TITLE: Is a quotient of a reductive group reductive?
QUESTION [19 upvotes]: Is a quotient of a reductive group reductive?
Edit [Pete L. Clark]: As Minhyong Kim points out below, a more precise statement of the question is: 
Is the quotient of a reductive linear group by a Zariski-closed normal subgroup reductive?

REPLY [18 votes]: It is important in answering this question that one can extend scalars to a perfect (e.g., algebraically closed) ground field, as was implicit in many of the other answers even if not said explicitly.  Indeed, if $k$ is an imperfect field then it always happens that there exist many examples of pairs $(G,H)$ with $G$ a smooth connected affine $k$-group  containing no nontrivial smooth connected unipotent normal $k$-subgroup and $H$ a smooth connected normal $k$-subgroup in $G$ such that $G/H$ contains a non-trivial smooth connected unipotent normal $k$-subgroup.
This can even happen when $G$ and $H$ are perfect (i.e., equal to their own derived groups), which is really disorienting if one is accustomed to working with reductive groups.  
For a simple commutative example, let $k'/k$ be a purely inseparable extension of degree $p = {\rm{char}}(k)$ and let $G$ be the Weil restriction ${\rm{Res}} _{k'/k}(\mathbf{G} _m)$, which is just "${k'}^{\times}$ viewed as a $k$-group".  This contains a natural copy of $\mathbf{G}_m$, and since $k'/k$ is purely inseparable this $k$-subgroup $H$ is the unique maximal $k$-torus and the quotient $G/H$ is unipotent of dimension $p-1$ (over $\overline{k}$ it is a power of $\mathbf{G} _a$ via truncation of $\log(1+x)$ in degrees $< p$, as one sees using the structure of the $\overline{k}$-algebra $\overline{k} \otimes_k k'$).  The main point then is that $G$ itself contains no nontrivial smooth connected unipotent $k$-subgroups, which is true because we are in characteristic $p > 0$ and $G$ is commutative with $G(k_s)[p] = {k'_s}^{\times}[p] = 1$! Note: the unipotent quotient $G/H$ is an example of a smooth connected unipotent $k$-group (even commutative and $p$-torsion) which contains no $\mathbf{G}_a$ as a $k$-subgroup (proof: commutative extensions of $\mathbf{G}_a$ by $\mathbf{G}_m$ split over any field, due to the structure of ${\rm{Pic}}(\mathbf{G}_a)$ and a small calculation); that is, $G/H$ is a "twisted form" of a (nonzero) vector group, which never happens over perfect fields.  
Making examples with perfect $G$ and $H$ is less straightforward; see Example 1.6.4 in the book "Pseudo-reductive groups".  
As for the suggestion to use Haboush's theorem (whose proof I have never read), I wonder if that is circular; it is hard to imagine getting very far into the theory of reductive groups (certainly to the point of proving Haboush's theorem) without needing to already know that reductivity is preserved under quotients (a fact that is far more elementary than Haboush's theorem, so at the very least it seems like killing a fly with a sledgehammer even if it is not circular).  
Finally, since nobody else has mentioned it, look in any textbook on linear algebraic groups (Borel, Springer, etc.) for a proof of the affirmative answer to the original question.  For example, 14.11 in Borel's book.  Equally important in the theory is that for arbitrary smooth connected affine groups, formation of images also commutes with formation of maximal tori and especially (scheme-theoretic) torus centralizers; see corollaries to 11.14 in Borel's book.<|endoftext|>
TITLE: Finite extension of fields with no primitive element
QUESTION [34 upvotes]: What is an example of a finite field extension which is not generated by a single element?
Background: A finite field extension E of F is generated by a primitive element if and only if there are a finite number of intermediate extensions. See, for example, [Lang's Algebra, chapter V, Theorem 4.6].

REPLY [2 votes]: If you're willing to work with commutative rings instead of just fields, you can gain some Galois-theoretic insight by base-changing to an algebraic closure $\bar{K}$ (or even a suitably large finite field extension).  Tensor product takes any primitive extension of your original field $K$ to a $\bar{K}$-algebra generated by a single element, so it suffices to find an extension that base changes to a $\bar{K}$-algebra generated by at least 2 elements.
If we take Alon Amit's minimal example $K = F(x,y), L = F(x^{1/p},y^{1/p})$, we can tensor it up to $\bar{K}[r,s]/(r^p-x,s^p-y)$.  By a suitable change of coordinates, we see that the spectrum of $\bar{K} \otimes_K L$ is a trivial torsor under $\alpha_p \times \alpha_p$ (and $\mu_p \times \mu_p$ and various other groups).  This description descends to $L$, but the torsor becomes nontrivial over $K$.  However, we do see that there is a 2-dimensional tangent space, and this gives us the derivations in KConrad's answer.  Furthermore, intermediate fields are fixed by subgroup schemes, and since we are in the the height one setting, these are cut out by the infinitely many $K$-rational slopes in the tangent space.
It's been almost 5 years since JBorger's comment to KConrad's answer, so I'm not sure if he's still interested.  Anyway, there is a Galois theory with group schemes that can be applied to purely inseparable extensions, worked out more or less in Chase, S.U.: Infinitesimal Group Scheme Actions on Finite Field Extensions Am. J. Math., Vol. 98, No.2, pp. 441-480 (1976).  Basically, you can corepresent the functor $A \mapsto \text{Aut}_A(L \otimes_K A)$ by a Hopf algebra over $K$, whose spectrum is called the Galois group scheme of $L/K$.  As far as I can tell, in our case, it is the Weil restriction of $\alpha_p \times \alpha_p$ (or $\mu_p \times \mu_p$, etc. - these are not canonical) along $L/K$, and the coordinate ring has dimension $(p^2)^{p^2}$.  I think there is a Galois correspondence involving subgroup schemes and descent data, but I haven't actually read the paper thanks to the paywall.<|endoftext|>
TITLE: Can $N^2$ have only digits 0 and 1, other than $N=10^k$?
QUESTION [47 upvotes]: Pablo Solis asked this at a recent 20 questions seminar at Berkeley. Is there a positive integer $N$, not of the form $10^k$, such that the digits of $N^2$ are all 0's and 1's?
It seems very unlikely, but I don't have a proof. It's easy to see that such a number must end in 1 or 9, and then easy to see that it must end in 01, 49, 51 or 99, and you can continue recursively for as long as you like, determining possible "suffixes". Using this, I had a computer check for me that there are no such N up to about $10^{24}$.
If you pretend that the digits of $N^2$ are randomly distributed, and $N$ has $n$-digits, there's a $(2/10)^{2n}$ chance of satisfying this condition. There are only $10^n$ $n$-digit numbers, so you might expect a $(4/10)^n$ chance of having a some $n$-digit number. This suggests we shouldn't expect to find anything.
(If you try the same problem in other bases, where the probabilities are better, you do find a few:  in base 5, 222112144, 22222111221444 and 100024441003001 work.)

REPLY [8 votes]: Let me redo the probabilistic argument. We know that the $k$ last digits of $n^2$ are determined by the $k$ last digits of $n$. For instance, $n^2$ ends with a $1$ if and only if $n$ ends with a $1$ or a $9$. By induction, one proves that there are $2^k$ $k$-uplets of digits $\bar a=a_1\ldots a_k$ such that the square of a number $n$ ending with $\bar a$ ends with $k$ ones and zeroes. For instance if $k=3$, the triplets are $001,249,251,499,501,749,751, 999$. 
If $n$ is a solution to the problem, and $k$ is the number of digits of $n$, then $n$ must be an $\bar a$ as above. Take the square of such an $\bar a$. We know that its $k$ last digits are ones and zeroes. In order that it provides a solution, the $k$ first digits have to be ones and zeroes, a property whose probability is $5^{-k}$. Therefore, the probability that there be a solution $n$ of length $k$ is approximately $(2/5)^{-k}$. Since the series
$$\sum_{k=1}^\infty\left(\frac{2}{5}\right)^k$$
converges, I bet that there are only finitely many solutions. But if you do not find a short one, I guess there will be no solution at all.<|endoftext|>
TITLE: Is there a good version of Artin-Wedderburn for semisimple algebra objects?
QUESTION [8 upvotes]: Artin-Wederburn says that if you have a semisimple algebra then it is a product of matrix algebras over division rings.
Suppose that $C$ is a fusion category over the complex numbers (if you want to assume pivotal or similar things, that's fine, but don't assume symmetric or braided).  Suppose that $X$ is an algebra object in $C$.  That is $X$ is an object in $C$ together with a multiplication map $X \otimes X \rightarrow X$ and a unit etc.  We call $X$ semisimple if the category of $X$-module objects in $C$ is semisimple.  Is there some good analogue of Artin-Wedderburn?

REPLY [4 votes]: I have a paper on that topic:

Finite, connected, semisimple, rigid tensor categories are linear, Math. Res. Lett. 10 (2003), 411-421, doi:10.4310/MRL.2003.v10.n4.a1, arXiv:math/0209256.<|endoftext|>
TITLE: Can a vector space over an infinite field be a finite union of proper subspaces?
QUESTION [62 upvotes]: Can a (possibly infinite-dimensional) vector space ever be a finite union of proper subspaces?
If the ground field is finite, then any finite-dimensional vector space is finite as a set, so there are a finite number of 1-dimensional subspaces, and it is the union of those. So let's assume the ground field is infinite.

REPLY [4 votes]: Anton Geraschenko's comment prompted me to write a new version of this short answer. I'm leaving the old version to make Anton's comment clearer (and also to increase the probability of having at least one correct answer).
NEW VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each (affine) line. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.
Indeed, as pointed out by Anton, the $K$-valued functions on $A$ which are polynomial on each line form obviously a ring $R$. This ring is a domain, because if $f$ and $g$ are nonzero elements of $R$, then there is a line on which none of them is zero, and their product is nonzero on this line.
OLD VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each finite dimensional affine subspace. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.
Indeed, we can assume that $A$ is finite dimensional, in which case the result is easy and well known.<|endoftext|>
TITLE: Non-quasi separated morphisms
QUESTION [10 upvotes]: What are some examples of morphisms of schemes which are not quasi separated?

REPLY [8 votes]: Here is some intuitive propaganda for Anton's answer...
We know that a  qsep (quasi-separated) scheme (over $\mathbb{Z}$) is precisely one where the intersection U∩V of any two open affines, U=Spec(A) and V=Spec(B), is quasi-compact.  Looking at compliments gives a different perspective: that their differences U\V and V\U are cut out by finitely many elements in A,B respectively, meaning that these differences are "easy to see".  I'd say this justifies the following credo:


A quasi-separated scheme is one where any two open affines are "easy to distinguish". 
A non-qsep scheme is one containing some "subtle distinction" between open affines.


The two copies of $\mathbb{A}^\infty$ in Anton's answer differ only by the origin, which is "hard to see" in that it cannot be cut out by finitely many ring elements, and I'd say using infinitely many variables to cut out one point is about the most natural way to achieve this.  Thus, I like to characterize non-qsep schemes as containing "(infinitely) subtle distinctions" such as this one. 
Further tinkering yields a similar way to think about a qsep morphism $f:X\to Y$.  I'd say the corresponding credo is that:


A quasi-separated morphism is one which preserves the existence of "subtle distinctions".
A non-qsep morphism is one which destroys some "subtle distinctions".


This helps intuitivize theorems like:
(1) " Any map from a qsep scheme is qsep ", because it has no subtle distinction that can be destroyed.
(2) " If $Y$ is qsep, then $f:X\to Y$ is qsep iff $X$ is qsep ", since $f$ destroys subtle distinctions iff $X$ has them.
(3) " If $g\circ f$ is qsep, then $f$ is qsep ", since if $f$ destroyed some subtle distinction, then $g$ could not recover it.
Here is a coarse and a fine justification for this credo in each direction...
Coarse version: By 1971 EGA I 6.1.11, for any cover of Y by qsep opens $V_{i}$, $f$ is qsep iff each preimage $f^{-1}(V_i)$ is qsep.  Thus, $f$ is non-qsep iff there is some qsep open $V\subseteq Y$ such that $f^{-1}(V)$ is non-qsep, meaning it contains some subtle distinction which is lost after application of by $f$.
Fine version: Suppose $f$ is qsep.  By 1971 EGA I 6.1.9, fibre products and compositions of qsep morphisms are qsep, and any universal injection is qsep (for example any immersion).  Now suppose 
$S\hookrightarrow X$ 
$T\hookrightarrow Y$ 
are any universal injections such that $f|_S$ factors through $T$, for example if $T$ is the scheme-theoretic image of $S$.  Then $T$ qsep $\Rightarrow$ $S$ qsep, hence $S$ non-qsep $\Rightarrow$ $T$ non-qsep, meaning $f$ preserves the existence of subtle distinctions in passing from any such $S$ to $T$.<|endoftext|>
TITLE: If $\Omega_{X/Y}$ is locally free of rank $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$, is $X\rightarrow Y$ smooth?
QUESTION [9 upvotes]: Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?
The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\dim X-\dim Y$?
Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{char}\,k=2$.

REPLY [5 votes]: I think Ishai's example is close, but one must be a little careful; the normalization of the node is a good example, but the normalization of the cusp is ramified, and the sheaf of relative differentials in that case is not even locally free.
The differential-wise condition you want is this: for the morphism morphism $f: X \to Y$ to be smooth, you need that the sequence
$$0 \to f^* \Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0$$
be  exact and locally split (I can't find a reference that says this is sufficient, so it may not be). In the special case when $\dim X = \dim Y$, 
 $\Omega_{X/Y}$ is 0 if and only if $f$ is unramified.  But in this case $f^* \Omega_Y \to \Omega_X$ can still fail to be injective.<|endoftext|>
TITLE: Which are the best mathematics journals, and what are the differences between them?
QUESTION [92 upvotes]: Suppose you have a draft paper that you think is pretty good, and people tell you that you should submit it to a top journal. How do you work out where to send it to?
Coming up with a shortlist isn't very hard. If you look for generalist journals, it probably begins:

Journal of the American Mathematical Society
Annals of Mathematics
Inventiones
...?

How do you begin deciding amongst such a list, however? I know that you can look up eigenfactors and page counts, and you should also look for relevant editors and perhaps hope for fast turn around times. Depending on your politics, you might also ask how evil the journal's publisher is.
But for most people thinking about submitting to a good journal, these aren't really the right metrics. What I'd love to hear is something like "A tends to take this sort of articles, while B prefers X, Y and Z." This sort of information is surprisingly hard to find on the internet.

REPLY [29 votes]: Look at the "big name" of your field, and if they are still alive, they are probably part of the editorial board of goods journals, to which you should submit your paper.<|endoftext|>
TITLE: Weil divisors on non Noetherian schemes
QUESTION [6 upvotes]: Let X be an integral scheme that is separated (say over an affine scheme). Define a Weil divisor as a finite integral combination of height 1 points of X, where the height of a point of X is the dimension of its local ring.  Let Z be a closed subscheme of X, Z not equal to X.  
Is there an example of such an X and Z where Z contains infinitely many Weil divisors?

REPLY [3 votes]: Let R be the integral closure of $\mathbb{Z}$ in the algebraic closure of $\mathbb{Q}$.
Then $\dim (R)=1$ and there are infinitely many prime ideals lying above a prime ideal
$p\mathbb{Z}$. Hence the closed set $V(pR)$ has the required property.<|endoftext|>
TITLE: What is the universal property of normalization?
QUESTION [31 upvotes]: What is the universal property of normalization? I'm looking for an answer something like

If X is a scheme and Y→X is its
  normalization, then the morphism
  Y→X has property P and any other
  morphism Z→X with property P
  factors uniquely through Y.

REPLY [3 votes]: While this comes pretty late I wanted to add that normalization has a somewhat stronger universal property than what is mentioned in the other answers:
(N) Let $f: Z \to X$ be a morphism between integral, excellent schemes with $Z$ normal such that the image contains a normal point of $X$. Then $f$ lifts uniquely to the normalization $Y \to X$.
If the cited normal point is the generic point of $X$, one gets back the criterion demanding dominance (ND), as in Giulia's answer. However it also encompasses situations like a normal curve $Z$ on a singular surface $X$ not contained in the singular locus, which also lifts uniquely to the normalization. 
To justify (N) one uses (ND) to reduce to $Z$ being the normalization of a closed integral subscheme $R \subset X$ whose generic point $\eta$ is normal in $X$. Denote by $R'$ the reduced closure of $\{\eta\}$ in the normalization of $R \times_X Y$. Because $\eta$ is a normal point of $X$, this is the unique component mapping dominantly to $R$. In the resulting diagram
$\require{AMScd}$
$$\begin{CD}
R' @>>> R\times_X Y@>>> Y\\
@VVV @VVV @VVV\\
Z @>>> R@>>> X
\end{CD}$$
we get the left vertical arrow again by (ND) from $R' \to R$. However, as $X$ is excellent, the map $Y \to X$ is finite, hence $R' \to Z$ is a finite birational map to a normal scheme, so an isomorphism. This gives a lift $Z \to R \times_X Y$ in the above diagram. Any other such lift also induces a section of $R' \to Z$ by (ND) which implies uniqueness.<|endoftext|>
TITLE: Supersingular elliptic curves
QUESTION [14 upvotes]: I've read that an elliptic curve is supersingular if and only if its endomorphism ring is an order in a quaternion algebra.  Does anyone have a simple explanation of this (or a good reference)?

REPLY [4 votes]: I'd like to add a question to this discussion: the endomorphism ring of a supersingular elliptic curve is not just an order in a quaternion algebra, it is a maximal order in such an algebra.  Is there a simple explanation for this?<|endoftext|>
TITLE: Is there a universal property for Witt vectors?
QUESTION [34 upvotes]: Do the Witt vectors satisfy a universal property?

REPLY [47 votes]: As Morten Brun said, the big Witt vector functor is the right adjoint of the forgetful functor from the category of lambda-rings to the category of rings (commutative). But this answer is not completely satisfying in that Witt vectors usually come up in number theory, in contexts that have little direct connection to K-theory. It's also not clear what the analogue of that statement for the p-typical Witt vectors is. (p is a prime here. The p-typical Witt vectors are the usual "non-big" Witt vectors as defined by Witt and which come up in the theory of local fields, for instance.)
To me, the most satisfying answer to this question is that Witt W(A) gives the universal way of equipping your ring A with lifts of Frobenius maps. For simplicity, let's look at the p-typical Witt vectors. Then W(A) has a ring endomorphism F which is congruent to the p-th power map modulo the ideal pW(A). In other words, W(A) has a lift of the Frobenius endomorphism of W(A)/pW(A). There is also a ring map W(A)-->A given by projection on the the first component.
Now, in what sense is W(A) universal? Suppose B is another ring equipped with a ring map B-->A and an endomorphism F:B-->B lifting the Frobenius endomorphism of B/pB. Then, assuming B is p-torsion free, there exists a unique ring map B-->W(A) commuting with the two maps F and the two maps to A. (This is a theorem called "Cartier's Dieudonne-Dwork lemma" in classical exposition of the Witt vectors, but is essentially true by definition in some more recent ones.) Thus, ignoring the issue of p-torsion, W(A) is the universal ring mapping to A with a Frobenius lift.
How do we deal with torsion? First, if A itself is p-torsion free, then so is W(A) -- it is actually a subring of an infinite product of copies of A. So then W is the right adjoint of the forgetful functor from the category of p-torsion-free rings equipped with a Frobenius lift to the category of p-torsion-free rings. Now it will one day be clear that the most important uses of W(A) are when A is torsion free, but certainly the most important existing applications are when A is an F_p-algebra, where everything is p-torsion. So it would be nice to have a universal property that works whether there is p-torsion or not. 
Probably the most straightforward way to do this is to use a better definition of "Frobenius lift". If F is a Frobenius lift on B as above and B is p-torsion free, then d(x)=(F(x)-x^p)/p is a well-defined operator on B. The condition that F be a ring endomorphism can be of course be expressed in terms of slightly complicated identities on d. The key point, then, is that by the magical properties of binomial coefficients modulo p, these identities have integral coefficients -- there are no p's in the denominators! Then we can define a d-ring structure on any ring to be an operator d satisfying these conditions. Then you can show by reduction to the p-torsion-free case that W is the right adjoint of the forgetful functor from the category of d-rings to the category of rings. The point of all this is to eliminate the existential quantifier hidden in the word "lift" by specifying a y such that F(x)-x^p is p times y, rather than just saying some such element y exists.
Pretty much everything is the same when dealing with more than one prime, except that the Frobenius lifts are required to commute. The big Witt vectors are what you get when you have commuting Frobenius lifts at all primes.
I think this point of view was first discovered by Joyal. You can also see the first section of my paper "Basic geometry of Witt vectors", which is on the archive. Unlike mine, Joyal's papers on this are wonderfully short. I don't have their precise details, but you can see the references in my paper.<|endoftext|>
TITLE: Points of a weakly locally separated algebraic space
QUESTION [5 upvotes]: If X is a quasi-separated algebraic space and Spec k -> X is an etale presentation, then X is isomorphic to Spec k' for a field k'.  (This is also true if X is Zariski locally quasi-separated.)  The examples of algebraic spaces I know where this fails have diagonals that are not immersions.
Is this also true for weakly locally separated algebraic spaces?
An algebraic space is locally weakly separated if the diagonal is an immersion (not necessarily quasi-compact).  In the literature, the term locally separated seems to have been reserved for quasi-compact immersions.
(This question is asked by Johan in the stacks project (currently Remark 32.4.8))

REPLY [2 votes]: Coincidently, I thought about this a few weeks ago (without any conclusion). I think that I can prove that a "weakly locally separated" algebraic space X with an étale cover Spec(k)->X is of the form Spec(k') if X lives over a field k_0 such that k/*k*_0 is algebraic. If X is not locally separated, this condition does not always hold (take A^1/Z where Z acts by translation and restrict this action to the generic point).
Let K be the algebraic closure of k. Let R=Spec(K) x_ X Spec(K). By assumption
j : R -> Spec(K) x_{k_0} Spec(K)
is an immersion and it is enough to show that this is a closed immersion (since fpqc morphisms descend closed immersions).
We can replace k_0 with its perfect closure. This follows from the observation that R is reduced.
Now, the right-hand scheme is a group scheme over Spec(K). Indeed, it is the fundamental group scheme \pi_0(k_0). It is totally disconnected and all its residue fields are K and the group of K-points is the pro-finite group Gal(K/*k*_0).
R is also totally disconnected and all its points have residue field K. The map
j(K) : R(*K*) -> Gal(K/*k*_0)
is injective and locally closed. Since R(*K*) => K(*K*) is an equivalence relation, it follows that j(K) identifies R(*K*) with a subgroup of Gal(K/*k*_0).
Lemma: A locally closed subgroup H of a topological group G is closed.
pf: The closure of H is a subgroup so we can assume that H is open. It is then easily seen that the complement of H is open.
Thus, R(*K*) is a closed subgroup of Gal(K/*k*_0). In particular, j is a closed immersion.
Remark: If K/*k*_0 is not algebraic, then (if K is algebraically closed) we still have a group structure on the K-points of the fiber product of K over k_0. R(*K*) will be a closed subgroup of this group but it is not clear whether this implies that j is closed.<|endoftext|>
TITLE: Can a coequalizer of schemes fail to be surjective?
QUESTION [18 upvotes]: Suppose $g,h:Z\to X$ are two morphisms of schemes. Then we say that $f:X\to Y$ is the coequalizer of $g$ and $h$ if the following condition holds: any morphism $t:X\to T$ such that $t\circ g=t\circ h$ factors uniquely through $f$. The question is whether it is possible for a coequalizer $f:X\to Y$ to fail to be surjective.
Remark: $f$ must hit all the closed points of $Y$. To see this, suppose $y\in Y$ is a closed point that $f$ misses. Then $f$ factors through the open subscheme $Y\smallsetminus\{y\}$. It is easy to check (using the fact that $Y$ is the coequalizer) that $Y\smallsetminus\{y\}$ satisfies the universal property of the coequalizer. But coequalizers are unique, so we get $Y=Y\smallsetminus\{y\}$.
Background: A categorical quotient of a scheme $X$ by a group $G$ is the same thing as a coequalizer of the two maps $G\times X\rightrightarrows X$ (given by $(g,x)\mapsto x$ and the action $(g,x)\mapsto g\cdot x$) in the category of schemes. In Geometric Invariant Theory, Mumford defines the notion of a geometric quotient of a scheme by a group (Definition 0.6), which is stronger than the notion of a categorical quotient (Proposition 0.1). Part of the definition is that a map $f:X\to Y$ must be surjective in order to be a geometric quotient. In subsequent pages, he suggests strongly (but doesn't explicitly state) that a categorical quotient need not be surjective.

REPLY [16 votes]: Let $k$ be a field. Take $Y=\mathrm{Spec}\,k[[t]]$, and take for $X$ the disjoint sum of the closed subschemes $X_n:=\mathrm{Spec}\,k[[t]]/(t^n)$ ($n>0$). Put $Z=X\times_Y X$ with the two obvious maps to $X$. A coequalizer is just a direct limit of the system $X_1\hookrightarrow\dots X_n\hookrightarrow  X_{n+1}\hookrightarrow\dots$ in the category of schemes (look at the definition!).  Clearly, $Y$ is a direct limit in the category of affine schemes, hence also in the category of schemes since the $X_n$'s are one-point schemes and every compatible system of morphisms $(X_n\to T)_{n>0}$ must factor through an affine open subscheme of $T$. 
So, $X\to Y$ is a coequalizer of $pr_1, pr_2 :Z\to X$, but its set-theoretic image is the closed point.<|endoftext|>
TITLE: What's an example of a function whose Taylor series converges to the wrong thing?
QUESTION [24 upvotes]: Can anyone provide an example of a real-valued function f with a convergent Taylor series that converges to a function that is not equal to f (not even locally)?

REPLY [24 votes]: Another thing to note is that there are smooth functions whose Taylor series do not converge to the function in a neighborhood of ANY point!  An easy example of this can be found here:
http://web.archive.org/web/20141230224759/http://www.math.niu.edu/~rusin/known-math/99/nowhere_analy
The example looks as follows:
$$F(x) = \sum_{n=0}^{\infty} \frac {\exp(2^n i x)} {n!},
\quad F^{(k)}(x) = \sum_{n=0}^{\infty} (2^n i)^k \frac {\exp(2^n i x)} {n!}$$
For every $k$, the above series converges absolutely for real $x$, so the function $F$ is smooth.
On the other hand, if $x=a/2^N$ for some integers $a$ and natural number $N$, then for $k\in 4\mathbb{Z}$ we have
$$|F^{(k)}(x)| \geq \frac {2^{Nk}} {N!} - \sum_{n<N} \frac {2^{nk}} {n!} > \frac {2^{Nk-1}} {N!}$$
provided that $2^k > 2N$. Therefore, the Taylor series of $F$ has convergence radius $0$ at $x$. Since the set on which $F$ is analytic is open, this means that $F$ is nowhere analytic on $\mathbb{R}$.<|endoftext|>
TITLE: Stack with affine stabilizers but not quasi-affine diagonal
QUESTION [8 upvotes]: Give an example of a stack X with affine stabilizer groups and separated but not quasi-affine diagonal.
Remarks:
1) If X has finite stabilizer groups then the diagonal is quasi-finite and separated, hence quasi-affine (Zariski's MT).
2) If we drop the condition that the diagonal of X is separated, it is easy to find examples.
3) The stabilizer groups of X are affine if and only if they are quasi-affine.

REPLY [9 votes]: Here is an example:
In X13 of "Faisceaux amples sur les schemas en groupes", Raynaud provides an example of a group scheme G -> S in chacteristic 2 where

S is a local regular scheme of dimension 2. 
G -> S is smooth, separated and quasi-compact.
The fibers of G ->S are affine and the generic fiber is connected.

such that G -> S is not quasi-projective.
Therefore, the classifying stack BG has affine stabilizers but does not have a quasi-affine diagonal.  
On the other hand, in VII 2.2, Raynaud proves that if G -> S is a smooth, finitely presented group scheme such that 

S is normal.
G -> S has connected fibers.
The maximal fibers are affine.

then G -> S is quasi-affine.
Question: Is the above statement true if (2) is weakened to require that the number of connected components over a fiber s \in S be prime the characteristic of the residue field k(s)?
Of course, one would really like to know if the statement is true if G->S is not necessarily flat so that one could apply it to the inertia stack.
On a related note, Raynaud also provides an example in VII3 of a smooth quasi-affine group scheme G -> A^2 over a field k with connected fibers but which is not affine.  The classifying stack BG gives an example of stack with affine and connected stabilizers but with non-affine inertia stack.  In the example of a scheme with non-affine diagonal, the inertia is of course affine.  It's also easy to provide examples of non-affine group schemes with affine but non-connected fibers  (eg. the group scheme obtained by removing the non-identity element over the origin from Z/2Z -> A^2).<|endoftext|>
TITLE: Can the valuative criteria for separatedness/properness be checked "formally"?
QUESTION [9 upvotes]: Suppose f:X→Y is a morphism of finite type between locally noetherian schemes. The valuative criterion for separatedness (resp. properness) says roughly that f is a separated (resp. proper) morphism if and only if the following condition holds:

For any curve C in Y and for any lift
  of C-{p} to X, there is at most one
  (resp. exactly one) way to extend this
  to a lift of C to X.

More precisely,

If C is the spectrum of a DVR with
  closed point p (a very local version
  of a curve: the intersection of all
  open neighborhoods of p on an "honest"
  curve), C→Y is a morphism, and
  C-{p}→X is a lift of that
  morphism along f, there is at most one
  (resp. exactly one) way to complete it to
  a lift C→X.

Does it suffice to check the valuative criteria on an even more local kind of object: the spectrum of a complete DVR? This would be quite nice because the only complete DVRs over a field k are rings of the form L[[t]] where L is an extension of k.
More generally, if you drop the hypotheses that f is of finite type and X and Y are locally noetherian, the usual valuative criteria must be verified for arbitrary valuation rings. Is it enough to check them for complete valuation rings?

REPLY [7 votes]: Yep, a quasi-compact morphism of schemes (resp. locally noetherian schemes) is universally closed if and only the existence part of the valuative criterion holds for complete valuation rings (resp. complete DVRs) with algebraically closed residue field.
This is in EGA (see II.7.3.8 and the remark II.7.3.9).  Note that the separated hypothesis is not necessary there; for the valuative criterion of properness one needs to require that the morphism is quasi-separated.  
This holds more generally for Artin stacks if one allows a field extension of the fraction field of the valuation ring (see LMB 7.3).<|endoftext|>
TITLE: What is interesting/useful about Castelnuovo-Mumford regularity?
QUESTION [15 upvotes]: What is interesting/useful about Castelnuovo-Mumford regularity?

REPLY [19 votes]: Here's how I think about Castelnuovo-Mumford regularity.  It's an invariant of an ideal (or module or sheaf) which provides a measure of how complicated that ideal (or module or sheaf) is.  This invariant is related to free resolutions, and thus it measures complexity from that perspective.
Why is it interesting?  One answer is that it can be used to provide an effective bound for two famous theorems.  The first theorem I have in mind is that the Hilbert function of a graded ideal (or a finitely generated graded module) over the polynomial ring eventually agrees with the Hilbert polynomial of that ideal (or module).  The second theorem I have in mind is Serre vanishing, which says that, given a coherent sheaf $\mathcal F$ on $\mathbb P^n$, there exists $d$ such that $H^i(\mathbb P^n, \mathcal F(e))=0$ for all $i>0$ and all $e>d$. These two theorems are related: if $M$ is a graded module of depth $> 0$, and $\mathcal F$ is the associated sheaf of $M$, then the Hilbert function of $M$ in degree $e$ equals $H^0(\mathbb P^n,\mathcal F(e))$.
An example where Castelnuovo-Mumford is particularly useful comes from the construction of the Hilbert scheme (I have heard that this is related to Mumford's original use, though I have no reference.)  The basic point is that you can parametrize the set of ideals with a given Hilbert function by considering subloci of certain Grassmanians satisfying determinantal criteria, whereas it's less clear (at least to me) how to parametrize ideals with a given Hilbert polynomial.
Another great example where  Castelnuovo-Mumford is useful is presented in Eisenbud "The Geometry of Syzygies" chapter 4, where he solves the interpolation problem for points in affine space.<|endoftext|>
TITLE: How to approximate a solution to a matrix equation?
QUESTION [8 upvotes]: Suppose a matrix equation Ax = b has no solution (b is not in the column space of A)
How can I find a vector x' so that Ax' is the closest possible vector to b?

REPLY [3 votes]: Andrew has a correct answer in the pseudoinverse $A^+$, which is characterized by the property that $x = A^+b$ is the shortest vector that solves $A^TAx = A^Tb$ (equivalently, $x$ has zero nullspace(A) component).  Computationally, it is typically found using a singular value decomposition: if
$\displaystyle A = U \Sigma V^T = \left[ \begin{array}{cc} U_{col} & U_{null} \end{array} \right] \left[ \begin{array}{cc} \Sigma_{pos} & 0 \\\ 0 & 0 \end{array} \right] \left[ \begin{array}{cc} V_{row} & V_{null} \end{array} \right]^T,$
then the pseudoinverse is $A^+ = V_{row} \Sigma_{pos}^{-1} U_{col}^T$.
I'd like to mention that the pseudoinverse is both rather computationally expensive and unstable in the presence of noise.  In particular, if $b$ is given by taking real-world data with limited accuracy, and $A$ is ill-conditioned (e.g., singular), the output of least-squares can vary wildly with the error.  One common approach to rectify this is Tikhonov regularization, which typically means minimizing $\Vert Ax - b \Vert^2 + \alpha \Vert x \Vert^2$ for some small $\alpha$.  This generically yields a nonsingular optimization problem, which can be computed quickly by Gaussian elimination, and as $\alpha$ approaches zero, the solution approaches the pseudoinverse solution.  It will not in general yield an exact solution, but there are error-minimizing heuristics (e.g., using the Discrepancy Principle) for choosing $\alpha$ based on knowledge about the size the noise.
Reference: Strang, Computational Science and Engineering<|endoftext|>
TITLE: How can you find small denominators inside triangles?
QUESTION [7 upvotes]: Darsh asked over at the 20 questions seminar:
Take a triangle in R^2 with coordinates at rational points. Can we find the smallest denominator point in the interior? (Consider denominator of an element of Q^2 to be the lcm of the denominators of the coordinates.) (Hint: you can do the 1-d version using continued fractions.)

REPLY [3 votes]: One way to interpret the problem is as an integer programming problem in 3-dimensions. If one has 3 points $(a_i/c_i,b_i/c_i) \in \mathbb{R}^2, i=1,2,3$
with $gcd(a_i, b_i, c_i) =1$ and $c_i \geq 1$, then take the three lattice points $P_i=(a_i, b_i, c_i) \in \mathbb{Z}^3$.
Take the cone C spanned by positive integer combinations of $P_i$ (this is the projective region sitting above the triangle). Then one wants to find the lattice point of $\mathbb{Z}^3$ inside the interior of this cone with smallest z-coordinate. This may be interpreted as an integer linear programming problem. However, I don't know enough about integer programming to know if this will help (Scarf seems to have thought about precisely this problem, but doesn't address the computational complexity). The following gives one possible approach, but might not be any better than Nikokoshev's approach except in certain regimes. 
These vectors generate a sublattice $\Lambda \subset \mathbb{Z}^3$ of index D, where D is the determinant of the matrix $[P_1,P_2,P_3]$, and is the volume of the parallelepiped F spanned by these vectors. One can see that a lattice vector in this cone with minimal z-coordinate must lie in this F, since F is a fundamental domain for the action of $\Lambda\cap C$. A brute-force approach is to compute the finite abelian group $\mathbb{Z}^3/\Lambda$, finding a basis (which will consist of at most 2 vectors since $P_i$ is primitive). Then translate these vectors into the fundamental domain F, and take enough positive linear combinations to generate all coset representatives of $\mathbb{Z}^3/\Lambda$ inside C. Then subtract elements from the positive semigroup $\Lambda\cap C$, until you find all coset representatives in the fundamental domain F, and find the one with minimal z-coordinate. This approach should be pretty effective when D is small or the group $\mathbb{Z}^3/\Lambda$ is cyclic, but I'm not sure how the size of D correlates with the size of the final solution. For example, if $D=1$, then the minimal vector will be $P_1+P_2+P_3$ with denominator $c_1+c_2+c_3$. If $D\geq 2$, the minimal denominator will be $<(c_1+c_2+c_3)/2$ by the symmetry of F.<|endoftext|>
TITLE: Is there an example of a scheme X whose reduction X_red is affine but X is not affine?
QUESTION [15 upvotes]: For Noetherian schemes this follows from Serre's criterion for affineness by a filtration argument.

REPLY [18 votes]: No, if X is any algebraic space such that X_red is an affine scheme, then X is an affine scheme. This follows from Chevalley's theorem. For X noetherian scheme/alg. space this theorem is in EGA/Knutson. As you noted, this can also be showed using Serre's criterion for affineness or by an even simpler argument (see EGA I 5.1.9, first edition).
For X non-noetherian, the following general version of Chevalley's theorem is proved in my paper "Noetherian approximation of algebraic spaces and stacks" (arXiv:0904.0227):
Theorem: Let W->X be an integral and surjective morphism of algebraic spaces. If W is an affine scheme, then so is X.
Recall that any finite morphism is integral, in particular X_red -> X. As a corollary, it follows that under the same assumptions, if W is a scheme then so is X.<|endoftext|>
TITLE: How to understand character sheaves
QUESTION [12 upvotes]: There's a well-known series of articles by Lusztig about Character Sheaves. They have important connections to many things in (geometric) representation theory, e.g. 0904.1247
How to understand these for a person with less than excellent representation theory background?

REPLY [9 votes]: That's a pretty vague question.  The vague answer is that all the operations (like induction from subgroups) that can be done for representations and characters can be done for sheaves, and doing these results in a category of sheaves on a group (like GL_n over a finite field), which are close enough to the characters of representations to tell us something about them, but which also have more structure, since they are sheaves, not just functions.
Here's a somewhat more precise description: if you have a variety X which a group G acts on, then you can take the action of G on the cohomology of X.  Better yet, you can get a sheaf on the group G, whose stalk over a group element g is the cohomology of the fixed points of g on X.  The function sheaf correspondence sends this sheaf to the character of the representation on the cohomology of X (this follows from Lefschetz).  Deligne and Lusztig defined certain varieties (the set of flags over F_q in a given relative position to their conjugates by Frobenius) on which GL(n,F_q) acts (actually, this works for any split simple algebraic group), and the corresponding sheaves (or rather the simple perverse constitutuents) are called character sheaves, and roughly capture the structure of the corresponding representations.<|endoftext|>
TITLE: What is the "right" hermitian structure on tensor products of quantum group representations?
QUESTION [5 upvotes]: This is pretty specific, but there are some experts around.
So, in Chari & Pressley, it's explained that in the standard *-structure, every irreducible, finite-dimensional representation of a quantum group (at a generic parameter) is unitary.  Is it written somewhere what the "right" unitary structure on a tensor product of these representations is?  
I ask because if one categorifies such representations, one gets a unitary structure essentially for free, so it would extremely useful if someone had already written down one I could match up with.

REPLY [4 votes]: I know these are all about the root of unity case, but you might look at this paper by Kirillov, and this one by Wenzl.<|endoftext|>
TITLE: What do epimorphisms of (commutative) rings look like?
QUESTION [113 upvotes]: (Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)
What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
  $$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$

Also, if $A\to B_i$ is a finite collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.
Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?

REPLY [14 votes]: A special case where epimorphisms are surjective is the category of finite-dimensional commutative $k$-algebras where $k$ is a field. See for example this page in the Stacks Project. 
This may come in handy on occasion; I was trying to convince myself this morning that monomorphisms between cocommutative $k$-coalgebras are those whose underlying functions are injective, and needed the result above as a lemma (first check the result on finite-dimensional cocommutative $k$-coalgebras by taking linear duals on the result above, and then use the fact that every coalgebra is the directed colimit of the system of finite-dimensional subcoalgebras and inclusions between them).<|endoftext|>
TITLE: What is the exact statement of "there are 27 lines on a cubic"?
QUESTION [21 upvotes]: I think there was a theorem, like 

every cubic hypersurface in $\mathbb P^3$ has 27 lines on it. 

What is the exact statement and details?

REPLY [7 votes]: There is a nice recent proof in a paper by Galkin and Shinder. These authors prove a relation in the Grothendieck of varieties involving the classes of a smooth cubic hypersurface, its variety of lines, and its symmetric square. The proof of this formula is based on the following elementary geometric principle: take a point on a cubic hypersurface and a line passing through that point; either the line is contained in the hypersurface, or it cuts the hypersurface in three points, the given one and a pair of points. (They prove a similar formula in the singular case, as well.)
Given the classical computations of the Euler-Poincaré characteristic/Hodge-Deligne polynomial of a hypersurface, and of its symmetric square, this gives a formula for the Hodge-Deligne polynomial of the variety of lines of the given cubic hypersurface. 
In the case of a surface, one gets exactly 27: there are 27 lines on a smooth cubic surface.<|endoftext|>
TITLE: Is there a good computer package for working with complexes over non-commutative rings?
QUESTION [8 upvotes]: I'm interested in doing computations with certain non-commutative rings, most of which involve taking derived tensor products.  Does anyone know of a computer algebra package which will find projective resolutions of complexes of modules over a finite-dimensional non-commutative ring, tensor with a bimodule, and do it all over again?

REPLY [6 votes]: I don't know whether Magma can handle all you ask for, but if I remember my coding for Magma correctly, at least the projective resolutions of modules over a non-commutative ring should be covered by that - for nice enough non-commutative rings. It's all been developed there as part of Jon F. Carlson's work on computing group cohomology rings.
If there is a system that does all you ask for, and does it efficiently, it is probably been written in connection to a group cohomology computation effort - which narrows the candidates down significantly: Magma and GAP do group cohomology rings, and SAGE now with the work of Simon King and David Green.
In contrast, I'm reasonably certain that Macaulay only does commutative things, and Singular doesn't have resolutions as a naturally occuring object at all.
Bergman might be able to deal with what you ask for, though.
To conclude: I'd recommend you to take a look at the homological algebra modules in Magma, GAP, SAGE and Bergman - I'd be highly surprised to see any other packages deal with the case you describe, and I'm not entirely convinced either of these do it well either.<|endoftext|>
TITLE: Linearity of the inner product using the parallelogram law
QUESTION [31 upvotes]: A norm on a vector space comes from an inner product if and only if it satisfies the parallelogram law.  Given such a norm, one can reconstruct the inner product via the formula:
$2\langle u,v\rangle = |u + v|^2 - |u|^2 - |v|^2$
(there are minor variations on this)
It's straightforward to prove, using the parallelogram law, that this satisfies:

$\langle u,u\rangle \ge 0$ for all $u$, and $\langle u,u\rangle = 0$ iff $u = 0$
$\langle tu,tu\rangle = t^2 \langle u,u\rangle$;
$\langle u,v\rangle = \langle v,u\rangle$;
$\langle u,v+w\rangle = 2\langle u/2,v\rangle + 2\langle u/2,w\rangle$;

From 4 with the special case $w=0$ one quickly deduces that $\langle u,v+w\rangle = \langle u,v\rangle + \langle u,w\rangle$.
The usual method of proving $\langle u,tv\rangle = t\langle u,v\rangle$ is to use 4 with induction to prove that $\langle u,nv\rangle = n\langle u,v\rangle$, then deduce $\langle u,tv\rangle = t\langle u,v\rangle$ for $t$ rational, and finally appeal to continuity to extend to the reals.
Is there any way to avoid this last bit?  In particular, is there a more geometric view of why $\langle u,tv\rangle = t\langle u,v\rangle$ for all real $t$?  Pictures would be great!
If not, is there a different way to express the condition that a norm comes from an inner product that does make all the conditions obviously geometrical?
Clarification added later: My reason for asking this is pedagogical.  I teach a course which introduces, in quick succession, metric spaces, normed vector spaces, and inner product spaces.  The properties of metrics and norms are very easy to motivate from intuitive properties of distances and lengths.  I'd like to do the same for inner products in terms of angles.  Thus by "geometric" I mean "geometric intuition" rather than geometry as geometers understand it.  Since the inner product is introduced after the norm, I argue that using the cosine law one can define the notion of "angle" between two vectors using any norm.  However, unless the norm is "special", that notion of angle doesn't behave how we would expect it to do so.  In particular, in order for angles to add properly, one needs the norm to satisfy the parallelogram law.  Here "add" means that (modulo a pi or two), the angle from $u$ to $v$ plus the angle from $v$ to $w$ should be the angle from $u$ to $w$.  Once one has the parallelogram law then the fact that it comes from an inner product follows via the route above.  However, the properties of an inner product are not particularly obvious from thinking about properties of angles.  So the easier they are to deduce from the parallelogram law, the easier they are to motivate.  I consider the route to $\langle u,\lambda v\rangle = \lambda\langle u,v\rangle$ to be a little long.  I was hoping someone could shorten it for me.
Alternative, there may be a different starting point than that angles "add".  Perhaps some other property, say similarity of certain triangles, that could be used.  However, I'd like a single property that would do the lot.  I don't want "add" for some properties and "something else" for others.  That's too complicated.

REPLY [37 votes]: To me continuity is more geometric and intuitive than the rest of the argument (which is purely algebraic manipulation). So I take the liberty to mis-read you question as follows:

Is it possible to derive linearity of the inner product from the parallelogram law using only algebraic manipulations?

By "only algebraic" I mean that you are not allowed to use inequalities. (It is triangle inequality that allows one to use continuity. In fact, one can derive continuity using only the inequality $|u|^2\ge 0$ and the parallelogram law.) Also, an algebraic argument must work over any field on characteristic 0.
The answer is that it is not possible. More precisely, the following theorem holds.
Theorem. There exists a field $F\subset\mathbb R$ and a function $\langle\cdot,\cdot\rangle: F^2\times F^2\to F$ which is symmetric, additive in each argument (i.e. $\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$), satisfies the identity $\langle tu,tv\rangle = t^2\langle u,v\rangle$ for every $t\in F$, but is not bi-linear.
Note that the above assumptions imply that the "quadratic form" $Q$ defined by $Q(v)=\langle v,v\rangle$ satisfies $Q(tv)=t^2Q(v)$ and the parallelogram identity, and the "product" $\langle\cdot,\cdot\rangle$ is determined by $Q$ in the usual way. [EDIT: an example exists for $F=\mathbb R$ as well, see Update.]
Proof of the theorem. Let $F=\mathbb Q(\pi)$. An element $x\in F$ is uniquely represented as $f_x(\pi)$ where $f_x$ is a rational function over $\mathbb Q$. Define a map $D:F\to F$ by $D(x) = (f_x)'(\pi)$. This map satisfies

$D(1) = 0$;
$D(\pi)=1$;
$D(x+y) = D(x)+D(y)$;
$D(xy) = x D(y) + y D(x)$.

Define $P:F\times F$ by $P(x,y) = xD(y)-yD(x)$. From the above identities it is easy to see that $P$ is additive in each argument and satisfies $P(tx,ty)=t^2 P(x,y)$ for all $x,y,t\in F$. Finally, define a "scalar product" on $F^2$ by
$$
 \langle (x_1,y_1), (x_2,y_2) \rangle = P(x_1,y_2) + P(x_2,y_1) .
$$
It satisfies all the desired properties but is not bilinear: if $u=(1,0)$ and $v=(0,1)$, then $\langle u,v\rangle=0$ but $\langle u,\pi v\rangle=1$.
Update. One can check that if $\langle\cdot,\cdot\rangle$ is a "mock scalar product" as in the theorem, then for any two vectors $u,v$, the map $t\mapsto \langle u,tv\rangle - t\langle u,v\rangle$ must be a differentiation of the base field. (A differentiation is map $D:F\to F$ satisfying the above rules for sums and products.) Thus mock scalar products on $\mathbb R^2$ are actually classified by differentiations of $\mathbb R$.
And non-trivial differentiations of $\mathbb R$ do exist. In fact, a differentiation can be extended from a subfield to any ambient field (of characteristic 0). Indeed, by Zorn's Lemma it suffices to extend a differentiation $D$ from a field $F$ to a one-step extension $F(\alpha)$ of $F$. If $\alpha$ is transcedental over $F$, one can define $D(\alpha)$ arbitrarily and extend $D$ to $F(\alpha)$ by rules of differentiation. And if $\alpha$ is algebraic, differentiating the identity $p(\alpha)=0$, where $p$ is a minimal polynomial for $\alpha$, yields a uniquely defined value $D(\alpha)\in F(\alpha)$, and then $D$ extends to $F(\alpha)$. The extensions are consistent because all identities involved can be realized in the field of differentiable functions on $\mathbb R$, where differentiation rules are consistent.
Thus there exists a mock scalar product on $\mathbb R^2$ such that $\langle e_1,e_2\rangle=0$ but $\langle e_1,\pi e_2\rangle=1$. And I am sure I reinvented the wheel here - all this should be well-known to algebraists.<|endoftext|>
TITLE: How can you tell if a space is homotopy equivalent to a manifold?
QUESTION [37 upvotes]: Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)?  Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient.  Are there any other special homotopical properties of manifolds?

REPLY [6 votes]: Sean: this gives a Poincare space which is not homotopy equivalent to a closed manifold. the idea is that the Spivak fibration of the $5$ dimensional Poincare space doesn't lift to a stable vector bundle. One can prove this as follows: let $X^5$ be as in Madsen and Milgram. Then $X$ fibers over $S^3$ with fiber $S^2$. Call this fibration $\xi$, and let the projection $X \to S^3$ be $p$.
Note that $p$ has a section, call it $s$.
It's not hard to prove that the Spivak fibration of $X$ in this case is just $p^*\xi$. One then needs to verify that $p^*\xi$ doesn't lift to a stable vector bundle. But if it did, then
so would $$\xi = (p \circ  s)^*\xi = s^*p^* \xi\ .$$  But it's very easy to see that $\xi$ doesn't lift
($\xi$ is given by the nontrivial element of $\pi_2(F) = \mathbb{Z}_2$, where $F$ classifies spherical fibrations with section, whereas $\pi_2(O)$ is trivial).<|endoftext|>
TITLE: Critical points on a fiber bundle
QUESTION [7 upvotes]: Consider a (smooth) bundle E→_B_, and a (smooth) function f: E → R on the total space.  Then it makes sense to talk about the derivatives of f along the fibers.  Let C be the subspace of E consisting of all points for which all fiber-wise derivatives of f vanish, so that upon intersecting with any fiber C consists of the critical points of the restriction of f to the fiber.  If the fiber is n-dimensional, then C is carved out by n equations, and so generically has codimension n in E.
Let's say that c is a point in C so that the second derivative of f in the fiber is nondenegerate (i.e. has non-degenerate Hessian; i.e. f restricts to a Morse function on the fiber through c).  Does it follow that the projection C→_B_ is a local diffeomorphism near c?
The answer is yes when everything is finite-dimensional (and I believe the statement is iff).  I am interested in the case when B is finite-dimensional but the fibers of E are infinite-dimensional.
Edit:  This is a response to Andrew's question below (since answering in the comments proves difficult).
I'm using the word "Morse" loosely, largely because I don't actually know Morse theory.  I would suggest that the definition I give is better than what's traditionally used.  What I actually mean is this:
Let M be a smooth manifold and f:M→R a smooth map.  What type of object is the second derivative f(2)?  In general, you should not talk about it by itself, because it does not transform as a tensor, although the pair (f(1),f(2)) is a vector in the 2-jet bundle over M.  But if c is a critical point of f, then f(2)(c) is naturally a symmetric bilinear form TcM x TcM → R.  Thus it is a map TcM→T*cM.  All I ask is that this map have zero kernel.
But if this condition is too weak, whereas a reasonable stronger condition works, I'd love to hear it.

REPLY [2 votes]: For completeness, I should post a follow-up to Chris's answer above.  Chris's argument shows that the map dp: TcC → Tp(c)B is an injection, where p is (the restriction of) the projection E → B.  But the conditions I originally proposed do not force this to be a surjection except in the finite-dimensional case (when a (co)dimension count shows that dim C is at least dim B), as the following example shows.
Let N be a compact connected manifold, B = N x N, E = Smooth Maps ( [0,1] → N ), and the projection is π(e) = (e(0),e(1)).  Pick a function L : TN → R, and define f : E → R to by f(_e_) = ∫[0,1] L(_e_'(t),e(t)) dt.  Then the set C consists of solutions to the Euler-Lagrange equations dt[\partialv L] = \partialq L.  Generically, this is a non-degenerate second-order differential equation, and so dim C = 2 dim N = dim B.  This is in fact the application I am interested in.
On the other hand, let's pick a one-form b and a function c on N, and let's suppose that the exterior derivative d_b_ is nondegenerate, so that d_b_ is a symplectic structure on N.  Let L(v,q) = _b_v + c.  Then one can check (it's straightforward) that the Euler-Lagrange equations are a non-degenerate first order differential equation on N, equivalent to the Hamilton equation for the sympectic manifold (N,d_b_) with Hamiltonian c (or perhaps -c depending on your conventions).  Thus a solution is determined by its initial location, and so dim C = dim N.
On the other hand, the Hessian H defined in Chris's answer is now a nondegenerate first-order linear differential equation, and the condition says that the only solution φ to this equation with φ(0) = 0 = φ(1) is the trivial solution φ = 0.  For a general Lagrangian, the Hessian is a second-order operator, and this condition is nontrivial, but when the Lagrangian is first-order, the Hessian necessarily has no kernel — a solution is determined by a single value.
Thus the condition that Chris thought I was imposing — that C be a manifold with dimension the same as B — cannot be dropped.
A final remark is that in finite dimensions, C is cut out by dim F = dim E - dim B equations, and so dim C is at least dim B.  The point is that this dimension count fails when dim F = ∞.<|endoftext|>
TITLE: Atiyah-MacDonald, exercise 2.11
QUESTION [29 upvotes]: Let $A$ be a commutative ring with $1$ not equal to $0$.  (The ring A is not necessarily a domain, and is not necessarily Noetherian.)  Assume we have an injective map of free $A$-modules $A^m \to A^n$.  Must we have $m \le n$?
I believe the answer is yes.  For instance, why is there no injective map from  $A^2 \to A^1$?  Say it's represented by a matrix $(a_1, a_2)$.  Then clearly $(a_2, -a_1)$ is in the kernel.  In the $A^{n+1} \to A^{n}$ case, we can look at the $n \times (n+1)$ matrix which represents it; call it $M$.  Let $M_i$ denote the determinant of the matrix obtained by deleting the $i$-th column.  Let $v$ be the vector $(M_1, -M_2, ..., (-1)^nM_{n+1})$.  Then $v$ is in the kernel of our map, because the vector $Mv^T$ has $i$-th component the determinant of the $(n+1) \times (n+1)$ matrix attained from $M$ by repeating the $i$-th row twice.
That almost finishes the proof, except it is possible that $v$ is the zero vector.  I would like to see either this argument finished, or, even better, a nicer proof.
Thank you!

REPLY [12 votes]: Every few years this question re-appears in my life (as it just did this week) and I have to rediscover the Euler Characteristic proof which I half-remember. Let me just write my thoughts down here so I don't ever have to do this again. 
Definition. Let $A$ be a non-zero commutative ring. A finite free resolution (or FFR) of an $A$-module $M$ is an exact sequence
$$0\to F_n \to F_{n-1} \to \cdots \to F_1 \to F_0 \to M\to 0$$
where each $F_i$ is a free $A$-module, of rank $r_i$. 
If $M$ has an FFR then we define the Euler Characteristic $\chi(M)$ of $M$ to be $\sum_i(-1)^ir_i$. This turns out to be a well-defined invariant of $M$ (i.e. independent of the resolution) because a relatively straightforward induction on $n$ shows that if $G_i$ is another resolution then $F_0\oplus G_1\oplus F_2\oplus\cdots\oplus (F\ \mbox{or}\ G)_n$ is isomorphic to $G_0\oplus F_1\oplus G_2\oplus\cdots$. For more details on this part of the argument see Lemma 4 in section 19 of Matsumura's book "Commutative Ring Theory".
The fundamental fact is:
Theorem. If $M$ has an FFR then $\chi(M)\geq0$.
As a consequence, not only does an injection $A^m \to A^n$ imply $m\leq n$ (apply the theorem to the cokernel), but if $0\to A^m\to A^n \to A^r$ is exact then $m+r\geq n$ and so on and so on. Do the other solutions also generalise to prove this?
The theorem is Theorem 19.7 of Matsumura's "commutative ring theory". Here's a sketch. Localising at a minimal prime one can assume that $A$ is local with nilpotent maximal ideal. It suffices to prove that in this case $M$ is free, as then the exact sequence splits and the Euler Characteristic is just the rank of $M$. It suffices to prove that the cokernel of $F_n\to F_{n-1}$ is free, because then one is done by induction on $n$. The trick now is to replace $F_n$ and $F_{n-1}$ by a minimal free resolution, look at the image of $F_n$ and see that its coordinates must lie in the maximal ideal, and it's then not difficult to concoct a non-zero element of $A$ which kills this finite set of coordinates, and hence $F_n$ is a free module annihilated by a non-zero element of $A$ and must hence be zero.
This last part of the argument feels very similar to Georges Elencwajg's argument above, which reduces to the Noetherian case and uses lengths, but Matsumura explicitly avoids this reduction.<|endoftext|>
TITLE: Can one check formal smoothness using only one-variable Artin rings?
QUESTION [9 upvotes]: Let $f:X\rightarrow Y$ be a morphism of schemes over a field $k$. Can one check that $f$ is formally smooth using only Artin rings of the form $k^{\prime}\left[t\right]/t^{n}$, where $k^{\prime}$ is also a field?
Considering cuspidal curves one can show that you do at least need arbitrarily large $n$.

REPLY [3 votes]: I'm not sure what you mean by "using" but I think the answer is no.  X can be nonempty and singular over Y with X(k) empty.<|endoftext|>
TITLE: are deformations of torsion modules always torsion?
QUESTION [5 upvotes]: Let's say I have a field $\mathbb{K}$ and a flat family of $\mathbb{K}[t]$-modules $M$ over the formal disk $Spec \mathbb{K}[[h]]$.  
Now, assume that $M/hM$ is torsion as a $\mathbb{K}[t]$-module (but NOT finitely generated).  Can I conclude that $M[h^{-1}]$ is torsion as a $\mathbb{K}((h))[t]$-module?

REPLY [4 votes]: Let M = k[[h]][x] = \bigoplus_{i=0}^{\infty}{ k[[h]]x^i }.  We make this into a flat family of k[t] modules by setting t x^i = h x^{i+1}.  Or in other words, p(h,t) \in k[[h]][t] acts by multiplication by p(h,xh) (wrt the natural ring structure on k[[h]][x]).  Clearly M/hM = k[x] with the action by t equal to zero.  Consider M[h^{-1}] = k((h))[x].  Since k((h))[x] is without zero divisors, and p(h,hx) is nonzero so long as p(h,t) was nonzero, we see that M[h^{-1}] is torsion free as a k((h))[t] module.<|endoftext|>
TITLE: Model category structures on categories of complexes in abelian categories
QUESTION [12 upvotes]: Section 2.3 of Hovey's Model Categories book defines a model category structure on Ch(R-Mod), the category of chain complexes of R-modules, where R is a ring. Lemma 2.3.6 then essentially states (I think) that taking projective resolutions of a module corresponds to taking cofibrant replacements of the module, at least in nice cases (e.g. when the projective resolution is bounded below). There is of course also a "dual" model category structure which gives the "dual" result for injective resolutions and fibrant replacements (Theorem 2.3.13).

I think the results in Hovey are proven for not-necessarily-commutative rings. Do things become nicer if we restrict our attention to commutative rings only?
Do these results generalize? For example, is there an analogous model category structure and an analogous result for Ch(OX-Mod), the category of chain complexes of OX-modules, where X is a scheme? More generally, how about for Ch(A), where A is an abelian category?

If the answers to these questions are known, then I assume they would be "standard", but I don't know a reference.

I've re-asked my question in a different form here.

REPLY [10 votes]: I don't think the existence of the dual "injective" model structure merits an "of course," since its generators are much less obvious to construct.  However, it turns out that injective model structures actually exist in more generality than projective ones, for instance they exist for most categories of sheaves.  I believe this was originally proven by Joyal, but it was put in an abstract context by Hovey and Gillespie.
The basic idea is that model structures on Ch(A) correspond to well-behaved "cotorsion pairs" on A itself.  The projective model structure comes from the (projective objects, all objects) cotorsion pair (which is well-behaved for R-modules, but not for sheaves), and the injective one comes from (all objects, injective objects).  There is also e.g. a flat model structure coming from (flat objects, cotorsion objects) which is monoidal and thus useful for deriving tensor products.  A good introduction, which I believe has references to most of the literature, is Hovey's paper Cotorsion pairs and model categories.<|endoftext|>
TITLE: When do fibre products of smooth manifolds exist?
QUESTION [17 upvotes]: Harold asks what conditions on $f:M\to L$ and $g:N\to L$, both smooth maps of smooth manifolds, ensures the existence of the fibre product $M \times_L N$.

REPLY [24 votes]: It suffices for one of the two maps to be a submersion for the fibre product to exist.  In fact, if $f$ and $g$ are maps from $X$ to $Y$, the fibre product is the inverse image of the diagonal in $Y \times Y$ under the map $f \times g : X\times X \to Y \times Y$.  So a sufficient condition to have a nice fibre product is that $f \times g$ be transverse to the diagonal.  
The next best thing to transverse intersection is clean intersection, as Ben has pointed out.  Another definition of clean intersection of $A$ and $B$ in $X$ (more easily checked than the one about the local normal form) is that the intersection of $A$ and $B$ is a submanifold $C$, and that the tangent bundle to $C$ is the intersection of the tangent bundles to $A$ and $B$.<|endoftext|>
TITLE: How do you see the genus of a curve, just looking at its function field?
QUESTION [26 upvotes]: Yuhao asked in the 20-questions seminar:
The genus of a curve is a birational invariant; the function field of a curve determines it up to birational equivelance.
How do you see the genus directly from the field?

REPLY [2 votes]: See C. Chevalley, Introduction to the theory of Algebraic functions of one variable. That book develops the whole theory of algebraic curves using just its function field.<|endoftext|>
TITLE: Is there a category in which finite limits and directed colimits *don't* commute
QUESTION [7 upvotes]: Andrew Critch asks at the 20-questions seminar:
In Set and AbGrp (the categories of sets and abelian groups, respectively), finite limits commute with directed colimits. As an example, if you're working with sheaves of sets, you can take kernels (a type of limit) and stalks (a type of directed colimit) safely.
Is there an example of a category where they don't commute?
(Depending on how you choose to talk about topoi, this condition is sometimes an axiom, capturing the idea of "looking sufficiently like sets".)

REPLY [5 votes]: Here is a non exotic example:
In the category of topological spaces, the functor
$Y \mapsto X \times Y$ does not commute with directed colimits in general.
However, if $X$ is locally compact it does.<|endoftext|>
TITLE: When does the sequence of iterates of a rational function converge?
QUESTION [9 upvotes]: Darsh asks at the 20-questions seminar:
Let $f:P^1 \rightarrow P^1$ be rational function.
Can you say when the sequence $\{ f^n(x)\}_n=\{ x,f(x),f(f(x)),\cdots\} $ converges? What about the sequence of averages $\left\{ \frac{1}{n+1}\sum_{i=0}^n f^i (x)\right\}_n$? Anything else?

REPLY [3 votes]: The question about convergence of the sequence of iterates is simple, as mentioned above: This happens if and only if the sequence converges to a fixed point of $f$. This means that the sequence converges if and only if one of the following holds:

$z$ belongs to the basin of attraction of an attracting or parabolic fixed point, or
$f^n(z)$ is a fixed point for some $z$.

Every rational map has a repelling fixed point in the Julia set. So the set of points in 2., while countable, is dense in the Julia set. 
Now let us turn to the convergence of averages. For points in the Fatou set, this average will always converge. Indeed, there are the following possibilities:

$z$ belongs to the basin of an attracting or parabolic periodic point, and convergence of the averages is trivial.
$z$ belongs to a rotation domain, where the map is analytically conjugate to an irrational rotation. It follows from the properties of rotations that the averages converge (to the integral of the conjugacy over the circle, with respect to Lebesgue measure). 

This leaves us with points in the Julia set. We already saw that there is a dense set of points where the orbit itself converges. On the other hand, the map is topologically chaotic on the Julia set, and it follows easily that there are orbits where the averages do not converge. Indeed, consider an orbit that follows one periodic orbit for a long time, then moves on to another, and switches back again. (I suppose that technically speaking I should justify that there are periodic orbits with different averages - this should not be difficult, but I won't go into details. It should be clear there is no reason to expect otherwise.) Again, once there is one such point, there is a dense set.
So we cannot expect any statement about all points in the Julia set. To make other statements, we should ask about ergodic invariant measures on the Julia set. This is a huge field, and in many cases the existence of such measures is known; then Birkhoff's ergodic theorem implies the convergence of the averages.
The simplest question one may ask is about convergence almost everywhere. It is known since work of Mary Rees in the 1980s that there is a positive measure set in the space of rational functions of any degree (at least $2$) where the map is ergodic with respect to Lebesgue measure on the sphere. So here the averages converge almost everywhere. It is reasonable to conjecture that the set of rational maps of any degree $\geq 2$ where the average converges for almost every point has full measure.
On the other hand, it seems reasonable to expect that there are cases where the Julia set has positive measure, but the averages do not exist almost everywhere. For the logistic family (and one-dimensional Lebesgue measure), I believe that it is known that there are maps without physical measures. I am not sure if it has been done for rational maps.
For quadratic polynomials, Julia sets of positive measure were proved to exist by Buff and Chéritat. It is considered likely that these maps are not ergodic with respect to Lebesgue measure, though as far as I know this has not yet been formally proved. Of course, this does not necessarily mean that the averages do not converge.<|endoftext|>
TITLE: Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1?
QUESTION [14 upvotes]: Background: the Hochschild homology of an associative algebra is the homology of the complex 
$$ \ldots \longrightarrow A \otimes A \otimes A \longrightarrow A \otimes A \longrightarrow A$$
where the last two differentials are $$a \otimes b \otimes c \mapsto ab \otimes c-a \otimes bc + ca \otimes b$$ and $a \otimes b \mapsto ab-ba$, and you can guess the rest. More generally, it's "derived coinvariants": take a projective resolution of your algebra, then coinvariants of that.
For $k[t]$, the Hochschild homology is concentrated in degrees 0 and 1, and in both of those degrees it's $k[t]$. I know that I can go look up Loday (\S 3.2.2) and find a calculation, but I'd like a better explanation.
I know that the zero-th Hochschild homology $\operatorname{HH}_0(k[t])$ must just be $k[t]$, because the zero-th Hochschild homology is just coinvariants, and $k[t]$ is commutative.
What I'd like is a "good" explanation for $\operatorname{HH}_1(k[t])$.
Edit: Ben has a simple explanation below. Let me also rephrase the question, hoping for more. Here are a few things: If $A$ is semisimple, then $\operatorname{HH}_{\ast}(A)$ is concentrated in degree 0. Is there something about $k[t]$ that ensures it's concentrated in degrees 0 and 1? Conversely, can I conclude anything about $A$ from the fact that $\operatorname{HH}_{\ast}(A)$ is zero above $\ast=1$?

REPLY [9 votes]: Is there something about k[t] that ensures it's concentrated in degrees 0 and 1.
Yes: it's a free associative algebra - that's why you only have 0th and 1st homology.<|endoftext|>
TITLE: What's the sense in which A_\infty algebras are "deformable"?
QUESTION [10 upvotes]: I realise this is a very vague question! I've heard people say that A∞ algebras are the right homotopy-theoretic generalization of usual associative algebras, because you can deform them. What exactly does this mean?
This roughly makes sense -- if you "deform" an associative algebra, it's generically going to stop being associative, but it will be "associative up to homotopy" in exactly the sense A∞ algebras are.

REPLY [3 votes]: Well here's my shot: (skip to the punchline at the bottom if you want)
Take an associative algebra A and a k-local ring R (the formal power series over k, or the infinitesimal ring will do nicely).
The algebra A is naturally a homotopy algebra and so may be given by a degree -1 square-zero coderivative on the free coassociative coalgebra on A[1].  We write this coalgebra BA, the bar resolution.  Note that in homotopy theory it often makes life easier if we forget any unit elements; BA is non-unital.
An A-infty R-deformation of A is now a square-zero coderivative on the coalgebra BA⊗R, such that the "obvious" diagram commutes (I could post this as an image when I'm permitted).  The condition could alternatively by phrased as the following:
"such that it extends the original coderivative on BA".
So far this has all been definitions, my answer to your question comes next:
Consider now the cobar functor applied to the morphism BA⊗R→BA,
Ω(BA⊗R) ≅ (ΩBA)⊗R → ΩBA.
This is a proper algebra deformation, nothing infinity about it!  Except... ΩBA is homotopy equivalent to A.
The short and snappy answer:
Infinity deformations are homotopy invariant, classical algebra deformations are not.
Edit: I should have added, if you would like me to expand on anything, I'm more than willing.<|endoftext|>
TITLE: Zeta-function regularization of determinants and traces
QUESTION [18 upvotes]: The short answer to my question may be a pointer to the right text.  I will give all the background I know, and then ask my questions in list form.
Let A be an operator (on an infinite-dimensional vector space).  You might as well assume that its spectrum is all real and positive.  In fact, I only care when the spectrum is discrete and grows polynomially, but I hear that this stuff works more generally.
In general, A is not trace-class (the sum of the eigenvalues converges) or determinant-class (the product of the eigenvalues converges) — if the nth eigenvalue grows as np for some p>0, then it won't be.  But there is a procedure to try to define a "trace" and "determinant" of A nevertheless.
Let us hope that for large enough s, the operators A-s (=exp(-s log A), and log A makes sense if the spectrum of A is positive) are trace-class.  If so, then we can define ζA(s) = tr(A-s); it is analytic for Re(s) large enough.  Let's hope that it has a single-valued meromorphic continuation and that this function (which I will also call ζA(s)) is smooth near s=0 and s=-1.  All these hopes hold when the eigenvalues of A grow polynomially, whence ζA(s) can be compared to the Riemann zeta function.
Then we can immediately define the "regularized trace" TR A = ζA(0) and the "regularized determinant" DET A = exp(-ζA'(0)), where by ζA'(s) I mean the derivative of ζA(s) with respect to s.  (If the eigenvalues λn are discrete, then ζA(s) = Σ λn-s, and so one would have TR A = Σ λn and DET A = Σ (log λn) λn-s |s=0,  if they converged.)  If A is trace- (determinant-) class, then TR A = tr A (DET A = det A).
So, here are my questions:

Is it true that exp TR A = DET exp A?
Let A(t) be a smooth family of operators (t is a real variable).  Is it true that d/dt [ log DET A(t) ] = TR( A-1 dA/dt )?  (I can prove this when A-1dA/dt is trace-class.)
Is DET multiplicative, so that DET(AB) = DET A DET B?  (I can prove this using 1. and 2., or using the part of 2. that I can prove if B is determinant-class.)
Is TR cyclic, i.e. TR(AB) = TR(BA)?
Is TR linear, i.e. TR(A + B) = TR A + TR B?

None of these are even obvious to me when A and B (or dA/dt) are simultaneously diagonalizable (except of course cyclicity), but of course in general they won't commute.

REPLY [7 votes]: I will answer some of my questions in the negative.
3.
First consider the case of rescaling an operator A by some (positive) number λ.  Then ζλA(s) = λ-sζA(s), and so TR λA = λ TR A.  This is all well and good.  How does the determinant behave?  Define the "perceived dimension" DIM A to be logλ[ (DET λA)/(DET A) ].  Then it's easy to see that DIM A = ζA(0).  What this means is that DET λA = λζA(0) DET A.
This is all well and good if the perceived dimension of a vector space does not depend on A.  Unfortunately, it does.  For example, the Hurwitz zeta functions ζ(s,μ) = Σ0∞(n+μ)-s (-μ not in N) naturally arise as the zeta functions of differential operators — e.g. as the operator x(d/dx) + μ on the space of (nice) functions on R.  One can look up the values of this function, e.g. in Elizalde, et al.  In particular, ζ(0,μ) = 1/2 - μ.  Thus, let A and B be two such operators, with ζA = ζ(s,α) and ζB = ζ(s,β).  For generic α and β, and provided A and B commute (e.g. for the suggested differential operators), then DET AB exists.  But if DET were multiplicative, then:
DET(λAB) = DET(λA) DET(B) = λ1/2 - α DET A DET B
but a similar calculation would yield λ1/2 - β DET A DET B.
This proves that DET is not multiplicative.
1.
My negative answer to 1. is not quite as satisfying, but it's OK.  Consider an operator A (e.g. x(d/dx)+1) with eigenvalues 1,2,..., and so zeta function the Reimann function ζ(s).  Then TR A = ζ(-1) = -1/12.  On the other hand, exp A has eigenvalues e, e2, etc., and so zeta function ζexp A(s) = Σ e-ns = e-s/(1 - e-s) = 1/(es-1).  This has a pole at s=0, and so DET exp A = lims→0 es/(es-1)2 = ∞.  So question 1. is hopeless in the sense that A might be zeta-function regularizable but exp A not.  I don't have a counterexample when all the zeta functions give finite values.
5.
As in my answer to 3. above, I will continue to consider the Hurwitz function ζ(s,a) = Σn=0∞ (n+_a_)-s, which is the zeta function corresponding, for example, to the operator x(d/dx)+a, and we consider the case when a is not a nonpositive integer.  One can look up various special values of (the analytic continuation) of the Hurwitz function, e.g. ζ(-m,a) = -Bm+1(a)/(m+1), where Br is the _r_th Bernoulli polynomial.
In particular,
TR(x(d/dx)+a) = -ζ(-1,a)/2 = -a2/2 + a/2 - 1/12
since, for example (from Wikipedia):
B2(a) = Σn=02 1/(n+1) Σk=0 n (-1)k {n \choose k} (a+_k_)2 = a2 - a + 1/6
Thus, consider the operator 2_x_(d/dx)+a+_b_.  On the one hand:
TR(x(d/dx)+a) + TR(x(d/dx)+b) = -(a2+b2)/2 + (a+_b_)/2 - 1/6
On the other hand, TR is "linear" when it comes to multiplication by positive reals, and so:
TR(2_x_(d/dx)+a+_b_) = 2 TR(x(d/dx) + (a+_b_)/2) =  -(a2+2_ab_+b2)/4 + (a+_b_)/2 - 1/6
In particular, we have TR(x(d/dx)+a) + TR(x(d/dx)+b) = TR( x(d/dx)+a + x(d/dx)+b ) if and only if a=_b_; otherwise 2_ab_ < a2+b2 is a strict inequality.
So the zeta-function regularized trace TR is not linear.
0./2.
My last comment is not so much to break number 2. above, but to suggest that it is limited in scope.  In particular, for an operator A on an infinite-dimensional vector space, it is impossible for A-s to be trace-class for s in an open neighborhood of 0, and so if the zeta-function regularized DET makes sense, then det doesn't.  (I.e. it's hopeless to say that det A = DET A.)  Indeed, if the series converges for s=0, then it must be a finite sum.
Similarly, it is impossible for A to be trace class and also for A-s to be trace class for large s.  If A is trace class, then its eigenvalues have finite sum, and in particular cluster near 0 (by the "divergence test" from freshman calculus).  But then the eigenvalues of A-s tend to ∞ for positive s.  I.e. it's hopeless to say that tr A = TR A.
My proof for 2. says the following.  Suppose that dA/dt A-1 is trace class, and suppose that DET A makes sense as above.  Then
d/dt [ DET A ] = (DET A)(tr dA/dt A-1)
I have no idea what happens, or even how to attack the problem, when dA/dt A-1 has a zeta-function-regularized trace.<|endoftext|>
TITLE: Definition of Hochschild (co)homology of a (dg or A-infinity) category
QUESTION [12 upvotes]: How do you define the Hochschild (co)homology of a dg category or an A-infinity category? I've only seen it defined when the category is equivalent to a category of modules over a dg algebra; then the Hochschild (co)homology is just that of the algebra. But more generally?

REPLY [3 votes]: Maybe this answer comes five years too late, but I have recently put an elementary construction of the Hochschild (co)homology of $A_\infty$-algebras on the arXiv:
http://arxiv.org/abs/1601.03963<|endoftext|>
TITLE: Ribbon graph decomposition of the moduli space of curves
QUESTION [11 upvotes]: What is a ribbon graph? What is the ribbon graph decomposition of the moduli space of curves? What are some good references for this material?

REPLY [3 votes]: Kiyoshi Igusa constructs the category of fat graphs (ribbon graphs) both in his paper Graph cohomology and Kontsevich cycles and his book Higher Franz-Reidemeister Torsion and are excellent references, containing lots of details in a clearly-written manner. There you can find, for example, a proof of the fact that the homology of the category of fat graphs is rationally isomorphic to the homology of the mappping class groups of marked surfaces. Details relating ribbon graphs to the Miller-Morita-Mumford classes are found there, too.<|endoftext|>
TITLE: When is fiber dimension upper semi-continuous?
QUESTION [28 upvotes]: Suppose $f\colon X \to Y $ is a morphism of schemes. We can define a function on the topological space $Y$ by sending $y\in Y$ to the dimension of the fiber of $f$ over $y$.

When is this function upper semi-continuous?

I have the following "concrete" application in mind. If an algebraic group $G$ acts on a scheme $X$, I'm pretty sure the stabilizer dimension is an upper semi-continuous function on $X$ (i.e. it can jump up on closed sub-schemes), but I don't know a proof. The stabilizers of points are the fibers of the map $\text{Stab}\to X$ in the following Cartesian square:
\begin{equation}
\require{AMScd}
\begin{CD}
\text{Stab} @>>> G \times X \\
@VVV @VV{\alpha}V \\
X @>{\Delta}>> X \times X.
\end{CD}
\end{equation}
where $\alpha\colon G\times X\to X\times X $ is given by $(g,x) \mapsto (g\cdot x,x)$, and $\Delta\colon X\to X\times X $ is the diagonal map $x\mapsto (x,x)$. It would be nice to have a condition satisfied by $\alpha\colon G\times X \to X\times X$ that would guarantee the upper semi-continuity of fiber dimension.

REPLY [21 votes]: I just discovered that Shavarevich (second edition) has a wrong answer to this question. In Section I.6.3, after Theorem 7 (which is correct), he gives the following Corollary. This quotation combines the Theorem and the Corollary.

Let $f: X \to Y$ be a regular map between irreducible varieties. Suppose that $f$ is surjective ... The sets $Y_k := \{ y \in Y: \dim f^{-1}(y) \geq k \}$ are closed in $Y$.

Note that this differs from the true EGA IV 13.1.5 by replacing "closed" with "surjective". I figured I'd record a counterexample here, which is slightly more public than just creating a handout for my class.
Our map is a composition $X \subset \mathbb{A}^3 \to \mathbb{A}^3 \to Y \subset \mathbb{A}^4$. We'll call the two $\mathbb{A}^3$'s $A$ and $B$ respectively.
$X$ is the quasi-affine variety $A \setminus \{ (0,\ast,0) \}$. We map $A \to B$ by $(x,y,z) \mapsto (x, xy, z)$. We map the $B$ to $\mathbb{A}^4$ by $(p,q,r) \mapsto (p(p-1), p^2(p-1), q,r)$. $Y$ is the affine variety $\{ (a,b,c,d) : a^3 = b(b-a) \}$. In other words, $Y$ is the product of a nodal cubic with $\mathbb{A}^2$.
To see surjectivity, note that $X$ hits every point of $B$ where $p$ is nonzero. The points of $B$ where $p \neq 0$ map to the points of $Y$ where $(a,b) \neq (0,0)$. The points $(0,0,c,d)$ in $Y$ are the images of $(1,c,d) \in B$, which are in turn the images of $(1,c,d)$ in $X$.
Now, let's look at $\dim f^{-1}(0,0,0,r)$. When $r \neq 0$, this is the union of $(1,0,r)$ and $(0, \ast, r)$, so one dimensional. When $r = 0$, the line $(0, \ast, 0)$ is deleted, so the preimage is only a point.
This suggests that something nicer may happen if we insist that fibers are irreducible, or that $Y$ is normal (perhaps Zariski's Main Theorem gets involved?) but I don't have a proposed statement to make.<|endoftext|>
TITLE: Is there an example of a formally smooth morphism which is not smooth?
QUESTION [15 upvotes]: A morphism of schemes is formally smooth and locally of finite presentation iff it is smooth. 
What happens if we drop the finitely presented hypothesis? Of course, locally of finite presentation is part of smoothness, so implicilty I am asking for the flatness to fail.

REPLY [26 votes]: Here's an elementary example. For any field $k$, consider the ring $k[t^q|q\in\mathbb Q_{>0}]$, which I'll abbreviate $k[t^q]$. I claim that the natural quotient $k[t^q]\to k$ given by sending $t^q$ to $0$ is formally smooth but not flat, and therefore not smooth.
First let's show it's formally smooth. Let $A$ be a ring with square-zero ideal $I\subseteq A$, and suppose we have maps $f:k[t^q]\to A$ and $g:k\to A/I$ making the following square commute (I drew it backwards because you're probably thinking of Spec of everything)
$$
\begin{array}{ccc}
 A/I & \xleftarrow g & k \\
 \uparrow & & \uparrow\\
 A & \xleftarrow f & k[t^q]
\end{array}
$$
We'd like to show that there's a map $k\to A$ filling the diagram in. For any $q\in \mathbb Q_{>0}$, note that $f(t^q)\in I$ by commutativity of the square, so $f(t^{2q})\in I^2=0$. But every $q$ is of the form $2q'$ for some $q'$, so we've shown that $f(t^q)=0$ for all $q\in \mathbb Q_{>0}$. So $f$ factors through $k$, as desired.
Now let's show that $k$ is not flat over $k[t^q]$. Consider the exact sequence 
$$0\to (t)\to k[t^q]\to k[t^q]/(t)\to 0.$$
When you tensor with $k$, you get 
$$0\to k\to k\to k\to 0,$$
which is obviously not exact. So $k$ is not flat over $k[t^q]$.<|endoftext|>
TITLE: How do you show that $S^{\infty}$ is contractible?
QUESTION [42 upvotes]: Here I mean the version with all but finitely many components zero.

REPLY [8 votes]: A very nice proof is via classifying spaces of categories. It goes as follows: Take the “walking isomorphism“ category $J$, that is the unique category with two isomorphic objects and four morphisms in total. The geometric realization of its simplicial nerve is exactly the infinity-sphere! (The non-degenerate $n$-simplices are given by the functors $[n] \to J$ starting at either $0$ or $1$ and then going back and forth using the isomorphism to produce upper and lower hemisphere of the standard cell-decomposition.) Then you conclude by observing that this category is equivalent to the terminal category and thus, as nerve and realization send natural transformations to simplicial, respectively topological homotopies, $S^\infty$ is contractible.<|endoftext|>
TITLE: Is there an example of a variety over the complex numbers with no embedding into a smooth variety?
QUESTION [17 upvotes]: Is there an example of a variety over the complex numbers with no embedding into a smooth variety?

REPLY [8 votes]: A really neat well known example is as follows: 
Choose a conic $C_1$ and a tangential line $C_2$ in $\mathbb{P}^2$ and asssociate to a point $P$ on $C_1$ the point of intersection $Q$ of $C_2$ and the tangent line to $C_1$ at $P$. This gives a birational isomorphism from $C_1$ to $C_2$. Identify the curves by this map to get the quotient variety $\phi:\mathbb{P}^2\rightarrow{X}$ with $C:=\phi(C_1)$. Now if there was an embedding of $X$ in a smooth scheme then, there would surely exist an effective line bundle on $X$, say $L$ whose pull back to $\mathbb{P}^2$ will obviously be effective. Let us see why this is a contradiction. Let $L'$ be the pullback of $L$ to $\mathbb{P}^2$. Note that the degrees of $L'|C_1$ and $L'|C_2$ both coincide with the degree of $L|C$ and are therefore equal. But $L'\cong\mathcal{O}(k)$ and therefore the degrees in question are $2k$ and $k$ respectively for $C_1$ and $C_2$. Therefore $k=0$ and $L'\cong\mathcal{O}$, which is non-effective! A contradiction!     
In view of VA's comment, I give a complete proof here for constructing $X$ as a scheme.
In our special case it is a trivial pushout construction: Here I am thinking of $Y$ as $C_1\amalg{C_2}$, $Y'=C$ (the quotient by the birational isomorphism above), and $Z=\mathbb{P}^2$, but the argument is more general provided any finite set of closed points in $Z$ is contained in an affine open set. $X$ will denote the quotient.  
Claim: Suppose $j:Y\rightarrow{Z}$ is a closed subscheme of a scheme $Z$, and $g:Y\rightarrow{Y'}$ is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram (which I don't know how to draw here, but one visualize it easily): $Y\xrightarrow{j}{Z}$, $Y\xrightarrow{g}Y'$, $Y'\rightarrow{X}$, $Z\xrightarrow{h}X$ where $X$ is a scheme, $h$ is finite and induces monomorphisms on coordinate rings and $Y'\rightarrow{X}$ is a closed immersion. 
Proof: First assume that $Z$ is affine, in which case $Y,Y'$ are both affine too. Let $A,A/I,B$ be their respective coordinate rings. Then $B\subset{A/I}$ in a natural way. Let's use $j$ again to denote the natural map $A\rightarrow{A/I}$. Put $A'=j^{-1}(B)$ and $Spec(A')=X$. The claim is clear for $X$. Also if $Z$ is replaced by an open subset $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$, $X$ would be replaced by $U'=h(U)$ which is an open subset. 
Now this guarantees the existence of $X$ once it has been shown that $Z$ can be covered by affine open subsets $U$ such that $g^{-1}g(U\cap{Y})=U\cap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of $X$ that it is actually reduced and irreducible. QED. 
I hope this is satisfactory.<|endoftext|>
TITLE: Can a quotient ring R/J ever be flat over R?
QUESTION [25 upvotes]: If R is a ring and J⊂R is an ideal, can R/J ever be a flat R-module? For algebraic geometers, the question is "can a closed immersion ever be flat?"
The answer is yes: take J=0. For a less trivial example, take R=R1⊕R2 and J=R1, then R/J is flat over R. Geometrically, this is the inclusion of a connected component, which is kind of cheating. If I add the hypotheses that R has no idempotents (i.e. Spec(R) is connected) and J≠0, can R/J ever be flat over R?
I think the answer is no, but I don't know how to prove it. Here's a failed attempt. Consider the exact sequence 0→J→R→R/J→0. When you tensor with R/J, you get

0→ J/J2→R/J→R/J→0

where the map R/J→R/J is the identity map. If J≠J2, this sequence is not exact, contradicting flatness of R/J.
But sometimes it happens that J=J2, like the case of the maximal ideal of the ring k[tq| q∈Q>0]. I can show that the quotient is not flat in that case (see this answer), but I had to do something clever.
I usually think about commutative rings, but if you have a non-commutative example, I'd love to see it.

REPLY [10 votes]: Just a small point.  There are rings over which every module is flat.  These are the absolutely flat rings in the checked answer above, but they are also called von Neumann regular rings.  An infinite product of fields is the standard example of a commutative von Neumann regular ring that is not semisimple.  But there are lots and lots of noncommutative von Neumann regular rings.   Note that a Noetherian von Neumann regular ring is semisimple, but every von Neumann regular ring is coherent.<|endoftext|>
TITLE: Questions about Quivers
QUESTION [5 upvotes]: Hi,
The definition I have for a Path Algega of a quiver Q is that it is the algebra whose basis is formed by the oriented paths in Q, including the trivial ones.  Apparently multiplication is given by concatenation of paths, and those that can't be concatenated are considered zero.  That part I think I understand, but I am wondering if there is a non-formal interpretation of the addition operation on two elements of a path algebra someone could provide?

REPLY [7 votes]: Maybe it is useful to add that a path algebra is a special case of a tensor algebra of a bimodule. Namely, the path algebra of a finite quiver $Q$ is the tensor algebra $T_RV$ of a finite dimensional bimodule $V=V_Q$ over a split semisimple finite dimensional commutative algebra $R=\oplus_{i\in I}k$ over a field $k$. More precisely, if $M_{ij}$ are the simple (1-dimensional) $R$-bimodules, then $V_Q=\oplus n_{ij}(Q)M_{ij}$, where $n_{ij}(Q)$ is the number of edges in $Q$ going from $i$ to $j$.<|endoftext|>
TITLE: Equivalent statements of the Riemann hypothesis in the Weil conjectures
QUESTION [18 upvotes]: In the cohomological incarnation, the Riemann hypothesis part of the Weil conjectures for a smooth proper scheme of finite type over a finite field with $q$ elements says that: the eigenvalues of Frobenius acting on the $H^i_\mathrm{et}(X, Q_\ell)$ are algebraic integers with complex absolute value $q^{i/2}$.
For smooth proper curve $ C$ over $F_q$ of genus g, the Riemann hypothesis is often stated in another way as $|N- q - 1| \leq 2g \sqrt(q)$  where $N$ is the number of rational points on $ C$.
Why are these two statements equivalent?  Is (are) there any corresponding inequality(ies) which are equivalent to the Riemann hypothesis in higher dimensions?

REPLY [6 votes]: Regarding the question about higher dimensions, Scholl showed that RH in all dimensions follows from the statement:

If $X$ is a smooth hypersurface in $\mathbb{P}^{d+1}$ over $\mathbb{F}_q$, then $X(\mathbb{F}_{q^n}) - \mathbb{P}^d(\mathbb{F}_{q^n}) = O(q^{dn/2})$.

Here the $O(\ )$ is as $n \to \infty$, with the constant depending on $X$.<|endoftext|>
TITLE: Is the long line paracompact?
QUESTION [10 upvotes]: A manifold is usually defined as a second-countable hausdorff topological space which is locally homeomorphic to Rn. My understanding is that the reason "second-countable" is part of the definition is to make sure that the space is paracompact, which you want so that you get locally finite partitions of unity. Once you have locally finite partitions of unity, basically anything you can multiply by a function can be constructed locally (any presheaf that is a module over the sheaf of functions is automatically a sheaf), a property you want manifolds to have.
But do we unnecessarily throw out some paracompact topological spaces which "should" be manifolds by requiring second-countability. A boring example is an uncountable disjoint union of manifolds, but there are other more interesting spaces that kind of look like they should be manifolds.
In particular, is the long line paracompact? Should I consider it a manifold?

REPLY [3 votes]: Let me give another proof that the long line is not paracompact. A theorem of Tamano states that a completely regular space $X$ is paracompact if and only if $X\times\beta X$ is normal where $\beta X$ denotes the Stone-Cech compactification of $X$.
Suppose that $L$ is the long line. To make our lives easier, let $X=L\cup\{0\}$ be the totally ordered set with least element $0$. Then $X$ is a manifold with boundary $0$. Then $\beta X=X\cup\{\infty\}$ where $\infty\not\in X$ and $X\cup\{\infty\}$ is given the one-point compactification topology.
Then $X\times\beta X=X\times(X\cup\{\infty\})$. Let $C=\{(x,x)|x\in X\}$ and let $D=\{(x,\infty)|x\in X\}$. Then by an easy application of Fodor's lemma, it can be shown that if $U,V$ are open sets with $C\subseteq U,D\subseteq V$, then $U\cap V\neq\emptyset$. Therefore $X\times\beta X$ is not normal, so $X$ is not paracompact.
Before I go, I should mention that the result and its proof stating that every paracompact connected manifold is second countable follows from the more general fact that a locally compact space is paracompact if and only if it is the disjoint union of open $\sigma$-compact sets. In particular, every locally compact connected space is paracompact if and only if it is $\sigma$-compact.<|endoftext|>
TITLE: How does one think about the "off-diagonal" part of the $R$-matrix?
QUESTION [5 upvotes]: The universal $R$-matrix of a quantized universal enveloping algebra is typically written as the product of two terms, one only involving elements of the Cartan, and one only involving elements of the upper and lower triangular pieces of $U_q(g)$.
How do I think about commuting these individual pieces past things like coproducts?  Are calculations like this written somewhere?

REPLY [4 votes]: Another way to think about this, which I learned from Mark Haiman's class, is that there are actually $4$ natural coproducts on the quantum group.  $\Delta\left(E\right)$ could be $E\otimes K + 1 \otimes E$, or you could move the $K$ to the other factor, or you could invert $K$, or you could do both.  The individual parts of the $R$-matrix move between these four different coproducts.  In particular, the quasi-$R$-matrix (iirc) turns the $K$ into a $K^{-1}$.<|endoftext|>
TITLE: A $k$-component link defines a map $T^{k}\rightarrow \mathrm{Conf}_{k} S^{3}$. Does the homotopy type capture Milnor's invariants?
QUESTION [8 upvotes]: A $k$-component link defines a map $T^{k}\rightarrow \mathrm{Conf}_{k} S^{3}$. Does the homotopy  type of this map capture the Milnor invariants?
Some special cases:

$k=2$, no, it's null homologous, but you can look instead at the map $T^{2}\rightarrow \mathrm{Conf}_{2} R^{3}$, which captures  linking number.
$k=3$, Melvin et al. proved it does.

REPLY [8 votes]: It's been a while since I've thought about this but I think Koschorke answered much of your question back in 1997  "A generalization of Milnor's mu-invariants to higher-dimensional link maps" Topology 36 (1997), no 2. 301--324.  Scanning through the paper I see he recovers many of the mu invariants but not all.  He lists it as an open question (6.3) if the homotopy class of the map T^k --> C_k R^3 is a complete link homotopy invariant of the link.
Related: 
Brian Munson put these Koschorke "linking maps" into the context of the Goodwillie calculus in a recent arXiv paper.  I've wondered for a while if you could use these types of maps to create a direct construction of the Cohen-Wu correspondence between the homotopy groups of S^2 and their corresponding simplicial quotient object made from the brunnian braid groups.<|endoftext|>
TITLE: Has anyone tabulated 2-knots? Would anyone like to try?
QUESTION [10 upvotes]: I'd love to have a list of 'small' 2-knots, for some sense of small. It's not clear what one should filter by, but there are two obvious candidates

Write a movie presentation, and count the frames.
Project the 2-knot to R^3, and count the triple points.

Does anyone know if this has been attempted? Such a list could be quite useful.

REPLY [6 votes]: Another complexity filter could be the number of 4-dimensional simplices in a triangulation of the complement.
2-knots complements have simple ideal triangulations.  The proof is very similar to the proof for 1-knots, viewed through the lens of Morse theory on manifolds with boundary.  ie: the Wirtinger presentation is viewed as the fundamental group computation coming from the cell decomposition of a Morse height function (in the stratified sense) on the knot complement. 
It'd be nice if there was a SnapPea-like algorithm for triangulating 2-knot complements.  As far as I know those kinds of details haven't been worked out by anyone.  But it should be reasonably doable.<|endoftext|>
TITLE: Why and how are moduli spaces of (semi)stable vector bundles well-behaved?
QUESTION [16 upvotes]: The slope of a vector bundle $E$ is defined as $\mu(E) = \deg(E)/\mathrm{rank}(E)$. Then a vector bundle $E$ is called semistable if $\mu(E') \leqslant \mu(E)$ for all proper sub-bundles $E'$. It is called stable if $\mu(E') < \mu(E)$.
I've heard that moduli spaces of stable and semistable vector bundles are somehow well-behaved, but I don't know why this is, nor do I know exactly what well-behaved should mean in this context. What goes wrong if we try to consider moduli of more general vector bundles? Moreover the definitions of slope and (semi)stable seem a bit artificial -- where do they come from?
I've also only seen the above definitions made in the context of vector bundles over curves. Why just curves? Does something stop working in higher dimensions or in greater generality?

REPLY [8 votes]: From a topological viewpoint, I believe the idea is that one wants to have a Hausdorff quotient space.  In other words, consider the space of all holomorphic structures on a fixed (topological) vector bundle on a curve.  Holomorphic structures can be viewed as differential operators on sections of the bundle, such that a section is holomorphic if and only if this operator evaluates to zero on the section.  (See, for example, sections 5 and 7 of Atiyah and Bott's "The Yang-Mills equations over Riemann surfaces.")  This makes the space of holomorphic structures (i.e. the space of bundles with a fixed topological type) into an affine space.  The group of complex automorphisms of the bundle acts on this space, and the quotient is the moduli space of holomorphic bundles.  If you don't restrict to stable bundles, this quotient space fails to be Hausdorff.  Atiyah and Bott reference this to Mumford's 1965 GIT book.  Actually, they just say that the moduli space of stable bundles is Hausdorff, due to the fact that the orbits of stable bundles are closed.  (Hmmm... that really just says points are closed in the quotient...)  I don't know how much of this is spelled out in Mumford; in particular, I don't know whether there's a proof in the literature that the full quotient space fails to be Hausdorff.<|endoftext|>
TITLE: Compact Kaehler manifolds that are isomorphic as symplectic manifolds but not as complex manifolds (and vice-versa)
QUESTION [10 upvotes]: What are some examples of compact Kaehler manifolds (or smooth complex projective varieties) that are not isomorphic as complex manifolds (or as varieties), but are isomorphic as symplectic manifolds (with the symplectic structure induced from the Kaehler structure)? Elliptic curves should be an example, but I can't think of any others. I'm sure there should be lots...
In the other direction, if I have two compact Kaehler manifolds (or smooth complex projective varieties) that are isomorphic as complex manifolds (or as varieties), then are they necessarily isomorphic as symplectic manifolds?
And one last question that just came to mind: If two smooth complex (projective, if need be) varieties are isomorphic as complex manifolds, then they are isomorphic as varieties?

REPLY [4 votes]: If $M \to X$ is smooth and proper, and $M$ is K\"ahler, then the fibers are all symplectomorphic. (Proof: the Levi-Civita connection generates symplectomorphisms.) The family of elliptic curves was already mentioned, but another interesting one has every general fiber being $F_0$ and the special fiber $F_2$ (Hirzebruch surfaces).
A curious example is the family $\{ xy = t \}$ of hypersurfaces in ${\mathbb C}^2$ as $t$ varies (away from $0$). There, the fibers are all holomorphic, and symplectomorphic, but not by the same diffeomorphism (their unique closed geodesics are of varying length).<|endoftext|>
TITLE: When are Hilbert schemes smooth?
QUESTION [13 upvotes]: I know that Hilbert schemes can be very singular. But are there any interesting and nontrivial Hilbert schemes that are smooth? Are there any necessary conditions or sufficient conditions for a Hilbert scheme to be smooth?

REPLY [4 votes]: For some more examples of smooth HS see A.P.Staal: The ubiquity of smooth Hilbert schemes, arxiv AG 31.Jan. 2017.<|endoftext|>
TITLE: Are abelian non-degenerate tensor categories semisimple?
QUESTION [7 upvotes]: A pivotal monoidal category is called non-degenerate if the inner product $\left(x,y\right) = Tr\left(xy^{*}\right)$ (where $y^{*}$ is the dual map) is non-degenerate.  As a rule of thumb non-degenerate is closely related to semisimplicity.  For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension $0$).  Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.
If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

REPLY [2 votes]: I believe this is Proposition 5.7 in Deligne's “La Categorie des Representations du Groupe Symetrique S_t, lorsque t n’est pas un Entier Naturel”. See also this question.<|endoftext|>
TITLE: does a line bundle always have a degree
QUESTION [19 upvotes]: For curves there is a very simple notion of degree of a line bundle or equivalently of a Weil or Cartier divisor.  Even in any projective space $\mathbb P(V)$ divisors are cut out by hypersurfaces which are homogeneous polynomials of a certain degree.  
Is there a more general notion of degree that applies to schemes with less structure?
Also, say you have a nice enough scheme $X$ so line bundles correspond to Cartier divisors under linear equivalence.  In whatever the most general setting is so that the degree of a line bundle makes sense, is there an example of a line bundle $L \ne O_X$ that is degree 0 and has $h^0(L$) = 1?

REPLY [3 votes]: I have a silly answer to your second question.  Take a disjoint union of an elliptic curve with any other curve, and set L to be a nontrivial degree 0 bundle on the elliptic curve and trivial on the other curve.<|endoftext|>
TITLE: Non-zero sheaf cohomology
QUESTION [5 upvotes]: Let $\mathbb{R}$ denote the real line with its usual topology. Does there exist a sheaf $F$ of abelian groups on $\mathbb{R}$ whose second cohomology group $H^{2}\left(\mathbb{R},F\right)$ is non-zero? What about $H^{j}\left(\mathbb{R},F\right)$ for integers $j\ge 2$ ?
(Here cohomology means derived functor cohomology as in, say, Hartshorne or EGA. Anyway this cohomology coincides with Cech cohomology since $\mathbb{R}$ is paracompact.)

REPLY [10 votes]: The sheaf cohomology Hi(X,F) of a (topological) manifold X of dimension n vanishes for i > n. This is a topological version of Grothendieck's vanishing theorem above. You can find this result in Kashiwara-Schapira's "Sheaves on manifolds" proposition III.3.2.2.<|endoftext|>
TITLE: Questions about ordering of reals and irrationals
QUESTION [12 upvotes]: Three problems from G.Rosenstein "Linear orderings" (from the end of Chapter 2 and beginning of Chapter 4):
1) Is there a nondecreasing function from irrationals onto reals?
2) Is there a nondecreasing function from reals onto irrationals?
3) Is there an increasing function from reals into irrationals? (In other words, are reals a subordering of irrationals?)
Any hints would be appreciated.
(Please tag the question set-theory order-theory)

REPLY [13 votes]: Here is a simple proof of (3), without continued fractions.
For any real number x, let f(x) be the real obtained by interleaving the binary digits of x with the binary digits of pi, a fixed irrational number. This is clearly order-preserving, and f(x) is irrational since the digits will not repeat. QED<|endoftext|>
TITLE: What is a TMF in topology?
QUESTION [22 upvotes]: What is a topological modular form? How are they related to 'normal' (number-theoretic) modular forms?

REPLY [18 votes]: If you know about classical modular forms, and you want to understand what they have to do with tmf, it is good to contemplate something more classical: the "derived functors of modular forms".
Modular forms of weight $n$ are $H^0(M, \omega^n)$, where $\omega$ is a certain line bundle over $M=$compactified moduli stack of elliptic curves.  There is non-trivial cohomology in
$H^s(M,\omega^n)$     
for $s>0$.  This comes in two flavors:
* free abelian group summands in $H^1(M,\omega^n)$ for $n\leq-10$.  These correspond by a kind of "Serre duality" to the usual modular forms in $H^0(M,\omega^{-n-10})$.
* finite abelian groups (killed by multiplication by $24$) in $H^s(M,\omega^n)$ for arbitrarily large $s$.
There is a spectral sequence 
$H^s(M,\omega^{t/2}) \Rightarrow \pi_{t-s} Tmf$,
and the edge of the spectral sequence gives a map $\pi_{2n}Tmf \rightarrow H^0(M,\omega^n)$.  This is essentially the map Chris describes in his answer.  
Slightly confusingly, the gadget I've called Tmf (which is the global sections of a sheaf of spectra over $M$) is not the same as TMF (the $576$-periodic guy, which is sections on $M'=$stack of smooth elliptic curves), or tmf (the connective cover of Tmf, which I don't believe is known to be sections on anything).  
I recommend Goerss's Bourbaki talk for learning more about this, especially section 4.6.<|endoftext|>
TITLE: What is the Theorem of the Cube?
QUESTION [10 upvotes]: What is the "theorem of the cube" for abelian varieties? What is the statement and how should I think about it?

REPLY [9 votes]: I'm a bit late to the party, but since these question are clearly still getting views, I'll answer the second question a bit. Given a nice category, one can form the pointed category $C$, and consider functors $F:C\to Ab$. There are cannonical maps $\beta:F(X_0\times\dots \times X_n)\to \prod_i F(X_0\times \dots \times X_{i-1}\times X_{i+1}\times \dots \times X_n)$ and $\alpha:\prod_i F(X_0\times \dots \times X_{i-1}\times X_{i+1}\times \dots \times X_n)\to F(X_0\times\dots \times X_n)$. We have $F(X_0\times\dots \times X_n)=Ker(\beta)\oplus Im(\alpha)$, and we say that $F$ is of order $n$ if either $Ker(\beta)=0$ or alternatively, if $\alpha$ is surjective. Now if you have an exact sequeunce of functors, $T_1\to T_2\to T_3$, and $T_1, T_3$ are of order $n$, then $T_2$ is by the Snake Lemma. Additionally, if $T$ is of order $n$, then it is of order $m$ for $m>n$. Lastly, we note that $H^n(X; \mathcal{F})$ is of order $n$ by the Kunneth theorem.
Now note that the Theorem of a Cube is just the statement that $Pic$ is of order $2$. Now in the complex case, the exponential exact sequence $0\to \mathbb{Z}\to \mathcal{O}_X\to \mathcal{O}_{X}^*\to 0$ gives an exact sequence $H^1(X; \mathbb{Z})\to H^1(X; \mathcal{O}_{X}^*)\to H^2(X;\mathcal{O}_X)$. Now the left-hand and right-hand term are both of order $2$, so we have derived the theorem of the cube in the complex case.
So when thinking of the theorem of the cube, I usually think that it expresses the fact that $\mathcal{O}_X^*$ is resolved with two sheaves, in some sense, and this generalizes to the non-complex case in spirit.<|endoftext|>
TITLE: Elliptic curve over spectra?
QUESTION [11 upvotes]: Filling the gaps in my knowledge to understand the tmf question.
So, what is the analogue of elliptic curve over the category of spectra?

REPLY [9 votes]: First you should know what a "derived scheme" (or "spectral scheme") is. Roughly, this is the same as an ordinary scheme, except instead of locally being the Spec of an ordinary commutative ring, it's locally the Spec of an E-infinity ring spectrum, or just E-infinity ring for short. An E-infinity ring is not the same as a ring spectrum; a ring spectrum is something that is "a ring up to homotopy", and an E-infinity ring is something that is "a ring up to coherent homotopy".
As a topological space, Spec of an E-infinity ring A is defined to be the Spec of pi0(A), which is an ordinary commutative ring. The difference is that the sheaf of functions is no longer a sheaf of rings but a sheaf of E-infinity rings. This sheaf of E-infinity rings is (analogously to the structure sheaf for ordinary affine schemes) defined by Uf -> A[f-1], where Uf is a distinguished open subset, and the localization A[f-1] is now taken in the category (or rather infinity-category) of E-infinity rings (see section 2.2 of Lurie's survey for a characterization of this localization).
There is a natural functor from derived schemes to ordinary schemes. It is (X, OX) -> (X, pi0(OX)), where pi0(OX) denotes the sheafification of the presheaf U -> pi0(OX(U)).
Then Definition 4.1 of Lurie's survey article defines an elliptic curve over an E-infinity ring A: it is a commutative A-group E -> Spec A such that (E, pi0(OE)) -> Spec pi0(A) is an ordinary elliptic curve over pi0(A).
I think one of the punchlines is that there is a derived Deligne-Mumford moduli stack of oriented derived elliptic curves, which becomes the ordinary Deligne-Mumford moduli stack of ordinary elliptic curves after hitting it with pi0. Taking global sections on this derived moduli stack is more or less tmf. I don't have a good short explanation of what an orientation is, but it's in the beginning of section 3 of the survey.
Unfortunately I think a lot of the details of this stuff are still not available. (I think it is supposed to be in "DAG VII: Spectral Schemes"; there may be some hints and related material in "DAG V: Structured Spaces".)<|endoftext|>
TITLE: Beilinson-Bernstein and Koszul duality
QUESTION [7 upvotes]: For geometric representation theorists down here.
Consider the Beilinson-Bernstein theorem: 

Functor of global sections establishes
  the correspondence between twisted
  D-modules with fixed twist θ on the
  flag variety and g-representations
  with fixed central character. These
  are modules over the same algebra D[θ]
  = U /(Z −χ). This correspondence respects the  structure of abelian
  category. It takes K-equivariant
  D-modules to (g, K)-admissible
  modules.

Why do people refer to its derived version as the Koszul duality?
How is this related to Soergel's conjecture?

REPLY [5 votes]: Some general discussions related to Soergel's conjecture and Beilinson-Bernstein can also be found in the introductions to my two papers with David Nadler on the arXiv,  0706.0322 (which is about to be split into two much improved papers) and 0904.1247.<|endoftext|>
TITLE: Are there interesting monoidal structures on representations of quantum affine algebras?
QUESTION [6 upvotes]: Is there a good monoidal structure on a category of integrable representations of a quantum affine algebra?  In the ordinary affine Kac-Moody case, there is the usual tensor product (symmetric, adds charges) and a fusion structure (braided, comes from G-bundles on curves, preserves central charge).  In the quantum case, there is the usual tensor product (braided meromorphic-braided$^\ast$), but all I see in the literature about fusion is vague comments that it can't exist.  I guess my question should be "what is the major malfunction?"
$^\ast$ Edit: The meromorphic property (in the sense of Soibelman's Meromorphic tensor categories) seems to be a first hint at problems, and I should have paid better attention to it.

REPLY [4 votes]: The poles of the R-matrices for quantum affine algebras are the price to pay for the abovementioned simplification - the braiding becomes symmetric under q-deformation. 
If there were no poles, the category would have been symmetric and hence would be a representation category of a (usual) group. In fact, the poles of the R-matrices contain much of the information about the structure of this category. 
Also, I'd like to add that there is an interesting paper by Hernandez, math/0504269, where he discusses a new "fusion" tensor product on representations of quantum affine algebras.<|endoftext|>
TITLE: Eigenvalues of Laplacian
QUESTION [6 upvotes]: What's the most natural way to establish the asymptotics of $\Delta$  on a compact Riemannian manifold $M$ of dimension $N$? The asymptotics should be 
$$ \#\{v < A^2\} = \mathrm{const}\ast\mathrm{vol}(M)\ast A^n + o(\mathrm{something}) $$
(Perhaps one could consider first the case of a Kähler manifold? The Laplacian is particularly simple there.)

REPLY [2 votes]: Wener Müler - Weyl laws...
http://www.math.uni-bonn.de/people/mueller/papers/weyllaw.pdf 
has a pretty good introduction. 
Edit: I see now that only the second section addresses you specific question and not the intro:(<|endoftext|>
TITLE: Dualizing sheaf on singular curves
QUESTION [7 upvotes]: I am trying to understand the stabilization map, which takes a prestable curve (a curve with some marked points, and at worst nodal singularities) and returns a stable curve (a curve with some marked points, at worst nodal singularities, and finite automorphisms). Essentially what the stabilization map does is it contracts all of the unstable components of the curve, that is, those with infinite automorphisms.
There is furthermore supposed to be a map from the moduli of prestable curves to the moduli of stable curves, and therefore we need a nice description of the stabilization map which works well in flat families.
One possible such description is supposed to be: Take the dualizing sheaf omegaC of the prestable curve C; then take L := omegaC(x1 + ... + xn), where the xi are the marked points; then some large power of L will be generated by global sections, and the stabilization of C is the image of C under the corresponding map to projective space.
In particular, if C' is an unstable component of C, then L restricted to C' should be OC', as the image of C' should be a point.
I have several dumb questions:

Why is omegaC an invertible sheaf? Does it follow from, e.g., Hartshorne III.7.11?
Why is some large power of L generated by global sections?
How do we know that this power can be chosen to be constant in families?
How do we see that L restricted to unstable C' is OC'?

My more general question is: Given a singular curve, or at least a prestable curve, what explicit information can we deduce about the dualizing sheaf? For example, is there a way to figure out what the dualizing sheaf looks like when restricted to an irreducible component, or at least a smooth irreducible component?

REPLY [5 votes]: The dualizing sheaf looks like the sheaf of 1-forms, with logarithmic singularities at nodes (of the form f(z)dz/z, f regular at 0) such that the residues on each component of a node add to zero.  The answer to the second part of number 1 is "yes" but you can also just calculate using the explicit description.<|endoftext|>
TITLE: Spectrum of the Grothendieck ring of varieties
QUESTION [31 upvotes]: Here's a problem that may ultimately require just simple algebraic-geometry skills to be solved, or perhaps it's very deep and will never be solved at all. From the comments, some literature and my memory it appears this was posed by Grothendieck as part of the big program of motives.
Consider classes of complex algebraic varieties X modulo relations 
    [X] - [Y] = [X\Y], 
    [X x Y] = [X] x [Y], 

Also, if you're familiar with taking inverse of an affine line, let's do that too: 
$$ \exists \mathbb A^{-1}\quad  \text{such that}\quad [\mathbb A] \cdot [\mathbb A^{-1}] = [\mathbb A^0].$$ 
(+ if you want, you can also take idempotent completion and formal completion by A^-1).
It's not hard to see that you can add (formally) and multiply (geometric product as above) those things, so they form a ring. Let's denote this ring  Mot (It's actually very close to what Grothendieck called baby motives.) 
And for things that form a ring you can study their Spec. For example, you can talk about  points of the ring — each point is by definition a homomorphism to complex numbers.

Question: what are the properties of Spec Mot? How to describe its points?

For example, one point is Euler characteristics $\chi \in \text{Spec}\,\mathbf{Mot}$, since it's additive and multiplicative (it's even integral!) Any other homomorphism to complex numbers is thus sometimes called generalized Euler characteristics.
There's also a plane there given by mixed Hodge polynomials (that is, polynomials whose coefficients are weighted Hodge numbers $h^{p,q}_k$), since Hodge polynomial at a given point satisfies those relations too (see the references below). 
As Ben says below, things would become even more interesting if we considered this ring for schemes over $\mathbb Z$, because then each $q$ would give a generalized Euler characteristic $\chi_q$ that counts points of $X(\mathbb F_q).$
Are there any other points? Any more information?

REPLY [4 votes]: This ring is very important for motivic integration; so it might be useful for you to read surveys on this subject.
Yet I would say that this ring is too large and complicated. A reasonable factor-ring of it is K_0 of Chow motives. If you take Chow motives with rational coefficients then as a group it (conjecturally!) would be a free abelian group with generators being isomorphism classes of indecomposable numerical motives.
You could also be interested in weight complexes: see  H. Gillet, C. Soule, Descent, motives and K-theory, J. Reine Angew. Math. 478 (1996)
or my own paper
https://arxiv.org/abs/math/0601713<|endoftext|>
TITLE: What is Koszul duality?
QUESTION [67 upvotes]: Okay, let's make sure I'm on the same page with those who know homological algebra.
What is Koszul duality in general?
What does it mean that categories are Koszul dual (I guess representations of Koszul dual algebras are the examples?) What are examples of "categories which seem to a priori have no good reason to be Koszul dual actually are" [Koszul dual] other than (g, R)-admissible modules?

REPLY [59 votes]: I've spent many years researching Koszul duality in its various versions.  To me, Koszul duality is a fundamental homological phenomenon which has many manifestations, e.g.

the relation between the homotopy groups of a topological space and its (co)homology groups;
the relation between an augmented algebra A and its Ext-algebra $\text{Ext}_{A}(k,k)$ (and between modules over these);
the relation between the ring of differential operators and the de Rham DG-algebra of differential forms (and between modules over these).

Version 1. is obviously complicated and one can approach it from various angles, like rational homotopy theory or stable homotopy theory.  The former is related to the duality between commutative and Lie DG-algebras (Quillen), while the latter is more akin to module duality, in that it is additive.  Morphisms from the sphere spectrum to the Eilenberg-MacLane spectrum are very simple to describe, while endomorphisms of the latter spectrum are more complicated and endomorphisms of the former one are much more complicated.  One can view the sphere spectrum as a kind of projective generator and the Eilenberg-MacLane spectrum as a kind of irreducible module.
Version 2. is better to approach by splitting it in two branches, namely
2a. the relation between a conilpotent coalgebra $C$ and its Ext-algebra $\text{Ext}_{C}(k,k)$; and
2b. the relation between an augmented algebra $A$ and its Tor-coalgebra $\text{Tor}^{A}(k,k)$.  Then one can generalize 2a. to DG-coalgebras, and 2b. to DG-algebras, which makes these two correspondences inverse to each other.  Subsequently one can notice that the augmentation assumption or structure is largely irrelevant for 2b., and generalize this duality even further, obtaining a correspondence between conilpotent CDG- (curved DG-) coalgebras and (nonaugmented) DG-algebras.
Version 3. is a relative one, with the ring of functions playing the role of the basic field.  The functions are not central in the differential operators and the de Rham differential is not linear over the functions, which makes such a relativization of Koszul duality highly nontrivial and interesting.
With both 2. and 3., there is a major problem that the duality functors do not preserve acyclicity of complexes.  For example, a nonacyclic DG-algebra is sometimes assigned to an acyclic DG-coalgebra, and a nonacyclic complex of D-modules is assigned to an acyclic DG-module over the de Rham complex.  And for curved DG-structures, the notion of quasi-isomorphism does not even make sense.  The general solution is that one has to introduce an equivalence relation more delicate than quasi-isomorphism, particularly on the coalgebra side of the story.  The relevant references include Hinich's paper, Lefevre-Hasegawa's thesis and Keller's exposition of some results contained therein, and my recent preprint, which is supposed to contain state of the art.  Concerning DG-modules over the de Rham complex, there were earlier approaches by Kapranov and Beilinson-Drinfeld.<|endoftext|>
TITLE: Is every functor a composition of adjoint functors?
QUESTION [23 upvotes]: My understanding of Ben's answer to this question is that even though associated graded is not an adjoint functor, it's not too bad because it is a composition of a right adjoint and a left adjoint.
But are such functors really "not that bad"? In particular, is it true that any functor be written as the composition of a right adjoint and a left adjoint?

REPLY [19 votes]: Here's a really trivial way to see that the answer is "no": a functor from the empty category to a nonempty category is never a composite of adjoints (since a functor from the empty category to a nonempty category is never an adjoint).<|endoftext|>
TITLE: Why are torsion points dense in an abelian variety?
QUESTION [8 upvotes]: Hi everyone, 
let $A$ be an abelian variety of dimension $g$ over an algebraically closed field $k$ of characteristic $p\geqslant 0$. I'm trying to prove that the subgroup $A'$ which is the union of all torsion points $a\in A(k)$ of order prime to $p$ is Zariski dense in $A$.
The statement would follow if the Zariski closure $B$ (which by construction is a group variety) of $A'$ in $A$ would again be an abelian variety of dimension $d$, because assuming $d<g$, the $\ell$-primary part of $B$ would still be $(\mathbf Q_\ell/\mathbf Z_\ell)^{2 g}$, while it SHOULD be of rank $2d<2g$, contradiction.
However, I fail to see why $B$ should be irreducible. Does anyone see a way to salvage the argument, or a different, (simpler) argument?

REPLY [16 votes]: Let $C$ be the connected component of the identity in $B$. Then $C$ is a projective group variety, hence an abelian variety; let it have dimension $d$. Let $B/C=G$, a finite group. Then the number of $\ell$-torsion points of $B$ is at most $\left|G\right|\cdot\ell^{2d}$. For large $\ell$, this is less than $\ell^{2n}$ if $d$ is not $n$.<|endoftext|>
TITLE: A reading list for topological quantum field theory?
QUESTION [61 upvotes]: Can you suggest a reading list, or at least a few papers that you think would be useful, for a beginner in topological quantum field theory? I know what the curvature of a connection is, know basic algebraic topology, and have some basic background in quantum field theory.
Perhaps others with different backgrounds will also be interested in a reading list on TQFTs, so feel free to ignore my background and suggest material at a variety of levels.

REPLY [6 votes]: This recent paper of 2016 contains a useful introduction to the new development in TQFT for strongly coupled condensed matter system and topological quantum matter in 3-dimensions, 4-dimensions and any dimension. 
The authors point out the relations between quantum Hamiltonian lattice models, the continuum TQFTs and group cohomology/cobordism theory.
https://arxiv.org/abs/1612.09298
Annals of Physics 384C, 254-287 (2017)
DOI: 10.1016/j.aop.2017.06.019<|endoftext|>
TITLE: Motivation for algebraic K-theory?
QUESTION [92 upvotes]: I'm looking for a big-picture treatment of algebraic K-theory and why it's important.  I've seen various abstract definitions (Quillen's plus and Q constructions, some spectral constructions like Waldhausen's) and a lot of work devoted to calculation in special cases, e.g., extracting information about K-theory from Hochschild and cyclic homology.  As far as I can tell, K-theory is extremely difficult to compute, it yields deep information about a category, and in some cases, this produces highly nontrivial results in arithmetic or manifold topology.  I've been unable to piece these results into a coherent picture of why one would think K-theory is the right tool to use, or why someone would want to know that, e.g., K22(Z) has an element of order 691.  Explanations and pointers to readable literature would be greatly appreciated.

REPLY [5 votes]: Grothendieck's original motivation for K-theory was to give a natural setting for the intersection theory on algebraic varieties.<|endoftext|>
TITLE: Derived categories and homotopy categories
QUESTION [18 upvotes]: There are two constructions that look quite similar to me: the derived category of an abelian category, and the homotopy category of a model category. Is there any explicit relationship between these two constructions? (This question is related to, and indeed the inspiration for, one of my previous questions.)

REPLY [3 votes]: Both give rise to derivators, and indeed thinking about homotopy theories as non-abelian derived categories is what led Grothendieck to introduce then (note that Heller and Franke independently came up with derivators, but I'm not sure they had the same motivation)<|endoftext|>
TITLE: Deligne's conjecture (the little discs operad one)
QUESTION [16 upvotes]: Deligne's conjecture states that the Hochschild cochain complex of an A-infinity algebra is an algebra over the operad of chains on the topological little discs operad.
Of course the conjecture has been proven, but proofs aside -- what are some moral reasons why we should believe in Deligne's conjecture?

REPLY [17 votes]: Here is a super simple way to think about it: The Hochshcild cochains on $A$ is derived $Hom$ over $A \otimes A^{op}$ of $A$ into$ A$. This has two multiplications, given by composition and multiplication in the target respectively.
These commute, or intertwine, so the result has an action of the tensor product of two $A$-infinity operads, which is (at least if you chose the right $A$-infinity operads) an $E_2$ operad.<|endoftext|>
TITLE: Motivating the Laplace transform definition
QUESTION [66 upvotes]: In undergraduate differential equations it's usual to deal with the Laplace transform to reduce the differential equation problem to an algebraic problem.
The Laplace transform of a function $f(t)$, for $t \geq 0$ is defined by $\int_{0}^{\infty} f(t) e^{-st} dt$.
How to avoid looking at this definition as "magical"? How to somehow discover it from more basic definitions?

REPLY [5 votes]: There is in fact a very good paper of 2011 which addresses this question exactly:
The Laplace Transform: Motivating the Definition
by Howard Dwyer
Abstract:

Most undergraduate texts in ordinary
  differential equations (ODE) contain a
  chapter covering the Laplace transform
  which begins with the definition of
  the transform, followed by a sequence
  of theorems which establish the
  properties of the transform, followed
  by a number of examples. Many students
  accept the transform as a Gift From
  The Gods, but the better students will
  wonder how anyone could possibly have
  discovered/developed it. This article
  outlines a presentation, which offers
  a plausible (hopefully) progression of
  thoughts, which leads to integral
  transforms in general, and the Laplace
  transform in particular.<|endoftext|>
TITLE: Deformation theory and differential graded Lie algebras
QUESTION [33 upvotes]: There is supposed to be a philosophy that, at least over a field of characteristic zero, every "deformation problem" is somehow "governed" or "controlled" by a differential graded Lie algebra. See for example http://arxiv.org/abs/math/0507284
I've seen this idea attributed to big names like Quillen, Drinfeld, and Deligne -- so it must be true, right? ;-)
An example of this philosophy is the deformation theory of a compact complex manifold: It is "controlled" by the Kodaira-Spencer dg Lie algebra: holomorphic vector fields tensor Dolbeault complex, with differential induced by del-bar on the Dolbeault complex, and Lie bracket induced by Lie bracket on the vector fields (I think also take wedge product on the Dolbeault side).
I seem to recall that there is a general theorem which justifies this philosophy, but I don't remember the details, or where I heard about it. The statement of the theorem should be something like: 

Let k be a field of characteristic zero. Given a functor F: (Local Artin k-algebras) -> (Sets) satisfying some natural conditions that a "deformation functor" should satisfy, then there exists a dg Lie algebra L such that F is isomorphic to the deformation functor of L, which is the functor that takes an algebra A and returns the set of Maurer-Cartan solutions (dx + [x,x] = 0) in (L^1 tensor mA) modulo the gauge action of (L^0 tensor mA), where mA denotes the maximal ideal of A.

Furthermore, I think such an L should be unique up to quasi-isomorphism.
Does anyone know a reference for something along these lines?
Any other nice examples of cases where this philosophy holds would also be appreciated.

REPLY [2 votes]: Some precise statements with proofs can be found in this paper of Manetti: http://arxiv.org/abs/math/9910071<|endoftext|>
TITLE: What's the right way to think about "anomalies" in 3d TQFTs?
QUESTION [17 upvotes]: 3d TQFTs constructed from modular tensor categories don't in general give an honest 3d TQFT, instead they have an "anomaly."  My vague understanding from Kevin Walker's talks and from skimming Freed-Hopkins-Lurie-Teleman is that what's really going on is that its a 4d TQFT that's almost boring on the 4d part and that's what the "anomaly" means.  Does anyone know how to make this more precise?

REPLY [8 votes]: Here is my understanding from physics point of view.
A quantum field theory is anomalous if it lacks of a UV completion. In other words there is no lattice theory in the same dimension, whose continuum limit produces the quantum field theory.
However, even an anomalous quantum field theory can be produced at the boundary of a topological ordered state on lattice in one higher dimension. 
In other words, for a given quantum field theory, there is always a gapped lattice theory in one higher dimension, whose continuum limit at the boundary produces the quantum field theory. Thus 
the types of anomalies in quantum field theories = types of topological orders in one higher dimension
This leads to a classification of gravitional anomalies in terms of topological orders in one higher dimension.
Also
the types of anomalies in topological quantum field theories （ie gapped quantum field thoeries）= types of topological orders with gappable boundary in one higher dimension （ie topological quantum field theories in one higher dimension that have state sum realization ）
I have two papers discussing those points in details:
arXiv:1303.1803
    Classifying gauge anomalies through SPT orders and classifying gravitational anomalies through topological orders (for gauge anomaly and group cohomology)
    Xiao-Gang Wen 
arXiv:1405.5858
    Braided fusion categories, gravitational anomalies, and the mathematical framework for topological orders in any dimensions (for gravitational anomaly and n-category, as well as the notion of H-type and L-type anomalies)
    Liang Kong, Xiao-Gang Wen 
Kapustin also has a paper
arXiv:1404.3230
    Anomalies of discrete symmetries in various dimensions and group cohomology (for gauge anomalies)
    Anton Kapustin, Ryan Thorngren<|endoftext|>
TITLE: Reading list for basic differential geometry?
QUESTION [80 upvotes]: I'd like to ask if people can point me towards good books or notes to learn some basic differential geometry. I work in representation theory mostly and have found that sometimes my background is insufficient.

REPLY [5 votes]: My take on it is like this:
Smooth manifolds

Loring Tu, Introduction to manifolds - elementary introduction,
Jeffrey Lee, Manifolds and Differential geometry, chapters 1-11 cover the basics (tangent bundle, immersions/submersions, Lie group basics, vector bundles, differential forms, Frobenius theorem) at a relatively slow pace and very deep level.
Will Merry, Differential Geometry - beautifully written notes (with problems sheets!), where lectures 1-27 cover pretty much the same stuff as the above book of Jeffrey Lee

Basic notions of differential geometry

Jeffrey Lee, Manifolds and Differential geometry, chapters 12 and 13 - center around the notions of metric and connection.

Will Merry, Differential Geometry - lectures 28-53 also center around metrics and connections, but the notion of parallel transport is worked out much more thoroughly than in Jeffrey Lee's book.

Sundararaman Ramanan, Global calculus - a high-brow exposition of basic notions in differential geometry. A unifying topic is that of differential operators (done in a coordinate-free way!) and their symbols.


What I find most valuable about these books is that they try to avoid using indices and local coordinates for developing the theory as much as possible, and only use them for concrete computations with examples.
However, the above books only lay out the general notions and do not develop any deep theorems about the geometry of a manifold you may wish to study. At this point the tree of differential geometry branches out into various topics like Riemannian geometry, symplectic geometry, complex differential geometry, index theory, etc., not to mention a close sister, differential topology.
I will only mention one book here for the breadth of topics discussed

Arthur Besse, Einstein manifolds - reviews Riemannian geometry and tells about (more or less) the state of the art at 1980 of the differential geometry of Kähler and Einstein manifolds.<|endoftext|>
TITLE: "A gentleman never chooses a basis."
QUESTION [43 upvotes]: Around these parts, the aphorism "A gentleman never chooses a basis," has become popular.

Question. Is there a gentlemanly way to prove that the natural map from $V$ to $V^{**}$ is surjective if $V$ is finite-dimensional?

As in life, the exact standards for gentlemanliness are a bit vague. Some arguments seem to be implicitly picking a basis. I'm hoping there's an argument which is unambiguously gentlemanly.

REPLY [18 votes]: At the price of being too categorical for the question, one can follow up Todd's answer as follows.
Consider any closed symmetric monoidal category $\mathcal{V}$ with product $\otimes$ and unit object $k$, such as vector spaces over a field $k$. Let $V$ be an object of $\mathcal{V}$ and let $DV = Hom(V,k)$.  Just from formal properties of $\mathcal{V}$, there are canonical maps $\iota\colon k\to Hom(V,V)$ and $\nu\colon DV\otimes V\to Hom(V,V)$, which are the usual things for vector spaces.  Say that $V$ is dualizable if there is a map  $\eta\colon k\to V\otimes DV$ such that $\nu \circ \gamma \circ \eta = \iota$, where $\gamma$ is the commutativity isomorphism.  Formal arguments show that $\nu$ is then an isomorphism and if $\epsilon\colon DV\otimes V \to k$ is the evaluation map (there formally), then, with $W=DV$, $\eta$ and $\epsilon$ satisfy the conditions Todd stated for $e$ and $f$.  This is general enough that it can't have anything to do
with bases.  But restricting to vector spaces, we can choose a finite set of elements $f_i\in DV$
and $e_i\in V$ such that $\nu(\sum f_i\otimes e_i) = id$.  Then it is formal that $\{e_i\}$ is 
a basis for $V$ with dual basis $\{f_i\}$.  This proves that $V$ is finite dimensional, and the 
converse is clear as in Todd's answer. There is a result in Cartan-Eilenberg called the dual basis theorem that essentially points out that the precisely analogous characterization holds for finitely generated projective modules over a commutative ring $k$, with the same proof.
Still in a general symmetric monoidal category, if $V$ is dualizable, then a formal argument also shows that the canonical map $V \to V^{**}$ (again defined formally) is an isomorphism.  Also, in answer to Peter Samuelson, while the name ``dual basis theorem'' dates from long before my time, it does have some justification.  When  $\mathcal{V}$ is modules
over a commutative ring $k$, if one takes a dualizable $V$ and constructs the free module $F$ on basis $\{d_i\}$ in 1-1 correspondence with the $e_i$ in my previous post, then $\alpha(v) = \sum f_i(v) d_i$
specifies a monomorphism $\alpha\colon V\to F$ such that $\pi\alpha = id$, where $\pi(d_i) = e_i$.  This completes the proof that dualizable implies finitely generated projective, with a relevant basis in plain sight.<|endoftext|>
TITLE: Does a scheme have a "separification"?
QUESTION [38 upvotes]: Background: 
(1) If C and D are categories and there is a forgetful functor U:C→D, then a C-ification functor F:D→C is an adjoint to U. For example, the (left) adjoint to the forgetful functor from groups to monoids is "groupification" of a monoid, given by formally adjoining inverses. The (left) adjoint to the forgetful functor from presheaves to sheaves is the usual "sheafification" functor.
Note that whenever you have a (left adjoint) C-ification functor F (whenever you have an adjunction, for that matter), you get a universal property. For any object X∈D, there is a canonical morphism (called the unit of adjunction) εX:X→U(F(X)) with the property that any morphism f:X→U(Y) factors as f=U(g)\circ εX for a unique morphism g:F(X)→Y in C.
(2) A scheme X is separated if the diagonal morphism X→XxX is a closed immersion. It is enough to check that the image of the diagonal is closed. Being separated is the algebro-geometric analogue of being hausdorff, which nothing in algebraic geometry ever is. 

My question is whether there exists a "separification" functor adjoint to the forgetful functor U from the category of separated schemes to the category of schemes. Note that the forgetful functor U does not respect colimits (you can glue together separated schemes to get a non-separated scheme), so it has no hope of having a right adjoint. But U does respect limits (it's enough to show that an arbitrary product of separated schemes is separated and that fiber products of separated schemes are separated), so it might have a left adjoint.
To put it another way, given a scheme X, is there a canonically defined separated scheme Xs and a morphism X→Xs so that any morphism from X to a separated scheme factors uniquely through X→Xs?

Related questions I'd like to know the answer to:

Is there a "relative separification" functor. That is, does an arbitrary morphism of schemes f:X→Y admit a canonical factorization through a separated morphism fs:X'→Y. This would be analogous to Stein factorization, which I regard as "relative affinification". An arbitrary (quasi-compact and quasi-separated) morphism f:X→Y canonically factors through the affine morphism SpecY(f*OX)→Y
Is there a separification functor for algebraic spaces? Is it possible that the separification of a scheme is naturally an algebraic space?
Is there a separification functor for algebraic stacks? (An algebraic stack is separated if the diagonal is proper.)

REPLY [3 votes]: Doesn't a separification exist for trivial reasons?
Given a scheme $X$, consider the category of separated schemes $Y$ together with a morphism $X \to Y$. View this as a diagram, and take the inverse limit. This means we take the product of all schemes $Y$, and set $X^s$ equal to the intersection of the closed graphs of each arrow in this category.
Then every morphism from $X$ to a separated scheme factors through $X^s$. However this is unique because, were there two different morphisms, there equalizer would be a nonempty closed subscheme of $X^s$ which itself is separated and admits a map from $Y$, which is impossible because $X^s$ would have to lie in a map of this graph.
Then we get functoriality the obvious way.
The only issue is set-theoretic. However somehow I think we only need consider maps to separated schemes of bounded cardinalities. 
There's likely some obvious flaw in this thinking. If not, I think the examples David and Gro-Tsen have come up with conclusively show that there is no geometrically natural construction of the separification.<|endoftext|>
TITLE: How is tropicalization like taking the classical limit?
QUESTION [33 upvotes]: There is a folk — I can't call it a theorem — "fact" that the mathematical relationship between Complex and Tropical geometry is analogous to the physical relationship between Quantum and Classical mechanics.  I think I first learned about this years ago on This Week's Finds.  I'm wondering if anyone can give me a precise mathematical statement of this "fact".  Or to the right introduction to tropical mathematics.
I can do the beginning.  In classical mechanics, roughly (lower down I will mention some ways what I am about to say is false), when a system transitions from one configuration to another, it takes the route that minimizes some "action" (this idea dates at least to Maupertuis in 1744, and Wikipedia gives ancient-Greek analogs).  Thus, for a system to transition from state $A$ to state $C$ in two seconds, after one second it is in the state $B$ that minimizes the sum of the action to get from $A$ to $B$ plus the action to get from $B$ to $C$.  For comparison, quantum mechanics assigns to the pair $A,B$ an "amplitude", and the amplitude to go from $A$ to $C$ in two seconds is the sum over all $B$ of the amplitude to go from $A$ to $B$ times the amplitude to go from $B$ to $C$.  (This is the basic principle of Heisenberg's matrix mechanics.)  Anyway, we can understand both situations within the same language by considering the a matrix, indexed by states, filled with either the actions or the amplitudes to transition.  In the quantum case, the matrix multiplication is the one inherited from the usual $(\times, +)$ arithmetic on $\mathbb{C}$.  In the classical case, it is the $(+, \min)$ arithmetic of the tropical ring $\mathbb{T}$.
Let's be more precise.  To any path through the configuration space of your system, Hamilton defines an action $\operatorname{Action}(\mathrm{path})$.  The classically allowed trajectories are the critical paths of the action (rel boundary values), whereas if you believe in the path integral, the quantum amplitude is
$$\int \exp\left(\frac{i}{\hbar} \operatorname{Action}(\mathrm{path}) \right) \mathrm{d}(\mathrm{path}),$$
where the integral ranges over all paths with prescribed boundary values and the measure $\mathrm{d}(\mathrm{path})$ doesn't exist (I said "if").  Here $\hbar$ is the reduced Planck constant, and the stationary-phase approximation makes it clear that as $\hbar$ goes to $0$, the integral is supported along classically allowed trajectories.
Of course, the path integral doesn't exist, so I will describe one third (and more rigorous) example, this time in statistical, not quantum, mechanics.  Let $X$ be the space of possible configurations of your system, and let's say that $X$ has a natural measure $\mathrm{d}x$.  Let $E : X \to \mathbb{R}$ be the energy of a configuration.  Then at temperature $T$, the probability that the system is in state $x$ is (unnormalized) $\exp(-(kT)^{-1} E(x))$, by which I mean if $f : X \to \mathbb{R}$ is any function, the expected value of $f$ is (ignoring convergence issues; let's say $X$ is compact, or $E$ grows quickly and $f$ does not, or...):
$$\langle f \rangle_T = \frac{\int_X \exp(-(kT)^{-1} E(x)) f(x) \mathrm{d}x}{\int_X \exp(-(kT)^{-1} E(x)) \mathrm{d}x}$$
Here $k$ is Boltzmann's constant. It's clear that as $T$ goes to $0$, the above integral is concentrated at the $x$ that minimize $E$.  In tropical land, addition and hence integration is just minimization, so in the $T \to 0$ limit, the integral becomes some sort of "tropical" integral.
Here are some of the issues that I'm having:

When you slowly cool a system, it doesn't necessarily settle into the state that globally minimizes the energy, just into a locally-minimal state.  And good thing too, else there would not be chocolate bars.
The problems are worse for mechanics.  Classically allowed paths are not necessarily even local minima, just critical points of the Action function, so the analogy between quantum and warm systems isn't perfect.  Is there something like $\min$ that finds critical points rather than minima?
More generally, the path integral is very attractive, and definitely describes a "matrix", indexed by the configuration space.  But for generic systems, connecting any two configurations are many classically-allowed trajectories.  So whatever the classical analogue of matrix mechanics is, it is a matrix valued in sets (or sets with functions to the tropical ring), not just in tropical numbers.
Other than "take your equations and replace every $+$ with $\min$ and every $\times$ with $+$", I don't really know how to "tropicalize" a mathematical object.

REPLY [12 votes]: The analogy I've worked out from pieces here and there goes like this:
using the logarithm and exponential, we define for two real numbers x and y the following binary operation  $x §_h y := h .ln( e^{x/h} + e^{y/h} )$  which depends on some positive real parameter $h$.  Then we observe that as $h \rightarrow 0$ the number $x §_h y$ tends to $max(x,y)$. (Proof: assume without loss of generality that $x>y$, so $(y-x)/h <0$. But since  $h. ln( e^{x/h} + e^{y/h} ) = h. ln( e^{x/h} . (1+e^{(y-x)/h} )$ as $h \rightarrow 0$ we tend to  $h. ln (e^{x/h} . (1+0) ) = x = max(x,y)$. QED.)
Now, in quantum mechanics the canonical commutation relations between positions and momenta operators read  $[x_u,p_v] = i \hbar \delta_{uv}$ and in the limit $\hbar \rightarrow 0$ those commutators thus tend to $0$, which says that we recover classical mechanics where everything commutes. And in quantum mechanics what matters are wavefunctions which are superpositions of things of the form $A.e^{iS/\hbar}$ where $A$ is some amplitude and $S$ some phase (the action of the path).
Going back to $§_h$ we can rewrite $e^{(x §_h y)/h } = e^{x/h}+ e^{y/h}$, and so there is your analogy: the tropical mathematics operation max(,) is some kind of classical limit of the (thereby quantum) operation +.<|endoftext|>
TITLE: What is an example of a smooth variety over a finite field F_p which does not embed into a smooth scheme over Z_p?
QUESTION [15 upvotes]: Such an example of course could not be projective and would not itself lift to Z_p. The context is that one can compute p-adic cohomology of a variety X over a finite field F_p via the cohomology of an embedding of X into a smooth Z_p scheme. 
This is similar in spirit to my questions here and here (but a different question than the second link).

REPLY [7 votes]: A theorem of Wlodarczyk in "Embedding varieties in toric varieties" says that any smooth variety such that any two points are contained in an affine open set can be embedded in a smooth toric variety. Toric varieties can be lifted to Z_p so any variety over F_p with the above property can be embedded in a smooth scheme over Z_p.
Unfortunately, not all smooth varieties have this property; the example in Hartshorne of a smooth proper 3-fold which is not projective appears to be one where this fails (though for suitable choices these could lift to Z_p).<|endoftext|>
TITLE: Does homology have a coproduct?
QUESTION [60 upvotes]: Standard algebraic topology defines the cup product which defines a ring structure on the cohomology of a topological space.  This ring structure arises because cohomology is a contravariant functor and the pullback of the diagonal map induces the product (using the Kunneth formula for full generality, I think.)
I've always been mystified about why a dual structure, perhaps an analogous (but less conventional) "co-product", is never presented for homology.  Does such a thing exist?  If not, why not, and if so, is it such that the cohomology ring structure can be derived from it?
I am aware of the intersection products defined using Poincare duality, but I'm seeking a true dual to the general cup product, defined via homological algebra and valid for the all spaces with a cohomology ring.

REPLY [14 votes]: $\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$I noticed these excellent answers were missing an explicit example of a space $X$ and a class $\eta$ in $H_{\ast}(X, \ZZ)$ such that the image of $\eta$ in $H_{\ast}(X \times X)$ (under the diagonal map) is not in the image of $H_{\ast}(X) \otimes H_{\ast}(X) \to H_{\ast}(X \times X)$. So, for the record, this occurs for the fundamental class of $\mathbb{RP}^3$. 
Verification We write points of $\RR \PP^3$ in homogenous coordinates as $(w:x:y:z)$. For $t \in [0,1]$, define 
$$D_t := {\Big \{} ((w_1:x_1:y_1:z_1), (w_2:x_2:y_2:z_2)) \in \RR \PP^3 \times \RR \PP^3 : $$
$$w_2 x_1 = t w_1 x_2,\ x_2 y_1 = t x_1 y_2,\ y_2 z_1 = t y_1 z_2,$$
$$w_2 y_1 = t^2 w_1 y_2,\ x_2 z_1 = t^2 x_1 z_2,\ w_2 z_1 = t^3 z_1 w_2 {\Big \}}$$
and set $D = \bigcup_{t \in [0,1]} D_t$. 
$D_1$ is the diagonal and, for $t \in (0,1]$, we have $D_t \cong \RR \PP^3$, the equations defining $D_t$ simply say that $(w_2:x_2:y_2:z_2) = (t^3 w_1: t^2 x_1: t y_1: z_1)$. The space $D_0$ is a bit more interesting: the equations are $w_2 x_1 = x_2 y_1 = y_2 z_1 = w_2 y_1 = z_2 x_1 = z_2 w_1=0$, and considering cases shows that
$$D_0 = (\RR \PP^3 \times \RR \PP^0) \cup (\RR \PP^2 \times \RR \PP^1) \cup (\RR \PP^1 \times \RR \PP^2) \cup (\RR \PP^0 \times \RR \PP^3).$$
Here $\RR \PP^2 \times \RR \PP^1$, for example, is all points of the form $(\ast: \ast: \ast : 0) \times (0 : 0 : \ast : \ast)$. 
$D$ is an oriented manifold with corners, whose boundary is $D_1 - D_0$. So $D_1$ is homologous to $D_0$. 
Put the standard CW-structure on $\RR \PP^3$, writing $C_i$ for the cell of dimension $i$, and write $C_{ij}$ for $C_i \times C_j$. It is easy to compute in this structure that 
$$H_3 = \ZZ \cdot [C_{30}] \oplus \ZZ \cdot [C_{03}] \oplus (\ZZ/2\ZZ) \cdot ([C_{21}]+[C_{12}]).$$
The image of $\bigoplus_{i+j =3 } H_i(\RR \PP^3) \otimes H_j(\RR \PP^3)$ is the first two summands, and $[D_0] = [C_{30}] + [C_{21}] + [C_{12}] + [C_{03}]$, so $[D_0]$ is not in the image of $\bigoplus_{i+j =3 } H_i \otimes H_j$. (Recall that $H_1(\RR \PP^3) = H_2(\RR \PP^3) = 0$. )

I'll point out two things which confused me at first. First of all, why isn't $[D_0]$ the image of $\sum_{i+j=3} [\RR \PP^i] \otimes [\RR \PP^j]$? The answer is that $\RR\PP^2$ is not orientable, so it does not define a class in $H_2(\RR \PP^3)$.
Once we realize that, what does $[D_0]$ mean, since the $\RR \PP^1 \times \RR \PP^2$ and $\RR \PP^2 \times \RR \PP^1$ components are also not orientable? The answer is that the orientation on the interior of $D$ (which is $(0,1) \times \RR \PP^3$, so orientable) gives rise to orientations on the interiors of the components of $D_0$, but these orientations are discontinuous across the codimension $2$ strata where these components meet. Nonetheless, this is enough to give us a map from an oriented closed cell to each component, whose boundaries cancel out, so we can speak of the $H_3$ class of $[D_0]$ even though we can't speak of the classes of its individual components.<|endoftext|>
TITLE: Existence of (smooth) models
QUESTION [15 upvotes]: Hi everyone,
let X be a variety over a field k, S an integral scheme such that the function field K of S is contained in k. An S-scheme X is called model of X/k if X x_S k = X, i.e. if the generic fiber of X over S is isomorphic to X.

Are there general conditions on X, S, k, such that X exists? 
If X is smooth and projective, what are the conditions, such that there is a smooth model X?
Any good references that go into models and reduction in general, and not only in the case of curves?

REPLY [17 votes]: I'm happy to present my example of a smooth projective surface $X$ over
$K=\mathbb{Q}_p$ ($p$ prime) such that $X(K)\neq\emptyset$, whose
$l$-adic cohomology groups are unramified (for all primes $l$) and which still
has bad reduction : there is no smooth $\mathbb{Z}_p$-scheme whose generic
fibre is $X$.  (The method works for any finite extension of $\mathbb{Q}_p$ and was worked out a few years ago.)
The surface $X$ is going to be a conic bundle over $\mathbb{P}_1$ with four
degenerate fibres, so it is a rational surface in the sense of being $\bar
K$-birational to $\mathbb{P}_2$.  It will be clear that the example is not
isolated.
If $p$ is odd, let $d\in\mathbb{Z}_p^\times$ be a unit which is
not a square, and take $d=5$ if $p=2$, so that $K(\sqrt{d})|K$ is the unramified quadratic extension.
Let $e_1, e_2$ be two distinct units of $K$.  We take $X$ to be the surface in
$\mathbb{P}({\cal O}(2)\oplus{\cal O}(2)\oplus{\cal O})$ (coordinates $y:z:t$)
over $\mathbb{P}_1$ (coordinates $x:x'$) defined by the equation
$$
y^2-dz^2=xx'(x-e_1x')(x-e_2x')t^2.
$$
I claim that this $X$ has all the properties stated above, if $v_p(e_1-e_2)>0$.
First, $X(K)\neq\emptyset$ because each degenerare fibre is a pair of
intersecting lines conjugated by $\mathrm{Gal}(\bar K|K)$.
Secondly, the $l$-adic cohomology is unramified because the action of
$\mathrm{Gal}(\bar K|K)$ on the Picard group $\mathrm{Pic}(\bar{X})$ of $\bar
X=X\times_K\bar K$ factors via the quotient $\mathrm{Gal}(K(\sqrt{d})|K)$.
Finally, $X$ has bad reduction because its Chow group $A_0(X)_0$ of $0$-cycles of degree
$0$ is $\mathbb{Z}/2\mathbb{Z}$ (cf. prop. 1 of arXiv:math/0302156), and a theorem of Bloch (th. 0.4, On the Chow groups of certain
rational surfaces, Annales scientifiques de l'École Normale Supérieure, Sér.
4, 14 no. 1 (1981), p. 41-59, available at Numdam) asserts that if a conic bundle has good
reduction, then its Chow group of $0$-cycles of degree $0$ is $0$.
Addendum  (in response to a question in an email I received).  One can show moreover that no smooth projective surface $Y$ over $\mathbf{Q}_p$ which is $\mathbf{Q}_p$-birational to $X$ can have good reduction.  This follows from the facts recalled above and the theorem of Colliot-Thélène and Coray (which can be found in Fulton's Intersection theory) : $A_0(Y)_0$ is isomorphic to $A_0(X)_0$.<|endoftext|>
TITLE: What is an example of a smooth variety over a finite field F_p which does not lift to Z_p?
QUESTION [9 upvotes]: Somebody answered this question instead of the question here, so I am asking this with the hope that they will cut and paste their solution.

REPLY [17 votes]: Examples are also in my paper "Murphy's Law in Algebraic Geometry", which you can get from my preprints page 
Here is a short (not quite complete) description of a construction, with two explanations of why it works.  I hope I am remembering this correctly!
In characteristic $>2$, consider the blow up of $\mathbf{P}^2$ at the $\mathbf{F}_p$-valued points of the plane.  Take a Galois cover of this surface, with Galois group $(\mathbf{Z}/2)^3$, branched only over the proper transform of the lines, and the transform of another high degree curve with no $\mathbf{F}_p$-points. 
Then you can check that this surface violates the numerical constraints of the Bogomolov-Miyaoka-Yau inequality, which holds in characteristic zero; hence it doesn't lift.  (This is in a paper by Rob Easton.)   Alternatively, show that deformations of this surface must always preserve that Galois cover structure, which in turn must preserve the data of the branch locus back in $\mathbf{P}^2$, meaning that any deformation must preserve the data of those $p^2+p+1$ lines meeting $p+1$ to a point, which forces you to live over $\mathbf{Z}/p$.
The two papers mentioned above give more exotic behavior too (of different sorts in the two papers), e.g. you an find a surface that lifts to $\mathbf{Z}/p^{10}$ but still not to $\mathbf{Z}_p$.

REPLY [3 votes]: There is an article by Illusie in Fundamental Algebraic Geometry: Grothendieck's FGA Explained that contains a construction of a non-liftable surface.  The article is online (pdf
- see section 6) and the construction is rather complicated.  There are additional references to other nonliftability results (section 5F).<|endoftext|>
TITLE: When does a transitive action of a profinite group have an infinite orbit?
QUESTION [5 upvotes]: That is: suppose G is a profinite group acting 1-transitively (but maybe not regularly) on a set X.  Is there a reasonable criterion for when there is a g in G and a point a in X such that the g-orbit of a is infinite?
I wonder if it's enough to have a family (g_i, a_i) of pairs in G times X such that the g_i-orbit of a_i has size at least i.
Also, does anybody study these things much?  A google search for "profinite group action" yields only a few hits; "profinite permutation group(s)" yields none.

REPLY [2 votes]: The answer depends on whether the action map GxX -> X is continuous (where I'm assuming X has the discrete topology).  If so, then I think transitivity implies X is finite.  If not, then you might as well view G as some abstract infinite group.  If X is not discrete, e.g., given by a profinite system of sets, then I think you can have more interesting actions.<|endoftext|>
TITLE: Homological algebra for commutative monoids?
QUESTION [26 upvotes]: Homological algebra for abelian groups is a standard tool in many fields of mathematics. How much carries over to the setting of commutative monoids (with unit)? It seems like there is a notion of short exact sequence. Can we use this to define ext groups which classify extensions? What works and what doesn't work and why?

REPLY [5 votes]: Jaret Flores's PhD thesis,

Jaret Flores, Homological Algebra for Commutative Monoids, arXiv:1503.02309.

treats precisely this topic.<|endoftext|>
TITLE: E_\infty spectrum corresponding to Z_p
QUESTION [5 upvotes]: First of the questions about derived algebraic geometry from a noobie.
The way I understand it, every discrete ring R corresponds to some ring spectrum whose \pi_0 is R. Now consider p-adic numbers. They are a limit of discrete rings — what should correspond to them? How to generalize?
(+ what could be good tags for derived algebraic geometry? I was considering: e-infinity, infty-structures, math-0703204, derived-alggeom, derived-spaces, infty-topology, a-infinity-algebras (2 currently tagged))

REPLY [10 votes]: I would argue that looking for a ring spectrum is not the right thing to do.  What you should be looking for is the category of modules over that possibly non-existent ring spectrum, as an infinity-category or just as a triangulated category.  If you think about it this way, an obvious answer presents itself.  
Begin with the category of HZ-modules, or the derived category of Z, or its infinity-category version.  Now take the Bousfield localization with respect to the object Fp (thought of as a complex in degree 0).  This is not a smashing localization, so this category is not equivalent to modules over HZp .  As a triangulated category, it is compactly generated, but by Fp itself.  The sphere is not small.  So this category is equivalent to modules over a DGA, the endomorphism DGA of Fp, but it is not commutative.  It is more like a DG Hopf algebra, I suspect, so that its homotopy category has a tensor product even though it is not commutative.  I have always thought this example needs more investigation, though it might be in Dwyer-Greenlees-Iyengar somewhere.  It is a toy version of the K(n)-local stable homotopy category.  
Mark<|endoftext|>
TITLE: Sums of cubes and more
QUESTION [9 upvotes]: It's well-known that every natural number can be written as a sum of 4 squares of integers.
Has there been any recent progress about the similar problem for the cubes, 4-th powers and so on? I believe this was proven to be representable using some N — that depends on the power — and what's the story about it?

REPLY [8 votes]: Arguably, the problem "what is the least g=g(k) such that every integer can be written as the sum of g k-th powers" is less interesting than the version that ignores the random stuff that's happening with a finite number of special cases. 
Namely, the "real" question should be "what is the least G=G(k) such that for some N, every integer greater than N can be represented as the sum of G k-th powers". 
For example, every number is the sum of 19 4th powers, but every number greater than 13,792 is actually the sum of just 16 4th powers. The "16" was known for quite some time; the verification that 13,792 is the last offender is quite recent (I found the value on Wikipedia, btw).
Evaluating G(k) is harder than evaluating g(k), and most of the actual values are still not known. I don't think there was tremendous recent progress on this front, although there certainly is progress on things like bounds, number of representations, etc. 
You should look at Wooley's survey here (I haven't read it yet).<|endoftext|>
TITLE: Advice on doing mathematical research
QUESTION [29 upvotes]: Please share any general tips or advice you have on doing mathematical research.
How do you identify good problems to work on or to think about? What do you do when you get stuck on a problem? Etc.

REPLY [6 votes]: Often, I find that the good problems come from errors in my work -- I write a proof of something that (at the time) I think is trivial and later find a counter-example.  Suddenly, the problem is non-trivial and interesting.
When I get stuck on a problem?  I work on another problem, a related problem, a broader problem, a more specific problem, etc.  Anything to try to understand why it is difficult to solve this problem.<|endoftext|>
TITLE: Characterizing the Radon transforms of log-concave functions
QUESTION [7 upvotes]: $f:\mathbf{R}^d\to \mathbf{R}_{\ge 0}$ is log-concave if $\log(f)$ is concave (and the domain of $\log(f)$ is convex).
Theorem: For all $\sigma$ on the sphere $\Bbb S^{d-1}$ and $r\in \mathbf{R}$,
$$
g_\sigma(r) := \int\limits_{\sigma\cdot x=r}f(x)\,\mathrm{d}S(x)
$$
is a log-concave function of $r$. (Note: $g$, as a function of $\sigma$ and $r$, is the Radon transform of $f$.)
Question: does this characterize log-concavity? That is, if $g_\sigma(r)$ is log-concave as a function of $r$ for all $\sigma$, is $f$ log-concave?

REPLY [6 votes]: The following answer exploits a wider context, where non-negative functions $f$, viewed as densities of absolutely continuous measures $f(x)dx$, are replaced by non-negative measures $\mu$. Because a concave function unbounded from above is $\equiv+\infty$, a log-concave measure $\mu$ is naturally an absolutely continuous measure with log-concave density. 
Therefore the Lebesgue measure over the $2$-dimensional sphere $S^2$ is not log-concave over ${\mathbb R}^3$. Nevertheless, its Radon transform in a direction $\sigma$ is
$$g_\sigma=2\pi\chi_{(-1,1)},$$
which is log-concave for every $\sigma$.<|endoftext|>
TITLE: What is an example of a function on M_g?
QUESTION [7 upvotes]: It feels bad talking about a space without knowing a single function on it, hah?
So what is a function on the moduli space of curves, from the geometric point of view?
From the functorial point of view, it should be invariants of family of curves, which is natural w.r.t. pullback of families. The j-invariant maybe one for M_1. But does anybody have a concrete example for higher genus?
I do have two guesses of sources of functions:

There are some geometrically defined divisors, maybe take one, and then pick two sections of the line bundle it defines, and take their quotient? 
Maybe there are some "natural" differential form, whose integral over the whole curve is a function on M_g?

REPLY [4 votes]: The period mapping lets you embed $M_g$ in the quotient of the Siegel upper half-space by the integral symplectic group; so any function on this locally symmetric space restricts to a function on $M_g$.<|endoftext|>
TITLE: References for homotopy colimit
QUESTION [16 upvotes]: (1) What are some good references for homotopy colimits?
(2) Where can I find a reference for the following concrete construction of a homotopy colimit?  Start with a partial ordering, which I will think of as a category and also as a directed graph (objects = vertices, morphisms = edges).  Assume we have a functor $F$ from the graph into (say) chain complexes.  We will construct a big chain complex (the homotopy colimit) in stages.
Stage 0: direct sum over all vertices $v$ of $F(v)$
Stage 1: direct sum over all edges $e$ of the mapping cylinder of $F(e)$, with the ends of the mapping cylinder identified with the appropriate parts of stage 0.
Stage 2: direct sum over all pairs of composable edges $(e_1, e_2)$ of a higher order mapping cylinder, with appropriate identifications to parts of stage 1.  This implements a relation between the three stage 1 mapping cylinders corresponding to $e_1, e_2$ and $e_1*e_2$.
Stage 3: direct sum over all triples of composable edges $(e_1, e_2, e_3) \dots$

REPLY [3 votes]: I learned today that Pascal Lambrechts also wrote a short primer on homotopy limits and colimits, filled with good exercises. It is online here (for now).<|endoftext|>
TITLE: Elements of infinite order in a profinite group
QUESTION [12 upvotes]: Say G is a profinite group with elements of arbitrarily large order.  Do elements of infinite order exist (A) if we assume G is abelian? (B) in general?
A start for (A): we can ask the same question for the closure of the torsion subgroup of G (a subgroup since G is abelian), so WLOG we can assume the torsion subgroup is dense in G.

REPLY [7 votes]: I would like to add that the restricted Burnside problem is equivalent to the fact that a finitely generated profinite group of finite exponent is finite. Now, this was of course proved by Zelmanov. But he also proved a stronger result: every finitely generated compact Hausorff torsion group is finite, see  Zelʹmanov, E. I., On periodic compact groups, Israel J. Math. 77 (1992), no. 1-2, 83--95. In particular, every finitely generated torsion profinite group is finite, i.e. the Burnside problem is true for profinite groups. 
BTW, Zelmanov has a more general result regarding when a pro-p group is finite (I don't remember now the exact formulation, Ignore: it should be something like all generators have finite order and the associated graded Lie algbera satisfies an identity), however, he only published a sketch of the proof which I think is in E. Zelmanov, Nil Rings and Periodic Groups, The Korean Mathematical Society Lecture Notes in Mathematics, Korean Mathematical Society, Seoul, 1992.
Edit: I was asked about this recently. So I really had to search my memory and the literature. This is what I have found: I think I read about the result in Shalev’s chapter in New Horizons in Pro-$p$ Groups (Theorem 2.1 and Corollary 2.2). However, the original reference is Zelmanov’s paper in Groups ’93 Galway St. Andrews Volume 2, LMS Lecture Note Series 212.
Here it is: 
Let $G$ be a group. Write $(x_1,x_2,\ldots,x_i)$ for the left normalized commutator of the elements $x_1,x_2,\ldots,x_i \in G$. Let $D_k$ be the subgroup of $G$ generated by $(x_1,x_2,\ldots,x_i)^{p^j}$, where $ip^j \geq k$ and we go over all $x_1,x_2\ldots,x_i \in G$. Let $L_p(G)$ be the Lie subalgebra generated by $D_1/D_2$ in the Lie algebra  $\oplus_{i \geq 1} D_i/D_{i+1}$. We say that $G$ is Infinitesimally PI (IP) if $L_p(G)$ satisfies a polynomial identity (PI). Zelmanov proved the following theorem:
Theorem: If $G$ is a finitely generated, residually-$p$, IP, and periodic group, then $G$ is finite. 
The proof is based on the following theorem for which he only sketched the proof (according to Shalev):
Theorem: Let $L$ be a Lie algebra generated by $a_1,a_2,\ldots,a_m$. Suppose that $L$ is PI and every commutator in $a_1,a_2,\ldots,a_m$ is ad-nilpotent. Then $L$ is nilpotent. 
The question whether a torsion profinite group is of finite exponent is still open as far as I know and is considerd very difficult. (Burnside type problems seem to be very difficult.)
Edit: Zelmanov recently published his results in:
Zelmanov, Efim; Lie algebras and torsion groups with identity, J. Comb. Algebra 1 (2017), no. 3, 289–340. 
Here is the abstract: 
"We prove that a finitely generated Lie algebra $L$ such that (i) every commutator in generators is ad-nilpotent, and (ii) $L$ satisfies a polynomial identity, is nilpotent. As a corollary we get that a finitely generated residually-$p$ torsion group whose pro-$p$ completion satisfies a pro-$p$ identity is finite."<|endoftext|>
TITLE: What are the Schur functions of the eigenvalues of a non-negative integer matrix counting?
QUESTION [14 upvotes]: Let A be a non-negative integer square matrix with eigenvalues x1, x2, ... xn.  Any symmetric function of these eigenvalues with integer matrices is an integer.  I'm aware of the following results regarding the combinatorial interpretation of these integers:

If A is the adjacency matrix of a finite directed graph G, the power symmetric functions of the eigenvalues count closed walks on A with a distinguished starting point.
Similarly, the complete homogeneous symmetric functions of the eigenvalues count non-negative integer linear combinations of aperiodic closed walks on A.  
(Gessel-Viennot-Lindstrom) If Aij is the number of paths from source i to sink j on, say, a 2-D lattice where the only permissible moves are to the right and up, then the elementary symmetric functions of the eigenvalues count the number non-intersecting k-tuples of paths from the sources to the sinks.  In particular det A is the number of non-intersecting n-tuples of paths.

Do these results generalize to give a nice combinatorial interpretation of the value of the Schur function associated to an arbitrary partition evaluated at x1, x2, ... xn in terms of some combinatorial object attached to A?  What conditions need to be placed on A so that the Schur functions are always non-negative?
Feel free to either talk about the GL(n) perspective or to frame your discussion entirely in terms of tableaux.

REPLY [6 votes]: This is moving far enough afield from my own understanding that I'm not sure how useful it is, but have you looked at the Ph.D. thesis of Andrius Kulikauskas?  He seems to be giving some sort of combinatorial interpretation in terms of Lyndon words for the Schur functions.  
http://www.selflearners.net/uploads/AndriusKulikauskasThesis.pdf<|endoftext|>
TITLE: understanding Steenrod squares
QUESTION [73 upvotes]: There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called Steenrod squaring: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$.  (Coefficient group suppressed from here on out.)  Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product).  A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.
I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing.  I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$.  This is nice, but still somewhat mysterious to me.  Does anyone have any good ideas or references for how to think about these maps?

REPLY [11 votes]: If I interpret the request a bit differently, I would say that the Steenrod operations in the cohomology of a spectrum tell you about the attachments of the cells.  If $Sq^1 x = y$, then a cell dual to $y$ is attached by a map of degree 2 mod 4 to a cell dual to $x$.  Similarly, $Sq^2 x = y$ tells us the attaching map is $\eta$, $Sq^4$ detects $\nu$ and $Sq^8$ detects $\sigma$.    This doesn't go very far, but may help with the need to 'get a real grip on what they're doing'.
Next, let's assume you're really interested in homotopy, not just (co)homology.  A class dual to a homology class in the image of the Hurewicz homomorphism must be indecomposable under the action of the Steenrod algebra, by naturality w.r.t. the  map $S^n \longrightarrow X$.   This limits the homotopy of $X$ which can be detected by the homomorpism $\pi_* X \longrightarrow H_* X$:  the homomorphism $H^* X \longrightarrow H^* S^n$ can only map indecomposables non-trivially, since all classes in degrees below $n$ must go to $0$.
Then there are the relations.  The fact that $Sq^n$ is decomposable when n is not a power of two tells us that if $y = Sq^n x$, there must be other classes between $x$ and $y$.  EG, $Sq^3 x = y \neq 0$ tells us that $Sq^2 x \neq 0$ also, since $Sq^3 = Sq^1 Sq^2$.  So our spectrum can't have just two cells, dual to $x$ and $y$, but must have a three cell subquotient with top cell attached by 2 (mod 4) to a cell attached by $\eta$ to the bottom cell.
Or, if $Sq^2 Sq^2 x = y \neq 0$ then we must also have nonzero classes $Sq^1 x$ and $Sq^2 Sq^1 x$, since $Sq^2 Sq^2 = Sq^1 Sq^2 Sq^1$, and vice versa, if $Sq^1 Sq^2 Sq^1 x = y \neq 0$ then $Sq^2 x \neq 0$ as well. This leads to an easy proof that the mod 2 Moore spectrum $M$ isn't a ring spectrum, since $2 \pi_0M = 0$ but $\pi_2 M = Z/4$, by looking at the obstruction to attaching the top cell of a putative spectrum with nonzero cohomology spanned by $x$, $Sq^1 x$, $Sq^2 Sq^1 x$, and $Sq^1 Sq^2 Sq^1 x$.   More, the fact that you can only add such a top cell if you also have a class $Sq^2 x$ so that the top cell can be attached by the sum of $Sq^1$ on $Sq^2 Sq^1 x$ and $Sq^2$ on $Sq^2 x$ shows that $\eta^2$ (corresponding to the path $Sq^2 Sq^2$ from bottom to top, must lie in the Toda bracket $\langle 2, \eta, 2\rangle$, corresponding to the path $Sq^1$, $Sq^2$, $Sq^1$ from bottom to top. 
Similarly, $y = Sq^1 Sq^2 x$ tells us that homotopy supported on a cell dual to $x$ can be acted on by $\text{v}_1$ to get $y$, literally if we have a $ku$-module and multiply by $\text{v}_1 \in ku_2$, or as the Toda bracket $\langle 2, \eta, -\rangle$ more generally.   The key fact here is that $\text{v}_1 \in ku_2$ is in $\langle 2, \eta, 1_{ku} \rangle$, where $1_{ku} : S \longrightarrow ku$ is the unit. 
Likewise, $Sq^2 Sq^1 Sq^2 x = y$ corresponds to multiplication by the generator of $ko_4$, literally for $ko$-modules, or as a bracket $\langle \eta, 2, \eta, - \rangle$ more generally.   Here you have to be in a situation where $2 \nu = 0$ to form the bracket, since $\langle \eta, 2, \eta \rangle = \{ 2\nu, 6 \nu\}$.   This hints that the role of $\nu$ is non-trivial in real K-theory, despite going to $0$ under the homomorphism $\pi_* S \longrightarrow \pi_* ko$ and despite the cohomology of $ko$ being induced up from the subalgebra $A(1)$ generated by $Sq^1$ and $Sq^2$.   The Adem relation $Sq^2 Sq^1 Sq^2 = Sq^1 Sq^4 + Sq^4 Sq^1$ shows that $Sq^4$ must act nontrivially if $Sq^2 Sq^1 Sq^2$ does.  Also, the fact that $A(1)//A(0)$ is spanned by $1$, $Sq^2$, $Sq^1 Sq^2$, and $Sq^2Sq^1Sq^2$ tells us (with a bit more work) that we can build $HZ$ as a four cell $ko$-module. 
A good way to organize all this information is the Adams spectral sequence, which tells you that the mod $p$ cohomology of $X$ gives a decent first approximation, $\text{Ext}_{A}(H^*X,F_p)$, to the homotopy of the $p$-completion of $X$.<|endoftext|>
TITLE: What is the expected number of maximal bicliques in a random bipartite graph?
QUESTION [5 upvotes]: Maximal Biclique: A complete bipartite subgraph, that isn't a subgraph of another complete bipartite subgraph.
Given a bipartite graph $G=(V_{1}\cup V_{2}, E)$ where $|V_{1}|=|V_{2}|$ with probability $p$ of there being an edge from any $a\in V_{1}$ to any $b\in V_{2}$, what is the expected number of maximal bicliques?
What I have worked out is the upper and lower bounds:
Lower Bound: 1 or 2. If $|E|$ is divisible by $n$, then $\frac{|E|}{n}$ nodes can be connected to completely to $\frac{|E|}{n}$ other nodes, making one maximal biclique. Otherwise, connect $\frac{|E|}{n}$ nodes completely to $\frac{|E|}{n}$ nodes and connect one node to $|E|(mod\ n)$ nodes. 
Upper Bound: There are $2^n$ unique subsets, the empty set and the entire set not included, leaves $2^n-2$ subsets. Therefore, there can be at most $2^n-2$ maximal bicliques. This upper bound is achievable when there are $n^2-n$ edges (I can prove this if anyone wants me to).
Both of these results are also easily extended to bipartite graphs where $|V_{1}|\neq |V_{2}|$.
The upper and lower bounds are both fairly trivial for the most part and it's the expected number of maximal bicliques that I've had the most trouble with. I've done a little work analyzing simple cases and brute forcing the expected number for small values of $n$ (I suppose I could write a program to brute force larger values of $n$), but it hasn't amounted to anything worth saying.
I'd appreciate any suggestions for methods of attacking the problem or references that I might find useful.

REPLY [5 votes]: One approach might be as follows (I will assume your initial graph had n vertices on each side):
Given a subset S of size s on one side, and a subset T of size t on the other side, what is the probability it is a maximal clique?  We need two things to happen.

It must be a clique.  This occurs with probability p^{ab}.  
No vertex outside the clique is connected to all the vertices inside the clique on the appropriate side.  For any given vertex, this occurs with probability (1-p^{n-a}) or (1-p^{n-b}), depending on which side it needs to fail to connect to.  The handy thing is that everything in sight is independent here, so we can just multiply over all vertices outside the clique.   

Multiplying, the probability SXT is a maximal clique is p^{ab} (1-p^{n-a})^b (1-p^{n-b})^a.  Now by linearity of expectation you just need to add up over all subsets, and (possibly) work out the asymptotics as n tends to infinity.<|endoftext|>
TITLE: Can adjoint linear transformations be naturally realized as adjoint functors?
QUESTION [23 upvotes]: Last week Yan Zhang asked me the following: is there a way to realize vector spaces as categories so that adjoint functors between pairs of vector spaces become adjoint linear operators in the usual sense?
It seems as if one needs to declare an inner product by fiat for this to work out.  An obvious approach is to take the objects to be vectors and hom(v, w) to be the inner product (so the category should be enriched over C).  But I don't see how composition works out here, and Yan says he tried this and it didn't work out as cleanly as he wanted.  In this setup I guess we want the category to be additive and the biproduct to be vector addition, but I have no idea whether this actually happens.  I think John Baez's ideas about categorified linear algebra, especially categorified Hilbert spaces, are relevant here but I don't understand them well enough to see how they work out.  
Anyone who actually knows some category theory care to clear things up?

REPLY [4 votes]: Not exactly an answer to the question as posed, but it's worth noting that adjoint linear maps and adjoint functors can both be realized as instances of the same thing, namely morphisms in a Chu construction.
On one hand, a vector space $V$ with an inner product (or indeed any bilinear form) $B_V:V\otimes V \to \Bbbk$ can be regarded as an object of $\mathrm{Chu}(\mathrm{Vect},\Bbbk)$, call it $V'$.  A morphism $V'\to W'$ in $\mathrm{Chu}(\mathrm{Vect},\Bbbk)$ is then a pair of linear maps $f:V\to W$ and $g:W\to V$ such that $B_V(v,g w) = B_W(f v,w)$, i.e. an adjoint pair of transformations in the linear sense.
On the other hand, any category $C$ has a hom-functor $\hom_C : C^{\mathrm{op}}\times C\to \mathrm{Set}$ and can thereby be regarded as an object of the 2-categorical Chu construction $\mathrm{Chu}(\mathrm{Cat},\mathrm{Set})$, as discussed here; call this object $C'$.  Then a morphism $C'\to D'$ in $\mathrm{Chu}(\mathrm{Cat},\mathrm{Set})$ is a pair of functors $f:C^{\mathrm{op}}\to D^{\mathrm{op}}$ (or equivalently $C\to D$) and $g:D\to C$ together with an isomorphism (this is why the Chu construction has to be 2-categorical here) $\hom_C(c,g d) \cong \hom_D(f c,d)$, i.e. an adjoint pair of functors in the categorical sense.<|endoftext|>
TITLE: Can the valuative criteria be checked "on a dense open"?
QUESTION [5 upvotes]: The valuative criterion for separatedness (resp. properness) says that a noetherian scheme X is separated (resp. proper) if and only if

for any DVR R, with fraction field K,
  any map Spec(K)→X extends in at most one way (resp.
  extends uniquely) to a map
  Spec(R)→X.

If U⊆X is a dense open subscheme, is it sufficient to check the valuative criteria only in the cases where the map Spec(K)→X lands in U?
Intuitively, the valuative criteria are checking if it is possible to fill in a "missing point" on a curve on X (and if it is possible to fill it in in multiple ways). If there is a curve in Z=X\U with a missing point, it feels like it should be possible to find a curve in U that should have the same limit point. Also, I think it's common to verify that a compactified moduli space is separated by checking the valuative criterion using only families in the original moduli space (not "on the boundary").

REPLY [2 votes]: I'm probably not in the good decade to answer, but Lemma 4.1.1 in this article looks like what you want.<|endoftext|>
TITLE: Finite groups with the same character table
QUESTION [20 upvotes]: Say I have two finite groups G and H which aren't isomorphic but have the same character table (for example, the quaternion group and the symmetries of the square). Does this mean that the corresponding categories of finite dimensional complex representations are isomorphic (ignoring the forgetful functor to vector spaces), or just that the corresponding representation rings are?

REPLY [29 votes]: This is a great question, and the answer leads to one of the best arguments for why category theory should be studied at all!
Every undergraduate mathematician should discover for themselves that character tables alone don't determine finite groups --- and then, just as their faith in the beauty of mathematics is about to shatter, they should be reassured that character tables are just a 'shadow' of the group's compact monoidal category of representations, and that DOES determine the group (or in general, groupoid).
The procedure for reconstructing a groupoid, up to equivalence, from its category of unitary complex representations, is stunningly beautiful: if G is our groupoid and Rep(G) is its representation category, then construct the groupoid which has objects given by symmetric monoidal functors Rep(G)-->Rep(1), and morphisms given by monoidal natural transformations between them. Here, Rep(1) is just the category representations of the trivial group --- in other words, just the category of finite-dimensional Hilbert spaces, with monoidal structure given by tensor product. This is known as "Doplicher-Roberts style" reconstruction, and the best reference is Muger's appendix to this paper. It's more elegant than "Tannakian" reconstruction, as there's no need to start with a given fiber functor (i.e., a specified functor Rep(G)-->Rep(1)).
This should remind you strongly of the way you recover a compact topological space from the commutative C*-algebra of functions from that space into the complex numbers ... and there are indeed deep connections!<|endoftext|>
TITLE: What is interesting/useful about big Witt Vectors?
QUESTION [32 upvotes]: $p$-typical Witt vectors are (among other things) a canonical way of associating to a perfect ring $A$ of characteristic $p$ a complete DVR of characteristic $0$ with residue ring $A$ generalizing $\mathbb{Z}_p$ and $\mathbb{F}_p$. 

What is interesting/useful about other flavors of Witt rings, in particular the big Witt ring?

I am particularly interested in knowing their original motivation and applications.

REPLY [9 votes]: I am particularly interested in knowing their original motivation and applications.

Kummer theory says that every degree-$n$ cyclic extension $L|k$ of any field $k$ containing a primitive $n$-th root $\zeta$ of $1$ is of the form $L=k(\root n\of D)$ for some order-$n$ cyclic subgroup $D\subset k^\times/k^{\times n}$, and conversely.
(Something can be said even when $\zeta\notin k$ but when $n$ is invertible in $k$.  Look up a certain exercise in Schoof's book on Catalan's Conjecture.)
This leaves out degree-$p$ cyclic extensions $L|k$ of a characteristic-$p$ field.  Artin-Schreier proved that $L=k(\wp^{-1}(D))$ for some ${\mathbb F}_p$-line $D\subset k/\wp(k)$, where $\wp:k\to k$ is the endomorphism $x\mapsto x^p-x$ of the additive group $k$, and conversely.
What about degree-$p^m$ cyclic extensions $L|k$ of a characteristic-$p$ field ?  Many complicated constructions for particular cases were given in the 1930s (by people such as Albert) before Witt introduced the ring $W_m(k)$ of $p$-typical Witt vectors of length $m$ and the endomorphism $\wp:W_m(k)\to W_m(k)$ of the additive group, and proved that $L=k(\wp^{-1}(D))$ for some order-$p^m$ cyclic subgroup $D\subset W_m(k)/\wp(W_m(k))$, and conversely. 
There were many other papers in the same volume of Crelle 176 (1937) applying Witt vectors to other outstanding problems.  My favourite is Hasse's characterisation of those $\alpha$ in a finite extension $K$  of ${\mathbb Q}_p$ containing a primitive $p^m$-th root of $1$ for which the extension $K(\root p^m\of\alpha)|K$ is unramified ($p^m$-primary numbers; see for example the book by Fesenko and Vostokov, freely available on Fesenko's homepage).
See also Harder, Wittvektoren,
Jahresber. Deutsch. Math.-Verein. 99 (1997), no. 1, 18--48.
An English translation of this paper has appeared in Ernst Witt, Gesammelte Abhandlungen, Springer, Berlin, 1996.<|endoftext|>
TITLE: Rings over which every module is free
QUESTION [19 upvotes]: We know that modules over skewfields are free. Is the converse true? In other words, is it true that a nontrivial ring over which every module is free is a skewfield?
If the ring A is commutative, then writing that for any proper ideal I of A, the A-module A/I is free yields the result. What about the general case?

REPLY [14 votes]: Caution: the zero ring is not a division ring, but all of its modules are free!  But other arguments already given here show that any nonzero ring over which every right module is free is a division ring.
I'll pipe in with one more approach, a personal favorite.  Assume that every right R-module is free and that R ≠ 0.  Let S be a simple right R-module (at least one exists because R is nonzero: choose any maximal right ideal M of R and let S = R/M).  Then S is free, so choose a basis for S.  Pick any element x of the basis.  Because S is simple and x ∈ S∖{0} we have S = xR, and because x was a basis element we have xR ≅ R as right R-modules.  In particular, RR ≅ SR is simple, from which it follows that R is a division ring.<|endoftext|>
TITLE: what is the connection between D-modules and coordinate bundles?
QUESTION [8 upvotes]: Fix $n$ and a field $k$ of characteristic zero.  Let $G$ be the pro-algebraic group of automorphims of $k[[x_1,...x_n]]$.  Let $G_0$ be the subgroup of automorphisms preserving the closed point  (note that for general $T$, $G_0(T)$ can be a proper subgroup of $G(T)$).  Let $X$ be a regular variety over $k$ and let $P$ be the principal $G$ bundle of formal coordinate systems, naturally a $G$ torsor over $X$.  I hear that there is a connection between $P$ and $D_X$-modules.  what is this connection?

REPLY [4 votes]: Assume X is n-dimensional and regular.  Then there is a functor from G-modules V to DX-modules, given by an associated bundle construction.  Take the trivial (ind-)bundle on P with fiber V, and quotient by the action of G on P and V.  If you replace G with G0 and P with the canonical G0-torsor, the same construction yields an OX-module.  The extra structure of a G-action lets you identify infinitesimally nearby fibers.<|endoftext|>
TITLE: Limit Linear Series
QUESTION [8 upvotes]: A linear series on a curve C is a line bundle L together with a subspace V of the global sections of L.  Eisenbud and Harris develeoped a theory of limit linear series which explans how (L, V) degenerate when C does.  I was under the impression that one of the nice things about limit linear series is it allows you to prove statements by induction on the genus of a curve.  What are some examples of such statements that you can prove like this?

REPLY [8 votes]: Not sure off the top of my head whether any of the Eisenbud-Harris papers literally used induction. One example is my paper "Linked Grassmannians and crude limit linear series" which gives a simple inductive proof of the Brill-Noether theorem using limit linear series. One can give similar arguments for existence of certain maps with prescribed ramification in characteristic p (see Theorem 7.1 of version 1 on the arxiv of "Linear series and existence of branched covers). 
But more broadly, one might say that the "inductive" structure of limit linear series is encapsulated by the fact that one can describe limit linear series component by component (in contrast to the situation for higher rank vector bundles, where one has to worry about gluing maps). Many arguments can either be phrased in terms of induction or by degenerating immediately to a maximally degenerate case (e.g., a comb curve, or a chain of elliptic curves), and are essentially the same either way. But the point either way is that the machinery reduces to studying linear series on the individual components of the degeneration, which have smaller genus.<|endoftext|>
TITLE: How small can a group with an n-dimensional irreducible complex representation be?
QUESTION [18 upvotes]: More precisely, what is the smallest exponent e such that, for every n, there exists a group of size at most Cn^e for some absolute constant C and with an n-dimensional irreducible complex representation?  
I know that e ≤ 3.   For various special values of n we can attain the lower bound e ≥ 2.  For example, if n+1 = q is a power of a prime then the group of affine linear transformations ax + b on Fq is doubly transitive, so the linearization of this group action decomposes into the trivial representation and an irreducible representation.
Edit:  Oops.  Reading Noah's comment, I probably should've justified the upper bound.  I hope this example is correct: let zeta be an nth root of unity and consider the group generated by the diagonal matrix (zeta, zeta, ... zeta), the diagonal matrix (1, zeta, zeta^2, ... zeta^{n-1}), and a cyclic permutation matrix of order n.  It can be verified that this is a group of order n^3 and that its elements span M_n(C) , so the given representation is irreducible.

REPLY [2 votes]: Note that for every odd prime $p$, the finite group ${\rm PSL}(2,p)$ has an irreducible complex character of degree $p$, and has order $\frac{p(p^{2}-1)}{2}$.  Hence if $n = p_{1}^{a_{2}}\ldots p_{r}^{a_{r}}$ where each $a_{i} \in \mathbb{N}$ and the $p_{i}$ are distinct odd primes, then there is a finite group $G$ of order less than 
$\frac{n^{3}}{2^{t}}$, where $t$ is the number of odd prime divisors of $n$ (counting multiplicities), which has a faithful complex irreducible character of degree $n$. The group $G$ is a direct product of groups of the form ${\rm PSL}(2,p_{i})$
($a_{i}$ factors ${\rm PSL}(2,p_{i})$ for each $i$). If $n$ is as above, and $h = 2^{a}n$ for some positive integer $a$, then we may take the direct product of $a$ copies of ${\rm SL}(2,2)$ with the previous group to obtain a group of order less than $\left(\frac{3}{4}\right)^{a} \frac{h^{3}}{2^{t}}$ with an irreducible character of degree $h$ (and if $h = 2^{a}$, a case which, strictly speaking, we have not yet covered, we obtain a group of order $\frac{2^{3a}3^{a}}{4^{a}}$ with a faithful complex irreducible character of degree $2^{a}$). Actually, since $S_{4}$ has an irreducible character of degree $4$, 
we can find a group of order $(24)^{\frac{a}{2}}$ with an irreducible character of degree $2^{a}$ when $a \geq 2$ is even, which improves $6^{a}.$ Likewise, we can find a group of order $ 6 \times (24)^{\frac{a-1}{2}}$ with an irreducible character of degree $2^{a}$ when $a \geq 1$ is odd, which improves $6^{a}$ for $a \geq 3$ So this improvement could also be used to improve cases when $h$ above is even).
This is only a marginal improvement of the bound obtained by the construction in the question, and does not impact on the optimal choice of $e$ asked for.<|endoftext|>
TITLE: Largest hyperbolic disk embeddable in Euclidean 3-space?
QUESTION [26 upvotes]: Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$.  On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder).  What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?
Edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere.  First, the pseudosphere is parametrized by the region $$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ on the upper half-plane model of $H^2$.  Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.
Next, Euclidean circles drawn on in the upper half-plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$.  I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $\mathrm{PS}$.  This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.
Surely one can do better?
Edit 2: fixed mistakes in formulas above (didn't affect the bound).  Here're some pictures:

REPLY [23 votes]: I didn't see the exact answer to your question in the Borisenko paper, since section 2.4 only seems to address immersions of subsets of ℍ2 into ℝ3. However, a perturbation of the pseudosphere, Dini's surface, which is an isometrically embedded one-sided tubular neighborhood of a geodesic in the hyperbolic plane (see https://mathoverflow.net/a/149884/1345), seems to do the trick since it contains arbitrarily large disks in the hyperbolic plane. See Dini's Surface at the Geometry Center.<|endoftext|>
TITLE: Why are subfactors interesting? 
QUESTION [24 upvotes]: I get asked this question a lot, and am not very happy with any of the answers. 
Vaguely I think of subfactor theory as a generalization of representation theory of groups. That is, if you have a group/subgroup subfactor, and look at the fusion category, you get a category that has all of the induction and restriction data for the subgroup (I think). So maybe that is my first question, what group theoretic information can you extract from the fusion category of a subgroup subfactor? You can do a similar thing with representations of quantum groups, and you also get "sporadic" subfactors, which don't come from groups or quantum groups. This is interesting because it looks a little like the classification of finite groups, you have a bunch of families and then the fun unexpected sporadic groups. I would love to hear someone else's opinion, if anyone can make the subfactors/group theory analogy more formal. 
The other explanation as to the interest in subfactors that I hear is, "they have a lot of surprising connections to other topics and show up all over the place." Can someone tell me about such an instance? What are some other topics where subfactors unexpectedly showed up? I am vaguely aware of something to do with random matrices ...
Of course, the easy answer is that they are really beautiful and cool in their own right, and no one has to convince me of that.

REPLY [2 votes]: The title of this talk suggests that there is a kind of refinement of Galois theory associated with subfactors, but I can't watch it here. Do you know where one can read about that?<|endoftext|>
TITLE: Specializations of Schur functions at consecutive integers
QUESTION [6 upvotes]: Given a partition λ = (λ1, λ2, ..., λn) denote with sλ the associated Schur function.
There exists a nice product formula for the principal specializations:
sλ(1, q, q2, ..., qn-1) = Πi<j (qλi+n-i - qλj+n-j) / (qj-1 - qi-1).
Is a similar evaluation known for specializations of the type sλ(1, 2, ..., n)?

REPLY [6 votes]: Let's take the strictest meaning of "similar evaluation", i.e., a product formula.  The by the remark, the answer is probably no, because for instance the Stirling number s(9,3) = 118124 = 2*2*29531.
A different meaning is whether there is some explicit combination of nested sums and products for a Schur evaluation at consecutive integers.  I don't know the answer to that.  Unlike for a straight product formula, it is difficult to look for such formulas or rule them out.
The loosest possible meaning is whether there is an efficient way to evaluate sλ(1, 2,...,n).  The answer to that is an easy yes, because you can use the Jacobi-Trudi determinant.<|endoftext|>
TITLE: Does finite mathematics need the axiom of infinity?
QUESTION [34 upvotes]: A statement referring to an infinite set can sometimes be logically rephrased using only finite sets/objects.  For example, "The set of primes is infinite" <-> "There is no largest prime".  Pleasantly, the proof of this statement does not seem to need infinity either (assume a largest prime, contradiction).  
What reason is there, other than convenience or curiosity, to adjoin infinite sets to our universe by axiomatically declaring that one exists?
Specifically:

What is an example of a theorem in ZF or ZFC which 1) does not refer to infinite sets, but 2) cannot be proven if the axiom of infinity is excluded?

(See Zermelo–Fraenkel set theory for the axiom of infinity in context.)

REPLY [22 votes]: Note that this question is part of Hilbert's program. (http://en.wikipedia.org/wiki/Hilbert%27s_program)
The first part is to formalize mathematics.  Some argue that the development of ZFC, together with the work of Russell/Whitehead and Bourbaki and others has done this, or at least shown that it is possible.
The second part is to prove that the resulting axiomatization is complete.  Godel's results show that this can't be done.
The third part is to show (finitistically) that formalized mathematics is consistent.  Godel's results show that standard finitistic mathematics (like Peano arithmetic) can't even prove itself consistent, much less the full formalized mathematics, but of course it doesn't rule out the possibility of finitistic systems like PA+Con(ZFC).  Hilbert didn't anticipate this sort of result because he didn't realize that every theory even of finitistic mathematics would be incomplete.
The fourth part is what you ask here, to show that full formalized mathematics is conservative over the finitistic part.  Since ZFC proves Con(PA), but PA doesn't, Godel's results already show that this was impossible.  Paris-Harrington, and Goodstein's theorems, and other examples mentioned above are nice additions only because they show that relatively straightforward statements are independent.  (All three of these results are just statements that every natural number has some particular property, though the property used in Con(PA) is the property of not being the Godel code of a proof of a contradiction from the axioms of PA, which is more complicated than the properties used in the others.)
The fifth part of Hilbert's program was to give a decision procedure for all mathematical statements.  This of course is far more ambitious even than the 10th problem and various other problems that have turned out to be impossible.  But again, Godel's results already showed that this part was doomed, because any decision procedure could be used to generate a complete axiomatization of mathematics.<|endoftext|>
TITLE: What is the affinization of M_g?
QUESTION [8 upvotes]: This question is inspired by What is an example of a function on M_g? . Consider Mg, the moduli space of genus g curves, NOT compactified. When g is 3 or greater, this is not affine. Does anyone know a good description of the ring of global functions on Mg, or of the spectrum of this ring?

REPLY [10 votes]: By exercise 2.10 of Moduli of Curves by Harris and Morrison there exists a complete curve through any two points of M_g (this follows from the fact that the `boundary of the Hodge theory' compactification of M_g has codimension at least 2 and that M_g is quasi-projective). It follows the only global functions on M_g are constant.

REPLY [8 votes]: Doesn't M_g have codimension 2 in the Satake compactification for g > 2?<|endoftext|>
TITLE: Deformation theory of representations of an algebraic group
QUESTION [29 upvotes]: For an algebraic group G and a representation V, I think it's a standard result (but I don't have a reference) that

the obstruction to deforming V as a representation of G is an element of H2(G,V⊗V*)
if the obstruction is zero, isomorphism classes of deformations are parameterized by H1(G,V⊗V*)
automorphisms of a given deformation (as a deformation of V; i.e. restricting to the identity modulo your square-zero ideal) are parameterized by H0(G,V⊗V*)

where the Hi refer to standard group cohomology (derived functors of invariants). The analogous statement, where the algebraic group G is replaced by a Lie algebra g and group cohomology is replaced by Lie algebra cohomology, is true, but the only proof I know is a big calculation. I started running the calculation for the case of an algebraic group, and it looks like it works, but it's a mess. Surely there's a long exact sequence out there, or some homological algebra cleverness, that proves this result cleanly. Does anybody know how to do this, or have a reference for these results? This feels like an application of cotangent complex ninjitsu, but I guess that's true about all deformation problems.
While I'm at it, I'd also like to prove that the obstruction, isoclass, and automorphism spaces of deformations of G as a group are H3(G,Ad), H2(G,Ad), and H1(G,Ad), respectively. Again, I can prove the Lie algebra analogues of these results by an unenlightening calculation.
Background: What's a deformation? Why do I care?
I may as well explain exactly what I mean by "a deformation" and why I care about them. Last things first, why do I care? The idea is to study the moduli space of representations, which essentially means understanding how representations of a group behave in families. That is, given a representation V of G, what possible representations could appear "nearby" in a family of representations parameterized by, say, a curve? The appropriate formalization of "nearby" is to consider families over a local ring. If you're thinking of a representation as a matrix for every element of the group, you should imagine that I want to replace every matrix entry (which is a number) by a power series whose constant term is the original entry, in such a way that the matrices still compose correctly. It's useful to look "even more locally" by considering families over complete local rings (think: now I just take formal power series, ignoring convergence issues). This is a limit of families over Artin rings (think: truncated power series, where I set xn=0 for large enough n).
So here's what I mean precisely. Suppose A and A' are Artin rings, where A' is a square-zero extension of A (i.e. we're given a surjection f:A'→A such that I:=ker(f) is a square-zero ideal in A'). A representation of G over A is a free module V over A together with an action of G. A deformation of V to A' is a free module V' over A' with an action of G so that when I reduce V' modulo I (tensor with A over A'), I get V (with the action I had before). An automorphism of a deformation V' of V as a deformation is an automorphism V'→V' whose reduction modulo I is the identity map on V. The "obstruction to deforming" V is something somewhere which is zero if and only if a deformation exists.
I should add that the obstruction, isoclass, and automorphism spaces will of course depend on the ideal I. They should really be cohomology groups with coefficients in V⊗V*⊗I, but I think it's normal to omit the I in casual conversation.

REPLY [5 votes]: About what Anton said at the end about deformations of a group.    Let $m_0$ be the standard multiplication.  Then I want to consider a deformation of the form $m:(G \times \epsilon \mathfrak{g}) \times (G \times \epsilon \mathfrak{g}) \to G \times \epsilon \mathfrak{g}$ where $m(g_1, g_2) = m_{0}(g_1,g_2) + \epsilon m_1 (g_1,g_2)$.  When you write out the associativity condition $m\circ (m \times 1) = m \circ (1 \times m)$ it seems that you find that 
$(g_1,g_2) \mapsto (m_{1}(g_{1},g_{2}))(g_{1}g_{2})^{-1}$ is a group cohomology cocycle for G acting on $\mathfrak{g}$ by the adjoint representation.  Now one has to identify $H^{2}(G,Ad)$ with $H^{2}(BG,Ad)$ (taking care of the topology somehow).<|endoftext|>
TITLE: Cauchy-Schwarz and pigeonhole
QUESTION [25 upvotes]: I've occasionally heard it stated (most notably on Terry Tao's blog) that "the Cauchy-Schwarz inequality can be viewed as a quantitative strengthening of the pigeonhole principle." I've certainly seen the inequality put to good use, but I haven't seen anything to make me believe that statement on the same level that I believe that the probabilistic method can be used as a (vast) strengthening of pigeonhole. 
So, how exactly can Cauchy-Schwarz be seen as a quantitative version of the pigeonhole principle? And for extra pigeonholey goodness, are there similarly powered-up versions of the principle's other generalizations? (Linear algebra arguments [particularly dimension arguments], the probabilistic method, etc.)

REPLY [2 votes]: The contrapositive of PH is: If you put at most one pigeon per hole, then you have at most $n$ pigeons. Letting $a = (1,\dots,1)$ and $b_i$ be the number of pigeons in hole $i$ with $b_i \in \{0,1\}$, then Cauchy-Schwarz is
$$ \text{number of pigeons} = \sum_i b_i \leq \sqrt{n} \|b\| . $$
We have $\|b\| \leq \sqrt{n}$ so this relaxes to say there are at most $n$ pigeons. So, maybe this is a good generalization of PH:
Theorem. Suppose that pigeon species $i$ weighs $a_i$ pounds. If the total weight of your pigeons exceeds $\|a\| \|B\|$ where $B \geq \vec{0}$, then there exists a species $i$ for which you have more than $B_i$ pigeons.
Proof. By contrapositive, if you have $b_i \in [0,B_i]$ pigeons of each species, then the total weight is $\sum_i a_i b_i \leq \|a\| \|b\| \leq \|a\| \|B\|$.<|endoftext|>
TITLE: Reading for finite Fourier analysis
QUESTION [9 upvotes]: Can anyone recommend some good reading for Fourier analysis (and the Fourier transform) over finite abelian groups? I've found it given brief descriptions in both books on representation theory and on regular Fourier analysis (the best so far in the chapter in Tao and Vu's Additive combinatorics), but can't find a dedicated, detailed exposition of this area proper.

REPLY [3 votes]: Belatedly, considering that the question was asked 13 years ago: I have some notes on Fourier analysis on finite abelian groups, as an early part of a repn theory course. Just a few pages, but trying to be forward-looking.
https://www-users.cse.umn.edu/~garrett/m/repns/notes_2014-15/01_finite_abelian.pdf<|endoftext|>
TITLE: Differentials in the Lyndon-Hochschild spectral sequence
QUESTION [6 upvotes]: The Lyndon-Hochschild(-Serre) spectral sequence applies to group extensions in a manner analogous to the Serre-Leray spectral sequence applied to a fibration.
Does anyone know of a good description (or reference) of the transgression maps in the Lyndon-Hochschild spectral sequence? MacLane describes them in terms of an additive relation, but I don't find this helpful in computing them.
More generally, I don't know how to calculate the differentials in this spectral sequence. In the Serre spectral sequence, I can see how an exact couple arises and the differentials are straightforward to see if not easy to calculate. But the LHSS arises from a double complex and I'm not sure how to get an exact couple from this.

REPLY [8 votes]: If you only need the first couple of transgressions, there is a nice description of them in the paper
MR0641328 (83a:18021)
Huebschmann, Johannes
Automorphisms of group extensions and differentials in the Lyndon\mhy Hochschild\mhy Serre spectral sequence.
J. Algebra 72 (1981), no. 2, 296--334.
For concrete calculations, this description is sometimes easier than the abstract nonsense description from the double complex.<|endoftext|>
TITLE: Logarithmic structures on moduli of elliptic curves over Z
QUESTION [9 upvotes]: I've heard it stated that if you take the moduli of elliptic curves with some level structure imposed (as a moduli scheme over Spec(Z)), there is a logarithmic structure that you can impose at the cusps so that the natural projection maps obtained by forgetting the level structure are log-etale (at least away from primes dividing the order of your level structure).
I can have some rough intuition about how this happens over a field of characteristic zero, but not integrally.  Can anybody explain this or give me a reference for this structure?
Additionally, has anybody worked out the appropriate integral ring of modular forms with logarithmic structure in some cases, similar to the Deligne-Tate calculation of modular forms over Z?

REPLY [4 votes]: I thing Kato's log purity theorem gives you this.  See, for instance, Theorem B in Mochizuki's "Extending Families of Curves over Log Regular Schemes."  I think all you need is that the cusps form a normal crossings divisor on X(1) [if you're worried about X(1) being a stack rather than a scheme, you can start with a bit of extra level structure coprime to the primes you're interested in] and then your map Y(N) -> Y(1) is tamely ramified, which tells you that the normalization X(N) of X(1) in Y(N) carries a canonical log-structure in which the map X(N) -> X(1) is log-etale.<|endoftext|>
TITLE: Ideals in Factors
QUESTION [13 upvotes]: One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $III$ factors are algebraically simple (any 2-sided ideal must contain a projection. All projections are comparable in a factor, so you can show 1 is in the ideal). Ideals in $B(H)$ ($\dim(H)=\infty$, $H$ separable) have been studied extensively. What about ideals in $II_\infty$ factors?
One might think, since every $II_\infty$ factor $M$ can be written as $N\overline{\otimes} B(H)$ for $N$ a $II_1$ factor, if $I\subset B(H)$ is an ideal, then $N\otimes I$ is a 2-sided ideal. This is false. One needs to take the ideal generated by $N\otimes I$. What does that mean from a von Neumann algebra viewpoint? Is it the same as taking the norm closure? 
We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2=\{x\in M | tr(x^\ast x)<\infty\}$$ and the trace class operators: 
$$I_1=\{x\in M | tr(|x|)<\infty\}=I_2^\ast I_2 =\left\{\sum^n_{i=1} x_i^\ast y_i | x_i, y_i\in I_2\right\}.$$ 
What is the relation of $I_j$ to $N\otimes L^j(H)$ for $j=1,2$ (where $L^2(H)$ is the Hilbert-Schmidt operators and $L^1(H)$ is the trace class operators in $B(H)$)? Is $I_j$ the norm closure of $N\otimes L^j(H)$?

REPLY [4 votes]: Regarding your first question, $N\otimes I$ is the subalgebra of "diagonal" operators, which is already norm-closed, and it is by no means an ideal. The ideal you are looking for is the ideal generated by the finite projections, which is $N\otimes K(H)$ (that is, the norm closure of the set of "finite matrices" with entries in N). 
And I think it is correct that $I_j$ is the norm closure of $N\otimes L^j(H)$, $j=1,2$.<|endoftext|>
TITLE: "synthetic" reasoning applied to algebraic geometry
QUESTION [16 upvotes]: A hyperlinked and more detailed version of this question is at 

nLab:synthetic differential geometry applied to algebraic geometry. 

Repliers are kindly encouraged to copy-and-paste relevant bits of their reply here into that wiki page.


The axioms of synthetic differential geometry are intended to pin down the minimum abstract nonsense necessary for talking about the differential aspect of differential geometry using concrete objects that model infinitesimal spaces.
But the typical models for the axioms – the typical smooth toposes – are constructed in close analogy to the general mechanism of algebraic geometry: well-adapted models for smooth toposes use sheaves on C ∞Ring op (the opposite category of smooth algebras) where spaces in algebraic geometry (such as schemes) uses sheaves on CRing op.
In fact, for instance also the topos of presheaves on k−Alg op, which one may think of as being a context in which much of algebraic geometry over a field k takes place, happens to satisfy the axioms of a smooth topos (see the examples there).
This raises some questions.
Quesions:
To which degree do results in algebraic geometry depend on the choice of site CRing op or similar?
To which degree are these results valid in a much wider context of any smooth topos, or smooth topos with certain extra assumptions?
In the general context of structured (∞,1)-toposes and generalized schemes: how much of the usual lore depends on the choice of the (simplicial)ring-theoretic Zariski or etale (pre)geometry (for structured (∞,1)-toposes), how much works more generally?
More concretely:
To which degree can the notion of quasicoherent sheaf generalize from a context modeled on the site CRing to a more general context. What is, for instance, a quasicoherent sheaf on a derived smooth manifold? If at all? What on a general generalized scheme, if at all?
Closely related to that: David Ben-Zvi et al have developed a beautiful theory of integral transforms on derived ∞-stacks.
But in their construction it is always assumed that the underlying site is the (derived) algebraic one, something like simplicial rings.
How much of their construction actually depends on that assumption? How much of this work carries over to other choices of geometries?
For instance, when replacing the category of rings /affine schemes in this setup with that of smooth algebra / smooth loci, how much of the theory can be carried over?
It seems that the crucial and maybe only point where they use the concrete form of their underlying site is the definition of quasicoherent sheaf on a derived stack there, which uses essentially verbatim the usual definition QC(−):Spec(A)↦AMod.
What is that more generally? What is AMod for A a smooth algebra? (In fact I have an idea for that which I will describe on the wiki page in a moment. But would still be interested in hearing opinions.)
Maybe there is a more intrinsic way to say what quasicoherent sheaves on an ∞-stack are, such that it makes sense on more general generalized schemes.

REPLY [4 votes]: I think I found the answer to the question: it's easy using the basic insight from Lurie's Deformation Theory.
I have written up what I think the answer is here: oo-vector bundle.<|endoftext|>
TITLE: Non-conjugate words with the same trace
QUESTION [10 upvotes]: Let n>=2, p a large prime, G = SL_n(Z/pZ).
If n=2, there are words that, while not conjugate in the free group, do have identical trace in G. For example, tr(g h^2 g^2 h)= tr(g^2 h^2 g h) for all g, h in SL_2(Z/pZ).
Question: does the same happen for n>=3?
Could it even be possible that there are words w_1, w_2 that are not conjugate even in G=SL_n(Z/pZ), yet always have the same trace: 
tr(w_1(g_1,g_2,..,g_k)) = tr(w_2(g_1,g_2,...,g_k)) for all g_1,...,g_k in SL_n(Z/pZ)?
This would be extremely helpful.
Actually, I just need a weaker statement:
Wild guess.- Let g, h be elements of SL_n(K), n>2. There is a constant
k (which may depend on n but not on K)
such that there are two elements a, b of the ball
({g,h,g^{-1},h^{-1},e})^k for which
(1) tr(a)=tr(b) and
(2) a is not conjugate to b (for g and h generic).
Does anybody have a clue as to whether this is or isn't true?

REPLY [12 votes]: I don't think you asked the question you intended. Even though this question is old, I'll say a bit:
For any fixed $n$ and $p$, there are a finite number of homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$.
Hence, there are only a finite number of possibilities for the trace of a group element as a function of the representation.  Therefore, there are many duplicates.
I think a better question is to ask whether there are pairs $(g_1, g_2)$ of non-conjugate elements of $F_2$ such that for all primes $p$ and all homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$ the traces are equal. In this form, you can pass to a limit $p \rightarrow \infty$ and conclude if nonconjugate elements can be distinguished by traces for an infinite sequence of $SL_n(Z/pZ)$, they could be distinguished by a  homomorphisms to $SL_n(\mathbb C)$. Conversely, if there are no trace identities in $SL_n(\mathbb C)$, then there are no trace identities true in all $SL_n(Z/pZ)$, by finding $Z/pZ$ quotients of the ring of coefficients.
Martin Kassabov has been interested in the latter question, and in discussions he and I have had, we've both come to the opinion that there are probably no trace identitites for $SL_n(C)$ when $n > 2$, but it's not easy to find a proof.  One possible strategy is to first characterize all trace identities in $SL_2$, and then construct representations in $SL_3$ where they break down.  This is interesting to me in any case because of its meaning in 2 and 3-manifold topology -- the trace identities give collections of distinct elements in $\pi_1$ that are forced to have the equal length in any hyperbolic structure.
A weaker question is whether the characteristic polynomials for representations in $SL_n$ can distinguish conjugacy classes in $F_2$.<|endoftext|>
TITLE: Linear Algebra Over $F_{2}$
QUESTION [5 upvotes]: Suppose we call a subset S of $F^{n}$ ($F$ is the field with two elements) good if for any $x$ and $y$ (possibly $x=y$) we have $[x,y]=1$ where $[ , ]$ denotes the obvious bilinear form on F. What's the best bound as a function of $n$ on the size of $S$? In particular, do we have $|S|$ less than or equal to $n$ (or just a linear function in $n$)?

REPLY [6 votes]: Andy's example is actually within a constant factor of optimal.  To see this, imagine adding on an additional coordinate equal to 1 to each element of S.  These new elements now satisfy [x',y']=0.  
Now let W be the subspace of F^{n+1} spanned by the augmentation of S.  The above relation is equivalent to saying that W is contained in its own orthogonal complement.  Since the dimension of a space and that of its complement add to n+1, we know the dimension of W is at most (n+1)/2, so W contains at most 2^{(n+1)/2} vectors.  
This result (and proof) is one I've heard referred to as the Eventown theorem and attributed to Berlekamp.
As an interesting side note, if we replaced the condition [x,y]=1 for all x,y by one saying that [x,y]=1 if and only if x does not equal y, then we'd have a bound of n on the size of S (this is Berlekamp's oddtown theorem).<|endoftext|>
TITLE: What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf?
QUESTION [25 upvotes]: There is a standard way to construct the sheafification of a presheaf on a Grothendieck topology which involves matching families. Details may be found here:
http://ncatlab.org/nlab/show/matching+family
In short, there is a functor + sending presheaves to separated presheaves and then separated presheaves to sheaves. So P^++ is always a sheaf.
Gelfand/Manin's Methods of Homological Algebra has a wrong proof that P^+ is a sheaf, and I have seen in several places a proof that P^++ is a sheaf. However, it seems that for any presheaf P I run into, P^+ is already a sheaf.
Does anyone know an example of a presheaf P where P^+ is not a sheaf i.e. where you actually need to apply the functor + twice to get a sheaf?

REPLY [7 votes]: An example is given in MacLane's "Sheaves in Geometry and Logic." Consider the constant presheaf on a space $X$ with $P(U) = S$ where $S$ is a set with more than one element and restriction maps are identities. The plus construction doesn't change anything except that $P(0) = 0$, and one can show easily that this is not a sheaf.
In general it is true that the plus construction turns separated presheaves into sheaves and any presheaf into a separated presheaf; hence ++ = sheafification.<|endoftext|>
TITLE: Two finite groups with the same identical relations?
QUESTION [5 upvotes]: An identical relation on a group G is a word w in Fr, the free group on r elements (for some r), such that evaluating w on any r-tuple of elements of G yields the identity (this just means substituting elements of g for the variables in w, and evaluating the product).  
Does the complete set of identical relations characterize a finite group? That is, are there two finite groups with precisely the same set of identical relations?

REPLY [5 votes]: Two finite-dimensional prime algebras (in a certain extended sense, what includes the usual algebras with binary multiplication over a field) with the same identities are isomorphic over the algebraic closure of the ground field (Yu.P. Razmyslov,  Identities of Algebras and Their Representations, AMS, 1994, p. 30 onwards). This suggests that the proper condition one should impose on finite groups to guarantee that the same identities imply isomorphism, would be something related to primeness. 
Unfortunately, Razmyslov's reasonings seem not be extendible to a broader class of algebraic systems, in particular, to groups: the linear structure is crucial there (a relatively free algebra in the corresponding variety is enlarged, via the action of a suitable extension of the ground field, to an algebra which is isomorphic to an extension of the initial algebra). Also, it is not clear what the group-theoretic analog of primeness should be in this context.
However, there is an old result saying that two finite simple groups with the same identities are isomorphic (H. Neumann, Varieties of Groups, p. 166, Corollary 53.35). I believe that the machinery developed in that book (critical groups, etc.) would allow to answer this question for any reasonably defined class of finite groups.<|endoftext|>
TITLE: A complex manifold which is quasiprojective in two different ways
QUESTION [13 upvotes]: Does there exist a complex manifold M which is a quasiprojective variety in two "essentially" different ways?  Let me be more specific.  I'm looking for a complex manifold M together with two projective varieties X and Y such that M is a locally closed Zariski dense subset of both X and Y.  This gives two different notions of "algebraic" objects on M, and I want them to be different.  For instance, can the resulting sheaves of regular function on M be different?  Or the Picard groups?  etc...

REPLY [3 votes]: This may be relevant, although I'm not sure if it answers your question directly. The Russell cubic x+x^2y+t^2+z^3=0 is diffeomorphic to affine space A^3. But, it is known not to be algebraically isomorphic to A^3. I'm afraid I don't know the references for this.<|endoftext|>
TITLE: Quotient of a category by a free group action
QUESTION [6 upvotes]: Let Cat denote the 1-category of small categories.  The functor Mor : Cat -> Set which assigns to a category its set of morphisms (aka Hom([• -> •], -)) does not commute with most colimits.  Does it commute with quotients by free group actions?  In other words, if C is a small category and G is a group acting on C such that the action of G on the objects of C is free, does Mor(C/G) = (Mor C)/G?

REPLY [2 votes]: This came up in something I'm working on, and this question was the first hit on google.
It seems to me that this has nothing to do with G acting freely, and that the answer is always yes.  One can always define a quotient category C/G with objects (Ob C)/G and morphisms (Mor C)/G, with composition and identities defined in a more-or-less obvious manner (use the action of G to match up two arrows $f: C\to D$ and $h: gD\to E$ whose domains are equal in C/G, and then compose; note that $h\circ g(f)$ and $g^{-1} (h) \circ f$ are equivalent morphisms in the quotient).  I was a little surprised that this makes sense, but it seems to.  
I think it is elementary to check that C/G satisfies the universal property defining the colimit of the functor $BG \to$ Cat.  
Am I missing something?  Edit: YES!  See the comments below.  If you care to see what I was actually thinking about (a version of this which is actually correct) look at my answer to  this subsequent question.<|endoftext|>
TITLE: What is (co)homology, and how does a beginner gain intuition about it?
QUESTION [93 upvotes]: This question comes along with a lot of associated sub-questions, most of which would probably be answered by a sufficiently good introductory text.  So a perfectly acceptable answer to this question would be the name of such a text.  (At this point, however, I would strongly prefer a good intuitive explanation to a rigorous description of the modern theory.  It would also be nice to get some picture of the historical development of the subject.)
Some sub-questions: what does the condition that d^2 = 0 means on an intuitive level?  What's the intuition behind the definition of the boundary operator in simplicial homology?  In what sense does homology count holes?  What does this geometric picture have to do with group extensions?  More generally, how does one recognize when homological ideas would be a useful way to attack a problem or further elucidate an area?

REPLY [57 votes]: I think the most intuitive way to look at topology is as a way to make precise the following idea. A warning: the idea by itself does not define homology, but something much scarier. Homology is what you get when you give up studying the scary but intuitive thing, and try to get something similar, but which you can calculate.
Consider a manifold. Homology is meant to count its submanifolds, up to cobordism. In other words, as out "chains of dimension $n$", take the formal sums of submanifolds of dimension $n$, where the submanifolds might have boundary. The boundary operation $\partial$ just takes the boundary. Notice that the boundary of any manifold is a manifold without boundary, so it's clear that $\partial^2=0$.
Now, the homology of this chain complex counts submanifolds without boundary; two submanifolds are considered different if they are not boundaries of the same higher-dimensional submanifold. If you think of the higher-dimensional submanifolds as a way to "move" one of the submanifolds to another, it makes sense that you might want to think of them as the "same". If a submanifold does not surround a "hole", it is the "same" as the "empty submanifold", this is a sense in which this homology counts holes.

Again, if you try to make this precise, you will run into all kinds of trouble. You'll have to define what a "submanifold" is, whether they can have self-intersections, etc. Then, you'll find that the above kind of "homology" is not really a homotopy invariant, and terribly difficult to calculate.
However, you should compare the above idea to the definition of simplicial homology. You'll see that the cycles you get in simplicial homology are similar to submanifolds, and all the wonderful algebraic machinery will show you that you can actually calculate homology of anything you'd like.
One algebraic topology book that seems to have this approach in it is Bredon's "Topology and Geometry".
The above intuition is especially useful in differential topology. There is a way to make the above idea (called cobordism theory), but you need to know how to use homology to do it. For a taste of it (that doesn't require homology), look in the last chapter of Milnor's "Topology from the Differential Viewpoint".

Three unrelated comments: 

Never expect any intuition from singular (co)homology, and never calculate anything with it; it is merely a tool for showing that other kinds of homology (that you actually care about) are the same and invariant under homotopy.
A completely different (and very precise) way to intuitively think of cohomology is as solutions to a certain differential equation. This is the approach of the "Calculus to Cohomology" book and Bott and Tu's "Differential Forms in Algebraic Topology", and is called De Rham cohomology.
Don't be surprised if there are some mistakes in any of the above; wise people - feel free to point them out and clarify!<|endoftext|>
TITLE: Stalks of sheaf-hom
QUESTION [14 upvotes]: Let $F$ and $G$ be sheaves on $X$. Under what conditions is the natural map from the stalk at $p$ of $\mathcal{H}\kern{-1pt}\mathit{om}(F,G)$ to $\mathrm{Hom}(F_p, G_p)$ an isomorphism?

REPLY [6 votes]: The result in Hartshorne if I recall correctly only really uses the fact that affine locally a coherent sheaf on a scheme has a locally free resolution by finite rank projectives and that one can compute stalks affine locally. In particular, as pointed out by David in the comments we only ready need the first two steps so that one can use the exactness properties of Hom/SheafHom.
So the right condition on F as in the comments is that it be finitely presented.<|endoftext|>
TITLE: Homological algebra and calculus (as in Newton)
QUESTION [21 upvotes]: This question reminded me of a possibly stupid idea that I had a while back.
On page 2 of this paper, while discussing Euclid's axioms of plane geometry and spatial geometry, Manin makes an extremely interesting comment:

Euclid misses a great opportunity here: if he stated the principle
“The extremity of an extremity is empty”,
he could be considered as the discoverer of the
BASIC EQUATION OF HOMOLOGICAL ALGEBRA: d^2 = 0.

Ever since I read this, I've had a suspicion that the equation "d^2 = 0" of homological algebra is somehow related to the equation "epsilon^2 = 0" of (first-order) calculus (as in Newton)*, since the latter equation can be interpreted as saying "a very very small quantity is zero" which at least superficially seems similar to "the extremity of an extremity is empty". I once explained my suspicion to Dan Erman over beers, and he responded by asking another question: Can we do some sort of homological algebra using the equation d^n = 0 rather than d^2 = 0? Perhaps if d^2 = 0 can be related to first-order calculus, then d^3 = 0 can be related to second-order calculus, and so on...
I don't really have a specific question to ask -- I just thought I might put this idea out there. Maybe someone can tell me why this idea is stupid, or why it is not stupid.

*or the ring of dual numbers k[epsilon]/(epsilon^2) if you're an algebraist or an algebraic geometer.

REPLY [3 votes]: There is a whole theory of calculus of functors started by Goodwillie. There are Taylor approximations of functors and so on. Here's the wikipedia page which contains over references:
http://en.wikipedia.org/wiki/Calculus_of_functors<|endoftext|>
TITLE: Can one make Erdős's Ramsey lower bound explicit?
QUESTION [13 upvotes]: Erdős's 1947 probabilistic trick provided a lower exponential bound for the Ramsey number $R(k)$. Is it possible to explicitly construct 2-colourings on exponentially sized graphs without large monochromatic subgraphs?
That is, can we explicitly construct (edge) 2-colourings on graphs of size $c^k$, for some $c>0$, with no monochromatic complete subgraph of size $k$?

REPLY [14 votes]: Finding explicit constructions for Ramsey graphs is a central problem in extremal combinatorics. Indeed, computational complexity gives a way to formalize this problem. Asking for a graph which can be constructed in polynomial time is a fairly good definition although sometimes the definition is taken as having a log-space construction.
Until quite recently the best method for explicit construction was based on extremal combinatorics. The vertices of the graphs were certain sets (say k-subset of an n element sets) and the edges represented pairs of sets with presecibed intersection. The best result was by Frankl and Wilson and it gives a graph with n vertices whose edges are colored by 2 colors with no monochromatic clique of size $\exp (\sqrt{(\log n))}$. (I think this translates to $k^{\log k}$ in the way the question was formulated here.) Using sum-products theorems Barak Rao Shaltiel and Wigderson improved the bound to $\exp (\log n^{o(1)})$.
Payley graphs are conjectured be explicit examples for the correct behavior. But proving it is much beyond reach.
Update(Nov 11, 2015):  Gil Cohen found an explicit construction with no monochromatic cliques of size $2^{(\log \log n)^L}$. An independent construction which applies also to the bipartite case was achieved by Eshan Chattopadhyay and David Zuckerman<|endoftext|>
TITLE: What is a cup-product in group cohomology, and how does it relate to other branches of mathematics?
QUESTION [34 upvotes]: I have a few elementary questions about cup-products.
Can one develop them in an axiomatic approach as in group cohomology itself, and give an existence and uniqueness theorem that includes an explicitly computable map on cochains?  Second, how do they relate to cup-products in algebraic topology?  In general, are there connections between cup-products and other mathematical constructions that may provide more intuition into them?

REPLY [3 votes]: You should definitely take a look at Lang's "Topics in Cohomology of Groups", chapter 4. There a general notion of cup-products on delta-functors is introduced. This may look like a lot of abstract nonsense to most people, but I like it :)<|endoftext|>
TITLE: Handling arXiv feeds to avoid duplicates
QUESTION [30 upvotes]: I subscribe to feeds from the arXiv Front for a number of subject areas, using Google Reader. This is great, but there is one problem: when a new preprint is listed in several subject categories, it gets listed in several feeds, which means I have to spend more time reading through the lists of new items, and due to my slightly dysfunctional memory, I often download the same preprint twice. Is there a way to get around this problem, by somehow merging the feeds, using a different arXiv site, or using some other clever trick?
(Hope this is not too off-topic, I think a good answer could be useful to a number of mathematicians. Also, I would like to tag this "arxiv" but am not allowed to add new tags.)

REPLY [2 votes]: I wrote a BASH script for automated downloading of selected arXiv RSS feeds.
That content is deduplicated, then parsed into one of two documents:

keyword-matched articles of interest;
the remaining articles.

The script can be scheduled to run daily via crontab, or manually executed.
By example, today (Jan 10, 2019) among five arXiv RSS feeds my RSS reader provided 690 entries (including duplicated, cross-posted entries) while my script returned 344 de-duplicated (unique) articles.

script | files: https://github.com/victoriastuart/arxiv-rss
accompanying research blog post: https://persagen.com/2019/06/10/arxiv-rss.html<|endoftext|>
TITLE: Finiteness conditions on simplicial sheaves/presheaves
QUESTION [10 upvotes]: Could someone give an overview, or just some examples, of "finiteness conditions" for simplicial sheaves/presheaves and/or simplicial schemes? Any answer or comment about this would be interesting, but I am interested in particular in the following two things:
1) I once heard Toen say something about this, and that one can express some kind of condition on simplicial sheaves/presheaves/schemes in terms of convergence of some power series, or something along these lines. What kind of power series is this, and what are the definitions/statements?
2) As explained for example in Deligne's classical papers on Hodge theory, a cohomology theory for varieties (defined as hypercohomology of a complex of sheaves, say) can be extended to simplicial varieties using a spectral sequence. I suspect (but am not sure) that this works well in some sense only under some kind of condition on the simplicial variety, but what would such a condition be? 
Some background: I am interested primarily in simplicial sheaves/presheaves on a site coming from algebraic geometry, which would typically be some category of schemes equipped with the Zariski/Nisnevich/etale/flat topology. A simplicial sheaf/presheaf should be thought of a "generalized space", some of the most important examples being the "motivic spaces" in A1-homotopy theory, and stacks in the sense of Toen.

REPLY [16 votes]: I don't know what Toën was talking about, but I suspect that it was about finiteness conditions for Artin stacks: the problem is that the usual finiteness conditions we look at for schemes (like the notion of constructibility for l-adic sheaves) do not extend to stacks in a straightforward way, which gives some trouble if one wants to count points
(i.e. to define things like Euler characteristics). Some notions of finiteness are developed to define Grothendieck rings of Artin stacks (e.g. in Toën's paper arXiv:0509098 and in Ekedahl's paper arXiv:0903.3143), which can be realized by our favourite cohomologies (l-adic, Hodge, etc), but the link with a good notion of finiteness for categories of coefficients over Artin stacks (l-adic sheaves, variation of mixed Hodge structures) does not seem to be fully understood yet, at least conceptually (and by myself).
As for finiteness conditions for sheaves (in some homotopical context), the kind of properties we might want to look at are of the following shape.
Consider a variety of you favourite kind X, and a derived category D(X) of sheaves over
some site S associated to X (e.g. open, or étale, or smooth subvarieties over X etc.).
For instance, D(X) might be the homotopy category of the model category of simplicial sheaves, or the derived category of sheaves of R-modules.
Important finiteness properties can be expressed by saying that for any U in the site S,
we have
(1) hocolimᵢ RΓ(U,Fᵢ)= RΓ(U,hocolimᵢ Fᵢ)
where {Fᵢ} is a filtered diagram of coefficients. If you are in such a context, then
you can look at the compact objects in D(X), i.e. the objects A of D(X) such that
(2) hocolimᵢ RHom(A,Fᵢ)= RHom(A,hocolimᵢ Fᵢ)
for any filtered diagram {Fᵢ}. In good situations, condition (1) will imply that the category of compact objects will coincide with constructible objects (i.e. the smallest subcategory of D(X) stable under finite homotopy colimits (finite meaning: indexed by finite posets) which contains the representable objects).
Sufficient conditions to get (1) are the following:
a) For simplicial sheaves (as well as sheaves of spectra or R-modules...), a sufficient condition is that the topology on S is defined by a cd-structure in the sense of Voevodsky (see arXiv:0805.4578). These include the Zariski topology, the Nisnevich topology, as well as the cdh topology (the latter being generated by Nisnevich coverings as well as by blow-ups in a suitable sense), at least if we work with noetherian schemes of finite dimension. Note also that topologies associated to cd structures define what Morel and Voevodsky call a site of finite type (in the language of Lurie, this means that, for such sites, the notion of descent is the same as the notion of hyperdescent: descent for infinity-stacks over S can be tested only using truncated hypercoverings; this is the issue discussed by David Ben Zvi above).
In practice, the existence of a cd structure allows you to express (hyper)descent using only Mayer-Vietoris-like long exact sequences (the case of Zariski topology was discovered in the 70's by Brown and Gersten, and they used it to prove Zariski descent for algebraic K-theory).
b) For complexes of sheaves of R-modules, a sufficient set of conditions are
i) the site S is coherent (in the sense of SGA4).
ii) any object of the site S is of finite cohomological dimension (with coefficients in R).
The idea to prove (1) under assumption b) is that one proves it first when all the Fᵢ's are concentrated in degree 0 (this is done in SGA4 under assumption b)i)).
This implies the result when the Fᵢ's are uniformly bounded. Then, one uses the fact, that, under condition b)ii), the Leray spectral sequence converges strongly, even for unbounded complexes (this done at the begining of the paper of Suslin and Voevodsky "Bloch-Kato conjecture and motivic cohomology with finite coefficients").
This works for instance for étale sheaves of R-modules, where R=Z/n, with n prime to the residual characteristics. Note moreover that, in the derived category of R-modules, the compact objects (i.e. the complexes A satisfying (2)) are precisely the perfect complexes.
The fact that the six Grothendieck operations preserves constructibility can then be translated into the finiteness of cohomology groups (note however that the notion of constructiblity is more complex then this in general: if we work with l-adic sheaves
(with Ekedahl's construction, for instance), then the notion of constructiblity does not agree with compactness anymore). However, condition (1) is preserved after taking the Verdier quotient of D(X) by any thick subcategory T obtained as the smallest thick subcategory which is closed under small sums and which contains a given small set of compact objects of D(X) (this is Thomason's theorem). This is how such nice properties survive in the context of homotopy theory of schemes for instance.
Note also that, in a stable (triangulated) context, condition (2) for A implies that we have the same property, but without requiring the diagrams {Fᵢ} to be filtering.
For your second question, the extension of a cohomology theory to simplicial varieties is automatic (whenever the cohomology is given by a complex of presheaves), at least if we have enough room to take homotopy limits, which is usually the case (and not difficult to force if necessary). The only trouble is that you might lose the finiteness conditions, unless you prove that your favorite simplicial object A satisfies (2). The fact that Hironaka's resolution of singularities gives the good construction (i.e. gives nice objects for open and/or singular varieties) can be expained by finiteness properties related to descent by blow-ups (i.e. cdh descent), but the arguments needed for this use strongly that we work in a stable context (I don't know any argument like this for simplicial sheaves). The fuzzy idea is that if a cohomology theory satisfies Nisnevich descent and homotopy invariance, then it satisfies cdh descent (there is a nice very general proof of this in Voevodsky's paper arXiv:0805.4576 (thm 4.2, where you will see we need to be able to desuspend)); then, thanks to Hironaka, locally for the cdh topology, any scheme is the complement of a strict normal crossing divisor in a projective and smooth variety. As cdh topology has nice finiteness properties (namely a)), and as
any k-scheme of finite type is coherent in the cdh topos, this explains, roughly, why we get nice extensions of our cohomology theories (as far as you had a good knowledge of smooth and projective varieties). If we work with rational coefficients, the same principle applies for schemes over an excellent noetherian scheme S of dimension lesser or equal to 2, using de Jong's results instead of Hironaka's, and replacing the cdh topology by the h topology (the latter being obtained from the cdh topology by adding finite surjective morphisms): it is then sufficient to have a good control of proper regular S-schemes.<|endoftext|>
TITLE: Simplicial objects
QUESTION [38 upvotes]: How should one think about simplicial objects in a category versus actual objects in that category? For example, both for intuition and for practical purposes, what's the difference between a [commutative] ring and a simplicial [commutative] ring?

REPLY [8 votes]: This is a good place to mention the notion of a "Grothendieck test category". This is a small A category which has the property that presheaves of sets on A, with an appropriate class of weak equivalences, models the homotopy theory of spaces.  So the simplicial indexing category Δ is a test category, but there are others, such as the category which indexes cubical sets.  I would guess that whenever one needs to use simplicial rings (say), you could replace simplicial with any test category (though I don't know that anyone has worked this sort of thing out).
These notions are developed in some papers of Cisinski, and a good introduction (which gives the definition of test category) is the paper of Jardine.  Of course, this doesn't really answer your question on how to think about simplicial objects, but perhaps it puts it in a broader context.<|endoftext|>
TITLE: Euler characteristic of a manifold and self-intersection
QUESTION [22 upvotes]: This is probably quite easy, but how do you show that the Euler characteristic of a manifold M (defined for example as the alternating sum of the dimensions of integral cohomology groups) is equal to the self intersection of M in the diagonal (of M × M)?
The few cases which are easy to visualise (ℝ in the plane, S1 in the torus) do not seem to help much.
The Wikipedia article about the Euler class mentions very briefly something about the self-intersection and that does seem relevant, but there are too few details.

REPLY [5 votes]: There is also a nice proof of this fact (which I first saw in Milnor-Stasheff's Characteristic Classes) which involves decomposing the class $\eta(\Delta) \in H^*(M \times M)$ obtained as the Poincare dual of the diagonal $\Delta \subset M \times M$. 
I'll assume $M$ is a compact (not necessarily oriented) $n$-manifold and use $\mathbb{Z}/2$-coefficients. If $M$ is oriented one can use coeffecients in any field. In addition I will assume $n$ is even. It simplifies the argument, and by another theorem in that book, the Euler characteristic of an odd dimensional compact manifold is zero -- so we won't miss out on much. 
Theorems 11.10 and 11.11 on p. 128 show: for each basis $b_1, \dots, b_r$ for $H^*(M)$ there is a dual basis $b_1^\vee, \dots, b_r^\vee$ characterized as follows: if $\mu \in H_n(M)$ is the fundamental class of $M$, then 
$$
\begin{equation}
\label{eq:1}
\tag{1}
\langle b_i \smile b_j^\vee , \mu \rangle = 
\begin{cases}
1, & \text{  if  } i = j \\
0, & \text{  otherwise  }\\
\end{cases}
\end{equation}
$$
Note that $\deg b_i^\vee = n - \deg b_i$  In terms of the $b_i, b_i^\vee$, we have 
$$
\begin{equation}
\label{eq:2}\tag{2}
\eta(\Delta) = \sum_{i=1}^r (-1)^{\deg b_i}b_i \times b_i^\vee \in H^n(M \times M)
\end{equation}
$$
Here's how we use the fact that the diagonal is the diagonal: If $\tau: M \times M \to M \times M$ sends $(x, y) \mapsto (y, x)$, and $\tau^* : H^*(M \times M) \to H^*(M \times M)$ is the induced map on cohomology, one can show $\tau^* \eta(\Delta) = \eta(\Delta)$. Substituting in the above formula, we have 
$$
\eta(\Delta) = \tau^*\eta(\Delta) = \tau^*(\sum_{i=1}^r (-1)^{\deg b_i}b_i \times b_i^\vee) = \sum_{i=1}^r (-1)^{\deg b_i} (-1)^{\deg b_i \cdot \deg b_i^\vee} b_i^\vee \times b_i
$$
$$
\begin{equation}
\label{eq:3} \tag{3}
= \sum_{i=1}^r (-1)^{\deg b_i (n - \deg b_i + 1)} b_i^\vee \times b_i = \sum_{i=1}^r   b_i^\vee \times b_i 
\end{equation}
$$
where I've used that $(-1)^{\deg b_i (n - \deg b_i + 1)} = 1$ since $n$ is even and $\deg b_i \cdot (1 - \deg b_i)$ is always even. 
Now when we self-intersect the diagonal, substitute formula \eqref{eq:2} for one copy of $\eta(\Delta)$ and formula \eqref{eq:3} for the other:
$$
\begin{equation}
\label{eq:4} \tag{4}
\eta(\Delta) \smile \eta(\Delta) = (\sum_{i=1}^r (-1)^{\deg b_i}b_i \times b_i^\vee) \smile (\sum_{i=1}^r  b_i^\vee \times b_i ) 
\end{equation}
$$
$$
\begin{equation}
\label{eq:5} \tag{5}
= \sum_{i, j} (-1)^{\deg b_i} b_i \times b_i^\vee \smile b_j^\vee \times b_j
\end{equation}
$$
$$
\begin{equation}
\label{eq:6} \tag{6}
= \sum_{i, j} (-1)^{\deg b_i} (-1)^{\deg b_i^\vee \deg b_j^\vee} b_i \smile b_j^\vee \times b_i^\vee \smile b_j
\end{equation}
$$
$$
\begin{equation}
\label{eq:7} \tag{7}
= \sum_{i, j} (-1)^{\deg b_i}  (-1)^{\deg b_i^\vee \deg b_j^\vee} (-1)^{\deg b_i^\vee \deg b_j} b_i \smile b_j^\vee \times b_j \smile b_i^\vee
\end{equation}
$$
Long overdue simplification of the factors of $(-1)$: the factor of $(-1)$ on the $i, j$ term of the sum is 
$$
(-1)^{s} \text{  where  } s = \deg b_i + \deg b_i^\vee \cdot (\deg b_j + \deg b_j^\vee) = \deg b_i  + \deg b_i^\vee \cdot n
$$
Since $n$ is even, $(-1)^s = (-1)^{\deg b_i}$, and we've shown 
$$
\eta(\Delta) \smile \eta(\Delta) = \sum_{i, j} (-1)^{\deg b_i} (b_i \smile b_j^\vee) \times (b_j \smile b_i^\vee)
$$
Finally, pairing with the fundamental class $\mu \times \mu$ of $M \times M$ and recalling equation \eqref{eq:1}, we see that 
$$
\langle \eta(\Delta) \smile \eta(\Delta), \mu \times \mu \rangle = \sum_{i, j} (-1)^{\deg b_i} \langle b_i \smile b_j^\vee , \mu \rangle \cdot \langle b_j \smile b_i^\vee , \mu \rangle
$$
$$
\label{eq:8} \tag{8}
 = \sum_i (-1)^{\deg b_i} = \sum_i (-1)^i \dim H^i(M)
$$<|endoftext|>
TITLE: Reference for the `standard' Tate curve argument.
QUESTION [14 upvotes]: I'd like a reference (e.g. something published somewhere that I can cite in a paper) for the proof of the following: 

Let $E$ be an elliptic curve over $\mathbb Q$ with minimal discriminant $\Delta$, let $p$ be a prime, at which $E$ has potentially multiplicative reduction and let $\ell$ be a prime different than $p$. Then the mod $\ell$ representation is unramified at $p$ iff $\ell$ divides the valuation of $\Delta$ at $p$.

This is used for instance in the proof of Fermat's last theorem. In On modular representations of ${\rm Gal}(\overline{\mathbb Q}/\mathbb Q)$ arising from modular forms by Ken Ribet he cites Serre's (awesome) paper Sur les représentations modulaires de degré $2$ de ${\rm Gal}(\overline{\mathbb Q}/\mathbb Q)$, which (4.1.12) says this follows immediately from the theory of Tate curves.
It is pretty easy: the Tate curve gives you a explicit description the field obtained by adjoining the $\ell$-torsion points to $\mathbb Q_p$, and one can just check directly that the divisibility condition implies that this field (and thus the mod $\ell$ representation on the $\ell$-torsion points) is unramified at $p$. 
Nonetheless I'm curious to know if anyone writes this down explicitly anywhere in the literature.

REPLY [10 votes]: I think most people just mentally have in mind the argument you give.
In my thesis I actually wrote this down semi-carefully (including the case l = p, in which case what you want to say is that E[l] is finite over Zp, where E is now the Neron model of your elliptic curve over Q_p.) Or rather I wrote down the direction "l divides Delta => unramified" in Corollary 1.2 of the short version of my thesis.  The goal of the thesis, by the way, was to extend this assertion to abelian varieties with real multiplication; the point being that it's not obvious what's supposed to play the role of Delta.<|endoftext|>
TITLE: Mirror symmetry for noncompact Calabi-Yau manifolds
QUESTION [8 upvotes]: In analogy with the Hodge diagram for ordinary de Rham cohomology, we should have some kind of diagram for Alexander-Spanier cohomology. Doing all the relevant duality stuff and assuming that now our space is a noncompact Calabi-Yau manifold, we get a reduced Hodge diamond, to which mirror symmetry probably applies. 
Unfortunately, I don't know anything about mirror symmetry. Do we still get meaningful geometric information (deformations, etc.)? I'd like to know what all the subtle obstructions are to defining things in the above way.

REPLY [7 votes]: There is a version of mirror symmetry, called "local mirror symmetry", for certain non-compact Calabi-Yaus, for example the total space of the canonical bundle of P^2 (exercise: show this is CY). The mirror (or rather one possible mirror) of this non-compact Calabi-Yau is an affine elliptic curve in (C^*)^2. I don't think that there is as yet a version of mirror symmetry for more general non-compact CYs, though I don't know too much about this story. In all of the papers that I've looked at on this stuff at least, the only non-compact CYs that have been considered in mirror symmetry so far are total spaces of vector bundles over compact (probably Fano) things.
I guess we should probably get some sort of "Hodge diamond symmetry" in local mirror symmetry, but the story becomes more complicated. One immediate thing to notice is that, at least in the example I've given, the dimensions of the manifolds aren't the same! So things will have to be modified.
The Hodge diamond symmetry in mirror symmetry for compact Calabi-Yaus should really be thought of as coming from a correspondence between certain deformations of a Calabi-Yau and certain deformations of its mirror. In the non-compact case, the deformations that we should consider will be somewhat different from the deformations that we should consider in the compact case.
A somewhat recent point of view, due to Kontsevich, is that this correspondence between deformations can be gotten from homological mirror symmetry. In homological mirror symmetry for compact Calabi-Yau manifolds, we consider a derived category of coherent sheaves on one side and a Fukaya category on the other side. Then we should have an equivalence of categories, plus an equivalence of a certain structure on their Hochschild cohomologies -- in particular their Hochschild cohomologies should be equivalent at least as vector spaces. These Hochschild cohomologies should be thought as the appropriate deformation spaces (or maybe rather the tangent spaces to the appropriate deformation spaces?), and an appropriate identification of the Hochschild cohomologies (plus the extra structure that I mentioned) should give in particular the Hodge diamond symmetry. There should also be homological mirror symmetry for non-compact Calabi-Yau manifolds, but we must define the analogues of derived category and Fukaya category in this situation appropriately. Then there should be an analogous story on the Hochschild cohomologies of the categories, and an appropriate analogue of the Hodge diamond symmetry. See the paper "Hodge theoretic aspects of mirror symmetry" by Katzarkov-Kontsevich-Pantev for more details.<|endoftext|>
TITLE: Formal consequences of Riemann-Roch (multiple answers welcome)
QUESTION [22 upvotes]: This question aims to pin down what Riemann-Roch can tell us about a divisor on a curve, without any "geometric thinking".  It can be annoying to wonder if there is some clever trick you're missing out on in a problem, and it would be nice to know the limits of Riemann-Roch formalism the way we know we can't solve a system of 2 linear equations in 3 variables.
If someone prefers a different formalization of this question, I'd be happy to get an answer to that one instead.  Here is mine: 
Situation:
S1) Div is a free abelian group generated by an infinite set of letters P. [like points] 
 S2) D ∈ Div is "effective" if all its coefficients are non-negative.
 S3) deg: Div -> Z is the sum of coefficients map.
 S4) Prin is a distinguished subgroup of ker(deg). [like principal divisors]
 S5) l: Div/Prin -> N is a function to the non-negative integers. [like the dimension of global sections]
 S6) K is an element of Div. [like a cannonical divisor]
Relations (g:=l(K)):
R1) l(D)-l(K-D) = deg(D) + 1 - g. [Riemann Roch]
 R2) l(D+P) = l(D) or l(D)+1 for any generator P.
 R3) l(D)>0 iff D is equivalent mod Prin to an effective divisor.
 R4) If l(D)>0 and deg(D)=0 then D ∈ Prin.

Question A (hopefully manageable): What exactly can be inferred here about one of l(D) or deg(D), given the other?  

Maybe someone already knows the answer to this, from experience with solving RR-related problems, or from literature.  
Awesomely, many other concepts can be reformulated in this context, and we can ask more...
Optional definitions:
O1) D is "free" if l(D-P) = l(D)-1 for any generator P.
O2) D is "very ample" if l(D-P-Q) = l(D)-2 for any generators P,Q (not nececesarily distinct)
O3) D is "ample" if nD is very ample for some n>0.
O4) D is "big" if for some c>0 and all large n, l(nD) ≥ cm^n

Question B (partial answers welcome): What exactly can be inferred here about l(D), deg(D), effectiveness, freeness, (very) ampleness, and bigness of D given information about the others?  

Some examples (see Hartshorne chapter IV): 

l(0)=1
deg(K) = 2g-2
If D is very ample then deg(D)>0
If deg(D) ≥ 2g then D is free
If deg(D) ≥ 2g+1 then D is very ample 
D is ample iff deg(D)>0

So, yeah!  What's the deal with Riemann-Roch?

REPLY [4 votes]: Dino Lorenzini has a preprint in which he considers "Riemann-Roch structures", roughly as you've laid them out. http://www.math.uga.edu/~lorenz/RRNovember11.dvi
Namely, he considers the situation of 
(S1) The free abelian group $\mathbf{Z}^n$ (okay, so not infinite... Riemann-Roch also works over a finite field)
(S2) $R\in \mathbf{Z}^n$ an "effective" vector or divisor with coprime integer entries
(S3) degree of a divisor $D$ with respect to $R$ is simply the dot product $R\cdot D$
(S4) the "principal divisors" are chosen to be a lattice $\Lambda$ inside of $\Lambda_R$, the lattice of vectors or divisors perpendicular to $R$
So that if (S6) a canonical divisor exists, there exists a function (S5) $h: \mathbf{Z}^n/\Lambda \to \mathbf{Z}_{\ge 0}$ which satisfies some of your relations and optional requirements called a "Riemann-Roch structure".(This is proposition 2.4)
Of interest and proved by Baker and Norine is that if we take $\Lambda$ to be the image of the LaPlacian matrix of a graph with $n$ vertices and $m$ edges, we get such a structure.<|endoftext|>
TITLE: Is the Fourier transform of $\exp(-\|x\|)$ non-negative?
QUESTION [7 upvotes]: Is the $n$-dimensional Fourier transform of $\exp(-\|x\|)$ always non-negative, where $\|\cdot\|$ is the Euclidean norm on $\mathbb{R}^n$? What is its support?

REPLY [2 votes]: These questions are closely related to the so-called stable distributions. In particular, 
the cauchy distribution on the real line has the characteristic function e^{-|x|}. 
Go to the wikipedia page, and in the definition section set: 
mu=0      (this is the drift parameter) 
alpha=0   (this is the skewness parameter) 
To get the same thing in higher dimensions, take independent copies in each coordinate.  
Take note:  These distributions are not square integrable--otherwise the 'universal' Central 
Limit Theorem would hold.  The cauchy distribution is only weakly integrable.<|endoftext|>
TITLE: What is the Tutte polynomial encoding?
QUESTION [15 upvotes]: Pretty much exactly what it says on the tin. Let G be a connected graph; then the Tutte polynomial T_G(x,y) carries a lot of information about G. However, it obviously doesn't encode everything about the graph, since there are examples of non-isomorphic graphs with the same Tutte polynomial.
My question is, what information exactly does the Tutte polynomial encapsulate? I'm aware of a few answers to this question, but I don't find any of them particularly satisfying. For instance, T_G(x,y) can be characterized as "the universal Tutte-Grothendieck invariant," but the definition of Tutte-Grothendieck invariants is just as unintuitive as the definition of the Tutte polynomial (because it's essentially the same definition!) One can also define the coefficients as counting certain spanning trees of G, but this doesn't make apparent the fact that the Tutte polynomial specializes to the chromatic polynomial, or the notion that it carries most of the information one can obtain via linear algebra methods.
So is there a nice way of thinking about what data about G the Tutte polynomial encodes?
ETA: Okay, here's a very rough conjecture. Suppose that there's some "computationally simple" (i.e., testing membership is in NP) class of graphs such that there are two connected graphs G, H with the same Tutte polynomial, and G is in the class and H is not. Then there are spanning trees S, T of G, H respectively, such that S is in the class and T is not.
This would mean, in a sense that I can't make entirely rigorous, that the information about a graph G not encoded in the Tutte polynomial is just information about the structure of spanning trees of G. (Update: As Kevin Costello points out in a comment, this idea appears to be severely limited by the existence of certain pairs of co-Tutte graphs. In particular, we would need to count spanning trees with multiplicity for it to have even a chance of being true.)
As stated, the above conjecture is false for trivial reasons. But is there a way of making it true, perhaps by requiring the property to be, in some sense, natural? Is there a broad notion of "graph properties" for which it is true? Can we at least state a conjecture along these lines which does seem to be true?

REPLY [4 votes]: The Tutte polynomial of a graph $G$ is the convolution of the modular flow polynomial and the modular tension polynomial of $G$, c.f. [1,2,3]. This leads to a combinatorial interpretation of the values of the Tutte polynomial at every integer point in the plane [4, Theorem 3.11.7].

W. Kook, V. Reiner, and D. Stanton, A convolution formula for the Tutte polynomial,
J. Combin. Theory Ser. B 76 (1999), no. 2, 297–300.
V. Reiner, An interpretation for the Tutte polynomial, European J. Combin. 20
(1999), no. 2, 149–161.
F. Breuer and R. Sanyal, Ehrhart theory, Modular ow reciprocity, and the Tutte polynomial, Mathematische Zeitschrift, to appear. arXiv:0907.0845
F. Breuer, Ham Sandwiches, Staircases and Counting Polynomials, PhD thesis, Freie Universität Berlin, 2009. Available here and here.

Disclaimer: the last two of the cited references are my own.<|endoftext|>
TITLE: algebraic group G vs. algebraic stack BG
QUESTION [25 upvotes]: I've gathered that it's "common knowledge" (at least among people who think about such things) that studying a (smooth) algebraic group G, as an algebraic group, is in some sense the same as studying BG as an algebraic stack. Can somebody explain why this is true (and to what extent it is true)? I can get as far as seeing that quasi-coherent sheaves on BG are the same as representations of G, but it feels like there's more to it.
In particular, Scott Carnahan mentioned here that deformations of BG as an algebraic stack should correspond exactly to deformations of G as an algebraic group. I assume this means that any deformation of BG must be of the form BG', where G' is a deformation of G (as a group). It's clear to me that such a BG' is a deformation, but why should these be the only deformations?

REPLY [12 votes]: The stack $BG$ only recovers $G$ up to inner automorphisms, not canonically (as suggested by blah) - this can lead to serious issues in families or equivalently over a nonalgebraically closed field, as Shenghao's comment points out. One way to say this is the following: the loops in $BG$ are $G/G$, the adjoint quotient of $G$. On the other hand, if you give a map
$pt \to BG$ then the based loop space (fiber product of $pt$ with itself over $BG$) is $G$,
so you recover the group canonically.<|endoftext|>
TITLE: An "existence contra partition of unity" statement for integer matrices?
QUESTION [11 upvotes]: While reading a blog post on partitions of unity at the Secret Blogging Seminar the following question came into my mind.
Let $n$ be a positive integer and let $B_1$ and $B_2$ be $n \times n$ matrices with integer entries. Is it true that exactly one of the following two statements is true?

There is a vector $v \in \mathbb{Q}^n \backslash  \mathbb{Z}^n$ such that both $B_1v$ and $B_2v$ are in $\mathbb{Z}^n$.
There are matrices $A_1$ and $A_2$ with integer entries such that $A_1B_1+A_2B_2=I$. 

Here, $I$ denotes the $n \times n$ identity matrix. The case $n=1$ is Bézout's identity.

REPLY [5 votes]: I believe that what you say is true. I'll sketch an argument.
Let f:Zn ---> Z2n be the map of free Z-modules given by the matrices B1, B2 put in column (i.e. the direct sum of the morphisms given by B1 and B2). Now we rephrase conditions (1) and (2) in a slightly more abstract way:

(1) fails to hold if, and only if there exists p:Z2n ---> Zn such that, together with f, fit in a short exact sequence
0 ---> Zn ---> Z2n ---> Zn ---> 0                               (*)

Indeed, the failure of (1) means that any v in Qn such f(v) in Z2n must be integral (i.e. v in Zn). In particular, this implies that f is injective. Moreover, take w in Z2n representing a nonzero torsion element in the cokernel of f. As w represents a torsion element, Nw belongs to the image of f for some big enough positive integer N, so there is v in Zn such that f(v) = Nw. But now f(1/N v) = w, and this means, by the failure of (1), that 1/N v is integral, so w is in the image of f and the cokernel of f has no torsion. As a finitely generated torsion-free Z-module is free, we get an exact sequence like (*) above. This argument can easily be reversed, to show the equivalence between the existence of this exact sequence and the failure of (1).  

(2) holds if, and only if there exists a morphism of Z-modules r:Z2n ----> Zn such that rf = id.

Let r be represented by a matrix (A1,A2). Then gf has matrix A1B1 + A2B2, and gf = id if, and only if (2) holds.
Now, the proof of what you asked for is easy. (1) fails if, and only if we can form the exact sequence (*), but such an exact sequence is always split because Z^n is projective, so we can form such exact sequence if, and only if there exists a splitting r:Z2n ----> Zn, which is exactly condition (2).<|endoftext|>
TITLE: "Points" in algebraic geometry: Why shift from m-Spec to Spec?
QUESTION [37 upvotes]: Why were algebraic geometers in the 19th Century thinking of
m-Spec as the set of points of an affine variety associated to the
ring whereas, sometime in the middle of the 20 Century, people started to think Spec was more appropriate as the "set of points".
What are advantages of the Spec approach? Specific theorems?

REPLY [6 votes]: The reason why $\operatorname{Spec} A$ is an important notion is because it solves the following problem for a commutative ring $A$:
Find a local ring $\mathcal O$ together with a localisation morphism $A \to \mathcal O$ such that every other localisation morphism $A \to B$ to a local ring $B$ factors as a local morphism over $A \to \mathcal O$, i.e. one looks for a kind of universal localisation of $A$.
Stated as above, this problem has no solution at least as long one is not willing to leave the world of rings in the category of sets. However it has a solution in the following more general setting:
There is a topos $X$ endowed with a local ring object $\mathcal O$ and a localisation morphism $A \to \mathcal O$ such that for every other topos $Y$ together with a local ring object $B$ and a localisation morphism $A \to B$ there is a pair of a geometric morphism $f\colon Y \to X$ and a morphism $f^* \mathcal O \to B$ of local rings, which is unique up to a unique natural isomorphism, such that $A \to B$ is given by the composition of $f^* \mathcal O \to B$ and $A \to f^* \mathcal O$.
In fact, the solution to this problem is the topos $X$ of sheaves on $\operatorname{Spec} A$ together with the structure sheaf $\mathcal O_X$ as a local ring object. Now if you replace $\operatorname{Spec} A$ by the max-spectrum, the locally ringed sheaf topos you get will not solve the universal localisation problem in general.
This means that the usual definition of $\operatorname{Spec} A$ with prime ideals is a correct one (as long as one is working in classical logic with the axiom of choice) but it  does not mean that it is the only correct definition: You can, for example, replace $\operatorname{Spec}$ by any other topological space or, more generally, by any other site such that the sheaf topos over it is still equivalent to $X$.<|endoftext|>
TITLE: Co-induction understanding
QUESTION [14 upvotes]: Hi,
I am studying coinduction(not induction) as part of a class on static analysis. Rummaging around the internet, I am simply not finding a clear, concise description of:

How coinduction actually proves something(it seems that coinduction is like waving a magic  hand in the treatments I've read)
What propositions require coinductive proof
How to operate a coinductive proof

I have reviewed Wikipedia and a tutorial on co-induction/co-algebras.
My best understanding:
Coinduction is used in propositions such as:
f(x) = ... f(x) ...

Where ... denotes "stuff here that doesn't rely on f(x)". 
Then, f(x) is assumed true, and the ... is proved. If ... holds, then the proposition f(x) holds, out to and including infinity. 
The treatments I've read include a functional function Q that takes f(x) and returns some f'(x), and somehow that makes it all better.
At an abstract level, I read that coinduction operates over a coalgebra, which is dual to induction/algebras. 
This (1) seems awfully like circular reasoning and (2)seems awfully like voodoo. I suspect part of what I'm missing is a careful and clear description of coalgebras, plus how we jump from algebra into a coalgebra and back again. The professor declined to go into that part of the subject.
Note:
I'm not an algebraicist by study- I write software and my thesis is over low-level software operations. So this is really unfamiliar to me and I'm trying to de-jargon this into my head.
(semi-cross-posted from stackoverflow.com)

REPLY [9 votes]: Here's an informal example of a proof by coinduction.
The extended natural numbers E = {0, 1, 2, ..., ∞} are the final coalgebra for the functor F(X) = 1 + X.  (If we have an F-coalgebra X, i.e. a set X with a map f : X -> 1 + X, and an element of X, what we can do is repeatedly apply f until we get the element of 1 rather than a new element of X.  Counting the number of times we got new elements of X gives us an element of E.  This is the unique F-coalgebra map X -> E.)  In Haskell we would write
data Nat = Z | S Nat   -- possible values are Z, S Z, S (S Z), ..., S (S (S ...))

We can define addition:
add :: Nat -> Nat -> Nat
add Z b = b
add (S a1) b = S (add a1 b)

I'll prove that add a Z = a by coinduction.  We perform case analysis on a.  If a = Z then it follows from the first equation.  If a = S a1 then add a Z = add (S a1) Z = S (add a1 Z) = S a1 = a where the next-to-last step used the coinductive hypothesis.
Why isn't that circular reasoning?  We were allowed to apply the coinductive hypothesis because we did so inside an application of the "constructor" S.  In other words, even if you didn't believe the coinductive hypothesis, you would still conclude that add a Z and a were both of the form Z or both of the form S x--in other words, add a Z and a agree for to one observation.  Using that statement where we used the coinductive hypothesis, you find that add a Z and a agree to two observations, etc.  Since an element of E is determined by all the finite sequences of observations we can make, it follows that add a Z = a.
(Note that this argument works even when a = S (S (S ...))!)
Exercise: Give a formal version of this argument, using the definition of E as the final F-coalgebra.<|endoftext|>
TITLE: What do models where the CH is false look like?
QUESTION [6 upvotes]: Additionally, is there any intuitive way to visualize the cardinalities that result?

REPLY [5 votes]: I have an objection to the way in which the question and some of the answers are phrased.  To me it seems meaningless to ask whether the Continuum Hypothesis is "true" or "false".  It is of course meaningful to ask whether CH is true or false in a given model. 
But unless you can reveal to me some objective world of sets, I'll have no idea what you mean by CH being "true" or "false" in an absolute sense.  
It's like asking whether the parallel postulate is "true" or "false".  There are geometrical systems in which it's true, and geometrical systems in which it's false; end of story.  Similarly, we know that from any model of ZFC, one can build (i) another model of ZFC in which CH is true, and (ii) another model of ZFC in which CH is false. 
I know what the original question is asking, but I'm surprised that so many people continue to use this language.  The same goes for the Axiom of Choice; it amazes me when people argue about whether it's "true" or "false".<|endoftext|>
TITLE: References for syntomic cohomology
QUESTION [14 upvotes]: Could anyone point to good readable references for learning about syntomic cohomology?

REPLY [2 votes]: Here is a diploma thesis on syntomic cohomology: https://www.mathi.uni-heidelberg.de/~venjakob/diplom/kuemmerer.pdf<|endoftext|>
TITLE: Does there exist a sequence of groups whose representation theory is described by plane partitions?
QUESTION [6 upvotes]: More precisely, does there exist a sequence $G_1 < G_2 < \cdots$ of finite groups such that the irreducible representations of $G_n$ are parameterized by the plane partitions of total size $n$?

REPLY [10 votes]: Not if you want the direct analogue of the branching rule to hold: namely, if V is the representation of Gn corresponding to a plane partition A of n, then the restriction of V to Gn-1 is the direct sum of one copy of the representation corresponding to each plane partition of n-1 contained in A.  That would allow you to compute the dimension of the representation corresponding to A as the number of paths in the containment poset of plane partitions from the empty partition to A.  Some computation then shows that the order of G3 would be 1+4+4+1+4+1=15, but there's only one group of order 15, the abelian one, which doesn't work.
You could imagine some variations of the branching rule, though, such as "if B is obtained from A by replacing k by k-1 then the irrep corresponding to A contains k copies of the irrep corresponding to B", and maybe something like that would work.<|endoftext|>
TITLE: Undergraduate Level Math Books
QUESTION [47 upvotes]: What are some good undergraduate level books, particularly good introductions to (Real and Complex) Analysis, Linear Algebra, Algebra or Differential/Integral Equations (but books in any undergraduate level topic would also be much appreciated)?
EDIT: More topics (Affine, Euclidian, Hyperbolic, Descriptive & Diferential Geometry, Probability and Statistics, Numerical Mathematics, Distributions and Partial Equations, Topology, Algebraic Topology, Mathematical Logic etc)
Please post only one book per answer so that people can easily vote the books up/down and we get a nice sorted list. If possible post a link to the book itself (if it is freely available online) or to its amazon or google books page.

REPLY [2 votes]: I would also recommend the book entitled Analysis on Manifolds by James Munkres. I think that this is a good undergraduate textbook in mathematics for any student wishing to pursue multivariable calculus in greater depth. My only complaint is that Munkres often chooses to include details which can be seen easily after a little bit of thought. Perhaps this can be viewed as an effort to show the student how to "properly do analysis": doing analysis, just like doing any other branch of mathematics, requires you to carefully apply definitions and theorems, and it is important for the student to appreciate this early in his/her mathematical learning. 
That said, the book is an excellent text overall for "advanced calculus". The student will need to be familiar with single variable analysis and perhaps some linear algebra. (Even a rudimentary knowledge of linear algebra will do since Munkres develops most of the necessary theory from scratch.) 
Roughly speaking, the book splits into two parts. The first part covers most of the results students see when doing multivariable calculus that are stated "without proof" in their texts. For example, "the equality of the mixed partials", "double integrals can be done in any order", "a bounded function is Riemann integral if and only if it is continuous almost everywhere", "the change of variables theorem" etc., are (very) imprecise forms of some of the results Munkres establishes.
In the second part of the book, manifolds and their theory are introduced. Thus, for example, a rudimentary introduction to tensors is given, and this is supplemented by the basic theory of differential forms, the De Rham groups (of the punctured plane), Stokes' theorem etc. 
I think that the exposition could be tightened: if you actually pick up the book and really make an effort to read it, it is quite possible to finish the first half of the book in the space of a week (that is, approximately 200 pages in a week) simply because certain topics are explained in more detail (at least in my opinion) than necessary. (One example is Munkres' proof of the linearity, monotonicity, additivity etc. of the Riemann integral. This is proved in three contexts separately: the case of the integral over a rectangle, that over a bounded set, and that of improper integrals, when essentially the proofs can be left as relatively easy exercises in some cases.)
As the above comments suggest, I think that this is an excellent book for undergraduate students, but perhaps less so for graduate students. (Spivak's Calculus on Manifolds is good for both undergraduate and graduate students, in my opinion, but some people may suggest that it is too hard for undergraduates.) And after reading this book, you should have more than enough preparation to read more advanced texts such as William Boothby's An Introduction to Differentiable Manifolds and Riemannian Geometry.<|endoftext|>
TITLE: Exhibit an explicit bijection between irreducible polynomials over finite fields and Lyndon words.
QUESTION [21 upvotes]: Let $q$ be a power of a prime.  It's well-known that the function $B(n, q) = \frac{1}{n} \sum_{d | n} \mu \left( \frac{n}{d} \right) q^d$ counts both the number of irreducible polynomials of degree $n$ over $\mathbb{F}_q$ and the number of Lyndon words of length $n$ over an alphabet of size $q$.  Does there exist an explicit bijection between the two sets?

REPLY [5 votes]: The correspondence invented by Golomb relies on the choice of a primitive element a in the field of order q^n. Then, to each Lyndon word L=(l_0,l_1,...,l_{n-1}) one assigns the primitive polynomial having as a root the element a^{m(L)} where m(L) is the integer sum of l_i*q^i over i=0,1,...,n-1.<|endoftext|>
TITLE: What is an example of a ring in which the intersection of all maximal two-sided ideals is not equal to the Jacobson radical?
QUESTION [7 upvotes]: What is an example of a ring in which the intersection of all maximal two-sided ideals is not equal to the Jacobson radical? Wikipedia suggests that any simple ring with a nontrivial right ideal would work, but this is clearly false (take a matrix ring over a field, for instance).
Benson's Representations and Cohomology I, on the other hand, claims that the Jacobson radical is in fact the intersection of all maximal two sided ideals. He defines the Jacobson radical as the intersection of the annihilators of simple R-modules, which are precisely the maximal two-sided ideals. Since this is the same as the intersection of the annihilators of the individual elements of the simple modules, then this is the same as the intersections of the maximal left (or right) ideals.
I don't see the flaw in Benson's reasoning, but I seem to recall hearing somewhere else that the Jacobson radical is not always the intersection of the maximal two-sided ideals. Who is correct here?

REPLY [7 votes]: A very important example is the quotient of $U(\mathfrak{g})$ (where $\mathfrak{g}$ is a simple complex Lie algebra) by the central elements killing a finite dimensional representation.  This has a unique maximal ideal (the annihilator of the finite dimensional module), but its Jacobson radical is trivial, since the annihilator of the simple highest weight module in this block whose highest weight is in the anti-dominant Weyl chamber is actually faithful.  There's actually an extremely interesting poset (by inclusion) of primitive ideals sitting in between.<|endoftext|>
TITLE: Hochschild/cyclic homology of von Neumann algebras: useless?
QUESTION [9 upvotes]: Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. There are plenty of other connections to homological algebra. If we use cyclic homology, there are connections to geometry and topology involving the Chern character.
Von Neumann algebras are complex algebras, so we can take their Hochschild and cyclic homologies. When I have asked experts in the fields of von Neumann algebras and non-commutative geometry about what you get, I usually hear some approximation of the following: "There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful."
Although this answer makes some sense, I find it very displeasing and cryptic. Why shouldn't it tell you something? Is there some way to make "it doesn't tell you anything" quantitative? Is there an example of a von Neumann algebra with nontrivial Hochschild or cyclic homology (different from that of the complex numbers)?
EDIT: After reading the responses so far, I should specify that I really want to know if there is a $II_1$-factor with nontrivial Hochschild or cyclic (co)homology.

REPLY [4 votes]: There is important work by Alain Connes and Dimitri Shlyakhtenko (see $L^2$-homology for von Neumann algebras (MSN)). They come up with a definition of $\ell^2$-homology for finite von Neumann algebras and define numerical invariants called $\ell^2$-Betti numbers for finite von Neumann algebras. This approach builds on the more classical theory of $\ell^2$-invariants developed by Atiyah, Cheeger–Gromov and also Lück. So far, there are no really interesting computations.
However, it seems that this homology group (or some variant of it) is more likely to be able to detect the differences among free group factors. Of course, this is only speculation. There is also a cohomological picture (see Kadison, Liu, and Thom - A note on commutators in algebras of unbounded operators (MSN)) which boils down (in dimension one) to a study of derivations with values in the algebra of affiliated operators. Unfortunately, this more algebraic approach has not been very successful so far.<|endoftext|>
TITLE: Examples of rational families of abelian varieties.
QUESTION [15 upvotes]: I'd like to know examples of non-trivial families of abelian varieties over rational bases (e.g. open subschemes of the projective line P^1).
One can generate many examples as Jacobians of rational families of curves (e.g. the hyperellitpic family, plane curves, complete intersections). Prym varieties are another example.
Are there any examples which are not obviously Jacobians of a family of curves? I would like to know both principally polarized and non principally polarized examples.

REPLY [3 votes]: There are nice examples of Gross and Popescu of Calabi-Yau threefolds fibered over $\mathbb P^1$ with generic fibers being abelian surfaces. See  arXiv:math/000108 and arXiv:0904.3354.<|endoftext|>
TITLE: Describing the universal covering map for the twice punctured complex plane
QUESTION [21 upvotes]: As is well known, the universal covering space of the punctured complex plane is the complex plane itself, and the cover is given by the exponential map.
In a sense, this shows that the logarithm has the worst monodromy possible, given that it has only one singularity in the complex plane. Hence we can easily visualise the covering map as given by the Riemann surface corresponding to log (given by analytic continuation, say).
Seeing how fundamental the exponential and logarithm are, I was wondering how come I don't know of anything about the case when two points are removed from the complex plane.
My main question is as follows: how can I find a function whose monodromy corresponds to the universal cover of the twice punctured complex plane (say ℂ∖{0,1}), in the same way as the monodromy of log corresponds to the universal cover of the punctured plane.
For example, one might want to try f(*z*) = log(*z*) + log(*z*-1) but the corresponding Riemann surface is easily seen to have an abelian group of deck transformations, when it should be F2.
The most help so far has been looking about the Riemann-Hilbert problem; it is possible to write down a linear ordinary differential equation of order 2 that has the required monodromy group. 
Only trouble is that this does not show how to explicitly do it: I started with a faithful representation of the fundamental group (of the twice punctured complex plane) in GL(2,ℂ) (in fact corresponding matrices in SL(2,ℤ) are easy to produce), but the calculations quickly got out of hand.
My number one hope would be something involving the hypergeometric function 2F1 seeing as this solves in general second order linear differential equations with 3 regular singular points (for 2F1 the singular points are 0, 1 and ∞, but we can move this with Möbius transformations), but I was really hoping for something much more explicit, especially seeing as a lot of parameters seem to not produce the correct monodromy. Especially knowing that even though the differential equation has the correct monodromy, the solutions might not.
I'd be happy to hear about any information anyone has relating to analytical descriptions of this universal cover, I was quite surprised to see how little there is written about it. 
Bonus points for anything that also works for more points removed, but seeing how complicated this seems to be for only two removed points, I'm not hoping much (knowing that starting with 3 singular points (+∞), many complicated phenomena appear).

REPLY [4 votes]: I am sorry to revive such an old topic. But it really needs to be mentioned that this universal cover of the twice punctured plane was constructed by PICARD, and as one of the comments says, this was used in the proof of little Picard (and also big Picard). This is one of Picard's most famous constructions, so you will excuse my emphasizing it. This is explicitly constructed in Ahlfors' book on complex analysis.<|endoftext|>
TITLE: Teaching statements for math jobs?
QUESTION [77 upvotes]: What is the purpose of the "teaching statement" or "statement of teaching philosophy" when applying for jobs, specifically math postdocs?  I am applying for jobs, and I need to write one of these shortly.
Let us assume for the sake of argument that I have a teaching philosophy; I am not asking you to tell me what my teaching philosophy should be.  I would like to know how those responsible for hiring view teaching statements, especially in the case of new PhD's who don't necessarily have extensive teaching experience.  
(I believe this is appropriate for mathoverflow because it is of interest to "a person whose primary occupation is doing mathematics", as I am.)

REPLY [21 votes]: I found this MO question while writing my own teaching statement. I'd like to point out that the Notices of the AMS published results of a survey which identified quite a few themes in successful teaching statements. I will paraphrase the list, beginning with the most-cited comments:

Specific examples linking philosophy to practice.
Dedication to teaching.
Good writing.
Thoughtful reflection on teaching.
Student-centered.
Match between individual and department.

Here's a handy rubric from the University of Michigan.<|endoftext|>
TITLE: triangulated vs. dg/A-infinity
QUESTION [32 upvotes]: Someone recently said "derived/triangulated categories are an abomination that should struck from the earth and replaced with dg/A-infinity versions".
I have a rough idea why this is true ("don't throw away those higher homotopies -- you might need them some time in the future"), but I would be interested to have a more informed/detailed understanding of why triangulated categories are so abominable.
So, my question: Where are some good places to read about this "Down with triangulated!" philosophy?
I'm interested in this because in my research I have come across some categories which, very surprisingly, have a (semi-)triangulated structure.  Perhaps I should be looking for an underlying A-infinity structure.

REPLY [33 votes]: The analogy that Nadler and I like is that if (oo,1)-categories
are like vector spaces, then model categories are like vector spaces with a fixed 
basis and homotopy categories are like vector spaces mod isomorphism - i.e., dimensions
of vector spaces. Which of the three would you rather work with?
In more details I'll take the low road and quote my paper with Francis and Nadler:
The theory of triangulated categories is inadequate
to handle many basic algebraic and geometric operations. Examples include 
the absence of a good theory of gluing or of descent, of functor categories, 
or of generators and relations. The essential problem is that
passing to homotopy categories discards essential information (in particular,
homotopy coherent structures, homotopy limits and homotopy colimits).
This information can be captured in many alternative ways, the most common 
of which is the theory of model categories. Model structures keep weakly equivalent objects distinct but retain the extra
structure of resolutions which enables the formulation
of homotopy coherence. This extra structure can be very useful for calculations but makes 
some functorial operations difficult. In particular, it can be hard to 
construct certain derived functors because the given resolutions are inadequate.
There are also fundamental difficulties with the consideration of functor categories between model categories. 
Anyway, in characteristic zero, the theories of dg categories, Aoo categories and stable oo,1 categories are all the same, and provide a "promised land" between the two extremes.
(The problems with localization of the Fukaya category are very serious, but are issues in geometry - the existence of instanton corrections - and not in the theory of A_oo categories). Perhaps the most useful result in this context that doesn't have analogs for dg or Aoo categories is Jacob Lurie's oo-Barr-Beck theorem, which has lots and lots of consequences - such as descent theory or the generation result that Ben mentioned. Another is having a good theory of tensor products of categories and internal hom of categories, neither of which works in the two extreme theories.<|endoftext|>
TITLE: Embedding abelian categories to have enough projectives
QUESTION [6 upvotes]: Is it true that the pro-objects of an abelian category form a category with enough projectives? 
In general, given an abelian category A, is there a canonical way to embed it a bigger abelian category A' with enough projectives (or injectives) and such that A' is universal with respect to this property?

REPLY [3 votes]: It seems that Pro(A) does not have enough projectives in general. In Kashiwara-Schapira's book "Categories and Sheaves" they prove (corollary 15.1.3) that Ind(k-Mod) does not have enough injectives. This means, taking opposite categories, that Pro(k-Mod^{op}) does not have enough projectives.
I don't know of any universal way of adding enough projectives.<|endoftext|>
TITLE: Is there a version of the valuative criteria for separateness/properness for varieties?
QUESTION [11 upvotes]: What I had in mind was something like the following:
X is separated/proper iff for all curves C and all maps f : C \ c -> X, f extends to C in at most/exactly one way.
Is there a good reason why this cannot possibly be true? 
Here X denotes a reduced scheme of finite type of a field k (I guess people usually call this prevariety).  I am mostly interested in the case where k is algebraically closed.

REPLY [5 votes]: I would like to answer my own question after nearly 12 years.  I am teaching algebraic geometry this year and in the process I found the statement that I wanted as Corollary 15.10 in Gortz-Wedhorn (an excellent book).
Theorem
Let $Y $ be a locally Noetherian scheme and let $ f : X \rightarrow Y $ be a morphism of finite type.  Then $ f $ is proper if and only if for all normal absolute curves $C $ which are schemes over $Y $, for every closed point $ c \in C$, and for every $ Y $-morphism $ g : C \setminus \{c \} \rightarrow X $, there exists a morphism $ \bar g : C \rightarrow X $ such that $ \bar g|_{C \setminus \{c\}} = g $.
When I asked my question above, I was wondering if there was a more geometric version of the valuative criterion and that is provided by this theorem.  In the special case where $ Y = Spec k $, then $ C $ will be a regular curve over $ k $.<|endoftext|>
TITLE: Arithmetic progressions without small primes
QUESTION [16 upvotes]: The following question came up in the discussion at How small can a group with an n-dimensional irreducible complex representation be? :
Is it known that there are infinitely many primes p for which the least prime q which is 1 mod p is > c p^2 (for some positive constant c, independent of p)?
Wikipedia's article on primes in arithmetic progressions says that the expected bound for the least prime is p^{2 + \epsilon}, given various strengthenings of the Riemann Hypothesis, but it doesn't say much about lower bounds.
By the way, for the applications in the linked post, it would be even better if the same lower bound applied to finding a prime power q which is 1 mod p.

REPLY [3 votes]: For a fixed a>0 it is known that there are infinitely many moduli q such that the least prime in the arithmetic progresion a mod q is at least >> q ln (q) lnln (q) ln ln ln ln (q) / (ln ln ln(q) )^2. This can be found in:
Prachar, K. Über die kleinste Primzahl einer arithmetischen Reihe. (German) J. Reine Angew. Math. 206 1961. 
To the best of my knowledge this is the world record.<|endoftext|>
TITLE: Algorithms for semistable reduction of families of curves
QUESTION [6 upvotes]: This is a somewhat vague question which came up MSRI a few days ago: Suppose I have a family of curves over a one dimensional base, given in a computationally explicit way. For example, maybe I have a family F_t(x,y,z) of homogenous polynomials, whose coefficients are polynomials in t, and which cut out smooth curves for t \neq 0. 
Is there an algorithmically practical way to write down the limit of this family in \overline{M}_g?

REPLY [4 votes]: If your curves are in P^n (specifically in P^2 - as in your example), I think there is something you can do: project your curves from a general P^{n-2} to P^1. This means that you
are now looking for a limit in a Hurwitz scheme. This can be broken into two problems:

looking for the limit on the underlying M_{0,n}
tracing the ramification structure.

Here is an example: find the limit of F+t Q^2 where F is a plane quartic, and Q is a plane quadric. 
Project from your favorite random point. You can verify that the limit of the ramification points on the family are

the eight intersection points of F and Q
twice on each of ramification points of the projection of p from Q.

From here you can continue in a variety of ways (e.g. you have a pencil of g^1_4 s on the limit curve which break through a map from the limit curve to a plane conic, which has 8 ramification points)<|endoftext|>
TITLE: Do homotopy pullbacks commute with homotopy orbits (in spaces)?
QUESTION [6 upvotes]: Suppose we are given a diagram $X \to Z \gets Y$ of $G$-spaces ($G$ a discrete group). Let $(- \times^h -)$ denote homotopy pullback. Is $(X \times^h_Z Y)_{hG}$ weakly equivalent to $X_{hG} \times^h_{Z_{hG}} Y_{hG}$?

REPLY [7 votes]: Yes.  A sketch:
Taking products with the free $G$-space $EG$ commutes with the pullback diagram (because product is also a limit) and so you can assume they're free, and one of the maps is a fibration.  
Having done this, there is a natural long exact sequence of homotopy groups
$\to \pi_* (U) \to \pi_*(U_{hG}) \to \pi_*(BG) \to \dots$
and applying this to the pullback diagram you can deduce (from the 5-lemma) that the natural map from the orbit of pullbacks to the pullback of the orbits is a weak equivalence.<|endoftext|>
TITLE: Is any representation of a finite group defined over the algebraic integers?
QUESTION [23 upvotes]: Apologies in advance if this is obvious.

REPLY [14 votes]: This is not really an answer, but is too long for a comment. The proof given by Moonface above is given in more or less that form in the 1962 book of Curtis and Reiner. As far as I know, it is still open whether all irreducible representations of a finite group $G$ can be realized over 
$\mathbb{Z}[\omega]$, where $\omega$ is a complex primitive $|G|$-th roots of unity, though I think the paper of Cliff,Ritter and Weiss settles the questions for finite solvable groups. The paper
of Serre ( the three letters to Feit) give counterexamples to a slightly different question:
they show (among other things) that a representation of a finite group can be realised over 
some number fields, but might not be able to be realised over the ring of algebraic integers
of that field. Brauer's characterization of characters/Brauer's induction theorem show that 
all representations of the finite group $G$ may be realised over $\mathbb{Q}[\omega]$ for 
$\omega$ as above ( $|G|$ can be replaced by the exponent of $G$ if desired). As I said, 
realizability over $\mathbb{Z}[\omega]$ is a different matter.<|endoftext|>
TITLE: What is inter-universal geometry?
QUESTION [41 upvotes]: I wonder what Mochizuki's inter-universal geometry and his generalisation of anabelian geometry is, e.g. why the ABC-conjecture involves nested inclusions of sets as hinted in the slides, or why such inclusion structures should be simpler if they are between categories
, how that relates to F_1. It seems to me that his basic idea is that algebraic geometry has in general a kind of semantic feedback-loop, what sounds very beautifull, if it were true. His view of Grothendieck/Deligne's idea of using the section conjecture for indirect proving finiteness statements seems to me as if he would say "The first part of that is just the first jump into the feedback-loop". 
Edit: A nice link was jut given in: Mochizuki's proof and Siegel zeros
Edit: A relatively new survey
Edit: Mochizuki' report on the current review status. Someone told me that there may be a seminar in Harvard on this next year. 
Edit: New short survey by Ivan Fesenko :" This text is expected to help its readers to gain a general overview of the theory and a certain orientation in it, as well as to see various links between it and existing theories. Together with several mathematicians, we hope to organise a workshop in Europe, as well as an international conference in Kyoto in the summer of 2016. Feel free to contact me if you are interested to seriously study this work."

REPLY [10 votes]: In a research statement, he says:
"The essence of arithmetic geometry lies not in the various specific schemes that occur in a specific arithmetic-geometric setting, but rather in the abstract combinatorial patterns, along with the combinatorial algorithms that describe these patterns, that govern the dynamics of such specific schemes."
Regarding this, he then talks about how his main motivations are monoids, Galois categories, and dual graphs of degenerate stable curves, which leads him to talking about his geometry of categories stuff, and then to "absolute anabelian geometry."  He then links to a bunch of papers that I would assume elaborate a bit on it.  He then goes on to talk about extending Teichmuller Theory.
Generally, his research statement is fairly readable (and consider that I'm very much a nonspecialist in arithmetic anything) and seems to link to things with more details.<|endoftext|>
TITLE: Etale covers of the affine line
QUESTION [26 upvotes]: In characteristic p there are nontrivial etale covers of the affine line, such as those obtained by adjoining solutions to x^2 + x + f(t) = 0 for f(t) in k[t].  Using an etale cohomology computation with the Artin-Schreier sequence I believe you can show that, at least, the abelianization of the absolute Galois group is terrible.
What is known about the absolute Galois group of the affine line in characteristic p?  In addition, can spaces which are not A^1 (or extensions of A^1 to a larger base field) occur as covers?

REPLY [10 votes]: As stated by David Speyer and Clark Barwick, the awnser to the second question is the following:
Any smooth projective curve $C$ defined over a field $k$ of positive characteristic $p$ can be realized as a finite cover of the projective line only ramified above one point.
Here is a short constructive proof only based on Riemann-Roch theorem. It can be considered as an illustration of Kedlaya's proof, only dealing with curves.

First of all, there exists a generically étale finite cover $C\to\mathbf P^1$, induced by a rational function $f\in k(C)$ (in fact, any element of $k(C)-k(C)^p$ will do the job).
Denote by $R\in$ Div$(C)$ the (reduced) ramification divisor of the above cover (i.e. the ramified points are couted without multiplicity). From Riemann-Roch theorem, for large $n$, there exist a rational function $g\in k(C)$ having a pole of order $n$ at each point of the support of $R$. We may take $n$ strictly greater than $\frac{\deg(f)}p$.
Then, the rational function $h=f+g^p$ induces a cover $C\to\mathbf P^1$ only ramified above infinity.<|endoftext|>
TITLE: Higher vanishing cycles
QUESTION [6 upvotes]: The generalisation of the vanishing cycle formalism in SGA 7 is apparently since the 1970's an issue, Morava mentioned a connection with Bousfield localization. I find the Morava's remarks un-understandable - would someone explain?

REPLY [8 votes]: My apologies if this is too much or too little; leave a comment and I can try and correct it.  He's talking about a specific issue in homotopy theory that we'd like a better understanding of.
The stable homotopy category (implicitly localized at a prime p) has a stratification into "chromatic" layers, which correspond to a connection to formal group laws.  We geometrically think of the stable homotopy category as some kind of category of sheaves on a moduli stack X which has a sequence of open substacks X(n) - these are the "E(n)-local categories", and there are Bousfield localization functors taking a general element M to its E(n)-localization LE(n) M, which you can think of as restricting to the open substack.  (A general Bousfield localization will take some notion of "equivalence" and construct a universal new category where those equivalences become isomorphisms, but in an appropriately derived way.)
The difference X(n) \ X(n-1) between two adjacent layers is a closed substack of X(n), which in our language is the "K(n)-local category".  There is also a Bousfield localization functor that takes an element M to its K(n)-localization LK(n) M.  Bousfield localization is pretty general machinery and in the previous "open" situation it acted as restriction; in this "closed" situation it acts as a completion along the closed substack.
We have some general understanding of the K(n)-local categories.  They act a lot like some kind of quotient stack of some Lubin-Tate space classifying deformations of a height n formal group law by the group scheme of automorphisms of said formal group law, which is the n'th Morava stabilizer group Sn.  Geometrically we think about it as a point with a fairly large automorphism group (even though this is, of course, the wrong way to think about things).  These are places where you can get dirty and do specific computations and examine one chromatic layer at a time.
There are two remaining pieces of data we need, then, to understand M itself from its localizations LK(n) M: we need to understand how they are patched together into the E(n)-localizations, and we need to understand the limit of the LE(n)M.  The latter is a "chromatic convergence" question and not immediately relevant to the point under discussion.
In general there is a "patching" diagram, which is roughly something like the data you'd usually associate with a recollement.  (My favorite reference for data in this kind of situation is Mazur's "Notes on etale cohomology of number fields".)  We have a (homotopy) pullback diagram

LE(n) M  ->  LE(n-1) M
     |             |
     V             V
LK(n) M  ->  LE(n-1) LK(n) M

that tells us that a general E(n)-local object is reconstructed from a K(n)-local object (something concentrated on the closed stack), an E(n-1)-local object (concentrated on the open stack), and patching data (a map from the object on the open stack to the restriction of the complete object to the open stack).  This roughly follows because the K(n)-localization of any E(n-1)-local object is trivial.
The functor that takes an object concentrated near the closed substack and restricts it (in a derived way) to the open substack is what Morava considers.  Here, in the language of Bousfield localization, it is E(n-1)-localization applied to K(n)-local objects.  What he seems to be proposing is that this general Bousfield localization setup should be one way of thinking about the vanishing cycles functors (and I concur with his dislike for the "vanishing" terminology) in which we can, in a fully derived way, view sheaves on a large stack as coming from patching data on an open-closed pair.
Just to close the loop, what we don't really understand at all in this picture is what this "trans-chromatic-layer" stuff really does.  We have, for example, two stabilizer groups connected to formal group laws of adjacent heights, and we don't really understand what the specialization functor is really doing in this case.<|endoftext|>
TITLE: Short Introduction to Planar Algebras
QUESTION [13 upvotes]: Are there any good short expositions of planar algebras out there? I am interested primarily in seeing the main definition and some explicit examples.

REPLY [4 votes]: You should first read the new Wikipedia page: Planar algebra.
Next, a neat course by Vijay Kodiyalam is available on YouTube, see here (HD).<|endoftext|>
TITLE: Most interesting mathematics mistake?
QUESTION [222 upvotes]: Some mistakes in mathematics made by extremely smart and famous people can eventually lead to interesting developments and theorems, e.g. Poincaré's 3d sphere characterization or the search to prove that Euclid's parallel axiom is really necessary unnecessary.
But I also think there are less famous mistakes worth hearing about. So, here's a question:

What's the most interesting mathematics mistake that you know of?

EDIT: There is a similar question which has been closed as a duplicate to this one, but which also garnered some new answers. It can be found here:
Failures that lead eventually to new mathematics

REPLY [2 votes]: A posthumous book on number theory by Dirichlet appeared in 1859. It stated that Euclid's proof of the infinitude of primes was by contradiction, starting with an assumption that only finitely many primes exist and then deducing a contradiction.
Euclid's actual proof, recast in modern language, was that if $S$ is any finite set of primes (with no assumption that it constains the smallest $n$ primes nor that it contains all primes), then the prime factors of $1+\prod S$ are not members of $S;$ hence there are always more primes than what one already has.
For example, if $S=\{5,7\}$ then $1+\prod S=36=2\times2\times3\times3$ and the new primes are $2$ and $3.$
This requires no assumption that $S$ contains all primes.
Only the assumption that $S$ contains all primes could justify the conclusion that $1+\prod S$ has no prime factors, and so "is therefore itself prime", to quote G. H. Hardy (no relation to me, as far as I know) on pages 122–123 of the 1908 edition of A Course of Pure Mathematics (but not in the posthumous 10th edition).
The erroroneous belief that $1+\prod S$ is prime whenever $S$ is the set of the smallest $n$ primes for some $n$ has been held by some conscientious persons. The smallest (but not the only) counterexample is $1+(2\times3\times5\times7\times11\times13) = 59\times509.$
Catherine Woodgold and I examined in some detail the error of thinking that this proof is by contradiction in "Prime Simplicity", Mathematical Intelligencer, volume 31, number 4, Fall 2009, pages 44–52.<|endoftext|>
TITLE: Why is the Euler characteristic of powers of a line bundle a polynomial in the power?
QUESTION [8 upvotes]: Mumford's book Abelian Varieties asserts that for a line bundle L on a projective variety (if necessary, you can assume it is as nice as possible), the Euler characteristic $\chi(L^k)$ of tensor powers of $L$ is a polynomial in $k$.  If L is very ample, this is just the Hilbert polynomial, and this can be proven by an induction argument twisting the short exact sequence $0 \to \mathcal O(-1) \to \mathcal O \to K \to 0$.  More generally, if $L$ (or $L^*$) is an ideal sheaf, the same argument should work.  Why does the result still hold for arbitrary $L$?
Edit: I'd be particularly interested in an elementary proof that does not involve proving an entire Riemann-Roch theorem--Mumford is using this result to prove Riemann-Roch for abelian varieties!

REPLY [6 votes]: See EGA III$_1$ 2.5.3 and EGA IV$_2$ 5.3. The elegant generalization there which incorporates an auxiliary coherent sheaf opens the door to using Grothendieck's unscrewing lemma (EGA III$_1$ 3.1.2) to vary that and get the result in considerably more generality: any line bundle on any proper scheme over a field (or artin local ring, using length in place of dimension).  The method is via Chow's Lemma to reduce to projective case and ultimately use slicing methods in the style of DeLand's answer to get back to the very ample case.<|endoftext|>
TITLE: Points and lines in the plane
QUESTION [22 upvotes]: Does a positive real number $k\geq1$ exist such that for every finite set $P$ of points in the plane (with the property that no three points of $P$ lie on a common line and $|P|\geq3$), one can choose a subset $Q$ of $P$ with
$|Q| \geq |P|/k$
points and with the property that there exist two different points $a$ and $b$ in $Q$ such that no line $\overline{(p,q)}$ through two different points $p,q$ of $Q\backslash\{a,b\}$ crosses the interior of the segment $(a,b)$?
If such a number exists, what is the smallest integer $k$, fulfilling the property?
If you take a set $P=\{a,b,c\}$ with three points, then you can set $P=Q$ since no line crosses the interior of the segment $(a,b)$. However, a counterexample for $k=1$ is given below by Reid Barton.

REPLY [3 votes]: I have started with the perturbed lattice suggested in Dmitri's answer and using it I think I can show such a $k$ cannot exist. If it does fix $k$. We choose $n$ very large we take $n$ times $n$ grid of points in a square grid since we cannot have three points in a line we perturb each amount by a small quantity small enough not to cause any point within a triangle of other points in the grid to be outside the triangle or on the boundary in the new grid. The rest of this proof will be about points on a lattice when we reach a certain point we will use the above to get a contradiction in the perturbed set of points.
For any pair of points $A$ and $B$ the number of points whose distance from the line containing the points is less than $nk^{2}$ is less than $\sqrt2n^{2}k^{2}$ (we bound the number of points by the number of points whose distance from the  line segment with greatest length which is that connecting two diagonal points. This line segment has length $\sqrt2n$ multiplying by the distance gives  $\sqrt2n^{2}k^{2}$. So we have $n^2k-\sqrt2n^{2}k^{2}$ points whose distance from the line containing the two points is greater than $nk^{2}$. Now I can use Pick's theorem to show that there must be at least $nk^{2}/2$ points in the triangle or twice as many on the border of the triangle. I use the distance of the point to the line containing the line segment as the altitude an assume the line segment has at least length one to get a lower bound on the area of the triangle.
Now if any of these points are inside the triangle then the line connecting this point and the point of the triangle not equal to the two points of the segment must pass through the line segment but we cannot have this. If any of the lattice points lie on the line segment AB then the area will increase by an amount greater than one for each point and there will be more points that have to lie on the other two sides or the interior. So we can ignore these points in our argument
Now I choose $n$ large enough so that $n^{2}k-\sqrt2n^{2}k^{2}/16$ satisfies the following property the number of integers whose prime factor  are all less than $k^{-100}$ is less than $(n^{2}k-\sqrt2n^{2}k^{2})k^{100}/16$. I can do this because for any set of primes the percentage of numbers divided by members of those primes only goes to zero as n goes to infinity.
So we have $n^{2}k-\sqrt2n^{2}k^{2}$ points whose distance from the line containing the two points is greater than $nk^{2}$ we divide them into 16 sets depending on whether their $x$ and coordinates are greater than the $x$ and $y$ coordinates of the two points. So we will have one set of $n^{2}k-\sqrt2n^{2}k^{2}/16$ points whose relation with the coordinates of $x$ and $y$ is always the same we can use the above property of $n$ to get the the number of points whose greatest prime factor in the distance $y$ from the first point is greater than $k^{-100}$ and whose greatest prime factor in the difference from the $y$ coordinate of the second point is greater than $k^{-100}$. To see this note that the number of points which don't satisfy this property with respect to the first coordinate is $(n^{2}k-n^{2}k^{102}/16$ and the number of points which don't satisfy this property with respect to the second coordinate is $1/16(n^{2}k-\sqrt2n^{2}k^{102}$
assuming the worst case that the sets are disjoint and subtracting from $n^{2}k-\sqrt2n^{2}k^{2}/16$ we get $n^{2}k-\sqrt2n^{2}k^{2}/16-(n^{2}k-\sqrt2n^{2}k^{102}$.Then we do the same with the $x$ coordinate ending up with $1/16(n^{2}k-n^{2}k^{2}-2\sqrt2n^{2}k^{102})$.
We want to limit the number of points on the the line $PA$ by getting a large prime to divide the different of the $x$ coordinates of the two points or the difference of the $y$ coordinates of the two points and also the same for $PB$. However for this to work the large prime cannot divide both coordinates. 
So we have to filter out all points $P$ where the difference of the $x$ coordinate of $P$ from the $x$ coordinate of $A$ is divisible by a prime $q$ greater than $k^{-100}$ and the difference of the $y$ coordinate of $P$ from the $y$ coordinate of $A$ is divisible by the same prime $q$. Fortunately we can bound this by $n^{2}$ multiplied by the series $1/x^{2}$ with $x$ greater than $k^{-100}$ which sums to roughly $k^{100}$ and $2(k^{100})n^{2}$ is much less than $1/16(n^{2}k-n^{2}k^{2}-2\sqrt2n^{2}k^{102})$ and an error term. 
The error term can be viewed as extra points near the edge of the grid of points whose $x$ coordinate and $y$ coordinate differ from the $x$ coordinate and $y$ coordinate of $A$ by a multiple of $q$. The point is that some of the points near the edge of the square may not be counted. These can be bounded by the set of points whose $x$ and $y$ coordinates differ from $A$(or $B$, we are doing this for both points) by a multiple of $q$ and one of those differences is the largest multiple of $q$ at which such a difference occurs. Thus these points lie on the edges of a square we bound the error as follows for a certain value of $n/q$ this square will have less points on it than inside so we just add another copy of the series and that takes care of them. Other wise we have $n/q$ is small and the ratios of the  multiples of $q$ are limited and we can cover them by taking all the grid points that lie on a certain set of lines.
For $q$ such that $n/q$ is greater than 10 the error will be less than the series as the points at the edge of the square will bound the error. So for both $x$ and $y$ we take an additional copy of the original series.If we have $n/q$ less than 10 than the error will be bound by the border of the square and the points on the border of the square will have certain ratios as $n/q$ is less than 10 the ratio two numbers less then 10 so we can remove in addition all lines through $A$(and $B$ we have to do this for each point)whose slope is a ratio of two numbers less than 10. There will be less than a 100 such lines so we will subtract at most $100n$. This will still leave the remaining number of points greater than zero as for large $n$ $100n$ will be much less than a quadratic function of $n$ and  $2(k^{100})n^{2}$ is much less than $1/16(n^{2}k-n^{2}k^{2}-2\sqrt2n^{2}k^{102})$ as mentioned above.
So we have at least one point $P$ whose distance from the line containing $A$ and $B$ is greater than $nk^{2}$ satisfying the above properties and hence the distance from the two points is greater than this amount for each of the two points in particular there must be one coordinate of each point whose distance is $1/\sqrt2$ this amount that difference coordinate of these coordinates distance from $P$ must be divisible by a large prime greater than $k^{-100}$ and the other coordinate is not. but that limits the points on the line segments from $P$ to $A$ and $B$ to the distance times $k^{100}$ for the other point we get a similar limitation but by Pick's theorem we need at least $2n^{2}k^{2}$ points on the edges of the triangle to prevent points from forming in the interior and giving a contradiction. The factor of $k^{100}$ prevents this and forces a point in the interior which remains in the interior of the perturbed points and thus the line from this point and $P$ will cross the line segment between $A$ and $B$ forcing a contradiction.<|endoftext|>
TITLE: Integer division: the length of the repetitive sequence after the decimal point 
QUESTION [6 upvotes]: When dividing two integers, there may be an infinite sequence of digits after the decimal point (e.g. in the cases of 1/3, 1/7 etc).
As far as I know, if the two numbers divided are integers, this infinite sequence will be, from some point, a repetition of a some finite sequence (this may be also true to dividing rational numbers and not just integers, but I'm keeping it simple for now).
Take for example 1/7, it is equal to 0.14285714285714285714285714285714... 
The sequence repeating itself is 142857, a sequence of length 6.
Another example is 1/31, which is equal to 0.032258064516129032258064516129032...
The sequence repeating itself is 032258064516129, a sequence of length 15.
Some divisions do not immediately start with the repeating sequence, e.g. 3/14 which equals 0.21428571428571428571428571428571... (starting with 2, then followed by repetitions of 142857, a sequence of length 6).
Finally, after this long speech, here comes a question that bothers me for a long time: is it possible to find out the length of the repeating sequence using the 2 numbers being divided?
I'm looking for a function that receives a numerator and a denominator as parameters and returns the length of the repetitive sequence. Examples for input+output:
F(1,7) = 6
F(1,31) = 15
F(3,14) = 6
F(1,3) = 1

Does anyone know if this is possible? And if so - how to achieve it?

REPLY [2 votes]: The length of the period of a fraction is indifferent to the numerator unless the numerator and the denominator are not in lowest terms (in which case, reduce them to apply the following method).
However, when they are in lowest terms, take the denominator and factor out all factors of the base you're in, as they change nothing (like 2s and 5s in base 10, 3s and 7s in base 21, etc).  Then, the length of the period is equal to n, where n is the smallest integer such that your denominator divides bn-1 (where b is your base).  So, for example, 1/7 = 0.142857..., and the smallest 10n-1 such that 7 is a factor is 7 is 106-1 = 999999, and 999999/7 = 142857, which is, incidentally, your repeating portion.  The same thing even works for, say, 1/2 -- the smallest n is 0, because 100 - 1 = 0, and all non-zero integers divide 0.  This is one of the methods for factoring repunits.  Note that n will always be less than the denominator.
Similarly, 1/147 (which is 1/11 in base 10) = 0.0431162355..., and the smallest n such that 7n-1 is a multiple of 11 is 10 (710 - 1 = 282475248, and 282475248 / 11 = 25679568).
Multiplying by a constant does nothing to change the period, though it will change the numbers themselves.  Interestingly, though, when the period n is equal to 1 less than the denominator (called "max" periodic length), multiplying the fraction by a constant (other than a multiple of the denominator itself) will actually produce a cycle of the period.  2/7 = 0.285714..., 3/7 = 0.428571..., 4/7 = 0.571428..., etc.
Edit: When you divide by factors of the base, let the whole number by which you divide be f.  Then, $\lceil \log_{base}f\rceil$ (ceiling function) is equal to the number of non-repeating places before the repeating portion of the decimal.  Thus, 1/6, divided by 2, yields 1/3 (1/6 has the same repeating length as 1/3), and we divided by 2.  $\lceil\log_{10}2\rceil$ = 1, and indeed, 1/6 = 1.1(6...), with 1 non-repeating decimal.
Furthermore, for fraction m/n, $\lfloor\log_{base}n\rfloor$ is the number of leading zeroes in the repeating portion of the decimal.<|endoftext|>
TITLE: Why the search for ever larger primes?
QUESTION [15 upvotes]: I understand why primes are useful numbers and also why the product of large primes are useful such as for application in public key cryptography, but I am wondering why it is useful to continue the search for larger and larger prime numbers such as in the GIMPS project. It would seem to me since that since it already proven that there are an infinite number of primes, I am not quite sure why working to finding bigger and bigger really matters!? Is this a "Climbing Mt Everest because it is there" kind of thing, or is the search and finding bigger results somehow furthering mathematics in some kind of way?

REPLY [2 votes]: While it is true that primes are useful such as for application in public key cryptography, you won't be using the largest known primes for such tasks -- that is much too impractical.
I have discovered some large primes in the past (see my prime search profile: http://primes.utm.edu/bios/page.php?id=776), which made a cute little addition to my CV.
It's probably also a popular pursuit because it's so easy to understand the concept of primality, that there's always a larger prime, etc.  Lots of people can understand "I found the X-th largest known prime!"<|endoftext|>
TITLE: Is there a high-concept explanation for why characteristic 2 is special?
QUESTION [74 upvotes]: The structure of the multiplicative groups of $\mathbb{Z}/p\mathbb{Z}$ or of $\mathbb{Z}_p$ is the same for odd primes, but not for $2.$ Quadratic reciprocity has a uniform statement for odd primes, but an extra statement for $2$. So in these examples characteristic $2$ is a messy special case.
On the other hand, certain types of combinatorial questions can be reduced to linear algebra over $\mathbb{F}_2,$ and this relationship doesn't seem to generalize to other finite fields. So in this example characteristic $2$ is a nice special case.
Is anything deep going on here? (I have a vague idea here about additive inverses and Fourier analysis over $\mathbb{Z}/2\mathbb{Z}$, but I'll wait to see what other people say.)

REPLY [11 votes]: Here is my computational reason (instead of a high-concept explanation) why the primes 2 and 3 are special (hypotheses of many theorems on algebraic groups, linear or projective exclude both these primes). 
The numbers 2 and 3  got into the Primes Club by 'dubious' means!
Given $p$, to certify it as a prime a number, we  need to check no number $d $ with $1 < d \leq [ \sqrt p ]$ divides it. For 2 and 3 this condition is vacuously true as there are no integers in that interval, wheres 5 onwards they really needed to pass the test!<|endoftext|>
TITLE: a general theory of configurations?
QUESTION [5 upvotes]: Once I found by accident an article by MacPherson: "Classical projective geometry and modular varieties", in "Algebraic analysis, geometry, and number theory" (Baltimore, MD, 1988), whose introduction thrilled me and I'm still curious about how that developed and if now a general theory of configurations as continuation of classical geometry exists. Do you know something about it?  
copy from the article: 
"Classical projective geometry was a beautiful field in mathematics. It died, in our opinion, not because it ran out of theorems to prove, but because it lacked organizing principles by which to select theorems that were important. Also, it was isolated from the rest of mathematics. Much of what we do may be regarded as direct continuation of nineteenth century synthetic geometry. In fact, we hope the new motivation of studying C-complexes will provide projective geometry with one organizational principle, and with one relation tying it to "mainstream" mathematics. We note … representable  matroids, arrangements of hyperplanes, and motivic cohomology. A large part of this paper's exposition is motivated by this dream of continuing classical projective geometry."
Edit: Mnev's theorem, that every scheme over Z "is" a moduli space for point-configurations in the plane, makes me ask about applications  and if versions for other number rings exist?

REPLY [4 votes]: We model theorists have been studying such things for the past couple of decades, so I know something about this.
Suppose that (S, cl) is a matroid -- i.e. a set S endowed with a closure operator satisfying a couple of natural axioms; canonical examples are where S is a vector space and cl(X) = Span(X), or when S a "projective space" over some field (i.e. the set of all 1-dimensional subspaces of a vector space, and cl is the closure operator induced by linear span).  In addition to the standard matroid axioms, let's assume:

cl(emptyset) = emptyset;
cl({a}) = {a} for any a in S;
(nontriviality) There is a subset X of S such that cl(X) is not the union of {cl(a) : a is in X};
(modularity) For any finite-dimensional closed subsets X and Y of S, dim(X union Y) = dim(X) + dim(Y) - dim(X intersect Y);
dim(S) is at least 4, and any 2-dimensional subset contains at least 3 elements.

Then it turns out that (S,cl) is isomorphic to a projective space over some skew field.
I believe this was first proved in Emil Artin's book Geometric Algebra (1957).
In the context of model theory, there has been a lot of research recently on matroids that arise in the models of certain theories (only we call them "pregeometries") and how properties of these matroids are related to definable group actions and vector spaces, in analogy to Artin's result above.  See, for instance, Hrushovski's "stable group configuration theorem," which says that there is an infinite definable group whenever you have a certain finite configuration in a model of your theory.
http://en.wikipedia.org/wiki/Pregeometry_(model_theory)<|endoftext|>
TITLE: Operator Valued Weights
QUESTION [5 upvotes]: One of the basic tools in subfactors is the conditional expectation. If $N\subset M$ is a $II_1$-subfactor (or an inclusion of finite factors), then there is a unique trace-preserving conditional expectation of $M$ onto $N$. This should be thought of as a (Banach space) projection of norm 1. In fact, it is the restriction of the Jones projection $e_N$ on $L^2(M)$ to $M$. In the finite index case, we get another conditional expectation (Jones projection...) from the basic construction $M_1=\langle M, e_N\rangle$ onto $M$.
In his thesis, Michael Burns showed that if we iterate the basic construction in the infinite index case, we only get half the conditional expectations (we only get the odd Jones projections). The other half of the time, we get a generalization of the conditional expectation called an operator valued weight, originally defined by Haagerup.
Given an inclusion of semifinite von Neumann algebras $(N, tr_N)\subset (M, tr_M)$, there is a unique normal, faithful, semi-finite trace-preserving operator valued weight $T\colon M_+\to \widehat{N_+}$, where we must take the "extended part" of the positive cone $N_+$ of $N$. 
Edit as per @Dmitri's answer:
Let
$$
n_T=\{x\in M| T(x^\ast x)\in N_+\}
$$
and set
$$
m_T=n_T^\ast n_T=span\{x^\ast y| x,y\in n_T\}.
$$
There is a natural extension of $T$ to $m_T$. Is there an example of a normal, faithful, semifinite operator valued weight such that 

$N$ is not contained in $T(m_T)$, and/or
$1\notin T(M_+)$?

What about when $M$ and $N$ are factors ($M$ is $II_\infty$)?

REPLY [6 votes]: Since $T(m_T)$ is a non-trivial two sided ideal in $N$, it follows from spectral calculus that $T(m_T)$ contains a non-zero projection.  $T(m_T)$ then contains all subprojections, hence there exists $x \in m_T$ such that $p = T(x)$ is a projection with trace Tr$(p) = 1/n$ for some natural number $n$.  
We may assume that $pxp = x$, and by considering $(x + x^*)/2$ we may assume that $x$ is self-adjoint.  Since $m_T$ is spanned by its positive part it follows that we can write $x$ as $x_1 - x_2$ where $x_j \in m_T$ are positive.  Hence $T(x_1) \geq p$ and so by considering the element $(pT(x_1)^{-1/2}p)x_1(pT(x_1)^{-1/2}p) \in m_T$ we may assume that $x \geq 0$.
If $N$ is a factor, then since Tr$(p) = 1/n$ we may find a (finite, if $N$ is finite) sequence of partial isometries $v_k$ such that $v_k^*v_k = p$ for each $k$ and $\Sigma_k v_kv_k^* = 1$.  Then $x_m = \Sigma_{k = 1}^m v_k x v_k^*$ is a bounded increasing sequence in $m_T$ and we have that $T(x_m)$ increases to $1$.  Since the weight is normal we have that $x_\infty = \Sigma_k v_k x v_k^* \in m_T \cap M_+$ and $T(x_\infty) = 1$.  In particular, since $T(m_T)$ is an ideal we have $N \subset T(m_T)$.
If $N$ has infinite dimensional center then as Dmitri explained there are many examples (even bounded ones) for which $1 \not\in T(M_+)$.  I imagine that for finite dimensional center the argument above should still work.<|endoftext|>
TITLE: Where are mathematics jobs advertised if not on mathjobs (e.g. in Europe and elsewhere)?
QUESTION [102 upvotes]: My impression is that in the US, there is a canonical place for finding math jobs, namely mathjobs.org. For those of us who live and apply for jobs elsewhere, life is more complicated, and searching for advertised academic mathematics jobs for example in Europe can be a real hassle, with loads of different sites, different systems, and some jobs apparently advertised only on the web page of the hiring institution, or one some obscure mailing list.
So, where are academic math jobs advertised when they for some reason are not or cannot be on mathjobs.org? Of course I know of a few such places, but I am sure there must be many more.
All answers welcome, this would help me and probably many others.
Edit : All answers are 10 years old. Some of them might not be working. Any new answers (or corrections to old answers) would be useful.

REPLY [9 votes]: Another site for jobs in the UK & Ireland is http://www.lms.ac.uk/jobs/index.html
Jobs at Oxford and Cambridge are often advertised in the Oxford Gazette http://www.ox.ac.uk/gazette/ and the Cambridge reporter http://www.admin.cam.ac.uk/reporter/<|endoftext|>
TITLE: Analogue to covering space for higher homotopy groups?
QUESTION [47 upvotes]: The connection between the fundamental group and covering spaces is quite fundamental.  Is there any analogue for higher homotopy groups?  It doesn't make sense to me that one could make a branched cover over a set of codimension 3, since I guess, my intuition is all about 1-D loops, and not spheres.

REPLY [2 votes]: Since Peter Scholze in his answer used his condensed framework to consider higher covering spaces treating the topological and homotopic information separately, perhaps an alternative approach to doing this should also be mentioned, alluded to in comments there, namely, cohesive homotopy types. The constructions are very similar, see sections '5.2.5 Universal coverings and geometric Whitehead towers' and '6.3.12 Universal coverings and geometric Whitehead towers' of Urs Schreiber's 'Differential cohomology in a cohesive ∞-topos'.<|endoftext|>
TITLE: Fourier transform of $\exp(-\|x\|_p)$: more general question
QUESTION [11 upvotes]: David Corfield asked the following questions yesterday: Is the
$n$-dimensional Fourier transform of $\exp(-\|x\|)$ always non-negative,
where $\|\cdot\|$ is the Euclidean norm on $\Bbb R^n$?  What is its support?
I want to ask a more general question: what happens when $\|\cdot\|$ is the
$p$-norm, for arbitrary $p\in [1, 2]$?
David's question is here: Is the Fourier transform of $\exp(-\|x\|)$ non-negative?

REPLY [11 votes]: Okay, I think I do have an answer now.  I'm borrowing arguments from the proof of Lemma 2.27 in the book "Fourier Analysis in Convex Geometry" by A. Koldobsky (apparently not available online at all).  That lemma states that the Fourier transform of the function (on $\Bbb R$) $\exp(-|x|^p)$ is positive everywhere for $p \in (0,2]$.
The central tool is a theorem of Berstein, which in particular implies that if $s$ is in $(0,1]$ then $\exp(-z^s)$ is the Laplace transform of some finite positive measure $\mu$ on $[0,\infty)$; that is,
$$
\exp(-z^s) = \int \exp(-uz) d\mu(u).
$$
Applying this with $s=1/p$ and $z=\|x\|_p^p$ yields
$$
\exp(-\|x\|_p) = \int \exp(-u \|x\|_p^p) d\mu(u).
$$
Now calculate the Fourier transform on $\Bbb R^n$ of this.  Using Fubini you get an integral wrt $\mu$ of a product of Fourier transforms of $\exp(-|x|^p)$, and you can now apply the one-dimensional lemma.  (The one-dimensional lemma is proved by using the same theorem of Bernstein to reduce to the case $p=2$.)<|endoftext|>
TITLE: Looking for cubic, bipartite graphs with girth at least six and no cycles of length 8.
QUESTION [6 upvotes]: Aside from the Desargues graph, are there nice (at least vertex-transitive), small (say, less than 60 vertices), cubic, bipartite graphs with girth at least 6 and no 8-cycles? (or, even better, no cycles of length congruent to 0 mod 4.) 
Bonus: any idea how to use Mathematica to pull these out of its list of graphs it's got data for in Mathematica 7?

REPLY [5 votes]: Desargues graph, also known as the generalized Petersen graph $G(10,3)$ has girth 6 and also contains cycles of length 8. There exist three 10-cages, smallest cubic graphs of girth 10. They have 70 vertices and none of them is vertex-transitive. For more information about the motivation
see here. 
Therefore it is quite surprising that Gordon's example is so small. By the way, it is also arc-transitive. One may consider it as a Levi graph of a self-dual, flag-transitive 3-configuration of 19 points and 19 lines. Configuration contains triangles but has no quadrangles.
Here is a geometric realization of this graph.
Configuration with triangles but no quadrangles

A natural question is, whether this is the smallest graph with required properties. 
Another related question is, whether there exist smaller bipartite graphs if the vertex-transitivity condition is dropped.<|endoftext|>
TITLE: Does Ribet's level lowering theorem hold for prime powers?
QUESTION [10 upvotes]: I often use the following theorem (that one can state more generally) in my research.
Let E/Q be an elliptic curve of conductor N corresponding to a modular form f(E), l a prime of good or multiplicative reduction for E,  and \rho(l) the 2 dimensional mod l Galois representation given by the action on the l-torsion points. Suppose that the torsion subscheme E[l] extends to a finite flat group scheme over Z_l, and let p be a prime of multiplicative reduction for E such that \rho(l) is unramified at p (e.g. the number field (Q(E[l]) generated by the coordinates of the l-torsion points is unramified at p). Then there exists a modular form f of conductor N/p  such that f is congruent to f(E) mod l (when f has Fourier coefficients over Z then this means that all but finitely many of the coefficents are congruent mod l); one can `lower the level' from N to N/p.
Does such a result hold for powers of primes? E.g. if this holds for the mod l^n representation (instead of the mod l) does one get a congruence mod l^n?

REPLY [3 votes]: One case that you can say a bit more is if the congruence number at level $N/p$ is coprime to $l$. Specifically, if $f(E)$ is congruent to a unique modular form $g$ at level $N/p$ modulo $l$, then you can say that $f(E)$ is congruent to $g$ modulo $l^n$.<|endoftext|>
TITLE: Algorithm for finding the volume of a convex polytope
QUESTION [41 upvotes]: It's easy to find the area of a convex polygon by division into triangles, but what is the optimal way of finding the volume of higher-dimensional convex bodies? I tried a few methods for dividing them into simplices, but gave it up and went with a Monte Carlo estimation scheme instead. Bonus question: How to find the surface area of those same convex bodies?
EDIT: To answer David's question: the data set is a Voronoi tessellation of an n-dimensional volume (n usually 4) with a periodic boundary (like a torus). So I have the coordinates of the vertices of the convex bodies as well as the connectivity of all the facets, faces, etc. For the Monte Carlo I mentioned, I did convert everything to half-spaces, so I think that was not very difficult.

REPLY [6 votes]: An implementation of randomized algorithms using hit-and-run and comparison with exact software presented in Bueler, Enge and Fukuda is presented here.
It works efficiently in high dimensions (say few hudrend) but it only computes an estimation, not the exact volume.
The software is available on github.<|endoftext|>
TITLE: Rational maps with all critical points fixed
QUESTION [13 upvotes]: What can be said about rational self-maps of $\mathbb P^1$
for which all critical points are also fixed points ?
If all but one of the fixed points are critical, there is
a characterization in http://arxiv.org/abs/math/0411604v1 
( see Corollary 1 and the discussion just after the statement ).
Still assuming that all critical points are fixed: Is it possible to bound the degree of the rational map if all but two of the fixed points are critical ?
I think that the answer is probably no, but I would really
love to hear the contrary.

Motivation. The question is motivated by a rather specific problem I like to think about from time to time. It concerns the 
classification of some special arrangements of lines on the projective plane. More specifically, I would like to classify arrangements
of $3d$ lines(or rather hyperplanes through the origin of $\mathbb C^3$) invariant by degree $d$ homogeneous polynomial vector fields on $\mathbb C^3$. Given one arrangement like that one can produce a degree $d$ rational map having all its critical points fixed. 

Update. (08/28/2013) The paper On the classification of critically fixed rational maps by Cordwell, Gilbertson, Nuechterlein, Pilgrim, and Pinella classifies rational maps for which all critical points are fixed. In particular, for every degree $d\ge 3$ there exists a rational  such that all but two of the fixed points are critical.

REPLY [6 votes]: Theorem: for each n, there are up to Mobius conjugacy a finite, nonzero, number of rational maps of degree n such that all of the critical points are fixed.
Proof: Nonempty is clear: z^n.  For finiteness: the degree bounds the number of critical points. Given a bound on the degree and the size of the postcritical set, with the usual set of flexible Lattes exceptions (which have non-fixed critical points), the set of such rational maps up to Mobius conjugacy is finite; this is a consequence of Thurston's Rigidity Theorem [Douady, Adrien; Hubbard, John H. A proof of Thurston's topological characterization of rational functions. Acta Math. 171 (1993), no. 2, 263–297.]; cf. also section 3 of: [E. Brezin, R. Byrne, J. Levy, K. Pilgrim, and K. Plummer). Conf. Geom. & Dynam. Sys. (electronic) 4(2000), 35-74. ]<|endoftext|>
TITLE: What is lambda calculus related to?
QUESTION [15 upvotes]: So I'm not much of a math guy but I've really enjoyed programming in Lisp and have become interested in the ideas of lambda calculus which it is based.
I was wondering if anyone had a suggestion where I should go from here if I'm interested in learning about similar fields. If would be nice if I could relate it back to programming, but not necessarily a prerequisite.
Thanks.

REPLY [3 votes]: I suggest learning Coq.  It's pretty much the most powerful type system with a complete implementation right now.<|endoftext|>
TITLE: algebraic K-theory and tensor products
QUESTION [11 upvotes]: Algebraic K-theory defines a functor K taking commutative rings to E_\infty ring spectra.  I'm interested in which pushouts (tensor/smash products) K preserves.  For example, if R is a regular noetherian ring then (I believe) K(R[t, t^{-1}]) = K(R) / ΣK(R) = K(R) /\K(Z) K(Z[t, t^{-1}]).  On the other hand, K(Q) = K(Q ⊗ Q) is not the same as K(Q) /\K(Z) K(Q) as you can check by computing π1.
Are there useful conditions under which K-theory preserves pushouts?
Edit: I'm equally interested in more general positive answers and more geometric counterexamples.  For example, what is an example of smooth schemes X and Y over Spec k such that K(X) /\K(k) K(Y) -> K(X xk Y) is not an equivalence?
Also, what if I only cared about K0?  Is the product map more often an isomorphism then?
More generally, is there a spectral sequence to compute the K-theory of a fiber product of schemes?

REPLY [13 votes]: For smooth varieties $X$ and $Y$ over a field $k$, it is pretty rare to have that the product map $K_0(X)\otimes K_0(Y)\to K_0(X\times_kY)$ is an isomorphism (or surjective). 
For instance, suppose $X$ is smooth and proper over $k$, and the product map above is surjective for $Y=X$. Consider the class of the diagonal $\Delta_X$ in $K_0(X\times_kX)$ (this makes sense because $K_0$ is the same for vector bundles and coherent sheaves). Then the class of $\Delta_X$ is expressible as a finite sum $\sum_ia_i\otimes b_i$ with $a_i,b_i\in K_0(X)$. 
View elements $\alpha\in K_0(X\times_kX)$ as correspondences, i.e., as endomorphisms of $K_0(X)$, given by $x\mapsto (p_2)_*(p_1^*(x)\cdot\alpha)$, where $\cdot$ is the multiplication in $K_0$ (the direct image $(p_2)_*$ exists because $X$ is proper). The class of the diagonal gives the identity endomorphism, while the class of an element which is the image of $a\otimes b$ is rather special: it is of the form $x\mapsto \chi(x\cdot a)b$, where $\chi:K_0(X)\to {\mathbb Z}$ is the Euler characteristic ($\chi$ is the direct image map $K_0(X)\to K_0(k)$).  Using this, one can deduce that $K_0(X)$ must be free abelian of finite rank, and $a_i$ (and also $b_i$) form a $\mathbb Z$-basis (and in fact, for the non-degenerate pairing $(x,y)\mapsto \chi(x\cdot y)$, they form dual bases). This argument, in some form, is well-known in some circles; this ``dual bases'' formulation appears in work of Ivan Panin.
Thus it is easy to give examples of such $X$ for which $K_0(X)\otimes K_0(X)\to K_0(X\times_kX)$ is not surjective, e.g., a smooth projective curve of positive genus over an algebraically closed field, or (more tricky) a complex Enriques surface ($K_0$ is finitely generated, but has torsion). 
I do not know a counterexample to the following:
let $X$ be a smooth (say, proper) variety over a field $k$ for which $K_0(X)\otimes K_0(X)\to K_0(X\times_kX)$ is an isomorphism; then for any smooth variety $Y$, the product map $K_0(X)\otimes K_0(Y)\to K_0(X\times_kY)$ is an isomorphism.<|endoftext|>
TITLE: Applications and Natural Occurrences of Prime Numbers
QUESTION [13 upvotes]: I'm fascinated by prime numbers, and over the years, I've found multiple applications and natural occurrences for them.  But can anyone suggest some alternatives that aren't in my list?
Applications of Prime Numbers

Public key cryptography algorithms
The lengths of hash tables (not recommended)
Prime polynomials can be used for hash functions and CRC algorithms
Programs that search for large prime numbers can be used as “torture tests” for compilers, multitasking operating systems, memory, processors, etc.
Pseudo-random-number generators
Hard disk interleaving
Error-correction codes (quadratic residue codes)
Since no known natural process generates prime numbers, extraterrestrials might use them at the beginning of a radio transmission so we can distinguish it from a natural process (as in Carl Sagan's Contact).
Generating organically tiling images

Applications of Coprimality

If the number of teeth on a sprocket and the number of links in a chain are coprime, then the sprocket-chain system will experience even wear (cyclical wear will be minimized).
Nanotech symmetrical sleeve bearings, in which the outer sleeve has m-fold rotational symmetry and the inner sleeve has n-fold rotational symmetry, would in theory function most smoothly when m and n are coprime. (Nanosystems, K. Eric Drexler, p. 286)

Natural Occurrences of Prime Numbers

The 13- and 17-year life cycles of periodical cicadas (genus magicicada) may be an evolutionary advantage (minimizing exposure to predators and competing broods).

REPLY [3 votes]: Maybe this one is too frivolous, but sometimes, at Chinese restaurants, I get 5 shumai or 7 peking ravioli, instead of 6 or 8, and I've always wondered if that was a deliberate ploy to entice people to order more servings!<|endoftext|>
TITLE: Example where you *need* non-DVRs in the valuative criteria
QUESTION [9 upvotes]: The valuative criterion for separatedness (resp. properness) says that a morphism of schemes (resp. a quasi-compact morphism of schemes) f:X→Y is separated (resp. proper) if and only if it satisfies the following criterion:

For any valuation ring R (with K=Frac(R)) and any morphisms Spec(R)→Y and Spec(K)→X making the following square commute

Spec(K) ---> X
  |          |
  |          | f
  v          v
Spec(R) ---> Y

there exists at most one (resp. exactly one) morphism Spec(R)→X filling in the diagram.

But if Y is locally noetherian and f:X→Y is of finite type, then this condition only needs to be verified for discrete valuation rings. Does anybody know of an example where it is not sufficient to use DVRs? In other words, does there exist a morphism of schemes f:X→Y which is not separated (resp. proper), but does satisfy the valuative criterion for DVRs?
Reference: EGA II, Proposition 7.2.3 and Theorem 7.3.8

REPLY [14 votes]: You can probably just take Y to be the spectrum of a valuation ring A which is not a DVR,
for example the integral closure of C[[t]] in an algebraic closure of C((t)). In this case
any homomorphism from A to a DVR R has to factor through the quotient field or the residue field of A.
For an explcit example, Let X be two copies of Spec(A) glued along the complement of the closed point and let X --> Y be the map which is the identity on both copies. The fibres of this map are both proper so it satisfies the valuative criterion using only DVRs by the observation above but the map itself fails to be proper.<|endoftext|>
TITLE: Motivation/interpretation for Quillen's Q-construction?
QUESTION [24 upvotes]: This question has been on my mind for a while. As I understand it, the Q-construction was the first definition for higher algebraic K-theory. Some details can be found here.
http://en.wikipedia.org/wiki/Algebraic_K-theory
I have always wondered what train of thought led Quillen to come up with this definition. Does anyone know an interpretation of the Q-construction that makes it seem natural?

REPLY [11 votes]: There is an altogether different motivation different from the ones discussed above that appears in a paper by Graeme Segal 
("K-homology and algebraic K-theory," LNM 575 K-theory and Operator Algebras, Athens Georgia 1975, pp. 113–127).
The $Q$-construction there is motivated by considering
self-adjoint Fredholm operators on Hilbert space.
More, precisely Segal shows that the homotopy type of the classifying space of Q-construction of the category of finite dimensional vector spaces over the reals or complex numbers is the same as that of the space $Saf(H)$ consisting of self-adjoint Fredholm operators on infinite dimensional Hilbert space $H$:
$$
BQC \simeq Saf(H) .
$$
A map $V\to V'$ in the $Q$-construction on the category $C =$ Vect of finite dimensional vector spaces is represented as a triple $(W_+,W_-;\alpha)$ in which $\alpha: W_+\oplus V\oplus W_- \to V'$  is an isomorphism of vector spaces. 
According to Segal,
the idea is supposed to be that a Fredholm operator is determined up to contractible choice by its kernel and cokernel, which are a pair of finite dimensional vector spaces. When a Fredholm operator is deformed continously, its kernel and cokernel can jump but only by adding isomorphic pieces to each. 
In the self adjoint case, the operator is determined by its kernel up to contractible choice. The kernel then corresponds to an object of the $Q$-construction.
When the operator is deformed, the kernel jumps in such a way that the part added to 
it is a direct sum of the part on which the operator was positive and a part on which it was negative. These correspond to the terms $W_\pm$ appearing above. 
So the heuristic motivation in a nutshell is this: the objects of the $Q$-construction correspond to self-adjoint Fredholm operators and the morphisms correspond to deformations of such operators. The passage is given by taking operator kernels.
(note: I think Segal wants to consider the above $C$ as a topological category).<|endoftext|>
TITLE: What is the right way to think about / represent general  tilings?
QUESTION [11 upvotes]: For periodic/symmetric tilings, it seems somewhat "obvious" to me that it just comes down to working out the right group of symmetries for each of the relevant shapes/tiles, but its not clear to me if that carries over in any nice algebraic way for more complicated objects such as a penrose tiling
 and just following wikipedia just leads to the statement that groupoids come into play, but with no references to example constructions! Moreover, at least naively thinking about, it seems any such algebraic approach should naturally also apply to fractals.

what references am I somehow not able to find that do a good job talking about this further?
is my "intuition" that the mathematical structure for at least some classes of fractals and quasicrystals being equivalent correct?

REPLY [14 votes]: Aperiodic tilings can be thought of (in a sometimes useful way) as leaves of laminations; the groupoid in question (as in Emily's answer) is then the holonomy groupoid of the lamination.
There is a standard description of the Penrose tiles in this way; think of an irrational plane (i.e. an $R^2$) in $R^n$ for some $n>2$, and consider the set of 2-dimensional faces of the $Z^n$ lattice in $R^n$ that intersect a (uniform) thickened tubular neighborhood of your plane. Project each such 2-dimensional face perpendicularly down to your plane; the result is an aperiodic tiling. If the irrational plane happens to be chosen with extra symmetries (eg it could be an eigenspace of a finite order element in $GL(n,Z)$) one gets quite a tile set with extra "partial symmetries". The Penrose tiling is of this kind: think of $Z/5Z$ permuting the coordinate axes in $R^5$. This fixes the vector $(1,1,1,1,1)$ and has two perpendicular irrational eigenspaces on which it acts as an order 5 rotation; translates of these eigenspaces give rise to the "standard" Penrose tilings.
The lamination in this case is the "irrational foliation" of the torus $R^5/Z^5$ by planes with slope equal to the slope of the $R^2$ (and one can easily imagine generalizations).<|endoftext|>
TITLE: Intro to automatic theorem proving / logical foundations?
QUESTION [14 upvotes]: Is there any web-based course or materials about logic / automatic theorem proving? (I checked MIT's OpenCourseWare and I only found a vaguely related AI course)

REPLY [4 votes]: John Harrison "Handbook of Practical Logic and Automated Reasoning" - superbook<|endoftext|>
TITLE: "Fermat's last theorem" and anabelian geometry?
QUESTION [20 upvotes]: Do I remember a remark in "Sketch of a program" or "Letter to Faltings" correctly, that acc. to Grothendieck anabelian geometry should not only enable finiteness proofs, but a proof of FLT too? If yes, how?
Edit: In this transcript, Illusie makes a remark that Grothendieck looked for a connection between "FLT" and "higher stacks". BTW, here a note on (acc. to Illusie) Grothendieck's favored landscape.

REPLY [6 votes]: As Minhyong Kim points out in one of his papers on unipotent fundamental group- it is not quite clear how the section conjecture would imply Faltings theorem. The nature of implication (i.e. section conjecture implies Faltings or FLT) may be known to some experts but I don't know if it is explicitly written in literature.<|endoftext|>
TITLE: Explicit Direct Summands in the Decomposition Theorem 
QUESTION [6 upvotes]: Let f:X→Y be a semismall resolution of singularities. Then the pushforward of the constant sheaf on X is a semisimple perverse sheaf on Y. Under these conditions, I know how to calculate the direct summands of the pushforward f*ℚX[dim X].
My question is as follows: What more general statements are there that enable us to explicitly calculate the direct summands of the pushforward? I'm thinking especially of the case where f:X→Y is as above (so in particular semismall), but we replace the constant sheaf ℚX with an arbitrary perverse sheaf (of geometric origin).

REPLY [8 votes]: I agree with Ben that the question is a little confusing.
There are two possible questions:
How do you calculate direct summands of the direct image when we start with an IC (not necessarily the constant sheaf) on the source?
In this case I think the direct image need not be perverse. In which case you are in the general situation of describing the direct summands of a semi-simple complex, for which you need to know the characters of all IC's, or be very clever.
How do you calculate the local systems occurring in the direct image of the constant sheaf?
Ben describes what happens above. Semi-small means that the fibre over each stratum of the base has dimension bounded by half the codimension of the stratum (a number I will call d). The local system is then given by local system of 2d^th cohomology of the fibre.
In the semi-small situation this is beautiful: the 2d^th cohomology of the fibre is zero if the stratum isn't relevant, and has a basis given by fundamental classes of irreducible components if the stratum is relevant.
Note that the fundamental group of the stratum acts on the irreducible components of the fibre via monodromy, and this is precisely the local system that you get. (As an aside, this explains why the local systems are semi-simple, even though the fundamental group might be infinite: the representation always factors through the permutation group on the irreducible components.)
I first learnt this in the beautiful article "The Hard Lefschetz Theorem and the topology of semismall maps" by de Cataldo and Migliorini.<|endoftext|>
TITLE: decidability of group homomorphism existence
QUESTION [7 upvotes]: Fix a finitely-presented group $G$ with distinguished non-identity element $g$.  For any finitely-presented group $H$ with element $h$, is it decidable whether there is a homomorphism $h: G \rightarrow H$ such that $h(g) = h\ ?$
If we know $G$ is cyclic, the question is undecidable by reduction from the Word Problem. But what if we don't know anything about $G$? What if we know $g$ has finite order in $G$?

REPLY [8 votes]: The problem is decidable if and only if there exists a homomorphism from $G$ to the infinite cyclic group $\mathbb Z$ taking $g$ to 1 (the generator of $\mathbb Z$). Clearly, if such a homomorphism exists, the answer to your question is "yes" for every $H,h$. Suppose that there is no such homomorphism.  Consider the signature (group operations, nullary operation) of pairs $(H,h)$ where $H$ is a group, $h\in H$. Let $X$ be the set of generators and $R$ be the set of defining relations of $G$, suppose $g$ is represented by a word $w$ in $X$. Then you are asking, whether a formula $\theta=\exists X (\& R \& (w=h))$ is true for the pair $(H,h)$. But the negation $\neg\theta$ is a Markov property. Indeed, there exists a pair, say, $({\mathbb Z},1)$ which satisfies $\neg\theta$ because there is no homomorphism $G\to \mathbb Z$ that takes $g$ to $1$, and there exists a pair, say, $(G,g)$ which cannot be embedded into any pair $(G',g)$ which satisfies $\neg\theta$. The proof that Markov properties for pairs $(H,h)$ are undecidable is the same as the proof of the Adian-Rabin theorem for groups .<|endoftext|>
TITLE: What is the size of the category of finite dimensional F_q vector spaces? 
QUESTION [11 upvotes]: The size of a finite skeletal category C in the sense of Leinster is defined as follows: Label the objects of C by integers 1,2,...,n and let aij be the number of morphisms from i to j (for i and j between 1 and n). The size (or Euler characteristic) of C is defined as the sum of the entries of the inverse of the nxn matrix A=(aij), if the inverse exists.  
Let Fq be a finite field with q elements. For every natural number i, there is up to isomorphism exactly one Fq-vector space Vi of dimension i. The number of linear maps from Vi to Vj is equal to qij. We ignore the zero dimensional vector space V0. Consider the infinite matrix
Q=(qij)
where rows and and columns are indexed by positive integers 1,2,3,... From now on let us treat q as a formal parameter, don't care about convergence issues, and set v=q-1.
Is there a notion of an inverse of Q? (The entries will probably be formal power series in v.) If the answer is yes, what is a closed form for the sum of the entries of the inverse (as a formal power series in v), i.e. the size of the category of finite dimensional Fq-vector spaces?
At least every truncation Qn of Q to an upper left nxn corner has an inverse for every positive integer n, since Qn is a Vandermonde matrix. What is the limit of the sum of the entries of Qn-1 as n goes to infinity? I believe the answer is a power series in v. Is there an explicit form?
How can you interpret the answer? Is it the Euler characteristic of some moduli space? Is it equal or related to a sum of 1/Gl(Vi)? Does something interesting happen at q=1?

REPLY [5 votes]: Following the observations made in the comments one can compute the sum of the entries of Qn-1. It turns out that Kevin Costello's formula is true for every n.
Let (a1, a2, ..., an) be the the transpose of the kth column vector of Qn-1. (Of course, this vector depends on k, but we omit the index k.) Qiaochu Yuan suggested to consider the polynomial
A(x) = a1x + a2x2 + ... + anxn.
The degree of A is n, therefore A is determined by values at n+1 points. But we know that A(0)=0 and that
A(qi) = deltaik for i = 1, 2, ... , n. 
By Lagrange interpolation, A(x) is equal to
x(x-q)(x-q2)...(x-qk-1)(x-qk+1)...(x-qn) / qk(qk-q)(qk-q2)...(qk-qk-1)(qk-qk+1)...(qk-qn).  
The sum a1+a2+...+an is equal to A(1). Let us work with quantized integers. We use the notation [k] = (1-qk)/(1-q) = 1+q+...+qk-1. (Note that people from quantum groups sometimes use a different convention.) Furthermore, let [n choose k] be the quantized binomial coefficient. Then, A(1) is equal to 
(-1)k-1 [n choose k] qk(k-1)/2-kn.
We sum A(1) over all k. A variant of the quantum binomial theorem gives that the sum of the entries of Qn-1 is equal to 1 - (1-1/q)(1-1/q2)...(1-1/qn).<|endoftext|>
TITLE: Why is it so cool to square numbers (in terms of finding the standard deviation)?
QUESTION [47 upvotes]: When we want to find the standard deviation of $\{1,2,2,3,5\}$ we do
$$\sigma = \sqrt{ {1 \over 5-1} \left( (1-2.6)^2 + (2-2.6)^2 + (2-2.6)^2 + (3-2.6)^2 + (5 - 2.6)^2 \right) } \approx 1.52$$. 
Why do we need to square and then square-root the numbers?

REPLY [4 votes]: Here is a simple-minded explanation: The standard deviation as a "measure of dispersion" is the natural partner of the arithmetic mean as a "central statistic". 
Suppose we are given $n+1$ measurements (say lengths) $x_0 \le x_1 \le \cdots \le x_{n},$ and wish to choose a single value $x^*$ to represent them. We need a metric for how good a particular $x^*$ is. Then we choose the value which minimizes the "agregate discrepancy".

If our metric is $\sum |x_i-x^*|,$ then it is best to take $x^*=x_{n/2}$ (the median) for even $n$ and any $x_{(n-1)/2} \le x^* \le x_{(n+1)/2}$ for odd $n$. It is perhaps unfortunate that only one or two of the $x_i$ actually matter. 
Of course for $\sum(x_i-x^*)^2$ the unique minimum occurs for the familiar arithmetic mean $x^*=\frac{\sum x_i}{n+1}.$ We prefer to use the metric $\sqrt{\sum(x_i-x^*)^2}$ Since the "dispersion" is the same for measuring in inches as in feet (and the units are correct). There are also reasons to divide by $n+1$ or by $n,$  but none of this changes the minimizing value and the question was about the squaring.
For $\sum|x_i-x^*|^p$ with varying $p$ we get the standard median as $p \rightarrow 1^+$ and $\frac{x_0+x_n}2$ as $p \rightarrow \infty.$
I suppose the mode would result from calling the discrepancy $0$ or $1$ according as $x_i = x^*$ or $x_i \neq x^*.$
Would $\sum \ln|x_i-x^*|$ (equivalently, $e^{\sum \ln|x_i-x^*|}$) give the geometric mean $\sqrt[n+1]{\prod{x_i}}?$
It might not be hard to find other metrics which yield the  harmonic mean $$\frac1{\sum \frac1{x_i}},$$ and perhaps even the AGM.<|endoftext|>
TITLE: Splitting Pythagorean triples
QUESTION [19 upvotes]: Can one partition the set of positive integers into finitely many Pythagorean-triple-free subsets? If so, what is the smallest number of such subsets?  Taking a wild guess, I would
be least surprised if the answer were 3.
Notice that the 2 subsets of integers such that highest power of 5 that divides them is
a) even
b) odd
manage to split most primitive triples, plus all the multiples of those.
Notice also that Schur proved the positive integers cannot be split into any finite number of sum-free subsets (i.e. no finite partition can split all power-of-1 Fermat triples), while Fermat's theorem proves that all power-of-n (n>2) triples can be split by the trivial partition into 1 set.
Edit: Since this turns out to be a known open problem, we're adding the tag [open-problem] and converting this question to community wiki. The idea is to have a separate answer for each possible approach to solving this problem. If you have some additional insight or a reference to contribute to an answer, you only need 100 rep to do so. We're still figuring out exactly how to handle open problems on MO. The discussion is happening on this tea.mathoverflow.net thread.

REPLY [17 votes]: This problem appears in Croot and Lev's 2007 "Open Problems in Additive Combinatorics" (http://people.math.gatech.edu/~ecroot/E2S-01-11.pdf ), where it is attributed to Erdos and Graham (the latter of whom offers $250 for its solution).  
Other references may include Cooper and Poirel's "Notes on the Pythagorean Triple System" (http://www.math.sc.edu/~cooper/pth.pdf ), which mentions that even the case of whether 2 colors is enough is open.  They also exhibit a 2-coloring of the integers from 1 to 1344 without any monochromatic Pythagorean triples.
UPDATE (5/4/16): A new preprint of Heule, Kullmann, and Marek (to appear in the SAT 2016 conference) answers the question for $2$ subsets in the negative: If $\{1,2,\dots,7825\}$ is split into two subsets, one subset must contain a pythagorean triple.  The $7825$ here is tight.  The proof is heavily computer-assisted, and the key ideas seem to be to cast the problem in the language of satisfiability, and set up a divide-and-conquer method so that the (enormous) resulting instance is handleable by state-of-the-art satisfiability solvers.<|endoftext|>
TITLE: What m minimizes E(|m-X|^3) for a random variable X?
QUESTION [10 upvotes]: Let X be a random variable.  Then E(|m-X|^1) is minimized when (as a function of m) when m is the median of X, and E(|m-X|^2) is minimized when m is the mean of x.
A couple weeks ago in a technical stretch of a proof involving the Lyapunov condition for the central limit theorem I ended up with the expression E(|m-X|^3).  Does this statistic have a name, or any nice properties?
Edit: Earlier versions of this question had |m-EX| where |m-X| was; this isn't what I meant.

REPLY [5 votes]: The minimizer $m$ is the nearest point projection of $X$ onto the subspace of $L^p$ formed by the constant functions ($p=3$ in your case). This $m$ is sometimes called the $p$-prediction or $p$-predictor of $X$. Apparently, this terminology began with Andô and Amemiya. Some of later papers are Landers and Rogge (who wrote a few other papers, e.g. this one), and Cuesta and Matrán. The term "generalized (conditional) expectation" also appeared.<|endoftext|>
TITLE: When does Cantor-Bernstein hold?
QUESTION [87 upvotes]: The Cantor-Bernstein theorem in the category of sets (A injects in B, B injects in A => A, B equivalent) holds in other categories such as vector spaces, compact metric spaces, Noetherian topological spaces of finite dimension, and well-ordered sets.
However, it fails in other categories: topological spaces, groups, rings, fields, graphs, posets, etc.
Can we caracterize Cantor-Bernsteiness in terms of other categorical properties?
[Edit: Corrected misspelling of Bernstein]

REPLY [3 votes]: The Schroeder-Bernstein Theorem holds in categories where every endomorphism is an isomorphism. More precisely, if every endomorphism is an isomorphism and there are morphisms $f:A\rightarrow B,g:B\rightarrow A$, then $A$ and $B$ are isomorphic.
It turns out that the category of ultrafilters has no non-trivial endomorphisms and hence the Schroeder-Bernstein Theorem holds for the category of ultrafilters.
To be precise, the objects in the category of ultrafilters are pairs of the form $(X,\mathcal{U})$ where $X$ is a set and $\mathcal{U}$ is an ultrafilter on the set $X$. If $(X,\mathcal{U}),(Y,\mathcal{V})$ are objects in the category of ultrafilters, then a premorphism from $(X,\mathcal{U})$ to $(Y,\mathcal{V})$ is a function $f:X\rightarrow Y$ such that if $R\subseteq Y$, then $R\in\mathcal{V}$ iff $f^{-1}[R]\in\mathcal{U}$. Now let $\mathcal{A}_{(X,\mathcal{U}),(Y,\mathcal{V})}$ be the set of all premorphisms from $(X,\mathcal{U})$ to $(Y,\mathcal{V})$. Then define an equivalence relation $\simeq$ on $\mathcal{A}_{(X,\mathcal{U}),(Y,\mathcal{V})}$ so that $f\simeq g$ iff $\{x\in X|f(x)=g(x)\}\in\mathcal{U}$. Then the morphism between $(X,\mathcal{U})$ and $(Y,\mathcal{V})$ are the equivalence classes in $\simeq$. The composition of morphisms is ordinary composition of functions. Every endomorphism in the category of ultrafilters is the identity mapping.
In fact, the fact that the category of ultrafilters satisfies the Schroeder-Bernstein theorem is one reason why the Rudin-Keisler ordering on ultrafilters is well known while the actual category of ultrafilters is less well known:
Take note that $\mathcal{U}\leq_{RK}\mathcal{V}$ if and only if there is a morphism from $\mathcal{V}$ to $\mathcal{U}$. Since the category of ultrafilters satisfies the Schroeder-Bernstein Theorem, if $\mathcal{U}\leq_{RK}\mathcal{V}\leq_{RK}\mathcal{U}$, then $\mathcal{U}$ and $\mathcal{V}$ are isomorphic ultrafilters.
Therefore the Rudin-Keisler ordering on ultrafilters is a satisfactory ordering on the class of ultrafilters because the category of ultrafilters satisfies the Schroeder-Bernstein Theorem (and because the ultrafilters have no non-trivial endomorphisms also).
The Rudin-Keisler ordering has been generalized to other kinds of objects such as filters and Boolean ultrapowers. However, the generalized Rudin-Keisler orderings on the category of filters and the category of Boolean ultrapowers respectively does not satisfy the Schroeder-Bernstein Theorem (presumably), so the generalization of the Rudin-Keisler ordering are less satisfactory in these categories (and yes filters and Boolean ultrapowers do form categories).<|endoftext|>
TITLE: Definition of infinite permutations
QUESTION [7 upvotes]: I've been trying to find a definition of an infinite permutation on-line without much success. Does there exist a canonical definition or are there various ways one might go about defining this?
The obvious candidate I guess would be a bijection p : {1,2,...} -> {1,2,...} between the natural numbers. One might also try to use the Robinson-Schensted correspondence between permutations of length n and pairs of standard Young tableaux of size n. Then one would need a definition of infinite Young tableaux.
Another correspondence that might be used is between permutations and permutation matrices.

REPLY [2 votes]: You might also be interested in the "juggling patterns" of Knutson, Lam, and Speyer, defined here: http://arxiv.org/abs/1111.3660.
Also known as bounded affine permutations, these are a subset of the affine permutations that Lusztig introduced mentioned by Igor Pak. Namely, they are the affine permutations $\Sigma_n$ (so $\pi$ a permutation $\mathbb{Z} \to \mathbb{Z}$ with $\pi(i+n) = \pi(i)+n$) that also satisfy $i \leq \pi(i) \leq i + n$. They are related to total positivity and the positroid stratification. There is a nice way to visualize them as juggling patterns, where the number of balls being juggled is equal to the average $\frac{1}{n}\sum_{i=1}^{n}(f(i) -i)$.<|endoftext|>
TITLE: Order of an automorphism of a finite group
QUESTION [48 upvotes]: Let G be a finite group of order n.  Must every automorphism of G have order less than n?
(David Speyer: I got this question from you long ago, but I don't know whether you knew the answer.  I stil don't!)

REPLY [49 votes]: Yes every automorphism has order bounded by |G|-1, provided G is not the trivial group. A reference is
M V Horoševskiĭ 1974 Math. USSR Sb. 22 584-594
which can be found at
http://www.iop.org/EJ/abstract/0025-5734/22/4/A08/
It is even shown that the upper bound is reached only for elementary abelian groups.<|endoftext|>
TITLE: Motivation for coherence axioms
QUESTION [9 upvotes]: The pentagon and hexagon axioms in the definition of a symmetric monoidal category are one example that I was thinking of here; the axioms of a weak 2-category are another.  I understand that it can be checked laboriously that these few coherence axioms are sufficient to show, e.g. in the first case, that all coherence conditions we want on associativity and commutativity to hold do, but this is rather tedious.  Is there some other motivation for the choice of coherence axioms?

REPLY [10 votes]: I agree with Viritrilbia: there is in general no "nice" or canonical choice of coherence axioms.  By "in general" I mean for an arbitrary 2-dimensional theory, such as the theory of weak 2-categories, monoidal categories, braided monoidal categories with duals, etc.  This is true for the same reason that there is in general no "nice" or canonical presentation of a given group.
Nevertheless, there are some famous patterns in the coherence axioms for certain commonly encountered theories.  The associahedra, or Stasheff polytopes, give the higher associativity coherence axioms.  (There is one such polytope of each dimension.  The pentagon is the associahedron of dimension 2.)  The hexagon in the definition of symmetric monoidal category is also part of a general pattern.  
Topologists often sweep units under the carpet.  In the definition of monoidal category, the pentagon is not the only coherence axiom: there is also the triangle, which has to do with the unit coherence isomorphisms $X \otimes I \to X$.  People talk much less about the triangle, even though it's a crucial part of the definition.  I don't know of a general pattern that it fits into.
Incidentally, when the definition of monoidal category was first given (by Mac Lane?), there were one or two extra coherence axioms that were later shown to be redundant.  This is an indication of how non-obvious the choice of axioms is.<|endoftext|>
TITLE: Boolean network as a gauge field
QUESTION [5 upvotes]: Consider a set of N binary-state nodes at "time" t, each of which is a (boolean) transition function of two nodes in the set, evaluated at time t-1. Thus there are N of these boolean functions of two arguments. The nodes undergo an evolution (ie, follow a trajectory in state space) driven by the transition functions.
Further assume, initially, that the set of transition functions can change arbitrarily from one time-step to the next; the evolution is unconstrained and non-deterministic as possible, given the grounding assumptions.
Now, consider that the node states have names or labels (eg, "0" and "1"). The labeling is arbitrary, so that a description of such a system has a dual, obtained by globally exchanging state label "0" for "1" and "1" for "0", and each transition function by its dual, given that the functions are specified in terms of the labels. For example, AND(x,y) means a mapping {0,0:0; 0,1:0; 1,0:0; 1,1:1}, and OR(x,y) means a mapping {1,1:1; 1,0:1; 0,1:1; 0,0:0}. Thus, an exchange of label in the given mappings amounts to an exchange of AND for OR and vice versa.
We have a global symmetry of the description, reflecting the purely conventional choice of labeling for the node states. This symmetry does not imply any constraint on the dynamics. Suppose we would like to replace this global symmetry by a local symmetry, that allows local or "selective" relabeling of node states, both across the node collection, and through "time". How does this constrain the dynamics, and what is or should be invariant in this dynamics? For that matter, what is required to make the problem well-posed?
My question: Does the above sketch correspond to a developed area of mathematical investigation? In asking this question, I have in mind the observation that the set of N boolean transition functions can be regarded as an abstract simplicial complex; the binary-state nodes are vertices, and the transition functions are edges. I am therefore implicitly asking a question about the dynamics of this simplicial complex when bound to the dynamics of the node states in the way I have described. (This is the real motivation for my interest in this question.)

REPLY [2 votes]: This is a bad idea.  First, causal sets are, to put it politely "not well-respected" in theoretical physics.  These kinds of discrete structures tend to break a significant amount of the structure of the theories; enough that they typically can't reduce to anything that looks close to any existing theory in any limit.
In particular, they break Lorentz invariance pretty directly, which is tied up with energy and momentum conservation.  The standard "hope" is that that this invariance is really "approximate" somehow, and becomes exact in some limit.  But this can't really happen, because, in some sense the "errors" cannot average out, they can only add together.  
In particular, you no longer have properties like $E = m$ in a particle's rest frame, because this formula is a consequence of local Lorentz invariance.  So, what you end up with, is having to fine-tune because making this approximate means saying something like "$E=m+\delta$" for a particle.  But what about when you have $10^{23}$ particles?  Well, now, that $\delta$ term has to be fine-tuned by hand to be $10^{-23}$ to keep the microscopic deviations of order one (which is already way too big).  Then you can ask what happens if you boost a collection of particles (e.g., a lead ion) to near the speed of light, etc....
There are many other problems with discreteness, too, but this is an immediate and fatal one.
Incidentally, due to the seriousness of making spacetime discrete, a lot of the people who try to do these things (causal sets, dynamical triangulations, LQG, etc) like to claim that their theories do not actually do this.  But this is basically a lie; there's no politer way to put it.  If you ever see someone claim they do not make spacetime discrete, but do sometime like that, walk away, because they're probably a crackpot.  
Also, the idea that spacetime has to be "discrete" somehow is a huge (but, sadly, common) misunderstanding of what quantum mechanics means.  There is nothing "inherently" discrete, or any need to introduce any discreteness into physics other than as a computational tool (e.g., lattice field theory) or as a physical system whose initial conditions make this a convenient approximation (e.g., a regular solid structure, crystals, etc).
The initial idea behind thinking about these kinds of things wasn't so bad, though.  It comes from the thinking of, in GR, interpreting the abstract points on a spacetime manifold as corresponding to "events," and the geodesic structure as connecting events together (see, for example, the first few chapters of Misner, Thorne, and Wheeler's GR book).  This lead to trying to think of, instead of the geometric structure, the structure of "events", not in the sense of the topological structure of the manifold, but as some other structure inherited by how one can connect things together with geodesics.  This naturally partitions things into "all things that could have effected the point $x$" and "all things that the point $x$ can effect," with the hope that by studying the structure of these kinds of causal relations, one can reformulate GR in a new (but equivalent!) way that may fit together nicely with the "observables" formulation of quantum mechanics.
It turns out, this doesn't work so well.  The structures that you get are so uncontrolled and pathological and unconstrained, that the only thing you can do to get anything sensible is by just doing GR.  AFAIK, almost all physicists (and certainly all of the top people in GR) abandoned this approach by the '80s.
You can find some references for this kind of stuff if you look, but it does not really lead anywhere.
Unfortunately, in the past 10 or 15 years, a handful of people came along and misunderstood what these people were doing, and started building crazy discrete models analogous to these, sometimes in combination with taking "lattice GR" ideas way too seriously, and produced a bunch of nonsense.
And, really, the only thing stopping most physicists from calling these ideas outright crackpot nonsense, is the involvement of actual GR bigshots in the ancestors to these ideas decades ago!  Although, there are certainly a few well known physicists out there who are famous for getting very angry when these ideas are brought up ;).
If you're really determined to discover in detail why these ideas are so wrong, other than working out stuff for yourself, or learning why things are the way they are in GR and QFT (which is tough!--but you should!) you may be in some trouble.  Physicists aren't really in the habit of writing papers about theories that they think are obviously, fundamentally broken.  We just tend to make fun of them when talking to our colleagues and otherwise ignore them!  So you can't find much in the way of refutations of specifics in published literature.  But if you search carefully you can find the occasionally angry rant published somewhere, and there are a few unpublished papers on the arxiv about why these theories are all broken.  The only author I remember offhand doing this is Peeters about LQG's many problems, but there are others if you look.<|endoftext|>
TITLE: What is an example of a topological space that is not homotopy equivalent to a CW-complex?
QUESTION [8 upvotes]: It would also be nice if someone can explain this comment appearing on the Wikipedia page on CW-complexes:
"The homotopy category of CW complexes is, in the opinion of some experts, the best if not the only candidate for the homotopy category."

REPLY [10 votes]: My favorite example of a space which is not homotopy equivalent to a CW complex is the Long Line. All it's homotopy groups vanish (exercise 1) but the long line is not contractible (exercise 2). It's too long!<|endoftext|>
TITLE: Commutativity in K-theory and cohomology
QUESTION [13 upvotes]: The Chern classes give a map $f : BU \to \prod_n K(\mathbb{Z},2n)$, which is a rational equivalence. However, it is not an equivalence over $\mathbb{Z}$ because the cohomology of $BU$ is just a polynomial algebra and has no Steenrod operations. In particular, the generators of the homotopy groups $\pi_{2n}(BU)=\mathbb{Z}$ will not map to generators of the homotopy groups $\pi_{2n}(K(\mathbb{Z},2n))=\mathbb{Z}$. Another way to say this is that the duals of the Chern classes in homology are not in the image of Hurewicz; only certain multiples of them are. What multiples you have to take is determined by the order of the k-invariants of $BU$, which are certain Steenrod operations of the fundamental classes of $K(\mathbb{Z},2n)$.
Steenrod operations can be understood as obstructions to the cup product on ordinary cohomology being strictly commutative.  On the other hand, the fact that $f$ is not an equivalence can also be understood as an obstruction to addition in K-theory being strictly commutative. Indeed, any space with a strictly commutative group structure is a product of $K(\pi_n,n)$'s under the map given by a right inverse of the Hurewicz map.
So in some sense you could say that products in cohomology are only homotopy-commutative and sums in K-theory are only homotopy-commutative "for the same reason". Is there some deeper story behind this? I don't know exactly what I'm asking for, but I'd like to get a better understanding of what's going on in this picture.

REPLY [13 votes]: Perhaps the deeper story you want involves the notion of "$E_{\infty}$ product".  The cup product in cohomology, and the sum (and for that matter, the product) in K-theory are commutative (and associative, and unital) not merely up to homotopy, but "up to all possible higher homotopies".  You can make this precise by saying that the appropriate binary operation on the representing space (the $K(\mathbb{Z},n)$'s, or $\mathbb{Z}\times BU$), is part of an $E_{\infty}$ algebra structure on that space.
It seems that most "naturally occuring" sums or products in topology turn out to be $E_{\infty}$ (addition for any generalized cohomology theory, multiplication for many nice ones such as ordinary cohomology, K-theories, bordism, elliptic cohomology).  As you observe, having a strictly commutative operation is very special, and basically forces the representing spaces to be a product of $K(A,n)$'s, by the Dold-Thom theorem.  
To me, the mystery here is that "stricly commutative" is so much more special than "$E_{\infty}$ commutative". Associative products don't behave this way: "strictly associative" turns out to be no more special than "$A_{\infty}$ associative", that is, any $A_{\infty}$ product on a space can be "rigidified" to a strictly associative product on a weakly equivalent space.<|endoftext|>
TITLE: easy(?) probability/diff eq. question
QUESTION [9 upvotes]: I've been wondering about this ever since I was a little kid and I used to ride in the back of the car and my mom would speed like hell towards a green light, only to slam on the brakes when she realized she wasn't going to make it.
Here is my question, loosely phrased: Given that we want to make it across the intersection before the light is up, what should be our position function?  To make things precise, we could specify some initial conditions, put a cap on the car's acceleration/deceleration (or its speed or its jerk or whatever), fix a probability distribution on when the light will change, and assign some point values to our happiness if we make the light vs. don't make the light vs. end up entering the intersection after the light has changed and get a ticket for it.  And then of course we could throw in the extra curveball of a yellow light warning you that it's about to turn red...
A similar question arises if you're approaching a red light but you think it might turn green soon.  Ideally you'd like to enter the intersection at the highest possible speed just after the light has turned green, but then again you don't want to enter when it's still red.
I'm sure a computer could solve such problems easily, but it seems like there should be some better way to think about this than just asking a machine to do it for me.  For the first question, it seems like the answer will just be either "hit the gas and go for it" or "cruise to a stop and don't plan on going through", depending on the parameters.  (Of course, there might be something in the "go just fast enough that you can slow down and not enter on a red" plan if the cost of a ticket is high enough.)  On the other hand, the second question seems to admit much more interplay between probability and differential equations.  The real issue here is that I know almost nothing about either of these two fields.  Any ideas?

REPLY [4 votes]: The general type of  extremal' that Amit talks about has a name within the field of optimal control.Most' control strategies tend to be  `bang-bang'. For a control
(here an acceleration) to bebang-bang' means that you operate at the max (of A above) or min (as in slamming on the brakes). The whole problem boils down to figuring out the `switching times' : when to switch from one strategy to the other.
There is a LARGE mathematical and engineering literature on this type of problem, some of the most useful works having been done by Pontrjagin,Gamkrelidze, and Boltjanskii and going under the name `Maximum principle'.  LC Young wrote a beautiful book on it. 
The general set-up for the maximum principle  is an ODE $\dot x = f(x,u, t)$
where $u$ is called the control'.  The simplest version of the optimal control
problem is to choose $u = u(t)$  tosteer' the vector or scalar  $x(t)$  from $x_0$ to $x_1$.
The constraints are of the form $a < u < A$ (if $u$ is 
vector-valued , then $u$ varies over some convex set). 
 The thing to maximize might be the time the steering takes (`optimal time')
or some integral $\int L (x, u, t) dt$.  
If instead one looks for $u = u(x)$ one is talking about ``optimal feedback control''.<|endoftext|>
TITLE: What's a groupoid? What's a good example of a groupoid?
QUESTION [60 upvotes]: Or more specifically, why do people get so excited about them?  And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?

REPLY [7 votes]: The  holonomy groupoid of  a  foliation is  another example of  a useful groupoid
it is described here:
http://www.ams.org/journals/bull/2005-42-01/S0273-0979-04-01036-5/S0273-0979-04-01036-5.pdf
For  a singular foliation see http://users.uoa.gr/~iandroul/AS-holgpd-final.pdf<|endoftext|>
TITLE: What are the higher $\mathrm{Ext}^i(A,\mathbf{G}_m)$'s, where $A$ is an abelian scheme?
QUESTION [13 upvotes]: Let $S$ be a base scheme, let $A/S$ be an abelian scheme, and let $\mathbf{G}_m/S$ be the multiplicative group; consider $A$ and $\mathbf{G}_m$ as objects in the abelian category of sheaves of abelian groups on the fppf site of $S$, and take $\mathrm{Ext}$'s between them. We know that $\mathrm{Ext}^1(A,\mathbf{G}_m)$ is the dual abelian scheme; but what is $\mathrm{Ext}^i(A,\mathbf{G}_m)$ for $i>1$?
Here is an argument why these higher $\mathrm{Ext}$'s contain no important information: the dual abelian scheme captures already all the data of $A/S$, since applying $\mathrm{Ext}^1(\cdot, \mathbf{G}_m)$ one more time recovers $A$; therefore the higher $\mathrm{Ext}$'s cannot hold any more information. Still, we should know explicitly what they are.

REPLY [11 votes]: It seems that here you are talking about the local exts, i.e. about the sheaves $\underline{Ext}^{i}(A,G_m)$ on $S$. The question is rather tricky actually. The problem is that before we try to answer it, we should first specify what we mean by $\underline{Ext}^{i}(A,G_m)$. We could mean  exts in the category of sheaves of abelian groups in the flat topology on $S$, or we could mean exts in the category of commutative group schemes. The latter is not an abelian category so you have to do something before you can define exts. It is possible to do this though. The standard lore is to use Yoneda exts. This is carried out in detail in the LNM 15 book by Oort. Among other things Oort checks that if $S$ is the spectrum of an algebraically closed field, then ext sheaves are all zero for $i \geq 2$.
Over general base schemes the situation is more delicate. First of all there are examples of Larry Breen showing  that the ext sheaves in the category of sheaves of abelian groups are strictly larger than the ext sheaves in the category of commutative group schemes. In his thesis
Breen, Lawrence
Extensions of abelian sheaves and Eilenberg-MacLane algebras.
Invent. Math. 9 1969/1970 15--44. 
Breen also showed that over a regular noetherian base schemes the global ext groups (in either category) are torsion if $i \geq 2$. Later in 
Breen, Lawrence
Un théorème d'annulation pour certains $E{\rm xt}\sp{i}$ de faisceaux abéliens.
Ann. Sci. École Norm. Sup. (4) 8 (1975), no. 3, 339--352. 
he strengthened his result to show that the higher exts sheaves are always zero for $1 < i < 2p-1$, where $p$ is a prime which is smaller than the (positive) residue characteristic of any closed point in $S$.<|endoftext|>
TITLE: Does Cantor-Bernstein hold for classes?
QUESTION [11 upvotes]: In Bonn, we've been have a discussion on the topic in the title: 

Suppose that A and B is are classes and that there are injections from A to B and fom B to A. Does it follow that there is a bijection between A and B?

Example: Let A the class of sets of cardinality one and let B be the class of sets of cardinality two. There is an injection 
A -> B sending a to {a, emptyset},
B-> A sending b to {{b}}.
Does it follow that there is a bijection between A and B?

REPLY [21 votes]: Since the question has arisen whether the standard
arguments really do work with classes, let me post this
answer giving some fuller details about one comparatively
concrete way to do it. Other methods are also possible.
I shall work in the theory of Goedel-Bernays GB set theory
(without global choice), a general setting for treating
classes, which forms a conservative extension of ZF. This
theory includes as a special case the traditional treatment
of classes as definable collections in ZF, since every
model of ZF, when augmented with its definable classes,
forms a model of GB. Thus, the GB context seems the most
comprehensive way to answer (and if you prefer ZF, then
just imagine that all classes here are definable from
parameters).
Suppose that $A$ and $B$ are classes and that we have class
functions $F:A\to B$ and $G:B\to A$ which are injective, as
in the question. Let $A_0=A-\operatorname{ran}(G)$, which is a class,
since it is definable from $A$ and $G$. Let us say that a
sequence $\langle x_0,x_1,x_2,\ldots\rangle$ is a
back-and-forth-iteration sequence if $x_0\in A$ and
$x_{2n+1}=F(x_{2n})$ and $x_{2n+2}=G(x_{2n+1})$ for all natural numbers $n$. Let $A_n$
be the elements $a_{2n}$ that appear at the even
coordinates of a back-and-forth iteration sequence with
$a_0\in A_0$. This notion is definable from $F$ and $G$, and the point is that we have a uniform presentation of the $A_n$ in a single class $\{(n,a)\mid a\in A_n\}$.
Let $A^+=\bigcup_n A_n$, which is a class definable from
$F$ and $G$.
Let $H$ be the function $(F\upharpoonright
A^+)\cup(G^{-1}\upharpoonright A-A^+)$. I claim that this
is the desired bijection between $A$ and $B$. First, it is
clearly a class that is definable from $F$ and $G$. Second,
it is a function from $A$ to $B$. Note that if $a\in A_n$,
then $G(F(a))\in A_{n+1}\subset A^+$, and so $F(a)$ is not
$G^{-1}(a')$ for any $a'\in A-A^+$. Thus, the function $H$
is injective. Secondly, if $b\in B$ and $G(b)\in A_n$, then
it must be that $n\geq 1$ and so $b=F(a')$ for some $a'\in
A_{n-1}\subset A^+$, putting $b\in\operatorname{ran}(H)$. Otherwise,
$G(b)\notin A^+$, and so again $b\in\operatorname{ran}(H)$. So $H$ is a
bijection. QED
Some of the other proofs can also be formalized for
classes, if one simply uses the sequences as I did here for
iterating.<|endoftext|>
TITLE: Category theory sans (much) motivation?
QUESTION [21 upvotes]: So I have a friend (no, really) who's taking algebra and is struggling to gain intuition for it. My story is as follows: I used to hate abstract algebra, with pretty much a burning passion, until I started to learn about the categorical way of thinking.
I think that the deal is as follows: One begins to gain all sorts of intuition about, for instance, groups when one realizes that it doesn't pay to think about elements of a group nearly as much as it does to think about morphisms to/from a group. The category-theoretic point of view is a tool that lets you gain intuition by moving up and down the hierarchy of abstraction. Maybe I'm not articulating this that clearly, but hopefully you've had similar experiences.
The problem is, I don't know any way to get a handle on the categorical way of thinking without learning category theory, and I don't know any way of learning category theory without wading through tons of abstract nonsense before you can begin to understand why it's valuable. (This is an even worse problem if, like my friend, you don't have any interest in the motivating examples from topology or geometry.)
So, what can I recommend that might help my friend start thinking categorically without drowning him in a sea of abstraction? Or is the "algebra sucks" phase a necessary stage of mathematical development?
ETA: Just to be clear, this is mostly undergrad-level stuff we're dealing with, so while I'm not opposed to easy ways to motivate category theory or get someone hooked, the fewer prerequisites the better...

REPLY [2 votes]: Maybe you could construct an explicit exact sequence which describes something irl about which your friend has already an intuition?
Start with a tiny category (1,2,3 elements)
 (source: Wayback Machine)
which is theoretically simple, and by a sequence of algebra-respecting surjections how a complex model of the real-world thing can be broken down (at different levels)  into smaller bits with an easier logic to them. (Alternatively you could branch out in different directions and show how chain 1 captures this aspect of the model, whilst chain 2 captures that aspect of the model.) I don't have a ready example for that right now but I'll edit this answer if I think of one.
You could also spend time talking about "congruence"—just like it would be idiotic—pedantic beyond the worst human pedant—to treat two different hand-writings of the letter a as having different meanings, we want to be able to use the word "the same" in mathematics as we use it in normal language—like, "OK, not the same same, but, you know, basically the same". (And this needs some precise definition in order to give us, in mathematics, the freedom-of-movement we get gratis in eg English.) I like the familiar toddlers' play blocks as a metaphor for "The same like how?".
 (source: Wayback Machine)
I'm not sure if this applies to your friend, but the simplest functor I can think of is between $(o,e,\mathbb{Z}_+) \longleftrightarrow (pos,neg,\mathbb{R}_{\times})$. Anyone who remembers grade-school arithmetic can follow the technical aspects of that relationship. (And you cover "the arrows changing along with the objects"—the relationship wouldn't stay the same if I substituted positives for evens without changing + to ×—and I think this is intuitively clear to anyone whose brain has been sufficiently jogged.)

I'm not sure if that captures what took you out of algebraic doldrums, but that's how I would explain category without busting out a Rube Goldberg of definitions first.<|endoftext|>
TITLE: Sheaf cohomology and injective resolutions
QUESTION [55 upvotes]: In defining sheaf cohomology (say in Hartshorne), a common approach seems to be defining the cohomology functors as derived functors. Is there any conceptual reason for injective resolution to come into play? It is very confusing and awkward to me that why taking injective stuff into consideration would allow you to "extend" a left exact functor.

REPLY [22 votes]: This is an interesting discussion to someone raised in the 60's.  It illustrates how lack of motivation creeps into books unnoticed.  Back when Hartshorne was being written everyone was steeped in the then standard derived functor formalism of Cartan Eilenberg and Grothendieck, axiomatizing constructions of cohomology via complexes.  So the pattern of this formalism explained so nicely by Anton and Andrew above was taken for granted. As the subject evolved, it was understood that acyclic resolutions could replace the more categorically natural injective ones.  This is the gist of Evan's answer. Perhaps also there is a tradition in mathematics books of giving definitions without historical background.  I have always had difficulty with any unmotivated definitions, such as abstract "residues" and modern Riemann Roch theorems, so I keep stressing to my students the value of learning the original version of Riemann, even if their goal is to understand cohomological and arithmetic versions.  Anyway, excellent question.<|endoftext|>
TITLE: Graded local rings versus local rings
QUESTION [31 upvotes]: A lot of times I see theorems stated for local rings, but usually they are also true for "graded local rings", i.e., graded rings with a unique homogeneous maximal ideal (like the polynomial ring). For example, the Hilbert syzygy theorem, the Auslander-Buchsbaum formula, statements related to local cohomology, etc.
But it's not entirely clear to me how tight this analogy is. I certainly don't expect all statements about local rings to extend to graded local rings, so I'd like to know about some "pitfalls" in case I ever decide to make an "oh yes, this obviously extends" fallacy. What are some examples of statements which are true for local rings whose graded analogues are not necessarily true? Or another related question: what kind of intuition should I have when I want to conclude that statements have graded versions?
There is a notion of "generalized local ring" due to Goto and Watanabe which includes graded local rings and local rings: a positively graded ring that is finitely generated as an algebra over its zeroth degree part, and its zeroth degree part is a local ring, so one possibility is just to see if this weaker definition is enough to prove the statement. Of course the trouble comes when the proofs cite other sources, and become unmanageable to trace back to first principles.

REPLY [11 votes]: I will try to provide some geometric intuition, why there should be an analogy between local rings and graded rings with unique homogeneous maximal ideal. Maybe this also helps to guess whether a statement true for local rings should still hold in the graded case.
A graded k-Algebra can be thought of as an affine space with $k^*$-action. Homogeneous prime ideals correspond to invariant closed sub-varieties. So your sort of algebras corresponds to spaces with exactly one fixpoint. For example in the case of the polynomial ring its $k^n$ with the obvious action of the multiplicative group and $0$ is the fixpoint. From this example we see, that the action can be used to "contract" the space to the fixed point. Hence the local nature of the space.<|endoftext|>
TITLE: Atiyah-Singer index theorem
QUESTION [54 upvotes]: Every year or so I make an attempt to "really" learn the Atiyah-Singer index theorem.  I always find that I give up because my analysis background is too weak -- most of the sources spend a lot of time discussing the topology and algebra, but very little time on the analysis.  Question : is there a "fun" source for reading about the appropriate parts of analysis?

REPLY [2 votes]: There is a new and attractive book on Index theory and applications to Physics by Booss and Bleecker  which covers all the necessary analysis background. To quote from its preface : 

In order to enjoy reading or even work through Parts I-III, we expect the readerto  be  familiar  with  the  concept  of  a  smooth  function  and  a  complex  separableHilbert  space.   Nothing  more.) 

Note that this is different from the 1984 book 'Topology and Analysis : ASIT and Gauge theoretic Physics' by the same authors which has been mentioned above.<|endoftext|>
TITLE: Given a sequence defined on the positive integers, how should it be extended to be defined at zero?
QUESTION [7 upvotes]: This question is inspired by a lecture Bjorn Poonen gave at MIT last year.  I have ideas of my own, but I'm interested in what other people have to say, so I'll make this community wiki and post my own thoughts later.  Here are some examples of what I'm talking about:

Why does a^0 = 1?
Why does 0! = 1?
If the Fibonacci number Fn+1 counts the number of ways to tile a board of length n with tiles of length 1 and 2, why does F1 = 1?
What is the determinant of a 0x0 matrix?
What is the degree of the zero polynomial?
What is the direct product of zero groups?
What is the zeroth homotopy group of a space?

I want to be very precise about exactly what I'm asking for here.
Question 1:  What general principles do you apply in a situation like this?  Can they be stated as theorems, or do they only exist at the level of intuition?
Question 2:  Do you know of any examples where there are two different ways to extend a sequence to zero, both of which are reasonable from the perspective of some principle?  
Feel free to answer at any level of sophistication.

REPLY [2 votes]: The determinant of an endomorphism f of a free R-module of dimension n (R commutative) is the $d \in R$ such that $\bigwedge^n f$ is the homothety of ratio d.
Our case corresponds to $n=0$, and $\bigwedge^0 f$ is the identity of R, so d=1.
The reasons, already given, why 0^0=1 (m^n is the number of functions from a set of cardinality n to a set of cardinality m) and 0!=1 (n! is the number of bijections of a set of cardinality n), are illustrations of Baez's ideas on counting as decategorification.<|endoftext|>
TITLE: Pontryagin product from an operad
QUESTION [6 upvotes]: For a topological group G, we have a Pontryagin product in homology by multiplying representative cycles. This gives the homology the structure of an associative graded algebra. Am I correct in thinking we can prove this by seeing a topological group as an algebra over the Ass-operad and applying homology everywhere?

REPLY [6 votes]: You can do that.  If an operad O acts on a space X, then the structure maps
    O(n) x Xn -> X
induce homology operations
H*O(n) ⊗ H*(X)⊗n -> H*(X).  In particular, any path component in O(2) produces a multiplication on H*X, if it's in the same path component as its own image under the symmetric group action it's commutative, if the two composites of it are in the same path component of O(3) it's associative, et cetera.
In particular if O is the associative operad (so O(n) are discrete) then this structure reduces to the Pontrjagin ring structure.<|endoftext|>
TITLE: How to partition R^3 into pairwise non-parallel lines?
QUESTION [7 upvotes]: Problem. How to partition R^3 into pairwise non-parallel lines?
A possible solution is to stack infinitely many ``concentric'' hyperboloids, by increasing radius and decreasing slope. And don't forget the line on the $z$ axis at the center. The prototype hyperboloid looks like this.
I heard a talk to which I didn't understand a lot ; a solution was given using Hopf fibration. I'm not familiar to these notions, and at the end it went like ``Tadaa! And here is our partition!''. The speaker could not describe what the partition looks like.
I would be very glad to: (1) understand the math he did (article, book?), (2) see what his solution looks like, and (3) know what kind of solutions exist.
Thanks in advance!

REPLY [2 votes]: Each foliation of 3-dimensional Euclidean space by pairwise nonparallel lines extends to a foliation of real projective 3-space by nonintersecting projective lines. Hence lifts to a foliation of the 3-sphere by nonintersecting great circles. So the complete classification of all foliations of 3-dimensional Euclidean space by pairwise nonparallel lines is the classification of great circle fibrations of the 3-sphere, which is worked out very clearly in Gluck, Warner, Great circle fibrations of the 3-sphere, Duke Math. J.,  Volume 50, Number 1 (1983), 107-132.
To give some idea of the method: each great circle spans an unique oriented 2-plane through the origin of $\mathbb{R}^4$, which is spanned by an oriented orthonormal basis $u,v$. The 2-vector $\zeta=u\wedge v$ is uniquely determined. It splits into a self-dual and an anti-self-dual part, say $\zeta_+,\zeta_-$, each of the same length, which we rescale to be unit length.
For a great circle fibration, every choice of $\zeta_-$ occurs uniquely, i.e. for a unique circle fiber, so $\zeta_+$ can be written as a function of $\zeta_-$. This functions turns out to be smooth, a smooth map from the unit sphere in the anti-self-dual 2-vectors to the unit sphere in the self-dual. This map is strictly contracting in the usual metric on those spheres. Moreover, any strictly contracting map arises in this way. All of this is proven in great detail in the paper.<|endoftext|>
TITLE: The large sieve for primes
QUESTION [9 upvotes]: Let $\Lambda(n)$ be the von Mangoldt function, i.e., $\Lambda(n) = \log p$ for $n$ a prime power $p^k$ and $\Lambda(n) = 0$ for all $n$ that not prime powers. Let 
$$S(\alpha) = \sum_{n \leq N} \Lambda(n) e(\alpha n).$$
Now, using, say, Lemma 7.15 in Iwaniec-Kowalski (or the same result in Montgomery), we get
$$\sum_{q \leq q_0} \sum_{a \pmod{q}: \gcd(a,q)=1} \lvert S(a/q)\rvert^2 
    \leq \frac{(N + Q^2) N \log N}{\sum_{\substack{q\leq Q \text{ squarefree} \\ \gcd(q,P(q_0))=1}} \phi(q)^{-1}},$$
where $Q$ is arbitrary and $P(z):=\prod_{p \leq z} p$. 
In practice, we would choose $Q$ slightly smaller than $\sqrt{N}$, and obtain
$$\sum_{q \leq q_0} \sum_{\substack{a \pmod{q} \\ \gcd(a,q)=1}} \lvert S(a/q) \rvert^2 \leq (1+\epsilon) 2 e^\gamma N^2 \log q_0,$$
where gamma is Euler's constant $0.577\cdots$ and $\epsilon$ is very small.
Now, the 2 in the bound $\leq (1+\epsilon) 2 e^\gamma N^2$ is due to the parity problem,
and thus should be next to impossible to remove (except for very small $q_0$).
However, the factor of $e^\gamma$ clearly has no right to exist. The true asymptotic should be simply $N^2 \log q_0$.
Can we remove that nasty $e^\gamma$? That is, can you prove a bound of type
$$\sum_{q \leq q_0} \sum_{\substack{a \pmod{q} \\ \gcd(a,q)=1}} \lvert S(a/q)\rvert^2 \leq (1+\epsilon) 2 N^2 \log q_0 ?$$
Harald

REPLY [10 votes]: Heh, I think I know why you are interested in this question, Harald, as Ben and I thought about essentially the same problem for what I suspect to be the same reason :-)
Anyway, we were able to get rid of the e^gamma factor.  One way to proceed is to work not with the exponential sums, but rather the inner product of Lambda with Dirichlet characters.  Using  a Selberg sieve majorant (the same one used to prove, say, the Brun-Titschmarsh inequality) and the TT^* method, one gets a nice l^2 bound on these inner products (losing only the 2 from the parity problem, or not even that if one deletes the principal character), and then one can do some Fourier analysis on finite groups to pass from Dirichlet characters back to exponentials.  It helps a little bit to smooth the sum, but it's not too essential here.  (See also my paper with Ben on the restriction theory of the Selberg sieve for some related results.)
Our argument isn't written up (or checked, for that matter) yet, but we can talk more if you want to know more details.<|endoftext|>
TITLE: Are submersions of differentiable manifolds flat morphisms?
QUESTION [40 upvotes]: Let $\pi \colon M\to N$ be a smooth map between real smooth manifolds. Then $C^\infty(M)$ forms a module over $C^\infty(N)$ (via pullback). Is this module flat when $\pi$ is a submersion?
Recall that the usual definition of flatness is equivalent to the following equational condition: 
whenever $ h_1 \ldots h_k\in C^\infty(N) $ and $g_1 \ldots g_k\in C^\infty(M)$ are such that:
$$h_1 g_1 + \ldots + h_k g_k = 0$$ (as functions on $M$)
then there are functions $G_1 \ldots G_r\in C^\infty(M)$ and $a_{i,j}\in C^\infty(N)$ such that:
$$g_i= \sum_j  a_{i,j}G_j \; \forall i $$ 
and 
$$\sum_i h_i a_{i,j}= 0 \; \forall j$$
Some remarks:

It's known that the inclusion of an open subset $U\subset N$ is a flat morphism since smooth functions on $U$ are obtained from smooth functions on $N$ by localizing w.r.t. functions vanishing nowhere on $U$.
It's also known that smooth flat maps have to be open. Proofs of both of these facts can be found for example in the book: Gonzales, Salas, $C^\infty$-differentiable spaces, Lecture notes in Mathematics, Springer 2000.
I've asked some of the experts including Malgrange and the above authors and it seems that the answer is not known.
I gave the equational condition of flatness since it seems like the most reasonable thing to use here. But considering already the simplest situation here's what gets me stuck: suppose you want to check flatness of the standard projection $\mathbb{R}^2 \to \mathbb{R}, (x,y)\mapsto x$, and take the case of just one $h\in C^\infty(\mathbb{R})$ and one $g\in C^\infty(\mathbb{R}^2)$ with $hg=0$. If you pick $h(x)$ to be strictly positive for $x<0$ and $0$ for $x\geq 0$, then the flatness condition translates into: 

Any smooth function $g(x,y) \in C^\infty (\mathbb{R}^2)$ that vanishes on the half plane $x\leq 0 $ admits a "factorization":
$$g(x,y)= \sum_j a_j (x)G_j (x,y)$$
where the $a_j\in C^\infty(\mathbb{R})$ all vanish on $x\leq 0$ and the $G_j\in C^\infty(\mathbb{R}^2)$ are arbitrary.
Anyone has an idea how to prove this "simple" case, or sees a counter example?
(Edit: George Lowther beautifully proves this "simple" case, and also comes closer to the full result in his second answer. If you also think he deserves some credit consider up-voting his second answer since the first one turned community wiki.)
Motivation
My personal interest is that a positive answer would allow me to finish a certain proof, which trying to explain here would take this too far afield. But I may try to put the question into context as follows: the notion of flat morphism plays an important role in algebraic geometry where it is basically the right way to formalize the notion of parametrized families of varieties (fibers of such a morphism being these families). One may also say that it is the right "technical" notion allowing one to do all the things one expects to do with such parametrized families (correct me if I'm wrong). 
Now I've been taught that differential topology may also be seen as a part of commutative algebra (and that taking such a point of view might even be useful at times). For example: a manifold itself may be recovered completely from the algebra of smooth functions on it, and any smooth map between manifolds is completely encoded by the corresponding algebra morphism. Other examples: vector fields are just derivations of the algebra, vector bundles are just finitely generated projective modules over the algebra etc. Good places to learn this point of view are: Jet Nestruev, Smooth manifolds and observables, as well as the above mentioned book.
Now in differential topology there is a well know notion of smooth parametrized families of manifolds, namely smooth fiber bundles. Hence from this algebraic point of view it would be natural to expect that fiber bundles are flat morphisms.

REPLY [4 votes]: In George Lowther's CW answer taking care of the "simple case" there are a Lemma 1, 2 and 3 concerning sets of function $U$ and $V$ about which it is stated "they could possibly be standard results, but I've never seen them before." I wanted to add that, by performing an inversion, those results become instead statements about rapidly decaying functions and Schwartz functions. In the latter form, they are easier to find in the literature. Really the present answer would be better as a comment, but I think there will not be space for that.
Use the map $x \mapsto 1/x$ to put functions on $(0,1]$ into bijection with functions on $[1,\infty)$. This correspondence puts the functions $f$ on $(0,1]$ with $\lim_{x \to 0^+} x^{-n} f(x) = 0$ for all $n$ into bijection with the functions $f$ on $[1,\infty)$ with $\lim_{x \to \infty} f(x) x^n = 0$ for all $n$, i.e. the functions of rapid decay. It also puts the functions $f$ on $(0,1]$ for which putting $f(x)=0$ for $x\leq 0$ yields a smooth extension into bijection with the Schwartz functions on $[1,\infty)$.
So George Lowther's Lemma 1 is equivalent to asking whether, for every function which decays rapidly at $\infty$, there exists a Schwartz function which decays more slowly. Lemmas 2 and 3 can be given similar restatements. Stated in this revised form, Lemma 1 is arguably easier to prove.

Claim: Suppose $f$ is a rapidly decaying function on $[1,\infty)$, then there exists a Schwartz function $g$ on $[1,\infty)$ with $g \geq |f|$.
Proof: Without loss of generality, $f$ is postive-valued and decreasing, or else replace it by $x \mapsto \sup_{y \geq x} f(y)$. Let $\varphi$ be a nonnegative $C^\infty$ bump function with support in $[0,1]$ and $\int_\mathbb{R} \varphi =1$. Then it is simple to check that the convolution $g= \varphi * f$ is a Schwartz function with $g \geq f$ (the main point is that rapidly decaying functions are closed under convolution, which implies the convolution of a Schwartz function and a rapidly decaying function is Schwartz).

The Schwartz VS rapid decay versions of the facts seem to be easier to find in the literature as well. See this expository note of Paul Garret: Schwartz-function envelopes for rapidly decreasing functions and note that the three bullet points of the theorem in Garrett's note correspond more or less to the three Lemmas in Lowther's answer. This post by Abdelmalek Abdesselam points to additional references in the literature.

K. Miyazaki, Distinguished elements in a space of distributions, Lemma 1.
H. Petzeltová; P. Vrbová Factorization in the algebra of rapidly decreasing functions on $R_n$.
J. Voigt, Factorization in some Frechet algebras of differentiable functions.<|endoftext|>
TITLE: Is ${\rm S}_6$ the automorphism group of a group?
QUESTION [72 upvotes]: The automorphism group of the symmetric group $S_n$ is $S_n$ when $n$ is not $2$ or $6$, in which cases it is respectively $1$ and the semidirect product of $S_6$ with the (cyclic) group of order $2$. (For this famous outer automorphism, see for instance wikipedia or Baez's thoughts on the number $6$.)
On the other hand, $S_2$ is the automorphism group of $Z_3$, $Z_4$ and $Z_6$ (and only those groups among finite groups). Hence my question: is $S_6$ the automorphism group of a group? of a finite group?

REPLY [7 votes]: There is a whole array of results, going back to G. Birkhoff at 1930s saying that every group is an automorphism group of some universal algebra (or some universal algebra inside some class).

(This really should be merely a comment to the previous answer, but I am still not reputable enough to leave comments).<|endoftext|>
TITLE: How to write math well?
QUESTION [35 upvotes]: Let's learn about writing good mathematical texts. 
For some people it could be especially interesting to answer about writing texts on Math Overflow, though I personally feel like I've already mastered a certain level in writing online answers while being hopelessly behind the curve in writing papers. So,

What is your advice in writing good mathematical texts, online or offline?

REPLY [8 votes]: One useful strategy (after a few years' experience) might be to read your own papers from a few years before. Do you still understand your own proofs and explanations ( or would you, if you were reading them from "scratch" without your own prior knowledge)? If the answer is "no", or "with difficulty", you might ask yourself why, and what could be done to improve that in future writing.<|endoftext|>
TITLE: Ways to characterize supersingular primes?
QUESTION [7 upvotes]: I've read the definition, and it basically says p is a supersingular prime iff 
the fundamental domain of a group generated by \Gamma(p) and a matrix ((0, 1), (-p, 0)) is rational.
And there's a finite list of those, so it's kind of stupid to try to characterize a finite set.
But, anyway, what is the deep meaning of supersingular primes? What are different ways to characterize them?
This question actually arose when I posted a different question about one of the characterizations, the one related to Monster finite group. I hope to collect all possible answers from number theory here.

REPLY [7 votes]: Supersingular primes are those primes p for which all supersingular elliptic curves over an algebraic closure of Fp have j-invariant in Fp.  There is a theorem of Deuring that implies the j-invariant always lies in Fp2 for any prime p, so supersingular primes form a rather distinguished class.  From the standpoint of probabilistic heuristics, you should expect these primes to be rather small, since there are exactly (p-1)/24 supersingular elliptic curves over an algebraic closure of Fp, weighted by automorphisms.<|endoftext|>
TITLE: Where stands functoriality in 2009?
QUESTION [27 upvotes]: Robert Langlands is famous in number theory for making famous and deep conjectures about very abstract things called automorphic forms, somewhere in the 60s.
There's a very interesting article by Langlands called Where stands functoriality today which describes the development of the subject from Langlands' point of view as of 1990s.
But if somebody was to write an overview of the current state of Langlands functoriality, what would it say?

REPLY [9 votes]: Perhaps the best answer to the question "Where stands functoriality today?" is given by Langlands himself in his informal write-up of two recent lectures (March 2011) given at the IAS, available here: 
http://publications.ias.edu/sites/default/files/functoriality.pdf<|endoftext|>
TITLE: cardinality of final coalgebras in Top
QUESTION [11 upvotes]: Let P be a polynomial functor from Top to Top, by which I mean a functor of the form P(X) = ∐i ≥ 0 Si × Xi where the Si are finite sets, all but finitely many of which are empty.  Corresponding to P there is an ordinary polynomial p(x) = Σi ≥ 0 |Si| xi.  I believe that there is a final P-coalgebra XP in Top; the structure map XP -> P(XP) is a homeomorphism; the underlying set of XP is the final P-coalgebra in Set; and a basis for the open sets of XP is given by the preimages of the points of the targets of the maps XP -> P(n)(XP) -> P(n)(•) as n ranges over all nonnegative integers.
Examples are the Cantor space, corresponding to p(x) = 2x, and the space of binary trees, corresponding to p(x) = x^2 + 1.
My question is: if P and Q are two polynomial functors, such that neither p(x) nor q(x) is of the form x + k, and the spaces XP and XQ are homeomorphic, does it follow that the polynomials p(x) - x and q(x) - x have a common root?
More generally, consider all the spaces that can be formed starting with the collection of spaces XP (p(x) not of the form x + k) by taking disjoint unions and products.  Can I assign an algebraic number to each of these spaces in a way which is a homeomorphism invariant and commutes with sums and products?

REPLY [2 votes]: I don't know, but one could begin to investigate your first question by considering polynomials of the form ax + k, where a is a natural number greater than 1.  Maybe you already know that the answer to the question is "yes" in that case?  (If so, I'd be interested to see your answer.)  
A point related to this, and to the general idea about algebraic numbers, is discussed here:
http://golem.ph.utexas.edu/category/2007/04/reportback_on_bmc.html#c009149
There are also close relations with the theory described in arXiv:math.DS/0411343 (an improved version of which appears in Real and Complex Singularities, ed. Paunescu et al, World Scientific 2007).  But there are differences too, e.g. you allow products but I didn't.<|endoftext|>
TITLE: What does "supersingular" mean?
QUESTION [7 upvotes]: Are supersingular primes and supersingular elliptic curves related?
(this was essentially a subquestion in my earlier question, but still looks sufficiently different to me to deserve a separate post)

REPLY [21 votes]: There are many equivalent ways to define supersingularity for an elliptic curve over a characteristic $p$ field.  One of them is that the $p$-torsion of the curve is connected, i.e., it is a purely infinitesimal group scheme of order $p^2$.  As Jonah mentioned, supersingular means very special, and is not a statement about smoothness.  There is a theorem of Deuring that implies the j-invariant of a supersingular elliptic curve always lies in $\mathbb{F}_{p^2}$, and as a consequence, all such curves are defined over a finite degree extension of $\mathbb{F}_p$.  
There are two notions of supersingular prime: one is relative to a fixed elliptic curve over $\mathbb{Q}$, and one is absolute.  For any elliptic curve $E/\mathbb{Q}$, a prime $p$ is supersingular for $E$ if $E$ has good supersingular reduction at $p$.  Such primes are known to be asymptotically density zero, but infinite in number (by a theorem of Elkies).  Lang and Trotter conjectured that the number of supersingular primes less than $N$ limits to a constant times $\sqrt{N}/\log(N)$ as $N$ gets large.
Supersingular primes in the absolute sense are those primes $p$ for which all supersingular elliptic curves over an algebraic closure of $\mathbb{F}_p$ have $j$-invariant in $\mathbb{F}_p$ instead of just $\mathbb{F}_{p^2}$.  These happen to be the primes that divide the order of the monster simple group, and they are also the primes for which the normalizer of $\Gamma_0(p) \in SL_2(\mathbb{R})$ acts on the complex upper half plane with a genus zero quotient. For general $p$, this normalizer contains $\Gamma_0(p)$ as an index 2 subgroup, with the nontrivial coset called the "Fricke involution" (a special case of Atkin-Lehner involultion).  There is a standard order 2 representative, taking $z \mapsto -1/pz$.  The quotient curve classifies unordered pairs of elliptic curves with dual degree p isogenies between them.  I do not know any canonical relations between these characterizations of supersingularity.
Edit: Thanks to Emerton for pointing out the connection.  I'll try to expand on it a bit.  The moduli problem of generalized elliptic curves with $\Gamma_0(p)$-structure has a coarse moduli space that is a smooth irreducible curve away from p, but has mod p fiber given by taking a disjoint union of two copies of $X(1)$ (a genus zero curve) and gluing along supersingular points (this description is more or less in Katz-Mazur, chapter 13).  A geometric point describing an elliptic curve with j-invariant $\alpha \in \mathbb{F}_{p^2}$ is glued to a geometric point on the other irreducible component describing an elliptic curve with j-invariant $\alpha^p$.  The Fricke involution switches the components, so the quotient of $X_0(p)$ by this involution is a genus zero curve glued to itself at finitely many supersingular points.  The quotient has arithmetic genus zero if and only if all supersingular geometric points are glued to themselves - otherwise, the flat modular deformation to characteristic zero yields a smooth curve of higher genus.  In other words, it is necessary and sufficient that all supersingular geometric points have no Frobenius conjugates, i.e., that the j-invariants of all supersingular curves lie in $\mathbb{F}_p$.
More Edit: I should give a more honest reply to Mariano's question, which was originally raised by Ogg in the mid 1970s (and he famously offered a bottle of Jack Daniels to anyone who could solve it).  Half of the question has an answer.  If we combine the results of Borcherds's paper Monstrous moonshine and monstrous Lie superalgebras with the results of the paper Modular equations and the genus zero property of moonshine functions by Cummins and Gannon, we get the following fact:

Let G be a finite group acting faithfully on a conformal vertex algebra V by conformal symmetries, and suppose V has central charge 24 and character $\operatorname{Tr}(q^{L_0-1}|V) = j(\tau)-744$.  Then for any element $g \in G$, the series $\operatorname{Tr}(gq^{L_0-1}|V)$ is the q-expansion of a modular function that is holomorphic on the upper half plane, invariant under a discrete group $\Gamma \subset PSL_2(\mathbb{R})$ satisfying $\Gamma \supset \Gamma_0(N)$ for some $N$, and generates the function field of the quotient curve $\Gamma \backslash \mathbf{H}$.  In particular, the quotient curve is genus zero.

The monster simple group arises in this context because I. Frenkel, Lepowsky, and Meurman constructed a conformal vertex algebra satisfying the above hypotheses, whose group of conformal automorphisms is the monster.  The fact given above implies that for each prime p dividing the order of the monster, the quotient $X_0^+(p) = X_0(p)/\langle w_p \rangle$ is genus zero, where $w_p$ is the Fricke involution.  In particular, each prime dividing the order of the monster is necessarily supersingular.
I know some people who would like there to be a conceptual (read: non-enumerative) explanation for why all supersingular primes divide the order of the monster.  So far, the best I've heard is that the monster is really big, while there aren't that many supersingular primes.<|endoftext|>
TITLE: Alternatives to pi day
QUESTION [15 upvotes]: If you don't already know, pi day happens on March 14 (3-14) every year. Festivities include reciting digits of pi and eating pies. I understand that it's all in good fun, but I've always felt that pi day is bad PR for mathematics. To non-mathematicians, it gives the impression that mathematics is about voodoo numerology, memorizing (or computing) digits of pi, and bad puns. The bad puns part is pretty accurate, but I don't care for the others.
So I propose we come up with some alternatives to pi day that send a better message about what mathematicians do. If you've got an idea, post it here. If possible, include the following information about your proposed holiday:

when is it celebrated?
what are the festivities?
what kind of food would be associated with this day?

Please post only one proposed holiday per post. If you have more than one, please post multiple answers.

REPLY [4 votes]: Pi Approximation Day on July 22nd. (ie, 22/7)
I celebrate it every year, by going out in the summer and getting cake, instead of pie like we do in March. It's approximately pie :)<|endoftext|>
TITLE: erfc lower bound
QUESTION [10 upvotes]: I've seen the following lower bound for the complementary error function (erfc) but I haven't been able to prove it.  Does anyone know how to establish the following?
$$erfc(x) > \frac{ x \exp(-x^2) }{ \pi(1 + 2x^2) }$$

REPLY [10 votes]: Paradoxically, it is quite easier to prove stronger bounds.
Let $X$ be a random variable with gaussian distribution and density $$ f(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2).$$
Now let, for any $k\in\mathbb{R}^+$, 
$$A_k = \sqrt{2\pi}\;\exp(k^2/2)\;\mathbb{P}[X>k] = \sqrt{\frac{\pi}{2}}\;\exp(k^2/2)\;\operatorname{Erfc}\left(\frac{k}{\sqrt{2}}\right).$$
Since $\mathbb{E}\left[\left(X-\mathbb{E}[X]\right)^2\right]\geq 0$, $\mathbb{E}[X^2]\geq\mathbb{E}[X]^2$, and the same holds for the conditional expected values, under the hypothesis $X>k$. That gives, in terms of $A_k$:
$$ (1-k\; A_k)^2 \leq A_k\;\left(-k+(1+k^2)\; A_k\right), $$
that can be restated as:
$$(\heartsuit)\quad A_k^2+k\;A_k-1\geq 0.$$
From:
$$ A_k \geq \frac{2}{k+\sqrt{k^2+4}}, $$
we immediately have a lower bound for the $\operatorname{erfc}$ function:
$$ e^{k^2}\;\operatorname{erfc}(k) \geq \frac{2}{\sqrt{\pi}}\left(\frac{1}{k+\sqrt{k^2+2}}\right).$$
An interesting fact is that the "reverse inequality"
$$(\spadesuit)\quad (1-k\; A_k)^2 \geq \frac{2}{\pi}\;A_k\;\left(-k+(1+k^2)\; A_k\right), $$
equivalent to:
$$(\spadesuit_2)\quad\pi\left(\int_{k}^{+\infty}(x-k)\;e^{-x^2/2}\;dx\right)^2\geq 2\int_{k}^{+\infty}e^{-x^2/2}\;dx\int_{k}^{+\infty}(x-k)^2\;e^{-x^2/2}\;dx,$$
holds, too. Since:
$$A_k =\int_{0}^{+\infty}\exp\left(-x^2/2-kx\right)\,dx, $$
by using the Fubini-Tonelli theorem and switching to polar coordinates one can see that $(\spadesuit_2)$ is equivalent to:
$$\int_{0}^{+\infty}\rho^3 e^{-\rho^2/2}\int_{0}^{\pi/4}(\pi\cos(2\theta)-2)\; e^{-k\rho\sqrt{2}\cos\theta}\;d\theta \; d\rho\geq 0, $$
also equivalent to (by integrating by parts):
$$\int_{0}^{+\infty}\rho^3 e^{-\rho^2/2}\int_{0}^{\pi/4}(\pi\sin\theta\cos\theta-2\theta)\;k\rho\sqrt{2}\sin\theta\; e^{-k\rho\sqrt{2}\cos\theta}\;d\theta \; d\rho\geq 0, $$
that is trivial since over $[0,\pi/2]$, by the concavity of the sine function, we have $\sin\phi\geq\frac{2\phi}{\pi}$. Following the line of the previous proof, the $(\spadesuit)$-inequality can be used to have tight upper bound for the $e^{k^2}\;\operatorname{erfc}(k)$ - function. Moreover, from the $(\spadesuit_2)$-inequality the convexity of $\frac{1}{A_k}$ follows.<|endoftext|>
TITLE: A learning roadmap for algebraic geometry
QUESTION [122 upvotes]: Unfortunately this question is relatively general, and also has a lot of sub-questions and branches associated with it; however, I suspect that other students wonder about it and thus hope it may be useful for other people too.  I'm interested in learning modern Grothendieck-style algebraic geometry in depth.  I have some familiarity with classical varieties, schemes, and sheaf cohomology (via Hartshorne and a fair portion of EGA I) but would like to get into some of the fancy modern things like stacks, étale cohomology, intersection theory, moduli spaces, etc.  However, there is a vast amount of material to understand before one gets there, and there seems to be a big jump between each pair of sources.    Bourbaki apparently didn't get anywhere near algebraic geometry.
So, does anyone have any suggestions on how to tackle such a broad subject, references to read (including motivation, preferably!), or advice on which order the material should ultimately be learned--including the prerequisites?  Is there ultimately an "algebraic geometry sucks" phase for every aspiring algebraic geometer, as Harrison suggested on these forums for pure algebra, that only (enormous) persistence can overcome?

REPLY [4 votes]: The following seems very relevant to the OP from a historical point of view: a pre-Tohoku roadmap to algebraic topology, presenting itself as a "How to" for "most people", written by someone who thought deeply about classical mathematics as a whole. The source is 

Ernst Snapper:  Equivalence relations in algebraic geometry. Bulletin of the American Mathematical Society,
  Volume 60, Number 1 (1954), 1-19.

However, I feel it is necessary to precede the reproduction I give below of this 'roadmap' with a modern, cautionary remark, taken literally from http://math.stanford.edu/~conrad/:

That said:
A 'roadmap' from the 1950s.


It is interesting, and indicative of how much knowledge is required in algebraic geometry, that Snapper recommends Weil's 'Foundations' at the end of this "How to get started"-section.<|endoftext|>
TITLE: Geometric calculations using Grassmann variables
QUESTION [9 upvotes]: Physicists seem to get huge computational value by introducing Grassmann variables and Grassmann integration into differential geometric calculations.  
See: http://en.wikipedia.org/wiki/Grassmann_number
Can someone here motivate these techniques mathematically, and include the simplest  pure-math example where their use and value can be illustrated.  I have thought they were invented for computing volumes of constrained moduli spaces, although physicists did not originally describe them this way.  Do any mathematicians use them rigorously to actually solve problems?

REPLY [4 votes]: Grassman variables are a neat and handy way to talk about exterior algebras and the geometric objects built on them, like forms and spinors. The physics notation in particular gets a lot of mileage out of Grassman calculations that look formally like Gaussian integrals and behave remarkably like them.  The best intro to that, and the bet purely mathematical application I know, is Mathai and Quillen's paper where they get an explicit representation of the Thom form that is quite powerful.  They can be a little intimidating because physicists speak about them in language grating to mathematicians and tend to use them in the vicinity of highly nonrigorous thinking, but they themselves can be worked with mathematically without difficulty.<|endoftext|>
TITLE: L_p norm balls for 1<p<2 - is it always similar to an L_q norm ball for some q>2?
QUESTION [8 upvotes]: The L_1 ball in 2D is shaped like a diamond (L_1 is also known as the Manhattan norm). The L_∞ ball is shaped like a square (L_∞ is also known as the supremum norm). They are similar, i.e. have same shape. The L2 ball is shaped like a circle.
Hypothesis: For all p in the interval (1,2), there is q>2 such that the q-ball and the p-ball are similar. One further hypothesis is that this occurs precisely when p,q are Hölder conjugates.
I wasn't sure how to tag this.

REPLY [10 votes]: $\ell_p^n$ is isometric to $\ell_r^n$ with $1\le p < r\le \infty$ iff $n=1$ or ($n=2$ and $p=1$ and $r=\infty$ and the scalars are real). While known before L. Dor's 1976 paper in the Israel J. Math. 24, 260-268, it can be deduced from Theorem 2.1 in that paper, which classifies when $\ell_s^n$ isometrically embeds into $L_t(0,1)$.<|endoftext|>
TITLE: Explicit convergence of Baker-Campbell-Hausdorff
QUESTION [6 upvotes]: Let g be a finite dimensional simple Lie algebra over C. The Baker-Campbell-Hausdorff series defines a (multivariable) analytic function from a neighborhood of 0 in g \times g \to g. What is the domain of (absolute) convergence of this function?

REPLY [5 votes]: You might try On the convergence and optimization of the Baker–Campbell–Hausdorff formula, via google.<|endoftext|>
TITLE: Is there a matrix whose permanent counts 3-colorings?
QUESTION [14 upvotes]: Actually, I suppose that the answer is technically "yes," since computing the permanent is #P-complete, but that's not very satisfying. So here's what I mean:
Kirchhoff's theorem says that if you take the Laplacian matrix of a graph, and chop off a row and a column, then the determinant of the resulting matrix is equal to the number of spanning trees of the graph. It would be nice to have some analogue of this for other points of the Tutte polynomial, but this is in general too much to ask: the determinant can be computed in polynomial time, but problems such as counting 3-colorings are #P-hard.
However, if we use the permanent instead of the determinant, we don't run into these complexity-theoretic issues, at least. So, given a graph G on n vertices, can we construct a matrix of size around nxn whose permanent is the number of 3-colorings of G?
(The secret underlying motivation here is a vague personal hope that we can extend the analogy between the Laplacian matrix and the Laplacian operator [no, the naming isn't a total coincidence] to analogies between other matrices and general elliptic operators, and then prove some sort of "index theorem," which could [even more speculatively, here] help us understand why graph isomorphism is hard, prove or construct a counterexample to the reconstruction conjecture, prove the Riemann hypothesis, and achieve world peace forever.)

REPLY [6 votes]: "The number of edge 3-colorings of a planar cubic graph as a permanent" by David E. Scheim gives a permanent formula for the number of edge-3-colorings of planar cubic graphs (i.e. the number of 3-coloring of graphs which are line graphs of cubic planar graphs.)
This is generalized (using some results of Ellingham and Goddyn) to the case of n-colorings of n-regular planar graphs in "Colorings and orientations of matrices and graphs" by Uwe Schauz. This paper interprets Ryser's permanent formula as a statement about colorings and gives a "matrix form" of a theorem of Alon and Tarsi.
This doesn't answer your question but I hope you find the above references interesting. On the other hand about the fact that the Laplacian matrix generalizes to the Laplace operator on graphs, I wanted to mention that, in turn, it generalizes to the Laplacian on vector bundles on graphs. I learned about this generalization in the talk that Kenyon gave this year at the JMM. This new approach generalizes Kirkhoff's theorem from spanning trees to cycle-rooted-spanning-forests.<|endoftext|>
TITLE: How many embeddings are there of super-Virasoro into n Fermions?
QUESTION [8 upvotes]: What is the space of N=1 super-Virasoro vertex superalgebras inside the c=n/2 free fermion vertex superalgebra?  [Said differently, how many Neveu-Schwartz vectors are there in n fermions?]  Answers in terms of VOAs, CFTs, or nets all welcome, and any interpretation of "N=1 super" that makes you happy is okay.

REPLY [13 votes]: I would be very surprised if this were not known, but I cannot find a suitable reference, so I did the calculation as the question seemed interesting to me.  I believe that this is now the full answer to the question.
Let $V$ be an $n$-dimensional real vector space with a symmetric inner product $\langle-,-\rangle$.  Formulae are easier to write if we choose a basis $(e_i)$ for $V$ and let
$$g_{ij} = \langle e_i,e_j\rangle.$$
Let $\psi_i$ denote the corresponding free fermion, with OPE
$$\psi_i(z) \psi_j(w) = \frac{g_{ij} 1}{z-w} + \mathrm{reg}$$
The standard Virasoro vector is then
$$T = 1/2 g^{ij} \partial \psi_i \psi_j,$$
where the product is the normal-ordered product, which in my conventions associates to the left, so that $ABC = (A(BC))$.  $T$ obeys the standard Virasoro OPE with $c = n/2$:
$$T(z) T(w) = \frac{\frac{n}4 1}{(z-w)^4}  + 2 \frac{T(w)}{(z-w)^2} + \frac{\partial T(w)}{z-w} + \mathrm{reg}$$
The fields $\psi_i(z)$ are primary with weight $\frac12$ relative to $T$.
Then you are asking about the existence of a field $G$ which is primary of weight $\frac32$ relative to $T$ and whose OPE is
$$G(z) G(w) = \frac{\frac{n}3 1}{(z-w)^3} + \frac{2 T(w)}{z-w}$$
The most general such $G$ takes the form
$$G = \frac16 A^{ijk} \psi_i \psi_j \psi_k + B^i \partial \psi_i.$$
Calculating the relevant OPEs one sees that:

for $G$ to be primary, $B^i=0$, and
for $2T$ to appear in the first-order pole of $GG$, $A$ must satisfy two conditions which I will now rephrase.

First of all, $A$ defines an alternating bilinear map $[-,-]: V \times V \to V$ by
$$[e_i,e_j] = A_{ijk} g^{kl} e_l.$$
The existence of $G$ translates into the following conditions on this map:

$\langle[x,y],z\rangle$  is totally skew
$\langle[x,y],[z,w]\rangle$ is an algebraic curvature tensor
and
$$\mathrm{Tr}~ \mathrm{ad}_x \mathrm{ad}_y = 2 \langle x,y\rangle$$
where $\mathrm{ad}_x$ is the skewsymmetric endomorphism defined by $\mathrm{ad}_x(y) = [x,y].$

The algebraic curvature tensor condition means that the fourth rank tensor obeys the algebraic Bianchi identity:
$$\langle[x,y],[z,w]\rangle + \mathrm{cyclic~in}~(x,y,z) = 0,$$
which using the invariance of the inner product under $\mathrm{ad}_x$ becomes
$$\langle[[x,y],z],w\rangle + \mathrm{cyclic} = 0,$$
which in turn is equivalent to the Jacobi identity for the bracket.  (Thanks for Paul de Medeiros for the observation.)  Finally the last condition says that the Killing form, being twice the inner product, is nondegenerate, whence the Lie algebra is semisimple.
In summary, the solutions are in one-to-one correspondence with real semisimple Lie algebras.  If you further require the inner product to be positive-definite, then these are the compact semisimple Lie algebras.
Interestingly, in this case one can embed an affine Lie algebra in such a way that the Virasoro vector coincides with the Sugawara construction.  As I said above, I'm sure that this is standard, but cannot locate the reference right now.
EDIT This result is indeed known and can be found in: Goddard and Olive's Kac-Moody algebras, conformal symmetry and critical exponents, Nuclear Physics B, Volume 257, 1985, Pages 226-252, in the very last section of the paper.  (Thanks to Matthias Gaberdiel for the suggestion to look there.)
EDIT: I just noticed that the question asked "how many", whereas my answer just showed that as many as "semisimple Lie algebras".  If, for the sake of simplicity, we take the inner product to be positive-definite, then it is easy to write down a partition function and evaluating this gives a numerical answer to the question, depending on n.  The first 100 values, starting at n=1, are the following:
0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 3, 5, 3, 4,
8, 4, 5, 8, 7, 8, 11, 10, 11, 12, 13, 15, 19, 16, 21, 24, 21, 24, 32,
27, 34, 43, 37, 39, 53, 47, 54, 65, 65, 68, 79, 80, 90, 98, 102, 114,
129, 122, 138, 160, 157, 172, 207, 193, 211, 247, 244, 262, 306, 305,
329, 363, 378, 399, 448, 460, 505, 548, 554, 601, 675, 669, 739, 822,
826, 877, 990, 999, 1068, 1184, 1227, 1288, 1419, 1458, 1554, 1693, 1765
(A previous version of this edit had only taken into account the A-series... apologies.)<|endoftext|>
TITLE: Gromov-Witten theory and compactifications of the moduli of curves
QUESTION [13 upvotes]: Why, from a string theory perspective, is it natural to consider the Deligne-Mumford (resp. Kontsevich) compactification of the moduli of curves (resp. maps [from curves to a target space X]) rather than some other compactification? 
In any case, what other compactifications of the moduli of curves have been studied? Similarly, what other compactifications of the moduli of maps have been studied? Do any of these other compactifications lead to an interesting "Gromov-Witten theory"?

REPLY [4 votes]: I would prefer to leave this as a comment to Tony Pantev's answer, but I don't have enough reputation to do so.  Anyway, as Tony mentions the Laumon compactification provides a semismall resolution of singularities of the Drinfeld compactification.  However, to the best of my knowledge, the Laumon compactification is special to the case where G = SL_n, whereas the Drinfeld compactification can be defined for any reductive algebraic group.  In Kuznetsov's paper, he says in the introduction that he would like to study the resolutions of the Drinfeld compactification for groups besides SL_n in a future paper.  As far as I know, no such paper was ever written.  Does anyone know if Kuznetsov ever wrote such a paper and, if not, if anyone else has ever worked on this?
One partial solution to this question seems like it might be buried in the paper by Braverman, Finkelberg, Gaitsgory, and Mirkovic where they compute the IC sheaves of the Drinfeld compactification.  This is related because, assuming a semismall resolution of singularities existed, its cohomology would compute the intersection cohomology of the Drinfeld compactification.  However, given their use of finite-dimensional Zastava spaces to model the singularities of the Drinfeld compactification (and hence to compute the IC sheaf), it does not seem to me that a semismall resolution can be found in their paper (although maybe this means that if a semismall resolution existed it should also provide a resolution of each of the Zastava spaces?).
If appropriate, I would be happy to start this as its own topic.<|endoftext|>
TITLE: Automatically updating PDF reader for Windows
QUESTION [13 upvotes]: So, when doing LaTeX, it is absolutely necessary for ones sanity to using a preview program which updates automatically every time you compile.  Of course, any previewer designed for DVIs will do this, but as far I can tell, Adobe Acrobat not only does not automatically update, but will not let you change the PDF with it open.  
On a Mac one can get around this by using Skim, and on *nix by using xpdf, but what should one do on Windows?

REPLY [2 votes]: gsview can display pdf files, automatically updates them when the file changes, and does not lock the file so you can modify your files and compile them without closing the pdf file.<|endoftext|>
TITLE: analog of principle of inclusion-exclusion
QUESTION [16 upvotes]: When I teach elementary probability to my finite math students, a common error is to mix up the concepts of disjointness and independence.  At some point I thought that it might be helpful to some students to draw the analogy between the two concepts implied by the following pair of statements:

To compute the probability of the union of disjoint events, you add the probabilities of the events.
To compute the probability of the intersection of independent events, you multiply the probabilities of the events.

I also teach them is that when events are not disjoint, you can still compute the probability of their union by applying the principle of inclusion-exclusion.  Hence the question: Is there a useful analog of the principle of inclusion-exclusion for computing the probability of the intersection of non-independent events?
Edit:
I am incorporating the following clarification that I made in a comment responding to the answer of Anna Varvak:

In inclusion-exclusion, one alternately adds and subtracts intersections. Intersections measure the degree to which disjointness fails. Can we write the right-hand side of Bayes Theorem as alternate multiplications and divisions of something, where "something" measures the degree to which independence fails?

REPLY [3 votes]: In belief propagation there is a notion of inclusion-exclusion for computing the join probability distributions of a set of variables, from a set of factors or marginals over subsets of those variables.  For example, suppose {X,Y,Z} is your set of variables, and you know the marginal probabilities for pX,Y(x,y) and pY,Z(y,z).  If these two are compatible, then the marginal pY can be computed in either of the ways
pY(y) = integral pX,Y(x,y) dx = integral pY,Z(y,z) dz
Then a maximum entropy guess at the full joint distrubution is given by inclusion-exclusion over subsets of variables
pX,Y,Z(x,y,z) = pX,Y(x,y) pY,Z(x,y) / pY(y)
You might take a look at "Belief propagation" on wikipedia, or the more techinical article "Constructing Free Energy Approximations and Generalized Belief Propagation Algorithms" by Yedidia, Freeman and Weiss, which uses inclusion-exclusion in the form of 'counting numbers'.<|endoftext|>
TITLE: Representablity of Cohomology Ring
QUESTION [9 upvotes]: I know that the individual cohomology groups are representable in the homotopy category of spaces by the Eilenberg-MacLane spaces. Is it also true that the entire cohomology ring is representable? If so, is there a geometric interpretation of the cup product as an operation on the representing space?

REPLY [8 votes]: The total cohomology of spaces should be thought of as a graded ring, or even more precisely as a graded E* algebra where E* is the cohomology of a point.  It is representable in the sense that there is a graded E* algebra object in hTop representing it.
Let's unpack that a little.
First, you have to understand about group objects and so forth.  The full story is in Lawvere theories, but the basic idea is that a, say, group object in a category 𝒞 is an object, say C, of 𝒞 together with all the structure needed to ensure that the contravariant hom-functor 𝒞(-,C) lifts from a functor to Set to a functor to Group.  Providing 𝒞 has enough products of C, it's simple to write down a few morphisms that C must have, together with some diagrams that must commute.  Lawvere theories are the correct way of thinking about these things in general, but in a specific instance it can be instructive to just write everything down.
Now, that works fine for groups, abelian groups, modules, rings, and so forth, but cohomology isn't any of those things.  Cohomology is a graded ring.  So we need graded objects in our category.  To talk about graded objects we need a grading set, say Z.  This doesn't have to have any structure whatsoever.  A Z-graded object in 𝒞 is just a functor Z → 𝒞 where Z is viewed as a discrete category.  So it's a family of objects in 𝒞, indexed by the elements (objects) of Z.  A Z-graded object in 𝒞 represents a functor from 𝒞 into SetZ, the category of Z-graded sets.  While we can send a Z-graded set to a set either by its coproduct or its product, we shouldn't do so.  We should keep the labelling.
That's because we now want to mix these two things and talk of graded rings, or more generally graded theories, also called many-sorted theories.  In a single-sorted theory, such as groups or rings, the standard set-based (or more generally 𝒞 based) groups or rings consist of a set (object) together with certain functions (morphisms), called operations, from certain n-fold products of that set (object) to itself.  In a graded theory, the operations go from products of components of the graded set (object) to other components.  Thus in a graded abelian group, say A, we have operations A(z) × A(z) → A(z) but not A(z) × A(z') → A(z'').  That is, we can only add terms in the same component.
So back to cohomology.  Cohomology is a representable functor into the category of graded E* algebras, where the grading set is ℤ.  So for each integer n we have a space En in hTop and for each operation of a graded E* algebra we have an operation En × Em → El.  In particular, multiplication corresponds to a map (technically, a homotopy class of maps as we're in hTop) En × Em → En+m.  For ordinary cohomology, these spaces are the Eilenberg-Maclane spaces.
Several remarks are in order here.
Sometimes, the operations in a graded object come in families.  Then it can be severely tempting to put the graded pieces together into one single thing, either by taking the coproduct or the product.  However, this destroys information.  Just because the primary operations come in families doesn't mean that all of the operations come in families.  For example, the primary structure of a graded E* algebra come in suitable families and it can be tempting to put the pieces of the cohomology together.  (One has to choose what method to use: product or coproduct.  Product generally works better because in a product one doesn't have to worry about things remaining finite, which is good, because you have no control over this.)  But cohomology theories don't just have their primary structure operations, they also have a whole raft of other operations.  The main division of these two is into stable and unstable operations.  By putting your theory together into a single object, you are effectively saying that you will only consider stable operations and will ignore unstable ones.  While this is basically okay for cohomology, for K-theory it is a disaster because we don't know what the stable ones are!.
Secondly, to correct something Chris said, the spaces En do keep track of the suspension isomorphisms.  The fact that there are suspension isomorphisms in cohomology theories is encoded in the fact that that there is an equivalence ΩEn ≃ En-1.  Also, the Mayer-Vietoris maps are part of the fact that cohomology is representable (you get MV from the long exact sequence for pairs, and that's needed for representability).
Of course, knowing that ΩEn ≃ En-1 allows you to construct a spectrum from the spaces En, even an Ω-spectrum, and those operations that come in families, namely the stable ones, become morphisms of spectra.  Once you have that, you can define the associated homology theory and lots of other wonderful things.  But the point is that there's a lot that you can do before going to spectra and, up to a point, the language of spectra is merely a way of keeping track of the grading.
Finally, this probably isn't geometric enough for the cup product in ordinary cohomology, but there you're asking a difficult question.  To give a truly geometric interpretation, you would first need to find good geometric models of all the K(π,n)s.  While this can be done for a few, I don't know of a good family for all of them ("good" in the sense of "easy to think about" rather than anything technical).  One can start out fairly well, say with π = ℤ, with ℤ, S1, ℂℙ∞, but then it gets a bit sticky.  Another model for K(ℤ,2) is the projective unitary group on a Hilbert space and this acts on the general linear group for the space of Hilbert-Schmidt operators on that same Hilbert space so you could make K(ℤ,3) as the quotient of GL(H⊗H)/ℙ(H) and this is essentially the start of gerbe theory, but, as I'm sure you'll agree, it's hard to look at that and say "Aha! Now I understand K(ℤ,3)." in quite the same way as one can look at ℂℙ∞ and understand the connection between H2(X;ℤ) and line bundles.<|endoftext|>
TITLE: What are examples of good toy models in mathematics?
QUESTION [46 upvotes]: This post is community wiki.
A comment on another question reminded me of this old post of Terence Tao's about toy models.  I really like the idea of using toy models of a difficult object to understand it better, but I don't know of too many examples since I don't see too many people talk about them.  What examples are common in your field, or what examples do you personally think are very revealing?  Here's what I've got so far, starting with Terence Tao's example.  Feel free to modify any of these examples if I'm not stating them correctly and to elaborate on them in answers if you want.

$F_p[t]$ is a toy model for $\mathbb Z$.
$F_p[[t]]$ is a toy model for $\mathbb Z_p$.
Simplicial complexes are a toy model for topological spaces.
$\mathbb Z/n\mathbb Z$ is a toy model for $\mathbb Z$ (for the purposes of additive number theory).
The DFT is a toy model for the Fourier transform on the circle.

Which properties of the original objects carry over to your toy model, and which don't?  As usual, stick to one example per post.

REPLY [5 votes]: The Noncommutative torus serves as a toy model for noncommutative differentiable manifolds.<|endoftext|>
TITLE: Functorial characterization of open subschemes?
QUESTION [19 upvotes]: Given a morphism of schemes f: U → X, can one determine when f is an isomorphism of U onto an open subscheme of X in terms of some induced functors between the categories of quasicoherent modules Qcoh(U) and Qcoh(X)?

To begin, it might be helpul to simply assume that U ⊆ X is an open subscheme, and consider some properties of the resulting functors.  I can think of one interesting functor: there is an exact functor Qcoh(X) → Qcoh(U) given by restriction of sheaves.  I assume this is a special case of some more general construction (direct or inverse image functor?) that probably has an adjoint on some side.
Can we continue to list enough functors and properties of these functors to the point where we have determined precisely when the above map f is an isomorphism onto an open subscheme?

REPLY [21 votes]: The abelian category of quasicoherent sheaves on a schemes determine the scheme. This is an old result of Gabriel ("des categories abeliennes" 1962), proved in full generality by Rosenberg. This means that, $\operatorname{QCoh}(X)$ does not only tell you the open subschemes of $X$ but also gives you the structure sheaf! I've known this result for some time but I had never looked at it in detail until today. I'll sketch what I have just learned hoping not to make big mistakes...
An abelian subcategory $B$ of an abelian category $A$ is said to be a thick subcategory if it is full and for any exact sequence in $A$
$$0\to M'\to M \to M''\to 0,$$
$M$ belongs to $B$ if and only if $M'$ and $M''$ do.
If $B$ is a thick subcategory of $A$ there is a well defined localization $A/B$, which is again an abelian category. $A/B$ has the same objects as $A$ and a morphism $f:M\to N$ in $A/B$ is an isomorphism if and only if $\ker f$ and $\operatorname{coker} f$ belong to $B$.
Let $T\colon A\to A/B$ be the localization functor. Then $B$ is said to be a localizing subcategory if $B$ is thick and $T$ has a right adjoint. The condition of being localizing can be rephrased only in terms of $A$ and $B$. see Gabriel's thesis above (proposition 4 in chapter III).
Finally, if $M$ is an object of $A$, we denote by $\langle M\rangle$; the smallest localizing subcategory containing $M$. 
Now let $X$ be a scheme, $j\colon U \to X$ an open embedding and $i\colon Y\to X$ its closed complement. Then there are a bunch of adjunctions between the categories of quasicoherent sheaves of $U,X,Y$: $i_*\colon \operatorname{QCoh}(Y)\to \operatorname{QCoh}(X)$ has a left adjoint $i^*\colon \operatorname{QCoh}(X)\to \operatorname{QCoh}(Y)$ and a right adjoint $i_!\colon \operatorname{QCoh}(X) \to \operatorname{QCoh}(Y)$. On the other hand, the functor $j^*\colon \operatorname{QCoh}(X)\to \operatorname{QCoh}(U)$ has a left adjoint $j_!\colon \operatorname{QCoh}(U)\to \operatorname{QCoh}(X)$ and a right adjoint $j_*\colon \operatorname{QCoh}(U)\to \operatorname{QCoh}(X)$. This is sometimes called a recollement.
Let's assume that $X$ is Noetherian and let $A = \operatorname{QCoh}(X)$. We have an exact sequence of abelian categories
$$0 \to \operatorname{QCoh}(Y) \to A \to \operatorname{QCoh}(U) \to 0$$
in the sense that the category $\operatorname{QCoh}(Y)$ happens to be a localizing subcategory of $A$ and its quotient is identified with $\operatorname{QCoh}(U)$. The first map in the exact sequence is $i_*$ and the second $j^*$. Moreover, I think that $\operatorname{QCoh}(Y)$ is the smallest localizing subcategory of $\operatorname{QCoh}(X)$ containing $i_*O_Y$. Gabriel proves that there are no more such localizing subcategories, that is closed subschemes of $X$ correspond exactly to localizing subcategories $\langle M\rangle$ generated by a single coherent sheaf (i.e. Noetherian object in $A$). Moreover, irreducible closed subsets (the points in the underlying topological space of $X$) are given by localizing subcategories $\langle I\rangle$ for $I$ an indecomposable injective. We have described the points of $X$ and its closed sets in terms of only the category $A$, so we can recover the underlying topological space of $X$ from $A$.
In particular, an open subscheme $U$ of $X$ gives a complementary closed subscheme $Y$, which is in correspondence with a localizing subcategory $\langle M\rangle$ and, moreover, $\operatorname{QCoh}(U) = A/\langle M\rangle$. So, responding to the queston above, for any $f\colon U\to X$, $U$ is an open subscheme if and only if the kernel of $f^*\colon \operatorname{QCoh}(X) \to \operatorname{QCoh}(U)$ is a localizing subcategory of the form $\langle M\rangle$ for a coherent sheaf $M$.
Regarding the structure sheaf $O_X$ there is an isomorphism between $O_X(U)$ and the ring of endomorphisms of the identity functor on $\operatorname{QCoh}(U)$ (which happens to be $A/\operatorname{QCoh}(Y)$), so the structure sheaf can be recovered only in terms of the category $A$.
Finally, just say that there are other results in the spirit of reconstructing a scheme from some category of sheaves on it. This is the starting point for using such categories of sheaves as a definition of noncommutative scheme. There is more information on this entry in nlab.<|endoftext|>
TITLE: R2 and S3 for rings.
QUESTION [15 upvotes]: For a noetherian ring R, Serre's criterion for normality states that R is normal if and only if R satisfies conditions R1 and S2, where R1 is regularity in codimension one, and S2 is Serre's condition that every prime P of codimension at least 2 satisfies depth R_P \geq 2.  Likewise, there's a similar condition for whether or not R is reduced:  R is reduced iff R satisfies R0 and S1.  Following the pattern, it seems like Rn and S(n+1) should be equivalent to some desirable property for rings.  
Now, there's a nice and canonical way to take any ring and create a reduced (or normal) ring out of it, and this is something we should not expect to extend to higher dimensions, since we know that resolving singularities is (a) hard and (b) not very canonical.  That aside, surely we ought to be able to say something nontrivial about the next case, ie, R2 and S3?

REPLY [4 votes]: If you weaken birational to just proper and generically finite ("alterations"), then you can do this thanks to de Jong's theorems. In fact, he allows you to construct generically etale alterations. Also, if you just try to make things Cohen-Macaulay (S_k for all k) while forgetting about regularity, then there is a theorem of Kawasaki that says essentially everything admits a proper birational map from something Cohen-Macaulay. It seems hard to do better: a 3-dimensional normal non-CM singularity in characteristic 0 does not admit a finite map from something CM (the local cohomology of singularity is a summand of that of the cover for any normal ring in char 0).<|endoftext|>
TITLE: Regular languages and the pumping lemma
QUESTION [11 upvotes]: Let's say that I want to prove that a language is not regular.
The only general technique I know for doing this is the so-called "pumping lemma", which says that if $L$ is a regular language, then there exists some $n>0$ with the following property.  If $w$ is a word in $L$ of length at least $n$, then we can write $w=xyz$ (here $x$, $y$, and $z$ are subwords) such that $y$ is nontrivial and $xy^{k}z$ is an element of $L$ for all $k>0$.
This lemma basically reflects the trivial fact that in any directed graph, there is some $n$ such that any path of length at least n contains a loop.
Question: are there any other general techniques for proving that a language is not regular?

REPLY [5 votes]: The following lemma, which I saw in some old book, can handle most textbook examples of pumping lemma usage and is usually much easier to apply when it does apply.  Let $L$ be a language, and suppose there are strings $\alpha_1,\alpha_2,\ldots $ and $\beta_1,\beta_2,\ldots$ with the property that $\alpha_i\beta_j\in L$ iff $i=j$.  Then $L$ is not regular (easy proof using a DFA).
Note that the $\alpha$s and $\beta$s are just strings, they don't need to be in $L$ and usually aren't.
Examples:

$\alpha_k:=a^k$ and $\beta_k:=b^k$ shows $\lbrace a^kb^k : k\ge 0\rbrace$ is not regular. 
$\alpha_k:=a^k$ and $\beta_k:=ba^k$ shows the set of palindromes is not regular.
$\alpha_k:=ba^k$ and $\beta_k:=ba^k$ shows $\lbrace ww : w\in\Sigma^*\rbrace$ is not regular.<|endoftext|>
TITLE: Is there a "universal group object"? (answered: yes!)
QUESTION [21 upvotes]: I want to say that a group object in a category (e.g. a discrete group, topological group, algebraic group...) is the image under a product-preserving functor of the "group object diagram", $D$.  One problem with this idea is that this diagram $D$ as a category on its own doesn't have enough structure to make the object labelled $``G\times G"$ really the product of $G$ with itself in $D$.

Is there a category $U$ with a group object $G$ in it such that every group object in every other category $C$ is the image of $G$ under a product-preserving functor $F:U\rightarrow C$, unique up to natural isomorphism? 

(It's okay with me if "product-preserving" or "up to natural isomorphism" are replaced by some other appropriate qualifiers, like "limit preserving"...)

REPLY [2 votes]: I would really suggest looking at Steve Awodey's lecture notes on Categorical Logic., found here http://www.andrew.cmu.edu/user/awodey/catlog/notes/  The category you are looking for is called "the theory of groups".  I find these notes much more digestible than Lawvere's original papers on the subject. 
Essentially, you can form a category in which all objects are products of a single object G, and the only morphisms between them are those morphisms you get out of the basic definition of a group object.  Then a group in any category is a product preserving functor from this category.  Actually, in this case, it is easy to see that the appropriate category is just the opposite of the full subcategory 1, F(1), F(2), ... , F(n), ... where F(k) is the free group on k generators.<|endoftext|>
TITLE: Global fields: What exactly is the analogy between number fields and function fields?
QUESTION [28 upvotes]: Note: This comes up as a byproduct of Qiaochu's question "What are examples of good toy models in mathematics?"
There seems to be a general philosophy that problems over function fields are easier to deal with than those over number fields. Can someone actually elaborate on this analogy between number fields and function fields? I'm not sure where I can find information about this. Ring of integers being Dedekind domains, finite residue field, RH over function fields easier to deal with, anything else? Being quite ignorant about this analogy, I am actually not even convinced that why working over function fields "should" give insights about questions about number fields.

REPLY [2 votes]: Here
 is an interesting survey by Buium on the surprising analogies conc. differential equations, here a very interesting article by Esnault on char 0/char p analogies and how to make use of them.<|endoftext|>
TITLE: Isomorphisms of Banach Spaces 
QUESTION [21 upvotes]: Suppose $X$ and $Y$ are Banach spaces whose dual spaces are isometrically isomorphic.  It is certainly true that $X$ and $Y$ need not be isometrically isomorphic, but must it be true that there is a continuous (not necessarily isometric) isomorphism of $X$ onto $Y$?

REPLY [6 votes]: One may also remark that $c_{0}$ $\times$ (ℓ∞$/c_{0}$) is a predual of (ℓ∞)$^*$.                      
However, $c_{0}$ $\times$ (ℓ∞$/c_{0}$) is not isomorphic to ℓ∞.<|endoftext|>
TITLE: Is there a natural measures on the space of measurable functions?
QUESTION [34 upvotes]: Given a set Ω and a σ-algebra F of subsets, is there some natural way to assign something like a "uniform" measure on the space of all measurable functions on this space?  (I suppose first it would be useful to know if there's even a natural σ-algebra to use on this space.)
The reason I'm asking is because I'd like to do the following.  Let Ω be the (2-dimensional) surface of a sphere, with the uniform probability distribution.  Let F be the Borel σ-algebra, and let G be the sub-algebra consisting of all measurable sets composed of lines of longitude.  (That is, S is in G iff S is measurable and for all x in S, S contains all points with the same longitude as x.)  Let A be the set of all points with latitude 60 degrees north or higher (a disc around the north pole).
Let f be a G-measurable function defined on Ω such that the integral of f over any G-measurable set B equals the measure of (A\cap B).  (This is a standard tool in defining the conditional probability of A given G-measurable sets.)  It's not hard to show that for any such function f, for almost-all x, f(x) will equal the unconditional measure of A.
What I'd like to be able to say is that for any x, for almost-all such functions f, f(x) will equal the unconditional measure of A.  However, I can't say "almost-all" on the functions unless I have some measure on the space of functions.
Clearly I can do this by concentrating all the measure on the single constant function in this set.  But I'd like to be able to pick out this most "generic" such function even in cases where A isn't so nice and symmetric.
Maybe there's some other, simpler question I should be asking first?

REPLY [3 votes]: Kirk Sturtz proves that the category $Meas$ is monoidal closed in the paper Categorical probability theory.<|endoftext|>
TITLE: Morphisms of (quasi-)projective varieties
QUESTION [14 upvotes]: This is another "homework help" question, which is still hopefully of at least pedagogical interest to working mathematicians.
So, I'm currently taking an intro algebraic geometry class, and one thing I've had some trouble with is grokking what a morphism of projective or quasi-projective varieties should be. I know at least one definition by heart, which is Hartshorne's, which is the one about locally pulling back regular functions to regular functions. 
The problem is, I don't have the Grothendieckian superpower of being able to grasp these abstract ideas without playing around with some concrete examples. And Hartshorne's definition isn't all that conducive to actually checking in practice whether a given map is actually a morphism. So my question has, I guess, three parts:

Is there a more concrete definition of a morphism of projective/quasi-projective varieties that I can use in practice to check if something's a morphism?
What are some of the motivating examples of morphisms between varieties, that give one a sense of what they should be? What's an example of something that isn't a morphism, that gives one a sense of what's too much to ask?
Most abstractly, is there a big-picture explanation that makes the definition of morphism "intuitively obvious," as is the case (for instance) for groups, or even for affine varieties?

REPLY [2 votes]: It is impossible to give a concrete description of a map between two abstract varieties without first giving some concrete way of representing those varieties.  That is why Hartshorne’s definition of morphism is in terms of abstract properties. If you tell how the varieties are represented, then it is possible to describe a morphism in those same terms.  To check Hartshorne's definition, you must have a way of representing both the morphism and the regular rational functions.  Thus concrete descriptions of maps of projective varieties must use projective models.  This answer is hence in the spirit of those of David and Charles.
Examples:

The first step in visualizing an algebraic morphism of a projective variety is visualizing an algebraically embedded set, e.g. a plane curve C of degree d.  E.g. imagine a cubic in the plane, or any curve of degree d.  Then an example of a morphism defined on that curve is projection π of the curve C from a point p to a line L.

For each point x of the curve C, π(x) is the intersection of the line through p and x with L.  The fiber π^-1(y) of π over a point y of L is the intersection of the line through p and y with the curve.  Thus fibers of π correspond to lines through p.  This suggests that the number of points on different fibers of π is a constant equal to d, and that the fibers vary in a “linear family”.   Lines through p tangent to C meet C in fewer than d points, so π represents C as a “branched cover” of degree d over L, which is a covering space except over images of those tangential intersections.  
Algebraically we are taking a polynomial in x and y, fixing y, and representing the fiber over y as the solutions of the resulting polynomial in x.  That polynomial generically has degree d in x, and branching occurs when it has multiple roots.  Rational functions of y which are regular at y0 pull back to (the same) rational functions on C, regular at points (x,y0) over y0, as in Hartshorne’s definition of morphism.

To define a map f from the plane curve C into 3 space P^3, one can choose a linear family spanned by 4 auxiliary plane curves more general than lines, such as any 4 independent conics passing through two specific points p,q of the plane not on C.  This will map the entire plane, except the two points p,q where the map f is not defined, to a surface Q in P^3, and maps C onto this surface.  Since the linear coordinate functions on P^3 pull back under f to the quadratic polynomials defining the family of plane conics, a plane in P^3 meets f(C) in a set of points which pull back to the 6 common points of the cubic C with one of these conics.  Thus f(C) is a sextic curve in P^3.   Rational functions a/b on P^3 which are regular at f(x) pull back via substitution to rational functions a(f)/b(f) on C which are regular at x.  Indeed all morphisms of plane curves can be given by rational functions.
Given a surface S of degree d, and a line L in P^3, one can define an algebraic projection map π from S to L by choosing an auxiliary line M, and considering all planes passing through M.  Given a point x of S, π(x) = the point where L meets the plane spanned by x and M.  Then for a point y of L, the fiber π^-1(y) is the intersection curve of S with the plane spanned by y and M.  As this family of planes revolves around M, they sweep out on S the pencil of curve fibers of π.  Since the line L also meets S in a finite set of d points, the map π is not defined at those d points.  Except for these base points, an algebraic map of S to a line again amounts to fibering the source S into a linear family of algebraic subsets of codimension one, i.e. into a linear family of “divisors”.  If L is the z axis, and S is defined by a polynomial G(x,y,z) = 0, the fiber π^-1(z) is the plane curve of solutions of G=0 in x,y obtained by fixing z.  Again rational functions of z regular at π(p), pull back to rational functions regular at p on S, away from the d points common to L and S.

Remark: It helps to recall that Hartshorne actually taught and wrote the chapters on curves and surfaces before the abstract chapters.  He indeed gives a very nice description of projective morphisms of curves on pages 309-314, and the later discussion of cubic surfaces is equally concrete.  Indeed every morphism from one projective variety to another is obtained as in these examples, first mapping into a high dimensional projective space by rational functions, then projecting down to a lower one.<|endoftext|>
TITLE: What does the property that path-connectedness implies arc-connectedness imply?
QUESTION [5 upvotes]: A space X is path-connected if any two points are the endpoints of a path, that is, the image of a map [0,1] \to X. A space is arc-connected if any two points are the endpoints of a path, that, the image of a map [0,1] \to X which is a homeomorphism on its image. If X is Hausdorff, then path-connected implies arc-connected.
I was wondering about the converse: What properties must X have if path-connected implies arc-connected? In particular, what are equivalent properties?

REPLY [3 votes]: It suffices that $X$ be Hausdorff: the path is then a compact metric image of [0,1] and as such arc-wise connected (do Problem 6.3.11 of Engelking's General Topology).<|endoftext|>
TITLE: What's the "best" proof of quadratic reciprocity?
QUESTION [106 upvotes]: For my purposes, you may want to interpret "best" as "clearest and easiest to understand for undergrads in a first number theory course," but don't feel too constrained.

REPLY [3 votes]: A nice proof was given by Dirichlet (see Dirichlet P.G.L., Lectures on number theory). Poisson summation proves that (see also Davenport, Multiplicative number theory)
$$S(q)=\sum_{k=1}^qe(k^2/q)=\dfrac{1+i^{-q}}{1+i^{-1}}\cdot\sqrt{q}.$$
Together with multiplicative property
$$S(pq)=\left(\dfrac{p}{q}\right)\left(\dfrac{q}{p}\right)S(p)S(q)$$
it proves the law:
$$\left(\dfrac{p}{q}\right)\left(\dfrac{q}{p}\right)=\frac{1+i^{-pq}}{1+i^{-p}} \cdot\frac{1+i^{-1}}{1+i^{-q}}=(-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}.$$<|endoftext|>
TITLE: Existence of proper regular models for varieties over Q and other global fields
QUESTION [5 upvotes]: What is known about regular proper models for smooth projective varieties over Q? Results for other global fields would also be interesting, as well as general comments and suggested references for integral models.
This is a followup to this question on smooth models, and here is part of what I wrote as an answer to the previous question: Nekovar's survey article on the Beilinson conjectures from the early 90s mentions some results for varieties over Q. He says in section 5.3 that given a smooth projective variety over Q, there always exists a proper flat model over Z, but that a regular such model is rarely known to exist. However, in the published version of the same survey, there is an added note at the very end of the article saying that "Spivakovsky recently announced a general result on resolution of singularities, which implies that a regular proper flat model of X mentioned in 5.3 always exist". However, I have never seen this result of Spivakovsky mentioned anywhere else, so I doubt that it is true. Does anyone else know more about this?
The survey is available here. For the published version, google "Serre Jannsen Motives", click at the Google Books link, and then search for "Spivakovsky" within the book.

REPLY [3 votes]: Regular proper models over Z are only known to exist for curves (by a theorem of Abhyankar). Spivakovsky's claims were never substantiated; as far as I know, no preprint was ever circulated.
For many purposes, de Jong's theorem on alterations suffices: For a proper variety X over a number field K there exists a variety Y with a proper generically finite map to X such that Y has a regular proper model over the ring of integers of K. (The actual theorem is more general.)<|endoftext|>
TITLE: Examples and intuition for arithmetic schemes
QUESTION [13 upvotes]: How should a beginner learn about arithmetic schemes (interpret this as you wish, or as a regular scheme, proper and flat over Spec(Z))? What are the most important examples of such schemes? Good references? What kind of intuition do people have for such schemes?

REPLY [3 votes]: One example I always found useful was that if you consider an elliptic curve like (the projective model of) y^2=x^3+1, then this equation gives an elliptic curve not only over the complex numbers but over any field of characteristic not 2 or 3. So in fact it's giving an elliptic curve over Spec(Z[1/6]), which will look like a scheme with a map to Spec(Z[1/6]) such that the fibre over the point corresponding to the prime ideal p>3 is just the curve over Z/pZ, and the point corresponding to the fibre over the generic point (0) is just the curve over Q. I think that checking these statements formally gave me some sort of intuition as to what was going on at the time. Note that Spec(Z[1/6]) is just the open subscheme of Spec(Z) that you get by throwing away the two closed points (2) and (3).<|endoftext|>
TITLE: Why is homology not (co)representable?
QUESTION [28 upvotes]: This is in the same vein as my previous question on the representability of the cohomology ring. Why are the homology groups not corepresentable in the homotopy category of spaces?

REPLY [22 votes]: While it's true that there are lots of internal things that a corepresentable homology functor wouldn't support, I think it's also enlightening to see that you wouldn't get the nice sorts of dualities that homology and cohomology theories have.  After all, we've already agreed that cohomology theories ought to be somehow representable, so maybe we should start there.  Instead of using stable maps $X \to E_n$ to produce $n$-degree $E$-cohomology classes of $X$, you can think of these instead as elements in the stable homotopy groups $\pi_{-*}^S F(X, E)$, where $F(X, E)$ denotes the function spectrum of maps $X \to E$.  This presentation makes the right choice for defining $E$-homology somehow much more obvious: the functor $F(X, -)$ has an adjoint, called the smash product (this is the whole point of the smash product -- it plays the role of "tensor product" for spaces!), and so for homology we think about maps $S^n \to E \wedge X$ instead.  That homology and cohomology are not (usually) exact duals in a linear algebraic sense is somehow measuring the twist introduced by this adjunction.  This does actually turn out to be the right definition for homology; (extraordinary) homology theories in the traditional sense are in fact modeled by functors of the form $\pi_*^S (E \wedge -)$.
This construction has a number of attractive features -- for instance, it means that we can (under some flatness and ringy conditions) think about "homology cooperations" associated to a spectrum, and they look like $E_* E$, a pleasant mirror of cohomology operations living in $E^* E$.  We also always get a pairing $E^* X \times E_* X \to E_* E$ of cohomology and homology classes that lands in homology operations, by composing as $S^n \to E \wedge X \to E \wedge E$.  (This pairing even gets used occasionally, though I'd be hard-pressed to come up with an obvious citation.)
For the most familiar homology theory, singular homology with $\mathbb{Z}/p$-coefficients, this flatness business does hold, the operations and cooperations even turn out to be $\mathbb{Z}/p$-vector space duals, and the coaction and action line up in the way you'd expect from Milnor's work.
(This belongs as a comment on Lawson's answer, I think, but it looks like I'm too new here to make that happen.)<|endoftext|>
TITLE: Freyd-Mitchell for triangulated categories?
QUESTION [16 upvotes]: Is there a nice analog of the Freyd-Mitchell theorem for triangulated categories (potentially with some requirements)?  Freyd-Mitchell is the theorem which says that any small abelian category is a fully faithful, exact embedding into the module category of some ring.
Therefore, I'd like a theorem like this:
Any small triangulated category is a fully faithful, triangulated subcategory of the unbounded derived category of modules on some ring.
My guess is that this fails to be true, for similar reasons to a triangulated category not always being the derived category of its core.  Is there a simple example of this?  -and- Is there a set of properties that do imply the above theorem?

REPLY [8 votes]: Don't know if this is germane, but Freyd actually proved that every small triangulated category has an abelian envelope.  That is, you can embed a triangulated category into an abelian category, in fact a Frobenius category, where the triangulated category is the category of projectives=injectives.  I believe he introduced this construction in the same paper, in 1965 or so, in some La Jolla conference proceedings, in which he introduced the generating hypothesis in stable homotopy theory.  It also appears in Amnon Neeman's book on triangulated categories, I am pretty sure.  
Mark<|endoftext|>
TITLE: Euclidean volume of the unit ball of matrices under the matrix norm
QUESTION [24 upvotes]: The matrix norm for an $n$-by-$n$ matrix $A$ is defined as $$|A| := \max_{|x|=1} |Ax|$$ where the vector norm is the usual Euclidean one. This is also called the induced (matrix) norm, the operator norm, or the spectral norm. The unit ball of matrices under this norm can be considered as a subset of $\Bbb R^{n^2}$. What is the Euclidean volume of this set? 
I'd be interested in the answer even in just the $2$-by-$2$ case.

REPLY [15 votes]: Building on the nice answer of Guillaume: The integral
$$ \int_{[-1,1]^n} \prod_{i < j} \left| x_i^2 - x_j^2 \right| \, dx_1 \dots dx_n $$
has the closed-form evaluation
$$ 4^n \prod_{k \leq n} \binom{2k}{k}^{-1}.$$
This basically follows from the evaluation of the Selberg beta integral Sn(1/2,1,1/2).
Combined with modding out by a typo, we now arrive at the following product formula for the volume of the unit ball of nxn matrices in the matrix norm:
$$ n! \prod_{k\leq n} \frac{ \pi^k }{ ((k/2)! \binom{2k}{k})} .$$
In particular, we have:

2/3 π2 for n=2
8/45 π4 for n=3
4/1575 π8 for n=4<|endoftext|>
TITLE: Can we categorify the equation (1 - t)(1 + t + t^2 + ...) = 1?
QUESTION [19 upvotes]: Polynomials in $\mathbb Z[t]$ are categorified by considering Euler characteristics of complexes of finite-dimensional graded vector spaces.  Now, given a rational function that has a power series expansion with integer coefficients, it seems natural to consider complexes of (locally finite-dimensional) graded vector spaces.
Are there nice examples of this in nature?

REPLY [7 votes]: I don't know whether or not this satisfies the criterion of "categorification" (what that?), but the equation (1 - t)(1 + t + t2 + ...) = 1 and its relation to vector spaces is well-known to differential topologists and geometers.  We use it all the time in K-theory and index theory where it becomes the identity Λ-1V ⊗ S1V = ℂ.  Here, Λ-1V denotes the alternating sum of the exterior powers of V whilst S1V is the sum of the symmetric powers.
Of course, the sum of the symmetric powers isn't a class in K-theory as it is an infinite sum.  To get round this, we work in K[[t]] and allow parameters, whereupon the equation becomes Λ-tV ⊗ StV = ℂ.  Here, the t means formally multiply the kth exterior or symmetric power by tk.
Where this breaks out of mere formalism and becomes very powerful is in equivariant K-theory.  Then the parameter becomes a way of measuring the action of the group, which is usually S1 or a finite cyclic group for index theory calculations.  In particular, in Witten's original adaptation of index theory to loop spaces, we end up with positive energy representations of S1 which become vector bundles over the original (finite dimensional) manifold with circle actions preserving the fibres.  One can decompose these according to the circle action whereupon one has a vector bundle for each k ∈ ℤ.  The positive energy criterion means that these are trivial below a certain integer and are always finite dimensional.  However, as there are an infinite number of them then the total dimension can be infinite dimensional.  Then the identity Λ-tV ⊗ StV = ℂ has real meaning as the power of the t parameter indicates how the circle acts on that component of the vector bundle.  That is, if the circle action on V is the standard action then it is multiplication by tk on Λk V and on Sk V.<|endoftext|>
TITLE: Are curves with `fractional points' uniquely determined by their residual gerbes?
QUESTION [17 upvotes]: One makes precise the vague notion of "curve with a fractional point removed" (see for instance these slides) using stacks -- one should really consider Deligne-Mumford stacks whose coarse spaces are curves, and the "fractional points" correspond to the residual gerbes at the stacky points.
One example: let a,b,c > 1 be coprime integers and let S be the affine surface given by the equation  x^a + y^b + z^c = 0. Then there is a weighted Gm action on S (t sends (x,y,z) to (t^bc x, t^ac y, t^ab z) and one can check that the stack quotient [S-{0}/Gm] has coarse space P^1 and, that since the action is free away from xyz = 0 but has stabilizers at those 3 points, one gets a stacky curve with three non-trivial residual gerbes.

Question: Are two 1-dimensional DM stacks with isomorphic coarse spaces 
  and residual gerbes themselves isomorphic?

I have an idea for how to prove this when the coarse space is P^1, but in general don't know what I expect the answer to be. Also, one may have to restrict to the case when the coarse space is a smooth curve.
This might be analogous to the statement that the `angle' of a node of a rational nodal curve with one node doesn't affect the isomorphism class of the curve.
Also, the recent papers of Abromivich, Olsson, and Vistoli (on stacky GW theory) may be relevant.

REPLY [7 votes]: I love this question so much that I can't stop thinking about it. Since this is a separate line of thought, I'll post it as a separate answer. Here's a general way to distinguish between DM stack whose coarse spaces are isomorphic with all the same residual gerbes.
Suppose a DM stack X contains a BG. Then the tangent space to the stack at that point is a representation of G. To get this representation, pull the sheaf of differentials of X back to BG. This is a coherent sheaf on BG, which is the same thing as a representation of G. Alternatively, there is a structure theorem for DM stacks (see Theorem 38.20 of my notes from Martin Olsson's course) that says that if G is the stabilizer of a geometric point x∈X, then there is an etale cover U→X of a neighborhood of x with a G action so that the neighborhood  of x is the stack quotient [U/G]. I'm pretty sure that the action of G on the tangent space of the point over x is the same representation as in the other construction (or maybe it will be the dual representation).
Now the idea is that if two DM stacks have the same coarse space and the same residual gerbes, then you might still be able to tell them apart if they have different "residual tangent representations". In the three examples I gave in my other answer, the residual representations were (1) the sign representation of Z/2, (2) a two-dimensional representation (sign plus trivial) of Z/2, and (3) the trivial representation of Z/2. However, there is the annoying bit that you only recover G up to isomorphism, so if two representations differ by an automorphism of G, you can't conclude that the stacks are different.

Using this kind of thinking, we can construct two DM stacks which are both smooth, separated, and have any other kind of nice properties you can imagine, and have isomorphic coarse spaces with the same residual gerbes, but are non-isomorphic. Unfortunately, the coarse space is a surface.
Consider the action of Z/6 on A2 given by 1⋅(x,y)=(x,ζy), where ζ is a primative 6-th root of unity. This action is generated by pseudo-reflections, so by the Chevalley-Shephard-Todd theorem (the easy half of it), the coarse space of the quotient stack [A2/(Z/6)] is A2, with a residual B(Z/6) at the origin.
On the other hand, consider the action of Z/6 on A2 given by 1⋅(x,y)=(ζ2x,ζ3y). Again, the action is generated by pseudo-reflections, so the coarse space of the quotient stack is A2, with a residual B(Z/6) at the origin.
But the two residual representations of Z/6 are non-isomorphic, even if you twist by automorphisms of Z/6. The first representation has a vector invariant under the action (namely (1,0)), but the second does not.<|endoftext|>
TITLE: Cohomology and fundamental classes
QUESTION [70 upvotes]: Let X be a real orientable compact differentiable manifold. Is the (co)homology of X generated by the fundamental classes of oriented subvarieties? And if not, what is known about the subgroup generated?

REPLY [2 votes]: According to an article in The Geometry and Topology of Submanifolds and Currents, edited by Weiping Li and Shinshu Walter Wei
The lack of suitable representatives in smooth homology by submanifolds was 
'partial motivation' for Federer to introduce rectifiable homology by defining 
• the chain groups as:
$Z_{i}(A,B):=\{T\in R_{m}(\mathbb{R}^{n},\mathbb{Z}): \partial T\in R_{m-1}(\mathbb{R}^{n},\mathbb{Z}),spt(T)\subset A$ and $spt(\partial T)\subset B\}$
• the boundary groups as 
$B_{i}(A,B):=\{\partial T:T\in Z_{i+1}(A,A)\}$
And the homology groups as usual by $H_{i}(A,B):=Z_{i}(A,B)/B_{i}(A,B)$
(Here, $A$ & $B$ are both compact Lipschitz neghbourhood retracts (of $\mathbb{R}^{n}$) and we have $B\subset A$). 
He showed that it satisfied the Eilenberg-Steenrod axioms, and hence is isomorphic to singular homology; and that every homology class in this homology contains a mass-minimising rectifiable representative. Thus we should see Federers notion of rectifiable currents as an appropriate generalisation of submanifold which allows us to identify homology classes with special sub-manifolds.<|endoftext|>
TITLE: Software for rigorous optimization of real polynomials
QUESTION [9 upvotes]: I am looking for software that can find a global minimum of a polynomial function over a polyhedral domain (given by, say, some linear inequalities) in $\mathbb R^n$. The number of variables, $n$, is not more than a dozen. I know it can be done in theory (by Tarski's elimination of quantifiers in real closed fields), and I know that the time complexity is awful. However, if there is a decent implementation that can handle a dozen variables with a clean interface, it would be great. I have tried builtin implementations in Mathematica and Maple, and they do not appear terminate on 4-5 variable instances.
If the software can produce some kind of concise "certificate" of its answer, it would be even better, but I am not sure how such a certificate should look like even in theory.
Edit: Convergence to the optimum is nice, but what I am really looking for is ability to answer questions of the form "Is minimum equal to 5?" where 5 is what I believe on a priori grounds to be the answer to optimization to be (in particular, it is a rational number). That also explains why I want a certificate/proof of the inequality, or a counterexample if it is false.

REPLY [2 votes]: Certificates for this sort of questions are discussed in my paper
H. Schichl and A. Neumaier, Transposition theorems and qualification-free optimality conditions, Siam J. Optimization 17 (2006), 1035-1055.
They can be obtained numerically by semidefinite programming.<|endoftext|>
TITLE: When do PROP-morphisms induce adjunctions?
QUESTION [9 upvotes]: If (C,tensor,1) is a symmetric monoidal category and f:A-->B is a morphism of PROPs (or monoidal cats = colored PROPs), one gets a forgetful functor f^*:B-Alg(C)-->A-Alg(C) (where B-Alg(C)=tensor-preserving functors from B to C) defined by precomposing with f.
Does anyone conditions on A,B,C under which this functor has a left or a right adjoint?
(e.g. if C has the monoidal structure coming from products, it has a left adjoint, is there more to say?)

REPLY [3 votes]: Paul-André Melliès has quite an interesting paper on this topic:
http://hal.archives-ouvertes.fr/docs/00/33/93/31/PDF/free-models.pdf
...but phrased in the more general terms of T-algebras of a pseudomonad. The idea is that a pseudomonad on a 2-category (especially Cat), let you put algebraic structures on categories the same way monads let you put them on objects of a category, like sets. This is motivated by the need to put PROPs, PROBs, PROs, Lawvere theories, etc. all under one roof.
He begins by talking about how a T-algebra homomorphism (a monoidal functor in the case where the T-algebras are monoidal categories) j : A -> B induces a forgetful functor U_j from Models(B,C) to Models(A,C) in the way you mentioned. Looking for left adjoint to U_j amounts to looking for a way to push some functor backwards along j in a suitably natural way. As Tom already mentioned, this is the left Kan extension. This process is functorial, and usually written Lan_j : [A,C] -> [B,C]. Furthermore, Lan_j -| U_j.
But if we were done there, all PROPs would have free algebras, which we know is not true in general (cf. bialgebras). The hard part is proving the Lan_j is a T-algebraic left Kan-extension. In the case of Lawvere theories, this is easy, because the product structure guarantees all natural transformations of cartesian functors are cartesian, but in the monoidal case, this stuff all needs to be checked.
This is where the story starts to get more complicated. It seems quite tricky to come up with suitably weak conditions under which Lan_j is T-algebraic. Mellies phrases these in terms of distributers (aka profunctors, modules, depending on who you ask and what country you are in :-P). If functors are like functions, this are a bit like relations. The nice thing about them is they always come in adjoint pairs f_* and f^* for any functor f.
So, thm 1 in the paper is (roughly) this. If j and j^* are T-algebraic in the suitable 2-categories, C is (T-algebraically) complete and co-complete, and for any model f : A -> C, f_* o j^* factors through the up-star of the Yoneda embedding y : C -> Psh(C), then U_j has a left adjoint computed as Lan_j that is indeed the free functor.
This is quite heavy-duty (pro-arrow equipment, ends, etc.), but it seems to get the job done. It would be nice to see more concrete/specific examples of this.<|endoftext|>
TITLE: Does projectiveness descend along field extensions?
QUESTION [6 upvotes]: Background: Properness is a much more robust notion than projectiveness. For example, properness descends along arbitrary fpqc covers (see, for example, Vistoli's Notes on Grothendieck topologies, fibered categories and descent theory, Proposition 2.36). This is far from true for projectiveness. In fact, projectiveness doesn't even descend along Zariski covers! The standard example of a proper non-projective morphism is locally projective over the base. (pictured in an appendix of Hartshorne; page 443, unfortunately not on Google books)

So how delicate is projectivity? Suppose X is a scheme over a field k, and suppose K is a field extension of k such that XK is projective over K. Does it follow that X is projective over k?
The obvious thing to do (I think) is to pick a very ample line bundle L on XK and try to descend it to X. If K is a finite extension of k, then the norm of L will be an ample line bundle on X (if I haven't misunderstood EGA II, section 6.5 and Corollary 6.6.2). But could it be that there is an infinite extension K of k such that XK is projective over K, but X is not projective over k?

REPLY [10 votes]: The entirety of sections 8--12 (apart from 10) of EGA IV is devoted to fleshing out in remarkably exhaustive and elegant generality the entire yoga of "spreading out and specialization" of which this question is but a special case.  Highly recommended reading; it is used (implicitly, if not explicitly) all the time when people prove theorems in algebraic geometry by specialization.  This includes proving results over $\mathbf{C}$ by "reduction to the case of positive characteristic or finite fields" (e.g., Mori, Deligne-Illusie) as well as construction of moduli spaces of stable curves by digging out subschemes of Hilbert schemes, etc. 
Here is a nifty little exercise to test one's understanding of the EGA formalism:  if $X$ and $Y$ are schemes of finite type over a field $k$ and if there is an extension field $K/k$ such that there is a $K$-morphism $f:X _K \rightarrow Y _K$ with any "reasonable" property $\mathbf{P}$ then there is such a morphism with $K/k$ a finite extension; here, "reasonable" can be lots of things: isomorphism, surjective, open immersion, closed immersion, finite flat of degree 42, a semistable curve fibration, smooth, proper and flat with geometric fibers having 12 irreducible components which intersect according to such-and-such configuration and dimensions, and so on.  The point is that the initial $f$ is certainly not descended to a finite subextension of the initial 
$K/k$, and if you made the construction over such an extension and extended scalars back up to the original $K$ then it has absolutely nothing to do with the original $f$. 
On the topic of specialization for morphisms, I can't resist mentioning a useful fact which is not a formal consequence of that general stuff:  if $A$ and $B$ are abelian varieties over a field $k$ then there exists a finite (even separable) extension $k'/k$ such that (loosely speaking) "all homomorphisms from $A$ to $B$" are defined over $k'$.  This means that if $K/k'$ is any extension field whatsoever, then every $K$-homomorphism $A_K \rightarrow B_K$ is defined over $k'$.  (Quick proof: the locally finite type Hom-scheme has finitely generated group of geometric points, and is unramified by functorial criterion, so it is \'etale since we are over a field.)  There is nothing like this for general (even proper smooth) varieties; just think about automorphisms of projective space.<|endoftext|>
TITLE: What is the geometric meaning of integral closure?
QUESTION [32 upvotes]: More precisely, how does one characterize integrally closed finitely generated domains (say, over C) based on geometric properties of their varieties?  Given a finitely generated domain A and its integral closure A' (in its field of fractions), what's the geometric relationship between V(A) and V(A')? 
If you can, phrase your answer in terms of complex affine varieties.

REPLY [28 votes]: The property you are interested in is known as being normal.  For affine varieties, the definition of normal is just that the coordinate ring is integrally closed, and the operation on varieties that corresponds to taking the integral closure of the coordinate ring is known as normalization (a general variety is said to be normal if it is locally isomorphic to a normal affine variety.)  So far I've just restated your question; but there are a number of things known.  For this we need the notion of smoothness: for varieties over $\mathbb{C}$, this should be equivalent to being a smooth manifold (the general definition is a bit technical).

Any smooth variety is normal.
The set of singular points of a normal variety has codimension $\geq 2$.

Corollary:  For curves, normal $\iff$ smooth.
Shafarevich's Basic Algebraic Geometry vol. 1 is a good reference for this from the varieties point of view (and deals with smoothness more rigorously).
As regards the relationship between the varieties corresponding to $A$ and $A'$:  I mostly just have intuition for curves, so I'll stick to talking about them.  For curves, $A'$ is a version of $A$ with the singularities "resolved":  more specifically, $V(A')$ is a smooth variety equipped with a surjective morphism of varieties from $A' \to A$  which is an isomorphism away from the preimages of the singular points of $A$.  (This should be true in higher dimensions too I think: it's definitely true if one is talking about schemes, but I think it's also true that for affine varieties the map $A \to A'$ induces a map $V(A') \to V(A)$.) The two basic examples to keep in mind here are the cuspidal cubic $C_1: y^2 = x^3$ and the nodal cubic $C_2: y^2 = x^3 + x^2$.
In the case $C_1$: the coordinate ring $\mathbb{C}[x, y]/(y^2 - x^3)$ has integral closure isomorphic to $\mathbb{C}[t]$, and the map of varieties here is the map from the affine line to $C_1$ given by $t \mapsto (t^2, t^3)$.  In this case the map is a bijection as sets (but not an isomorphism of affine varieties! because the inverse map cannot be expressed as a polynomial map), and the "cusp" of $C_1$ that is visible at the point $(0,0)$ is no longer evident.
In the case $C_2$: the coordinate ring also has integral closure isomorphic to $\mathbb{C}[t]$: this time the map is a bit more complicated, but it's $t \mapsto (t^2 -1 , t(t^2-1))$.  How did I find that?  In this case, looking at the curve one sees that it has a self-intersection at the origin.  This means that there should be two distinct points in the normalization that have been sent to the same point in $C_2$.  Another way of stating that is that because the curve appears to have two tangent lines at the origin, there really should be two different points there, one on each tangent line.  How to tell them apart?  Well, as one approaches the origin from one direction, the ratio $y/x$ tends to $1$ in the limit, whereas if one approaches it from the other direction, the ratio $y/x$ tends to $-1$: so at one of our two points, $y/x=1$, and at the other one, $y/x = -1$.  Since $y/x$ is well-defined everywhere else on the curve, this suggests that we want $t = y/x$ to belong to our coordinate ring at the origin.  Indeed, $t^2 = x +1$, so $t$ is integral, and we can solve for $x$ and $y$ in terms of $t$ to get the original answer.  So in this case we have a surjective map from the affine line to a self-intersecting curve which is injective everywhere except at the preimage of the singular point at the origin.<|endoftext|>
TITLE: Universal covers of domains in complex projective space
QUESTION [8 upvotes]: The Uniformization Theorem states that the universal cover of a Riemann surface is biholomorphic to the extended complex plane, the complex plane or the open unit disk. Each of these three is a domain in the extended complex plane. In particular, then, the universal cover of a domain in the extended complex plane is biholomorphic to a domain in the extended complex plane. This leads to an analogous question in higher dimensions: Is the universal cover of a domain in complex projective space biholomorphic to a domain in complex projective space? More precisely, I am asking for a counterexample. Many results in one complex variable break in several complex variables, and the Uniformization Theorem is fairly delicate, so it seems reasonable to expect it to break. Perhaps there is a counterexample that one can see just by topology?

REPLY [10 votes]: Consider a tubular neighborhood of three generic lines on P^2. 
The fundamental group is Z. The universal covering will contain
an infinite chain of P^1's, and in particular two
disjoint   P^1's. Thus it cannot be a domain in P^2.<|endoftext|>
TITLE: Is there a Murnaghan-Nakayama Rule for GL(n,q)?
QUESTION [18 upvotes]: The Murnaghan-Nakayama rule for S_n is a combinatorial rule to compute the irreducible characters of the symmetric group. Is there a q-analogue of this rule for GL(n,q) to compute the irreducible characters? For example, exhibiting that the value of the unipotent characters of GL(n,q) on a unipotent class is given by the cocharge Kostka-Foulkes polynomials, and showing other special cases.

REPLY [7 votes]: I presume you're talking about characters over C. In which case Green's classic paper "The characters of the finite general linear groups" is the only thing I know on this subject. He defines a whole load of polynomials (that have since come to be known as Green polynomials), out of which he's able to extract a list of the irreducible characters.
Green's paper is the mathematical ancestor of the Deligne-Lusztig theory of irreducible characters.<|endoftext|>
TITLE: Unstable Vector Bundles
QUESTION [10 upvotes]: As a follow up to me other question, what can be said about unstable vector bundles? I know this is rather open ended, but what sorts of horrible things does having a subbundle of strictly greater slope imply?

REPLY [3 votes]: Another reason behind the notion of stability is the "jumping phenomenon."  Namely, you can construct a family of vector bundles parametrized by the disk where all the fibers apart from the origin are mutually isomorphic, but not isomorphic to the fiber at the origin.  Concretely, you can realize this by scaling an extension class.  In the Atiyah-Bott-Kempf-Ness picture, this corresponds to non-closed orbits of the complex gauge group. When the base manifold has dimension greater than one, the pieces in the Harder-Narasimahan filtration may not be subbundles, so the limiting object is in general a torsion-free sheaf, and not necessarily a bundle.<|endoftext|>
TITLE: Point singularity of a Riemannian manifold with bounded curvature
QUESTION [21 upvotes]: Suppose you have an incomplete Riemannian manifold with bounded sectional curvature such that its completion as a metric space is the manifold plus one additional point. Does the Riemannian manifold structure extend across the point singularity?
(Penny Smith and I wrote a paper on this many years ago, but we had to assume that no arbitrarily short closed geodesics existed in a neighborhood of the singularity. I was never able to figure out how to get rid of this assumption and still would like someone better at Riemannian geometry than me to explain how. Or show me a counterexample.)
EDIT: For simplicity, assume that the dimension of the manifold is greater than 2 and that in any neighborhood of the singularity, there exists a smaller punctured neighborhood of the singularity that is simply connected. In dimension 2, you have to replace this assumption by an appropriate holonomy condition. 
EDIT 2: Let's make the assumption above simpler and clearer. Assume dimension greater than 2 and that for any r > 0, there exists 0 < r' < r, such that the punctured geodesic ball B(p,r'){p} is simply connected, where p is the singular point. The precludes the possibility of an orbifold singularity.
ADDITIONAL COMMENT: My approach to this was to construct a differentiable family of geodesic rays emanating from the singularity. Once I have this, then it is straightforward using Jacobi fields to show that this family must be naturally isomorphic to the standard unit sphere. Then using what Jost and Karcher call "almost linear coordinates", it is easy to construct a C^1 co-ordinate chart on a neighborhood of the singularity. (Read the paper. Nothing in it is hard.)
But I was unable to build this family of geodesics without the "no small geodesic loop" assumption. To me this is an overly strong assumption that is essentially equivalent to assuming in advance that that differentiable family of geodesics exists. So I find our result to be totally unsatisfying. I don't see why this assumption should be necessary, and I still believe there should be an easy way to show this. Or there should be a counterexample.
I have to say, however, that I am pretty sure that I did consult one or two pretty distinguished Riemannian geometers and they were not able to provide any useful insight into this.

REPLY [7 votes]: Once we considered a similar problem but around infinity,
try to look in our paper "Asymptotical flatness and cone structure at infinity".
Let us denote by $r$ the distance to the singular point.
If dimensions $\not= 4$ then the same method shows that at singular point we have Euclidean tangent cone even if curvature is "much less" than $r^{-2}$ (say if $K=O(\tfrac{1}{r^{2-\varepsilon}})$ for some $\varepsilon>0$, but one can make it bit weaker).
In dimension 4 there might be some funny examples: Your singular point has tangent space $\mathbb R^3$,
the $r$-spheres around this point are  Berger spheres, so its curvature is very much like curvature of $r\cdot(S^2\times \mathbb R)$, the size of Hopf fibers goes to $0$ very fast. 
However if you know that dimension of the tangent space is $4$ then it has to be Euclidean.
All this can happen if curvature grows slowly.
If it is bounded then one can extend Bishop--Gromov type inequality for balls around singular point.
It implies that the dimension of the tangent space is $4$. That will finish the proof.<|endoftext|>
TITLE: Categorifying the Reals via von Neumann Algebras?
QUESTION [15 upvotes]: So one way to categorify the natural numbers is to replace them with vector spaces. Then the dimension of the vector space reproduces the natural number. More generally you can categorify integers to graded vector spaces. Also a linear monoidal category (assuming some finiteness conditions) will lead to an algebra defined over the natural numbers and so can be viewed as a categorification of the this algebra. 
Now one thing that I found intriguing when I learned about factor von Neumann algebras is that the type II_1 factor has modules which have "dimensions" which land in the (positive) real numbers. Has anyone ever seen a categorification of some real number quantity by using these sorts of modules? It seems that graded modules (or complexes of modules) would then give all real numbers.
Is there some sort of categorification of real algebras to a monoidal category enriched over II_1-modules? Or some other type of categorification of the reals which I am not even guessing at?

REPLY [3 votes]: In a sense we already have a great categorification of the complex numbers, and that's given by FdHilb, the *-category of finite-dimensional Hilbert spaces and continuous maps. Results like the Doplicher-Roberts theorem give us good reasons to believe this. So from this perspective, we don't need to go as far as looking at fancy von Neumann algebras to get what you want.
One point I'm implicitly making here is that categorification isn't necessarily the inverse process to taking isomorphism classes!<|endoftext|>
TITLE: Quotients of Schemes by Free Group Actions
QUESTION [26 upvotes]: I've often seen people in seminars justify the existence of a quotient of a scheme by an algebraic group by remarking that the group action is free. However, I'm pretty sure they are also invoking something else. So my question is: when you can you quotient a scheme by a free action and get a scheme? In particular, when do the coset spaces of a subgroup of an algebraic group exist as a scheme? And in these cases, how do you construct the quotient?

REPLY [17 votes]: In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, proposition 1.8.
In particular, if $X$ is a closed subscheme of projective space, then for any finite set of  points of $X$ (in particular, any orbit of a single point) we can find a hyperplane not containing any of those points.  The complement will be an affine open of $X$ containing the orbit.  This covers the existence of the quotient in the cases that $X$ is projective (and by a similar argument, quasi-projective).   
Here is a sketch of the argument in SGA I:
If the condition is satisfied, one first shows that this implies that $X$ is covered by $G$-invariant open affines, then by taking invariants, constructs the quotient of each affine by $G$, and then glues together to get the global quotient $X/G$. 
Conversely, if $X/G$ is a scheme then for any open affine $V \subseteq X/G$ the inverse image of $V$ under the morphism $X\longrightarrow X/G$ is an open affine of $X$ stable under $G$.  Since we can cover $X$ by such opens, the orbit of any point of $X$ is contained in some affine open set.
As in George McNinch's answer there is no need for the $G$-action to be free.  (This discussion was limited to quotients by finite groups though!)<|endoftext|>
TITLE: Can you construct a mapping space from local data? (looking for reference)
QUESTION [7 upvotes]: I'd to know if/where there is a reference for the following construction.
Let C_*(maps(M, T)) denote the singular chains on the space of continuous maps from an n-manifold M to some target space T.  Our goal is to construct C_*(maps(M, T)) out of local information.  More specifically, let B be any n-manifold homeomorphic to the n-ball, and let c: \boundary(B) -> T be some fixed map.  Define L_*(B, c) to be the singular chains on the space of all maps B -> T which restrict to c on the boundary of B.  We want to construct C_*(maps(M, T)) (up to homotopy) out of {L_*(B, c)}, where B ranges through all n-balls and c ranges through all boundary conditions.
This can be done as follows.  Let D be the category of all decompositions of M.  An object x of D is decomposition of M into n-balls.  There is a (unique) morphism x -> y if and only if x is a refinement of y (i.e. the balls of x are subdivisions of the balls of y).  Let D_T be a similar category, where the objects are decompositions of M into balls with the additional structure of a map from the (n-1)-skeleton of the decomposition to T, and the morphisms (anti-refinements of decompositions) are required to respect this additional structure.
We can define a functor F from D_T to the category of chain complexes.  Define F(x) to be the tensor product of all L_*(B, c), where B ranges over all n-balls of x and c is determined by the map of the (n-1)-skeleton of x to T.  Then (theorem) C_*(maps(M, T)) is homotopy equivalent to the homotopy colimit of F.
I doubt the above construction is new, but I haven't come across it anywhere.  Hence the question in the first sentence above.
EDIT: It looks like the special case, where T is n-connected and we use the undecorated category D instead of D_T, exists in some form(s) in the literature (see Oscar Randal-Williams' answer below).  So I'm particularly interested in the case where no assumptions about the connectivity of T are made.

REPLY [3 votes]: This seems to be Theorem 3.8.6 in Lurie's DAG-VI, which he says is also in Paolo Salvatore, "Configuration spaces with summable labels", Cohomological Methods in Homotopy Theory. Progress in Mathematics 196, 2001, 375-396. They work on the space level instead of chains.
It is worth noting that Lurie has a condition for the theorem to hold: that T is dim(M)-connected, and he says that it does not hold without this assumption.<|endoftext|>
TITLE: Pairs of shortest paths
QUESTION [16 upvotes]: It is known that the binomial coefficient $2n \choose n$ is equal to number of shortest lattice paths from $(0,0)$ to $(n,n)$. The Catalan number $\frac{1}{n+1} {2n\choose n}$is equal to the number of shortest lattice paths that never go above the diagonal. Here, the diagonal may be viewed as a path from $(0,0)$ to $(n,n)$.
Is there a formula for the number of pairs $(P_{1},P_{2})$ where each $P_{i}$ is a shortest lattice path from $(0,0)$ to $(n,n)$ such that $P_{1}$ never goes above $P_{2}\ ?$ Here, "$P_{1}$ never goes above $P_{2}$" means that $P_{1}$ lies inside or on the boundary of the region determined by $P_{2}$, the $x$-axis, and the line $x=n$.

REPLY [7 votes]: If you ask the analogous question about p noncrossing paths in an m x n rectangle, you are essentially counting plane partitions (if you write in each box the number of paths it is above, then all rows and columns are weakly decreasing). There is then a beautiful formula by MacMahon for the number of these: it is the product of (i+j+k-1)/(i+j+k-2) over all 1<=i<=m, 1<=j<=n, 1<=k<=p.<|endoftext|>
TITLE: When is a scheme a zero-set of a section of a vector bundle?
QUESTION [14 upvotes]: Are there any general results on when a closed subscheme X of a quasi-projective smooth scheme M can be written as the zero-set of a section of a vector bundle E on M?
To put it in a diagram: When is X the fiber product of M -> E <- M , where one arrow is the zero section and the other arrow is the section I'm looking for. 
If this is not possible, can X be written as a degeneracy locus?

REPLY [7 votes]: Are you assuming that the rank of $E$ equals the codimension of the subscheme? You don't say so explicitly. If not, the answer is that every closed subscheme is a zero section, since it is the intersection of finitely many hypersurfaces.<|endoftext|>
TITLE: Is there a free digraph associated to a graph?
QUESTION [8 upvotes]: A little bit of background: A graph G is, of course, a set of vertices V(G) and a multiset of edges, which are unordered pairs of (not necessarily distinct) vertices. We say that two vertices v_1, v_2 are adjacent if {v_1, v_2} is an edge. A directed graph, or digraph is essentially the same thing, except that the edges are now ordered pairs. Digraphs have a rather nice categorical interpretation.
A graph homomorphism is exactly what it should be, categorically, if you think of graphs as "sets with an adjacency structure." It's a map f: V(G) \rightarrow V(H) is a homomorphism if v_1, v_2 \in V(G) adjacent implies f(v_1), f(v_2) adjacent. The notion of homomorphism for directed graphs is essentially the same; if there's an edge from v_1 to v_2, then there's an edge from f(v_1) to f(v_2). Taking these definitions as morphisms, we can define categories of graphs and of digraphs. 
Oftentimes in graph theory (particularly when linear algebra methods come into play), it's easier to work with a digraph than a graph, and so we usually orient the edges arbitrarily. But this isn't really natural...
There's a forgetful functor from the category of digraphs to the category of graphs. Does this functor have an adjoint? (I forget which is left and which is right.) Now that I think about it, is the obvious thing (replace each edge by a pair of directed edges, one in each direction) an adjoint, and if so is there a way to fudge the categories so that simple graphs are taken to simple digraphs?

REPLY [6 votes]: I like to use the following definitions, which give a nonstandard definition of undirected graph but produce particularly nice categories.

A directed graph is a pair of sets V and E together with two maps s, t : E -> V.  Call the category of these DirGraph.
An undirected graph is a pair of sets V and E together with two maps s, t : E -> V plus a map r : E -> E such that r^2 = 1, sr = t and tr = s.  Call the category of these UndirGraph.

The interpretation of directed graphs should be obvious.  For an undirected graph, an edge is an orbit under r of an element of E.  Note that there are two kinds of loops--orbits of size 1 or 2.
We can represent the categories of directed graphs and undirected graphs as the categories of presheaves on two categories Dir and Undir respectively (each has two objects corresponding to V and E and a small number of morphisms).  There's an inclusion functor i : Dir -> Undir which induces a restriction functor UndirGraph -> DirGraph; this is your doubling functor.  It has adjoints on both sides (left and right Kan extension).  The left adjoint is your "forgetful" functor DirGraph -> UndirGraph.  The right adjoint is another functor DirGraph -> UndirGraph which roughly speaking sends a directed graph G to the undirected graph G' with the same vertices where an edge between v and w in G' is a pair of an edges in G, one from v to w and one from w to v.
So with these definitions, not only does the "forgetful" functor have a right adjoint, but its right adjoint also has a right adjoint.<|endoftext|>
TITLE: Are generalized cohomology theories a homotopy category of some category of invariants?
QUESTION [14 upvotes]: I was taught to think of generalized cohomology theories as the homotopy category of (symmetric) spectra. But is there also a category of 'invariants', that is, some category of contravariant functors from a suitable category of topological spaces to a suitable category of algebraic objects, which has a model category structure such that the homotopy category gives the category of generalized cohomology theories, without referring to spectra?
If such a thing exists, why do people prefer to use spectra?

REPLY [12 votes]: An even more compelling reason is that stable natural transformations between generalized cohomology theories on spaces form the wrong category, i.e. are not the same as the morphisms of the stable category. Let $E,F$ be generalized cohomology theories. Let $Z_n$ be the terms of the $\Omega$-spectrum of a CW model of $E$. Then, more or less by definition, stable natural transformations from $E$ to $F$ are the inverse limit of $F^nZ_n=F^0\Sigma^{\infty-n}Z_n$. This is the correct Hom set in the stable homotopy category when $\lim^1=0$, but this may not be the case (for example when $E=S$, $F$ is the wedge of $\Sigma^n HZ/2$ over $n\in \mathbb{Z}$).<|endoftext|>
TITLE: Is 8 the largest cube in the Fibonacci sequence?
QUESTION [26 upvotes]: Can you prove that 8 is the largest cube in the Fibonacci sequence?

REPLY [4 votes]: In Siegel's article Zum Beweise des Starkschen Satzes. Inventiones mathematicae (1968), Siegel reduces the class number one problem for imaginary quadratic fields to the determination of Fibonacci cubes. The latter problem he solves in an elementary way. However, he does not mention Fibonacci numbers.<|endoftext|>
TITLE: K(F_1) = sphere spectrum?
QUESTION [21 upvotes]: I repeatedly heard that K(F_1) is the sphere spectrum. Does anyone know about the proof and what that means?

REPLY [10 votes]: Here is another heuristic, related to what Randal-Williams said above. The sphere spectrum is the unit object in nice categories of spectra. That is, ring spectra are algebras over the sphere spectrum. Now, to every scheme X you can associate a K-theory ring spectrum K(X), and this is contravariant. So, in the usual theory there is a morphism K(Z)->K(X) for all schemes X. So, finding F_1 also means finding something (its K-theory spectrum) that maps to the (homotopy) limit of all K-theory spectra. That this should be the unit object of the category of spectra doesn't seem very surprising.<|endoftext|>
TITLE: Homotopy type of stabilizers
QUESTION [6 upvotes]: Let X be a contractible metric space and G a topological group acting transitively on X (i.e. given any two points x,y \in X, there exists g \in G such that gx=y).
My question is the following: is it true that given any x \in X its stabilizer 
Stab(x)={ g \in G : gx=x } and the whole group G have the same homotopy type?
If the answer is "no", I'd like to know some "mild" hypothesis that could be add to have an affirmative response.
For instance, I know that whenever G is a Lie group and H < G is a closed subgroup such that G/H is contractible, then G and H are homotopically equivalent (in this case H can be seen as the stabilizer of the coset H under the natural G-action on G/H). However, to assume that G is a Lie group seems to be too restrictive. In fact, I'd like to apply this "result" to some groups which are not locally compact.

REPLY [5 votes]: The short: no.  E.g. let X be a topological group and G be the underlying discrete group of X, acting on X by left translation.
One standard hypothesis is the existence of slices.  For some (hence any) point x in X, if there is an open neighborhood U of x and a section s:U -> G such that s(u)*x = u for all u in U, then the map p:G -> X given by p(g) = g*x is a fiber bundle with fiber Stab(x).
This hypothesis automatically implies that Stab(x) is weakly equivalent to G by the long exact sequence of the fibration.  If G and Stab(x) have the homotopy type of CW-complexes then they are homotopy equivalent by the Whitehead theorem.<|endoftext|>
TITLE: Finite type/finite morphism
QUESTION [22 upvotes]: I am not too certain what these two properties mean geometrically. It sounds very vaguely to me that finite type corresponds to some sort of "finite dimensionality", while finite corresponds to "ramified cover". Is there any way to make this precise? Or can anyone elaborate on the geometric meaning of it?

REPLY [22 votes]: I definitely agree with Peter's general intuitive description.
In response to some of the subsequent comments, here are some implications to keep in mind: 
Finite ==> finite fibres (1971 EGA I 6.11.1) and projective (EGA II 6.1.11), hence proper (EGA II 5.5.3), but not conversely, contrary to popular belief ;)
Proper + locally finite presentation + finite fibres ==> finite (EGA IV (part 3) 8.11.1)
When reading about these, you'll need to know that "quasi-finite" means "finite type with finite fibres."  Also be warned that in EGA (II.5.5.2) projective means $X$ is a closed subscheme of a "finite type projective bundle" $\mathbb{P}_Y(\mathcal{E})$, which gives a nice description via relative Proj, whereas "Hartshorne-projective" more restrictively means that $X$ is closed subscheme of "projective n-space" $\mathbb{P}^n_Y$.
When the target (or "base" scheme) is locally Noetherian, like pretty much anything that comes up in "geometry", a proper morphism is automatically of locally finite presentation, so in that case we do have 
finite <==> proper + finite fibres
Regarding "locally finite type", its does not imply finite dimensionality of the fibres; rather, it's about finite dimensionality of small neighborhoods of the source of the map.  For example, you can cover a scheme by some super-duper-uncountably-infinite disjoint union of copies of itself that is LFT but not FT, since it has gigantic fibres.<|endoftext|>
TITLE: Can Walsh-Hadmard transform be used for convolution ?
QUESTION [5 upvotes]: The Walsh-Hadamard transform is very fast to compute. 
Can it be used to compute the convolution of two functions as it can be done with Fourier transform ?

REPLY [4 votes]: Not in the sense I think you mean it.  First of all, the Walsh-Hadamard transform is a Fourier transform - but on the group (Z/2Z)^n instead of on the group Z/NZ.  That means you can use it to compute convolutions with respect to the space of functions (Z/2Z)^n -> C.  Unfortunately, unlike the case with Z/NZ you can't use this to approximate a compactly supported convolution on Z, at least not directly.<|endoftext|>
TITLE: Can one calculate the (co)homology of the loopspace of a Lie group from its Lie algebra?
QUESTION [7 upvotes]: Compact connected simply-connected Lie groups have so much structure that you can calculate their cohomology from their Lie algebras using Lie algebra cohomology (certain Ext-groups) and similarly their homology from their Lie algebras using Lie algebra homology (certain Tor-groups). 
Is there similar theorem that gives the (co)homology of the loop space of a Lie group in terms of its Lie algebra?

REPLY [2 votes]: Yes.
At least, rationally.
The result that you want starts on p68 of "Loop Groups" by Pressley and Segal.  There, they prove surjectivity of H*(L𝔤;ℝ) → H*(LG;ℝ).  The basic idea of the argument is as follows: for reasonably simple reasons, the cohomology of LG is easily obtainable from that of G.  This yields specific formulae for generators of the de Rham cohomology of LG.  Although these forms are not themselves left invariant, they are cohomologous to rational multiples of left invariant forms, and thus come from the cohomology of the Lie algebra, L𝔤.  The proof of surjectivity is thus not hard.
The proof that it is an isomorphism is a little trickier and they defer that to section 14.6 (p299).  That this is quite some way through the book is a good indication of how much you need to absorb to understand it.
Amazingly, parts of Loop Groups (including pages 68, 69, and 299, but not 300) are available on Google books.  It says that it is a "Limited preview" but whether or not that is just limited in pages or limited in time, I do not know.  However, Loop Groups is a great book for anyone interested in Lie Groups and infinite dimensional "stuff".
(Incidentally, this is all for G simply connected.)<|endoftext|>
TITLE: When should I expect a quiver with potential to be rigid?
QUESTION [19 upvotes]: This question is pretty technical, but there are some very smart people here.
Fix a quiver Q, WITH oriented cycles. Let k[[Q]] be the completed path algebra. (Like the path algebra, but we allow formal infinite sums.) My question is about what behavior I should expect when I equip Q with a generic choice of potential.
A potential, denoted S, is a formal sum of (nontrivial) closed cycles in k[[Q]]. I'm interested in the case where the coefficients of that sum are chosen generically. Let J(S) denote the Jacobian ideal of S. This is a two sided ideal which is, roughly speaking, generated by the derivatives of S. (See Derksen-Weyman-Zelevinsky, Quivers with potentials and their representations I: Mutations, for the precise definitions.) [DWZ] show that the space of deformations of (Q,S) is k[[Q]]/(J(S) + [,]) where [,] is the vector space (NOT usually an ideal) spanned by commutators. They define (Q,S) to be rigid if this vector space is spanned by the empty cycle.
Heuristic arguments seem to suggest that a generic potential is always rigid. But example 8.6 in [DWZ] shows that this is false. Can someone give me a better heuristic or, better yet, a theorem?
EDITED to fix errors involving cycles of length 0.

REPLY [6 votes]: As I understand it, the quotient C(Q) = k[[Q]]/[,] is the (completed) vector space of formal linear combinations of oriented circuits in Q.  Define a detour in Q to be an edge e and an oriented path p with the same source and sink as e.  For every detour (e,p), there is a detour operator D(e,p).  Given a circuit s, D(e,p)(s) is a sum of terms for each occurrence of e in s; each term is obtained by replacing that occurrence of e by p.  Then the image I(S) of J(S) appears to be the span of all D(e,p)(S).
If I have all of that straight, then I can't think of a significant condition to guarantee that there is a rigid potential S.  However, I can think of a significant condition to guarantee that there isn't one.  Suppose for simplicity that the quiver Q has no parallel edges.  Consider further just the shortest non-trivial circuits in C(Q); they span C(Q)g, where g is the oriented girth of Q.  For these circuits, the only detour operators D(e,p) that matter are the trivial ones with p = e.  In other words, you have no choice but to replace an edge e with itself.  But often C(Q)g is bigger than the space of edges, i.e., Q could have more shortest cycles than it has edges.  In this case, just for dimension reasons, I(S) can't contain C(Q)g.
What bothers me is that if this is correct, then Derksen, Weyman, and Zelevinsky worked a little harder than necessary to make their Example 8.6.

Now that I read on a bit more of the paper, the authors give some constructions of quivers with rigid potentials.  For instance, they establish that the existence of a rigid potential is a mutation-invariant property of quivers.  They also provide an example of a rigid quiver which does not mutate to an acyclic quiver.  You can go a little further:  In studying the cyclic part of the path algebra of a quiver, you can restrict attention to its strongly connected components.  So you can start with any collection of strongly connected, rigid quivers, then connect them any way you like acyclically, and then mutate that.
It sounds like the status quo of the question is that there are several constructions of quivers that do not have rigid potentials, and several constructions of quivers that have rigid potentials, but that finding a good characterization of the dichotomy is an open problem.
Also, clearly the quivers that do not have rigid potentials are recursively enumerable.  (I.e., there is an algorithm to confirm that a quiver does not have a rigid potential, but it might not terminate if it does have one.)  Maybe a good remaining question is whether quivers with rigid potentials are also recursively enumerable, i.e., that the dichotomy is recursive.

One way to confirm that a quiver has a rigid potential is to find a J(S) that contains all paths of some length.  I(S) then automatically contains all loops of that length or greater.  Moreover, it does not matter whether S includes any terms beyond the cutoff length.  If a rigid potential always has this property, then determining whether there is one is recursive.<|endoftext|>
TITLE: Can you do surgery on framed tangles?
QUESTION [6 upvotes]: Surgery on framed links give orientable compact closed connected 3-manifolds. Can you do surgery on framed tangles? Would Fenn-Rourke moves be invariant?

REPLY [3 votes]: You can cut out a collection of 3-balls (a regular neighbourhood of the tangle), and glue them back in a different way. Such modifications are a part of the Montesinos trick. Montesinos uses them to prove that any compact oriented connected 3-manifold is a 3-fold branched cover of the 3-sphere. In this case, the "surgery on the tangle" downstairs lifts to honest Dehn surgery upstairs.
In the context of branched covering spaces, your second question generalizes to the open question "does the Kirby theorem hold for surgery in an orbifold?".<|endoftext|>
TITLE: What is the difference between a homogeneous stochastic process and a stationary one?
QUESTION [8 upvotes]: Hello.
I am studying stochastic process.
here,
I don't know what is difference of
"the process is homogeneous"
and
"the process is stationary"
I feel confusing. It seems to similar to me.

REPLY [5 votes]: A process is (strictly) stationary if any sequence of n consecutive points has the same distribution as any other sequence of n consecutive points. There are weaker definitions, for example weak stationarity is based only on the first two moments.
A (discrete valued) process is homogeneous if the transition probability between two given state values at any two times depends only on the difference between those times. 
However, some references uses "homogeneous" rather loosely and confuse the two concepts.<|endoftext|>
TITLE: Number of metric spaces on N points
QUESTION [15 upvotes]: Given $X = \{x_1, ..., x_n\}$, how many collections $C$ of subsets of $X$ are there such that $C$ is the listing of all open balls of some metric space?
The first nontrivial example is $n=3$; let's call the points $x, y$ and $z$. Also, let $a = d(x, y), b = d(y, z), and c = d(z, x)$. For any collection $C$ to be a listing of all the open balls, it must contain all the singleton sets  and the whole set $X$. Let $C_0 = \{\{x\},\{y\},\{z\},X\}$. If $x, y$ and $z$ are equidistant, these are the only open balls. If, say, $a < b < c$, then we get $C = C_0 \cup \{\{x,y\},\{y,z\}\}$. Through careful case enumeration, we can answer this for small $n$, but the process quickly becomes unwieldy. Has anyone ever looked at this before, and is there a recursive formula or a generating function for this? What about if the points are unlabeled? For $n=3$, I count $7$ possibilities for $C$ if the points are labeled, and $3$ if the points are unlabeled. It's already somewhat time-consuming to count when $n=4$.

REPLY [5 votes]: My analysis leads to different answers from the above, and to some references.  Edit: I get different answers because I changed the question without realizing it.  I'm working from the collection of pointed metric balls, i.e., metric balls with a distinguished center.  Gabe asked the question about the collection of unpointed metric balls, noting that the same set can be a metric ball with two different centers.  That seems more complicated, although I would suggest working from the pointed solution.
Let's let $\{1,2,\ldots,n\}$ be the points in $X$.
The distances from $i$ to the other points in the set $X$ induce a strict weak ordering of those points by their distance from $i$.  This has the same information as the set of metric balls with center $i$.  Thus the information in the metric is given by all of the comparisons between the distances $x_{(i,j)} = d(i,j)$ and $x_{(i,k)} = d(i,k)$.  You can express these relations by a hyperplane arrangement in $\mathbb{R}^{n(n-1)/2}$, where the hyperplanes are given by the equations $x_{(i,j)} = x_{(i,k)}$.
You might also think about the triangle inequalities satisfied by all of the distances, and the fact that the distances are all positive numbers.  The set of feasible distance vectors is called the "metric cone".  Although the metric has an interesting combinatorial structure, it looks like it matters for nothing in this particular question:  The hyperplanes all meet at an interior point of the cone in which the distances are all equal.  In other words, you can always add a constant distance $h \gg 0$ to all of the distances without changing any of the metric balls, so that the triangle inequality and positivity of distance become irrelevant.
If a hyperplane arrangement has the property that all hyperplanes are given by setting two coordinates equal, then the arrangement is called "graphical".  The coordinates correspond to the vertices of a graph $G$, and the hyperplanes correspond to the edges.  The hyperplane arrangement gives you a partially ordered set of chambes and other faces, ordered by inclusion.  This poset has a lot of properties and there are techniques to compute the number of faces from $G$.  In our case, $G$ is the line graph $T_n$ of the complete graph $K_n$, which is sometimes confusingly called a triangular graph.
For $n=3$, my answer is that there are 13 different types of metrics, not 7, corresponding to the hyperplane arrangement $x=y$, $y=z$, $x=z$ in $\mathbb{R}^3$.    In other words, I count 13 types of triangles:  6 scalene, 3 short-base isosceles, 3 long-base isosceles, and 1 equilateral.
Some of the types of metrics lie in chambers, meaning generic metrics in which $d(i,j) \ne d(i,k)$ for all $i$, $j$, and $k$.  It is a theorem that the number of chambers of the graphical arrangment $A(G)$ of a graph $G$ is $|\chi_G(-1)|$, where $\chi_G$ is the chromatic polynomial of $G$.  (See this excellent review by Richard Stanley.)  So I asked Maple to compute 
ChromaticPolynomial(LineGraph(CompleteGraph(n)),q)

for small values of $n$, and I got the following answers:
$$\chi_{T_3}(q) = q(q-1)(q-2)$$
$$\chi_{T_4}(q) = q(q-1)(q-2)(q^3-9^2+29q-32)$$
$$\chi_{T_5}(q) = q(q-1)(q-2)(q-3)(q-4)(q^5-20q^4+170q^3-765q^2+1804q-1764).$$
Evaluating at $q=-1$, I get that there are 1, 6, 426, and 542880 generic types of metrics on 2, 3, 4, and 5 points.  This sequence is not in the Encyclopedia of Integer Sequences, although possibly it should be.  I think that you can obtain the total number of faces of a graphical arrangement $A(G)$ from the Tutte polynomial of $G$, and therefore the total number of types of metrics, but I did not do the calculation.<|endoftext|>
TITLE: Graded or stacky Serre duality
QUESTION [12 upvotes]: I am considering the following situation. $A$ is a finitely generated ring over a field $K$ with non-negative grading and $A_0=K$ of Krull dimension n+1, but I don't necessarily assume A is generated in degree 1. Then $X=\mathrm{Spec} A$ carries an action of the multiplicative group $\mathbb{G}_m$, which is really what the grading means to me. Also, I want to assume that $X$ has a unique singularity at the `origin'
0 corresponding to the maximal ideal of positive elements of $A$, so that $U=X\setminus 0$ is smooth.
I am interested in (the derived category of) coherent sheaves on the quotient stack $[U/\mathbb{G}_m]$
or equivalently in $\mathbb{G}_m$-equivariant coherent sheaves on $U$. I'd like to have Serre duality
in this category. I think one should be able to state this in the form
$\operatorname{Ext}^k(F,G) \simeq \operatorname{Ext}^{n-1}(G,F \otimes \omega_U)^*$ 
where $\omega_U$ is the canonical sheaf of $U$ and $*$ is the graded dual, so that taking $\mathbb{G}_m$-invariants (degree 0) produces the desired Serre duality on $[U/\mathbb{G}_m]$. 
I am willing to assume the singularity of $X$ is Cohen-Macaulay or even Gorenstein. I think such a statement could be deduced from local duality if $A$ were local rather than graded. But I don't understand these things well enough to see right away if there is a graded version. Also, I'm not sure what reference to consult. It would be helpful to have both
a geometric and an algebraic reference.

REPLY [5 votes]: Fabio Nironi wrote a paper on Serre Duality on Deligne-Mumford stacks, http://arxiv.org/abs/0811.1955.<|endoftext|>
TITLE: Why is the exterior algebra so ubiquitous?
QUESTION [59 upvotes]: The exterior algebra of a vector space V seems to appear all over the place, such as in

the definition of the cross product and determinant,
the description of the Grassmannian as a variety,
the description of irreducible representations of GL(V),
the definition of differential forms in differential geometry,
the description of fermions in supersymmetry.

What unifying principle lies behind these appearances of the exterior algebra?  (I should mention that what I'm really interested in here is the geometric meaning of the Gessel-Viennot lemma and, by association, of the principle of inclusion-exclusion.)

REPLY [2 votes]: I think that what unifies some of the different examples of when the exterior algebra occurs is that it is the structure that transforms the action of a commutative ring on a module (or, more concretely, the action of several commuting linear operators on a vector space) into a chain complex. The ubiquitous structure is really commutative rings and modules over commutative rings.  The exterior algebra actually encodes commutativity, in a sense. (One might then ask why is it that mathematicians love so much commutative rings, commuting operators, etc.)
When tensoring a commutative action with the graded algebra one gets the Koszul complex, and the anti-commutative nature of the graded algebra is precisely what, when coupled with the commutative action of the ring, makes possible the definition of a dirac operator $d$ with $d^2=0$.
I got these ideas after thinking about Joseph Taylor's paper, "A joint spectrum for several commuting operators", J. Funct. Anal. , 1970.<|endoftext|>
TITLE: Can algebraic varieties be rigidified by finite sets of points?
QUESTION [32 upvotes]: For an algebraic variety X over an algebraically closed field, does there always exist a finite set of (closed) points on X such that the only automorphism of X fixing each of the points is the identity map? If Aut(X) is finite, the answer is obviously yes (so yes for  varieties of logarithmic general type in characteristic zero by Iitaka, Algebraic Geometry, 11.12, p340). For abelian varieties, one can take the set of points of order 3 [added: not so, only for polarized abelian varieties]. For P^1 one can take 3 points. Beyond that, I have no idea.
The reason I ask is that, for such varieties, descent theory becomes very easy (see Chapter 16 of the notes on algebraic geometry on my website).

REPLY [25 votes]: I get that the answer is "no" for an abelian variety over the algebraic closure of Fp with complex multiplication by a ring with a unit of infinite order. Since you say you have already thought through the abelian variety case, I wonder whether I am missing something.
More generally, let X be any variety over the algebraic closure of Fp with an automorphism f of infinite order. A concrete example is to take X an abelian variety with CM by a number ring that contains units other than roots of unity. Any finite collection of closed points of X will lie in X(Fq) for some q=p^n. Since X(Fq) is finite, some power of f will act trivially on X(Fq). Thus, any finite set of closed points is fixed by some power of f.
As I understand the applications to descent theory, this is still uninteresting. For that purpose, we really only need to kill all automorphisms of finite order, right?<|endoftext|>
TITLE: What do gerbes and complex powers of line bundles have to do with each other?
QUESTION [16 upvotes]: We all know how to take integer tensor powers of line bundles.  I claim that one should be able to also take fractional or even complex powers of line bundles.  These might not be line bundles, but they have some geometric life.  They have Chern classes, and one can twist differential operators by them.  How should I think about these?  What do they have to do with gerbes?

REPLY [5 votes]: If L is any line bundle on a space (scheme, whatever) X, A is any (additive) abelian group, and a an element of A, there is a natural construction of an A-gerbe $L^a$ as follows.  By definition, $L^a$ should be a "sheaf of categories", or stack (not algebraic) on X, and here are its categories of sections.  Identify L with its total space, which is a $\mathbb{G}_m$-bundle on X, and for any open set U in X, let $L^a(U)$ be the category of all A-torsors on $L|_U$ whose monodromy about each fiber of $L|_U \to U$ is a.
One can check that this really is a gerbe: it is locally nonempty, since if L is trivial over U, you can write $L|_U = \mathbb{G}_m \times U$ and then pull back the unique A-torsor on $\mathbb{G}_m$ with monodromy a.  It has a natural action of A-torsors on X, given by pulling up along the bundle map $L \to X$ and tensoring.  And this action is free and transitive, since the difference of two a-monodromic torsors on $L|_U$ has trivial monodromy on each fiber and therefore descends to X.
Why do I call this $L^a$?  Suppose that $L = \mathcal{O}_X(D)$ for a divisor D, where for simplicity let's say that D is irreducible of degree n; then L gets a natural trivialization on $U = X \setminus D$ having a pole of order n along D.  As shown above, this induces a trivialization $\phi$ of $L^a$ on U, and if we pick a small open set V intersecting D and such that D is actually defined by an equation f of degree n, then we get a second (noncanonical) trivialization $\psi$ of $L^a$ on V.  You can check that the difference $\psi^{-1} \phi$, which is an automorphism of the trivial gerbe on $U \cap V$, is in fact described by the A-torsor $\mathcal{T} = f^{-1}(\mathcal{L}_a)$, where $f \colon U \cap V \to \mathbb{G}_m$ and $\mathcal{L}_a$ is the A-torsor of monodromy a.  Since f has degree n, $\mathcal{T}$ has monodromy na about D.  Thus, it is only reasonable to say that the natural trivialization $\phi$ has a pole of order na, which is consistent with the behavior of the trivialization of L itself on U, when raising to integer powers.
What does this have to do with twisting of differential operators?  Suppose we have some kind of sheaves (D-modules, locally constant sheaves, perverse sheaves; technically, they should form a stack admitting an action of A-torsors).  On the one hand, one could mimic the above construction of $L^a$ to describe a-monodromic sheaves on L, and this is what is often called twisting.  On the other hand, there is a natural way to directly twist sheaves by the gerbe $L^a$ without mentioning L at all (that is, you can twist by any A-gerbe).  The procedure is as follows: a twisted sheaf is the assignment, to every open set U in X, of a collection of sheaves on U parameterized by the sections of $L^a(U)$, and compatible with tensoring by A-torsors.  Of course, since if $L^a(U)$ is nonempty this is the same as giving just one sheaf, this is sort of overkill, but the choice of just one such sheaf is noncanonical whereas this description is canonical.  These collections should be compatible with the restriction functors $L^a(U) \to L^a(V)$ when $V \subset U$.  It is an exercise to reader to check that this is the same as the other definition of twisting :)
Man, you asked the right question at the right time.  My thesis is all about this stuff.<|endoftext|>
TITLE: Free, high quality mathematical writing online?
QUESTION [76 upvotes]: I often use the internet to find resources for learning new mathematics and due to an explosion in online activity, there is always plenty to find.  Many of these turn out to be somewhat unreadable because of writing quality, organization or presentation.
I recently found out that "The Elements of Statistical Learning' by Hastie, Tibshirani and Friedman was available free online: http://www-stat.stanford.edu/~tibs/ElemStatLearn/ .  It is a really well written book at a high technical level.  Moreover, this is the second edition which means the book has already gone through quite a few levels of editing.
I was quite amazed to see a resource like this available free online.
Now, my question is, are there more resources like this?  Are there free mathematics books that have it all: well-written, well-illustrated, properly typeset and so on?
Now, on the one hand, I have been saying 'book' but I am sure that good mathematical writing online is not limited to just books.  On the other hand, I definitely don't mean the typical journal article.  It's hard to come up with good criteria on this score, but I am talking about writing that is reasonably lengthy, addresses several topics and whose purpose is essentially pedagogical.
If so, I'd love to hear about them.  Please suggest just one resource per comment so we can vote them up and provide a link!

REPLY [2 votes]: Within the framework of the project retro.seals.ch, scientific journals are retrodigitized and made available via internet. The project contains the following mathematical journals:

Commentarii Mathematici Helvetici
Elemente der Mathematik
Elemente der Mathematik (Beihefte zur Zeitschrift)
L'Enseignement Mathématique<|endoftext|>
TITLE: How should I think about B-fields?
QUESTION [15 upvotes]: So, physicists like to attach a mysterious extra cohomology class in H^2(X;C^*) to a Kahler (or hyperkahler) manifold called a "B-field."  The only concrete thing I've seen this B-field do is change the Fukaya category/A-branes: when you have a B-field, you shouldn't take flat vector bundles on a Lagrangian subvariety, but rather ones whose curvature is the B-field.  How should I think about this gadget?

REPLY [3 votes]: Let me try to add a different point of view on B-fields and mirror symmetry. Ideally in mirror symmetry, given a Calabi-Yau manifold X, you would like to "construct" its mirror X', where the symplectic form on X should give you the complex structure on X'. As already mentioned, classes of symplectic forms have moduli of real dimension $h^{1,1}(X)$ and complex structures on X' have moduli of complex dimension $h^{2,1}(X') = h^{1,1}(X)$. So the kahler class is not enough to determine all complex structures on X'. In the context of the Strominger-Yau-Zaslow conjecture there is a nice interpretation of the B-field. Suppose X = $T^*B / \Lambda$, where B is a smooth manifold and $\Lambda$ is locally the span over the integers of 1-forms $dy_1$, ..., $dy_n$ (here $y_1$, ..., $y_n$ are coordinates which 
change with affine transformations from one chart to the other). Then $X$ has a standard symplectic form. We can consider $X'= TB / \Lambda'$, where  $\Lambda'$ is the dual lattice. 
Then X' has a natural complex structure defined as follows. In standard coordinates on TB, given by $(y,x)$ --> $x \partial_y$, the complex coordinates on X' are $z_k = e^{2\pi i(x_k + i y_k)}$, which are well defined due to the nature of the coordinates x and y. But the above complex coordinates can be twisted locally (on a coordinate patch) by  $z_k (b) =  e^{2\pi i(x_k + b_k  + i y_k)}$, where $b = (b_1, \ldots, b_n)$ is some local data. But since on overlaps $U_i \cap U_j$ the coordinates have to match, we must have $b(i) - b(j) \in \Lambda$. It turns out that by putting $b_{ij} = b(i) - b(j)$ on overlaps, we get a cohomology class in $H^{1}(B, \Lambda)$, this is the B-field. The cohomology group $H^{1}(B, \Lambda)$ shoud coincide (in some cases at least) with $H^2(X, R/Z)$, which is what Kevin Lin mentioned. The elliptic curve case (mentioned by Kevin) can be seen from this point of view. 
This point of view is also called "mirror symmetry without corrections" and it only approximates what happens in compact Calabi-Yaus. I have learned this in papers by Mark Gross (such as "Special lagrangian fibrations II: geometry") or the book "Calabi-Yau manifolds and related geometries" by Gross, Huybrechts and Joyce. 
I would be interested to know how this interpretation connects to the other ones which have been described.<|endoftext|>
TITLE: What is the constant of the Coppersmith-Winograd matrix multiplication algorithm
QUESTION [14 upvotes]: Or at least it's order of magnitude.
I've only ever heard it described as "huge", and a google search turned up nothing.
Also, given that the Strassen algorithm has a significantly greater constant than Gaussian Elimination, and that Coppersmith-Winograd is greater still, are there any indications of what constant an O(n^2) matrix multiplication algorithm might have?

REPLY [8 votes]: In your second question, I think you mean "naive matrix multiplication", not "Gaussian elimination".
Henry Cohn et al had a cute paper that relates fast matrix multiply algorithms to certain groups.  It doesn't do much for answering your question (unless you want to go and prove the conjectured results =), but it's a fun read.
Also, to back up harrison, I don't think that anyone really believes that there's an $O(n^2)$ algorithm.  A fair number of people believe that there is likely to be an algorithm which is $O(n^{2+\epsilon})$ for any $\epsilon > 0$.  An $O(n^2 \log n)$ algorithm would fit the bill.
edit: You can get a back-of-the-envelope feeling for a lower bound on the exponent of Coppersmith-Winograd based on the fact that people don't use it, even for n on the order of 10,000; naive matrix multiplication requires $2n^3 + O(n^2)$ flops, and Coppersmith-Winograd requires $Cn^{2.376} + O(n^2)$.  Setting the expressions equal and solving for $C$ gives that the two algorithms would have equal performance for n = 10,000 (ignoring memory access patterns, implementation efficiency, and all sorts of other things) if the constant were about 627.  In reality, it's likely much larger.

REPLY [3 votes]: In answer to the second part of your question, I think the conventional wisdom is that there isn't a O(n^2) algorithm; analogously to the case for integer multiplication, you shouldn't be able to do better than about O(n^2 log n). (Raz has shown that this is a lower bound in the arithmetic circuits with bounded coefficients model.)
What's the implied constant there? Probably just "huge." As far as I know, the reason that people believe that we can achieve close to O(n^2) is basically by analogy with integer multiplication, so if you want some grasp on the constants it might be worthwhile to look at the constants in FFT multiplication.
Incidentally, has the appropriate volume of Art of Computer Programming been released, or will it be soon? I know Knuth's a stickler for including these kinds of details, so that might be the most obvious reference apart from the original paper...<|endoftext|>
TITLE: Can anyone give me a good example of two interestingly different ordinary cohomology theories?
QUESTION [65 upvotes]: An answer to the following question would clarify my understanding of what a cohomology theory is. I know it's something that satisfies the Eilenberg-Steenrod axioms, and I know that those axioms allow you to work out quite a lot. But what sort of thing is not determined by the axioms? In particular, can someone give me a simple example of a space that has different cohomology groups with respect to two different theories? Obviously a trivial answer would be to take coefficients in different rings, so let me add the requirement that the coefficient rings should be the same. And if there's some other condition needed to make the question non-trivial, then add that in too.

REPLY [5 votes]: (Some inaccurate info appeared in the Answers above).
Different co/homology theories may behave differently on infinite CW-complexes. If two E-S theories are the same for the singleton then they are equivalent for all spaces homotopically dominated by finite polyhedra. It seems to me that for any other topological space one can construct two E-S theories (equal for singletons) which will give different groups (let's consider completely regular $T_1$-spaces only).
Surely dimension of the Hilbert cube   $C$   is infinite, while   $H^n(C)=0$   for every   $n > 0$. In the case of any Hausdorff compact space   $X$   the (topological) covering dimension, say   $n = \dim(X)$,   is equal to the cohomological dimension, meaning that there exists a closed subset   $A$   of   $X$   such that   $H^n(X\ A) \ne 0$,   but there is no such closed subset   $B$   that   $H^m(X\ B)\ne 0$   for any   $m > n$   (a few more words should be devoted to the infinite-dimensional case).
Intuitively the singular and the Cech co/homology view topological spaces in a drastically different way. What is connected for the cruder Cech theory might be not connected for the more sensitive singular theory; the former one is much more tolerant. One can say that the Cech theory applies the shape theory as its foundation rather than just homotopy theory. More precisely, all co/homology functors split into a composition of the homotopy functor and the rest; while the Cech functor does more, it splits into composition of the shape functor and the rest (while the shape functor is a composition of the homotopy functor and the rest).<|endoftext|>
TITLE: Asymptotics of the number of compositions whose summands are the divisors of a number?
QUESTION [6 upvotes]: Let $n$ be a natural number.  Let $dc(n)$ be the number of compositions of $n$ where the summands are required to be in the set of divisors of $n$.  Standard lore in analytic combinatorics yields the following formula for $dc(n)$:
$$dc(n) = n\text{th Taylor coefficient of }\frac{1}{1- \sum_{m\in\text{divisors of n}} z^m}.$$
But what are the asymptotics of $dc(n)$?  Here's a plot that I made (the y-axis is $dc(n)$ on a log scale and the x-axis is $n$):

I would like to understand the "fanning", which presumably has something to do with whether numbers have lots of small divisors or not, and I would also like to understand why these fans seem to be so close to exponentials.  The solid fit line at the top is $2^n$, which is the number of unrestricted compositions.
If that's too much to ask for, I guess I'd like to know how one might use known facts about the number of divisors, etc. to say something about this, as this rather artificial construction ought to be governed by some more fundamental number theoretic functions.

REPLY [5 votes]: The number of compositions of n with all parts equal to 1 or 2, for example, is the (n+1)st Fibonacci number, which grows like φn where φ = (1+√5)/2.
The number of compositions of 2p for p a prime, where all parts are equal to 1, 2, p, or 2p,  isn't all that much bigger.  In particular I believe it has the same exponential growth rate.  I suspect this is one of the lines you see.
Replacing 2 with any integer, you can do the same sort of thing; this is what causes your points to fall, for the most part, along straight lines.<|endoftext|>
TITLE: Kodaira-Spencer Theory and moduli of curves
QUESTION [9 upvotes]: I was looking at a paper of Farkas and the following confusing point came up.
Let $\mathscr{M}_g$ be the moduli stack of smooth genus $g$ curves and let $\pi: \mathscr{C} \to \mathscr{M}_g$ be the universal curve. Let $\mathscr{F}$ be $\Omega^1_\pi \otimes \Omega^1_\pi$, where $\Omega^1_\pi$ is the sheaf of relative differentials of $\pi$. Then the pushforward $\pi_* \mathscr{F}$  is isomorphic $\Omega^1_{\mathscr{M}_g}$.
Why is this true? Farkas says this follows from Kodaira-Spencer theory. I googled for a while and asked a few students, but couldn't figure this out.

REPLY [7 votes]: By standard deformation theory (see e.g., Hartshorne III Ex 4.10, but there are probably better references), the tangent sheaf of $\mathscr{M}_g$ is $R^1\pi_{\ast}(\mathscr{C}, T_{\mathscr{C}/\mathscr{M}_g})$, which is Serre dual to $\pi_{\ast}\mathscr{F}$. The tangent sheaf is dual to what you wanted.<|endoftext|>
TITLE: For which spaces is homology (or cohomology) determined by the Eilenberg-Steenrod axioms
QUESTION [36 upvotes]: This is a spinoff of Can anyone give me a good example of two interestingly different ordinary cohomology theories? .  By an ordinary homology theory, I mean a functor on topological spaces which satisfies the usual set of axioms (e.g., excision, homotopy, infinite coproducts, & dimension axiom (with integer coefficients)).
It is well-known to the sort of people who know these sorts of things that any two such homology theories are isomorphic on any space with the homotopy type of a CW-complex.  (If I remember correctly, one proves that there is a natural map from singular homology to another theory defined on all spaces, and that this map is an isomorphism for spaces homotopy equivalent to a CW-complex.)
The question is: are spaces with the homotopy type of CW-complexes the largest class of spaces for which this is so?  A similar question is: given a space X not homotopy equivalent to a CW-complex, are there two homology theories which have different values on X?
(I'd like to add the tag "shape-theory" to this, since that's the area where an answer is mostly likely to be found.  But I can't create new tags!)

REPLY [6 votes]: Clark Barwick suggested to me that this should be true for any Δ-generated space, and that this is a strictly larger class than spaces homotopy equivalent to a CW-complex.  I haven't attempted to verify either of these statements.<|endoftext|>
TITLE: Lifting bases for (Z/pZ)^n to Z^n
QUESTION [8 upvotes]: The following question came up in my research.  I suspect that it has a slick answer,
but I can't seem to find it.
Fix an integer n>=2 and a prime p.  Define X(n) to be the set of primitive
vectors in the Z-module Z^n and Y(n,p) to be the set of "lines" in the vector space (Z/pZ)^n (ie the spans of non-zero vectors).  There is a natural surjective map f:X(n)-->Y(n,p) ("reduce mod p and take the span").
Question : Does there exist a map g:Y(n,p)-->X(n) with the following two properties.

f(g(L))=L for all L in Y(n,p).
If {L_1,...,L_n} \subset Y(n,p) spans the vector space (Z/pZ)^n, then {g(L_1),...,g(L_n)} is a basis for the Z-module Z^n.

Of course, I expect that the answer is no except in certain simple situations (for instance,
it is yes for n=p=2), but I can't seem to find a proof.
EDIT : Oops!  I phrased the question incorrectly.  Above is a corrected version.

REPLY [4 votes]: Here's a unified argument based on my comments to Scott's post that doesn't use quadratic reciprocity in any form.  Suppose n=2 and p >= 5, and lift each line of slope i in Y(2,p) to a point (ai+pbi, iai+pci).
Since each pair of lifts should give a basis of Z2 and thus a matrix with determinant \pm 1, taking each pair from among i=1,2,k+2 (with 1 <= k <= p-3) gives us conditions
a1a2 = \pm 1 (mod p)
k*a2ak+2 = \pm 1 (mod p)
(k+1)*a1ak+2 = \pm 1 (mod p).
Combining the first two gives ka22*a1ak+2 = \pm 1, or a22 = \pm(1+1/k) (mod p).
But for k=1 this gives us a22 = \pm 2, and for k=2 we get a22 = \pm (1 + (p+1)/2) = \pm (p+3)/2, so either (p+3)/2 = 2 (mod p) or (p+3)/2 = -2 (mod p).  These imply p=1 and p=7, respectively, so already the only possible solution is p=7.  But if p=7 then k=3 gives a22 = \pm 6, which is not \pm 2 (mod 7), so that doesn't work either.  Thus a lift with n=2 can only possibly exist if p is 2 or 3.<|endoftext|>
TITLE: How do you keep your research notes organized?
QUESTION [118 upvotes]: One of the things I struggle with most in doing research is keeping my notes organized. Since research tends to do a lot of branching, keeping notes in a linear fashion seems useless to me. On the other hand, this means that I end up with several notebooks that have ideas and dead-ends everywhere. Then if I want to piece parts together, or if I eventually want to go back and re-investigate what looked like a dead-end at the time -- perhaps because I have learned some new tool -- it takes me a long time to find what I am looking for. How do other people surmount these obstacles?

REPLY [2 votes]: I use a tablet notebook running Windows 7, so my way is probably not what you're looking for. 
I take handwritten notes (my handwriting is rather neat) using Windows Journal. I can save it directly to PDF (if I need to share the notes), or save it in its original format for convenience of editing. Almost all my notes now reside on my hard disk (portability).
I use Mindjet MindManager to create a mindmap of how the notes pertaining to a particular broad topic are related (in content, and time-wise, as further notes are taken). I can easily put in direct (hyper)links to these files right in the mindmap, which is a nice feature since the file is a click away on the map. I can add more notes, web links etc. to the mindmap, as required. I can also link to preprints, journal articles, ebooks (even ebook pages/sections, as I can link to ebook bookmarks on the map) on my notebook, so the note- taking and keeping become quite dynamic, which is nice. With MindConnect, you can also sync the maps online, and even share them with others, or brainstorm with other people on a map online. 
I can easily hand-annotate PDF files using PDF Annotator. With a library of ebooks, imagine the web of information you can create right on your computer, in a mindmap. It's quite fantastic. Lots of possibilities there!<|endoftext|>
TITLE: Does "finitely presented" mean "always finitely presented"?  (Answered: Yes!)
QUESTION [61 upvotes]: Precisely, if an R-module M has a finite presentation, and Rk → M is some unrelated surjection (k finite), is the kernel necessarily also finitely generated?  

Basically I want to believe I can choose generators for M however I please, and still get a finite presentation.  I have reasons from algebraic geometry to believe this, but it seems like a very basic result, so I would like to understand it directly in terms of the commutative algebra, which I just can't seem to figure out...
(Here R is an arbitrary commutative ring, with no other hypotheses.)
Edit: All maps here are maps of R-modules.  Also, the reason this is not the same as "does finite presentation imply coherent?" is that I am only asking for finite type kernels of surjections Rk → M.  That the hypotheses assume surjectivity is a common misreading of the general definition of "coherent".
If the answer to the above is "yes", then coherent will mean "finite type, and all finite type submodules are finite presentation"

REPLY [6 votes]: Here's a further generalization of the generalization from YCor's answer.
Claim: Let $\mathcal K$ be a locally $\kappa$-presentable category containing a coequalizer diagram $A \rightrightarrows B \twoheadrightarrow C$ where $B,C$ are $\kappa$-presentable. Then there exists a coequalizer diagram $A' \rightrightarrows B \twoheadrightarrow C$ with $A'$ $\kappa$-presentable and a map $A' \to A$ making the obvious diagram commute.
It follows, for example, that if $B$ is any finitely-presentable module and $C$ is any finitely-presentable quotient, then the kernel is finitely-generated (we need not assume that $B$ is free). The result will still be most useful in "algebraic" categories $\mathcal K$ (the context YCor described already discussed) where there is a good supply of coequalizer diagrams. Nevertheless, it's true in this more general context.
Proof: Write $A = \varinjlim A_i$ as the $\kappa$-filtered colimit of $\kappa$-presentable objects. We have resulting coequalizer diagrams $A_i \rightrightarrows B \twoheadrightarrow C_i$, and $C = \varinjlim_i C_i$. In fact, the isomorphism $C = \varinjlim_i C_i$ lifts to the coslice category $\mathcal K_{B/}$. Because $C$ is $\kappa$-presentable (even regarded as an object of $\mathcal K_{B/}$), the identity map $C = C$ factors through some stage of the colimit, $C \rightarrowtail C_i \twoheadrightarrow C$ This is true in the coslice category, so that $(B \twoheadrightarrow C_i) = (B \twoheadrightarrow C \rightarrowtail C_i)$. Because this map is epi, it follows that $C \rightarrowtail C_i$ is epi. But it is also split mono, and hence iso. Thus $C = C_i$ is the coequalizer of $A_i \rightrightarrows B$.<|endoftext|>
TITLE: Non-free projective modules for a Universal Enveloping Algebra?
QUESTION [6 upvotes]: Let g be a finite dimensional Lie algebra over k, and let U be its universal enveloping Lie algebra.  Is there a left module M of U which is projective but not free?  That is, is the Quillen-Suslin theorem still true for enveloping algebras?
Quillen-Suslin says this there are no non-free projectives for S(g), the associated graded algebra of U.  Thus, if the associated graded module of a projective is projective, then it is free (and so the original module was also free).  Therefore, this question is equivalent to the question "Is the associated graded module of a projective U-module always projective?"
My guess is no, because the Weyl algebra has non-free projectives, even though it's associated graded algebra is a polynomial algebra.  However, the tricks I know that work for the Weyl algebra don't work for Lie algebras.  I would love a simple example of a non-free projective U-module.

REPLY [5 votes]: In this paper Stafford shows that whenever g is a finite-dimensional non-abelian Lie algebra the enveloping algebra has non-free but stably free (and therefore projective) right ideals. He also shows how to construct them.<|endoftext|>
TITLE: Does every morphism BG-->BH come from a homomorphism G-->H?
QUESTION [14 upvotes]: Given a homomorphism f:G→H between smooth algebraic groups, we get an induced homomorphism of algebraic stacks Bf:BG→BH, given by sending a G-torsor P over a scheme X to the H-torsor PxGH, whose (scheme-theoric) points are {(p,h)|p∈P,h∈H}/∼, where (pg,h)∼(p,f(g)h).
Is every morphism of algebraic stacks BG→BH of the form Bf? If not, what is an example of a morphism not of this form?

REPLY [4 votes]: As a complement to the answers above : it is kind of well-known (at least I thought it was) that the natural morphism 
$$\operatorname{\mathbf{Hom}}_{gr} (G,H) \to \operatorname{\mathbf{Hom}}(BG,BH)$$
is an $H$-torsor (where $H$ operates on $\operatorname{\mathbf{Hom}}_{gr} (G,H)$ by conjugacy). In other words, the internal Hom stack $\operatorname{\mathbf{Hom}}(BG,BH)$ identifies with the quotient stack
$$\left[\operatorname{\mathbf{Hom}}_{gr} (G,H) \right/H ] \simeq \operatorname{\mathbf{Hom}}(BG,BH)$$
of the Hom sheaf $\operatorname{\mathbf{Hom}}_{gr} (G,H)$ by the conjugacy action of $H$.
This of course answers the original question completely (globally, no; locally, yes).
The proof is straightforward enough. For instance here is a sketch of the local surjectivity: let $\alpha : BG\to BH$. If $s_G$ is the canonical section of $BG$, we get a morphism  $G=\operatorname{\mathbf{Aut}}(s_G)\to \operatorname{\mathbf{Aut}}(\alpha(s_G))$. Since $\alpha(s_G)$ is isomorphic to $s_H$ locally, this defines locally a morphism $G\to H$, up to conjugacy by $H=\operatorname{\mathbf{Aut}}(s_H)$ of course.
I have learnt this from Angelo Vistoli, and since this is nice and quite useful, I am happy to share it back with you. For the context: this isomorphism is a frequent guest in non-abelian cohomology, and in tannaka duality.<|endoftext|>
TITLE: Learning new mathematics
QUESTION [20 upvotes]: While many of us have had the experience of learning mathematics informally by osmosis or more formally in classes, there are times when we have to sit down and systematically learn, without the benefit of a class, large amounts of mathematics.  For instance, there might be a technique that we need from a field we are not familiar with.
When you find yourself in such a situation, what are your best tricks for teaching yourself new mathematics?

REPLY [9 votes]: There are many good answers here already, but I'll just add an observation on my own technique once I have decided on a book or paper to read. I like to read fast at first to get an idea of the lay of the land, so to speak, and keep on reading until I am lost due to not having absorbed previously introduced concepts. Then I backtrack and try to tease out the essential definitions before proceeding. Also, rather than reading proofs line by line I take a quick look and then try to fill in the details myself. That's time consuming, though, so it is reserved for proofs I really want to learn. This method is probably not suited for everybody, though. Everyone has to find their own method for absorbing new material.<|endoftext|>
TITLE: What are some examples of coarse moduli spaces?
QUESTION [11 upvotes]: It took me some effort to work out Gerashenko's nice simple example Can a singular Deligne-Mumford stack have a smooth coarse space? of a DM stack non-equisingular with its coarse moduli space, which means I must improve my understanding of coarse moduli spaces. 
What are your favourite examples of coarse moduli spaces? One per answer, please, so we can rank them.

REPLY [4 votes]: The moduli space of semi-stable vector bundles with trivial determinant over a genus $g$ curve. If the rank is 2 then the coarse space is isomorphic to $\mathbb{P}^3$!!<|endoftext|>
TITLE: How can I prove that a sequence of squares of graph norms is never cyclotomic?
QUESTION [9 upvotes]: The norm of a graph is the largest eigenvalue of the adjacency matrix. I'll write ||G|| for the norm of G.
Now, fix some graph G with a chosen vertex *, and consider the family of graphs G_k obtained by adding a chain of k edges to *.
For many such examples, the sequence {||G_k||^2}_k appears to be never cyclotomic; I'd like some ideas as to how I might try to prove such statements for particular graphs G.
I know how to show individual algebraic integers aren't cyclotomic -- modulo any prime not dividing the discriminant, the minimal polynomial of a cyclotomic integer must factor into factors with uniform degree. This approach seems very hard to make work for a family of numbers, although I'm aware of the work of Asaeda-Yasuda in which they did this for the graph

        o-o-o
       /
*-o-o-o
       \
        o-o-o

(with the exception of k=4, where the norm-square is in fact cyclotomic). If anyone has ideas about how one should attack such a question, or examples of similar problems, please let me know!
Finally -- the application here is to subfactors; Etingof-Nikshych-Ostrik proved that the index of a subfactor must be a cyclotomic integer, and the index is just the norm square of the principal graph. When we look for possible new examples of subfactors, we tend to get results constraining the principal graph to lie in such a sequence {G_k}.

REPLY [3 votes]: This is a vague thought: is there some simple recurrence for the characteristic polynomials of the charctertistic polynomials of the corresponding matrices. For example, if you look at the A_n chains, the polynomials are the Chebyshev polynomials, whose roots are cyclotomic, and which obey a simple resursion.
Even if you had a recursion, I do not know how to show that the roots are not cyclotomic, but the problem feels more tractable.<|endoftext|>
TITLE: HOMFLY and homology; also superalgebras
QUESTION [16 upvotes]: My understanding is that an analogy along the following lines is (roughly) true: 
"The Alexander polynomial is to knot Floer homology is to gl(1|1)
as the Jones polynomial is to Khovanov homology is to sl(2)
as a-lot-of-other-specializations-of-HOMFLY are to Khovanov-Rozansky homology are to sl(n)."
1) To what extent is it possible to add another line that starts with the (unspecialized) HOMFLY polynomial?  I think there is a triply-graded complex that I can put here (and that maybe this is what I should be calling Khovanov-Rozansky homology? or at least is also due to them?), but is there an analogous object to put in place of the Lie (super-)algebras appearing above?
2) Why is gl(1|1) here? That seems weird.

REPLY [10 votes]: Lots of good answers above, but the one that seems to be missing is what belongs in the third column for unspecialized HOMFLY.  (As Geordie Williamson points out, Khovanov and Rozansky wrote two related papers, one of which does the unspecialized case.)  There's a one parameter family of categories that looks like the category of representations of a Lie superalgebra, called $gl(\alpha)$.  When $\alpha$ is an integer, you can kill off the representations that have dimension 0 and get the usual category of representations of $gl(n)$.  (You can also kill off fewer representations and get the representations of $gl(n+k|k)$ for any $k$.)  Many people have written about this, but perhaps the coolest place to look is in Deligne's La série exceptionelle de groupes de Lie, C. R. Acad. Sci. Paris, Série I 322 (1996) 321—326, where he also conjectures a similar structure specializing to all the exceptional groups.  (There are also several follow-up papers.)<|endoftext|>
TITLE: Helm's improvement to Beck-Fiala theorem
QUESTION [5 upvotes]: Beck-Fiala theorem states that if $X$ is a finite set and $H$ is any family of subsets of $X$, in which every vertex occurs in at most $d$ sets of $H$, then there is a function $f\colon X\to\{1,-1\}$ such that for every set $S$ in $H$ we have $|\sum_{x\in S} f(x)|\le 2d-2$. In combinatorics parlance, one formulates this as "every hypergraph of maximal degree $d$ has discrepancy at most $2d-2$". The theorem is striking since the bound on discrepancy depends only on $d$, but not on the sizes of $X$ and $H$.
There were two papers that improve the bound of $2d-2$. The first is due to Bednarchak and Helm, which replaces $2d-2$ by $2d-3$ for $d\ge 3$. Their argument is short and sweet. The later improvement is due to Helm to $2d-4$. However, I have been unable to follow the paper. I also tried to contact the author, but I could not locate him. Has anyone been able to follow the paper, or at least understood the algorithm to find $f$ well enough to explain it in pseudo-code?

REPLY [2 votes]: I still do not understand Helm's paper, and in particular what his algorithm for finding a coloring is. I found an argument that improves on Helm's result. Some of the ideas in the new argument are similar to what Helm seems to have used. The paper is available at http://arxiv.org/abs/1306.6081. (Sorry for answering my own question with a reference to my own paper.)<|endoftext|>
TITLE: Determinant of a pullback diagram
QUESTION [9 upvotes]: Suppose that X and Y are finite sets and that f : X → Y is an arbitrary map. Let PB denote the pullback of f with itself (in the category of sets) as displayed by the commutative diagram 
PB → X 
↓      ↓
X   → Y
Terence Tao observes in one the comments on his weblog that the product of |PB| and |Y| is always greater than or equal to |X|2. (This is an application of the Cauchy-Schwarz inequality.) This fact may be rephrased as follows: If we ignore in the above diagram all arrows and replace the sets by their cardinalities we obtain a 2x2 matrix with a non-negative determinant.
The question is whether this is a general phenomenon. Suppose that n is a positive integer and that X1, X2, ... ,Xn are finite sets; furthermore we are given maps f1 : X1 → X2, f2 : X2 → X3, ... , fn-1 : Xn-1 → Xn. We construct a pullback diagram of size nxn. The diagram for n=4 is shown below.
PB → PB → PB → X1
↓       ↓       ↓     ↓ 
PB → PB → PB → X2
↓       ↓       ↓     ↓ 
PB → PB → PB → X3
↓       ↓       ↓     ↓ 
X1 →  X2 →  X3 → X4
Here, the maps between the Xi in the last row and column are the corresponding fi and the PBs denote the induced pullbacks. (Of course, although they are denoted by the same symbol, different PBs are different objects.) The PBs can be constructed recursively. First, take the pullback of X3 → X4 ← X3; it comes with maps X3 ← PB → X3. Having constructed this, take the pullback of X2 → X3 ← PB and so forth.
Ignore all arrows and replace sets by their cardinalities. Is the determinant of the resulting nxn matrix always non-negative?

REPLY [10 votes]: Write Xn = {x1, ..., xk}.  For each 1 ≤ i ≤ k let wi be the vector whose jth component is the cardinality of the inverse image of xj in Xi.  Then your matrix is the sum w1w1T + ... + wkwkT, a sum of positive semidefinite matrices, so it is positive semidefinite and in particular has nonnegative determininant.<|endoftext|>
TITLE: Orbits of real groups, canonical forms of matrices
QUESTION [9 upvotes]: There are a lot of results in textbooks concerned with canonical forms of matrices under certain complex groups of transformations, e.g. GL(n|C), O(n|C),... 
Could anybody give me references where the canonical forms of real matrices under the action of SO(p,q|R) were found. Of most interest is the canonical form of antisymmetric matrices, i.e. that of the adjoint representation. Other related results, e.g. on canonical forms under SU(p,q), SP(p,q) are also appreciated. Thanks!

REPLY [5 votes]: Another reference is: D. Djokovic, J. Patera, P. Winternitz, H. Zassenhaus, Normal forms of elements of classical real and complex Lie and Jordan algebras, J. Math. Phys. 24 (1983), N6, 1363-1374 http://dx.doi.org/10.1063/1.525868 .<|endoftext|>
TITLE: Extending Functions on Closed Submanifolds of C^n
QUESTION [5 upvotes]: Functions on an algebraic subvariety X of A^n are the same as functions on A^n restricted to X. So the statement that functions on X extend to all of A^n follows by the definition. My question is: does the analogous statement hold for C^n and closed complex submanifolds (maybe even closed analytic subvarieties), and if so, how is this proved?

REPLY [4 votes]: Yes, this is true. It follows from "Cartan's Theorem B" which says that H^1 of any coherent analytic sheaf on a closed submanifold of C^n is 0; the same result is also true for analytic subspaces. Look up any book on several complex variables for a proof. (It is quite possible that there is a more elementary proof.)
(One uses the theorem as follow: Let X be the submanifold or analytic space and consider the exact sequence of sheaves on C^n
0 --> I --> O_{C^n} --> O_X --> 0
where I is the ideal sheaf of X. The vanishing of H^1(C^n,I) implies that the map H^0(C^,O_{C^n}) to H^0(X,O_X) is surjective, which is what you want.)<|endoftext|>
TITLE: Why do zeta functions contain so much information?
QUESTION [26 upvotes]: Is there some intuitive explanation why Dedekind zeta functions contain so much information about their number field?
For example the residue at the pole $s=1$ relates several invariants of the number field. the class number formula.

REPLY [16 votes]: This question is a major open problem. One theoretical framework has been proposed that would explain why all the various zeta and L-functions (the ones attached to number fields, function fields, modular forms, algebraic varieties, Hecke characters, ...) have the same nice properties (analytic continuation, special values that "contain so much information," ...). It is known as the theory of motives. It is somewhat stymied by the fact that no one can (with proof) define the right category of motives, nor has anyone been able to for some 40 years. Milne's lecture notes "What is a motive?" are a great introduction to the theory; http://www.jmilne.org/math/xnotes/mot.html
For a particular answer to your question, the so-called Equivariant Tamagawa Number Conjecture in the theory of motives encompasses every one of the "information-packed" formulas we have about special values of L-functions. Of course, no one expects a proof any time soon (since such a proof would at once prove long outstanding questions about the category of motives, Birch and Swinnerton-Dyer's conjecture and the Stark conjectures).
Here's one more concrete idea. One important gadget in algebraic geometry is etale cohomology. Most L-functions that we know of can be defined in terms of the eigenvalues of various Frobenius maps (global and local) acting on etale cohomology groups. In many cases, by clever arguments using analogues of classical Betti cohomology theorems (Poincare duality, Lefschetz fixed point theorems, ...), we can express the results on special values of L-functions in terms of etale cohomology. In other words, the conceptual reason that L-functions defined from different objects behave so similarly to one another is that all these objects are algebro-geometric, and etale cohomology ought to be an awful lot like Betti cohomology. Of course, this is more believable for a variety over the complexes than it is for Spec Z, but it's a start.<|endoftext|>
TITLE: Infinite matrices and the concept of "determinant"
QUESTION [22 upvotes]: Suppose we have an infinite matrix A = (aij) (i, j positive integers). What is the "right" definition of determinant of such a matrix? (Or does such a notion even exist?) Of course, I don't necessarily expect every such matrix to have a determinant -- presumably there are questions of convergence -- but what should the quantity be? The problem I have is that there are several ways of looking at the determinant of a finite square matrix, and it's not clear to me what the "essence" of the determinant is.

REPLY [28 votes]: There is a class of linear operators that have a determinant.  They are, for some strange reason, known as "operators with a determinant".
For Banach spaces, the essential details go along these lines.  Fix a Banach space, X, and consider the finite rank linear operators.  That means that T: X → X is such that Im(T) is finite dimensional.  Such operators have a well-defined trace, tr(T).  Using this trace we can define a norm on the subspace of finite-rank operators.  If our operator were diagonalisable, we would define it as the sum of the absolute values of the eigenvalues (of which only finitely many are non-zero, of course).  This norm is finer than the operator norm.  We then take the closure in the space of all operators of the space of finite-rank operators with respect to this trace norm.  These operators are called trace class operators.  For such, there is a well-defined notion of a trace.
(Incidentally, these operators form a two-sided ideal in the space of all operators and are actually the dual of the space of all operators via the pairing (S,T) → tr(ST).)
Now trace and determinant are very closely linked via the forumula etr T = det eT.  This means that we can use our trace class operators to define a new class of "operators with a determinant".  The key property should be that the exponential of a trace class operator should have a determinant.  This suggests looking at the family of operators which differ from the identity by a trace class operator.  Within this, we can look at the group of units, that is invertible operators.
So an "operator with a determinant" is an invertible operator that differs from the identity by one of trace class.
For more details, I recommend the book "Trace ideals and their applications" by Barry Simon (MR541149) and the article "On the homotopy type of certain groups of operators" by Richard Palais (MR0175130).
But defining the determinant of an arbitrary operator is, of course, impossible.  One can always figure out a renormalisation for a particular operator but there just ain't gonna be a system that works for everything: obviously det(I) = 1 but then det(2I) = ?
(I should also say that I picked Banach spaces for ease of exposition.  One can generalise this to locally convex topological spaces, but that involves handling nuclear materials so caution is advised.)<|endoftext|>
TITLE: Describe a topic in one sentence.
QUESTION [50 upvotes]: When you study a topic for the first time, it can be difficult to pick up the motivations and to understand where everything is going. Once you have some experience, however, you get that good high-level view (sometimes!) What I'm looking for are good one-sentence descriptions about a topic that deliver the (or one of the) main punchlines for that topic.
For example, when I look back at linear algebra, the punchline I take away is "Any nice function you can come up with is linear." After all, multilinear functions, symmetric functions, and alternating functions are essentially just linear functions on a different vector space. Another big punchline is "Avoid bases whenever possible."
What other punchlines can you deliver for various topics/fields?

REPLY [6 votes]: Geometric representation theory: keep translating the problem until you run into Hard Lefschetz, then you are done.

REPLY [5 votes]: Etale cohomology - you can apply fixed-point theorems from algebraic topology to Galois actions on varieties.<|endoftext|>
TITLE: What can't be described by categories?
QUESTION [5 upvotes]: I've been reading some "introduction to categories" type materials and have been impressed with the all-encompassing nature, but the skeptic in me wonders: is there any mathematical object that categories can't describe? 
To be quite specific, I'd be interested any of these:
a.) Objects that can be described by categories that have properties that can't.
b.) Category equivalents of set-theoretic type limits, like how "the set of all sets" causes problems.
c.) Some type of mathematics so pathological it foils, say, associativity. It doesn't need to be a mathematics that's useful in any sense, just one designed specifically to be impossible to describe with categories.

REPLY [11 votes]: Someone (Halmos?) said that category theory can only describe the most trivial parts of any subject, and that this a valuable service because it's good to clearly mark which parts are trivial.<|endoftext|>
TITLE: is amalgamation of groups associative
QUESTION [5 upvotes]: Given groups $G_1, G_2, G_3$ and injections $A_1 \to G_1$ and $A_1 
\to G_2$ , from $A_2 \to G_2$ and $A_2 \to G_3$, let $G_1 *_{A_1} *G_2 *_{A_2} G_3$ be the amalgam formed these groups and maps. 
Then is it true that  $G_1 *_{A_1} *G_2 *_{A_2} G_3$ is the same as (G_1 *_{A_1} G_2 ) *_{A_2} G_3. If yes, how do we see this?

REPLY [7 votes]: Amalgamation of groups is a categorical construction known as a "pushout": http://en.wikipedia.org/wiki/Pushout_(category_theory)
By general category theory, pushouts are associative up to unique isomorphism, i.e., the two things you wrote are isomorphic in a unique way (subject to commuting with the inclusions from A_i, etc.)<|endoftext|>
TITLE: What are dessins d'enfants?
QUESTION [48 upvotes]: There was an observation that any algebraic curve over Q can be rationally mapped to P^1 without three points and this led Grothendieck to define a special class of these mappings, called the Children's Drawings, or, in French, Dessins d'Enfants (his quote was something like "things as simple as the drawings...").
I'm not an expert in this field, so could somebody please write more about those dessins, and  what things they are related to? What's their importance? How does the cartographic group act on these?

REPLY [21 votes]: Given a compact Riemann surface $X$, every holomorphic function on $X$ is constant. This is obvious if you think about holomorphic functions as locally conformal mappings, that is, transformations of $X$ into a plane which locally preserve angles ("infinitesimal similarities"). Such a transformation is continuous, thus has a compact image, but the image is also open, so it must be constant: collapses $X$ into a point. In other words, the ring $\mathcal{O}(X)$ of holomorphic functions on $X$ reduces to the constants $\mathcal{O}(X) = \mathbb{C}$.
That's why, in the theory of compact Riemann surfaces, one is interested in meromorphic functions, i. e., those with at least one pole. These functions constitute a field denoted by $\mathcal{M}(X)$, and can be seen as branched coverings $X \rightarrow \mathbb{S}^2$ of the Riemann sphere/complex projective line.
One can show that there always exist non-constant meromorphic functions on $X$ (Riemann's existence theorem). This means that every compact Riemann surface is a branched covering of the sphere. This can be used to show that the field $\mathcal{M}(X)$ is a finite field extension of $\mathcal{M}(\mathbb{S}^2) =\mathbb{C}(z)$ (this last field is the field of rational functions = polynomial fractions).
Moreover, one can show that each finite extension of $\mathbb{C}(z)$ gives rise to a unique compact Riemann surface, up to isomorphisms, with the given extension as its field of meromorphic functions (Dedekind-Weber theory of algebraic function fields in one variable). This compact Riemann surface can always be realized as an algebraic curve in the complex projective space $\mathbf{P}^3(\mathbb{C})$ (without singularities, naturally). This means that it is the set of zeroes of some homogeneous polynomials with complex coefficients, in projective space.
One interesting question is to ask when a compact Riemann surface $X$ can be given by an equation with coefficients in $\overline{\mathbb{Q}}$ (the algebraic closure of rationals). It is known that this is the case if and only if there is a meromorphic function $X \rightarrow \mathbb{S}^2$ with at most three critical values (Belyi's theorem). A function of this kind is a covering of the sphere which is branched over three points (or less).
Thus, the study of compact Riemann surfaces/smooth plane algebraic curves over the algebraic numbers $\overline{\mathbb{Q}}$ is reduced to the study of coverings of the sphere branched over three points, which we can assume to be the points $0, 1, \infty$.
These branched coverings $f: X \rightarrow \mathbb{S}^2$ can be given a geometric representation. The fiber of $0$, is a finite set of points in $X$, which can be marked as black points. Points in the fiber of $1$ are usually colored white. The preimage of the interval $[0,1]$ (as a curve joining $0$ and $1$) is given by a set of curves joining black and white points, alternatively. This graph on $X$, formed by black points, white points and curves, is the dessin d'enfant associated to the branched covering $f: X \rightarrow \mathbb{S}^2$.
It is a remarkable fact that the branched covering $f: X\rightarrow \mathbb{S}^2$ determines the Riemann surface structure of $X$ (by pullback of the same structure in $\mathbb{S}^2$), and so, compact Riemann surfaces over algebraic numbers (with a specified meromorphic function branched over at most three points) are just orientable compact surfaces with certain graphs in them (the graphs associated with branched coverings).
Now, Grothendieck was interested in the Galois group of $\overline{\mathbb{Q}}$ over $\mathbb{Q}$, which acts on the coefficients of the equation of an algebraic curve over $\overline{\mathbb{Q}}$, giving another curve of the same type. Considering the dessins associated to those curves, that group then transforms a dessin into another dessin, and so, dessins can be used to obtain a geometric interpretation of the absolute Galois group.<|endoftext|>
TITLE: Properties of monodromy of a fibration?
QUESTION [7 upvotes]: Sorry for a loaded question.
I'm not an expert on those things, but I do know that a fibration gives rise to the representations of pointed fundamental group of the base on the cohomology of the fiber.
What are the properties of this map for different classes of fibrations? I think it's known what the image of this map can be. And the local properties are governed, at least in the complex case, by what type the manifold is.
And, most importantly, there is something about uppertriangularity. What exactly is that?

REPLY [16 votes]: A small clarification on bhargav's answer: in algebraic geometry we only have quasi-unipotency of the local monodromy in one-parameter families (which is what bhargav is talking about); or in multi-parameter families but only near a normal crossing point of the discriminant. Global monodromies are reductive and local monodromies near bad points of the discriminant can be more general.
For concreteness look at a projective morphism $f : X \to B$, where $X$, $B$ are smooth complex projective varieties. Let $D \subset B$ be the discriminant divisor of $f$, i.e. the divisor where the differential of $f$ is not surjective. The global monodromy of the smooth fibration $f : X - f^{-1}(D) \to B - D$ is always reductive by a theorem of Borel. That is: the Zariski closure of the monodromy in the linear automorphisms of the cohomology of the marked fiber is a complex reductive group. If we take a small analytic ball $U \subset B$ centered at some point of $D$, and if we know that $D\cap U$ is a normal crossings divisor in $U$, then the monodromy of the local fibration $f : f^{-1}(U-D) \to U-D$ is quasi-unipotent as bhargav explained. Note that the normal crossings condition implies that the fundamental group of $U - D$ is abelian, so the quasi-unipotency condition makes sense here. 
If however $U\cap D$ does not have normal crossings, then $\pi_{1}(U-D)$ need not be abelian and the monodromy of $f : f^{-1}(U-D) \to U-D$ need not be quasi-unipotent. An easy example is to look at a generic projective plane $\mathbb{P}^{2}$ in the $9$ dimensional projective space of cubic curves in $\mathbb{P}^{2}$. This plane parametrizes a family of cubics which degenerates along a discriminant curve $D \subset \mathbb{P}^{2}$ and under the genericity assumption $D$ has only nodes and cusps. The cuspidal points of $D$ correspond to cuspidal cubics, and near a cusp of $D$ the local fundamental group of $U - D$ is the amalgamated product of $\mathbb{Z}/4$ and $\mathbb{Z}/6$ over $\mathbb{Z}/2$ and so is isomorphic to $SL_{2}(\mathbb{Z})$ the local monodromy representation near the cusp, i.e. the representation of $\pi_{1}(U-D,u_{0})$ to the linear automorphisms of the first integral cohomology of the cubic corresponding to $u_{0} \in U - D$, is an inclusion, i.e. has image $SL_{2}(\mathbb{Z})$. In particular it is not quasi-unipotent.<|endoftext|>
TITLE: Space of $(1,0)$-holomorphic forms on a Riemann surface
QUESTION [6 upvotes]: In a complex analysis course I have been given the following definition:
Let $X$ be a Riemann surface, denote by $H(1,0)$ the space of all $(1,0)$-holomorphic forms on $X$ and consider the quotient vector space (over $\mathbb{C}$) of $H(1,0)$ by $$\{f \in H(1,0) \mid f = d(\phi) \text{ for some } \phi \in C(X)\}.$$ The dimension of this vector space is called the genus of the surface.
Does anyone know of any good book that deals with this?
Thank you.

REPLY [9 votes]: There is the introductory graduate-level text Riemann Surfaces by Otto Forster which approaches the subject from just the angle suggested by the definition you were given. If you read French there is the book Quelques Aspects des Surfaces de Riemann by Eric Reyssat, a gentle introduction with a broad outlook. Rather more demanding is Compact Riemann Surfaces by Raghavan Narasimhan, a modern treatment that is not overly long but covers considerable ground. Actually, there are many good books on Riemann surfaces, not all from an algebraic geometry viewpoint. If you can get your hands on them, the wonderful Columbia University notes of Lipman Bers show Riemann surfaces from a complex analysis/PDE/differential geometry angle that you should not miss. They date from 1957, so inevitably some things are not there (I don't recommend them for the specific purpose you had in mind). If your taste is towards analysis, there is also Compact Riemann Surfaces by Jürgen Jost.
I think Forster's book is my best response to your question. Or perhaps even more useful if you are in a hurry; Chapter 9 of the second edition of Complex Analysis in One Variable by Narasimhan may be all you need!<|endoftext|>
TITLE: What is the Cayley projective plane?
QUESTION [46 upvotes]: One can build a projective plane from $\Bbb R^n$, $\Bbb C^n$ and $\Bbb H^n$ and is then tempted to do the same for octonions. This leads to the construction of a projective plane known as $\Bbb OP^2$, the Cayley projective plane.
What are the references for the properties of the Cayley projective plane? In particular, I would like to know its (co)homology and homotopy groups.
Also, what geometric intuition works when working with this object? Does the intuition from real projective space transfer well or does the non-associativity make a large difference? For example, I would like to know why one could have known that there is no $\Bbb OP^3$.

REPLY [6 votes]: There's a very pretty construction using the exceptional Jordan algebra of dimension 27, described in the book "On Quaternions and Octonions" by Conway and Smith, as well as in Baez's article on the Octonions (see http://math.ucr.edu/home/baez/octonions/node12.html). To summarize, you take Hermitian 3 by 3 matrices over the Octonions, with Jordan product $A \circ B = (AB + BA)/2$. That gives you the exceptional Jordan algebra. If you further restrict to matrices which are of unit trace and idempotent, you get $\mathbb{OP}^2$. You say point $P$ lies on line $L$ if $P \circ L = 0$.
As some of the answers above have pointed out, the usual construction of higher dimensional projective spaces doesn't work because the Desargues theorem holds automatically in these (see Courant and Robbins, "What is Mathematics", pg. 171, for a nice illustration and quick proof), and would imply that $\mathbb{O}$ is associative, which it isn't. Note that the plane $\mathbb{OP}^2$ is non-Desarguian.<|endoftext|>
TITLE: What are some reasonable-sounding statements that are independent of ZFC?
QUESTION [272 upvotes]: Every now and then, somebody will tell me about a question. When I start thinking about it, they say, "actually, it's undecidable in ZFC."
For example, suppose $A$ is an abelian group such that every short exact sequence of abelian groups $0\to\mathbb Z\to B\to A\to0$ splits. Does it follow that $A$ is free? This is known as Whitehead's Problem, and it's undecidable in ZFC.
What are some other statements that aren't directly set-theoretic, and you'd think that playing with them for a week would produce a proof or counterexample, but they turn out to be undecidable? One answer per post, please, and include a reference if possible.

REPLY [4 votes]: "There exists a complete metric space $(X,d)$ and a Borel probability measure $\mu$ on $X$ with non-separable support."
This has been discussed elsewhere on MO (I forget where, sorry), but is shown to require a large cardinal axiom in Fremlin's Measure Theory, section 438.<|endoftext|>
TITLE: Bimodules in geometry
QUESTION [27 upvotes]: Grothendieck's approach to algebraic geometry in particular tells us to treat all rings as rings of functions
on some sort of space.  This can also be applied outside of scheme theory (e.g., Gelfand-Neumark theorem
says that the category of measurable spaces is contravariantly equivalent to the category
of commutative von Neumann algebras).  Even though we do not have a complete geometric description
for the noncommutative case, we can still use geometric intuition from the commutative case effectively.
A generalization of this idea is given by Grothendieck's relative point of view, which says that a morphism of rings f: A → B
should be regarded geometrically as a bundle of spaces with the total space Spec B fibered over the base space Spec A
and all notions defined for individual spaces should be generalized to such bundles fiberwise.
For example, for von Neumann algebras we have operator valued weights, relative L^p-spaces etc.,
which generalize the usual notions of weight, noncommutative L^p-space etc.
In noncommutative geometry this point of view is further generalized to bimodules.
A morphism f: A → B can be interpreted as an A-B-bimodule B, with the right action of B given by the multiplication
and the left action of A given by f.
Geometrically, an A-B-bimodule is like a vector bundle over the product of Spec A and Spec B.
If a bimodule comes from a morphism f, then it looks like a trivial line bundle with the support being
equal to the graph of f.
In particular, the identity morphism corresponds to the trivial line bundle over the diagonal.
For the case of commutative von Neumann algebras all of the above can be made fully rigorous
using an appropriate monoidal category of von Neumann algebras.
This bimodule point of view is extremely fruitful in noncommutative geometry (think of Jones' index,
Connes' correspondences etc.)
However, I have never seen bimodules in other branches of geometry (scheme theory,
smooth manifolds, holomorphic manifolds, topology etc.) used to the same extent as they are used
in noncommutative geometry.
Can anybody state some interesting theorems (or theories) involving bimodules in such a setting?
Or just give some references to interesting papers?
Or if the above sentences refer to the empty set, provide an explanation of this fact?

REPLY [4 votes]: The paper
Adam Nyman. The Eilenberg-Watts theorem over schemes, available at arXiv
studies the connection between cocontinuous functors $Qcoh(Y) \to Qcoh(X)$, which are there called bimodules, and $Qcoh(X \times Y)$ in detail.<|endoftext|>
TITLE: Why is the Hodge class of \bar{M_g} big and nef?
QUESTION [8 upvotes]: Let pi: \bar{Mg,1} \to \bar{M_g} be natural projection of compactified moduli stacks of curves and let omega be the relative dualizing sheaf. Then the Hodge class \lambda of \bar{M_g} is the first chern class of the pushforward \pi_*(ω). Among other things the hodge class, together with the boundary divisors, freely generates the Picard group of \bar{M_g}.

Question: Why is lambda big and nef?

REPLY [5 votes]: Some multiple of lambda is defined on the coarse moduli space and this is the pullback of an ample bundle on \bar{A_g}, the Satake-Baily-Borel compactification of A_g. Since \bar{M_g} maps birationally onto its image in \bar{A_g}, it follows that lambda is nef and big, in fact also semi-ample (some multiple is base point free) on the coarse moduli space. 
(The map to \bar{A_g } contracts the boundary divisor corresponding to irreducible nodal curves so lambda is not ample.)<|endoftext|>
TITLE: Separable and finitely generated projective but not Frobenius?
QUESTION [7 upvotes]: Let R be a commutative ring, and $A$ an $R$-algebra (possibly non-commutative). Then $A$ is separable if it is finitely generated (f.g.) projective as an $(A \otimes_R A^{\mathrm{op}})$-algebra. Suppose further that $A$ is f. g. projective as an $R$-module. Does this imply that $A$ is a (symmetric) Frobenius algebra?
There are lots of equivalent definitions of a Frobenius algebra. One (assuming $A$ is a f.g. projective R-module) is that there exists an $R$-linear map $\mathrm{tr}: A\to R$, such that $b(x,y) := \mathrm{tr}(ab)$ is a non-degenerate.
I know that the answer is yes when $R$ is a field. What about other rings?
I am not an expert on algebras, but this question is related to understanding obstructions for extended TQFTs, and so I am very interested in knowing anything I can about it.

REPLY [4 votes]: Theorem 4.2 of On separable algebras over a commutative ring says that the answer is always yes.<|endoftext|>
TITLE: Maximal localizations of von Neumann algebras
QUESTION [9 upvotes]: Suppose M is a von Neumann algebra.
Denote by L its maximal noncommutative localization,
i.e., the Ore localization with respect to the set of all left and right regular elements,
i.e., elements whose left and right support equals 1.
Denote by A the set of all closed unbounded operators with dense domain
affiliated with the standard representation of M on a Hilbert space, i.e., L^2(M),
also known as the standard form of M.
Von Neumann proved that if M is finite, then L and A are canonically isomorphic.
What can we say about the relationship of L and A when M has type III?
I am also interested in the properly infinite semifinite case.

REPLY [4 votes]: I think the question is not well-posed or has a negative answer.
One first has to deal with the question whether the left-right-regular elements satisfy the Ore condition, or equivalently, we have to ask: Can we find common denominators? If one is not in the finite case, this is not possible. 
For $B(H)$, let us take injective bounded operators $T$ and $S$ such that the images are dense but intersect only in the zero vector. In order to find a common denominator, we need to find an operator $R$ (bounded, injective, dense image) and bounded operators $X$ and $Y$ such that $R = TX$ and $R= SY$. This cannot possibly work since $T$ and $S$ have disjoint image.
Since $B(H)$ sits inside any type $III$-factor, no Ore localization in the above sense exists.<|endoftext|>
TITLE: Does the cohomology ring of a simply-connected space X determine the cohomology groups of ΩX?
QUESTION [19 upvotes]: One could try to apply the Eilenberg-Moore spectral sequence to the pullback diagram • → X ← •, obtaining a spectral sequence TorH•(X, R)(R, R) => H•(ΩX, R), but could there be differentials or extension problems which differ for different spaces X with the same cohomology ring?

REPLY [8 votes]: Complementing the other answers in this thread: while the cohomology ring of a simply connected space does not determine the cohomology of the loop space, the rational cohomology viewed as an $A_\infty$-algebra does.
Namely, the cohomology of any $A_\infty$-algebra $A$ over $\mathbf{Q}$ (in particular, of any differential graded algebra) carries an $A_\infty$-structure such that there is an $A_\infty$ map $H^\ast(A)\to A$ inducing the identity in cohomology; this $A_\infty$ structure is unique up to a non-unique isomorphism. See e.g. Keller, Introduction to A-infinity algebras and modules, 3.3 and references therein. By taking $A$ to be the rational singular cochains of a topological space $X$ we get an $A_\infty$-structure on $H^\ast(X,\mathbf{Q})$.
To each $A_\infty$ algebra $H$ there corresponds a bar construction, which is a free differential coalgebra on $H$ shifted by 1 to the left (see e.g. 3.6 of Keller's paper mentioned above). It is an old result of Kadeishvili (see MR0580645) the that if $H$ is the cohomology of a simply-connected space $X$ with the above $A_\infty$-structure, then the cohomology of the bar construction is the cohomology of $\Omega (X)$.
This also explains why we should expect a negative answer to the question as it is stated: all components of the $A_\infty$ structure on the cohomology participate in the bar construction, and not just the product.<|endoftext|>
TITLE: Dyck paths on rectangles
QUESTION [15 upvotes]: The number of Dyck paths in a square is well-known to equal the catalan numbers:
http://mathworld.wolfram.com/DyckPath.html
But what if, instead of a square, we ask the same question with a rectangle? If one of its sides is a multiple of the other, then again there is a nice formula for the number of paths below the diagonal, but is there a nice formula in general? What is the number of paths from the lower-left corner of a rectangle with side lengths a and b to its upper-right corner staying below the diagonal (except for its endpoint)? I am also interested in asymptotics.

REPLY [10 votes]: Since that Mirko Visontai told me that the answer is ${a+b\choose a}/(a+b)$ if $\gcd(a,b)=1$. The proof is the following (with k=a and l=b):
The number of 0--1 vectors with $k$ 0's and $l$ 1's is ${k+l\choose k}$, so we have to prove that out of these vectors exactly $1/(k+l)$ fraction is an element of $L(k,l)$. The set of all vectors can be partitioned into equivalence classes. Two vectors $p$ and $q$ are equivalent if there is a cyclic shift that maps one into the other, i.e., if for some $j$, $p_i = q_{i+j}$ for all $i$. We will prove that exactly one element from each equivalence class will be in $L(k,l)$. This proves the statement as each class consists of $k+l$ elements because $gcd(k,k+l)=1$.
We can view each 0--1 sequence as a walk on $\mathbb R$ where each 0 is a $-l/(k+l)$ step and each 1 is a $+k/(k+l)$ step. Each $(k,l)$ walk starts and ends at zero and each walk reaches its maximum height exactly once, otherwise $ak + bl = 0$ for some $0 < a +b < k+l$ which would imply $\gcd(k,l) \neq 1$.
If we take the cyclic shift that ``starts from the top'', we stay in the negative region throughout the walk, which corresponds to remaining under the diagonal in the lattice path case. Any other cyclic shift goes above zero, which corresponds to going above the diagonal at some point.<|endoftext|>
TITLE: Hopf Algebra Reference
QUESTION [17 upvotes]: I was talking this morning to a colleague who thinks about combinatorial Hopf algebras. He mentioned several rings, which are of interest in combinatorics, for which he didn't know whether a Hopf structure existed. I was able to rule out several by the following result:
If A is a finitely generated commutative algebra over a field of characteristic 0, and A has a Hopf structure, then A is a regular ring.
So, two questions:
(1) The only reference I know for this is Tate's article on group schemes in "Modular Forms and Fermat's Last Theorem." Does anyone know a version which targeted towards a reader who likes algebra better than geometry? (So, for example, "Hopf algebra" is a friendlier term than "group scheme".)
(2) Are there any useful generalizations that take out "commutative" or "finitely generated"?

REPLY [6 votes]: Oort has an elementary proof that group schemes in char. 0 are reduced -- see MR0206005.<|endoftext|>
TITLE: Is there a complex structure on the 6-sphere?
QUESTION [83 upvotes]: I don't know who first asked this question, but it's a question that I think many differential and complex geometers have tried to answer because it sounds so simple and fundamental. There are even a number of published proofs that are not taken seriously, even though nobody seems to know exactly why they are wrong.
The latest published proof to the affirmative: http://arxiv.org/abs/math/0505634 Even though the preprint is old it was just published in Journ. Math. Phys. 56, 043508-1-043508-21 (2015)

REPLY [18 votes]: There was a workshop on this problem at Univ. Marburg, in March 2017:
https://www.mathematik.uni-marburg.de/~agricola/Hopf2017/
It resulted in a special issue of the Journal of Differential Geometry and its Applications (April 2018):
https://www.sciencedirect.com/journal/differential-geometry-and-its-applications/vol/57/suppl/C
Most of the papers from that issue are on the ArXiv.  The introductory one is a good historical overview: 

Ilka Agricola, Giovanni Bazzoni, Oliver Goertsches, Panagiotis Konstantis, Sönke Rollenske, On the history of the Hopf problem, arXiv:1708.01068

The pdf of that paper has arxiv links to most of the other papers (or see below).  It mentions the papers by Etesi and Atiyah, saying about them that "the community of experts does not seem to find unity" and linking these MO threads.
By all appearances though, the problem is still open ;-).

The rest of the papers are:

Panagiotis Konstantis, Maurizio Parton, Almost complex structures on spheres, arXiv:1707.03883
Cristina Draper, Notes on $G_2$: The Lie algebra and the Lie group, arXiv:1704.07819
Ilka Agricola, Aleksandra Borówka, Thomas Friedrich, $S^6$ and the geometry of nearly Kähler $6$-manifolds, arXiv:1707.08591
Ana Cristina Ferreira, Non-existence of orthogonal complex structures on the round 6-sphere, arXiv:1906.02062
Boris Kruglikov, Non-existence of orthogonal complex structures on 6-sphere with a metric close to the round one, arXiv:1708.07297
Daniele Angella, Hodge numbers of a hypothetical complex structure on $S^6$, arXiv:1705.10518
Christian Lehn, Sönke Rollenske, Caren Schinko, The complex geometry of a hypothetical complex structure on $S^6$, ResearchGate (request only)
Aleksy Tralle, Markus Upmeier, Chern's contribution to the Hopf problem: an exposition based on Bryant's paper, arXiv:1708.02904<|endoftext|>
TITLE: Is a smooth closed surface in Euclidean 3-space rigid?
QUESTION [14 upvotes]: Classical theorem of Cohn-Vossen: A closed convex surface in Euclidean 3-space cannot be deformed isometrically.
Robert Connelly found an example of a polyhedral surface that can be deformed isometrically. A metal hinged model of it can be found at IHES.
But what about an arbitrary not-necessarily-convex smooth closed surface? Is it necessarily rigid? Or maybe it might be possible to make a smooth version of Connelly's example? It's easy to make smooth "hinges". The real challenge is finding a smooth model of the vertices, which is where two or more hinges meet.

REPLY [3 votes]: Apparently the question is still open for smooth enough surfaces and deformations (that is, at least $C^2$). 
Mike Anderson wrote a preprint claiming to prove local rigidity of smooth enough surfaces, but it was later withdrawn. 
Idjad Sabitov and his collaborators have been working on this question, developing for instance a theory of higher-order isometric deformations, see e.g. Sabitov, I. Kh. Local theory of bendings of surfaces [MR1039820 (91c:53004)]. Geometry, III, 179–256, Encyclopaedia Math. Sci., 48, Springer, Berlin, 1992. He conjectures that local rigidity holds for analytic surfaces.<|endoftext|>
TITLE: Why is the gradient normal?
QUESTION [80 upvotes]: This is a somewhat long discussion so please bear with me.  There is a theorem that I have always been curious about from an intuitive standpoint and that has been glossed over in most textbooks I have read.  Quoting Wikipedia, the theorem is:

The gradient of a function at a point is perpendicular to the level set of $f$ at that point.

I understand the Wikipedia article's proof, which is the standard way of looking at things, but I see the proof as somewhat magical.  It gives a symbolic reason for why the theorem is true without giving much geometric intuition.
The gradient gives the direction of largest increase so it sort of makes sense that a curve that is perpendicular would be constant.  Alas, this seems to be backwards reasoning.  Having already noticed that the gradient is the direction of greatest increase, we can deduce that going in a direction perpendicular to it would be the slowest increase.  But we can't really reason that this slowest increase is zero nor can we argue that going in a direction perpendicular to a constant direction would give us a direction of greatest increase.
I would also appreciate some connection of this intuition to Lagrange multipliers which is another somewhat magical theorem for me.  I understand it because the algebra works out but what's going on geometrically?
Finally, what does this say intuitively about the generalization where we are looking to: maximize $f(x,y)$ where $g(x,y) > c$.
I have always struggled to find the correct internal model that would encapsulate these ideas.

REPLY [3 votes]: Imagine two surfaces: $S_1 : f(x,y,z) = c_1$ and $S_2 : f(x,y,z) = c_2$ where $c_1 < c_2 $. Now let's see what happens in a region $ x \in (a,a+\Delta x),y \in (b,b+\Delta y) $ in which $S_1$ is approximately flat. Let's take the perpendicular to $S_1$ lines that extend from the corner points of $S_1$ to $S_2$. In the region that these lines define, $S_2$ is going to have a point $P_{min}$ with minimal distance from $S_1$ and a point $P_{max}$ with maximal distance from $S_1$. Let's define as $D$ the difference between their respective distances from $S_1$. Now we can arbitrarily choose a $D' < D$ and a surface of the same form will exist between $S_1$ and $S_2$ with a point $P'_{max}$ with maximal distance $D'$ from $S_1$. This is allowed because $f(x,y,z)$ is continuous and the surfaces cannot intersect. Let $P'_{min}$ be the point with minimal distance from $S_1$. Then the difference of their respective 
distances will be less than $D'$. If we choose a $D'$ small enough the resulting surface $S' : f(x,y,z) = c'$ will be approximately flat and 
parallel to $S_1$ in the specified region. Now the gradient $\nabla f$ points to the direction of the biggest increase of $f(x,y,z)$. Inside the region, the shortest path from any point on $S_1$ to $S'$ is through the perpendicular line. Since the same will happen for all surfaces between $S_1$ and $S'$, the gradient will be perpendicular to the surface.<|endoftext|>
TITLE: Regulators of Number fields and Elliptic Curves
QUESTION [19 upvotes]: There is supposed to be a strong analogy between the arithmetic of number fields and the arithmetic of elliptic curves.  One facet of this analogy is given by the class number formula for the leading term of the Dedekind Zeta function of K on the one hand and the conjectural formula for the leading term of the L function associated to an elliptic curve defined over a number field on the other.  One can look at the terms that show up in these two expressions and more or less 'pair' them off as analogous quantities.  
One of these pairs consists of the respective regulators.  The regulator of a number field K is defined by taking a basis for the free part of the units of the ring of integers and then using the embeddings of K into C, taking logs of absolute values, etc.  The regulator of an elliptic curve is defined by taking a basis for the free part of the K points of the curve and then computing the determinant of a symmetric matrix built out of this basis using a height pairing.  
My question is: is there some way to view the number field regulator as coming from some kind of symmetric pairing on the units of K?  Alternatively, just give some reasoning why these constructions appear so different.

REPLY [16 votes]: Let me first give you a heuristic "reason", why the regulator in the class number formula looks different from the regulator in the Birch and Swinnerton-Dyer conjecture. It is often more convenient (and more canonical) to combine the regulator and the torsion term in the Birch and Swinnerton-Dyer conjecture: one chooses a free subgroup $A$ of the Mordell-Weil group of $E(K)$ with finite index and looks at the quantity $R(A)/[E(K):A]^2$, where $R(A)$ is the absolute value of the determinant of the Néron-Tate height pairing on a basis of $A$. As you can easily check, the square in the denominator insures that this quantity is independent of the choice of $A$. Now, in the class number formula, the torsion term is replaced by the number of roots of unity in $K$ and that term is not squared. So for such a canonical formulation to be possible, it is reasonable to expect that not the regulator of the number field should be defined through a symmetric pairing on the units, but its square. Then you could make the same definition as in the elliptic curves case, and it would be independent of the choice of finite index subgroup.
Now, that we have established this, there are several ways to bring the two situations closer together.
You could do the naïve thing: take the matrix $M=(\log|u_i|_{v_j})$, where $u_i$ runs over a basis of the free part of the units (or more generally $S$-units for any set of places $S$ which includes the Archimedean ones), and where $v_j$ runs over all but one Archimedean place (or more generally all but one place in $S$), v_0, say. The absolute values have to be suitably normalised (see e.g. Tate's book on Stark's conjecture). Now take the symmetric matrix $MM^{tr}$ and define this to be the matrix of a new pairing on the units. In other words, you would define your symmetric pairing as
$$
(u_1,u_2) = \sum_{v\in S\backslash\{v_0\}} \log|u_1|_v\log|u_2|_v.
$$
Then, it's clear that you have a symmetric pairing and that the determinant of that pairing with respect to any basis on the free part of the units is $R(K)^2$. As I mentioned above, the square was expected.
Depending on what you want to do, this pairing might not be the best one to consider. For example if now $F/K$ is a finite extension and you consider the analogous pairing on the $S$-units of $F$ and restrict it to $K$, then it's not clear how to compare it to the pairing on $K$. In the elliptic curves case by contrast, the former is $[F:K]$ times the latter. To fix this, we can make the following definition:
$$
(u_1,u_2) = \sum_{v\in S} \frac{1}{e_vf_v}\log|u_1|_v\log|u_2|_v,
$$
where $e$ and $f$ are the absolute ramification index and residue field degree respectively. With this definition, the compatibility upon restriction to subfields is the same as in the elliptic curves case. On the other hand, the relationship with the actual regulator is slightly less obvious. It is however quite easy to show (and I have done it in http://arxiv.org/abs/0904.2416, Lemma 2.12, in case you are interested in the details) that the determinant of this pairing is given by
$$
\frac{\sum e_vf_v}{\prod e_vf_v}R(K)^2,
$$
with both the sum and the product again ranging over the places in $S$.
So I guess, the moral is that one shouldn't seek analogies between the BSD-formula and the class number formula but rather between the BSD and the square of the class number formula (note that also sha, which is supposed to be the elliptic curve analogue of the class number, has square order whenever it is finite). There is also a corresponding heuristic on the analytic side.<|endoftext|>
TITLE: p-power roots of unity in local fields
QUESTION [8 upvotes]: Let $K$ be a number field and suppose $K$ contains no $p$-power roots of unity. Let $\mathcal{P}$ be a prime of $K$ above the rational prime $p$. Can someone prove or disprove the assertion that the local field $K_{\mathcal{P}}$ will contain no $p$-power roots of unity?

REPLY [6 votes]: Another counterexample, along the same lines as the one given by the other David S. but perhaps more "standard", is that $\mathbb{Q}_p(\zeta_p) = \mathbb{Q}_p((-p)^{1/(p-1)})$; so $K = \mathbb{Q}((-p)^{1/(p-1)})$ will do.  As "unknown" has mentioned, Krasner's lemma explains why you would expect this to be false.<|endoftext|>
TITLE: If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a nice proof of this?
QUESTION [80 upvotes]: There is a standard problem in elementary probability that goes as follows.  Consider a stick of length 1.  Pick two points uniformly at random on the stick, and break the stick at those points.  What is the probability that the three segments obtained in this way form a triangle?
Of course this is the probability that no one of the short sticks is longer than 1/2.  This probability turns out to be 1/4.  See, for example, problem 5 in these homework solutions.
It feels like there should be a nice symmetry-based argument for this answer, but I can't figure it out.  I remember seeing once a solution to this problem where the two endpoints of the interval were joined to form a circle, but I can't reconstruct it.  Can anybody help?

REPLY [6 votes]: Here's how I explained it in my blog post a few years ago, as a step in solving a related problem:
If we break the stick at two random points, $x$ and $y$, the three resulting pieces will have lengths $x$, $(y - x)$, and $(1 - y)$ if $x$ is to the left of $y$

and $y$, $(x - y)$, and $(1 - x)$ if $x$ is to the right of $y$

The three pieces form a triangle if none of the pieces is greater than half the length of the stick. In other words, if
$(y > 1/2) AND (x < 1/2) AND (y - x) < 1/2$
when point $x$ is to the left of point $y$ (first image above), and
$(x > 1/2) AND (y < 1/2) AND (x - y) < 1/2$
when $x$ is to the right of $y$ (second image above). If we plot all six of these inequalities we get the area that represents the proportion of triangles formed from our broken stick.

The shaded regions representing the conditions that form a triangle add up to 1/4 of the total area, or a 0.25 probability of forming a triangle.<|endoftext|>
TITLE: Can the Category of Schemes be Concretized?
QUESTION [12 upvotes]: If not, are there any interesting subcategories that can be concertized? If I am not mistaken, the category of reduced finite type varieties over the complex numbers would be an example, where the forgetful functor to sets would be given by looking at the underlying map of points.

REPLY [22 votes]: Apparently, there is an abstract nonsense argument that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof. 
We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.<|endoftext|>
TITLE: Definition and meaning of the conductor of an elliptic curve
QUESTION [44 upvotes]: I never really understood the definition of the conductor of an elliptic curve.
What I understand is that for an elliptic curve E over ℚ, End(E) is going to be (isomorphic to) ℤ or an order in a imaginary quadratic field ℚ(√(-d)), and that this order is uniquely determined by an integer f, the conductor, so that End(E) ≅ ℤ + f Oℚ(√(-d)) (where O just means ring of integers).
However I feel that this is not very convenient; this definition does not say anything about elliptic curves without complex multiplication.
The other definition I have come across gives the conductor as the product of primes at which the elliptic curve does not have good reduction:
N = ∏ pfp
where fp = 0 if E has good reduction at p, fp = 1 if the reduction is multiplicative, fp = 2 if it is additive and p ≠ 2 or 3, and fp = 2 + δ if p = 2 or 3, where δ is some (seemingly complicated) measure of how bad the reduction is.
I've never been able to make much sense of the second definition, nor have I seen any relation with the first. How did the idea initially appear, and why is this particular definition more useful (or "natural") than other similar definitions?

REPLY [41 votes]: Let me complete a little bit the story. The conductor mesures the ramification of the Galois group of the local field on the Tate module of the elliptic curves. The formal definition is given in Serre's book as said Jordan, in Buhler's text in the link given by Rob, and also in Sliverman's second volume on elliptic curves. 
The conductor is a non-negative integer involved in the $L$-function of $E$. Its $p$-part $f_p$ vanishes iff the Tate module is unramified (the inertia group of ${\mathbb Q}_p$ acts trivially). Otherwise it has two part, a tame one, which depends solely on the reduction type of the Néron model of $E$ (I think this is proved in Serre-Tate's paper ''Good reduction of Abelian varieties''). 
The second part (the wild one) in the conductor is the Swan conductor. It is the most headhach one. It vanishes if and only if the $p$-Sylow acts trivially on the Tate module. In very simple cases, it can be computed directly. In general, it is related to the invariants of $E$ given by Tate's algorithm: the conductor $f_p$ is given by Ogg's formula:
$$ f_p=\nu_p(\Delta) - n +1 $$
where $\Delta$ is the discriminant of a minimal Weierstrass equation of $E$, and $n$ is the number of geometric irreducible components of the fiber at $p$ of the minimal regular projective model of $E$ over $\mathbb Z$ (the fiber at $p$ is a projective, possibly reducible curve over $\mathbb F_p$, when $n$ is computed over the algebraic closure of $\mathbb F_p$). In Buhler's text, ''geometric'' is missing. 
Tate's algorithm gives $\Delta$ and $n$ and computer can find them very quickly. So everybody is happy. 
But, Ogg's formula, stated in his late 60's paper, was not fully proved. He checked the equality by case by case analysis. In residue characteristic 2, he said ''for the sake of simplicity, we will work in equal characteristic'' ! We know that equal characteristic is kind of limit of mixed characteristic (when the absolute ramification index tends to infinity), of course, this hypothesis simplifies a lot the computation, but does not give any crew for the mixed characteristic case (e.g. $\mathbb Q_p$). While this formula was widely used in computer programs, and often used as a definition of the conductor (!), some people were awared of the incompleteness of the proof. For example, Serre said this in seminars. This was also pointed out in the paper of Lockhart, Rosen and Silverman bounding conductor of abelian varieties (J. Alg. Geometry).  
This situation is repaired in 1988 in a magistral paper of Takeshi Saito. Let $R$ be a d.v.r. with perfect residue field, let $C$ be a projective smooth and geometrically connected curve of positive genus over the field of fractions of $R$ and let $X$ be the minimal regular projective model of $C$ over $R$. One defines the Artin conductor ${\rm Art}(X/R)$ which turns out to be $f+n-1$ with the same meaning as above ($f$ is the conductor associated to the Jacobian of $C$). Saito proved that 
$${\rm Art}(X/R)=\nu(\Delta)$$
where $\Delta\in R$ is the ''discriminant'' of $X$ which mesures the defect of a functorial isomorphism which involves powers of the relative dualizing sheaf of $X/R$. When $C$ is an elliptic curve, one can prove that $\Delta$ is actually the discriminant of a minimal Weierstrass equation over $R$, and le tour est joué ! This paper of Saito was apparently not very known by the number theorists. Some more details are given in a text (in French). 
So Ogg's formula should be called Ogg-Saito's formula. That some people do.<|endoftext|>
TITLE: Does the fiber product of two normal varieties remain normal?
QUESTION [8 upvotes]: Suppose $k$ is an algebraically closed field, and $X$, $Y$ are two normal varieties over $k$. Is the product $X \times Y$ necessarily still normal?

REPLY [9 votes]: The answer is yes.
In general one can define a normal morphism of schemes $f:X \rightarrow Y$ to be a flat morphism such that for every $y \in Y$ the fibre over $y$ is geometrically normal.
Then we have the following theorem on normality and base change (see EGA Ch 2 IV 6.14.1)
Let $g: Y' \rightarrow Y$ be a normal morphism of locally noetherian schemes. Then for every normal $Y$-scheme $X$ the fibre product $X \times_Y Y'$ is normal.
Over an algebraically closed field flatness and geometric normality reduce to just being normal so the result follows.<|endoftext|>
TITLE: Why are functional equations important?
QUESTION [39 upvotes]: People who talk about things like modular forms and zeta functions put a lot of emphasis on the existence and form of functional equations, but I've never seen them used as anything other than a technical tool.  Is there a conceptual reason we want these functional equations around?  Have I just not seen enough of the theory?

REPLY [7 votes]: It seems worth mentioning another application of the functional equations.  To fix ideas, I will suppose that $E$ is an elliptic curve over ${\mathbb Q}$, and let $L(E,s)$ be the $L$-function of $E$.  There is then a functional equation relating $L(E,s)$ and $L(E,2-s)$;
one feature of this functional equation is that it has a sign, i.e. has the form
$L(E,s-2) = \pm (\text{ a positive constant })L(E,s),$ where $\pm$ is some fixed sign, depending on $E$, and easily computed (say using the modular form corresponding to $E$).
Thus one find that the order of vanishing of $L(E,s)$ at $s = 1$ is odd (even) precisely
if the sign is $-1$ ($+1$).  
Since the order of vanishing is conjectured to coincide with the rank of $E({\mathbb Q})$
(the BSD conjecture), this is pretty important arithmetic information which is obtained pretty easily from the functional equation.  In particular, if the sign is $-1$, we always expect there to be a rational point of infinite order, and arranging things so that the sign is $-1$ (by twisting by a well-chosen character, say) is a common way of forcing the existence of rational points in various situations.  (There is a large body of research focused around this expected relationship between signs of functional equations and the
existence of rational points on elliptic curves; for example, it lies at the heart of the study of Heegner points by Gross and Zagier and Kolyvagin. Some important recent contributions are by Nekovar, Mazur and Rubin, and T. and V. Dokchitser.)<|endoftext|>
TITLE: How do I describe a fusion category given a subfactor?
QUESTION [14 upvotes]: I felt like following up on Kate's question.  There were some good motivational answers there.
Given a pair of factors M < N, there is a standard way to construct a 2-category whose objects are M and N, whose morphism categories are the categories of bimodules, and whose composition is described by some kind of Connes product.  If I restrict to the endomorphism category of M, I get a monoidal category structure, but I don't know how to say anything about it.  Here's a barrage of questions:

When people talk about fusion categories coming from subfactors, are they referring to the endomorphism category of one of the factors?
How are the endomorphism categories of M and N related?  Are they equivalent?  Are they Koszul dual?
Does the Jones index say something concrete about the category, like Frobenius-Perron dimension?  (How does one compute Jones index, anyway?)
How do people go about constructing exotic subfactors?  Do they just arise in nature? I'm totally okay with pointers to references here.
(bonus) I should get a braided tensor structure from a net of factors on a circle.  Is this the center of the fusion category, and is it in the literature?

Edit: Based on the (fantastically illuminating) responses, it seems that my bonus question doesn't make sense, because the M-M bimodule fusion category depends on the choice of N in an essential way.  Maybe the phrase "conformal defect" should be used somewhere.  If I come up with a suitable replacement, I'll present it as a separate question.

REPLY [2 votes]: There are lots of great answers above, I'm just going to try for as succinct of answers as I can to all the questions.

In the II_1 case, the category is the subcategory of the category of bimodules over N tensor generated by M.  (Similarly you get another category by reversing the roles of N and M).  For III_1 factors you can use endomorphisms instead of bimodules.  This tensor category is fusion iff the subfactor is finite depth.  You can consider either algebraic bimodules or L^2 bimodules and you end up with the same category by a preprint of Vaughan's.
Morita equivalent with the equivalence given by an appropriate bimodule category of N-M bimodules.
The Jones index measures the FP dimension not of the whole category, but of your particular favorite object M.
There are no deeply satisfying constructions yet.  All the constructions are by hands and somewhat messy.  The descriptions in Emily Peters thesis and related papers at least give a nice description of the category by generators and relations, but the proof that they exist is still very computational.  Everyone would love if someone managed to give a really conceptual construction of one of these beasts.<|endoftext|>
TITLE: Non-finitely generated ring of regular functions
QUESTION [11 upvotes]: It is remarked in Shafarevich's Basic Algebraic Geometry 1 that Rees and Nagata constructed examples of quasiprojective varieties such that the ring of regular functions is not finitely generated, but I cannot find the source he is referring to. Can anyone give such examples here? Does that mean we can't really say anything about the ring of regular functions of a quasi-projective variety?

REPLY [5 votes]: "Does that mean we can't really say anything about the ring of regular functions of a quasi-projective variety?" 
Since every variety contains an open affine, the ring of regular functions is always a subring of a finitely generated ring. (I assume that you consider varieties to be integral.) This is a nontrivial restriction. Also, the ring of regular functions will be noetherian, since any infinite ascending chain of ideals would give an infinite descending chain of subschemes. Wrong, see below.<|endoftext|>
TITLE: Surjective maps given by words, redux
QUESTION [8 upvotes]: I asked some time ago:
Let $w(X,Y)$ be a word in $X$ and $Y$ (i.e., an element in the free group on $X$ and $Y$). Let the variables $x$ and $y$ now range among elements of $SL_n(K)$, $K$ an algebraically closed field.
For which $w$ is it the case that, for $x$ generic, the image of the map given by
$y \rightarrow w(x,y)$
is not Zariski-dense?
It seems that the map has Zariski-dense image for most w one can try. At the same time, as I said back when I first asked the question, the image of the map $y \rightarrow w(x,y)$ is not Zariski-dense for $w(x,y) = y x y^{-1}:$ the image of the map is contained in the conjugacy class of $x$.
By the same reasoning, the image of the map $y \rightarrow w(x,y)$ is not Zariski-dense for $x$ generic when $w$ is of the form
$w(x,y) =  v(x, u(x,y)^{-1} x u(x,y)),\                            (*)$
where $v$ and $u$ are some words. The question can then be made precise: are all examples of this form? That is: is it the case that, for all words $w(x,y)$ not of the special form $(*)$, the image of the map $y\rightarrow w(x,y)$ is Zariski-dense for $x$ generic?
I would be extremely interested in the correct answer, even in the case $n=3$. I suspect that the answer is "yes", at least for $n=2$. At the same time, it would be rather nice if it were "no".

[The most obvious approach may be to take derivatives at the origin. However, while this often proves that a map is surjective, it does not prove that a map is not surjective -
and in this problem it leaves too many candidates of possible words w for which the map $y\rightarrow w(x,y)$ might not be surjective but probably really is.
For those who have asked about the characteristic: if the characteristic is finite, you can assume it's much greater than the sum of the absolute values of the exponents of the word we are considering. In other words, let us not consider things like $y\rightarrow y^p, p = char(K)$.

REPLY [3 votes]: You need some conditions on K to make sense of the question.  For instance, y ⟼ y2 is not surjective when K = ℝ and n = 2, because you cannot reach [[-1,0],[0,-2]]. Of course, you might have meant surjectivity in the sense of algebraic groups rather than in the sense of set-theoretic groups.  But I think that that just amounts to asking for K to be algebraically closed.  There is another problem when K has characteristic p.  In this case y ⟼ yp is not surjective; its image is only the diagonalizable matrices.
The conjecture as stated seems plausible when K = ℂ.  Or if K has positive characteristic and is algebraically closed, you could perhaps ask for a Zariski-dense image.
Updates:
Re Tom's comment: I think that Harald's question is clear enough, even though I agree that one has to work a little to see what he means.  He must mean that w is an element of the free group in the letters x and y, of course.  The second objection is also not essential.  The question can be phrased as:  For each w, what are the Zariski topology properties of the set of all x for which y ⟼ w(x,y) is surjective?  I believe that Chevalley's theorem on constructible images tells you that the set of such x is constructible, so it either contains a Zariski open or it is disjoint from a Zariski open.
Re Harald's comment concerning if-and-only-if conditions.  You have probably thought of this, but here goes anyway.  If your derivative is non-singular for any z, then the w-map is Zariski dense.  Of course, there are interesting maps in algebraic geometry that are Zariski dense but not surjective.  You seem to suggest that that cannot happen here, but I do not know how to prove it.<|endoftext|>
TITLE: Stein Manifolds and Affine Varieties
QUESTION [26 upvotes]: When is a Stein manifold a complex affine variety? I had thought that there was a theorem saying that a variety which is Stein and has finitely generated ring of regular functions implies affine, but in the comments to my answer here, Serre's counterexample was brought up.  I'm guessing that the answer is that the ring of regular functions must be nontrivial somehow, like it must separate points, but I'm curious about what the exact condition is.

REPLY [26 votes]: Charlie, it is funny answering this way but here it is. 
The criterion you are thinking about is a criterion that is relative to an embedding. It says that if $X$ is a quasi-affine complex normal variety, whose associated analytic space $X^{an}$ is Stein, then $X$ is affine if (and only if) the algebra $\Gamma(X,\mathcal{O}_{X})$ is finitely generated. This is a theorem of Neeman. 
You can reformulate the requirement of $X$ being quasi-affine as a separation of points property: for any point $x \in X$ consider the subset $S_{x} \subset X$ defined as the set of all points $y \in X$ such that all regular functions on $X$ have equal values at $x$ and $y$. Then by an old theorem of Goodman and Hartshorne $X$ is quasi-affine if $S_{x}$ is finite for all $x$. So you can say that $X$ is affine if it satisfies: 1) $X^{an}$ is Stein; 2) $S_{x}$ is finite for all $x \in X$; 3) $\Gamma(X,\mathcal{O}_{X})$ is finitely generated.<|endoftext|>
TITLE: What is the volume of a $\delta$-ball in the orthogonal group $O(n)$? Is there a (simple) lower bound?
QUESTION [10 upvotes]: The volume in the orthogonal group is measured by the Haar measure, which is the up to scaling unique measure that is invariant under the group operation. I consider the usual metric that is induced by the spectral norm $|M| = \max |Mx|$ where $x$ ranges over all vectors of length 1 and the vector norm is the Euclidean one. A $\delta$-ball is the set of all orthogonal matrices that have distance less or equal $\delta$ to a fixed matrix $M$. Because of the invariance of the Haar measure, for a fixed $\delta$, all $\delta$-balls have the same volume.

REPLY [6 votes]: The volume of the delta-ball of the special orthogonal group can be computed exactly by applying the Weyl integration formula: (Without loss of generality, we assume that the delta-ball is around the unit group element).
a. One notices (Again due to the invariance under the Haar measure) that the characteristic function of the delta ball is a class function. Thus upon the application of the Weyl integration formula we are left only with the radial part on the eigenvalues which is a $\lfloor N/2\rfloor$-dimensional integral for $\mathrm{SO}(N)$. Here, the radial integral is described explicitely.
b. The eigenvalues of an orthogonal matrix of dimension $N=2m+1$ are $1$ and $m$ pairs $\exp(i \phi_ m)$ and $\exp(-i \phi_ m)$, $0\leq\phi_ 1 \leq\ldots\leq\phi_m \leq\pi$. In the case of even dimensions, the unit eigenvalue is absent.
c. The delta-ball condition on the eigenvalues becomes:
$$
|\exp(i\phi_k)-1|\leq\delta ,
$$
which implies:
$$\phi_k\leq 2 \arcsin\sqrt{\delta/2}.$$
d. Applying the Weyl integration formula, we obtain for the odd case $\mathrm{SO}(2m+1)$:
$$
\mathrm{Vol}(\delta\mathrm{-ball}) = \frac{2^{m^2}}{\pi^m m!} \int_{\phi_1\leq\ldots\leq\phi_m \leq 2 \arcsin\sqrt{\delta/2}} 
\prod_{1\leq j < k \leq m} (\cos\phi_k-\cos\phi_j)^2 \prod_{l=1}^m \sin^2(\phi_l/2) d\phi_1 \cdots d\phi_m.
$$
e. For the even dimensional case, the only changes are $2^{m^2}$ is replaced $2^{(m-1)^2}$ and the sine terms are absent.<|endoftext|>
TITLE: Adapting families of diffeomorphisms to an open cover
QUESTION [5 upvotes]: Has anyone seen the following result in the literature?  I've asked a few experts but so far I've come up with nothing.
Given a manifold M and an open cover {U_i} of M, we want to see how families of diffeomorphisms of M can be adapted to {U_i}.  We will think of families of diffeomorphisms as generators of C_*(Diff(M)), where C_*() denotes singular chains.
Def: A k-parameter family of diffeomorphisms f: P^k \times M -> M is supported on V \subset M if, for all y not in V, we have f(p, y) = f(q, y) for all p, q \in P.  In other words, f is independent of the parameters P outside of V.
Define A_k \subset C_k(Diff(M)) to be the subcomplex generated by all k-parameter families (k-chains) of diffeomorphisms f such that f is supported on a union of at most k of the U_i's, and such that (inductively) the boundary of f is in A_{k-1}.
Claim: A_* is homotopy equivalent to C_*(Diff(M)).
There is a similar result if we replace Diff(M) with Maps(M -> T), where T is some topological space.  It is used in the proof of the claim in this question.

REPLY [3 votes]: This is not a full answer, but there is a once-well-known technique in geometric topology that could help clarify this result.  Kirby and Siebenmann, in their old work on triangulability of manifolds, have a technique called handle straightening.  It is a very nice way to push around pieces of a diffeomorphism without disturbing other pieces.
To use this technique, give the manifold M a handle decomposition (which in your case should refine your open cover) and collar all of the handles.  Then you can isotop the 0-handles with without disturbing the 1-handles.  How?  You should shrink back the higher handles along the collars of the 0-handles, then isotop the 0-handles in a region that is that contains the original 0-handles but is disjoint from the trimmed higher handles.  Then you can extend the higher handles back again so that they attach to the 0-handles.  You can repeat this for the k-handles in turn, provided that the requested isotopy of the k-handles is the identity in a neighborhood of the (k-1)-skeleton.
I don't know if your whole result is standard, but this part, if you can use it in your construction, is standard.<|endoftext|>
TITLE: Interesting families of sparse graphs?
QUESTION [9 upvotes]: I'm interested in graph families which are sparse, and by sparse I mean the number of edges is linear in the number of vertices. |E| = O(|V|).
Besides non-trivial minor-closed families of graphs (these turn out to be sparse), I don't know any other families. Can anyone suggest any interesting graph families (which are not minor-closed) which are sparse?
Please don't suggest the trivial family ("the family of sparse graphs").
EDIT: Thanks to the first few people who replied, I realized that bounded degree graphs (max degree < k) also form an interesting and large class of sparse graphs. So perhaps I'll refine my question to exclude those too. Any interesting sparse graph families where the max degree isn't bounded? For example the family of star graphs is sparse and not bounded degree. (But they're minor-closed.)

REPLY [3 votes]: Let me recommend the book J. Nešetřil and P. Ossona de Mendez, Sparsity: 
Graphs, Structures, and Algorithms, Springer 2012. It is the first systematic study of sparse graphs and related topics.<|endoftext|>
TITLE: Is there a coalgebraic characterisation of the hyperfinite II_1 factor?
QUESTION [9 upvotes]: Peter Freyd showed that the real interval [0, 1] is a final coalgebra for a functor on sets equipped with two points, which sends such a set to the 'wedge' of two copies of itself, identifying the second point of the first copy with the first point of the second copy.
The hyperfinite II_1 factor has trace values in [0, 1] and arises from a process of completing a union of finite subalgebras by a form of doubling discussed here.
Is there then a similar coalgebraic characterisation of the factor?

REPLY [2 votes]: I thought about this question some yesterday.  As I was saying in the related post, von Neumann algebras are a non-commutative or quantum generalization of measurable spaces, $C^*$ algebras are a non-commutative/quantum generalization of compact Hausdorff spaces, and both generalizations are contravariant.  Your question has a covariant spirit, which is a bit misaligned but not an essential point for this particular question.
The more germane issue is that the interval is a topological space in Peter Freyd's construction, while the hyperfinite factor, like any von Neumann algebra, behaves as a measurable space.  It is true that the hyperfinite factor is a non-commutative analogue of an interval, and you can construct it as the closure of a certain class of operators on $L^2([0,1])$.  But Freyd's construction does not work for measurable maps, only continuously.  In fact, as a measurable space, the unit interval doesn't have endpoints.  It has points, sort-of, but not in any useful way.  (To be precise, the measurable model of the interval here is the class of measures on it which are Lebesgue absolutely continuous, not all Borel measures.)
Maybe there is a $C^*$-algebra with a Freyd-type property, and which generates the hyperfinite factor.  You would have to decide whether its "endpoints" are a classical bit or a qubit.  You could try to work in the category of $C^*$-algebra homomorphisms, which tends to be short of morphisms from non-commutative objects. Or you could try to work in the category of completely positive maps on $C^*$-algebras, which has plenty of morphisms but also has other complications.  Certainly you would want to reverse arrows in passing from topology to $C^*$-algebras.  I don't know a whole lot about making $C^*$-algebras; I couldn't come up with anything.<|endoftext|>
TITLE: What does a projective resolution mean geometrically?
QUESTION [50 upvotes]: For R a commutative ring and M an R-module, we can always find a projective resolution of M which replaces M by a sequence of projective R-modules. But as R is commutative, we can consider the affine variety X=Spec R and the sheaf of modules associated to M. What is the projective resolution doing geometrically to this sheaf?
Projectives are locally free sheaves, so if M itself is not projective then it must have some sort of "sharp twisting" or "pinching". In some way a projective resolution is "un-pinching" M. Geometrically, is this the same "un-pinching" that happens in a resolution of a singularity of a variety? Is there an example in low dimensions where one can actually draw a picture of this happening for modules?

REPLY [4 votes]: I think this important question deserves one more answer, even after all the excellent ones already given. 
One can think of a free resolution as "approximating the module by the ring". This tautology has some corollaries: 
If you want the resolutions to reveal geometric properties, you need the ring to be nice to start with. Examples of this have been explained in all the other answers. In fact, you get the most information if your ring is "smooth" (regular local or polynomial rings). 
Flipping the argument, this means that "good module + bad ring = bad resolution". For example, let $M$ be a field $k$, viewed as a module over the ring $R =k[x,y]/(x^2,xy)$. Then a resolution of $M$ is always bad, the ranks of the free modules will increase exponentially, and no nice information about $M$ can be learned from those numbers. In fact, the information now flows the other way, it tells you how bad $R$ is. For example, it follows that $R$ can not be a complete intersection, because if it was, a result by Eisenbud implies that the Betti numbers would have had polynomial growth.<|endoftext|>
TITLE: Density of a subset of the reals
QUESTION [7 upvotes]: The rationals are clearly dense in the real number system, i.e. for every pair a < b of real numbers there exists a rational number p/q s.t. a < p/q < b. I conjecture the same to be true with p and q both primes. Any idea of how one could prove it? It should depend on some strong result on the distribution of prime numbers.

REPLY [3 votes]: You don't need the prime number theorem.  Suppose the result is false, i.e. for fixed a and b there are only finitely many q such that there is a prime between qa and qb.  Then the nth prime pn grows at least as fast as (b/a)^n; in particular, sum 1/pn converges, which we know to be false (and which is totally elementary).  (This argument is slightly thorny to make rigorous, but it's just a matter of handling constants.)<|endoftext|>
TITLE: Good introductory references on algebraic stacks?
QUESTION [64 upvotes]: Are there any good introductory texts on algebraic stacks?
I have found some readable half-finsished texts on the net, but the authors always seem to give up before they are finished. I have also browsed through FGA explained (Fantechi et al.). Although I find the level good, it is somewhat incomplete and I would want to see more basic examples. Unfortunately I don't read french.

REPLY [2 votes]: Daniel Halpern-Leistner is teaching a foundational course on moduli theory based on stacks at Cornell. Very nice addition to the literature in my opinion.<|endoftext|>
TITLE: Transformation of the Black-Scholes PDE into the diffusion equation - shift of coordinate system
QUESTION [5 upvotes]: The aim of transforming the Black-Scholes PDE is of course to find a form where an relatively easy solution exists. Most of the steps seem to be straightforward - please use this reference:
https://planetmath.org/AnalyticSolutionOfBlackScholesPDE
...all but one, actually the last one where a convection-diffusion equation is being transformed into the basic diffusion equation. 
[In the article you find it here: "Under the new coordinate system (z), we have the relations amongst vector fields ... leading to the following transformation of equation..."]
The u_x-term vanishes by some magic coordinate transformation. When you look at the derivatives they even seems wrong to me because they state that tau=s and then derive delta/delta tau = delta/delta s + some extra term (delta/delta y * -(r-1/2 sigma^2).
I simply don't get it: first how it works and second how they find that kind of transformation.

REPLY [5 votes]: I'm afraid the Planetmath page put my browser into an infinite reload loop, so I can't help you with the formalism there.
I would recommend instead looking at the change of variables in the Wikipedia article.  The last time I checked it, it seemed to work.
Edit:  Okay, I have a formulation that works.  I'll write s for sigma, so the equation is initially:
Vt + (1/2)s2S2VSS = rV - rSVS.
Since S follows a lognormal random walk (in particular the stochastic diff eq governing S involves a logarithmic derivative), it is natural to change to x = log S, or S = ex, so the log price x follows normal Brownian motion.  This yields the equation:
Vt + (1/2)s2(Vxx - Vx) = r(V - Vx).
Black-Scholes is a final-value problem, i.e., we know the value of the option at time T, and diffusion works backwards.  It is therefore natural to negate the time variable (and multiply by a suitable scalar to make things neater).  tau = (1/2)s2(T-t).  Then we get:
(1/2)s2(Vxx - Vx - Vtau) = r(V - Vx).
Finally we rescale the value function to remove exponential growth effects.  u = eax + b(tau)V for undetermined coefficients a and b.  We can substitute, multiply the equation by 2eax+b(tau)/s2, and we get:
uxx + (something)ux + (something else)u = utau.
(something) is a degree one polynomial in a and is independent of b.  (something else) is a degree one polynomial in b, so we can choose a and b to kill those terms.  This yields the diffusion equation.
Hope that helps.<|endoftext|>
TITLE: Young's lattice and the Weyl algebra
QUESTION [18 upvotes]: Let L be the lattice of Young diagrams ordered by inclusion and let Ln denote the nth rank, i.e. the Young diagrams of size n.  Say that lambda > mu if lambda covers mu, i.e. mu can be obtained from lambda by removing one box and let C[L] be the free vector space on L.  The operators
U lambda = summu > lambda mu
D lambda = sumlambda > mu mu
are a decategorification of the induction and restriction operators on the symmetric groups, and (as observed by Stanley and generalized in the theory of differential posets) they have the property that DU - UD = I; in other words, Young's lattice along with U, D form a representation of the Weyl algebra. 
Is this a manifestation of a more general phenomenon?  What's the relationship between differential operators and the representation theory of the symmetric group?  
Edit:  Maybe I should ask a more precise question, based on the comment below.  As I understand it, in the language of Coxeter groups the symmetric groups are "type A," so the Weyl algebra can be thought of as being associated to type A phenomena.  What's the analogue of the Weyl algebra for types B, C, etc.?

REPLY [3 votes]: Sasha Kleshchev's book "Linear and Projective Representations of Symmetric Groups" is the reference I'd suggest. Chapter 1 contains the connection with Young's lattice, and the subsequent chapters develop the functors that Ben described above. The second half of the book develops the theory for spin representations of symmetric groups which is an honest type B analogue (The functors $E_m$ and $F_m$ in Ben's answer satisfy the Serre relations for the Kac-Moody algebra of type $B_\infty$).
To add a little more detail to Ben's answer, the right level of generality to think about these questions is the affine Hecke algebra (either the degenerate or nondegenerate varieties). I'll describe the degenerate case:
Let $F$ be an algebraically closed field of characteristic $p$. As a vector space the (degenerate) affine Hecke algebra is a tensor product of a polynomial algebra with the group algebra of the symmetric group: $H_d=F[x_1,\ldots,x_d]\otimes FS_d$. Multiplication is defined so that each tensor summand is a subalgebra, and $H_d$ satisfies the mixed relations $s_ix_j=x_js_i$ for $j\neq i,i+1$ (here $s_i=(i,i+1)$), and $s_ix_i=x_{i+1}s_1-1$. 
Note that in addition to being a subalgera, $FS_d$ is also a quotient of $H_d$ obtained by mapping $s_i\mapsto s_i$ and $x_1\mapsto 0$ (so that the $x_i$ map to Jucys-Murphy elements). 
The polynomial subalgebra forms a maximal commutative subalgebra, so given a finite dimensional $H_d$-module $M$, we may decompose $$M=\bigoplus_{(a_1,\ldots,a_d)\in F^d}M_{(a_1,\ldots,a_d)},$$ where $$M_{(a_1,\ldots a_d)}=\lbrace m\in M|(x_i-a_i)^Nm=0,\mbox{ for }N\gg0\mbox{ and }i=1,\ldots,d \rbrace$$ is the generalized $(a_1,\ldots,a_d)$-eigenspace for the action of $x_1,\ldots,x_d$. Let $I=\mathbb{Z}1_F\subset F$ and $Rep_IH_d$ be the category of finite dimensional $H_d$-modules which are integral in the sense that if $M\in Rep_IH_d$, and $M_{(a_1,\ldots,a_d)}\neq 0$, then $(a_1,\ldots,a_d)\in I^d$.
Now, let $K_d=K_0(Rep_IH_d)$, and $K=\bigoplus_d K_d$. Then, the categorification statement is that parabolic induction and restriction give $K$ the structure of a bialgebra, and as such $$K\cong U_{\mathbb{Z}}(\mathfrak{n}).$$ In the above statement,  $\mathfrak{n}\subset \mathfrak{g}$ is the maximal nilpotent subalgebra of the Kac-Moody algebra $\mathfrak{g}$ generated by the Chevalley generators $e_i$, where, if $char F=p$, then $\mathfrak{g}=\hat{sl}(p)$, and if $char F=0$, $\mathfrak{g}=\mathfrak{gl}(\infty)$. In both cases $U_{\mathbb{Z}}(\mathfrak{g})$ denotes the Kostant-Tits $\mathbb{Z}$-subalgebra of the universal enveloping algebra. Note here that the Chevalley generators are indexed by $I$.
Now, for each dominant integral weight $\Lambda=\sum_{i\in I}\lambda_i\Lambda_i$ ($\Lambda_i$ the fundamental dominant weights) for $\mathfrak{g}$, define the polynomial $$f_\Lambda=\prod_{i\in I}(x_1-i)^{\lambda_i}.$$ Then, the algebra $H_d^\Lambda=H_d/(H_d f_\Lambda H_d)$ is finite dimensional. In the case $\Lambda=\Lambda_0$, $H_d^\Lambda\cong FS_d$. 
One can form $K_d(\Lambda)$ and $K(\Lambda)$ as above corresponding to the category $H_d^\Lambda-mod$. Then, the categorification statement is $$K(\Lambda)\cong V_{\mathbb{Z}}(\Lambda)$$ as $\mathfrak{g}$-modules, where $V(\Lambda)$ is the irreducible $\mathfrak{g}$-module of highest weight $\Lambda$ generated by a highest weight vector $v_+$, and $V_{\mathbb{Z}}(\Lambda)=U_\mathbb{Z}(\mathfrak{g})v_+$ is an admissible lattice. The action of the Chevalley generators on $K$ are analogues of the functors in Ben's answer. The action of the Weyl module corresponds to the action of $D=\sum_{i\in I}e_i$ and $U=\sum_{i\in I}f_i$ (in characteristic 0, this is defined in the completion $\mathfrak{a}(\infty)$ of $\mathfrak{gl}(\infty)$.
One can generalize this story to $\hat{\mathfrak{sl}}_\ell$ by working with the (nondegenerate) affine Hecke algebra $H_d(t)$, where $t$ is a primitive $\ell$-th root of unity. In this case, the finite dimensional quotients are Hecke algebras of complex reflection groups. The hyperoctohedral group corresponds to the highest weight $\Lambda=2\Lambda_0$. Then $V(\Lambda)$ is a level 2 representation,  hence the central element acts by $2\cdot Id$ as in Sammy's comment).
In the second half of Kleshchev's book, the Hecke algebra is replaced by the so-called Hecke-Clifford (or Sergeev) algebra, and $\mathfrak{g}$ is of type $B_\infty$ or $A_{2\ell}^{(2)}$ depending on the ground field (or one can work in the non-degenerate case so that $\ell$ needn't be prime.
The algebras introduced by Khovanov-Lauda and Rouquier generalize this story to arbitrary symmetrizable Kac-Moody algebra. These algebras are graded, so one gets a categorification of the quantum group $U_q$, where $q$ keeps track of the grading . . .<|endoftext|>
TITLE: Lower bound on volume of minimal hypersurface contained in a unit ball with curvature bounds
QUESTION [5 upvotes]: I was just wondering, if I have a geodesic ball of radius one in a manifold M whose sectional curvature lies between -epsilon and epsilon for epsilon small, and the injectivity radius of my manifold is two say, and I have a SINGULAR minimal hypersurface in M passing through the centre of my unit ball, can I find a lower bound, away from zero, on the volume of the singular minimal hypersurface contained in the ball, in terms of only the dimension of the ambient manifold M? Thanks!

REPLY [3 votes]: @Deane, when M is Euclidean space, this is a consequence of monotonicity, which is one of the most fundamental facts about minimal surfaces.
As for the original question, I believe that the answer is YES if the hypersurface S is minimizing in M.  Think of S as sitting in the Euclidean ball determined by geodesic coordinates.  Of course, S is no longer minimizing in the Euclidean ball, but the curvature bounds should give you control on the ratio of areas computed wrt to the ambient metric vs areas computed wrt the Euclidean metric.  So although any "competitor" surface can have Euclidean area less than that of S, it must have area greater than |S|/C for some big constant C.  Using a modification of the standard monotonicity argument, this is sufficient to derive a lower bound on the area of S.  (For this part, consult arxiv 0705.1128 Section 5 for details.)
As for when S is not necessarily minimizing, I suspect(?) that the answer is still yes.  One idea is to look at the proof of Allard's monotonicity formula and account for what happens in the presence of curvature.
I also suspect that all of this is written up in full generality and rigor... somewhere.<|endoftext|>
TITLE: What is the Beilinson regulator?
QUESTION [8 upvotes]: Trying to understand answer to this question.
What is the (Beilinson) higher regulator of a number field?

REPLY [27 votes]: Many, many years ago I happened to be in the room when somebody asked exactly the same question --- "What is the Beilinson regulator?" --- of Marc Levine.  Marc went straight to the blackboard and drew the key diagram that makes everything clear.  I knew it had to be preserved for posterity, so I ran across the street to buy a disposable camera.  I am delighted to be able to share this:

(Click picture for larger version.)<|endoftext|>
TITLE: Learning about Galois representations
QUESTION [8 upvotes]: My goal was to learn about l-adic representations on some example — I'm a newbie in these topics.
Thus take pt = Spec F_q, G=\pi_1(pt) and consider lisse schemes over pt. My understanding is that such a scheme always comes from Galois extension with some group, e.g. H, a subgroup of G and that the fibers are, by Galois theory, parametrized by classes G/H.
So in this case, as with any Galois covering, I get the action of Galois group on the cohomology of the fiber, that is G acts on H^*(f_*QQ_l). Now this representation may be not irreducible.

(1) Is is true that irreducible l-adic rep is called geometric iff it's part of H^*(f_*QQ_l) for some f?

(my understanding is that the above construction gives the representations with kernel H)

(2) Is it true that I get all representations with open kernel that way?

I think (2) is very similar to  a classical theorem of algebraic number theory.

(3) Did we just prove a Brauer theorem? Or did we, on the contrary, somehow use it?

And, finally, I hope that this example is related to more complicated Galois representations.

(4) What does the above teach us about more complicated Galois representations?

REPLY [3 votes]: Regarding the first question (about the term geometric): the notion of geometric I know of is about p-adic representations of the Galois group of Q (or a number field). In this case, a rep is called geometric if (1) it is unramified at almost all places, (2) at places above p, it is potentially semistable (in the sense of p-adic Hodge theory). This is to be distinguished from representations "coming from geometry" which are those that occur as subquotients of the etale cohomology of some smooth projective variety over the number field. The Fontaine-Mazur conjecture is that "geometric" representations "come from geometry". (The terminology is somewhat confusing). I suggest Bellaïche's clay math lecture notes (link text) on the subject. But perhaps, this is not what you were getting at with this question.<|endoftext|>
TITLE: Cartographic group and flat stringy connection
QUESTION [8 upvotes]: There's a literature about dessins d'enfants (including my previous question here), and one amazing thing about them is that absolute Galois group Gal Q acts on cartographic group which, I believe, is isomorphic to  letters_2 = <<A, B>> (group, freely generated by two noncommuting letters).
The funny thing about the latter group is that there is a flat connection coming from string theory defined on its group algebra, C[letters_2], which I think has the name of Knizhnik-Zamolodchikov. So, it that latter connection somehow related to Galois group?

REPLY [12 votes]: Yes, this is the so called Grothendieck-Teichmuller theory. The reference is the paper of Drinfeld, "On quasitriangular quasi-Hopf algebras and a group closely related to $Gal(\bar Q/Q)$".
Following Grothendieck, the idea is the following: let $M_{0,n}$ be the moduli space of Riemann spheres with $n$ marked points. The absolute Galois group acts on the profinite completion $\hat T_n$ of the fundamental group of $M_{0,n}$. In particular, $M_{0,4}$ is isomorphic to $\mathbb{P}_1-${$0,1,\infty$}' whose fundamental group is the free group $F_2$. This is the first action you mention.
The point is to look at the action of $Gal(\bar{Q}/Q)$ on the whole "tower" of the $\hat T_n$, i.e. simultaneous actions on the $\hat T_n$ compatible with morphisms induced by natural geometric operations like adding or removing marked points.
So far I know, KZ equations are not quite related to string theory but rather to conformal field theory. But anyway, they leads to "universal" representations of the braid groups which are closely related to the $\hat T_n$. Drinfeld gives an algebraic description of these representations by introducing objects called "associators", which satisfy complicated equation expressing somehow a notion of compatibility with these natural geometric operations. He uses this machinery to define a rather "explicit" group (i.e. defined by explicit but very complicated algebraic equations) over which the set of associators is a torsor, called the Grothendieck-Teichmuller group, which is a subgroup of $Aut(\hat F_2)$ containing the image of $Gal(\bar Q/Q)$ through the morphism induced by its action on the $\hat T_n$. It is well known that this map is injective, and so far I know it is a plausible conjecture that these two groups are actually equal.<|endoftext|>
TITLE: Solving "a, b, a+b have given divisors" problem
QUESTION [6 upvotes]: I've read an interesting article, math.NT/0409456 where you're just trying to solve a simple problem:

For a given (finite) set of primes S find all solutions to an equation a + b = c with the condition that all prime divisiors of integers a, b, c must be in S.

and this problem turns out to be very geometric. It turns out (and I tell that in comments below) you're actually dealing with sections of certain projective morphism of schemes R --> Spec ZZ \ S. The article then proves that the number of solutions to the equation is finite by proving that the number of these sections is finite.
Is there kind of general theory or other methods to prove things about sections of these maps? What is the intuition used here? Would there be a way to count these solutions?

REPLY [2 votes]: This sounds to me like it is related to some recent work of Lagarias and Soundararajan:
http://arxiv.org/pdf/0911.4147
Their paper has some relations to the ABC conjecture and to GRH. Hope this helps.<|endoftext|>
TITLE: A single paper everyone should read?
QUESTION [250 upvotes]: Different people like different things in math, but sometimes you stand in awe before a beautiful and simple, but not universally known, result that you want to share with any of your colleagues. 
Do you have such an example?
Let's try to go in the direction of papers that can actually be read online or accessible with little effort, e.g. in major libraries, so that people could actually follow your advice and read about it immediately.
And as usual let's do one per post and vote freely, vote a lot.

REPLY [4 votes]: PDE as a Unified Subject   by Sergiu Klainerman.
An essay on partial differential equations written by a leading expert in the field, I strongly recommend to anyone who aspires to know more on the subject as well as to those who are not interested strictly in PDE's, but would like to get a grasp of interactions between Mathematics and Physics. There are also many interesting references.<|endoftext|>
TITLE: What's the "Yoga of Motives"?
QUESTION [73 upvotes]: There are some things about geometry that show why a motivic viewpoint is deep and important. A good indication is that Grothendieck and others had to invent some important and new algebraico-geometric conjectures just to formulate the definition of motives.
There are things that I know about motives on some level, e.g. I know what the Grothendieck ring of varieties is or, roughy, what are the ingredients of the definition of motives.
But, how would you explain the Grothendieck's yoga of motives? What is it referring to?

REPLY [7 votes]: I think a deep sense of this Yoga is that "Grothendieck's vision of motives was as a universal cohomolgy theory but also as higher dimensional version of Galois theory". You can see this for examples by its 0-dimensional examples, the so called Artin motives and how motives are "understood" via Motivic Galois groups. In André book, there are plenty of Galoisian remarks on motives.
In the same way as there are schemes of all dimensions, there are fibrations of all relative dimensions above a scheme $S$, and not only the relative dimension 0 of the coverings. When $S=Spec(F)$, these are ‘varieties’ defined by algebraic equations in more than one variable. In this vertical direction, Grothendieck also gave a partial generalization of Galois theory, the "$l$-adic cohomology" of fibrations. The cohomology, or rather the homology, of a topological space had been invented by Poincaré, and as soon as the 1940’s, André Weil was interested in adapting it to algebraic geometry. After pioneering work by Jean-Pierre Serre, Grothendieck realized this adaptation, associating to every fibration of a scheme $S$ the $l$-adic cohomology spaces that are continuous linear representations of the fundamental group $π_S$; we call these Galois representations of $S$. We would have a complete generalization of Galois theory if we could have moved back up from these to algebraic varieties; this is the object of Grothendieck’s theory of ‘motives’, which, even today, remains conjectural. Outside of relative dimension 0, we know only the case of the varieties called ‘Abelian’ conjectured by John Tate and proved by Gerd Faltings in 1983: when two Abelian varieties have the same $l$-adic cohomology, each parametrizes the other. But if it is true that the category of fibrations, or rather of ‘motives’, over a base scheme $S$ is equivalent to that of the Galois representations of $S$, determining these representations and their mutual relations is crucial. Galois theory and Arithmetic, Lafforgue and Florence.
My own reading of Grothendieck's Récoltes et semailles suggested to me that was the Motivic Galois groups the are at the core of the yoga and not motives themselves. The Galois nature of this yoga is important since the very quest for the most profound "invariante de la forme", Grothendieck says: Ainsi, le motif m'apparait comme le plus profond "invariant de la forme" qu'on a su associer jusqu'a présent a une variété algébrique, mis a part son "groupe fondamental motivique". L'un et l'autre invariant représentent pour moi comme les "ombres" d'un "type d'homotopie motivique" qui resterait a décrire, Récoltes et semailles". Because of this yoga is about the most profound galoisian invariant it is for me that not Motives as cohomology theory but Galois theory of a homotopy theory or a higher relative is the nature of this yoga, it is still missing a motivic theory for Grothendieck's homotopy developed in Lawrence Breen letters. 
To conclude I would like to cite the following: "Grothendieck's broken dream was to develop a theory of motives, which would, in particular, unify Galois theory and topology". A mad day's work, Cartier.
Autour de la «théorie de l'ambiguite«, de Galois a nos jours, Y André
Groupes de galois motiviques et periodes, Y André.
Note: Some edits will appear later, there are some things yet to be explained in detail, but it is enough for today.<|endoftext|>
TITLE: Exactness of filtered colimits
QUESTION [17 upvotes]: Are filtered colimits exact in all abelian categories?
In Set, filtered colimits commute with finite limits.  The proof carries over to categories sufficiently like Set (i.e. where you can chase elements round diagrams), in particular A-Mod where A is a commutative ring.  This implies that filtered colimits are exact in A-Mod.
I am aware of a vague principle that things that are true in A-Mod are true for all abelian categories, but I have never seen a precise statement of this principle so I am not sure if it applies in this case.

REPLY [4 votes]: A counterexample which is non-trivial is given in Chapter 6 of Neeman's book Triangulated Categories. The category in question is the full subcategory of additive functors Cat(S^{op}, Ab) where S satisfies some hypotheses (e.g. is an essentially small triangulated category) and we take those functors which are product preserving for small enough products (so as contravariant functors S-> Ab they send sufficiently small coproducts to products). This category is complete and cocomplete but has neither exact filtered limits or exact filtered colimits. More precisely it satisfies [AB4] and [AB4*] but neither [AB5] nor [AB5*].<|endoftext|>
TITLE: What are the most important instances of the "yoga of generic points"?
QUESTION [19 upvotes]: In algebraic geometry, an irreducible scheme has a point called "the generic point." The justification for this terminology is that under reasonable finiteness hypotheses, a property that is true at the generic point is actually generically true (i.e. is true on a dense open subset).
For example, there is a result called "generic flatness" (EGA IV (2), Theorem 6.9.1). Suppose Y is locally noetherian and integral, with f:X→Y a morphism of finite type and F is a coherent OX-module. If F is flat over the generic point of Y (a condition which is always satisfied, since anything is flat over a point), then there is a dense open subscheme U⊆Y such that F is flat over U.
I'm sure that there are lots of instances of this "yoga of generic points", but whenever I try to come up with one, it's kind of lame (in my example above, the condition at the generic point is vacuous). What are the main examples of the yoga of generic points?

REPLY [2 votes]: There is a spectacular proof of the ACC for log canonical thresholds in $\mathbb C ^N$ due to Ein, de Fernex and Mustata https://arxiv.org/abs/0905.3775 that relies on taking generic limits of polynomials / power series (see also the beautiful paper by Koll\'ar especially \S 4 of https://arxiv.org/pdf/0805.0756.pdf). The idea is that the use of generic points allows us to take the "correct" limit of a sequence of polynomials.<|endoftext|>
TITLE: How much theory works out for "almost commutative" rings?
QUESTION [16 upvotes]: I've been reading about D-modules, and have seen a proof that D_X, the ring of differential operators on a variety, is "almost commutative", that is, that its associated graded ring is commutative.  Now, I know that with the Ore conditions, we can localize almost commutative rings, and so we get a legitimate sheaf D to do geometry with.  But how far does the analogy go? What theorems are true for commutative rings but can't be modified reasonably to be true for (nice, say, left and right noetherian or somesuch) almost commutative rings?

REPLY [3 votes]: Check out the paper differential operator on noncommutative ring, Rosenberg-Lunts define the noncommutative version of Grothendieck differential operator. This framework works perfect for arbitrary noncommutative ring. As an application. They developed quantum D-module theory as a special case of Noncommutative D-module theory. They treated the much more general case in Differential Calculus in noncommutative algebraic geometry
As Greg mentioned above, considering example of Weyl algebra. In commutative geometry, there is not good notion for closed subscheme(at least for diagonal). So,the crucial step is to find what is the correct notion for closed subscheme for noncommutative ring which led to the definition of noncommutative subscheme(naturally goes to the content of noncommutative algebraic geometry). In the paper I mentioned above, they gave the correct notion for noncommutative subscheme and correct definition for diagonal. Then they defined noncommutative differential operator as certain kind of differential bimodule and developed localization theory for differential operators. Then one can localize differential operators.(Which means (noncommutative)D-modules automatically becomes noncommutative schemes)
Notice
These framework works for any noncommutative ring, either noncommutative D-module(very noncommutative, such as ring of quantized differential operators) or usual commutative D-module(itself noncommutative but almost commutative)<|endoftext|>
TITLE: Curves with negative self intersection in the product of two curves
QUESTION [15 upvotes]: I wonder if the following is known: Are there two compact curves C1 and C2 of genus>1 defined over complex numbers, such that their product contains infinite number of irreducible curves of negative self-intersection and arbitrary large genus?  It we aks the same question replacing "negative" by "zero", the answer will be yes, moreover there will be examples of with C1 and C2 of any genus. These examples can be obtained as ramified covers ExE where E is an elliptic curve. 
(PS. The answers given to this question in 2009 did not solve it)

REPLY [7 votes]: Disclaimer. The answer below is a variation of Bogomolov's argument and it would not come to be without Dmitri's answer. If you feel like upvoting this, please upvote his answer.
Curves on  products of isogeneous  elliptic curves. 
As already suggested in the body of the question, if we start with a pair of elliptic 
curves, say $E_1$ and $E_2$, admitting a non-constant morphism $f : E_1 \to E_2$ then
given any point $p \in X=E_1 \times E_2$  we have infinitely many elliptic curves with self-intersection zero on $X$ passing through $p$. It suffices to consider translates of the graphs of endomorphisms of $E_2$ (there are at least $\mathbb Z$ of them) composed with $f$.
If we blow-up $p$ then we get a surface $S$ containing infinitely many (elliptic) curves
with negative self-intersection.
Jacobians of genus $2$ curves. As pointed out in Dmitri's answer the 
natural morphism
$$
\mathrm{Sym}^2 C \to \rm{Pic}^2(C) \cong \rm{Jac}(C)
$$
identifies $\mathrm{Sym}^2 C$ with the blow-up of $\rm{Jac}(C)$ at a point. 
Thus if 
we have a genus $2$ curve with Jacobian isogenous to the square of an elliptic curve
then the discussion in the previous paragraph shows that $C^2$ has infinitely many
curves of negative self-intersection since we can pull-back the 
negative curves on $\mathrm{Sym}^2 C$ through the natural morphism $C^2 \to \rm{Sym}^2 C$. 
Notice also that the negative curves have unbounded intersection with the diagonal $\Delta \subset C^2$. It is not hard to verify that the pull-backs of the negative elliptic curves to $C^2$ will have unbounded genus. 
Explicit example.
If $C$ is a genus $2$ curve admitting a morphism
$\pi : C \to E$ to an elliptic curve $E$ then $\rm{Jac}(C)$ is isogeneous to the product of $E$ with another
elliptic curve $E'$ ( the connected component  of the kernel of $\pi$ through zero). 
Automorphisms of $C$ act naturally on $\rm{Jac}(C)$. If there is an element
$\varphi \in \mathrm{Aut}(C)$ with induced action on $\rm{Jac}(C)$ not preserving $E'$ then $E$ is isogeneous to $E'$ since $$\pi_* \circ \varphi_* : \rm{Jac}(C) \to \rm{Jac}(E)\cong E$$ restricted to $E'$ is an isogeny. Therefore $\rm{Jac}(C)$ is isogeneous to the square of $E$.
To have a concrete example we can take $C = \lbrace y^2 = x^6 - 1\rbrace$ which maps to $E =\lbrace y^2 = x^3 -1\rbrace$ and has automorphism group isomorphic to $\mathbb Z_3 \rtimes D_8$ (which is not the automorphism group of any elliptic curve). From the discussion above
it follows that $C^2$ has infinitely many curves of negative self-intersection and unbounded genus. 
Question. Suppose  $C$ is genus $2$ curve such that $C^2$ contains infinitely many
curves of negative self-intersection. Is the Jacobian of $C$  isogeneous 
to the square of an elliptic curve ?<|endoftext|>
TITLE: How to think about model categories?
QUESTION [34 upvotes]: I've read about model categories from an Appendix to one of Lurie's papers.
What are the examples of model categories? What should be my intuition about them?
E.g. I understand the typical examples come from taking homotopy of something — but are all model categories homotopy categories?

REPLY [16 votes]: Model categories are 1-categorical presentations of (∞,1)-categories, which you can just think of as categories enriched in topological spaces, such as the category of spaces itself.  (Actually, there are conditions on (∞,1)-categories that come from model categories--most importantly they must have all homotopy limits and colimits.)  They're particularly convenient for computations, as with any kind of presentation.<|endoftext|>
TITLE: Variation on a matrix game
QUESTION [19 upvotes]: The original problem appeared on last year's Putnam exam:
"Alan and Barbara play a game in which they take turns filling entries of an initially empty 2008×2008 array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?"
It's not hard to see that Barbara can win this game by reflecting Alan's moves over a vertical line. (In fact, you might say she "wins with multiplicity 1004".) My question is, what if the goals were reversed? That is, suppose Alan (the first player) wants the determinant to be zero, and Barbara wants it to be nonzero. Now who has the winning strategy?
If you expect the result to rely solely on parity, then you should note that Alan wins in the 2×2 case, because he can force a row or column to have only zeroes. Unfortunately, it's not at all clear (to me, anyway) that he can do anything similar to a 4×4 matrix, let alone a 2008×2008 one.

REPLY [3 votes]: Let me expand upon my comment. Let us say we are dealing with a 4x4 matrix Alan is restricted to integers, Barbara can use any rational numbers. Alan moves first and wants to force the matrix to have an integral determinant. Alan moves first Barbara moves second and her entry is 1/2 in the same row as Alan the look at the matrix with the row Alan moved first in and the column Barbara moved first in deleted. Suppose Alan moves next in that matrix then Barbara moves in the same row and then the row Alan moved in and the column Barbara moved in is deleted or Alan doesn't move in the matrix still Barbara moves in the derived matrix and again her entry is 1/2 in the same row as Alan. In either case the row Alan moved in and the column Barbara moved in are deleted and we get a new derived matrix in which the process is repeated until there are 4 elements in the matrix with entry 1/2 whose product is in the determinant adding or subtracting 1/16. Barbara makes sure that the rest of her entries are integral as a result the 1/16 cannot be canceled out since the all other contributions to the determinant can be expressed as fractions with denominator 8 or less and the determinant is not integral. This idea can be extended in various ways depending on the restiction on Alan's moves.<|endoftext|>
TITLE: Manifolds distinguished by Gromov-Witten invariants?
QUESTION [25 upvotes]: What is a simplest example of a manifold $M^{2n}$ that admits two symplectic structures with isotopic almost complex structures, and such that  Gromov-Witten invariants of these symplectic structures are different? 
(unfortunately I don't know any example...) If we don't impose the condition that almost complex structures are isotopic, such examples exist in dimension 6. 
Added
Refined question. Is there a manifold $M^{2n}$ with two symplectic forms $\omega_1$, $\omega_2$,
such that the cohomology classes of $\omega_1$ and $\omega_2$ are the same and
the corresponding almost complex structures are homotopic, 
but at the same time the Gromov-Witten invariants are different?

REPLY [11 votes]: Here is an answer to the REFINED question given to me by Richard Thomas.
In this refined version one looks for an example such that the cohomology 
classes of two symplectic forms coincide.
In a later paper 1996, Duke Vol. 83 
TOPOLOGICAL SIGMA MODEL AND DONALDSON TYPE
INVARIANTS IN GROMOV THEORY, Ruan proved that such refined examples exist.
He says in this paper that for product examples $V\times S^2$ 
from the paper in JDG  1994 
(cited by Mike Usher) he does not know whether the classes of 
constructed symplectic forms can coincide as well. In fact this does not seem very plausible.
These refined examples are two $3$-dimensional Calabi-Yau manifolds,
constructed by Mark Gross. The construction is described in the paper
of Mark Gross (1997): "The deformation space of Calabi-Yau $n$-folds with canonical singularities can be obstructed". One $3$-dimensional Calabi-Yau
is a smooth anti-canonical section of $\mathbb CP^1\times \mathbb CP^3$ and the over is a smooth anti-canonical section of the projectivsation of  the bundle 
$O(-1)+O+O+O(1)$ over $\mathbb CP^1$. 
The construction of Gross is recalled on pages 47-48 of 
http://xxx.soton.ac.uk/PS_cache/math/pdf/9806/9806111v4.pdf
Using Wall's theorem Ruan proves that these Calabi-Yau 
manifolds are differomorphic. Then he studies the quantum cohomology
rings of these Calabi-Yaus and proves that they are different.<|endoftext|>
TITLE: What is the relationship between integrable systems and toric degenerations?
QUESTION [21 upvotes]: Given an integrable system on a Kahler manifold X, is there a way to associate a toric degeneration of X whose Milnor fibers are related to the fibers of the integrable system?
An integrable system is, at least, a map from X to R^n whose coordinate functions Poisson commute.  The moment map of a Hamiltonian torus action will have this property, but there are other examples.  For instance, the flag variety GL(n,C)/B has a famous integrable system and a famous toric degeneration, both of which are related to the same polytope--a Gelfand-Tsetlin polytope.  (Famous but I don't know the original references for these constructions.)
Given a toric degeneration Y --> C, you can try to construct an integrable system on a general fiber Y1 by flowing along a gradient vector field from Y1 to Y0 (the special fiber, a toric variety) and projecting to R^n via the moment map of the torus action on Y0.  I heard that this doesn't work on the nose, but that it does work well enough that you can at suitable points identify the fibers of e.g. the Gelfand-Tsetlin integrable system with the Milnor fibers of the Gelfand-Tsetlin toric degeneration.  Possibly starting with an integrable system and trying to construct a toric degeneration is easier and more algebraic.
P.S.  Some references after all:  Guillemin and Sternberg, "The Gelfand-Cetlin system and quantization of the complex flag manifolds,"  and Gonciulea and Lakshmibai, "Degenerations of Flag and Schubert varieties to toric varieties."

REPLY [10 votes]: Dear David, you may find the article "Integrable systems, toric degenerations and Okounkov bodies", arXiv:1205.5249, useful.<|endoftext|>
TITLE: How do quantum knot invariants change when I pick a funny ribbon element?
QUESTION [6 upvotes]: So, there's a construction of Reshetikhin and Turaev which extracts knot invariants from ribbon monoidal categories, which are (usually) the representation category a Hopf algebra with a choice of ribbon element.
How do these knot invariants change if I pick a different ribbon element in the same Hopf algebra?  In particular, will something strange happen with 3-manifold invariants?

REPLY [2 votes]: Did you look at prop 5.21 in the paper with Peter?  I think that should answer your question.
There are two slightly different questions you could ask.  First how does the framing-dependent invariant change.  Here it is just (\pm 1)^#L where # is the number of components.  Second how does the framing-corrected invariant change?  Here it's (\pm 1)^#L (\pm 1)^writhe.  In both cases the \pm 1 just measures whether you've changed the FS indicator of your rep V.
If you want to think about things labelled with components labelled by more than one irrep it'll get yuckier.<|endoftext|>
TITLE: Can I finitely color Z^2 such that (x,a) and (a,y) are different for every x,y,a?
QUESTION [6 upvotes]: I ran into this obstacle in a harmonic analysis problem; I know epsilon about coloring problems.
Is it possible to finitely color Z^2 so that the points (x,a) and (a,y) are differently colored for every x, y and a in the integers (excepting, of course, the trivial cases x=y=a)?

REPLY [11 votes]: Nope.
Suppose this were possible. For z \in Z, let Rz be the colors that appear in the row {(x,z) : x≠z \in Z}, and let Cz be the colors that appear in the column {(z,y) : y≠z \in Z}. The coloring condition is that each Cz and Rz is disjoint. Since there are finitely many possibilities for each set, find distinct z and z' where Rz = Rz', and Cz = Cz'. What color is (z,z')? Well, it's in Rz' (and not Cz'), and it's in Cz (but not Rz). So it's both in Rz and not: a contradiction.
(Ninja'd. Drat. Well, different proof at least.)<|endoftext|>
TITLE: Covering the primes by 3-term APs ?
QUESTION [11 upvotes]: Hello, the Green-Tao theorem says infinitely many k-term Arithmetic Progressions exist for any integer k.  
My question is: can we actually partition the primes into 3-term APs only (or is there a simple reason why it cannot be expected) ? And if it were possible, then what would it mean for the set of primes ?
For example fix a large integer (say M=10,000) then take the primes (except 2): 3, 5, 7, 11, 13... and remove the longest possible AP with common difference less than M as you go along. It provides the partition:
3    5    7 -- 11    17    23    29 -- 13    37    61 -- 19    31    43 -- 41    47    53    59 -- 67    73    79 -- 71    89    107 -- 83    131    179    227 -- 97    103    109 -- 101    137    173 -- ... Numerical data for the first 10,000 primes with that M shows that the average length of APs so defined is 3.2 (which is thus quite close to 3, so at least numerically there exists partitions where 3-terms APs cover a large fraction of the primes, hence the question).

REPLY [8 votes]: Using the greedy algorithm, this would follow if for any fixed prime q, there exist infinitely prime "pairs" of the form p and 2p-q. This follows from standard (difficult) conjectures if q is odd (for example, the case q = -1 corresponds to "Sophie Germaine Primes"). On the other hand, it would be an implication of such a partition that for each odd prime q, there either:
(i) exists at least one prime pair (p,2p-q),
(ii) 2q is the sum of two primes,
(iii) exists at least one prime pair (p,q-2p).
Almost all proofs showing that there exist primes of a certain form also prove that there are infinitely many such primes. Thus, I suspect, one could not prove this result without also proving the difficult conjectures alluded to above.<|endoftext|>
TITLE: A comprehensive overview of finite fields
QUESTION [12 upvotes]: I've read numerous introductions to finite fields, but I feel like my intuition about them is fairly lacking.  Considering that finite fields are the the most "inert" objects in algebraic geometry, I think I could use a serious surge of perspective.
What I would like to read now is a comprehensive overview that tells me "everything I need to know" about how finite fields and their algebraic closures work, algebraically.  I don't mind working out the proofs on my own if they are terse or absent; I'm just looking for quality and quantity of results.  Hopefully some intense reading will help steep out some of my insecurities about characteristic p.
Can anyone recommend a single source for such an overview?
Thanks!

REPLY [10 votes]: Finite Fields by R. Lidl and H Niederreiter (CUP). Probably as comprehensive as it gets.
The ams review calls it the ``the Bible of finite fields''. You can find it (the review)here.<|endoftext|>
TITLE: Characterize P^NP (a.k.a. Delta_2^p)
QUESTION [11 upvotes]: What can you say about the complexity class $\text{P}^{\text{NP}}$, i.e. decision problems solvable by a polytime TM with an oracle for SAT? This class is also known as $\Delta_2^p$.
Obviously $\text{P}^{\text{NP}}$ is in $\text{PH}$ somewhere between $\text{NP} \cup \text{coNP}$, and $\Sigma_2^p \cap \Pi_2^p$. What else is known about that complexity class?

REPLY [2 votes]: András Salamon's answer to a similar question on cs.stackexchange.com named some additional $P^{NP}$-complete problems.
Krentel also gave another problem besides the one mentioned by Ryan O'Donnell:

Input: Weighted graph $G$, integer $k$.
Question: Is the length of the shortest TSP tour in $G$ divisible by $k$?

Before him, Papadimitriou had already found:

Input: An instance of the TSP (that is, an $n \times n$ symmetric distance matrix of nonnegative integers)
  Question: Is there a unique shortest TSP tour?

According to Krentel, this was the only known $P^{NP}$-complete problem before his work.

Mark W. Krentel, The Complexity of Optimization Problems, JCSS 36 490–509, 1988. doi:10.1016/0022-0000(88)90039-6
Christos H. Papadimitriou, On the Complexity of Unique Solutions, JACM 31:2, 1984. doi:10.1145/62.322435<|endoftext|>
TITLE: Iwasawa Decomposition
QUESTION [14 upvotes]: Does anyone know where I can find a proof of the Iwasawa decomposition for reductive groups? I know that there are a couple of related results that are called the Iwasawa decomposition, but I am interested in the following statement:
Let G be a complex reductive group, let O be Taylor series with complex coefficients, and let K be Laurent series with complex coefficients. The G(K) = G(O) * T(K) * U(K), where T is a maximal torus and U is a maximal unipotent subgroup of G. 
I am interesting in finding the proof because this is how one shows that the semi-infinite cells in the affine Grassmannian cover the entire space. I have been using this fact for quite a while now and am becoming uncomfortable about not knowing where to find the proof. A proof in the case where K is a p-adic field and O is its ring of integers would also be great since I am sure a proof would carry over to the above case.

REPLY [8 votes]: The original reference for this is the paper of Bruhat-Tits (available on NUMDAM), see Prop. 4.4.3. Another reference, probably easier to read, is the book of Macdonald, "Spherical functions on a group of p-adic type", Theorem 2.6.11.
There is a nice proof of this fact using the geometry of buildings, which goes as follows. You can think only about trees (eg for G=SL(2)) to get the main ideas.
Consider the affine building X associated to G(K). The buildings at infinity of X (which is also the space of flags) can be seen as equivalence classes of sectors in X. Then U(K) is the union of fixators of sectors in such a class \xi. The group T(K) is the group of translation in some apartment A, and B:=T(K)U(K) is the stabiliser of the equivalence class of sector. G(O) is the stabiliser of some vertex o.
The main point is that the building is the union of all apartments containing a sector pointing towards \xi. It follows that, for every x in X, there is an element u of U(K) such that u.x is in A.
Let g in G. Applying this to the element x=g.o, we see that the vertex ug.o is in A. By transitivity of the action of T(K) on vertices of A, there is an element t in T such that tug.o=o. Thus tug is in G(O), which gives the decomposition of g.<|endoftext|>
TITLE: Which commutative rigs arise from a distributive category?
QUESTION [8 upvotes]: A rig is an algebraic object with multiplication and addition, such that multiplication distributes over addition and addition is commutative. However, instead of requiring that the set forms an abelian group under addition, we require only that it forms an abelian monoid. A commutative rig is a rig in which multiplication induces an abelian monoid.
A distributive category is a small category with finite products and coproducts, which I'll denote * and + respectively, such that the canonical morphism X * Y + X * Z \rightarrow X * (Y+Z) is an isomorphism. Essentially, taking products distributes over taking coproducts.
There's an apparent formal similarity between the definitions, and in fact you can get a commutative rig out of a distributive category C by taking the objects of the rig to be isomorphism classes of objects in the category (equivalently, considering the skeletal category) and letting coproducts correspond to addition and products to multiplication. For instance, if you start with the category of finite sets, you can skeletonize to obtain a commutative rig isomorphic to the rig of natural numbers.
So the question is, does every commutative rig arise this way from some distributive category? I suspect the answer is "no," and I can even think of some likely counterexamples (the nonnegative real numbers, Z), but I don't see an obvious way to prove that they are counterexamples.
Supposing that my hunch is correct, is there a nice way to classify the rigs that do arise in this manner? Is there a way to classify the commutative rings that arise when you add negatives to these rigs (analogously to the Grothendieck group)?

REPLY [6 votes]: Reid gave half the answer I was going to give — that for your rig to be arise from a category (what Schanuel calls the "Burnside rig"; by the way, the Schanuel paper in question is MR1173024, and is very good) a necessary condition is that there be no nontrivial solutions to 0 = x + y.  Similarly, there can be no nontrivial solutions to 1 = x * y.  Indeed, these conditions do not require any distributivity, so it's highly unlikely that they are sufficient conditions.
Since Categorization is popular, as Reid mentions there are constructions to get around these difficulties.  Schanuel shows that a number of natural categories have Burnside rig N[x]/(x=2x+1); the cancelative quotient of this rig is Z.  Baez and Dolan like the 2-category of finite groupoids as a prototypical example in which division can be introduced.  Indeed, they define "groupoid cardinality" to take values in the non-negative rationals.  These examples can be combined to some category of finite orbifolds.  But all these maps are very lossy.
Let me quote a bit from Schanuel's paper:

The Burnside rig (of isomorphism classes of objects, added by coproduct and multiplied 
  by product) of a distributive category has some special features, the first of which we have 
  already seen. 

If a + b = 0, then a = b = 0. 
If Σ ai = Σ bj, then there exist cij such that Σj cij = ai and Σi cij = bj. 
If a is connected (a is not 0, and a = b + c implies b = 0 or c = 0), then a is 
  cancellable (a + x = a + y implies x = y). 
1 is cancellable (whether it is connected or not). 
If ab=1, then a=b=1. 

Properties (1) and (3) follow from (2), which follows easily from the observation that 
  coproduct decompositions A = ΣI Ai correspond to maps A → 1 + 1 + ... + 1 (I terms). I don't know what additional properties characterize Burnside rigs of distributive categories. 

I'll conclude by mentioning two homomorphisms out of any rig, which may be of interest in trying to classify Burnside rigs.  First, there's the cancelative quotient R/~, where a~b if there is any x with a+x=b+x; because of Schanuel's geometric picture, he calls this "Euler Characteristic".  Second there's what Schanuel calls "dimension" R/~ where the relationship ~ is generated by 1+1~1.<|endoftext|>
TITLE: ubiquitous quantum cohomology
QUESTION [12 upvotes]: Manin stressed that every projective scheme should have a quantum-cohomology structure. I'd like to know more about that. And since the varieties considered in texts about monodromy resp. vanishing cycles which I have read are projective, I am curious about the behaviour of quantum cohomology under monodromy.
Edit: A coming seminar: "Quantum motives: realizations, detection, applications", incl. a lecture "Quantum motives: review (of) the classical idea of how to linearize algebraic geometry with an eye to utilizing it in the quantum setup." and a minicourse "Geometric Langlands and quantum motives: a link".  
Edit: The slides on Manin's talk on the concept of classical motives and it's relation to  quantum cohomology are here. 
Edit: Manin's and Smirnov's interesting computations and thoughts on e.g. a "membrane quantum cohomology" (+ russian videos of a talk in june 2011 on the program "... elucidation of this "self-referentiality" of quantum cohomology has just begun. The talk tries to outline the contours of this huge program and the first steps of it.": part 1, part 2). BTW, has anyone the text by Kapranov which is mentioned in the paper above and in Hacking's introduction?

REPLY [6 votes]: About monodromy and quantum cohomology: while everybody knows that Gromov-Witten invariants are invariant under deformations, the immediate consequence that they are invariant under the monodromy action for any smooth family of varieties is sometimes forgotten. In other words, they are invariant under the mapping class group. This can be extremely useful in computations when the mapping class group is big.
To elaborate a bit: when we are saying "GW invariants are invariant under small deformations", then we are implicitly already using the Gauss-Manin connection: If $X \to T$ is a family, say over the disc, then it doesn't make sense to identify Gromov-Witten invariants of $X_{t_1}$ and $X_{t_2}$ unless you know which GW-invariants to compare; in other words, given cohomology classes $\gamma_1, \dots, \gamma_n$ and a homology class $\beta$ on $X_{t_1}$, we need to find corresponding classes on $X_{t_2}$. Well, fortunately, this is not a problem, as $X_{t_1} \cong X_{t_2}$ as smooth manifolds, and thus $H^*(X_{t_1}) \cong H^*(X_{t_2})$.
This identification is nothing but the Gauss-Manin connection. In particular, when the bases $T$ is more complicated, you need to choose a path from $t_1$ to $t_2$ to obtain the identification; and it will depend on the homotopy class in the path. In particular, when $T$ is not simply-connected, we get a representation of $\pi_1(T)$ on $H^*(X_0)$, i.e. the monodromy action. Since it is pieced together out of identifications $H^*(X_0) \cong H^*(X_{t_1}) \cong H^*(X_{t_2}) \cong \dots$ as above, GW-invariants are invariant under this group action:
$
\langle \gamma_1, \dots, \gamma_n \rangle_{\beta}^{g,n} =
\langle \Phi(\gamma_1), \dots, \Phi(\gamma_n) \rangle_{\Phi(\beta)}^{g,n} 
$
for any $\Phi$ in the image of $\pi_1(T) \to \mathrm{Aut} H^*(X, \mathbb{Q})$.
(The mapping class group is basically the biggest possible group $\pi_1(T)$ you can obtain this way, i.e. the fundamental group of the moduli space of varieties diffeomorphic to $X$.)<|endoftext|>
TITLE: Why do I find Category Theory mostly just a way to make simple things difficult?
QUESTION [14 upvotes]: I have a basic working knowledge of category thoery since I do research in programming languages and typed lambda-calculus.  Indeed, I have refereed many papers in my area based on category theory.
But, in doing this refereeing, and in reading many important categorial papers in my area, I simply find the terminology and presentation style extremely opaque compared to style that I prefer which instead emphasizes logic inference rules and extensions of the lambda calculus.
Given the (extended) Curry-Howard Isomorphism between programming languages, logics, and categories, it's clear that I can understand the concepts I need by mapping category theory into the other two.  But, am I missing something in the process, or are the papers I'm refereeing just making things more difficult than they need to be?

REPLY [19 votes]: I think the other answers miss one aspect of this question.  Mathematicians vary in how they do math. Some are "syntactic thinkers" (maybe you), some are "conceptual", and some are "geometric" in the way they think.  That is the way Leone Burton's book Mathematicians as Enquirers: Learning about Learning Mathematics., Kluwer, 2004, analyzes it.  Others take geometric and conceptual to be variations of the same category, and different names are used for the categories, too.  
People are different, and in how they think about abstract ideas they are different in a very deep way.  That is my own experience in both my research career in and teaching.  I took logic from Joe Shoenfield (which gives me a respectable background!) and did work in abstract algebra and then discovered category theory and thought: Way to go!  That is because I think primarily conceptually.
Mike Barr said that to a person with a hammer, everything looks like a nail.  I keep translating problems into categorical language.  You go the other way.  These differences run deep, and should be taken into account when reading other people's stuff.<|endoftext|>
TITLE: Pseudorandom generators
QUESTION [8 upvotes]: Has there been any progress about constructing strong pseudorandom generators?
I'm not an expert on this topic, basically everything I know is a definition of a pseudorandom generator, the idea that they are related to one-way functions, as well as other standard parts of complexity theory. 
I'll appreciate any related information, e.g. what it is equivalent to.

REPLY [11 votes]: Ilya,
It's possible that I'm misinterpreting what you're asking (since complexity theorists, applied cryptographers, and combinatorialists all tend to use slightly different definitions of "pseudorandom"), but from a theory standpoint, I think the question is essentially solved and has been for a while.
As you mentioned, PRNGs are related to one-way functions. In particular it's easy to see that a PRNG immediately provides you with a one-way function; just plug in some initial parameters and run the generator! (Whence the implication to P \neq NP; a one-way function is clearly hard-on-average, and a hard-on-average function is clearly hard in the worst case.) It was a longstanding open problem whether the converse was true, whether you could build a PRNG from a one-way function. About a decade ago, Hastad, Impagliazzo, Levin and Luby answered this question in the affirmative in a massive and technically challenging paper. The HILL construction is hugely inefficient, but from a theory standpoint, it shows that the question of the existence of PRNGs is equivalent to the existence of one-way functions.
If you want a PRNG strong enough to derandomize BPP, the question actually becomes a bit easier, and certainly less technical. On the assumption that some problem in E requires exponential-size circuits, Impagliazzo and Wigderson construct such a generator. (The I-W paper is fantastic -- essential reading for anyone interested in derandomization.) It's also known that this is essentially the best we can do, in that derandomizing BPP requires either proving circuit lower bounds for E, or showing that one of several things no one really believes is true. (E.g., P = NP.) This is a result of I think Kabanets and someone in {Impagliazzo, Wigderson, Goldreich}, although I can't remember or find the paper.<|endoftext|>
TITLE: Use of n-transitivity in finite group theory
QUESTION [14 upvotes]: Hello,  apparently finite groups which are n-transitive with n>5 are only the permutation groups Sn or the alternating groups An+2, see e.g. page 226 this book by Isaacs http://books.google.fr/books?id=pCLhYaMUg8IC&pg=PA226
Is this characterization useful at all?  For instance, are there famous proofs (maybe in a geometric context), which use it to, say, show that a certain group is in fact some An ?

REPLY [8 votes]: [Edited to add the final field-theoretic paragraph]
Just noticed this in a list of "Related" problems.  I've seen applications to the computation of Galois groups $G$ of explicit polynomials $P \in k[X]$.  See for example Abhyankar's survey paper

[A] Abhyankar, Shreeram S., with an appendix by J.-P. Serre: Galois theory on the line in nonzero characteristic, *Bull. AMS (N.S.) 27 #1 (July 1992), 68–133.

(Note the dedication to "Walter Feit, J-P. Serre, and e-mail"!)
Let $N = \deg P$.  If $G = S_N$ or $A_N$ then this can be proved by showing that $G$ is $n$-transitive for $n$ large enough (depending on $N$), at which point $G=A_N$ if ${\rm disc}(P\phantom.)$ is square in $F$, and $G=S_N$ if not.  [This last assumes $2 \neq 0$ in $k$; there's a pseudo-discriminant criterion that works in characteristic 2.]  As T. Sauvaget notes in his question, $n>5$ is always enough, but in fact $n=4$ suffices except for $N=11,12,23,24$ (Mathieu groups), and even $n=3$ brings it down to a usually-manageable list of possibilities (see [A, pages 86–87]).
This approach can be useful because showing (say) 4-transitivity amounts to proving that a few polynomials are irreducible.  Indeed $P$ itself is irreducible iff $G$ is 1-transitive; in this case, we may adjoin to $k$ a root $X_0$ of $P$, and then the point stabilizer in $G$ is the Galois group of the degree-$(N-1)$ polynomial $P_1 := P(X)/(X-X_0)$ over $k_1 := k(X_0)$, so $G$ is 2-transitive iff $P_1$ is irreducible over $k_1$, in which case we can adjoin a second root, etc.; if $P_3$ is irreducible then $G$ is 4-transitive, and you're done (except in the four Mathieu cases where you must go one or two steps further).  Again see [A], in particular Section 4 "Throwing away roots" (p.69).
This also has the following amusing consequence.  Let $P\phantom.$ be an irreducible separable polynomial over $k$, and define $P_1, P_2, \ldots, P_n$ as before, as long as $P_m$ is irreducible for each $m<n$.  Then:

i) If each of $P_1,P_2,P_3$ is irreducible but $P_4$ is reducible then $\deg P \phantom. \in \lbrace 6, 11, 23 \rbrace$.
ii) If each of $P_1,P_2,P_3,P_4$ is irreducible but $P_5$ is reducible then $\deg P \phantom. \in \lbrace 7, 12, 24 \rbrace$.
iii) Suppose $n \geq 5$.  If $P_m$ is irreducible for each $m \leq n$, but $P_{n+1}$ is reducible, then $\deg P \phantom. = n + 3$.
Moreover, each of the allowed possibilities for $\deg P\phantom.$ in (i), (ii), and (iii) occurs for suitable $k$ and $P$.

This statement does not explicitly mention finite groups at all, but given Galois theory it is equivalent to the classification of 4-transitive permutation groups.<|endoftext|>
TITLE: Iwasawa and Cartan Decompositions.
QUESTION [7 upvotes]: Consider the tome of Bruhat and Tits: Groupes réductifs sur un corps local : I. Données radicielles valuées. Publications Mathématiques de l'IHÉS, 41 (1972), p. 5-251. (available on NUMDAM). I am interested primarily in the statements of Propositions 4.4.3 and 4.4.4 (and maybe also 7.3.1).
In the swathe of notation and technical conditions present, I find it hard to read exactly what the precise statements of these two propositions are. My question is, can anyone give
(a) a precise version of the statements of 4.4.3 and 4.4.4?
(b) if the request in (a) is too much, some sort of simplified version that is easy to state/comprehend and hopefully still reasonably general.
or
(c) an alternative reference covering at least the case of a split reductive group?
(this question is related to, but more general than Dinakar's question)

REPLY [2 votes]: I figured this exact same stuff a while ago. I will look at my notes sometime soon and post something more precise, but here is what I can say off the top of my head.

"bon" or "good" is pretty much by definition specifically the capability of $K$ to make true the 'Iwasawa decomposition', which most people nowadays would write $G=P\cdot K$ (not direct) where $K$ is one of these maximal compact-open subgroups and $P$ is a minimal parabolic (a Borel subgroup, essentially by definition, if the group is quasi-split). In fact, let's just say it's a Borel subgroup $B$ and be done with it.
that horrible symbol $\hat{\mathfrak{B}}^0$ is really the unipotent radical $U$ of the $B$
that $\hat{V}$ is something like the cocharacter group of the maximal torus $T\subset B$ interpreted as a subgroup of $B$ by evaluation at the uniformizer of the base field, i.e. $X_{*}(T)\rightarrow B : \mu \mapsto \mu(\pi)$

(so "$\hat{G}=\hat{\mathfrak{B}}^0 \hat{V} K$" really just means $G=BK$)

you have to be very careful in this book with what kind of thing $K$ is. There are four (I think) closely related groups that are differentiated only by cryptic combinations of hats, superscript zeros, and primes. The can be pointwise fixers or merely stabilizers of facets in the whole group or some subgroup, and under various hypotheses on $G$ some of them turn out to be equal to others.<|endoftext|>
TITLE: Elliptic Curves, Lattices, Lie Algebras
QUESTION [12 upvotes]: I've recently started to look at elliptic curves and have three basic questions: 

Is it correct to say that elliptic curves $E$ in the projective plane are in bijective correspondence with lattices $L$ in the complex plane via $E$ <--> $C/L$. 
If so, is there an explicit expression of the lattice generators in terms of the equation defining the curve? Or, at least, is there a simple example of a curve and its corresponding lattice?
Since every elliptic curve is a Lie group, it must have a corresponding Lie algebra. Is there an explicit expression of the Lie algebra in terms of the equation or lattice? Or, again, a simple example of a curve and its Lie algebra (or, even better, an example of a curve, its lattice, and its Lie algebra).

REPLY [6 votes]: For a bit more info on question 3: if you are interested in the elliptic curve only as a complex Lie group, then when you identify it with C/L for C the complex plane and L a lattice, the Lie algebra is canonically C and the exponential map is the reduction mod L.<|endoftext|>
TITLE: What is the field with one element?
QUESTION [100 upvotes]: I've heard of this many times, but I don't know anything about it.
What I do know is that it is supposed to solve the problem of the fact that the final object in the category of schemes is one-dimensional, namely $\mathop{\text{Spec}}\mathbb Z$.
So, what is the field with one element? And, what are typical geometric objects that descend to $\mathbb F_1$?

REPLY [10 votes]: One of the many resources pointed out above linked to the unpublished preprint by Kapranov and Smirnov called Cohomology determinants and reciprocity laws: the number field case. It's posted page by page in jpgs, but it is definitely worth looking at. They work out the details of vector spaces over F_1^n in detail and also relate the classical power residue symbol to determinants of morphisms of these vector spaces.<|endoftext|>
TITLE: Hamiltonian $S^1$ actions with isolated fixed points
QUESTION [9 upvotes]: I have in mind the following question for some time. Is there an example of a compact symplectic manifold with a Hamiltonian $S^1$-action with isolated fixed points, that does not admit a compatible $S^1$-invariant Kahler strucutre? One would say, of course there should be such an example. But I have not seen any...
Added, 2009. Apparently this is an open problem.
PS, 2019. Not anymore!

REPLY [2 votes]: Nick Lindsay and have  just proved that such a manifold indeed exists. And surprise, surprise, this is Tolman's manifold. See Theorem 1.3 and Corollary 1.4 of our paper: https://arxiv.org/abs/1912.02785<|endoftext|>
TITLE: When does the converse to Schur's Lemma hold?
QUESTION [26 upvotes]: Let $R$ be a commutative ring, let $A$ be an $R$-algebra, and let $M$ be an $A$-module.  If $M$ is simple, then End$_{A-mod}(M)$ is a division ring.
A common use is when $R$ is the complex numbers $\mathbb{C}$, and $M$ is such that End$_{A-mod}(M)$ is finite dimensional.  Then End$_{A-mod}(M) = \mathbb{C}$.
Under what circumstances (regarding $R$ and/or $A$) is the converse true, that the endomorphism ring being a division ring, or being just $R$ itself, implies that $M$ is simple?

REPLY [27 votes]: This CSL ("converse of Schur's Lemma") condition on a ring is actually a topic of interest in ring theory lately.  This basically means that there is not yet any simple answer to the question.  But there is some interesting progress toward partial answers.  For instance, a recent paper by G. Marks and M. Schmidmeier shows that the converse of Schur's Lemma holds in the category of right R-modules of finite length if and only if all extensions of simple right R-modules are split.  This holds, in particular, over any commutative ring.  
Those authors also show that a semiprimary ring R (that is, R has a nilpotent Jacobson radical J, and R/J is semisimple) satisfies CSL on the right if and only if all extensions of simple right R-modules are split, if and only if R is a finite direct product of matrix rings over local rings.  (Examples of semiprimary rings include one-sided artinian rings, such as finite dimensional algebras over fields.)
The same paper cites a number of other sources if you are interested in further exploring the topic.  For instance, there are references for the following result, similar to the one above: A one-sided noetherian ring has CSL on the right if and only if it is a finite direct product of matrix rings over local perfect rings (which must be one-sided noetherian, hence one-sided artinian).<|endoftext|>
TITLE: Where can I learn about (the asymptotics of) Toeplitz operators?
QUESTION [8 upvotes]: Toeplitz operators provide a natural language with which to do geometric quantization.  I don't want to really understand them, and I don't need them in full generality.  I'm looking for some references that will provide formulas with which to compute products of Toeplitz operators, and specifically formulas for the asymptotics of such products as Planck's constant h → 0.  I will give some background (also so the experts can correct any errors I might have), and then sketch the type of calculation I would like to perform.
Background
A quantization of a commutative Poisson algebra A (a commutative algebra A is Poisson if it comes equipped with a bilinear bracket {,}: A ⊗ A → A that is a derivation in each coordinate (i.e. Leibniz rule) and a Lie bracket (i.e. antisymmetry and Jacobi)) is a smooth bundle of algebras over the real line (or some open subinterval thereof), with certain properties.  Think of the bundle as a family Ah of (noncommutative) algebras, where h is my real variable.  The conditions are that A0 = A; that we have given linear isomorphisms  φh : A0 → Ah; and that limh→0 φh-1( h-1 [ φh(a) , φh(b) ] ) = {a,b}, where [,] is the commutator bracket and {,} is the Poisson bracket.  Quantizations are not determined by their Poisson algebras: for example, take any smooth map R → R that sends 0 to 0 and has derivative 1 at 0, and pull your bundle of algebras back along this map.
(A typical example of a quantization is the universal enveloping algebra of a finite-dimensional Lie algebra.  Indeed, if G is a finite-dimensional Lie algebra (sorry, I don't know how to make fraktur letters here), then the symmetric algebra S_G_ is the algebra of polynomials on the dual space G*, and inherits a Poisson bracket by {f,g}(p) = <p,[df(p),dg(p)]>, where <,> is the pairing G* ⊗ G → R, and since G* is a vector space, I can canonically identify T*pG* = G; [,] is the Lie bracket on G.  Anyway, let Gh be the Lie algebra G with the rescaled bracket [a,b]h = h[a,b].  Then the universal enveloping algebra U_G_h is a quantization of the Poisson algebra S_G_.)
((Another parenthetical: most people actually do quantizations of C, and then ask that complex conjugation deform well.  They also sometimes decorate their formulas with _i_s.  This has to do with the ability to find good representations noncommutative algebras in terms of bounded operators on Hilbert spaces.  I'll skip such parts of the definition.))
Quantizations were originally invented to describe quantum mechanics on Rn, and this is the situation I'm trying to understand.  Let A be an algebra of functions on the cotangent bundle T*Rn = R2n.  If we take the algebra of polynomials on R2n, it is generated by {p1,..., pn, q1,..., qn}, and the Poisson bracket is defined by {pi,qj} = δij, the Kronecker delta.  Here are two quantizations:

The QP quantization.  For h nonzero, let Ah be the algebra of differential operators on the polynomial ring R[q1,..., qn].  Construct the maps φh by sending qi∈A to (multiplication by) q in Ah, and send pi∈A to the partial derivative operator δi.  For a more complicated monomial, first write it with all qs to the left and all ps to the right, and then apply the above maps letter-by-letter.
The Z Z-bar quantization.  Complexify A, and change variables so that zj = qj + ipj and wj = qj - ipj.  Write every monomial with the zs to the left and the ws to the right, and let the monomials act on the polynomial algebra C[z1,..., zn] analogous to in the QP case.  I believe that you can reincorporate by real structure by recording the action of "complex conjugation".

So far, I've been playing with the QP quantization.  This has a natural extension to the algebra of functions on  T*Rn that are polynomial in the p variables but smooth in the q variables (i.e. C∞Rn ⊗ R[p1,..., pn]).
What I've been told is that Toeplitz operators quantize (at least) the algebra of analytic functions on T*Rn, (actually, they work much more generally to quantize symplectic manifolds, but I don't need them to), and that the quantization corresponds to the Z Z-bar quantization above.
The calculation I'd like to do
Starting with a Lagrangian on Rn, it's relatively straightforward to write down the formal Feynman path integral (a power series in h), and I know some things about the derivatives of this formal power series with respect to the physical variables.  This series is a putative solution to Schroedinger's equation.  When the Lagrangian is quadratic in velocity, I can compute the Hamiltonian explicitly, and sure enough Schroedinger's equation is satisfied.  When the Lagrangian is not quadratic, the Hamiltonian is still well-defined, but I don't have explicit enough formulas, and in general it is not polynomial in momentum.  I do have various differential equations satisfied by the Hamiltonian, and I'm hoping that with these and the right formulas for the asymptotics of products of Toeplitz operators, I can check the Schroedinger equation (asymptotically) without knowing precisely what the Schroedinger operator is.

REPLY [2 votes]: You probably need some conditions to restrict the kind of Schroedinger operators you want to work with.  Have a look at the work of Charles and Vu-Ngoc on Toeplitz operators and the semiclassical limit, in particular theorem 1 of this http://people.math.jussieu.fr/~charles/Articles/BerToep.pdf   and  section 1.3 of this http://people.math.jussieu.fr/~charles/Articles/Half1.pdf   and for spectral asymptotics in more concrete examples (a non-degenerate potential well) see this http://hal.archives-ouvertes.fr/docs/00/06/73/18/PDF/birkhoff.pdf<|endoftext|>
TITLE: 2-adic Coefficients of Modular Hecke Eigenforms
QUESTION [18 upvotes]: Suppose that $N$ is prime, and consider the normalized cuspidal Hecke eigenforms of weight 2 and level $\Gamma_0(N)$.
For such an eigenform $f$, the coefficients generate (an order in) the ring of integers of a totally real field $E$. The corresponding simple abelian variety quotient $A_f$ of  $J_0(N)$ has dimension $[E:\mathbb{Q}]$. Let $C(N)$ denote the maximal degree $[E:\mathbb{Q}]$ amongst all such eigenforms. It is not too hard to prove that $C(N)$ is unbounded, and not much harder to show that the $\lim \inf$ of $C(N)$ as $N\to \infty$ is infinite. (I don't want to mention the argument here because I don't think it will apply to my question below.)
Is the same result true over $\mathbb{Q}_2$? Namely, fix an embedding of $\bar{\mathbb{Q}}$ into $\bar{\mathbb{Q}}_2$. Then, for any $d$, is it true that for all sufficiently large prime $N$ there exists a cuspidal eigenform of weight 2 and level $\Gamma_0(N)$ whose coefficients generate some field $E/\mathbb{Q}_2$ with $[E:\mathbb{Q}_2] > d$?
This would be interesting to know even for $d = 1$.

REPLY [12 votes]: A natural way to approach this question is to ask for an eigenform $f$ with coefficients in a local field $K$ with residue field degree $\ge 2$. Or, asking for slightly more, with coefficient ring $\mathcal{O}$ admitting a surjective map to a field $\mathbf{F}$ of order divisible by $4$ (this is slighly more since the rings $\mathcal{O}$ can typically be non-trivial orders in $\mathcal{O}_K$). By Serre's conjecture, this is equivalent to asking for the existence of an irreducible Galois representation
$$\rho: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{F})$$
with Serre weight and conductor $(2,p)$ (note, no oddness condition!). It's hard to construct such representations, but one natural way is to use induced ($=$ projectively dihedral) representations. 
If $E$ is the corresponding quadratic extension, and $F/E$ the cyclic extension, then the Serre weight and level will be $(2,p)$ providing that $F/E$ is totally unramified and $E = \mathbf{Q}(\sqrt{-1})$, $\mathbf{Q}(\sqrt{p})$, or $\mathbf{Q}(\sqrt{-p})$.
(EDIT: In a previous version of this post, I only thought about extensions that were totally unramified at $2$ for some unknown reason.)
One is in good shape providing that the odd part of the class group of one of the fields $\mathbf{Q}(\sqrt{\pm p})$ is not $(\mathbf{Z}/3 \mathbf{Z})^n$. Since one can't really say much about real quadratic fields, let's think about the imaginary quadratic fields.
It's a theorem, proved by (amongst other people)  Lillian Pierce, and Jordan Ellenberg and Akshay Venketesh, that 
the 3-torsion part of the class group is a negligible part of the class group. Thus we are done if $p$ is big providing that the $2$-part of the class group is not so big. The $2$-part is trivial if $p$ is $3$ mod $4$, and has order $2$ if $p$ is $5$ mod $8$. However, it is quite possible that the class group of $\mathbf{Q}(\sqrt{-p})$ is $\mathbf{Z}/2^n \mathbf{Z}$, and very rough heuristics suggest that this might happen infinitely often.
[EDIT: LaTeX issues now mostly sorted in the sequel]
If $\mathbf{T}$ denotes the Hecke algebra tensored with $\mathbf{Z}_2$, then
$\mathbf{T}$ is a free $\mathbf{Z}_2$-module of rank $n = \mathrm{dim}(S_2(\Gamma_0(p)))$. 
It is also a semi-local ring. For each maximal ideal $m$ of $\mathbf{T}$, we are asking that 
$$\mathbf{T}_{m} \otimes \mathbf{Q}$$ 
is not a number of copies of the 2-adic numbers. If $m$ is the $2$-adic Eisenstein ideal, then in a paper with Matthew Emerton, I prove that 
$$\mathbf{T}_{m}/2$$
has rank $2^{e-1}-1$, where the $2$-part of the class group of
$\mathbf{Q}(\sqrt{-p})$ (we may assume that $p$ is $1$ modulo $8$) has order $2^{e}$.
If $p$ is $9$ modulo $16$ then $\mathbf{T}$ localized at $m$ is also a domain, and so we are done in this case if $2^{e-1} -1 \ge d$. Even if $p$ is $1$ modulo $16$, this shows that the bulk of $\mathbf{T}/2$ must come from non-Eisenstein primes $m$, since 
$2^e \le h \ll p^{1/2 + \epsilon} \ll p/12 \simeq n$.
So, in the case $d = 1$, one may assume that the odd part of the class group is
$3$-torsion, and then hope that for the corresponding $m$, one can prove that
the corresponding rings $\mathbf{T}_{m}/2$ can't be too large.
It turns out that the relevant question becomes: Given a $\overline{\rho}$ corresponding to an $S_3$ extension, and given a finite flat deformation of
$\overline{\rho}$ to $\mathrm{SL}_2(A)$ for a local ring $A$ killed by $2$, can one bound the rank of $A$ as an $\mathbf{F}_2$-module? For example, can one prove that $A$ has rank $p^{1 - \epsilon}$? The example of Eisenstein $m$ suggests that one can not neccesarily do better than $p^{1/2 + \epsilon}$. 

Here is an answer that shows works for d = 1, and works for general d assuming GRH.
Step I. Let $C$ be the class group of $\mathbf{Q}(\sqrt{-p})$. The group $C$ decomposes as 
$C_{odd} \oplus C_{even}$.
Step II. Suppose that $C_{odd}$ is not annihilated by $2^{2d} - 1$. Then there exists a dihedral representation induced from a character of $C_{odd}$ which is finite flat at $2$, ordinary at $p$, and has which has image in 
$$\mathrm{SL}_2(\mathbf{F})$$
for a field $\mathbf{F}$ of degree $>d$. 
Thus we are done unless $C_{odd} = C_{odd}[2^{2d} - 1]$.
Step III. If $d = 1$, then then the $2^{2d} - 1 = 3$-torsion of $C_{odd}$ has small order, namely, of order at most
$$p^{1/2 - \delta}$$
 for some explicit $\delta > 0$. For general $d$, the $2^{2d} -1$-torsion has order at most $p^{\epsilon}$, assuming GRH. 
Step IV: If $p$ is $-1$ modulo $4$ or $5$ modulo $8$, then $C_{even}$ has order $\le 2$.
This implies by Steps II and III that $C$ has order $\le p^{1/2 - \delta}$, which contradicts the estimate $h \gg p^{1/2 - \epsilon}$.
Step V: We may assume that $p$ is $1$ modulo $8$. Let $|C_{even}| = 2m$. Using the estimate $h \gg p^{1/2 - \epsilon}$, we deduce that
$m \gg p^{\delta - \epsilon}$ for some explicit $\delta > 0$.
Step VI: Let $\mathbf{T}$ denote the localization of the Hecke algebra at the Eisenstein prime of residual characteristic $2$. It is a consequence of one of the main theorems of FC-ME that one has $\mathbf{T} = \mathbf{Z}_2[x]/f(x)$, where:
$$f(x) \equiv x^{m - 1} \ \mathrm{mod} \ 2,$$
$$\mathbf{Z}_2/f(0) \mathbf{Z}_2 \simeq \mathbf{Z}_2/n \mathbf{Z}_2,$$
and $n$ is the numerator of $(p-1)/12$. (Compare the discussion in Mazur's Eisenstein ideal paper, section 19, page 140.)
Step VII: If the coefficients of all forms of level $\Gamma_0(p)$ define extensions of degree $\le d$, then the normalized $2$-adic valuation of every root of $f(x)$ is at least $1/d$. We deduce that
$$\frac{m - 1}{d} \le v_2(n),$$
Since $v_2(n) \le \log_2(n) < \log_2(p)$, we deduce that
$$ m \le d \log_2(p).$$
This contradicts the previous estimate $m \gg p^{\delta - \epsilon}$ for sufficiently large $p$.
Remark: For $d = 1$, it should be easy to use the unconditional results to give an explicit lower bound on possible $p$.

To summarize, for any $d$, and sufficiently large $p$, there exists either:
(i) A modular form $f$ of level $\Gamma_0(p)$ with coefficients in an extension of $\mathbf{Q}_2$ which contains an unramified extension of degree $> d$, and whose residual representation is dihedral,
(ii) A modular form $f$ of level $\Gamma_0(p)$ which coefficients in an extension of $\mathbf{Q}_2$ which contains a ramified extension of degree $> d$, and whose residual representation is Eisenstein.

Extra: For any $p$, the coefficients of an $f$ of level sufficiently divisible by $p$ contains   the totally real subfield of  the $p^n$th roots of unity.
For any $p$, and sufficiently large $n$, this contains a large extension of the $2$-adic numbers.
Reason why this argument still sucks: Doesn't work at all for primes $> 2$, and uses GRH for $d > 1$.<|endoftext|>
TITLE: Most harmful heuristic?
QUESTION [167 upvotes]: What's the most harmful heuristic (towards proper mathematics education), you've seen taught/accidentally taught/were taught? When did handwaving inhibit proper learning?

REPLY [6 votes]: The "size" of a finite-dimensional vector space is proportional to its dimension.

In fact, the "size" of a finite-dimensional vector space is almost always better thought of as being exponential in its dimension. This is easiest to see for (finite-dimensional) vector spaces over finite fields, which have finite cardinality. But it's a better heuristic even for vector spaces over infinite fields.
Internalizing the correct intuition makes it clear why forming the (algebraic) direct product of two vector spaces causes their dimensions to add, and not to multiply as you might naively expect based on the fact that taking the direct product of groups multiplies their orders.
Another confusing point to which this misconception leads regards the advantage that quantum computers give over classical ones. The difference is sometimes stated as "quantum computers have a state space that's exponentially large in the number of qubits," but this is highly misleading, because classical computers also have a state space that's exponentially large in the number of bits. The better intuition is: since quantum computers have a state space whose dimension is exponentially large in the number of qubits, the state space itself is actually doubly exponential in the number of qubits, while the state space of a classical computer is only singly exponential in the number of bits.
The reason why this misconception is so widespread is that early courses in linear algebra almost always begin with vectors spaces over infinite fields (usually $\mathbb{R}$ or $\mathbb{C}$), which have infinite cardinality, so the dimension is the only finite number available. This practice leads to misleading intuition for general vector spaces.<|endoftext|>
TITLE: Any work on the Adams-Watters triangle?
QUESTION [5 upvotes]: Does anyone know whether any arithmetical or asymptotic results have been obtained about the Adams-Watters triangle?

REPLY [3 votes]: There seems to be no response to this, but perhaps somebody knows something about it in another terminology.
Franklin T. Adams-Watters defined a triangle similar to Pascal's, but where the latter has 
c = a + b between and under a and b, Adams-Watters takes c = (a+b)/gcd(a,b). The first few 
rows look like this:
                      1
                    1   1
                  1   2   1
                1   3   3   1
              1   4   2   4   1
            1   5   3   3   5   1
          1   6   8   2   8   6   1
        1   7   7   5   5   7   7   1
      1   8   2  12   2   12  2   8   1

Adams-Watters summed the rows and obtained a sequence a(n) which is numbered A125606 in the OEIS database. He conjectures that log(a(n))/n tends to log(2*zeta(3)/zeta(2)). That was what I meant when I asked about asymptotic information. (The analogous limit is log(2) for Pascal's triangle, of course)
I am actually more interested in whether the Adams-Watters triangle has any nontrivial arithmetic structure. When you add a and b, the sum is trivially divisible by gcd(a,b), 
so the operation (a+b)/gcd(a,b) removes this trivial information. Pascal's triangle has a lot of arithmetic structure, but I did not see any when I factored the entries in the A-W triangle out to the fiftieth row. Perhaps dividing by the gcd leaves nothing, or maybe I didn't think about it the right way.
Two caveats: The little piece above is not enough to see what happens. For example, the 
twelfth row is the first that has both odd and even entries (apart from the 1s). Also there seems no reason to think that the A-W triangle has any combinatorial significance.<|endoftext|>
TITLE: In a Banach algebra, do ab and ba have almost the same exponential spectrum?
QUESTION [21 upvotes]: Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $x\in A$ by $$e(x)= \{\lambda\in\mathbb{C}: x-\lambda1 \notin G_1(A)\},$$ where $G_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity.

Is it true that $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$?

Equivalently, is it true that 
$1-ab$ is in $G_1(A)$ if and only if $1-ba$ is in $G_1(A)$, for all $a,b \in A$?
Note: The usual spectrum has this property.
Just an additional note:
We have $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$ if
1) The group of invertibles of $A$ is connected, because then the exponential spectrum of any element is just the usual spectrum of that element.
2) The set $Z(A)G(A) = \{ab: a \in Z(A), b\in G(A)\}$ is dense in $A$, where $Z(A)$ is the center of $A$. (One can prove this). In particular, we have $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$ if the invertibles are dense in $A$.
3) $A$ is commutative, clearly.
But what about other Banach algebras? Can someone provide a counterexample?

REPLY [5 votes]: Just to update this: a negative solution was recently given by Klaja and Ransford. See arXiv 1510.08109.<|endoftext|>
TITLE: The ants-on-a-ball problem
QUESTION [28 upvotes]: Suppose I put an ant in a tiny racecar on every face of a soccer ball. Each ant then drives around the edges of her face counterclockwise. The goal is to prove that two of the ants will eventually collide (provided they aren't allowed to stand still or go arbitrarily slow).
My brother told me about this result, but I can't quite seem to prove it. Instead of a soccer ball, we should be able to use any connected graph on a sphere (provided that there are no vertices of valence 1). We may as well assume there are no vertices of valence 2 either, since you can always just fuse the two edges.
I (and some people I've talked to) have come up with a number of observations and approaches:

Notice that if two ants are ever on the same edge, then they will crash, so the problem is discrete. You can just keep track of which edge each ant is on, and let the ants move one at a time. Then the goal is to show that there is no way for the ants to move without crashing unless some ant only moves a finite number of times.
You can assume all the faces are triangles. If there is a face with more than three edges, then you can triangulate it and make the ants on the triangles move in such a way that it looks exactly the same "from the outside". If there is a 1-gon, it's easy to show the ants will crash. If there is a 2-gon, it's easy to show that you can turn it into an edge without changing whether or not there is a crash.
One approach is to induct on the number of faces. If there is a counterexample, I feel like you should either be able to fuse two adjacent faces or shrink one face to a point to get a smaller counterexample, but I can't get either of these approaches to work.
If you have a counterexample on a graph, I think you get a counterexample on the dual graph. Have the dual ant be on the next edge along which a (non-dual) ant will pass through the given vertex.
It feels like there might be a very slick solution using the hairy ball theorem.

REPLY [7 votes]: A quick comment on the idea of "a very slick solution using the hairy ball theorem".  Any such very slick proof will surely only use the fact that the Euler characteristic is non-zero, and so should apply just as well to surfaces of higher genus (at least two).  For instance, if I understand it correctly, then Reid's answer above would work just as well on a higher-genus surface, by the Poincare--Hopf Theorem.
But the theorem does not hold on surfaces of higher genus.
What follows would probably be better with pictures, but I'll try to describe it without and hope for the best.
For instance, it's easy to divide a torus into two rectangles such that there are "traffic schedules" with no crashes.  (OK, the Euler characteristic of a torus is zero, but bear with me.)  (Also, what is to ants as a traffic schedule is to cars?)
Now consider the surface of genus two as an octagon with sides identified, as usual.  You can divide this into two rectangles and two pentagons in a way that mimics two copies of the torus picture in the previous paragraph: the surface is the union of two tori with boundary, and each torus is divided into a rectangle and a pentagon.  Schedule each torus as before (stretching one of the ants' route over the fifth side of the pentagon).  As only one ant from each torus traverses the fifth side of the pentagon, it is easy to arrange that the ants from different tori do so at different times.
Does that make sense, or does anyone want a picture?
EDIT:
Oh, and of course it follows that the theorem is also false on any orientable surface of even higher genus, as they all cover the surface of genus two.<|endoftext|>
TITLE: Number of subdivisions of an n-gon
QUESTION [5 upvotes]: Suppose I have a regular n-gon. I want to draw some noncrossing diagonals to subdivide it into smaller polygons. In how many ways can I do this? The vertices are unlabeled, so I don't distinguish between rotations or reflections of a given subdivision.
A triangle has 1 subdivision (do nothing!); a square has 2, a pentagon has 3, and a hexagon has at least 9 -- I'm not certain that I haven't missed any.
In fact, what I would really like is not just a count, but an algorithm for generating such subdivisions. There are obvious algorithms that generate some subdivisions multiple times, but what I'd really like is an algorithm that only generates distinct subdivisions, and that generates all of them.

REPLY [4 votes]: I think this could be A001004 in Sloane's Encyclopedia.  It's hard to be sure without checking the references given there; the sequence is defined as "Number of symmetric dissections of a polygon.", which may or may not be what you mean.  (In particular, the OEIS claims the next term is 20 and I'm afraid to try to check that.)<|endoftext|>
TITLE: Why does non-abelian group cohomology exist?
QUESTION [37 upvotes]: If $K$ is a non-abelian group on which a group $G$ acts via automorphisms, we can define 1-cocycles and 1-coboundaries by mimicking the explicit formulas coming from the bar resolution in ordinary group cohomology, and thus we have a reasonable notion of $H^1(G, K)$.
It turns out we have a part of the expected long exact sequence, until this construction breaks down for building $H^i$ when $i > 1$, where the long exact sequence stops. There are other analogues to ordinary group cohomology as well. The only proof I've ever seen of any of this is by hand. Is there some deeper explanation of why non-abelian group cohomology exists (and then ceases to exist)?

REPLY [22 votes]: A concrete and arithmetically useful way to interpret it without appeal to explicit cocycle formulas is to express everything in the language of torsors.  More specifically, for arithmetic purposes if the group G is Gal(F'/F) for a Galois extension F'/F and if the group K is H(F') for an F-group scheme H of finite type and K is equipped with the evident left G-action then ${\rm{H}}^1(G,K)$ is the set of isomorphism classes of H-torsors over F which split over F' (i.e.,, admit an F'-rational point).  The low-degree exact sequence can then be expressed entirely in such terms, using pushouts and pullbacks with torsors. (Implicit in the argument is effectivity of Galois descent for H-torsors, which uses that H is quasi-projective over F.)
This is useful in settings as varied as H an abelian variety and H a linear algebraic group, and even the non-smooth case. In fact, when using non-smooth H it is rather restrictive to use Galois cohomology (but not unnatural if studying Tate-Shafarevich sets with coefficients in an Aut-functor, such as for a projective variety), and in such cases the "right" variant that is often more useful is to work with torsors for the fppf topology over F. The torsor viewpoint also gives a useful perspective when working over richer base rings than fields, such as rings of S-integers in a global field, even in the case of a smooth coefficient group (over the ring of S-integers), for which the etale topology is "enough". See section 5.3 in Chapter I of Serre's book on Galois cohomology for the Galois case, Milne's "Etale cohomology" book for generalization with flat and 'etale topologies, and Appendix B in my paper on "Finiteness theorems for algebraic groups over function fields" for a concrete fleshing out of the dictionary between the torsor and Galois languages (where I work with affine group schemes, due to the context of that paper).
Some papers of Mazur and Grothendieck (not as co-authors...) on abelian varieties make creative use of the torsor viewpoint when working with Tate-Shafarevich groups.  The exact sequence for Brauer groups in global class field theory also has a useful interpretation via torsors; see Grothendieck's papers on Brauer groups for more in that direction (and  somewhere in there he also discusses Tate-Shafarevich). Beware that when the base is not a field (or even when it is a field but we relax "quasi-projective" to "locally finite type" on $H$, such as for Aut-schemes of projective or proper varieties) then effectivity of descent for torsors is not at all clear, even with quasi-projective hypotheses, and so the torsors often need to be understood to be taken in the category of algebraic spaces (for which fppf descent is always effective).  In the Néron Models book they have a discussion (somewhere in Chapter 6, I think) on effectivity of descent for torsors if one wishes to avoid algebraic spaces (under suitable hypotheses on the "coefficient group"), but working with algebraic spaces isn't so bad once one gets used to them and it is a more natural setting due to their better general behavior with respect to descent.<|endoftext|>
TITLE: Solving polynomial equations when you know in which number field the solutions live
QUESTION [16 upvotes]: Suppose I have a bunch of polynomial equations with coefficients in a number field, and suppose further that I'm guaranteed a priori that they have a solution in that number field.  Can I leverage that knowledge into a technique for solving the equations more easily?  The cases we care about are massively overdetermined systems of linear and quadratic equations.

REPLY [2 votes]: One minor comment: if you can ever reduce to solving equations in a single variable, and looking for rational roots, then you can use the rational root theorem. You can reduce to the case where your number field is the rationals easily enough. (For example, if you started out looking for roots in Q(i), write everything as a+bi.) But I don't know that there is any good way to reduce to equations in a single variable.<|endoftext|>
TITLE: Abel's equation for the dilog
QUESTION [16 upvotes]: Abel's identity for the dilogarithm (see the  wikipedia page about polylogarithms)
plays a role in web geometry as it is one of the abelian relations of the
first example of exceptional web (Bol's 5-web) to appear in the literature. 
I have heard it is important in other domains (cohomology of SL(3,C), algebraic K-theory, motives ). 
I would like to learn more about it. 
I am asking for:

Insights on why Abel's identity is relevant in this or that field;
References where it plays a role.


Edit.
I have just learned from this blog about Bridgeman's orthospectrum identity.
Those interest in the question above might want to take a look at it.

REPLY [8 votes]: There is a remarkable article, "The remarkable dilogarithm," J. Math. Phys. Sci. 22 (1988),   131--145, by Don Zagier, which was recently reprinted and updated as "The dilogarithm function" (63 pages!) in one of the collections by Springer Verlag.<|endoftext|>
TITLE: Strong Law of Large Numbers for weakly dependent random variables
QUESTION [11 upvotes]: Let Xi be a sequence of identically-distributed random variables with finite-range dependence (i.e. there exists I such that if |i-i'| ≥ I, then Xi and Xi' are independent), and a finite moment-generating function (i.e. EerXi < ∞ for all r ∈ R).
It's not too hard to show that Xi satisfies a strong law of large numbers, and I've got a proof written.  However, I'm sure that this is a standard theorem in the probability literature, and I'd rather just cite it in the paper I'm writing.  Do you have a good reference for this result?
Here are two follow-up generalizations:  what if Xi instead has only a finite moment condition?  Or what if Xi has exponential correlation decay (i.e. EXiXi' ≤ Ce-c|i-i'| for some positive c, C)?

REPLY [7 votes]: This question sounds like an exercise: Split the sequence into I sequences of iid random variables. Apply the classical SLLN to each sequence.  Recombine.  
Tom: 
Of course it is true with exponential decay of the corellation function, but it is not easy. 
The essential difficulty is that one wants to reduce the SLLN to an exponentially growing subsequence of N's.  In the classical case, this is done by a martingale inequality. (Prob of a supremum of a martingale is dominated by the probability at end of the martingale.)  
Once one moves away from an implicit martingale structure, then tricks have to be employed---of which the most obvious is that if the sequence of random variables is bounded, then obviously you can reduce to an exponentially growing subsequence.  This point is much of the content of the paper of Lyons cited already.  
Not sure that this would appear in a text book however. My sense is that these considerations are well-known.<|endoftext|>
TITLE: Sheaf description of $G$-bundles
QUESTION [29 upvotes]: Now, among algebraic geometers, at least, it is well known that there is an equivalence between locally free $\mathcal{O}_X$-modules of rank $n$ and vector bundles of rank $n$.  So, equivalently, principal $\mathrm{GL}(n,\mathbb{C})$-bundles are given by locally free sheaves of rank $n$.
So...what about other groups?  I guess that $\mathrm{SL}(n,\mathbb{C})$ bundles are then locally free sheaves of rank $n$ with top exterior power trivial, but can we phrase everything in terms of the properties of a sheaf and a group?
My guess is that in this context, if we can do it, we'll end up with something that's not quite locally free sheaves of rank n for $\mathrm{GL}(n,\mathbb{C})$, but which will be equivalent.
Note: I'm aware that we could just say something like "the sheaf of local sections of a $G$-bundle" but I'm looking for something intrinsic, a set of properties of the sheaf without reference to the geometric bundle, which can be reconstructed from the sheaf description.

REPLY [3 votes]: Adding it so that it's easily found.  The thing I was looking for, which is generally not written out except in the case of vector bundles, is that the sheaf of sections of an F-bundle with fiber F is a sheaf of sets that is locally isomorphic in the etale topology to the sheaf hom(-,F) ranging over small enough open subsets of X.<|endoftext|>
TITLE: Making an l_2 distance out of l_1 distance
QUESTION [12 upvotes]: If we think of the l1 distance as a grid-distance between points, then we can think of l2 distance as what we get when we "shortcut" the grid by going "inside" a cell. 
Making the grid finer doesn't change the l1 distance, so there's no obvious sense in which the l2 distance can be seen as a limiting version of the l1. 
So here's my question: is there any way to take an l1-like distance and extract an l2 like distance from it (possibly as a limiting case). I'm asking because i have a distance defined in a discrete space that has an l1-like behaviour, and I'd like to generalize it as the discrete space gets finer and finer, but I want to end up with a distance that goes "directly" through the space like l2. 
apologies if this is way too vague.

REPLY [4 votes]: In some cases, you can recover L_2 information in the limit by considering random walks, or said differently, basically by counting paths. When you say "discrete space with L_1-like behavior," I imagine a big grid of points (something like Z^d) that are connected to their nearest neighbors by edges, with the distance between two points being defined in the natural graph-theoretic way as the length of the shortest path between them. If that's the case, you can consider a uniform random walk along the edges of the graph, and then successively refine the grid. This is important: if you refine the grid, say, by a factor of two (so each step of the random walk becomes half as big), then you have to refine the walk by taking steps four times as fast. (In general, when you refine by a factor F, you have to take steps F^2 times as fast.) As you keep refining, the probability of the walk ending up near a particular point at a fixed time in the future may stabilize to something that looks like a function of L_2 distance from the starting point. 
If this construction works in your case, then you can think of the logarithm of the probability of ending up in a small neighborhood as being roughly proportional to the volume of the neighborhood and the square of its L_2 distance from the starting point, but except in very special cases, the approximation probably won't be good for all pairs of points, especially if they aren't very close to one another. Note that computing such probabilities basically amounts to counting paths of particular lengths between points, as opposed to just finding the length of the shortest path (which would roughly corresponding to the L_1 distance). 
How much that computation will actually resemble an L_2-like distance depends on many things. The construction will work basically as stated when you successively refine a grid like Z^d because of the central limit theorem. If your "discrete space" isn't very similar to that, then the construction might not work at all, or it might give you something that isn't quite right but is "close enough for government work," so to speak. You'll have to be the judge of that.<|endoftext|>
TITLE: Best algebraic geometry textbook? (other than Hartshorne)
QUESTION [224 upvotes]: I think (almost) everyone agrees that Hartshorne's Algebraic Geometry is still the best.
Then what might be the 2nd best?
It can be a book, preprint, online lecture note, webpage, etc.
One suggestion per answer please. Also, please include an explanation of why you like the book, or what makes it unique or useful.

REPLY [7 votes]: Next to the classics, I enjoy Introduction to schemes by Ellingsrud-Ottem very much (available here).
The best part about this book is that many insightful examples are explicitly computed accompanied with nice, colorful pictures. E.g. already chapter 5 (just after introducing schemes!) spans 18 pages of examples about gluing (from $\mathbb{P}^n$ to hyperelliptic curves to Hirzebruch surfaces $\mathbb{F}_r$). Prof. Ottem himself has mentioned that writing out these examples are important to him, and it really shows!<|endoftext|>
TITLE: Line bundles on moduli spaces
QUESTION [10 upvotes]: This is perhaps too broad or vague (or silly) a question, but here it is anyway: why should I care about constructing line bundles on a moduli space?  This comes up all of the time, but I seem to be missing the (probably obvious) motivation.
Ideally, it would be nice to attach to this question a particular moduli space (vector bundles on a curve, instantons, etc.), but I think I will leave the task of finding an efficient yet instructive example to someone with more knowledge.
Thanks.

REPLY [4 votes]: In general Cohomology is a tool to linearize global properties (just like calculus is a tool to linearize local properties). 
Line bundles are elements in the (probably) most important cohomology group you have on the space - the Picard group. Depending how you construct them you can know a lot of things about the NEF cone of the space and it's Kodaira dimension. 
So, assuming that moduli space are interesting (they are because you don't have so many ways to construct concrete objects), and that the invariants I mentioned are (they are because they are almost the only ones we know for general spaces), it is interesting.<|endoftext|>
TITLE: What is Drinfeld's manuscript "Best Dream" (in Russian!) about?
QUESTION [6 upvotes]: I would like to know what Drinfeld's scanned manuscript "Best Dream" is about: the title makes me curious.   
It's in Russian.

REPLY [7 votes]: From the first line it appears to about D-modules on stacks.
The next topic is an attempt to construct a "duality" (called F) for derived category of (quasicoherent) D-modules on Bun_G (the space/stack of G-bundles). The problem is to find a "reverse Langlands transform" (?) by defining a suitable scalar product on those.
In the middle a single question  appears, "What is the Eisenstein fuctor?" (indeed, what is it?) Next, there's a question about Hecke functors.
Part two starts with the self-explanatory
        D (Bun_G)  =?=  O(LocSys)

The Eisenstein functor for GL(2) appears in the discussion, with the standard definition. Some constructions relevant to the formula above appear. Pages 59–60 then are in English. Then the derived categories and Eisentein functor continue.
Page 76 contains some specific questions ("I don't understand!") again along the topics above which continue until the end of paper.<|endoftext|>
TITLE: Minkowski sum of small connected sets
QUESTION [20 upvotes]: Suppose that the convex hull of the Minkowski sum of several compact connected sets in $\mathbb R^d$ contains the unit ball centered at the origin and the diameter of each set is less than $\delta$. If $\delta$ is very small (this smallness may depend on $d$ but on nothing else), does it follow that the sum itself contains the origin?

REPLY [13 votes]: I finally figured it out. My solution is here. I would repost it on mathoverflow but until LaTeX is enabled, it is quite hard for me to communicate such things here...<|endoftext|>
TITLE: "isotropic" subspaces of a simple Lie algebra
QUESTION [21 upvotes]: Let $\bf g$ be a finite-dimensional real simple Lie algebra of compact type and let $\left<-,-\right>$ denote the positive-definite inner product induced from the negative of the Killing form.   Let $\Omega$ denote the trilinear map defined by
$$\Omega(X,Y,Z) = \left<[X,Y],Z\right> .$$
It is easy to see that it is alternating, because of the ad-invariance of the Killing form.  Let us call a subspace $S\subset{\bf g}$ isotropic if $\Omega$ vanishes identically when restricted to $S$; that is, if
$$\Omega(X,Y,Z) = 0, \forall X,Y,Z \in S.$$
In other words, $S$ is isotropic iff $[S,S] \subset S^\perp$, where ${}^\perp$ means the perpendicular complement relative to the Killing form.
Furthermore we say that an isotropic subspace is maximal if it is not properly contained in an isotropic subspace.  It is not hard to show that $S$ is maximal isotropic if and only if $[S,S] = S^\perp$.
The question is how to characterise the maximal isotropic subspaces of $\bf g$.
It is easy to see that the maximally isotropic subalgebras are precisely the Cartan subalgebras, but I am interested in subspaces which are not necessarily subalgebras.
The only examples I know are those for which $S = {\bf k}^\perp$ and ${\bf k} < {\bf g}$ a subalgebra, whence 
$${\bf g} = {\bf k} \oplus S$$
is a symmetric decomposition corresponding to the compact riemannian symmetric space $G/K$.
Question: Are there any other maximal isotropic subspaces?

REPLY [4 votes]: The answer is yes, there are other maximal isotropic subspaces for at least some real Lie algebras of compact type.  I thought of a dimension-counting argument that avoids an explicit construction.  A subspace of symmetric type is rigid; it has no free parameters under than conjugation in the Lie algebra $\mathbf{g}$.  In the dimension sense, there may not be enough of them to contain all of the isotropic subspaces.
For instance, suppose that $\mathbf{g} = \mathrm{su}(3)$.  It is 8-dimensional and it has two types of symmetric subspaces of symmetric type, $W_5 = \mathrm{so}(3)^\perp$ which is 5-dimensional, and $W_4 = (\mathrm{su}(2) \oplus \mathrm{u}(1))^\perp$, which is 4-dimensional.  Each such isotropic subspace $W_n$ lies in a conjugacy class which is $n$-dimensional.
On the other side, suppose that we build an isotropic space as a flag $V_1 \subset V_2 \subset \cdots \subset V_k$. then there are 8 parameters for $V_1$, 7 for $V_2$, at least 4 for $V_3$ (because the kernel of the map to $\bigwedge^2 V_2^*$ is at least 5-dimensional), and at least 1 for $V_4$ (by the same kernel argument).  Then we have to subtract the dimension of the flag variety of $V_4$, which is 6.  That makes a moduli space of $V_4$s which is at least 12-dimensional, and possibly exactly that.  That is bigger than the manifold of $W_4$s, and it is also bigger than the manifold of $W_5$s times the Grassmannian of 4-planes in each $W_5$.
I haven't checked much larger cases than this one.  You could try to refine this argument by explicitly identifying when $V_j$ for some small value of $j$ knocks out some $W_n$ that would have contained it.  I conjecture that there are arbitrarily large examples, but I don't know if you need such a refinement to get them.<|endoftext|>
TITLE: What are the "special" strata of Sym^n(C^2)?
QUESTION [5 upvotes]: The affine variety $Sym^n(\mathbb{C}^2)$ has a natural quantization as a spherical rational Cherednik algebra.  Thus, any primitive ideal of the rational Cherednik algebra has an corresponding ideal in $\mathbb{C}[X_1,Y_1..,X_n,Y_n]^{S_n}$.  By a theorem of Ginzburg, the zero-set of this ideal is the closure of the elements where the partition given by looking at which points coincide is a particular fixed one.  
My question: Which paritions can come up this way?

REPLY [4 votes]: Just by coincidence, I was catching up on my arxiv reading and noticed that Losev's paper appears to answer your question.  See part (3) of Theorem 4.3.1 there.  He says it was "known previously".
Edit
To be kinder to the reader: the Cherednik algebra is really a family of algebras depending on a parameter $c$; it looks like Losev's result is that at $c=k/m$ for relatively prime integers $k$ and $m$ the varieties we want correspond to partitions of $n$ which are of the form $(m,m,\dots,m,1,1,\dots,1)$ for some number of $m$'s and some number of $1$'s.
2nd Edit
Having looked more carefully at Losev's paper, it seems the idea of the proof is this: in the spirit of Bezrukavnikov-Etingof, having fixed a partition $\lambda$ Losev gives (his Theorem 1.3.1) an inductive description of primitive ideals with associated variety corresponding to $\lambda$.  The partition $\lambda$ corresponds to a "parabolic" subgroup $W$ of $S_n$ consisting of those permutations fixing a generic point of the corresponding variety, and the set of primitive ideals corresponding to $\lambda$ is in bijection with the set of primitive ideals of finite codimension in the Cherednik algebra attached to $W$ at the same parameter $c=k/m$.  But $W$ is a product of symmetric groups, so the classification due to Berest-Etingof-Ginzburg of finite dimensional modules for the type $S_n$ Cherednik algebra (they exist exactly when $c$ has denominator $n$) implies the result.
It appears that ideas from Losev's work on finite $W$-algebras also work for symplectic reflection algebras.  It would therefore be nice to understand the finite dimensional modules outside of the symmetric group case a little better.  (And now I'm more motivated to learn about $W$-algebras!)<|endoftext|>
TITLE: What is an Oper?
QUESTION [14 upvotes]: Given a curve C, and a reductive group G, there is a moduli stack Loc_G(C), the stack of G-local systems.  I keep reading that there's a substack of "opers" but am having trouble locating a definition.  So what's an oper, and how should I think about them?

REPLY [2 votes]: For reviews, one can try the articles of Frenkel and Teschner.
In the physics context, a fairly recent article which uses opers is this one, where Gaiotto and Witten define an oper for $G = \text{SU}(2)$ to be a flat rank two complex bundle $E$ over a Riemann surface $C$, with structure group $SL(2, \mathbb{C})$, together with a holomorphic line sub-bundle $L$ in $E$ such that $L$ is nowhere invariant under parallel transport by the connection $\mathcal{D}_z$ on $E$.
An oper which is endowed with a covariantly constant reduction of its structure group to a Borel subgroup (ie. the group of upper triangular matrices) is called a Miura oper.  This notion is also described in detail in the above article by Frenkel.<|endoftext|>
TITLE: Which computer algebra system should I be using to solve large systems of sparse linear equations over a number field?
QUESTION [9 upvotes]: This is related to Noah's recent question about solving quadratics in a number field, but about an even earlier and easier step.
Suppose I have a huge system of linear equations, say ~10^6 equations in ~10^4 variables, and I have some external knowledge that suggests there's a small solution space, ~100 dimensional. Moreover, the equations are sparse; in fact, the way I produce the equations gives me an upper bound on the number of variables appearing in each equation, ~10. (These numbers all come form the latest instance of our problem, but we expect to want to try even bigger things later.) Finally, all the coefficients are in some number field.
Which computer algebra system should I be using to solve such a system? Everyone knows their favourite CAS, but it's often hard to get useful comparisons. One significant difficulty here is that even writing down all the equations occupies a big fraction of a typical computer's available RAM.
I'll admit that so far I've only tried Mathematica; it's great for most of our purposes, but I'm well aware of its shortcomings, hence this question. A previous slightly smaller instance of our problem was within Mathematica's range, but now I'm having trouble.
(For background, this problem is simply finding the "low weight spaces" in a graph planar algebra. See for example Emily Peter's thesis for an explanation, or our follow-up paper, with Noah Snyder and Stephen Bigelow.)

REPLY [4 votes]: If Magma can do this, you may well look at Sage, which is open source, remarkably powerful, and with support for sparse linear algebra.<|endoftext|>
TITLE: Mathematical podcasts/audio
QUESTION [40 upvotes]: Just to ask if anyone is aware of any interesting math podcasts?  I am particularly interested in podcasts describing mathematics in the wider world; but interesting academic podcasts would also be useful.
Interesting mathematical audio other than podcasts is also welcome.
Summary of Podcast Links 

More or Less
Travels in a Mathematical World
Mathematical Moments
Math Mutation
Math Factor
In Our Time
Strongly Connected Components
inSCight
The Science of Better

REPLY [2 votes]: History of mathematics, but very very good is:
Opinionated history of mathematics<|endoftext|>
TITLE: Is there a constructive description of type in the p-local stable homotopy category?
QUESTION [13 upvotes]: The title pretty much sums it up - but let me give a little bit of background first.
In the p-local stable homotopy category (basically one localizes away the torsion spectra which are not p-torsion) the  Morava K-theories  are a non-negative integer indexed family of spectra (morally they are residue objects or "homological fibre functors") which classify the thick subcategories of SH^{(fin)_p the p-local stable homotopy category of finite spectra. This classification assigns to each p-local finite spectrum a type (namely the smallest thick subcategory it occurs in).
As far as I know the only definitions of type are directly in this way via the Morava K-theories or in terms of periodic self maps which are still really in terms of Morava K-theory.
Is there a more "constructive" definition of type? For instance can one determine the type of a p-local finite spectrum in terms of how bad the obstruction to it generating the whole of SH^{(fin}}_p is? Really I would be happy with any answer which was somewhat more constructive or to find out that the question is open/ridiculous.
Now let me explain somewhat the motivation for this question and how it came about. One can view the Morava K-theories (as I mentioned above) as residue objects/homological fibre functors (the term homological fibre functor has less algebraic geometry bias and sounds cooler) in the sense that their behaviour is analogous to residue fields of points in the derived category of quasi-coherent modules on a scheme (to be safe one should really take the derived category of O_X-modules with quasi-coherent cohomology) and with \kappa-modules in modular representation theory. Namely they all give tensor functors to some flavour of graded vector space category which classify thick subcategories.
The mod n Moore spectra are `Koszul objects' which we want to view as analogues of the usual Koszul complexes on a scheme and of Carlson modules in modular rep theory. In other words a Koszul object is a cone on some (possibly graded and maybe also twisted) element of the endomorphism ring of the tensor unit of our category.
Now one can associate some geometry to a biexact tensor product (by biexact tensor product I mean symmetric monoidal structure which is exact in each variable, there are no decency assumptions or extra axioms regarding compatibility with the triangulation required) on an essentially small triangulated category. It is possible to cook up a locally ringed space associated to such a category with tensor product (this is work of Paul Balmer). In the two algebraic cases, derived categories of schemes and stable categories in modular rep theory, one gets (with some mild hypothesis in the algebraic geometry case) back the scheme or recovers the projective support variety. This comes down in some sense to the fact that the Koszul objects determine the topology, or equivalently that they determine the thick subcategories in some sense.
This fails for the stable homotopy category of finite spectra (both globally and p-locally). The mod n Moore spectra are not enough - one needs the Morava K-theories. The locally ringed space one gets is not a scheme (nor an algebraic space). One can then ask if there is some "global" reason that this happens (even though it is not at all a surprise) other than the fact that the Morava K-theories are just there generating subcategories p-locally. This is motivated partially by trying to understand the failure of a certain comparison map to be injective (which I didn't mention - it would be interesting to have a good criterion for its injectivity and I currently only have quite hard to check ones) and to try to get some feel for what properties can cause the associated locally ringed space to fail to be algebraic (one really needs more examples computed for this and I haven't found the time yet unfortunately).
So basically I feel that if there were some definition of type that made clear the failure to be able to reduce the type by taking triangles and suspensions (or something other than just the residue objects being there) it might be quite enlightening. In particular, in general one wouldn't expect to produce an algebraic gadget from a topological triangulated category (in the sense of  Schwede ). That is why I mentioned the extension problem as if topologicalness provided some global obstruction to Koszul objects being enough that would be very interesting.
That got very long - I hope it is interesting/useful.

REPLY [8 votes]: My theory on this, which could be wrong, is that if the stable homotopy of the sphere (whatever the words "sphere" and "stable homotopy" mean in your situation) is Noetherian commutative, then type should be determined by prime ideals in the stable homotopy of the sphere.  
For example

D(R).  Stable homotopy is homology, sphere is R. If R is Noetherian commutative, the type of X is determined by ann H_* X.  This is the Hopkins thick subcategory theorem.
Stmod(kG).  This one is a little confusing, because stable homotopy is really Tate cohomology, which is never Noetherian.  But group cohomology is Noetherian, and Stmod(kG) is just a localization of D(kG).  The sphere is k.  So we should expect the type of M to be determined by ann [k,M] in D(kG), which is something like the Benson-Carlson-Rickard theorem.
Stable homotopy category.  Here the homotopy of the sphere is far, far from Noetherian, so you cannot expect the type to be determined by the homotopy of the sphere.  There are "embedded primes"; prime ideals in the category not visible in the homotopy of the sphere.  You kill p, and a new prime ideal pops up that was not previously visible.<|endoftext|>
TITLE: What is Eisenstein series?
QUESTION [13 upvotes]: There are several related questions here, the latter being especially interesting. We know the classical Eisenstein series.

What are the Eisenstein series on a group G and why they are interesting?
What's the geometric analogue of Eisenstein series? I think I saw the definition, but what's its importance?

REPLY [4 votes]: I highly recommend Drinfeld's lectures on Eisenstein series, notes available at
http://www.math.utexas.edu/users/benzvi/GRASP/lectures/drinfeldEisen.pdf<|endoftext|>
TITLE: Sheaves on Bun_G
QUESTION [5 upvotes]: What's the background I need to know to understand the conjectural
 D (Bun_G)  =?=  O(LocSys)

from this question. I know the LHS is about the derived category of D-modules on the space (stack?) of some (stable?) bundles for some preselected group G. What's the RHS? I've seen the physics articles that say something about his topic, but how would you explain it?
I know this goes under the name of Geometric Langlands. Why?

REPLY [2 votes]: The right hand side of the equation describes the derived category of coherent sheaves on the stack of LG torsors with connection.  In the traditional Langlands correspondence, one side is described by certain homomorphisms of a Galois group into the complex reductive group that is Langlands dual to G.  In the function field setting, the (unramified) Galois group is the etale fundamental group of a proper algebraic curve, and conjugacy classes of homomorphisms to L G are in natural bijection with LG torsors with connection.  These torsors are in turn in bijection with certain skyscraper sheaves on the corresponding moduli stack.  Passing to coherent sheaves is a natural expansion of the category (but I imagine there are better motivations that escape me at the moment).
There is a more symmetric quantized version of this conjecture, due to Feigin, E. Frenkel, and Stoyanovsky.  Coherent sheaves on the right hand side are replaced by twisted D-modules on the stack of LG bundles.  There is a discussion of it in the introduction to Gatsgory's paper on the twisted Whittaker model (on the arXiv).<|endoftext|>
TITLE: Homotopy theory of schemes examples
QUESTION [22 upvotes]: Is it possible give an example of (or explain) how the Voevodsky et al.'s homotopy theory of schemes computes higher Chow groups?

REPLY [6 votes]: If X is not smooth, then it is possible for the Chow groups and the A^1-represented motivic cohomology theory to disagree.
For instance, if we take X to be two copies of A^1 identified at a point then CH^0(X) has rank 2, but the sheaf represented by X in the A^1 category is contractible, so H^0(X,Z(0)) has rank 1.
To see this last point, we consider X as the colimit of a diagram A^1 <- * -> A^1. Since the maps in this diagrams are monomorphisms of schemes, they are cofibrations. The colimit of the diagram is therefore equivalent to the homotopy colimit, which is invariant under pointwise equivalences of diagrams. Since A^1 is contractible, we are left with the colimit of * <- * -> *, a point.
Morally, A^1 is contractible, so we can shrink down the A^1s without changing the A^1 homotopy type.<|endoftext|>
TITLE: What are Gromov-Witten invariants in terms of physics?
QUESTION [26 upvotes]: What do Gromov-Witten invariants (of say a Calabi-Yau 3-fold) represent, or what are they supposed to represent, in terms of string theory? When I compute GW invariants, am I actually computing some interactions between some particles, or what ...?
Please be gentle, and use only undergraduate-level physics words, if possible. (Perhaps this is too much to ask! Ok well, I'd rather get a response that involves fancier physics words than no response at all.)
I suppose this question is more physics than math -- I hope that's ok.

REPLY [33 votes]: Here is a very rough answer.
The Gromov-Witten invariants show up in a few a priori different 
contexts within string theory.  Let me focus on one particular place they show up that is 
directly related to conventional physics, as opposed to topological
quantum field theory.
Type IIA string theory is formulated on a spacetime "background" 
which is, in the simplest setup, just a Lorentzian 10-manifold.  The 
equations of motion of the theory require (at least in their leading 
approximation) that the metric on this 10-manifold should be Ricci-flat.
A popular thing to do is to take this 10-manifold of the form 
X x R^{3,1}, where X is a compact Calabi-Yau threefold.  
We can simplify matters by taking X to be very small --- 
smaller than the Compton wavelength of any of the particles we are able 
to create.  (Remember that in quantum mechanics particles have a 
wavelike character, with wavelength inversely related to their energy; 
since we only have limited energy available to us, we can't make 
particles with arbitrarily short wavelength.)  A little more precisely, 
let's take X such that the first nonzero eigenvalue of the Laplacian is 
larger than the energy scale we can access.
In this case we low-energy 
observers will not be able to detect X directly in any experiments.  To 
us, spacetime will appear to be R^{3,1}.
What will be the physics we see on this R^{3,1}?  We will see 
various different species of particle.  Each species of particle that 
we see corresponds to some zero-mode of the Laplacian of X.
In particular, there are particles corresponding to classes in H^{1,1}(X).
The genus 0 Gromov-Witten invariants are giving
information about the interactions between these particles.  (So if you want to calculate what will come out when you 
shoot two of these particles at each other, one of the inputs to that calculation
would be the genus 0 Gromov-Witten invariants.)  The higher genus Gromov-Witten 
invariants are giving information about interactions which involve these particles 
together with other particles related to the gravitational interaction.<|endoftext|>
TITLE: Elementary $\mathrm{Ext}^1$ intuition
QUESTION [17 upvotes]: $\DeclareMathOperator{\Hom}{\operatorname{Hom}}\DeclareMathOperator{\Ext}{\operatorname{Ext}}$I am wondering what sort of basic basic intuitive meaning $\Ext^1(M,N)$ has.
As a base case: if $M$ and $N$ are say, (finite-dimensional) vector spaces (with a compatible group/algebra action), and $M$ and $N$ are indecomposable inequivalent (so $\Hom(M,N)={0}$), can I somehow conclude that $\Ext^1(M,N)$ is zero?
All I can get from Weibel/Wikipedia is that $\Ext^1(M,N)$ is a group under the Baer sum operation, and is in bijection with the set of solutions $X$ to the short exact sequence $0\to N\to X\to M\to 0$.  I don't know how to use this second meaning, but it seems the most hands-on.
Full disclosure: If this sounds like a homework exercise, it (almost) was -- past tense.  Although, the problem/text/instructor had no desire for use of $\Ext$ or $\Hom$, I just want to know how to use these functors (better).  I could give character references (even from some past Berkeley grads) to allay fears....
I would be happy to have a good reference to look this up myself.  I hear Rotman's first book was good, but I've only read negative responses to the new edition (and the old one isn't for sale anywhere I've seen), and Weibel is apparently too abstract for me, in some way.  I'll post a separate question for that, in fact.

REPLY [5 votes]: It is fruitful to think of Ext^1 as a generalization of Hom.  If A and B are objects of an abelian category, the set Hom(A,B) has the structure of an abelian group.  Here is one way to define the group law: any two maps A --> B induce a map A x A --> B x B.  This induces a sequence
A --> A x A --> B x B --> B

where the first map is the diagonal and the last is the codiagonal (the addition map).  The composition is the sum.  (What we are using here is that finite products and finite direct sums coincide in an abelian category; in fact, this can be taken as the definition of an abelian category.)
We have used three properties of Hom to construct the addition map: Hom is covariant in the second variable, contravariant in the first, and comes with a canonical map 
Hom(A,B) x Hom(A',B') --> Hom(A x A', B x B')

(plus some compatibilities which I have suppressed).  If we define Ext^1(A,B) to be the category of extensions 
0 --> B --> E --> A --> 0

we see that it also has these properties (using base change for the variable A and pushout for the variable B).  Thus we get a group law on the category Ext^1(A,B).  Passing to isomorphism classes gives the usual group law on the set Ext^1(A,B).
The zero object of Ext^1(A,B) also corresponds nicely to the zero object of Hom(A,B).  In the latter case it is constructed by pushing forward the unique map A --> 0 via the unique map 0 --> B.  For Ext^1(A,B), we push forward the unique extension of A by 0 
0 --> 0 --> A --> A --> 0

by the unique map 0 --> B and get
0 --> B --> B x A --> A --> 0.

So an extension of A by B is zero if and only if it admits a splitting.  Thus in the category of vector spaces over a field, Ext^1(A,B) is always zero because can always be split.<|endoftext|>
TITLE: Homological Algebra texts
QUESTION [51 upvotes]: I would like to hear the communities' ideas on good Homological Algebra textbooks / references.  The standard example is of course Weibel (which I'll leave for someone else to describe).
As usual, the rule is one reference per post.  Please include some description which distinguishes it from other texts.

REPLY [4 votes]: Appendix 3 of Eisenbud's "Commutative Algebra" is the best short treatment I know. I find it fantastic. It clearly and concisely covers a surprising number of topics in homological algebra.<|endoftext|>
TITLE: "Albanese" schemes: When does an "initial abelian scheme" exist under a given scheme? 
QUESTION [13 upvotes]: For a variety V, its Albenese variety Alb(V) is a variety with a map V → Alb(V) which factors uniquely into any map from V to an abelian variety.  Can we say something similar for an arbitrary scheme?  When do we know there exists an "Albanese" scheme Alb(X)?  That is,

Under what conditions on a scheme X does there exist a morphism X → Alb(X) which factors uniquely into any map from from X to an abelian scheme?

REPLY [2 votes]: The recent paper "Para-abelian varieties and Albanese maps" by Bruno Laurent and Stefan Schröer (https://arxiv.org/abs/2101.10829) revived the topic and seems to give an almost optimal existence result which I'll leave you discover in the article itself.<|endoftext|>
TITLE: Real-world applications of mathematics, by arxiv subject area?
QUESTION [196 upvotes]: What are the most important applications outside of mathematics of each of the major fields of mathematics? For concreteness, let's divide up mathematics according to arxiv mathematics categories, e.g. math.AT, math.QA, math.CO, etc.
This is a community-wiki question, so please edit and improve pre-existing answers: let's keep it to a single answer for each subject area.
(This is inspired by Terry Tao's recent post about a periodic table of the elements listing commercial applications. He suggested it might be fun to have such a summary for either the MSC top-level subjects or the arxiv subjects.)
I'd like to propose that for areas in which the applications are either numerous, non-obvious, or generally worthy of discussion, someone volunteers to open up a new question specifically about that subject area, and takes care of providing a summary here of the best answers produced there.

REPLY [9 votes]: math.MP Mathematical Physics: this subject already uses several of the other already mentioned categories, like group theory, functional analysis, applied to physical theories like quantum field theory. But since the intention is to see concrete applications to the real world taken from a mathematically rigorous framework, then we could mention

In the first place Noether's theorem: "every differentiable symmetry of the action of a physical system has a corresponding conservation law".
Onsager reciprocal relations: They "express the equality of certain ratios between flows and forces in thermodynamic systems out of equilibrium, but where a notion of local equilibrium exists". Onsager's contribution was to demonstrate that not only is $L_{\alpha \beta}$ positive semi-definite (the Onsanger matrix of phenomenological coefficientes), it is also symmetric, except in cases where time-reversal symmetry is broken. 
The issues about proving thermodynamical statements strictly from statistical mechanics. This became one the biggest discussions in mathematical physics on its time, between Ludwig Boltzmann and Ernst Zermelo (see the book of the colleted works of Ernst Zermelo, volume II, edited by Herausgegeben von, Heinz-Dieter Ebbinghaus and Akihiro Kanamori, Springer, 2013).
The existence of anti-particles by P.A.M. Dirac, which was first a mathematical result. Dirac realised that his relativistic version of the Schrödinger wave equation for electrons predicted the possibility of antielectrons, themselves discovered four years latter, by Carl D. Anderson.
The Universality of the Feigenbaum constants, proven by Landorf ( Lanford III, Oscar (1982). "A computer-assisted proof of the Feigenbaum conjectures". Bull. Amer. Math. Soc 6 (3): 427–434.), with some corrections by Eckmann and Wittwer (Eckmann, J. P.; Wittwer, P. (1987). "A complete proof of the Feigenbaum conjectures". Journal of Statistical Physics 46 (3–4): 455).
The very concept of Universality, together with scaling, introduced by Kadanoff (see for example, Physica A 163 (1990) 1-14 "Scaling and Universality in Statistical Physics"). Roughly speaking, scaling is about the description of changes in the behavior of physical phenomena, in terms of adimensional constants, and how do they scale with them, as is the case of the Reynolds number. Universality concerns the invariance of properties for different dynamical systems, independently of some other physical details. For example, (see the aforementioned paper of Kadanoff), any perturbation which does not drive away a Hamiltonian near a critical point is deemed irrelevant, and all the Hamiltonians with any such kind of perturbations are said to belong to the same Universality class.
In the speculative region (as of today), the prediction of a possible second Island of Stability, where some new stable chemical elements might be found, with possibly interesting physical properties (see, for example, Zeitschrift für Physik 1969, Volume 228, Issue 5, pp 371-386 "Investigation of the stability of superheavy nuclei around Z=114 and Z=164", by Jens Grumann, Ulrich Mosel, Bernd Fink, Walter Greiner).
Even more speculative, but still valid mathematical results, some solutions from Einstein's General Relativity equations, allowing for closed timelike curves (i.e. time machines to the past), such as Gödel Spacetime (see, Review of Modern Physics, volume 21, Number 3, July 1949, "An example of a new type of cosmological solutions of Einstein's field equations of gravitation", Kurt Gödel).  
Roger Penrose's arguments against the possibility of Strong Artificial Intelligence (the actual construction of a physically based intelligence, not based on a brain like ours, but on an algorithmic machine), based on the  insolubility of the halting problem and Gödel's incompleteness theorem preventing the existence of an algorithmically based system of logic from reproducing such traits of human intelligence as mathematical insight. 
Not yet published as mathematically solved, the strict mathematical derivation of turbulence, from the Navier-Stokes equation, or otherwise (turbulence does not have a complete mathematical theory, to the extent of my knowledge). Navier-Stokes does not even have a theorem for the following problem: "Prove or give a counter-example of the following statement: In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations."
Along similar lines for mathematical completeness, is the Yang-Mills existence and mass gap: "Prove that for any compact simple gauge group $G$, a non-trivial quantum Yang–Mills theory exists on $\mathbb{R}^4$ and has a mass gap $\Delta > 0$. Its importance comes from the fact that it is the simplest Quantum Field Theory available (as far as I know), and it does not need to assume the existence of quarks.<|endoftext|>
TITLE: Examples of applications of the Borel-Weil-Bott theorem?
QUESTION [33 upvotes]: In "Quantum field theory and the Jones polynomial" (Comm. Math. Phys. 1989 vol. 121 (3) pp. 351-399), Witten writes:

A representation Ri of a group G should be seen as a quantum object.  This representation should be obtained by quantizing a classical theory.  The Borel-Weil-Bott theorem gives a canonical way to exhibit for every representation R of a compact group G a problem in classical physics, with G symmetry, such that the quantization of this classical problem gives back R as the quantum Hilbert space.  One introduces the "flag manifold" G/T, with T being a maximal torus in G, and for each representation R one introduces a symplectic structure ωR on G/T, such that the quantization of the classical phase space G/T, with the symplectic structure ωR, gives back the representation R.  Many aspects of representation theory find natural explanations by thus regarding representations of groups as quantum objects that are obtained by quantization of classical physics. [page 372; emphasis added]

I'm fascinated by this idea — I haven't seen it before, but it seems natural, in that classical objects should not be linear, whereas quantum objects should be.  I'm most interested in the last sentence: what examples can y'all come up with of representation-theoretic facts that can be "explained" by "physics" on G/T?  (Besides, of course, Witten's application in the paper I quoted from.)
More generally, I've read the Wikipedia discussion of the Borel-Weil-Bott theorem, and done some random googling, but I haven't found an elementary description of the symplectic structure Witten refers to.  Anyone want to pedantically spell out Witten's comment, please?

REPLY [2 votes]: A belated response, but appropriate nevertheless, I think: the locus classicus for this general theme about the deep connection between quantum particles and representation theory is given by E Wigner, "On unitary representations of the inhomogeneous Lorentz group," Annals of Mathematics 40 (1): 149–204. Wigner is the one who gave us Wigner's classification -- viz., the classification of most nonnegative energy irreducible unitary representations of the Poincare group. Since Wigner, it has become quite standard simply to define elementary particles as these irreducible representations.<|endoftext|>
TITLE: Silly Question about "opposite groups"
QUESTION [7 upvotes]: Given a group G, we can define Gop to be a new group whose underlying set is the same as that of G but with the new multiplication g.h = hg, i.e. multiply as if you were in G but reverse the order.  
Is there anything interesting to say about this construction?  
When is a group isomorphic to its opposite group?
One sees the "opposite ring" of a non-commutative ring showing up in a crucial way in number theory in the study of Brauer groups, so this is not necessarily an artificial question.  This is just a question that has been poking around my brain for a little while. . . it seems like a potentially neat little problem.

REPLY [22 votes]: Every group is isomorphic to its opposite group, the map $g \mapsto g^{-1}$ being an isomorphism.
In ring theory this is not the case, and the opposite ring is an important construction.  For instance, if $A$ is a central simple algebra over a field (or an Azumaya algebra over a ring), then the classes of $A$ and $A^{\operatorname{op}}$ are inverse in the Brauer group.
I don't know of anything similarly interesting to say about opposite groups.<|endoftext|>
TITLE: Are there any recordings of Grothendieck online?
QUESTION [18 upvotes]: Illusie mentions tape recordings of Grothendieck explaining his trace formula and more. Are they or similar recordings online? I guess, even if (what I doubt) everything he thought about that is somewhere in print, it would give an interesting insight in his way of thinking.

REPLY [9 votes]: Now you can access the audio lectures that Grothendieck gave at Buffalо in 1973.
Lectures on Topoi
Algebraic geometry
Algebraic groups
Thèmes pour une Harmonie
In the section 1973.
The page is still work in progress.<|endoftext|>
TITLE: Can any countably generated k-algebra occur as the ring of global sections of some variety?
QUESTION [8 upvotes]: In the answer to  this  question we saw that  there exists  a nonsingular quasi-projective threefold over a field with non-finitely generated global sections.
I was talking about this previous question today and the following question came up - given any countably generated noetherian k-algebra R which is an integral domain and whose field of fractions has finite transcendence degree over k, where k is a field does there exist some quasi-projective variety X (by variety I mean an integral separated scheme of finite type over k) such that the ring of global sections of X is R?
It is possible one needs more hypotheses to make this work - if this is false I think it would be interesting to know the class of algebras which can occur.

REPLY [10 votes]: No. Take k[t] and invert countably many relatively prime polynomials. This obeys all of your adjectives (localization preserves noetherianness, the others are obvious.)
However, a ring of global sections must be a subring of some finitely generated k-algebra. (I pointed this out in the last discussion.) Hence, its unit group must be a subgroup of the group of units of a finitely generated k-algebra. If A is any finitely generated k-algebra, then Units(A)/Units(k) is a finitely generated abelian group, and thus can't contain the countably generated group of the above example.
I can't find a reference for the fact about Units(A)/Units(k) at the moment, Tevelev describes this as well known in

Compactifications of subvarieties of tori, American Journal of Mathematics 129 no. 4 (2007) pp. 1087–1104, doi:10.1353/ajm.2007.0029, arXiv:math/0412329

REPLY [2 votes]: No, but for somewhat trivial reasons.  Let R be the polynomial ring in countably many variables, with no relations.  This is a countably generated k-algebra, and it can't be the ring of functions on a quasi-projective.  Any quasi-projective has function field of finite transcendence degree over the base field, because they are birational to hypersurfaces in P^n.  Now, if you add the hypothesis that R is countably-generated, reduced, noetherian k-algebra, it might be true, though both reduced and noetherian are necessary hypotheses.<|endoftext|>
TITLE: Intuition about the cotangent complex?
QUESTION [63 upvotes]: Does anyone have an answer to the question "What does the cotangent complex measure?"
Algebraic intuitions (like "homology measures how far a sequence is from being exact") are as welcome as geometric ones (like "homology detects holes"), as are intuitions which do not exactly answer the above question.
In particular: Do the degrees have a meaning? E.g. if an ideal $I$ in a ring $A$ is generated by a regular sequence, the cotangent complex of the quotient map $A\twoheadrightarrow A/I$ is $(I/I^2)[-1]$. Why does it live in degree 1?

REPLY [36 votes]: First a correction: the cotangent complex of a local complete intersection embedding is concentrated in degree -1, not in degree 1.
In general, the cotangent complex of an algebraic space can be supported in arbitrary non-positive degrees.  The cotangent complex of an Artin stack can be nonzero in degree 1.  The degrees in which the cotangent complex is concentrated imply various things about a morphism of schemes:
it is perfect in degree 0 if and only if the map is smooth;
it is perfect in $[-1,0]$ if and only if the map is lci;
$H^1 = 0$ if and only if it is a DM stack;
$H^0 = H^1 = 0$ if and only if it is an etale local immersion.
Other people have already said some things about the relationship to deformation theory.  The cotangent complex actually has two immediate relationships to deformation theory: one to the deformations of morphisms and one to the deformation of spaces.
In what's written below, $L_X$ is the absolute cotangent complex and $L_{X/S}$ is the relative cotangent complex.
If $f : S \to X$ is a map of schemes and $S'$ is a square-zero extension of $S$ with ideal $J$, there is an obstruction to extending $f$ to $S'$ in the group $Ext^1(f^\ast L_X, J)$.  If this obstruction vanishes, such extensions have a canonical structure of a torsor under $Ext^0(f^\ast L_X, J)$.
If $p : X\to S$ is a morphism, $S'$ is a square-zero extension with ideal $J$, and $p^\ast J \rightarrow I$ is a homomorphism of quasi-coherent sheaves on $X$, then the problem of finding a square-zero extension $X'$ with ideal $I$ and a map $X' \to S'$ extending $X \to S$ compatible with the given map on ideals is obstructed by a class in $Ext^2(L_{X/S}, I)$.  If this class is zero, isomorphism classes of solutions form a torsor under $Ext^1(L_{X/S}, I)$ and isomorphisms between any two solutions form a torsor under $Ext^0(L_{X/S}, I)$.<|endoftext|>
TITLE: Least number of charts to describe a given manifold
QUESTION [30 upvotes]: Hello,  I'm wondering if there is a standard reference discussing the least number of charts in an atlas of a given manifold required to describe it.  
E.g. a circle requires at least two charts, and so on (I couldn't manage to get anything relevant neither on wikipedia nor on google, so I guess I'm lacking the correct terminology).
Edit: in the case of an open covering of a topological space by n+1 contractible sets (in that space) then n is called the Lusternik-Schnirelman Category of the space, see Andy Putman's answer.  The following book seems to be the standard reference http://books.google.fr/books?id=vMREfNN-L4gC&pg=PP1
Great, now I'm still interested by the initial question: does anybody know of another theory without this contractibility assumption (hoping that it allows more freedom)? e.g. would it lead to different numbers say for genus-g surfaces? 
Final edit: yes different numbers for genus-g surfaces (see answers below), but not sure there is a theory without contractibility.   Right, really lots of interesting literature on the LS category nevertheless, hence the accepted answer.  For example there are estimates for non-simply connected compact simple Lie groups like PU(n) and SO(n) in Topology and its Applications, Volume 150, Issues 1-3, 14 May 2005, Pages 111-123.

REPLY [12 votes]: After "dimension" this is the most basic numerical invariant of a manifold and the least explored. I found this reference some years ago: I. Bernstein, "On Imbedding Numbers of Differentiable Manifolds", Topology, Vol. 7, pp. 95-109.<|endoftext|>
TITLE: Longest Element of an Affine Weyl Group
QUESTION [9 upvotes]: I know that the Weyl groups of affine Lie algebras don't have a longest element, but are there any good substitutes for w_0. In particular, is there any good substitute for a reduced decomposition of the longest element?

REPLY [5 votes]: Here is another notion of a reduced decomposition for the long word, as used for example in a recent preprint of Baumann, Kamnitzer and Tingley.
Recall the theorem from the finite type case that reduced decompositions of w0 are in bijection with convex orderings on the set of positive roots. The latter notion is easier to generalise, and can be considered an appropriate analogue.
Definition: A convex order on the set of all positive roots Φ+ is a preorder ≤ such that
i) For all α, β in Φ+, we have α ≤ β or β ≤ α (OR not XOR).
ii) If α ≤ β and α+β is a root, then α ≤ β+α ≤ β.
iii) If α ≤ β and β ≤ α then α and β are proportional.
Voila.<|endoftext|>
TITLE: Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups
QUESTION [23 upvotes]: In a forthcoming paper with Venkatesh and Westerland, we require the following funny definition.  Let G be a finite group and c a conjugacy class in G.  We say the pair (G,c) is nonsplitting if, for every subgroup H of G, the intersection of c with H is either a conjugacy class of H or is empty.
For example, G can be the dihedral group of order 2p and c the class of an involution.
The case where c is an involution is o special interest to us.  One way to construct nonsplitting pairs is by taking G to be a semidirect product of N by (Z/2^k Z), where N has odd order, and c is the conjugacy class containing the involutions of G.  Are these the only examples?  In other words:
Question 1:  Is there a nonsplitting pair (G,c) with c an involution but where the 2-Sylow subgroup of G is not cyclic?
Slightly less well-posed questions:
Question 2:  Are there "interesting" examples of nonsplitting pairs with c not an involution?  (The only example we have in mind is G = A_4, with c one of the classes of 3-cycles.)
Question 3:  Does this notion have any connection with anything of pre-existing interest to people who study finite groups?
Update:  Very good answers below already -- I should add that, for maximal "interestingness," the conjugacy class c should generate G.  (This eliminates the examples where c is central in G, except in the case G = Z/2Z).

REPLY [2 votes]: If your $c$ is a conjugacy class of elements of odd prime order $p$, then (using the classification
of finite simple groups), it is still the case that $G = O_{p^{\prime}}(G)C_{G}(x)$ for 
each $x \in c$, where $O_{p^{\prime}}(G)$ denotes the largest normal subgroup of $G$ of order prime to $p$. This might be considered as an odd analogue of Glauberman's $Z^{\ast}$-theorem.
If anyone could come up with a classification-free proof of such a result, it would be of
considerable interest (it is relatvely easy to prove this directly when $G$ is solvable
(or, more generally, $p$-solvable)). Incidentally, your question seems to be related to the
old concept of pronormality (which is treated, for example, in Gorentstein's book
"Finite Groups"). Back to the case $p = 2$, interesting generalizations of Glauberman's
$Z^{\ast}$-theorem were given by D. Goldschmidt and also by E. Shult. One result that Shult
proved which you might find interesting is in a Bull AMS paper (circa 1966), in which he showed
that an element of order $p$ in a finite group which commutes with none of its other conjugates
AND centralizes every $p^{\prime}$-group it normalizes, is central in $G$.<|endoftext|>
TITLE: opposite Banach space
QUESTION [9 upvotes]: I heard this from Haskell Rosenthal many years ago.
If V is a complex vector space, say the opposite of V is the complex vector space with the same elements, the same operations except switch scalar multiplication to scalar multiplication by the complex conjugate scalar.   Of course this definition applies in particular to a complex Banach space.  Question: Is every complex Banach space isomorphic to its opposite?  (An isomorphism is a complex-linear homeomorphism.)
[ prompted by question Silly question about opposite groups ]

REPLY [9 votes]: Does this paper of Kalton do the trick? (disclaimer: I haven't read through the details)<|endoftext|>
TITLE: Is a map that is locally fiberwise equivalent to a product a Hurewicz fibration?
QUESTION [5 upvotes]: The following is a result I feel like I've seen some form of before, but can't figure out how to prove or find a reference for.  Suppose you have a map p:E \to B, with B paracompact, and suppose that every point in B has a neighborhood U such that there is a map p^{-1}(U) \to U \times F over U which is a fiber homotopy equivalence.  Does it follow that p is a Hurewicz fibration?  The converse, if B is locally contractible, is standard: a Hurewicz fibration is locally equivalent to a product.

REPLY [4 votes]: The answer is no; Allen Hatcher sent me the following:
An example where this fails is the projection of the letter L onto its horizontal base, which I'll call B.  The deformation retraction of L onto B is a fiberwise homotopy equivalence.  The homotopy lifting property fails:  Map a point to the left endpoint of B, then lift this to a point of L - B and take a homotopy that moves the left endpoint of B to the right endpoint.<|endoftext|>
TITLE: Whitehead for maps
QUESTION [21 upvotes]: I made the following claim over at the Secret Blogging Seminar, and now I'm not sure it's true:
Let $f: X \to Y$ and $g: X \to Y$ be two maps between finite CW complexes. If f and g induce the same map on $\pi_k$, for all k, then f and g are homotopic.
Was I telling the truth?
EDIT: Since I didn't say anything about basepoints, I probably should have said that f and g induce the same map 
$[S^k, X] \to [S^k, Y]$.
This will also deal better with the situation where X and Y are disconnected. I'd be interested in knowing a result like this either with pointed maps or nonpointed maps. (Although, of course, if you work with pointed maps you have to take X and Y connected, because $[S^k, -]$ can't see anything beyond the number of components in that case.)

REPLY [6 votes]: Another class of examples comes from group cohomology: if $G$ is any group with interesting higher cohomology $H^n(G, A), n \ge 2$ then there are non-nullhomotopic maps $BG \to B^n A$ for some $n \ge 2$. But the source and target don't have nonzero homotopy groups in any shared degree so any such map induces the zero map on homotopy groups. If these cohomology groups all vanished then, among other things, groups would have no nontrivial central extensions and so all finite $p$-groups would be elementary abelian! 
(Edit: admittedly, the above examples don't usually involve finite complexes. But with the right $G$ we can take $BG$ to be an aspherical manifold and then we can truncate $B^n A$.) 
This is the precise analog, in topology, of the fact in homological algebra that in most interesting abelian categories $\text{Ext}^i$ can be nontrivial for $i \ge 1$, which reflects the existence of maps in derived categories between two objects which don't have nonzero homology groups in any shared degree. 
Group cohomology turns out to be a pretty good model for the more general question: 

What is a complete set of obstructions for a map $f : X \to Y$ (let's say pointed, between pointed CW complexes) to be nullhomotopic? 

There is an obstruction theory coming from trying to lift $f$ up through the stages of the Whitehead tower
$$\dots \to Y_2 \to Y_1 \to Y_0 \cong Y$$
of $Y$. Here $Y_k$ is the $(k-1)$-connected cover of $Y$, obtained from $Y$ by killing $\pi_1, \pi_2, \dots \pi_{k-1}$ (so the index tells you the lowest degree in which it can have a nontrivial homotopy group). In particular $Y_1$ is the connected component of the basepoint and $Y_2$ is the universal cover. 
At each stage, suppose we've lifted $f$ to a map $f_n : X \to Y_n$. Now, $Y_n$ has lowest homotopy group $\pi_n(Y_n) \cong \pi_n(Y)$, and hence there is a natural map
$$k_n : Y_n \to B^n \pi_n(Y)$$
inducing an isomorphism on $\pi_n$. The pullback of this map along $f_n$ gives a cohomology class 
$$k_n \in H^n(X, \pi_n(Y))$$
which I'll also call $k_n$, and $f_n$ lifts to a map $f_{n+1} : X \to Y_{n+1}$ iff this cohomology class vanishes. (In general, $k_n$ is only well-defined once we've chosen a lift $f_n$.) The punchline now is that $f$ is nullhomotopic iff all lifts $f_n$ exist iff all classes $k_n$ vanish. If $X$ has finite cohomological dimension then this is in principle only finitely many conditions to check. 
Example. Let $Y = BO$ be the classifying space of stable real vector bundles, let $X$ be a smooth manifold, and let $f : X \to BO$ be the classifying map of the stable tangent bundle. Then $f$ is nullhomotopic iff $X$ is stably parallelizable. The characteristic classes $k_n$ are Bott periodic. Here are the first few:
$k_1$ is the first Stiefel-Whitney class $w_1$. It vanishes iff $f$ lifts to a map $f_2 : X \to BSO$ iff $X$ is orientable. Here $BSO$ is $Y_2$.
$k_2$ is the second Stiefel-Whitney class $w_2$. It vanishes iff $f_2$ lifts to a map $f_3 : X \to BSpin$ iff $X$ has a spin structure. Here $BSpin$ is $Y_3 \cong Y_4$.
$k_3$ automatically vanishes. $k_4$ is the first fractional Pontryagin class $\frac{p_1}{2}$. It vanishes iff $f_3$ lifts to a map $f_5 : X \to BString$ iff $X$ has a string structure. Here $BString$ is $Y_5 \cong Y_6 \cong Y_7$.

REPLY [5 votes]: A simple example with finite complexes would be the collapse map $q:X\times Y\to X \wedge Y$.  Since the inclusion $X \vee Y \hookrightarrow X\times Y$ is surjective on $\pi_*$,
$q_* = 0$.  But $q$ is generally nonzero on homology; for example if $X = S^n$, $Y=S^m$, 
then $q_*$ is an isomorphism on $H_{n+m}$.<|endoftext|>
TITLE: Abelianization of Lie groups
QUESTION [6 upvotes]: If G is a group, its abelianization is the abelian group A and the map G → A such that any map G → B with B abelian factors through A.  Abelianization is a functor, and in general a very lossy operation.  The map G → A is always a surjection/quotient, because we can construct A by dividing G by the minimal normal subgroup that contains all conjugations ghg-1h-1 for g,h∈G.
If V is a finite-dimensional (super)vector space over a field K, then the abelianization of GL(V) is isomorphic to the multiplicative group K* of non-zero numbers in K.  Indeed, the determinant exhibits the desired isomorphism.
Here are two questions I'm curious about:

What can be said about the abelianizations of other (finite-dimensional) Lie groups?
If V is an infinite-dimensional vector space, what can be said about the abelianization of GL(V)?  Most infinite-dimensional vector spaces have some analytic structure, e.g. topological vector spaces, and so it's reasonable to ask that the operators in GL(V) should preserve that structure; you are welcome to take your favorite type of infinite-dimensional vector space and your favorite type of GL(V), if you want.

REPLY [3 votes]: (In some sense, this is just a restatement of what Eric said above....)
For compact groups, quite a lot can be said.  Every compact group H' has a finite cover H which is Lie group isomorphic to $T^{k} \times G$, where $G$ is compact and simply connected.
Then, one can easily show that [H,H] = {$e$}$\times G$ and hence that the abelianization of $H$ (which, as in the finite case is H/[H,H]) is $T^{k}$.
The same holds true of the original group $H'$: the abelianization is $H'/[H',H']$ and is isomorphic to $T^{k}$.<|endoftext|>
TITLE: Formalism of homotopy theory of schemes
QUESTION [12 upvotes]: I have some vague knowledge about the philosophy that schemes should be thought of as similar to topologic spaces, and we should divide everything by homotopy, and that the space should be actually sheaf in a correct topology. 
Could somebody provide a concise and modern introduction that would allow me to work with statements like (from an answer to question about motivic cohomology)

K(Z(0),0) is simply the constant sheaf Z.

REPLY [23 votes]: Let $E(S)$ be the category of Nisnevich sheaves on the site of smooth schemes over some base $S$. Then Morel and Voevosy's homotopy category $\mathrm{H}(S)$ is obtained as a localization of the category $sE(S)$ of simplicial objects in $E(S)$. The localization functor $\mathrm{loc}:sE(S)\to \mathrm{H}(S)$ can always by constructed in a way that it is the identity on objects, so that you can always think of the objects as genuine simplicial sheaves. A statement as above reads simply as: the object $K(\mathbb Z(0),0)$ is isomorphic to the image by $\mathrm{loc}$ of the constant sheaf $Z$. Now, if you want to use such a statement to compute something, you will need to know what are the maps in $\mathrm{H}(S)$. For this, you will need quite a bit of general homotopy theory (here the homotopy theory of simplicial sheaves as well as the theory of left Bousfield localization), and, of course, at some point, some geometry. However, the theory of Bousfield localization applies here in a rather gentle way, if you admit the homotopy theory of simplicial sheaves.
Let $\mathrm{Ho}(sE(s))$ be the localization of $sE(S)$ by the class of local weak equivalences (i.e. maps inducing weak equivalences of simplicial sets stackwise). Then, for any smooth $S$-scheme $X$, you have a derived global section functor $R\Gamma(X,?)$ (with values in Kan complexes). Morel and Voevodsky's homotopy category $\mathrm H(S)$ is obtained by inverting some maps in $\mathrm{Ho}(sE(S))$, so that we have a localization functor $\mathrm{loc}:\mathrm{Ho}(sE(S))\to H(S)$. As this comes from a left Bousfield localization at the level of the underlying model categories, this latter localization functor has a right adjoint $i:\mathrm H(S)\to\mathrm{Ho}(sE(S))$ which is fully faithful (almost by construction/definiton). Hence, we can understand $\mathrm H(S)$ as the full subcategory of $\mathrm H(sE(S))$. Moreover, we can understand the essential image of $i$ in a rather simple way: it consists of the objects $F$ such that, for any smooth $S$-scheme $X$, the map
$R\Gamma(X,F)\to R\Gamma(X\times \mathbb A^1,F)$
is a weak equivalence of Kan complexes.
This means that, whenever you have your favourite cohomology theory $F$, if it satisfies Nisnevich descent (hence is representable in $\mathrm Ho(sE(S))$, then it is representable in $\mathrm H(S)$ if and only if it is homotopy invariant. If it only satisfies Nisnevich descent, then you still have a universal way to force $\mathbb A^1$-homotopy invariance (by applying the functor loc).
for instance, $K(\mathbb Z(0),0)$ is really the Nisnevich cohomology with coefficients in $\mathbb Z$.
However, in the latter case, you might have some trouble to compute what you get. 
For the higher $K(\mathbb Z(n),2n)$, there is an explicit description in terms of complexes which is obtained as follows. Let $F$ be a sheaf of abelian groups. Consider the cosimplicial scheme $\Delta^n$; defined as the spectrum of the (sheaf of) ring(s) $\mathcal O[t_0,...,t_n]$ modulo the relation $t_0+...+t_n=1$ ($\mathcal O$ is the sheaf of functions on S). Taking the internal Hom's, you get a simplicial sheaf of abelian groups $\mathrm{Hom}(\Delta^n,F)$ (letting $n$ vary; you can also play with the Dold-Kan correspondance to get a complex if you prefer a hypercohomology point of view). Applying this for the object which represents $K(\mathbb Z(n),2n)$ (as explained in there),
if $S$ is smooth over a field, one of the deepest and less trivial result of Voevodsky is that we obtain an simplicial sheaf which satisfies Nisnevich descent and is $\mathbb A^1$-homotopy invariant, so that it represents motivic cohomology both in $\mathrm H(S)$ and in $\mathrm{Ho}(sE(S))$.
If I may suggest an exercise: apply Morel and Voevodsky's construction to describe usual algebraic topology: instead of the Nisnevich site of smooth $S$-scheme, consider the site of smooth analytic manifold on C (with the usual topology) (and replace the affine line by the disk $\mathbb D^1$). Then Morel and Voevodsky theory gives a category which is canonically equivalent to the usual homotopy theory of topological spaces (this is due to the fact that any smooth complex manifold is locally constractible, so that after trivializing $\mathbb D^1$, only locally constant invariants remain). Then, for instance, Poincaré lemma says that the de Rham complex is $\mathbb D^1$-homotopy invariant. In this precise sense, this shows that complex de Rham cohomology is very well defined on any homotopy type.<|endoftext|>
TITLE: Some intuition behind the five lemma?
QUESTION [31 upvotes]: Slightly simplified, the five lemma states that if we have a commutative diagram (in, say, an abelian category)
$$\require{AMScd}
\begin{CD}
A_1 @>>> A_2 @>>> A_3 @>>> A_4 @>>> A_5\\
@VVV     @VVV     @VVV     @VVV     @VVV\\
B_1 @>>> B_2 @>>> B_3 @>>> B_4 @>>> B_5
\end{CD}
$$
where the rows are exact and the maps $A_i \to B_i$ are isomorphisms for $i=1,2,4,5$, then the middle map $A_3\to B_3$ is an isomorphism as well.
This lemma has been presented to me several times in slightly different contexts, yet the proof has always been the same technical diagram chase and no further intuition behind the statement was provided. So my question is: do you have some intuition when thinking about the five lemma? For instance, particular choices of the $A_i, B_i$ which make it more transparent why the result should be true? Some analogy, heuristic, ...?

REPLY [4 votes]: Variant of what other people already said, but I find the following statement quite intuitive:

If G is a group and $f: X \to Y$ is a $G$-equivariant map inducing a bijection on orbit sets $X/G \to Y/G$ and isomorphism of all stabilizer groups $G_x \to G_{f(x)}$, then $f$ must be a bijection.

Now in the setting of the 5-lemma, specialize to $X = A_3$, $Y = B_3$, and $G = A_2 \cong B_2$ acting by addition on $X$ and $Y$.  By exactness of the rows, all maps of stabilizers are identified with the isomorphism $\mathrm{Im}(A_1 \to A_2) \cong \mathrm{Im}(B_1\to B_2)$ induced by the leftmost square in the diagram, and the map of orbit sets is identified with the isomorphism $\mathrm{Ker}(A_4 \to A_5) \cong \mathrm{Ker}(B_4 \to B_5)$ induced by the rightmost square.
From this point of view it is also clear that "abelian group" is more structure than necessary.  For example, the rightmost square need only be given in the category of pointed sets.<|endoftext|>
TITLE: An intuitive reason why the "Rule 30" CA is random/pseudorandom?
QUESTION [13 upvotes]: I'm a little bit hesitant to ask this here, so please notice the tag.  My hope is that someone will have a more satisfying answer than what I've heard before...
A long time ago I read (perhaps 'browsed' is a better word) Wolfram's "A New Kind of Science".  There are many many references to the "Rule 30" CA - http://mathworld.wolfram.com/Rule30.html.  However, no intuitive reasoning for it's random/pseudorandom behavior is provided prior to a digression to the importance of the discovery.  I was recently reminded of this when I heard that Mathematica (a program I use quite frequently) uses certain outputs from this CA as its random number generator.  
So my question is - beyond 'numerical phenomenology' is there an intuitive understanding why Rule 30 should behave in this random/pseudorandom manner?

REPLY [2 votes]: As I understand it Mathematica in fact uses rule 30 to generate pseudorandom numbers by just reading down a single column: so what it's exploiting is not just that rule 30 behaves pseudorandomly but that columns look more or less like strings of independent uniform random bits.  If we label the four cells on which the rule is defined

a b c
  d

it seems a lot easier to achieve this if (b,d) takes on all its possible values equally often, as rule 30 does.  
In fact, the stronger property is true that rule 30 can be run leftwards.  a is a (total) function of (b,c,d), therefore any two columns of cells can be completed uniquely to a halfplane extending leftwards satisfying rule 30.  This is suggestive at the very least -- if more things could appear to the left of some columns than others, the columns probably wouldn't be nice and uniform and all that.<|endoftext|>
TITLE: cosimplicial algebras to dg-algebras
QUESTION [7 upvotes]: The normalized Moore complex functor is usually considered taking simplicial abelian groups to chain complexes. But there is an obvious dual version that takes cosimplicial abelian groups to N-graded cochain complexes.
Moreover, applied on a cosimplicial group that has the structure of a cosimplicial ring, the Moore cochain complex yields a cochain complex that has the structure of a dg-algebras (the details are at monoidal Dold-Kan correspondence).
This seems obvious and useful enough, but I find surprisingly little literature on this dual monoidal Dold-Kan correspondence. In fact the only relevant reference that I am aware of is Castiglioni-Cortinas, Cosimplicial versus dg-rings: a version of the Dold-Kan correspondence. 
They consider not the Moore cochain functor but its right adjoint, and show that its left derived functor is an equivalence of homotopy categories. 
But it would seem that instead considering directly the Moore cochain complex functor on cosimplicial rings would be at least as interesting. 
I can dream up some of its properties myself, but I keep feeling I must be missing the canonical literature on this, which must exist. Does anyone have more references on the Moore cochain complex functor on cosimplicial rings/algebras?

REPLY [3 votes]: This construction is closely related to de Rham cohomology, and used to compute the continuous cohomology of Lie groups.
This is part of the tools to compute Beilinson's regulator, as explained in
J. I. Burgos Gil, The regulators of Beilinson and Borel. CRM Monograph Series, 15. American Mathematical Society, Providence, RI, 2002.
See Chapter 7 for the abstract results, and Chapter 8 for its applications. (It seems that there is an online version here.)
The same kind of constructions is considered by Larry Breen and Bill Messing in
Combinatorial differential forms, Advances in Mathematics, vol. 164 (2001), 203-282
and used in this paper of Larry Breen arXiv:0802.1833 to study differential geometry on gerbes ("non-abelian de Rham cohomology").<|endoftext|>
TITLE: Beilinson conjectures
QUESTION [21 upvotes]: Continuing an amazingly interesting chain of answers about motivic cohomology, I thought I should learn about the Beilinson conjectures, referred there.
I have found some references, and they seem to present the conjectures from different sides, e.g. there's the statement about vanishing but then there are also connections to motivic polylogarithms. 
What I miss from these articles in a general picture that would allow us to start somewhere natural. So,

how would you describe an introduction into Beilinson conjectures in motivic homotopy?

Sorry for such a loaded question — I really don't know how to make it fit MathOverflow format better. One could theoreticlly post lost of specific questions on the topic, but to ask the right questions in this case you might need to know more than I do. Also, I know there are some technical developments, e.g. the language of derived stacks, and my hope would be that somebody could make a connection to these conjectures using some  clear and suitable language.

REPLY [2 votes]: Beilinson's conjecture, together with Tate's conjecture (on algebraic cycles on smooth projective varieties over $\mathbf F_p$) and with Soulé's conjecture (on the pole orders of zeta functions for schemes of finite type over $Spec \mathbf Z$) can be unified into one statement: subject to certain "standard" conjectures on motivic t-structures, they are (jointly) essentially equivalent to the following: for a geometric motive $M$ over $Spec \mathbf Z$ (i.e., a compact object in the triangulated category $DM(Spec Z)$), there is a natural pairing
$$Hom(1, M) \otimes Hom(M, \hat 1(1)[2]) \to Hom(1, \hat 1(1)[2]).$$
Here, 1 is the monoidal unit (i.e., the motive of $Spec Z$ itself), and $\hat 1$ is the fiber of the map $1 \to H_D$, where $H_D$ is the spectrum representing Deligne cohomology. This object was introduced in joint work of Andreas Holmstrom and myself (Arakelov motivic cohomology I, Journal of Algebraic Geometry (2015), Vol. 24, pp. 719–754).
Conjecturally (See my paper Special L-values of geometric motives in Asian Journal of Mathematics (2017), Vol. 21 (2) pp. 225–264):

The pairing is perfect (of finite-dimensional R-vector spaces, provided that motivic (co)homology is with real coefficients).
The special L-value of $M$ can be read off from the pairing by considering $\mathbf Q$-structures on the above $\mathbf R$-vector spaces.

This conjectural duality is somewhat in the spirit of Artin-Verdier duality, or also of Poincaré duality.<|endoftext|>
TITLE: Links between Riemann surfaces and algebraic geometry
QUESTION [32 upvotes]: I'm taking introductory courses in both Riemann surfaces and algebraic geometry this term. I was surprised to hear that any compact Riemann surface is a projective variety. Apparently deeper links exist.
What is, in basic terms, the relationship between Riemann surfaces and algebraic geometry?

REPLY [8 votes]: To me, excellent as the others are, engelbrekt's is the most direct answer to your question. I.e. 
1) Every projective plane curve is a compact Riemann surface, essentially because of the implicit function theorem. 
2) Conversely, every compact Riemann surface [immerses as] a projective plane curve because it has [enough] non constant meromorphic functions which almost embed it in the plane. Also every meromorphic function is the pullback of a rational function in the plane.  So all the analytic structure is induced from the algebraic structure. 
In higher dimensions, complex projective algebraic varieties are a special subcategory of compact complex spaces, namely those that admit [holomorphic is sufficient] embeddings in projective space.  
More precisely, an n dimensional compact complex variety has a field of meromorphic functions that has transcendence degree ≤ n, and projective algebraic varieties are a subcategory of those (Moishezon spaces) for which the transcendence degree is n.  Indeed I believe Moishezon proved the latter are all birational modifications of projective varieties.
I would also add something about the impact of Riemann surfaces on algebraic geometry. Namely it was Riemann's introduction of the topological and analytic points of view, showing that path integrals and differential forms could be profitably used to study projective algebraic curves, that deepened and revolutionized algebraic geometry forever.<|endoftext|>
TITLE: Can isomorphisms of schemes be constructed on formal neighborhoods?
QUESTION [6 upvotes]: Let (A,m) be a complete local Noetherian ring and let X and Y be two schemes of finite type over A (and flat over A). Let Xn and Yn be the reductions of X and Y mod mn+1.

Question: Suppose there is a compatible system of isomorphisms between Xn and Yn (for all n). Does there necessarily exist an isomorphism between X and Y over A?

In other words, suppose the formal schemes \hat{X} and \hat{Y} are isomorphic; are X and Y isomorphic? 
Remark: The answer is `no' if we drop flatness (you can just stick an extra component over the generic fiber) or finite type (A[t] vs. A{t} = the completion of A[t]).

REPLY [4 votes]: In David S.'s example, I don't see a compatible sytem of isomorphisms $X_n\to Y_n$. Here is a similar example with the required properties. 
Let $A=k[[t]]$, let $Y={\rm Spec} A[x]]$ be the affine line over $A$ and $X=Y\setminus \{ y_0 \}$, where $y_0$ is the closed point of $Y$ corresponding to the ideal $(1-tx)A[x]$ (the quotient ring is $A[1/t]=k((t))$). 
Then the canonical inclusion $X\to Y$ induces clearly a compatible system of isomorphisms $X_n\to Y_n$ 
for all $n\ge 1$ (they are actually identities), and of course $X$ is not isomorphic to $Y$. 
The reason behind is that the point $y_0$ is invisible in all the formal neighborhood of the closed fibler of $Y\to {\rm Spec } A$. 
In general, a compatible system of isomorphisms $X_n\to Y_n$ induces an isomorphism between the formal schemes $\hat{X}\to \hat{Y}$. When $X, Y$ are projective, one can use GAGA to show that this isomorphism comes from an isomorphism $X\to Y$. This gives an alternative proof to the use of ${\rm Hom}(X, Y)$.<|endoftext|>
TITLE: Strong Bertrand postulate
QUESTION [7 upvotes]: Is it known that for every epsilon there is N_0 such that all intervals of the form [N, (1+\epsilon)*N], where N > N_0, contain prime numbers?

REPLY [8 votes]: If one wants an explicit bound on N0, apparently this can be gleaned from a Ph.D. thesis by Pierre Dusart (in French) which contains the result that for all x > 3275 there is a prime between x and x(1 + 1/(2 ln2 x)). So we can take N0 to be max(3275,exp((2 epsilon)−1/2)).<|endoftext|>
TITLE: Intrinsic characterization of a star shaped domain
QUESTION [6 upvotes]: Let A be a closed (compact no boundary), embedded (no self intersections), smooth surface of R^3. We say that the interior of A is star shaped if there exists a point p in A, such that for any point q in A, the line segment joining p and q lies entirely within A. My questions is this:
Let (S^2,g) be the sphere with a given smooth Riemannian metric (may not be the constant curvature one), and assume there exists a smooth isometric embedding f: S^2 -> R^3. Does there exist a condition on the Gaussian curvature of g that ensures the interior of f(S^2) (the image of S^2 under f) is star shaped?

REPLY [3 votes]: Here is an example of a smooth sphere in R^3 that is not star shaped, and such that both positive and negative cuvature parts are connected. Take in R^3 a thin rotation-invariant torus. Its positive and negative curvature parts are long thin annuli. Now cut a half of it by a plane going through the axe of rotation. Take one of the halthes (a long, thin, curved tube) and cup its two little holes with two little half-spheres -- so you get a sort of banana. This can be done in a smooth way so that positive curvature part is connected.<|endoftext|>
TITLE: A learning roadmap for Representation Theory
QUESTION [70 upvotes]: As Akhil had great success with his question, I'm going to ask one in a similar vein.  So representation theory has kind of an intimidating feel to it for an outsider.  Say someone is familiar with algebraic geometry enough to care about things like G-bundles, and wants to talk about vector bundles with structure group G, and so needs to know representation theory, but wants to do it as geometrically as possible.
So, in addition to the algebraic geometry, lets assume some familiarity with representations of finite groups (particularly symmetric groups) going forward.  What path should be taken to learn some serious representation theory?

REPLY [10 votes]: All of these recommendations are very good, and I'd like to add that the book D-Modules, Perverse Sheaves, and Representation Theory (which you can download at the provided link if you have institutional access; otherwise you can get it from, say, Amazon) contains some very good introductory chapters (chapters 9, 10, and 11) on the various sorts of things one would want to know in representation theory and algebraic geometry. The whole book is quite good if you're interested in the D-modules/perverse sheaves side of the story, but even if you're not interested in that, those particular chapters might be of interest.<|endoftext|>
TITLE: Generators for congruence subgroups of SL_2
QUESTION [22 upvotes]: For positive integers $n$ and $L$, denote by $SL_n(Z,L)$ the level $L$ congruence subgroup of $SL_n(Z)$, i.e. the kernel of the homomorphism $SL_n(Z)\rightarrow SL_n(Z/LZ)$.
For $n$ at least $3$, it is known that $SL_n(Z,L)$ is normally generated (as a subgroup of $SL_n(Z)$) by Lth powers of elementary matrices.  Indeed, this is essentially equivalent to the congruence subgroup problem for $SL_n(Z)$.
However, this fails for $SL_2(Z,L)$ since $SL_2(Z)$ does not have the congruence subgroup property.
Question : Is there a nice generating set for $SL_2(Z,L)\ ?$  I'm sure this is in the literature somewhere, but I have not been able to find it.

REPLY [5 votes]: I was at a workshop on noncongruence modular forms recently (a noncongruence subgroup of $SL_2(\mathbb{Z})$ is a subgroup of finite index not containing any $\Gamma(N)$ ) when this question came up. I believe that the consensus was that although one can, in principle, compute a generating set for $\Gamma(N)$, the algorithm is not terribly effective for $N$ large (where by large I mean greater than $13$ or so). The algorithm involves computing the Farey symbol associated to $\Gamma(N)$ and getting coset representatives. The difficulty is that the index of $\Gamma(N)$ increases very rapidly, making the calculation of the Farey symbol very lengthy. 
Ling Long and Chris Kurth have written a paper about the algorithm (it references the paper of Kulkarni that algori mentioned), which is available here.<|endoftext|>
TITLE: Cofinality of Theta if sharps exist
QUESTION [6 upvotes]: If ℝ# exists then why is cof(θL(ℝ)) = ω? Also I have the same question for the L(Vλ+1) generalization (if it's actually a different proof; I presume it isn't), i.e. if θ is defined as the sup of the surjections in L(Vλ+1) of Vλ+1 onto an ordinal, then if Vλ+1# exists why is cof(θL(Vλ+1)) = ω?

REPLY [4 votes]: Scott, the best way to think of sharps is via mice. Think of x^# as a mouse over x with one measure which is iterable. R^# is a mouse over R with one measure which is iterable. Things become very easy ones you make the move from sharps as reals or sets of reals or etc to sharps as mice.<|endoftext|>
TITLE: Proof of no rational point on Selmer's Curve $3x^3+4y^3+5z^3=0$
QUESTION [21 upvotes]: The projective curve $3x^3+4y^3+5z^3=0$ is often cited as an example (given by Selmer) of a failure of the Hasse Principle: the equation has solutions in any completion of the rationals $\mathbb Q$, but not in $\mathbb Q$ itself. 
I don't think I've ever seen a proof of the latter claim — is someone able to provide an outline? What are the necessary tools?

REPLY [9 votes]: This problem is in Cassels' book "Local Fields" and I wrote up a solution once along those lines, for an algebraic number theory class.  See 
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf,
but I should advise that it comes out seeming pretty tedious.  Solutions that involve elliptic curves are more conceptual.  Others have already provided pointers to references for that approach.<|endoftext|>
TITLE: "Understanding" $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$
QUESTION [96 upvotes]: I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers $G = \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. What do people mean when they say this? The Kronecker-Weber theorem gives a good idea of what the abelianization of $G$ looks like. But in one of Richard Taylor's MSRI talks, Taylor said that he's never heard of anyone proposing a similar direct description of $G$ and that to understand $G$ one studies the representations of $G$.
I know that there is a strong interest in showing the Langlands reciprocity conjecture [Edit: What I had in mind in writing this is evidently Clozel's conjecture, not the Langlands reciprocity conjecture - see Kevin Buzzard's post below] - that $L$-functions attached to $\ell$-adic Galois representations coincide with $L$-functions attached to certain automorphic representations. And I've heard people refer to the Tannakian philosophy which I understand as (roughly speaking) asserting that $G$ is determined by all of its finite dimensional representations.
Here is a representation of $G$ understood not to be a representation of $G$ as an abstract group but as a group together with a labeling of some of the conjugacy classes of G by rational primes (the Frobenius elements)?
When people talk about "understanding $G$" do they mean proving [Edit: Clozel's conjecture] (in view of the Tannakian philosophy)? If not, what do they mean? If so, this conceptualization seems quite abstract to me. Is this what people mean when they say "understand $G$"? Can [Edit: Clozel's conjecture] be used to give more tangible statements about $G$? 
Something that I have in mind as I write this is the inverse Galois problem (does every finite group occur as a Galois group of a normal extension of $\mathbb{Q}$?) and Gross' conjecture (mostly proven by now) that for each prime $p$ there exists a nonsolvable extension of $\mathbb{Q}$ ramified only at $p$. But I am open to and interested in other senses and respects in which one might "understand" $G$.

REPLY [15 votes]: There are already 8 good answers to this old question, but surprisingly there is something which I believe is true and important and that none of the answers says explicitly: Understanding Gal$(\overline{\mathbb Q}/\mathbb Q)$ as a group is not a major goal of number theory, and actually is not even a problem that really belongs to number theory. This is not to say this problem is not interesting. But it is a problem that belongs to algebra, field theory if you like, not to number theory. For example, the conjecture of Shafarevich mentioned by JSE (that's a natural subgroup of tat group is free profinite), while interesting, is not really number theory.
What number theory is concerned about is understanding  Gal$(\overline{\mathbb Q}/\mathbb Q)$, together with its family of subgroups $D_p$ (for $p$ a prime or infinity), the decomposition groups well defined up to conjugacy, and isomorphic to Gal$(\overline{\mathbb Q}_p/\mathbb Q_p)$ for $p$ finite, and to $\mathbb Z/2$ when $p=\infty$. I think that this what most people mean, but sometimes forget to say, when they say that understanding is a major goal of Number Theory. Indeed, if we "understand" sufficiently well Gal$(\overline{\mathbb Q}/\mathbb Q)$ with its decomposition subgroup, we understand in principle a lot of finite Galois groups which can be obtained with elementary operations from the above, and that can be interpreted say as various class groups of number fields. In turn those groups governs a lot of classical problems in number theory, like classifying quadratic forms over $\mathbb Z$ (a problem considered since Legendre), understanding which prime can be represented by a give quadratic forms, understanding all kind of Selmer groups, etc.<|endoftext|>
TITLE: Intuition for Integral Transforms
QUESTION [45 upvotes]: It is well known that the operations of differentiation and integration are reduced to multiplication and division after being transformed by an integral transform (like e.g. Fourier or Laplace Transforms).
My question: Is there any intuition why this is so? It can be proved, ok - but can somebody please explain the big picture (please not too technical - I might need another intuition to understand that one then, too ;-)

REPLY [4 votes]: Simply because the exponential function $\exp(xy)$ as a function of $x$ is an eigenfunction of the derivative operator and also the integration operator so we have:
$$ \frac{\mathrm{d}}{\mathrm{d}x} \exp(xy)=y \exp(xy)$$
If we think of integration as the inverse of the derivative operation we have:
$$\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{-1} \exp(xy)=\frac{1}{y} \exp(xy)$$
The situation is likely as we are working in a "continuous" basis $\exp(xy)$ indexed by the continuous parameter $y$, the derivation an integration being the diagonal matrices $diag(y)$ and $diag(1/y)$ respectively. So  because the family  $\exp(xy)$ with $y=-i\omega$ in the case of the fourier transform, is a "basis" for functions, the operations of differentiation and integration are reduced to multiplication and division for functions admitting such decompositions.<|endoftext|>
TITLE: Equivalent forms of the Grand Riemann Hypothesis
QUESTION [21 upvotes]: I have long been curious about equivalent forms of the Riemann hypothesis for automorphic L-functions.
In the case of the ordinary Riemann hypothesis, one gets a very good error term for the prime number theorem, one has the formulation involving the Mobius  mu function which is a result to the effect of the parity of prime factors in a square free number having a distribution related to that of flips of an unbiased coin, and one also has the reformulation in terms of Farey fractions.
I know that for L-functions attached to Dirichlet characters, one gets a very good error term for the prime number theorem for primes in arithmetic progressions. Presumably if one focuses on Dedekind zeta functions and Hecke L-series one gets a very strong effective Chebotarev density theorem or something like that.
But for L-functions attached to Hecke eigenforms for GL(2), or more abstract things like symmetric n-th power L-functions attached to automorphic forms or automorphic representations, it seems quite unclear to me what the significance of the Riemann hypothesis for these L-functions is. I think that I remember something about a zero free region to the left of the boundary of the critical strip being related to the Sato-Tate conjecture, so I have a vague impression that one might be able to get a good bound on the speed of convergence to the Sato-Tate distribution as an equivalent to the Riemann hypothesis for some of these L-functions.
What are some interesting equivalents to the Riemann hypothesis for automorphic L-functions that you know? I'm particularly interested in statements that have qualitative interpretations.
P.S. I've blurred the distinction between an equivalent of the Riemann hypothesis for a single L-function and equivalents to the Riemann hypothesis for a specified family of L-functions. I am interested in both things
P.P.S. I am more interested in equivalents than in consequences of the Riemann hypotheses for these L-functions in so far as equivalents "capture the essence" of the statement in question to a greater extent than consequences do. Still, I would would welcome references to interesting consequences of the Riemann hypothesis for automorphic L-functions, again, especially those with qualitative interpretations.

REPLY [13 votes]: Well, suppose pi is a cuspidal automorphic representation of GL(n)/Q.  This has the structure of a tensor product, indexed by primes p, of representations pi_p of the groups GLn(Qp).  The Satake isomorphism tells us that at almost all primes, each pip is determined by a conjugacy class A(p) in GLn(C).  In this language, the Riemann hypothesis for the L-function associated to pi says that the partial sums of tr(A(p)) over p < X show "as much cancellation as possible," and are of size sqrt(X).  But if n>1, we are dealing with very complicated objects, and the local components of these automorphic representations vary in some incomprehensible way...
You are right, there are certainly special cases.  If we knew GRH for L-functions associated to Artin representations then the Cebotarev density theorem would follow with an optimal error term.  Likewise, GRH for all the symmetric powers of a fixed elliptic curve E implies (and is in fact equivalent to; see Mazur's BAMS article for a reference) the Sato-Tate conjecture for E with an optimal error term.  But in general, reformulations like this simply don't exist.
There are many interesting consequences of GRH for various families of automorphic L-functions.  I recommend Iwaniec and Kowalski's book (Chapter 5), the paper "Low-lying zeros of families of L-functions" by Iwaniec-Luo-Sarnak, and Sarnak's article at http://www.claymath.org/millennium/Riemann_Hypothesis/Sarnak_RH.pdf<|endoftext|>
TITLE: 3/4-Lie superalgebras: how much of a theory can one develop?
QUESTION [11 upvotes]: Let $\mathfrak{s} = \mathfrak{s}_0 \oplus \mathfrak{s}_1$ be a real Lie superalgebra.  (The ground field does not matter much, but at least one formula will not work as written if the characteristic is 2 or 3.)  Recall that this means that there is a bilinear 2-graded bracket $[-,-]$ with three components
(a)  $\mathfrak{s}_0 \times \mathfrak{s}_0 \to \mathfrak{s}_0$   (skewsymmetric)
(b)  $\mathfrak{s}_0 \times \mathfrak{s}_1 \to \mathfrak{s}_1$
(c)  $\mathfrak{s}_1 \times \mathfrak{s}_1 \to \mathfrak{s}_0$   (symmetric)
satisfying the Jacobi identity, which splits into 4 components, which can be paraphrased as
(1) $\mathfrak{s}_0$ is a Lie algebra under (a)
(2) $\mathfrak{s}_1$ is an $\mathfrak{s}_0$-module under (b)
(3) the map in (c) is $\mathfrak{s}_0$-equivariant
(4) $[[x,x],x] = 0$ for all $x \in \mathfrak{s}_1$
The fact that the first three components can be written using words, whereas the fourth is easiest via a formula, suggests that they should perhaps be treated differently.
Indeed, over time I have come across many examples of superalgebras where the first three components of the Jacobi identity are satisfied but not the fourth.  I'd like to call them 3/4-Lie superalgebras.  I would like to know how far can this notion be pushed and in particular how much of the theory of Lie superalgebras still works in the 3/4 case.
To motivate this seemingly random question, let me end by pointing out one generic example where they arise.  There are others, but they are lengthier to describe.
Let $\mathfrak{g}$ be a metric Lie algebra; that is, a Lie algebra with an ad-invariant inner product $(-,-)$ and let $V$ be a symplectic $\mathfrak{g}$-module; that is, one possessing a $\mathfrak{g}$-invariant symplectic form $\langle-,-\rangle$.  Now let $\mathfrak{s} = \mathfrak{g} \oplus V$.  Then maps (a) and (b) are obvious: given by the Lie bracket on $\mathfrak{g}$ and the action of $\mathfrak{g}$ on $V$, respectively.  Map (c) is the transpose of map (b) using the inner products of both $\mathfrak{g}$ and $V$; in other words, if $x,y \in V$ then $[x,y] \in \mathfrak{g}$ is defined by
$$([x,y],a) = \langle a\cdot x,y\rangle$$
for all $a \in \mathfrak{g}$.
Then it is easy to see that $[x,y] = [y,x]$ and that (1)-(3) are satisfied, whereas in general (4) is not satisfied and instead defines a subclass of symplectic $\mathfrak{g}$-modules.

REPLY [5 votes]: At least in the semisimple case, it doesn't seem like there can be a theory of 3/4-Lie superalgebras other than a fairly predictable set of examples within Cartan-Weyl theory.  I don't know how to do all of the relevant calculations, but it is easy to see roughly how they would go.
Suppose that the even part $s_0$ is a semisimple Lie algebra $g$.  Then $s_1$ is can be a self-dual representation $V$ of $g$.  Unless possibly $g$ is $E_6$, I don't think that $V$ can be anything other than a self-dual representation.  So then the form is a symmetric element of $\mathrm{Inv}(g \otimes V \otimes V)$, which is a vector space whose dimension can be computed.  The dimension is at most the rank of $g$ when $V$ is irreducible.  Indeed it is usually the rank of $g$, when the highest weight of $V$ is far away from the walls of the Weyl chamber.  The dimension of the symmetric part is smaller, but it is often non-zero.  (This is the part that I don't know how to compute off the top of my head.)
So inevitably there will be a classification of these 3/4 Lie algebras using these branching rules.  Besides the classification, the only theory that springs to mind is representations of these 3/4 Lie algebras.  A representation $W$ would first off be a representation of $g$ (and I suppose a super vector space?).  Then there would be a $g$-invariant map $V \otimes W \to W$ which represents the action of $V$.  I don't see a rationale for imposing restrictions on this map other than $g$-invariance, for otherwise $g \oplus V$ would not be the adjoint representation of itself.  So once again, there is some Cartan-Weyl classification that says what $W$ can be, and I'm not sure what more you could say.<|endoftext|>
TITLE: What are conditions on real coefficients for zeros of a polynomial to be on the unit circle?
QUESTION [11 upvotes]: My complex analysis is decades in the rear view mirror.  Perhaps someone here can help.  I am looking for necessary and sufficient conditions on the coefficients of of a real polynomial of one complex variable (i.e. an element of ℝ[z]) so that all of its zeros will be on the unit circle.  Clearly it is necessary that the polynomial is palindromic, but that is not sufficient.  My question arose, after I discovered that if r,s are both real numbers then the polynomial p(z) = 

has all its roots on the unit circle.  Since I had not seen anything like this before, I wondered if there were perhaps just some conditions on the coefficients one could check.

REPLY [4 votes]: See paper
Palindrome-Polynomials with Roots on the Unit Circle by John Konvalina and Valentin Matache<|endoftext|>
TITLE: Infinite Ramsey theorem with infinitely many colours
QUESTION [6 upvotes]: Clearly, it is possible to colour the edges of an infinite complete graph so that it does not contain any infinite monochromatic complete subgraph. Now what about the following?

Let $G$ be the complete graph with vertex set the
  positive integers. Each edge of $G$ is then coloured c with probability $\frac{1}{2^c}$, for
  $c = 1, 2, \dots$ What is the probability
  that G contains an infinite
  monochromatic complete subgraph?

It is unclear for me if the answer should be $0, 1$, or something in between.

REPLY [10 votes]: Every countably infinite random graph is almost surely the Rado graph which contains all finite and countably infinite graphs as induced subsets. So each color class almost surely contains the Rado graph and hence a infinite monochromatic subgraph. See the following for more details here:
http://en.wikipedia.org/wiki/Random_graph
There are also links in the article to other articles including one on the Rado graph.<|endoftext|>
TITLE: One Point Compactification
QUESTION [6 upvotes]: Suppose X is a path-connected, locally compact, Hausdorff space and Y is its one-point compactification.  Let G be the fundamental group of X and H be the fundamental group of Y.  Is it true that the embedding of X into Y always induces a monomorphism G->H?  More generally, what is the relationship between G and H?

REPLY [12 votes]: More generally, if A and X0 are any finite CW complexes, and f : A → X0 is any map, let Y be the mapping cone of f, and let X be Y with the cone point removed; then Y is the one-point compactification of X, and the inclusion X0 → X is a homotopy equivalence.  (David's example is the case A = X0 = S1, f = id.)  So any map X → Y which is the mapping cone of something is homotopy equivalent to a one-point compactification.
I don't think you can realize any map of groups as the induced map on π1 of a mapping cone (0 → Z/2Z?) but you can realize (G → 0, 0 → a free group, ...)<|endoftext|>
TITLE: Prime numbers and strings of symbols
QUESTION [6 upvotes]: Suppose you have N symbols (e.g. "1","2",...,"N" or "a","b",...,"$") and a string of these symbols (say, the first trillion digits of pi).  Then does there exist a prime number whose N-ary representation contains that string of digits?

REPLY [5 votes]: If the sequence of digits ends in a digit that is coprime to N, one can make the stronger statement that there exists a prime number whose final digits are exactly that sequence; this is just Dirichlet's theorem about primes in arithmetic progression.
If the last digit isn't already coprime to N, just tack on an additional "1".

REPLY [4 votes]: I think there's an easier way (well, easier if you assume Dirichlet's Theorem - I think it's simpler than the strong form of Betrand's Postulate). Interpret the string as an integer M in base N; if it happens to be relatively prime to N, you're done - use Dirichlet to find a prime congruent to M modulo N^k where k is the length of the string. David's method places the given string as the most-significant digits of the prime, whereas this approach places it as the least-significant ones. 
If M happens to have a common divisor with the base N, you can just pad it with a "1" at the end (e.g. replace 31415 with 314151 in base 10). The padding replaces M by NM+1 which is obviously relatively prime to N.<|endoftext|>
TITLE: How does one intersect non-transverse divisors on Mg-bar.
QUESTION [7 upvotes]: Let Mg-bar be the Deligne-Mumford compactification of genus g curves, and let δ1 be the divisor of degenerate curves of the form `genus 1 meeting a genus g-1 transversely". 

Question 1: What is the n-fold self intersection of δ1?

I've seen mentioned in the literature that the answer for n = g is rational curves with g many genus 1 tails. This certainly seems reasonable -- this cycle has the right dimension and isvisually' obtained by g-1 many degenerations.

Question 2: Let C be a curve in Mg-bar contained in δ1. How do I calculate the intersection of C and δ1?

I'm interested in the case where C is a family of rational curves with g many fixed elliptic tails (so that each elliptic tail is the same elliptic curve), each with at most two rational components. The answer is in the literature, but without proof. Does anyone know a reference or a quick way to do this calculation?

REPLY [3 votes]: Both questions reduce to showing the 2-fold intersection of D_1 is a the closure of the locus of
genus g-2 curve with two elliptic tails.
The first question follows from this claim by induction and using the map D_1 -> M(g-1)^bar x M(1,1)^bar.
The second is a direct application of the claim.
Which brings us to the claim: Instead of working on Mg^bar - which is very hard, you can rigidify the question, and work on some compactified Hurwitz scheme. The Hurwitz scheme H(g,d) is the scheme of degree d covers of P^1 by genus g curves, with simple ramficiation only. The crucial facts here are:

Given branch points on P^1, there are only finitely many curves with these branch points.
You can compactify this space: when branch points colide you add a "buble": a copy of P^1, connected to the original P^1 where these points colided, and marked ramification points on the buble. You now have a node connecting the two P^1s; the rule is that the over this nodes, the ramification pattern is identical for the curves lying over the buble and the original P^1. Needless to say, you can add as many bubles as you want.

The pullback of D1 to H(g,d) is the closure of the locius of a pair of P^1's, such that: there are 2(g-2+d)-1 marked branch points on one of them, and 2d-1 marked branch points on the other. This is a purely combinatorial condition. Intersecting the condition with itself you get the closure of the loci of a chain of 3 P^1s, with 2d-1, 2(g-3+d) -2, 2d-1 points. Which is a pullback to H(g,d) of the expected class.
Note that we worked with pullbacks, so we have an equality of classes, not of the underlying reduced schenes.<|endoftext|>
TITLE: Non trivial colouring of the edges of an infinite complete graph
QUESTION [6 upvotes]: Can you build a probabilistic scheme for colouring each edge (independently of all other edges) of the complete graph G on the positive integers such that the probability that G contains an infinite monochromatic complete subgraph is neither 0 nor 1?

REPLY [9 votes]: It seems like this would go against the Kolmogorov 0-1 law..  If we let Xi denote the  coloring of all of the edges from i to integers larger than i, wouldn't the existence of an infinite monochromatic subgraph be a tail event?<|endoftext|>
TITLE: Unitary representations of SL(2, R)
QUESTION [19 upvotes]: I've heard that irreducible unitary representations of noncompact forms of simple Lie groups, the first example of such a group G being SL(2, R), can be completely described and that there is a discrete and continuous part of the spectrum of L^2(G).

How are those representations described?
Do all unitary representations come from L^2(G)?
How are those related to representation of compact SO(3, R)? 
What happens in the flat limit between SL(2, R) and SO(3, R)?

Also, is it possible to answer the questions above simultaneously for all Lie groups, not just SL(2, R)?

REPLY [2 votes]: Regarding classification of irreducible unitary representations of arbitrary Lie groups, Michel Duflo showed (I guess in the early 80's) that at least for algebraic Lie groups the classification can be reduced to the case of reductive Lie groups. See Vogan's Unitary representations of reductive Lie groups for more information.<|endoftext|>
TITLE: When does the sheaf direct image functor f_* have a right adjoint?
QUESTION [11 upvotes]: Say f: X → Y is a morphism of schemes.  The sheaf direct image functor f★ always has a left adjoint, namely the sheaf inverse image functor f★ (with tensoring).

Under what (sufficient) conditions do we know that f★ has a right adjoint?  What is it?

Answer to a related question (edit): If f★ preserves quasicoherence, then its restriction to quasicoherents f★: QCoh(X) → QCoh(Y) has a right adjoint when f is affine (in particular, any closed immersion or finite morphism will do).  The basic idea is to globalize the affine case (see Eric Wofsey's answer below); thanks to Pablo Solis for pointing me to page 6 of Ravi Vakil's notes explaining this.
In this question, I'm not restricting to the quasi-coherent categories.  One reason for working with non-quasicoherents is that j! , the "extension by zero" right adjoint to j★ for an open immersion j, doesn't take qcoh to qcoh.

REPLY [4 votes]: For your question. 
i agree with the answer from Greg, but in a different formalism. He used derived category. I did not. If f:X--->Y is a morphism of scheme. Then if f is affine morphism. Then direct image functor f_* :Cx---->Cy, has right adjoint functor f^!. Where Cx and Cy are category of sheaves or in particular, category of quasi coherent sheaves. So, in general, what we need is only the scheme is quasi compact and quasi separated.(I believe the quasi compact can be dropped, but I need some time to check globalization, I believe the flag variety of affine Kac-Moody algebra which is not quasi compact lies in this case).
The reference is 
M.Kontsevich and A.Rosenberg
Noncommutative spaces and flat descent. MPIM preprint
There is another related question. In category of quasi coherent sheaves. Say, if we have scheme morphism X---->Y, we always can get inverse image functor f^*: QcohY--->QcohX. But the direct image functor does not always exist. But if the scheme we are talking about is quasi compact and quasi separated. It exists. There is of course weaker condition. For this case, one can see the following papers:
1 D.Orlov Quasi coherent sheaves in commutative and noncommutative geometry
2 M.Kontsevich, A.Rosenberg. Noncommutative stack MPIM preprint 
3 SGA 6<|endoftext|>
TITLE: Actions of finite permutation groups on hereditarily finite sets.
QUESTION [7 upvotes]: Model theorists have a lot to say about so-called definable imaginary elements of a structure. One way to formulate imaginaries is the following: Suppose $\mathcal{M}$ is a structure with universe $M$, and let $G$ be the automorphism group of $\mathcal{M}$. 
Define
$$V_0(M) = M, \quad V_{\alpha+1}(M) = V_\alpha\cup P(V_\alpha(M)), \quad V_\lambda = \bigcup_{\alpha<\lambda}V_\alpha(M)$$
when $\lambda$ is a limit ordinal. Let
$$V(M) = \bigcup_{\alpha<\infty}V_\alpha(M).$$
Then $G$ has a natural action on $V(M)$ defined inductively by
$$g(x) = \{g(y):y\in x\}.$$
If $x\in V(M)$ and $S\subseteq M$, then $S$ supports $x$ if $G(S)\subseteq G(\{x\})$, where $G(S)$ is the pointwise stabilizer of $S$ and $G(\{x\})$ is the setwise stabilizer of $x$. The imaginaries are those $x\in V(M)$ which have finite support.
Now, for a finite structure $\mathcal{M}$, it makes sense to work with $HF(M)$ instead of $V(M)$ (stop the construction at the first countable infinite ordinal). 
I would like to know if anyone has done any work regarding the action of a group of permutations of a finite set $X$ on the hereditarily finite sets above $X$. Ideally, I'd like to get results "off the shelf" if they're out there.

REPLY [4 votes]: Well, since there have been no answers in a month, let me at least point out the easy fact that if M is finite, then every set in HF(M),
and indeed, every set in V(M), is imaginary over M.
(I assume here that M is taken as urelements in the
definition of V(M), as I mentioned in my comments to the
question above, since otherwise there are problems with the
action of G on V(M) and even HF(M) being well-defined.)
Theorem. If M is finite, then every object in HF(M),
and indeed every object in V(M), is an imaginary element.
Proof: Since M is finite, we may take S=M. If pi fixes
every element of M, then it is easy to see by transfinite
induction that the action of pi on V(M) is the identity.
Namely, if pi fixes every element of V_alpha(M), then it
clearly also fixes every element of V_{alpha+1}(M). And so
it fixes every element of V(M), including HF(M)=V_omega(M).
QED
OK. What this answer really shows is that the question is not about the imaginaries over M, but rather, about gaining a greater understanding of the actiom of G on V(M). Perhaps it would be helpful to define the parameter-free version of imaiginary, where we might say that X in V(M) is pure
imaginary over M if whenever pi is a permutation of M,
then pi(X)=X, under the induced action of pi on V(M). For
example, the set M itself has this property, as does the
power set P(M), the set {M} and {emptyset,M}, and so on. In
addition, any set whose transitive closure includes no
urelements from M will be pure imaginary. The question would be to characterize the pure-imaginary sets
over M.
This question shares many similarities with the various
forcing arguments showing the consistency of the negation
of the Axiom of Choice. Specifically, in the pre-forcing
days, set theorists built what are called the symmetric
models of set theory, by taking an infinite set of
urelements M and restricting to the elements of V(M) having
finite support. One can show that this is a model of
ZF-with-urelements having no wellordering of M. The forcing
proofs of the consistency of not-AC have exactly the same
flavor, where one adds an infinite set of mutually generic
Cohen reals, and then considers the sets that have names
with finite support over this set. This is precisely how
Cohen produced a model of ZF+not-AC, without urelements.
So one of the good reasons to study the imaginary elements over a set M is that they form a model of the set theory ZF-with-urelements. When M is infinite, however, then there can be no linear order of M in the pure imaginaries, since swapping elements outside the support of this set will not fix the order. In particular, M will not be well-orderable in this model of set theory, and so AC will not hold. For finite M, of course, there are linear orders of M having support M, and one can show that V(M) satisifes ZFC-with-urelements. But if one considers only the collection of pure-imaginaries, as I defined them above, then one will not even get ZF-with-urelements, unless M has only one element, since one will lose the Comprehensive (subset) axiom when there are parameters from M. For example, no proper nonempty subset of M can be pure imaginary. From this perspective, the pure imaginary sets are not so nice as the imaginary sets.<|endoftext|>
TITLE: Equivalence of derived categories which is not Fourier-Mukai
QUESTION [6 upvotes]: D. Orlov proved that any equivalence of bounded derived categories F:Db(X) -> Db(Y) is a Fourier-Mukai transform, when X and Y are smooth projective varieties. Is there any example of such equivalence, which is not a Fourier-Mukai transform (it is not an integral transform)?

REPLY [9 votes]: Schlichting gave an example of two categories of singularities which are derived equivalent but whose K-groups are not isomorphic. Dugger and Shipley (arXiv:0710.3070) expanded on this example and noted that it gives two dga's which are derived equivalent but not by an integral transform. 
Otherwise, Lunts and Orlov's results on uniqueness of enhancements give a large class of triangulated categories for which one might lift exact functors to dg-functors and apply Toen's result.<|endoftext|>
TITLE: Cohomology and Eilenberg-MacLane spaces
QUESTION [17 upvotes]: This question is related to this question from Dinakar, which I found interesting but don't yet have the background to understand at that level.
Unless I'm mistaken, the rough statement is that $H^n(X;G)$ (the $n$-dimensional cohomology of $X$ with coefficients in $G$) should somehow correspond to (free?) homotopy classes of maps $X \to K(G,n)$. I want to understand this better, in relatively elementary terms. Here are some questions which (I hope) will point me in the right direction.

What category are we working in? My guess is that $X$ should just be a topological space, the cohomology is singular cohomology, and our maps $X \to K(G,n)$ just need to be continuous.
Does this carry over if we give $X$ a smooth structure, take de Rham cohomology, and require our maps $X \to K(G,n)$ to be smooth?
How does addition in $H^n(X;G)$ carry over?
How does the ring structure on $H^*(X;G)$ carry over? (This has probably been adequately answered to Dinakar already.)

REPLY [17 votes]: We are working in the homotopy category of topological spaces where morphisms are homotopy classes of continuous maps.  More accurately, we tend to work in the based category where each object has a distinguished base point and everything is required to preserve that base point.  The non-based category can be embedded in the based category by the simple addition of a disjoint base point, so we often pass back and forth between the two without worrying too much about it.  The cohomology theory itself is slightly more interesting.  For CW-complexes, it doesn't matter which one you choose as they are all the same.   However, outside the subcategory of CW-complexes then the different theories can vary (as was mentioned in another question).  So what we do is the following: using Big Theorems we construct a topological space, which we call $K(G,n)$, which represents the $n$th cohomology group with coefficients in $G$ for CW-complexes.  So whenever $X$ is a CW-complex, we have a natural isomorphism of functors $\tilde{H}^n(X;G) \cong [X, K(G,n)]$, where the right-hand side is homotopy classes of based maps.  For arbitrary topological spaces, we then define cohomology as $[X, K(G,n)]$.  If this happens to agree with, say, singular cohomology then we're very pleased, but we don't require it.
Depends what you mean by "smooth structure".  Certainly in the broadest sense, you will get different answers if you insist on everything being smooth.  But for smooth manifolds, continuous maps are homotopic to smooth maps (and continuous homotopies to smooth homotopies) so the homotopy category of smooth manifolds and smooth maps is equivalent to the homotopy category of smooth manifolds and continuous maps.  However, you need to be careful with the $K(G,n)$s as they will, in general, not be finite dimensional smooth manifolds.  However, lots of things aren't finite dimensional smooth manifolds but still behave nicely with regard to smooth structures so this shouldn't be seen as quite the drawback that the other answerants have indicated.
Addition in $\tilde{H}^n(X;G)$ translates into the fact that $K(G,n)$ is an $H$-space.  The suspension isomorphism, $\tilde{H}^n(X;G) \cong \tilde{H}^{n+1}(\Sigma X, G)$ implies the stronger condition that $K(G,n)$ is the loop space of $K(G,n+1)$ and so the $H$-space structure comes from the Pontrijagin product on a (based!) loop space.  But the basic theorem on representability of cohomology merely provides $K(G,n)$ with the structure of an $H$-space.
As for the ring structure, that translates into certain maps $K(G,n)\wedge K(G,m) \to K(G,n+m)$.  I don't know of a good way to "see" these for ordinary cohomology, mainly because I don't know of any good geometric models for the spaces $K(G,n)$ except for low degrees.  One simple case where it can be seen is in rational cohomology.  Rational cohomology (made 2-periodic) is isomorphic to rational $K$-theory and there the product corresponds to the tensor product of vector bundles.

(It should be said, in light of the first point, that $K$-theory should only be thought of as being built out of vector bundles for compact CW-complexes.  For all other spaces, $K$-theory is homotopy classes of maps to $\mathbb{Z} \times BU$.)

REPLY [5 votes]: Here's a precise statement: reduced singular cohomology $H^n(X;G)$ is naturally isomorphic to homotopy classes of pointed maps from $X$ to $K(G,n)$, for any pointed topological space $X$ having the homotopy type of a CW complex.  Explicitly, the identity map $G = \pi_n(K(G,n)) = H_n(K(G,n); \mathbb{Z}) \to G$ gives an element $i_n$ of $H^n(K(G,n);G)$, and the isomorphism is given by taking a map $f : X \to K(G,n)$ to the class $f*(i_n)\in H^n(X;G)$.
By Yoneda, the additive and multiplicative structure on $H^*(X;G)$ come from certain (homotopy classes of) maps $K(G,n) \times K(G,n) \to K(G,n)$ and $K(G,n) \times K(G,m) \to K(G,m+n)$, respectively. The addition map is actually quite easy to see: $K(G,n)$ is the loopspace of $K(G,n+1)$, so it has a natural binary operation $K(G,n) \times K(G,n) \to K(G,n)$ given by concatenating loops. Since $K(G,n)$ is actually the double loopspace of $K(G,n+2)$, the Eckmann-Hilton argument (the same argument that shows higher homotopy groups are abelian) shows that this operation is commutative up to homotopy. I don't know of a good way to see the multiplication map.
As for your second question, the answer should be yes whenever it makes sense.  For any good notion of a smooth structure, it should be true that smooth maps up to smooth homotopy are the same as continuous maps up to continuous homotopy (at least, it is true for smooth manifolds). However, as far as I know there is rarely a natural smooth structure to put on $K(G,n)$, so this doesn't make sense (though I may be wrong!). In particular, to do de Rham cohomology you presumably want $G$ to be $\mathbb{R}$ or $\mathbb{C}$, and then $K(G,n)$ is really monstrous geometrically. You may want to take a look at this question.

REPLY [2 votes]: Indeed, the statement is that homotopy classes of continuous maps of pointed spaces $[X, K(G, n)]$ are in 1-1 correspondence with the elements of singular homology $H^n(X, G)$ for a CW-complex $X$.
The simplest example would be $G = \mathbb{Z}$, $n = 1$. Then you have $K(\mathbb{Z}, 1) = S^1$ and you can use a first cohomology class $c \in H^1(X, \mathbb{Z})$ to map the 1-skeleton of $X$ to $S^1$ (edge $e$ will make $c(e)$ loops around $S^1$). It's not hard to check that it gives the equivalence.
You also see that the choice of basepoint is irrelevant since you can shift it without affecting homotopy.
The example also helps to see that the additivity doesn't become obvious just from the things you wrote. To add properly, you need some kind of addition map on your target, that is, $K(G, n) \times K(G, n) \to K(G, n)$. How you prove this depends on your definition of Eilenberg-Mac Lane space, e.g. by universality. The ring structure comes from the wedge product (from an answer to this question).
If $X$ is smooth then the de Rham cohomology is the same as singular cohomology, but the space $K(G, n)$ has so little chance of being smooth (there was a question on MathOverflow explaining this) that the smooth maps are not really relevant, as expected for homotopy theory.<|endoftext|>
TITLE: A group action of the Heisenberg group with special symmetries
QUESTION [11 upvotes]: Suppose we look at the Heisenberg group $H_{d}$ as a matrix group of upper triangular matrices over the ring $\mathbb{Z}/d\mathbb{Z}$.  You can even choose $d$ to be prime if you want.  A natural irrep of $H_{d}$ acting on $\mathbb{C}^{d}$ maps the group elements into the "shift" and "phase" operators, plus roots of unity.  More specifically, the two natural generators map the orthonormal basis vectors from $j \to j+1\mod d$, and the Fourier transform of that operation, plus overall phases by roots of unity.  The question is this:
Can you find a unit vector $v$ such that $|(v,U_g v)| = c$ for all g not in the center of $H_{d}\ ?$  One can solve for the constant: $c=\frac{1}{\sqrt{d+1}}$.
Numerics suggests that these vectors exist in all the dimensions $< 67$, hence they may exist in every dimension, but the form of the vectors contains no (obvious) hint as to how to prove this.
This problem seems extremely truculent and any help is greatly appreciated!

REPLY [4 votes]: I just wanted to point out this paper to anyone who is interested. The authors report on a massive computational test of Zauner's conjecture. Don't be intimidated by the length; there are 18 pages of math, the rest is all tables of data.
I also want to use the word SIC-POVM, as that is what anyone searching this site for references will probably look for.<|endoftext|>
TITLE: Computations in Knot Homology Theories
QUESTION [6 upvotes]: 1) Relative to one another, how computable are the various knot homology theories?  For example, how many crossings can we allow a knot and still hope to compute its Khovanov homology, versus its knot Floer or Khovanov-Rozansky homologies?  (The latter two seem to be generally unlisted at the Atlas; KF at least has a page about how it can be computed, but KR seems totally absent. If I'm just not seeing the right link, feel free to let me know)
2) Do the algorithms by which these invariants are computed share common features, or are they really very specific to the particular homology being computed?  
For example, people computing KF homology draw square pictures that look very different from the pictures drawn by people doing Khovanov homology.  On a less superficial level, KF algorithms (as far as I can tell) appear to be 'global' in an essential way, while Khovanov calculations can be done locally, breaking a knot into smaller pieces and then working with the pieces.   So I'm led to believe the computations involved are different in a really fundamental way, but would be interested to see some combinatorial connection (as opposed to a topological relationship).  I have no idea how KR homology is computed, so have no idea how closely related the computations involved are to, say, ordinary Khovanov homology.

REPLY [6 votes]: Khovanov homology can be computed quite well, by Dror Bar-Natan's algorithm, as Ben says. It's nowhere near as good as algorithms for computing the Jones polynomial however.
The basic idea is that the speed of the computation depends largely on the "girth" of the link diagram. This is greatest number of intersections you see with a horizontal line. Within each girth, the algorithm applies to be a small degree polynomial in the number of crossings. However changing the girth dramatically affects the algorithm. Girth 14 is critical -- we can do up to about 80 crossings. See my paper with Freedman, Gompf and Walker on the smooth 4-d Poincare conjecture for details. Girth 12 and below is 'easy', and yes, 100 crossings is plausible, but more likely in weeks than seconds! Girth 16 and above seems to be out of range of current computers.
It's important to point out that memory constraints are the real issue for large Khovanov homology calculations. There's only a limited degree to which you can parallelize the computation, and so you end up wanting a single big computer with tonnes of RAM. The latest implementation of Dror's algorithm (my partial rewrite of Jeremy Green's program) tries to cache a lot of stuff of disk, but nevertheless we've done runs on 32gb machines and wished we had more memory!
As Ben says, Khovanov-Rozansky is much harder, and the purely combinatorial calculations you do for Khovanov homology get replaced by lots of commutative algebra. Ben at one point had a program that was doing this -- I'm not sure what the status of that is.
Knot Floer homology is a different matter. Even though there is now a nice 'cube of resolutions' picture for Knot Floer homology, it's only directly for knots, rather than tangles, so Dror's divide and conquer strategy doesn't apply. Dylan Thurston and others have a local description for tangles, but I don't know that anyone has tried implementing it. Computers aren't yet good at A_\infty tensor products!
If anyone is ever interested in writing better programs for Khovanov homology (or indeed trying to generalize to cover some of the other cases), please contact me, as I have some ideas! I'm no longer interested in writing code myself, but I'm happy to talk about it and assist. Particularly next semester at MSRI, if anyone is interested in doing some link homology programming they should talk to me.<|endoftext|>
TITLE: What is Floer homology of a knot?
QUESTION [9 upvotes]: I've heard that there are different theories providing knot invariants in form of homologies.  My understanding is that if you embed knot in a special way into a space, there is a special homology theory called Floer homology.

Question: what's the definition and properties of a Floer homology of a knot? How is it related to other knot homology theories?

REPLY [16 votes]: I can say something about this for Heegaard Floer homology.  Given a 3-manifold Y, you can take a Heegaard splitting, i.e. a decomposition of Y into two genus g handlebodies joined along their boundary.  This can be represented by drawing g disjoint curves a1,...,ag and g disjoint curves b1,...,bg on a surface S of genus g; then you attach 1-handles along the ai and 2-handles along the bi, and fill in what's left of the boundary with 0-handles and 3-handles to get Y.
The products Ta=a1x...xag and Tb=b1x...xbg are Lagrangian tori in the symmetric product Symg(S), which has a complex structure induced from S, and applying typical constructions from Lagrangian Floer homology gives you a chain complex CF(Y) whose generators are points in the intersection of these tori and whose differential counts certain holomorphic disks in Symg(S).  Miraculously, its homology HF(Y) turns out to be independent of every choice you made along the way.  We can also pick a basepoint z in the surface S and identify a hypersurface {z}xSymg-1(S) in Symg(S), and we can count the number nz(u) of times these disks cross that hypersurface: if we only count disks where nz(u)=0, for example, we get the hat version of HF, and otherwise we get more complicated versions.
Given two points z and w on the surface S of any Heegaard splitting we can construct a knot in Y: draw one curve in S-{ai} and another in S-{bi} connecting z and w, and push these slightly into the corresponding handlebodies.  In fact, for any knot K in Y there is a Heegaard splitting such that we can construct K in this fashion.  But now this extra basepoint w gives a filtration on CF(Y); in the simplest form, if we only count holomorphic disks u with nz(u)=nw(u)=0 we get the invariant $\widehat{HFK}(Y,K)$, and otherwise we get other versions.  The fact that this comes from a filtration also gives us a spectral sequence HFK(Y,K) => HF(Y).
This was constructed independently by Ozsvath-Szabo and Rasmussen, and it satisfies several interesting properties.  Just to name a few:

for knots K in S3 it has a bigrading (a,m), and the Euler characteristic $\sum_m (-1)^m HFK_m(S^3,K,a)$ is the Alexander polynomial  of K;
there's a skein exact sequence relating HFK for K and various resolutions at a fixed crossing;
the filtered chain homotopy type of CFK tells you about the Heegaard Floer homology of various surgeries on K;
the highest a for which HFK*(S3,K,a) is nonzero is the Seifert genus of the knot;
If Y-K is irreducible and K is nullhomologous, then HFK(Y,K,g(K)) = Z if and only if K is fibered (proved by Ghiggini for genus 1 and Ni in general, and later by Juhasz as well).

For knots in S3 it is also known how to compute HFK(K) combinatorially: see papers by Manolescu-Ozsvath-Sarkar and Manolescu-Ozsvath-Szabo-Thurston.
The relation to other knot homology theories isn't all that well understood, but there are some results comparing it to Khovanov homology.  For example, given a knot K in S3:

Just as Lee's spectral sequence for Khovanov homology gave a concordance invariant s(K), the spectral sequence from HFK(K) to HF(S3) gives a concordance invariant tau(K), and both of these provide lower bounds on the slice genus of K.  (Hedden and Ording showed that these invariants are not equal.)
There's a spectral sequence from the Khovanov homology of the mirror of K to HF of the branched double cover of K.
For quasi-alternating knots, both Khovanov homology and HFK are determined entirely by the Jones and Alexander polynomials, respectively, as well as the signature; this can be proven using skein exact sequences for both (Manolescu-Ozsvath).

Anyway, that was long enough that I've probably made several mistakes above and still not been anywhere near rigorous.  There's a nice overview that's now several years old (and thus probably missing some of the things I said above) on Zoltan Szabo's website, http://www.math.princeton.edu/~szabo/clay.pdf, if you want more details.<|endoftext|>
TITLE: Examples where Kolmogorov's zero-one law gives probability 0 or 1 but hard to determine which?
QUESTION [26 upvotes]: Inspired by this question, I was curious about a comment in this article:

In many situations, it can be easy to
  apply Kolmogorov's zero-one law to
  show that some event has probability 0
  or 1, but surprisingly hard to
  determine which of these two extreme
  values is the correct one.

Could someone provide an example?

REPLY [8 votes]: I was just looking through a book which proves many interesting and rather difficult results on Brownian motion (pdf link, website link), and it seems that the Kolmogorov zero-one law applies to most of these.
Using Fourier transforms, a standard Brownian motion Xt on the range 0≤t≤1 can be decomposed as
$$
X_t = At + \sum_{n=1}^\infty\frac{1}{\sqrt{2}\pi n}\left(B_n(\cos 2\pi nt - 1)+C_n\sin 2\pi nt\right)
$$
where A, Bn, Cn are independent normals with mean 0 and variance 1.
It follows that any property of the Brownian motion which is unchanged under addition of a linear combination of sines, cosines and linear terms is a tail event and, by Kolmogorov's zero-one law, has probability zero or one. Eg, Brownian motion is known to be nowhere differentiable (with probability 1).
It gets more interesting if you look at the modulus of continuity of Brownian motion. For any time t, the Law of the iterated logarithm says that
$$
\limsup_{h\downarrow 0}\frac{|X_{t+h}-X_t|}{\sqrt{2h\log\log (1/h)}}=1
$$
with probability 1. From this, you can say that, with probability one, Brownian motion satisfies this limit almost everywhere (but not everywhere - there are exceptional times).
More generally, they show that with probability one, the following are true at all times,
$$
\limsup_{h\downarrow 0}\frac{|X_{t+h}-X_t|}{\sqrt{2h\log(1/h)}}\le 1,\ 
\limsup_{h\downarrow 0}\frac{|X_{t+h}-X_t|}{\sqrt{h}}\ge 1.
$$
With probability one, these bounds are achieved. Times where the left inequality is an equality are fast times and slow times are when the right one is an equality. In the book I linked, they calculate lots of stuff about these fast and slow times, such as their fractal dimensions.
According to my decomposition of Brownian motion above, all of these definitions and statements are about tail events and we know that they must have probability 0 or 1 of being true. In fact, the sets of slow and fast times are defined in terms of tail events, so any measurable statement about these sets must be either always true or always false with probability one, and any measurable function of them, such as their fractal dimensions, must be deterministic constants with probability one, even though it is hard to calculate what they are. The same goes for many of the other properties of Brownian motion in the book I linked - they are tail events and therefore always true or false with probability one.<|endoftext|>
TITLE: Is the Fukaya category "defined"?
QUESTION [23 upvotes]: Sometimes people say that the Fukaya category is "not yet defined" in general.
What is meant by such a statement? (If it simplifies things, let's just stick with Fukaya categories of compact symplectic manifolds, not Fukaya-Seidel categories or wrapped Fukaya categories or whatever else might be out there)
What should a "correct" definition of the Fukaya category satisfy? (--- aside from, perhaps, "it makes homological mirror symmetry true")
What are some of the things which make defining the Fukaya category difficult?
What are some cases in which we "know" that we have the "correct" Fukaya category?

REPLY [10 votes]: Mike has given a good answer (basically - bubbling), but perhaps I can elaborate/add.
In general, due to disk bubbling one gets a curved or obstructed A-infinity category. The reason that's bad is that in this case m1 does not square to zero - no Floer homology. In the case when m0 is a multiple of fundamental class, then for CF(L,L) the A-infinity relations still say m1 squares to zero, and more generally for L_1 and L_2 if it's the same multiple \lambda. So you get completely disjointed categories for each \lambda. This happens for fibers of toric symplectic manifolds, for example (and on their mirror Landau-Ginsburg model we get the categories of singularities D^b sing, \lambda corresponding to the value of the superpotential). As Mike said, the business of bounding cochains is understanding when A-infinity relations can be modified to get m0 to be zero (unobstructed L) or multiple of fundamental class (weakly unobstructed L).
As for finding the "right" definition of Fukaya category, my understanding it is tricky business. Form the point of view of mirror symmetry, the current definition is a cheat, where we use passing to derived category to sweep under the rug all the problems - one of which is not dealing with immersed Lagrangians, and another is perhaps ignoring the suggestion of Kapustin-Orlov to include coisotropics. As I understand people are studying ways to do Floer theory for both of these new types of objects. This is however, a somewhat different question from the one you asked.<|endoftext|>
TITLE: Analogues of the Weierstrass p function for higher genus compact Riemann surfaces
QUESTION [9 upvotes]: There was a previous post on the correspondence between Riemann surfaces and algebraic geometry.  I want to ask a related but more detailed question.
BACKGROUND:
Engelbrekt gave an overview of how you start with a compact Riemann surfaces and map them into projective space
Links between Riemann surfaces and algebraic geometry
In the case of a genus 1 surface X there's a very explicit construction.  Namely X can be realized as ℂ/L for a lattice L ≅ ℤxℤ.  From here the Weierstrass p function and its derivative can be constructed
http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions
and these give you a map ℂ/L --> ℙ^2 via z |--> [p(z), p'(z), 1] which realizes X as a degree three curve in ℙ^2
QUESTION:
Say now X is a compact Riemann surface of genus g > 1.  As has been pointed out below I should restrict to say g = 1/2(d-1)(d-2) where d>3, because otherwise there is no hope to realize X as a nonsingular curve in ℙ^2.    
Is there   
1) a complex manifold Y that is a covering space of X such that X ≅ Y/G where G is the covering group of Y over X  
2) holomorphic functions f₁, f₂, f₃ from Y/G to ℂ∪∞  
such that z |--> [f₁,(z), f₂(z), f₃(z)] realizes X as a projective variety of dimension 1 in ℙ^2?
I'm told a good choice for Y would be the hyperbolic plane because then the 4g-gon representation of a genus g surface tiles the plane.

REPLY [15 votes]: For $X = \Delta/\Gamma$ a compact Riemann surface of genus $g>1$ a good analogue of the Weierstrass $p$-function is the Poincaré series $f_a(z) = \sum_{\gamma \in \Gamma} \gamma^*(q_a)$, where $q_a = dz^2/(z-a)$ is a quadratic differential with a simple pole at $a \in \Delta$.  This series gives a meromorphic quadratic differential on $X$ with a simple pole at any desired point and no other singularities.  If $b$ is close enough to $a$ then $f_b/f_a$ gives a meromorphic function on $X$ with a simple zero at $a$, providing local coordinates.  By compactness one can then choose finitely many $a_1,\ldots,a_n$ such that the corresponding $f_1,\ldots,f_n$ give an embedding of $X$ into $P^{n-1}$.<|endoftext|>
TITLE: Localization(s) of Categories
QUESTION [8 upvotes]: I've been trying to read a paper by Krause and came across a strange (to me, of course) notion of localization.  After looking around for a long time, and then finding this on his site, I see that there are two notions for localization, both with significant usage online.  These are namely Verdier localization and Bousfield localization.  Is there a strong motivation to use one over the other?  A little bit of context:
I see that Bousfield localization is defined for model categories, and this includes the notion of modules over a ring, among many many others.  I don't see a similar restriction for the Verdier localization.
Verdier localization uses the (standard for ''localization'') idea of a multiplicative set S of maps which are formally inverted by a functor Q from a category T to a new category denoted T/S.  Hartshorne's Residues and Duality is a reference for this.  (BTW, where does the assumption that the pullback of a multiplicative map is multiplicative come from?)
Bousfield localization is stated in several places (such as the Krause reference above) as a Verdier localization composed with a right adjoint for Q, which I understand to mean a functorial way of choosing objects in the isomorphism classes, and maps in the multiplicative subsets of each Hom(A,B).  It is also stated in the generality of model categories as needing three distinguished collections of morphisms: namely quasi-isomorphisms and (co)fibrations.  What bothers me more is the definition as given in Krause: an exact functor L from a triangulated category T to itself for which there exists a natural transformation η:Id-->L which commutes with L (ηL=Lη) and for which ηL is invertible.  As a second, smaller, question, what is encoded by the commutative condition (what would be lost without it?)?  I can come up with contrived examples (using the automorphisms of the objects LX) of course, but in what precise way does η really just encode L as a natural transformation?

REPLY [4 votes]: The conceptual home of all these notions is best understood by realizing that whenever we have a category with weak equivalences it is best viewed as an (oo,1)-category: a category which has an oo-groupoid (aka Kan complex) of morphisms between each pair of objects.
An ordinary localization of a category at a collection of morphisms is a geometric embedding: a faithful inclusion functor that has a left exact left adjoint.
This abstract-nonsense statement has -- and that's the advantage of abstract nonsense -- a straightforward generalization to (oo,1)-categories: a localization of an (oo,1)-category is just a faithful inclusion into it that has a left exact left adjoint -- only that now all these terms are interpreted in the corresponding higher category theory context. For instance adjoint functor now means adjoint (oo,1)-functor and so on.
So this tells us what localization should be conceptually. To actually do something with it we usually pick concrete presentations of these higher structures by model categories.
Under this presentation, the abstractly-defined localization of (oo,1)-categories is presented by the Bousfield localization  of the presented model categories.<|endoftext|>
TITLE: Tools for Organizing Papers?
QUESTION [46 upvotes]: Much like a previous question on keeping research notes organized, my question is how people keep their pile of papers organized. I've got a stack of about 100 in my office, most of them classifying as "want to read", a couple "have read", and lots in between.

REPLY [3 votes]: There is a new tool called colwiz, which seems to have some advantages for mathematicians. In particular, it has MathSciNet support, unlike Mendeley. I wrote a summary of why I like it here.<|endoftext|>
TITLE: Modular forms reference
QUESTION [5 upvotes]: If f is a weight 2 newform on $\Gamma_1(N)$ then there exists an abelian variety Af whose endomorphism algebra is isomorphic to the field generated by the coefficients of f.
I've seen this proven in Shimura's book, but was wondering if anyone knows of a different reference (perhaps one that is a bit more readable...).
Thanks.

REPLY [2 votes]: The Anterp conference volumes, "Modular functions in one Variable - I, II, III, .... " might contain what you want. I am not sure though, as I am unable to verify it by looking into all the volumes.<|endoftext|>
TITLE: Great mathematical figures and/or diagrams?
QUESTION [48 upvotes]: Most math papers have few figures, if any, although sometimes a well-chosen figure can be a tremendous help in understanding mathematical concepts. Does anyone have any examples of notable uses of figures in mathematical writing and/or texts that make great use of figures/diagrams/illustrations?

REPLY [3 votes]: In my opinion, a book that sets a new standard, not only in the quality of the pictures, but in overall design is A singular mathematical promenade by Étienne Ghys. A free PDF of the book is available here:
http://perso.ens-lyon.fr/ghys/promenade/
The pictures are in fact more vivid in the PDF version than on paper.<|endoftext|>
TITLE: What is a rigorous statement for "linear time-invariant systems can be represented as convolutions"?
QUESTION [7 upvotes]: In Signal Processing books, a fundamental theorem is that linear time invariant systems can be represented as a convolution with a distribution.  Could you give a mathematically rigorous statement of this theorem, or refer a book that includes it?
Edit: For example, would the following be a correct statement?
"Let S' be the space of tempered distributions.  If L is a linear operator on S' that commutes with translations, then there exists a distribution h in S' such that Lf = f*h for all f in S'"

REPLY [2 votes]: The Schwartz kernel theorem seams relevant here. You might recall from your signal processing books that in the linear but non-time-invariant case we still get the output as a integral $\int K(x,y) f(y)dy$ where $f$ is the input. The kernel theorem makes this rigorous as I recall where, $K$ then can be a distribution. 
Once you have that theorem it is probably easy to get the statement you want.<|endoftext|>
TITLE: D-modules supported on the nilpotent cone
QUESTION [13 upvotes]: I am reading some papers which involve D-modules on a Lie algebra g, which are supported on the nilpotent cone n.  They are equivariant for the action of G.  (In particular, I consider g=sl_n).
It was explained to me (the statement, not the proof) that the category of such D-modules is semisimple, and that the simple objects are given by (the intermediate extension) of the constant sheaf sheaf of functions on each g-orbit on n, so they are in bijection with Jordan decompositions with all zeroes, so just partitions of n.
I'm not strong with D-modules (I'm learning!).  My question is this:  Since g is affine, D(g) is an associative algebra, namely the Weyl algebra on the vector space g.  How can I describe the D(g)-module M corresponding to partition \lambda explicitly as a module over D(g)?

REPLY [2 votes]: This is a very old question, but for future visitors, I would like to point out that the answer is described in section 3 of:
T. Levasseur, "Equivariant D-modules attached to nilpotent orbits in a semisimple Lie algebra", 1998.
It isn't clear to me that all of the simple objects are covered by the constructions in this paper, but at least it seems that the type you're considering are - i.e. those arising from minimal extension of functions on a nilpotent orbit.<|endoftext|>
TITLE: Intuitionistic Lowenheim-Skolem?
QUESTION [13 upvotes]: Is there a version of the Löwenheim-Skolem theorem in intuitionistic logic?  I'm particularly interested in the "downward" form.  The standard proof I know uses the Tarski-Vaught test for elementary substructures, which in turn relies on the fact that "forall" is equivalent to "not exists not", and that fails intuitionistically.

REPLY [6 votes]: Is the question about the Löwenheim–Skolem theorem for classical models in intuitionistic metatheory, or about the Löwenheim–Skolem theorem for intuitionistic (Kripke) models in classical metatheory? The latter certainly holds. One can prove it easily by realizing that a Kripke model can be represented by a suitable two-sorted classical model in such a way that satisfaction of any intuitionistic formula in the original model is first-order definable in the representation, and then applying the classical Löwenheim–Skolem theorem.<|endoftext|>
TITLE: Derived functors vs universal delta functors
QUESTION [19 upvotes]: I would like to understand the relationship between the derived category definition of a right derived functor $Rf$ (which involves an initial natural transformation $n: Qf \rightarrow (Rf)Q$, where $Q$ is the map to the derived category) and the "universal delta functor" definition given in Hartshorne III.1.
I already know that $R^if(A) = H^i(Rf(A))$.  What I want to know most is:

What is the role of the natural transformation n in this comparison?

I guess it can be thought of as a natural map from a injective resolution of $f(A)$ to $f$(an injective resolution of $A$), but I'm not sure what is the significance of this... Does anyone know a good reference explaining such things?

REPLY [10 votes]: I haven't checked all the details, but I think the story could go like this. (I have to apologize: it's a bit long.)
(1) Let  $F:\mathsf A\rightarrow \mathsf B$  be an additive left exact functor between two abelian categories. Take an injective resolution of an object  $A$  in  $\mathsf A$:
$$0\rightarrow A \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots $$
Let us call  $i: A \rightarrow I^0$  the first morphism. Apply $F$ to this exact sequence:
$$0\rightarrow FA \rightarrow FI^0 \rightarrow FI^1 \rightarrow \cdots $$
Now, the total right derived functor of $F$ applied to $A$ (thought as a complex concentrated in degree zero) is the complex
$$\mathbb RF(A) = [ FI^0 \rightarrow FI^1 \rightarrow FI^2 \rightarrow \cdots ]$$
and the classical right derived functors of $F$ are its cohomology:
$R^nF(A) = H^n(\mathbb RF(A)) = H^n(FI^)$.
These ${R^nF}_n$ are a universal cohomological delta-functor and we have a natural transformation of functors
$$qF \Rightarrow (\mathbb RF)q$$
which is essentially
$$Fi: FA \rightarrow \mathbb RF(A)$$ 
(here we have extended  $F$ degree-wise to the category of complexes, and this is the degree zero of the natural transformation, because  $\mathbb RF(A)^0 = FI^0$ ).
(2) Now, let  $T^n : \mathsf A \rightarrow \mathsf B$  be a cohomological delta-functor and  $f^0 : F \Rightarrow T^0$ a natural transformation. We have to extend this $f^0$  to a unique morphism of delta-functors  ${ f^n : R^nF \Rightarrow T^n }$.
To do this, observe that, in general, given two right-derivable functors between two, say, model categories $$F,G: \mathsf C \rightarrow  \mathsf D$$, and a natural transformation between them $t: F \Rightarrow G $, we have a natural transformation between the total right derived functors $\mathbb Rt : \mathbb RF \Rightarrow \mathbb RG$ because of the universal property of the derived functors:
Indeed, if $f : qF \Rightarrow (\mathbb RF)q$  and  $g : qG \Rightarrow (\mathbb RG)q$
 are the universal morphisms of the derived functors, then we have a natural transformation
$$gt : F \Rightarrow (\mathbb R G)q$$
and, so, because of the universal property of derived functors, a unique natural transformation $\mathbb R t : \mathbb R F \rightarrow \mathbb R G$ such that $(\mathbb R t)qf = g$.
(3) So, take our  $f^0 : F \Rightarrow T^0$ , extend it to a natural transformation between the degree-wise induced functors between complexes. Passing to the derived functors, we obtain
$$\mathbb R f^0 : \mathbb R F \Rightarrow \mathbb R T^0.$$
Taking cohomology, for each $n$, we get
$$H^n(\mathbb R f^0) : H^n (\mathbb R F) \Rightarrow H^n (\mathbb R T^0).$$
But these are the classical right derived functors, so we have natural transformations
$$R^nf : R^n F \Rightarrow R^nT^0$$
and because the classical right derived functors are universal delta-functors, we have unique natural transformations
$$i^n : R^nT^0 \Rightarrow T^n$$
which extend the identity 
$$i^0 : R^0T^0 = T^0.$$
The composition
$$i^n \circ R^f : R^F \Rightarrow T^n$$
is, I think, the required morphisms of delta-functors that we need.<|endoftext|>
TITLE: Cone shaped solutions to wave equation
QUESTION [10 upvotes]: When I studied physics, we learned how to write down planar waves and spherical waves. But, when I turn on my flashlight, I see a cone of light. How can I see that there is a solution to the wave equation which describes a wave in a conical region, dropping off sharply outside that cone? And am I right that the wave equation cannot describe a cylindrical beam?

REPLY [17 votes]: At human scales, the wavelength of visible light is so tiny (or equivalently, the frequency is so high) that the wave equation can be modeled by geometric optics (this is the high frequency limit of the wave equation).  The cone you see from the flashlight is then nothing more than the shadow that the casing of the light casts from the light bulb. 
At larger wavelengths, diffractive effects become more pronounced (when was the last time you saw a water wave propagating along a cone?).  The asymptotic behaviour of any localised wave disturbance (in odd dimensions) is then an outgoing spherical wave, modulated by a scattering amplitude that depends only on the direction of propagation (and which is basically a kind of Radon transform of the initial data); this can be seen from asymptotics of the fundamental solution.  (It's slightly more complicated in even dimensions due to the failure of the sharp Huygens principle.)<|endoftext|>
TITLE: Is there any meaning to a "nice bijective proof?"
QUESTION [16 upvotes]: From Zeilberger's PCM article on enumerative combinatorics:

The reaction of the combinatorial enumeration community to the involution principle was mixed. On the one hand it had the universal appeal of a general principle... On the other hand, its universality is also a major drawback, since involution-principle proofs usually do not give any insight into the specific structures involved, and one feels a bit cheated. [... O]ne still hopes for a really natural, "involution-principle-free proof."

The quest for bijective proofs really is, at heart, the quest to categorify an equation proved via "decategorified" arithmetic, to provide an explicit family of isomorphisms in the category of finite sets (or the category of finite sets and bijections, I don't think it matters). So categorifying multiplication and addition is easy -- they just correspond to product and coproduct of sets, respectively. Categorifying subtraction is trickier, but that's what the involution principle does for us. (On a tangential note, I know that categorifying division is [in]famously hard in the category of sets -- is it anything like as hard when we talk about finite sets? ETA: Looking at Conway's paper, the answer seems to be "yes and no." No, because with finite sets it's kosher to use a bijection between a set of size n and {0, 1, ..., n-1};  yes, because this is sort of like a "finitary Axiom of Choice," and in particular it's not canonical.)
So is there any real meaning, in a categorical sense, to the enumerative-combinatorial dream of "really nice proofs?" Or will there necessarily be identities that can only be proved bijectively by categorifying, in a general and universal way, their "manipulatorics" proofs?
Edit: So philosophically this is a category-theory question, and it'd be nice to have it as a "real" category-theory question. Here's my (very rough) attempt at phrasing it as such.
Let T be a topos where we can categorify addition, multiplication, and subtraction of natural numbers. Then, are there functors between T and FinSet that preserve these decategorifications? If so, then I think maybe we can ask the question in terms of topos theory, although maybe not -- I don't really know much topos theory, so I'm certainly having trouble.

REPLY [3 votes]: Igor Pak wrote a paper in which he defines the notion of an "asymptotically stable" partition bijection, then shows there is no such bijection proving the Rogers-Ramanujan identities.<|endoftext|>
TITLE: Division Algebras as Algebraic Groups
QUESTION [7 upvotes]: If I'm given a division algebra D with Z(D)=F, then how can I view Dx as an algebraic group defined over F?  I'd like to see first how Dx can be given the structure of a variety defined over F, and then to see how the group law on Dx is defined over F.

REPLY [11 votes]: Choose an F-basis of D. The multiplication is described by certain quadratic functions, with respect to this basis; D* is given by the nonvanishing of a polynomial function (the norm). 
So the multiplication can be understood as defining an algebraic group structure on the complement of a hypersurface in an affine space.<|endoftext|>
TITLE: Number theoretic sequences and Hecke eigenvalues
QUESTION [14 upvotes]: What are some number theoretic sequences that you know of that occur as (or are closely related to) the sequence of Fourier coefficients of some sort of automorphic function/form or the sequence of Hecke eigenvalues attached to a Hecke eigenform?
I know many such sequences, but am always looking for more.
Some examples
(1) The sequence a(n) deriving from the traces a(p) of the Frobenius elements in a Galois representation (Langlands reciprocity conjecture)
(2) Number of representations of a natural number as a sum of k squares (theta functions)
(3) The sum of powers of divisor functions (Eisenstein series)
(4) The central critical values of L-functions attached to all quadratic twists of a Hecke eigenform (Kohnen, Waldspurger)
(5) Intersection numbers of certain subvarieties of Hilbert modular surfaces (Hirzebruch-Zagier)
I'll end with a question that is ill-posed but nevertheless very interesting (at least to me personally): why do so many familiar and yet diverse sequences appear in this fashion? Note that many of them have a history of study that precedes the recognition that they are essentially coefficients of automorphic functions/forms.

REPLY [6 votes]: Characters of rational vertex operator algebras tend to yield modular functions.  This is due to the space of torus partition functions in a chiral conformal field theory being a complex moduli invariant.  The standard example is the monster vertex algebra, whose character is j-744.  Other examples come from lattice CFTs (presumably describing a bosonic string propagating in a torus), and have the form of a theta function divided by a power of eta.  The characters are never Hecke eigenforms, because of the pole at infinity, but traces arising from higher-weight vectors may be.  In some cases, the vertex algebra structure is supposed to arise from geometry of a target space, so this phenomenon may be related to Hirzebruch-Zagier (number 5).
Characters of highest weight representations of affine Kac-Moody algebras yield modular forms.  One can reasonably argue that this is a special case of the previous paragraph, since (I think) they come from Wess-Zumino-Witten.  The Weyl-Kac character formula for such representations is one way to get Macdonald identities, and the smallest case (trivial rep of affine sl2) yields the Jacobi triple product.<|endoftext|>
TITLE: Characterisation for separable extension of a field
QUESTION [8 upvotes]: Can someone verify this for me.. or tell me what reference shows me this... is this true:

Let $k$ be a field. Then a field extension $K$ of $k$ is separable over $k$ iff for any field extension $L \supseteq k$ the Jacobson radical of the tensor product $K\otimes_k L$ is trivial.

I got this idea by looking at some definitions of separable algebras (which is not my field of research.. but somehow this definition got me intruiged). Anyone knows if this is true and why so? or maybe a reference or two about it?

REPLY [8 votes]: Let $k\subset K$ be a completely arbitrary extension of fields. This extension is said to be separable if equivalently
a) For all extensions $k\subset L$, the ring  $K \otimes _k L$ is reduced.  [A ring is reduced if $x^n=0 \Rightarrow x=0$] 
or
b) For some algebraically closed extension $k \subset \Omega$ , the ring $K\otimes _k \Omega$ is reduced.
If the extension $k\subset K$ is algebraic the above are equivalent to the more elementary definition
c) Every $x\in K$ is separable, i.e. the minimum polynomial of $x$ over $k$ is separable. [Which means that its roots are distinct in an algebraic closure $k^a$ of $k$].
This can all be found in Bourbaki's Algebra, chapter V.<|endoftext|>
TITLE: Examples of well-displayed mathematics on the internet
QUESTION [5 upvotes]: I'm interested in hearing of examples of mathematical (or, at a pinch, scientific) websites with serious content where the design of the website actually makes it easy to read and absorb the material.  To be absolutely clear, the mathematical content of the website should be on the website itself and not in an electronic article (so meta-sites that make it easy to find material, like MathSciNet or the arXiv, don't count).
Edit: I'm extending this to non-internet material.  I want examples where the design of the document/website/whatever actually helped when reading the material.
As a little background, I know that LaTeX is meant to help us separate content from context and concentrate on each one in turn, but I often feel when reading an article that the author has concentrated solely on the content and left all of the context to TeX.  This is most obvious with websites where there are some really well-designed websites to compare with, but holds as well with articles.

REPLY [2 votes]: Hi, 
For french speaking readers, there's an example of discussion forum 
using latex2html to generate a nice display of mathematical forumlas:
http://www.les-mathematiques.net/phorum/addon.php?0,module=recent_messages 
It is realized using the Phorum php framework ( www.phorum.org ), 
and is easily customizable (the addition of a latex2html module 
 is part of this customization).
But I don't know if something similar exists for english speaking users.
Best regards,
Eric<|endoftext|>
TITLE: Errata database?
QUESTION [26 upvotes]: Some authors do a really great job by collecting errors and comments to their books and putting a list on their websites. I wonder if there is some (perhaps wiki-style) website where errata are collected. Does anybody know?

REPLY [4 votes]: I have created a web page recently. If you know about any Errata (official or unofficial) leave a comment here after checking it in mathbookserrata.<|endoftext|>
TITLE: Tools for collaborative paper-writing
QUESTION [115 upvotes]: I personally use a revision control system (git) to manage my own paper writing, back things up, and synchronize between different machines.  However, I've found most programmer's revision control systems to require a bit too much training to try and push on a co-author for the purposes of working on a joint paper.
Are there any people out there using software tools to help handle joint writing?  How successful have they been?

REPLY [13 votes]: Overleaf.com (previously WriteLaTeX) is a service that lets you edit & compile  latex in the browser, with real-time edit merging.
It's comparable to ShareLaTeX.com.  Notable differences (some unchecked since 2013):

ShareLaTeX is mostly open source.  Easy to self-host on Sandstorm.
Overleaf doesn't require signup => Less friction when convincing collaborators to try it, just send them the URL and they can edit.
Overleaf's embedded preview uses images (fast but blurry), ShareLaTeX uses PDF.js.  Both allow PDF download, of course.
both save history, can show diffs.

Overleaf free plan doesn't save full history, only when you explicitly create a version ?
ShareLaTeX has a Track Changes mode letting you later reject/accept them.

both have some spell checking and auto-complete.
ShareLaTeX.com on paid plans has 2-way Dropbox sync allowing some offline/local editing (though only for one user; I tried it and wasn't impressed).  WriteLaTeX only has 1-way backup to dropbox.
ShareLaTeX.com on paid plans has [sync with Github](https://www.sharelatex.com/blog/2015/02/10/sharelatex-github-sync.html; Overleaf supports direct Git access.
Overleaf has "rich text" mode somewhat resembling LyX.
Constructs such as sections, bullet lists, math are typeset inside the source.
Both have UI for adding/replying/closing comments.  In Overleaf those are part of the source so the UI only applies in rich text mode.
both embedded chat, with math rendering.
both have forward/reverse search (SyncTeX)
Overleaf seems to work better on Android & iOS.
UPDATE: unclear, Android typing in Overleaf is also problematic as of 2016.  I think the only way I ever got it to work acceptably on Android was with external keyboard?<|endoftext|>
TITLE: Is there a topology on growth rates of functions? 
QUESTION [29 upvotes]: I've often idly wondered one can say about the collection of "growth rates". By growth rate, let's say we mean an equivalence class of functions (0,infty) \to (0,\infty), where two functions f_1,f_2 are equivalent if f_1/f_2 and f_2/f_1 are bounded away from 0 and infinity. You can add, and multiply them, and they form a poset under the pordering where f_1 <= f_2 if f_1/f_2 is bounded above.  
So, in loose terms, does the sequence xlog(1+x), xlog(log(10+x)), xlog(log(log(100+x))), ... converge to x in some natural way? With a little thought you can construct a growth rate which is strictly greater than x and strictly less than all growth rates in that sequence, so it probably no. Still is there any sort of natural "topology"? Can you find a directed set of growth rates which are linearly ordered, and eventually smaller than anything larger than x? 
There's probably a better way to look at this (which is why I ask). =)

REPLY [17 votes]: There is some fascinating work in the subject of cardinal
characteristics of the continuum in set theory that
directly relates to the concept of growth rates of
functions. I believe that it is the ideas in this subject
that are ultimately fundamental to your question. I explain
a little about the general subject of cardinal
characteristics in my answer
here.
Much of the interest of your question is already present
for functions on the natural numbers. The two main orders on such functions that
one considers in cardinal characteristics are

almost-less-than, where $f \lt^\ast g$ means that
$f(n) \lt g(n)$ for all $n$ except finitely often, and
domination, where $f \lt g$ means that $f(n) \lt g(n)$ for
all $n$.

A family $F$ of functions is said to be unbounded if there
is no function $g$ that has $f\lt^\ast g$ for all $f\in F$. That is, an unbounded family is a family that is not
bounded with respect to $\lt^\ast$. A family $F$ is
dominating if every function $f$ is dominated by some
function $g\in F$. The corresponding cardinal characteristics
of these two types of families are:

The bounding number $\frak{b}$ is the size of the smallest
unbounded family.
The dominating number $\frak{d}$ is
the size of the smallest dominating family.

It is easy to see that $\frak{b}\leq\frak{d}$, simply
because any dominating family is also unbounded. Also, both
$\frak{b}$ and $\frak{d}$ are at most the continuum $\frak{c}$,
the size of the reals. It is not difficult to see that both
of these numbers must be uncountable, since for any
countable family of functions $f_0,f_1, f_2,\ldots$, we can build the function $g(k) = \sup_{n\leq k}f_n(k)+1$, which eventually exceeds any particular
$f_n$. In other words, any countable family of
functions is bounded with respect either to
almost-less-than or with respect to domination. Thus,
$\omega_1\leq\frak{b},\frak{d}\leq\frak{c}$. 
It follows from these simple observations that if the
Continuum Hypothesis holds, then both the bounding number
and the dominating number are equal to $\omega_1$, which under CH is the same as the continuum $\frak{c}$. 
Now, the amazing thing is that ZFC independence abounds
with these concepts. First, it is relatively consistent
with ZFC that the Continuum Hypothesis fails, and both the
dominating number and the bounding number are as large as
they could possibly be, the continuum itself, so that  $\frak{b}=\frak{d}=\frak{c}$. 
Second, it is
also consistent that both are strictly intermediate between
$\omega_1$ and the continuum $\frak{c}$, but still
equal. Next, it is also consistent with $\text{ZFC}+\neg\text{CH}$ that
the bounding number $\frak{b}$ is as small as it could be,
namely $\omega_1$, but the domintating number is
much larger, with value $\frak{c}$. The tools for proving all
these results and many others involve the method of
forcing.
Now, let me get to the part of my answer that directly
relates to the idea of rates-of-growth. A slalom is
defined to be a sequence of natural number pairs
$(a_0,b_0), 
(a_1,b_1), \ldots$ with $a_n\lt b_n$. Each slalom s corresponds to the collection
of functions $f:\omega\to\omega$ such that $f(n)$ is in the
interval $(a_n,b_n)$ for all but
finitely many $n$. That is, imagine an olympic athlete on
skiis, who must pass through (all but finitely many of) the
slalom posts. An $h$-slalom is a slalom such that
$|b_n-a_n|\leq h(n)$. 
Thus, a slalom is a growth rate of functions, in a very
precise sense. With suitably chosen (countable collections
of) slaloms, it is possible to express the concept of
growth rate that you mentioned in your question.
The set theory gets quite interesting. For example, a major
question is: how many slaloms suffice to cover all the
functions? This is particularly interesting when one
restricts the size of the slaloms by considering $h$-slaloms.
A fat slalom is a $2^n$-slalom, where the
$n^{\rm th}$ interval has size at most $2^n$. 
It turns out that this is connected with ideas involving
meagerness, otherwise known as category. For example,
Bartoszynski
proved that every set of reals of size less than $\kappa$ is
meager if and only if for every function $h$ and every family
of $h$-slaloms $F$ of size less than $\kappa$, there is a
function $g$ eventually missing every slalom in $F$. In other
words, the possibility of a family of fewer than $\kappa$
many $h$-slaloms covering all the functions is equivalent to
every set of size less than $\kappa$ being meager.
And so on. There is a large amount of work on these and
similar ideas. An article particularly focused on slaloms
would be
this.
And there is a survey article by
Brendle on
cardinal characteristics.<|endoftext|>
TITLE: Simplicial Model of Hopf Map?
QUESTION [21 upvotes]: The Hopf fibration is a famous map S3 --> S2 with fiber S1, which is the generator in pi_3(S2). We can model this map in terms simplicial sets by taking the singular simplicial sets of these spaces and the induced map of simplcial sets. But this model is HUGE and isn't really useful for doing calculations. Does anyone know a nice small model for this map in terms of simplicial sets? Something suitable for computations? This map is also the attaching map used to build CP2 out of S2, so I would equivalently be interested in a small combinatorial model for CP2.

REPLY [5 votes]: There is a small simplicial set description of the Hopf map in Clemens Berger's thesis, Exemple 1.19, pp. 45-47.<|endoftext|>
TITLE: Constructing Affine Kac-Moody Groups
QUESTION [9 upvotes]: Does anyone know a simple construction for Affine Kac-Moody groups? There is a book by Kumar ("Kac-Moody groups, their flag varieties, and representation theory") that does the construction for the general Kac-Moody case, but I find the presentation dense. There is also a section that constructs a one-dimensional extension of the loop group by loop rotation, which is a fairly transparent definition. However, I don't know how to add on the final central extension.
Even if the answer to my question is "There is no simpler construction," could someone also tell me about a fruitful way to get my hands on Affine Kac-Moody groups?

REPLY [6 votes]: As you said, the main thing is to construct the central extension. 
The story is relatively straightforward for groups of type $A$ and gets more complicated in the general case. 
First, let $\mathcal K=k((t)), \mathcal O= k[[t]]$.
As usual, let the affine Grassmannian $Gr_G$ be the quotient $G(\mathcal K)/G(\mathcal O)$. Then in order to construct a central extension of $G(\mathcal K)$ it is enough to construct a line bundle $\mathcal L$ on $Gr_G\times Gr_G$ which is
a) $G(\mathcal K)$-equivariant (note that this is the same as to specify a $G(\mathcal O)$-equivariant line bundle on $Gr_G$)
b) Has the property that for any $x_1,x_2,x_3\in Gr_G$ we have an isomorphism $\mathcal L(x_1,x_3)=\mathcal L(x_1,x_2)\otimes \mathcal L(x_2,x_3)$ satisfying obvious associativity relation (of course, all this is actually an additional structure - a careful way to say it is to identify two lifts of $\mathcal L$ to
$Gr_G\times Gr_G\times Gr_G$ such that two identifications on $Gr_G^4$ coincide, but this is standard).
Indeed given such $\mathcal L$ we can define the central extension $\hat{G}$ as the set of pairs $(g,\alpha\in \mathcal L(1,g)), \alpha\neq 0$ (here I identify elements of $G$ with their image in $Gr_G$) and the multiplication is easy from b) above. 
Now we need to construct $\mathcal L$ as above. This is easy if $G=GL(n)$ (or $G=SL(n)$). Namely, in this case $Gr_G$ is the same as the space of lattices $\Lambda \subset \mathcal K^n$ and for any two such lattices $\Lambda_1,\Lambda_2$ we can set $\mathcal L(\Lambda_1,\Lambda_2)=\det(\Lambda_1/\Lambda_1\cap \Lambda_2)\otimes \det(\Lambda_2/\Lambda_1\cap\Lambda_2)^{-1}$. Properties a), b) are obvious.
For general $G$ the story is more complicated. First, canonically central extensions are in one-two-one correspondences with even invariant bilinear forms
on the Cartan of $Lie(G)$. For simple $G$ there is a minimal such form and it is enough to construct the extension corresponding to this minimal form (all others are "powers" of it in the appropriate sense). For $G=SL(n)$ the above construction gives exactly the minimal extension. For general $G$ you can easily adapt the above construction once you choose a representation $V$ of $G$. The problem is that when $G$ is not of type $A$, there is no $V$ that gives the minimal form (any $V$ defines an even invariant form on the Lie algebra of $G$ but usually it is not minimal). So, in this way you are going to get powers of the correct line bundle (and thus powers of the correct central extension). For instance, if you take $V$ to the be the adjoint representation, you get the $2h^{\vee}$-power of the minimal bundle, where $h^{\vee}$ is the dual Coxeter number. Constructing minimal $\mathcal L$ for general $G$ is a relatively tricky business - the best treatment of this that I know was given by Faltings in 

Gerd Faltings, Algebraic loop groups and moduli spaces of bundles. J. Eur. Math. Soc. 5 (2003), 41-68. doi: 10.1007/s10097-002-0045-x.<|endoftext|>
TITLE: How exactly is Hochschild homology a monad homology?
QUESTION [14 upvotes]: Many texts which praise the generality of the bar construction associated to a monad, say that Hochschild homology is an example of this. 
What exactly is in this case the underlying endofunctor of the monad, on which category is it an endofunctor, what are the monad structure maps and, most important (since I think my confusion lies here), why do then the face maps look as on the wikipedia page?

REPLY [4 votes]: Assuming that the base ring $k$ is a field (or that the algebra is projective over $k$), a functor whose monad computes Hochschild homology is the one which maps $k$-modules $M$ to the $A$-bimodule $A\otimes M\otimes A$, which is adjoint to the forgetful functor in the other direction.
The bar complex corresponding to this adjunction is not exactly the one used by Hochschild to define his cohomology, but it is easily seen to give naturally isomorphic results, since it more or less evidently constructs gives projective resolutions of bimodules (not only of $A$, as is the case with the monad coming from one-sided extension of scalars)<|endoftext|>
TITLE: Most helpful heuristic?
QUESTION [13 upvotes]: What's the most useful piece of mathematical "folk wisdom" you've encountered? I'm talking here about things that aren't theorems, or even conjectures, or even shadows of conjectures -- just broad analogies or slogans that nonetheless tend to be incredibly useful for formulating conjectures, understanding when a line of attack on a problem will or won't work, etc.
The first example that leaps to my mind is the Cramer probabilistic model of the primes, which tends to give astoundingly good predictions for all sorts of things despite being horribly simple.
Oh, right: Community wiki, one tidbit of wisdom per post, please!

REPLY [2 votes]: Every set or function you can build in a concrete fashion starting from (countablye many) other measurable sets or functions is measurable.
There are some counterexample to this. But it gets you a feel for measurability like we have a feel for what is continuous and what is not.<|endoftext|>
TITLE: simplicial deRham complex and model category structure
QUESTION [6 upvotes]: To every simplicial manifold is associated its simplicial deRham complex.
Is there any literature that discusses explicitly to which extent this classical construction, regarded as a (contravariant) functor from simplicial manifolds to dg-algebras, is homotopical?
For instance: simplicial manifolds naturally embed into the category of simplicial presheaves on the category of manifolds, on which we have the standard local model structure on simplicial presheaves. On dg-algebras there is the standard model structure on dg-algebras.
Is there any literature that discusses explicitly the respect of the simplicial deRham complex operation of the respective weak equivalences?

REPLY [7 votes]: There is a projective model structure on the category of (pre)sheaves with value in any reasonnable model category (e.g. simplicial sets, complexes of abelian groups, commutative k-dg-algebras, where k is some field of char. 0, or k-dg-algebras for any commutative ring k); see for instance  def. 4.4.33 and 4.4.40 and cor. 4.4.42 in Ayoub's book (Astérisque 315), whose online version is here (there is also a paper of Barwick which does the job if you want descent à la Lurie (to appear in HHA soon, I think)).
If you have a site C and a left Quillen functor F:M->M', you get a left Quillen functor Sh(C,M)->Sh(C,M') between model categories of sheaves. The fibrant objects will always have the good taste of being exactly the termwise fibrant sheaves which satisfy (hyper)descent, and, if you have enough points, the weak equivalences are defined stalk-wise.
If M=SSet and M'=Complexes of R-vector spaces, the usual adjunction SSet<->Comp(R) gives a Quillen adjunction:
Sh(C,Sset) <-> Sh(C,Comp(R))
If you consider the projective model structure on Sh(C,Sset), any simplicial sheaf X such that, for each n, X_n is a sum of representables, is cofibrant (and any cofibrant object is weakly equivalent to such a thing). Hence, if C={differential real manifolds}, if you allow your manifolds to be stable under small sums, the simplicial manifolds are just your favourite cofibrant objects. You can thus apply all the machinery of model categories (e.g. the homotopy category of cofibrant objects is equivalent to the whole Ho(M) for any model category M). As the de Rham complex satisfies descent on manifolds, then it is a fibrant object (for the projective model structure on sheaves on complexes), and using the above Quillen adjunction, you will get that the derived sections of the de Rham complex over a simplicial manifold behaves like maps from a cofibrant object to a fibrant object in any model category: quite well. From there, you should be able to prove that de Rham cohomology of simplicial manifolds is in fact the explicit description of the total left derived functor of the left Quillen functor DR:Sh(manifolds,Sset)->cdga^op.
[Added comments] You can also consider the left Bousfield localization L_R of the projective model category of simplicial sheaves on the category of manifolds which consists to invert the maps XxR->X for any manifold X (where R denotes the real line). The result is a model category which is Quillen equivalent to the model category of simplicial sets. Using the Poincaré lemma (which gives you the homotopy invariance of de Rham cohomology), the functor
DR:Sh(manifolds,Sset)->cdga^op induces a functor of shape
Ho(Sset)=Ho(L_R Sh(manifolds,Sset))->Ho(cdga)^op
which is simply (isomorphic to) the functor of Quillen-Sullivan.<|endoftext|>
TITLE: Bound on cardinality of a union
QUESTION [6 upvotes]: Suppose I have n finite sets A1 through An contained in some fixed set S, and I am given non-negative integers N and N1 through Nn such that each Ai has cardinality N, and each k-tuple intersection has cardinality less than or equal to Nk.
Can I use this to construct a good lower bound on the cardinality of the union of the Ai?

REPLY [9 votes]: I don't know your reason for asking this question, so it's unlikely that what I'm about to write will be helpful. Nevertheless, there's an easy method I like a lot for deducing a lower bound just from the knowledge that N_2 is small. It may be contained in what has been said above -- I haven't checked.
The idea is to think of each set A_ i as a 01-valued function on a set of size M. We then look at the ell_ 2 norm of sum_i A_i. The square of the ell_ 2 norm is   sum_ {i,j}|A_ i cap A_ j|,   which by assumption is at most nN + n(n-1)N_ 2. But we also have a lower bound: the ell_ 1 norm of the sum is nN, from which it follows that the square of the ell_ 2 norm is at least (nN)^2/M.
Putting these two bits of information together gives a lower bound for M of nN/(1+(n-1)N_ 2/N). So, for example, if N_ 2 = cN for some smallish positive constant c, then the lower bound is roughly N/c.<|endoftext|>
TITLE: What are the higher homotopy groups of Spec Z ?
QUESTION [55 upvotes]: The homotopy groups of the étale topos of a scheme were defined by Artin and Mazur.  Are these known for Spec Z?  Certainly π1 is trivial because Spec Z has no unramified étale covers, but what is known about the higher homotopy groups?

REPLY [46 votes]: $Spec(\mathbb{Z})$ should only be considered as $S^3$, if you "compactify" that is add the point at the real place. This is demonstrated by taking cohomology with compact support.
The étale homotopy type of $Spec(\mathbb{Z})$ is however contractible (indeed what do you get by removing a point form a sphere?)  to see this (all results apper in Milne's Arithmetic Dualities Book)
(let $X=Spec(\mathbb{Z})$ )

$H^r_c(X_{fl},\mathbb{G}_m)=H^r_{c}(X_{et},\mathbb{G}_m) = 0$ for $r \neq 3$.

$H^3_c(X_{fl},\mathbb{G}_m)=H^3_{c}(X_{et},\mathbb{G}_m) = \mathbb{Q}/\mathbb{Z}$

by 2+1, we have:


$H^3_c(X_{fl},\mu_n)= \mathbb{Z}/n$
$H^r_c(X_{fl},\mu_n)= 0$ for $r \neq 3$.

since we have a duality $$H^r(X_{fl},\mathbb{Z}/n)\times  H^{3-r}_c(X_{fl},\mu_n) \to \mathbb{Q}/\mathbb{Z} $$
we have

$H^0(X_{fl},\mathbb{Z}/n) = H^0(X_{et},\mathbb{Z}/n) = \mathbb{Z}/n,$


$H^r(X_{fl},\mathbb{Z}/n) = H^r(X_{et},\mathbb{Z}/n) = 0$, $r >0$

Now  since $\pi_1$ is trivial we have by the Universal Coefficient Theorem, the Hurewicz Theorem and the profiniteness theorem for 'etale homotopy that all homotopy groups are zero.<|endoftext|>
TITLE: Can the "physical argument" for the existence of a solution to Dirichlet's problem be made into an actual proof?
QUESTION [15 upvotes]: Caveat: I don't really know anything about PDEs, so this question might not make sense.
In complex analysis class we've been learning about the solution to Dirichlet's problem for the Laplace equation on bounded domains with nice (smooth) boundary. My sketchy understanding of the history of this problem (gleaned from Wikipedia) is that in the 19th century everybody "knew" that the problem had to have a unique solution, because of physics. Specifically, if I give you a distribution of charge along the boundary, it has to determine an electric potential in the domain, which turns out to be harmonic. But Dirichlet's proof was wrong, and it wasn't until around 1900 that Hilbert found a correct argument for the existence and uniqueness of the solution, given reasonable conditions (the boundary function must be continuous, and the boundary really has to be sufficiently smooth). 
Is the physical heuristic really totally meaningless from a mathematical point of view? Or is there some way to translate it into an actual proof?

REPLY [3 votes]: The electrostatic intuition does lead to a correct mathematical formulation of the Dirichlet problem.
Let's consider an electric charge distribution of two thin layers (one layer is positive and the other is negative) located along a closed surface $S\subset\mathbb R^3$. Assume that $d>0$ is the distance  between charges along the normal $n_p$ to the surface at point $p$. Let $\rho\in C(S,\mathbb R)$ denote the distribution's density. 
A pair of two opposing charges $+Q=\rho/d$ and $-Q=-\rho/d$ creates an electric field. The limit of the field when $d\to 0$ is known as the dipole. For any $x\in \mathbb R^3$, the dipole potential at the point $p\in S$ has the form
$$\frac{\rho}{d}\Phi(x-(p+n_pd/2))-\frac{\rho}{d}\Phi(x-(p-n_pd/2))=\rho\frac{\partial \Phi(x-p)}{\partial n_p} +o(1)\quad{\rm as\ \ } d\to0,\qquad(1)$$
where $\Phi(x)=-(4\pi|x|)^{-1},\quad x\in \mathbb R^3,$
is the fundamental solution of Laplace's equation. (1) gives the dipole potential of a single dipole at the point $p\in S$ and the integral
$$u(x)=\int_{S}\rho(p) \frac{\partial \Phi(x-p)}{\partial n_p} dp,\quad x\in\mathbb R^3,$$
is the potential of the whole distribution.
Now, a simple computation shows that $u(x)$ is a harmonic function when $x$ is not on the surface. It has a jump when $x$ passes through $S$:
$$u_{-}(x_0)=u(x_0)-2\pi \rho(x_0),\quad u_{+}(x_0)=u(x_0)+2\pi \rho(x_0),\quad x_0\in S,\qquad\qquad\qquad\qquad (2)$$
where $u_{-}(x_0)$ ($u_{+}(x_0)$) is the limit from the interior (exterior) of the surface.
Relations (2) are integral equations w.r.t. the unknown density (assuming that the potential on the surface is known). The equations can be solved using the Fredholm approach.  The function $u(x)$ then gives a solution to the Dirichlet problem.
Edit. See a nice little textbook by Arnold where he shows how to make the physical intuition rigorous in this problem.<|endoftext|>
TITLE: Construction of maps $f:S^3 \to S^2$ with arbitrary Hopf invariant?
QUESTION [8 upvotes]: The well-known Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$ has explicit constructions involving the geometry of $C^2$ and intersections of complex lines with the $3$-sphere.  They don't seem to generalize easily to "higher" Hopf maps from $S^3 \rightarrow S^2$ with Hopf invariant not equal to one.  Are there any simple expressions for those maps?

REPLY [4 votes]: Actually, yes, there is a construction involving complex projective line. 
Consider all points (x1, x2, x3, x4) on a 3-sphere in the 4-dimensional space. Our goal is to map them to $S^2$  which is the same as $CP^1$
To do this, take a quaternion 

$$x_1+x_{2}i+x_3j+x_4k$$

raise it to the $n$-th power (this is that group law on a 3-sphere) and decompose back into two complex numbers $z_1+z_2j$ . Now $z_i:z_i$
is a point of a complex projective line, that's it!<|endoftext|>
TITLE: Prestacks and fibered categories
QUESTION [10 upvotes]: It seems to be a well-known fact that there is a "one-to-one correspondence'' between prestacks and fibered categories. Here a prestack (called a pseudo-functor in SGA1) means a contravariant lax functor $F$ on a small category taking values in the $2$-category of small categories in which the structure natural transformation $F(f)\circ F(g)\Rightarrow F(gof)$ is invertible.
For example, Vistoli says in this note that "the theory of fibered categories is equivalent to the theory of pseudo-functors" at the end of section $3.1$. 
Is this "equivalence" an equivalence of 2-categories? If so, where can I find a proof?

REPLY [4 votes]: The proof of the equivalence of 2-categories between the 2-category of "prestacks"(whose meaning is a pseudofunctor in the context of this question) and the 2-category of fibered categories is mentioned in the theorem 2.2.3. in the paper 

Fosco Loregian, Emily Riehl, Categorical notions of fibration, Expositiones Mathematicae (Available online 14 June 2019) doi:10.1016/j.exmath.2019.02.004, arXiv:1806.06129.

Though my answer is posted after a decade, but I felt this information may help some future readers. 
Thank you.<|endoftext|>
TITLE: Do orbits and stable loci of group actions have natural scheme structures?
QUESTION [14 upvotes]: Suppose G is an algebraic group with an action G×X→X on a scheme. Then many of the usual constructions you make when you talk about group actions on sets can be made scheme-theoretically. For example, if x∈X is a point (thought of as a map x:∗→X, where ∗ is Spec of a field or the base scheme), then the stabilizer Stab(x) is naturally a scheme because it is the fiber product

Stab(x) ----> G×X    (g,y)
  |            |       _
  |            |       |
  v   (x,x)    v       v
  ∗ --------> X×X   (gy,y)


Does the orbit of a point have a natural scheme structure?
Does the fixed locus (the set of points x∈X fixed by all of G) have a scheme structure?

For (1), if everything is sufficiently nice, then the morphism G×∗→X, given by g→g⋅x has a scheme-theortic closed image, and the actual image is constructible and invariant under the G-action, so the actual image is an open subset of its closure. Thus, the orbit gets the structure of an open subscheme of a closed subscheme of X. But this construction doesn't feel very natural.
For (2), you can obviously define the functor Fix(T)={t∈X(T)|t is fixed by every element of G(T)}. Is this functor always representable?
Edit: Given that Scott has given such an excellent (negative) answer to question (1) but not said anything about question (2), I've asked (2) as a separate question.

REPLY [8 votes]: Question number 1: Let your base S be Spec k[x] (say k is an algebraically closed field), let X be Spec k[x,y], and let G be Ga,S, with action over the point s given by gs(xs) = (sgs) + xs.  This action is transitive away from zero, so the orbit of the zero section is a plane with a slit.  This is not an open subscheme of a closed subscheme of X, because it is not a scheme.<|endoftext|>
TITLE: When does Tannakian theory work over affine schemes besides fields?
QUESTION [10 upvotes]: By 'work' I would like the correspondence between fiber functors (to finitely generated projective modules) and algebraic groups to be the same as in the field case.
Specifically, if $A$ is an affine ring, and if $\operatorname{Proj}(A)$ is the category of finitely generated projective A-modules, when can we say that a fiber functor $w:\mathcal{T}\to\operatorname{Proj}(A)$ corresponds to an algebraic group over $A$, where $\mathcal{T}$ is an $A$-linear tensor category.

REPLY [4 votes]: There is a nice recent paper by Michael Broshi on the arxiv which is related to this theme when the base is a Dedekind scheme (such as Dedekind domain, or regular proper curve over a field).<|endoftext|>
TITLE: How do I compare the different notions of Fourier transform for sheaves?
QUESTION [27 upvotes]: There is a close but not perfect relationship between algebraic D-modules on C^n, constructible sheaves on C^n in the analytic topology, and \ell-adic sheaves on an n-dimensional vector space over a field of characteristic p:

Forming the sheaf of not-necessarily-algebraic solutions to the algebraic system of differential equations carries a class of D-modules (regular holonomic D-modules) to constructible sheaves, and in an appropriately derived setting this is an equivalence of categories.  This is the Riemann-Hilbert correspondence.  
Less functorially, you can try to find an \ell-adic model for your constructible sheaf on affine space over not C but something like the p-adic integers (which you embed in C), and reduce mod p.   

Many things can go wrong, but there are comparison theorems.  For instance there is a good notion of "constant sheaf on a subvariety" in all three settings, and taking the cohomology of this sheaf gives a similar answer in the three cases: this is the comparison theorem between de Rham, Betti, and \ell-adic cohomology.
In all three settings there is an operation called Fourier transform.  An algebraic D-module on C^n is a module over the Weyl algebra C[x,y,...,d/dx,d/dy,...], and its Fourier transform is the pullback along the change of variables x --> -d/dx, d/dx --> x.  In the topological setting you have the Fourier-Sato/Kashiwara-Schapira transform, whose target is sheaves on the real dual vector space to C^n.  And in characteristic p you have the Fourier-Deligne transform, which involves the Artin map x^p - x somehow.
For both D-modules and \ell-adic sheaves, there is no restriction on the kind of sheaf you can transform.  But it may take a D-module with regular singularities to one without regular singularities, or take an \ell-adic sheaf without wild ramification to one with wild ramification.  In the topological setting, you can only take the Fourier transform of sheaves that are constant along rays from the origin (usually because these sheaves are C^*-equivariant), but then the new thing is as well-behaved as the old.
I would like to understand better how these things are related, or what can go wrong.  Maybe they just have misleading names--I am pretty sure that the so-called Fourier-Mukai transform is a red herring, here.  But I have seen them used in the same way in Springer theory: to construct representations of Weyl groups on the cohomology (Betti or \ell-adic) of Springer fibers.  Are there any comparison results between the different Fourier transforms?
PS Ben's suggestion is that there should be very strong comparison theorems if we work with C^* or G_m-equivariant objects in all settings.  So, specifically, the Riemann-Hilbert correspondence should commute with the Fourier transform on C^*-equivariant holonomic D-modules and constructible sheaves.  Is this a well-known and referencable result?

REPLY [4 votes]: Malgrange has written a book called "Equation differentielles a coefficients polynomiaux". The whole book is about the compatiblity of the topological Fourier transform and the Fourier-Laplace transform of D-modules in dimension 1 (even for irregular holonomic D-modules). 
In the G_m equivariant, i.e. monodromic, setting, the topological Fourier transform is the Fourier-Sato transform and the compatiblity is considered well known but I don't know any reference for it.   
PS: This article of L. Daia generelasizes Malgrange's results to higher dimensional vector spaces.<|endoftext|>
TITLE: Non-commutative geometry from von Neumann algebras?
QUESTION [16 upvotes]: The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^*$-algebras with unital $*$-homomorphisms to the category of compact Hausdorff spaces with continuous maps. Hence the study of $C^*$-algebras is sometimes referred to as non-commutative topology.
All diffuse commutative von Neumann algebras acting on separable Hilbert space are isomorphic to $L^\infty[0,1]$. Hence the study of von Neumann algebras is sometimes referred to as non-commutative measure theory.
Connes proposed that the definition of a non-commutative manifold is a spectral triple $(A,H,D)$. From a $C^*$-algebra, we can recover the "differentiable elements" as those elements of the $C^*$-algebra $A$ that have bounded commutator with the Dirac operator $D$. 
What happens if we start with a von Neumann algebra? Does the same definition give a "differentiable" structure? Is there a way of recovering a $C^*$-algebra from a von Neumann algebra that contains the "differentiable" structure on our non commutative measure space? This would be akin to our von Neumann algebra being $L^\infty(M)$ for $M$ a compact, orientable manifold (so we have a volume form). Or are von Neumann algebras just "too big" for this?
One of the reasons I am asking this question is Connes' spectral characterization of manifolds (arXiv:0810.2088v1) which shows we get a "Gelfand theory" for Riemannian manifolds if the spectral triples satisfy certain axioms. Connes starts with the von Neumann algebra $L^\infty(M)$ instead of the $C^*$-algebra $C(M)$.

REPLY [2 votes]: Alain Connes proves in the paper "A unitary invariant in Riemannian Geometry" ( http://arxiv.org/abs/0810.2091 )
that all information on a Riemannian geometry can be decoded from the spectrum of a Dirac operator and the relative position of two commutative von Neumann algebras. 
I don't know how you would translate this into a statement about noncommutative spaces however, I'll have to look at the paper in more detail.<|endoftext|>
TITLE: (∞, 1)-categorical description of equivariant homotopy theory
QUESTION [24 upvotes]: I'm trying to learn a bit about equivariant homotopy theory.  Let G be a compact Lie group.  I guess there is a cofibrantly generated model category whose objects are (compactly generated weak Hausdorff or whatever) topological spaces with G-action and whose morphisms are G-maps, in which the generating cofibrations are maps of the form G/H x Sn-1 → G/H x Dn (n ≥ 0, H a closed subset of G) and the generating acyclic cofibrations are the obvious analogous thing.  Apparently the weak equivalences in this category are those maps which induce weak equivalence on H-fixed points for every closed subgroup H of G.  I assume the corresponding (∞,1)-category is presentable.  (My preliminary question is, does anyone know a good source for this paragraph?)
My real question is: Can you give an (∞,1)-categorical description of this category, say via a universal property, or built somehow from the category of spaces?  For instance, what is an explicit presentation as a localization of a category of presheaves of spaces?  (An example of the kind of answer I am looking for is "functors from BG to Spaces", but that describes a model category of G-spaces whose weak equivalences are simply weak equivalences of the underlying spaces.)
(My next question would be asking for an analogous description of the equivariant stable homotopy category.  I imagine this would be easy if I knew how to answer the first question, but if something special happens in the stable situation, I would like to know about it.)

REPLY [12 votes]: I have been trying to hold myself back from answering this, because I am not entirely sure my view on this is accurate.  To me, it seems like a G-spectrum should be a spectrum with an action of G on it, full stop.  Obviously you have to specify your notion of spectrum, and obviously if you want to include topological groups G it had better be a symmetric monoidal category of spectra that is enriched over topological spaces.  So you could take S-modules of EKMM or orthogonal spectra (my personal favorite) or symmetric spectra based on topological spaces.  With any of these categories, there is a notion of a G-spectrum, by which I mean a spectrum with an action of G.  
I can hear you objecting--you must be being too naive--what about complete G-universes?  I take the point of view that picking a universe corresponds to picking a model structure on the one God-given category of G-spectra.  Picking a smaller universe just means localizing the model structure.  So the complete universe is the "initial" one, and every other universe is a localization of the complete universe.  The naive universe is the "terminal" one, in the sense that it is a localization of every other universe.  There are lots of universes corresponding to model structures in between these.  
I cannot now remember how these model structures are supposed to go, but I believe that both Neil Strickland and Tony Elmendorf have separately written something about this approach.  Tony's might be part of a joint paper, I can't remember.  I think it is just a different way to look at things,  but it goes so much against the prevailing viewpoint that it has not gotten so much traction.  
Again, I have to confess that I am working from memory from something I probably did not completely understand.  Possibly Mike Shulman or someone else will be able to convince me I am completely wrong.<|endoftext|>
TITLE: Non-Lie Subgroups
QUESTION [12 upvotes]: A result of Borel and Lichnerowicz states that the holonomy group of a connection on a principal $G$-bundle is a Lie subgroup of $G$ (Cartan had earlier asserted this, but apparently without proof).
This restriction, that it be a Lie subgroup, allows for a lot of poorly-behaved subgroups, for example a line with irrational slope on a torus.  This subgroup comes from a perfectly fine immersion of the Lie group $\mathbb{R}$, but it's not closed in the induced topology of the torus.  
As an example of something that's not a Lie subgroup, let $G= \mathbb{R}$, consider an uncountable set of $\mathbb{Q}$-independent points, none of which are rational, and consider the subgroup they generate.  If this were a Lie subgroup it would be the image of an uncountable discrete space (there can't be anything $1$-dimensional, since we left out the rationals), which wouldn't be second countable, hence not a manifold and not a Lie group.  
This seems like a pretty contrived example, and I suspect there is more content to 'being a Lie subgroup' than having countably many components.  However, I can't seem to pin down something that would illustrate this.  Can anyone give me an example of a connected subgroup of a Lie group that is not a Lie subgroup?

REPLY [4 votes]: Apparently, there is no consensus in the literature on how to define a Lie subgroup of a Lie group. One should be careful not to mix different definitions. The result of Yamabe (1950) states that every arcwise connected subgroup of a real Lie group is Lie subgroup (in the sense of Yamabe). Bourbaki uses the notion of an integral subgroup (Lie Groups and Lie Algebras, Chapter III, §8, Exercise 4). Onishchik/Vinberg use the notion of a virtual Lie subgroup (Foundations of Lie theory in Lie groups and Lie algebras I, p. 39, Theorem 2.4). Note that Onishchik/Vinberg also use the (stronger) notion of a Lie subgroup and present a counterexample not fulfilling their definition on page 14. For more on the technical details one can refer to Bourbaki (Chapter III, §6 and §8) or Onishchik/Vinberg.<|endoftext|>
TITLE: Presheaves as limits of representable functors?
QUESTION [16 upvotes]: If I remember correctly, I read that given a presheaf $P:\mathcal{C}^{op} \to Set$, it is possible to describe it as a limit of representable presheaves. Could someone give a description of the construction together with a proof?

REPLY [2 votes]: I will give two easier-seeming facts and proofs, and then show that the "colimit of representables" idea is obtained by the essentially same arguments in a more-general setting.

If $R$ is a ring, then any $R$-module $M$ is a quotient of a free module.

Proof: Choose generators for $M$ by some intricate procedure, or by taking every element of $M$ to be a generator.  For each generator $m$, build a map $R^1 \to M$ specified by sending $1 \mapsto m$.  This assignment extends to at most one map of modules since $1$ generates $R^1$ as a module, and moreover, it does extend, since the formula $r \mapsto mr$ constructs a valid map of right modules.
By summing these maps over the generating set, we obtain a map from a large free module to $M$.  Every generator is in the image of this map since it is hit by its corresponding basis vector.  Since the image is a submodule of $M$ that contains a generating set, the map is a surjection, and so $M$ is a quotient of a free module.

Any $R$-module $M$ has a presentation by generators and relations.

Proof: Witness $M$ as a quotient of a free module, which we will call the module of generators.  The kernel of the quotient map is again an $R$-module (actually, it is a submodule of the module of generators), and so is itself a quotient of a free module (the module of relations).  There is a natural map from the module of relations to the module of generators given by the quotient map followed by the inclusion, and the cokernel of this map is $M$.  This shows that $M$ has a presentation by generators and relations.
Ok, and here are the corresponding facts and proofs for categories.  Set $\mathcal{D} = \mathcal{C}^{op}$.  By an "element" of $F \colon \mathcal{D} \to \mathrm{Set}$ we mean a pair $(d \in \mathcal{D}, x \in Fd)$.  Essentially, an "element of F" is an element of the disjoint union of the various $Fd$.  If $f \colon d \to d'$ is a morphism of $\mathcal{D}$, then we write $(d,x) \cdot f = (d', (Ff)(x))$.  This gives a right action of $\mathcal{D}$ on the elements of $F$.

If $\mathcal{D} = C^{op}$ is a category, and $F : \mathcal{D} \to \mathrm{Set}$ is a functor to sets, then $F$ is a quotient of a disjoint union of representable functors.

Proof: Choose enough elements of $F$ so that every other element may be obtained from these by use of the action of $\mathcal{D}$.  For example, one may choose every element of $F$ at this stage.  Call this collection of elements the "generators".  Then, for each generator $(d, x)$, build a map
$$ \mathcal{D}(d,-) \to F $$
specified by $(d, 1_d) \mapsto (d, x)$ and therefore, by naturality we would have at a general element $(d', f \colon d \to d') \mapsto (d,x) \cdot f$.  As this formula does give a natural transformation, the desired map exists.
By taking the coproduct of these maps over the generating set, we obtain a map from a large disjoint union of representables to $F$.  Every generator $(d,x)$ is in the image of this map since it is hit by the element $(d, 1_d)$ in the corresponding summand of the disjoint union.  Since the image is a subfunctor that contains a generating set, it is a surjection, and so $F$ is a quotient of a disjoint union of representables.

Any functor is a the coequalizer of a map from one disjoint union of representables to another.

Proof: By the lemma, write $F$ as a quotient of a disjoint union of representables $X$, so that $X$ the $\mathcal{D}$-set of generators.  The set of pairs of elements of $X$ that map to the same element of $F$ gives a subfunctor of $X \times X$.  Write this subfunctor as a quotient of a disjoint union of representables $Y$, the relations.  We obtain two natural transformations $Y \to X$ by the quotient map followed by each projection.  The coequalizer of these two maps identifies all pairs of elements of $X$ that map to the same thing in $F$, so the coequalizer is isomorphic to $F$.
Remarks
For a good example to work by hand, I recommend setting $R=\mathbb{Z}[x,y]$ to be a graded ring with $|x|=|y|=1$, and taking $M$ to be a monomial ideal.  The objects of $\mathcal{D}$ are the grades of $R$, and there can be some nice combinatorics in the construction of the relations.
In commutative algebra, it is often useful to extend a presentation to a free resolution.  The analog would be finding a simplicial resolution by coproducts of representables.<|endoftext|>
TITLE: "Philosophical" meaning of the Yoneda Lemma
QUESTION [131 upvotes]: The Yoneda Lemma is a simple result of category theory, and its proof is very straightforward.
Yet I feel like I do not truly understand what it is about; I have seen a few comments here mentioning how it has deeper implications into how to think about representable functors.
What are some examples of this? How should one think of the Yoneda Lemma?

REPLY [4 votes]: This perspective seems to be absent so far, even though this is a very old question.
To properly credit the source for this idea, the perspective seems to be taken for granted in MacLane and Moerdijk's Sheaves in Geometry and Logic, which was where I was introduced to it.
The observation is the following: Categories are generalizations of monoids, and functors are the correct analog of representations.
The analogy:
This goes as follows. If $M$ is a monoid, we can define a one object category, which I'll also denote by $M$ which has a single object $*$ and $M(*,*)=M$ with the composition coming from the multiplication in the monoid. 
Then if $M=G$ is a group for example, then a functor $F$ from $G$ to $\newcommand\Set{\mathbf{Set}}\Set$ is a choice of set $X=F(*)$ together with mappings $F(g):X\to X$ for all morphisms $g\in G$. If we write $g\cdot x$ for $F(g)(x)$, then the equation $F(g)\circ F(h) = F(gh)$ becomes $g\cdot (h\cdot x) = (gh)\cdot x$ for all $x\in X$, $g,h\in G$. This is exactly what it means for $X$ to be a $G$-set.
A natural transformation between two functors $F,F' : G\to \Set$ is a map $\phi : F(*)\to F'(*)$ such that 
$g\cdot \phi(x) = \phi(g\cdot x)$ for all $g\in G$ and $x\in X$. So morphisms of functors are $G$-equivariant maps, as they should be. 
Replacing the category of sets with any category you care to think about gives the expected notion of $G$-representation in that category.
Back to categories and functors
In this perspective, a covariant functor $\newcommand\C{\mathcal{C}}F:\C\to \Set$ is a choice of set $F(c)$ for all objects $c\in\C$ and a function $F(f) : F(c)\to F(d)$ for each morphism $f:c\to d$ in $\C$. Subject to the requirement that $F(fg) = F(f)F(g)$ when $f$ and $g$ are composable arrows. 
We can think about this as a family of sets $X_c$ for all $c\in \C$ such that for $f:c\to d$, and $x\in X_c$, $f_*x\in X_d$ and $g_*f_*x=(gf)_*x$ when $g$ and $f$ are composable.
A natural transformation between two representations $X_c$, $Y_c$ is a family of maps 
$\phi_c:X_c\to Y_c$ such that 
$\phi(f_*x)=f_*\phi(x)$ whenever this makes sense.
Now the Yoneda lemma becomes the following observation.
The functor $\C(a,-)$ is "freely generated" as a $\C$-representation by $1_a$.
What I mean by this is that a natural transformation $\C(a,-)\to F-$ is determined by the image of $1_a$ and there are no restrictions on the choice of image (except that of course it must lie in $Fa$).
This is because for any morphism $f$, we have 
$$\phi(f)=\phi(f_*1_a)=f_*\phi(1_a).$$
The contravariant version is identical, except now we think of functors as right $\C$-representations because contravariance becomes the rule $x|_f |_g = x|_{fg}$, where $|_f$ denotes the action of $f$ on $x$ when this makes sense.
This philosophical perspective is related to Sridhar Ramesh's answer, based on what I can see, but I'm not really familiar with the algebraic theory perspective, and I think this is a bit more of an elementary algebraic viewpoint. 
The point here is that you should think of Yoneda functors $\C(a,-)$ or $\C(-,a)$ as the free objects in a single variable supported at an object $a$.<|endoftext|>
TITLE: Are proper linear subspaces of Banach spaces always meager?
QUESTION [22 upvotes]: Let X be a Banach space, and let Y be a proper non-meager linear subspace of X. If Y is not dense in X, then it is easy to see that the closure of Y has empty interior, contradicting Y being non-meager. So Y must be dense. If Y has the Baire property, then it follows from Pettis Lemma that Y is open and hence closed (since the complement of Y is the union of translates of Y), contradicting Y being proper. Thus, Y must be dense and not have the Baire property.  
My question is: is there a Banach space X with a proper non-meager linear subspace Y? Such a Y must be dense and not have the Baire property. Any such Y must be difficult to construct since all Borel sets and even all continuous images of separable complete metric spaces have the Baire property.
More info:
1. Meager is just another word for first category, i.e. the countable union of nowhere dense sets.
2. A set A in a topological space has the Baire property if for some open set V (possibly empty) the set (A-V)U(V-A) is meager.
3. The collection of sets with the Baire property form a sigma-algebra. All open sets trivially have the Baire property, thus all Borel sets have the Baire property. All analytic sets also have the Baire property.
4. Pettis Lemma: Let G be a topological group and let A be a non-meager subset of G with the Baire property. Then the set A*A^{-1} (element-wise multiplication) contains an open neighborhood of the identity. This is an analog to a similar theorem about Lebesgue measure: If A is a Lebesgue measurable subset of the reals with positive Lebesgue measure, then A - A (element-wise subtraction) contains an open set around 0.

REPLY [23 votes]: I am afraid that Konstantin's accepted answer is seriously flawed.
In fact, what seems to be proved in his answer is that $\ker f$ is of second category, whenever $f$ is a discontinuous linear functional on a Banach space $X$. This assertion has been known as Wilansky-Klee conjecture
and has been disproved by Arias de Reyna under Martin's axiom (MA). 
He has proved that, under (MA), in any separable Banach space there exists a discontinuous linear functional $f$ such that $\ker f$ is of first category.
There have been some subsequent generalizations, see Kakol et al.
So, where is the gap in the above proof?
It is implicitly assumed that $\ker f = \bigcup A_i$. Then $f$ is bounded on $B_i=A_i+[-i,i]z$. But in reality, we have only $\ker f \subset \bigcup A_i$ and we cannot conclude that $f$ is bounded on $B_i$.
And finally, what is the answer to the OP's question?
It should not be surprising (remember the conjecture of Klee and Wilansky) that the answer is:
in every infinite dimensional Banach space $X$ there exists a discontinuous linear form $f$ such that $\ker f$ is of second category.
Indeed, let $(e_\gamma)_{\gamma \in \Gamma}$ be a normalized Hamel basis
 of $X$. 
Let us split $\Gamma$ into countably many pairwise disjoint sets $\Gamma =\bigcup_{n=1}^\infty \Gamma_n$, each of them infinite. We put $X_n=span\{e_\gamma: \gamma \in \bigcup_{i=1}^n \Gamma_i\}$. It is clear (from the definition of Hamel basis) that $X=\bigcup X_n$. Therefore there exists $n$ such that $X_n$ is of second category. Finally, we define $f(e_\gamma)=0$ for every $\gamma \in \bigcup_{i=1}^n\Gamma_i$ and $f(e_{\gamma_k})=k$ for some sequence $(\gamma_k) \subset \Gamma_{n+1}$. We extend $f$ to be a linear functional on $X$. It is clearly unbounded, $f\neq 0$, and $X_n \subset \ker f$. Hence $\ker f\neq X$ is dense in $X$ and of second category in $X$.<|endoftext|>
TITLE: Is the fixed locus of a group action always a scheme?
QUESTION [12 upvotes]: Suppose G is an algebraic group with an action G×X→X on a scheme. Does the fixed locus (the set of points x∈X fixed by all of G) have a scheme structure? You can obviously define the functor Fix(T)={t∈X(T)|t is fixed by every element of G(T)}. Is this functor always representable?
(This question was "broken off" of a compound question of mine after Scott Carnahan answered the other part so wonderfully that I had to accept his answer.)

REPLY [28 votes]: The question gives the "wrong" definition of Fix(T), hence the resulting confusion. 
A more natural definition of the subfunctor X^G of "G-fixed points in X" is
(X^G)(T) = {x in X(T) | G_T-action on X_T fixes x}
               = {x in X(T) | G(T')-action on X(T') fixes x for all T-schemes T'}.
(Of course, can just as well restriction to affine T and T' for "practical" purposes.) 
By way of analogy with more classical situations, if the base is a field k then a moment's reflection with the case of finite k shows that
{x in X(k) | G(k) fixes x}
is the "wrong" notion of (X^G)(k), whereas
{x in X(k) | G-action on X fixes x}
is a "better" notion, and is what the above definition of (X^G)(k) says. 
From this point of view, if (for simplicity of notation) the base scheme is an affine Spec(k) for a commutative ring k then the "scheme of G-fixed points" exists whenever G is affine and X is separated provided that k[G] is k-free (or becomes so after faithfully flat extension on k).  So this works when k is a field, or any k if G is a k-torus (or "of multiplicative type").  See Proposition A.8.10(1) in the book Pseudo-reductive groups.<|endoftext|>
TITLE: What are the Benefits of Using Algebraic Spaces over Schemes?
QUESTION [18 upvotes]: I have heard that algebraic spaces have better formal properties than schemes. What are these benefits? Also, is there a natural way to go straight from affine schemes to algebraic spaces bypassing the locally ringed space construction?

REPLY [4 votes]: In addition to being closed under taking quotients by etale equivalence relations, algebraic spaces are also closed under taking quotients by arbitrary finite group actions (need not be free, so the induced relation need not be etale). I only realized this recently, but I think the following argument is correct.
Suppose G is a finite group acting on an algebraic space X. Then the stack quotient [X/G] is a Deligne-Mumford stack (it has an etale cover by X). By the Keel-Mori theorem, DM stacks have coarse spaces. Let [X/G]→Y be the coarse space of [X/G]. Then Y is an algebraic space and the map [X/G]→Y is universal for maps from [X/G] to algebraic spaces. This means that Y is a categorical quotient of X by G. In fact, I think Y might actually be a geometric quotient (or at least a good quotient) of X by G, but I haven't unraveled all the definitions yet.<|endoftext|>
TITLE: Does any method of summing divergent series work on the harmonic series?
QUESTION [53 upvotes]: It's sort of folklore (as exemplified by this old post at The Everything Seminar) that none of the common techniques for summing divergent series work to give a meaningful value to the harmonic series, and it's also sort of folklore (although I can't remember where I heard this) that the harmonic series is more or less the only important series with this property.
What other methods besides analytic continuation and zeta regularization exist for summing divergent series?  Do they work on the harmonic series?  And are there other well-known series which also don't have obvious regularizations?

REPLY [6 votes]: As other answers have mentioned, the most "natural" value for the harmonic series seems to be the Euler-Mascheroni constant $\gamma$. The article Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru says

We believe that Euler constant is not just the "renormalized" value of
  the Riemann zeta function in 1. In a sense that we shall clarify it is
  in fact the normal and natural value of zeta of 1. In this paper we
  first propose a limit definition of a function whose values coincide
  everywhere with those of the Riemann zeta function, save in 1, where
  our limit definition yields the Euler constant. Since in the
  literature one can find more than one way to regularize the value of
  the zeta function at s=1, we give asymptotic expansions where, by dint
  of some extended analogies, Euler constant appears to be the true
  "renormalized" value. As a striking example of such analogies, we
  propose an expansion of the logarithm function based on Euler constant
  and on all values of the zeta function at odd positive integers, in
  which all these presumably irrational numbers are accompanied by
  Harmonic numbers of corresponding orders. The other aim of this paper
  is to show how sequences of rationals, often the same, arise in
  computations related to Dirichlet L-functions. Here, a connection with
  the Liouville lambda function appears to have been found. Thus we
  raise the question about the possible usefulness of an extension of
  the Liouville lambda function to rationals.

One example given in the article (on page 4, equation 11) is
$$\ln \Gamma(x) = -\ln x - \gamma x + \sum_{k \geq 2} \frac{\zeta(k)}{k}(-x)^k$$
where $\Gamma$ is the gamma function and $\zeta$ is the zeta function. If we let $\zeta(1) = \gamma$, this can be simplified to
$$\sum_{k \geq 1} \frac{\zeta(k)}{k} x^k = \ln (-x)!$$
which is, I think, supremely elegant. More generally, we seem to have
$$\sum_{k \geq 1} \zeta(k) k^n x^k = \sum_{k=0}^n \left\{{n+1 \atop k+1}\right\} (-x)^{k+1} \psi^{(k)}(1-x)$$
where $\psi^{(k)}$ is the polygamma function and the brackets indicate Stirling numbers of the second kind.<|endoftext|>
TITLE: What kind of geometric operations "scale up" cohomology?
QUESTION [13 upvotes]: There's an obvious operation on the category of graded rings, given by "scaling up," multiplying the grading of every element by some fixed constant.
Does anyone know of an operation on the level of spaces that will do this to the cohomology rings?

REPLY [16 votes]: How about we change the question to this: for what kind of spaces X are there a family of  continuous maps ψk: X->X which act on cohomology by "scaling up"?  Here is a famous and bewildering example of something sort of like this:
As you can read in the notes of Sullivan's 1970 MIT course (see esp. chapter 5), if you have an algebraic variety X defined over Q(the rationals), then you get an action of G(=absolute Galois group of Q) on a space Xet=the etale homotopy type of X.
If X is a Grassmanian variety, then
* Xet has the homotopy type of the usual complex Grassmannian (up to a "profinite completion"),
* G acts on the profinite cohomology of Xet through its abelianization Gab=Z*(=the units of the profinite integers),
* this action of Gab on cohomology is by scaling.
I don't know what the motivation of the original question is, so I don't know if this kind of thing has any relation for what you want.  (I really posted this to advertise the Sullivan notes, which everybody should read.)<|endoftext|>
TITLE: Controlling Ultrapowers
QUESTION [8 upvotes]: Say I start with some  a transitive model of a large fragment of ZFC (say enough to run Łoś' Theorem externally) and a specific set x∈M. Now let's say I'm going to pick some M-ultrafilter U on x. By M-ultrafilter, I mean that U measures the subsets of x which are in M.
My question is, by varying U, how much can I affect the ultrapower of M by U? Let's say I limit myself to a U which are countably complete, so that the ultrapower will be wellfounded.  If this question is too vague or broad, I'd welcome any interesting examples of things that are possible or impossible.

REPLY [4 votes]: As much as you wish. Lowenheim Skolem give you such situations and then you can affect it too much. For instance, you can construct situations where the critical point is singular in the ultrapower. You cannot do much if U is amenable to M. Then it is like a real ultrafilter. 
I don't know what you are asking actually.
An interesting question is whether you can have an ultrafilter on kappa such that the powerset of kappa^+ is in the ultrapower and James Cummings solved this by showing that you can. I don't know if it is interesting to look for ultrafilters that can code powerset(kappa^++) into the ultrapower. probably his proof already gives that.<|endoftext|>
TITLE: Finiteness of Obstruction to a Local-Global Principle
QUESTION [6 upvotes]: Say that a projective variety V over Q satisfies the local-global principle up to finite obstruction (#) if there are only finitely many isomorphism classes of projective varieties over Q that are not isomorphic to V over Q despite being isomorphic to V over every completion of Q..
In section 7 of Barry Mazur's 1993 article titled On the Passage From Local to Global in Number Theory, Mazur describes his attempt to prove that (#) for abelian varieties over Q implies (#) for all projective varieties over Q, and a partial result that he, Yevsey Nisnevich and Ofer Gabber achieved in this direction. Has there been further progress in this direction since 1993?
My understanding is that an effective version of (#) for genus 1 curves (an effective bound on certain Tate-Shafarevich groups) gives a finite algorithm (of a priori bounded running time) for determining whether a genus 1 curve has a rational point, and also that such an effective bound on Tate-Shafarevich groups is expected.
Is an effective version of (#) for general projective varieties over Q expected? If so, how does this relate to Hilbert's 10th problem over Q (which Bjorn Poonen has conjectured to be undecidable)?

REPLY [4 votes]: "Has there been further progress in this area since 1993?"
So far as I know, there has been no direct progress.  I feel semi-confident that I would know if there had been a big breakthrough: Mazur was my adviser, this is one of my favorite papers of his, and I still work in this field.  Also, I just checked MathReviews and none of the citations to this paper makes a big advance on the problem, although two are somewhat relevant:
MR1905389 Thăńg, Nguyêñ Quoc On isomorphism classes of Zariski dense subgroups of semisimple algebraic groups with isomorphic $p$-adic closures.  Proc. Japan Acad. Ser. A Math. Sci.  78  (2002),  no. 5, 60--62.
MR2376817 (2009f:14040) Borovoi, M.; Colliot-Thélène, J.-L.; Skorobogatov, A. N. The elementary obstruction and homogeneous spaces. Duke Math. J. 141 (2008), no. 2, 321--364. 
I'm not sure what you mean by an effective bound on Shafarevich-Tate groups (henceforth "Sha").  It is certainly expected that the Sha of any abelian variety over a global field is finite.  If this is true, then in any given case one can, "in principle", give an explicit upper bound on Sha by the method of n-descents for increasingly large n.  (In practice, even for elliptic curves reasonable algorithms have been implemented only for small values of n.)  I really can't imagine any algorithm having to do with Sha that has "a priori bounded running time".  What do you have in mind here?
As to the final question, let me start by saying that it seems reasonable at least that the set of "companion varieties" (i.e., Q-isomorphism classes of varieties everywhere locally isomorphic to the given variety) of a projective variety V/Q is finite: as above, we believe this for abelian varieties, and Barry Mazur proved in this paper a lot of results in the direction that the conjecture for abelian varieties implies it for arbitrary varieties.  (For instance, quoting from memory, I believe he proved the implication for all varieties of general type.)
Here is a key point: suppose you are given a variety V/Q and you are wondering whether it has rational points.  If V is itself a torsor under an abelian variety (e.g. a genus one curve), then if you can compute Sha of the Albanese abelian variety of V, you can use this to determine whether or not V has a Q-rational point.  In general, the connection between computation of sets of companion varieties of V and deciding whether V has a Q-rational point is less straightforward.  If V is a curve, then there are theorem in the direction of the fact that finiteness of Sha(Jac(V)) implies that the Brauer-Manin obstruction is the only one to the existence of rational points on V.  In particular, people who believe this (including Bjorn Poonen, I think), believe that there is an algorithm for deciding the existence of rational points on curves.  But nowadays we know examples of varieties where the Brauer-Manin obstruction is not sufficient to explain failure of rational points.
So, in summary, it is a perfectly tenable position to believe that companion sets are always finite, even effectively computable, but still there is no algorithm to decide the existence of Q-points on an arbitrary variety.<|endoftext|>
TITLE: Subgroups of a group generated by a free semigroup
QUESTION [8 upvotes]: Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, and suppose that $G$ (as a group) is generated by $F$. The most simple such situation would be when $G$ is a free group, but there are lot of groups, besides free ones, which could occur in this situation (for example, $G$ could be solvable). Is it possible that $G$ contains a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$ (the direct product of two copies of the infinite cyclic group $\mathbb{Z}$)?
Update: thanks to all the people for a very interesting discussion. Sorry that I cannot award a few "accepted answers", so the only one goes to Greg who supplied the most lucid and explicit example.

REPLY [3 votes]: It would be schizophrenic to try to slice this into comments on the various answers, so:
Brandon's condition (P) does not imply that no two elements of the semigroup lie in the same coset. Let F=F2; from Richard's and Henry's answers we have the example of G = [F,F], which has (P), while nevertheless we have ab=ba. The same issue showed up in Henry's first solution.
However, continuing the construction nevertheless we obtain the universal 2-step solvable group Γ=F/[[F,F],[F,F]]. Since we know from Greg's and Henry's answers that free semigroups exist in 2-generated 2-step solvable groups, it follows from universality that the semigroup in Γ generated by a and b is free. Since Γ contains [F,F]/[[F,F],[F,F]], an infinite-rank free abelian group, we see that Brandon's construction does work, at least in this case.
(Please direct your upvotes to the people who came up with these examples, not me.)<|endoftext|>
TITLE: Is no proof based on "tertium non datur" sufficient any more after Gödel?
QUESTION [11 upvotes]: There are many proofs based on a "tertium non datur"-approach (e.g. prove that there exist two irrational numbers a and b such that a^b is rational).
But according to Gödel's First Incompleteness Theorem, where he provides a constructive example of a contingent proposition, which is neither deductively (syntactically) true nor false, we know that there can be a tertium.
My question: Are all proofs that are based on that principle useless since now we know that a tertium can exist?

REPLY [3 votes]: I think that many of the responses here are overlooking the key part of your question, "Are all proofs that are based on that principle useless?".  The question of whether LEM is not merely formally consistent but actually useful, i.e., pertaining to reality, is a deep question that has been debated for at least a century (since Brouwer).  One partial answer to your question is that no, not all such proofs are useless, because often they involve LEM on propositions which are indeed decidable.  In constructive mathematics, $\varphi \vee \neg \varphi$ is precisely what it means for a proposition $\varphi$ to be decidable, and making use of this fact in a proof amounts to calling a decision procedure.<|endoftext|>
TITLE: When is a locally convex topological vector space normal or paracompact?
QUESTION [26 upvotes]: All locally convex topological vector spaces (LCTVS) are completely regular, since their topology is given by a family of semi-norms.  I'm interested in conditions that imply that a LCTVS is paracompact or normal.
Some background: I have a particular space in mind.  It is a directed colimit (union) over an uncountable family of nuclear Frechet spaces.  It is complete but not metrisable nor separable nor nuclear.  I have a very concrete description of it and can describe the bounded (and compact) subsets.  This space is fairly badly behaved in other ways so I'm half anticipating a negative result, thus answers along the lines of "If you can find a subset that looks like X then it can't be normal" could be just what I'm looking for.
So in particular, I'm interested in the general question.  To forestall a couple of "easy" answers: as my space is not Frechet, it is not metrisable so I can't directly use theorems on metrisability (however, as it is a colimit there may be some scope for indirect use).  And, of course, paracompact would imply normality since it is completely regular.
To forestall another possible comment, I'm not going to tell you what the particular space is.  Partly because I'm more interested in the general situation, this space is merely focussing my attention on the question, and partly because it's a more useful question if it's general.  A bit of intelligent searching would reveal what space it is anyway so it's no great hardship.
Edit: There is a simple example of a space that would be very interesting to know about: the sum (i.e. coproduct) of an uncountable number of copies of $\mathbb{R}$, or even more specifically $\sum_{\mathbb{R}} \mathbb{R}$.  This isn't the specific space that I'm interested in, but is close enough that I think that an answer either way for this space will tell me what to do for my space.

REPLY [8 votes]: Thanks to your other question, I was on a LCTVS kick.  I did find one general criterion that implies that a locally convex space is paracompact.  According to the Encyclopedia of Mathematics, if it is Montel (which means that it is barrelled and the Heine-Borel theorem holds true for it), then it is paracompact.  Although this criterion is important, it is of no use to your specific question.
I thought I had a proof for half of your question, which I wrote up as the first version of this answer, but I made a mistake and proved something different.  My thinking is based on the fact that the normality axiom for a topological space is equivalent to the Tietze extension theorem.  (Tietze extension follows from normality.  In the other direction, if $A$ and $B$ are the two closed sets, you obtain disjoint open neighborhoods from a continuous function that is 0 on $A$ and 1 on $B$.)  However, in my argument I conflated the locally convex direct sum of spaces with the topological direct sum.  For a countable direct sum of copies of $\mathbb{R}$, they are the same topology, and they agree with the box topology.  But Waelbroeck, LNM 230, § I.3 Note 4 points out that they are different in the uncountable case.
Let $\alpha$ be an ordinal, for instance an ordinal of cardinality $2^{\aleph_0}$.  Then $\mathbb{R}^\alpha$ in the topological direct sum topology satisfies Tietze extension.  Let $A \subset \mathbb{R}^\alpha$ be a closed set and let $f:A \to \mathbb{R}$ be a continuous function.  For $\beta < \alpha$, let $A_\beta$ be the intersection of $A$ and with $\mathbb{R}^\beta$.  Suppose that $\alpha = \beta+1$ is a successor ordinal.  If $\alpha$ is finite, then the conclusion is standard.  Otherwise, by induction, there is an extension $f_\beta$ of $f$ to $\mathbb{R}^\beta$.  Moreover, by induction in a different sense, we have already proved that $\mathbb{R}^{\beta+1}$ is normal, since $\beta$ and $\beta+1$ have the same cardinality.  So there exists an extension $f_\alpha$ to $\mathbb{R}^\alpha$.  If instead $\alpha$ is a limit ordinal, then the extensions all the way up to $\alpha$ work just because they work; that's the behavior of topological direct limits.
Having failed to normality for the locally convex direct sum, I can't say much about paracompactness either.  :-)  However, there is an interesting result called the Michael selection theorem which seems to do for paracompactness what the Tietze theorem does for normality.  If the Tietze theorem is useful for your spaces, then maybe the Michael selection theorem is too.<|endoftext|>
TITLE: internal homs and adjunctions?
QUESTION [7 upvotes]: This is probably an easy question. Let C be a category with (finite) products.
 An internal hom in C category is an object uhom(X, Z) which represents the functor:
Y   |----->  hom(Y x X, Z)
here "uhom" is for "underlined hom" as that is how it is commonly denoted. Many example of categories with internal homs satisfy an a priori stronger for of adjunction:
uhom(Y x X. Z) = uhom(Y, uhom(X,Z))
Is this automatic for categories with internal homs? Is there an easily understood counter example?
(*) This might not be the most general/best definition of internal hom, but it is valid for many examples.

REPLY [2 votes]: As Dimitri ans bhargav said, this is automatic for categories with internal homs. And you don't need  x  to be the product: you have the same result for any closed monoidal category (in your case it is cartesian closed). You can find it in "Basic concepts of enriched category theory", by G.M. Kelly, page 14, which you can get for free here: http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html .<|endoftext|>
TITLE: Is there a stable algorithm for polynomial division (in several variables)?
QUESTION [8 upvotes]: Suppose you have a homogeneous ideal $I$ inside the algebra $\mathbb{C}[x_1,...,x_d]$  of complex polynomials in $d$-variables. Can one find a basis for $I$, say $\{f_1,...,f_k\}$, such that every $h \in I$ can be written as 
$$h = a_1 f_1 + ... + a_k f_k $$
where the coefficients appearing in each summand $a_i f_i$ are not much bigger then the coefficients appearing in $h$?
More specifically, given that  $\{f_1,...,f_k\}$  is a Groebner basis for $I$, can one modify the standard division algorithm so that one gets 
$h = a_1 f_1 + ... + a_k f_k$
with controlled terms?
Added 13.11.09 - By controlled I mean that the coefficients of the terms $a_i f_i$ are bounded in a non-exponential manner by the coefficients of $h$. There is no problem with degree of the $a_i$'s.
I will share that I found this possible in some special cases, for example when $d=2$ or when $I$ is generated by monomials, and I am now interested in the general question.
Note: My question begins after a basis has been found, I am not concerned here with the terrible complexity of actually computing a Groebner basis.
Another note (added 12.11.09): The answers and links that I am getting suggest that this problem has not been considered before. So I re-eamphasize my note from above: assume that a Groebner basis, even a universal Groebner basis, has already been found for the ideal. What can be said about the stability of certain variants of the division algorithm now?

REPLY [3 votes]: Revised June 25, 2010: I feel I need to close this circle. Thanks for the answers given but they have all been quite off the mark. My partial results on this problem (there is much room for improvement) as well as an application to operator theory can be found in this link:
http://arxiv.org/abs/1003.0502<|endoftext|>
TITLE: Intuition for the last step in Serre's proof of the three-squares theorem
QUESTION [47 upvotes]: Serre's A Course in Arithmetic gives essentially the following proof of the three-squares theorem, which says that an integer $a$ is the sum of three squares if and only if it is not of the form $4^m (8n + 7)$ :  first one shows that the condition is necessary, which is straightforward.  To show it is sufficient, a lemma of Davenport and Cassels, using Hasse-Minkowski, shows that $a$ is the sum of three rational squares.  Then something magical happens:
Let $C$ denote the circle $x^2 + y^2 + z^2 = a$.  We are given a rational point $p$ on this circle.  Round the coordinates of $p$ to the closest integer point $q$, then draw the line through $p$ and $q$, which intersects $C$ at a rational point $p'$.  Round the coordinates of $p'$ to the closest integer point $q'$, and repeat this process.  A straightforward calculation shows that the least common multiple of the denominators of the points $p'$, $p''$, ... are strictly decreasing, so this process terminates at an integer point on $C$.
Bjorn Poonen, after presenting this proof in class, remarked that he had no intuition for why this should work.  Does anyone have a reply? 
Edit:  Let me suggest a possible reformulation of the question as follows.  Complete the analogy:  Hensel's lemma is to Newton's method as this technique is to _____________________.

REPLY [5 votes]: The Lemma in the aswer by Bjorn Poonen can be sharpened to give an exact relationship
between $\mathrm{den}(x)$ and $\mathrm{den}(x')$.

Lemma 1.
  $~$Let $f=f_2+f_1+f_0\in\mathbb{Z}[X]=\mathbb{Z}[X_1,\ldots,X_n]$ ($n\geq 1$),
  with each $f_i$
  homogeneous of degree $i$.
  Let $y,v\in\mathbb{Z}^n$, where $v$ is primitive and $f_2(v)\neq0$,
  and define $F=AT^2+BT+C:=f(y+Tv)\in\mathbb{Z}[T]$.
  Suppose that the two zeros $t$ and $t'$ of $F$ are rational,
  so that the two rational points $x=y+tv$ and $x'=y+t'v$ are zeros of $f$.
  If $x=a/b$ and $x'=a'/b'$ are the reduced representations (with $b,b'>0$),
  and $x\neq y$ (which is certainly true if $x\notin\mathbb{Z}^n$),
  then
  \begin{equation*}
b' \,=\, \mathrm{sgn}(A)\cdot
  \frac{f_2(x-y)}
   {\mathrm{gcd}(A,B,C)\,\mathrm{gcd}(a-by)^2}\cdot b~; \tag{1}
\end{equation*}
  exchanging $x$ and $x'$ gives the analogous identity (provided $y\neq x'$).

Remark.
$~$For any $u=(u_1,\ldots,u_n)\in\mathbb{Z}^n$ we write $\mathrm{gcd}(u):=\mathrm{gcd}(u_1,\ldots,u_n)$.
Proof. $~$Since $tv=x-y=(a-by)/b=(c/b)v$,
where $c=\pm\,\mathrm{gcd}(a-by)$ and $\mathrm{gcd}(c,b)=1$,
we have $t=c/b$, and similarly $t'=c'/b'$ with $c'=\pm\,\mathrm{gcd}(a'-b'y)$
and $\mathrm{gcd}(c',b')=1$. The leading coefficient of $F$ is $A=f_2(v)=(b/c)^2f_2(x-y)$
($c\neq 0$ because $a-by=b(x-y)\neq 0$).
By Gauss lemma $F=d(bT-c)(b'T-c')$ with $d=\mathrm{sgn}(A)\cdot\mathrm{gcd}(A,B,C)$.
Then $dbb'=A=(b/c)^2f_2(x-y)$ gives us $b'$ expressed
as in the lemma.$~$
Done.
The Lemma in Bjorn Poonen's post is an immediate consequence.
Lemma 1 is about the geometric background of Davenport-Cassels lemma:
it relates the reduced representations of two rational zeros of $f$
that lie on an integral line $L=y+\mathbb{Q}v$
    whose direction vector $v$ is an anisotropic vector of the quadratic form $f_2$.
(An integral line is an affine line in $\mathbb{Q}^n$
    that contains an integral point and hence infinitely many integral points.)
It is not required that one or the other of the two zeros is non-integral,
the lemma says something interesting even when both zeros are integral.
Also, the two zeros may coincide, in which case the line $L$
    is tangent to the quadric $\{f=0\}$.
If we actually write out the other identity mentioned in the lemma
and then compare the two identities,
we obtain the identity (supposing $y\neq x,x'$)
\begin{equation*}
f_2(x-y)f_2(x'-y)
 \,=\, \bigl(\mathrm{gcd}(A,B,C)\,\mathrm{gcd}(a-by)\,\mathrm{gcd}(a'-b'y)\bigr)^2~.\tag{2}
\end{equation*}
It is not in the least surprising that $f_2(x-y)f_2(x'-y)$ is a square:
if $q$ is any quadratic form on a vector space $V$ over some field $K$,
and $u\in V$ and $\lambda,\mu\in K$,
then it is trivial that $q(\lambda u)q(\mu u)=\bigl(\lambda\mu q(u)\bigr)^2$.
It may seem slightly surprising that $f_2(x-y)f_2(x'-y)$ is a square of an integer,
but this is
a consequence of $A=dbb'$: in the situation of the lemma
the trivial identity satisfied by a general quadratic form reads
\begin{equation*}
f_2(x-y)f_2(x'-y) \,=\, \Bigl(\frac{c}{b}\,\frac{c'}{b'}f_2(v)\Bigr)^2~,
\end{equation*}
and substituting $f_2(v)=A=dbb'$ yields
\begin{equation*}
f_2(x-y)f_2(x'-y) \,=\, (dcc')^2 \,=\, C^2 \,=\, f(y)^2
\end{equation*}
(this time without the restriction $y\neq x,x'$).
Let $L_{\mathbb{Z}}$ denote the set of all integral points on the line $L$:
$L_{\mathbb{Z}}:=L\cap\mathbb{Z}^n=y+\mathbb{Z}v$.
For any $z\in L_{\mathbb{Z}}$
the identity (2) still holds when $y$ is replaced by $z$,
but we must be careful and write it as
\begin{equation*}
f_2(x-z)f_2(x'-z)
 \,=\, \bigl(\mathrm{gcd}(A,B_z,C_z)\,\mathrm{gcd}(a-bz)\,\mathrm{gcd}(a'-b'z)\bigr)^2~,
\end{equation*}
because the coefficients $B_z$ and $C_z$ of $F_z=f(z+Tv)$ depend on $z$.
However,
note that the coefficient $A_z=A=f_2(v)$,
as well as the greatest common divisor of the coefficients of $F_z$
(the content of $F_z$),
$\mathrm{gcd}(A,B_z,C_z)=\left|A\right|/bb'=\left|d\right|$, do not depend on $z$.
For $z\in L_{\mathbb{Z}}$ we define $c(z),c'(z)\in\mathbb{Z}$
by $x-z=\bigl(c(z)/b\bigr)v$ and $x'-z=\bigl(c'(z)/b'\bigr)v$.
Since $\mathrm{gcd}(a-bz)=\left|c(z)\right|$ and $\mathrm{gcd}(a'-b'z)=\left|c'(z)\right|$,
we have
\begin{equation*}
f_2(x-z)f_2(x'-z) \,=\, \bigl(d\,c(z)\,c'(z)\bigr)^2 \,=\, C_z^2 \,=\, f(z)^2~,
     \qquad\quad z\in L_{\mathbb{Z}}\,.
\end{equation*}
Remark. $~$Idiot me!
This is just a very special case of the general power-of-a-point theorem,
which does not rely on specific factorization properties of integers
and is almost trivial to prove:

Let $K$ be a field,
  let $f\in K[X] = K[X_1,\ldots,X_n]$ be of degree $m$ (where $m, n\geq 1$),
  and denote by $f_m$ the homogeneous component of $f$ of degree $m$.
  Let $L$ be an affine line in $K^n$ with a direction vector $v$, where $f_m(v)\neq 0$.
  Let $y\in L$, and define $F_y := f(y+Tv)\in K[T]$, a polynomial of degree $m$.
  Suppose that $F_y$ has $m$ zeros (counting multiplicities) $t_1$, $\ldots$, $t_m$ in $K$.
  Then the points $x_i=y+t_iv\in L$, $1\leq i\leq m$, are zeros of $f$,
  the multiset of the $x_i$'s does not depend on the choice of $y\in L$,
  and
  \begin{equation*}
f_m(y-x_1)f_m(y-x_2)\cdots f_m(y-x_m) \,=\, f(y)^m~.
\end{equation*}

The independence is easy: if $z=y+sv$,
then $F_z$ has the zeros $t_i-s$, whence $z+(t_i-s)v = y+t_iv = x_i$.
The leading coefficient of $F_y$ is $f_m(v)$ and its constant term is $f(y)$.
From $F_y=f_m(v)(T-t_1)\cdots(T-t_m)$ we get $f(y)=(-1)^m t_1\cdots t_m f_m(v)$,
whence $f_m(y-x_1)\cdots f_m(y-x_m) = \bigl((-t_1)\cdots(-t_m)f_m(v)\bigr)^m = f(y)^m$.
We digress.
Let's return to the situation in Lemma 1.
We regard the point $y$ as fixed, serving as an origin of $L_{\mathbb{Z}}$.
Let us determine $c(z)$ for a general point $z=y+kv\in L_{\mathbb{Z}}$, $k\in\mathbb{Z}$:
from
\begin{equation*}
\frac{c(y+kv)}{b}\,v \,=\, x-(y+kv) \,=\, (x-y)-kv \,=\, \frac{c(y)-kb}{b}\,v
\end{equation*}
we see that
\begin{equation*}
c(y+kv) \,=\, c(y) - kb~. \tag{3}
\end{equation*}
For any $z\in L_{\mathbb{Z}}$ we have
\begin{equation*}
f_2(x-z) \,=\, \frac{c(z)^2}{b^2}f_2(v)
 \,=\, \frac{c(z)^2}{b^2}\,dbb'
 \,=\, \frac{db'}{b}\,c(z)^2
 \,=\, \frac{e_0}{b_0}\,c(z)^2~, \tag{4}
\end{equation*}
where $b_0=b/\mathrm{gcd}(b,db')$ and $e_0=db'/\mathrm{gcd}(b,db')$.
Since $\mathrm{gcd}\bigl(b,c(z)\bigr) = 1$, and hence $\mathrm{gcd}\bigl(b_0,c(z)\bigr) = 1$, it follows that
\begin{equation*}
\mathrm{den}\bigl(f_2(x-z)\bigr) \,=\, b_0 \qquad\quad \text{for every $z\in L_{\mathbb{Z}}$}\,.
\end{equation*}
Combining (3) and (4) we obtain
\begin{equation*}
f_2\bigl(x-(y+k)v\bigr) \,=\, \frac{e_0}{b_0}\bigl(c(y)-kb\bigr)^2~, \qquad\quad k\in\mathbb{Z}\,.
\end{equation*}
In the special case $x=x'$, when the line $L$ is a tangent of the quadric $\{f=0\}$,
we have $b=b'$, whence
\begin{equation*}
f_2(x-z) \,=\, d\,c(z)^2~, \qquad\quad z\in L_{\mathbb{Z}}\,,
\end{equation*}
thus $f_2(x-z)$ is an integer for every integral point $z$ in $L$.
The discussion above has demonstrated that the identity (1) and its brethren have uses
unrelated to Davenport-Cassels lemma.
Now we return to applications of (1) to Davenport-Cassels lemma and (a little way) beyond it.
The assumptions $x\in\mathbb{Q}^n\setminus\mathbb{Z}^n$ and $0<\left|f_2(x-y)\right|<1$
imply the premises $f_2(v)\neq 0$ and $x\neq y$ of Lemma 1
    and then yield the instantaneous result $b'<b$.
But besides the $f_2(x-y)$ in the numerator on the right hand side of (1)
there are also the factors $\mathrm{gcd}(A,B,C)$ and $\mathrm{gcd}(a-by)^2$ in the denominator
whose product can be greater than $1$ and can help make $b'$ smaller than $b$
    even when $\left|f_2(x-y)\right|\geq 1$.
This leads to the idea of walking with Aubry 'on the far side':
when there is no integral point $y$ satisfying $\left|f_2(x-y)\right|\geq 1$
we may still find an integral point $y$
so that we can make a step along the line $L$
    from a zero $x=a/b$ of $f$ to a zero $x'=a'/b'$ of $f$ with $b'<b$.
The folowing two examples attest that this idea actually works.
But first, a definition.
A quadratic form $q$ on $\mathbb{Q}^n$ is said to be Euclidean
if for every $x\in\mathbb{Q}^n\setminus\mathbb{Z}^n$
there exists $y\in\mathbb{Z}^n$ such that $0 < \left|q(x-y)\right| < 1$.
For the first example let $q(X_1,X_2,X_3)=X_1^2+X_2^2+5X_3^2$
and $f(X)=q(X) - m$, where $m$ is a positive integer.
The positive-definite quadratic form $q$ is not Euclidean on $\mathbb{Q}^3$,
since for any $x\in\mathbb{Z}^3+\bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigr)$
and any $y\in\mathbb{Z}^3$ we have $q(x-y)\geq 7/4$.
Consider the set $F$ of all points $x$ in the cube
$\bigl\{(x_1,x_2,x_3)\in\mathbb{R}^3 \bigm| 0\leq x_1,x_2,x_3\leq \frac{1}{2}\bigr\}$
at which $q(x)\geq 1$;
the set $F$ ('the far side') is shown as the darker shaded part of the cube
    in the following figure:

The function $q(1-x_1,1-x_2,1-x_3)$ of a point $(x_1,x_2,x_3)$ in the set $F$
attains its largest value $8-2\sqrt{5}$ at the point $P=\bigl(0,0,1/\sqrt{5}\bigr)$.
$\quad$Suppose that $x\in\mathbb{Q}^3\setminus\mathbb{Z}^3$ is a zero of $f$,
and let $y:=\mathrm{round}(x)$.
If $q(x)<1$ (we certainly have $q(x)>0$) then fine,
we make a step to the next zero of $f$ with smaller denominator
this side of the Euclidean horizon.
Otherwise $q(x-y)\geq 1$, we are on the far side, and must tread more carefully.
Let $x=a/b=(a_1,a_2,a_3)/b$ be the reduced representation.
We claim that since $q(x)=m$ is an integer, the denominator $b$ is odd.
Suppose that $b$ is even;
then at least one of $a_1$, $a_2$, $a_3$ is odd,
and so in $q(x)=q(a)/b^2$ the numerator $q(a)$ is congruent to $1$, $2$, or $3$ modulo $4$,
while the denominator is divisible by $4$, contradiction.
Note that $b$ being odd implies $0 \leq \left|x_i-y_i\right| < \frac{1}{2}$, $i=1,2,3$.
Now we choose an integral point $z$, close to the integral point $y$.
If $a_1-by_1$ is even, then we set $z_1:=y_1$.
If $a_1-by_1$ is odd, we let $z_1$ be a second closest integer to $x_1$
(there are two possible choices for $z_1$ iff $x_1$ is an integer,
    and we may choose either of them);
then $z_1=y_1\pm1$, $a_1-bz_1$ is even, and $\left|x_1-z_1\right|=1-\left|x_1-y_1\right|$.
In either case we have $\left|x_1-z_1\right|\leq1 - \left|x_1-y_1\right|$.
The coordinates $z_2$ and $z_3$ are chosen analogously.
We step to the next zero $x'=a'/b'$ of $f$ along the line laid through the points $x$ and $z$.
Since $\bigl(\left|x_1-y_1\right|,\left|x_2-y_2\right|,\left|x_3-y_3\right|\bigr)\in F$,
it follows that
$$q(x-z)\leq q\bigl(1-\left|x_1-y_1\right|,1-\left|x_2-y_2\right|,1-\left|x_3-y_3\right|\bigr)
 \leq 8-2\sqrt{5} < 4~.$$
By the choice of the point $z$ all three coordinates of $a-bz$ are even,
thus $\mathrm{gcd}(a-bz)\geq 2$,
and Lemma 1 tells us that $b'<b$.
$\quad$Done.
For the second example we consider $q(X_1,X_2,X_3)=X_1^2+X_2^2+2X_3^2$
and $f(X)=q(X)-m$ with $m$ a positive integer.
The quadratic form $q$ is barely non-Euclidean:
for every $x\in\mathbb{Q}^3\setminus\mathbb{Z}^3$ there is $y=\mathrm{round}(x)\in\mathbb{Z}^3$
such that $q(x-y)\leq 1$.
The problem is that there exist points $x$ for which $q(x-y)=1$ is the best we can do:
if $x\in M := \mathbb{Z}^3+\bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigr)$,
then $q(x-y)\geq 1$ for any $y\in\mathbb{Z}^3$.
Note that $q(x)$ is an odd integer for every $x\in M$,
thus there do exist integers $m$, all of them odd,
so that $f$ has rational zeros,
but woe, it also has a rational zero,
with all three coordinates precisely halfway between consecutive integers,
at which we get stuck, because there is no Euclidean step from it to another zero.
On the other hand, if $m$ is even we never get stuck,
there is always a Euclidean step from a non-integral zero;
that is, though $f$ is not Euclidean, it is 'conditionally' Euclidean
on the set of its zeros.
$\quad$Now suppose that $m$ is odd and that $f$ has rational zeros,
and that we walked ourselves into a point $x=a/2\in M$.
In this case the trick we have used in the preceding example does not work,
because $\mathrm{gcd}(a-2y)^2\leq f_2(x-y)$ for any $y\in\mathbb{Z}^3$;
we must seek help from the other factor $\mathrm{gcd}(A,B,C)$
in the denominator on the right hand side of (1).
Note that we can round the point $x$ to any of the eight integral points
$x+\frac{1}{2}(\delta_1,\delta_2,\delta_3)$, where $\delta_1,\delta_2,\delta_3\in\{-1,1\}$;
let $y$ be one of these eight points.
We have $v=(\delta_1,\delta_2,\delta_3)$,
and $F(T) = f(y+Tv) = q(v)T^2 + 2\langle y,v\rangle T + q(y) - m$,
where $\langle\text{-},\text{-}\rangle$
is the bilinear form associated with the quadratic form $q$,
$\langle X,Y\rangle = X_1Y_1+X_2Y_2+2X_3Y_3$.
The leading coeffient $A=q(v)=4$ is even,
the next coefficient $B=2\langle y,v\rangle$ is also even,
thus it remains to make $C=q(y)-m$ even.
But this is easy: choose $\delta_1$ and $\delta_2$ so
that one of $y_1$, $y_2$ is even and the other one is odd.
Such a choice makes $\mathrm{gcd}(A,B,C)\geq 2$,
and since $q(x-y)=1$, we can step to a zero $x'=a'/b'$ of $f$ with $b'<2$,
that is, to an integral zero (whence $\mathrm{gcd}(A,B,C)=2$).
$\quad$Done.<|endoftext|>
TITLE: Which rings are subrings of matrix rings?
QUESTION [20 upvotes]: In this question, all rings are commutative with a $1$, unless we explicitly say
so, and all morphisms of rings send $1$ to $1$.
Let $A$ be a Noetherian local integral domain. Let $T$ be a non-zero $A$-algebra
which, as an A-module, is finitely-generated and torsion-free.
Can one realise $T$ as a subring of the (not necessarily commutative)
ring $End_A(A^n)$ for some $n \ge 1$?

REPLY [6 votes]: Hello, I just want to add a few minor comments here, since this is a topic very close to my heart:
1) Finite MCM modules are not known to exist in dimension 3. What Hochster proved for equicharacteristic case and also in general dimension 3 (based on Ray Heitmann result) is that non-finitely generated MCM modules exist.  
2) If R is a N-graded domain over a perfect field of char p > 0 and R is locally Cohen-Macaulay on the punctured spectrum, then R admits a finite MCM. A proof can be found in:
http://www.math.utah.edu/vigre/minicourses/algebra/hochster.pdf 
3) A possible candidate for a counter-example is the local ring at the origin of the cone over some abelian surfaces. 
4) Many module-theoretic consequence of existence of finite MCM can be deduced from existence of non-finitely generated MCM and other approaches to the homological conjectures, so it may be helpful to what you want to do.
Cheers,<|endoftext|>
TITLE: How hard is it to compute the Euler totient function?
QUESTION [23 upvotes]: Are there any efficient algorithms for computing the Euler totient function? (It's easy if you can factor, but factoring is hard.)
Is it the case that computing this is as hard as factoring?
EDIT: Since the question was completely answered below, I'm going to add a related question. How hard is it to compute the number of prime factors of a given integer? This can't be as hard as factoring, since you already know this value for semi-primes, and this information doesn't seem to help at all. Also, determining whether the number of prime factors is 1 or greater than 1 can be done efficiently using Primality Testing.

REPLY [36 votes]: For semiprimes, computing the Euler totient function is equivalent to factoring.  Indeed, if n = pq for distinct primes p and q, then φ(n) = (p-1)(q-1) = pq - (p+q) + 1 = (n+1) - (p+q).  Therefore, if you can compute φ(n), then you can compute p+q.  However, it's then easy to solve for p and q because you know their sum and product (it's just a quadratic equation).
If you believe factoring is hard for semi-primes, then so is computing the Euler totient function.
Update!  Factoring and computing the Euler totient function are known to be equivalent for arbitrary numbers, not just semiprimes.  One reference is "Riemann's hypothesis and tests for primality" by Gary L. Miller.  There, the equivalence is deterministic, but assumes a version of the Riemann hypothesis.  See also section 10.4 of "A computational introduction to number theory and algebra" by Victor Shoup for a proof of probabilistic equivalence.<|endoftext|>
TITLE: What's a reasonable category that is not locally small?
QUESTION [37 upvotes]: Recall that a category C is small if the class of its morphisms is a set; otherwise, it is large.  One of many examples of a large category is Set, for Russell's paradox reasons.  A category C is locally small if the class of morphisms between any two of its objects is a set.  Of course, a small category is necessarily locally small.  The converse is not true, as Set is a counterexample.
Now, I can construct categories that are not locally small.  However, what's the most common or most reasonable such category?

REPLY [9 votes]: The category of Grothendieck topoi and (equivalence classes of) geometric morphisms. For example, if $A$ is the classifying topos for abelian groups, the geometric morphisms from $Set$ to $A$ are in bijection with isomorphism classes of abelian groups, which is certainly not a set.<|endoftext|>
TITLE: Hopf algebra structure on the universal enveloping algebra of a Leibniz algebra?
QUESTION [14 upvotes]: A Leibniz algebra L may be thought of as a noncommutative generalisation of a Lie algebra.  One drops the requirement that the bracket be alternating and substitutes the Jacobi identity for the Leibniz identity
$$ [x,[y,z]] = [[x,y],z] + [y,[x,z]] $$   for all $x,y,z \in L$
Remark. What is being defined above is a left Leibniz algebra.  There is also the notion of a right Leibniz algebra where the Leibniz identity now says that it is the right multiplication $[-,x]$ which is a derivation, instead of the left multiplication $[x,-]$ as in the equation above.
Since the bracket is no longer alternating, left- and right-multiplications are no longer related simply by a sign as in the case of Lie algebras, and this means that representations in general admit two actions of $L$: one on the left and one on the right, satisfying some identities which are explained, for example, in a paper of Loday (who seems to have introduced the concept) and Pirashvili (Math. Ann. 296 (1993) pp. 139–158).  In that paper they also define the universal enveloping algebra $U(L)$ of a Leibniz algebra $L$ and show that the there is a categorical equivalence between representations of $L$ and left modules over $U(L)$.  (Right modules of $U(L)$ correspond to the notion of corepresentation.)
Also notice that in that paper they work with right Leibniz algebras, so everything there is the mirror image to what I'm saying here.  One difference with the case of a Lie algebra is that $U(L)$ is a quotient of the tensor algebra of $L\oplus L$, to take into account the two actions of $L$ on a representation.
My question is whether there is a Hopf algebra structure on $U(L)$.
My interest in this question is that in some recent work on the deformation theory of n-Leibniz algebras, I studied cohomology with values in a representation $M $of a Leibniz algebra L and also with values on $End(M)$.  The action of $L$ on $End(M)$ follows from the formalism and one can check that it is indeed a representation, but it does not follow in any obvious way from the action of $L$ on $M$.  In Lie theory, we are used to the fact that if $M$ is a (finite-dimensional) representation of a Lie algebra $G$, then we have an isomorphism $End(M) \cong M \otimes M^*$ as representations of $G$, where to determine the action of $G$ on $M \otimes M^*$ we use the Hopf algebra structure on $U(G)$.  Hence my question.
EDIT:  I am adding more details about $U(L)$, as requested in the comment below by Theo Johnson-Freyd.
To motivate it, let us first define a representation $M$ of a (left) Leibniz algebra $L$ to be a vector space admitting two actions of $L$:
$$
(x,m) \mapsto [x,m]  \textrm{  and  }  (m,x) \mapsto [m,x], \forall m \in M \textrm{ and } x \in L
$$
satisfying three identities, which are obtained from the Leibniz identity above by replacing $x,y,z$ in turn by $m$; that is,
$$
[m,[x,y]] = [[m,x],y] + [x,[m,y]] \\
[x,[m,y]] = [[x,m],y] + [m,[x,y]] \\
[x,[y,m]] = [[x,y],m] + [y,[x,m]]
$$
To define $U(L)$ we start with the tensor algebra $T(L\oplus L)$ of $L \oplus L$.  Let $l_x = (x,0)$ and $r_x = (0,x) \in L \oplus L$.  Then $U(L)$ is the quotient of $T(L+L)$ by the two-sided ideal generated by the following elements (which can be read off from the conditions defining a representation):
$$
r_{[x,y]} - r_y r_x - l_x r_y \\
l_x r_y - r_y l_x - r_{[x,y]} \\
l_x l_y - l_y l_x - l_{[x,y]}
$$
for all $x, y \in L$, and where I have omitted the $\otimes$'s.
Notice that adding the first two, we can substitute one of them by the simpler
$$
  r_y (l_x + r_x) = 0 
$$
I don't know what the coalgebra structure is, though.  That's part of the original question.

REPLY [2 votes]: Let $L$ be the abelian Leibniz algebra of dimension $n$. Then $U(L)$ is the polinomial algebra $k[X_1, \cdots, X_n, Y_1, \cdots, Y_n]$ subject to the relations: 
$$
X_i X_j = X_j X_i, \quad X_j Y_i = Y_i X_j = - Y_j Y_i
$$
for all $i$, $j$. Then $U(L)$ is not a Hopf algebra (whith the usual structures, i.e. $X_i$ and $Y_i$ are primitives) since the above two-sided ideal is not a coideal.<|endoftext|>
TITLE: Are there good programs to create mathematical pictures in svg format?
QUESTION [13 upvotes]: I've recently started my personal wiki to organize my notes and thoughts. I use the wiki program instiki which I believe is the same as the n-lab uses. Instiki can upload svg's. I want to be able to create nice looking pictures of some (not to complicated) geometric objects, e.g. knots, pair of pants, etc. My question is twofold.

Do people draw these things using svg format? For example people who do diagram algebra, do you use svg format?
If so, what are some good free/open source programs for creating these pictures?

REPLY [2 votes]: i wrote small script to paint math functions in SVG, please feel free to use it
http://webdev.ts9.ru/#/step-4<|endoftext|>
TITLE: Does the super Temperley-Lieb algebra have a Z-form?
QUESTION [17 upvotes]: Background  Let V denote the standard (2-dimensional) module for the Lie algebra sl2(C), or equivalently for the universal envelope U = U(sl2(C)).  The Temperley-Lieb algebra TLd is the algebra of intertwiners of the d-fold tensor power of V.

TLd = EndU(V⊗…⊗V)

Now, let the symmetric group, and hence its group algebra CSd, act on the right of V⊗…⊗V by permuting tensor factors.  According to Schur-Weyl duality, V⊗…⊗V is a (U,CSd)-bimodule, with the image of each algebra inside EndC(V⊗…⊗V) being the centralizer of the other.
In other words, TLd is a quotient of CSd.  The kernel is easy to describe.  First decompose the group algebra into its Wedderburn components, one matrix algebra for each irrep of Sd.  These are in bijection with partitions of d, which we should picture as Young diagrams.  The representation is faithful on any component indexed by a diagram with at most 2 rows and it annihilates all other components.
So far, I have deliberately avoided the description of the Temperley-Lieb algebra as a diagram algebra in the sense that Kauffman describes it.  Here's the rub:  by changing variables in Sd to ui = si + 1, where si = (i i+1), the structure coefficients in TLd are all integers so that one can define a ℤ-form TLd(ℤ) by these formulas.

TLd = C ⊗ TLd(ℤ)

As product of matrix algebras (as in the Wedderburn decomposition), TLd has a ℤ-form, as well:  namely, matrices of the same dimensions over ℤ.  These two rings are very different, the latter being rather trivial from the point of view of knot theory.  They only become isomorphic after a base change to C.

There is a super-analog of this whole story.  Let U = U(gl1|1(C)), let V be the standard (1|1)-dimensional module, and let the symmetric group act by signed permutations (when two odd vectors cross, a sign pops up).  An analogous Schur-Weyl duality statement holds, and so, by analogy, I call the algebra of intertwiners the super-Temperley-Lieb algebra, or STLd.
Over the complex numbers, STLd is a product of matrix algebras corresponding to the irreps of Sd indexed by hook partitions.  Young diagrams are confined to one row and one column (super-row!).  In that sense, STLd is understood.  However, idempotents involved in projecting onto these Wedderburn components are nasty things that cannot be defined over ℤ

Question 1:  Does STLd have a ℤ-form that is compatible with the standard basis for CSd?
Question 2:  I am pessimistic about Q1; hence, the follow up:  why not?  I suspect that this has something to do with cellularity.
Question 3:  I care about q-deformations of everything mentioned:  Uq and the Hecke algebra, respectively.  What about here?  I am looking for a presentation of STLd,q defined over ℤ[q,q-1].

REPLY [2 votes]: I would define this algebra by a presentation. This algebra is over $\mathbb{Z}[\delta]$ but you can specialise to $\mathbb{Z}[q,q^{-1}]$ by taking $\delta\mapsto q+q^{-1}$.
The generators are $u_1,\ldots ,u_{n-1}$ (I have $n$ where OP has $d$). Then defining relations are
$$u_i^2=\delta u_i$$
$$u_iu_j=u_ju_i\qquad\text{if $|i-j|>1$}$$
$$u_iu_{i+1}u_i-u_i=u_{i+1}u_iu_{i+1}-u_{i+1}$$
$$u_{i-1}u_{i+1}u_i(\delta-u_{i-1})(\delta-u_{i+1})=0$$
$$(\delta-u_{i-1})(\delta-u_{i+1})u_iu_{i-1}u_{i+1}=0$$
Does this count?
If you want to move this to nLab that's fine by me.<|endoftext|>
TITLE: Can a discrete set of the plane of uniform density intersect all large triangles?
QUESTION [43 upvotes]: Let S be a discrete subset of the Euclidean plane such that the number of points in a large disc is approximately equal to the area of the disc. Does the complement of S necessarily contain triangles of arbitrarily large area?
(Added in July 2021:) I have recently learned that this is Danzer's problem. A Danzer set
(https://en.wikipedia.org/wiki/Danzer_set) is a counterexample. It is apparently not so well-known although the beginning of Gowers answer suggests that he has heard of it.

REPLY [3 votes]: I believe that quite a lot of research has been put into related questions, with much of the attention given to the unit square (and even more, the unit cube in higher dimensions).  Some key references are: 
[1] H.Niederreiter, Random Number Generation and Quasi-Monte Carlo Methods (SIAM, 1992) 
[2] J.Beck and W.Chen, Irregularities of distribution (CUP, 1987) 
[3] B.Chazelle, The Discrepancy Method: Randomness and Complexity (CUP, 2000). 
All I can do is quote you various results from these.  (Some of this is covered in Wikipedia-discrepancy.)  The results are typically expressed in terms of "discrepancy": for N points in the unit square, the discrepancy DN can be defined as supJ |A(J,N) - N λ(J)|where J is all subintervals of rectangles aligned with the axes, A(J,N) is the counting function (that is, the number of points inside J), and λ(J) is the Lebesgue measure.  This definition has an extra factor of N compared to [1], but I think this is useful when you scale the unit square to area N: a rectangle of area λ then "should" contain λ points; and the discrepancy is the difference between that and the number it actually contains.
There are results for the best possible behaviour of DN as a function of N.  For the axis-oriented rectangles, it is Ω(log N), and this can be achieved by the Halton-Hammersley sequence.  When J is changed to rectangles with any rotation, discrepancy is Ω(N1/4), and O(N1/4 √log N) can be achieved by jittered sampling.  When J is all convex subsets of the unit square, you've moved to "isotropic discrepancy" JN -- harder to study, but DN ≤ JN ≤ 8√(N DN).  Also, JN is Ω(N1/4).All of this is related to discrepancy of shapes that can contain some points, and the question is whether they contain the "correct" number.  If you start with a shape j1 with n1 points, and add an empty region (while staying in the set J), you can show that the empty region must have area O(2DN), but I can't see a guarantee that an empty shape obeys the Ω limit.
I'm not sure how close this is to an answer.  I certainly agree that it's a nice area, and I recommend those references -- they can all be read, if not fully understood, by someone with limited mathematical education, like me.<|endoftext|>
TITLE: Are there two groups which are categorically Morita equivalent but only one of which is simple
QUESTION [26 upvotes]: Can you find two finite groups G and H such that their representation categories are Morita equivalent (which is to say that there's an invertible bimodule category over these two monoidal categories) but where G is simple and H is not.  The standard reference for module categories and related notions is this paper of Ostrik's
This is a much stronger condition than saying that C[G] and C[H] are Morita equivalent as rings (where C[A_7] and C[Z/9Z] gives an example, since they both have 9 matrix factors).  It is weaker than asking whether a simple group can be isocategorical (i.e. have representation categories which are equivalent as tensor categories) with a non-simple group, which was shown to be impossible by Etingof and Gelaki.
Matt Emerton asked me this question when I was trying to explain to him why I was unhappy with any notion of "simple" for fusion categories.  It's of interest to the study of fusion categories where the dual even Haagerup fusion category appears to be "simple" while the principal even Haagerup fusion category appears to be "not simple" yet the two are categorically Morita equivalent.

REPLY [12 votes]: Categorically Morita equivalent groups were studied by Deepak
Naidu in Categorical Morita equivalence for group-theoretical categories. He obtained there a complete description  of
Morita equivalent groups. It is also shown 
that simple groups are categorically Morita rigid.<|endoftext|>
TITLE: Total Spaces of Quasicoherent Sheaves
QUESTION [15 upvotes]: You can construct a total space of a quasicoherent sheaf on an scheme by taking relative spec of the symmetric algebra of the dual sheaf. For locally free sheaves, you get vector bundles, and every vector bundle arises this way. What about sheaves that are not locally free? Are there any other sheaves for which the total space is a useful construction?

REPLY [2 votes]: We can back up a step and ask about the relative spec of the symmetric algebra of any coherent sheaf. Then there are lots of nice examples. If $X=\mathrm{Spec}\ k[t]$ and $E$ is the skyscraper sheaf at $0$, then we get $\mathrm{Spec}\ k[t,u]/(tu)$. If $X$ is $\mathrm{Spec}\ k[x,y]$ and $E$ is the ideal sheaf of the origin, then I get the relative spec of $\mathrm{Sym}_X(E)$ is $\mathrm{Spec}\ k[x,y,t,u]/(xu-ty)$. This has a one dimensional fiber over the points of $X$ other then $(0,0)$, and a two dimensional fiber over $(0,0)$. It looks like the fiber over $x \in X$ is the dual vector space to $E_x \otimes_{\mathcal{O}_x} k(x)$, but I am not sure whether this is always true.
If you insist that $E$ be the dual sheaf to some sheaf $F$, then you run into the problem that non-locally-free dual sheaves are hard to come by in dimensions 1 and 2. See my question, to which there are several good answers.<|endoftext|>
TITLE: Recovering a monoidal category from its category of monoids
QUESTION [13 upvotes]: What kind of additional properties and/or structures one needs to impose on the category
of (commutative or noncommutative) monoids of some monoidal category
so that one can recover the original monoidal category from this data?
What kind of additional properties and/or structures one needs to impose on a category
to ensure that it is the category of monoids of some monoidal category?
The example I have in mind is the category of (commutative or noncommutative) C*-algebras (or von Neumann algebras).
Can we obtain one of these categories as the category of monoids of some monoidal category?

REPLY [18 votes]: Here is a characterization of categories of commutative monoids.  I don't know the answer in the non-commutative case.
Let C be a category.  Then C is the category of commutative monoids in some symmetric monoidal category if and only if C has finite coproducts.
For suppose that C = CMon(M) for some symmetrical monoidal category M = (M, @, I).  Then one can show that the tensor product @ of M also defines a tensor product on C --- and that this is, in fact, binary coproduct in C.  (Example: if M is the category of abelian groups then C is the category of commutative rings, and the tensor product of commutative rings is the coproduct.)  Similarly, the unit object I of M is a commutative monoid in a unique way, and is in fact the initial object of C.  So C has finite coproducts.
Conversely, suppose that C has finite coproducts.  Then (+, 0) defines a symmetric monoidal structure on C, and with respect to this structure, every object of C is a commutative monoid in a unique way.  Thus, C = CMon(C).<|endoftext|>
TITLE: What is known about K-theory and K-homology groups of (free) loop spaces?
QUESTION [15 upvotes]: Calculating the homology of the loop space and the free loop space is reasonably doable. There exists the Serre spectral sequence linking the homology of the loop space and the homology of the free loop space. Furthermore, for finite CW complexes the James product construct a homotopy approximation to the loop space of a suspension.
One wonders how to extend these calculations to generalized (co)homology theories, like K-theory and K-homology. For which spaces do we know the K-theory and/or K-homology groups? How can one calculate these?

REPLY [6 votes]: I take the liberty to reference to a paper of mine (which is hopefully no bad style): Spectral Sequences in String Topology. Here, the K-homology of the free loop spaces of spheres and complex projective spaces is computed (up to extensions when torsion occurs). The method is actually pretty straightforward: one first finds concrete manifold generators for the homology of the free loop spaces of these spaces, which one can show to have a (stably) almost complex structure. This shows that the Atiyah-Hirzebruch spectral sequence for $MU$ degenerates and K-homology is determined by complex cobordism.
Additionaly, I want to comment that calculating the ordinary homology of free loop spaces is in general not that easy, too. For example, in the paper of Cohen, Jones and Yan, where the compute the Chas-Sullivan product for spheres and projective spaces, they reference for the computation of the additive homology of free loop spaces of projective spaces to a paper by Ziller, where he uses a Morse homology method, which, at least to my mind, is not altogether easy. Indeed, there are very few computations of homologies of free loop spaces, as far as I know. Spectral Sequences are a nice tool, but if they don't collapse calculations tend to be not straightforward.<|endoftext|>
TITLE: How can I generate random permutations of [n] with k cycles, where k is much larger than log n?
QUESTION [19 upvotes]: I've been thinking a lot lately about random permutations.  It's well-known that the mean and variance of the number of cycles of a permutation chosen uniformly at random from Sn are both asymptotically log n, and the distribution is asymptotically normal.
I want to know what a typical permutation of [n] with k(n) cycles "looks like" (in terms of cycle structure), where k(n)/(log n) → ∞ as n → ∞.  The special case I have in mind is permutations of [n] with n1/2 cycles, since I've come across such permutations in another context, but I'm also curious about the more general problem.  In order to do this I would like an algorithm that generates permutations of n with k cycles uniformly at random -- that is, it generates each one with probability 1/S(n,k) where S(n,k) is a Stirling number of the first kind -- so that I can experiment on them.  (I'd be willing to settle for a Markov chain that converges to this distribution if it does so reasonably quickly.)
Unfortunately the only way I know to do this is to take a permutation of [n] uniformly at random (this is easy) and then throw it out if it doesn't have k cycles.  If k is far from log(n) this is very inefficient, since those permutations are rare.
A few references I've come across that are related: This paper of Granville looks at permutations with o(n1/2-ε) cycles or Ω(n1/2+ε) cycles and shows that their cycle lengths are "Poisson distributed", but right around n1/2 is a transitional zone.  And this paper of Kazimirov studies "the asymptotic behavior of various statistics" under the distribution I've claimed, but I haven't read it yet because I can't read Russian and I'm waiting for the English translation.  Finally, the algorithm I'm looking for might be in one of the fascicles of volume 4 of Knuth, but our library doesn't have them.

REPLY [3 votes]: One can sample from the uniform distribution on permutations on $n$ elements with $k$ cycles in expected time $O(n\sqrt{k})$. 
For large $n$ and $k$ this may be more feasible than 
the method Herb Wilf refers to in his answer, which,
if I understand right, requires the generation of the Stirling cycle numbers $S(m,r)$ 
for $m\leq n$ and $r\leq k$. 
The idea is as follows: consider a Poisson-Dirichlet($\theta$) partition of $[n]$. The
blocks of such a partition have the distribution of the cycles of a random permutation
of $[n]$ from the distribution which gives weight proportional to $\theta^r$ to any permutation with $r$ cycles. In particular, conditioned on the number of cycles, the permutation is uniformly distributed. (Once one has a partition into blocks corresponding to the cycles, one can just fill in the elements of $[n]$ into the positions in the cycles uniformly at random). 
Choose $\theta$ in such a way
that the mean number of cycles is around $k$. 
One can sample a PD($\theta$) partition of $[n]$ in time $O(n)$ (see below). 
Keep generating independent samples of the partition until you get one
with exactly $k$ blocks. 
The variance of the number of cycles will be $O(k)$ (see below) and the 
probability that the number of cycles is precisely $k$ will be on the order of $1/\sqrt{k}$,  so 
one will need to generate
$O(\sqrt{k})$ such samples before happening upon one with precisely $m$ cycles.
So this is really not so far from what you suggested in the question 
(generate random permutations until you find one that fits) with the twist that
instead of generating from the uniform distribution (which corresponds to PD(1))
you choose a better value of $\theta$ and generate from PD($\theta$).
Here are two nice ways to sample a PD($\theta$) partition of $[n]$:
(1) "Chinese restaurant process" of Dubins and Pitman.
We add elements to the partition one by one. Element 1 starts in a block on its own. Thereafter, when we add element $r+1$,
suppose there are currently $m$ blocks whose sizes are $n_1, n_2, ... n_m$. Add element $r+1$ to block $i$ with probability 
$n_i/(r+\theta)$, for $1\leq i\leq m$, and put element $r+1$ into a new block on its own with probability $\theta/(r+\theta)$. 
(2) "Feller representation". Generate independent Bernoulli random variables $B_1, \dots, B_n$ with $P(B_i=1)=\theta/(i-1+\theta)$. 
Write a string of length $n$ divided up into blocks, with the rule that we start a new block before position $i$ whenever $B_i=1$. 
So for example if $n=10$ with $B_1=B_5=B_6=B_9=1$ and the other $B_i$ equal to 0, then the pattern is
(a b c d)(e)(f g h)(i j).
(Note that always $B_1=1$). Then assign the elements of $[n]$ to the positions in the blocks uniformly at random.
The expected number of blocks is $\sum_{i=1}^n \mathbb{E}B_i$, which is $\sum_{i=1}^n \theta/(i+\theta)$,
which is approximately $\theta(\log n-\log \theta)$. If this is equal to $k$ with $1 << k << n$, then
the number of blocks will be approximately normal with mean $k$ and variance $O(k)$. 
For details of some of the things mentioned here 
to do with Poisson-Dirichlet partitions, random permutations etc, see e.g. Pitman's lecture notes from
Saint-Flour: http://bibserver.berkeley.edu/csp/april05/bookcsp.pdf<|endoftext|>
TITLE: Two commuting mappings in the disk
QUESTION [183 upvotes]: Suppose that $f$ and $g$ are two commuting continuous mappings from the closed unit disk (or, if you prefer, the closed unit ball in $R^n$) to itself. Does there always exist a point $x$ such that $f(x)=g(x)$? 
If one of the mappings is invertible, then it is just a restatement of Brouwer's fixed point theorem but I do not know the answer in the general case and would not even dare to guess what it must be. Also, the answer is well-known to be "Yes" in dimension $1$.

REPLY [11 votes]: There is also a result due to Shields which implies a common fixed point (and hence a coincidence point) assuming that $f$ and $g$ are both analytic on the disk. 
A. L. Shields, On fixed points of commuting analytic functions, Proc. Amer. Math. Soc. 15(1964), 703-706. MR 29 #2790.<|endoftext|>
TITLE: Equivariant Derived Categories via their properties.
QUESTION [5 upvotes]: There are some ways to define equivariant derived categories of all sorts. But all the ways i know of involve giving a concrete construction. Is the other way around possible? Is there some universal property that characterizes them?
More explicitly i'm thinking of a bifibration D -> T where T is for example some nice subcategory of Top and D are the derived k-sheaves. For every group object G in T there should be a equivariant bifibration D_G -> T^G and these fibrations should again be bifibered over the category of group objects in T.
So here's the question: What properties do we actually want in such a situation?
For example i have in mind the "induction equivalence" and "quotient equivalence" as described in the book by Bernstein and Lunts.

REPLY [2 votes]: A bifibration is presumably equivalent to the data of a functor from T to categories, with the property that the map between categories induced by a map between spaces has an adjoint.  So we can understand your bifibration D -> T as a contravariant functor that takes a topological space to its category of sheaves, and a continuous map to the pullback operation.  
If by "category of sheaves" we mean the 1960's style triangulated category, this functor has bad properties--e.g. there is no simple way to recover D(X) from the categories D(U), where U runs through an open cover of X.  But by now there is better technology: we can replace the triangulated category of sheaves by a suitable infinity-category.  In this setting, D(X) will be an inverse limit in the infinity-categorical sense of the categories D(U) belonging to an open cover.  This is a nice way of formulating what is sometimes called cohomological descent.  We actually have descent for more general kinds of covers, for instance coming from free group actions.  If Y is a principal G-bundle over X, then D(X) is the inverse limit of the cosimplicial diagram of infinity-categories D(Y x G x ... x G).  This is the kind of thing that makes the equivariant theory of sheaves work.
So we can try to answer your question like this.  A candidate theory of equivariant sheaves is a contravariant functor D_G from T^G to infinity-categories.  It must agree with the usual theory of sheaves on T, which we identify with the full subcategory of T^G consisting of free G-spaces.  And it must satisfy a strong enough kind of descent.  After pinning down that last condition, I am pretty sure this will uniquely determine D_G.
You could translate all this back into the language of bifibrations (well, infinity bifibrations), but that point of view isn't so easy for me.
(Another way to put this is that D is itself a sheaf of categories on T in a suitable Grothendieck topology, and that this extends in a unique way to sheaf of categories on T^G.  But I'm worried that there is some technical problem with this statement.)<|endoftext|>
TITLE: Sources for exact triangles in triangulated categories.
QUESTION [6 upvotes]: The other day I came across the statement that in the triangulated category $\mathfrak{KK}$ (of C*-algebras with KK-groups as morphism sets) "there are many other sources of exact triangles besides extensions". Except for mapping cone triangles I don't know what is meant. What can you come up with?
I would also appreciate answers focussing on triangulated categories in general.

REPLY [4 votes]: This question has kind of been bothering me since I started thinking about it - I am far from an expert on KK-theory but I thought I'd throw something out there and maybe someone else will see it and come along and agree with me or correct me.
I think this statement is a way of thinking rather than something precise. Indeed by definition every distinguished triangle in KK is an isomorph of an extension triangle (although in the equivariant case I believe life is not so simple). Alternatively one can define the triangulation by taking distinguished triangles to be those candidates triangle (i.e. $X\to Y\to Z \to \Sigma X$ where each pair of composites vanishes) which are isomorphic to mapping cone triangles. So this is really all one has. The same story is true in the derived category of an abelian category for instance.
But (here is the punchline) one does not generally build the triangles one needs to prove things by considering short exact sequences of chain complexes! Indeed, one of the virtues of triangulated categories is that one has the ability to produce lots of new objects and triangles starting with very little. Often it is hard/impossible to do this in any explicit way - in fact one generally just knows that some collection of triangles doing the job exists and has no idea what they look like. So even though every triangle one might construct is (up to KK equivalence) an extension there is a very good chance that one didn't obtain it by writing down an explicit extension.
I guess there is also the fact that a triangle which is just isomorphic in KK to an extension triangle is not itself literally an extension of $C^*$-algebras. I don't know the stuff well enough to know whether or not one can produce interesting triangles via other constructions where there is a guarantee that some extension exists to make it distinguished. This is entirely possible (and in my opinion viewing such a construction as a different source of triangles is a worthwhile psychological distinction).<|endoftext|>
TITLE: Is the set of primes "translation-finite"?
QUESTION [32 upvotes]: The definition in the title probably needs explaining. I should say that the question itself was an idea I had for someone else's undergraduate research project, but we decided early on it would be better for him to try adjacent and less technical questions. So it's not of importance for my own work per se, but I'd be interested to know if it easily reduces to a known conjecture/fact/counterexample in number theory.
Apologies if the question is too technical/localized/unappealing/bereft of schemes.
Given a subset $X$ of the natural numbers $N$, and given $n \in N$, we write $X-n$ for the backward translate of $X$, i.e. the set {$\{x-n : x\in X\}$}.
We say that $X$ is translation-finite if it has the following property: for every strictly increasing sequence n1 < n2 < in $N$, there exists k (possibly depending on the sequence) such that
$(X-n_1) \cap (X-n_2) \cap \dots\cap (X-n_k)$
is finite or empty.
Thus every finite set is trivially translation-finite: and if the elements of $X$ form a sequence in which the difference between successive terms tends to infinity, then $X$ is translation-finite and we can always take k=2. Moreover:

if $X$ contains an infinite arithmetic progression, or if it has positive (upper) Banach density, then it is NOT translation finite;
there exist translation-finite sets which, when enumerated as strictly increasing sequences, grow more slowly than any faster-than-liner function.
there exist translation-finite sets containing arbitrarily long arithmetic progressions.

These resultlets suggest the question in the title, but I don't know enough about number theory to know if it's a reasonable question. Note that if, in the definition, we were to fix k first (i.e. there exists k such that for any sequence (nj)...) then we would get something related to Hardy-Littlewood conjectures; but I was hoping that this might not be necessary to resolve the present question.
EDIT (2nd Nov) It's been pointed out below that the question reduces in some sense to a pair of known, hard, open problems. More precisely: if the answer to the question is yes, then we disprove the Hardy-Littlewood k-tuples conjecture; if the answer is no, then there are infinitely many prime gaps bounded by some absolute constant, and this is thought to be beyond current techniques unless one assumes the Eliott-Halberstam conjecture.
 Added in 2013: Stefan Kohl points out that the latter is Yitang Zhang's famous recent result. However, as Will Sawin points out in comments, a negative answer to the main question would imply there are 3-tuple configurations occurring infinitely often in the primes, and (see thelink in Will's comment) this is thought to be out of reach even if we assume the EH conjecture holds.

REPLY [23 votes]: As you mention, this is related to the Hardy-Littlewood k-tuple conjecture.  In particular, if their conjecture is true, then the primes are not translation-finite.  Indeed, it is possible to find an increasing sequence n1 < n2 < n3 < ⋯ so that for every k, the first k nis form an admissible k-tuple.  (For example, I think ni = (i+1)! works.)  Then, by the k-tuple conjecture, infinitely many such prime constellations exist and so for all k, (X-n1) ∩ (X-n2) ∩ ⋯ ∩ (X-nk) is infinite.  (Here and below, X is the set of primes.)
However, maybe we can prove that the primes are not translation finite by some other means.  Unfortunately, the technology is not quite good enough to do that.  Proving that the primes are not translation finite would, in particular, prove that there exist n1 < n2 such that (X-n1) ∩ (X-n2) is infinite.  In particular, this implies that the gap n2-n1 occurs infinitely often in primes, and so pn+1-pn is constant infinitely often.  (The standard notation pn indicates the nth prime.)
The best known upper bound for the size of small gaps in primes is that lim infn→∞ (pn+1-pn)/log pn = 0.  This was established by Goldston and Yildirim around 2003 and the proof was later simplified.  To the best of my knowledge, the best conditional result is by the same authors; they show that given the Elliott-Halberstam conjecture, the prime gap is infinitely often at most 20 or so.<|endoftext|>
TITLE: What is the most compelling reason to believe Church's thesis?
QUESTION [12 upvotes]: Church's thesis states that the Turing machine is a universal model of computation.  What is the most compelling argument supporting this assertion?

REPLY [6 votes]: There is a good amount of 'experimental evidence', I guess. There have been lots of other models proposed, and they've all been proven to be computationally equivalent to Turing machines. This is true even for quantum computers. That is, they can all do the exact same things that Turing machines can. But there is also the question of whether this equivalence is a polynomial-time equivalence. This leads to the 'strong Church-Turing thesis'. It is believed by many that quantum computers are strictly better than classical computers in terms of time complexity, though as far as I know this has not been proven yet. If this were proven, then the strong Church-Turing thesis would be false. Then the natural thing would be to assert a quantum version of the strong Church-Turing thesis...

Edit: I see that this topic has been closed. I agree that it is perhaps too philosophical for Math Overflow. But still, let me make a few more mathematical comments, and one philosophical comment. Consider the lambda calculus. This is a very simple model of computation; its functionality is essentially restricted to a notion of a function and a notion of evaluation of functions which allows for recursion. The interesting thing is that even with this seemingly very limited functionality, the lambda calculus is still Turing-equivalent. What this shows is that it is relatively easy for a "reasonable model of computation" to be at least as powerful as Turing machines; once it has a notion of functions and a notion of evaluation of functions, it will be at least as powerful as the lambda calculus, and thus Turing machines. The more philosophical question arises when we consider the converse situation, namely, why should we expect "reasonable models of computation" to not be strictly more powerful than Turing machines? There is (probably) no good mathematical answer to this, because there is (probably) no good mathematical definition of "reasonable".

REPLY [3 votes]: The usual answer is that all proposed models of computation have turned out to be equivalent.<|endoftext|>
TITLE: What kind of operations does the Tall-Wraith monoid encode?
QUESTION [7 upvotes]: According to the nLab page, for an algebraic theory V a Tall–Wraith V-monoid is "the kind of thing that acts on V-algebras". Well, it certainly does act on V-algebras, but in which sense is it "the kind of thing"?
Can one express explicitly (in terms of elements and operations of the V-algebra) what is an action of the kind taken into account by TW-V-monoids?
In general TW-V-monoids do not form a variety, and they probably are not related to the structure of the set of endomorphisms of a V-algebra. But is it true that, for V a commutative theory, TW-V-monoids form a variety again, with operations given by those of V (applied pointwise on endomorphisms) plus the monoid structure coming from composition - i.e. the structure of the set endomorphisms?
(This plethory of questions remained after looking into Stacey/Whitehouse's "Hunting Hopf"-paper and the google-browsable parts of the Bergman/Hausknecht book, are there other references for the general V-algebra case?)

REPLY [3 votes]: But is it true that, for V a commutative theory, TW-V-monoids form a variety again, with operations given by those of V (applied pointwise on endomorphisms) plus the monoid structure coming from composition - i.e. the structure of the set endomorphisms?

Yes, this is true.  And the simplest example of a TW-V-monoid is where V is abelian groups whereupon TW-V-monoids are simply rings (unital, not necessarily commutative).
What led us to write our paper is that we couldn't find an accurate description of what we wanted (the full structure on unstable operations in a "generator-relation" type description).  After a paper chase through near rings and composition rings, we finally came across Tall and Wraith's 1970 paper in PLMS and the Borger-Wieland paper.  Neither was quite what we wanted, though, since the algebras from cohomology theories are more complicated than "mere" algebras.  The Hunting of the Hopf Ring paper (we were sure that a referee was going to complain about the title, by the way) concentrates on those technicalities rather than trying to explain the simple details of TW V-monoids.  We're currently writing another paper which tries to lay out the simpler ideas and which will include a proof of the commutative case (not that this is difficult).
The slogan "TW V-monoids are that-which-acts-on-V-monoids" is intended as an adaptation of the slogan "Rings are that-which-acts-on-abelian-groups".  The story is more complicated because it is not automatic that Hom(V,V) is a TW V-monoid.  One simple case where it is is when V is a finite ring, but in general one needs a Kunneth-type formula to hold.  Nonetheless, it is quite easy to find lots of examples of TW V-monoids and the Tall-Wraith paper, and the Borger-Wieland paper contain plenty of examples (as will ours).  The point about TW V-monoids being the thing that acts is that they are the representable things that act.  The general idea is that whenever you have a set-with-structure acting on a V-algebra then there is an associated Tall-Wraith V-monoid.  Of course, one might wonder why bother with that bloated gadget, but then one might wonder why bother with group rings when you already have the group.
I feel that this isn't really a sufficient answer, though.  My problem is that we came at this with examples in hand that we already knew about and wanted some algebraic way of encoding the structure that we already had.  As we couldn't find such a description, we invented one (and we hope that Tall and Wraith don't object to the approbation of their names!).  So for us, the intuition is all in cohomology operations and not in "bare algebra".  If you can expand on what you are looking for a little, I may be able to refine my answer somewhat.<|endoftext|>
TITLE: Quantum Computing Complexity?
QUESTION [5 upvotes]: After reading a recent post on Church's Thesis, I ran into Turing-Church's Strong Thesis, that may be potentially disproven by advances in Quantum Computing. Does anyone know of a good resource that gets into the potential of quantum computing on complexity theory? And how complexity calculations would be done in that environment?
I am also not well read in complexity theory in general, so I may have made an ill-posed question here.

REPLY [12 votes]: First, Alon's summary of applications of quantum computing is not complete.  What Shor's algorithm (or more precisely, Simon-Shor-Kitaev) really does is that it fully analyzes any finite abelian group in polynomial time (polynomial in log |G|), provided that inverses and the group law are available in black-box form, and elements have unique names.  Thus, you can factor N by analyzing the multiplicative group of Z/N.  You can analyze elliptic curves, other abelian varieties, D-modules, etc.  You can find the cardinality, find orders of elements, compute discrete logarithms, etc.
There are other quantum algorithms that do black-box things that certainly look like they could be useful.  What makes Shor's algorithm special is that there are lots of obvious ways to replace the black box by a "white box", i.e., by an explicit computational problem.
These algorithms provide strong evidence that the quantum polynomial time class, BQP, is larger than P, polynomial time (and BPP, randomized polynomial time).
And that is the original question about the strong Church-Turing thesis.  The strong Church-Turing thesis posits that all natural models of polynomial-time computation are equivalent.  At the moment, it looks like there are two natural models, P (which is conjectured to equal BPP) and BQP.  As Alon says, factoring is not known to be NP-complete, but there is nothing unfortunate about that.  It would surprise a lot of people if NP turned out to be a natural class, and it is a standard conjecture that BQP does not contain NP.<|endoftext|>
TITLE: Relationship between universal coefficient theorem and $[K(\mathbb{Z},n), K(G,n)]$?
QUESTION [11 upvotes]: In short, I'm wondering whether the universal coefficient theorem can be understood/reinterpreted by using maps of Eilenberg-MacLane spaces.  This is a wishy-washy idea and I don't have evidence to back it up, but it would be very nice if the "freebie" cohomology classes in $H^n(X; G)$ we get when we're changing our coefficients from $\mathbb{Z}$ to $G$ (i.e., those that come simply from tensoring with the new group) corresponded to elements of $[K(\mathbb{Z}, n), K(G, n)]$. Then, the other classes that arise from $\operatorname{Ext} /\operatorname{Tor}$ would correspond to elements of $[X, K(G,n)]$ which don't factor through $K(\mathbb{Z},n)$. Is anything like this even remotely true?
This question is in part motivated by the responses to an earlier question of mine, which mentioned that viewing $H^n(X; G)$ as $[X, K(G,n)]$ helps us understand cohomology operations (in that case, Steenrod squaring). It seems as if the representability of cohomology is probably only useful for studying honest cohomology operations, but I don't think I understand exactly what that means well enough to deduce whether changing coefficients qualifies...

REPLY [2 votes]: Just a minor addition to the other answers, since you seem particularly interested in cohomology theories and operations on them.  We often concentrate on operations of a particular cohomology theory, but the theory works equally well for operations between cohomology theories.  What you describe is a particular instance of that.  Where the cohomology theories are particularly well-behaved, all this operational structure is a sort-of souped-up version of rings and modules.  You can think of the operations of a single cohomology theory as like a ring, and operations between cohomology theories as like a bimodule.  (This is an analogy, they aren't rings and bimodules, they're Tall-Wraith monoids and bimodules for such.)<|endoftext|>
TITLE: A few questions on model theory, especially model theory of rings
QUESTION [8 upvotes]: I have never really read anything proper about model theory, so I have a few questions:
Someone told me that a school of logicians managed to give a very short proof of Falting's Theorem using model theory (and apparently "elimination of quantifiers"); I have not been able to find any reference about this, any ideas?
I also remember being given some very interesting information about the model theory of some rings. I do not recall the specifics, but it did give precise insights about how something like the complex numbers is easy to work with, whereas for the integers it is much harder (with problems such as Matiyasevich's Theorem/Hilbert's Tenth Problem). I have not managed to find anything similar to what I remember about it.
Does anyone have any information about all this, or a good source to learn from (which would, hopefully, mention these examples, at least the second one as it seems like it must be fairly standard)?

REPLY [11 votes]: As far as I know, there is no model theoretic proof of Faltings' Theorem itself.  Hrushovski's proof applies only to algebraic varieties over function fields and fails for varieties over number fields. 
On the other hand, one of Abraham Robinson's last works, a joint paper with Roquette published posthumously, Robinson, A.; Roquette, P. On the finiteness theorem of Siegel and Mahler concerning Diophantine equations, J. Number Theory 7 (1975), 121--176, contains a proof of Siegel's theorem on the finiteness of the number of integral points on curves of positive genus via nonstandard analysis.<|endoftext|>
TITLE: Are there two non-diffeomorphic smooth manifolds with the same homology groups?
QUESTION [6 upvotes]: I know that there definitely are two topological spaces with the same homology groups, but which are not homeomorphic. For example, one could take $T^{2}$ and $S^{1}\vee S^{1}\vee S^{2}$ (or maybe $S^{1}\wedge S^{1}\wedge S^{2}$), which have the same homology groups but different fundamental groups. But are there any examples in the smooth category?

REPLY [5 votes]: Serre has shown with the help of two embeddings phi and psi of a quadratic number field into C that there exist two projective surfaces V(phi)and V(psi) over C which have non isomorphic fundamental groups (and so are non homeomorphic) but have isomorphic Betti numbers.
The comparison theorem between étale cohomology and singular cohomology ( which didn't exist when Serre wrote his article )  even proves thar these surfaces have the same singular cohomology with value in any finite abelian group or over Q_l(l-adics) for any prime l.
I don't know if these surfaces have the same homology and so I don't answer your question in the strict sense (anyway, now you have Andy's and Eric's most satisfying solutions); but these remarks in an algebraic geometry context might interest you.
Serre's article is 
Exemples de variétés projectives conjuguées non homéomorphes,  C.R. Acad.Sci.Paris 258 (1964), 4194-4196
It is of course reproduced in his Collected Papers.<|endoftext|>
TITLE: Mayer-Vietoris homotopy groups sequence of a pull-back of a fibration
QUESTION [5 upvotes]: This must be an elementary question: could somebody tell me a reference for the Mayer-Vietoris homotopy groups sequence of a pull-back of a fibration?
I'm working in the category of pointed simplicial sets. So I've a pull-back of a (Kan) fibration of pointed simplicial sets, and I've read that in this situation you have an associated Mayer-Vietoris sequence relating the homotopy groups of the simplicial sets of the pull-back that looks like the classical Mayer-Vietoris sequence for the singular homology of a pair of open sets covering a topological space.
I've been searching in May's "Simplicial objects in Algebraic Topology" and Goerss-Jardine's "Simplicial Homotopy Theory", but I couldn't find it.

REPLY [3 votes]: Here is a lower-tech version of Reid's answer.  In general, if one has a Serre fibration p: E->B, and a map f: X->B, then there is a Mayer-Vietoris type sequence that comes from weaving together the long exact sequences in homotopy associated to p and to f*(p): f*(E)->X.  These exact sequences agree on every third term, because the fibers of f*(p) and p are homeomorphic.  The weaving process I'm talking about is an exercise in Hatcher's book (Exercise 38 in Section 2.2).
The relation between this exact sequence and the usual Mayer-Vietoris sequence in cohomology is explained here: Mathematically mature way to think about Mayer–Vietoris<|endoftext|>
TITLE: Surfaces that are 'everywhere accessible' to a randomly positioned Newtonian particle with an arbitrary velocity vector
QUESTION [9 upvotes]: Consider an idealized classical particle confined to a two-dimensional surface that is frictionless.  The particle's initial position on the surface is randomly selected, a nonzero velocity vector is randomly assigned to it, and the direction of the particle's movement changes only at the surface boundaries where perfectly elastic collisions occur (i.e. there is no information loss over time).
My question is - Does there exist such a bounded surface where the probability of the particle visiting any given position at some time 't', P(x,y,t), becomes equal to unity at infinite time?  In other words, no matter where we initialize the particle, and no matter the velocity vector assigned to it, are there surfaces that will always be 'everywhere accessible'?
(Once again, I welcome any help asking this question in a more appropriate manner...)

REPLY [2 votes]: I think the answers are not to the question asked (at least as it is asked). The ergodicity of the geodesic flow (which, by the way holds for all negatively curved surfaces -- a fact surprising not mentioned in any of the above answers) does not mean that a fixed geodesic will hit every point on the surface eventually, but merely that it will become dense (well, more than that, but less than hitting every point). The OP asks for every point to be hit.<|endoftext|>
TITLE: Is a subgroup of a free abelian group free abelian?
QUESTION [9 upvotes]: It's well-known that that a subgroup of a free group is free. Is a subgroup of a free abelian group (may not be finitely generated) always a free abelian group?

REPLY [25 votes]: A variety of groups $V$ is said to have the Schreier property if every subgroup of a free group in the variety is free. It is a classical theorem of Peter Neumann and James Wiegold that the only varieties of groups with the Schreier property are: the (absolutely) free groups, the free abelian groups, and the free exponent $p$ abelian groups for $p$ prime.<|endoftext|>
TITLE: How should we define "locally small"?
QUESTION [14 upvotes]: Let U be a Grothendieck universe, and U+ its successor universe (assume Grothendieck's universe axiom).
Everybody agrees that a U-small category is a category whose sets of objects and morphisms are both elements of U.  For the "next larger size" of categories which are not necessarily even locally small, call them just U-categories, there are two possible definitions:

a category whose set of objects and Hom-sets are all subsets of U;
a category whose set of objects and Hom-sets are all elements of U+ (U+-small categories).

I quite prefer the second notion, so that the category of U-categories is cartesian closed and we can form localizations.  This is the usage of Dwyer-Hirschhorn-Kan-Smith, "Homotopy Limit Functors on Model Categories and Homotopical Categories".  I think the first more closely corresponds to non-Grothendieck universe-based treatments of category theory using sets and classes, but I might be wrong about that.
For U-locally small categories there are again two possible definitions:

a category whose set of objects is a subset of U and whose Hom-sets are elements of U,
a category whose set of objects is an element of U+ and whose Hom-sets are elements of U.

I don't see a strong reason to prefer one over the other, except that the second is more parallel with my preference for U-categories.  DHKS uses the first.  As an example of the difference between them, if I have a U-locally small category C, I can form the category (poset) of full subcategories of C; this is U-locally small under the second definition, but not the first.  Is this a good thing or a bad thing?  Or are there no theorems I would care about that are affected by this difference?  Does anyone have an opinion about these two definitions?

REPLY [8 votes]: You are correct that the first notion of U-category corresponds more
closely to non-Grothendieck-universe-based treatments, e.g. using NBG
or MK set-class theory.  To be precise, if U is a universe, defining
"set" to mean "element of U" and "class" to mean "subset of U" gives a
model of MK set-class theory (and hence also NBG, which is weaker than
MK).  A comparison of the relationships between different set-theoretic
treatments of large categories can be found in my expository paper
"Set theory for category theory."
Here is an example of one theorem that can (maybe) tell the difference
between the two notions of U-locally-small categories.  Let C be a
U-category whose hom-sets are in U (i.e. a "U-locally-small category"
by your second definition).  Then C has a Yoneda embedding C →
[Cº,Set] where Set is the U-category of U-small sets.  Note that
[Cº,Set] is only a U-category by your second definition (i.e. a
U⁺-small category).  We say that C is lex-total if this Yoneda
embedding has a left adjoint which preserves finite limits.  It is a
theorem of Freyd, which can be found in Ross Street's paper "Notions
of topos," that if C is lex-total and also U-locally-small according
to your first definition (its set of objects is a subset of U), then C
is a Grothendieck topos (i.e. the category of U-small sheaves on some
U-small site).  The converse is not hard to prove, so this gives a
characterization of Grothendieck toposes.  As far as I know, it is
unknown whether there can be lex-total U-categories with very large
object sets that are not Grothendieck toposes.
I would personally be inclined to use your second definition of
"U-locally small," because as you say it matches your preferred
definition of large category relative to U (which I would prefer to
just call a "U⁺-small category", since its definition makes no reference
to U), and also because the term "U-locally small" sounds as if it
only imposes a smallness condition locally.  Street uses "moderate"
for a category with at most a U-small set of isomorphism classes of
objects, so if one wants to state a theorem (such as the above) about
U-locally-small categories according to your first definition, one can
instead say "U-locally-small and U-moderate."<|endoftext|>
TITLE: Countable subgroups of compact groups
QUESTION [14 upvotes]: What is known about countable subgroups of compact groups? More precisely, what countable groups can be embedded into compact groups (I mean just an injective homomorphism, I don't consider any topology on the countable group)? In particular, can one embed S_\infty^{fin} (the group of permutations with finite support) into a compact group? Any simple examples of a countable group that can't be embedded into a compact group?

REPLY [20 votes]: As a complement to Reid's answer: a finitely generated group is maximally almost periodic if and only if it is residually finite. Indeed, if a group is residually finite, it embeds into its profinite completion, which is compact. Conversely, if a finitely generated group $G$ embeds into a compact group $K$, then using first that homomorphisms $K\rightarrow U(n)$ separate points of $K$, second that finitely generated linear groups are residually finite (Mal'cev theorem), we conclude that $G$ is residually finite.<|endoftext|>
TITLE: spiral of Theodorus
QUESTION [14 upvotes]: A long time ago when I was in college I read about making a spiral out of right triangles with sides 1 and $\sqrt{N}$. (A google search seems to indicate that this is called the Spiral of Theodorus.)

I spent a long time trying to prove that the series of points approximated a spiral $R = K\theta + \varphi$, by trying to show the limit of the difference $\varphi = \sqrt{N+1} - K \sum_{i=1}^N \arctan(1/\sqrt{i})$ existed for some $K$. I think I managed to do it but it was confusing and can't find my papers. (and I'm still an amateur mathematician!)
Is this a known problem, and is there a closed-form solution to $K$ and $\varphi$?

REPLY [15 votes]: Here's a sketch of a proof that the constant you want exists, and how to find it.
Let 
$$
f(n) = \arctan(1) + \arctan(1/\sqrt{2}) + \arctan(1/\sqrt{3}) + \ldots + \arctan(1/\sqrt{n}).
$$
You want to show that $f(n) = \sqrt{n} + C + o(1)$ for some constant $C$.  (If you're not familiar with the $o$-notation, think of $o(1)$ as representing some function which goes to $0$ as $n$ goes to infinity.)  
Then take the power series expansion of $\arctan(1/\sqrt{k})$; this is
$$
(*) ~~~~~~~k^{-1/2} - \frac{1}{3} k^{-3/2} + \frac{1}{5} k^{-5/2} + \ldots
$$
So summing over $1$ to $n$, we should get
\begin{align*}
f(n)  = & (1^{-1/2} + 2^{-1/2} + ... + n^{-1/2}) \\\
  - \, \frac{1}{3} &(1^{-3/2} + 2^{-3/2} + ... + n^{-3/2}) \\\
 + \, \frac{1}{5}& (1^{-5/2} + 2^{-5/2} + ... + n^{-5/2}) - \ldots 
\end{align*}
Now, $1^{-1/2} + 2^{-1/2} + \ldots + n^{-1/2}$ has the asymptotic form
$$
2 \sqrt{n} + \zeta(1/2) + O(n^{-1/2}) 
$$
where I cheated a bit and asked Maple, and $\zeta$ is the Riemann zeta function.  And $1^{-j/2} + 2^{-j/2} + \ldots + n^{-j/2}$ has the asymptotic form
$$
\zeta(j/2) - O(n^{-j/2 + 1})
$$
where, if you're not familiar with the $O$-notation, $O(n^{-j/2+1})$ should be thought of as a function that goes to zero at least as fast as $n^{-j/2 + 1}$ as n goes to infinity.  So, assuming that we can rearrange series however we like,
$$
f(n) = 2 \sqrt{n} + (\zeta(1/2) - \frac{1}{3} \zeta(3/2) + \frac{1}{5} \zeta(5/2) - \ldots) + o(1).
$$
Since $\zeta(s)$ is very close to $1$ when $s$ is a large real number, that alternating series should converge.  Again cheating and using Maple, I claim it converges to about $−2.157782997$.  This is the constant you call $\varphi$, and what you called $K$ is equal to $2$.  (An easier way to see that your $K$ is $2$ is to note that $\arctan(1/\sqrt{n})$ is about $1/\sqrt{n}$, and approximate the sum by an integral.<|endoftext|>
TITLE: Tannakian Formalism
QUESTION [39 upvotes]: The Tannakian formalism says you can recover a complex algebraic group from its category of finite dimensional representations, the tensor structure, and the forgetful functor to Vect. Intuitively, why should this be enough information to recover the group? And does this work for other base fields (or rings?)?

REPLY [63 votes]: A very nice and very general version of Tannakian formalism is in Jacob Lurie's paper,
Tannaka duality for geometric stacks, arXiv:math/0412266. 
I like to think of Tannaka duality as recovering a scheme or stack from its category of coherent (or quasicoherent) sheaves, considered as a tensor category. 
From this POV the intuition is quite clear: having a faithful fiber functor to Vect (or more generally to R-modules) means your stack is covered (in the flat sense) by a point (or by Spec R). This is why you get (if you're over an alg closed field) that having a faithful fiber functor to Vect_k means you're sheaves on the quotient BG of a point by some group G, i.e. Rep G.. Over a more general base, you only locally look like a quotient of Spec R (or Spec k for k non-alg closed) by a group ---- ie you're a BG-bundle over Spec R, aka a G-gerbe. Even more generally, the kind of Tannakian theorem Jacob explains basically says that any stack with affine diagonal can be recovered from its tensor category of quasicoherent sheaves..
Actually the construction of the stack from the tensor category is just a version of the usual functor Spec from rings to schemes. Recall that as a functor, Spec R (k) = homomorphisms from R to k. So given a tensor category C let's define Spec C as the 
stack with functor of points Spec C(k) = tensor functors from C to k-modules (for any ring k, or algebra over the ground field etc). The Tannakian theorems then say for X reasonable
(ie a quasicompact stack with affine diagonal), we have X= Spec Quasicoh(X)  ---
so X is "affine in a quasicoherent-sheaf sense". Again, the usual Tannakian story is the case X=BG or more generally a G-gerbe.

REPLY [4 votes]: Some time ago I was also puzzled by this same question, and only now, after seeing yours, I start to think maybe I understand the idea. These are my own thoughts though, so you're encouraged to re-check them.
Consider for simplicity an affine algebraic group (on the opposite side would be a complex compact one). Then we know that the regular representation k[G] (respectively, L2(G)) should decompose as \sum R^*\otimes R.
Now what does knowing all the whole tensor category of representation imply? This means we can reconstruct the space k[G] as a G-representation by the above formula, so we know the functions on G! The multiplicatication of functions on G should be embedded in the product above, though I can't yet figure exactly how. 
E.g. for a linear group G there is some representation V which contains linear functions. Then functions of the form x^n can be obtained, up to conjugacy, as the highest vectors in V\otimes ... \otimes V, taken n times.
By the correspondence between functions and spaces, this would allow us to reconstruct the original space of group G. Now the action on the space of functions translates into the action of G.
So, I think if on this road one tries to define what the point of G is, one would arrive exactly at the defintion of it as the automorphisms of the fiber functor, the way it is in the Tannakian formalism.
I wonder if somebody could give a reference to a text explaining this point of view?<|endoftext|>
TITLE: Are there two non-isomorphic number fields with the same degree, class number and discriminant?
QUESTION [27 upvotes]: If so, do people expect certain invariants (regulator, # of complex embeddings, etc) to fully 'discriminate' between number fields?

REPLY [5 votes]: I recently heard a fascinating talk by Drew Sutherland on work he's done on this sort of question. Rather than trying to summarize his talk, here is a link to the slides for his talk. He also indicated that he'll be posting an article soon on the ArXiv, but it's not quite ready yet.<|endoftext|>
TITLE: Do convolution and multiplication satisfy any nontrivial algebraic identities?
QUESTION [35 upvotes]: For (suitable) real- or complex-valued functions $f$ and $g$ on a (suitable) abelian group $G$, we have two bilinear operations: multiplication -
$$(f\cdot g)(x) = f(x)g(x),$$
and convolution -
$$(f*g)(x) = \int_{y+z=x}f(y)g(z)$$
Both operations define commutative ring structures (possibly without identity) with the usual addition. (For that to make sense, we have to find a subset of functions that is closed under addition, multiplication, and convolution. If $G$ is finite, this is not an issue, and if G is compact, we can consider infinitely differentiable functions, and if $G$ is $\mathbb R^d$, we can consider the Schwarz class of infinitely differentiable functions that decay at infinity faster than all polynomials, etc. As long as our class of functions doesn't satisfy any additional nontrivial algebraic identities, it doesn't matter what it is precisely.) 
My question is simply: do these two commutative ring structures satisfy any additional nontrivial identities? 
A "trivial" identity is just one that's a consequence of properties mentioned above: e. g., we have the identity
$$f*(g\cdot h) = (h\cdot g)*f,$$
but that follows from the fact that multiplication and convolution are separately commutative semigroup operations. 
Edit: to clarify, an "algebraic identity" here must  be of the form $A(f_1, ..., f_n) = B(f_1, ..., f_n)$," where $A$ and $B$ are composed of the following operations: 

addition
negation
additive identity (0)
multiplication
convolution

(Technically, a more correct phrasing would be "for all $f_1, ..., f_n$: $A(f_1, ..., f_n) = B(f_1, ..., f_n)$," but the universal quantifier is always implied.) While it's true that the Fourier transform exchanges convolution and multiplication, that doesn't give valid identities unless you could somehow write the Fourier transform as a composition of the above operations, since I'm not giving you the Fourier transform as a primitive operation. 
Edit 2: Apparently the above is still pretty confusing. This question is about identities in the sense of universal algebra. I think what I'm really asking for is the variety generated by the set of abelian groups endowed with the above five operations. Is it different from the variety of algebras with 5 operations (binary operations +, *, .; unary operation -; nullary operation 0) determined by identities saying that $(+, -, 0, *)$ and $(+, -, 0, \cdot)$ are commutative ring structures?

REPLY [3 votes]: Late to the thread, but I wanted to quickly mention an identity that shows up for separable functions. Although this is a close cousin of your trivial identity and hardly theoretically deep, it turns out to be very useful in practice.
Let's take $\mathrm{R}^2$ as an example. If $f(x,y) = f_1(x)\ f_2(y)$ and $g(x,y) = g_1(x)\  g_2(y)$ then
$$f * g = (f_1\ f_2) * (g_1\ g_2) = (f_1 * g_1)\ (f_2 * g_2).$$
I abused notation a little to highlight the resemblance to distributivity. This identity finds use in a folklore trick of image processing that is described here:
http://www.stereopsis.com/shadowrect/
Edit by anonymous user: This is not true in general. Take $$f_1=f_2=g_1=g_2=\Pi,$$ where $\Pi$ denotes the rectangular function, which is one in the interval $[-1/2,1/2]$ and zero elsewhere. Then:
$$(f_1\ f_2)*(g_1\ g_2)=(\Pi^2)*(\Pi^2)=\Pi*\Pi=\Delta\neq \Delta^2=(\Pi*\Pi)^2=(f_1* g_1)(f_2*g_2).$$
Here, $\Delta$ is the triangle function.<|endoftext|>
TITLE: Characterizations of non-wellfounded models?
QUESTION [6 upvotes]: My question is whether there are any characterizations of non-wellfounded models of set theory. A wellfounded model is one that does not have any \epsilon-descending infinite sequences. I'm not asking about models that satisfy ZF-Foundation, but rather ones that satisfy ZF, but are not wellfounded in V. For example, taking the ultrapower by an ultrafilter which is not countably complete produces such a model. 
I would assume that the reason one does not work with non-wellfounded models is the inability to collapse them, and therefore the inability to study their structure in relation to V. And perhaps there is a theorem which says that "anything is possible" when it comes to these models, and therefore it's a hopeless cause to understand them.

REPLY [11 votes]: There is a large body of work studying ill-founded models of set theory. The goal is to provide a robust model theory for models of set theory, usually focussing on the countable models. Much of this theory grows out of the study of nonstandard models of arithmetic, 
and many tools and theorems from models of arithmetic generalize to the study of models of ZFC. 
Let me give a few examples. If M is a model of ZF, one can form the standard system of M by looking at the trace on the standard ω of all the reals of M. It is easy to see that Ssy(M) is a Boolean algebra, closed under Turing reducibility and if T is an infinite binary tree coded in Ssy(M), then there is a path through T coded in Ssy(M). Any set of reals with those three properties is called a Scott set, in honor of Dana Scott, who proved the following amazing characterization:
Theorem.(Scott) If ZFC is consistent, then every countable Scott set arises as the standard system of a model of ZFC.
Scott's theorem is usually stated for models of PA, but the proof for ZFC is identical. It remains a big open question whether Scott's theorem holds for all uncountable Scott sets. The answer is known for Scott sets of size ω1, and so under CH the problem is solved, but it remains open when CH fails. 
A key definition is that a model M of ZFC is computably saturated if it realizes every finitely consistent computable type over M. All such models are ω-nonstandard. It turns out that M is computably saturated if and only if it is (isomorphic to) a model that is an element of an ω-nonstandard model of ZFC. Furthermore, these models have interesting closure properties.
Theorem. Any two countable computably saturated models of ZFC with the same standard system and same theory are isomorphic. 
Theorem. Every countable computably saturated model M of ZFC is isomorphic to a rank initial segment Vα of itself. 
Much of the analysis of models of PA, such as that in the book by Jim Schmerl (UConn) and Roman Kossak (CUNY) extends to models of ZFC. Ali Enayat has also done a lot of interesting work along these lines. 
Here is another interesting theorem having to do with nonstandard ZFC models. Let ZFC* be any finite fragment of ZFC. If there is a (very small) large cardinal, then one can use full ZFC in this theorem (e.g. it suffices is there some uncountable θ with Lθ satisfying ZFC). The theorem is interesting in the case that there are nonconstructible reals. 
Theorem. Every real x is an element of a model of ZFC*+V=L. Furtheremore, one can find such a model whose ordinals are well-founded at least to α, for any desired countable ordinal α.
Proof. First, the statement of the theorem is definitely true in L, since every real in L is in some large Lθ. Second, the complexity of the statement is Σ11 in x and a real coding α. Thus, by Shoenfield's Absoluteness theorem, it is true in V. QED
Thus, even when x is non-constructible, it can still exist in a model of V=L! This is quite remarkable. (This theorem was shown to me by Adrian Mathias, but I'm not sure to whom it is originally due.)<|endoftext|>
TITLE: Do torsion-free groups give projectionless group ($C^\ast$) algebras?
QUESTION [13 upvotes]: One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the von Neumann algebras they generate must have nontrivial projections (unless it's just the complex numbers, of course). A good example of this is the reduced group $C^\ast$-algebra of any free group $F_n$. If $n=1$, then $C_r^\ast(Z)\cong C(S^1)$ via the Gelfand transform, which is clearly projectionless. If $n\geq 2$, the proof is fairly complicated. See Davidson's book for a proof when $n=2$.
If $G$ is a torsion-free group, is the reduced group $C^\ast$-algebra of $G$ projectionless? This $C^\ast$-algebra always contains the group algebra $C[G]$, so a simpler question is whether $C[G]$ is projectionless if $G$ is torsion-free.
Note that torsion-free is a necessary condition as one gets a projection from summing up the elements in the cyclic group generated by a torsion element and dividing by the order of the element.
EDIT: changed typestting. still some bugs... help please?

REPLY [11 votes]: The book Introduction to the Baum-Connes Conjecture, by Alain Valette, begins with a discussion of this problem.<|endoftext|>
TITLE: Are the asymptotics of Fourier coefficients to periodic solutions of ODE known?
QUESTION [5 upvotes]: The Van der Pol equation, given by
$$x'' + x = g x' (1 - x^2),$$
has periodic solutions $x(t)$, with the period $T(g)$ depending on the parameter. Thus, one can expand $x(t)$ as a Fourier series with coefficients $a_n(g)$ also depending on $g$.
Question: Can one find an asymptotic formula for $a_n(g)$, as $g \to\infty$?
For example, the asymptotic formula for the period is well known: 
$$T(g) \sim g [ (3 - \log 4) + O(g^{-4/3}) ].$$

REPLY [2 votes]: I am quite sure I found a perturbative solution to this during graduate school, what you are asking for is the Fourier representation of that. Are you familiar with the Method of Dominant Balance ?<|endoftext|>
TITLE: Decomposition of k[G]
QUESTION [12 upvotes]: There's a well-known decomposition of $L^2(G)$, a regular representation of compact complex group Lie $G$, called Peter-Weyl theorem.
Turns out for some reason I automatically  think that there is a similar theorem that decomposes regular representation $k[G]$ of algebraic group $G$:
$$k[G] = \bigoplus_R \ R^* \otimes R$$
where sum goes over representations to $GL(n, k)$. For this to work I think we need $G$ to be a linear reductive group over, say, algebraically closed field $k$ of characteristic 0. Also, perhaps we need $\pi_1(G) = 1$. 
But perhaps this is not true — the search hasn't given me a reference yet, but I wasn't able to provide a counterexample either.
Consider, for example, the multiplicative group $\mathbb G_m$. Then $k[\mathbb G_m] = k[x, x^{-1}]$ where each summand $k\cdot x^n$ is a separate representation of $\mathbb G_m$ into $\mathbb G_m = GL(1, k)$, specifically the one given by $a \mapsto a^n$. So the identity works.
So, is there such a theorem? What's a reference or a counterexample?

REPLY [6 votes]: This statement is false in general for algebraic groups. It's true in characteristic 0, but it is not in general true in positive characteristic. Instead, one has a weaker statement in positive characteristic (cf Proposition 4.20 on page 213 in Jantzen's "Algebraic Groups"): 

Let $G$ be a reductive linear algebraic group over an algebraically closed field of positive characteristic $k$. Then $k[G]$ has an increasing filtration whose subquotients are of the form $H(\lambda) \otimes H(-w_0 \lambda)$, where $\lambda$ runs over the dominant weights for $G$ and the $H(\lambda)$ are the modules arising as global sections of line bundles on the flag variety of $G$ (the so-called costandard modules for $G$). 

Moreover, this is true when $k[G]$ is considered as a $G\times G$-module.
Note that unlike in characteristic 0, these modules $V$ are not in general irreducible. (It's worth noting that the category of modules over a reductive algebraic group is not in general a semisimple category — this is only true in characteristic 0).<|endoftext|>
TITLE: Is there a categorification of the integers in terms of "graded sets"?
QUESTION [11 upvotes]: One way to categorify the non-negative integers is to consider the category FinSet whose objects are finite sets and whose morphisms are functions.  The isomorphism classes of objects in FinSet can be labeled "sets of cardinality 0, sets of cardinality 1," and so forth, so are a natural way of talking about the non-negative integers.  
What I want is a good definition of a category FinSSet (where SSet stands for "super set") whose isomorphism classes can be thought of as "sets of cardinality k" for all integers k in a natural way.  A natural candidate for the set of objects is the set of "Z2-graded sets," i.e. pairs of sets (S0, S1).  What we want is to define the cardinality of such a set as card(S0) - card(S1), so we want the isomorphism classes of objects to only depend on this number.
Unfortunately, I can't find a definition of the morphisms that actually accomplishes this.  What should it be?  For what it's worth, I've read "From Finite Sets to Feynman Diagrams" and I think John Baez gives up a little too early on the integers.

REPLY [3 votes]: For many years I am planning to try the following construction but never had enough time/energy to look at it carefully.
Take the object classifier $\mathscr E:=Set^{set_{fin}}$, and let $I$ be the generic object (the embedding of finite sets into sets). There is a triple of adjoint endofunctors $I\times(\_)\dashv(\_)^I\dashv\nabla_I(\_)$, induced by $(\_)+1:set_{fin}\to set_{fin}$; in particular these give an essential geometric endomorphism $E:\mathscr E\to\mathscr E$.
We now can either look at the $E$-completion or at the $E$-localization of $\mathscr E$, i. e. either at the colimit or at the limit of
$$
\cdots\to\mathscr E\xrightarrow[]{E}\mathscr E\xrightarrow[]{E}\mathscr E\to\cdots;
$$ 
what we get must contain an additive inverse of $(\_)+1$ in one sense or other...<|endoftext|>
TITLE: What is the symbol of a differential operator?
QUESTION [69 upvotes]: I find Wikipedia's discussion of symbols of differential operators a bit impenetrable, and Google doesn't seem to turn up useful links, so I'm hoping someone can point me to a more pedantic discussion.
Background
I think I understand the basic idea on $\mathbb{R}^n$, so for readers who know as little as I do, I will provide some ideas.  Any differential operator on $\mathbb{R}^n$ is (uniquely) of the form $\sum p_{i_1,\dotsc,i_k}(x)\frac{\partial^k}{\partial x_{i_1}\dots\partial x_{i_k}}$, where $x_1,\dotsc,x_n$ are the canonical coordinate functions on $\mathbb{R}^n$, the $p_{i_1,\dotsc,i_k}(x)$ are smooth functions, and the sum ranges over (finitely many) possible indexes (of varying length).  Then the symbol of such an operator is $\sum p_{i_1,\dotsc,i_k}(x)\xi^{i_1}\dotso\xi^{i_k}$, where $\xi^1,\dotsc,\xi^n$ are new variables; the symbol is a polynomial in the variables $\{\xi^1,\dotsc,\xi^n\}$ with coefficients in the algebra of smooth functions on $\mathbb{R}^n$.
Ok, great.  So symbols are well-defined for $\mathbb{R}^n$.  But most spaces are not $\mathbb{R}^n$ — most spaces are formed by gluing together copies of (open sets in) $\mathbb{R}^n$ along smooth maps.  So what happens to symbols under changes of coordinates?  An affine change of coordinates is a map $y_j(x)=a_j+\sum_jY_j^ix_i$ for some vector $(a_1,\dotsc,a_n)$ and some invertible matrix $Y$.  It's straightforward to describe how the differential operators change under such a transformation, and thus how their symbols transform.  In fact, you can forget about the fact that indices range $1,\dotsc,n$, and think of them as keeping track of tensor contraction; then everything transforms as tensors under affine coordinate changes, e.g. the variables $\xi^i$ transform as coordinates on the cotangent bundle.
On the other hand, consider the operator $D = \frac{\partial^2}{\partial x^2}$ on $\mathbb{R}$, with symbol $\xi^2$; and consider the change of coordinates $y = f(x)$.  By the chain rule, the operator $D$ transforms to $(f'(y))^2\frac{\partial^2}{\partial y^2} + f''(y) \frac{\partial}{\partial y}$, with symbol $(f'(y))^2\psi^2 + f''(y)\psi$.  In particular, the symbol did not transform as a function on the cotangent space.  Which is to say that I don't actually understand where the symbol of a differential operator lives in a coordinate-free way.
Why I care
One reason I care is because I'm interested in quantum mechanics.  If the symbol of a differential operator on a space $X$ were canonically a function on the cotangent space $T^\ast X$, then the inverse of this Symbol map would determine a "quantization" of the functions on $T^\ast X$, corresponding to the QP quantization of $\mathbb{R}^n$.
But the main reason I was thinking about this is from Lie algebras.  I'd like to understand the following proof of the PBW theorem:

Let $L$ be a Lie algebra over $\mathbb{R}$ or $\mathbb{C}$, $G$ a group integrating the Lie algebra, $\mathrm{U}L$ the universal enveloping algebra of $L$ and $\mathrm{S}L$ the symmetric algebra of the vector space $L$.  Then $\mathrm{U}L$ is naturally the space of left-invariant differential operators on $G$, and $\mathrm{S}L$ is naturally the space of symbols of left-invariant differential operators on $G$.  Thus the map Symbol defines a canonical vector-space (and in fact coalgebra) isomorphism $\mathrm{U}L\to\mathrm{S}L$.

REPLY [47 votes]: One way to understand the symbol of a differential operator (or more generally, a pseudodifferential operator) is to see what the operator does to "wave packets" - functions that are strongly localised in both space and frequency.
Suppose, for instance, that one is working in $\mathbb R^n$, and one takes a function $\psi$ which is localised to a small neighbourhood of a point $x_0$, and whose Fourier transform is localised to a small neighbourhood of $\xi_0/\hbar$, for some frequency $\xi_0$ (or more geometrically, think of $(x_0,\xi_0)$ as an element of the cotangent bundle of $\mathbb R^n$).  Such functions exist when $\hbar$ is small, e.g. $\psi(x) = \eta( (x-x_0)/\epsilon ) e^{i \xi_0 \cdot (x-x_0) / \hbar}$ for some smooth cutoff $\eta$ and some small $\epsilon$ (but not as small as $\hbar$).
Now apply a differential operator $L$ of degree $d$ to this wave packet.  When one does so (using the chain rule and product rule as appropriate), one obtains a bunch of terms with different powers of $1/\hbar$ attached to them, with the top order term being $1/\hbar^d$ times some quantity $a(x_0,\xi_0)$ times the original wave packet.  This number $a(x_0,\xi_0)$ is the principal symbol of $a$ at $(x_0,\xi_0)$.  (The lower order terms are related to the lower order components of the symbol, but the precise relationship is icky.)
Basically, when viewed in a wave packet basis, (pseudo)differential operators are diagonal to top order.  (This is why one has a pseudodifferential calculus.)  The diagonal coefficients are essentially the principal symbol of the operator.  [While on this topic: Fourier integral operators (FIO) are essentially diagonal matrices times permutation matrices in the wave packet basis, so they have a symbol as well as a relation (the canonical relation of the FIO, which happens to be a Lagrangian submanifold of phase space).]
One can construct wave packets in arbitrary smooth manifolds, basically because they look flat at small scales, and one can define the inner product $\xi_0\cdot(x-x_0)$ invariantly (up to lower order corrections) in the asymptotic limit when $x$ is close to $x_0$ and $(x_0,\xi_0)$ is in the cotangent bundle.  This gives a way to define the principal symbol on manifolds, which of course agrees with the standard definition.<|endoftext|>
TITLE: Intuitive Example of a Jacobson Radical
QUESTION [6 upvotes]: Can anyone explain what a Jacobson radical is using an intuitive example? I can't quite understand Wikipedia's explanation.

REPLY [5 votes]: Lucky timing, I just worked out a couple of examples for the class I'm teaching this semester.  Let A be a commutative local ring, m the maximal ideal, and k=A/m the residue field.  Let's consider two rings: M = M2(A) (2-by-2 matrices with entries in A), and I = the Iwahori order consisting of 2-by-2 matrices with entries in A whose lower-left entry is in m, i.e. matrices of the form
[ A A ]
[ m A ]
We'll compute rad(M) and rad(I) using the fact that rad = intersection of annihilators of simple left modules = { x : 1 - xy is a unit for all y }.
First let's compute rad(M).  M acts naturally on k^2 by left multiplication (via M2(A) ->> M2(k)), and this is a simple M-module.  The annihilator of k^2 is M2(m) = 2-by-2 matrices with entries in m, so rad(M) is contained in M2(m).  On the other hand every element x in M2(m) has the property that 1-xy is a unit in M2(A) for all y in M2(A) (since xy is in M2(m), the determinant of 1-xy is a unit); therefore rad(M) contains M2(m), and we've shown rad(M) = M2(m).
Next let's compute rad(I).  Now k is a simple left I-module in two ways: first by multiplication by the upper-left entry (mod m), and second by multiplication by the lower-right entry (mod m).  (Check that if x,y are in I then the upper-left entry of xy is congruent mod m to the product of the upper-left entries of x and y, and similar for the lower right.)  The annihilator of the first I-module is therefore matrices of the form
[ m A ]
[ m A ]
and the second is matrices of the form
[ A A ]
[ m m ]
and the intersection of these annihilators is the ring of matrices of the form
[ m A ]
[ m m ]
which therefore contains the Jacobson radical.  On the other hand every matrix x of that form has the property that det(1-xy) is a unit in A for y in I, and therefore the above collection of matrices is equal to the Jacobson radical.<|endoftext|>
TITLE: What is Chern-Simons theory?
QUESTION [42 upvotes]: What is Chern-Simons theory? I have read the wikipedia entry, but it's pretty physics-y and I wasn't really able to get any sense for what Chern-Simons theory really is in terms of mathematics.
Chern-Simons theory is supposed to be some kind of TQFT. But what kind of TQFT exactly? When mathematicians say that it is a TQFT, does this mean that it's a certain kind of functor from a certain bordism category to a certain target category? If so, what kind of functor is it? What kind of bordism category is it? What kind of target category is it? How exactly is the functor defined?
Also, from attending talks of Michael Freeman, I know that Chern-Simons theory is supposed to describe some aspects of the fractional quantum Hall effect. How does this work? How do I take some sort of Chern-Simons computation on a 3-(or 4-?)manifold and extract from that some kind of physical prediction about some 2d electron gas? I've also heard that Witten has interpretted various knot invariants like the Jones polynomial in terms of Chern-Simons theory. So does this mean that the Jones polynomial of a knot has a physical interpretation? If so, what is it?

REPLY [2 votes]: I don't know about mathematicians but when physicist say that it is a TQFT, they just mean that the theory doesn't depend on the metric choice. You can easily see that the Chern-Simons action is metric free. Thus, it is a TQFT.
About your fractional quantum hall effect question, 
you can look at Zee's "Quantum Field Theory in a Nutshell", Part VI - Field Theory and Condensed Matter. It is a nice introduction.
About you last question, 
Witten showed that Wilson loop expectation values of Chern-Simons theory are given by link invariant polynomials. In condensed matter field theory, Wilson loop is related to conductivity. So, that is how link invariants are related to physical observables.
There are a few good books on these subjects. I recommend Altland & Simons's "Condensed matter field theory" and Fradkin's "Field Theories of Condensed Matter Physics".<|endoftext|>
TITLE: Uniformization theorem in higher dimensions
QUESTION [16 upvotes]: Let $M$ be a 4-manifold with a complex structure.
Does there exist a finite list of simply connected complex 4-manifolds $M_1, ... , M_n$ such that M is the quotient of some $M_i$ by the action of a group acting discretely on $M$?
This would be an analog of the Poincare-Koebe uniformization theorem in (real) dimension 2. People who I've asked this question to think that it's unlikely that there is such a list, but haven't been able to offer an argument or a reference.

REPLY [4 votes]: As a complement to Georges Elencwajg's answer, there are lots of different simply connected hypersurfaces in projective space. The second volume of Shafarevich's introduction to algebraic geometry has a discussion of just this question of uniformization of complex manifolds.<|endoftext|>
TITLE: distinguished triangles and cohomology
QUESTION [19 upvotes]: Start with A an abelian category and form the derived category D(A).
Take a triangle (not necessarily distinguished) and take it's cohomology. We obtain a long sequence (not necessarily exact). If the triangle is distinguished it is exact. How about the converse: if the long sequence in cohomology is exact does it follow that the triangle is distinguished? (My guess is no, but I can't find a counter-example).

REPLY [3 votes]: You might be interested in the paper: Vaknin A., Virtual Triangles// K-Theory, 22 (2001), no. 2, 161--197.<|endoftext|>
TITLE: When are probability distributions completely determined by their moments?
QUESTION [62 upvotes]: If two different probability distributions have identical moments, are they equal?  I suspect not, but I would guess they are "mostly" equal, for example, on everything but a set of measure zero.  Does anyone know an example of two different probability distributions with identical moments?  The less pathological the better.
Edit: Is it unconditionally true if I specialize to discrete distributions?
And a related question: Suppose I ask the same question about Renyi entropies.  Recall that the Renyi entropy is defined for all $a \geq 0$ by
$$ H_a(p) = \frac{\log \left( \sum_j p_j^a \right)}{1-a} $$
You can define $a = 0, 1, \infty$ by taking suitable limits of this formula.  Are two distributions with identical Renyi entropies (for all values of the parameter $a$) actually equal?  How "rigid" is this result?  If I allow two Renyi entropies of distributions $p$ and $q$ to differ by at most some small  $\epsilon$ independent of $a$, then can I put an upper bound on, say, $||p-q||_1$ in terms of $\epsilon$?  What can be said in the case of discrete distributions?

REPLY [3 votes]: I stumbled across this post while googling a question about moment determinacy. Then I also found this survey (Recent Developments on the Moment Problem, Gwo Dong Lin), which summarizes pretty much everything said here on the moment problem and gathers a lot more checkable conditions for moment (in)determinacy of probability distributions. I found it to be a great resource.<|endoftext|>
TITLE: Are there any interesting examples of random NP-complete problems?
QUESTION [26 upvotes]: Here's an example of the kind of thing I mean. Let's consider a random instance of 3-SAT, where you choose enough clauses for the formula to be almost certainly unsatisfiable, but not too many more than that. So now you have a smallish random formula that is unsatisfiable. 
Given that formula, you can then ask, for any subset of its clauses, whether that subset gives you a satisfiable formula. That is a random (because it depends on the original random collection of clauses) problem in NP. It also looks as though it ought to be pretty hard. But proving that it is usually NP-complete also seems to be hard, because you don't have the usual freedom to simulate. 
So my question is whether there are any results known that say that some randomized problem is NP-complete. (One can invent silly artificial examples like having a randomized part that has no effect on the problem -- hence the word "interesting" in the question.)

REPLY [28 votes]: Important update (Oct. 6, 2010): I'm pleased to say that I gave the "random 3SAT" problem in the OP to Allan Sly, and he came up with a simple NP-hardness proof.  I've posted the proof to my blog with Allan's kind permission.

Sorry that I'm extraordinarily late to this discussion!
Regarding Tim's specific question: if we stick enough clauses in our instance that every 3SAT instance (satisfiable or not) occurs as a subformula with high probability, then certainly the resulting problem will be NP-hard.  Indeed, it would suffice to have a large enough set of clauses that we can always find a subformula that can serve as the output of one of the standard reductions.  On the other hand, I don't know of any techniques for proving NP-hardness that are tailored to the setting Tim has in mind -- it's a terrific question!
Since I can't answer the question he asked, let me talk for a while about a question he didn't ask (but that I, and apparently some of the commenters, initially thought he did). Namely, what's known about the general issue of whether there are NP-complete problems that one can show are NP-hard on average (under some natural distribution over instances)?
(1) The short answer is, it's been one of the great unsolved problems of theoretical computer science for the last 35 years!  If there were an NP-complete problem that you could prove was as hard on average as it was on the worst case, that would be a huge step toward constructing a cryptosystem that was NP-hard to break, which is one the holy grails of cryptography.
(2) On the other hand, if you're willing go above NP-complete, we know that certain #P-complete problems (like the Permanent over large finite fields) have worst-case/average-case equivalence.  That is to say, it's exactly as hard to compute the permanent of a uniform random matrix as it is to compute the permanent of any matrix, and this can be proven via an explicit (randomized) reduction.
(3) Likewise, if you're willing to go below NP-complete, then there are cryptographic problems, such as shortest lattice vector (mentioned by Rune) and discrete logarithm, that are known to have worst-case/average-case equivalence.  Indeed, this property is directly related to why such problems are useful for cryptography in the first place!  But alas, it's also related to why they're not believed to be NP-complete.  Which brings me to...
(4) We do have some negative results, which suggest that worst-case/average-case equivalence for an NP-complete problem will require very new ideas if it's possible at all.  (Harrison was alluding to these results, but he overstated the case a little.)  In particular, Feigenbaum and Fortnow showed that, if there's an NP-complete problem that's worst-case/average-case equivalent under randomized, nonadaptive reductions, then the polynomial hierarchy collapses.  (Their result was later strengthened by Bogdanov and Trevisan.)  There are analogous negative results about basing crytographic one-way functions on an NP-complete problem: for example, Akavia, Goldreich, Goldwasser, and Moshkovitz (erratum).  At present, though, none of these results rule out the possibility of a problem being NP-complete on average under the most general kind of reductions: namely, randomized adaptive reductions (where you can decide what to feed the oracle based on its answers to the previous queries).
(5) Everything I said above implicitly assumed that, when we say we want an average-case NP-complete problem, we mean with a distribution over instances that's efficiently samplable.  (For example, 3SAT with randomly generated clauses would satisfy that condition, as would almost anything else that arises naturally.)  If you allow any distribution at all, then there's a "cheating" way to get average-case NP-completeness.  This is Levin's universal distribution U, where each instance x occurs with probability proportional to $2^{-K(x)}$, K(x) being the Kolmogorov complexity of x.  In particular, for any fixed polynomial-time Turing machine M, the lexicographically-first instance on which M fails will have a short description (I just gave it), and will therefore occur in U with large probability!
(6) If you're willing to fix a polynomial-time algorithm M, then there's a beautiful result of Gutfreund, Shaltiel, and Ta-Shma that gives an efficiently-samplable distribution over NP-complete problem instances that's hard for M, assuming $NP \nsubseteq BPP$.  The basic idea here is simple and surprising: you feed M its own code as input, and ask it to find you an instance on which it itself fails!  If it succeeds, then M itself acts as your sampler of hard instances, while if it fails, then the instance you just gave it was the hard instance you wanted!
(7) Finally, what about "natural" distributions, like the uniform distribution over all 3SAT instances with n clauses and m~4.3n variables?  For those, alas, we generally don't have any formal hardness results.<|endoftext|>
TITLE: Are there two non-homotopy equivalent spaces with equal homotopy groups?
QUESTION [28 upvotes]: Could someone show an example of two spaces $X$ and $Y$ which are not of the same homotopy type, but nevertheless $\pi_q(X)=\pi_q(Y)$ for every $q$? Is there an example in the CW complex or smooth category?

REPLY [2 votes]: Consider $X=S^1\vee S^3$ and its double cover $X_2$ i.e, attach two copy of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) =\mathbb{Z} = \pi_1(X_2)$. And covering map induced isomorphism in $\pi_n$ for all $n\geq 2$. But they are not homotopically equivalent since their Eular Characteristics are different.

REPLY [2 votes]: The  simplest examples I know  are the $3$-dimensional lens spaces $L(p,q)$. They display many oddities.
Consider the  lens spaces  $L(p,q_0)$ and $L(p,q_1)$, $\gcd(p,q_0)=\gcd(p,q_1)=1$. Their fundamental groups are isomorphic to the   cyclic group $\newcommand{\bZ}{\mathbb{Z}}$ $\bZ/p\bZ$. Since  both these lens spaces have the same universal cover $S^3$,  their  higher homotopy groups are also isomorphic.
A theorem of Franz-Rueff-Whitehead (see Theorem 2.60 of these notes)  shows that $L(p,q_0)$ and $L(p,q_1)$ are homotopy equivalent if and only 
$$q_1\equiv \pm \ell^2 q_0\bmod p,  $$
for some $\ell\in\bZ$.   This reduces  the problem to  a number theoretic one. For example,  $L(5,1)$ is not homotopy equivalent to $L(5,2)$ since $\pm 2$ is not a quadratic residue mod $5$.
On the other hand, the lens spaces $L(7,1)$ and $L(7,2)$ are homotopy equivalent, but they are not homeomorphic.<|endoftext|>
TITLE: Algebraically closed fields of positive characteristic
QUESTION [47 upvotes]: I'm taking introductory algebraic geometry this term, so a lot of the theorems we see in class start with "Let k be an algebraically closed field." One of the things that's annoyed me is that as far as intuition goes, I might as well add "...of characteristic 0" at the end of that.
I know the complex numbers from kindergarten algebra, so I have a fairly good idea of how at least one algebraically closed field of characteristic 0 looks and feels. And while I don't have nearly the same handle on the field of algebraic numbers, I can pretty much do arithmetic in it, so that's two examples.
But I've never (really) seen an algebraically closed field of characteristic p > 0! I can build one just fine, and if you put a gun to my head I could probably even do some arithmetic in it, but there's no intuition of the sort that you get with the complex numbers. So: does anyone know of an intuitive description of such a field, that it's possible to get a real sense of in the same way as C?

REPLY [8 votes]: It is a little unfair to pass this off as an answer to your question, but it also seems too interesting to be buried as a comment.  (I can say that because I am citing other people's work!)
The comments following Ben's answer point out some nice senses in which ‘any’ calculation over the algebraic closure of $\mathbb F_p$ is really a calculation over a suitably large finite extension of $\mathbb F_p$, but I didn't see a precise statement in the comments.  I think it is very much worthwhile to note that, not just algebraically-closed-positive-characteristic computations, but even algebraically-closed-characteristic-$0$ calculations can be viewed this way.  For example, this is one way to prove that an injective, polynomial self-map of $\mathbb C^n$ is bijective.
See Serre's lovely article and Tao's lovely summary of it for more details on this point of view.<|endoftext|>
TITLE: Where are some interesting places where the axiom of choice crops up in category theory?
QUESTION [11 upvotes]: The two that come to mind are splitting epics in Set and taking the Skel of a category. Surely there are lots of other interesting (and maybe upsetting) places where this comes up.

REPLY [3 votes]: If $s:\mathcal{X} \to \mathcal{C}$ is a fibered category and $\varphi:C \to D$ is a morphism in $\mathcal{C}$, for each object $y$ in the fiber $\mathcal{X}_D$ the axiom of choice allows us to specify exactly one pullback $f:y_C \to y$ (i.e., a cartesian arrow $f$ with $s(f)=\varphi$).  The choice of such a collection is called a 'cleavage', and a cleavage always exists by the axiom of choice.  
This enables us to define the 'change of base functor' $\varphi^*:\mathcal{X}_D \to \mathcal{X}_C$.<|endoftext|>
TITLE: Colloquial catchy statements encoding serious mathematics
QUESTION [62 upvotes]: As the title says, please share colloquial statements that encode (in a non-rigorous way, of course) some nontrivial mathematical fact (or heuristic).  Instead of giving examples here I added them as answers so they may be voted up and down with the rest.
This is a community-wiki question: One colloquial statement and its mathematical meaning per answer please!

REPLY [2 votes]: Adding an axiom saying that something's provable
  doesn't help you prove it.

A characterization for Lob's theorem from reddit user noop_noob.<|endoftext|>
TITLE: Polynomials that are sums of squares
QUESTION [14 upvotes]: Is any algorithm known for determining whether or not a multivariate polynomial with integer coefficients can be written as a sum of squares of such polynomials? 
By way of background, if we one replaces "polynomial" by "rational function" then there is such an algorithm, because a rational function is a sum of squares iff it is positive definite. But this equivalence fails for polynomials.

REPLY [3 votes]: Determining whether or not a polynomial is a sum of squares can be done in polynomial time; the problem is equivalent to semidefinite programming.  As Jose Capco mentioned above, Pablo Parrilo is an authority on this subject, and has written a number of papers explaining the relationship.
Parrilo has a particularly short, simple, and self-contained exposition of this relationship in Parrilo, Pablo A. "Sum of squares programs and polynomial inequalities." SIAG/OPT Views-and-News: A Forum for the SIAM Activity Group on Optimization. Vol. 15. No. 2. 2004.<|endoftext|>
TITLE: An inverse problem: Number fields attached to elliptic curves over Q
QUESTION [5 upvotes]: If I understand FC's remark under the post "Very strong multiplicity one for Hecke eigenforms," in the course of Faltings's proof of the Tate conjecture, Faltings proves the following statement: let E/Q and F/Q be elliptic curves and write Q(E[p]) (resp. Q(F[p])) for the number field obtained by adjoining the x and y coordinates of the p-torsion of E to Q. Then if Q(E[p]) = Q(F[p]) for infinitely many primes p, E/Q and F/Q are isogenous.
Learning of this result prompted me to wonder: suppose P is a finite set of primes. Then do there exist E/Q and F/Q such that Q(E[p]) = Q(F[p]) for each p in P with E/Q and F/Q not isogenous?
If not in general, what is known about the particular P for which the above question does (or doesn't) have an affirmative answer?

REPLY [8 votes]: Most people would say no.  Indeed, there's a conjecture, most commonly ascribed to Frey, that for p a SINGLE large enough prime, E[p] isomorphic to F[p] implies that E and F are isogenous.  I believe p=37 is thought to be large enough, but don't hold me to it.<|endoftext|>
TITLE: Mathematicians who were late learners?-list
QUESTION [112 upvotes]: It is well-known that many great mathematicians were prodigies. 
Were there any great mathematicians who started off later in life?

REPLY [31 votes]: Preda Mihailescu is a good example http://en.wikipedia.org/wiki/Preda_Mihăilescu.
He received his PhD with 42, and proved the Catalan conjecture 5 years later.
The Catalan conjecture was open for 160 years.
He proposed in 2009 a proof of the Leopoldt conjecture, but I am not sure about the status of this.<|endoftext|>
TITLE: Is there a simple way to compute the number of ways to write a positive integer as the sum of three squares?
QUESTION [22 upvotes]: It's a standard theorem that the number of ways to write a positive integer N as the sum of two squares is given by four times the difference between its number of divisors which are congruent to 1 mod 4 and its number of divisors which are congruent to 3 mod 4.  Alternatively, there are no such representations if the prime factorization of N contains any prime of form 4k+3 an odd number of times.   If the prime factorization of N contains all such primes an even number of times, then we have
r2(N) = 4(b1 + 1)(b2 + 1)...(br+1)
where b1, ..., br are the exponents of the primes congruent to 1 mod 4 in the factorization of N.
For example, 325 = 52 × 13 can be written in 4(2+1)(1+1) = 24 ways as a sum of squares.  These are 182 + 12, 172 + 62, 152 + 102, and the representations obtained from these by changing signs and/or permuting.
Is there an analogous formula in the three-square case?  I know that an integer can be written as the sum of three squares if and only if it is not of the form 4m(8n+7).  There is a simple argument that shows that the number of ways to write all integers up to N as a sum of three squares is asymptotically 4πN3/2/3 -- representations of an integer less than N as a sum of three squares can be identified with points in the ball in R3 centered at the origin with radius N1/2.   Differentiating, a "typical" integer near N should have about 2πN1/2 representations as a sum of three squares.   From playing around with some data it looks like 
limn → ∞ #{ k ≤ n and r3(k)/k1/2 ≤ x} / n 
might be a nonzero constant.  That is, for each positive real x, the probability that a random integer k can be written in no more than x k1/2 ways approaches some constant in the open interval (0, 1) as k → ∞. 
One way to prove this (if it is in fact true) would be if there were some formula for r3(k), in terms of the prime factorization, which is why I'm curious.
(I apologize if this is something that is well-known to number theorists, although I'd appreciate a pointer if it is.  I am not a number theorist, I just play around with this sort of thing every so often and generate amusing conjectures.)

REPLY [26 votes]: I put a fair amount of effort into this, just about as the recent duplicate question was being closed. So I am moving it.
I wanted to include the viewpoint of Burton Wadsworth Jones, given in his little book "The Arithmetic Theory of Quadratic Forms." The theorem, with many cases, is that the number of "primitive" or "proper" representations $R_{0}(n)$ of a number by $x^2 + y^2 + z^2,$ (meaning $\gcd (x,y,z) = 1$) is a multiple of the class number of binary quadratic forms of discriminant $-4n,$ but the multiple changes depending on congruence properties of $n.$ Also there are "ground" cases, here $n=1,$ which are done separately anyway.To get the actual number of representations for a number that is not squarefree it is  necessary to take a sum.
Let's see, if $n$ is a multiple of 4 there are no primitive representations, as $x^2 + y^2 + z^2 \equiv 0 \pmod 4$ means that $x,y,z$ are all even. But that is fine, because this also means that the number of representations of $4n$ is exactly the same as the number of representations of $n.$ Also, if
 $ n \equiv 7 \pmod 8$ there are no representations at all.
For $n > 1$ and $ n \equiv 1 \pmod 8,$  $\; \; R_{0}(n) = 12 h(-4n).$
For $ n \equiv 3 \pmod 8,$  $ \; \; R_{0}(n) = 8 h(-4n).$
For $ n \equiv 5 \pmod 8,$  $ \; \; R_{0}(n) = 12 h(-4n).$
For $ n \equiv 2 \pmod 8,$  $ \; \; R_{0}(n) = 12 h(-4n).$
For $ n \equiv 6 \pmod 8,$  $ \; \;R_{0}(n) = 12 h(-4n).$
Just to include something that is not entirely about proper representations, from the Hecke eigenform method one gets, with p an odd prime,
$$ R(p^2 n) = (p + 1 - (-n|p) ) \; \; R(n) - \; \; p \; R( n / p^2) $$
where $R(n)$ is the number of representations including both proper and improper, the Jacobi symbol $(-n|p)$ is taken to be 0 if $p | n,$ while $R(n/p^2)$ is taken to be 0 if $p^2$ does not divide $n.$ This appears in an article by Hirschhorn and Sellers called, and I think this is clever, "On representations of a number as a sum of three squares" which appeared about 1999 in a journal with the word "Discrete" in the title. I just have a preprint here.<|endoftext|>
TITLE: Nonprojective Surface
QUESTION [11 upvotes]: Let k be an algebraically closed field.  It's well known that every complete curve, period, is projective.  Also, that every smooth surface is, and that there are smooth 3-folds which are not, and people go to reasonable lengths to include these examples all over the place, so they're easy to find.  However, Hartshorne does say that singular complete surfaces are not all projective.  Is there a simple example? A complete normal surface that is not projective? Is there some "least singular" possible such surface? I suspect that normality is too much to hope for, but I can't quite phrase why I think this, so is every normal complete surface projective?

REPLY [2 votes]: There is also an example in an Exercise from Hartshorne's Algebraic geometry involving infinitessimal extensions which I am trying to understand.
Let me recall some definitions and properties in the first place:

Infinitessimal lifting property: given an algebraically closed field $k$, a finitely generated $k$-algebra $A$ with $X=\mbox{Spec } A$ non-singular, and an exact sequence $0\rightarrow \mathcal{I} \rightarrow B' \rightarrow B \rightarrow 0$, where $B,B'$ are k-agebras and $I\subset B'$ is an ideal such that $\mathcal{I}^2=0$, any k-algebra homomorphism $A\rightarrow B$ lifts to a h-algebra homomorphism $A\rightarrow B'$, and two such homomorphism differ by a k-derivation of A into $\mathcal{I}$, namely an element in $Hom_A(\Omega_{A/k},\mathcal{I})$.
An infinitessimal extension of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ is a pair $(X',\mathcal{I})$ where $X'$ is a k-scheme and $\mathcal{I}$ is a sheaf of ideals on $X'$ with $\mathcal{I}^2=0$ and such that we have isomorphisms $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$ (as k-schemes) and $\mathcal{I}\cong \mathcal{F}$ (as $\mathcal{O}_X$-modules). For instance, the trivial extension of $X$ by $\mathcal{F}$ is given by the pair $(X,\mathcal{F})$, where the $X$ has structure sheaf $\mathcal{O}_X'=\mathcal{O}_X\oplus \mathcal{F}$ with product $(a\oplus f)\cdot (a'\oplus f')=aa'\oplus (af'+a'f)$, so that $\mathcal{F}$ becomes an ideal sheaf in $X$.
If $X=\mbox{Spec }A$ is affine and $\mathcal{F}$ is a coherent sheaf, then any extension is isomorphic to the trivial one: we just use the previous lifting property to construct a splitting of an appropriate short exact sequence.
There is a correspondence between isomorphism classes of infinitessimal extensions of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ and the cohomology group $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ where $\mathcal{F}_X$ is the tangent sheaf of $X$. If $(X',\mathcal{I})$ is an infinitessimal extension of $X$ by $\mathcal{F}$ and and $\{U_i\}$ is an affine open cover of $X$ (so that sheaf cohomology is isomorphic to Cech cohomology) then on every open affine set the extension is trivial, namely of the form $(U_i,\mathcal{I}_{|U_i}=\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i})$. It is easy to see from the construction of the trivialisation (choosing a lift, and noting that the difference of two lifts gives an element of $Hom_A(\Omega_{A/k},\mathcal{I})$) that this gives a cocyle in $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$. Conversely, given a cocyle $\xi=(\xi_{ij})\in \check{H}^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ and an open affine cover $\{U_i\}$, on each $U_i$ we have a trvial extension $(U_i,\mathcal{F}_{|U_i|}$ with $\mathcal{O}_{|U_i}'\cong\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i}$ and we can glue them all via $\xi=(\xi_{ij})$ to give an extension of $X$ by $\mathcal{F}$.

Then Hartshorne suggests that we perform the following computation:
let $X=P_k^2$ and consider the sheaf of differential 2-forms
$\omega_X$; then $H^1(X,\Omega_X^1)\cong H^1(X,\omega_X\otimes
\mathcal{T}_X)$ and a nontrivial extension $X'$ of $X$ by $\omega_X$
is given by the cocylce $\xi \in H^1(X,\omega_X^1)$ given over
$U_{ij}=U_i\cap U_j$ (where the $\{U_i\}$ are the standard open
subsets covering $P_k^2$) by
$\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$. This is our
target proper non-projective surface and in order to see that it is
indeed non-projective we shall prove that it has no ample invertible
sheafs (in fact no invertible sheaf at all, namely $Pic X'=0$). We
have a short exact sequence
$0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast}
\rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long
exact cohomology sequence $\cdots \rightarrow
\underbrace{H^1(X,\omega_X)}_0 \rightarrow
\underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow
\underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)}
\stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k
\rightarrow \cdots$ 
and in order to see that $Pic X'==$ it suffices
to prove that $\delta$ is injective and nonzero. Since $Pic X\cong
\mathbb{Z}$, any invertible sheaf is of the form
$\mathcal{L}=\mathcal{O}_X(d)\cong \mathcal{O}_X(1)^{\otimes d}$ and
it suffices to see that $\delta(\mathcal{O}_X(1))\neq 0$. I am
confused as to how to carry out this computation since I guess I
still do not understand very well the correspondence between
infinitessimal extensions and the cohomology group. What I intend to
do is to compute $\delta$ explicitly in the standard way, namely via
the diagram
$\begin{array}{ccccccccc} 0 & \rightarrow & \check{C}^1(U,\omega) &
\rightarrow & \check{C}^1(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow &
\check{C}^1(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0
\\ && \downarrow && \downarrow && \downarrow && \\ 0 & \rightarrow & \check{C}^2(U,\omega) &
\rightarrow & \check{C}^2(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow &
\check{C}^2(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \end{array}$
The cycle corresponding to $\mathcal{O}_X(1)$ in
$\check{C}^1(U,\mathcal{O}_X^{\ast})$ is
$\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$. How
does it map down to $\check{C}^2(U,\omega)$?
Thanks in advance for any insight.<|endoftext|>
TITLE: What does ramification have to do with separability?
QUESTION [10 upvotes]: Does ramification have anything to do with inseparability? It feels like an extension of Q in which p ramifies should somehow correspond to an extension of F_p(t). Does totally ramified <--> purely inseparable?
In fact, saying an irreducible polynomial f(x) is inseparable is the same as saying that f(x) ramifies when we extend Q[x] to L[x], where L is the splitting field of f(x).
By correspond, I generally mean taking an extension of Q defined by a root of p(x)=r, where r is a rational making the extension nontrivial, and then extending F_p(t) by a root of p(x)=t. It's interesting because then this extension of F_p(t) corresponds to a number of extensions of Q (this is the same thing when you do Galois theory by looking at fundamental groups of branched coverings of C. Then do you look at etale fundamental groups of objects associated to these function fields over finite fields?).

REPLY [3 votes]: Here's a candidate definition for "total ramification" which jibes quite well with pure inseparability: 
A finite map f:X -> Y is totally ramified at y if the scheme theoretic fibre X_y -> y is a universal homeomorphism. 
If k is a field, and A is a finite k-algebra, then A is totally ramified over k in the above sense if and only if a) A is local, and b) the last condition holds after all base changes on k. If k has characteristic 0, this is equivalent to requiring that A be local with residue field k. This means that in the case X and Y are curves over a field of characteristic 0, this gives the usual notion. On the other hand, a finite extension L/K of fields is totally ramified in the above sense iff L/K is geometrically connected iff L/K is purely inseparable.<|endoftext|>
TITLE: When do six operations work?
QUESTION [11 upvotes]: This question comes (heavily edited) from my notes, thus slightly unusual structure.

We know that algebraic maps have very strict structure, and in many settings the operations f_*, f_!, their adjoints f^*, f^!, bioperations ⊗ and =>  as well as duality D behave well. They satisfy (whenever defined) some good identities, especially for proper morphisms.
There are specific subtleties in the following cases:

case Z: constructible sheaves := (finite) local systems (finitely) glued  ...
case O: coherent sheaves := finitely generated O-modules ...
case D: (D-modules) holonomic := 'number of equations is just right' ...


Question: I wonder if there are other sheaves of non-commutative algebras for which we can define operations and duality? That is, is it possible to continue this list with another "case ?".

REPLY [5 votes]: This article of Fausk, Hu and May adds nothing to the list but gives a very nice analysis of the categorical situations in which you have six operations.<|endoftext|>
TITLE: Cubical vs. simplicial singular homology 
QUESTION [78 upvotes]: Singular homology is usually defined via singular simplices, but Serre in his thesis uses singular cubes, which he claims are better adapted to the study of fibre spaces. This young man (25 years old at the time) seemed to know what he was talking about and has had a not too unsuccessful career since.
So my (quite connected) questions are 
1) Why do so few books use this approach ( I only know Massey's) ?
2) What are the pro's and con's of both approaches ?
3) Does it matter ? After all the homology groups obtained are the same.

REPLY [6 votes]: A disadvantage of using cubical singular homology comes form the following example (copied from Natalia Hajlasz's homework):

Example. The circle $c(t)=(\cos 2\pi t,\sin 2\pi t)$, $t\in [0,1]$ is not the
  boundary  of any  singular cubical chain.

Proof. For a singular chain $\sum_i a_i c_i$, where $a_i\in\mathbb{Z}$ and $c_i$ is a singular cube, define its `length' by $\sum_i a_i$. Since opposite sides of a singular cube have signs $\pm 1$, it follows that the length of $\partial c_i$ equals zero and hence
the length of $\partial(\sum_i a_i c_i)$ equals zero. However, the length of  $c(t)=(\cos 2\pi t,\sin 2\pi t)$ equals $1$ so it cannot be represented as a boundary of a singular chain. $\Box$
If you consider the singular cube: $b=(s\cos 2\pi t, s \sin 2\pi t)$, $s,t\in [0,1]$, then $\partial b$ is a difference of two singular cubes: one is $c$ and one is a constant cube (mapping to the center). It is not possible to avoid such degenerate cubes. 
As I understand this problem is what Tyler Lawson had in mind when he wrote in his answer: The main disadvantage that comes solely from the point of view of homology is that you have this irritating normalization procedure: you have to take the cubical chains on X and take the quotient by degenerate cubes.<|endoftext|>
TITLE: Making D-modules on affine varieties more explicit
QUESTION [10 upvotes]: This is a follow up to my question about D-modules supported on the nilpotent cone.  I got some good answers there but now I have a more basic question.
Consider an affine algebraic variety X, a closed subvariety i:Y-->X, and the intermediate extension of the structure sheaf on Y to all of X (do I denote this i!*OY ?  For that matter, explaining either the * or ! extension instead would be a helpful start if its easier).
My question is this:  Since X is affine, D(X) is just an associative algebra, generated by O(X) and Vect(X) by the usual construction. My question is how can I understand i!*OY as a module the associative algebra D(X), supposing I understand D(X)?
In other words, what is the vector space underlying i!*OY, how do functions in O(X) and vector fields act?
I probably need to invest some serious time with a textbook to answer this question myself, but any help getting started would be most appreciated!

REPLY [2 votes]: Since Y is closed, the !, *, and !* extensions coincide, and there should be a straightforward (but confusing for me) way of doing what you want.  If Y is smooth, then I wonder if you can identify the vector space you're talking about with something like sections of the conormal bundle to Y in X, but this is not quite right (e.g. when Y = X).  When Y is singular, there is a more complicated story, having to do with the fact that the obvious definition of D-module on a singular variety is no good.
When Y is only locally closed, describing the !* extension is a difficult problem.  Vilonen's thesis, the only reference I know about this, is available here: 
http://gdz.sub.uni-goettingen.de/en/dms/load/img/?PPN=PPN356556735_0081&DMDID=dmdlog12
There is even something nontrivial to say about the ! and * extensions--the characteristic cycles of such D-modules were computed by Schmid and Vilonen.  Maybe there is a more elementary answer to your question about these extensions, though.<|endoftext|>
TITLE: Faithfully flat descent over Hopf algebras in terms of comodule structures
QUESTION [10 upvotes]: Let $A$ be a (finite-dimensional graded cocommutative) Hopf algebra over a field $k$, $E$ be a Hopf subalgebra, and $R=A \otimes_E k$.  Then the comultiplication on $A$ induces a coalgebra structure on $R$.  Furthermore, $R$ is a coalgebra in the monoidal category of $A$-modules, with $A$ acting on $R \otimes R$ diagonally via the comultiplication.  Define an internal $R$-comodule to be an object $M$ which is simultaneously an $A$-module and an $R$-comodule such that the structure map $M \to R \otimes M$ is a map of $A$-modules, for the diagonal $A$-module structure on the tensor product.
$A$ itself is naturally an internal $R$-comodule, via the comultiplication $A \to A \otimes A \to R \otimes A$.  For any $E$-module $N$, $A \otimes_E N$ then inherits an internal $R$-comodule structure from $A$.  Conversely, if $M$ is an internal $R$-comodule, $N={m:d(m)=1 \otimes m}$ is an $E$-module, where $d:M \to R \otimes M$ is the structure map.
Is it true (possibly under some reasonable niceness hypotheses) that these two functors between E-modules and internal $R$-comodules are inverse?  In particular, I'd like to interpret this in terms of faithfully flat descent: $A$ is faithfully flat over $E$, and I want to say that for an $A$-module $M$, there is a natural bijection between descent data that allows us to identify $M=A \otimes_E N$ for an $E$-module $N$ and internal $R$-comodule structures $M \to R \otimes M$.
Sorry if I'm getting some things wrong about what hypotheses are needed for this to make sense; I'm trying to understand this in a specific example and don't know much of the general theory.

REPLY [4 votes]: A very small example where the answer is no:
Suppose $k$ has characteristic not two and $A=k\langle x,y:x^2=1, y^2=0\rangle$ with $\Delta(x)=x\otimes x$, $\Delta(y)=y\otimes 1+x\otimes y$, $\varepsilon(x)=1$ and $\varepsilon(y)=0$; this is the Sweedler Hopf algebra. Let $E$ be the subHopf algebra generated by $x$, which has $\{1,x\}$ as a basis. Then $R=k\otimes_EA$ has $\{\overline 1=1\otimes 1,\overline y=1\otimes y\}$ as a basis, and its coalgebra structure is given by $\Delta(\overline 1)=\overline 1\otimes\overline 1$, $\Delta(\overline y)=\overline y\otimes\overline1+\overline1\otimes\overline y$, $\varepsilon(\overline1)=1$ and $\varepsilon(\overline y)=0$.
Since $E\cong k\times k$ as an algebra, the category $\mathrm{Mod}_E$ is semisimple.
On the other hand, suppose $M\in\mathrm{Mod}_A^R$. One can check that the right $R$-comodule structure $\rho$ of $M$ is determined by a linear map $\phi:M\to M$ such that $\phi^2=0$ by the equation $$\rho(m)=m\otimes\overline 1+\phi(m)\otimes\overline y.$$ Similarly, the $A$-module structure on $M$ is easily seen to be such that $m\cdot y=0$ for all $m\in M$ and $\phi(m\cdot x)=\phi(m)\cdot x$ for all $m\in M$. It follows that one can identify an object $M$ of $\mathrm{Mod}_A^R$ with a $4$-tuple $(M^+,M^-,\phi^+,\phi^-)$ such that $M=M^+\oplus M^-$ is the decomposition of $M$ as direct sum of the eigenspaces of right multiplication by $x$ (the only possible eigenvalues are $1$ and $-1$, and it is diagonalizable) and $\phi^{\pm}:M^\pm\to M^\pm$ are the restrictions of the map $\phi$ to $M^+$ and $M^-$ (so in particular they square to zero).
Moreover, morphisms in $\mathrm{Mod}_A^R$ have the obvious description in terms of these $4$-tuples.
Now, it is very easy to see using this description that $\mathrm{Mod}_A^R$ is not semisimple: for example, the object $(k^2,0,\left(\begin{array}{cc}0&1\\\\0&0\end{array}\right),0)$ is not semisimple (in fact, the category is the direct sum of two copies of the module category over the quiver $\bullet\to\bullet$). It follows that $\mathrm{Mod}_E$ and $\mathrm{Mod}_A^R$ are not equivalent in this case.
(The answer is yes, though, in the two extreme cases where (i) $E=k$ or (ii) $E=A$ (the first one is the «fundamental theorem of Hopf algebras», the second one is trivial)<|endoftext|>
TITLE: What are the endofunctors on the simplex category?
QUESTION [14 upvotes]: Is there a 'classification' of the endofunctors F: Δ --> Δ where Δ denotes the simplex category with objects [n] and the weakly monotone maps [m] --> [n] as morphisms (Actually, I don't know if 'simplex category' is the right name)?
For instance there is a shift functor S: Δ --> Δ defined by S([n])=[n+1] on objects and S(d): [m+1] --> [n+1] being d on [m] and mapping m+1 to n+1 for a morphism d: [m]-->[n]. Hence for a simplicial set X one gets a path-object XoS.

REPLY [17 votes]: To carry Charles' train of thought further:
By an 'interval' let us mean a finite ordered set with at least two elements; let $Int$ be the category of intervals, where a morphism is a monotone map preserving both endpoints.
The simplicial set $\Delta^1$ can be viewed as a simplicial interval. That is, this functor $\Delta^{op}\rightarrow Set$ factors through the forgetful functor from $Int$ to $Set$. In fact, the resulting functor $\Delta^{op}\rightarrow Int$ is an equivalence of categories. 
This extra structure (ordering and endpoints) on $\Delta^1$ is inherited by Charles' $K_1=F^*\Delta^1$; it, too, is a simplicial interval.
There aren't that many things that a simplicial interval can be. Its realization must be a compact polytope with a linear order relation that is closed. That makes it at most one-dimensional, and makes each component of it either a point or a closed interval. Simplicially each of these components can be either a $0$-simplex, or a $1$-simplex with its vertices ordered one way, or a $1$-simplex with its vertices ordered the other way, or two or more $1$-simplices each ordered one way or the other and stuck together end to end. 
The three simplest things that a simplicial interval can be are: two points, a forward $\Delta^1$, and a backward $\Delta^1$. These arise as $F^*\Delta^1$ for three examples of functors $F:\Delta\rightarrow \Delta$, the only examples that satisfy $F([0])=[0]$, namely the constant functor $[0]$, the identity, and "op".
It's clear that any functor with $F([0])=[n]$ has the form $F_0\coprod\dots\coprod F_n$ where $F_i[0]=[0]$ for each $i$. This means that the corresponding simplicial interval can be made by sticking together those which correspond to the $F_i$. For example, the 'shift' functor mentioned in the question is $id\coprod [0]$; Reid mentioned $id\coprod id$ and $op\coprod id$. These correspond respectively to: a $1$-simplex with an extra point on the right, two $1$-simplices end to end, and two $1$-simplices end to end one of which is backward. As another example, the constant functor $[n]$ corresponds to $n+1$ copies of (two points) stuck together end to end, or $n+2$ points.
In short, every functor $F:\Delta\rightarrow \Delta$ is a concatenation of one or more copies of $[0]$, id, and op. I can more or less see how to prove this directly (without toposes or ordered compact polyhedra).<|endoftext|>
TITLE: Explanation for E_8's torsion
QUESTION [21 upvotes]: To study the topology of Lie groups, you can decompose them into the simple compact ones, plus some additional steps, such as taking the cover if necessary. After that, the structure of $SO(n)$'s is rather straightforward, but the exceptional groups are more interesting.
Any simple compact Lie group, by means of Hopf algebra theory, has the rational homology of a product $$S^a \times S^b \times \dots \times S^z$$ where the numbers are called exponents.  Other than that, their cohomology could also have torsion. Now the torsion for all groups is known:

Among classical groups, only 2-torsion is possible and only for $Spin(n)$
Exceptional groups can only have 2 and 3-torsion (most do), with the exception of:
$E_8$ which has 2-, 3-, and 5- torsion.

Well, this is bound to be related to $E_8$'s Coxeter number, which is 30, but are there any hints as to why? My reference would be math-ph/0212067 but it can't relate this to Coxeter number either. 
For the reference, exponents are known to be related to Coxeter number, see Kostant, The Principal Three-Dimensional Subgroup and the Betti Numbers of a Complex Simple Lie Group (google search).
Is this an open problem? Maybe yes, but maybe it's been explained, so I'm posting it as it is for now.

REPLY [4 votes]: Take a look at "Finite H-spaces and Lie Groups" by Frank Adams, particularly the
letter from E8 and the appendix which follows it.<|endoftext|>
TITLE: Stable homology of arithmetic groups
QUESTION [13 upvotes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$Suppose that $F/Q$ is a number field.
Using automorphic forms, Borel computed the ($R$-) stable cohomology of $\SL_n(O_F)$, and as a result, computed $K_i(O_F)\otimes Q$. Nowadays, using work of Suslin, Voevodksy, Rost, etc, one has an almost "complete" understanding of the actual integral groups $K_i(Z)$, say, modulo Vandiver's conjecture. This does not directly give the stable cohomology of $\SL_n$, because one set of groups is computing homotopy and the other is computing cohomology, but let us not worry about that distinction for the moment.
Borel also computed the ($R$-) stable cohomology of $\Sp_{2n}(O_F)$.
My question, loosely speaking, is whether one can describe the stable integral cohomology of $\Sp_{2n}(Z)$ in as detailed away as the algebraic K-groups of $Z$ "describe" the integral cohomology of $\SL_n(Z)$.

Let me summarize some of what I have found out (following up some of the answers below), mostly though emails from experts.
For affine objects, which certainly includes $Z$, K-theory is about the monoidal category
$P(A)$ of projective finitely generated A-modules, and
Hermitian K-theory is about the monoidal category
$P(A)_h$ of objects in $P(A)$ equipped with a non-degenerate
symmetric (or skew-symmetric) form. One of the issues with computing or working with such a theory over $Z$ is that irritating issues arise in characteristic $2$, as one might expect with quadratic forms present. It seems that one might be in good shape to understand the groups $K^h_i(Z[1/2])$. For usual K-theory, there is an excision formula relating $K_i(Z)$ to $K_i(Z[1/2])$ and $K_i(F_2)$. The latter group is "easy" (or at least was computed by Quillen).
Of interest to me in $K_i(Z)$ are the Soulé classes. The next thing I will be thinking about is whether Soule classes can give rise to elements in the stable cohomology of $\Sp_2n(Z)$.
Andy said some very interesting things, but I will probably be awarding the bounty to Oscar, since the paper he linked to was more directly relevant to what I was trying to find out. (Sorry Andy...but it looks like you have lots of reputation anyway!)

REPLY [4 votes]: The algebraic K-groups of Z are to the homology of SLn(Z) as the Hermitian K-groups of Z are to the homology of Sp2g(Z). There is a paper by Berrick and Karoubi here in which they discuss, and make some calculations of, the Hermitian K-theory of Z and Z[1/2].

REPLY [3 votes]: I don't know what $H_3(Sp_{2g}(Z))$ is, but I want to give a reference for a related fact.  Let $Mod_g$ be the mapping class group.  The action on $H_1$ of the surface preserves the algebraic intersection form and gives a representation $Mod_g\to Sp_{2g}(Z)$ which is well-known to be surjective.  Rationally, the stable cohomology of $Sp_{2g}(Z)$ injects into the stable cohomology of $Mod_{g}$ and gives "one half" of the stable cohomology of $Mod_{g}$.
Anyway, I do know a reference for some stable integral homology calculations for the mapping class group.  Namely, in The low-dimensional homotopy of the stable mapping class group, Ebert proves that in a stable range we have $H_3$ of the mapping class group equal to $Z/12Z$ and $H_4$ equal to $Z^2$.  He does this by using the fact that Madsen and Weiss's work identifies the infinite loop space $(Mod_{\infty})^+$ with a space whose first few homotopy groups are known.  Ebert then builds with his bare hands the first few stages of the Postnikov tower for this space.
Presumably this either has been done or could be done for $Sp_{2g}(Z)$ by a sufficiently motivated homotopy theorist.<|endoftext|>
TITLE: What are the automorphism groups of (principally polarized) abelian varieties?
QUESTION [24 upvotes]: What are the possible automorphism groups of a principally polarized abelian variety $(A,\lambda)$ of dimension $g,$ say an abelian surface ($g=2$) over the complex numbers or algebraic closure of a finite field? The fact that the moduli stack $A_g$ is of finite diagonal (over the integers) implies that the automorphism groups are all finite, but do we know more? Like the size.
When $g=1$ this is given in Silverman I, p.103.
Edit:  Let me make the question more specific. Let $(A,\lambda)$ be an $\mathbb F_q$-point of $A_g$ (i.e. an abelian variety $A$ over $\mathbb F_q$ of dimension $g$ and a principal polarization $\lambda$). We want to consider its automorphism group (over $\mathbb F_q$).
Let $\pi:A_{g,N}\to A_g$ be the natural projection, where $A_{g,N}$ is the moduli stack of p.p.a.v. of dimension $g$ with a level $N$ structure (a symplectic isomorphism $H^1(A,Z/N)\to(Z/N)^{2g}$). We always assume $q$ is prime to $N.$ Note that $\pi$ is a $G$-torsor, for $G=GSp(2g,Z/N),$ so it gives a surjective homomorphism $\pi_1(A_g)\to G.$ The sheaf $\pi_*\mathbb Q_l$ on $A_g$ is lisse (even locally constant), corresponding to the representation of $\pi_1(A_g)$ obtained from the regular representation $\mathbb Q_l[G]$ of $G$ and the projection $\pi_1(A_g)\to G.$ For any $\mathbb F_q$-point $x$ of $A_g,$ the local trace $\text{Tr}(Frob_x,(\pi_*\mathbb Q_l)_{\overline{x}})$ is either $|G|$ or 0, depending on $Frob_x\in\pi_1(A_g)$ is mapped to 1 in $G$ or not. 
We have isomorphisms $H^i_c(A_{g,N},\mathbb Q_l)=H^i_c(A_g,\pi_*\mathbb Q_l).$ By Lefschetz trace formula, applied to both $\mathbb Q_l$ on $A_{g,N}$ and $\pi_*\mathbb Q_l$ on $A_g,$ we have 
$$|A_{g,N}(\mathbb F_q)|=|G|\sum_{x\in S} 1/\#Aut(A_x,\lambda_x),$$
where $S$ is the subset of $[A_g(\mathbb F_q)]$ consisting of points $x$ such that all $N$-torsion points of the abelian variety $A_x$ are rational over $\mathbb F_q$ (i.e. $|A_x[N](\mathbb F_q)|=N^{2g}$), and $(A_x,\lambda_x)$ is the pair corresponding to $x.$ This equation gives some constraints (one for each $N$) that $|Aut(A,\lambda)|$ must satisfy. In particular, when $g=N=2$ and $q=3,$ we have $|A_{2,2}(\mathbb F_3)|=10$ and $|G|=720$ (in this case $G$ is the symmetric group $S_6$), and this becomes a puzzle of solving
$$
1/72 = \sum 1/n_i,
$$
and the $n_i$'s satisfy some additional conditions. Any idea on how to solve it? I'm considering the contributions of the two parts in $A_2,$ one for Jacobians of smooth genus 2 curves and one for Jacobians of stable singular ones $E_1\times E_2$. Any suggestion is appreciated.
Edit: Maybe it's easier to solve it over $\mathbb F_5,$ since the (orders of the) automorphism groups of smooth genus 2 curves over finite fields of characteristic 5 is known.

REPLY [10 votes]: If we start with the same problem over the complex numbers, then the
automorphism groups of principally polarised abelian varieties of dimension $g$
are exactly the stabilisers of $\mathrm{Sp}_{2g}(\mathbb Z)$ in its action on
the Siegel upper half space $\mathbb H_g$. These are finite groups and it is a
general fact that a finite subgroup of $\mathrm{Sp}_{2g}(\mathbb Z)$ fixes a
point of $\mathbb H_g$. Hence the answer in that case are that the automorphism
groups are exactly the maximal finite subgroups of $\mathrm{Sp}_{2g}(\mathbb
Z)$. Note that just as for $\mathrm{SL}_g(\mathbb Z)$ (which is a subgroup of
$\mathrm{Sp}_{2g}(\mathbb Z)$) the classification of such finite subgroups
quickly becomes untractable (for increasing $g$).
If we try to use this to get lower bounds for the case when the base field is
the closure of $\mathbb Z/p$ we can start with a pair $(A,G)$ of a p.p.a.v. and
a finite group $G$ of automorphisms over a field of characteristic. We can then
(after appropriately changing fields) assume that the base field is the fraction
field of a DVR $R$ of mixed characteristic with finite residue field of
characteristic $p$ and that $A$ has semi-stable reduction over $R$. The action
of $G$ extends and we then get $G$ as a subgroup of of the automorphism group of
a semi-abelian variety with a principal polarisation of the abelian part of
characteristic $p$. Note that if $G$ is large (for some definition of "large")
then the reduction is necessarily an abelian variety as the automorphism of a properly
semi-abelian variety is "smaller". This should give a lot of characteristic
$p$ examples. 
This however will not get everything as there are (already in the case of $g=1$)
pairs $(A,G)$ that don't lift to characteristic $0$. It does however give all
such pairs for which the order of $G$ is prime to $p$. In principle it should be
possible to resolve the problem with the approach suggested in Milne's
answer. However, I think there is a problem (apart from the fact that the algebraic problem
quickly becomes intractable) in that it is a somewhat tricky problem to figure
out which pairs consisting of a Rosati involution and a stable order in the
isogeny algebra are realisable by principal polarisations. Still there are some
constructions that can be used: One can take the product of two principally
polarised automorphism group. One can also tensor a principally polarised AV
with a positive definite unimodular hermitian form over the endomorphism ring
with the Rosati involution. The latter includes starting with a positive
definite integral unimodular form of rank $g$ and realising its automorphism
group as the automorphism group of the product of $g$ copies of any elliptic
curve. Starting instead with a supersingular elliptic curve probably give larger
examples.
The problem for finite fields instead of the algebraic closure of one is even
more unpredictable.<|endoftext|>
TITLE: Programming Languages Based on Category Theory
QUESTION [37 upvotes]: Since some computer scientists use category theory, I was wondering if there are any programming languages that use it extensively.

REPLY [2 votes]: The U.S. government has sponsored reseach in using category theory-based languages (e.g. https://www.kestrel.edu/home/projects/) with the hope specifying systems as the composition of theories using category theory to combine and refine these theories into executable robust code. The language used on some of the projects listed at the link above use the specification language Metaslang which has a subset which is executable. One side effect of this reseach led to our supporting the development of CRYPTOL, a good example of specifying a domain specific language using the functional language Haskell. In my use of Metaslang and the development environment (emacs/Specware), it is difficult to get "normal" programmers to use algebraic specification and functional programming. Despite being some of the nicest people I have met, Haskell programmers are definitely NOT normal. (;-)<|endoftext|>
TITLE: Cogroup Objects
QUESTION [16 upvotes]: Pretty much anyone who does algebra is familiar with group objects in categories, but what about cogroup objects? Most of what I've been able to find about them is that they "arise naturally in algebraic topology" (from wikipedia) and that, somehow, the n-sphere is one (nLab's meager entry).  Is there a reference for more on the stuff? Specifically wondering if the spaces of pointed maps from a topological space X to a pointed sphere are cogroups, and if anything is known about these "cohomotopy groups."

REPLY [6 votes]: This isn't a homotopy theory answer, but the most widely used type of cogroup in mathematics is an algebraic group.  An algebraic group is a cogroup object in the category of commutative rings, or in particular commutative algebras over a field.  What is important in this case is that people care a lot about the opposite category, the category of affine schemes, or in the field case affine algebraic varieties.  People care just as much about schemes and algebraic varieties in general, but if you want the algebraic groups analogue of connected semisimple Lie groups, then all examples are affine.
In my view, any time you endorse a co(something fancy) object in a category, you are really endorsing the opposite category in general.  Of course, at the formal level a category and its opposite category have exactly the same information, but conceptually it's a big leap.  For example, non-commutative geometry is largely the study of the opposite category of non-commutative rings.  Although in this case, people (correctly) use Hopf algebras as if they were cogroup objects, but they aren't!  The tensor product operation on non-commutative algebras is not a coproduct.<|endoftext|>
TITLE: Examples of algebraic stacks without coarse moduli space?
QUESTION [13 upvotes]: Keel-Mori's theorem says an algebraic stack with a finite diagonal over a scheme S has a coarse moduli space. What is an example of an algebraic stack without coarse moduli space?

REPLY [13 votes]: [A^1/Gm] is one example. You can check that any Gm invariant map from A^1 to a scheme is constant. Thus the map from [A^1/Gm]  to the point is universal for maps to schemes, but is not a bijection on geometric points (since [A^1/Gm] has two geometric points).
Check out Jarod Alper's thesis to learn more.<|endoftext|>
TITLE: When does "splits" imply "cosplits"?
QUESTION [7 upvotes]: In the category of groups, there are lots of "exact sequences", e.g. 4 → H → 2, that neither split nor cosplit, where H is the eight-element group of quaternions, and lots of sequences like 4 → D → 2 that split but do not cosplit, where D is the eight-element dihedral group.  By "2" and "4" I mean the cyclic groups of those orders.  By "exact sequence" A → B → C, I mean that A is the kernel of the quotient B → C (equivalently C is the cokernel of the subobject A → B).  A sequence A → B → C "splits" if there is a map C → B so that the compotision C → B → C is the identity; cosplitting is on the other side.
So in groups, a split exact sequence does not necessarily cosplit.  (In fact, I have a hard time thinking of any cosplit sequences.)  On the other hand, my friends who do ring theory state definitions like "A ring is semisimple if any short exact sequence of modules splits".  Why don't they ask for the sequence to cosplit?  Does that come for free?  (Or am I misremembering the definition?)
More generally, what conditions does one have to place on a category so that "splits" implies "cosplits"?

REPLY [5 votes]: So the fact that you had a hard time thinking of cosplit sequences of groups and the last question got me thinking (along the lines of Joel's comment actually)... what I came up with is probably standard (to people who well know it).
Suppose the sequence A-> B-> C is an exact sequence of groups. Then if it splits B is a quotient of the free product A*C (the coproduct in Grp) and if it cosplits B is a subgroup of the product AxC.
The idea (in the splitting case - this is enough since they are dual) is to use the universal property + the splitting to get a map from the coproduct. Then use the factorizations + exactness + element chase to check it is an epi in terms of right cancellation. There might be a slicker way to do this, but a way to get it for free from universal properties didn't occur to me.
This gives a philosophical explanation of why split sequences are easy to find, they are given by a presentation for B in terms of A and C (e.g. your dihedral group example). I think cosplitting seems a bit weirder since one is defining a group as a subgroup of a product - is this less natural to people who actually do group theory?
I haven't checked but I suspect that one cannot have a split and cosplit sequence in Grp - it seems like the fact that there is no biproduct should be an obstruction to this based on the above but I am not sure. The largest class of categories which springs to mind where Andrew's trick works would be quasi-abelian categories for strict exact sequences (so pretty much exact categories).
Finally I thought I'd point out some whackier examples where one has splitting iff cosplitting behaviour. In a triangulated category one has that every monomorphism is split and every epimorphism is split - in particular a triangle "splits" iff it "cosplits". The same is actually true for "exact sequences of triangulated categories". A fully faithful exact functor S -> T admits a right adjoint (cosplitting) iff T -> T/S admits a right adjoint (splitting).<|endoftext|>
TITLE: Does there exist a continuous function of compact support with Fourier transform outside L^1?
QUESTION [27 upvotes]: Let f be a complex-valued function of one real variable, continuous and compactly supported. Can it have a Fourier transform that is not Lebesgue integrable?

REPLY [2 votes]: Choose
$$f(x)=
\begin{cases}
\dfrac{\frac12 -x}{\log(x)},&0<x\leq1/2\\
0,&\text{otherwise}
\end{cases}$$
Then $$\operatorname*{Im} \int_{0}^{\infty} dk \ e^{-\varepsilon k}
\int_{-\infty}^{\infty} dx \ f(x) e^{-ikx} = \int_{0}^{\frac12}
dx \ \frac{\frac12 -x}{\log(x)} \cdot
\frac{-x}{\varepsilon^{2}+x^{2}} \to \infty\text{ for }
\varepsilon \downarrow 0.$$
Therefore the Fourier transform of $f$ is not in $L^{1}$ .<|endoftext|>
TITLE: What is new in MSC 2010?
QUESTION [6 upvotes]: Has anyone seen the new MSC 2010? I was browsing around and to my suprise there is another revision of MSC. Has anyone noticed any major changes in there? Do major journals already accept papers with MSC 2010 classification?

REPLY [3 votes]: You can get a more complete answer here:
www.msc2010.org<|endoftext|>
TITLE: Why do functions in complex analysis behave so well? (as opposed to functions in real analysis)
QUESTION [70 upvotes]: Complex analytic functions show rigid behavior while real-valued smooth functions are flexible. Why is this the case?

REPLY [14 votes]: Here is a related discussion from combinatorics and more: http://gilkalai.wordpress.com/2009/06/29/test-your-intuition-6/#comment-2057
A very nice explanation by John Baez (in two installments) can be found here
http://golem.ph.utexas.edu/category/2006/10/knowledge_of_the_reasoned_fact.html#c005571
and here
http://golem.ph.utexas.edu/category/2006/10/knowledge_of_the_reasoned_fact.html#c005623<|endoftext|>
TITLE: How hard is it to compute the number of prime factors of a given integer?
QUESTION [49 upvotes]: I asked a related question on this mathoverflow thread. That question was promptly answered. This is a natural followup question to that one, which I decided to repost since that question is answered.
So quoting myself from that thread:

How hard is it to compute the number of prime factors of a given integer? This can't be as hard as factoring, since you already know this value for semi-primes, and this information doesn't seem to help at all. Also, determining whether the number of prime factors is 1 or greater than 1 can be done efficiently using Primality Testing.

REPLY [83 votes]: There is a folklore observation that if one was able to quickly count the number of prime factors of an integer n, then one would likely be able to quickly factor n completely.  So the counting-prime-factors problem is believed to have comparable difficulty to factoring itself.
The reason for this is that we expect any factoring-type algorithm that works over the integers, to also work over other number fields (ignoring for now the issue of unique factorisation, which in principle can be understood using class field theory).  Thus, for instance, we should also be able to count the number of prime factors over the Gaussian integers, which would eventually reveal  how many of the (rational) prime factors of the original number n were equal to 1 mod 4 or to 3 mod 4.
Using more and more number fields (but one should only need polylog(n) such fields) and using various reciprocity relations, one would get more and more congruence relations on the various prime factors, and pretty soon one should be able to use the Chinese remainder theorem to pin down the prime factors completely.
More generally, the moment one has a way of extracting even one non-trivial useful bit of information about the factors of a number, it is likely that one can vary this procedure over various number fields (or by changing other parameters, e.g. twisting everything by a Dirichlet character) and soon extract out enough bits of information to pin down the factors completely.  The hard part is to first get that one useful bit...
[EDIT: The above principle seems to have one exception, namely the parity bit of various number-theoretic functions.  For instance, in the (now stalled) Polymath4 project to find primes, we found a quick way to compute the parity of the prime counting function $\pi(x)$, but it has proven stubbornly difficult to perturb this parity bit computation to find other useful pieces of information about this prime counting function.]<|endoftext|>
TITLE: The "miracle" of Heegard Floer.
QUESTION [13 upvotes]: Taking tori in symmetric products and "miraculously" proving that the Floer homology is independent of choices always seemed, well, miraculous.  Some time ago Max Lipyanski explained to me the origins of this construction from gauge theory on surfaces, a la Atiyah-Floer conjecture, which I have then forgotten. What is the origin of Heegard Floer?

REPLY [13 votes]: From Szabó's delightfully understated response (pdf) to receiving the Veblen prize:

The joint work with Peter Ozsváth
which is noted here grew out of our
attempts to understand Seiberg–Witten
moduli spaces over three-manifolds
where the metric degenerates along a
surface. This led to the construction of Heegaard Floer homology
that involved both
topological tools, such as Heegaard diagrams, and
tools from symplectic geometry, such as holomorphic
disks with Lagrangian boundary constraints.
The time spent on investigating Heegaard Floer
homology and its relationship with problems in
low-dimensional topology was rather interesting.

Of course, if one believes that Heegaard Floer homology is somehow the limit of monopole Floer homology as one degenerates the metric in some way that depends on the Heegaard diagram, then the independence of Heegaard Floer homology from the Heegaard diagram would fall out from the metric-independence of monopole Floer homology.  Unfortunately, I can't seem to find references that give any sort of precise picture of how Ozsváth and Szabó came to think that this should be the case  (though it might have been a baby analogue of the picture in this paperlink broken (pdf) by Yi-Jen Lee, written a few years later).
It perhaps bears mentioning that Heegaard Floer homology wasn't the first invariant that Ozsváth and Szabó constructed based on thinking about the interaction of the Seiberg-Witten equations with a Heegaard diagram—see The Theta Divisor and Three-Manifold Invariants and The theta divisor and the Casson–Walker invariant, which extract an invariant from the theta-divisor of the Heegaard surface, appear to have been based on thinking about what happens to the Seiberg–Witten equations when one has a neck $S\times [-T,T]$ ($S$ is the Heegaard surface) with the metric on $S$ at $t=-T$ itself having long cylinders over the compressing circles for one handlebody, while the metric on $S$ at $t=T$ has long cylinders over the compressing circles for the other handlebody.<|endoftext|>
TITLE: What is the geometric significance of Cartan's structure equations?
QUESTION [11 upvotes]: The Cartan structure equations for a connection and various associated 1-forms can be checked in a straightforward algebraic manner.  But is there a geometric or global significance to the equations- can one visualize the proof?

REPLY [6 votes]: There is a nice grand story behind all this. I don't know if you like thinking that way, but things do clarify when one looks at it from a more general perspective of oo-Lie algebroid valued differential forms with curvature.
I am still working on these entries, trying to expose some of my work with Jim Stasheff and Hisham Sati. If you bug me with questions, there is a good chance that I'll improve the exposition taylor-made for your needs.<|endoftext|>
TITLE: Introductory text on geometric group theory?
QUESTION [45 upvotes]: Can someone indicate me a good introductory text on geometric group theory?

REPLY [5 votes]: A book I completely appreciate is
"Geometric Group Theory", by Clara Loh.
It contains 250 exercises of varying difficulty including programming tasks.
She introduces the key notions from quasi-geometry, such as growth, hyperbolicity, boundary constructions and amenability.<|endoftext|>
TITLE: Maximal ideals in the ring of continuous real-valued functions on ℝ
QUESTION [25 upvotes]: For a compact space $K$, the maximal ideals in the ring $C(K)$ of continuous real-valued functions on $K$ are easily identified with the points of $K$ (a point defines the maximal ideal of functions vanishing at that point).
Now take $K=\mathbb{R}$. Is there a useful characterization of the set of maximal ideals of $C(\mathbb{R})$, the ring of continuous functions on $\mathbb{R}$? Note that I'm not imposing any boundedness conditions at infinity (if one does, I think the answer has to do with the Stone–Čech compactification of $\mathbb{R}$ — but I can't say I'm totally clear on that part either). Is this ring too large to allow a reasonable description of its maximal ideals?

REPLY [19 votes]: This isn't really an answer to your question, but I'd like to see it here next time I come looking, so I'll post it. The following result is basically Theorem 2.1 in C∞-differentiable spaces by Juan A. Navarro González and Juan B. Sancho de Salas.

Theorem: For any manifold M, the maximal ideals of C(M) whose residue field is ℝ is exactly in bijection with the points of M.

Proof: It's clear that points give you distinct maximal ideals with residue field ℝ, so we just need to show that every such ideal comes from a point. Suppose m is a maximal ideal in C(M) such that C(M)/m=ℝ and  ∩g∈m{g=0}=∅.
Choose a sequence of compact sets K1⊂K2⊂...⊂M such that Ki is in the interior of Ki+1 and M=∪Ki (you can do this since M is hausdorff and second countable). For each i, choose a function fi which is 0 on Ki but 1 outside of Ki+1, and define f=∑fi. Note that for any r∈ℝ, the set {x|f(x)=r} is a closed subset of some Ki, so it is compact.
Since we have a surjection C(M)→ℝ whose kernel is m, there is some r∈ℝ so that f-r∈m. Since ∩g∈m{g=0}=∅, the open sets {g≠0}g∈m is a cover of M, and in particular cover the compact set {f=r}. So there is some finite collection g1, g2, ..., gn∈m so that {g1=0}∩...∩{gn=0}∩{f=r}=∅. But then (g1)²+...+(g1)²+(f-r)²∈m is a nowhere vanishing function, so it is a unit, so m=C(M), a contradiction.<|endoftext|>
TITLE: Folding by Automorphisms
QUESTION [16 upvotes]: Background reading: John Stembridge's webpage.
The idea is that when you want to prove a theorem for all root systems, sometimes it suffices to prove the result for the simply laced case, and then use the concept of folding by a diagram automorphism to deduce the general case.
I have never seen an example of this in practice. So my question is: What are some (good) examples illustrating this technique?

REPLY [2 votes]: This is by now an old question, but a more recent example is given in a preprint by a current MIT graduate student
here.    I think the moral of the story is that there are quite a few directions in Lie theory where folding plays a significant role in getting from the simply-laced case to other cases.<|endoftext|>
TITLE: Motivation for BMO
QUESTION [13 upvotes]: At the moment, I don't have access to the early 1960's paper of John and Nirenberg that (from what I understand) introduced the space BMO (bounded mean oscillation). Why were John and Nirenberg interested in the space BMO? Were their interests motivated by PDEs? Did they initially make the connection with Hardy spaces? Any thoughts/input would be appreciated.

REPLY [4 votes]: I think the legend is that an official introduction to $BMO$ was given in the paper "Of Functions of Bounded Mean Oscillation" by F. John and L. Nirenberg , but the initial mention was in the paper by F. John "Rotation and Strain". F. John was looking at mappings and rotations. For example, the class of mappings $f(x)$ satisfying the following: 
\begin{equation*}
1-\epsilon '\leq \frac{f(y)-f(x)}{|y-x|} \leq 1+\epsilon '
\end{equation*}
for every $\epsilon '>\epsilon,~x\in\mathbb{R},~\exists \delta=\delta (x,\epsilon)$, and fixed $\epsilon <1$.
When looking at approximations for the derivatives of mappings in this class, F. John deduced some inequalities that look suspiciously like the definition for BMO. 
Both papers are available online via Communications in Pure and Applied Mathematics.<|endoftext|>
TITLE: Constructing a degeneration (as a group scheme) of G_m to G_a
QUESTION [13 upvotes]: SGA 3, expose 12, remark 1.6 says that one can easily construct a group scheme over a discrete valuation ring with generic fiber Gm and special fiber Ga. 
What is such an example?

REPLY [14 votes]: One more point of view :   if X is a 2x2 matrix, its centralizer in PGL(2) is -- for generic X -- isomorphic to Gm; however, it is isomorphic to Ga if X is nonzero and nilpotent.  This gives a typical naturally occurring family (you can, of course, restrict it to get a family over a dvr).<|endoftext|>
TITLE: question on sigma-fields
QUESTION [5 upvotes]: Let X,Y to be mappings from the sample space Ω to R and suppose Y is measurable with respect to σ(X), the smallest σ-field that makes X measurable. 
Does it follow that there exists some Borel-measurable function f: R → R such that 
Y=f(X)?

REPLY [2 votes]: It is trivially true but maybe worth noting that the converse is also true - if there exists such an f, then Y is σ(X)-measurable.
This and the question asked are theorem 20.1(ii) in Billingsley's Probability and Measure, 3rd edition.<|endoftext|>
TITLE: What does it mean to 'discharge assumptions or premises'?
QUESTION [21 upvotes]: When constructing proofs using natural deduction what does it mean to say that an assumption or premise is discharged? In what circumstances would I want to, or need to, use such a mechanism?
The reason I'm asking this question is that many texts on logic use this term as understood by the reader and don't take the time to adequately explain the technical sense in which they are using it.

REPLY [5 votes]: In the spirit of Kenny's observation, note also that we can formulate classical logic using a Peircian inference rule (equivalent to the usual theory in the presence of ex falso quodlibet) which clearly modifies the inferential properties of implication:

${{{A \rightarrow B} \atop \vdots} \atop A } \over A$

but in an odd way: is it an introduction rule?  But there is no logical structure in the conclusion.  Is it an elimination rule?  But the implication above the rule appears among the assumptions to the subderivation, not the conclusion.  It seems to be something like an implication elimination-eliminating rule, a kind of structurally double-negative introduction rule, where you can introduce logical structure by discharging it in the assumptions.
All the rules for adding classical-strength inference to the usual, well-behaved intuitionistic natural deduction involve inference rules that are eccentric in some way or another.  Parigot's lambda-mu calculus shows how the fundamental structural glue of (through a classical Curry-Howard correspondence) natural deduction can be tweaked to make it as good a fit for classical logic as the sequent calculus.
In chapter 3 of my DPhil dissertation (Stewart 2000), I give what I think is a successful reconstruction of Prawitz's inversion principle for the lambda-mu-based natural deduction, and show how this provides the basis for something we might call "classical constructivism", where the principle of the excluded middle is admitted as having constructive content by being a principle that provides a constructive, dialectical mediation between proofs and refutations.
Ref. Stewart (2000) On the formulae-as-types correspondence for classical logic.<|endoftext|>
TITLE: Discrete logs vs. factoring
QUESTION [25 upvotes]: One thing that I've never quite understood is why computing discrete logarithms (in the multiplicative group mod p) and factoring seem to be so closely related. I don't think that there's a reduction in either direction, at least when p is prime (although a quick literature search tells me that there's a probabilistic reduction in one direction), but there are a huge number of nontrivial similarities:

Neither of them are believed to have a classical polynomial-time algorithm, but essentially the same quantum algorithm works for both.
Both of them serve as an easy basis for public-key cryptography, while other problems (for instance, lattice-based problems) are much more difficult to turn into working cryptosystems, and some (graph isomorphism, computing the permanent) seem to have no hope whatsoever of being useful in crypto.
Many of the classical algorithms for one of them are closely related to a classical algorithm for the other -- number field sieve and function field sieve, Pollard's rhos, baby-step giant-step vs. Fermat-type methods, and so on.

Is there a simple reason why, even though in theory they aren't reducible to each other, in practice they behave as though they were?
Edit: To clarify further, as Rune points out, both factoring and discrete log can be seen as versions of the hidden subgroup problem over an abelian group. I understand why there are quantum algorithms for this, and some of the obstacles to quantum algorithms for nonabelian HSP, and even to a degree why HSP is good to base a public-key system around. My question is essentially why "they share a much closer bond that isn't seen with other abelian HSPs."

REPLY [5 votes]: Another relationship is that in both cases the fundamental structure is a ring of integers modulo some other integer m (prime or composite, as the case may be).  In such a ring there are distinguished "small prime" elements, 2,3,5, etc.  The reason that index calculus (and the Number Field Sieve, etc) works in both cases is that you can decompose (some) elements in the ring into these small prime elements, and then find relations between these decompositions.
On the other hand, on elliptic curves over (prime) finite fields there is no analogous "factor base," which is why the discrete log problem is so much harder in this case.
So I would say that the connection is not between discrete logs and factoring, but between the multiplicative group mod p and the multiplicative group mod N = pq.<|endoftext|>
TITLE: Are Q-curves now known to be modular?
QUESTION [18 upvotes]: I really should know the answer to this, but I don't, so I'll ask here.
A Q-curve is an elliptic curve E over Q-bar which is isogenous to all its Galois conjugates. A Q-curve is modular if it's isogenous (over Q-bar) to some factor of the Jacobian of X_1(N) for some N>=1 (here X_1(N) is the compact modular curve over Q-bar).
Has current machinery proved the well-known conjecture that all Q-curves are modular yet?
Remark: I know there are many partial results. What I'm trying to establish is whether things like Khare-Wintenberger plus best-known modularity lifting theorems are strong enough to give the full conjecture yet, or whether we're still waiting.

REPLY [18 votes]: Yes, this is a consequence of Serre's conjecture. The canonical reference is probably Corollary 6.2 of Ribet's paper on Q-curves:
http://math.berkeley.edu/~ribet/Articles/korea.pdf<|endoftext|>
TITLE: Infinite dimensional Newlander-Nirenberg theorem
QUESTION [7 upvotes]: The Newlander-Nirenberg theorem states that an almost complex structure is integrable if and only if the Nijenhuis tensor vanishes. 
I heard that this statement is not true in infinite dimensions, since for example the Loop space of a Riemannian 3-manifold is counterexample. (In fact, I think NN fails for Fréchet manifolds in general(?))
So my question is: 
Is the Newlander Nirenberg theorem valid for Banach- or Hilbertmanifolds? If not, is it possible to weaken the statement (or some conditions) such that it remains true for some class of infinite dimensional manifolds?
EDIT: Added the tag "open-problem", since NN for Hilbertmanifolds seems to be an open problem.

REPLY [5 votes]: As far as I understand, in a paper of Petyi [On the ∂-equation in a Banach space. Bull. Soc. Math. France 128 (2000), no. 3, 391–406.] it is shown that the NN theorem does not hold for Banach manifolds in general.  However, as you may know, the NN theorem has an easy proof when the almost complex structure is assumed to be real analytic.  In this case, the NN theorem is an easy consequence of the Frobenius theorem, which is true for Banach manifolds (reference? Smale?).  There is a paper by Daniel Beltita [http://arxiv.org/abs/math/0407395] which confirms that NN is true in this real-analytic case.  It would be interesting to see exactly where the NN proof breaks down for Banach manifolds, and exactly what one should assume to allow it to go through. But I am not aware of any such detailed work.<|endoftext|>
TITLE: When is a monic integer polynomial the characteristic polynomial of a non-negative integer matrix?
QUESTION [27 upvotes]: Suppose $P(x)$ is a monic integer polynomial with roots $r_1, ... r_n$ such that $p_k = r_1^k + ... + r_n^k$ is a non-negative integer for all positive integers $k$.  Is $P(x)$ necessarily the characteristic polynomial of a non-negative integer matrix?
(The motivation here is that I want $r_1, ... r_n$ to be the eigenvalues of a directed multigraph.)
Edit:  If that condition isn't strong enough, how about the additional condition that 
$$\frac{1}{n} \sum_{d | n} \mu(d) p_{n/d}$$
is a non-negative integer for all d?

REPLY [33 votes]: This question is completely answered, and the result is that the condition involving
the Moebius inversion you mention is both necessary and sufficient! See

K. H. Kim, N. Ormes, F. Roush. The spectra of nonnegative integer matrices via formal power series, J. Amer. Math. Soc. 13 (2000) 773–806. https://doi.org/10.1090/S0894-0347-00-00342-8

This is really a remarkable and
beautiful theorem.<|endoftext|>
TITLE: Minimal surface in a ball
QUESTION [22 upvotes]: Assume a minimal surface $\Sigma$ has boundary on the unit sphere in the Euclidean space
and $r$ is the distance from $\Sigma$ to the center of the ball.
Is it true that 
$$\mathop{\rm area} \Sigma\ge \pi\cdot(1-r^2).$$
Comments:

The problem is solved in all dimensions and codimension, see "Area bounds for minimal..." by Brendle and Hung in 2016. (Thanks Rbega for the ref.)
If $r=0$, the statement follows directly from the monotonicity formula.
If $\Sigma$ is topological disc the answer is YES, see answer of Oleg Eroshkin below.
The general question is formulated as a conjecture in 1975 --- see comment of Ian Agol.
There is an analog in all dimension and codimension for area minimizing surfaces, see Alexander, H.; Hoffman, D.; Osserman, R. Area estimates for submanifolds of Euclidean space. 1974.

REPLY [7 votes]: This has just been solved (in full generality) by Brendle and Hung using the first variation formula together with a clever (if mysterious) choice of vector field.<|endoftext|>
TITLE: Tate uniformization of nonsplit semistable elliptic curves
QUESTION [12 upvotes]: Let $E/\mathbf{Q}_p$ be an elliptic curve having split multiplicative reduction. Then Tate uniformization gives a surjective homomorphism of $p$-adic analytic groups $G_m \to E$, with infinite cyclic kernel. Is there an analogue of this fact for $E$ having nonsplit multiplicative reduction, perhaps replacing Gm with a nonsplit torus? E.g., can one uniformize $E$ over the quadratic extension where the reduction splits, and then somehow descend?
(My intuition was as follows. Take $E/\mathbf{Q}_p$ with nonsplit multiplicative reduction, and let $K/\mathbf{Q}_p$ be quadratic so that $E$ becomes split semistable over $K$, and let $E'$ be the $K$-twist of $E$ (which has split multiplicative reduction). Then one has a short exact sequence
$$0 \to Z \to \mathbf{G}_m \to E' \to 0$$
(where $Z$ is the constant analytic group of integers). Extending scalars to $K$ then applying Weil restriction of scalars, we get
$$0 \to X \to T \to A \to 0$$
where $X$ is an etale-locally-constant analytic group, $T$ is a torus, and $A$ is an abelian variety, each of rank $2$ in the appropriate sense. The latter short exact sequence contains the former short exact sequence as a sub (direct factor?); the quotient sequence should be something like
$0 \to Z' \to \mathbf{G}_m' \to E \to 0$,
where ' still denotes twisting by $K/\mathbf{Q}_p$. Since $Z'$ has trivial $\mathbf{Q}_p$-points, then, one should have something like $\mathbf{G}_m'(\mathbf{Q}_p) = E(\mathbf{Q}_p)$, modulo any descent used in forming the quotient.
Does this sound sensical?
If anyone has access to Google Wave and wants to discuss, I've set up a wave here:
https://wave.google.com/wave/#restored:wave:googlewave.com!w%252BQCn6fZTuZ

REPLY [4 votes]: A form of this is contained in Silverman, second book, Chapter V, Corollary 5.4. I guess
that the image of Gm' in E (at the level of Q_p-points) may have index 2.

REPLY [2 votes]: Most of it makes sense.  Elliptic curves with non-split reduction can be analytically uniformized by the norm torus.  There is a "nice" picture of this using the Berkovich spectrum for a non-split torus.  I have my doubts about the statement concerning rational points - you should have a Galois cohomology exact sequence.<|endoftext|>
TITLE: Are there any interesting connections between Game Theory and Algebraic Topology?
QUESTION [14 upvotes]: I've been learning game theory on my own and was just curious how it connected with previous things I've learned.  So are there any interesting connections between Game Theory and Algebraic Topology? or possibly other branches of topology?

REPLY [4 votes]: I'm coming from the other side: familiar with game theory and learning about algebraic topology.

I know Kakutani's fixed point theorem was used to simplify Nash's seminal paper.
If you think about a simple normal-form game like Prisoner's Dilemma or Coordination Game,  (source: Wayback Machine)

the arrows between the payoff matrix are governed by $>_A$ and $>_B$ meaning the ordering of payoffs by players A and B. I know topological spaces aren't necessarily orderable, but this seems topological in flavour since distance doesn't matter. (Whether you lose £3 or get your head chopped off, I don't care as long as it doesn't affect me.) I think of this as like a "topological gravity" creating the Nash equilibria.<|endoftext|>
TITLE: What's a non-abelian totally ordered group?
QUESTION [32 upvotes]: Because I have heard the phrase "totally ordered abelian group", I imagine there should be non-abelian ones.  By this I mean a group with a total ordering (not to be confused with a well-ordering) which is "bi-translation invariant": a < b should imply cad < cbd.
Does anyone know any examples?  
Totally ordered abelian groups are easy to come up with: any direct product of subgroups of the reals, with the lexicographic ordering, will do.  Knowing some non-abelian ones would help reveal what aspects of totally ordered abelian groups really depend on them being abelian...
Edit: Via Andy Putman's answer below, I found this great summary of results about ordered and bi-ordered groups (i.e. groups with bi-translation invariant orderings) on Dale Rolfsen's site:
Lecture notes on Ordered Groups and Topology
He shows numerous examples of non-abelian bi-orderable groups, including a bi-ordering (bi-translation invariant ordering) on the free group with two generators.  As well, he mentions, due to Rhemtulla, that a left-orderable group is abelian iff every left-ordering is a bi-ordering, which I think really highlights the relationship between ordering and abelianity.

REPLY [4 votes]: Thompson's group $F$ is totally ordered.  See here for a description of all possible bi-orderings.  Indeed, all diagram groups are totally orderable (see here).  The pure braided Thompson group $BF$ also bi-orderable (see here).<|endoftext|>
TITLE: No simple duplication formula for factorials?
QUESTION [8 upvotes]: Many special functions including the gamma function have a duplication formula of some sorts.  In the case of the gamma function it reads:

Gamma(2z) = Gamma(z) Gamma(z+1/2) 22z-1/Gamma(1/2)

On the other hand, there is no algebraic relation between Gamma(2z) and Gamma(z) by themselves, meaning that is there is no nonzero polynomial f(x,y) such that f(Gamma(2z),Gamma(z))=0 for all complex z.  I can prove this by chasing poles and their order.
However, I'd be interested in a (simple) argument which shows that the following similar statement is true (which I believe it is):

There is no (nonzero) polynomial f(x,y) such that f((2n)!, n!)=0 for all integers n≥0.

Any ideas?  Thank you!

REPLY [5 votes]: Armin,
Let me try to solve your original problem differently.
First write the wanted polynomial in the form $f(x,y)=\sum_kx^kA_k(y)$ where
the leading polynomial $A_0(y)$ is not identically zero (otherwise we can
always replace $f(x,y)$ by $f(x,y)/x^\ell$ for a suitable $\ell$). Denote by $N$
the degree of the polynomial $A_0(y)$. For any prime $p>N!$ the numbers
$0$ and $(-1)^kk!$, where $k=0,\dots,N-1$, are distinct residues modulo $p$, 
so that $p!\equiv 0\pmod p$ and $(p-k)!=(p-1)!/\prod_{j=1}^{k-1}(p-j)\equiv(-1)^k(k-1)!^{-1}\pmod p$
are pairwise noncongruent modulo $p$ as well. Substituting $x=(2p-2k)!\equiv0\pmod p$ 
and $y=(p-k)!$ for each $k=0,1,\dots,N$ into $f(x,y)=0$ and reducing modulo $p$, we obtain
$N+1$ different solutions of the polynomial equation 
$A_0(x)\equiv0\pmod p$, so that all coefficients of $A_0(x)$ are divisible by $p$.
Since this is true for any prime $p>N!$, the polynomial $A_0(x)$ is identically zero,
which contradicts our assumption.
Is it elementary enough?<|endoftext|>
TITLE: Logical endofunctors of Set?
QUESTION [20 upvotes]: What set-theoretic assumptions are necessary and sufficient to ensure the existence of a nontrivial (i.e. not isomorphic to the identity) endofunctor of the category Set which is logical (i.e. preserves finite limits and power objects—hence also finite colimits and exponentials)?
On the lower end, Andreas Blass proved ("Exact functors and measurable cardinals") that there exists a nontrivial exact endofunctor of Set (that is, preserving finite limits and colimits) iff there exists a measurable cardinal.  Since logical functors are a fortiori exact, the existence of a measurable cardinal is a necessary condition.  On the upper end, any nontrivial elementary embedding j:V→V surely induces a logical endofunctor of Set, so the existence of a Reinhardt cardinal is a sufficient condition.  But can it be pinned down more precisely?

REPLY [9 votes]: There is a set theoretic axiom due to Paul Corazza called the Wholeness Axiom, which is stated in the language of ZFC augmented by a single unary function symbol j. The axiom expresses, as a scheme, that j is a nontrivial elementary embedding from V to V. That is, we have the elementary axiom scheme, expressing "for all x, phi(x) iff phi(j(x))" and the nontriviality axiom, expressing "exists x, j(x) not=x" and the critical point axiom, expressing "there is a least ordinal kappa such that kappa < j(kappa)". 
Under this axiom, j really is an elementary embedding from the universe V to V, and so this presumably induces the kind of functor you want. 
The point is that the large cardinal consistency strength of this axiom is weaker than a Reinhardt cardinal. In fact, it is strictly below an I3 cardinal. 
But the situation with this axiom is not great, since the j you get will not be a definable class in the usual sense of ZF. Also, you will not have the Replacement Axiom in the full language with j. So to make use of the axiom, you in effect give up a little of what you mean by the existence of such a j.
There is an ambiguity, isn't there, in the question when you ask about the existence of a proper class object. What kind of existence is desired? The question is not directly formalizable in ordinary set theory, since the question is itself a quantification over proper classes (although Kelly Morse set theory would accommodate this). Do you want a definable class? Do you want a class in the sense of Goedel-Bernays? Having a relaxed attitude about this allows the large cardinal consistency strength of the answer to come down.<|endoftext|>
TITLE: Can Hom_gp(G,H) fail to be representable for affine algebraic groups?
QUESTION [16 upvotes]: Let $G$ and $H$ be affine algebraic groups over a scheme $S$ of characteristic 0 and let $\textbf{Hom}_{S,gp}(G,H)$ be the functor $T \mapsto \text{Hom}\_{T,gp}(G,H)$

Theorem (SGA 3, expose XXIV, 7.3.1(a)): Suppose G is reductive. Then 
  $\textbf{Hom}_{S,gp}(G,H)$ is representable by a scheme.

Can this fail if $G$ is not reductive? I worked out a few example with $G = \mathbb{G}_a$, but they were representable.

REPLY [17 votes]: There is a reasonable salvage, at least if the base scheme is a field: Hom(G,H) is a direct limit of representable subfunctors.  See Lemma A.8.13 in the book "pseudo-reductive groups" (where it is used to prove that the scheme-theoretic fixed locus for a linearly reductive group acting on a connected reductive group is always reductive (possibly disconnected) provided the base is a field.<|endoftext|>
TITLE: Crepant resolutions of toric varieties
QUESTION [9 upvotes]: Given a toric variety, is it easy to see if a crepant resolution exists?  If so, how can it be explicitly constructed?

REPLY [5 votes]: The first requirement is that the toric variety is $\mathbb Q$-Gorenstein, otherwise discrepancies and crepancy are not defined. Combinatorially, this means that for every cone of the fan $\Sigma$ the generators of the exceptional rays lie in a hyperplane. 
A resolution of such cone is crepant iff all new rays added lie on the aforementioned hyperplane for the cone of the original fan that they sit in. Nonsingularity is a statement that the finer fan is simplicial and unimodular.
While simpliciality is rather easy to satisfy, unimodularity is unlikely in dimension four or higher. It can be managed in dimension up to three by Pick's theorem.
One possible solution to this issue is to consider stacky resolutions which are smooth toric DM stacks that map birationally to the original toric variety. Then all one needs is a $\mathbb Q$-Gorenstein condition.<|endoftext|>
TITLE: How many dimensions is it safe to get drunk in?
QUESTION [17 upvotes]: In Michael Lugo's blog post Variations on the drunken-bird theorem, and real-world sightings he wonders (without coming to a conclusion) what the maximum 'safe' number of dimensions to get drunk in might be. 

So where's the line between "always finds his way home" and "gets lost forever"? It seems to me that this should be at 2+ε dimensions, since that's what we need to make the sum convergent, but I don't know if anyone's studied random walks on fractal spaces.

In the comments Matt Noonan recalls an unverified claim that the line is at 2.5 dimensions. 
Can anybody shed any further light on these musings?

REPLY [5 votes]: Indeed, the answer is 2+ϵ in most senses. For example, consider a wedge in Z3 which is all the points (x,y,z) such that |z|<f(x) for some specific increasing function f:N->N .
These wedges are transient already for pretty slowly growing functions. If you take f(x)=xϵ which correspond, in the sense of volume growth, to 2+ϵ dimensions, then it is transient for any ϵ>0.
One can get even more refined results by taking f(x)=log(x)a, in which case the wedge is transient if and only if a>1.
You can find more information in the book of Lyons with Peres.<|endoftext|>
TITLE: Questions about analogy between Spec Z and 3-manifolds
QUESTION [52 upvotes]: I'm not sure if the questions make sense: 
Conc. primes as knots and Spec Z as 3-manifold - fits that to the Poincare conjecture? Topologists view 3-manifolds as Kirby-equivalence classes of framed links. How would that be with Spec Z? Then, topologists have things like virtual 3-manifolds, has that analogies in arithmetics?
Edit: New MFO report: "At the moment the topic of most active interaction between topologists and number theorists are quantum invariants of 3-manifolds and their asymptotics. This year’s meeting showed significant progress in the field."
Edit: "What is the analogy of quantum invariants in arithmetic topology?", "If a prime number is a knot, what is a crossing?" asks this old report.
An other such question:
Minhyong Kim stresses
the special complexity of number theory: "To our present day understanding, number fields display exactly the kind of order ‘at the edge of chaos’ that arithmeticians find so tantalizing, and which might have repulsed Grothendieck." Probably a feeling of such a special complexity makes one initially interested in NT. Knot theory is an other case inducing a similar impression. Could both cases be connected by the analogy above? How could a precise description of such special complexity look like and would it cover both cases? Taking that analogy, I'm inclined to answer Minhyong's question
 with the contrast between low-dimensional (= messy) and high-dimensional (= harmonized) geometry. Then I wonder, if "harmonizing by increasing dimensions"-analogies in number theory or the Langlands program exist.    
Minhyong hints in a mail to "the study of moduli spaces of bundles over rings of integers and over three manifolds as possible common ground between the two situations". A google search produces an old article by Rapoport "Analogien zwischen den Modulräumen von Vektorbündeln und von Flaggen" (Analogies between moduli spaces of vector bundles and flags) (p. 24 here, MR). There, Rapoport describes the cohomology of such analogous moduli spaces, inspired by a similarity of vector bundles on Riemann surfaces and filtered isocrystals from p-adic cohomologies, "beautifull areas of mathematics connected by entirely mysterious analogies". (book by R., Orlik, Dat) As interesting as that sounds, I wonder if google's hint relates to the initial theme. What do you think about it? (And has the mystery Rapoport describes now been elucidated?)
Edit:
Lectures by Atiyah discussing the above analogies and induced questions of "quantum Weil conj.s" etc.
This interesting essay by Gromov discusses the topic of "interestung structures" in a very general way. Acc. to him, "interesting structures" exist never in isolation, but only as "examples of structurally organized classes of structured objects", Z only because of e.g. algebraic integers as "surrounding" similar structures. That would fit to the guesses above, but not why numbers were perceived as esp. fascinating as early as greek antiquity, when the "surrounding structures" Gromov mentions were unknown. Perhaps Mochizuki has with his "inter-universal geometry" a kind of substitute in mind?    
Edit: Hidekazu Furusho: "Lots of analogies between algebraic number theory and 3-dimensional topology are suggested in arithmetic topology, however, as far as we know, no direct relationship seems to be known. Our attempt of this and subsequent papers is to give a direct one particularly between Galois groups and knots."

REPLY [6 votes]: It seems the following remarks in M. Kapranov's paper http://arxiv.org/abs/alg-geom/9604018
page 64 bottom, has not been mentioned so far

According to the point of view going
  back to Y.I. Manin and B. Mazur, one
  should visualize any 1-dimensional
  arithmetic scheme X as a kind of
  3-manifold and closed points x ∈ X as
  oriented circles in this 3-manifold.
  Thus the Frobenius element (which is
  only a conjugacy class in the
  fundamental group) is visualized as
  the monodromy around the circle
  (which, as an element of the
  fundamental group, is also defined only
  up to conjugacy since no base point is
  chosen on the circle), Legendre
  symbols as linking numbers and so on.
  From this point of view, it is natural
  to think of the operators (algebra
  elements) af,x,d for fixed f and
  varying x, d as forming a free boson
  field Af on the “3-manifold” X; more
  precisely, for ±d > 0, the operator a
  ± f,x,d is the dth Fourier component
  of Af along the “circle” Spec(Fq(x)).
  The bosons a ± f,x,d and their sums
  over x ∈ X (i.e., the Taylor
  components of log Φ ± f (t)) will be
  used in a subsequent paper to
  construct representations of U in the
  spirit of [FJ].

It might be that recent paper by Kapranov and coauthors: 
http://arxiv.org/abs/1202.4073
The spherical Hall algebra of Spec(Z)
is somehow developing ideas quoted above.
The question which I heard from V. Golyshev and others many years ago is the following:
if Spec (Z) is analogous to 3-fold, what should be arithmetic analog of Chern-Simons theory ?<|endoftext|>
TITLE: Is Hodge theory somehow connected with a Galois group action Gal(C/R)?
QUESTION [20 upvotes]: I'm currently taking a course in Hodge theory ... and I wonder if all the splittings in $\{i,-i\}$ Eigenvalue pairs come from the Galois group action (of the extension $\mathbb{R}\rightarrow\mathbb{C}$) - it seems to me like that (and I couldn't find such a statement in my textbook).
Is this true? If yes, is this a good way to think of Hodge decomposition or does one need more data than just the Galois group? If not, what is my misconception?
I thought (if my assumption is true), this would be a way to generalize to other algebraic field extensions.. are there analogues of Hodge theory for any algebraic field extension? Does it involve the Galois group?
If this question isn't "researchy" enough, just close it ... I will come back asking questions in a year then :-)

REPLY [2 votes]: As far as I understand acc. to Gelfand/Manin "Homological Algebra" p. 140, the idea of Hodge structures comes from Galois representations in arithmetics and was then by Deligne's "yoga des poids" transfered to complex varieties. But there it is (or was? The book was written 20 years ago) unclear which symmetry group (called "Hodge symmetries" in the book) lurks behind it.<|endoftext|>
TITLE: Does every finitely generated group have a maximal normal subgroup?
QUESTION [12 upvotes]: Given an infinite group which is finitely generated, is there a proper maximal normal subgroup?

REPLY [2 votes]: So many answers! I'm completely lost. The paper of "B.H. Neumann, "Some remarks on infinite groups", Journal London Math. Soc, 12 (1937), 120-127" stated results for the existence of maximal subgroups, not maximal normal subgroup. Is this existence question of nontrivial normal subgroup still unsolved?<|endoftext|>
TITLE: a question on function fields
QUESTION [11 upvotes]: Consider the transcendental extension Q(t) of the field of rationals.
To Q(t) adjoin the root of the polynomial x^5+t^5=1. The resulting 
field Q(t)[x] is a radical extension of Q(t). Is it true that the 
only solutions to the equation X^5+Y^5=1 in the field Q(t)[x] are 
(0,1), (1,0), (t,x), (x,t), (1/t,-x/t) and (-x/t, 1/t)?
Comment: Using the ABC theorem one can prove that the Fermat
curve X^n+Y^n=1 does not have a non-trivial solution in Q(t) for n>2.
In particular in Q(t) the equation X^5+Y^5=1 does not have
non-trivial solutions.

REPLY [12 votes]: First, replace Q by the complex numbers C. 
Write A = C[x,y]/(x^n+y^n-1). Then the field you write down, call it K, is the fraction field of A, which is the function field of the Fermat Curve.
Finding a solution (X,Y) to x^n + y^n = 1 is equivalent to finding a map A --> K, where one sends x to X and y to Y. Any map to a field factors though A/P for some prime ideal P of A.
Since (x^n+y^n-1) is irreducible, the only primes in A are either (0) or maximal ideals m with A/m = C.
If A/P = C, then the map A --> K factors through C. Maps from A to C correspond to complex points on the Fermat curve.
If P = 0, then A --> K extends to a non-trivial map from K --> K. Giving a map between function fields is equivalent to giving a map of the corresponding smooth projective curves (in the opposite direction). Thus, the question becomes: what are the non-trivial maps from the Fermat curve to itself? For symmetry reasons, write the Fermat curve as x^n + y^n + z^n = 0 (over C).
Assume that n > 3. Since the genus of the Fermat curve is > 1, the Reimann-Hurwitz formula tells us that any non-trivial map must be an automorphism. On the other hand, the isomorphism group of the Fermat curve is the semi-direct product of (Z/nZ)^2 by S_3. (Explicitly, this corresponds to multiplying (x,y,z) by n-th roots of unity, and permuting the entries.)
In terms of the affine coordinates the S_3 action is generated by (x,y)->(y,x) and (x,y)->(1/y,-x/y).
To summarize, the K points are given by the (finitely many) automorphisms, and the C-points of the curve. You have noted all the automorphisms over Q, I would note that you are also invoking the non-trivial fact that the Fermat curve has no non-trivial points over Q, which (even for n = 5) that is harder than everything else in this answer (and is especially harder for general n > 2!). (The "ABC theorem" for polynomials only tells you that there are no non-constant solutions in C[t]. In general, a solution in C[t] will correspond to a map from P^1-->X, which can be non-trivial only if X has genus 0.)
Note that this argument used basically nothing about the Fermat curve itself --- any curve of genus > 1 has only finitely many automorphisms.
If g = 1, then there are some non-trivial maps from a curve of genus g to itself, but Riemann-Hurwitz tells us that they are all unramfied. X/C in this case is an elliptic curve, so there will be infinitely many non-trivial solutions in K, corresponding to the rational functions giving the multiplication by [n] map (or more if X has CM). In the Fermat case, this means that for n = 3 there will be many more solutions.
If g = 0, there are buckets of maps from P^1 to itself, as one knows in the Fermat case when n = 2 or 1.<|endoftext|>
TITLE: When can you desuspend a homotopy cogroup?
QUESTION [17 upvotes]: Any topological group $G$ has a classifying space, whose loopspace is a (homotopy) group which is homotopy equivalent to $G$ in a way that preserves the group structure.  More generally, if $G$ is an $A_\infty$-group (a space with a binary operation which satisfies the group axioms up to coherent homotopy), it similarly can be delooped to a classifying space.
Now suppose you have a cogroup in the category of pointed spaces.  If it is actually literally a cogroup, it's not hard to show it must be a point.  However, up to (coherent) homotopy, the suspension of any space is a cogroup.  Is the converse true?  Can you desuspend any $A_\infty$-cogroup?  Are there any examples of homotopy cogroups (possible not $A_\infty$) which are not suspensions?  More generally, are there any criteria that you can use to prove that a space does not have the homotopy type of a suspension?  The only one I know is that all cup products must vanish, but this also holds automatically for a homotopy cogroup (indeed, for any "co-H-space").

REPLY [9 votes]: Hopkins' result alluded to above gives a coordinate-free approach a lá Segal. The paper I wrote with Schwänzl and Vogt:
Comultiplication and suspension. Topology Appl. 77 (1997), no. 1, 1–18,
gives an actual co-operadic approach: we show that a 2-connected space which has a homotopy everything co-action by the Stasheff operad (i.e., a co-$A_\infty$-structure) is always a suspension.
This paper answers the first of the questions posed above.
There is an open question about this matter which I'd like to bring up. Hopkins claims his result holds for 1-connected spaces, but I and my co-authors could only get our result to work for 2-connected ones. Hopkins' proof, which was based on the Bousfield spectral sequence, never appeared (our proof involved the higher Blakers-Massey theorems). 
Hopkin's result is stated, but not proved, in: Formulations of cocategory and the iterated suspension. Algebraic homotopy and local algebra (Luminy, 1982), 212-226, Astérisque, 113-114, Soc. Math. France, Paris, 1984. Thus we don't know if the 1-connected case has really been settled or not.
Additional Note: Schwänzl, Vogt and I also show that if the space is highly connected with respect to its CW dimension (i.e., $(n-1)$-metastable), then the existence of a co-$A_n$-structure on it is tantamount to the existence a co-$A_\infty$-structure. It follows that it too desuspends.<|endoftext|>
TITLE: If Spec Z is like a Riemann surface, what's the analogue of integration along a contour?
QUESTION [35 upvotes]: Rings of functions on a nonsingular algebraic curve (which, over $\mathbb{C}$, are holomorphic functions on a compact Riemann surface) and rings of integers in number fields are both examples of Dedekind domains, and I've been trying to understand the classical analogy between the two.  As I understand it, $\operatorname{Spec} \mathcal{O}_k$ should be thought of as the curve / Riemann surface itself.  Is there a good notion of integration in this setting?  Is there any hope of recovering an analogue of the Cauchy integral formula?
(If I am misunderstanding the point of the analogy or stretching it too far, please let me know.)

REPLY [33 votes]: For a variety $X$ over a finite field, I guess one can take $\ell$-adic sheaves to replace differential forms. Then the local integral around a closed point $x$ (like integral over a little loop around that point) is the trace of the local Frobenius $\operatorname{Frob}_x$ on the stalk of sheaf, the so-called naive local term. Note that $\operatorname{Frob}_x$ can be regarded as an element (or conjugacy class) in $\pi_1(X)$, "a loop around $x$". The global integral would be the global trace map 
$$
H^{2d}_c(X,\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}(-d),
$$
and the Tate twist is responsible for the Hodge structure in Betti cohomology (or the $(2\pi i)^d$ one has to divide by). The Lefschetz trace formula might be the analog of the residue theorem in complex analysis on Riemann surfaces.
For the case of number fields, each closed point $v$ in $\operatorname{Spec} O_k$ still defines a "loop" $\operatorname{Frob}_v$ in $\pi_1(\operatorname{Spec} k)$ (let's allow ramified covers. One can take the image of $\operatorname{Frob}_v$ under $\pi_1(\operatorname{Spec} k)\to\pi_1(\operatorname{Spec} O_k)$, but the target group doesn't seem to be big enough). For global integral, there's the Artin-Verdier trace map $H^3(Spec\ O_k,\mathbb G_m)\to\mathbb{Q/Z}$ and a "Poincaré duality" in this setting, but I don't know if there is a trace formula. The fact that 3 is odd always makes me excited and confused.
So basically I think of trace maps (both local and global) as counterpart of integrals. Correct me if I was wrong.<|endoftext|>
TITLE: Relation between Lie Algebra Cohomology and Number of Relations of a Cyclic Module?
QUESTION [8 upvotes]: Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $k$, let $U$ be its enveloping algebra, and let $M$ be a $\mathfrak{g}$-module (not necessarily finite dimensional).  Call the invariant dimension of $M$ the largest $i$ such that $\operatorname{Ext}^i_U(k,M)\neq 0$.  This is the same as the degree of the largest non-vanishing Lie algebra cohomology group.
Here are two equivalent statements of my question.

If $M$ is a cyclic $\mathfrak{g}$-module (that is, its generated by a single element as a $U$-module), then is the number of relations of $M$ always greater than the invariant dimension minus $1$?
If $I$ is a left ideal in $U$, then is the number of generators of I always greater than the invariant dimension?

The reason why I would suspect such a thing is that it is true in the case of an abelian Lie algebra; that is, when $U$ is a polynomial ring.  In this case, the invariant dimension of an ideal $I$ coincides with the height of $I$, and so (2) becomes Krull's height theorem (and (1) follows immediately from (2)).

REPLY [3 votes]: This is FALSE as stated, even for the augmentation ideal. For example, consider the 3-dimensional Heisenberg Lie algebra g and the trivial module k. Then it's easy to see that the third (=top) cohomology is non-zero, but due to the noncommutativity of g, the left ideal I=annihilator of k may be generated by only two elements. The inequality proposed by Simon, in fact, goes the other way: the number of generators for I is at most the number of generators of gr I.
Admittedly, this is a cheap shot, since the right notion of the number of generators for a left ideal I of U(g) is the minimal dimension of an ad(g)-invariant subspace of U(g) generating I - in particular, this number survives the passage to the associated graded. I don't know whether the modified statement is true.<|endoftext|>
TITLE: One-parameter semigroups of bimodules
QUESTION [5 upvotes]: Suppose M is a von Neumann algebra.
Consider a monoidal category of bimodules over M.
Here a bimodule is a Hilbert space with two normal representations of M.
The monoidal structure is given by Connes' fusion.
Alternatively we can take right W*-correspondences (right Hilbert W*-modules over M with a normal left action of M)
together with the completed algebraic tensor product.
What can we say about one-parameter “semigroups” of such bimodules?
More precisely, consider a family E of bimodules parametrized by a real number t > 0
such that E_s ⊗ E_t = E_{s+t}.  Equality here denotes isomorphism.
Does it have an “infinitesimal generator”?
What is the “exponent” then?
Do we need any “continuity” conditions to guarantee good properties of such “semigroups”?
What kind of additional restrictions we can obtain on M provided that such a family exists?
It appears to me that such objects are known as “continuous tensor product systems”.
Any references on this matter will be appreciated.

REPLY [6 votes]: Such structures have been investigated at depth (welcome to the club!). Let me try to answer some of your questions.
1) First, let me suggest that you look at Bill Arveson's book on this subject, Noncommutative Dynamics and E-semigroups. Arveson is considered the pioneer of modern work on these structures, and you will enjoy his book.
2) An "infinitesemal generator" can mean various things, but in some sense, remarkably, it doesn't always exist. Arveson's book contains some treatment of this issue, known as the existence of type III examples, which are due to R. T. Powers and B. Tsirelson (there is also recet work of M. Izumi).
3) Do we need continuity conditions? Measurability conditions suffice. The condition is that the bundle {E_t}_{t>0} be isomorphic, as a measurable bundle of Hilbert spaces, to the trivial bundle (0,infinity) X H_0, with H_0 some fixed Hilbert space, plus compatibility of the measurable structure with addition, multiplication, etc.
4) Extensive research has been carried out also in the case where the E_t are Hilbert bimodules (C^*-correspondences). Search the ArXiv for works of Michael Skeide, or Paul Muhly and Baruch Solel.
5) I should also mention: these product systems arise naturally in the study of, give rise to, and are in a one-to-one correspondence with semigroups of *-endomorphisms on von Neumann algebras.<|endoftext|>
TITLE: Why are local systems on a complex analytic space equivalent to vector bundles with flat connection?
QUESTION [15 upvotes]: Let $X$ be a complex analytic space. It is a 'well known fact' that the categories of local systems on $X$ (i.e. locally constant sheaves with stalk $C^n$), and of (holomorphic) vector bundles on $X$ with flat connection, are equivalent. I've been looking for a proof of this, but every reference I can find merely says something like 'this is well known' without further argument. Does anyone know of a proof?

REPLY [10 votes]: You might also want to read Carlos Simpson's paper "Moduli of representations of the fundamental group of a smooth projective variety", parts I and II. He explains in great detail how to make the set of objects of each of these categories into the points of an algebraic variety, and why these algebraic varieties are analytically but not algebraically isomorphic. He doesn't use the category theoretic perspective much, as I recall, but he is invaluable for understanding how to work with concrete moduli spaces.<|endoftext|>
TITLE: Classification problem for non-compact manifolds
QUESTION [46 upvotes]: Background
It is well-known that the compact two-dimensional manifolds are completely classified (by their orientability and their Euler characteristic).
I'm also under the impression that there is also a classification for compact three-dimensional manifolds coming from the proof of the Geometrization Conjecture and related work.
Unfortunately for $n\ge4$ no similar classification is possible because it can be shown that it is at least as hard as the word problem for groups. Thus for higher-dimensional manifolds we instead focus on classifying all the simply-connected compact manifolds.
My question
Why in these "classification problems" are we only considering compact manifolds? Is there an easy reason why we restrict ourselves to the classification of compact manifolds? Does a classification of general (not necessarily compact) manifolds follow easily from a classification of compact manifolds?

REPLY [6 votes]: To illustrate both sides of all of the above comments (non-compact builds on compact, vs. non-compact is much harder than compact), I point out that ALL connected, separable, metric 2-manifolds (with or without boundary) meaning all connected, separable, metric spaces in which every point has an open neighborhood homeomorphic to the closed 2-disk have been classified.  See Brown, Edward M.; Messer, Robert: The classification of two-dimensional manifolds.  Trans. Amer. Math. Soc. 255 (1979), 377–402.  MR0542887 
The point is that the analysis is rather delicate and has some surprises, showing that the non-compact case does not follow easily from the compact.  Given the examples in dimension 3 by McMillan mentioned by algori, one of the surprises is that it could be done at all.<|endoftext|>
TITLE: Universal definition of tangent spaces (for schemes and manifolds)
QUESTION [44 upvotes]: Both schemes and manifolds are local ringed spaces which are locally isomorphic to spaces in some full subcategory of local ringed spaces (local models). Now, there is the inherent notion of the Zariski tangent space in a point (dual of maximal ideal modulo its square) which is the "right" definition for schemes and for $C^\infty$-manifolds (over $\mathbb{R}$ and $\mathbb{C}$). But for $C^r$-manifolds over $\mathbb{R}$ with $r<\infty$ this is not the correct definition. Here one has to take equivalence classes of $C^r$-curves through the point. Isn't there some general definition of tangent spaces which is always the right one?
I am also not completely sure what "right" means. So far, I think that one wants the dimension of the tangent space to be equal to the dimension of the point. This is for example the problem with the Zariski tangent spaces for $C^r$-manifolds. Can this failure be explained geometrically?

REPLY [6 votes]: Steven already explained a bit about the key to the answer: Synthetic differential geometry (cf. nLab where the hints on a higher categorical analogue are also present), but I would like to put it in a much broader perspective, though more in words and references than really explaining, mainly due to space, time and expertise limits. 
While there are mentions, in several of the answers above, of the (possibly relative) module of Kähler differentials leading to the algebraic version of cotangent space used in algebraic and analytic geometry; this partial notion predates a little bit the more fundamental work of Grothendieck, who invented an inherent geometrical way to found a differential calculus in geometry. Similarly to the differentiation in topological vector spaces, the basic idea is to approximate the maps with linear maps, but this time Grothendieck considered maps among sheaves of $\mathcal{O}$-modules over schemes; he described the linearization in the language of operations on sheaves in terms of infinitesimal neighborhoods of the diagonal $\Delta\subset X\times X$; these are described in terms of nilpotent elements in the structure sheaf; one can also define the related notion of infinitesimally close generalized points; the infinitesimal neighborhoods build up an increasing filtration, which induces a dual filtration on the hom-spaces, so called differential filtration. The union of the differential filtration is the differential part of the hom-bimodule, and its elements are regular differential operators. A "crystaline" variant of the picture related to divided powers leads to appropriate treatment of differential calculus in positive characteristics. The notion of crystal of quasicoherent sheaves is bases on the notion of infinitesimally closed generalized points; the geometric picture with pullbacks of sheaves, leads to a definition as sort of descent data, cf. 

P. Berthelot, A. Ogus, Notes on crystalline cohomology, Princeton Univ. Press 1978. vi+243. 
J. Lurie, Notes on crystals and algebraic $D$-modules, in Gaitsgory's seminar, pdf

These are a dual point of view on D-modules. The descent data in abelian context are equivalent to certain formally defined connection operator, called Grothendieck connection in this case. There are now abstract versions of the algebraic correspondence between descent data in abelian context and flat Kozsul connections for the associated "Amitsur" complexes (work of Roiter, T. Brzeziński and others, cf. connection for a coring). On the other hand, Grothendieck immediately came up with nonlinear version of crystals (crystals of schemes) which are dual point of view on what some now call D-schemes. 
Grothendieck's point of view on differential calculus has been soon after the discovery at the end of 1950s, introduced in works of Malgrange, Kodaira and Spencer in the development of obstruction and deformation theory for differential equations. Both works together in late 1960s motivated Lawvere, Kock and Dubuc to extend that geoemtric approach to differential calculus into differential geometry. Dubuc introduced $C^\infty$-schemes as yet another approach to manifolds, in the spirit of the theory of schemes. Lawvere did not look only at then recent work of Grothendieck (and Malgrange, Spencer...), but also at classical work on "synthetic geometry". This is a terminology which requires caution: in 19th and early 20th century, synthetic was viewed as differing from coordinatized, analytic, and pertained to either work from axioms, not referring to coordinate and even metric aspects, and some people in axiomatic descriptive geometry refer to their geometry as synthetic even now in that "clean", but less powerful sense. Another sense is that it is close to the engineering point of view that the path of a particle can be considered either as a point in the space of paths or as a map from interval into the space, what implies that the infinite-dimensional spaces of paths should exist and one should have the exponential law, i.e. we need to embed our category of spaces into closed monoidal category; there are many such embeddings of the category of manifolds available now, and some models of them offer the model $D$ of infinitesimals, which represents the functor of taking the tangent space in particular. This model has been shaped with having in mind the Grothendieck's field of dual numbers in algebraic geometry, but the language and multiplicity of models made it very flexible in the approach of the synthetic differential geometry of Kock and Lawvere. First of all they had an independent axiomatic approach as well as study of the topos theoretic models; including the study of Cahiers topos which is even more faithful to the Grothendieck's point of view. In all these models, they had nilpotent infinitesimals, like in scheme theory, but not like infinitesimals in nonstandard analysis. More recent approach of Moerdijk-Reyes offered both nilpotent and non-nilpotent infinitesimals (though possible variant related to nonstandard analysis is not known to me). For a differential geometer there are many attractive tools in synthetic differential geometry like infinitesimal simplices, enabling intuitive and effective of many quantities involving differential forms and geometry. 
On the other hand, the usual differential geometry is faithfully embedded into synthetic models, so one is bound to be conservative, i.e. not to get results about usual notions in manifolds theory which are inconsistent with the usual definitions. One just gets more intuitive and technical power. 
I should also mention that the Grothendieck's picture with the infinitesimal thickenings, aka resolutions of diagonals, leading to differential calculus, can be extended to noncommutative spaces rerpesented by abelian categories "of quasicoherent modules". This has been done in 1996 preprints

V. A. Lunts, A. L. Rosenberg, Differential calculus in noncommutative algebraic geometry I. D-calculus on noncommutative rings, MPI 1996-53 pdf, II. D-Calculus in the braided case. The localization of quantized enveloping algebras, MPI 1996-76 pdf

The resulting definition of the rings of regular differential operators on noncommutative rings has been used in the study toward the Beilinson-Bernstein correspondence for quantum groups in later published two articles, which however skip the geometric derivation of the definition of the differential operators used:

V. A. Lunts, A. L. Rosenberg, Differential operators on noncommutative rings, Selecta Math. (N.S.)  3  (1997),  no. 3, 335--359 (doi); sequel: Localization for quantum groups, Selecta Math. (N.S.) 5 (1999), no. 1, pp. 123--159 (doi).

Somewhat similar analysis in infinite-categorical setup of a $(\infty,1)$-version of the Cahiers Topos is in recent master diploma

Herman Stel, ∞-Stacks and their Function Algebras – with applications to ∞-Lie theory, Utrecht 2010, webpage, pdf

under the guidance of Urs Schreiber. This work leads to a correct theory of higher Lie algebroids.<|endoftext|>
TITLE: How to distinguish between natural and unnatural equivalences of categories
QUESTION [9 upvotes]: Some equivalences of categories are constructed by explicitly giving a pair of functors that are inverses up to isomorphism. For example, the equivalence between CRing^op and affine schemes is given by the pair (Spec, GlobalSections). I'd say these are "natural", since no choices are made.
Another equivalence of categories is between finite dimensional vector spaces and the category consisting of one vector space of each dimension. The functor in one direction is just the inclusion, but the inverse requires making a bunch of choices. I'd say this is "unnatural". 
But my definitions of "natural" and "unnatural" aren't precise. I suppose one of the triumphs of category theory has been the ability to make precise the definition of natural in some contexts. So my question is: how can I make this precise?

REPLY [4 votes]: On the other hand to Tom's answer, the distinction between strong and weak equivalences disappears if you use anafunctors instead of functors, which is arguably the "right" way to do category theory in the absence of choice.  Someone who takes this point of view would argue that the inverse (ana)functor from FDVect to the category of (say) Euclidean spaces actually doesn't involve any unnatural choices, because every time you have to choose something, all of the things you have to choose from are uniquely canonically isomorphic.
It's precisely analogous to how if a category C has binary products, you need to make a bunch of "choices" to define a product functor C × C → C, but in each case the "category of possible choices" is contractible and so you aren't really making any "contentful" choices at all.  And actually, depending on what foundational axiom system you use, your example of a "natural" equivalence might actually involve this sort of "contentless" choice as well.  The underying space of Spec R is some subset of P(R) (the prime ideals), but in a categorical set theory like ETCS, the power-object P(R) is only characterized up to isomorphism.
Thus, someone who takes this point of view might claim that the distinction you want to draw is actually an illusory one.<|endoftext|>
TITLE: Consequences of Geometric Langlands
QUESTION [43 upvotes]: So, lots of people work on the Geometric Langlands Conjecture, and there have been a few questions around here on it (admittedly, several of them mine).  So here's another one, tagged community wiki because there isn't really a "right" answer: what does GLC imply? Lots of big conjectures have well known consequences (Riemann Hypothesis and distribution of primes) but what about GLC? Are there any nice things that are known to follow from this equivalence of derived categories?
EDIT: The Geometric Langlands Conjecture says the following: Let $C$ be an algebraic curve (any field, though I think the formulation I know is only good in characteristic 0), $G$ a reductive algebraic group, $^L G$ its Langlands dual (the characters of G are cocharacters of $^L G$, if I recall correctly).  Then there's a natural equivalence of categories from the derived category of coherent sheaves on the stack of $G$-local systems to the derived category of coherent $\mathcal{D}$-modules on the moduli stack of principal $^L G$-bundles, such that the structure sheaf of a point is sent to a Hecke Eigensheaf (and I'm not going to sit down and define that on top of the rest here...the idea is that $G$-local systems on the curve are equivalent to eigensheaves for some collection of operators, but actually making it precise and having a hope of being true gets technical)
Edit 2: This paper states one version of the conjecture (for $GL(n)$ only) as 1.3, after defining the Hecke operators.

REPLY [59 votes]: OK, this is a very broad question so I'll be telegraphic.
There is a sequence of increasingly detailed conjectures going by the name GL -- it's really a "program" (harmonic analysis of $\mathcal{D}$-modules on moduli of bundles) rather than a conjecture -- and only the first of this sequence has been proved (and only for $GL_n$), but I don't want to get into this.
There are several kinds of reasons you might want to study geometric Langlands:

direct consequences. One application is Gaitsgory's proof of de Jong's conjecture (arXiv:math/0402184). If you prove the ramified geometric Langlands for $GL_n$, you will recover L. Lafforgue's results (Langlands for function fields), which have lots of consequences (enumerated eg I think in his Fields medal description), which I won't enumerate. (well really you'd need to prove them "well" to get the motivic consequences..)
In fact you'll recover much more (like independence of $l$ results). To me though this is the least convincing motivation..
Original motivation: by understanding the function field version of Langlands you can hope to learn a lot about the Langlands program, working in a much easier setting where you have a chance to go much further. In particular the GLP (the version over $\mathbb{C}$) has a LOT more structure than the Langlands program -- ie things are MUCH nicer, there are much stronger and cleaner results you can hope to prove, and hope to use this to gain insight into underlying patterns.

By far the greatest example of this is Ngo's proof of the Fundamental Lemma --- he doesn't use GLP per se, but rather the geometry of the Hitchin system, which is one of the key geometric ingredients discovered through the GLP. To me this already makes the whole endeavor worthwhile..

Relations with physics. Once you're over $\mathbb{C}$, you (by which I mean Beilinson-Drinfeld and Kapustin-Witten) discover lots of deep relations with (at least seemingly) different problems in physics. 

a. The first is the theory of integrable systems -- many classical integrable systems fit into the Hitchin system framework, and geometric Langlands gives you a very powerful tool to study the corresponding quantum integrable systems. In fact you (namely BD) can motivate the entire GLP as a way to fully solve a collection of quantum integrable systems. This has has lots of applications in the subject (eg see Frenkel's reviews on the Gaudin system, papers on Calogero-Moser systems etc).
b. The second is conformal field theory (again BD) --- they develop CFT (conformal, not class, field theory!) very far towards the goal of understanding GLP, leading to deep insights in both directions (and a strategy now by Gaitsgory-Lurie to solve the strongest form of GLP).
c. The third is four-dimensional gauge theory (KW). To me the best way to motivate geometric Langlands is as an aspect of electric-magnetic duality in 4d SUSY gauge theory. This ties 
in GLP to many of the hottest current topics in string theory/gauge theory (including Dijkgraaf-Vafa theory, wall crossing/Donaldson-Thomas theory, study of M5 branes, yadda yadda yadda)...

Finally GLP is deeply tied to a host of questions in representation theory, of loop algebras, quantum groups, algebraic groups over finite fields etc. The amazing work of Bezrukavnikov proving a host of fundamental conjectures of Lusztig is based on GLP ideas (and can be thought of as part of the local GLP). (my personal research program with Nadler is to use the same ideas to understand reps of real semisimple Lie groups). This kind of motivation is secretly behind much of the work of BD --- the starting point for all of it is the Beilinson-Bernstein description of reps as $\mathcal{D}$-modules.

There's more but this is already turning into a blog post so I should stop.<|endoftext|>
TITLE: Complexity of determining if two graphs have same cycle matroid?
QUESTION [5 upvotes]: Consider the following question:
Input: Two graphs G1 and G2
Question: Is the cycle matroid M(G1) isomorphic to the cycle matroid M(G2)
What is the complexity of this question?
It is well known that the two cycle matroids are isomorphic if and only if the graphs are "2-isomorphic" which means that there is a sequence of "Whitney flips" (where a graph is disconnected at a 2-vertex cutset and then reconnected with one of the pieces flipped) from one to an isomorph of the other.
This suggests that the complexity should be the same as graph isomorphism, but I cannot find a reference to this.

REPLY [4 votes]: The following paper seems to show that this problem is polynomial equivalent to graph isomorphism (see section 5):
http://arxiv.org/abs/0811.3859<|endoftext|>
TITLE: equivalence of Grothendieck-style versus Cech-style sheaf cohomology
QUESTION [95 upvotes]: Given a topological space $X$, we can define the sheaf cohomology of $X$ in 
I. the Grothendieck style (as the right derived functor of the global sections functor $\Gamma(X,-)$)
or
II. the Čech style (first by defining the Čech cohomology groups subordinate to an open cover, and then taking the direct limit of these groups over all covers).
When exactly are these two definitions equivalent? I'm unhappy with the explanation given by Hartshorne. Are they the same for any paracompact Hausdorff space? Or a locally contractible space?
And what is the relationship between these two sheaf cohomologies and singular cohomology?
Any elaboration on this circle of ideas related to the relationship between all the different cohomology theories would be appreciated.

REPLY [4 votes]: Here are some positive results and counterexamples for etale cohomology.
Definition: Let $X$ be a scheme. Say that $X$ has property "$AF_{n}$" if for every collection $x_{1},\dotsc,x_{n} \in X$ of $n$ points of $X$ there exists an affine open subscheme $U \subseteq X$ containing all the $x_{i}$. Let $$ a(X) $$ denote the supremum of the positive integers $n$ for which $X$ has $AF_{n}$. We say that $X$ is an "AF-scheme" (or "FA-scheme") if $a(X) = \infty$.
See Gross [4], Section 2.
The relevant theorem is:
Theorem (Artin 1971 [1]): Let $X$ be a quasi-compact AF-scheme. Then for any abelian sheaf $\mathscr{F}$ on the etale site of $X$, the map $$ \check{\mathrm{H}}^{i}(X,\mathscr{F}) \to \mathrm{H}_{\mathrm{et}}^{i}(X,\mathscr{F}) $$ is an isomorphism for all $i$.
Milne [6, III, Theorem 2.17] also discusses Artin's proof. This is more a claim about the etale topology on $X$, namely that if $U \to X$ is an etale cover, then any etale cover $V \to U \times_{X} \dotsb \times_{X} U$ may be refined by an etale cover of the form $U' \times_{X} \dotsb \times_{X} U'$ for $U' \to X$ which refines $U \to X$.
Later Schröer refined Artin's result as follows:
Theorem (Schröer 2003 [7]): Let $X$ be a Noetherian scheme and let $n$ be an integer such that $n \le a(X)$. Then for any abelian sheaf $\mathscr{F}$ on the etale site of $X$, the map $$ \check{\mathrm{H}}^{i}(X,\mathscr{F}) \to \mathrm{H}_{\mathrm{et}}^{i}(X,\mathscr{F}) $$ is an isomorphism for $i \le n$ and an injection for $i = n+1$.
Example: For an example of a scheme $X$ and an abelian sheaf $\mathscr{F}$ on the etale site of $X$ for which the map $$ \check{\mathrm{H}}^{2}(X,\mathscr{F}) \to \mathrm{H}_{\mathrm{et}}^{2}(X,\mathscr{F}) $$ is not an isomorphism, see the answers to this question. The current answers discuss the cases (1) $\mathscr{F}$ is a constant sheaf and (2) $\mathscr{F} = \mathcal{O}_{X}$.
(Gabber) For an example of a scheme $X$ for which the map $$ \check{\mathrm{H}}^{2}(X,\mathbb{G}_{m}) \to \mathrm{H}_{\mathrm{et}}^{2}(X,\mathbb{G}_{m}) $$ is not an isomorphism, let $R$ be a normal noetherian strictly henselian local ring of dimension $\ge 2$ whose punctured spectrum $U$ has nonzero Picard group (see e.g. this and this for examples of such $R$), and let $X$ be the gluing of two copies of $\operatorname{Spec} R$ along $U$. This is a local version of the counterexample to $\operatorname{Br} = \operatorname{Br}'$ due to Edidin, Hassett, Kresch, Vistoli [9].

Here are some conditions relevant to the AF-property:
Lemma (e.g. [8, 01ZY]): Let $X$ be a quasi-compact scheme admitting an ample line bundle. Then $X$ is an AF-scheme.
This is essentially the graded prime avoidance lemma.
The Chevalley conjecture, proved by Kleiman, states that for smooth proper varieties the converse is true:
Theorem (Kleiman 1966 [5]): Let $k$ be an algebraically closed field and let $X$ be a smooth proper $k$-scheme. If $X$ is an AF-scheme, then $X$ is projective over $k$.
Benoist proved the following generalization of Kleiman's result:
Theorem (Benoist 2013 [2]): Let $k$ be an algebraically closed field and let $X$ be a normal, finite type $k$-scheme. If $X$ is an AF-scheme, then $X$ is quasi-projective over $k$.
Theorem (Farnik 2013 [3, Theorem 2.2]) For every integer $n \ge 2$ there is a smooth proper variety $X$ with $a(X) = n$.

References:
[1] M. Artin, "On the joins of Hensel rings", Advances in Mathematics 7 (1971) pp 282–296, doi:10.1016/S0001-8708(71)80007-5, core.ac.uk.
[2] O. Benoist, "Quasi-projectivity of normal varieties", International Mathematics Research Notices, vol 2013, no 17 (2012) pp 3878–3885, doi:10.1093/imrn/rns163, arXiv:1112.0975.
[3] M. Farnik, "On strengthening of the Kleiman-Chevalley criterion", Proceedings of the AMS 141 no 11 (2013) pp 4005-4013, doi:10.1090/S0002-9939-2013-11695-3.
[4] P. Gross, "Tensor generators on schemes and stacks", Algebraic Geometry 4 (4) (2017) pp 501–522, doi:10.14231/ag-2017-026, arXiv:1306.5418.
[5] S. Kleiman, "Toward a numerical theory of ampleness", Annals of Mathematics 84 No. 3 (1966) pp 293–344, doi:10.2307/1970447.
[6] J.S. Milne, Etale Cohomology, Princeton University Press (1980) JSTOR (subscription needed).
[7] S. Schröer, "The bigger Brauer group is really big", Journal of Algebra  262 (2003) pp 210–225, doi:10.1016/S0021-8693(03)00026-7, arXiv:math/0108135.
[8] Stacks Project, https://stacks.math.columbia.edu/.
[9] Edidin, Hassett, Kresch, Vistoli, "Brauer groups and quotient stacks", American Journal of Mathematics 123 No. 4 (2001), doi:10.1353/ajm.2001.0024, JSTOR, arXiv:math/9905049.<|endoftext|>
TITLE: Eigenvalues of matrix sums
QUESTION [75 upvotes]: Is there a relationship between the eigenvalues of individual matrices and the eigenvalues of their sum? What about the special case when the matrices are Hermitian and positive definite?
I am investigating this with regard to finding the normalized graph cut under general convex constraints. Any pointers will be very helpful.

REPLY [4 votes]: Here is a trivial case with a simple solution. Applicable in Quantum Mechanics, for one.
Given two matrices of the form $A \otimes Id$, $Id \otimes B$, the eigenvalues of their sum are all combinations $a_i+b_j$, where $A\vec{a}_i=a_i\vec{a}_i$ and $B\vec{b}_i=b_i\vec{b}_i$. The eigenvectors are all tensor products of the individual eigenvectors of $A$ and $B$.
This is a one-liner to show:
$$(A \otimes Id + Id \otimes B) \,\vec{a}_i\otimes \vec{b}_j=(a_i+b_j)\,\vec{a}_i\otimes \vec{b}_j$$<|endoftext|>
TITLE: How should one approach tropical mathematics?
QUESTION [39 upvotes]: Let me preface this by saying that my background is pretty meagre (i.e. solid undergrad). However, a few months ago I came across Litvinov - The Maslov dequantization, idempotent and tropical mathematics: A brief introduction which presented an idea that struck me as really remarkable. One can develop a theory of analysis of functions taking values in an idempotent semiring (e.g. the max+ algebra), which happens to be naturally suited towards some traditionally nonlinear problems. Under this formulation (specialized to the case of max+), the integral of a good function corresponds to the supremum, the Fourier transform roughly corresponds to the Legendre transform (!), and it seems that one can develop a theory of "linear" (e.g. in terms of the max+ operations) PDE analogous to the traditional linear theory (!!). For example, the HJB equation is a nonlinear first order PDE, but linear in the max+ sense of the word. This all blew my mind, but after trying to read a few more papers on the subject I decided to put it on the back burner for later thought.
Then, a few days ago I was reading something written by Gian-Carlo Rota in which he makes a remark about developing an "algebra" for multisets. I guess distributive lattices model the "algebra" of sets well enough (in fact there is Birkhoff's theorem), but the quantitative aspect of multisets make this seem inappropriate. So just playing around a bit, I realize that if one models multisets on elements in X by functions from X to the nonnegative integers (the multiplicity), then multiset union corresponds to pointwise addition and multiset intersection corresponds to pointwise min. The min+ algebra on the nonnegative integers! Perhaps this points in the direction of why I think of tropical mathematics as something of interest to people in algebraic combinatorics (maybe this generalization is wrong).
Ok, sorry for that ramble. Essentially, I have 2 questions. First, aside from references to "dequantization", how should I envision the role of tropical mathematics? My lack of background makes it hard for me to get an idea of what is going on here (especially on the geometry side of things), but it seems like there are some big ideas lurking around.
Second, if I wanted to learn more about this stuff, what would be the best route to take? What expository papers should I look at / save for later? It's a bit intimidating that it seems like one needs background in algebraic geometry before one can seriously approach such ideas, but maybe that's just the way it is.

REPLY [2 votes]: From a completely different perspective, there is Jean-Eric Pin's article Tropical semirings, and then
Stéphane Gaubert's short introduction Methods and Applications of $(\operatorname{max},+)$ Linear Algebra, Report 3088, January 1997, INRIA and a longer set of lecture notes by the same author, Introduction aux Systèmes Dynamiques à Événements Discrets (these are in French but are excellent lecture notes). The point of view of Gaubert is to describe discrete event systems which are great fun and very approachable. Later when discussing the applications in that area some of the points that you made in the original question can be seen but with much greater emphasis on the applicability. See also Jean-Pierre Quadrat, Max–Plus Algebra and Applications to System Theory and Optimal Control, Proccedings of the International Congress of Mathematicians, Zurich 1994,  Birkhauser, 1995.
I learnt a lot from a talk by Jeremy Gunawardena many years ago, and for further information you could look at his Idempotent Semi-rings  and his website (which is easy to find).
Another useful website is http://www.maxplus.org/ linking into several groups working on the (max,+)-algebra.
Note none of this really needs the algebraic-geometric side as such, at least to start with, and is much more linked to ‘systems theory’,  and applications such as combinatorics, scheduling, and dynamic programming problems in Operational Research.  It is also great fun both to learn and to teach.<|endoftext|>
TITLE: Relating Category Theory to Programming Language Theory
QUESTION [70 upvotes]: I'm wondering what the relation of category theory to programming language theory is.
I've been reading some books on category theory and topos theory, but if someone happens to know what the connections and could tell me it'd be very useful, as that would give me reason to continue this endeavor strongly, and know where to look.
Motivation: I'm currently researching undergraduate/graduate mathematics education, specifically teaching programming to mathematics grads/undergrads. I'm toying with the idea that if I play to mathematicians strengths I can better instruct them on programming and they will be better programmers, and what they learn will be useful to them. I'm in the process (very early stages) of writing a textbook on the subject.

REPLY [5 votes]: If you want a very basic, entry level description of category theory from a Haskell programming point of view (not exactly what the OP asked for),

http://en.wikibooks.org/wiki/Haskell/Category_theory

is nice.<|endoftext|>
TITLE: Godel's 1st incompleteness theorem - clarification.
QUESTION [5 upvotes]: This should be a trivial question for people who know Gödel's 1st incompleteness theorem. I quote the statement the theorem from wikipedia: "Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory."
My question is: what is the meaning of 'true' in the last sentence? Let me elaborate: the only (introductory) proof of the theorem that I know starts with a specific model of the theory and constructs a sentence which is true in that model, but not provable from the theory.
So does true arithmetical statement' in the statement of the theorem meantrue in the (implicitly) given model', or true in EVERY model?

REPLY [6 votes]: It means true in the usual model. For 1st order logic we have Godel's Completeness theorem, which guarantees that if something is true in every model, then it is actually provable in the theory.

REPLY [4 votes]: As already pointed out here by Tom Leinster, "true" doesn't make sense without some sort of model. It is a technical word.<|endoftext|>
TITLE: Why is Milnor K-theory not ad hoc?
QUESTION [37 upvotes]: When Milnor introduced in "Algebraic K-Theory and Quadratic Forms" the Milnor K-groups he said that his definition is motivated by Matsumoto's presentation of algebraic $K_2(k)$ for a field $k$ but is in the end purely ad hoc for $n \geq 3$. My questions are:

What exactly could Milnor prove with these $K$-groups? What was his motivation except for Matsumoto's theorem?
Why did this ad hoc definition become so important? Why is it so natural?

REPLY [19 votes]: Also, about the motivations of Milnor, it is quite natural to try to understand the Witt ring of a field, classifying quadratic forms over this field (in char not 2). This ring has a natural filtration by the fundamental ideal, and it is natural to try to understand the associated graded ring, which is simpler than the Witt ring. One approach is to understand it by generators and relations. The relations defining Milnor's K-theory are elementary ones obviously satisfied in the graded Witt ring, and there are very few of them. Milnor's conjecture (now a theorem) says that Milnor K-theory mod 2 is isomorphic to that graded Witt ring. It is equivalent to the formulation with étale cohomology, but probably, an important part of Milnor's original motivation was about quadratic forms, as one can see in his original paper.
It is quite surprising that, a posteriori, this simple K-theory appears as such a fundamental object in intersection theory. There is a nice (and seminal) paper by Totaro explaining in rather elementary and geometric terms the simplest case of the connexion between Milnor K-theory and higher Chow groups, mentioned by Denis-Charles Cisinski. 
Milnor K-theory is the Simplest Part of Algebraic K-Theory, K-theory 6, 177-189, 1992<|endoftext|>
TITLE: Singularity of sparse random matrices
QUESTION [8 upvotes]: The following topic came up in conversation with my office-mate Lionel: Let $p$ be a fixed prime, $c$ a fixed positive real parameter and $n$ a large number. Consider a random $(0,1)$ matrix with entries in $Z/p$, where the probability of a $0$ is $1-\frac{c}{n}$ and that of a $1$ is $\frac{c}{n}$. As $n\rightarrow \infty$, what is the probability that this matrix is singular? UPDATE: As moonface points out, this probability is $1$. Furthermore, his argument points out that we should expect the corank to be something like $a*n$. So I'll ask more generally what we can say about the behavior of the rank as $n\rightarrow\infty$.
Actually, in our motivating example, the matrix is symmetric and the distribution on the diagonal is different than in the rest of the matrix. I left these details out, but please mention it if this point would strongly effect your answer.

REPLY [5 votes]: A few words of caution about random matrices over finite fields: it takes very little deviation from the zero distribution to force the spectral properties of a random matrix to look like one chosen with a uniform distribution.
For example, take the field to be Z/pZ and consider the distribution where each entry is 0 with probability 1 - (log n)^2/n and 1 with probability (log n)^2/n. The probability that the random matrix is non-singular then converges to (1 - 1/p)(1 - 1/p^2)(1 - 1/p^3)...(1-1/p^n), which is the same probability as a uniformly chosen matrix from M(n,Z/pZ) being non-singular.  In other words, the singularity probability will converge to a constant about 1/p, not to zero or to one, and the probability that the matrix has corank k is about 1/p^k.  This phenomena will hold as long as the distribution you choose is not concentrated more than about 1 - log n/n; otherwise, we do indeed expect the matrix to be almost surely singular as n -> infty, as is the case for the random matrix in the original post.
The key point seems to be bounding the probability that a row is all zeros or that the matrix has multiple equal rows.  If you can show that the probability of such an event goes to zero as n -> infty, then I would expect the asymptotics I mentioned above to take over.<|endoftext|>
TITLE: Curriculum vitae:  including grants you've applied for, not received (or not yet received).
QUESTION [17 upvotes]: I've heard from multiple sources now that one's CV should include grants you've applied for, even if you didn't receive them or won't find out if you've received them until after your CV goes out.  I haven't had much luck finding this in other people's CVs, though.  I'd like to have confirmation of this from someone who's been on a hiring committee before.  If this is the case, any advice on how to present this information?  (example appreciated as always).
General advice about CVs is appreciated too ...

REPLY [5 votes]: First, let's assume that you're applying for a position, where your research matters. If so, then the hiring committee wants to judge as best as it can whether you will do good research after you're hired. Also, let's assume that the hiring committee is not familiar with your specific area of research, so it is not able to judge directly the quality of your work by reading it.
Published and accepted papers, as well as grants awarded, are very useful for establishing the strength of your research ability. Other measures such as quality of journals and citation numbers can strengthen your case even further and are in fact quite important if you are applying to a strong department.
Submitted papers, preprints, and grant applications do not help in judging your research ability. But they do matter. In particular, they, along with the items above, show that you are committed not only to continuing your research but also documenting it in a way that your department, your university, and your peers can judge it properly. In short, it provides evidence that you're willing to and are continuing to "play the game". This stuff won't help you beat out someone who is viewed as a stronger mathematician than you, but it might help you beat out someone viewed as on the same level but does not demonstrate the same level of explicit effort.
You should provide all evidence of research activity, whether it represents something you've already accomplished or something you are still striving for.
And you should omit anything that a hiring committee might choose to interpret as a serious distraction to your research efforts.<|endoftext|>
TITLE: Level raising by prime powers
QUESTION [10 upvotes]: Suppose $f$ is a weight $2$ level $N$ cusp form.  When can we realize the mod-$\ell$ representation of $f$ in a form of weight $2$ and level $Np^3$, where $p$ is some prime not dividing $N$?  I assume that, if a simple criterion exists at all, it is a condition on the mod-$\ell$ representation of $f$ restricted to inertia at $p$, but I'm not sure what it would say...

REPLY [10 votes]: Presumably you want the form (let me call it g) of level Np^3 to be new at p, otherwise it's trivial.
Let me also assume ell isn't p.
If the form g is new at p, and has level Gamma0(p^3) at p, then the ell-adic representation attached to g will have conductor p^3. But this is a bit of a problem, because the conductor of the mod ell representation can't be that much lower than the conductor of the ell-adic representation. Indeed a theorem of Carayol and, independently, Livne, says that the p-conductor of the mod ell representation will be at least p if the p-conductor of the ell-adic representation is p^3 (the exponent can drop by at most 2). So if you're looking for Gamma_0(p^3) then you're in trouble. This is just a local calculation and isn't too deep.
Diamond and Taylor, in their second paper on the subject, give a list of the conductors of the newforms that can give rise to a given irreducible modular mod ell representation. You can see that Gamma0(p^3) is too much from the main theorem there. Of course the work in that theorem is realising everything that is possible, not ruling out everything that isn't.<|endoftext|>
TITLE: Two functors from Grp to Grp?
QUESTION [24 upvotes]: It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises.  Question 5 in section 1.3 asks you to find two different functors $\mathsf{T}: \mathsf{Groups} \to \mathsf{Groups}$ with object function $\mathsf{T}(\mathsf{G}) = \mathsf{G}$ for every group $\mathsf{G}$.  I have played with this for a long time, and none of the obvious choices end up working.  Was this a mistake on Mac Lane's part, or am I just missing something very obvious?
If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

REPLY [8 votes]: Since this question popped up again, I might as well sketch how far I got. The trivial morphism can be recognized because it factor through the trivial group, so the trivial morphism must be sent to itself. From this we deduce that the obvious inclusions and projections between $G$, $H$ and $G \times H$ are sent to themselves up to the sort of evil isomorphisms discussed in Reid's answer. For any group $F$ with maps $F \to G$ and $F \to H$, the product map $F \to G \times H$ is sent to the product map, again, up to evil isomorphisms. In particular, the diagonal embedding $G \to G \times G$ is sent to the diagonal up to isomorphisms.
If $G$ is abelian, then $\mu: (g_1, g_2) \mapsto g_1 g_2$ is a map $G \times G \to G$ and analyzing the composition of $\mu$ with the coordinate embeddings $G \to G \times G$ shows that $\mu$ goes to $\mu$ (up to isomorphism). Composing diagonals and $\mu$, one deduces that $g \mapsto g^n$ goes to itself (up to isomorphism) for any abelian $G$ and integer $n$. Using the classification of finitely generated abelian groups, one deduces that all maps of finite generated abelian groups go to themselves (up to isomorphism).
I believe I was able to argue that the mystery functor must commute with the abelianization function up to natural isomorphism, but I don't have notes from this anymore. My plan for getting beyond this was to chase the isomorphism down the derived series, argue is particular everything must be good for maps of free groups as the intersection of the derived subgroups is trivial there, and then somehow use that everything is a quotient of a free group to win. But I got stuck around this point.
This question has a very similar feel to this math.SE question: Is there a function $\mathcal{P}$ from connected topological spaces to groups such that $\mathcal{P}(X) \cong \pi_1(X)$? One can show that such a functor can't restrict to $\pi_1$ on pointed spaces, which is probably what the OP meant to ask, but it seems hard to rule out that there is some functor.<|endoftext|>
TITLE: Interesting applications of the pigeonhole principle
QUESTION [51 upvotes]: I'm a little late in realizing it, but today is Pigeonhole Day. Festivities include thinking about awesome applications of the Pigeonhole Principle. So let's come up with some. As always with these kinds of questions, please only post one answer per post so that it's easy for people to vote on them.
Allow me to start with an example:

Brouwer's fixed point theorem can be proved with the Pigeonhole Principle via Sperner's lemma. There's a proof in Proofs from The Book (unfortunately, the google books preview is missing page 148)

By the way, if you happen to be in Evans at Berkeley today, come play musical chairs at tea!

REPLY [4 votes]: I think the solutions of these questions are very interesting (by using pigeon-hole principle), first question is easy, but second question is more advanced:
1) For any integer $n$, There are infinite integer numbers with digits only $0$ and $1$ where 
they are divisible to $n$. 
2) For any sequence $s=a_1a_2\cdots a_n$, there is at least one $k$, such that $2^k$ begin with $s$.<|endoftext|>
TITLE: Is the wedge product of two harmonic forms harmonic?
QUESTION [18 upvotes]: Is the wedge product of two harmonic forms on a compact Riemannian manifold harmonic? I'm looking for a counter-example that the textbooks say exists. 
I would like to see a counter example that is on a complex manifold, Ricci-flat  (or Einstein) manifold or both, if it is at all possible. 
In general, I'm trying to understand the interaction between the wedge product, Hodge star and the Laplacian on forms and it's eigen-vectors, references will be much appreciated.

REPLY [8 votes]: Here is a cop-out answer (sorry in advance). Consider the 0-form $f(x)=x$ on $\mathbb{R}$. Clearly $f$ is harmonic, but $(f\wedge f)_{x}=x^2$ isn't. Since the wedge product is just the point-wise product on 0-forms, this is a valid example!<|endoftext|>
TITLE: A walk on a compact 2D surface embedded in 3-space that never returns home
QUESTION [5 upvotes]: At the risk of asking an uninformed question...
Imagine an ant on a compact two-dimensional surface embedded in 3-space.  The ant is placed at a point on the surface with random orientation.  Once placed and oriented, the ant will walk along a local geodesic path.
Are there examples of such 2D surfaces where we are guaranteed the ant will never return to some starting position?  
(Thanks to Henry Wilton for requesting clarification as per his comment below.)

REPLY [7 votes]: By Poincare recurrence theorem, on a compact manifold, almost every geodesic returns arbitrarily close to its starting point, infinitely many times.
So if you accept a slight error in the measure of the position (and direction) of the ant (say less than 1 picometer, and $10^{-9}$ degree), you are almost guaranted (w.r.t Lebesgue measure) that the ant returns to its initial position if you wait long enough.
The case of a non-compact manifold is even more interesting. 
If the volume is finite, there is always a geodesic that goes to infinity, yet the result still holds: a.e geodesics are recurrent.
There are also examples of infinite volume surfaces for which a.e. geodesics are recurrent. This is the case for a tubular neighborhood of the Cayley graph of $\mathbb{Z}^2\times \lbrace 0\rbrace$ in $\mathbb{R}^3$. 
Finally, there are examples of infinite volume surfaces for which there is a dense set of recurrent geodesics, yet almost all geodesics go to infinity. This is the case for a tubular neighborhood of the Cayley graph of $\mathbb{Z}^3$.<|endoftext|>
TITLE: f(f(x))=exp(x)-1 and other functions "just in the middle" between linear and exponential
QUESTION [70 upvotes]: The question is about the function f(x) so that f(f(x))=exp (x)-1.
The question is open ended and it was discussed quite recently in the comment thread in Aaronson's blog here http://scottaaronson.com/blog/?p=263
The growth rate of the function f (as x goes to infinity) is larger than linear (linear means O(x)), polynomial (meaning exp (O(log x))), quasi-polynomial (meaning exp(exp O(log log x))) quasi-quasi-polynomial etc. On the other hand the function f is subexponential (even in the CS sense f(x)=exp (o(x))), subsubexponential (f(x)=exp exp (o(log x))) subsubsub exponential and so on. 
What can be said about f(x) and about other functions with such an intermediate growth behavior? Can such an intermediate growth behavior be represented by analytic functions? Is this function f(x) or other functions with such an intermediate growth relevant to any interesting mathematics? (It appears that quite a few interesting mathematicians and other scientists thought about this function/growth-rate.) 
Related MO questions:

solving $f(f(x))=g(x)$
How to solve $f(f(x)) = \cos(x)$?
Does the exponential function has a square root
Closed form functions with half-exponential growth
$f\circ f=g$ revisited
The non-convergence of f(f(x))=exp(x)-1 and labeled rooted trees
The functional equation $f(f(x))=x+f(x)^2$
Rational functions with a common iterate
Smoothness in Ecalle's method for fractional iterates.

REPLY [3 votes]: First, by the kindness of Daniel Geisler, I have a pdf of the graph of this, along with $ y = x$ and $ y = e^x - 1,$ at:
    

(source)
The calculated function is analytic on the strictly positive real axis, analytic on the strictly negative real axis, and at least $C^1$ which applies only to behavior in the germ at the origin. Meanwhile, there is exactly one $C^1$ function that works, so this is it. The $C^1$ condition at $0$ is nothing more than the fact that the resulting function is between $x$ and $e^x-1,$ both for positive and negative $x.$ What I think is that the function is $C^\infty$ and the derivatives at $0$ are given by the formal power series solution. My hope is to prove, by basic methods, at least $C^8,$ as that is the approximation I use to graph near the origin. The method away from the origin is the same as that for sine, see
Does the formal power series solution to $f(f(x))= \sin( x) $ converge?
There are, however, a number of tweaks, as we actually need to work with the inverse function for $ x > 0,$ one must use a different branch of the logarithm for $x < 0,$ and so on. I am appending here a C++ program called abel.cc 
which is compiled with 
g++ -o abel abel.cc -lm 
and then run with
./abel 
#include <iostream.h>
#include <stdlib.h>
#include <fstream.h>
#include <sstream>
#include <list>
#include <set>
#include <math.h>
#include <iomanip.h>
#include <string>
#include <algorithm>
#include <iterator>
using namespace std;

// const int MINDISC =         27000;
// const int MAXDISC =         27000;

//   lines after double slashes are comments
//   also on a line with a command, anything after // is  commentary
//   on a Unix or Linux computer,  compile using line

//        g++ -o abel abel.cc -lm   

//  then run the program  with

//  ./abel  

double abel_positive(double x)
{
double eps = 0.000000001;
   eps /= 10000.0 ;    // satisfied at 10^{-14}
  double f = x ;
  double g = 1.0, g_old = 100.0, diff = 1.0 ;
 for( int n = 0; n <= 9000  && diff >= eps ; ++n)
 {
   g =  2 / f - log(f) / 3  + f / 36 - f * f / 540 - f * f * f / 7776 + 71 * f * f * f * f  /435456 - n ;
   diff = fabs(g - g_old);
//  cout.precision(16);
//   cout << n << "  " << x  << "  "  << f  << "  "  << g <<  "   " << diff << endl ;
   f = log ( 1.0 + f); 
   g_old = g;
 }
  return g;
}  // end  abel_positive


double abel_negative(double x)
{
double eps = 0.000000001;
   eps /= 10000.0 ;    // satisfied at 10^{-14}
  double f = x ;
  double g = 1.0, g_old = 100.0, diff = 1.0 ;
 for( int n = 0; n <= 1000  && diff >= eps ; ++n)
 {
   g =  2 / f - log(fabs(f)) / 3  + f / 36 - f * f / 540 - f * f * f / 7776 + 71 * f * f * f * f  /435456 + n ;
   g *= -1.0;
   diff = fabs(g - g_old);
//  cout.precision(16);
//   cout << n << " neg " << x  << "  "  << f  << "  "  << g <<  "   " << diff << endl ;
   f = exp (  f) - 1.0; 
   g_old = g;
 }
  return g;
}  // end  abel_negative

// alpha(x) = 2 / x - log(x) / 3  + x / 36 - x^2 / 540 - x^3 / 7776 + 71 * x^4 / 435456 - 8759 * x^5 / 163296000 - 31 * x^6 / 20995200


double inverse_abel_positive(double x)
{
  double eps = 0.000000001;
     eps /= 10000.0 ; //  satisfied at 10^{-14}
  double middle;
  if( x < -2.0001) return -5000000.0;
  else if ( x < -1.0001) return (  exp(inverse_abel_positive( x + 1.0)) - 1.0             );
  else
  {
    double left = 0.001, right = 200.0;
    middle = ( left + right) / 2.0; 
    double left_val = abel_positive(left) , right_val = abel_positive(right), middle_val = abel_positive(middle);
    while ( right - left > eps)
    {
      if (middle_val < x )
      {
        right = middle;
        middle = ( left + right) / 2.0;
         right_val = abel_positive(right);
        middle_val = abel_positive(middle);
      }
      else
      {
        left = middle;
        middle = ( left + right) / 2.0;
         left_val = abel_positive(left);
        middle_val = abel_positive(middle);
      }
    //  cout.precision(16);
    //  cout << middle << endl;
    } // while not accurate
  }  // else in range

  return  middle;
} //  end    inverse_abel_positive



double inverse_abel_negative(double x)
{
  double eps = 0.000000001;
     eps /= 10000.0 ; //  satisfied at 10^{-14}
  double middle;
  if( x < -2.0001) return -5000000.0;
  else if ( x < 1.04) return (  log(  1.0 + inverse_abel_negative( x + 1.0))             );
  else
  {
    double right = -0.01, left = -200.0;
    middle = ( left + right) / 2.0; 
    double left_val = abel_negative(left) , right_val = abel_negative(right), middle_val = abel_negative(middle);
    while ( right - left > eps)
    {
      if (middle_val > x )
      {
        right = middle;
        middle = ( left + right) / 2.0;
         right_val = abel_negative(right);
        middle_val = abel_negative(middle);    
      }
      else
      {
        left = middle;
        middle = ( left + right) / 2.0;
         left_val = abel_negative(left);
        middle_val = abel_negative(middle);
      }
   //   cout.precision(16);
   //   cout << middle << endl;
    } // while not accurate
  }  // else in range

  return  middle;
} //  end    inverse_abel_negative


double half_iterate(double x)
{
  if ( x > 0.1)  return inverse_abel_positive( -1/2.0 + abel_positive(x)  );
  else if ( x <= 0.1 && x >= -0.1 - 0.000000001 ) return x +  x * x / 4.0  +  x * x * x / 48.0   + x * x *  x * x * x / 3840.0  - 7 * x * x * x * x * x * x / 92160.0   +  x * x * x * x * x * x * x / 645120.0    + 53.0 * x * x * x * x * x * x * x * x / 3440640.0   ; // no x^4 term, it happens.
   else     return inverse_abel_negative( 1/2.0 + abel_negative(x)  );       
}  // half_iterate


//  gp pari :

// g = x + x^2 / 4 + x^3 /48 + x^5 / 3840 - 7 * x^6 / 92160 + x^7 / 645120 + 53 * x^8 /  3440640

// g + g^2 / 4 + g^3 / 48 + g^5 /  3840   - 7 * g^6 / 92160 + g^7 / 645120  + 53 * g^8 /  3440640

//  ...+ 488363/190253629440*x^13 + 5440363/713451110400*x^12 + 20071/1189085184*x^11 + 20971/825753600*x^10 + 971/46448640*x^9 + 1/40320*x^8 + 1/5040*x^7 + 1/720*x^6 + 1/120*x^5 + 1/24*x^4 + 1/6*x^3 + 1/2*x^2 + x


 //        g++ -o abel abel.cc -lm 

int main()
{

  for( double x = 5.4; x >= -3.45 ; x -= 0.01 )
  {

     cout.setf(ios::fixed, ios::floatfield);
     cout.precision(16);

//  cout << x << "    " << abel_positive( x) << "    "  << half_iterate( x) << "    "  << half_iterate(half_iterate( x)) << "    "  ;

// cout << x << "    " << abel_positive( x) << "    "  << half_iterate( x)  << "    "  ;


cout << x << "    "   << half_iterate( x)  << "    "  ;

cout.unsetf(ios::floatfield);
cout.unsetf(ios::fixed);
  cout.precision(4);
//cout << abel(log(1.0 + x)) - abel( x) - 1 << endl;
cout << half_iterate(half_iterate( x)) -  exp( x) + 1.0   << endl;

  }
  return 0 ;
}    //  end of main

 //        g++ -o abel abel.cc -lm   

Then I am appending the output, which is just the x value, the value of f(x), finally an error  term f(f(x)) - exp(x) + 1. Actually, I had to give fewer outputs, there seems to be a size limit on computer output in MO.
phoebus:~/Cplusplus> ./abel
5.4000000000000004    16.3650724302248491    1.331e-09
5.3000000000000007    15.7879819196927205    2.847e-10
5.2000000000000011    15.2243680870229294    3.423e-10
5.1000000000000014    14.6740278132216613    1.348e-09
5.0000000000000018    14.1367598155454832    5.125e-10
4.9000000000000021    13.6123646385552366    -1.41e-10
4.8000000000000025    13.1006446447147358    3.756e-11
4.7000000000000028    12.6014040044095186    -2.856e-10
4.6000000000000032    12.1144486866975196    7.443e-10
4.5000000000000036    11.6395864499040993    -1.007e-10
4.4000000000000039    11.1766268322195632    6.662e-11
4.3000000000000043    10.7253811422648102    -3.654e-11
4.2000000000000046    10.2856624497240059    2.202e-10
4.1000000000000050    9.8572855759604447    8.681e-11
4.0000000000000053    9.4400670849610382    4.856e-10
3.9000000000000052    9.0338252736914679    -1.539e-10
3.8000000000000052    8.6383801632549879    1.661e-11
3.7000000000000051    8.2535534890740454    1.432e-11
3.6000000000000050    7.8791686923891451    -1.033e-10
3.5000000000000049    7.5150509103819285    1.846e-10
3.4000000000000048    7.1610269672691711    -3.851e-10
3.3000000000000047    6.8169253652572603    -1.32e-10
3.2000000000000046    6.4825762748621774    -2.01e-11
3.1000000000000045    6.1578115260938269    -8.575e-11
3.0000000000000044    5.8424645990524926    -1.462e-11
2.9000000000000044    5.5363706146241647    1.361e-11
2.8000000000000043    5.2393663251515896    -2.929e-12
2.7000000000000042    4.9512901050766569    -4.294e-11
2.6000000000000041    4.6719819413692445    -1.235e-11
2.5000000000000040    4.4012834240498844    2.79e-11
2.4000000000000039    4.1390377362686763    -2.539e-11
2.3000000000000038    3.8850896444937177    -1.616e-13
2.2000000000000037    3.6392854882807599    4.231e-12
2.1000000000000036    3.4014731698489813    3.428e-12
2.0000000000000036    3.1715021431943606    6.262e-13
1.9000000000000035    2.9492234028327093    2.941e-12
1.8000000000000034    2.7344894719531982    1.509e-11
1.7000000000000033    2.5271543897887909    1.412e-13
1.6000000000000032    2.3270736982303024    1.964e-12
1.5000000000000031    2.1341044271723417    -9.805e-13
1.4000000000000030    1.9481050786134744    -4.962e-12
1.3000000000000029    1.7689356089101538    1.714e-13
1.2000000000000028    1.5964574088813634    -1.159e-12
1.1000000000000028    1.4305332811341520    -1.013e-12
1.0000000000000027    1.2710274138894109    7.816e-13
0.9000000000000027    1.1178053503667034    -1.945e-13
0.8000000000000027    0.9707339525168228    -3.206e-13
0.7000000000000027    0.8296813575533253    3.708e-13
0.6000000000000028    0.6945169252836292    2.309e-14
0.5000000000000028    0.5651111736539647    -1.235e-13
0.4000000000000028    0.4413356992547828    -3.93e-13
0.3000000000000028    0.3230630786173595    -1.743e-14
0.2000000000000028    0.2101667451936166    2.259e-14
0.1000000000000028    0.1025208358618962    8.632e-14
0.0000000000000028    0.0000000000000028    -8.327e-17
-0.0999999999999972    -0.0975208360134532    -1.577e-14
-0.1999999999999972    -0.1901667548369642    1.416e-14
-0.2999999999999972    -0.2780631873408490    5.551e-15
-0.3999999999999972    -0.3613363013525148    5.274e-14
-0.4999999999999972    -0.4401134277164546    -4.774e-14
-0.5999999999999972    -0.5145235016776626    -2.472e-13
-0.6999999999999972    -0.5846974891233467    1.36e-14
-0.7999999999999972    -0.6507687621527771    -4.419e-14
-0.8999999999999971    -0.7128733880168525    1.32e-13
-0.9999999999999971    -0.7711502997203856    -1.19e-13
-1.0999999999999972    -0.8257413246049203    3.442e-15
-1.1999999999999973    -0.8767910576263940    1.829e-13
-1.2999999999999974    -0.9244465773015511    -3.32e-14
-1.3999999999999975    -0.9688570130908518    -2.05e-13
-1.4999999999999976    -1.0101729822279837    -2.889e-14
-1.5999999999999976    -1.0485459210648198    2.558e-13
-1.6999999999999977    -1.0841273405174991    -2.204e-13
-1.7999999999999978    -1.1170680371664736    5.618e-14
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-2.0999999999999979    -1.2015263350700924    4.802e-15
-2.1999999999999980    -1.2253702540044693    3.331e-13
-2.2999999999999980    -1.2472896992571449    -3.854e-13
-2.3999999999999981    -1.2674156771038474    1.388e-13
-2.4999999999999982    -1.2858739130575336    -8.189e-13
-2.5999999999999983    -1.3027845214953375    -1.961e-13
-2.6999999999999984    -1.3182617679403474    -1.768e-14
-2.7999999999999985    -1.3324139189760200    -1.3e-14
-2.8999999999999986    -1.3453431728233207    -1.149e-14
-2.9999999999999987    -1.3571456621162299    -8.366e-14
-3.0999999999999988    -1.3679115197319236    7.921e-14
-3.1999999999999988    -1.3777249981938211    1.012e-14
-3.2999999999999989    -1.3866646333660464    2.682e-13
-3.3999999999999990    -1.3948034435566479    -6.27e-13
phoebus:~/Cplusplus><|endoftext|>
TITLE: Cohomology of fibrations over the circle
QUESTION [17 upvotes]: Are there any general results on the (integral) cohomology of manifolds that are fibrations over the circle? Any literature references much appreciated.

REPLY [17 votes]: This began as a comment, but it is interesting enough that I decided to make it an answer instead.  Let's consider bundles over the circle whose fibers are closed genus $g$ surfaces $\Sigma_g$.  
The diffeomorphism type of the total space of our bundle only depends on the isotopy class of the monodromy map.  Denote by $M_g$ the mapping class group, ie the group of isotopy classes of orientation-preserving diffeomorphisms of $\Sigma_g$.  For $f \in M_g$, denote by $B_f$ the surface bundle over the circle determined by $f$.
In a comment, Tom Church observed that the homology of $B_f$ will be the same as the homology of the trivial bundle if and only if $f$ acts trivially on $H_1(\Sigma_g)$.  The group of mapping classes that act trivially on $H_1(\Sigma_g)$ is known as the Torelli group.
One could demand more.  Namely, we could require that the cup-product structure on $H^{\ast}(B_f)$ be the same as the cup product structure on the trivial bundle.  As Tom observed, a beautiful theorem of Dennis Johnson gives a precise characterization of the subgroup of $I_g$ consisting of monodromies with this property.  One easy way of describing it is that it is the kernel of the (outer) action of $M_g$ on the second nilpotent truncation of $\pi_1(\Sigma_g)$ (the group $H_1(\Sigma_g)$ is the first nilpotent, ie abelian, trunctation).
The story does not end here.  A topological space has a "higher-order" intersection theory given by the so-called Massey products. They are sort of like generalized cup products. Anyway, Kitano generalized Johnson's work and gave a precise and beautiful description of the monodromies of surface bundles over the circle in which these higher intersection products (up to a certain level) are trivial.  Namely, all the degree at most $k$ Massey products of $B_f$ will be trivial if and only if $f$ acts trivially on the (k+1)st nilpotent truncation of $\pi_1(\Sigma_g)$.  
For the details of this plus references to Johnson's papers, see the following paper:
MR1381688 (97f:57014)
Kitano, Teruaki(J-TOKYTE)
Johnson's homomorphisms of subgroups of the mapping class group, the Magnus expansion and Massey higher products of mapping tori. (English summary)
Topology Appl. 69 (1996), no. 2, 165--172.<|endoftext|>
TITLE: Solvable class field theory
QUESTION [15 upvotes]: Is/should there be a theory of finite solvable extensions over a given base field?  Could it be based on/use class field theory?  Assume the base field isn't a local field.

REPLY [13 votes]: As FC says, since solvable extensions are built up out of abelian extensions, class field theory is certainly relevant and helpful in understanding the structure of solvable extensions.  On the other hand, to do this in a systematic way requires understanding class field theory of each number field in a tower "all at once".  The picture that one gets in this way seems quite blurry compared to the classical goal of class field theory: to describe and parameterize the finite abelian extensions L of a field K in terms of data constructed from K itself.  In the case of a number field, this description is in terms of groups of (generalized) ideal classes, or alternately in terms of quotients of the idele class group.  I'm pretty sure there's no description like this for solvable extensions of 
any number field.
What I can offer is a bunch of remarks:
1) Sometimes one has a good understanding of the entire absolute Galois group of a field K, in which case one gets a good understanding of its maximal (pro-)solvable quotient.  Of course this happens if the absolute Galois group is abelian.
2) Despite the OP's desire to exclude local fields, this is one of the success stories: the full absolute Galois group of a $p$-adic field is a topologically finitely presented prosolvable group with explicitly known generators and relations.
3) On the other hand, we seem very far away from an explicit description of the maximal solvable extension of Q. For instance, in the paper
MR1924570 (2003h:11135)
Anderson, Greg W.(1-MN-SM)
Kronecker-Weber plus epsilon. (English summary)
Duke Math. J. 114 (2002), no. 3, 439--475. 
the author determines the Galois group of the extension of Q^{ab} which is obtained by taking the compositum of all quadratic extensions K/Q^{ab} such that K/Q is Galois.  Last week I heard a talk by Amanda Beeson of Williams College, who is working hard to extend Anderson's result to imaginary quadratic fields.  
4) This question seems to be mostly orthogonal to the "standard" conjectural generalizations of class field theory, namely the Langlands Conjectures, which concern 
finite dimensional complex representations of the absolute Galois group.
5) A lot of people are interested in points on algebraic varieties over the maximal solvable extension Q^{solv} of Q.  The field arithmeticians in particular have a folklore conjecture that Q^{solv} is Pseudo Algebraically Closed (PAC), which means that every absolutely irreducible variety over that field has a rational point.  This would have applications to things like the Inverse Galois Problem and the Fontaine-Mazur Conjecture (if that is still open!).  Whether an explicit description of Q^{solv}/Q would be so helpful in these endeavors seems debatable.  I have a paper on abelian points on algebraic varieties, in which the input from classfield theory is minimal.
The two papers on solvable points that I know of (and very much admire) are:
MR2057289 (2005f:14044) Pál, Ambrus Solvable points on projective algebraic curves.  Canad. J. Math.  56  (2004),  no. 3, 612--637.
MR2412044 (2009m:11092) Çiperiani, Mirela; Wiles, Andrew Solvable points on genus one curves. Duke Math. J. 142 (2008), no. 3, 381--464.<|endoftext|>
TITLE: Pronunciation: Crapo
QUESTION [8 upvotes]: A similar question reminds me: When giving talks, I often want to refer to the work of Henry Crapo. I have asked several mathematicians, and none of them were sure how to pronounce his last name. Any help?

REPLY [2 votes]: I knew Henry quite well.  His last name is pronounced Cray-poe.<|endoftext|>
TITLE: What do decategorification and "compactification on a circle" have to do with each other? 
QUESTION [11 upvotes]: Some physicists have told me that if you think about an extended n-dimensional TQFT $F$, then the decategorification is given by $F'(X)=F(X\times S^1)$, which I believe they call "compactification on a circle."  Is there any way to make this statement precise?

REPLY [8 votes]: In a general extended TQFT Z, the assignment $Z(X x S^1)$ is the "dimension" of Z(X), in the following sense. Write the circle as an incoming arc followed by an outgoing arc. The incoming arc is a morphism (coevaluation) from the unit  (Z(empty set)) to Z(X) tensor its dual $Z(X^{op})=Z(X)^*$, followed by a morphism (evaluation) back to the unit.
In particular we learn Z(X) HAS a dual (is dualizable), and these are the two canonical maps that come in the definition of being a dual. The composition is an endomorphism of the unit Z(empty), which is very generally called the dimension of Z(X), or the Hochschild homology of Z(X). 
If Z(X) is a vector space, End Z(empty) = numbers and this is the usual dimension. If Z(X) is a category (or an algebra, or a 2-category, or....), this is what is usually known as its Hochschild homology. In particular Hochschild homology is where characters (or traces) of objects in Z(X) live, if it's a category. In simple situations this will be the same as the K-theory of Z(X) (in great generality there's a map from K-theory to Hochschild homology), if you want to compare this to another version of decategorification, which is taking K-groups.<|endoftext|>
TITLE: Is a torsion-free abelian group finitely generated, if all of its localizations at primes $p$ are finitely generated over $\mathbb{Z}_p$?
QUESTION [6 upvotes]: Background: When proving that the group of $k$-isogenies $\mathrm{Hom}_k(A,B)$ between two abelian varieties is finitely generated, one first shows that the Tate map $$\mathbb{Z}_\ell\otimes_{\mathbb{Z}} M \to \mathrm{Hom}_{\mathbb{Z}_\ell}(T_\ell A,T_\ell B)$$ is injective. Since each Tate module is free of finite rank over $\mathbb{Z}_\ell$, it follows that the localization $M_\ell$ is $\mathbb{Z}_\ell$-finite. One then uses a little trick to deduce the $\mathbb{Z}$-finiteness of $M$ itself. (See Silverman I, for example.)
The above proof needs only a single prime $\ell$,
but disregarding issues of the characteristic of the field (which are apparently surmountable) we actually have an injective Tate map at every prime. Thus...
Question: Can the $\mathbb{Z}$-finiteness of $M$ be deduced directly from the $\mathbb{Z}_\ell$-finiteness of  $M_\ell$ for all primes $\ell$?
One can consider this a question about general torsion-free abelian groups $M$. A non-counterexample to keep in mind is $M=\mathbb{Z}[1/p]$,
for which $M_\ell$ is $\mathbb{Z}_\ell$-finite for all $\ell\neq p$.
(A google search shows that there is actually quite a body of literature on torsion-free abelian groups, so perhaps the answer to this question is well-known, but I'm not sure where to look...)

REPLY [2 votes]: I had the same question when reading Silverman, and want to follow up on Jeremy's answer and explain "Step 4" of his proof in more detail. Corollary 1.6 in these notes was also very helpful.
So let's prove that if $N$ is a $\mathbb{Z}$-module such that

$N$ is torsion-free,
For every finitely generated $\mathbb{Z}$-submodule $M$ of $N$, the set $M^{\text{div}} := \{n \in N ~|~ \exists k \in \mathbb{Z} \text{ such that } kn\in M\}$ is a finitely generated $\mathbb{Z}$-module,
For one prime $\ell$ the $\mathbb{Z}_\ell$-module $N \otimes_{\mathbb{Z}} \mathbb{Z}_\ell$ is finitely generated,

then $N$ is finitely generated. (The rank estimate for $N=\text{Hom}(E_1,E_2)$ follows then precisely as in Silverman, Cor. III.7.5)
By 3. we know that $N\otimes_{\mathbb{Z}} \mathbb{Q}_\ell = (N\otimes_{\mathbb{Z}} \mathbb{Z}_\ell) \otimes_{\mathbb{Z}_\ell} \mathbb{Q}_\ell$ is a finite-dimensional $\mathbb{Q}_\ell$-vector space. But also $N\otimes_{\mathbb{Z}} \mathbb{Q}_\ell = (N\otimes_{\mathbb{Z}} \mathbb{Q}) \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$, thus the $\mathbb{Q}$-vector space $N\otimes_{\mathbb{Z}} \mathbb{Q}$ is finite-dimensional as well. Let $n_1,\dots,n_d$ be a basis of the vector space $N\otimes_{\mathbb{Z}} \mathbb{Q}$. By clearing denominators we may assume that $n_i \in N$ for all $i$. Let $M$ be the $\mathbb{Z}$-submodule of $N$ generated by $n_1,\dots,n_d$. We claim that $M^{\text{div}} = N$, which by 2. completes the proof.
Clearly $M^{\text{div}} \subset N$, so take $n \in N$. Then $n = \sum_i n_i \otimes \lambda_i$ for $\lambda_i \in \mathbb{Q}$. By clearing denominators of the $\lambda_i$ we see that some $\mathbb{Z}$-multiple of $n$ is a $\mathbb{Z}$-linear combination of the $n_i$ and thus lives in $M$, i.e. $n \in M^{\text{div}}$.<|endoftext|>
TITLE: Is the space of nondegenerate classical paths connected?
QUESTION [7 upvotes]: I have a fairly specific question.  My intuition says the answer is "yes", but there is a  natural generalizations in which I take out all the "physics", and then I think the answer is "no".
Edit number 2: the question without all the background
In response to Andrew's comments, here's the question I want to ask without all the infinite-dimensional preamble:
On $\mathbb R^d$ with its usual metric, pick a differential one-form $b$ and a smooth function $c$, and suppose that each has compact support.  Consider the following (nondegenerate, nonlinear, second-order) differential equation for a path $\gamma(t)$:
$$ \ddot \gamma = db \cdot \dot\gamma + dc $$
This is the Euler-Lagrange equation, and so I will abbreviate it as (EL).  In coordinates, it is:
$$ \ddot \gamma^i = (\partial\_i b\_j - \partial\_j b\_i) \dot\gamma^j + \partial\_i c $$
Since (EL) is nondegenerate and $b,c$ have compact support, every solution to (EL) extends to have domain all of $\mathbb R$, and the solutions are in bijection with the tangent bundle ${\rm T}\mathbb R^d = \mathbb R^{2d}$ by identifying $\gamma$ with $(\dot\gamma(0),\gamma(0))$.
For each $(v,q) \in {\rm T}\mathbb R^d$, define a second-order linear differential operator $h\_{(v,q)}$, given in coordinates by:
$$ h\_{(v,q)}[\eta]^j(t) = \ddot\eta^j(t) + \bigl(\partial\_i b\_j|\_{\gamma(t)} - \partial\_j b\_i|\_{\gamma(t)}\bigr) \dot\eta^i(t) + \bigl( \partial\_i \partial\_k b\_j|\_{\gamma(t)} \dot\gamma^k(t) - \partial\_i\partial\_j b\_k|\_{\gamma(t)} \dot\gamma^k(t) - \partial\_i\partial\_j c|\_{\gamma(t)}\bigr) \eta^j(t) $$
where $\gamma$ is the solution to (EL) with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$.
Let $C = {\rm T}\mathbb R^d \times \mathbb R\_{\>0}$.  For $(v,q,T) \in C$, consider the operator $h\_{(v,q)}$ as a map
$$ h\_{(v,q,T)} : \bigl\{ \eta: [0,T] \to \mathbb R^d \text{ s.t. } \eta(0) = 0 = \eta(T) \bigr\} \to \bigl\{ \eta: [0,T] \to \mathbb R^d \bigr\}$$
Define $C' \subseteq C$ to be the set $\{ (v,q,T) \in C \text{ s.t. } \ker h\_{(v,q,T)} = 0\}$.
Then I have the following questions:

Is $C'$ open (with the topology induced from $C$)?  I asserted that it was, because the coefficients of the second-order operator depend smoothly, and I think that kernels can only jump in dimension at closed regions.  But I'm not 100% sure.
Is $C'$ (path) connected?  This was my originally-posed question, and it definitely requires that $b,c$ have compact support.
Actually, for my research I need that for each $T > 0$, we have $C' \cap {\rm T}\mathbb R^d \times \{T\}$ is path connected.  Since $C'$ includes every $\gamma \in C$ with $\gamma([0,T])$ always outside the support of $b,c$, and since this set is path connected if the supports are compact, 3. implies 2., but perhaps 3. is stronger.  Also, perhaps 3. does not require that $b,c$ have compact support?

Bonus question: I used the metric exactly once in (EL) and exactly once in (HJ), to compare the folks with raised indices to the ones with lowered indices.  Does anything happen if I change the signature of the metric?
The rest is what I wrote before:
Background and definitions
On $\mathbb R^d$ (with its usual metric), pick a differential one-form $b$ and a smooth function $c$.  The tangent bundle $T\mathbb R^d$ is just $\mathbb R^{2d}$; define the Lagrangian $L: T\mathbb R^d \to \mathbb R$ by $L(v,q) = \frac12 |v|^2 + b(q)\cdot v + c(q)$, where $v$ is the fiber coordinate on $T\mathbb R^d$, $q$ is the base coordinate on $\mathbb R^d$, and $\cdot$ is the canonical pairing of a one-form with a vector.  A path of length $t$ is a smooth map $\gamma: [0,t] \to \mathbb R^d$; it has a canonical lift $(\dot\gamma,\gamma): [0,t] \to T\mathbb R^d$.  The action of a path $\gamma$ of length $t$ is the integral $A[\gamma] = \int_0^t L(\dot\gamma(\tau),\gamma(\tau))d\tau$.  By adjusting signs, one can include paths of negative length; a path of length $0$ is a point in $T\mathbb R^d$ and has zero action.
Consider the set $P$ of all paths (of arbitrary length); it is an infinite-dimensional smooth manifold.  There are various natural projections from $P$ to finite dimensions.  The "initial-value map" $P \to T\mathbb R^d \times \mathbb R$ takes a path $\gamma: [0,t]\to \mathbb R^d$ to the triple $(\dot\gamma(0),\gamma(0),t)$.  I will be more interested in the "boundary-value map" $P \to \mathbb R^d \times \mathbb R^d \times \mathbb R$ taking $\gamma \mapsto (\gamma(0),\gamma(t),t)$.  The fiber over a point in $\mathbb R^d \times \mathbb R^d \times \mathbb R$ is an affine space modeled on the space of Dirichlet paths $\gamma: [0,t] \to \mathbb R^d$ with $\gamma(0) = 0 = \gamma(t)$.
I like to think of the action $A$ as a Morse function on fibers of the boundary-value map.  Let $C \subset P$ be the set of classical paths, i.e. paths $\gamma$ so that $dA|_\gamma \cdot \xi = 0$ if $\xi$ is Dirichlet ($dA|_\gamma$ is the differential of the action at $\gamma$; $\cdot$ is the canonical pairing).  Equivalently, $\gamma \in C$ if $\gamma$ satisfies the Euler-Lagrange equations $\frac{\partial L}{\partial q}(\dot\gamma,\gamma) = \frac{d}{d\tau}\bigl[ \frac{\partial L}{\partial v}(\dot\gamma,\gamma) \bigr]$.  Since the Euler-Lagrange equations are second-order nondegenerate, the initial-value map restricts to a diffeomorphism of $C$ to an open subset of $T\mathbb R^d \times \mathbb R$ containing $T\mathbb R^d \times \{0\}$.
If I really want to think of $A$ as a Morse function, I should require that its critical points (the classical paths) be nondegenerate.  Let $\gamma$ be a (classical) path of length $t$, and $V$ the vector space of Dirichlet paths of length $t$.  Then the second derivative or Hessian of $A$ is well-defined as a map $H : V \to V^*$.  In fact, the Hessian makes sense as a second-order linear differential operator on the space of all paths of length $t$.  Let's say that a classical path is nondegenerate if $\ker H = 0$ (or, rather, does not intersect the space $V$ of Dirichlet paths).  The set $C'$ of nondegenerate classical paths is an open (I'm pretty sure) subset of $C$.
My question
Is the space $C'$ (path) connected?
Bonus question: what happens if you change the signature of the metric on $\mathbb R^d$?
Edit
The answer to my original question is "no".  Let $d = 1$, $b = 0$, and $c(q) = \frac12 q^2$.  Then a classical path of length $t$ is nondegenerate if and only if $t$ is not an integer multiple of $\pi$.  This is a very nongeneric Lagrangian (it is the harmonic oscillator, and is exactly solvable).  Also, I think with my definitions, paths of length $0$ are always degenerate.
So let me ask a more restricted question.  Let's suppose that $b$ and $c$ are only supported in a compact neighborhood.  Then classical paths that do not enter this neighborhood are precisely the straight lines, and they are all generic (provided $t \neq 0$).  Is it true that the space of classical nondegenerate paths with positive length is connected with the restriction that $b,c$ have compact support?

REPLY [2 votes]: If you consider a Riemannian manifold with the Lagrangian $L(\nu,q)=|\nu|^2$, where $q$ is a point in the manifold, $\nu$ is a tangent vector, and $|\nu|$ is defined using the metric at the point $q$, then the first variation gives you the geodesic equation and the second variation gives Jacobi fields. This case has been thoroughly studied (for example, it is one of the main topics of the beautiful book "Morse Theory" by J. Milnore). The space of paths $P$ between two fixed points (or the space of loops, which are Dirichlet paths) with the compact/open topology turns out to be homotopy equivalent to a cell complex with dimensions of cells determined by conjugate points (degenerate paths in your case).
In the case of your question, if we put $b=0$, we will get the equation
$$\ddot\eta^i=-R_{ij}\eta^j,$$
where 
$R_{ij}=\partial_{ij}c$. 
This is exactly the Jacobi equation along a fixed geodesic if we let $R$ be the curvature tensor with respect to the tangent vector of the geodesic. (It may be that in the case $b=0$ we can interpret our problem in the sense of the first paragraph of this answer for a certain Riemannian metric, but I am not sure. If this is true, then the harmonic oscillator should become the round sphere.) We should expect that the curvature will put certain restrictions on when conjugate points occur or do not occur. For example, if $R$ is positive and large, then by Bonnet-Myers Theorem, there exists certain $T_0$ such that for any $\nu$ and $q$, there exists $T\in (0,T_0)$ such that $(\nu,q,T)\not\in C$; therefore, $C$ cannot be connected. This is exactly what happens in the case of the harmonic oscillator and this would happen if we restricted our attention to solutions with bounded value of $|\nu|$, because in that case our solution will stay for time $T_0$ in a fixed compact region, where we could cook up $c$ so that $R$ is very large. However, if $|\nu|$ is too large, then $\gamma$ will escape from our compact region very fast and conjugate points will not have enough time to develop. (In other words, $(\nu,q,T)\in C$ for all $T>0$.) I think that in the nonnegative curvature case, the complement of $C$ should consist of a set of closed hypersurfaces that go to infinity as $|\nu|$ approaches a certain threshold; therefore, $C$ will still not be connected. On the other hand, if $R$ is always negative, then Hadamard-Cartan Theorem gives that all possible points lie in $C$ (thus it is connected).
Here is a particular outcome of the thoughts in the previous paragraph. For each nonnegative integer $k$, consider the set
$$
C_k=\{(\nu,q,T)\in C\mid N(T'\in (0,T)\mid (\nu,q,T')\not\in C)=k\};
$$
namely, we count how many degenerate points we had before time $T$. (We might need to introduce some multiplicities.) Then each $C_k$ is open; since $C_k$ form a nonintersecting cover of $C$, they should represent different connected components. The component $C_0$ is always nonempty; if we are able to prove that, say, $C_1$ is nonempty, then $C$ is not connected.<|endoftext|>
TITLE: Elementary solutions to f(z+1)-f(z)=g(z) in entire functions
QUESTION [22 upvotes]: Let g(z) be an entire function of a complex variable z. Does there exist an entire function f(z) such that f(z+1)-f(z)=g(z)? As I learned several years back, the answer to this is apparently 'yes', but I have not felt satisfied with the proof because it goes beyond my expertise.
I tried to find f using the power series expansion of g, for that works when g is a polynomial. But the results of partial inversions kept diverging. Representing g as an integral via Cauchy's formula, and doing inversion inside the integration led to similar problems. Perhaps I am overly optimistic, but a question this elementary should have equally elementary solution. Is there such a solution? If not, is there a reason to expect that no simple and elementary solution should exist?

REPLY [31 votes]: It took me some time to find a solution that satisfies both requirements:
a) If should be based on the power series expansion
b) It should use no tools heavier than contour integration.
So, let $g(z)=\sum a_ k z^k$. We know that $a_ k$ decay faster than any geometric progression. We want analytic functions $F_ k(z)$ such that $F_ k(z+1)-F_ k(z)=z^k$ and $F_ k(z)$ grows not faster than a geometric progression as a function of $k$ in every disk. Then $f=\sum a_ k F_ k$ is what we want. So, just choose any odd multiples $r_ k\in(k,k+2\pi)$ of $\pi$ and put
$$
F_ k(z)=\frac{k!}{2\pi i}\oint_ {|z|=r_ k}\frac {e^{z\zeta}}{e^\zeta-1}\frac{d\zeta}{\zeta^{k+1}}
$$
The key is that $|e^\zeta-1|\ge \frac 12$ on the circle and $r_ k^k\ge k!$, so $|F_ k(z)|\le 2e^{|z|(k+2\pi)}$ on the plane.<|endoftext|>
TITLE: Is there a theorem that says that there is always more than one way to "continue a finite sequence"?
QUESTION [6 upvotes]: I have come across a bit of folklore(?) which goes something like "given any finite sequence of numbers, there is more than one 'valid' way of continuing the sequence". For example see here. I would like to know if this is actually stated and proved rigorously, and if so, where can I find a statement and proof?
EDIT: The first couple of answers reflect my concerns exactly. But to quote from the book review I linked to, 

Wittgenstein's Finite Rule Paradox implies that any finite sequence of numbers can be a continued in a variety of different ways - some natural, others unexpected and surprising but equally valid.

I didn't use the terms "Wittgenstein's Finite Rule Paradox" and "Wittgenstein on rule-following" before as googling them turns up results which look more like philosophy and linguistics than mathematics. My background in logic is nonexistent, I'm looking for any logicians out there who may have seen this before.

REPLY [8 votes]: There is an area of computer science called grammatical inference.
Let $\Sigma$ be a finite alphabet, $\Sigma^*$ be the set of finite length strings over $\Sigma$ and a language be a set of strings. Let $F$ be a family of languages. The data for the problem is a sequence of words $w_0, w_1, \ldots$ from a language in $F$. A learner receives the data and generates a sequence of languages $H_0, H_1, \ldots$ called hypotheses. 
Typically the hypothesis $H_i$ includes all words provided till that point. The learner stabilizes if the hypotheses do not change after a point and is successful if the stable hypothesis is the language from which the words were drawn. A family of languages can be learnt in the limit if there is an algorithm that can successfully learn the source language. 
There is a theorem due to E. Mark Gold (Language Identification in the Limit, 1967) stating that if the family $F$ contains all languages of finite cardinality and at least one language of infinite cardinality, it cannot be learnt in the limit. 
This result may be one formalisation related to the original question, but does not tell the whole story. Dana Angluin (Inference of Reversible Languages, 1982) showed that there are families containing infinitely many languages that can be learnt in the limit. There is a recent text Grammatical Inference: Learning Automata and Grammars by Colin de la Higuera devoted to this area.<|endoftext|>
TITLE: How many of the true sentences are provable?
QUESTION [20 upvotes]: Is there a  natural  measure on the set of statements which are true in the usual model (i.e. $\mathbb{N}$) of Peano arithmetic which enables one to enquire if 'most' true sentences are provable or not? By the word 'natural' I am trying to exclude measures defined in terms of the characteristic function of the set of true sentences.

REPLY [21 votes]: It seems to me that the probability that a statement is provable and that it is undecidable should both be bounded away from 0, for any reasonable probability distribution.
Let $C_n$ be the number of grammatical statements of length $n$. For any statement $S$, the statement 

$S$, or $1=1$

is a theorem. So the number of provable statements of length $n$ is bounded below by $C_{n-k}$, where $k$ is the number of characters needed to tag on "or $1=1$".
On the other hand, let $G$ be an undecidable sentence, and $S$ any sentence. Then

Either $S$ and $1 \neq 1$, or else $G$

is undecidable. So the number of undecidable sentences of length $n$ is bounded below by $C_{n-\ell}$, for some constant $\ell$. For any reasonable grammar, the ratios $C_{n-k}/C_n$ and $C_{n- \ell}/C_n$ should both be bounded away from 0.
I am currently trying to figure out why my computation is seemingly incompatible with the paper of Calude and Jurgensen cited by Konrad. I suspect that the answer is hidden in the definition of prefix free, on page 4, but I am trouble understanding it. Any help?<|endoftext|>
TITLE: Where was/is Compensated Compactness used?
QUESTION [11 upvotes]: This last summer, I read up on Tartar's so called Method of Compensated Compactness (or at least how it applied to scalar conservation laws). I used this theory to prove the existence of $L^{\infty}$ solutions to the scalar conservation law $u_{t}+(f(u))_{x}=0$ with initial data in $L^{\infty}$. Here we do not need to have a strictly hyperbolic conservation law. Perhaps just assume $f\in C^{2}$. The sources I used for my presentation were from Yunguang Lu's book Hyperbolic Conservation Laws and the Compensated Compactness Method and Dafermos's book Hypberbolic Conservation Laws in Continuum Physics. 
I am aware that Compensated Compactness can be used to prove existence of solutions to $2\times 2$ systems of conservation laws. I haven't looked over this proof, but I believe Lu covers this.
My question is as follows: Where else is the method of Compensated Compactness used? If you can say, if possible, could you give a rough, brief sketch of how the method was used? Any input/thoughts would be appreciated.

REPLY [11 votes]: Compensated compactness helps when one needs to find the limit of $u_n \cdot v_n$, where the sequences of vector fields $u_n$ and $v_n$ converge weakly  in $L^2$:  $u_n\rightharpoonup u$,  $v_n\rightharpoonup v$. If none of the sequences converges strongly in $L^2$, the vector fields can still  possess some additional properties which would compensate for the lack of strong convergence. For instance, the following result and its versions are often used in the homogenization theory. 

Lemma. Let $u_n$, $v_n\in L^2(\Omega)$, where $\Omega$ is a bounded domain in $\mathbb R^d$ and let $u_n\rightharpoonup u$,  $v_n\rightharpoonup v$ in $L^2(\Omega)$. Assume that $$\mbox{curl }v_n=0\qquad  \mbox{and}\qquad \mbox{div } u_n\to w\quad \mbox{in }\ H^{-1}(\Omega).$$ Then $u_nv_n\to uv$ in the weak-* topology of $L^1(\Omega)$.

The lemma can be used to justify the convergence of solutions of the Dirichlet problem with 
strongly oscillating coefficients
$$\mbox{div }\left(a\left(\frac{x}{\varepsilon}\right)\nabla u_\epsilon\right)=f,\qquad 
\left.u\right|_{\partial\Omega}=0,$$
to the solutions of some averaged problem (see, e.g., "Homogenization of differential operators and integral functionals" by V. V. Jikov, S. M. Kozlov, O. A. Oleinik).<|endoftext|>
TITLE: What are the open subsets of $\mathbb{R}^n$ that are diffeomorphic to $\mathbb{R}^n$
QUESTION [64 upvotes]: I would like to know if there is a known necessary and sufficient
property on an open subset of $\mathbb{R}^n$ to be diffeomorphic to $\mathbb{R}^n$ :
For example :

Are all open star-shaped subsets of $\mathbb{R}^n$ diffeomorphic to $\mathbb{R}^n$ ?

Reciprocally, are all open subsets of $\mathbb{R}^n$ which are diffeomorphic to $\mathbb{R}^n$, star-shaped ?


Thank you for your answers and proofs

REPLY [2 votes]: Yet another proof in the case where $\Omega$ is star-shaped, in little details unless anyone is interested.
Assume for simplicity that $\Omega$ is star-shaped around $0$. Define the function $r$ mapping $\theta\in\mathbb S(\mathbb R^n)$ to the only $r(\theta)$ such that $\Omega\cap \mathbb R_+\theta = [0,r(\theta))\theta$ (possibly $r(\theta)=\infty$). Then $r:\mathbb S(\mathbb R^n)\to[0,+\infty]$ is lower semicontinuous and uniformly positive, so there exists a sequence $(r_k)_{k\geq0}$ of smooth uniformly positive functions such that $r=\sum_{k\geq0}r_k$ (e.g. use this post to write $r$ as an increasing limit of Lipschitz functions and reduce to the Lipschitz case, then iteratively fit a smooth function $r_{k+1}$ strictly between $(r-r_k)/2$ and $r-r_k$ by using a partition of unity). Without loss of generality, $r_0$ is actually a (positive) constant.
The idea is that we are looking for a diffeomorphism $\mathbb R^n\to\Omega$ such that the image of the sphere of radius $k$ is the surface $\lbrace |x|=r_k(x/|x|)\rbrace$. The final diffeomorphism will end up being a bit different, but I think it is a good image to keep in mind.
Find a collection of smooth functions $(\rho_k)_{k\geq0}$ from $\mathbb R_+$ to $[0,1]$ such that

for all $k>0$, $\rho_k$ starts at zero, then increases and finally stays constant equal to one, in such a way that $\rho_k'$ has support in $(k-1,k+2)$ and is (uniformly) positive over $[k,k+1]$;
$\rho_0$ starts at zero, then increases and finally stays constant equal to one, in such a way that $\rho_0'$ is one in a neighbourhood of zero, has support in $[0,2)$, and is (uniformly) positive over $[0,1]$.

Then $$\Phi:x=|x|\cdot\frac x{|x|}=r\cdot\theta\mapsto\sum_{k\geq0}\rho_k(r)r(\theta)\theta$$ is smooth at zero (it is a constant multiple of the identity) and everywhere else (it is locally a finite sum of smooth functions), it is injective (because it preserves rays and is strictly increasing on rays) and surjective with image $\Omega$ (because $r=\sum_{k\geq0}r_k$), and its differential is invertible (away from zero because it is strictly increasing with positive derivative along the rays and it preserves the angles). It means that $\Phi$ has an inverse $\Phi^{-1}$ that is locally smooth, so $\Phi$ is actually a diffeomorphism.<|endoftext|>
TITLE: Assumptions on the category C for sheafification of C-valued presheaves
QUESTION [11 upvotes]: For any category C and topological space X we have the notion of a C-valued presheaf on X.
What assumptions must be made about C in order that we have the notion of such a presheaf being a 'sheaf'? I understand the definition of the sheaf properties using an equalizer diagram which assumes C has products and a final object. Is this definition 'standard'?
Secondly, the definition of a sheafification of a presheaf in terms of the obvious universal property makes sense for any category C (for which the notion of sheaf makes sense). But what assumptions must be placed on C in order for such a sheafification to exist? For presheafs of sets I know the construction via the étale space of the presheaf (namely, the sheafification can be constructed as the sheaf of sections of the projection E->X of the etale space E onto X). This construction works in general right?

REPLY [2 votes]: Let PS(X,B) denote the category of presheaves on the category with values in category B. We denote by Sh((X,SX),B) the category of sheaves on the topos (CX,SX)(CX is category representing X and SX is pretopology). The assumption for category B is that 
B has filtered colimits which commutes with kernels of pairs of arrows. Let HX denote the endofunctor of PS(X,B) which assigns to every presheaf F: CX^op--->B of the presheaf HX(F) defined by 
HX(F)(N)=colim(Ker(F(M)-->-->F(M product M over N))),where N and M are objects of CX
where colimit is taken over Cov(N,t):(Cover of M, t is determined by SX mentioned above). We can easily check that Cov(N,t) is a filtered category. So the colimits we need is filtered colimits. 
We can also easily check that F(N)--->HX(F)(N) is a functor for each N belongs to ObCX
i.e. it defines a functor morphism F|--->HX(F)
This functor HX is called Heller functor. It is routine to check HX is left exact.
with following property 
1 Functor HX maps any presheaf to monopresheaf and maps any monopresheaf to sheaf. So sheafification functor is just HX compose HX.<|endoftext|>
TITLE: Exotic spheres and stable homotopy in all large dimensions?
QUESTION [14 upvotes]: Given that the kervaire invariant one problem has been solved in (almost) all dimensions....my question is whether there exists an exotic sphere in all sufficently lagre dimensions? Given the Kervaire-Milnor theory this "quasireduces" to determining whether coker J has nontrivial stable homotopy in all large dimensions. Has anyone looked at this? I've glanced at the work of Ravenel, etc. which doesn't seem to be sufficent.

REPLY [3 votes]: As Ryan Budney indicated, Mark Behrens gave a talk about joint work with Mike Hill, Mike Hopkins, and Mark Mahowald at the Milnor conference a few days ago.  Here's a link to the video.
The "reader's digest" version of this is that, by using periodic families of elements in the stable homotopy groups of spheres, they've established a large number of congruences where exotic spheres are always forced to exist, but it's not yet exhaustive.  In addition, they've done enough low-dimensional computations to know that in dimensions less than or equal to (at least) 63, there are exotic spheres in all dimensions except 1, 2, 3, 4, 5, 6, 12, and 61.  The first of these were known from work of Kervaire-Milnor in 1963 (as part of their enumeration of the number of exotic spheres that's indicated by the Wikipedia entry), but the nonexistence of exotic spheres in dimension 61 is a new result.<|endoftext|>
TITLE: Cures for mathematician's block (as in writer's block)
QUESTION [46 upvotes]: What kind of things do you find that help you get the "creative juices flowing," to use a tired cliche, when you're stuck or burnt out on a problem? I've read about some studies that suggest listening and playing music can stimulate mathematical thinking. Any particular style that someone finds helpful?
Other things that help?
(In case you haven't figured it out, I've been stuck on some things lately.)

REPLY [6 votes]: Here are a few that I haven't seen above yet:
1) Go for long walks (with a bit of math in mind). Don't worry about thinking about the math, just put it in your head and then walk. This is distinct from walking to clear your mind, as putting the math in there tends to get the juices flowing whether you like it or not!
2) Force yourself to write down questions. Lots and lots of questions...about whatever mathematics comes to mind. 
An advisor of mine said that there are uncountably many open math questions out there, some of which you have the tools to solve, in contrast to the finite list of `big fish' questions we usually focus on. Every hundred or so questions, you may be able to answer something or will at least feel a bit energized. If you asked the question yourself, chances are it is connected to something you've worked hard on...perhaps indirectly.<|endoftext|>
TITLE: Can we count isogeny classes of abelian varieties?
QUESTION [15 upvotes]: Let's fix a finite field F and consider abelian varieties of dimension g over F.  Can we say how many isogeny classes there are?  Is it even clear that there's more than one isogeny class?  For g=1, and some fixed F with characteristic not 2 or 3, we could probably just write down all the Weierstrass equations and count isogeny classes by brute force, but is there a cleaner way to do it?

REPLY [5 votes]: Let Fq be a finite field.  Two elliptic curves E1 and E2 defined over Fq are Fq-rationally isogenous iff # E1(Fq) = # E2(Fq).
(==>) Let \varphi be an isogeny, take l a prime which is prime to q*(degree of \varphi).  Then \varphi induces a Frobenius-equivariant isomorphism on the l-adic Tate modules, so the characteristic polynomials of Frobenius agree.  Since
E(Fq) = q+1 - trace(Frob),
this implies that the two elliptic curves have the same number of points.
(<==) Similarly, if the elliptic curves have the same number of rational points, their Frobenius traces aq are the same, so their characteristic polynomials of Frobenius are both T2 - aq T + q2.  It follows from Honda-Tate theory that they have the same number of points.
Since it is known exactly what the possibilities for #E(Fq) are in terms of q -- this the Hasse-Deuring-Waterhouse theorem -- it seems that it is known exactly how many Fq-rational isogeny classes there are.  For instance, when q = p is prime, the theorem asserts that any integer N satisfying the Weil bounds
p+1 - 2\sqrt{p} < N < p+1 + 2\sqrt{p}
occurs as #E(Fp) for some E/Fp, so it seems that there are exactly
1 + 2*floor(2\sqrt{p})
Fp-rational isogeny classes.
For abelian varieties of dimension g, to determine the Frobenius polynomial and hence the Fq-rational isogeny class, you need to know not just #A(Fq) but #A(Fqi) for 1 <= i <= g.
The answer for isogeny over the algebraic closure is different -- let me know if you're interested in that.<|endoftext|>
TITLE: What happens to Newtonian systems as the mass vanishes?
QUESTION [10 upvotes]: This question is closely related to another one I asked recently, and may be thought of as a warm-up to that one.
Consider $\mathbb R^n$ with its usual metric, and pick a one-form $b$ and a function $c$.  Let $m$ be a positive constant, and consider the second-order differential equation for a function $q(t)$
$$ m\ddot q = db \cdot \dot q + dc $$
where I have used the metric to identify vectors and covectors, $dc$ is the differential of $c$, and $db$ is the exterior derivative of $b$ (it is contracted with $\dot q$ to yield a covector).  In coordinates, and using Einstein's summation convention:
$$ m\ddot q^i = \left(\partial\_i b\_j - \partial\_j b\_i\right)\dot q^j + \partial\_i c $$
I am interested in the limit when $m\to 0$.  For example, when $m=0$ and $b=0$ (or anyway when $b$ is closed), then the differential equation forces the path $q(t)$ to stay within the set of critical points of $c$ (this set is generically discrete, so that the only solutions are constant).  At another (more generic) extreme, $db$ might be nondegenerate, and hence a symplectic form on $\mathbb R^n$.  Then the equation $0 = db \cdot \dot q + dc$ is a nondegenerate first-order differential equation, exactly equivalent to Hamilton's equations for the symplectic manifold $(\mathbb R^n,db)$ with Hamiltonian $-c$.  There is some gradation when $db$ is nonzero but has nontrivial kernel (as for example must happen if $n$ is odd).
So I basically get what happens when $m=0$.  But can we understand the limit $m\to 0$?  For example, if $m\neq 0$, then any initial value $(\dot q(0),q(0))$ determines a solution; for fixed initial values, how does this solution vary as $m\to 0$?  Alternately, we can try to solve the boundary value problem, in which we prescribe $q(0)$ and $q(1)$.  Then what happens to the solutions as $m$ shrinks?  Since when $m=0$ we cannot find solutions with arbitrary initial velocity, it is unlikely that anything is particularly well-behaved in the limit, but not impossible.
Very specifically, I would like to know about the asymptotics of the solutions to the boundary and initial-value problems — what do solutions look like when $m$ is a formal variable?  But more generally I'm happy with some statements about the regularity in the $m\to 0$ limit.

REPLY [2 votes]: This is not a real answer, but there is a branch of mathematics called "semiclassical analysis" which might be related. For example, consider a degenerate version of the problem above:
$$
(-h^2 \partial_x^2+V(x))u=0.
$$
Here $h^2=m$ and $V=dc$; we assume that $n=1$. Then the limit as $h\to 0$ is called "semiclassical limit". What should happen (if you prescribe some boundary conditions) is that the possible solutions $u$ should get "microlocalized" near the zero set $\{p=0\}$ of the semiclassical symbol
$$
p(x,\xi)=\xi^2+V(x).
$$
Here a function $u$ is "microlocalized" near a subset $K$ of the cotangent bundle if a certain norm $\|Au\|$ is small for any pseudodifferential operator $A$ with symbol $a$ supported outside of $K$.
A physical explanation of the above would be that our static Schr\"odinger equation governs the behaviour of a single quantum particle under the potential $V$ near the zero energy level; for small values of Planck constant $h$, this should correspond to the motion of a classical particle at this fixed energy level.
There are several sources to read about semiclassical analysis, including 
lecture notes by Evans-Zworski and a book by Dimassi and Sjostrand.<|endoftext|>
TITLE: Definitions of Hecke algebras
QUESTION [34 upvotes]: There is a definition of Iwahori-Hecke algebras for Coxeter groups in terms of generators and relations and there is a definition of Hecke algebras involving functions on locally compact groups. Are these two concepts somehow related? I think I read somewhere that the Hecke algebra with functions includes Iwahori-Hecke algebras, is that correct? Is there a good motivation for introducing and studying either type of algebras? How much is known about their representations?

REPLY [5 votes]: The motivation for studying Iwahori-Hecke algebras (the convolution type) is the study of the unramified principal series representations. The details can get lengthy, but it is all in a paper of Borel's (Armand Borel. Admissible representations of a semi-simple group over a local field with vectors fixed under an Iwahori subgroup. Invent. Math., 35:233–259, 1976.)
The relationship between the convolution algebra and the algebra using Coxeter groups was due to Iwahori and Matsumoto (N. Iwahori and H. Matsumoto. On some Bruhat decomposition and the structure of the Hecke rings of p-adic Chevalley groups. Inst. Hautes Etudes Sci. Publ. Math., (25):5–48, 1965.)<|endoftext|>
TITLE: Uniformization in algebraic/arithmetic geometry?
QUESTION [6 upvotes]: Jonah's question makes me wonder: What is with uniformization in algebraic/arithmetic geometry? E.g. this article by Faltings seems to be about that, the Shimura-Taniyama statement too, Mochizuki discusses a p-adic version of Fuchsian uniformization of hyperbolic curves. Do you know surveys or expositions of the theme? Or is 'uniformization' a mistaken concept in the context of arithmetic geometry (a remark in Faltings article sounds as if saying that)?

REPLY [3 votes]: Presumably, you want to look at uniformizations of curves, since blow-ups make it difficult to classify covers of higher dimensional varieties.  Scheme-theoretically, there doesn't seem to be a good notion of uniformization, but one can get good arithmetic results using analytification.
You already mentioned a couple references.  For the complex-analytic view, I think any book on modular forms (e.g., Shimura) should have some treatment.  Mochizuki's work in the nonarchimedean setting has a counterpart in the maximally degenerate case of Mumford curves.  There is a book by Gekeler and van der Put on uniformizations of these (essentially using the Bruhat-Tits building for SL2).  I think Darmon and Dasgupta have done some arithmetic work using this "upper half plane" uniformization.  It historically originates from Tate's work around 1960 on rigid analytic spaces (apparently, Grothendieck expressed doubt that such a theory would exist).<|endoftext|>
TITLE: What is the "intuition" behind "brave new algebra"?
QUESTION [37 upvotes]: Y.I. Manin mentions in a recent interview
the need for a “codification of efficient new intuitive tools, such as … the “brave new algebra” of homotopy theorists”. This makes me puzzle, because I thought that is codified in e.g. Lurie’s articles. But I read only his survey on elliptic cohomology and some standard articles on symmetric spectra. Taking the quoted remark as indicator for me having missed to notice something, I’d like to read what others think about that, esp. what the intuition on “brave new algebra” is.
Edit:
In view of Rognes' transfer of Galois theory into the context of "brave new rings" and his conference last year, I wonder if themes discussed in Kato's article (e.g. reciprocity laws) have "brave new variants".
Edit: I found Greenlees' introductions (1, 2) and Vogt's "Introduction to Algebra over Brave New Rings" for getting an idea of the topological background very helpful.

REPLY [15 votes]: re: Manin's comments, the article says that "Manin edited this translation for publication in the Notices", so it is not surprising the English and Russian versions are different.<|endoftext|>
TITLE: Is there an infinity × infinity lemma for abelian categories?
QUESTION [24 upvotes]: Many people know that there is a (3×3) nine lemma in category theory.  There is also apparently a sixteen lemma, as used in a paper on the arXiv (see page 24).  There might be a twenty-five lemma, as it's mentioned satirically on Wikipedia's nine lemma page.
Are the 4×4 and 5×5 lemmas true?  Is there an n×n lemma?  How about even more generally, if I have an infinity × infinity commutative diagram with all columns and all but one row exact, is the last row exact too?  For all of these, if they are true, what are their exact statements, and if they are false, what are counterexamples?
Note: There are a few possibilities for what infinity × infinity means -- e.g., it could be Z×Z indexed or N×N indexed.  Also, in the N×N case, there are some possibilities on which way arrows point and which row is concluded to be exact.

REPLY [18 votes]: Yes, there is an $n\times n$ lemma, and even an $\mathbb N\times\mathbb N$ lemma. The spectral sequence argument that Reid gives works. Another elementary proof uses the salamander lemma, a result of George Bergman's that I blogged about at SBS. It's exactly the same as the proof of the $3\times 3$ lemma I wrote up there.
Here's a counterexample to the $\mathbb Z\times\mathbb Z$ lemma. If you read about the salamander lemma, you'll understand how I came up with it. All non-zero maps are the identity
$$\require{AMScd}
\begin{CD}
0 @>>> 0 @>>> 0 @>>> 0 @>>> 0\\
@.    @VVV    @VVV   @VVV   @VVV\\
0 @>>> 0 @>>> 0 @>>> \mathbb{Z} @>>> 0\\
@.     @VVV   @VVV   @VVV            @VVV\\
0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\
@.     @VVV   @VVV             @VVV           @VVV\\
0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 @>>> 0
\end{CD}
$$
Extend the diagram by copies of $\mathbb Z$ down and to the left, and put $0$'s everywhere else. All columns are exact, and all rows except one (the one with a single $\mathbb Z$ in it) are exact.<|endoftext|>
TITLE: Maps to projective space determined by a line bundle
QUESTION [22 upvotes]: The following should be pretty standard for any algebraic geometer.
Let $X$ be a compact complex variety, and let $L$ be a line bundle on $X$.  We say $L$ is 'generated by global sections' if for every point $p$ in $X$, there is a global section of $L$ which doesn't vanish.  If this is true, then $L$ determines a map to a projective space in the following way.  The global sections of $L$ are finite dimensional, so choose a basis $(a_i)$.  Then send a point $p$ in $X$ to the projective point 
$$
[a_1(p):a_2(p):...:a_n(p)]
$$  
It should be noted that $a_i(p)$ is only a point in the fiber of $L$ over $p$, and not a complex number. By choosing an isomorphism from the fiber over $p$ to $\mathbb{C}$, the $a_i(p)$ can be identified with complex numbers.  The ambiguity introduced in choosing this isomorphism dissappears when taking the projective coordinates.
This constructions is remarkably ad hoc for something that ends up being foundational in algebraic geometry.  It requires the variety be over $\mathbb{C}$, and requires that some inconsequential choices be made enroute.
My question is, what are nicer, more intrinsically algebraic ways to construct this map?  Three ways this construction could be nicer:

It could work over other fields, or possibly even over $\mathbb{Z}$ (though then its less clear what a line bundle should be).
It could give a good intuitive justification of why this map is a natural and powerful thing to look at.
It could lend itself to generalizations in different directions.  For instance, a rank n vector bundle V with 'enough global sections' (in some sense) should determine a map from X to $Hom_{\mathbb{C}}(\Gamma(V),\mathbb{C}^n)//GL(n)$ (where this is a GIT quotient).

Aside.  Is there even a good name for this construction?  The "map to projective space determined by a line bundle" is a bit long-winded.

REPLY [11 votes]: This answer is meant as (1) the stacky answer Ben anticipated, and (2) an answer addressing the generalization to higher rank bundles. The basic thing I assume you know (or are willing to take as definition) is that a morphism from $X$ to a quotient stack $[Y/G]$ is equivalent to the data of a $G$-torsor $P\to X$, and a $G$-equivariant morphism $P\to Y$. For $BG=[*/G]$, it's just the data of a $G$-torsor, since there's only one possible choice of map to a point.
First, the stacky interpretation. A choice of a line bundle $\mathcal L$ on a scheme $X$ is equivalent to a morphism to the stack $\def\GG{\mathbb G} B\GG_m$; the $\GG_m$-torsor is the complement of the zero section in the total space $\mathbb V(\mathcal L)$. A choice of a line bundle together with $n$ sections is equivalent to a morphism $f:X\to\def\AA{\mathbb A} [\AA^n/\GG_m]$; the $\GG_m$ torsor is the complement of the zero section of $\mathcal L$, and the $n$ map to $\AA^n$ is given by the $n$ regular functions which are the pullbacks of the sections. The condition that $f$ misses the one stacky point of $[\AA^n/\GG_m]$ (i.e. the point where the $\GG_m$ doesn't act freely)—and therefore factors through the open subscheme $[(\AA^n\smallsetminus 0)/\GG_m]=\mathbb P^{n-1}$ —is precisely the condition that the sections do not all vanish at the same point.
Now the generalization. A rank $k$ vector bundle $\def\E{\mathcal E} \E$ on a scheme $X$ is equivalent to a morphism to the stack $BGL_k$; the $GL_k$-torsor is the sheaf $\def\O{\mathcal O} Isom(\O^k,\E)$. A vector bundle together with a choice of $n$ sections is equivalent to a morphism $f:X\to [(\AA^k)^n/GL_k]$; again, the $GL_k$-torsor is $Isom(\O^k,\E)$, and the $k\cdot n$ regular functions on the torsor are given by the pullback of the $n$ sections of $\E$, and the fact that the pullback of $\E$ is canonically identified with $\O^k$. Regarding $(\AA^k)^n$ as the space of $k\times n$ matrices, the stacky locus of $[(\AA^k)^n/GL_k]$ (i.e. the points with non-trivial stabilizers) is the locus where the rank of the matrix is less than $k$. So the condition that $f$ misses the stacky locus is equivalent to the condition that the $n$ given sections span the fiber of $\E$ at any point. The non-stacky locus is the open subscheme $[${$k\times n$ matrices of rank $k$}$/GL_k]$, the Grassmannian of $k$-planes in $n$-space, $Gr(k,n)$. We therefore have the following interpretation.

A vector bundle $\E$ on a scheme $X$, together with $n$ sections $s_1,\dots, s_n$ which span every fiber is equivalent to a morphism $X\to Gr(k,n)$.

If you don't want to dirty things up by choosing the sections, you can replace $(\AA^k)^n$ by $Hom(\Gamma(\E),\mathbb C^k)$ everywhere. The non-stacky locus is then the space of surjective linear maps. Then it looking at kernels, it looks like you naturally get $Gr(n-k,n)$ instead of (the isomorphic) $G(k,n)$. I'm sure this has something to do with a dualization involved in forming the total space of a locally free sheaf; it still regularly confuses me, so I'll just leave this thread loose.<|endoftext|>
TITLE: When is an algebraic space a scheme?
QUESTION [24 upvotes]: Sometimes general theory is "good" at showing that a functor is representable by an algebraic spaces (e.g., Hilbert functors, Picard functors, coarse moduli spaces, etc).  What sort of general techniques are there to show that an algebraic space is a scheme?  Sometimes it's possible to identify your algebraic space with something "else" (e.g. it "comes" from GIT as is the case for the moduli space of curves), but are there other methods?

REPLY [10 votes]: Another example: the Nakai-Moishezon theorem says that a divisor D on X is ample iff for every curve C on D, $D \cdot C > 0$ and $D^2 > 0$. This holds also for an algebraic space.
As an application, you can show for instance that the coarse space of $\bar{M_g}$, the Deligne-Mumford compactification of the moduli stack of smooth genus g curves, is represented by a projective variety. The point is that Artin's representibility theorem tells you that the coarse space exists as an algebraic space, and you can then use Nakai-Moishezon to show that it has an ample line bundle. This is cool because it avoids GIT.<|endoftext|>
TITLE: Analogue of Sperner's lemma for Lefschetz theorem?
QUESTION [9 upvotes]: Sorry if this is easy/well-known, I don't know much algebraic topology and I'm just curious about this question.
One of the easier proofs of the Brouwer fixed-point theorem (we'll say for n = 2 for concreteness in the terminology, although it generalizes) proceeds by establishing Sperner's lemma and noting that a continuous map gives us a Sperner labeling on a triangulation of our disk. All the "real work" in this proof is in establishing Sperner's lemma, which can be done completely combinatorially.
So I know that the Lefschetz fixed-point theorem generalizes Brouwer's theorem, and that it applies to more general spaces than Brouwer. Is there a (relatively) simple combinatorial statement, analogous to Sperner, that can be easily shown to imply the Lefschetz theorem, at least on some large class of topological spaces?

REPLY [7 votes]: Probably there aren't any such combinatorial statements in the literature, the difficulty being that the statement of Lefschetz theorem involves looking at the induced action on the homology. 
There is a combinatorial lemma - Tucker lemma - which implies the Borsuk-Ulam theorem. Both the Sperner Lemma and Tucker lemma had many generalizations (Ky Fan contributed many, there is a famous deep extension of Sperner lemma by Shapley), although there is less activity in this line of reaserch these days.<|endoftext|>
TITLE: Valuative criterion for properness
QUESTION [7 upvotes]: Let $f : X \rightarrow Y$ be a finite type morphism of Noetherian schemes. The valuative criterion for properness runs as follows. Suppose that for any DVR $R$ with fraction field $K$ that any $K$-valued point of $X$ lying above an $R$-valued point of $Y$ extends uniquely to an $R$-valued point of $X$. Then $f$ is proper.
Does it suffice to check instead that any $K$-valued point of $X$ lying above an $R$-valued point of $Y$ extends to an $R'$-valued point of $X$, where $R'$ is the integral closure of $R$ in a finite extension $K'$ of $K$?

REPLY [7 votes]: Yes.
I claim that, for any $K'$ point of $X$, if this point extends both to an $R'$ point and an $K$ point, then it extends to a $R$ point. This obviously proves the result. 
Let $x'$ be the closed point of $\mathrm{Spec}(R')$. Let $\mathrm{Spec}(A)$ be an affine neighborhood of the image of $x'$. Then the $R'$ point must factor through $\mathrm{Spec}(A)$. The $K$ point also factors through $\mathrm{Spec}(A)$, as it has the same image as the $K'$ point. 
We can now transform the question into algebra. The algebraic statement is that we have a map $A \to K'$ which factors through both $K$ and $R'$. But then the image of this map must lie in $K \cap R' = R$. So the map factors through $\mathrm{Spec}(R)$, and our $K$ point extends to an $R$ point.<|endoftext|>
TITLE: Can an algebraic space fail to have a universal map to a scheme?
QUESTION [14 upvotes]: Let $\mathcal{X}$ be an algebraic space. Can it happen that there does not exist a map $\mathcal{X} \to X$ with $X$ a scheme that is initial for maps from $\mathcal{X}$ to schemes? Are there reasonable conditions (e.g. finite type) that we can put on $\mathcal{X}$ so that there does exist such a universal map to a scheme?

REPLY [4 votes]: I think the answer is almost certainly yes, an algebraic space can fail to have a universal map to a scheme (a "schemification"). I don't have a proof, but I think I know the right place to look for one (besides David Rydh's immediate surroundings).
If we can find two maps of schemes which do not have a coequalizer in the category of schemes, but do have a coequalizer in the category of algebraic spaces, then the coequalizer algebraic space will not have a schemification.
Consider Hironaka's example of a non-projective proper variety (see page 15 of Knutson's Algebraic Spaces). It has an action of ℤ/2 for which there is an algebraic space quotient. But in the category of schemes, there is no geometric quotient for this action. The question is whether there is a categorical quotient in this case.<|endoftext|>
TITLE: Hironaka desingularisation theorem -- new proofs in literature?
QUESTION [19 upvotes]: I'm wondering what the landscape looks like for proofs of Hironaka's desingularisation theorem.

Are there many proofs in the literature?
Is there a commonly accepted simplest bare-knuckle proof out there
that might be considered a pleasant read for people outside of
algebraic geometry?
Would that still be Hironaka's original manuscript?

REPLY [3 votes]: Old thread, but let me point out the recent preprint Very fast, very functorial, and very easy resolution of singularities, by M. McQuillan and G. Marzo.
There is a also published version on GAFA, with McQuillan listed as the only author.<|endoftext|>
TITLE: Regularity of sparse Fourier transforms
QUESTION [5 upvotes]: Suppose $F$ has discrete Fourier transform $(a_n)$ where $a_n=0$ unless $n=2^k$ for some $k > 0$, in which case $a_n=1/k$ (or $a_n=1/k^2$ if you want: I'm happy with anything polynomial).  What sort of regularity conditions does $F$ have?  Is it Holder continuous, or not?
To be explicit:
$$
F(x)=\sum_{k=1}^\infty k^{-2} \exp(ix2^k)
$$
for example.
More generally, I'm interested in two dimensional (discrete) Fourier transforms: is there a good reference for this sort of thing?

REPLY [4 votes]: Gian Maria Dall'Ara's comment is the solution. This function that you describe is a (typical) example of a continuous function that's nowhere differentiable. In fact, suppose that you have an integrable function $F$ such that $\hat F(n) = a _ n $ whenever $n=\lambda _ k$ and zero otherwise, where we assume that the sequence $\lambda_k$ is lacunary in the sense of Hadamard (i.e ${\lambda _  {k+1}} / {\lambda _ k}\geq c$). If the function $F$ is differentiable at some point then $a _ {\lambda _k}=o(\frac{1}{\lambda _ k})$ (actually i have the impression that the proof of this fact uses the weaker assumption that $F$ is Lipschitz continuous at some point). More generally, if you replace differentiability of the function $F$ with $\alpha$-Hölder continuity (in a neighborhood of zero say) for $0<\alpha <1$ then you conclude that $a _ {\lambda _k}=O(\frac{1}{\lambda _ k ^\alpha})$. So your function is not $\alpha$- Hölder either.
Remark 1: The contrary is also true since $a _ {\lambda _k}=o(\frac{1}{\lambda _ k})$ implies that $F$ is differentiable at any point of the circle where the partial sums converge to the function. I have some doubts about the precise hypothesis needed here. I'm not sure if you need your function to have only positive spectrum, but your function here does anyways.
Remark 2: You can look in Grafakos book for example, or of course, in Zygmund's trigonometric series (that would be my first reference for this type of problems). Katznelson has also a lot of information. But I know that Grafakos book contains these results for sure.
Remark 3: So your function is nowhere differentiable and is not Hölder continuous either. However it has other nice properties. For example, it belongs to any $L^p$ for $1\leq p <\infty$ and the $L^p$ norm is comparable to the $L^2$ norm (here note that the lacunary gaps force the Littlewood-Paley pieces of the function to behave as independent random variables). On top of that, using kolmogorov's result on lacunary Fourier series you get an easy a.e convergence result of the partial sums to the function (something which is still true for $L^2$ functions in general, but several scales deeper and more difficult to prove).
Remark 4: Finally, your function has only positive frequences and belongs to $L^p$ on the circle, hence it belongs to the Hardy space $H^p$ on the circle. I don't know if you can use that in your problem, but it is a strong property.<|endoftext|>
TITLE: Explanation for Satake correspondence
QUESTION [9 upvotes]: Some time ago I was told there's an interesting classical Satake correspondence which I will write as 
$$[\mathop{\mathrm{disk}} \Rightarrow G] \,\backslash\, [\mathop{\mathrm{disk}^\times} \Rightarrow G] \,/\, [\mathop{\mathrm{disk}} \Rightarrow G] \,=\, X_*/W \,=\, G^\vee\mbox{-reps}$$
(where $G$ is reductive, $\mathrm{disk}$ could be the spectrum of either $\mathbb Z_p$ or $k[[t]]$ and $\Rightarrow $ denotes algebraic morphism) and its categorified/geometric version (equivariant perverse sheaves on affine grassmanian of $G$ form the tensor category of representations of $G^\vee$).
I think I'm missing the larger context here, though. I don't mean the context of perverse sheaves, geometric Langlands, etc. On the contrary, I feel like I miss any intuition for classical representation theory. Why would statements like this be interesting?
I wasn't able to find anything in wikipedia or nLab.
One thing I know is that the correspondence allows us to construct the Langlands dual group in a natural way. But still, it would be interesting to know if it's a part of larger picture and if  there are related results. 

Question: is there an intuition for Satake correspondence that would make its statement obvious?

REPLY [11 votes]: What you have written above isn't classical Satake; it's the generalized Bruhat decomposition.  Classical Satake is a much more interesting theorem, which says that the Hecke algebra of $G(\mathcal{K})$ over $G(\mathcal{O})$ (the compactly supported $G(\mathcal{O})$ bi-invariant functions on $G(\mathcal{K})$ with convolution multiplication) is isomorphic to the representation ring of $G^\vee$.  
Why is this interesting?  Because the Hecke algebra $L^2[G(\mathcal{O})\backslash G(\mathcal{K})/G(\mathcal{O})]$ is the endomorphism algebra of $L^2(G(\mathcal{K})/G(\mathcal{O}))$, so there's a bijection between $W$-orbits on $T^\vee$ and representations of $G(\mathcal{K})$ appearing in the $L^2$ above.<|endoftext|>
TITLE: Are supervector spaces the representations of a Hopf algebra?
QUESTION [14 upvotes]: Supervector spaces look a lot like the category of representations of $\mathbb{Z}/2\mathbb{Z}$ - the even part corresponds to the copies of the trivial representation and the odd part corresponds to the copies of the sign representation.  This definition gives the usual morphisms, but it does not account for the braiding of the tensor product, which I've asked about previously.
Since the monoidal structure on $\text{Rep}(G)$ comes from the comultiplication on the group Hopf algebra, which in this case is $H = \mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$, it seems reasonable to guess that the unusual tensor product structure comes from replacing the usual comultiplication, which is $\Delta g \mapsto g \otimes g$, with something else, or alternately from placing some extra structure on $H$.  What extra structure accounts for the braiding?

REPLY [4 votes]: The correct Hopf algebra is not the group algebra $kC_2$ but its dual $kC_2^*$. If the characteristic of $k$ is not $2$ then accidentally $kC_2\cong kC_2^*$. Otherwise, Greg answered the question as your new $R$-element is no longer trivial $1\otimes 1$ but $1 \otimes - 2\delta_x \otimes \delta_x$ where $\delta_x$ is the delta function on the generator of $C_2$.<|endoftext|>
TITLE: Classifying triangulated structures on a graded category
QUESTION [14 upvotes]: I know of several results to the effect that two triangulated categories are equivalent categories (usually one coming from algebra and one coming from topology).  However, it's never been clear to me how to classify the possible different triangulated structures lifting a given "graded" category (a category enriched in graded abelian groups).
For example, how many different triangulated structures can there be with underlying category the derived category of Z-modules?  All objects in this category are equivalent to direct sums of Z-modules concentrated in different degrees.
As another example, we have the category of "Z/2-graded abelian groups", which is equivalent to the derived category of differential graded modules over a Laurent polynomial ring on a generator in degree 2.  There's another, equivalent-looking, homotopy category of modules over the periodic complex K-theory spectrum KU.  These have equivalent underlying categories but I have never been clear on whether the triangulated structures are the same.

REPLY [3 votes]: In 2002, Paul Balmer wrote a nice two page note answering the same question.
He sent it to me because I had asked the same question in my paper ``The 
additivity of traces in triangulated categories'', where I had recalled the
standard fact (well-known in topological examples since the 1960's) that a 
triangulation usually differs from its negative.  I just looked, and it 
seems Paul never published his note.  I'll mostly use his notation.
Fix a category $\mathcal K$ with a triangulation $\mathcal T$ that differs 
from its negative $-\mathcal T$.  We will never change the suspension $\Sigma$.  His 
first example is trivial.  For any $n\geq 1$, the $n$-fold product $\mathcal{K}^n$
with the obvious suspension has at least $2^n$ distinct triangulations, choosing 
$\mathcal T$ or $-\mathcal T$ on each factor.  
Define a global automorphism $\alpha$ of $(\mathcal K,\Sigma)$ to be an automorphism of
the identity functor on $\mathcal K$ that commutes with $\Sigma$.  It is given by 
isomorphisms $\alpha\colon A\to A$ for all objects $A$ of $\mathcal K$ such that $\alpha f = f\alpha$
for all maps $f\colon A\to B$ in $\mathcal K$ and $\Sigma \alpha$ is the $\alpha$ for $\Sigma A$.
The simplest example is to take $\mathcal K$ to be (any version of) the derived category 
of a commutative ring $R$ and let $\alpha(u)\colon A\to A$ be multiplication by a unit $u\in R$.
For a global automorphism $\alpha$ of $\mathcal K$, define $\mathcal{T}_{\alpha}$ to be the
class of all triangles $(f,g,h)$ such that $(f\alpha,g,h)$ is a trangle in $\mathcal T$.  Then
$(\mathcal K,\mathcal{T}_{\alpha})$ is a triangulated category. 
Now return to the derived category $\mathcal K$ of a commutative ring $R$. Let $u$ be a unit of $R$.
Suppose there is an element $r$ such that $r$ is not a zero divisor and $1-u\notin rR$. 
Then the triangulations $\mathcal T$ and $\mathcal{T}_{\alpha(u)}$ are different. If 
we take $u=-1$, this very often proves that $\mathcal T \neq -\mathcal T$.  Now let 
$R = k[x]$ be a polynomial ring over a commutative ring $k$ with infinitely many units.
consider the units $u$ other than $1$ in $k\subset R$ and let $r=x$.  We have a 
triangulation $\mathcal{T}_{\alpha(u)}$ for each such $u$.  Clearly 
$(\mathcal {T}_{\alpha(u)})_{\alpha(v)} = \mathcal {T}_{\alpha(uv)}$.  Thus, if $\mathcal{T}_{\alpha(u)} = \mathcal{T}_{\alpha(v)}$,
then $\mathcal{T}= \mathcal{T}_{\alpha(u^{-1}v)}$.  Therefore $\mathcal K$ has infinitely
many distinct triangulations.<|endoftext|>
TITLE: Are there pairs of highly connected finite CW-complexes with the same homotopy groups?
QUESTION [36 upvotes]: Fix an integer n.  Can you find two finite CW-complexes X and Y which 
* are both n connected,
* are not homotopy equivalent, yet
* $\pi_q X \approx \pi_q Y$ for all $q$.
In Are there two non-homotopy equivalent spaces with equal homotopy groups? some solutions are given with n=0 or 1.  Along the same lines, you can get an example with n=3, as follows.  If $F\to E\to B$ is a fiber bundle of connected spaces such that the inclusion $F\to E$ is null homotopic, then there is a weak equivalence $\Omega B\approx F\times \Omega E$.  Thus two such fibrations with the same $F$ and $E$ have base spaces with isomorphic homotopy groups.
Let $E=S^{4m-1}\times S^{4n-1}$.  Think of the spheres as unit spheres in the quaternionic vector spaces $\mathbb{H}^m, \mathbb{H}^n$, so that the group of unit quarternions $S^3\subset \mathbb{H}$ acts freely on both.  Quotienting out by the action on one factor or another, we get fibrations
$$ S^3 \to E \to \mathbb{HP}^{m-1} \times S^{4n-1},\qquad S^3\to E\to S^{4m-1}\times \mathbb{HP}^{n-1}.$$
The inclusions of the fibers are null homotopic if $m,n>1$, so the base spaces have the same homotopy groups and are 3-connected, but aren't homotopy equivalent if $m\neq n$.
There aren't any n-connected lie groups (or even finite loop spaces) for $n\geq 3$, so you can't push this trick any further.  
Is there any way to approach this problem?  Or reduce it to some well-known hard problem?
(Note: the finiteness condition is crucial; without it, you can easily build examples using fibrations of Eilenberg-MacLane spaces, for instance.)

REPLY [3 votes]: There are quite a few finite CW-complexes $X$ and $Y$ that have homotopy equivalent loop spaces, thus have isomorphic homotopy groups, but themselves are not homotopy equivalent. Examples include taking $X=S^n\times S^n$ and $Y$ any simply connected space that has isomorphic cohomology ring to $X$, $X$ and $Y$ to be homotopically distinct closed simply connected $4$-manifolds with the same second Betti numbers, or more generally $(n-1)$-connected $2n$-manifolds at least when $n\neq 4,8$. This can be proved by obtaining homotopy decompositions of their loop spaces, showing the factors in the decomposititions of $\Omega X$ and $\Omega Y$ are the same.<|endoftext|>
TITLE: level sets of multivariate polynomials
QUESTION [11 upvotes]: Let $p:\mathbb R^n \rightarrow \mathbb R$ be a polynomial of degree at most $d$. Restrict $p$ to the unit cube $Q=[0,1]^n\subset\mathbb R^n$. We assume that $p$ has mean value zero on the unit cube $Q$:
$$\int_Q p(x) dx = 0.$$
For $\alpha>0$ consider the sublevel sets of $P$,
$$E_\alpha= \{x\in Q: |p(x)|\leq \alpha\}$$
There are several known estimates for the Lebesgue measure of this set which in some sense or another are uniform over some classes of polynomials. For example, we have that
$$|E_\alpha| \lesssim \min(pd,n) \frac{ \alpha^{1/d} }{ \|p\|_{ L^p(Q) }^{1/d} } $$
This particular estimate is due to Carbery and Wright and can be found here. 
I'm interested in studying the (induced Lebesgue) measure of the boundary of this set 
$$|\partial E_\alpha|=|\{x\in Q: |P(x)|=\alpha\}| $$
Consider first the easy case of dimension $n=1$. Then the set $E_\alpha$ is a finite union of closed intervals and the question is trivial. It is obvious that in this case there are at most $O(d)$ intervals so the $0$-dimensional measure of the boundary is $O(d)$.
Now in many variables things will be much more complicated. For example can we say that the set $E_\alpha$ has $O(d)$ connected components? Is there an estimate for the measure of the boundary $\partial E_\alpha $ in terms of $\alpha$, $d$ and $n$, assuming (say) that $\|p\| _ {L^1(Q)}=1$ ?
This question comes up naturally if one tries to study an oscillatory integral with phase $p$ and apply integration by parts (i.e Gauss theorem) imitating the one dimensional method of proving the van der Corput lemma (for example).

REPLY [3 votes]: An encouraging update: Daniel Kane posted his proof for the Gaussian case on arXiv yesterday. The proof is both very simple and brilliant. I won't be surprised if his technique can be modified to cover the cube case.<|endoftext|>
TITLE: How much "Morse theory" can be accomplished given only a continuous transformation of a space?
QUESTION [11 upvotes]: If $M$ is a Riemannian manifold and $f:M\to \mathbb{R}$ a Morse-Smale function (which is just a rigorous way to say "generic smooth function"), then Morse theory essentially recovers the manifold itself from relatively basic information about the gradient flow diffeomorphisms of $f$. To sketch briefly: for each pair of critical points $p$ and $q$ of $f$ (i. e., fixed points of the diffeomorphisms), we can consider the subset $S\_{p,q}$ of $M$ that is attracted to $p$ and repelled from $q$ under the diffeomorphisms. ("Repelled from $q$" just means attracted to $q$ under the inverse of the diffeomorphisms.) These $S\_{p,q}$ essentially constitute a decomposition of $M$ as a cell complex. If you just want the homology groups, then you can get away with just considering pairs of critical points whose indices differ by one, and if you just want the Euler characteristic, then you only need local information around each critical point (to define its index). The index of a critical point $p$ is the number of negative eigenvalues of the Hessian (which does not actually depend on the coordinates chosen or even the metric), and it is also the dimension of the submanifold of points in any small neighborhood of $p$ that are attracted to $p$ under the gradient flow diffeomorphisms. 
I want to know how much of that can be done if we don't have $f:M\to \mathbb{R}$, but just some transformation $F:M\to M$ homotopic to the identity, and if $M$ isn't necessarily even a manifold (but probably compact and metrizable). Given information about the fixed points of $F$ (or other dynamical information?), how much of the topology of $M$ can be recovered? (Can we still try to define the "index" of a fixed point of $F$ by looking at the set of points that are attracted to it as $F$ is iterated?) 
Some thoughts:

For some $M$, there might well be maps $F$ that have no fixed points at all. If the Euler characteristic can be recovered from the fixed points of $F$, then such $M$ would have to have an Euler characteristic of zero. (Is that the case??) So the fixed points of $F$ are not very useful in such cases, but are there more general dynamical features of $F$ that relate to the topology of $M$?
Some $M$ might admit perfectly continuous, even smooth, $F$ with chaotic dynamics. 
If $F$ has a unique fixed point $x\_0$ and for every $x\in M$, $F^n(x)\to x\_0$, then $M$ is contractible (recalling our assumption that $F$ is homotopic to the identity). 
Can we get better results by considering an even more restrictive class of transformations? Of course, I don't want to go as far as to say that $F$ belongs to some group of gradient flow diffeomorphisms on a manifold, but maybe we can try to relax that by supposing there exists $f:M\to \mathbb{R}$ such that $f\circ F \geq f$. (That condition makes sense even if $M$ is not a manifold.)

REPLY [3 votes]: Morse theory can be generalised in many directions.  The generalisation of which you speak would cover Morse-Novikov homology; this is where there is a function which is locally Morse, but it might not integrate to a full Morse function (there may be loops of 'flow').
Another generalisation it could cover is Morse-Bott homology and this will cover the situation where you don't have isolated fixed points, but the fixed points form nice enough shapes.
Both of these generalisations may be studied by taking cellular decompositions (the 'flow' out of a fixed point), although in the Novikov case things are more complicated as you have to take the limit of relative decompositions, but I wont try to make that precise here.  These cellular decompositions give a filtration (by index) of the cell complexes of the underlying space.  All of the Morse conditions are carefully chosen to make the associated spectral sequences nice (converging to the homology in the nicest case!).  So an approach to generalisation would normally involve relaxing these conditions and seeing what happens to the spectral sequence.<|endoftext|>
TITLE: Stable presentable categories as module categories
QUESTION [9 upvotes]: There is a theorem of Schwede and Shipley which classifies categories of modules over an A∞ ring spectrum as those stable presentable (∞,1)-categories with a compact generator.  Suppose I allow my A∞ rings to "have many objects", that is, I consider categories of the form FunSp(Iop, Sp) where Sp is the category of spectra, I is a small Sp-enriched category (in some appropriate sense) and Funsp denotes the category of Sp-enriched functors.  Is there a classification of which stable presentable categories can be obtained in this way?  Is it possible that all stable presentable categories are of this form?

REPLY [8 votes]: According to the abstract of http://arxiv.org/abs/math/0108143 (Schwede & Shipley, Classification of Stable Model Categories), they deal with the case of stable model categories (=stable presentable (∞,1)-categories, I suppose) which have a set of compact generators, and show they are the same as model categories of functors from spectral-enriched categories.
(Edited, in the light of Reid's comment, to include the hypothesis of compact generators.)<|endoftext|>
TITLE: K3 surfaces with good reduction away from finitely many places
QUESTION [15 upvotes]: Let S be a finite set of primes in Q.  What, if anything, do we know about K3 surfaces over Q with good reduction away from S?  (To be more precise, I suppose I mean schemes over Spec Z[1/S] whose geometric fibers are (smooth) K3 surfaces, endowed with polarization of some fixed degree.)  Are there only finitely many isomorphism classes, as would be the case for curves of fixed genus?  If one doesn't know (or expect) finiteness, does one have an upper bound for the number of such K3 surfaces X/Q of bounded height?

REPLY [8 votes]: Yves Andre has proved the finiteness of the number of K3 surfaces over a number field $K$ with a polarisation of fixed degree $d$ and having good reduction (as a polarised variety) outside a fixed finite set of primes. 
The reference is: "On the Shafarevich and Tate conjectures for hyper-Kähler varieties".Math. Ann. 305 (1996), no. 2, 205–248.<|endoftext|>
TITLE: Examples of left reversible semigroups
QUESTION [8 upvotes]: I am looking for concrete examples of cancellative, left reversible semigroups. Left reversible semigroups are also called "Ore semigroups". See this wikipedia page for the definition of a left reversible semigroup. Of course, commutative semigroups are automatically left reversible, and I am looking for non-commutative examples.
Please also mention if these semigroups arise in an interesting setting.

REPLY [3 votes]: The semigroup $S= \langle a,b\mid a^2=b^2\rangle$ is two-sided reversible.<|endoftext|>
TITLE: What's the right object to categorify a braided tensor category?
QUESTION [8 upvotes]: The yoga of categorification has gained a lot of popularity in recent years, and some techniques for it have made a lot of progress.  It's well-understood that, for example, a ring is probably categorified by a monoidal abelian (or triangulated) category.  
But I get a little more confused another step up.  If I have a braided tensor category, what sort of 2-category should I expect to categorify it?
Edit: I realized this wasn't the right question to ask; what I really wanted to know is What structure on a monoidal category would make its 2-category of module categories monoidal and braided?

REPLY [4 votes]: The free braided monoidal 2-category with duals on one self-dual object generator is the 2-category of 2-tangles. The original proof of Fisher had a gap, in part, because he relied on the original movie-move theorem. In our abstract, there was a slight mis-statement about the meaning of "movie" in that context. The error was fixed in CRS, and then Baez Langford gave a precise characterization of a free braided monoidal 2-category with duals. 
Ben's question might be along the lines of using Lauda-Khovanov construction to construct a representation theory of the categorification  of $u^+(sl(2))$ for example. Arron spoke about this in Riverside on Saturday morning. According to Arron's slides, there was hope that such representations can give a BM2Catw/D. 
Is anyone working on giving an explicit "other" braided monoidal 2-cat w/ duals?<|endoftext|>
TITLE: Number of paths equal less than equal to a certain length
QUESTION [5 upvotes]: Hey,
I need to count the number of paths from node $s$ to $t$ in a weighted directed acyclic graph s.t. the total weight of each path is less than or equal to a certain weight $W$. I have an algorithm to do it in $O(nW)$ using dynamic programming. Let $N_W(s,t)$ denote the number of such paths (with weight less than $W$) from $s$ to $t$.
It seems like doing it in polynomial time would be hard, since for instance if I take the strategy of calculating $N_W(s,u)$ for $u$ in the paths between $s,t$, then I would have to keep track of how many paths are there from $s\rightarrow p,p\in parents(u)$ for every weight in $\{1,\dots,W\}$ which alone would take $O(nW)$ time and space.
So my question is what is the complexity of this problem?
Thanks
Edit: Changed $O(W)$ to $O(nW)$ for the running time of the dynamic programming approach. Correction thanks to David Eppstein.

REPLY [5 votes]: The problem is and is not NP-hard.  If the length of the input is defined by writing the weights in binary, then it is indeed NP-hard and in fact #P-complete.  Say that the target weight $W$ is written in base 10.  Suppose further that the graph is a string, as in Reid's construction, with two edges from $i$ to $i+1$, one weight 0 and the other with weight some integer in base 10 with only 0s and 1s in its digits.  Also, choose all of the weights so that they cannot add together with any carries.  Then the $k$th digit $d$ of the target weight $W$ is a Boolean clause, which says that of those edges that have a non-zero $k$th digit, exactly $d$ are used used.  It is not hard to build general Boolean circuits with these clauses.
Also, to clarify a couple of points about this:  You can convert from weight at most $W$ to weight exactly $W$ by subtracting off weight at most $W-1$; so in this quick argument it is only #P-hard with a Turing reduction and not a Cook reduction. And of course I'm really showing that the knapsack counting problem is #P-hard.
On the other hand, if the length is defined by writing the weights in unary, to favor small integer values of the weights, then the dynamic programming algorithm given in the question is polynomial time.
Which complexity model is more appropriate depends on the context of the problem, which was not given.<|endoftext|>
TITLE: Littlewood–Richardson–Type Rule for Cohomology Ring of Grassmannians
QUESTION [12 upvotes]: $\DeclareMathOperator\GL{GL}$The ordinary Grassmannian of k-planes in n-space is a coset space for $\GL_n$.
It is $\GL_n$ mod a maximal parabolic. Here there is a nice basis given by Schubert varieties, which can be indexed by Young diagrams that fit in a $(k)\times(n-k)$ box. The structure constants for the cup product are then given by Littlewood–Richardson numbers.
My question: is there a similarly nice picture for Grassmannians of arbitrary simple groups?  Here the ordinary Grassmannian is replaced by $G/P$ where $G$ is a simple group and $P$ is a maximal parabolic. There are still Schubert varieties in this case, but I don't know how to say anything about the cup product.

REPLY [15 votes]: As yet, such a nice rule has only been formulated in the case that $G/P$ is minuscule or co-minuscule. See

Hugh Thomas, Alexander Yong, A combinatorial rule for (co)minuscule Schubert calculus, Adv. Math. 222 (2009), no. 2, 596–620, doi:10.1016/j.aim.2009.05.008, arXiv:math/0608276

for details.<|endoftext|>
TITLE: Is formal proof (formalized mathematics) interesting to practicing mathematicians? To educators? 
QUESTION [12 upvotes]: Formalizing mathematical proofs so that they can be checked for correctness and manipulated by computer is a recurrent proposal, most notably stated in the QED manifesto (1994).  The December 2008 issue of Notices of the AMS is entirely devoted to the state of the art in formal proof; an informal, general interest overview is provided by Cameron Freer's website vdash.org.
As practicing mathematicians, what's your perception of formal proof and interactive proof assistants?  Are you familiar with current systems?  Should they be used in mathematical education?
What are the obstacles to formal proof becoming a generally used tool, besides the effort required to build a useful library of current mathematical knowledge?

REPLY [12 votes]: Carlos Simpson, in the nineties wrote "Descente pour les n-champs" together with Andre Hirschowitz. Apparently they were quite ahead of their time: No one could be found to referee their paper. So to know whether it is correct or not, Simpson started developing an automatic theorem-checking program for category theoretical statements. Later it turned out that there was a small inaccuracy (they used covers where they needed hypercovers), but I don't know if it was found by help of the program...
Anyway it clearly shows how it can be useful.<|endoftext|>
TITLE: SL(2,Z/N)-decomposition of space of cusp forms for Gamma(N)
QUESTION [9 upvotes]: Since $\Gamma(N)$ is normal in $\mathrm{SL}(2,\mathbb{Z})$, the quotient group $\mathrm{SL}(2,\mathbb{Z}/N)$ acts on the spaces of cusp forms $S_k(\Gamma(N))$. How do these spaces decompose into irreducible representations? 
I can do the case $N=2$. I'm mostly interested in the case of $N$ a prime.

REPLY [7 votes]: As usual, once I spot a question on here I have anything useful to say about, somebody has already answered it.  
I can sum up that part of my thesis this way: let M be the induced representation of the character (-I) --> (-1)^k of the center up to all of SL(2,Z/NZ).  Then S_k(Gamma(N)) is roughly k/12 copies of M, plus some error term which can be given precisely, with some effort.  
When you ask instead about Hilbert modular forms over a totally real field K, the "1/12" becomes the absolute value of zeta_K(-1).<|endoftext|>
TITLE: Does some version of U_q(gl(1|1)) have a basis like Lusztig's basis for \dot{U(sl_2)}?
QUESTION [7 upvotes]: There's a non-unital algebra $\dot{U}$ formed from $U_q (sl_2)$ by including a system of mutually orthogonal idempotents $1_n$, indexed by the weight lattice.  You can think of this as a category with objects $\mathbb{Z}$ if you prefer.
Lusztig's basis $\mathbb{\dot{B}}$ for $\dot{U}$ has nice positivity properties:  structure coefficients are in $\mathbb{Z}[q,q^{-1}]$.
Has anyone tried to write down a similar type of basis for the algebra associated to $U_q (gl_{1|1})$?

REPLY [3 votes]: Kashiwara has developed some crystal theoretic methods for the Lie superalgebra $\mathfrak{q}(n)$. However, I think you should look at Khovanov, to get an idea of what it should look like.<|endoftext|>
TITLE: Lie Groups and Lie Algebras
QUESTION [5 upvotes]: What is the exact relationship between Lie groups and Lie algebras? I know it's not bijective because all commutative Lie groups have isomorphic Lie algebras.

REPLY [10 votes]: Up to isomorphism, there is one simply connected Lie group for every Lie algebra.  Indeed, there is also a homomorphism of simply connected Lie groups for every homomorphism of the corresponding Lie algebras so one gets an equivalence of categories this way.
This pans out nicely in yr commutative example: the simply connected abelian groups are just the Lie algebras under addition.  All other abelian Lie groups have a non-simply connected torus component and look like $T^k\times\mathbb{R}^{n-k}$.
To get the remaining Lie groups into the picture, recall that the universal cover of any Lie group is a simply connected Lie group.

REPLY [6 votes]: I'd suggest taking a look at "Lecture 23" of Fulton-Harris. In particular, given a complex Lie algebra, there's a unique connected, simply connected complex Lie group G with that Lie algebra. The other connected Lie groups with the same Lie algebra are quotients of G by discrete subgroups of its centre Z(G) (so G is the universal cover of these other groups). The representations of the Lie algebra are in correspondence with the representations of G. To see which representations of the Lie algebra are representations of one of the other Lie groups with the same Lie algebra, you just need to check which ones factor through the discrete subgroup of the centre that defines the other Lie group. If you're also interested in the real case, check out their "Lecture 26". Here the situation is more complicated.<|endoftext|>
TITLE: squares in stable homotopy
QUESTION [13 upvotes]: I noticed that the generator of the second stable stem b is the square of the generator of the first stable stem a, in the sense that if take two copies of a and smash product them together you get b out. I'm wondering if there are any ( many) other exmples of this. What are the elements in the stable homotopy of spheres which a squares in the above sense?

REPLY [11 votes]: Appendix 3 of Ravenel's Green Book, http://www.math.rochester.edu/u/faculty/doug/mu.html#repub, has a chart of stable homotopy groups including much of the multiplicative structure.  Figure A3.1 depicts some of this structure visually, while Table A3.3 lists the elements out by name and degree.
The next example of a square after \eta^2 is the element in the sixth stable stem, which is the square of the Hopf map \nu in the third stable stem.
Though stable homotopy is not multiplicatively finitely generated, you can consider Toda brackets, which are a form of higher multiplication in homotopy analogous to Massey products in cohomology, and it is known that the entire stable homotopy groups of spheres are generated by Toda brackets on the Hopf elements 2, \eta, \nu, and \sigma.<|endoftext|>
TITLE: Algorithm generalizing continued fractions for non-quadratic algebraic numbers
QUESTION [17 upvotes]: The continued fraction algorithm generates an integer sequence which terminates for a rational number, is periodic for the roots of irreducible integer quadratics, and is non-periodic for other algebraic numbers.  This sequence uniquely determines the number in a useful way, e.g. one can compute convergents and solve Diophantine equations.
Does there exist a corresponding algorithm for, say, roots of irreducible cubics which has similar properties?  What about other algebraic numbers?  What is known about this?  Or to save people time, what phrase should I google to find out the answer?

REPLY [6 votes]: Suppose that r is a positive real number with no nontrivial positive real Galois conjugates.  This case includes the case of nth roots of integers.  Then it is not difficult to work out the continued fraction of r algorithmically.
Suppose that f(x) is the minimal polynomial of r.  The greatest integer of r (which is a_0) is just the largest integer n such that f(n) is negative.  Then it's easy to work out the minimal polynomial of 1/(r-a_0) using the fact that the minimal polynomial of the reciprocal is given by reversing the coefficients.  Furthermore, 1/(r-a_0) again has the same property as r (no nontrivial positive real Galois conjugates).  Now compute a_1 as the largest integer such that f(a_1) is negative.  Wash, rinse, repeat.<|endoftext|>
TITLE: Why are powers of $\exp(\pi\sqrt{163})$ almost integers?
QUESTION [62 upvotes]: I've been prodded to ask a question expanding this one on Ramanujan's constant $R=\exp(\pi\sqrt{163})$.
Recall that $R$ is very close to an integer; specifically $R=262537412640768744 - \epsilon$ where $\epsilon$ is about $0.75 \times 10^{-12}$.  Call the integer here $N$, so $R = N - \epsilon$.
So $R^2 = N^2 - 2N\epsilon + \epsilon^2$.  It turns out that $N\epsilon$ is itself nearly an integer, namely $196884$, and so $R^2$ is again an almost-integer.  More precisely,
$$j(\tau) = 1/q + 744 + 196884q + 21493760q^2 + O(q^3)$$
where $q = \exp(2\pi i\tau)$.   For $\tau = (1+\sqrt{-163})/2$, and hence $q = \exp(-\pi\sqrt{163})$, it's known that the left-hand side is an integer.  Squaring both sides,
$$j(\tau)^2 = 1/q^2 + 1488/q + 974304 + 335950912q + O(q^2).$$
To show that $1/q^2$ is nearly an integer, we can rearrange a bit to get
$$j(\tau)^2 - 1/q^2 - 974304 = 1488/q + 335950912q + O(q^2)$$
and we want the left-hand side to be nearly zero.  $1488/q$ is nearly an integer since $1/q$ is nearly an integer; since q is small the higher-order terms on the right-hand side are small.  
As noted by Mark Thomas in this question, $R^5$ is also very close to an integer -- but as I pointed out, that integer is not $N^5$.  This isn't special to fifth powers.  $R$, $R^2$, $R^3$, $R^4$, $R^5$, $R^6$, respectively differ from the nearest integer by less than $10^{-12}$, $10^{-9}$, $10^{-8}$, $10^{-6}$, $10^{-5}$, $10^{-4}$, and $10^{-2}$.  But the method of proof outlined above doesn't work for higher powers, since the coefficients of the $q$-expansion of $j(\tau)^5$ (for example) grow too quickly.  Is there some explanation for the fact that these higher powers are almost integers?

REPLY [50 votes]: Another take on this:
As David Speyer and FC's answer shows, this question can be answered without any additional deep theory.
However, I'd like to explain a variant on their arguments that puts this in a little more context regarding modular forms.   It also means we can use a technique which makes it easier to see how good these approximations are in terms of the growth rate of the coeffients of the j-function.
The important fact here is that any modular function (for SL_2(Z)) with integer coefficients in its q-expansion takes on integer values at τ = (1+√(-163))/2 (and so q = exp(-π√163)).  This fact is in fact a consequence of the integrality of the j-value here, since any such function can be expressed as a polynomial in j with integer coefficients (although similar things are true in other contexts, such as modular functions of higher level, where there is not a canonical generator for the ring of such modular invariants).
This means that, just as we can use the integrality of
$j(\tau) = q^{-1} + 744 + O(q) $
to get an integer approximation to $q^{-1}$, if we have a modular function $f_n$ with power series of the form
$f_n(\tau) = q^{-n} + integer + O(q)$,
we can get an integer approximation to $q^{-n}$.  How good this approximation is will depend upon the size of the coefficients of the power series for the $O(q)$ part.
Fortunately for us, such a function $f_n$ always exists (and is unique up to adding integer constants).  How can we construct it?  One way is to take an appropriate polynomial in $j$, that is, take an appropriate linear combination of $j, j^2, \cdots , j^n$ to get a function with the right principal part.  This clearly works, and if one works out the details, it should turn out equivalent to FC's and David's approach.
However, now that we're in the modular forms mindset, we have other tools at our disposal.  In particular, another way to create new modular functions is to apply Hecke operators to existing modular functions, such as $j$.  This turns out to be an effective way to get modular functions of the type we need, since Hecke operators do predictable things to principal parts of q-series (for example, if $p$ is prime, $T_p j = q^{-p} + O(1)$).  I'll just explain how this works for $n = 5$, although the method should generalize immediately to any prime $n$ (composite $n$ might be a little trickier, but not much).
The theory of Hecke operators tells us that the function $T_5 j$ defined by
$$(T_5 j)(z) = j(5 z) + \sum_{i \ mod \ 5} j (\frac{z + i}{5})$$
is modular, with q-expansion given by
$$(T_5 j)(\tau) = q^{-5} + \sum_{n = 0}^{\infty} (5 c_{5n} + c_{n/5}) q^n$$.
where the $c_n$ are the coefficients in
$$j(\tau) = q^{-1} + \sum_{n = 0}^{\infty} c_n q^n$$
(and $c_n = 0$ if $n$ is not an integer).
So if as before we set $q = e^{- \pi \sqrt{163}}$, we find that
$q^{-5} + 6 c_0 + 5 c_5 q + 5 c_{10} q^2 + 5 c_{15} q^3 + 5 c_{20} q^4 + (c_1 + 5 c_{25}) q^5 + \dots$
is an integer.
Now, $q$ is roughly $4 \cdot 10^{-18}$, and looking up the coefficients for $j(z)$ on OEIS, we find that
$q^{-5} + 6 \cdot 744 + 5 \cdot (\sim 3\cdot 10^{11}) (\sim 4\cdot10^{-18})  + \text{clearly smaller terms}$
is an integer.  Hence $q^{-5}$ should be off from an integer by roughly $6 \cdot 10^{-6}$.  This agrees pretty well with what Wolfram Alpha is giving me (it wouldn't be hard to get more digits here, but I'm feeling back-of-the-envelope right now and will call it a night :-)<|endoftext|>
TITLE: Math History Question about the exponential function
QUESTION [10 upvotes]: While tutoring a student recently, I have come across the situation of explain logarithms by first introducing functions of the form $$f(x)= a^x$$ where $a \ge 0,x\in \mathbb{R}$. My student then asked me a seemingly innocent question, what if $a < 0.$ I explained to her, that it suffices to consider the basic case $$g(x)= (-1)^x$$ however I told her that the answer to that is beyond the scope of the material. I did tell her the answer though, namely that $$g(x) = (-1)^x = (e^{i \pi})^x = e^{i x \pi} $$ which is the unit circle in the complex plane. Of course this opens up a new can of worms.
This got me to thinking, in the course of history the most probable chronological sequence was that functions of the form of $f(x)$ were considered before complex numbers were formalized. Then did mathematicians of the past simply state that functions with $a<0$ were simply undefined? Or was it the case that complex numbers were considered first?

REPLY [4 votes]: I have stumbled upon this article, maybe it can help:

History of the Exponential and Logarithmic Concepts,
  Florian Cajori,
  The American Mathematical Monthly,
  Vol. 20, No. 7 (Sep., 1913), pp. 205-210.

By the way, already $(-1)^{1/2}$ is undefined/multivalued.<|endoftext|>
TITLE: Splitting in triangulated categories
QUESTION [7 upvotes]: Using the axioms for a triangulated category, is it possible to prove the following:

$A\stackrel{0}{\to}B\to A\oplus B\to$ is a distinguished triangle.

From the first axiom, the map 0:A-->B extends to its cone, but there is no guarantee I see that the direct sum fits into a triangle.  If it does, however, clearly they are (should be?) isomorphic.
I have tried using the universal properties of the direct sum, in that finite coproducts and finite products coincide in additive categories, so that I have two diagrams, and extended each of these diagrams into triangles in every way I can imagine, but I think I'm just getting lost in the plethora of sequences.  0:A-->B extends to a triangle A-->B-->X--> and so I can get things like $X\to A\oplus B\to \Sigma^{-1}X\cong X$, but by moving away from triangles, I have lost notions of exactness (so that this sequence of maps merely commutes..)
I ask this because the proof of Lemma 3.3(2) of this paper seems to use this without reference.

REPLY [16 votes]: So if I understand correctly the question you wanted to ask was:
Is it true that a triangle $$X \stackrel{u}{\to} Y \stackrel{v}{\to} Z \stackrel{w}{\to} \Sigma X$$ is split if and only if one of $u$, $v$, or $w$ is zero. The answer to this is yes.
It is clear (I think I can add details if someone wants) that if the triangle is split then one map must be zero (basically since we have an epi composing to zero). Conversely suppose that $w$ is zero, which is sufficient since we can always just rotate. Now we know by the axioms that $$Z \stackrel{-1}{\to} Z \to 0 \to \Sigma Z$$ and hence $$0 \to Z \stackrel{1}{\to} Z \to 0$$ are triangles (as an exercise check that any sequence of this form given by an isomorphism is necessarily a distinguished triangle), and $$X \stackrel{1}{\to} X \to 0 \to \Sigma X$$ is also a triangle. It is easy to check then that so is the direct sum $$ X \to X\oplus Z \to Z \stackrel{0}{\to} \Sigma X$$. The identity maps on $X$ and $Z$ then induce a map via [TR3] from this to the original triangle (since we have $w=0$ this trivially satisfies the necessary commutativity to apply [TR3]) and since two of the maps are isomorphisms so is the third. Hence any triangle as above with $w=0$ is isomorphic to one obtained by summing triangles on identity maps.
Proof of the claim that the direct sum of triangles is a triangle:
Let $X \to Y \to Z \to \Sigma X$ and $X' \to Y' \to Z' \to \Sigma X'$ be two distinguished triangles. We can complete the map $X\oplus X' \to Y \oplus Y'$ to a triangle $$X\oplus X' \to Y \oplus Y' \to Q \to \Sigma(X\oplus X')$$. We then get two diagrams whose rows are triangles by projecting to the factors $X, Y$ and $X', Y'$ respectively which we can complete to maps of triangles by the mapping axiom [TR3]. The maps $Q\to Z$ and $Q\to Z'$ such obtained induce a map $Q\to Z\oplus Z'$ by the universal property which gives a map from the triangle $X\oplus X' \to Y\oplus Y' \to Q$ to the pretriangle (a pretriangle is one where the maps compose to zero and it plays nicely with homological functors, that is they take it to a long exact sequence, this is clear from the fact that homological functors are additive and it is a sum of distinguished triangles) $X\oplus X' \to Y\oplus Y' \to Z\oplus Z'$. Two of the maps are isomorphims, namely the identities on the terms $X\oplus X'$ and $Y\oplus Y'$ so that the third must be also - this follows from the fact that $Hom(A,-)$ is a homological functor for any $A$, Yoneda, and the 5 lemma. So the triangle in question is isomorphic to a distinguished triangle and hence itself distinguished.
This works in the generality of arbitrary coproducts/products provided the coproducts/products in question exist and we consider a pretriangle to be one which homological/cohomological functors preserving the coproducts/products take to long exact sequences.
I'd recommend reading through the first chapter of Neeman's book Triangulated Categories - this is certainly covered in there as well as a bunch of other facts you might find useful in reading that paper. The reference for this result is Corollary 1.2.7 (it seemed lazy not to check since it is on my shelf) and the proof there is pretty much identical to the one here except that the facts I glossed over are proved earlier.

REPLY [2 votes]: The distinguished triangle is $B\to A\oplus B\to A \overset{0}{\to}$ is the exact triangle (this differs from what you have above by a shift).  I suspect that what you should should use is the map axiom (TR3) in Wikipedia's labeling to get a map from the cone  of the zero map to A (by applying this axiom for the identity on A and 0 on B) and to B (by doing the reverse) and the use the universal property of product (you might need to throw in the octahedral axiom for that).<|endoftext|>
TITLE: Braided Monoidal 2-categories with duals
QUESTION [7 upvotes]: Which categorifications give explicit braided monoidal 2-categories with duals? 
This question is in response to Ben Webster's questions in recent history. The point is that given a braided monoidal 2-category with duals (other than the 2-category of tangle surfaces) an invariant of knotted surfaces can be constructed. 
I've been told that Lurie's work gives examples, but I don't know where to look therein.

REPLY [5 votes]: Let $A$ be an $E_3$-algebra, so that $A$ is an $E_2$-algebra in the category of $E_1$-algebras by Dunn additivity. The functor
$$
    E_1-alg 
    \to
    Cat
$$
$$
    A 
    \mapsto
    A-mod
$$
is symmetric monoidal, so it will send a "banana" algebra in $E_1$-alg to a "banana" algebra in categories. In particular, the category of left $E_1$-modules over $A$ is an $E_2$-category; i.e., a braided monoidal (but not symmetric monoidal) category.
If you want the target to be 2-cats, rather than Cat, you can enhance by considering an $E_4$-algebra $A$, forget it to an $E_2$ algebra $A'$ and looking at the Morita 2-category of algebras over $A'$, or at the category of $E_2$-algebras over $A'$.<|endoftext|>
TITLE: Classification of homology 3-spheres?
QUESTION [18 upvotes]: Is there some general description of all homology 3-spheres?

REPLY [11 votes]: Another way to represent homology spheres is to take a Heegaard splitting
for $S^3$, cut and reglue by an element of the Torelli group. This is
not canonical, but any two Heegaard splittings are equivalent after some
number of stabilizations. If you wanted to enumerate every homology sphere, you could list elements of the Torelli group, and construct 3-manifolds, then throw away repeats by using some solution to the homeomorphism problem for 3-manifolds. This is not really feasible to carry out in practice, but is one way to give a "general description" of homology spheres at least in theory, by giving a recursive enumeration of them.<|endoftext|>
TITLE: What do higher Chow groups mean?
QUESTION [36 upvotes]: Let $z^i(X, m)$ be the free abelian group generated by all codimension $i$ subvarieties on $X \times \Delta^m$ which intersect all faces $X \times \Delta^j$ properly for all j < m. Then, for each i, these groups assemble to give, with the restriction maps to these faces, a simplicial group whose homotopy groups are the higher Chow groups CH^i(X,m) (m=0 gives the classical ones).
Does anyone have an intuition to share about these higher Chow groups? What do they measure/mean? If I pass from the simplicial group to a chain complex, what does it mean to be in the kernel/image of the differential?
Could one say that the higher Chow groups keep track of in how many ways two cycles can be rationally equivalent (and which of these different ways are then equivalent etc.)?
Finally: I don't see any reason why the definition shouldn't make sense over the integers or worse base schemes. Is this true? Does it maybe still make sense but lose its intended meaning?

REPLY [23 votes]: I believe Bloch's original insight was something like the following:
First, if $X$ is a regular scheme, you can filter $K_0$ by ``codimension of support''; that is, view $K_0(X)$ as the Grothendieck group of the category of all finitely generated modules and let $F^iK_0(X)$ be the part generated by modules with codimension of support greater than or equal to $i$.
Next, suppose you want to mimic this construction for $K_m$ instead of $K_0$.  The first step is to notice that if you patch two copies of $\Delta^m_X$ together along their "boundary" (i.e. the union of the images of the various copies of $\Delta^{m-1}_X$) and call the result $S^m_X$, then Karoubi-Villamayor theory tells you that $K_m(X)$ is a direct summand of $K_0(S^m_X)$.  (The complementary direct summand is $K_0(X)$.)
So it suffices to find a "filtration by codimension of support" on $K_0(S^m_X)$. 
The usual constructions don't work because $S^m_X$ is not regular (so that in particular, not all modules correspond to $K$-theory classes.)
But:  a cycle in $z^i(X,m)$ has a positive part $z_+$ and a negative part $z_-$ which, (if it is homologically a cycle) must agree on the boundary.  Therefore you can imagine taking $\Delta^m_X$-modules $M_+$ and $M_-$ supported on these positive and negative parts and patching them along the boundary to get a module on $S^m_X$.  If this module has finite projective dimension (which it ``ought'' to because of all the proper-meeting conditions, and as long as it has no bad imbedded components), then it gives a class in $K_0(S^m_X)$, hence a class in $K_m(X)$, and we can take the $i-th$ part of the filtration to be generated by the classes that arise in this way.
The Bloch-Lichtenbaum work largely bypasses this intuition, but this was (I think) the original intuition for why it ought to work.<|endoftext|>
TITLE: Which magazines should I read?
QUESTION [16 upvotes]: Suppose you are a mathematics student who has just graduated and you haven't yet come to graduate school, or maybe you are in your first year of graduate school. Which magazines should you read? I mean general interest magazines, not journals of a specific field, but the kind of magazine that in each volume contain at least a couple of articles that every mathematician could understand (or reasonably try to understand) and be interested in. I don't even know how to describe exactly what I mean. That's why I am asking.
Please post one suggestion per answer.

REPLY [9 votes]: I would get a copy of the Princeton Companion to Mathematics and read two or three articles a month.<|endoftext|>
TITLE: Erdős–Stone theorem type edge density estimates for bipartite graphs?
QUESTION [5 upvotes]: The Erdős–Stone theorem theory says that the densest graph not containing a graph H (which has chromatic number r) has number of edges equal to $(r-2)/(r-1) {n \choose 2}$ asymptotically.
However, this doesn't say much for bipartite graphs (since r=2). I wanted to know what are the best results known for the densest graphs not containing a particular bipartite graph H. I'm guessing this problem is still open and hasn't been completely resolved.
This problem is easy if H is a forest, since every graph with $|E| > k|V|$ contains every forest on k vertices as a subgraph. For even cycles, I know there is a result of Bondy and Simonovits which says:
"if $|E| \geq 100k|V|^{1+1/k}$  then G contains a $C_{2l}$ for every $l$ in $[k, n^{1/k}]$."
So can someone point me to the best known results now for bipartite cyclic graphs?

REPLY [3 votes]: To add to Konrad's answer:
No construction is known for $K_{4,4}$. Neither there is a construction for $C_{2k}$ for k other than 2,3,5. There is a construction by Lazebnik and Woldar that beats the probabilistic for $C_{2k}$, though. There are some non-trivial upper bounds on bipartite graphs of bounded degree and (more generally) of bounded degeneracy. These upper bounds (and some references) can be found in the survey on dependent random choice by Fox and Sudakov.<|endoftext|>
TITLE: Turaev-Viro extended TQFT
QUESTION [12 upvotes]: Hi I am looking for any papers which extends the Turaev-Viro TQFT to a 3-2-1 theory (i.e. allows manifolds with corners) . I know this construction is known, but I cannot find a source. Please help.
Thanks, Ben

REPLY [6 votes]: As Charlie pointed out in comments Balsam and Kirillov were working on this, and since his comment they posted a preprint to the arxiv: http://arxiv.org/abs/1004.1533
For another point of view see Kevin Walker's notes in progress at http://canyon23.net/math/  In theory Kevin's point of view should automatically lead to a theory extended all the way down to points, but it's phrased in a different language from the usual extended field theory language.<|endoftext|>
TITLE: Is there a cohomological criterion of nefness?
QUESTION [7 upvotes]: like serre's thm for ampleness?

REPLY [3 votes]: Well, kind of -- a line bundle is nef if and only if its tensor product with any ample line bundle is ample.<|endoftext|>
TITLE: How does one identify properties of objects with good "inheritance"?
QUESTION [8 upvotes]: When you are dealing with a very general object like a topological space or a ring, usually you impose an additional condition (such as compact Hausdorff or Noetherian) with the property that the subset of objects satisfying the additional condition is both flexible enough to be widely applicable and rigid enough that it is not too pathological.  Part of being flexible is having good inheritance properties.  For example, one reason the Noetherian condition is nice is Hilbert's basis theorem and one reason compact Hausdorff is nice is Tychonoff's theorem.
What is known about systematically identifying the properties of objects that will have good inheritance?  (I'm leaving this phrase vague because I am sure you know of more examples of such properties than I do.)  For example, perhaps one can say something general about properties that are defined using certain patterns of quantification.

REPLY [5 votes]: In first-order logic, some of the most natural inheritance properties have been studied.
First, you should recall what a substructure means in the sense of model theory -- basically just that you're closed under all the operations of the bigger structure, and the interpretation of any relational symbols in your language agrees on tuples common to the two structures.  (I'm using the "non-weak" sense in the link I provided.)
A first-order axiomatizable class of structures is closed under substructures if and only if it can be axiomatized by a set of universal sentences (of the form: $\forall x_1 \ldots \forall x_n \varphi(\overline{x})$, where $\varphi$ is quantifier-free).  Think of groups (in a language with a function symbol for inverses) or ordered groups.
A first-order axiomatizable class of structures is closed under superstructures if and only if it can be axiomatized by a set of existential sentences ($\exists x_1 \ldots \exists x_n \varphi(\overline{x})$, with $\varphi$ quantifier-free).
An axiomatizable class of structures is closed under unions of ascending chains of superstructures if and only if it can be axiomatized by a set of "AE-sentences," of the form $\forall x_1 \ldots \forall x_n \exists y_1 \ldots \exists y_m \varphi(\overline{x}, \overline{y})$ with $\varphi$ quantifier-free.  Think of fields in the language with only the symbols for 0, 1, and the two field operations: the axiom expressing that every element has a muliplicative inverse is AE.
Inheritance under products in first-order logic is trickier.  Any class that can be axiomatized by first-order Horn sentences is closed under products, but the converse is false, and I've never heard of a good syntactic characterization of such classes.  (I think this is why logicians are so fond of ultraproducts, which automatically preserve the truth of all first-order sentences!)
I'm not sure how useful these results actually are, since many (most?) properties that algebraists care about are not first-order axiomatizable.  E.g. the class of all Noetherian rings is not axiomatizable (by the compactness theorem).<|endoftext|>
TITLE: What precisely Is "Categorification"?
QUESTION [74 upvotes]: (And what's it good for.)
Related MO questions (with some very nice answers): examples-of-categorification; can-we-categorify-the-equation $(1-t)(1+t+t^2+\dots)=1$?; categorification-request.

REPLY [8 votes]: I believe that one of the oldest and most important motivations for / settings for categorification hasn't been explicitly mentioned yet (here or in the linked questions). It is due to Grothendieck (and Weil): the relation between cohomology and counts of points for varieties over finite fields. [This happened to be last week's lecture in a class I'm teaching, so a ready-made rant.]
As was explained categorification is understood in relation to DEcategorification, which can be made a well-defined mathematical process: replace a vector space or chain complex by a number (its dimension or Euler characteristic), an associative algebra or category by a vector space / chain complex (its cocenter / trace / Hochschild homology), a monoidal category by an associative algebra / category etc --- these (and many more) are captured by the notion of "dimension" of a dualizable ("finite dimensional") object in a symmetric monoidal [higher] category (which defines an endomorphism of the unit, taking the place of a "scalar").
One can use dimension in this sense as a general definition of decategorification. This has a natural motivation from topological field theory, where taking dimension corresponds to crossing with a circle -- $dim(Z(M))=Z(M\times S^1)$. (This sometimes give a more naive decategorification than passing to K-groups, though in examples the two often coincide or the simpler "dim" is often what we really want.)
But if you have a dualizable object you can talk not only about its dimension (trace of the identity), but about the trace of an endomorphism. For example in TFT you can study not $M\times S^1$ but the mapping torus of a diffeomorphism of $M$ to get the trace of the induced map on $Z(M)$.
In the setting of geometry over finite fields, there's a canonical choice of endomorphism to take, namely Frobenius, so we get a different version of decategorification by consistently studying $Tr(F)$ instead of $dim=Tr(Id)$, and correspondingly a different notion of categorification.
The Grothendieck-Lefschetz trace formula tell us that l-adic cohomology categorifies (in this general sense) counts of points over finite fields (as captured by zeta- and L-functions), a categorification just as revolutionary as the replacement of Euler characteristics by homology groups. But this is just the beginning. Note also that this categorification is richer in the sense that we can take traces of powers of Frobenius to find point counts over all finite extensions of our finite field.
Grothendieck's function-sheaf dictionary is a fundamental categorification, a relative version of the above: it suggests that interesting functions on sets of points of varieties over finite fields can be categorified by l-adic sheaves; or all together, interesting function spaces are categorified by categories of sheaves.
This idea is behind much (most?) of modern geometric representation theory, and in particular one of the most spectacular achievements in math: the entire representation theory of all the finite groups $G(\mathbb F_q)$ for $G$ reductive (like $GL_n, SL_n,SO_n,...$)  -- a list that includes almost all the finite simple groups -- was categorified in this sense in the collected works of Lusztig. This includes the Deligne-Lusztig construction of representations, the celebrated [first set of] Kazhdan-Lusztig papers, which can be interpreted as categorifying the [unipotent] principal series representations (closely related to Springer theory categorifying the representation theory of Weyl groups), and the theory of character sheaves, which roughly categorifies the entire character theory of these groups. In other words, a huge amount of what we know about this huge family of finite groups comes from categorification.
..and the entire Geometric Langlands program comes from this idea, starting with Drinfeld, taking the same philosophy from reductive groups over finite fields to reductive groups over local fields. The Geometric Satake correspondence categorifies the classical Satake correspondence, the basic mechanism behind the Langlands program, and is fundamental to our undrestanding of what local Langlands is about (ie how to organize representations of p-adic groups) thanks to Fargues-Scholze. V.Lafforgue's proof of the automorphic-->spectral direction of Langlands for function fields is inspired by the categorification idea, and the work of Arinkin-Gaitsgory-Kazhdan-Raskin-Rozenblyum-Varshavsky explicitly makes sense of this process, recovering spaces of [unramified] automorphic functions over function fields as trace of Frobenius on suitable categories of sheaves.<|endoftext|>
TITLE: Gelfand-Naimark from the category-theoretic point of view
QUESTION [13 upvotes]: I was thinking about the Gelfand-Naimark theorem asserting the isometric * isomorphism between a commutative C* algebra (with unit) A and the C* algebra of continuous complex-valued functions on its spectrum (via the Gelfand transform). Explicitly: let spec(A) denote the spectrum of A and C(X) the algebra of complex continuous functions on X. Then spec and C define contravariant functors from commC* alg to CompHausTop, which (correct me if i'm wrong) establish  an equivalence between the two categories. 
Gelfand-Naimark theorem has a non-commutative analogue, which is based on the so-called GNS construction and which shows that every non commutative C* algebras has a faithful isometric *-representation on a Hilbert space H. In this case I can't see an analogue of the preceding equivalence of categories, which is equally meaningful. Does it exist?

REPLY [15 votes]: The proper analogue is rather based on the characterization of the state space of a unital C*-algebra found in (sorry about the self-advertisement) E. Alfsen, H. Hanche-Olsen and F.W. Shultz: State Spaces of C∗-Algebras, Acta Math. 144 (1980) 267–305. So the category to replace CompHausTop would be the category of state spaces equipped with orientations on their facial 3-balls, and whose morphisms are certain affine maps between these compact convex sets.
In this context, a compact Hausdorff space X is represented by the set of probability Baire measures on X, which is in particular a Choquet simplex.<|endoftext|>
TITLE: Divisors, extensions of functions
QUESTION [6 upvotes]: This might be a silly/obvious question.  I know that we have the removable singularity theorem (of Riemann) on the complex line, and we also have the generalization of this to algebraic curves.  (Namely: if $U$ is an open subset of a Riemann surface and $a\in U$, and $f \in \mathcal{O}(U-a)$ is bounded in some neighborhood of a, then f can be extended uniquely to a function $F\in \mathcal{O}(U)$.)  In particular, if a function on a curve vanishes over a divisor, we can extend over it uniquely.
What do we know about extending of functions over divisors/hyperplanes in higher dimensions?

REPLY [7 votes]: To add to the excellent answers of jvp and David Speyer above: 
Actually Hartog's theorem does even better: Given a domain $U\subset\mathbb{C}^{n{\geq{2}}}$ and a compact $K\subset{U}$ such that $U\setminus{K}$ is just connected, any holomorphic function on $U\setminus K$ extends holomorphically to $U$. Note that we don't need the codimension condition. This is actually a simple consequence of Cauchy's formula in several variables. 
For algebraic geometry (resp. complex geometry) and dimension $\geq{2}$: The analog of Hartog's theorem is that any regular (resp. holomorphic) function on the complement of an algebraic (resp.  analytic) subset of codimension atleast $2$ in a normal algebraic(resp. analytic) variety, extends to the whole algebraic (resp. analytic) variety.
I would also like to point out the important difference between the affine (resp. Stein) case and the projective case as perhaps alluded to by jvp above. If $X$ is a projective algebraic variety over $\mathbb{C}$ and you find a holomorphic function $f$ on the complement of a divisor $D$ then you will not be able to extend the function to all of $X$ however hard you try(!) since there are no global non-constant functions on projective varieties (similar statement for compact analytic vars). So to add to jvp's answer (which might seem like hair-splitting but is important IMHO): His second paragraph must refer to a non-projective neighborhood of the contractible curve that he discusses, otherwise the argument is still true but only vacuously! This is because there is no non-constant function to be found on a projective surface on the complement of a contractible curve and hence there is nothing to extend!   
Moreover, it will not be possible in general to extend holomorphic functions from the complement of divisors whose examples can be easily constructed. So there is no hope in this direction without any local boundedness hypothesis.<|endoftext|>
TITLE: Is it possible to write a research statement which is too short?
QUESTION [14 upvotes]: We all know that it is possible to write a research statement which is so long it becomes counterproductive.  At some point people will give up on reading it.
But can you write one (let's say for job applications) which is so short that its length alone makes a bad impression on the reviewers?  I certainly suspect this to be the case.
It would be very useful if in answers people could specify which hiring environment (which country, which type of school) they have in mind.  It would be particularly interesting to hear a perspective from someone who's hired for tenure-track jobs at a liberal arts college in the US, but I'd welcome info about any sort of job or grant.

REPLY [6 votes]: I agree with JSE.  You should write your research statement so that someone who stops reading after the introduction still gets a lot out of it.  The people who make it to the end of a long (say 8-10 page) research statement are probably the ones who are considering hiring you.  Even if the work is strong, an overly short research statement might make me concerned that the candidate does not possess the requisite stamina and follow-through for an academic career.<|endoftext|>
TITLE: Existence of hyperelliptic curve with specific number of points in a family
QUESTION [9 upvotes]: Hi,
the following question was posed to me, it apparently has applications for linear codes. Let n>1, and $K = \rm{GF}(2^n)$. Let $k$ be coprime to $2^n-1$. Does there always exist $a \neq 0$ in $K$ such that the curve
$y^2+y = x^k+ax$
has exactly $2^n$ affine solutions? (I ran some computer checks [although only for quite small n and k] without finding a counterexample.)

REPLY [2 votes]: I think that this is equivalent to a known open question.  Here are
the details.  For $K:=\mathbb{F}\_{2^n}$, the function $f:y\mapsto
y+y^2:K\to K$ is $\mathbb{F}_2$-linear, and its kernel $\{0,1\}$ has
dimension 1. The image is therefore of dimension $n-1$, and for $z$ in
the image, the fiber $f^{-1}(z)$ has exactly 2 elements.  
Hence, to prove that $y^2+y=x^k+ax$ has exactly $2^n$ solutions
for some fixed $a\in K$, we have to show that $|\{x\in K|x^k+ax\in
\mathrm{Im}(f)\}|=2^{n-1}$.  
Since $\sigma:y\mapsto y^2$ is a generator of the Galois group of
$K/\mathbb{F}\_2$, Hilbert's Theorem 90 (in additive form) says that
$z\in \mathrm{Im}(f)$ if and only if $\mathrm{Tr}(z)=0$, where
$\mathrm{Tr}$ stands for the trace map from $K$ to $\mathbb{F}_2$. 
So the problem is equivalent to showing that there exists an $a\neq 0$ in $K$
such that $|\{x\in K|\mathrm{Tr}(x^k+ax)=0\}|=2^{n-1}$. In other
words, we would like to show that there exists a nonzero $a\in K$ such that
$$
S_k(a):=\sum_{x\in K}(-1)^{\mathrm{Tr}(x^k+ax)}=0.
$$
Apparently, this question was addressed in the coding community.  In
detail, in [1, p. 258], the following conjecture (of Helleseth) is
mentioned:
``Conjecture 3. For any $m$ and $k$ such that
$\mathrm{gcd}(2^m-1,k)=1$, the sum
$\sum_{x\in\mathbb{F}_{2^m}}(-1)^{\mathrm{Tr}(x^k+ax)}$ is null for at
least one nonzero $a$.'' (Note that $n$ in the current question is
$m$ in 1).
It seems that in [1, Corollary 1, p. 253], Conjecture 3 is proved for
even $m$ and for certain values of $k$ (the ``Niho exponents,''
defined on p. 252 of 1).
Interestingly, at least at a first glance it seems that 1 has
nothing to say on $k\in\{1,\ldots,2^{n-1}\}$, but to me it seems that
this case is trivial (am I missing something?):  Consider a normal
basis for $K/\mathbb{F}\_2$, that is, a basis $B$ consisting of an orbit of
an element $\gamma\in K$ under the Galois group of $K/\mathbb{F}_2$
(the $i$th element of $B$ is $b_i:=\gamma^{2^i}$ for
$i\in\{0,\ldots,n-1\}$).  
From the linearity of the trace and the fact that the trace is onto,
we must have $\mathrm{Tr}(b)=1$ for at least one element $b\in B$, and
from  $\mathrm{Tr}(b^2)=\mathrm{Tr}(b)$ we then have
$\mathrm{Tr}(b)=1$ for all $b\in B$. So the trace of an element in $K$
is just the modulo-2 sum of the coefficients in its decomposition
according to the basis $B$.  
Let $a$ be any element in the
trace-dual basis of $B$, say $\mathrm{Tr}(ab_i)=\delta_{i,0}$. Then
for $k=2^j$, if we write $x=\sum_i \alpha_i b_i$, we
get: $\mathrm{Tr}(ax)=\alpha_0$,
$\mathrm{Tr}(x^k)=\mathrm{Tr}(x)=\sum \alpha_i$ (sum in
$\mathbb{F}_2$). These agree for half of the $x\in K$, as required. 
That's about it.  I hope at least some of this makes sense :)  I also
hope that the original person asking this question didn't actually want
to solve the above open question by converting it to a question about
curves, for then this answer is useless.
1 P. Charpin, ``Cyclic codes with few weights and Niho exponents,''
Journal of Combinatorial Theory, Series A 108 (2004)
247--259.<|endoftext|>
TITLE: The Relationship between Complex and Algebraic Geomety
QUESTION [33 upvotes]: I have recently begun to study algebraic geometry, coming from a differential geometry background. It seems that there is a deep link between complex manifolds and complex varieties. For example, one often hears that they are two different ways of looking at the same thing. Can anybody give a precise statement of this relationship?

REPLY [33 votes]: No one seems to have mentioned that if a complex manifold is an algebraic variety, then it has a nice compactification. More precisely, by Hironaka it can be embedded in a compact manifold such that the boundary is a divisor with normal crossings. 
In dimension 1 this is the only condition: a Riemann surface is an algebraic curve if and only if it can be compactified by adding a finite number of points. In particular, bounded holomorphic functions on it are constant (so the complex upper half plane is not an algebraic curve because z-i/z+i is a bounded holomorphic function).
So the two main obstructions to a complex manifold being an algebraic variety (or, at least, an algebraic space) are that it may not have a nice compactification (in the above sense) and it may not have enough meromorphic functions.
To say that a complex manifold is an algebraic variety is a very strong statement.<|endoftext|>
TITLE: Where does the "easy" definition of a weak n-category fail?
QUESTION [5 upvotes]: Okay, I'm going to ask a naiive question that surely has an interesting answer. So, a first approximation of defining a (small) weak n-category probably goes something like this. Take a pre-n-category C of all the cells, source and target maps that do the right thing (i.e. are globular), and a composition defined for each r in {0,...,n}.
For C, define a family of coherent sets $(\Sigma\_1, \Sigma\_1, \ldots)$ as a family of sets $\Sigma_r$ of r-cells in C such that

$f : a \rightarrow b \in \Sigma\_r \Rightarrow \exists f' : b \rightarrow a \in \Sigma\_r$
$f, f' : a \rightarrow b \in \Sigma\_r \Rightarrow \exists \alpha : f \rightarrow f' \in \Sigma\_{r+1}$

Now, suppose C admits such a family of coherent sets and all r-cells have associators, uniters, and interchangers in $\Sigma\_{r+1}$, one might be tempted to say C  is an $\infty$-category. If for all $r \geq n+1$, $\Sigma\_r$ is only identites, one might say this defines an n-category.
So, the reason I say "one might be tempted to say" is that, if it were that easy, someone much smarter than me would have done it already. :) So, where does the above recipe fail? Or is this definition unsatisfactory because it doesn't express the structure using a finite generating set of commutative diagrams (cf. Mac Lane's coherence etc.)?

REPLY [9 votes]: If I understand correctly what you're getting at, I think the reason this fails is because for n>2, not every diagram of constraints can be expected to commute (even up to higher constraints) in a weak n-category.  For example, a braided monoidal category can be regarded as a weak 3-category with one 0-cell and one 1-cell, but then the "double twist" is a constraint isomorphism which is not equal to the identity (also a constraint isomorphism).
One way to get around this is, instead of talking about diagrams of constraints in some particular n-category, to talk about "formal" diagrams of constraints, i.e. diagrams of constraints in a free n-category.  I think that when one makes your idea precise using this corrected approach, one will end up with something very similar to Batanin's higher-operadic definition of weak n-category.<|endoftext|>
TITLE: Graphs with incidence matrices whose pseudoinverses are proportional to their transposes
QUESTION [7 upvotes]: When I was working on my PhD dissertation, I came across a physical situation involving nodes and flows between them. It turned out that I was working with a complete oriented graph $K_n$ (all nodes are connected to each other), and I needed to calculate the pseudoinverse of its incidence matrix T, i.e. the rectangular matrix N(V) x N(E) where N(E) = n is the number of edges and N(V) is the number of vertices, with matrix element 1 if the edge enters the vertex, -1 if the edge leaves the vertex, and 0 otherwise.
To my surprise the pseudoinverse turns out to be proportional to the transpose of the incidence matrix! Specifically
$T^{+} (K\_n) = \frac{1}{n} T^{\prime} (K\_n)$
where the prime denotes transposition.

My question is:
A) What other graphs $G$, if any, have this property, i.e. that
$T^+(G) \propto T^\prime(G)$,
or some suitable generalization thereof, and
B) How can I show that this result is invariant of orientation? (I determined empirically is certainly true for all possible orientations of the small complete graphs, and I haven't been able to find a counterexample, but I don't have a proof of this statement yet)
I'm not a professional mathematician, so any thoughts would be welcome.

One class of generalizations that is possible (but not so interesting IMO) are disconnected graphs where each subgraph is a complete graph, i.e.
$G = K_{n_1} \oplus K_{n_2} \oplus \dots \oplus K_{n_m}$
In this case one gets
$T^+(G) = \frac{1}{n_1} T^\prime(K_{n_1}) \oplus \frac{1}{n_2} T^\prime(K_{n_2}) \oplus \dots \oplus \frac{1}{n_m} T^\prime(K_{n_m})$
which is not really that exciting, but perhaps could point the way to a more interesting generalization.

P.S. There is a short proof of the first fact, which relies on the fact that the complete graph has a Laplacian of the form
$\Delta = n \mathbf{I} - \mathbf{1} \mathbf{1}^\prime$
where $\mathbf{I}$ is the n x n identity matrix and $\mathbf{1}$ is a column vector of ones.
With this fact, together with knowing that the column sums of $T$ are all zero, it is straightforward to show that $T^\prime(K_n)/n$ satifies all the Moore-Penrose conditions for the pseudoinverse.

P.P.S. If anyone is interested in the physical context, here is where it came from.

REPLY [5 votes]: Let the vertices be numbered $1,\ldots,n$. For a vertex $i$, let $\mathrm{deg}(i)$ be the total number of (unsigned) edges incident to $i$. If $A$ is the incidence matrix, consider $AA^TA$. An entry of $AA^TA$ is indexed by a vertex $i$ and an edge $e=(j,k)$. If I did this correctly, it's straightforward to show that $[AA^TA]_{i,(j,k)}$ is: 

$-\mathrm{deg}(i)$ if $i=j$, 
$\mathrm{deg}(i)$ if $i=k$,
the number of undirected edges $(i,k)$ minus the number of undirected edges $(i,j)$ otherwise. 

If $A^T$ is a scalar multiple of the pseudoinverse of $A$, then $AA^TA$ must be a scalar multiple of $A$. It's easy to see from the above that this is equivalent to requiring that each connected component of the graph is complete, each vertex has the same (undirected) degree, and within each component, all edges have the same (undirected) multiplicity. (For example, I think you could have two components $K\_3$ and $K\_5$, where there are two edges between each pair of vertices in the $K\_3$ component and one edge between each pair in the $K\_5$ component.) None of this depends on the orientations of the edges: flipping an edge of the graph is equivalent to negating a column of $A$, and if $AA^TA$ was a scalar multiple of $A$ before, then it still is now. Since $(AA^TA)^T = A^TAA^T$, if $AA^TA = cA$, then $A^TAA^T = cA^T$, so $(1/c)A^T$ is the Moore-Penrose pseudoinverse of $A$. ($AA^T$ and $A^TA$ are obviously symmetric.) This completely characterizes graphs with your property.<|endoftext|>
TITLE: On Euclid's proof of the infinitude of primes and generating primes
QUESTION [19 upvotes]: So looking at Euclid's proof he says 
1)take a finite family of primes (F) 
2)multiply all the primes and add one
3)this new number has at least 1 new prime factor
So I was wondering about what kind of primes you get by recursively feeding this process into it self.
Since the number you must factor grows exponentially, it's hard to get a lot of numerical evidence for what happens.
I calculated a few:
[2]-> [2,3]-> [2,3,7]->[2,3,7,43]->[2,3,7,43,13,139]->[2,3,7,43,13,139,3263443]
->[2,3,7,43,13,139,3263443,547,607,1033,31051]-> cannot factor 113423713055421844361000443
[5] (x5)-> [5,2,3,31,7,19,37,3343,79,193662529] -> cannot factor 234069798025176583891
Obviously quite a few primes are missing, 5,11,19,etc from the first list, but could show up later.
So my question is does a finite family of primes exist that eventually generates all the primes?  I figure this probably doesn't have an easy answer, but any information related to this process would be appreciated, or even why it can't be done.

REPLY [7 votes]: Going with what Qiaochu Yuan said about f(x), it follows that we will never get those primes unless we start, even if we don't include multiplicities.  Since we're starting with 'n', then we're taking the prime factors of 'n+1', then we're taking the prime factors of f(n+1), then f(f(n+1)), then f(f(f(n+1))) etc, even if we get, say, a 72 in there, our number is [f(f(f(n+1)))] / 7, which then goes into f(x).
So no, unless you start with the infinite product $p_1 = \prod_{n=1}^\infty 6n-1$, you will never get any of those numbers.
It's funny, though; I'd had a whole demonstration started to show that you'll never get a multiplicity when taking f(f(...f(2)...)), but this is simpler.
As for the Euclid-Mullin sequence, I have no idea.<|endoftext|>
TITLE: can you fool SnapPea?
QUESTION [16 upvotes]: A while back I thought I had some simple knots that "fooled" SnapPea.  But I no longer remember those examples, if I ever had them to begin with.  
What I'm looking for is a non-hyperbolizable knot or link in S^3 for which SnapPea thinks it finds a hyperbolic structure on the complement.  Do you have such an example?   I'm interested in the examples that work in SnapPea -- it's fine if Snap or the Harriet Moser criterion "knows" the gluing equations are not satisfied. 
edit: to make my question more rigid, can you find an unknot (or a trivial link) for which SnapPea thinks there is a hyperbolic structure?
In case this is all jargon to you, SnapPea is software used primarily for finding and exploring hyperbolic structures on 3-manifolds: http://www.math.uic.edu/~t3m/SnapPy/doc/

REPLY [3 votes]: Have you tried something like this (you probably have):
Take a diagram of a complicated hyperbolic knot and draw a diagram for the Whitehead double.  Then change a crossing in the clasp.
I'd like to test this, but my SnapPea can't handle it.
Edit: You could also try Haken's unknots. (Wayback Machine)<|endoftext|>
TITLE: Is there a compact group of countably infinite cardinality?
QUESTION [27 upvotes]: Apologies for the very simple question, but I can't seem to find a reference one way or the other, and it's been bugging me for a while now.
Is there a compact (Hausdorff, or even T1) (topological) group which is infinite, but has countable cardinality? The "obvious" choices don't work; for instance, $\mathbb{Q}/\mathbb{Z}$ (with the obvious induced topology) is non-compact, and I get the impression that profinite groups are all uncountable (although I might be wrong there). So does someone have an example, or a reference in the case that there are no such groups?

REPLY [2 votes]: It is a well known fact (see for example Hodel´s article on the Handbook of Set Theoretic Topology) that the cardinality of an infinite compact homogeneous space is always a power of $2$. Topological groups are homogeneous (as it is pointed out in Yemon´s answer) and therefore any infinite compact group has the size of the continuum or bigger.<|endoftext|>
TITLE: "Vector bundle" with non-smoothly varying transition functions
QUESTION [9 upvotes]: I'm working my way through Lang's Fundamentals of Differential Geometry, and when he introduces vector bundles, he states that for finite dimensional bundles, the third axiom is redundant.  I'm hoping someone can give a counterexample in infinite dimensions.
His axioms (for a $C^p$ bundle) are (1) local triviality, (2) transition maps are Banach space isomorphisms (linear homeomorphisms), and (3) the maps $x\mapsto A_x$ are $C^p$ (where $A_x$ is a transition map).
Essentially, I'm looking for a $C^p$ automorphism of $B\times V$ of the form $(x,v)\mapsto(x,A_xv)$ such that $x\mapsto A_x$ is not $C^p$.  Here, $B$ is open (say the unit ball) in some Banach space, $V$ is another Banach space, and the linear maps are given the operator norm.
All derivatives are Fréchet derivatives.

REPLY [12 votes]: I suspect that the issue here is actually to do with continuity, not differentiability.  Without more details on the exact definition of vector bundle as given, I can't be sure (and I don't have a copy of Lang's book to hand to check).  If the definition is a "top down" one, then continuity is certainly an issue.  By "top down" then I mean that a vector bundle consists of two smooth manifolds, $E$ and $B$, and a smooth map $p : E \to B$ satisfying the above conditions.  (The other approach is to build it up from the transition functions.)
I'll assume that this is so.
Then from local triviality and the fibrewise description, we obtain the transition functions $\psi : U \times F \to U \times F$.  Now, from the properties we can find a function $\theta : U \to GL(F)$ with the property that $\psi(x,v) = (x,\theta(x)v)$.  There is no difficulty in simply defining this function.
The problem is that continuity of $\psi$ (even higher differentiability) is not sufficient to guarantee the continuity of $\theta$ in infinite dimensions.  The map $\theta$ is continuous if $GL(F)$ is given the weak topology where $(A_n) \to A$ if $(A_nv) \to Av$ and $(A_n^{-1}v) \to A^{-1}v$ for all $v$.  But normally we ask for the strong (norm) topology on $GL(F)$.
In finite dimensions, this topology agrees with the standard topology but in infinite dimensions they are very different.  For example, if we take $\ell^2$ and let $P_n$ be the projection onto the first $n$-coordinates then $(P_n) \to I$ in the weak topology but not in the strong topology (this is an example in $L(H)$ but can be easily tweaked to give an example in $GL(H)$).
Further reading:

Topological Vector Spaces, Schaefer.  Contains lots about the different topologies.
A Convenient Setting of Global Analysis, Kriegl and Michor.  Contains lots about the intricacies of infinite dimensional differential topology.

Edit: I finally remembered the classic example of this: $L^2$-functions on a Lie group.  For simplicity, let's take $S^1$.  Then $S^1$ acts in the obvious way on $L^2(S^1,\mathbb{C})$ (hereinafter $L^2$).  The action $S^1 \times L^2 \to L^2$ is jointly continuous but the associated map $S^1 \to GL(H)$ is most assuredly not.  Indeed, if $\lambda \ne \mu \in S^1$ then $\|R_\lambda - R_\mu\| = 2$ so the image is discrete.  So if we have an $S^1$-principal bundle $P \to B$ then we can form a new space by taking the quotient $P \times_{S^1} L^2$.  This will be locally trivial, and the transition maps will be fibrewise linear as they are of the form $x \mapsto R_{\lambda(x)} : L^2 \to L^2$ where $x \mapsto \lambda(x)$ are the transition functions of the $S^1$-bundle.  But the associated map $x \to GL(H)$ is not continuous and so it won't be a genuine vector bundle in the sense Lang defines.
(What's particularly embarrassing about how long it took me to remember this example is that in a recent paper I go into great detail about the different "levels" one can require for continuity of the action of subgroups of $Diff(S^1)$ on various loop spaces.  The paper in question is this one in case anyone's interested.)<|endoftext|>
TITLE: Super-linear time complexity lower bounds for any natural problem in NP?
QUESTION [23 upvotes]: Do we know any problem in NP which has a super-linear time complexity lower bound? Ideally, we would like to show that 3SAT has super-polynomial lower bounds, but I guess we're far away from that. I'd just like to know any examples of super-linear lower bounds.
I know that the time hierarchy theorem gives us problems which can be solved in O(n^3) but not in O(n^2), etc. Thus I put the word "natural" in the question.
I ask for problems in NP, because otherwise someone would give examples of EXP-complete problems. 
I know there are time-space tradeoffs for some problems in NP. I don't know if any of them imply a super-linear time complexity lower bound though.
(To address a question below about machine models, consider either multitape Turing machines or the RAM model.)

REPLY [4 votes]: If you restrict space usage to be sublinear then you can prove superlinear time lower bounds on SAT.  See this article on Lipton's blog for a nice exposition.<|endoftext|>
TITLE: Categories which are not compactly generated
QUESTION [16 upvotes]: Do you know natural examples of triangulated categories (or [presentable] stable $\infty$-categories) which are not compactly generated? (ideally they'd be defined algebraically, but curious to hear any examples.. the ones I know are gotten as opposites of compactly generated categories or by slightly ad hoc geometric constructions)

REPLY [7 votes]: There is also, now a bunch of years later, a result of Hall, Neeman, Rydh that says over a field of char $p$,there are no non trivial compact objects in $D_{qc}(B\mathbb{G}_a)$. 
Moreover they prove for any algebraic stack ,$X$, which is quasi-compact, quasi-separated and "poorly stabilized" (roughly that there exists a point whose stabilizer contains a copy of $\mathbb{G}_a$) then $D_{qc}(X)$ is never compactly generated. 
https://arxiv.org/pdf/1405.1888.pdf<|endoftext|>
TITLE: What is the relationship between algebraic geometry and quantum mechanics?
QUESTION [17 upvotes]: The basic relationship in algebraic geometry is between a variety and its ring of functions.  Arguably a similarly basic relationship in quantum mechanics is between a state space and its algebra of observables.  In what sense is algebraic geometry a "classical" (i.e. commutative) phenomenon?  How does intuition from quantum mechanics influence how one should think about algebraic geometry and vice versa?  What modern areas of research study their interaction?
(This question is loosely inspired by a similar question about representation theory.)

REPLY [7 votes]: There is a very nice paper by Martin Schlichenmaier available at arXiv math/000528, where the author discusses the relationship between quantization on Kaehler manifolds and projective geometry. Basically, a (quantizable) Kaehler manifold M can be embedded into a complex projective space by the Kodaira embedding theorem. The quantum Hilbert space becomes the projective coordinate ring of M.<|endoftext|>
TITLE: Compact generation for modular representations
QUESTION [10 upvotes]: Are the derived categories of modular representations of algebraic groups compactly generated? (e.g. consider SL_2 in characteristic 2). Note modular reps of finite groups are compactly generated (by the regular representation) - that's an example of compact generation of modules for an algebra. But here we're asking about comodules for a coalgebra that's not dualizable so it's not immediately clear (to me).
This makes more specific my other question for any "nice" examples of non-compactly generated categories.

REPLY [8 votes]: It appears this question is resolved in a definitive fashion in today's preprint
Algebraic Groups and compact generation of
their derived categories of representations by Hall and Rydh. Their first theorem asserts that the quasicoherent derived category of the stack $BG$ for $G$ a group scheme of finite type over a field $k$ is compactly generated if and only if either the characteristic of $k$ is zero, or the component group of the reductive part of $G$ (after
base change to $\overline{k}$) is SEMI-ABELIAN - or equivalently, contains no additive groups. In particular semisimple groups in (any) positive characteristic are out. 
They then deduce (using another paper of theirs from today, with Neeman) that even for $G$ affine in these ``poor" cases the quasicoherent derived category differs from the derived category of representations 
(ie quasicoherent sheaves on $BG$), and other striking results.<|endoftext|>
TITLE: Fibered nots (non-geometric HNN extensions of free groups normally generated by the monodromy)
QUESTION [9 upvotes]: A fibered knot is a knot $K$ in the $3$-sphere whose complement is a surface bundle over a circle. If $S$ is the fiber, the fundamental group of $S$ is free (of even rank), and the fundamental group of the complement is an HNN extension 
$$\pi_1(S) \to \pi_1(S^3-K) \to Z$$
where the $Z$ is generated by the meridian of the knot.
Since surgery on $K$ recovers the $3$-sphere, the group $\pi_1(S^3-K)$ has the interesting property that it is normally generated by (the conjugacy class of) the meridian.
What I didn't realize until recently is that there are many examples of "non-geometric" automorphisms $\phi$ of free groups $F$ for which the associated HNN extension $F \to G \to Z$ is normally generated by the conjugacy class of the monodromy. One simple example is the case $F = \langle a,b,c \rangle$ and $\phi$ is the automorphism $a \to c^{-1}abac, b \to bac, c\to bc$. 
Is there any systematic way of generating such examples? Is there a classification? One reason to be interested is that such examples can be used to construct smooth $4$-manifolds which are topologically $S^4$ but not obviously diffeomorphically $S^4$.
Edit: a link to the construction is http://lamington.wordpress.com/2009/11/09/4-spheres-from-fibered-knots/ (this explains the construction in the case of a fibered knot, but the group-theoretic condition is the only important ingredient).

REPLY [5 votes]: This question seems to be interesting even for the rank 2 free group. In this case, it is well-known (since the commutator is preserved up to conjugacy and inverse) that the outer automorphisms are the same as the mapping class group of a 2-punctured torus. The only possible candidates for outer automorphisms satisfying your condition are determined homologically as homology circles, and correspond to the trefoil and figure eight knot complements. But of the automorphisms corresponding to those outer autmorphisms, I'm not sure which satisfy your condition. I think this amounts to asking which curves in the figure 8 & trefoil knot complements normally generate the fundamental group. If one considers peripheral curves, then this is answered by the knot complement problem (I think the attribution of this for the figure 8 knot is Thurston, who classified the Dehn fillings on the figure 8 - I don't know who originally did this for the trefoil).
Another obervation is that if a curve normally generates, then there are finitely many conjugates that generate the fundamental group. Jason Callahan has ruled out the case of a non-peripheral curve in the figure 8 knot complement such that two conjugates generate the fundamental group. There's probably some other 3-manifold machinery that may be brought to bear on this question in the rank 2 case.<|endoftext|>
TITLE: Stable w-length
QUESTION [12 upvotes]: Let $F$ be a free group, and $w$ an element of $F$. In any group $G$, a $w$-word is the image of $w$ or $w^{-1}$ under a homomorphism from $F$ to $G$. The subgroup of $G$ generated by $w$-words is denoted $G(w)$.
For any $g \in G(w)$, the $w$-length of $g$, denoted $l(g|w)$, is the minimum number of $w$-words in $G$ whose product is $g$, and the stable $w$-length of $g$, denoted $sl(g|w)$, is the limit $sl(g|w) = lim_{n \to \infty} l(g^n|w)/n$.
If $w$ is not in the commutator subgroup of $F$, the stable $w$-length of every element in any group is trivial. Otherwise, one has a universal inequality 
$$1/2 \le sl_F(w|w) \le 1$$
(where the subscript $F$ indicates that stable $w$-length is being calculated in the free group $F$ containing $w$ itself.)
The lower bound of $1/2$ is realized e.g. by the word $w=xyx^{-1}y^{-1}$ (i.e. a standard commutator) in $F_2$ but I don't know how to compute (or even approximate!) $sl(w|w)$ in (essentially) any other case.
What values are achieved by $sl(w|w)$? Are they all rational? Are they dense? Is $1$ ever achieved? Is $1/2$ ever achieved for a word other than $xyx^{-1}y^{-1}$? 
(Added:) After reading FC's answer, it is probably worth pointing out that the lower bound $1/2 \le sl_F(w|w)$ comes from the inequality $scl_G(g) \le sl_G(g|w)(scl_F(w)+1/2)$ for any $g$ in any $G$, so one gets a better lower bound on $sl_F(w|w)$ if one knows $scl_F(w)>1/2$ (the estimate $scl_F(w)\ge 1/2$ is always true). Upper bounds can be established by exhibiting identities (like FC's identity below). Does a blind computer search yield any interesting examples?

REPLY [4 votes]: In case any one is still thinking about this question, it turns out one can say a lot. For example, $sl(w|w)=1/2$ whenever $w$ is a word of the form $[a,b^n]$ (and several other examples), one has $2/3 \le sl(w|w) \le 4/5$ when $w=[a,b]^2$, and if $\gamma_n$ is the iterated commutator $[x_1,[x_2,\cdots[x_{n-1},x_n]\cdots]$ one has $sl(\gamma_n|\gamma_n)\le 1-2^{1-n}$. In fact, I would explicitly conjecture that $sl(w|w)<1$ (i.e. strict inequality) for every $w$.
Getting systematic lower bounds on $sl(w|w)$, other than $scl(w)/(scl(w)+1/2)$ seems difficult; one imagines that the (currently nonexistent) theory of nonabelian Bavard duality might do the trick.<|endoftext|>
TITLE: Exotic automorphisms of the fundamental group of a curve?
QUESTION [10 upvotes]: A while back, Jordan S. Ellenberg brought the following problem to my attention.
If $G$ is a residually finite group, let $\widehat G$ be its profinite completion.
Let $S$ be a closed surface of genus $g \geq 2$, and let $\pi$ be its topological fundamental group.
Let $\mathrm{Mod}(S)$ be the mapping class group of $S$.
There is homomorphism $\widehat{\mathrm{Mod}(S)} \to \mathrm{Out}(\widehat \pi)$.  Is this map surjective?
In other words, does the geometric fundamental group of the moduli space surject the outer automorphisms of the geometric fundamental group of the curve?
Edit:  As Jordan points out, the map is not surjective.  So, the question is: 
What's the closure of the image of $\mathrm{Mod}(S)$ in $\mathrm{Out}(\widehat \pi)$?
Or, more precisely, for Henry: 
As Jordan explains, there is a map $\mathrm{Out}(\widehat \pi) \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) \to \widehat{\mathbb{Z}}^\star$
Is the closure of the image of $\mathrm{Mod}(S)$ in $\mathrm{Out}(\widehat \pi)$ the preimage of 1 and -1?

REPLY [3 votes]: Just a non-thought-out thought:
For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $\mathrm{Out}(\widehat{\pi})$ acts on this set.  Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $\mathrm{Out}(\widehat{\pi})$ valued in $\pm 1$. 
[If $S$ is $\mathbb{Z}/p\mathbb{Z}$,  I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]
I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $\mathrm{Out}(\pi)$ is finitely generated, the restriction of $F$
to it must have finite image.  (Slight clarification: restrict the product over $S$ to nonabelian finite simple groups, since the abelian ones provide no new information.)<|endoftext|>
TITLE: How Does a Borel Subgroup Know Which Weights Are Dominant
QUESTION [14 upvotes]: Let $G$ be a simple group (say $SL_n$) and let $B$ be a Borel subgroup (say upper triangular matrices). Then all irreducible representations of $G$ are induced from one-dimensional representations of $B$, i.e characters of $B$. However, only some of the weights will induce to non-zero representations. Relative to the standard maximal torus of diagonal matrices, these weights appear to be the anti-dominant weights. However, $B$ contains other tori as well, and no one choice is canonical. 
My question is: since induction from $B$ didn't require a choice of torus, how are some characters of $B$ already deemed to be anti-dominant? Where is the symmetry broken?

REPLY [15 votes]: One of the sneaky tricks that Lie theorists play on students is that they tell them about Cartan subalgebras, and then at some point, they pull the rug out and say "just joking; really you should think about the abstract Cartan."  The abstract Cartan is a Borel mod its radical.  You might say "which Borel?" but it doesn't matter; all these quotients are canonically isomorphic (any two Borels are conjugate, and any two ways of making them conjugate differ by an element of the Borel).  
Note, a Cartan subgroup of your Lie algebra isn't canonically isomorphic to the abstract Cartan;  you have to choose a Borel containing that Cartan subgroup first.  That's where the symmetry is broken.
The abstract Cartan has a canonical notion of positive weight: you look at the action of $B/N$ on the quotient of the Lie algebra $n/[n,n]$, and the weights that appear there are your simple roots (all the positive roots can be obtained by taking the associated graded for the filtration $n\supset [n,n]\supset [n,[n,n]]$).  Again, any way of making Borels conjugate carries these weights to the corresponding ones for the other Borel.  So once you've picked a Borel, you can use the isomorphism of the abstract Cartan to your chosen one to get a root system on your chosen one.
EDIT: As Allen notes below: I learned most of this from the book of Chriss and Ginzburg.
You get fundamental coroots by taking minimal parabolics over your Borel (those obtained by adding one negative simple root).  There's a canonical map from the abstract Cartan of $P/[P,P]$ to that of $G$, and there's unique coweight of $P/[P,P]$ (which is $SL_2$ or $PSL_2$) so that the action on the unique simple root space has eigenvalue 2.  The image of that is the simple coroot.<|endoftext|>
TITLE: look into Delzant Polytope
QUESTION [6 upvotes]: A Delzant polytope in R^n by definition is a simple, rational, and smooth convex polytope in R^n (Ana Cannas da Silva's book for notions.) Do you guys have any insight of the definition, for example, anything we can say about the shape? They satisfy some rigidity conditions? All related comments are welcome!

REPLY [18 votes]: The standard model of a vertex which satisfies the Delzant condition is the positive "quadrant" $x_i \geq 0$ of $\mathbb{R}^n$ near the origin. In general a polytope is Delzant if and only if every vertex can be taken to this standard model by some element of $\mathrm{GL}(n, \mathbb Z)$.
The motivation for this definition, coming from symplectic geometry, is the following fact, due to Archimedes. Consider the standard radius-one sphere in 3-space enclosed in a cylinder of radius one. Project the sphere outwards onto the cylinder, orthogonal to the cylinder's axis. Archimedes' theorem says that this map preserves areas. In terms of Delzant polytopes, you should think of the sphere as 2-dimensional toric manifold. The corresponding "polytope" is the interval $[-1,1]$. Archimedes' theorem says that if you take the cylinder $S^1\times [-1,1]$ and collapse the circles $S^1\times\{\pm 1\}$ you obtain the sphere as a symplectic manifold. To see how to generalise this to higher dimensions, take the n-fold product of the cylinder $S^1\times [0,\infty)$ and consider the map to $\mathbb{R}^n$ given by forgetting the $S^1$-factors. The image is the positive "quadrant" mentioned in the first paragraph above. Archimedes' theorem tells us that if we collapse the preimage of the boundary of this quadrant in a controlled way we obtain something which admits a smooth symplectic structure. I.e., over the orgin we collapse the whole $T^n$-fibre, more generally, over each coordinate $r$-plane we contract the corresponding orthogonal copy of $T^{n-r}$ inside $T^n$. (To do this recall each $S^1$-factor corresponds to a direction in $\mathbb{R}^n$).
Now the meaning of Delzant's condition should be clear: given a Delzant polytope P we built a symplectic manifold by taking $P \times T^n$ and then collapsing parts of the boundary. To decide how, we map each vertex to the standard model described above and collapse as in that case. Archimedes theorem tells us the resulting object is a smooth symplectic manifold with a torus action.<|endoftext|>
TITLE: Minimize Perimeter(S)/Area(S) for S inside the unit square.
QUESTION [9 upvotes]: This is a very silly question.
For all regions S contained inside the unit square, what is the infimum of the quantity Perimeter(S)/Area(S)? This ratio being considered is not scale invariant, so it is only the constraint of being contained within the square which implies that this infimum is non-zero.
There are some "obvious" configurations to try, but I do not even know how to use a calculus of variations argument to show that these are local maxima.
Can one do better than $\displaystyle{2 + \sqrt{\pi}}$?
EDIT: One can probably show that an optimal S has $\mathbf{Z}/4\mathbf{Z}$ symmetry. Assume that S consists of line segments along part of boundary, leaving segments of length x at each corner, and then (four quarters of) a shape T in the corner. We can assume that T has volume Ax^2 and perimeter Px. Minimizing the quantity with respect to x, one obtains a minimum of:
$$2 +\sqrt{4 + \frac{(P/2-4)^2}{(A - 4)}},$$
which becomes $2 + \sqrt{\pi}$ when T is the circle, i.e., $A = \pi$ and $P = 2 \pi$.

REPLY [3 votes]: This is known as the Cheeger problem, you can have a look at the paper "An introduction to the Cheeger problem" (http://www.utgjiu.ro/math/sma/v06/a02.html) and references therein.<|endoftext|>
TITLE: Homotopy orbit spaces of representation spheres
QUESTION [7 upvotes]: Let $G$ be a finite group and $V$ be finite-dimensional real representation of $G$.  Write $S^V$ for the one-point compactification of $V$, with induced $G$-action, viewed as a pointed space, and consider the homotopy orbit space $(S^V)_{hG}$.  For instance, if $V$ is the regular representation of $G=\mathbb{Z}/2$, $(S^V)_{hG}$ works out to be the suspension of $B\mathbb{Z}/2$.  Is there some kind of general description of this space, presumably built out of classifying spaces of subgroups of $G$?  What if I only care about the stable homotopy type?  I'm most interested in the permutation representations of the symmetric groups.

REPLY [9 votes]: The most salient fact about (SV)hG is that it is the Thom space of a vector bundle over BG.  So this gives you a lot of control over the sorts of things seen by homology: e.g., a Thom isomorphism if the bundle is oriented with respect to your homology theory; if you put a CW-structure on BG, then (SV)hG has a CW-structure with cells in the same dimensions (shifted up by dim V).  Morally, it's a "twisted suspension" of BG, and that fact encodes most of what I know about it.
For G=Z/2, and L=sign rep, you have (SnL)hG = RP∞/RPn-1, and something similar holds for Z/p, and for the symmetric group Σp (and perhaps for every group with periodic tate cohomology?)
If you have a decomposition of the form BG = hocolim BH, then you can pull your bundle over BG back to all the terms in the hocolim to get (SV)hG = hocolim (SV)hH.<|endoftext|>
TITLE: Fundamental Examples
QUESTION [215 upvotes]: It is not unusual that a single example or a very few shape an entire mathematical discipline. Can you give examples for such examples? (One example, or few, per post, please)
I'd love to learn about further basic or central examples and I think such examples serve as good invitations to various areas. (Which is why a bounty was offered.)

Related MO questions:  What-are-your-favorite-instructional-counterexamples,
Cannonical examples of algebraic structures, Counterexamples-in-algebra, individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline, most-intricate-and-most-beautiful-structures-in-mathematics, counterexamples-in-algebraic-topology, algebraic-geometry-examples, what-could-be-some-potentially-useful-mathematical-databases, what-is-your-favorite-strange-function; Examples of eventual counterexamples ; 

To make this question and the various examples a more useful source there is a designated answer to point out connections between the various examples we collected.

In order to make it a more useful source, I list all the answers in categories, and added (for most) a date and (for 2/5) a link to the answer which often offers more details. (~year means approximate year, *year means a year when an older example becomes central in view of some discovery, year? means that I am not sure if this is the correct year and ? means that I do not know the date. Please edit and correct.) Of course, if you see some important example missing, add it!
Logic and foundations: $\aleph_\omega$ (~1890), Russell's paradox (1901),
Halting problem (1936), Goedel constructible universe L (1938), McKinsey formula in modal logic (~1941), 3SAT (*1970), The theory of Algebraically closed fields (ACF) (?),
Physics: Brachistochrone problem (1696), Ising model (1925), The harmonic oscillator,(?) Dirac's delta function (1927), Heisenberg model of 1-D chain of spin 1/2 atoms, (~1928), Feynman path integral (1948), 
Real and Complex Analysis: Harmonic series (14th Cen.) {and Riemann zeta function (1859)}, the Gamma function (1720), li(x), The elliptic integral that launched Riemann surfaces (*1854?), Chebyshev polynomials (?1854) punctured open set in C^n (Hartog's theorem *1906 ?)
Partial differential equations:  Laplace equation (1773), the heat equation, wave equation, Navier-Stokes equation (1822),KdV equations (1877),
Functional analysis: Unilateral shift, The spaces $\ell_p$, $L_p$ and $C(k)$, Tsirelson spaces (1974), Cuntz algebra,
Algebra: Polynomials (ancient?), Z (ancient?) and Z/6Z (Middle Ages?), symmetric and alternating groups (*1832), Gaussian integers ($Z[\sqrt -1]$) (1832), $Z[\sqrt(-5)]$,$su_3$ ($su_2)$,  full matrix ring over a ring, $\operatorname{SL}_2(\mathbb{Z})$ and SU(2), quaternions (1843), p-adic numbers (1897), Young tableaux (1900) and Schur polynomials, cyclotomic fields, Hopf algebras (1941) Fischer-Griess monster (1973), Heisenberg group, ADE-classification (and Dynkin diagrams), Prufer p-groups,
Number Theory: conics and pythagorean triples (ancient), Fermat equation (1637), Riemann zeta function (1859)  elliptic curves, transcendental numbers, Fermat hypersurfaces,
Probability: Normal distribution (1733), Brownian motion (1827), The percolation model (1957), The Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble, SLE (1999),
Dynamics: Logistic map (1845?), Smale's horseshoe map(1960). Mandelbrot set (1978/80) (Julia set), cat map, (Anosov diffeomorphism)
Geometry: Platonic solids (ancient), the Euclidean ball (ancient), The configuration of 27 lines on a cubic surface, The configurations of Desargues and Pappus, construction of regular heptadecagon (*1796), Hyperbolic geometry (1830), Reuleaux triangle (19th century), Fano plane (early 20th century ??), cyclic polytopes (1902), Delaunay triangulation (1934) Leech lattice (1965), Penrose tiling (1974), noncommutative torus, cone of positive semidefinite matrices, the associahedron (1961)
Topology: Spheres, Figure-eight knot (ancient), trefoil knot (ancient?) (Borromean rings (ancient?)), the torus (ancient?), Mobius strip (1858),  Cantor set (1883), Projective spaces (complex, real, quanterionic..), Poincare dodecahedral sphere (1904), Homotopy group of spheres, Alexander polynomial (1923), Hopf fibration (1931), The standard embedding of the torus in R^3 (*1934 in Morse theory), pseudo-arcs (1948), Discrete metric spaces, Sorgenfrey line, Complex projective space, the cotangent bundle (?), The Grassmannian variety,homotopy group of spheres (*1951), Milnor exotic spheres (1965)
Graph theory: The seven bridges of Koenigsberg (1735), Petersen Graph (1886), two edge-colorings of K_6 (Ramsey's theorem 1930), K_33 and K_5 (Kuratowski's theorem 1930), Tutte graph (1946), Margulis's expanders (1973) and Ramanujan graphs (1986), 
Combinatorics: tic-tac-toe (ancient Egypt(?)) (The game of nim (ancient China(?))), Pascal's triangle (China and Europe 17th), Catalan numbers (18th century), (Fibonacci sequence (12th century; probably ancient), Kirkman's schoolgirl problem (1850), surreal numbers (1969), alternating sign matrices (1982)
Algorithms and Computer Science: Newton Raphson method (17th century), Turing machine (1937), RSA (1977), universal quantum computer (1985)
Social Science: Prisoner's dilemma (1950) (and also the chicken game, chain store game, and centipede game), the model of exchange economy, second price auction (1961)
Statistics: the Lady Tasting Tea (?1920), Agricultural Field Experiments (Randomized Block Design, Analysis of Variance) (?1920), Neyman-Pearson lemma (?1930), Decision Theory (?1940), the Likelihood Function (?1920), Bootstrapping (?1975)

REPLY [2 votes]: 3264 is considered "emblematic" of a solution to an instance of the fundamental problem of constructing notions of 'random objects'/'generic object'/'objects in general position'.
In D. Eisenbud, J. Harris: 3264 and all that. Cambridge University Press (2016) one reads:

"[...] the determination, by Chasles,1 of the number of smooth conic plane curves tangent to five given general conics. The problem is emblematic of the dual nature of the subject. On the one hand, the number itself is of little significance: [...] But the fact that the problem is well-posed---that there is a Zariski open subset of the space of 5-tuples $(C_1,\dotsc,C_5)$ of conics for which the number of conics tangent to all five is constant, and that we can in fact determine that number---is at the heart of algebraic geometry. And the insights developed [...] [for] a rigorous derivation of the number [...] [e.g.] a new parameter space for plane conics, and the understanding why intersection products are well-defined for this space---are landmarks [...] the number 3264 [...] is as emblematic of enumerative geometry as the date 1066 [...] is of English history."

[emphasis added] The Zariski-openness means that there is a meaningful notion of 'general position'.
An introduction (in French):

Étienne Ghys:TROIS MILLE DEUX CENT SOIXANTE-QUATRE. Comment Jean-Yves a récemment précisé un théorème de géométrie. Images des Mathématiques. 2008

1  Joncquières is said to have obtained this result earlier, yet did not publish. None of the 'solutions' seems to be considered rigorous, and a proof, i.e. a correct deduction of '3264' from respected axioms, had to wait until the late 20th century.<|endoftext|>
TITLE: Explicit description of a free braided monoidal groupoid with inverses
QUESTION [5 upvotes]: Let G be a braided monoidal groupoid: it does no harm to suppose that the monoidal product on G is strictly associative, so I'll do that.  
"With inverses" means that for every object $X$ of G, there is an object $Y$ and an isomorphism $X\otimes Y\approx 1$ in G, where $1$ is the unit object.
I'd like to assume, without loss of generality, that inverses exist "on the nose", so that for every object $X$, there is an object $X^{-1}$ such that $X\otimes X^{-1} = 1$.  That is, the objects of G form a group.
First question: am I allowed to do this?
Now I can get a 2-category BG by "delooping" G (using the monoidal structure), so that BG has only one object *, and the category of morphisms BG(*,*) is exactly G.  This is a 2-groupoid with one object, and it carries some sort of additional structure encoding the braiding.
A connected 2-groupoid is exactly the same thing as a crossed module, which consists of data $(H,F, d: H\to F, \phi: F\to \mathrm{Aut}(H))$, where $H$ and $F$ are groups and $d$ and $\phi$ are homomorphisms.  In terms of G, F is the group of objects of G, while H is the set of 1-morphisms in G with unit object as domain.
Second question: what extra structure do I put on the crossed module to encode the braiding?
I want to understand such G which are free on some set S of objects.  In the translation to crossed modules, the group F will have to be the free group on S.  
Third question: how do you describe the group H in this crossed module?  (That's what I mean by "explicit".)  
There's an extensive literature on braided monoidal categories, so I bet someone has thought about this.  
(Oh, and I can deloop one more time to get a weak 3-groupoid B2G, and this thing will model a homotopy 3-type X.  If G is free, X is a wedge of 2-spheres.  Because of this, I know things like the image and kernel of $d: H\to F$.  But what is $H$ itself?)

REPLY [3 votes]: I think the answer to the first question is yes.  Let $C$ be a monoidal category whose monoid of isomorphism classes is a group, so that every object has an inverse.  For each object $X$ pick an object $X^{-1}$ and isomorphisms $l_X : X^{-1} \otimes X \to 1$, $r_X : X \otimes X^{-1} \to 1$ such that the composition $(r_X \otimes X)(X \otimes l_X) : X \to X \otimes X^{-1} \otimes X \to X$ is the identity.  (This should be no problem; by assumption we can find $X^{-1}$, $l_X$, $r_X$ satisfying all but the last identity, and we can alter $r_X$ to make the last identity hold.)  We also need to choose $1^{-1} = 1$, $l_1 = r_1 = \mathrm{id}$.
Now we build a strict monoidal category $D$ whose objects are the free group on the objects of $C$, where the monoidal structure $\cdot$ is strict and given by the multiplication in this group, and where the Hom sets are "pulled back" along the function $f : \mathrm{Ob}\,D \to \mathrm{Ob}\,C$ which sends a reduced word in objects of $C$ and their inverses to the tensor product of those objects and their chosen inverse objects.  This induces a fully faithful functor $F : D \to C$ which is obviously also surjective and thus an equivalence.  We need to make it a monoidal functor.  To do so, note that $F(A \cdot B) = F(A) \otimes F(B)$ unless there is cancellation between the words $A$ and $B$ in the free group.  Where there is cancellation, we use the isomorphisms $l_X$ and $r_X$ to build a map $F(A \cdot B) \to F(A) \otimes F(B)$.  The monoidal functor conditions are satisfied because of the conditions on $l_X$ and $r_X$.  Finally, the monoid of objects of $D$ is by definition a group.
As for the braiding, we should be able to pull it back along $F$ also, so that $F$ is braided monoidal.<|endoftext|>
TITLE: Notions of degree for maps $S^n \to S^n$?
QUESTION [12 upvotes]: In algebraic topology, we define a degree for a map $f: S^n \to S^n$ as where the induced map $f_*$ on the $n$-th homology group of $S^n$ sends $1$.
In differential topology, we have a different (same?) notion of degree for $f$.  You take a regular value $b \in S^n$, consider $f^{-1} (b)$ (which is finite by the inverse function theorem and some compactness argument), and take the difference between the number of points in the preimage where the Jacobian of $f$ is positive and the number of points in the preimage where the Jacobian of $f$ is negative.
Geometrically, I can see that they are the same, but I couldn't convince myself rigorously.  In Prop 2.30 of Hatcher, he mentions that the degree of $f$ is the sum of the local degrees of $f$ at each preimage point, and local degrees are either $\pm 1$. (Local degree is defined in the middle of page 136 in Hatcher.)
So, the final question is, must the sign of the local degree of $x \in f^{-1}(b)$ the same as the sign of the Jacobian of $f$ at $x$?

REPLY [7 votes]: I think what you need is the following lemma (usually called the "Stack of records" lemma):
Consider a smooth proper map of manifolds of the same dimension $f \colon M \to N$ and let $y \in N$ be a regular value of $f$.
Then there exists a neighbourhood $V \subset N$ of $y$ such that $f^{-1}(V) = \cup\_{i=1}^n U\_i$ with $U\_i \cap U\_j = \emptyset$ for $i \neq j$ and $f|\_{U_i} \colon U\_i \to V$ is a diffeomorphism for all $i$.

Now from this you can just sum up $\pm 1$ according to orientation on each $U_i$ to get the local degree of $f$ at $y$, and this works for both definitions of degree.<|endoftext|>
TITLE: Rational Group Cohomology
QUESTION [7 upvotes]: This is a general question about group cohomology.  I'm interested in the case when the coefficients are the rational numbers and hence I suppose when my groups are infinite.  The question splits into two:
1)  Are there any favoured examples that you would recommend a look at? (Recommended references would be just as welcome.)
And the main question:
2)  What sort of functors on the category of groups leave the rational cohomology unchanged?  In particular is there a projection onto a special subcategory of groups that is in some way the right category to study?
I have a feeling that someone with a good knowledge of rational homotopy theory would be able to answer this question with relative ease.

REPLY [3 votes]: One class of groups whose rational cohomology is particularly easy to handle are the finitely generated torsion-free nilpotent groups. This is because they admit refined Postnikov systems which can be localised, leading to a recipe for computing the Sullivan minimal model of cochains in terms of their central series. This means one can also read off cup and Massey products quite easily.
A good reference (with a crash course in rational homotopy theory included) is the paper
Oprea, John The category of nilmanifolds. Enseign. Math. (2) 38 (1992), no. 1-2, 27–40.<|endoftext|>
TITLE: What is induction up to epsilon_0?
QUESTION [26 upvotes]: This is a question asked out of curiosity, and because I can't understand the Wikipedia page.
I have often been told that PA cannot prove the validity of induction up to $\epsilon_0$, which has been expressed to me roughly as the claim that $\epsilon_0$ is well-ordered. I understand what ordinals are, and what $\epsilon_0$ is. I also understand first order logic and axiom schemes, so I understand how the induction axiom scheme formalizes the notion that $\omega$ is well-ordered. 
What I don't understand is how one could formulate the statement that $\epsilon_0$ is well-ordered as a first order sentence in arithmetic. Would someone mind spelling this out for me?

REPLY [2 votes]: The answer to your question can be found in Maria Hameen-Anttila's paper, "Nominalistic Ordinals, Recursion on Higher Types, and Finitism", Bulletin of Symbolic Logic,25(1), 101-124 (2019). Since the version I have immediate free access to is the Accepted author manuscript version found on the University of Helsinki research portal (www.researchportal.helsinki.fi/publications/nominalistic-ordinals-recursion-on-higher-types-and-finitism), I will be using that version's pagination rather than the published version's pagination when referring to the pages on which the answer to your question can be found.
To begin with, the answer to your question can be found on pp. 4-14 (these pages also provide the historical context regarding the answer--the actual answer can be found on pp. 12-14 in the Accepted author manuscript on the Web).  I hope this helps.
As regards the statement, "I have often been told that $PA$ cannot prove the validity of induction up to $\epsilon_0$, which has been expressed to me roughly as the claim that $\epsilon_0$ is well-ordered", you might consider Noah Schweber's nice answer to my mathoverflow question, "What does 'can almost be proven in $PA$' mean regarding Theorem 2 of Timothy Chow's expository article, 'The Consistency of Arithmetic' ":

We have an arithmetic statement of the form
$(*)$ $\forall$$x$$\varphi$($x$)
which while not provable in $PA$ has the property that for each natural number $n$ the instance
$(*)_{n}$ $\varphi$($\mathfrak n$)
is provable in $PA$ (where "$\mathfrak n$" is the numeral corresponding to the natural number $n$)...
...Note that we also have the same phenomenon with respect to consistency:  for each natural number $n$, $PA$ proves "there is no $PA$-proof of '0=1' of length $\lt$ $n$."  So $PA$ almost proves $Con$($PA$).

It should be noted that this same phenomenon holds for defining $\epsilon_0$ as well.  As is known, $\epsilon_0$ can be defined as follows:

{$\omega$, $\omega^{\omega}$, $\omega^{\omega^{\omega}}$,...} = $\epsilon_0$ for a countably infinite number of iterations of $\omega^{\alpha}$.

Since it is known that $PA$ proves that each of the elements of $\epsilon_0$ is well-ordered, but doesn't prove that the set of ordinals that is itself $\epsilon_0$ is well-ordered, one can see that the aforementioned set is well-ordered since each of its members is well-ordered.  Here we have a early concrete example of Godel's First Incompleteness Theorem before Paris-Harrington or Goodstein sequences.  I hope this clarifies things at least somewhat.<|endoftext|>
TITLE: Are there elementary-school curricula that capture the joy of mathematics?
QUESTION [33 upvotes]: UPDATE:  Wow, thank you everyone for the great insights!
A couple of months ago I stumbled across Paul Lockhart's essay A Mathematician's Lament and it made perfect sense to me.  I'm not meaning to argue this essay one way or the other, except to say that 12 years of what I did in math class really isn't mathematics as you -- and I, as an "enthusiastic amateur" -- enjoy it.
We're probably going to homeschool our daughter, who will be kindergarten age next fall.  I feel there's a place for knowing your times tables and the like, but there's also a place for knowing that mathematics is more than arithmetic and formulas: it's discovery, playing with ideas, etc.
Where I'm going is that I know enough to understand the difference, but I'm not quite confident enough to teach (or lead) this process effectively, since I'm not a professional mathematician.
Are there any curricula that would help provide some structure to facilitate this kind of learning, so that there is one less student who has a shortchanged opinion of the mathematics profession?
Or, alternatively, if you felt that your elementary school math education hit the mark, how was it done?
Thanks for your time!

REPLY [4 votes]: Here is something I've considered. I only have the first volume, though, and am not sure if it gets at the mathematics the way I'd like. One thing I can say is that the book is very funny to my 5 year old.
Another resource that I've heard is good but have not yet tried is Beast Academy, which was produced by the Art of Problem Solving. These present mathematics topics in a graphic novel format. Children purportedly have a hard time putting them down.<|endoftext|>
TITLE: Can one do without Riesz Representation?
QUESTION [15 upvotes]: In more detail, can one establish that the continuous linear dual of a Hilbert space is again a Hilbert space without appealing to the Riesz Representation Theorem?
For me, the Riesz Representation Theorem is the result that every continuous linear functional on a Hilbert space is of the form $v \mapsto \langle v, u \rangle$ for some $u$ in the Hilbert space.
Whilst I have no particular quarrel with the Riesz Representation Theorem itself, I wonder if it's possible to do without it.  My motivation is fairly flimsy, but consider the situation where you have an arbitrary inner product space, $V$.  Then its dual is a Hilbert space.  However, to use Riesz Representation to prove that, you first have to complete $V$ to a Hilbert space and then apply Riesz.  Completing metric spaces, and in particular showing that the completion of an inner product space is a Hilbert space, seems like a lot of just hard slog to me (and hard to motivate to students in particular) so I wondered if one could avoid it by proving directly that the dual was a Hilbert space.

REPLY [5 votes]: Well, there is a way, but I don't know if I like it: Let $V$ be a complex inner product space, and for any $f\in V^*$ and $\epsilon>0$ let $$V^f(\epsilon)=\{v\in V:\|v\|=\|f\|, f(v)\ge\|f\|^2(1-\epsilon)\}$$
Now, as $\epsilon\to0$ we expect the functionals induced by $v\in V^f(\epsilon)$ to converge to $f$. More precisely, let $\bar v$ be the functional $\bar v(u)=\langle u,v\rangle$. It's easy to estimate $|f(u)-\bar v(u)|$ when $u$ is parallel to $v$. It's just a tiny bit harder to do so when $u\perp v$: Assume first that $\|u\|=\|v\|=\|f\|$ and consider $f(\overline{f(u)}u+\overline{f(v)}v)$, compute the norm using Pythagoras and employ the definition of $\|f\|$ to find $$\|f(u)\|^2\le\|f\|^4-|f(v)|^2\le\|f\|^4(2\epsilon-\epsilon^2)=\|f\|^2\|u\|^2(2\epsilon-\epsilon^2)$$
I left out some details, but the end result is that $\|\bar v-f\|\to0$ when $v\in V^f(\epsilon)$ and $\epsilon\to0$. Now we can finally define the inner product on $H^*$ by $$\langle f,g\rangle=\lim\langle w,v\rangle$$ where $v\in V^f(\epsilon)$, $w\in V^g(\epsilon)$ and $\epsilon\to0$.
Pretty? I don't think so, but it seems to answer your question. As a bonus, the Riesz representation theorem is now of course just around the corner.<|endoftext|>
TITLE: Reference for iterated homotopy fixed points?
QUESTION [6 upvotes]: What are (good) references for results about iterated homotopy fixed points?  That is, suppose G is a topological group acting on a space (or spectrum) X, and H is a normal subgroup of G.  Then one would like to first compute the homotopy fixed points of X with respect to H, and use that as a stepping stone to compute the homotopy fixed points of X with respect to G.
(I am independently interested in both the space and spectrum versions, so am happy with pointers, comments regarding either.)

REPLY [8 votes]: The statement XhG = (XhH)hG/H is true for any G-object X of any complete (∞,1)-category C.  An object of C with a G-action is the same as a functor BG → C where BG represents the category (or (∞,1)-category if G is not discrete) with a single object with automorphism group G.  The G-fixed points are the homotopy limit of this functor, or equivalently its right Kan extension along the functor BG → •.  We can factor this latter functor as p: BG → B(G/H) followed by q: B(G/H) → •.  So
$X^{hG} = (qp)_* X = q_* p_* X = (p_* X)^{hG/H}$
It remains to compute the right Kan extension of X along p.  On the object • of B(G/H), it is given as the limit of the diagram X over the category • ↓ G, which is the translation groupoid of G acting on G/H, or equivalently BH.  So indeed $p_* X = X^{hH}$.  Identifying the action of G/H is left as an exercise for the reader. :)<|endoftext|>
TITLE: Solutions of the Quantum Yang-Baxter Equation
QUESTION [9 upvotes]: I am interested in finding non-constant solutions to the following Yang Baxter equation
$$R_{12}(x/y) R_{13}(x/z) R_{23}(y/z) = R_{23}(y/z)
R_{13}(x/z) R_{12}(x/y)$$
where $R(x)$ is an endomorphism of $V\otimes V$. Personally I happen to be interested in finding explicit solutions in the case where $V$ is of dimension 2.
I would like to know what families of solutions are known. For example, there is an example at the end of Ch 12 of Chari and Pressley related to quantum affine sl2. I would prefer answers that are as explicit as possible (eg writing $R$ as 4x4 matrix) and some context as to where they come from.

REPLY [3 votes]: The general theory (due to Jimbo) is that each irreducible finite dimensional representation
of the quantised enveloping algebra of a Kac-Moody algebra (not of finite type) gives a trigonometric R-matrix. There is substantial information on these representations but the R-matrices are not explicit. There is a special case which is explicit and is given by the
"tensor product graph" method (this was worked out by Niall MacKay and Gustav Delius).
I used this in my paper:
R-matrices and the magic square. J. Phys. A, 36(7):1947–1959, 2003.
and you can find the references there.
If you want to go beyond this special case and be explicit then you can use
"cabling" a.k.a "fusion".
The only papers which deal with R-matrices not covered by the tensor product graph method that I know of are
Vyjayanthi Chari and Andrew Pressley. Fundamental representations
of Yangians and singularities of R-matrices. J. Reine Angew. Math.,
417:87–128, 1991.
G´abor Tak´acs. The R-matrix of the Uq(d(3)4 ) algebra and g(1)2 affine
Toda field theory. Nuclear Phys. B, 501(3):711–727, 1997.
Bruce W. Westbury. An R-matrix for D(3) 4 . J. Phys. A, 38(2):L31–L34, 2005
Deepak Parashar, Bruce W. Westbury
R-matrices for the adjoint representations of Uq(so(n))
arXiv:0906.3419
The Chari & Pressley paper deals with rational R-matrices.
The last preprint was an incomplete attempt to try and find the trigonometric analogues of these R-matrices.<|endoftext|>
TITLE: Relation between higher algebraic K-groups and topological K-groups
QUESTION [12 upvotes]: Let X be a compact Hausdorff space. Swan's theorem provides an equivalence between the category of (say real) vector bundles on X and the category of finitely generated projective modules over the ring C(X,R) of continuous functions from X to the real numbers. This relates the topological K0 to the algebraic K0 of a ring, i.e. the group completion of the semiring with elements finitely generated projective modules.
Higher topological K-Groups are defined as K0 of suspensions, i.e. Kn(X)=K0(Sn(X)) and in particular Kn(X)=K0(C(Sn(X),R)). Higher algebraic K-groups are defined as certain homotopy groups of a space, given by topological constructions like the +-construction, Q-construction or Waldhausen's S-construction which remind one of the group completion.
Is there a kind of analogous statement to Swan's theorem for higher K-groups? To be more precise, is it possible to construct a functor Rn(-) from well behaved topological spaces to the category of rings such that Kn(X)=Kn(Rn(X))?

REPLY [5 votes]: I am quite sure that such a construction exists (for complex topological K-theory): Start with a space X, from this produce the C^*-algebra of continuous complex-valued functions A:=Cont(X,C) - there you already have a ring encoding your space and I am sure this is step one. 
The C^ *-algebra-K-theory of A is the topological K-theory of X. But the C^ *-K-theory is algebraic K-Theory made homotopy invariant (see Example 2.1.3 and what follows in the Notes of Cortinas you find here). Now I think one could enforce homotopy invariance somehow on a ring level, but can not recall how.
If you take the topological tensor product of A with K:=kernel of the map from the Toeplitz algebra to Cont(S^1,C) (see start of section 2.3 in Cortinas' notes), then algebraic and topological K-theory coincide on the result (Thm 3.4.1.3 in Cortinas' notes) - if this leaves topological K-theory unchanged for C^ *-algebras coming from spaces then that is what you do. If not then it is something else, which you might find in those notes. Actually I thought it sounded less complicated than what I wrote, something like, pass to infinite matrices over A...
There also is an article comparing algebraic and topological K-theory, from the handbook of K-theory, which might contain what you look for: here<|endoftext|>
TITLE: Richardson varieties over finite fields
QUESTION [8 upvotes]: Let me start with some background to set the notation before I ask my question.
Let G be a semisimple algebraic group over some algebraically closed field K, and suppose we have fixed a Borel subgroup B and an opposite Borel subgroup B'. Let P be a parabolic subgroup containing B and consider the partial flag variety X = G/P. The closures of the B-orbits on X are the Schubert varieties, and the closures of the B'-orbits on X are the opposite Schubert varieties. Their intersections are Richardson varieties.
As far as I am aware, all of this stuff can be defined over the integers (anyway I'm only interested in the case when G is a symplectic or orthogonal group), so we can ask about how many F_q-rational points these various varieties have. For a Schubert varieties Z I know how to do this: consider the poset of Schubert varieties contained in Z. It's graded, and specializing q into the rank generating function of this poset gives me the number of points. For opposite Schubert varieties it's similar. 
My question is what can one say about Richardson varieties? What I want to say is to just do the same thing: the Schubert varieties correspond to lower intervals in the Bruhat order, and the Richardson varieties correspond to arbitrary intervals. So it's tempting to say just take the rank generating function of this interval and specialize at q. Does this work? What if I assume something like G/P is minuscule? If that's not enough, I really only care about the case when the opposite Schubert variety is replaced by the open B' orbit on X.
A related question: do the Richardson varieties have nice cell decompositions like the Schubert varieties?
My feeling is that something like this is relatively easy and known (I just can't find it), or hopelessly complicated. Please inform me that I'm in the first case.

REPLY [7 votes]: The intersections of opposite Schubert cells have a very nice decomposition into products of tori and affine spaces due to Deodhar which, of course, induces such a decomposition of the Richardson.  This decomposition is defined over $\mathbb{Z}$ (actually it works in any building), so it lets you count points, and the strata are combinatorially described by special subwords of a reduced decomposition of one of the words.  I recommend reading the paper of Marsh and Rietsch.<|endoftext|>
TITLE: How have mathematicians been raised?
QUESTION [15 upvotes]: Many of us have -- or at some point want to have -- children, and wonder how we can do our best to fulfill the "nurture" component of helping them develop mathematical talent... not because we want them to be mathematical professionals, but we'd like them to have the choice, like we did.
However, this is not a question about how one "should" raise children, because of how fiercely debatable even the meaningfulness of such a question would be.  Instead,

What reputable statistics are known about how mathematicians or highly mathematically talented people have been raised, especially during their early, formative years when "potential" is thought to be more of a variable?

(This question is community wiki, so everyone can edit it.)
Edit: I changed the title from "How are mathematicians raised?", since I personally only intended to ask for descriptive, not prescriptive, answers.  The "However" paragraph above was inserted to further emphasize this.

REPLY [12 votes]: Buy the kid Duplo and Lego - the generic blocks, not the fancy kits.<|endoftext|>
TITLE: Pushouts in the Category of Schemes
QUESTION [29 upvotes]: When does it make sense to glue schemes together along subschemes?
In particular: is there a way to glue two schemes together along a closed point (say we're working over a field)? Can you glue two closed points of the same scheme together? 
Is it easier to glue in the category of algebraic spaces?

REPLY [4 votes]: Since it doesn't seem to have been mentioned yet, Ferrand proves in Theorem 7.1 of "Conducteur, descent, et pincement" that you can pushout closed immersions $Y' \to X'$ along various nice affine morphisms $g: Y' \to Y$. 
For example, if $g$ is finite, and every finite set of points of $X'$ and $Y$ are contained in an open affine (e.g., they are projective varieties) then the pushout exists (Theorem 5.4 of loc.cit.).<|endoftext|>
TITLE: Algebraic Geometry versus Complex Geometry
QUESTION [38 upvotes]: This question is motivated by this one.
I would like to hear about results concerning  complex projective varieties which 

have a complex analytic proof but no known  algebraic proof; or
have an algebraic proof but no known complex analytic proof.

For example, I don't think there exists an equivalent of Mori's bend-and-break argument that avoids reduction to positive characteristic. So the existence of rational curves on Fano varieties would be an example of 2.

REPLY [3 votes]: I am fairly certain that there is no complex analytic proof of the following theorem (but I would love to be proven wrong!). This is not strictly speaking an answer to the question, because the available proof is not exactly algebraic either; rather, it uses $p$-adic (analytic) methods.

Theorem. (Batyrev) Let $X$ and $Y$ be birational Calabi–Yau varieties (that is, smooth projective over $\mathbb C$ with $\Omega^n \cong \mathcal O$). Then $H^i(X,\mathbb C) \cong H^i(Y,\mathbb C)$.

The same methods were later refined to prove the following theorem:

Theorem. (Ito) Let $X$ and $Y$ be birational smooth minimal models (that is, smooth projective over $\mathbb C$ with $\Omega^n$ nef). Then $h^{p,q}(X) = h^{p,q}(Y)$ for all $p,q$.

Again, the proof goes through $p$-adic analytic methods, this time combined with $p$-adic Hodge theory (which I think counts as an algebraic method).
References.
V. V. Batyrev, Birational Calabi–Yau $n$-folds have equal Betti numbers. arXiv:alg-geom/9710020
T. Ito, Birational smooth minimal models have equal Hodge numbers in all dimensions. arXiv:math/0209269<|endoftext|>
TITLE: What are the possible images of a square under an area-preserving map?
QUESTION [7 upvotes]: Let S be the open unit square in R^2: the set of points (x,y) with 0 < x < 1 and 0 < y < 1.  Consider an area-preserving smooth map S --> R^2, that is, a map whose Jacobian has determinant 1 at every point.  What can the image of S look like?  
Can the image of the square have a smooth boundary?  I think you can smooth out the two corners on top with a transformation of the form (x,y) --> (x,y+f(x)), but this makes the other two corners sharper.
(I had added, and now removed, some nonsense about Gromov's nonsqueezing theorem, which does not hold in dimension 2, and doesn't say what I thought it said besides.)

REPLY [8 votes]: Your statement of Gromov's theorem is wrong. Of course you can map a square to any rectangle of the same area! Just by diagonal linear matrix with eigenvalues (k, 1/k) (i.e. shrink in one direction, stretch in another). 
Gromov nonsqueezing says $B^{2n}(1)$ can not be put inside $B(r)x\mathbb{C}^{n-1}$ for $B(r)$ - real 2-d ball in $\mathbb{C}$ with $r<1$ by a symplectomorphism. This means by a map that preserve symplectic form, which in turn can be viewed as measuring the "sum of the areas of projections of a 2-d object on the planes x1-y1, x2-y2 ....xn-yn". The non-squeezing in some sense is a statement about not being able to trade these "complex plane" directions for each other. It says nothing about squeezing inside those planes. 
I believe by a general volume preserving map you can take any contractible domain to any other of the same volume...<|endoftext|>
TITLE: Groupoid of moves on trivalent fatgraph
QUESTION [13 upvotes]: $\DeclareMathOperator\Out{Out}$Let $T$ be a finite trivalent fatgraph — i.e. a graph with a cyclic order of the edges at each vertex. Then there are certain basic "moves" we can perform on $T$: an embedded edge can be collapsed and then uncollapsed in a different way (a "rotation", or "2-2 move"), or the circular order of the three edges incident at a vertex can be reversed (a "flip").
Define a set $\mathcal{T}$ whose elements are trivalent fatgraphs $T'$ homotopic to $T$ with a labeling of the edges from $1$ to $n$ and a labeling of the vertices from $1$ to $m$ (note $m = 2n/3$). A "move" is a pair $(T,c)$ where $T \in \mathcal{T}$, and $c$ is an element of the set $e_1, e_2, \cdots, e_n, v_1, v_2, \cdots, v_m$. The move acts on the labeled fatgraph $T$, and turns it into a new labeled fatgraph $T'$ obtained from $T$ by performing a rotation on edge $e_i$ if $c=e_i$ or a flip on vertex $v_i$ if $c=v_i$. It is clear how a flip affects the labels (it doesn't). A rotation destroys one edge labeled by $e_i$ and creates a new edge, so label this new edge $e_i$.
Now define a marked fatgraph to be a labeled fatgraph (i.e. an element of $\mathcal{T}$) together with a homotopy class of homotopy equivalence to some fixed $K(\pi_1(T),1)$. The moves defined above generate a new groupoid on marked fatgraphs, by acting on the labeled fatgraph part. This groupoid — the groupoid acting on marked fatgraphs — I will denote by $V(T)$ (the notation $V(T)$ is suggested by the similarity to Thompson's group $V$).
This groupoid — or something like it — turns up in many different contexts, so as a preliminary question, it would be nice to know how it is referred to. (Or: does this construction even make sense?)
More substantially: what is known about the algebraic structure of $V(T)$? What can be said about the cohomology of its classifying space? What is the relation to the group $\Out(F)$, where $F$ is the (free) fundamental group of $T$? (note that $\Out(F)$ acts on marked fatgraphs in a way that commutes with $V(T)$, by acting by homotopy equivalences of the $K(\pi_1(T),1)$ and thereby changing the marking). Is there a good reference?

Since the answers I am getting are not really what I am after, I think I need to make the question more pointed. A marked fatgraph determines a certain amount of algebraic structure  on a free group (i.e. the fundamental group of $T$), namely a pair $(l,e)$ where $l$ is a length function, and $e$ is a bounded 2-cocycle. The first part of the data comes from the "thin" underlying graph, and is just the translation length of each element on its axis. The second part of the data comes from the fattening, and is an explicit cocycle representing the Euler class of the thickened surface. The first kind of move affects $l$, the second kind affects $e$. Crucially, both $l$ and $e$ are integer valued (this is the point of discussing discrete combinatorial objects, namely fatgraphs, instead of e.g. discrete faithful representations of $F$ into $\operatorname{PSL}(2,\mathbb{R}))$.
Many, many papers discuss length functions, and many, many papers discuss Euler classes, but I would like to have a (presumably homological) algebraic framework which treats the two components as a single object with, presumably, more structure. The question is: what is this structure? Is it something that is already well-studied? Is there a reference?

REPLY [2 votes]: This is a nice question and I find it very hard to make such things understandable so I'm impressed.
The relationship to Outer Space of course looks very strong, but you're just looking at the simplicies of maximal dimension (trivalent graphs), so to understand the relationship my first step would be to try to lift your moves to the whole of outer space.  
Your edge moves get factored into edge expansion and edge contraction, these are well understood and studied, see any reference on outer space.  As for the vertices this means having to alter fat vertices of higher degree and looking at the trivalent case I think that means that you have to allow moves between any two cyclic orderings.  A first glance seems to indicate that there is some kind of compatibility between edge and vertex moves: essentially it does look to me that you can lift your concept of moves to the whole of "fattened outer space".
Now we've done this we can study your object in the formalism of Kontsevich's graph complexes (well the associated (semi-)simplicial structures).  Fattened outer space becomes the associative graph complex but with some extra simplices (your vertex moves).  We can characterise these as certain simplices coming from the kernel of the map from the associative graph complex to the commutative graph complex.
As such I don't think that your groupoid will mean much beyond outer space, but it may encode some information from the associative complex, although I'm guessing only something along the lines of enumeration of the kernel.
I've run out of time and need to leave, hopefully what I've said makes some sense and the details are fill-in-able.  I'll address any comments tomorrow.<|endoftext|>
TITLE: Do quotients of representable sheaves represent quotients?
QUESTION [6 upvotes]: Here's the context for the question:  Proposition 4.6 of Freitag and Kiehl's book on etale cohomology shows that a sheaf (of sets) $\mathcal{F}$ (on the site Et(X)) is constructible if and only if it is the coequalizer of an etale equivalence relation $\mathcal{R}\rightrightarrows \mathcal{Y}$, where $\mathcal{R}$ and $\mathcal{Y}$ are representable sheaves.  Here an etale equivalence relation is defined exactly as you would expect: $\mathcal{R}\rightarrow \mathcal{Y}\times \mathcal{Y}$ is injective, and for every etale $U\rightarrow X$, $\mathcal{R}(U)\subset \mathcal{Y}(U)\times \mathcal{Y}(U)$ is an equivalence relation (of sets).
Now if the quotient $\mathcal{Y}/\mathcal{R}$ "should" be represented by the quotient $Y/R$ (where $Y$ represents $\mathcal{Y}$ and $R$ represents $\mathcal{R}$), well, it sounds like constructible sheaves should be algebraic spaces, or at least there should be some relationship.  On the other hand, I don't think this could be right.
So is the problem that you can't take sheafy quotients like this, or is it something more subtle?

REPLY [2 votes]: This is just a little side remark that in general it does make a difference whether you take quotients as representable functors or as schemes. There is a counterexample in section 4 of http://www.math.univ-toulouse.fr/~toen/cours1.pdf. 
The counter example is as follows. Let R be the real number and Q the rationals. Then Q acts on R by translation. The quotiens R/Q is nothing reasonably geometric. But if we take h_R to be the functor represented by R, then the quotient h_R / Q is an algebraic space!<|endoftext|>
TITLE: complex cobordism from formal group laws?
QUESTION [14 upvotes]: Reading Ravenel's "green book", I wonder about his question on p.15 "that the spectrum MU may be constructed somehow using formal group law theory without using complex manifolds or vector bundles. Perhaps the corresponding infinite loop space is the classifying space for some category defined in terms of formal group laws. Infinite loop space theorists, where are you?". What is the state of things on that now?

REPLY [3 votes]: There is a very easy theorem along much weaker but related lines in Adams's "Stable Homotopy and Generalized Homology."  I refer to Lemma 4.6 of section II.
It isn't written quite like this, but essentially it says that if E is a complex orientable spectrum together with a complex orientation $x \in E^{2}(\mathbf{C}P^{\infty})$ then there is a unique (up to homotopy) map of ring spectra 
$MU \rightarrow E$ taking the fixed (better fix one) complex orientation of $MU$ to the given complex orientation of $E$.
This doesn't build $MU$ out of formal group laws of course, but it shows $MU$ has this universal property for complex oriented cohomology theories, and this book was around of course when Ravenel wrote the green book.
It does seem like with the modern point of view these ideas should yield a construction out of complex oriented theories if not quite out of formal group laws.<|endoftext|>
TITLE: global fibrations of simplicial sheaves
QUESTION [12 upvotes]: I'm reading the classical Brown-Gersten's paper "Algebraic K-theory as generalized sheaf cohomology" and I'm stuck with their choose of global fibrations. Namely, a morphism of simplicial sheaves $p : E \longrightarrow B$ is a global fibration if for every inclusion of open sets $U\subset V$ the natural map $E(V) \longrightarrow B(V) \times_{B(U)} E(U)$ is a (Kan) fibration of simplicial sets.
My problem is: why these fibrations? As far as I can see, when they make use of this definition in constructing the factorizations of the model category structure, they could have chosen the fibrations to be defined open-wise: $p : E \longrightarrow B$ is a fibration if $p(V) : E(V) \longrightarrow B(V)$ is a (Kan) fibration of simplicial sets for every open set $V$ and apply as well the small object argument they use at this point.
In other contexts I understand this kind of fibrations. For instance, for the model structure of the category of diagrams $C^I$ of a model category $C$ when $I$ is a 'very small' category (Dwyer-Spalinski, "Homotopy theories"), or a Reedy category. In these cases, this kind of fibrations ensures that you can extend your liftings by induction. But I don't see if this is their role with a category of sheaves, since no induction seems to be at hand.
A colleague of mine has said to me thas this choice of fibrations is the consequence of choosing the cofibrations to be the monomorphism, following Joyal's "Letter to Grothendieck"; that is, these are precisely the fibrations if you choose monomorphisms as cofibrations and ask fibrations to have the RLP with respect to trivial cofibrations. But I couldn't find anywhere this famous Joyal's letter, so I would also be glad if someone could tell me where I can find it.
Thanks in advance for any hints.

REPLY [5 votes]: For model structures on simplicial sheaves, there is a difference between the Joyal-Jardine approach and the Brown-Gersten approach. This is well explained in Voevodsky's preprint: Homotopy theory of simplicial presheaves in completely decomposable topologies, available here. Briefly, the Brown-Gersten approach does not work for arbitrary sites, but it works for a class of sites defined in Voevodsky's paper - this class includes Noetherian finite-dimensional spaces. When the B-G approach works, the resulting model structure has better finiteness properties than the Joyal-Jardine model structure, which on the other hand can be defined for simplicial (pre)sheaves on any site.<|endoftext|>
TITLE: A coalgebraic description of the hyperfinite II_1 revisited
QUESTION [7 upvotes]: Back here I was asking for a coalgebraic characterisation of the hyperfinite $II_1$ factor. Recall the latter's construction by forming the inductive limit of a chain of matrix algebras $R \to M_2(R) \to M_{2^2}(R) \to ...$, where a matrix is sent to two copies of itself placed in the diagonal blocks, zero elsewhere. Then completion in the weak topology gives the hyperfinite factor.
The trace is halved each step along the chain. Traces for projections in the inductive limit are dyadic rationals in [0, 1], while in the hyperfinite factor they are the whole real interval [0, 1]. Now, the dyadic rationals are the initial algebra for the endofunctor on Bipointed Set, $X \mapsto X \vee X$, identifying the second point of the first copy with the first point of the second copy. The [0, 1] interval is the terminal coalgebra for the same functor and the Dedekind and Cauchy completion of the initial algebra. Perhaps results such as Adamek's Final Coalgebras are Ideal Completions of Initial Algebras may be extended here.
This put me on the quest of characterising the hyperfinite factor coalgebraically. So now the question: what relevant facts are known about the endofunctor on algebras over the reals: $X \to M_2(X)$? I believe the hyperfinite factor is a fixed point. Presumably there is a need to be clear over isomorphism versus Morita equivalence. Might it be that the hyperfinite factor is the greatest fixed point up to isomorphism? Maybe I should be looking in the category of a certain kind of algebra.

REPLY [6 votes]: This is an interesting question, but the motivation is a bit misaligned.  $C^*$ algebras are a non-commutative or quantum generalization of compact Hausdorff spaces and von Neumann algebras are a non-commutative or quantum generalization of (not too unreasonable) measurable spaces.  However, both of these generalizations are contravariant.  Your motivation is a covariant comparison between von Neumann algebras and topological spaces, which is problematic.
The von Neumann algebra or $C^*$-algebra $M_2(\mathbb{C})$ is now famously known as a "qubit"; it is a great non-commutative analogue of $\mathbb{C} \oplus \mathbb{C}$, which is of course the complex functional algebra of a classical bit.  The endofunctor that you ask about is a geometric product of $X$ and a qubit, and the morphism in your question is a geometric projection back to $X$ with a qubit fiber.  So the completion that you ask about is thus a geometric product with a quantum Cantor set.  I forget what the fiber is called in the $C^*$-algebra setting, but I remember that, unlike a classical Cantor set, its isomorphism type depends on the sizes of the matrices.  In the von Neumann case, this quantum Cantor set is interpreted as a measurable space, and then it is always the hyperfinite $II_1$ factor and does not depend on the matrix sizes.
I think that the hyperfinite factor is not the only fixed point of tensoring with a qubit.  Let $S$ be any set, let $F$ be the set of functions from $S$ to a bit (or any finite set), and then let $M$ be the von Neumann closure of the local operators on $\ell^2(F)$.  Here a local operator is one that affects only finitely many values of $f \in F$.  If $S$ is an infinite set of any cardinality, then $M$ goes to itself when you tensor it with a qubit.<|endoftext|>
TITLE: Abstract Relation between Presehaves and Simplicial Sets
QUESTION [6 upvotes]: Every presheaf (let's say on a topological space) comes with restriction maps. The open sets of a topological space are ordered by inclusion and these inclusions yield the restrictions. Now a sheaf satisfies a gluing condition: that you can glue along elements which coincide on common restrictions.
Every simplicial object (let's say a simplicial set) comes with face maps. The simplex category is ordered by faces & degeneracies and these maps yield simplicial maps. Now a Kan complex satisfies a gluing* condition: that you can glue along simplices which coincide on common faces.
Is there a deeper theoretical framework to relate these 2 notions? I guess that this is the case, and that it is rather trivial.
Side-question: Can we define "degeneracies" for presheaves?
Ideas?
*it's not a gluing condition, but "somehow similar" (see answers below)

REPLY [3 votes]: The Kan condition isn't exactly like the sheaf condition: the Kan condition allows you to "glue" (as you put it) in certain cases, but the result is not unique.
A better analogy to the Kan condition in sheaf theory might be the notion of a flasque sheaf: a sheaf F is flasque if for all subsets V of U, all sections of F over V extend to sections over U.<|endoftext|>
TITLE: Why is it a good idea to study a ring by studying its modules?
QUESTION [89 upvotes]: This is related to another question of mine.  Suppose you met someone who was well-acquainted with the basic properties of rings, but who had never heard of a module.  You tell him that modules generalize ideals and quotients, but he remains unimpressed.  How do you convince him that studying modules of a ring is a good way to understand that ring?  (In other words, why does one have to work "external" to the ring?)  Your answer should also explain why it is a good idea to study a group by studying its representations.

REPLY [3 votes]: If your friend knows well rings and ideals, he has certainly encountered quotient like $I/J$, hasn't he? What are they for him? sets? abelian groups? This point of view is wanting. 
You can formulate basic and very useful results, such as (for $A$ local, noetherian): if $I/mI$ has a set of $n$ generators, so has the ideal $I$ (Nakayama), only if you can see $I/J$ as a module...<|endoftext|>
TITLE: Derangements and q-variants
QUESTION [16 upvotes]: Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ derangements of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\frac1{(2)_q!}-\frac1{(3)_q!}+\cdots+(-1)^{n}\frac1{(n)_q!} \right)$ elements in $\mathrm{GL}(q,n)$ which do not have $1$ as an eigenvalue; here $q$ is a prime-power, $(k)_q!=(1)_q(2)_q\cdots(k)_q$ are the $q$-factorials, and $(k)_q=1+q+q^2+\cdots+q^{k-1}$ are the $q$-numbers.
Now, there are $D_n^+=\tfrac12\bigl(|D_n|-(-1)^n(n-1)\bigr)$ and $D_n^-=\tfrac12\bigl(|D_n|+(-1)^n(n-1))\bigr)$ even and odd derangements of $\{1,2,\dots,n\}$, as one can see, for example, by computing the determinant $\left|  \begin{array}{cccccc}    0 & 1 & 1 & \cdots & 1 & 1 \\    1 & 0 & 1 & \cdots & 1 & 1 \\    1 & 1 & 0 & \cdots & 1 & 1 \\    \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\    1 & 1 & 1 & \cdots & 0 & 1 \\    1 & 1 & 1 & \cdots & 1 & 0  \end{array}  \right|$ and looking at the result.
How should one define $D_n^+(q)$ and $D_n^-(q)$?

REPLY [5 votes]: Both of Reid's suggestions sort-of work and lead to the same formula.  However, it's easier to first change the question to a $q$-analogue of the difference
$$D^{\pm}_n = D^+_n - D^-_n = (-1)^n(n-1).$$
(Also, the formula for $D_n(q)$ is missing a factor of $q^{n(n-1)/2}$.)
You can find $D^{\pm}_n$ by the inclusion-exclusion formula just as you can find $D_n$.  In fact, it's even easier since only the last two terms of inclusion-exclusion survive.  This proof has a straightforward $q$-generalization, although the final formula isn't the same.  But I don't know how likely the latter is in the $q$-linear algebra context. The determinant of a matrix seems like the reasonable $q$-analogue of the sign of a permutation, yet it is not a perfect analogue.
Let $\chi$ be a complex character of the multiplicative group of $\mathbb{F}_q$.  Let $D^\chi(q)$ be the corresponding sum of $\chi(\det M)$, summed over deranged matrices $M$.  Then the $q$-analogue is that $D^\chi(q) = (-1)^nq^{n(n-1)/2}$ for all non-constant $\chi$.
The proof uses the Möbius function of the lattice of subspaces, just like for the ordinary enumeration of deranged matrices.  This time only the last term of the Möbius inversion survives, the term for the identity matrix.<|endoftext|>
TITLE: Algorithms for laying out directed graphs?
QUESTION [6 upvotes]: I have an acyclic digraph that I would like to draw in a pleasing way, but I am having trouble finding a suitable algorithm that fits my special case.  My problem is that I want to fix the x-coordinate of each vertex (with some vertices having the same x-coordinate), and only vary the y.  My aesthetic criteria are (in order of importance):

Ensure no two vertices are too close together
Minimize edge crossings and near misses
Make a reasonable use of the entire drawing space

I have tried several (modified) force-directed algorithms, but they haven't met my expectations on at least one of these - usually too many edge crossings.
Has anyone come across a problem like this, or can you point me to some good papers that deal with restrictions like this?

REPLY [2 votes]: The documentation for GraphViz (a software package that does this sort of thing) has a number of papers on the subject included.<|endoftext|>
TITLE: Visualizing how Cech cohomology detects holes
QUESTION [28 upvotes]: I think it's pretty intuitive how singular/simplicial cohomology detects "holes" in a space.

How can we directly visualize how and in what sense the Cech cohomology of a cover does this?

In case it's of any interest, here are two examples I've looked at with the constant sheaf $\mathbb{Z}$:
(1) The disk, covered "Venn diagram style" with three open patches $U_1, U_2, U_3$ overlapping near the center (like this, but with overlaps), and
(2) The restriction of this cover to the boundary circle of the disk: three opens $U_1, U_2, U_3$ with 3 double intersections $U_{12}, U_{13}, U_{23}$ and no triple intersection.
If you look at the Cech complex in (2), the $H^1=\mathbb{Z}$ "comes from" the fact that you can write down a triple of elements $(1,0,0)$ on $U_{12}, U_{13}$,$U_{23}$ which "would" disagree on the triple overlap in (1), but since it's "missing", $(1,0,0)$ gets counted as a cocycle, which is not a coboundary.  Even better, the presentation of this $H^1$ you get from the Cech complex is $\mathbb{Z}^3/\{(a,b,c)=(b,c,a)\}$, which is isomorphic to $\mathbb{Z}$ because you can "rotate" all the coordinates "around the missing intersection" into the first component.
I think a like minded analysis of higher dimensional analogues provides similar intuition.  Are there any formulations of the Cech complex to really make precise how this intuition should work?  What's going on here?

Follow up: Following Mariano's answer below, I started reading about Abstract simplicial complexes and their cohomology, which seem like just what I was looking for.  What helped me most were the ideas that

The (constant sheaf) Cech cohomology of a cover $\cal{U}$ of $X$ "is" the simplicial cohomology of its nerve  $N(\cal{U})$, an abstract simplicial complex,

The simplicial cohomology of an abstract simplicial complex "is" the singular cohomology of its geometric realization, and

The geometric realization of the nerve of a covering of $X$ is a "simple approximation" of $X$,


So in this sense, we can say precisely that

Cech (constant sheaf) cohomology on a cover detects holes in a "simple approximation" to $X$ defined by that cover.

In particular, seeing the faces of a simplicial complex encoded as formal wedge products of its vertices totally made my day :)

REPLY [4 votes]: This is explained in "Principles of Algebraic Geometry," by Griffiths & Harris, p42.
Given a simplicial decomposition of your space, associate each vertex $v_\alpha$ with its star $U_\alpha$, where the star means the union of interiors of the simplices that contain it. This gives an open cover $\{U_\alpha\}$ where the intersection of p open sets is nonempty precisely when the corresponding vertices span a p-simplex. So a p-cochain maps $(U_{\alpha_1},\dots,U_{\alpha_p})$ to a nonzero section of the coefficient sheaf only if $\{v_{\alpha_1},\dots,v_{\alpha_p}\}$ span a simplex.<|endoftext|>
TITLE: Side-Angle-Side Congruence and the Parallel Postulate
QUESTION [8 upvotes]: Is there a link between the side-angle-side congruence of triangles and the parallel postulate? Specifically, does it follow from Euclid's first four axioms alone? In fact, does it even follow from all five?

REPLY [4 votes]: You would probably do better to use either Hilbert's or Tarski's axioms, since Euclid's axioms aren't rigorous. In fact, non-Pasch geometries (see here) arguably satisfy Euclid's axioms. 
In other words, since Euclid had a muddled understanding of Euclidean geometry, attempting to use his axiomatic system without modifications will lead to you having a muddled understanding of Euclidean geometry as well.
They are separate axioms in two rigorous axiomatizations of Euclidean geometry.
In both of the (rigorous) axiomatic systems mentioned above, the SAS postulate (five-segment axiom in Tarski's system and the last congruence axiom in Hilbert's system) and the parallel postulate (axiom 10 of Tarski, and here in Hilbert) are different axioms. 
I don't know how much has been proven for Hilbert's axioms, but since Tarski's axiom system is simpler, most of the axioms have been proven to be independent of one another, including the SAS and the parallel postulate. (See e.g. here.)
There exist valid geometries satisfying SAS but not the parallel postulate.
Both Hilbert's and Tarski's axioms, which include SAS as one of the axioms, can also be used to create axiom systems for neutral geometry (by omitting the parallel postulate) and for hyperbolic geometry (by negating the parallel postulate). Since these axiom systems for hyperbolic geometry, which contain both SAS and the negation of the parallel postulate, are also consistent, this also shows that the parallel postulate is independent of SAS. (This is essentially the same argument made by Kristal Cantwell in another answer.)
There exist valid geometries satisfying the parallel postulate but not SAS.
Finally, in taxicab geometry, the parallel postulate does hold, but SAS doesn't. See here:

...the Taxicab Plane satisfies all of the usual axioms (rules) of the Euclidean plane except one of the congruence axioms... We have two triangles obeying SAS which are not congruent!

Since there exist consistent systems of geometry in which (1) the parallel postulate doesn't hold but SAS does, and (2) the parallel postulate does hold but SAS doesn't, we can conclude with certainty that the two are independent of one another.<|endoftext|>
TITLE: Lie Groups and Manifolds
QUESTION [12 upvotes]: I'm trying to get a better handle on the relation between Lie groups and the Manifolds they correspond to. Firstly, is the relationship injective? that is, does each Lie group correspond to a unique manifold? Or are all the manifolds corresponding to a particular group homeomorphic?
Also, what formal form does the relationship take? I can intuitively understand the relationship between, say, $SO(3)$ and $S^2$ by thinking about rotating the sphere into itself, but what how does this generalize to a more general group or manifold.

REPLY [2 votes]: Just to comment on the relation between S^2 and SO(3) : 
there is indeed, for symmetric spaces, a natural correspondence between them and Lie groups, which in particular gives the S^2-SO(3) pair. 
A symmetric space is roughly a connected manifold M with global symmetries s_x at each point (satisfying certain properties). 
Now define G(M) to be the group generated by even products of symmetries.
Then one can show (using Palais's theorem) that this is a Lie group, which is connected and acting transitively on the symmetric space. Obviously, if you are in the Riemannian context, this will be a Lie group of isometries. Further, it's the 'smallest' subgroup of the isometry group transitive and stable under the involution given by conjugation by symmetry at any base point, which is thus a sort of uniqueness.<|endoftext|>
TITLE: Functorial Whitehead Tower?
QUESTION [21 upvotes]: The Whitehead tower of a (pointed) space is a tower of spaces which successively kills the bottom homotopy groups. The first two spaces can be constructed functorially (at least for suitably nice spaces) as the connected component and the universal cover.
Can the remaining spaces be constructed functorially?
For the dual situation the answer is yes. I.e. for the Postnikov tower where we have a tower of spaces where the bottom homotopy groups are intact, but where we have killed off all the higher homotopy groups does have a functorial construction (again for nice spaces). The construction I know passes through simplicial sets. I'm wondering if something similar exists for the Whitehead tower?

REPLY [2 votes]: $W_{n+1}(X)$ is the homotopy fiber of the natural map $W_n(X)\to K(\pi_n(X),n)$, so all of them are functorial.<|endoftext|>
TITLE: Simplicial volume
QUESTION [9 upvotes]: Is there a finite dimensional closed manifold $M$ which is a $K(\pi,1)$, whose fundamental group is not word-hyperbolic, but which has a positive simplicial volume (ie "Gromov norm")?
(Added:) The answers of Jim and Richard are both excellent; another example is any closed, irreducible locally symmetric manifold (of non-positive curvature). But these examples are all CAT(0); I wonder if there is an example which is not CAT(0)? (of course then it is hard to see the example is a $K(\pi,1)$ . . .)

REPLY [2 votes]: Just take an irreducible, but not atoroidal, 3-manifold with the property that at least one of the pieces in the JSJ-decomposition is hyperbolic.
Such a manifold is nonpositively curved by Leeb's thesis, hence a $K(\pi,1)$. However the fundamental group is not hyperbolic because of the abelian subgroups coming from incompressible tori. Its simplicial volume is the sum of the simplicial volumina of the hyperbolic pieces, thus non-zero.<|endoftext|>
TITLE: Is the maximum domain to which a Dirichlet series can be continued always a halfplane?
QUESTION [8 upvotes]: Let $f(s)=\sum_n a_n n^{-s}$ be a Dirichlet series whose coefficients satisfy $\lvert a_n\rvert\leq n^{C}$. Then $f(s)$ converges absolutely in some halfplanes, and is conditionally convergent in (potentially different) halfplane. It might happen sometimes that $f(s)$ admits meromorphic continuation to a larger domain. Consider maximal such domain. Is this domain always a halfplane? For purposes of this question, I take it to mean that $\mathbb{C}$ is a halfplane.
The question is motivated by known results about analytic continuation of $L$-functions. It has vexed me: to what extent the analytic continuation is special to the algebraic world of $L$-functions, and how much its properties are common to the analytic world of Dirichlet series.

REPLY [6 votes]: The product 
$F(s) = \prod_{n=2}^{\infty}(1 - n^{-s})^{-1} = \sum_{n=1}^{\infty}c(n)n^{-s}$ is an explicit counterexample. The coefficient $c(n)$ is the number of ways of writing $n$ as a product of integers $\geq 2$ without regard to order, and the Dirichlet series has abscissa of convergence ${\sigma}_c = 1$. The function $F(s)$ has meromorphic continuation to $\sigma > 0$ minus the points $1,1/2,1/3,\ldots$ and has $\sigma = 0$ as a natural boundary. At the points $s = 1,1/2,1/3,\ldots$ it has essential singularities. One sees this by logarithmically differentiating the product. The essential singularities are obvious, but you have to do a little work to get the natural boundary.<|endoftext|>
TITLE: Measurable functions and unbounded operators in von Neumann algebras
QUESTION [5 upvotes]: How do you define unbounded measurable functions for a general von Neumann algebra?
For the commutative algebra $L^\infty(X,\mu)$, we can consider the space of all measurable functions that are almost everywhere finite. This set has certain nice properties: it is closed under multiplication and there is the notion of convergence almost everywhere.
For the the noncommutative algebra of bounded operators on a Hilbert space, we can consider the set of all closed unbounded operators with dense domain. These operators are quite important in PDE, since differential operators are always unbounded. It is not obvious to me, however, why the product of two unbounded operators will be again a nice operator or how to generalize convergence almost everywhere to this setting.
Is there any construction like the above ones for an arbitrary von Neumann algebra? Can one get any standard properties of measurable functions from this construction?
If the closure of the spectrum of an unbounded operator $T$ is not the whole $\mathbb C$, and $z$ does not lie in this set, then we can consider instead of $T$ the resolvent $(z-T)^{-1}$, which will be a bounded operator. However, it is not clear to me how to proceed when the spectrum is the whole $\mathbb C$ or whether there is a more conceptual way to define these objects.
Update: I must have phrased the question inaccurately; the answer to the original question would be given by affiliated operators, a construction beautiful and useful, but not quite what I had in mind. I am sorry for the confusion caused (by the way, does anyone know how I can mark two answers as accepted, one for the original question and one for the rephrased one?) The rephrased question is:
For each von Neumann algebra, define canonically a set $S$ such that:
* For the algebra $L^\infty(X)$, $S$ is the set of all measurable functions on $X$.
* For the algebra of bounded operators on a Hilbert space, $S$ is the set of all unbounded operators on the same Hilbert space.
Alternatively, explain why it impossible (or unreasonable to try) to define such a set for all von Neumann algebras. So, I wish to somehow see the set of objects that will somehow remind of unbounded operators, not to study some specific unbounded operators on given spaces. These objects need not be actual operators in any sense for an arbitrary algebra.

REPLY [2 votes]: This is a very old question but I would like to mention what I think is the most natural response, namely the fact that we can borrow the concept of rings of quotients from algebra.  In the  commutative case, this is transparent: if $f$ is measurable, then we can express it as the quotient of the two bounded, measurable functions $\frac f{1+|f|^2}$ and $\frac 1 {1+|f|^2}$.  Conversely, if $f$ and $g$ are bounded measurable functions and the set where $g$ vanishes is negligible, then $\frac f g$ is a measurable function (we use the usual sloppy notation to talk about equivalence classes of measurable functions).
The non-commutative case is, of course, more delicate but a standard result which is often used to extend the spectral theorem from the bounded to the unbounded case and which can be found in  Riesz-Nagy, states that if $T$ is a closed, densely defined, unbounded linear operator in Hilbert space, then both $T(I+T^\ast T)^{-1}$ and $(I+T^\ast T)^{-1}$ are bounded  and $T$ is their quotient (in the more subtle sense used for unbounded operators).  This suggests that a suitable definition for an unbounded operator $T$ to be associated with a von Neumann algebra $\cal A$ would be  for  these two operators to lie in $\cal A$.<|endoftext|>
TITLE: Basis of l^infinity
QUESTION [16 upvotes]: Is it possible to exhibit a (Hamel) basis for the vector space l^infinity, given by the bounded sequences of real numbers?

REPLY [16 votes]: The question is about the complexity of the simplest possible Hamel basis of l^infty, and this is a perfectly sensible thing to ask about even in a context where one wants to retain the Axiom of Choice. That is, we know by AC that there is a basis---how complex must it be? 
Such a question finds a natural home in descriptive set theory, a subject concerned with numerous similar complexity issues. In particular,
descriptive set theory provides tools to make the question
precise. 
My answer is that one can never prove a negative answer to the question, because it is consistent with the ZFC axioms of set theory that
Yes, one can concretely exhibit a Hamel basis of l^infty.
To explain, one natural way to measure the complexity of sets of reals
(or subsets of R^infty) is with the projective hierarchy.
This is the hierarchy that begins with the closed sets (in
R, say, or in R^omega), and then iteratively closes under
complements and projections (or equivalently, continuous
images). The Borel sets appear near the bottom of this
hierarchy, at the level called Delta11, and then the
analytic sets Sigma11 and co-analytic sets Pi11, and so
on up the hierarchy. Sets in the projective hierarchy are
exactly those sets that can be given by explicit definition
in the structure of the reals, with quantification only
over reals and over natural numbers. If we were to find a
projective Hamel basis, then it will have been
exhibited in a way that is concrete, free of
arbitrary choices. Thus, a very natural way of making the
question precise is to ask:
Question. Does l^infty have a Hamel basis that is
projective?
If the axioms of set theory are consistent, then they are
consistent with a positive answer to this question. This is
not quite the same as proving a positive answer, to be
sure, but it does mean that no-one will ever prove a
negative answer to the question.
Theorem. If the axioms of ZFC are consistent, then
they are consistent with the existence of a projective
Hamel basis for l^infty. Indeed, there can be such a basis
with complexity Pi13.
Proof. I will prove that under the set-theoretic assumption
known as the Axiom of Constructibility V=L, there is a
projective Hamel basis. In my answer to question about Well-orderings of the reals, I explained that
in Goedel's constructible universe L, there is a definable
well-ordering of the reals. This well-ordering has
complexity Delta12 in the projective hierarchy. From this
well-ordering, one can easily construct a well-ordering of
l^infty, since infinite sequences of reals are coded
naturally by reals. Now, given the well-ordering of
l^infty, one defines the Hamel basis as usual by taking all
elements not in the span of elements preceding it in the
well-order. The point for this question is that if the
well-order has complexity Delta12, then this definition
of the basis has complexity Pi13, as desired. QED
OK, so we can write down a definition, and in some
set-theoretic universes, this definition concretely
exhibits a Hamel basis of l^infty. There is no guarantee,
however, that this definition will work in other models of
set theory. I suspect that one will be able to find other
models of ZFC, in which there is no projective Hamel basis
of l^infty. It is already known that there might be no
projective well ordering of the reals (a situation that
follows from large cardinals and other set theoretic
hypotheses), and perhaps this also implies that there is no
projective Hamel basis. In this case, it would mean that
the possibility of exhibiting a concrete Hamel basis is
itself independent of ZFC. This would be an
interesting and subtle situation. To be clear, I am not
referring here merely to the existence of a basis requiring
AC, but rather, fully assuming the Axiom of Choice, I am
proposing that the possibility of finding a
projective basis is independent of ZFC.
Conjecture. The assertion that there is a projective
Hamel basis of l^infty is independent of ZFC.
I only intend to consider the question in models of ZFC, so
that l^infty has a Hamel basis of some kind, and the only
question is whether there is a projective one or not. In this situation, the fact that AD seems to imply that there is no Hamel basis is not relevent, since that axiom contradicts AC.
Apart from this, I also conjecture that there can never be
a Hamel basis of l^infty that is Borel. This would be a lower bound on the complexity of how concretely one could exhibit the basis.<|endoftext|>
TITLE: How to smootly interpolate between möbius transformations?
QUESTION [14 upvotes]: If you have two Möbius transformations represented as:
$f(z) = \frac{az + b}{cz + d}$
$g(z) = \frac{pz + q}{rz + s}$
where $a, b, c, d, p, q, r, s, z \in \mathbb{C}$
Is it possible to derive a third function $h(z, t)$, where $t \in \mathbb{R}$ and $0 \leq t \leq 1$, which "smoothly" interpolates between the transformations represented by $f(z)$ and $g(z)$?
Let me try to clarify the meaning of "smoothly".  I'm working in a the Poincaré Disc model of Hyperbolic geometry.  The functions $f$ & $g$ represent transformations which preserve congruence within the model, i.e. a Poincaré line segment $PQ$ will have the same Poincaré distance as $P'Q'$, where $P' = f(P)$ & $Q' = f(Q)$. I would like the interpolated transform to preserve this property as well.
Clarification:
The answers involving diagonalization of the matrix "almost" work.  Let me be clearer about how these mobius transformations are used, so that I can give some more context.
$f$ represents the position and orientation of a tile on the Poincare disc.  I use this to transform a tile centered at the origin. $g$ represents a neighboring tile.  I'd like to "animate" the $f$ tile moving it smoothly on top of the $g$ tile.  While animating, the center of $f$ should travel along the poincare line connecting the two tiles at rest position.
The problem with the diagonalization approach is that the centers of the two tiles will travel about a corner of one of the tiles, sometimes the "long" way around.
Of course it could just be a bug in my code...
P.S. This is for an iphone game I'm developing called Circull
http://www.youtube.com/watch?v=DiWijYb-xus

REPLY [11 votes]: I'd like to address two issues about how to implement these interpolations and what they mean.  The group that you ask about is actually equivalent to the group of real M\"obius transformations.  If you were in the upper half plane model, the coefficients would be real numbers.  You actually don't need to explicitly use the upper half plane model, but you do need the fact that comes from it that all of the motions are elliptic, hyperbolic, or parabolic.  The group is called $\mathrm{PSL}(2,\mathbb{R})$.
It is widely understood that a good way to make smooth motions in a Lie group is with geodesics.  This is an important principle in ordinary 3D computer graphics, where the Lie group is instead $\mathrm{SO}(3)$.  A standard method to find the geodesics in this comparison case is with quaternions.  As it happens, this introduces a wrinkle with doubled angles that also appears in your case.
I assume that you are using $f$ and $g$ as follows:  $z$ is a point in the standard bit map of your bird, and then $f(z)$ or $g(z)$ is the mapped position of the bird in your Poincare disk model of the hyperbolic plane.  If this is what you are doing, then a correct derivation of the geodesic is $h_t(f(z))$, where $h_t(z)$ is an exponential path from the identity to $g(f^{-1}(z))$.  I suppose that the swapped formula that other people have used, $f^{-1} \circ g$, is equivalent.
You can find the exponential path by diagonalizing the matrix of $h_1 = g \circ f^{-1}$.  However, there is the important wrinkle that the group of matrices is $\mathrm{SL}(2,\mathbb{R})$, which is twice as big as $\mathrm{PSL}(2,\mathbb{R})$.  Let $M$ be the matrix of $h_1$.  If $M$ is hyperbolic, then it has real eigenvalues.  In this case, you should first switch to $-M$ if the eigenvalues are negative.  In the basis in which $M$ is diagonal, you should then specifically use
$$M_t = \begin{bmatrix} \exp(ct) & 0 \\\\ 0 & \exp(-ct)\end{bmatrix}.$$
If you use some other choice, you will not follow the geodesic at a constant rate.  If $M$ is elliptic, then you should use
$$M_t = \begin{bmatrix} \exp(i\theta t) & 0 \\\\ 0 & \exp(-i\theta t)\end{bmatrix},$$
again in a (complex) basis in which $M$ is diagonal.  But here, how do you choose between $M$ and $-M$?  The total geometric rotation is $2\theta$, not $\theta$.  You should negate $M$ if after you diagonalized, $\theta$ is more than $\pi/2$; if you don't do that you might rotate by more than 180 degrees.  In the parabolic case (which you might not see because it requires a numerical coincidence),
$$M_t = \begin{bmatrix} 1 & t \\\\ 0 & 1\end{bmatrix}.$$
Finally, in computer graphics you might want a smooth trajectory accelerate from 0 at the beginning and decelerate to 0 at the end.  With the confidence that $t$ in the above formulas follows the geodesic at a uniform rate, you can do something like $t = \sin(s)^2$ and use $s$ as the time parameter instead.<|endoftext|>
TITLE: Flips in the Minimal Model Program
QUESTION [25 upvotes]: In order get a minimal model for a given a variety $X$, we can carry out a sequence of contractions $X\rightarrow X_1\ldots \rightarrow X_n$ in such a way that that every map contracts some curves on which the canonical divisor $K_{X_j}$ is negative.
Here we have, at least, the following technical problem: in contracting curves, the resulting variety $X_j$ might have become singular. In order to fix this issue, people consider a flip.
Here are my questions. What is the intuition to understand such a flip?
Are there examples of such things in other contexts of math or is it an ad hoc construction?

REPLY [10 votes]: This is a comment to Charles's answer, but I need more room than what comments allow.
"Glue back differently" means that the curve is "glued back" with its normal bundle "reversed". 
There is also an algebraic way to think about flips: If $f:X\to Y$ is a contraction, then $X$ can be considered as ${\rm Proj}_Y\sum_{m=0}^\infty f_*\mathcal O_X(-mK_X)$. Now if $f$ is small, then the flip of $f$ is given by the morphism $f^+: X^+={\rm Proj}_Y\sum_{m=0}^\infty f_*\mathcal O_X(mK_X)\to Y$. So, to prove the existence of a flip you "only" need to prove that the above algebra is finitely generated over $\mathcal O_Y$.
This might not seem an intuitive way right away, but remember that Proj comes with a relatively ample divisor, so what's happening is that we make an $f$-anti-ample divisor into an $f^+$-ample one without changing it on the locus where $f$ was an isomorphism. If $X$ and $Y$ are $3$-dimensional and $f$ is a small Mori-contraction then it contracts a single rational curve and being ample is equivalent to the degree of the divisor on the curve being positive. Now the (anti-)ampleness of the canonical class is then governed by the normal bundle of the curve and hence "flipping" the positivity of $K_X$ on this curve is essentially the same as "flipping" the normal bundle.<|endoftext|>
TITLE: "Requires axiom of choice" vs. "explicitly constructible"
QUESTION [9 upvotes]: I think I'm a bit confused about the relationship between some concepts in mathematical logic, namely constructions that require the axiom of choice and "explicit" results. 
For example, let's take the existence of well-orderings on $\mathbb{R}$. As we all know after reading this answer by Ori Gurel-Gurevich, this is independent of ZF, so it "requires the axiom of choice." However, the proof of the well-ordering theorem that I (and probably others) have seen using the axiom of choice is nonconstructive: it doesn't produce an explicit well-ordering. By an explicit well-ordering, I simply mean a formal predicate $P(x,y)$ with domain $\mathbb{R}\times\mathbb{R}$ (i. e., a subset of the domain defined by an explicit set-theoretic formula) along with a proof (in ZFC, say, or some natural extension) of the formal sentence "$P$ defines a well-ordering." Does there exist such a $P$, and does that answer relate to the independence result mentioned above?
More generally, we can consider an existential set-theoretic statement $\exists P: F(P)$ where $F$ is some set-theoretic formula. Looking to the previous example, $F(P)$ could be the formal version of "$P$ defines a well-ordering on $\mathbb{R}$." (We would probably begin by rewording that as something like "for all $z\in P$, $z$ is an ordered pair of real numbers, and for all real numbers $x$ and $y$ with $x\neq y$, $((x,y)\in P \vee (y,x)\in P) \wedge \lnot ((x,y)\in P \wedge (y,x)\in P)$, etc.) On the one hand, such a statement may be a theorem of ZF, or it may be independent of ZF but a theorem of ZFC. On the other hand, we can ask whether there is an explicit set-theoretic formula defining a set $P^\*$ and a proof that $F(P^\*)$ holds. How are these concepts related: 

the theoremhood of "$\exists P: F(P)$" in ZF, or its independence from ZF and theoremhood in ZFC;
the existence of an explicit $P^\*$ (defined by a formula) with $F(P^*)$ being provable. 

Are they related at all?

REPLY [2 votes]: Levy has a few interesting things to say on the definability of a set whose existence requires the axiom of choice, see pages 171- 175 Basic Set Theory, Perspectives in Mathematical Logic. In particular, he mentions Feferman's result on the unprovability in ZFC of the existence of definable well ordering of reals.
On your last question: Levy mentions an example (exceedingly trivial) of a formula F(x) for which one does have a definable set satisfying it in ZFC although the existential formula "for some x, F(x)" is unprovable in ZF.<|endoftext|>
TITLE: Is every matching of the hypercube graph extensible to a Hamiltonian cycle
QUESTION [11 upvotes]: Given that $Q_d$ is the hypercube graph of dimension $d$ then it is a known fact (not so trivial to prove though) that given a perfect matching $M$ of $Q_d$ ($d\geq 2$) it is possible to find another perfect matching $N$ of $Q_d$ such that $M \cup N$ is a Hamiltonian cycle in $Q_d$.
The question now is - given a (non necessarily perfect) matching $M$ of $Q_d$ ($d\geq 2$)  is it possible to find a set of edges $N$ such that $M \cup N$ is a Hamiltonian cycle in $Q_d$.
The statement is proven to be true for $d \in\{2,3,4\}$.

REPLY [15 votes]: This is a known open problem.  See "Matchings extend to Hamiltonian cycles in hypercubes" over at the Open Problem Garden.<|endoftext|>
TITLE: What's an example of a space that needs the Hahn-Banach Theorem?
QUESTION [43 upvotes]: The Hahn-Banach theorem is rightly seen as one of the Big Theorems in functional analysis.  Indeed, it can be said to be where functional analysis really starts.  But as it's one of those "there exists ..." theorems that doesn't give you any information as to how to find it; indeed, it's quite usual when teaching it to introduce the separable case first (which is reasonably constructive) before going on to the full theorem.  So it's real use is in situations where just knowing the functional exists is enough - if you can write down a functional that does the job then there's no need for the Hahn-Banach Theorem.
So my question is: what's a good example of a space where you need the Hahn-Banach theorem?
Ideally the space itself shouldn't be too difficult to express, and normed vector spaces are preferable to non-normed ones (a good non-normed vector space would still be nice to know but would be of less use pedagogically).
Edit: It seems wrong to accept one of these answers as "the" answer so I'm not going to do that.  If forced, I would say that $\ell^\infty$ is the best example: it's probably the easiest non-separable space to think about and, as I've learnt, it does need the Hahn-Banach theorem.
Incidentally, one thing that wasn't said, and which I forgot about when asking the question, was that such an example is by necessity going to be non-separable since countable Hahn-Banach is provable merely with induction.

REPLY [29 votes]: There is another logical tidbit here.
Hahn-Banach (HB) is strictly weaker than Axiom of Choice (AC), meaning--under the assumption of consistence as usual--in ZF, AC implies HB but not the other way around.  Another intermediate theorem of functional analysis is the Krein-Milman theorem: "A compact convex nonempty set in a locally convex space has an extreme point"  Call it KM.  In ZF, AC implies KM but not the other way around.  And the interesting point is that, taken both together, we do get AC.  So in ZF, HB+KM implies and is implied by AC.<|endoftext|>
TITLE: How to respond to "I was never much good at maths at school."
QUESTION [57 upvotes]: We've all heard it.  I even got it in Norwegian recently.  It's number 1 on the list of responses to the statement "I'm a mathematician.".  Does anyone have any good comebacks?  What other responses have you heard?
Standard community wiki rules (not that anyone seems to take any notice of them): one answer per post please.  Also, no snide comebacks, please - keep it nice.  Extra kudos for answers that would actually educate the other person.

REPLY [4 votes]: For our fantasts:
link text<|endoftext|>
TITLE: Theorems for nothing (and the proofs for free)
QUESTION [50 upvotes]: Some theorems give far more than you feel they ought to: a weak hypothesis is enough to prove a strong result.  Of course, there's almost always a lot of machinery hidden below the waterline.  Such theorems can be excellent starting-points for someone to get to grips with a new(ish) subject: when the surprising result is no longer surprising then you can feel that you've gotten it.
Let's have some examples.

REPLY [9 votes]: The only group with order $p$ a prime is $\mathbb{Z}/p\mathbb{Z}$<|endoftext|>
TITLE: What is a deformation of a category?
QUESTION [39 upvotes]: I have several naive and possibly stupid questions about deformations of categories. I hope that someone can at least point me to some appropriate references.
What is a deformation of a (linear, dg, A-infinity) category? Is it a "bundle of categories" over a scheme? How can you make such a notion rigorous? Maybe via stacks? Suppose we take some nice scheme $X$ and we consider $D^b\text{Coh}(X)$; if we deform $X$, then do we also get a corresponding deformation of the derived category? What does this corresponding deformation "look like"? Are we deforming the morphisms? The objects? Both?
Similarly, what about in the situation where we have a category of modules over an algebra $A$? If we deform the algebra, then do we also get a corresponding deformation of the category? Again, what does it "look like"?
And finally, in either of the above cases, are there deformations of the respective categories that don't correspond to deformations of $X$ or of $A$ respectively? I expect the answer to be "yes"; then my next question is: Are there any nice examples of such deformations that can still be described in an explicit or geometric way?
I am most of all interested in concrete examples, and less interested in general theory.

Edit 1: I probably should have mentioned this when I first posted this (almost 3 months ago now!), but somehow I forgot. Kontsevich has been at least implicitly talking about deformations of categories since at least 1994, in the original paper introducing homological mirror symmetry. The idea (or "philosophy") seems to be that the deformation theory of a category should have something to do with its Hochschild (co)homology. But I still do not understand this connection, at least in any sort of generality. Perhaps this is explained in some of the papers already listed in the answers below --- what I'd most like to see is how to relate "deformation of a (linear/dg/A-infinity) category", however one defines it, to Hochschild (co)homology.
Perhaps it's somehow obvious... but I'm pretty dense and would like to see it spelled out...
So I hope someone can explain this to me, or point me to a spot in a paper where it is explained.
I'm adding a bounty to this question just for the heck of it.
Edit 2: See my answer below.

REPLY [17 votes]: Maybe it is helpful to look at these lectures by Yekutieli: arXiv:0801.3233. There he discusses twisted deformations of algebraic varieties $X$ suggested by Kontsevich, i.e. deformations of the category of coherent sheaves on $X$. These deformations are classified by the second Hochschild cohomology of $X$, which by the Hochschild-Kostant-Rosenberg theorem is $H^0(\wedge^2T)\oplus H^1(T)\oplus H^2({\mathcal O})$, where ${\mathcal O}$ is the structure sheaf and $T$ is the tangent sheaf of $X$. 
Here, roughly speaking, the first summand corresponds to noncommutative deformations of the structure sheaf (global Poisson bivectors), the second summand corresponds to commutative deformations of this sheaf (i.e., formal deformations of X as a variety), and the third term corresponds to deformations of the category of coherent sheaves which do not arise from deformations of the structure sheaf as a sheaf of algebras (i.e., the algebra deformations exist only on local charts but do not glue into a sheaf; only their categories of modules glue into a sheaf of categories, called a gerbe).  
Another helpful reference on this may be van den Bergh's paper arXiv:math/0603200.<|endoftext|>
TITLE: When are there enough projective sheaves on a space X?
QUESTION [46 upvotes]: This question is being asked on behalf of a colleague of mine.
Let $X$ be a topological space.  It is well known that the abelian category of sheaves on $X$ has enough injectives: that is, every sheaf can be monomorphically mapped to an injective sheaf.  The proof is similarly well known: one uses the concept of "generators" of an abelian category.
It is also a standard remark in texts on the subject that on a general topological space $X$, the category of sheaves need not have enough projectives: there may exist sheaves which cannot be epimorphically mapped to by a projective sheaf. (Dangerous bend: this means projective in the categorical sense, not a locally free sheaf of modules.)  For instance, Wikipedia remarks that projective space with Zariski topology does not have enough projectives, but that on any spectral space, a space homeomorphic to $\operatorname{Spec}R$, there are enough projective sheaves.
Two questions:

Who knows an actual proof that there are not enough projectives on, say, $\boldsymbol{P}^1$ over the complex numbers with the Zariski topology?  What about the analytic topology, i.e. $\boldsymbol{S}^2$?
Is there a known necessary and sufficient condition on a topological space $X$ for there to be enough projectives?  


EDIT: I meant the question to be purely for sheaves of abelian groups.  Eric Wofsey points out that the results alluded to above on Wikipedia are not consistent when interpreted in this way, since $\boldsymbol{A}^1$ and $\boldsymbol{P}^1$, over an algebraically closed field $k$, with the Zariski topology are homeomorphic spaces: both have the cofinite topology.
I am pretty sure that when my colleague asked the question, he meant it in the topological category, so I won't try to change that. But the other case is interesting too. What if $(X,\mathcal{O}_X)$ is a locally ringed space. Does the abelian category of sheaves of $\mathcal{O}_X$-modules have enough projectives? I think my earlier warning still applies. Since for a scheme $X$ not every $\mathcal{O}_X$-module is coherent, it is not clear that "projective sheaves" means "locally free sheaves" here, even if we made finiteness and Noetherianity assumptions to get locally free and projective to coincide.

REPLY [7 votes]: I just saw this question now.
If $X$ is a Hausdorff space with a basis of compact open sets, then the category of sheaves of abelian groups on $X$ has enough projectives.  Let $R$ be the ring of locally constant functions $f\colon X\to \mathbb Z$ with compact support and pointwise operations.  $R$ is a unital ring iff $X$ is compact, but it is always a ring with local units (i.e., a directed union of unital subrings).  An $R$-module $M$ is unitary if $RM=M$.  This is equivalent to for all $m\in M$, there is an idempotent $e\in R$ with $em=m$.  It follows that the category of unitary $R$-modules has enough projectives (the direct sums of modules of the form $Re$ with $R$ idempotent do the job).
It is known that the category of sheaves of abelian groups on $X$ is equivalent to the category of unitary $R$-modules.  The equivalence takes a sheaf to its global sections with compact support. Thus the category of sheaves of abelian groups on $X$ has enough projectives.
I don't know a precise reference for this folklore. The compact totally disconnected case is easily deduced from Pierce's Memoir of the AMS on modules over commutative regular rings. The modification for the locally compact case is known to those studying representations of reductive groups over local fields.<|endoftext|>
TITLE: How does one handle two-body job searches?
QUESTION [28 upvotes]: One thing I've heard conflicting advice on is how to handle a job search for two people simultaneously; especially the strategy of things like: when do you mention your situation to the departments in question? In your cover letter?  When you're invited to an interview?  At the interview?
As usual, people should be specific about what kind of institutions and which nations they have in mind with their advice.  Also, presumably strategy differs a bit depending on whether the second person is a mathematician, or in a different discipline; it would be great to hear about both of those cases.

REPLY [9 votes]: It's important to know that if you want to discuss this with the chair of a department where you're interviewing, or have an offer, you will have to bring it up yourself; it's illegal (at least in the U.S., or at least in the states I'm familiar with) for a university to ask you about your marital status.  You'll find that most departments are eager to help to the extent they can, and that there's wide variation in what that extent is.
Update:  Based on the comments below, let me add that some chairs don't know what the law is, or don't care; so it's clearly not universally true that you won't be asked about your spouse.  
Maybe I'll also add that, if I remember right, I never brought up my spouse until I had an offer.  The interview stage is when you're trying to convince them that you should work there; the post-offer stage is the reverse.  The problem with bringing up spousal issues at the interview stage is that the department may start thinking "We're never going to find a job for her husband in post-Inca ethnography, so maybe we should make an offer to somebody we're more likely to get."  I think this would be somewhat unethical, but then again so is illegally asking you about your marital status, and it's sometimes done.
So the difficult question:  what to do if you don't want to talk about your two-body problem at the interview, and you're asked about it?  On general "keep the tone light" grounds I would advise against saying "You are breaking the law."  If you said, "I'd rather keep this just about me for the moment," I would be fine with it; but for the sake of honesty I should say I'd expect some chairs would find it weird.
Perhaps the best thing would be to concede that you have a spouse, and to say something about what field they're in, but to play down any sense of a two-body "problem."  If the spouse is an academic (which I think is the situation in the posted question), I think it is totally OK to create the impression that a TT job for your spouse would be a big draw, but not a necessity.  This allows the math department to get the ball rolling on an attempt to do a double hire, but reduces the risk that the department will find out there's no job for your spouse and give up on you, too.
Note that the advice above only applies if you're (illegally) asked about your spouse in the interview.  If you choose to bring it up yourself, I think you should be totally upfront about what will be required to hire you.  If you are deadset against considering anything other than a double TT offer (e.g. if you are already situated in a place where you have one TT and one non-TT offer) then you should go ahead and bring this up to avoid wasting anyone's time.<|endoftext|>
TITLE: Asymptotics of Power Series With Branch Singularities
QUESTION [8 upvotes]: I am wondering if there are analytic tools to find asymptotic formulae for the coefficients of a complex power series of a function with branch singularities.  For example, it is possible to show using elementary means that, for $q>1$, the coefficients of $\frac{1}{1-z} log(\frac{1}{1-qz})$ are asymptotic to $\frac{1}{1-q^{-1}} \frac{q^n}{n}$, but I would like to know if it is possible to see this from analytic properties of the function itself.
Motivation: There are nice asymptotic formulae for the coefficients of power series of meromorphic functions.  As a simple example, the coefficients of an entire function must be $O(\epsilon^n)$ for any $\epsilon$.  In general, the sum of the principal parts of the poles of smallest modulus will provide a very good first approximation, and contour integration can be applied to get more delicate bounds, see e.g. Wilf's Generatingfunctionology.

REPLY [8 votes]: The answer is "Quite often yes, but the error terms are seldom as good as in the meromorphic case". The reason the asymptotics for the meromorphic functions works is that we know the exact coefficients for $c_k(z-z_0)^{-k}$ and can always remove the singularity by subtracting a few terms of this kind thus increasing the radius of convergence and boosting the decay of the coefficients, so that the error is exponentially small compared to the main terms. 
When you have some branching singularity $S(z)$ (like $\log(z-z_ 0)$ or $\sqrt[]{z-z_ 0}$) for which you know the coefficients when it stands alone, you can still play the trick to get the asymptotics for a function $F(z)S(z)$ where $F$ is alnalytic in some larger disk by writing the sum as $F(z_ 0)S(z)+(F(z)-F(z_ 0))S(z)$ and noting that the second term is smoother than the first one on the boundary circle, so its Fourier coefficients decay a bit faster. If you take more terms in the Taylor series for $F$ at $z_ 0$, you can get any degree of smoothness in the remainder you wish, so the error terms can be made of size $n^{-k}$ times the leading terms with any given $k$, but that's about as far as you can go unless you want integral expressions instead of algebraic quantities.
In the latter case, you should make a radial cut starting at singularity and ending at some bigger circle and take the contour integral over that larger circle and 2 sides of the cut in the Cauchy formula. The integral over the larger circle will decay exponentially faster than the typical coefficient, so all asymptotics will come from the cut. The point is that the integral over the cut is not that of something oscillating but that of a fast decaying function, so you can use real variable methods to find its asymptotics in nice algebraic terms (or you can just leave it as is).<|endoftext|>
TITLE: Wick rotation in mathematics
QUESTION [12 upvotes]: In physics, esp. quantum field theory, Wick rotation (i.e. putting $t \mapsto i\tau$, imaginary time) is often used to simplify calculations, make things convergent or make connections between different models (e.g. quantum and statistical mechanics). Does the Wick rotation trick occur anywhere in mathematics not related to QFT (analysis, PDE etc.)?

REPLY [14 votes]: A general form of the Wick's rotation is the "Weyl's unitary trick". This construction allows to relate group actions of noncompact forms of a complex Lie group to those of the compact one by changing the signature of the Cartan-Killing form . Although, the representations of the compact and noncompact forms are different, the unitary trick introduces relations among their invariants and between the transition functions, hence the use in quantum field theory. Also, it introduces relations between their homogeneous spaces (see the example above of the sphere and the hyperboloid).<|endoftext|>
TITLE: Cocktail party math
QUESTION [58 upvotes]: Ok, hotshots. You're at a party, and you're chatting with some non-mathematicians. You tell them that you're a mathematician, and then they ask you to elaborate a bit on what you study, or they ask you to explain why you like math so much.

What are some engaging ways to do this in general? What are some nice elementary results, accessible to people with any mathematical background or lack thereof, that can be used to illustrate why math is interesting, and its depth, breadth, and beauty? Note the scenario: you're at a party so it should be a relatively quick and snappy kind of thing that doesn't require a blackboard or paper to explain.
Easy Mode: How do you explain what your particular sub-field of math is about in an accurate but still understandable and engaging way? Hard Mode: Assuming you work in a less applied area, how can you do this without mentioning any real-world applications? Again, note the scenario.

This question is inspired by this one, in particular Anton's answer.

REPLY [2 votes]: When trying to talk about specific results, I really like talking about Cantor's Theorem (or at least, the special case of $2^{\aleph_0}$), and then, if they're willing to accept that, talk about ordinals a bit. If the audience isn't taking it, I'll generally talk about some arbitrary graph problem that comes to my head.
But when asked, I typically try to approach it from the more philosophic perspective of "what mathematicians do"-- study abstract structure. (Or at least, that's the approach I take to math), and try to explain what that means, providing some examples here and there. The definition of group shows up a lot, since there are easy to understand examples of groups. I also view myself as a bit of an artist, so I tend to use analogies that deal with music and painting.<|endoftext|>
TITLE: $\omega$-topos theory?
QUESTION [10 upvotes]: I've been reading through Lurie's book on higher topos theory, where he develops the theory of $(\infty,1)$-toposes, which leads me to the following question: Is there any sort of higher topos theory on the more general $\omega$-categories, where we don't require all higher morphisms to be invertible?

REPLY [6 votes]: One should keep in mind that Jacob Lurie's book "only" (if I may use this word) discusses the (oo,1)-version of Grothendieck toposes/category of sheaves: the (oo,1)-toposes in Jacob Lurie's book are (oo,1)-categories of (oo,1)-sheaves/of oo-stacks.
This is less general than the "elementary (oo,1)-toposes" that one would eventually want to see, but it already goes a long way -- and it is more accessible.
Similarly, while a general theory of n-toposes for higher n is largely missing, there is a bit more known about (oo,n)-sheaves, i.e. of oo-stacks which are presheaves with values not just in oo-groupoids but in (oo,n)-categories. 
For instance Ross Street once proposed a notion  of descent for strict-omega-category-valued presheaves. Using a result by Verity this may be regarded as presenting descent for strict oo-groupoid valued presheaves, but I'd expect that with the required care exercised it goes further than that (and this seems to be what Street had in mind, though I can't tell that for sure). 
But a more developed general theory for descent of (oo,n)-category valued presheaves is developed notably in Hirschowitz-Simpson's Descente pour les n-champs. This yields at least part of a theory of Grothendieck-style n-toposes.<|endoftext|>
TITLE: How  many mathematicians are there?
QUESTION [83 upvotes]: Although we are not so numerous as other respected professionals, like for example lawyers, I wonder if we could come up with a reasonable estimate of our population.
Needless to say, the question  more or less amounts to the definition of"mathematician".
Since I should like to count only research mathematicians (and not, say, high-school teachers) some criterion of publishing should be applied. But it should not be too strict in order not to exclude  Grothendieck, for example, who has not published any mathematics for a long time.
An excuse for asking a question so soft as to verge on the flabby is that it might be considered an exercise in Fermi-type order of magnitude estimation.

REPLY [3 votes]: Only somewhat related, I think it was Frank Adams who commented on the varying percentage of a country's population in math(s) e.g Hungary and Roumania.<|endoftext|>
TITLE: Which mathematicians have influenced you the most?
QUESTION [98 upvotes]: There are mathematicians whose creativity, insight and taste have the power of driving anyone into a world of beautiful ideas, which can inspire the desire, even the need for doing mathematics, or can make one to confront some kind of problems, dedicate his life to a branch of math, or choose an specific research topic.
I think that this kind of force must not be underestimated; on the contrary, we have the duty to take advantage of it in order to improve the mathematical education of those who may come after us, using the work of those gifted mathematicians (and even their own words) to inspire them as they inspired ourselves.
So, I'm interested on knowing who (the mathematician), when (in which moment of your career), where (which specific work) and why this person had an impact on your way of looking at math. Like this, we will have an statistic about which mathematicians are more akin to appeal to our students at any moment of their development. Please, keep one mathematician for post, so that votes are really representative.

REPLY [5 votes]: Who: G. H. Hardy
When: As a high school student
Where: The book "Pure Mathematics" -- from which I learned real analysis.
Who: Serge Lang
When: As a college student
Where: At Columbia, Serge Lang was my mathematical mentor.  I took Math I C/II C from him (which I'd describe as freshman mathematics for prospective Ph.D.'s -- it was pretty much an undergraduate Abstract Algebra, plus Real Analysis plus more in two semesters).  His energy and love of mathematics was inspiring.  I know that nobody who met him felt neutral about him.  He was incredibly dedicated to his students.  If he liked you he would move mountains.
Who: Lipman Bers
When: As a college student
Where: At Columbia, Lipman Bers was my other inspiration.  I took Math III C/IV C from him -- sophomore mathematics for prospective Ph.D.'s.  Besides being a very lucid lecturer, with fantastic geometric intuition, he was sophisticated and kind -- a prince among men!  By example he showed how one could live a mathematical life (at perhaps a bit less than the frenetic pace of Serge Lang).<|endoftext|>
TITLE: "Dirty" proof that Eilenberg-MacLane spaces represent cohomology?
QUESTION [26 upvotes]: The standard approach to proving that $H^n(X; G)$ is represented by $K(G, n)$ seems to be to prove that $\text{Hom}(X, K(G, n))$ defines a cohomology theory and then use Eilenberg-Steenrod uniqueness. This is utterly spiffing, but as far as I can see gives little geometric intuition. In his treatment, Hatcher mentions that there is a more direct cell-by-cell proof, albeit a somewhat messy and tedious one. I haven't been able to find any such proof, but I'd really like to see one; I think it would help me solidify my mental picture of Eilenberg-MacLane spaces. Does anyone have a reference?

REPLY [14 votes]: I think that a nice write up can be found in the first chapter of Mosher and Tangora (a very nice book).<|endoftext|>
TITLE: Dirichlet series with integer coefficients as a UFD
QUESTION [9 upvotes]: I recall the following question from Ulam's book "Unsolved math problems": show that the ring of Dirichlet series with integer coefficients is a factorial ring. I believe that soon after Ulam wrote his book, this problem was solved; essentially, this ring is isomorphic to the ring of formal power series in infinitely many variables (one for each prime) with integer coefficients, and once it is reformulated this way it looks much less mysterious. However, my actual question is not about the proof, - I was always curious if a result like that (in its original formulation) can be really applied or interpreted from the Dirichlet series point of view. Has anyone heard/thought of any reasonable number-theoretic interpretation of this fact, or does the interpretation as a power series ring kill any hope of that sort?

REPLY [4 votes]: First, I want to nail down a reference to the solution to the problem alluded to above: the Dirichlet ring of functions f: Z+ -> Z with pointwise addition and convolution product is a UFD.  This was proved by L. Durst in his 1961 master's thesis at Rice University and independently by Cashwell and Everett.  References:
Deckard, Don; Durst, L. K.
Unique factorization in power series rings and semigroups.
Pacific J. Math. 16 1966 239--242. 
Cashwell, E. D.; Everett, C. J.
Formal power series.
Pacific J. Math. 13 1963 45--64. 
Note that the latter is the second paper of Cashwell and Everett on the subject of factorization in Dirichlet rings.  They also had a 1959 paper:
Cashwell, E. D.; Everett, C. J.
The ring of number-theoretic functions.
Pacific J. Math. 9 1959 975--985. 
Ulam makes reference to the 1959 paper in his statement of the problem in the first (1960) edition of his book.  In the 1964 paperback edition of his book, he announces the problem as solved by 1961 papers of Buchsbaum and Samuel.
I want to make the remark that since the original 1959 paper of Cashwell and Everett certainly makes the connection to formal power series rings, there is more to the solution of the problem than this.  (Indeed, the 1961 result of Samuel, that if R is a regular UFD, then R[[t]] is again a UFD, seems to be the key.)
There is a 2001 arxiv preprint of Durst which claims a more elementary proof of the result:
http://arxiv.org/PS_cache/math/pdf/0105/0105219v1.pdf
As to whether the result has number-theoretic significance: not as far as I know.  I should say that I first learned of the Cashwell-Everett theorem by reading an analytic number theory text -- Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, p. 26 -- but the author seems to mention it just for culture.  
Finally, remember that this is a result about factorization in the ring of formal Dirichlet series, an inherently algebraic beast.  So perhaps the following result is more relevant:
Bayart, Frédéric; Mouze, Augustin Factorialité de l'anneau des séries de Dirichlet analytiques. (French) [The ring of analytic Dirichlet series is factorial]  C. R. Math. Acad. Sci. Paris  336  (2003),  no. 3, 213--218<|endoftext|>
TITLE: When does a subgroup H of a group G have a complement in G?
QUESTION [29 upvotes]: Let H be a subgroup of G.  (We can assume G finite if it helps.)  A complement of H in G is a subgroup K of G such that HK = G and |H∩K|=1.  Equivalently, a complement is a transversal of H (a set containing one representative from each coset of H) that happens to be a group.
Contrary to my initial naive expectation, it is neither necessary nor sufficient that one of H and K be normal.  I ran across both of the following counterexamples in Dummit and Foote:

It is not necessary that H or K be normal.  An example is S4 which can be written as the product of H=⟨(1234), (12)(34)⟩≅D8 and K=⟨(123)⟩≅ℤ3, neither of which is normal in S4.
It is not sufficient that one of H or K be normal.  An example is Q8 which has a normal subgroup isomorphic to Z4 (generated by i, say), but which cannot be written as the product of that subgroup and a subgroup of order 2.

Are there any general statements about when a subgroup has a complement?  The Wikipedia page doesn't have much to say.  In practice, there are many situations where one wants to work with a transversal of a subgroup, and it's nice when one can find a transversal that is also a group.  Failing that, one can ask for the smallest subgroup of G containing a transversal of H.

REPLY [2 votes]: A lot can be said in the finitely generated abelian case, just by using the structure theorem. 
Call a group transversal if every subgroup has a complement; non-transversal of it has proper, nontrivial subgroups, none of which has a complement; and semi-transversal if it is not transversal but some proper, nontrivial subgroup has a complement.
Take a finite abelian group $G=\mathbb{Z}/p_1^{e_1}\mathbb{Z}\times\ldots\times\mathbb{Z}/p_k^{e_k}\mathbb{Z}$ with $r\geq 0$, distinct primes $p_1,\ldots,p_k$ and $e_i\geq 1$. 
Then $G$ is transversal when $e_i=1$ for all $i$, semi-transversal when $k>1$ and $e_i>1$ for some $i$, and non-transversal when $k=1$ and $e_1>1$.
For an infinite finitely generated abelian group $G$, $G$ is non-transversal if $G=\mathbb{Z}$, and semi-transversal otherwise.
Given a finitely generated abelian group $G$ and $S\subseteq G$, call $S$ independent if $0\not\in S$ and for all $x_1,\ldots,x_k\in S$, $r_1,\ldots, r_k\in\mathbb{Z}$, we have that $\sum_{i=1}^k r_ix_i=0$ implies $r_ix_i=0$ for all $i$. Call $S$ a basis of $G$ if it is independent and $G=\left< S\right>$.
Then if $H\leq G$, $H$ has a complement if and only if $H$ has a basis that can be expanded to a basis of $G$.  If $S$ is a basis for $H$ and $S\subseteq T$ for some basis $T$ of $G$, then the complement of $H$ is $\left< T\backslash S\right>$.
So if $G$ is a vector space over $\mathbb{Z}/p\mathbb{Z}$ then any subgroup has a complement, since any subspace has a basis that can be completed to the whole space. My definition of independence is the same as linear independence.
If $G$ is a free module over $\mathbb{Z}/p^n\mathbb{Z}$ then my definition of independence is weaker than linear independence. I would like to say that $H\leq G$ has a complement if and only if it has a module basis, but I can't prove the reverse direction of this.
I know less about the divisible abelian case, except that if $G$ is a divisible abelian group then saying that $H\leq G$ has a complement is the same as saying that $H$ is divisible. In particular $\mathbb{Q}$ and the Prufer $p$-group $\mathbb{Z}(p^\infty)$ are both non-transversal.<|endoftext|>
TITLE: The Importance of ZF
QUESTION [17 upvotes]: It seems as though many consider ZF to be the foundational set of axioms for all of mathematics (or at least, a crucial part of the foundations); when a theorem is found to be independent of ZF, it's generally accepted that there will never be a proof of the theorem one way or another.  My question is, why is this?  It seems as though ZF is flawed in a number of ways, since propositions like the axiom of choice and the continuum hypothesis are independent of it.  Shouldn't our axioms of set theory be able to give us firm answers to questions like these?  Yes, the incompleteness theorems say that we'll never develop a perfect set of axioms, and many of the theorems independent of our axioms will probably be quite interesting, but is ZF really the best we can do?  Is there hard evidence that ZF is the "best" set theory we can come up with, or is it merely a philosophical argument that ZF is what set theory "should" look like?

REPLY [3 votes]: This answer is essentially a Joel's version by another route.
ZF(C), possibly with appropriate large cardinal axioms, is one of the three most important formal axiomatisations in the foundations of mathematics, because it is foundationally complete (Friedman 1997):

The usual set theoretic foundations is very powerful, coherent, concise,
  successful, explanatory, impressive, and totally dominating at this time.
  Taken as a whole, with the major supporting classical developments, it is
  certainly one of the few greatest acheivments of the human mind of all time.
However, it also does not come close to doing everything one might demand
  of a foundation for mathematics. At the present time, there is no full
  blown proposal for scrapping it and replacing it with anything
  substantially different that isn't far more trouble than its worth. Present
  cures are far far far worse than any perceived disease.

...

Now before I remind everybody of some of the most vital features of the
  usual set theoretic foundations for mathematics, let me state a great,
  great, great, theorem in the foundations of mathematics:
THEOREM. Sets under membership form the simplest foundationally complete
  system.
There is one trouble with this result: I don't know how to properly
  formulate it. In particular, I don't know how to properly formulate
  "foundational completeness" or "simplest."<|endoftext|>
TITLE: Notions of convergence not corresponding to topologies
QUESTION [61 upvotes]: This question concerns the ramifications of the following interesting problem that
appeared on Ed Nelson's final exam on Functional Analysis some years ago:
Exam question: Is there a metric on the measurable functions on R such that a sequence $\langle F_n(x) \rangle$ converges almost everywhere iff $\langle F_n(x) \rangle$ converges in the metric? 
Answer: No.
Better Answer: Convergence ae does not even precisely correspond to a topology!
The later answer follows from the following (textbook) theorem:
Theorem: Let $\langle F_n(x) \rangle$ be a sequence of measurable functions on a measure space $X$. Then $\langle F_n(x) \rangle$ converges in measure iff every subsequence of $\langle F_n(x) \rangle$ has a subsequence converging almost everywhere.
In particular:
Corollary: If one places a topology $T$ on the measurable functions such that all the almost-everywhere convergence sequences converge in $T$, then all the convergent-in-measure sequences also converge in $T$.
Obvious questions are:

Are there other natural notions of convergence which don't exactly correspond to convergence in some topology?
Are there other pairs of natural convergences which have a similar topological relationship as convergence in measure and convergence ae?
Can one construct a nice theory of "convergences" different from the theory of topologies?  (Warning: This problem tortured me for some weeks some years ago.)

REPLY [4 votes]: Regarding your point 3 here is a list of some sources you might find interesting:

Dolecki & Mynard, "Convergence Foundations of Topology": topology from convergence theory point of view, basically this is what you are asking about
Schechter, "Hadbook of Analysis and its Foundations": good book on analysis that uses nets and filters in an essential way
Nel, "Continuity Theory"
Dolecki, "An invitation into Convergence Theory"

You can find more of them if you'll look for "convergence spaces" as this is a name of the object of the theory you asked for. Also you can look for "filters" and "nets".
As a side remark I can mention that there's a course on basic analysis that introduces limits using filter convergence (kind of): Zorich "Mathematical Analysis".
Also note, that the category of convergence spaces is quite general and there are some useful subcategories of it that are still more general than the topological spaces: pretopological and pseudotopological spaces.<|endoftext|>
TITLE: Why Drinfel'd-Jimbo-type quantum groups?
QUESTION [32 upvotes]: Hopf algebras are pretty easy to motivate, as a not-necessarily-commutative generalization of the ring of functions on an algebraic group (and there are many other ways in which they come up).  I like any situation where you take some interesting geometric construction, characterize it in terms of some structure on the corresponding ring of functions, and then ask what happens when you no longer require such algebras to be commutative.
However, most of the people I know who study Hopf algebras study a very particular class of them often called 'quantum groups'.  This term very often applies to much more general class of Hopf algebras, so for clarity I will call them Drinfel'd-Jimbo quantum groups, as wikipedia does.  I am referring to a very specific q-deformation of the universal enveloping algebra of a semi-simple Lie algebra $g$ (the definition is long, so I will defer to wikipedia).
My question is, what is so special about these Hopf algebras in particular?  Why not some other deformation of the Hopf algebra structure on $\mathcal{U}g$?  I have seen some cool things you can do with these, such as characterizing crystal bases of modules; but are these the reasons people started studying these in the first place?  Is there a natural reason why these are the 'best' deformations of $\mathcal{U}g$?

REPLY [9 votes]: I have seen some cool things you can do with these, such as characterizing crystal bases of modules; but are these the reasons people started studying these in the first place?

No. The original motivations came from quantum inverse scattering method, as they are the natural place where the quantum R-matrices live in abstract form (universal R-element); in particular godo representations of quantum groups supply examples of concrete R-matrices. Thus at least one restricts to quasitriangular or coquasitriangular Hopf algebras (cf. nlab entry).
Later it appeared that apart of some deformed integrable models (spin chains and alike), most physics applications belong to quantum groups at root of unity. For example, the Chern-Simons theory which somebody mentions has a Hamiltonian reduction to WZNW model which has roughly speaking a quantum group symmetry. But the representation theory at the root of unity is related to the representation theory of affine Lie algebras. There are many aspects of this, including the hints from deep work of Kazhdan and Lusztig mentioned in other answers, but also hypergeometric pairing
A.Varchenko, "Multidimensional Hypergeometric Functions and Representation Theory of Lie Algebras and Quantum Groups," Advanced Series in Mathematical Physics, Vol. 21, World Scientific (1995)
which can be interpreted as evaluating cohomology cocycles on certain homology cycles; this picture is later extended in many directions. Big step is a deep work of then extremely young Bezrukavnikov, summarized in the joint book Factorizable sheaves and quantum groups with Finkelberg and Schechtman. More recently this is further understood by joint work of Lurie and Gaitsgory, I know of the published part by
Dennis Gaitsgory, Twisted Whittaker model and factorizable sheaves, Selecta Math. (N.S.)  13  (2008),  no. 4, 617--659.
the newer part with even more exciting idea mentioned by Scott I did not see in any detailed form yet.
As far as constructing part of the quantum group (say just the Borel part) from Cartan data withouit explicit formulas but somehow a priori has with various degrees of success being attempted by various people. Luzstig, Nakajima and others take an appropriate variant of configuration spaces and look at certain perverse sheaves there (this kind of ideas is not unrelated to the business of Bezrukavnikov and others involving another category of (factorizable) sheaves to study quantum groups). Lyubashenko, Majid, Schauenburg and have tried starting with some simple braided tensor category and few other data to construct the quantum group; some categorical meaning of relations, though still complicated and with much input by hand is in Aguiar's thesis, allowing for some neat generalizations. Rosenberg (here) starts with only a braided tensor category and a finite family of distinguished objects to create a quantum group-like structure and claims without proof that the category of vector spaces with certain easy choice of braiding (imvolving formula q to the power involving Cartan matrix elements) gives Drinfel'd-Jimbo case; all the relations, including Serre relations are for free by general nonsense.<|endoftext|>
TITLE: Ramified covers of 3-torus
QUESTION [10 upvotes]: It is known that every orientable 3-manfiold can be obtained as a ramified cover of S3 with a ramification (of some order) at a link in S3. I am curious if there is a reasonable characterization of 3-manifolds that cover 3-torus? 
Added. Notice that such a manifold is enlargeble, so it does not admit a metric of positive scalar curvature, so for example a connected sum of n  copies of S2 x S1 does not admit a ramified cover of T3 (as far as I understand).

REPLY [12 votes]: Note that a branched covering induces an injection of rational cohomology rings, by transfer considerations. Therefore the cohomology of a manifold that is a branched covering of $T^3$ must contain three classes of degree 1 whose triple cup product is nontrivial. 
This condition on a manifold $M^3$ implies the existence of a map $M^3\to T^3$ of nonzero degree. Passing to a covering space of $T^3$ if necessary we obtain such a map that is also surjective on $\pi_1$. Assuming the resulting map is of degree $\ge 3$, the main result of [Edmonds, Allan L.Deformation of maps to branched coverings in dimension three. Math. Ann. 245 (1979), no. 3, 273--279.] implies that this map is homotopic to a branched covering.<|endoftext|>
TITLE: ubiquity, importance of path algebras
QUESTION [18 upvotes]: I work in planar algebras and subfactors, where the idea of path algebras on a graph (alternately known as graph algebras, graph planar algebras, etc.) is quite useful.  The particular result I'm thinking of is a forthcoming result of Jones and some others; it says that any subfactor planar algebra can be found inside the planar algebra of its principal graph.  If you're not into subfactors/planar algebras, the importance of this result is that it says you know a concrete place to begin looking for a particular abstract object.
At Birge Huisgen-Zimmermann's talk on quivers at the AMS meeting at Riverside last weekend, I encountered what seemed to be a similar result:  Gabriel's theorem, which says that any finite-dimensional algebra is equivalent (Morita equivalent I think?) to a path algebra modulo some relations.  (As far as I can tell, "quiver" is a fancy word for a directed finite graph).  I also know, though I don't know why, that path algebras are used in particular constructions in C*-algebras.
This got me thinking:

What are some other places that path algebras appear, and what are they used for?
Why is this idea so useful in these different fields?  Is it simply that path algebras are a convenient place to do calculations?  Or is there some philosophical reason path algebras are important?

REPLY [8 votes]: You also have the rather new field of Leavitt Path Algebras (in which I happen to be working right now), where you take a field $K$ and a directed graph $E$, generate its extended graph $E'$ (add to $E$ its own edges reversed, denoted as $e^*$ for every edge $e$), and compute the Leavitt path algebra of $E$, $L(E)$, as the path algebra $KE'$ modulo some relations called the Cuntz-Krieger relations, inherited from the $C^*$-algebras setting, concretely:
(CK1) $e^* f=\delta_{ef}$ for any two edges $e,f$ of $E'$.
(CK2) $\sum_{e\in s^{-1}(v)}ee^* = v$, for $v$ a vertex which emits a nonzero finite number of edges, and $s^{-1}(v)$ the set of those edges.
(One can look at (CK1) and (CK2) as an abstract generalization of the product of matrix units).
These associative algebras provide us simultaneously with a purely algebraic analog of $C^*$-algebras of graph and a generalization of the Leavitt algebras (some associative algebras which do not satisfy the IBN property).
The full matrix rings over $K$ of order $n$ then arise as the Leavitt path algebras of the graphs with $n$ (consecutive) vertices and $n-1$ arrows, one between every pair of consecutive vertices.
Another simple example of Leavitt path algebra is the ring of Laurent polynomials over $K$, $K[x,x^{-1}]$, which appears associated to the graph with one vertex and a single loop.
The theory of LPAs is useful, and even beautiful, because: 

They provide simple, visually attractive representations of well-known algebras.
They allow us to look at their algebraic properties by means of the combinatorial properties of their associated graphs. This happens to equip us with some rather powerful tools.
Conversely, they also enable "algebraic engineering", since they give us a straightforward, visual way to construct new algebras, customized with any algebraic or ring-theoretic properties we may desire. For example, we can show an algebra generated by five elements such that it is exchange but not purely innitely simple, by constructing a particular (small) graph with some (easy) graph-theoretic features.

Some references:

G. Abrams, G. Aranda Pino. "The Leavitt path algebra of a graph", J. Algebra 293 (2), 319-334 (2005). (Available at http://agt.cie.uma.es/~gonzalo/papers/AA1_Web.pdf).
P. Ara, M.A. Moreno, E. Pardo. "Nonstable K-Theory for graph algebras", Algebra Repr. Th. DOI 10.1007/s10468-006-9044-z (electronic).
(Available at http://www.springerlink.com/content/pu701474q5300m63/).
G. Abrams, G. Aranda Pino, F. Perera, M. Siles Molina. "Chain conditions for Leavitt path algebras".
(Available at http://agt.cie.uma.es/~gonzalo/papers/AAPS1_Web.pdf).
K.R. Goodearl. "Leavitt path algebras and direct limits", Contemp. Math. 480 (2009), 165-187.<|endoftext|>
TITLE: Math Vs Social Science
QUESTION [7 upvotes]: What is the impact of Mathematics in social science today?. That is to say, what are the mathematics that a social scientist is using and, from the point of view of a mathematician, what are the mathematics they should use more or start out using.
Let me give a particular example to start with: Social Inequality. Every day I can see governments trying to eradicate the social inequality in terms of wealth. Assuming wealth can be measure, Is there a theoretical result saying that it is possible (or impossible) to work out such a thing exactly (mathematically speaking)?

REPLY [2 votes]: There was a collaboration between the anthropologist Claude Lévi-Strauss and the Bourbarki's founder André Weil. The latter wrote a appendix to the first part of the former's book, "The Elementary Structures of Kinship".
In this appendix, named as "On the algebraic study of certain types of marriage laws (Murngin system)", André Weil shows how the Murgin's marriage laws can be interpreted algebraically with the tools of group theory. For more information, one can read this featured AMS article.<|endoftext|>
TITLE: The localisation long exact sequence in K-theory over an arbitrary base
QUESTION [10 upvotes]: If I work over a field k,write D for the formal disk k[[t]] and Dx for the formal punctured disk k((t)), then there is an associated long exact sequence in algebraic K-theory
... Kn+1(Dx) --> Kn(k) --> Kn(D) --> Kn(Dx) ...
I want to know, what happens if we replace the base k by a more general scheme?
(I am particularly interested in the map K2(Dx) --> K1(k) (which must be the tame symbol right?))

REPLY [12 votes]: I'm not sure that what I have to say really addresses the heart of your question, but it seems at least related.
Background
The general Localization Theorem (7.4 of Thomason-Trobaugh) states the following. Suppose $X$ a quasiseparated, quasicompact scheme, suppose $U$ a Zariski open in $X$ such that $U$ is also quasiseparated and quasicompact, and suppose $Z$ the closed complement. Then the following sequence of spectra is a fiber sequence:
$$K^B(X\textrm{ on }Z)\to K^B(X)\to K^B(U).$$
Here $K^B$ refers to the Bass nonconnective delooping of algebraic $K$-theory. One thus gets a long exact sequence
$$\cdots\to K_n^B(X\textrm{ on }Z)\to K_n^B(X)\to K_n^B(U)\to K_{n-1}^B(X\textrm{ on }Z)\to\cdots$$
(If one tries to work only with the connective version, then the exact sequence ends awkwardly, since $K_0(X)\to K_0(U)$ is not in general surjective; indeed, the obstruction to lifting $K_0$-classes from $U$ to $X$ is precisely $K_{-1}(Z)$ by Bass's fundamental theorem.)
The term $K^B(X\textrm{ on }Z)$ is the Bass delooping of the $K$-theory of the ∞-category of perfect complexes of quasicoherent $\mathcal{O}$-modules that are acyclic on $U$. Identifying this fiber term with $K^B(Z)$ is generally a delicate matter. Let me summarize one situation in which it can be done.
Suppose that $X$ admits an ample family of line bundles [Thomason-Trobaugh 2.1.1, SGA VI Exp. II 2.2.3], and suppose that $Z$ admits a subscheme structure such that the inclusion $Z\to X$ is a regular immersion (so that the relative cotangent complex $\mathbf{L}_{X|Z}$ is $I/I^2[1]$, where $I$ is the ideal of definition), and $Z$ is of codimension $k$ in $d$ in $X$. Then the spectrum $K^B(X\textrm{ on }Z)$ coincides with a nonconnective delooping of the Quillen $K$-theory of the exact category of pseudocoherent $\mathcal{O}_X$-modules of Tor-dimension $\leq k$ supported on $Z$. If now $Z$ and $X$ are regular noetherian schemes, then a dévissage argument now permits us to identify $K^B(X\textrm{ on }Z)$ with $K(Z)$.
Your case
Now I'm assuming that $K(D)$ refers just to the $K$-theory of the ring $k[[t]]$ (and not, for instance, the $K$-theory of the formal scheme $\mathrm{Spf}(k[[t]])$), then the discussion above applies to give you your desired localization sequence
$$K^B(X)\to K^B(X[[t]])\to K^B(X((t)))$$
for any scheme $X$ admitting an ample family of line bundles. If in particular $X$ is regular, then the negative $K$-theory vanishes, and we have a localization sequence
$$K(X)\to K(X[[t]])\to K(X((t)))$$<|endoftext|>
TITLE: Are quotients of polynomial rings almost UFDs?
QUESTION [5 upvotes]: If $K$ is a field then the polynomial ring $K[x_1,\ldots, x_n]$ is a UFD.  On the other hand, quotients of such a polynomial ring usually don't enjoy unique factorization: consider, for instance, $\Bbb R[x,y]$ modulo the ideal $(x^2+y^2-1)$. Then $x^2=(1-y)(1+y)$ and likewise $y^2=(1-x)(1+x)$. (Over the complex numbers we also have $1=(x+iy)(x-iy)$, and, as Georges points out, the quotient ring is in fact a UFD.)

My question is: are these (in some sense) the only examples which can be factored in different ways? Let me explain what I mean: The above quotient ring (over the reals), call it $A$, is Noetherian so every element can be factored into irreducible ones. I'd be interested to see further elements (not a multiple of the above ones) which don't factorize uniquely. Can something interesting be said about those elements?
What happens in more general quotient rings (let's assume they are domains)?

Thanks for any help and pointers (in particular, the ones already received!), as well as for your patience if this is trivial.

REPLY [8 votes]: (Edit:  I slightly misread your question.  In this answer "unique factorization" means "of ideals," not elements.)
I'll try to give a basic answer, although I'm still learning about this stuff myself.  The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$.  Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).  
More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish.  The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element.  For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization.  More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points. 
Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities.  On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.
(Again, I'm still learning about this stuff, so if I've misstated something please let me know!)

Or maybe you just wanted to know something about your specific case.  Make the substitution $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$; then for example $x^2 = (1 + y)(1 - y)$ can be written as $(1 - t^2)^2 = (1 + 2t + t^2)(1 - 2t + t^2)$.  So here the failure of unique factorization is quite simple: certain polynomials in $t$ are being treated as prime which "shouldn't be."  On the other hand any polynomial in $x, y$ which, when written as a rational function in $t$, avoids these anomalous primes, will have the usual prime factorization properties as a polynomial in $t$, but these prime factors will not necessarily always come from polynomials in $x, y$.

REPLY [6 votes]: As I interpret it, the answer to your question is "yes"; all ways that unique factorization fails come from writing an element of your ideal as the difference of two products.
The point the commenters above are making is that you should not think of this as an at all interesting or useful property.  In general, understanding all elements of an ideal is hard and all ways of writing them a product is way too much for a human brain to take in.<|endoftext|>
TITLE: Closed, complemented subspaces of $l^1(X)$ when $X$ is uncountable
QUESTION [6 upvotes]: ... are all isomorphic to $l^1$ on some other index set. At least, that much I "know" from 2nd-hand sources, since the original proof is apparently in a paper of Köthe from the 1930s 1960s (in German) that I can't get hold of have had trouble digesting. Since there are some Banach space specialists reading MO, I wondered if someone could sketch how the proof differs from the countable case, or point to a more recent text, preferably in English or French, that gives the proof? I hope this is a well-defined question for MO, since I'm not baiting with something where I know the answer.
(Some background for other readers: the analogous result when $X$ is countably infinite follows from combining two steps: one first uses a block basis argument to show that a closed, complemented subspace $V$ inside $l^1(\bf N)$ must either be finite-dimensional, or contain an infinite-dimensional, closed complemented subspace $W$ that is isomorphic to $\ell^1({\bf N})$. In the former case, $V$ is then obviously isomorphic to some finite-dimensional $\ell^1$. In the latter case, one applies Pelczynski decomposition. My impression is that it's the first step which might prove problematic if attempted for $\ell^1(X)$ when $X$ is uncountably infinite, but I could well be wrong and would welcome corrections.)

REPLY [7 votes]: Nice answer, Phil.
The case  of  $Z=\ell_p(X)$, 1 < p < infinity, and $Z=c_0(X)$ with $X$ uncountable are a bit easier because if $T$ is a bounded linear operator from $Y$ into $Z$ and the density character of $Y$ is smaller than $|X|$, then $T$ cannot be one to one.  So if $Y$ is a subspace of $Z$ with density character $|X|$, then a maximal set of disjointly supported unit vectors in $Y$ has cardinality $|X|$ (use  the obvious fact that a subspace $W$ of $Z$ is contained in $\ell_p(Q)$
[or $c_0(Q)$] with $|Q|$ equal to the density character of $W$).  So every subspace of $Z$ with density character $|X|$ has a norm one complemented subspace that is isometric to $Z$.  This and a second use of the ``obvious fact" gives the desired result.<|endoftext|>
TITLE: The class number formula, the BSD conjecture, and the Kronecker limit formula
QUESTION [24 upvotes]: If K is a number field then the Dedekind zeta function Zeta_K(s) can be written as a sum over ideal classes A of Zeta_K(s, A) = sum over ideals I in A of 1/N(I)^s. The class number formula follows from calculation of the residue of the (simple) pole of Zeta_K(s, A) at s = 1 (which turns out to be independent of A).
Let E/Q be an elliptic curve. One might try to prove the (strong) Birch and Swinnerton-Dyer conjecture for E/Q in an analogous way: by trying to define L-functions L(E/Q, A, s) for each A in the Tate-Shafarevich group, writing L(E/Q, s) as a sum of zeta functions L(E/Q, A, s) where A ranges over the elements of Sha, then trying to compute the first nonvanishing Taylor coefficient of L(E/Q, A, s) at s = 1.
Has there been work in the direction of defining such zeta functions L(E/Q, A, s)? If so, what are some references and/or what are such zeta functions called?
Also, taking K to be quadratic, there is not only a formula for the first nonvanishing Laurent coefficient of Zeta_K(s) (the class number formula), but there is a formula for the second nonvanishing Laurent coefficient of Zeta_K(s) (coming from a determination of the second nonvanishing Laurent coefficient of Zeta_K(s, A) - something not independent of K - this is the Kronecker limit formula). Does the Kronecker limit formula have a conjectural analog for the L-function attached to an elliptic curve over Q? 
A less sharp question : are there any ideas whatsoever as to whether any of the Taylor coefficients beyond the first for L(E/Q, s) expanded about s = 1 have systematic arithmetic significance?

REPLY [9 votes]: The following observation suggests that perhaps the analogy between
class groups and Tate-Shafarevich groups is not as close as one might 
think, and that, at least in the quadratic case, the right object is 
the group of ideal classes modulo squares. 
Let $Q_0$ denote the principal binary quadratic form with 
discriminant $d$, and let ${\mathcal P}: Q_0(X,Y) = 1$ be the 
associated Pell conic. For each prime power $q = p^r$, denote 
the number of ${\mathbb F}_q$-rational points on ${\mathcal P}$ 
by $q - a_q$; it is easily checked that $a_q = \chi(q)$, where
$\chi = (\frac{d}{\cdot})$ is the quadratic character with
conductor $d$.
Define the local zeta function at $p$ as the formal power series
$$ Z_p(T) = \exp\Big(\sum_{r=1}^\infty N_r \frac{T^r}r \Big), $$
where $N_r$ denotes the number of ${\mathbb F}_q$-rational points
on ${\mathcal P}$. A simple calculation shows that 
$$ Z_p(T) = \frac{1}{(1-pT)(1-\chi(p)T)}. $$
Set $P_p(T) = \frac1{1 - \chi(p)T} $ and define the global $L$-series as 
$$ L(s,\chi) = \prod_p P_p(p^{-s}). $$
This is the classical Dirichlet L-series, which played a major role in
Dirichlet's proof of primes in arithmetic progression, and was almost
immediately shown to be connected to the class number formula.
Class groups do not occur in the picture above; like their big brothers,
the Tate-Shafarevich group, they are related to the global object we
started with: the Pell conic. The integral points on the affine Pell
conic form a group, which acts (in a more or less obvious way - think 
of integral points as units in some quadratic number fields) on the 
rational points of curves of the form 
$$ Q(x,y) = 1, $$
where $Q$ is a primitive binary quadratic form with discriminant $d$.
This action makes $Q$ into a principal homogeneous space ("over the 
integers"), and the usual action of SL$_2({\mathbb Z})$ on quadratic 
forms respects this structure. The equivalence classes of such spaces 
form a group with respect to taking the Baer sum, which coincides with 
the classical Gauss composition of quadratic forms.
Principal homogeneous spaces with an integral point are trivial in the
sense that they are equivalent to the Pell conic ${\mathcal P}$. The
spaces with a local point everywhere (i.e. with rational points) form 
a subgroup Sha isomorphic to the group $Cl^+(d)^2$ of square classes. 
Defining Tamagawa numbers for each prime $p$ as $c_p = 1$ or $=2$ 
according as $p$ is coprime to $d$ or not, we find that the usual class
number (in the strict sense) is the order of Sha times the product of 
all Tamagawa numbers (the latter is twice the genus class number).
Now we can use Dirichlet's class number formula for proving the BSD 
conjecture for conics:
$$ \lim_{s \to 0} s^{-r} L(s,\chi) = \frac{2hR}{w} = 
   \frac{|Sha| \cdot R^+ \cdot \prod c_p}
   {| {\mathcal P}({\mathbb Z})_{tors}|}. $$
Observe that $R^+$ denotes the regulator of the Pell conic, i.e. the logarithm 
of the smallest totally positive unit $> 1$.
The proof of Dirichlet's class number formula uses the class group, which
is a group containing Sha as a quotient, and a group related to the Tamagawa 
numbers as a subgroup. It remains to be seen whether such a group exists in 
the elliptic case. 
It might be possible to make advance without having such a group:
in the case of Pell conics, the zeta functions of ideal classes are,
if I recall it correctly, closely related to the series defined by
summing over all $1/Q(x,y)^s$ for integers $x$, $y$. The question 
remains how to imitate such a construction for the genus 1 curves
representing elements in Sha. 
Remark: The Tamagawa numbers may be defined as certain $p$-adic integrals;
see an unpublished masters thesis (in Japanese) by A. Iwaomoto, Kyoto 2005.
For more info on the above, see 
here.
For ideas pointing in a different direction, see

D. Zagier, The Birch-Swinnerton-Dyer conjecture from a naive point of view,
Prog. Math. 89, 377-389 (1991)<|endoftext|>
TITLE: Does $\mathrm{Aut}(\mathrm{Aut}(...\mathrm{Aut}(G)...))$ stabilize?
QUESTION [163 upvotes]: Purely for fun, I was playing around with iteratively applying $\DeclareMathOperator{\Aut}{Aut}\Aut$ to a group $G$; that is, studying groups of the form
$$ {\Aut}^n(G):= \Aut(\Aut(...\Aut(G)...)) $$
Some quick results:

For finitely-generated abelian groups, it isn't hard to see that this sequence eventually arrives at trivial group.
For $S_n$, $n\neq 2,6$, the group has no center and no outer automorphisms, and so the conjugation action provides an isomorphism $G\simeq \Aut(G)$.
Furthermore, I believe that if $G$ is a non-abelian finite simple group, then $\Aut(\Aut(G))\simeq \Aut(G)$, though this is based on hearsay.
If one considers topological automorphisms of topological groups, then $\Aut(\mathbb{R})\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{R}$, and so $$\Aut(\Aut(\mathbb{R}))\simeq \Aut(\mathbb{Z}/2\mathbb{Z}\times \mathbb{R})\simeq\mathbb{Z}/2\mathbb{Z}\times \mathbb{R}.$$

When the sequence $\Aut^n(G)$ is constant for sufficiently large $n$, we will say the sequence stabilizes.  Despite my best efforts, I have been unable to find a group $G$ such that the sequence $\Aut^n(G)$ is provably non-stabilizing.  Is this possible?
A slightly deeper question is whether there are groups $G$ such that the sequence becomes periodic after some amount of time.  That is, $\Aut^n(G)\simeq \Aut^{n+p}(G)$ for some $p$ and for $n$ large enough, but $\Aut^n(G)\not\simeq \Aut^{n+m}(G)$ for $m$ between 0 and $p$.  A simple way to produce such an example would be to give two groups $G \neq H$ such that $\Aut(G)\simeq H$ and $\Aut(H)\simeq G$.  Does anyone know an example of such a pair?

REPLY [14 votes]: It seems that $Aut(G)$ is complete for any simple (not necessarily finite) group. The proof of this is in Rotman's book, but it is very short.<|endoftext|>
TITLE: What are the fibres of a representable simplicial sheaf (in the Nisnevich topology)
QUESTION [7 upvotes]: Let $k$ be a field and $X$ a smooth $k$-scheme. We consider now the pointed (constant) simplical Nisnevich-sheaf $X_{+}$ that is represented by $X$. Let now $\nu\in U$ be a point of a smooth scheme, then the fibre of $X_{+}$ at $\nu$ is defined as $\text{colim} \\ X_{+}(V)$ where the colimit is over all Nisnevich neighborhoods $V$ of $\nu$. 
What is this fiber now? Is this just the constant simplicial set associated to the set of all points of $X$ with residue field isomorphic to the one of $\nu$?
And as a further question; If I have a morphism $f:X\rightarrow Y$ between two smooths $k$-schemes, when is the associated morphism of simplicial Nisnevich sheaves $f_{+}:X_{+}\rightarrow Y_{+}$ a simplicial weak equivalence?  For example in the case where $Y$ is irreducible and $f$ an open immersion of a dense open subscheme $X$?

REPLY [2 votes]: Let $N:=\textrm{Neigh}\_{Nis}(U,\nu)$ denote the filtered category of Nisnevich neighborhoods of the point $\nu\in U$. Then $\textrm{colim}\_{V\in N^{op}} V\cong \textrm{spec}(\mathcal{O}^{h}\_{U,\nu})$, where $N^{op}$ denotes the opposite category of $N$ and $\mathcal{O}^{h}\_{U,\nu}$ is the henselization of the local ring $\mathcal{O}\_{U,\nu}$.
My guess now would be that the fiber is the constant simplicial set associated to the set      $\textrm{Hom}\_{k}(\textrm{spec}(\mathcal{O}^{h}\_{U,\nu}),X)$. 
Maybe someone can clarify this as well as your second question.<|endoftext|>
TITLE: Abstract nonsense versions of "combinatorial" group theory questions
QUESTION [15 upvotes]: In particular, I'm just curious whether there's a version of the Sylow theorems (which are very combinatorially-flavored) which allows horizontal and/or vertical categorification? Or at least can be stated in more category-theoretic terms?

REPLY [11 votes]: Sylow subgroups are an example of a type of object satisfying a sort of universal property.  Exploring other objects with similar properties gave birth to the modern theory of finite soluble groups in the 1960s.
If $X$ is a class of finite groups and $G$ is a finite group, then $P$ is an $X$-covering subgroup of $G$ if $P$ is in $X$, and whenever $P \le H \le G, N \unlhd H$, and $H/N$ in $X$, then $PN=H$.  In other words, $P$ covers every $X$-factor of $G$.  If $X$ is the class of finite $p$-groups, then $X$-covering subgroups of $G$ and Sylow $p$-subgroups of $G$ are the same thing.  Indeed, if $P$ is an $X$-covering group, and if $H$ is a Sylow $p$-subgroup containing $P$ and $N=1$, then we must have $H=P$.  If $P$ is a Sylow $p$-subgroup and $H$ contains $P$ with $N \unlhd H$ and $[H:N]$ a power of $p$, then $[H:NP]$ is a divisor of $[H:N]$ and $[H:P]$, so must be $1$.  Notice how the "containment" part of the Sylow theorems is replaced with a "covering" condition that behaves better with the normal structure of the group.
If $G$ is a finite solvable group and $X$ is the class of nilpotent groups, then there is a sort of "Sylow nilpotent subgroup", the $X$-covering groups or Carter subgroups.  They were studied by R.W. Carter who described them as self-normalizing nilpotent subgroups.  Like Sylow $p$-subgroups, there is exactly one conjugacy class of Carter subgroups, and they have some reasonable arithmetic properties.  People tried to determine which classes $X$ of groups are such that $X$-covering groups exist and are unique up to conjugacy.  Roughly speaking, this was the dawn of the modern theory of finite soluble groups, with Gaschütz's (et al.) classification of such $X$ as "saturated formations".
This shifts focus away from the subgroup $P$ to the class $X$.  If $X$ is sufficiently nice, then there will be a nicely embedded X-subgroup for any finite group.
Sylow $p$-subgroups also satisfy a dual condition, they are also $X$-injectors for the class $X$ of finite $p$-groups.  If $X$ is a class of finite groups, and $G$ is a finite group, then $P$ is an $X$-injector of $G$ if for every subnormal subgroup $N$ of $G, P\cap N$ is a maximal $X$-subgroup of $N$.  The dual definition of covering group (for $X$ a saturated formation) is that $P$ is an $X$-covering group iff $PN/N$ is a maximal $X$-subgroup of $G/N$ for every $N \unlhd G$.  If $X$ is the class of finite nilpotent groups, then $X$-injectors are called Fischer subgroups and again form a single, well-behaved conjugacy class of subgroups.  A Fischer subgroup of a finite soluble group is a nilpotent subgroup that contains every nilpotent subgroup that it normalizes.  This is similar to the idea that a Sylow $p$-subgroup contains every $p$-group that it normalizes.  $X$ such that $X$-injectors form a unique conjugacy class are called Fitting classes, due to their similarity to Fitting's lemma on subnormal nilpotent subgroups.
A very approachable introduction to these ideas is B.F. Wehrfritz's tiny textbook for a Second Course on Group Theory.  Some of these ideas are described in Robinson's textbook for a Course in the Theory of Groups, but I believe it spends very little time on general formations.  The standard textbook source for formations, especially in the soluble universe, is K. Doerk and T. Hawkes's book Finite Soluble Groups.  Doerk&Hawkes explains several of Gaschütz's arithmetically defined Xs, which you might find a good contrast.<|endoftext|>
TITLE: Simultaneous Equations Involving Power Sums
QUESTION [7 upvotes]: Let $\ell$ be a positive integer greater than 1. The problem is to find a set of $n$ real positive numbers $x_i$ and $n+1$ numbers $y_i$ such that
$$\sum_{i=1}^n x_i^k= \sum_{i=1}^{n+1} y_i^k$$
for $k=\ell,\cdots,2\ell-1$. These $2n+1$ numbers need to be upper/lower bounded by a constant independent of $\ell$ [thus $x_i,y_i=\Theta(1)$] and also I suspect that it is possible to do so with just $n=\ell$ or $n=O(\ell)$. [$\ell$ equations with $2\ell$ unknowns, why not!] An existential proof suffices but a constructive proof or a recipe would be really nice.
For me it is useful to find a bounded from below solution that scales in polynomially in the following sense: There exist positive $c$, and $s$ such that $c\le x_i,y_i$ and
$$\sum_i x_i+\sum_i y_i=O(\ell^s)$$. 
The problem is related to a follow up on this paper of mine: arxiv:0908.1526 .

REPLY [11 votes]: Actually Darsh gave an almost full solution. Let me fill in the minor technical details.
1) We need the following quantitative form of the inverse function theorem. Suppose that  $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_1$, that $\max_{Y\in B(X, \delta)}\|D^2F(Y)\|\le C_2$, and that $C_1C_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_1})$.  
2) Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_1,\dots,y_n)_k=\sum_{j=1}^n y_j^k$ where $k=1,2,\dots,n$. Take $X=(x_1,\dots,x_n)$ where $x_j=\frac{n+j}{n}$ for $j=1,\dots,n$.
3) Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.
4) Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,\dots,c_n)$ to the vector $p(x_1),\ldots,p(x_n)$ consisting of the values of the polynomial $p(x)=\sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.
5) Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.
6) In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.<|endoftext|>
TITLE: Picard-Fuchs equations
QUESTION [11 upvotes]: If I have the periods $$\pi_1(\lambda)=\int_0^1\frac{dx}{\sqrt{x(x-1)(x-\lambda)}}$$ and $\pi_2(\lambda)$ similarly defined of the cubic curve $$y^2z=z(x-z)(x-\lambda z)$$ Such functions will be holomorphic on $\lambda \in \mathbb{C}-\{0,1\}$. Then the sum $\pi(\lambda)=\pi_1(\lambda)+\pi_2(\lambda)$, as well as each $\pi_i$, satisfy the differential equation called  Picard-Fuchs equation $$0=\tfrac{1}{4}\pi+(2\lambda-1)\pi'+\lambda(\lambda-1)\pi''$$ derivation with respect to $\lambda$.
My questions are the following. Where does such a differential equation come from? 
why is it important for?
has it something to do with Gromov-Witten Theory? 
I'll appreciate any kind of comment related with such an equation.

REPLY [17 votes]: Here is a very rough outline:
Take your family $E$ of elliptic curves over $B := \mathbb{C} - \{0,1\}$. Then take the associated "(co)homology bundle" over $B$, whose fibre over $\lambda$ is the (singular) (co)homology of the elliptic curve $E_\lambda$. To be rigorous, the $i$-th cohomology bundle is $R^i \pi_\ast\mathbb{C}$, where $\pi$ is the map $E \to B$ and $\mathbb{C}$ is the constant sheaf (let us work in the analytic topology). Actually to be precise I should say that $R^i\pi_\ast\mathbb{C}$ is a (locally constant) sheaf of $\mathbb{C}$ vector spaces, and the corresponding vector bundle is $R^i\pi_\ast\mathbb{C} \otimes_\mathbb{C} \mathcal{O}\_B$. It is a fact that these cohomology bundles come with flat (Gauss-Manin) connections $\nabla$. One way to see that the vector bundles are flat is to observe that there are integral lattices $R^i\pi_\ast \mathbb{Z} \subset R^i\pi_\ast\mathbb{C}$.
Let $\omega$ be a 1-form on the family $E$. Note that the 1st cohomology of an elliptic curve is rank 2, so the cohomology bundle $R^1\pi_\ast \mathbb{C}$ is rank 2, thus if we have 3 sections, then they will be (fiber-wise) linearly dependent. So here are 3 sections: $\omega, \nabla_{d/d\lambda}\omega, (\nabla_{d/d\lambda})^2 \omega$. The Picard-Fuchs equation is essentially just the equation which expresses that these sections are linearly dependent. Your equation involving "$\pi$" (not the same as what I am calling "$\pi$") and its derivatives comes from taking this linear dependence equation and "plugging in" (i.e. integrating along) homology classes extended by parallel transport.
The story that I've described above generalizes to arbitrary families of smooth compact varieties.
Thomas Riepe's answer explains some of the more classical reasons why we might be interested in period integrals and Picard-Fuchs equations, so let me say a few words about the relation to Gromov-Witten theory.
The relation to GW theory arises from mirror symmetry, which is a duality between type IIA and type IIB string theories. One of the reasons why mathematicians first became interested in mirror symmetry was because of the prediction of the physicists Candelas-de la Ossa-Green-Parkes in the early 90s that the genus 0 GW invariants of a quintic threefold (type IIA theory) could be computed via an analysis of period integrals and Picard-Fuchs equations coming from a "mirror" family of Calabi-Yau manifolds (type IIB theory). This is a general principle of mirror symmetry: that GW invariants of certain manifolds can be computed via completely different methods on the "mirror manifold". Usually, studying the mirror manifold is "easier" than trying to study the GW theory of the original manifold directly; although by now our knowledge of GW theory has grown considerably, so this is less true than it used to be.
A very nice introductory paper on this material is "Picard-Fuchs equations and mirror maps for hypersurfaces" by David Morrison: http://arxiv.org/abs/alg-geom/9202026
If you're interested in reading further, you should check out the book "Mirror symmetry and algebraic geometry" by Cox-Katz, which covers all of this material in detail and explains the proofs (due to Givental and Lian-Liu-Yau) of the Candelas-et. al. prediction.<|endoftext|>
TITLE: Atiyah-MacDonald: exercise 5.29 - "local ring of a valuation ring"
QUESTION [5 upvotes]: The exercise is the following:

Let $A$ be a valuation ring, $K$ its field of fractions. Show that every subring of $K$ which contains $A$ is a local ring of $A$.

Does anyone know what is meant by "to be a local ring of a valuation ring"?

REPLY [4 votes]: Let B be the subring of K containing A. I am going to prove that B is the localization of A at a prime ideal $P\subset A$, which seems a reasonable interpretation of the statement that B is a local ring of A.
First B is a local ring [by Atiyah-MacDonald Prop 5.18 i)] ; let $M_B$ be its maximal ideal.
Similarly let $M_A$ be the maximal ideal of A. 
Define $P=A\cap M_B$. Then of course $P\subset M_A$ is prime and if we localize A at P, I claim that we get B:
a) If $\pi \in A-P$ then $\pi \notin M_B$ and so $\pi$ is invertible in B. Hence for all $ a \in A$ we have $a/ \pi \in B$.
This shows $A_P \subset B$. 
b) Now we see that B dominates $A_P$: this means we have an inclusion of local rings  $A_P \subset B$ and of their maximal ideals:  $PA_P \subset M_B$. But $A_P$ is a valuation ring of K because A is (This is the preceding exercise 28 : these guys know where they are heading!)
and valuations rings are maximal for the relation of domination (Exercise 27: ditto !). 
Hence we have the claimed equality   $B=A_P$.<|endoftext|>
TITLE: fonctions-faisceaux correspondence 
QUESTION [11 upvotes]: what are the standard references for this subject, i.e., the realization of functions as traces of Frobenius acting on sheaves?  Is there any motivation or philosophy coming from categorification, number theoretic analogies, or other directions?

REPLY [2 votes]: My preferred reference is Kiehl and Weissauer's book (very complete, and in English).
Of course, if you want motivation, you can read my blog post (this the companion to the one Thomas linked to).<|endoftext|>
TITLE: Is the Fell-Doran problem trivial in a topological setting?
QUESTION [5 upvotes]: The Fell-Doran problem is a problem in functional analysis. It goes as follows: Let $A$ be a complex unital algebra, $X$ a locally convex space, and $L(X)$ the algebra of all continuous endomorphisms of $X$. Suppose that we have a representation of $A$ on $X$, by which we simply mean an algebra homomorphism
$$
 T : A \rightarrow L(X)
$$
which is irreducible (no proper closed invariant subspace) and has trivial commutant (any bounded operator commuting with all the $T_a$ must be a multiple of the identity). The Fell-Doran problem is: Is $T(A)$ dense in $L(X)$ in the strong operator topology?  
My question is: Is this a problem having to do with the fact that we didn't require a topology on our algebra? In other words, what can be said about the case when $A$ is actually a 'topological algebra' and the map $T$ is required to be continuous in some sense? Does that make the problem trivial, i.e. is the answer then automatically yes?
By the way, I have heard that so far there is almost no progress on the Fell-Doran problem in general; not even for Hilbert spaces! The only thing that is known is that there exists a certain concrete space where the answer is affirmative.

REPLY [4 votes]: Actually, digging a bit deeper, I've found this paper by Zelazko: Colloquium Mathematicum which states the problem as: If $A\subseteq L(X)$ is topologically irreducible (that is, the orbit under $A$ of any non-zero vector is dense) and the commutant of $A$ is trivial, is $A$ totally irreducible.  This means that for each $n$, if $x=(x_1,\cdots,x_n)$ is a vector in $X^n$ with linearly independent co-ordinates, then $x$ should be cyclic for $A^n$ (where $A^n$ acts on $X^n$ is the obvious way).
Off hand, I don't see what, if anything, this has to do with $A$ being strong operator topology dense in $L(X)$.  Or is there more than one conjecture??<|endoftext|>
TITLE: Fourier-Mukai transform - a first example
QUESTION [7 upvotes]: Suppose $X$ and $Y$ are schemes of finite type over a field, and let $f: X\rightarrow Y$ be a morphism. Let $\Gamma_f$ be the closed subscheme of $X\times Y$, then the first example of the Fourier-Mukai transform says that
$$f_*()=p_Y{_*}(p_X^*()\bigotimes_{\mathcal{O}_{X\times Y}}\mathcal{O}_{\Gamma_f}),$$ similarly there is an expression for $f^*$. Has anyone checked this before? I am having some difficulties (at least on some commutative algebra) in verifying them. Does anyone know why those formulas are true?

REPLY [3 votes]: You should use the fact that $p_X^*(-)\otimes_{\mathcal{O}_{X\times Y}}\mathcal{O}_{\Gamma_f}\cong i_*(p_X|_{\Gamma_f})^*(-)$ where $i:\Gamma_f\to X\times Y$ is the inclusion. Then the statment becomes much easier.<|endoftext|>
TITLE: When does a pointwise CLT hold?
QUESTION [9 upvotes]: Let $X$ be a random variable with mean $0$ and variance $1$, and let $X_1, X_2, X_3, \dots$ be iid copies of $X$.  Under what conditions can we say that the density of $\frac{X_1+\dots+X_n}{\sqrt{n}}$ converges pointwise to $N(0,1)$?
In particular, when can I say that for any sequence $\epsilon_n \rightarrow 0$ we have 
$$\frac{P(|\frac{X_1+\dots+X_n}{\sqrt{n}}|<\epsilon_n)-P(|N(0,1)|<\epsilon_n)}{\epsilon_n} \rightarrow 0?$$
In flavor this is somewhat similar to what I've seen termed as "local limit theorems", except a little bit stronger; for example if $X$ is a Bernoulli variable the above would not hold (take $\epsilon_n=2^{-n}$).  My guess would be that a sufficient condition would be for the usual CLT to hold and $X$ to have bounded density functions, though I haven't seen this cited anywhere.

REPLY [2 votes]: Feller states the Berry-Esseen theorem in the following way. Let the $X_k$ be independent variables with a common distribution $F$ such that $$E[X_k]=0, E[X_k^2]=\sigma^2>0, E[|X_k|^3]=\rho<\infty,$$ and let $F_n$ stand for the distribution of the normalized sum $$(X_1+ \dots X_n)/(\sigma \sqrt{n}).$$ Then for all $x$ and $n$ $$|F_n(x)-N(x)| \leq \frac{3\rho}{\sigma^3 \sqrt{n}}.$$
The expression you are interested in is 
  $$\left|\frac{F_n(\epsilon)-F_n(-\epsilon)-N(\epsilon)+N(-\epsilon)}{\epsilon}\right|,$$ 
which is less than
  $$\left| \frac{F_n(\epsilon)-N(\epsilon)}{\epsilon} \right| + \left| \frac{F_n(-\epsilon)-N(-\epsilon)}{\epsilon} \right|,$$
which by Berry-Esseen is bounded by
  $$2\frac{3\rho}{\epsilon \sigma^3 \sqrt{n}}.$$
So, if $\epsilon\sqrt{n}$ goes to infinity, then you are good.
I realize this isn't what you asked, in that you wanted conditions on $X$, and this instead gives you conditions on $\epsilon_n$. Still, perhaps it'll help.
Reference: Feller, An Introduction to Probability Theory and Its Applications, Volume II, Chapter XVI.5.<|endoftext|>
TITLE: multi-index Dirichlet series
QUESTION [8 upvotes]: Hi,
I have recently got interested in multi-index (multi-dimensional) Dirichlet series, i.e. series of the form $F(s_1,...,s_k)=\sum_{(n_1,...,n_k)\in\mathbb{N}^k}\frac{a_{n_1,...,n_k}}{n_1^{s_1}...n_k^{s_k}}$. I found some papers suggesting that multi-index Dirichlet series are in fact a distinct subfield for itself within analytic number theory. So, I´m now looking for some 'basic' learning materials/books or similar on this subject.
Any suggestions are greatly appreciated!
efq
PS: I believe I have already checked most books on multi-dimensional complex analysis/several complex variables.

REPLY [2 votes]: See P. Deligne, Multizeta values, Notes d'exposes, IAS Princeton, for the deep mathematical aspects of this. 
Also for a general relevance philosophy, see Kontsevich and Zagier, Periods, Mathematics Unlimited(2001). An electronic version is available here.
There are various references, including those of Zudilin, Cartier, Zagier, Terasoma, Oesterle(On polylogarithms), Manin(iterated integrals and ....). Please look into mathscinet. 
There seem to be many papers by Dorian Goldfeld and collaborators, too.<|endoftext|>
TITLE: What are some interesting ways of making new metrics out of old metrics?
QUESTION [9 upvotes]: If $d(x,y)$ and $e(x,y)$ are metrics then $d(x,y)+e(x,y)$ and $\frac{d(x,y)}{1+d(x,y)}$ are metrics.
If $d_i(x,y)$ for $i=1,\dots,n$ are metrics then so is $\sqrt{\sum_{i=1}^n{d_i^2(x,y)}}$
Are there other interesting ways of constructing new metrics from old metrics?

REPLY [2 votes]: Associated to any metric is a length structure. The length of an absolutely continuous path is the integral of it's metric derivative. Associated to any length structure is a (pseudo)-metric. If this is an honest metric, it is also a path metric: for any $x, y\in X$ and any $\varepsilon>0$, there is some $z\in X$ such that $d(x, z)+d(z, y)\le d(x, y)+\varepsilon$.
Taken from Gromov, "Metric structures for Riemannian and non-Riemannian spaces".<|endoftext|>
TITLE: Relation between dendroidal and opetopic sets
QUESTION [6 upvotes]: To my shame I have to admit that I have as yet not looked much into opetopes and opetopic sets.
I am in the process of writing nLab entries on dendroidal sets and noticed that some remarks on the relation to opetopic sets should probably be in order. 
Now, I  know that I should just sit down and read the opetopic literature. But while I am busy doing that, and since the model structure on dendroidal sets wasn't around when most of it was written: does anyone know more about the relation?

REPLY [4 votes]: Urs, do you have a reason to think that there'll be much to say about this?  I can see that opetopic and dendroidal sets are both presheaf categories that arise in higher-dimensional category theory.  I can see that in both cases, the small category on which you're taking presheaves has a graphical or geometric interpretation, and there are some "face maps", and there's something tree-like going on.  But beyond that, I don't see what there is to say.  Do you have something in mind?
I just looked in Ittay Weiss's thesis, Dendroidal Sets.  "Opetope" is not in the index, nor is the relevant Baez--Dolan paper (Higher Dimensional Algebra III) cited.  So I guess he had no thoughts on the matter.<|endoftext|>
TITLE: Can Gröbner bases be used to compute solutions to large, real-world problems?
QUESTION [27 upvotes]: In particular, suppose I have an affine algebraic variety over $\mathbb{R}^n$ described by generators of a radical ideal $I$ and I want to find (perhaps not all of the) points in the variety.  Several important questions come up in practice:

are there versions of Buchberger's algorithm that work with inexact data?  For instance, suppose that the coefficients of the polynomials generating $I$ are known only to floating point precision.  Some CAS will try to find solutions assuming that these coefficients are exact.  Are there CAS that do something more intelligent (e.g., make certain guarantees given that the numerical coefficients are the truncation of exact coefficients)?
does a sparse system of polynomial equations yield a Gröbner basis with sparse elements?  In other words, if each polynomial in the original system has a small number of non-zero coefficients relative to $n$, do the basis elements also have this property?
what bounds are known for the size of a Gröbner basis in terms of size and sparsity of the original system?
are there more appropriate algorithms (than Buchberger's) if we just want to find a single point in the variety?  (Suppose that any such point is sufficient.)  More generally, which algorithms are better suited to address the kinds of issues mentioned above?

REPLY [5 votes]: Concerning your questions 1 and 4, you should also have a look at border bases:
http://www.risc.jku.at/Groebner-Bases-Bibliography/gbbib_files/publication_1140.pdf<|endoftext|>
TITLE: Change rates of the 2nd Chebyshev function
QUESTION [6 upvotes]: Let $\psi(x):=\sum_{n\leq x}\Lambda(n)$ denotes the 2nd Chebyshev function, where $\Lambda$ stands for the von Mangoldt function. Are there any known (and 'nice') estimates for the change rates $\psi(x+h)-\psi(x)$ for general or special $x$ and $h$?
Thanks in advance,
efq

REPLY [2 votes]: Also, I think Selberg's result asserts that for "almost all x"
we have $\psi(x + h) - \psi(x) \sim h$ for $h \asymp (\log x)^2$; 
this was shown to not be true "pointwise" by Mayer.<|endoftext|>
TITLE: Hodge Index Theorem for  Gr(n,k)
QUESTION [8 upvotes]: I'm trying to understand the Hodge-Index Theorem at the moment. What does it say explicitly for the case of the complex Grassmannian Gr($n,k$), and can this be established without recourse to the theorem?

REPLY [14 votes]: The answer is a happy surprise for me:  The Hodge index theorem for a Grassmannian matches a special case of John Stembridge's $q=-1$ phenomenon, that I also studied in an old paper.
First, some generalities about what is going on, and about when you do or don't "need" the Hodge Index Theorem.  I am following the Hodge Index theorem described in Claire Voisin's book, that says that the signature of a complex $n$-manifold is a certain alternating sum of Hodge numbers.  A Grassmannian is a type of partial flag variety, and flag varieties have Schubert cell decompositions.  That makes it easy to compute both the Hodge numbers, and the signature of the manifold separately so that the Hodge Index Theorem becomes an identity between two calculations.
More precisely, the Schubert cells are a basis for the cohomology: they make a CW complex with even-dimensional cells, which forces the CW differential to vanish.  For complex geometry reasons, the Schubert cells are all in $H^{k,k}$ for some $k$, i.e., all on the diagonal of the Hodge diamond.  The Hodge Index theorem then says that the signature is the alternating sum of these numbers.  At the same time, the intersection matrix of such a manifold is just a symmetric permutation matrix connecting dual Schubert cells.  So the signature is also the number of self-dual Schubert cells.  The Hodge Index theorem thus equates the number of self-dual cells with signed sum of all of the cells.
In the case of a Grassmannian $\mathrm{Gr}(a,a+b)$, the Schubert cells are labelled by partitions in an $a \times b$ rectangle.  The Hodge index formula is the $q$-enumeration of these partitions with $q = -1$.  On the other hand, the direct calculation is the number of self-complementary partitions, where you replace a partition with its complement in the rectangle and turn the rectangle upside-down.  It is easy to calculate the self-complementary partitions in the case.  The lattice path of the partition is symmetric about the middle of the rectangle, so these are just partitions in an $\lfloor a/2 \rfloor \times \lfloor b/2 \rfloor$ rectangle.  Or there are none if $a$ and $b$ are both odd: in this case the signature is trivial a priori because the middle dimension has odd degree.  This combinatorial equality is exactly the $q=-1$ phenomenon.
What I wonder now is whether every case of the $q=-1$ phenomenon, for the $q$-dimension of an irreducible representation of a complex simple or compact simple Lie group, can be explained by a Hodge Index theorem with coefficients, combined with some geometric model of the representation.  This would be sophisticated (to the point of overkill, if you just want the combinatorics), because the involution in general is something called "evacuation", which John related to canonical bases.<|endoftext|>
TITLE: Principal bundles, representations, and vector bundles
QUESTION [21 upvotes]: What is the exact relationship between principal bundles, representations, and vector bundles?

REPLY [7 votes]: Just to be pedantic, these notions are for finite dimensional groups and representations.  In infinite dimensions (either group or representation) one has to be a little bit more careful, see this question.<|endoftext|>
TITLE: How do I check if a functor has a (left/right) adjoint?
QUESTION [85 upvotes]: Because adjoint functors are just cool, and knowing that a pair of functors is an adjoint pair gives you a bunch of information from generalized abstract nonsense, I often find myself saying, "Hey, cool functor. Wonder if it has an adjoint?" The problem is, I don't know enough category theory to be able to check this for myself, which means I can either run and ask MO or someone else who might know, or give up.
I know a couple of necessary conditions for a functor to have a left/right adjoint. If it doesn't preserve either limits or colimits, for example, I know I can give up. Is there an easy-to-check sufficient condition to know when a functor's half of an adjoint pair? Are there at least heuristics, other than "this looks like it might work?"

REPLY [6 votes]: A special case I have found particularly tractable is when the categories involved are locally presentable.

Theorem: Suppose $\mathcal{C}$ and $\mathcal{D}$ are locally presentable categories, and $F : \mathcal{C} \to \mathcal{D}$ is a functor. Then:

$F$ has a left adjoint if and only if it preserves limits and is accessible
$F$ has a right adjoint if and only if it preserves colimits

(ref: nLab)

A lot of nice categories are locally presentable, and there are enough structure theorems that you can often quickly verify that a category is locally presentable.<|endoftext|>
TITLE: Is there a subfactor construction involving 2-groups?
QUESTION [9 upvotes]: I seem to recall that there is a straightforward subfactor construction that yields fusion categories given by G-graded vector spaces and representations of G, for finite groups G.  Is there an analogous construction for 2-groups?
Some background: A 2-group is a monoidal groupoid, for which the isomorphism classes of objects form a group.  Sinh showed that up to monoidal equivalence, these are classified by a group G (isom. classes of objects), a G-module H (automorphisms of identity), and an element of H3(G,H).  In the context of this discussion, we can limit our attention to G finite, H=Cx.  One notable feature is that when the action of G on H is trivial, the three-cocycle twists the associator in the G-graded vector space category.
I'm mostly curious about how to tell when two elements of H3(G,H) yield Morita-equivalent fusion categories, and am wondering if subfactors or planar algebras make it easy to detect this.

REPLY [6 votes]: This is a standard construction in Subfactor theory see the intro of http://arxiv.org/abs/0811.1084v2 for details.  The construction goes back a long long way (if I remember correctly both Vaughan Jones and Adrian Ocneanu's theses were related to this question, but I could be wrong there).
From a category theory perspective recall that a subfactor (N < M) is a unitary tensor category C (the N-N bimodules) together with a Frobenius algebra object A in C (M as an N-N bimodule with conditional expectation as trace).  In this case the tensor category C is the twisted category of G-graded vector spaces (where you use the 3-cocycle to change the associator), and the algebra object is a twisted version of the group algebra (or maybe just the group algebra? I'm getting confused, shouldn't group algebras be twisted by 2-cocycles?).<|endoftext|>
TITLE: Is there a natural way to give a bisimplicial structure on a 2-category? 
QUESTION [6 upvotes]: I mean by the nerve functor. 
Given a 2-category $\mathcal{C}$, if we forget the 2-category structure (just view $\mathcal{C}$ as a category), the nerve functor will give us a simplicial set $N\mathcal{C}$. However, $\mathcal{C}$ is a 2-category, thus for any two objects $x,y\in\mathcal{C}$, $Hom_{\mathcal{C}}(x,y)$ is a category, applying the nerve functor gives us a simplicial set $N(Hom(x,y))$.
My question is, can these two simplicial set structure compatible in some way, gives us a bisimplicial set $N_{p,q}(\mathcal{C})$, say? Or is there another way to give a bisimplicial structure on a 2-category?

REPLY [5 votes]: Yes. This is called the double nerve of a 2-category.
See in particular the first reference cited at that link.<|endoftext|>
TITLE: Generalization of the two bucket puzzle
QUESTION [14 upvotes]: The classic puzzle goes something like this: "You are standing in front of a lake with a 3 gallon bucket and a 5 gallon bucket, how can you get 4 gallons of water?"
Is there an easy way to generate the triple (A,B,C) where you can get C gallons of water using buckets of size A and B?

REPLY [4 votes]: Mathloger has a beautiful video about this, here.<|endoftext|>
TITLE: Local Joyal-simplicial presheaves?
QUESTION [5 upvotes]: It is well known that left Bousfield localizations of the global functor model category $Func(C^{op}, SSet_{standard})$ of functors with values in simplicial sets equipped with the standard model structure on simplicial sets yields the local model structure on simplicial presheaves that models hypercomplete oo-stacks.
What is known, though, about left Bousfield localizations of $Func(C^{op}, SSet_{Joyal})$, with simplicial sets equipped with the Joyal model structure?
If it exists, this should model $(\infty,2)$-sheaves on $C$. But possibly it doesn't exist.

REPLY [11 votes]: If $V$ is a reasonnable model category (i.e. combinatorial, etc), and $C$ a small category endowed with a Grothendieck topology $\tau$, there are $\tau$-local model structures on the category $Fun(C^{op},V)$ (a projective version as well as an injective version). You may have a look at this paper of Clark Barwick (see also section 4.4 of Joseph Ayoub's book for the hypercomplete versions). In the case $V$ is the standard model structure on simplicial sets, we get the usual homotopy theory of stacks in $\infty$-groupoids, while, if $V$ is the Joyal model structure, you get the homotopy theory of stacks in $(\infty,1)$-categories. If you consider the hypercomplete version, then these model structures will behave in the usual way; for instance, if moreover the topos of sheaves on $C$ has enough points, a morphism of simplicial presheaves on $C$ will be a weak equivalence for the hypercomplete $\tau$-local Joyal model structure if and only if its stalkwise is a weak equivalence for the Joyal model structure.
You may as well consider the case where $V$ is the Rezk model structure for $(\infty,n)$-categories to get the homotopy theory of stacks in $(\infty,n)$-categories (for $0\leq n\leq \infty$) to obtain the $(\infty,n+1)$-topos of stacks in $(\infty,n)$-categories (whatever this means). Alternatively, you may as well consider the case where $V$ is the model category of Segal $n$-categories, and get something rather close to Simpson and Hirschowitz theory of higher stacks. It is also possible to consider the case where $C$ is a simplicial category and $\tau$ is a Grothendieck topology on $C$ (in the sense of Toën-Vezzosi-Lurie, see HAG I and Lurie's book), and obtain stacks of $(\infty,n)$-categories over any $(\infty,1)$-topos.<|endoftext|>
TITLE: Is there any Grothendieck Riemman Roch theorem for general stack ?
QUESTION [8 upvotes]: It seems that there is no g.r.r for stack yet according to dejong. Does anyone know anything about it? But as you might know, there are some complex manifold which is not scheme having atiyah singer index theorem. So I was wondering if there exists some analogue of g.r.r for special stack.

REPLY [12 votes]: If you work with the naive Chow-groups and allow non-representable morphisms the GRR-Theorem does not hold! In the paper by Toen quoted above and in some of the papers by Joshua there are explicit counterexamples. They always involve non-representable morphisms. 
There are two ways to get around this. 
The first is to modify the definition of the Chow-groups. This is what Toen does. He takes Chow groups with coefficients in the characters of the stack, which is quite an involved definition. But it leads to a GRR-theorem for DM-stacks.
The second approach by Joshua is modify the topology to keep track of the stabilizer groups. He introduces a topology which he calls the isovariant etale site, which is motivated by ideas of Thomason. This gives a different kind of Chow groups. For this recall that you can define the Chow groups as cohomology of some sheaves using higher K-theory. For stacks this was done by Gillet. You can then get new kinds of Chow groups by calculating the cohomology of these sheaves in the isovariant etale topology. In a series of papers Joshua proves GRR-theorems in this context.<|endoftext|>
TITLE: Symmetric Powers, Tableau and Wreath Products
QUESTION [7 upvotes]: Let V and W be irreducible representations of $S_n$ and $S_m$ over a field of characteristic 0.  Then the Littlewood-Richardson coefficients allow us to compute the isomorphism type of the induced $S_{n+m}$-module V⊗W↑.  This induction comes from the inclusion
$S_n\times S_m \rightarrow S_{n+m}$.
Now suppose V=W.  Then V⊗V↑ is a $S_{2n}$-module.  But actually there's a symmetry coming from the symmetric monoidal category structure, so there is another induction up to an $S_{2n}$-module structure:
Extend the action of $S_n\times S_n$ on V⊗V by including the symmetry $c_V$, this naturally extends the group to the wreath product $S_n\sim S_2$.  Induction along
$S_n\sim S_2 \rightarrow S_{2n}$
gives the representation that I want:
$(V\otimes V)_{S_n\sim S_2}\uparrow^{S_{2n}} \hookrightarrow V\otimes V\uparrow^{S_{2n}}$.
Using the Littlewood-Richardson rules we know the structure of the last term in terms of semi-standard skew tableau.  My question is, how do we characterise the inclusion?

REPLY [2 votes]: My understanding was that induction from wreath products was supposed to correspond to plethystic substitution of symmetric functions.  But I don't really understand that, so I can't really say.<|endoftext|>
TITLE: Is there a Faa di Bruno-like formula for composition of three functions?
QUESTION [10 upvotes]: Faa di Bruno's formula (MathWorld, Wikipedia) gives the kth derivative of f(g(t)) as a sum over set partitions.  (I'm not sure how well-known the combinatorial interpretation is; for example see a 2006 paper of Michael Hardy).
Is there a similar formula for the kth derivative of f(g(h(t))), i. e. of a threefold composition?  One has, I think,
${d^2 \over dt^2} f(g(h(t))) = f^{\prime\prime}(g(h(t))) g^\prime(h(t))^2 (h^\prime(t))^2 + f^\prime(g(h(t))) g^{\prime\prime}(h(t)) h^{\prime}(t)^2$
$ + f^\prime(g(h(t))) g^\prime(h(t)) h^{\prime\prime}(t)$
but I am not even sure this is right (computation is tricky!) and I do not know how to generalize this.

REPLY [5 votes]: I would like to provide an analytically explicit answer. For simplicity, just take f,g,h smooth functions from $\mathbb R$ into itself. Then we have
\begin{align*}
\frac{(h\circ g\circ f)^{(p)}}{p!}=\sum_{\substack{p_1+\dots+p_q=p\\q\ge 1, p_j\ge 1}}
\frac{(h\circ g)^{(q)}}{q!}
\prod_{1\le j\le q}\frac{f^{(p_j)}}{p_j!}
\\
=
\sum_{\substack{p_1+\dots+p_q=p\\q\ge 1, p_j\ge 1}}
\sum_{\substack{q_1+\dots+q_r=q\\r\ge 1, q_k\ge 1}}
\frac{h^{(r)}}{r!}
\prod_{1\le k\le r}\frac{g^{(q_k)}}{q_k!}
\prod_{1\le j\le q}\frac{f^{(p_j)}}{p_j!},
\end{align*}
where of course the derivatives of $h$ (resp. $g$) are evaluated at $(g\circ f)(x)$ (resp. $f(x)$). Alphabetical order may help the reader to complete an expression for the higher order derivatives of the composition of $N$ functions.

REPLY [5 votes]: Similarly to Theo's answer, I think the best way to write such combinatorial formulas is in terms of trees (rather than multiindices or even set partitions). I explained that in my article "Feynman diagrams in algebraic combinatorics" in Sém Lothar. Combin.  49 (2002/04), Art. B49c.
See in particular Eq. 13 for composition of $n$ functions.<|endoftext|>
TITLE: Blackboard rendering of math fonts
QUESTION [42 upvotes]: I learned most of my math font rendering from watching others (for example, I draw ζ terribly).  In most cases it is passable, but I'm often uncomfortable using fonts like Fraktur on the board.  Does anyone know good resources for learning to write these passably?

REPLY [2 votes]: Concerning the greek letters, it is essential to know the following: there does not exist in the greek language a cursive style (joined-up writing, schreibschrift). To my knowledge, at least. And has never existed. As a further point, the greek z should be written as a 3. It is essentially the same symbol, I should say.<|endoftext|>
TITLE: Which quadratic forms on $\Lambda^2 V$ come from quadratic forms on $V$?
QUESTION [15 upvotes]: Let $V$ be a finite dimensional vector space, say over $\mathbf R$.  Let $g \in S^2 V^*$ be a quadratic form on $V$.  Then $g$ induces a quadratic form $\Lambda^2 g \in S^2 \Lambda^2 V^*$ on $\Lambda^2 V$ defined on simple vectors by $$\Lambda^2 g(u\wedge v, x\wedge y) := \det \begin{bmatrix} g(u,x) & g(u,y) \\\\ g(v,x) & g(v,y) \end{bmatrix}.$$
Is there a nice sufficient and necessary condition for a quadratic form on $\Lambda^2 V$ to be induced from a quadratic form on $V$?

REPLY [4 votes]: Your $\Lambda^2g$ is curvature operator for hypersurface with second fundamental form $g$.
Thus your question can be reformulated the following way

Find a nice algebraic condition for curvature operators which appear as a curvature operators of a hypersurface<|endoftext|>
TITLE: What is convolution intuitively?
QUESTION [176 upvotes]: If random variable $X$ has a probability distribution of $f(x)$ and random variable $Y$ has a probability distribution $g(x)$ then $(f*g)(x)$, the convolution of $f$ and $g$, is the probability distribution of $X+Y$.  This is the only intuition I have for what convolution means.
Are there any other intuitive models for the process of convolution?

REPLY [3 votes]: Although I like @TerryTao's answer, I would phrase a little differently. If you view $f$ as just a function and $g$ as a weight function with integral $1$ centered at $0$, then the convolution $f*g$ at $x$ is the weighted average of $f$ centered at $x$. This gives a "softened" approximation to $f$. If you rescale $g$ to maintain its integral at $1$ but concentrate it more and more at the origin, then it's clear the convolution will converge in some sene to $f$ itself. That's why it is often called an approximation of the identity.
If $f$ is a function of time, then the convolution of $f$ by $g$ can be viewed as a continuous analogue of a moving average of a sequence of measurements over time (except that the moving average usually has a time lag because it is using only values from the past). Both the moving average and convolution are used to soften the spikes that occur in the original sequence or function.<|endoftext|>
TITLE: Beamer printout
QUESTION [10 upvotes]: I have just created a presentation using beamer, and I want the
"one" command at the top of the file that creates a printable version. It is true that I can recompile having searched for all the \pause commands and percent them out, but I remember there is an elegant way of doing this.
For the record, I am using the Singapore style, and the talk will not be of much interest to mathematicians.

(This has now been closed as "off-topic" with one of the reasons being the existence of TeX-SX.  People interested in this question should therefore consider the question Is there a nice way to compile a beamer presentation without the pauses? on the TeX-SX site.)

REPLY [10 votes]: The first answer is the basic correct answer, but there are variants.

Don't change the background colour (waste of ink), rather use pgfpages to put a border around each frame (this isn't one of the standard page-type declarations, but it isn't hard and I can make mine available if anyone wants it).
It's possible to change the type of the output (between beamer, handout, trans, or article) without modifying the file.  The trick is to put the main document in one file, say geometry.tex but without the documentclass declaration.  Then you create a new file for each type with just the documentclass declaration.  For example, geometry.beamer.tex could contain:
\documentclass[12pt,t,xcolor=dvipsnames,ignorenonframetext]{beamer}
\input{geometry.tex}

whilst geometry.handout.tex might contain
\documentclass[12pt,xcolor=dvipsname,ignorenonframetext,handout,%
 notes=only%
]{beamer}
\input{geometry.tex}

and geometry.article.tex might be
\documentclass[a4paper,10pt]{article}
\usepackage[envcountsect]{beamerarticle}
\setjobnamebeamerversion{geometry.beamer}
\input{geometry.tex}

Not only does this make sure that you are always compiling the correct version of the document, it also means that if you use a version control system then it doesn't keep complaining about you modifying the file just because you change the output type.
If you are strong in the ways of beamer and TeX, you can go one step further.  I use beamer for lectures which means that one single file contains the beamer versions and the handout versions of nearly 30 lectures.  To produce a given version of a given lecture, I need to have a way of telling TeX what I want.  I could have 60 separate files all with variations on the above, but I've found a simpler way is to have TeX examine the jobname to determine this.  Then I just have to have 60 symlinks to the main file (and I can create all 60 symlinks with a single zsh command).  That is, lecture.beamer.2009-11-19.tex is a symlink to lectures.tex and when I run LaTeX on it then I get tomorrow's lecture in beamer format (well, I would if I'd written it yet).  Again, I'd be happy to share the code for this if anyone's interested.


Addendum 2012-11-13  Unsurprisingly, an early question on TeX-SX was essentially exactly this one.  Also unsurprisingly, I answered there as well and as that place is more suited to TeX answers than this one I incorporated this answer there.  That answer can be found at https://tex.stackexchange.com/a/1426/86 which also contains links to my code for the above.<|endoftext|>
TITLE: What are tame and wild hereditary algebras?
QUESTION [12 upvotes]: What are tame and wild hereditary algebras?
Are they related to hereditary rings? (Those are rings for which every left (resp. right) ideal is projective, equivalently, for which every left (resp. right) submodule of a projective module is again projective).
Googling them I can see they seem related to the "tame representation type", but this concept is also new to me.
I would also like to know what is their relation to path algebras, since sometimes they appear mentioned together.

Do you know any good (newbie) references for all this? (Or can you elaborate in any of the questions?)

REPLY [27 votes]: An $k$-algebra $A$ is tame (or, equivalently, it has tame representation type) if, for every dimension $d\geq0$, you can parametrize all isoclasses of indecomposable $A$-modules of dimension $d$, apart from a finite number of them, by a finite number of $1$-parameter families. On the other hand, a finite dimensional $k$-algebra $A$ is wild (or, equivalently, it has wild representation type) if in the category $\mathrm{mod}_A$ of finite dimensional modules contains a copy of the category $\mathrm{mod}_{k\langle x,y\rangle}$ of modules over the free $k$-algebra on two generators. It is an amazing theorem of Drozd that a finite dimensional algebra is either tame or wild; this is the so called dichotomy theory. One of the reasons that make this theorem so amazing is that one can show that if $A$ is wild then $\mathrm{mod}_A$ contains copies of the module categories of all finite dimensional algebras; in other words, wild algebras are really wild...
In particular, this concepts of tame and wild apply to hereditary algebras, which are those of global dimension $1$. 
Now,  a finite dimensional hereditary algebra is Morita equivalent to the path algebra $kQ$ on a quiver without oriented cycles. A well-known theorem of Gabriel and others tells us that such a path algebra $kQ$ is tame iff the quiver $Q$ is, when you forget the orientation of the arrows, an Dynkin or an extended-Dynkin diagram. In all other cases the parth algebra is wild.
Two great references on all of this are the (first volume of the) book by Assem, Skowroński and Simson, or the book by Auslander, Reiten and Smalø.<|endoftext|>
TITLE: Do the signs in Puppe sequences matter?
QUESTION [31 upvotes]: A basic construction in homotopy is Puppe sequences.  Given a map $A \stackrel{f}{\to} X$, its homotopy cofiber is the map $X\to X/A=X \cup_f CA$ from $X$ to the mapping cone of $f$.  If we then take the cofiber again, something remarkable happens: $(X/A)/X$ is naturally homotopy equivalent to the suspension $\Sigma A$ of $A$.  This isn't hard to see geometrically; a nice picture and discussion can be found in pages 397-8 of Hatcher.  If we iterate this, we end up with a sequence $$A \to X \to X/A \to \Sigma A \to \Sigma X \to \Sigma(X/A) \to \Sigma^2 A \to \cdots$$
in which each map is the homotopy cofiber of the previous map.  If we then apply a functor which sends cofiber sequences to exact sequences, we get a long exact sequence.  This can be understood as the origin of long exact sequences of cofibrations in (co)homology, using the fact that $H^n(X)=H^{n+1}(\Sigma X)$.
One subtlety of this construction is that under the natural identifications of $(X/A)/X$ and $((X/A)/X)/(X/A)$ with $\Sigma A$ and $\Sigma X$, the map $\Sigma A\to\Sigma X$ is not the suspension of the original map $f$, but rather its negative (i.e., $-1\wedge f: S^1\wedge A=\Sigma A \to \Sigma X=S^1\wedge X$, where -1 is a map of degree -1).  The geometric explanation for this can neatly be seen in Hatcher's picture, where the cones are successively added on opposite sides, so the suspension dimensions are going in opposite directions.
However, you usually don't need to worry about this sign issue.  First, since a map of degree -1 is a self-homotopy equivalence (even homeomorphism) of $\Sigma A$, we could just change our identification of $(X/A)/X$ with $\Sigma A$ by such a map and then we would just have $\Sigma f:\Sigma A \to \Sigma X$ (note though that then we are not changing how we identify the next space in the sequence with $\Sigma X$, which breaks some of the symmetry of the picture).  Alternatively, if we only care about the Puppe sequence because of the long exact sequences it gives us, we could note that an exact sequence remains exact if you change the sign of one of its maps.
My question is: is there any situation where these signs really do matter and have interesting consequences?  Might they be somehow connected to the signs that show up in graded commutative objects in topology?

REPLY [19 votes]: There is a specific sort of situation I know about where that sign matters.  Suppose you have $f:X \rightarrow Y$ and $g:Z \rightarrow W$ cofibrations (if the maps are not cofibrations, all the same things work - you just replace the quotient spaces by mapping cones).  You extend both maps to their Dold-Puppe sequences, so you get the sequences
$X \rightarrow Y \rightarrow Y/X \rightarrow \Sigma X \rightarrow \Sigma Y \ldots$
and 
$Z \rightarrow W \rightarrow W/Z \rightarrow \Sigma Z \rightarrow \Sigma W \ldots$
Now suppose you have maps $a: X \rightarrow W$ and $b: Y \rightarrow W/Z$ making the obvious square commute up to homotopy.  You can then extend these to make a commutative ladder from the first Dold-Puppe sequence to the second.  (Notice that the sequences are deliberately offset from each other by one spot.)    
Using the usual parameters and the obvious choices of homotopies you will get a square involving $Y/X, \Sigma X, \Sigma Z, \Sigma W$.    This square will commute if it includes the map $-\Sigma g: \Sigma Z \rightarrow \Sigma W$, but not generally with the map $\Sigma g$.  (To check all this, I recommend doing the Dold-Puppe sequences with mapping cones rather than quotient spaces but keeping the homotopy equivalences with the quotient spaces in mind, which is the only way I know to calculate what the right maps should be.)
At this point, if you were feeling stubborn, you could replace the map in your ladder $\Sigma X \rightarrow \Sigma Z$ with $-1$ times that map, and that would allow you to have used $\Sigma g$ in the square I mention in the above paragraph, but that creates other issues;  if you choose not to simply use suspensions of your original maps to go from one Dold-Puppe sequence to the other then you run into problem when you are mapping between Dold-Puppe sequences without the shift of this example.
I hope this helps unravel Greg's answer (which is correct - you need the sign to get good mapping properties).
Of course one sees exactly the same phenomenon in the category of chain complexes of abelian groups (where homotopy is chain homotopy) and other such categories.  I agree with Theo and Mark that one thinks about the suspension as "odd" (in the sense of parity not the sense of peculiar).
The published paper that Mark refers to that has an error of exactly this sort (which is unfortunately fundamental to the paper) is by Lin Jinkun in Topology v. 29, no. 4, pp. 389-407.  I read this paper in preprint form in 1988 and missed this error, but discovered it in 1992 when reading another paper by the same author with the same error.  In the Topology paper the error is made in diagram 4.4 on the right hand square (proof of Lemma 4.3).<|endoftext|>
TITLE: Cohomology map induced by the group actions on homogeneous vector bundles
QUESTION [8 upvotes]: Here is a topological question which seems quite elementary. The answer to this question may be useful e.g. in estimating the orders of the automorphism groups of some algebraic varieties and in computing the cohomology of some moduli spaces.
Let $K$ be a compact Lie group, $T$ a maximal torus of $K$ and $V$ a complex vector space of dimension $d$. (If one wishes, instead of $K$ and $T$ one can consider a complex reductive group and its Borel subgroup.) From a representation $R:T\to GL(V)$ we can construct a homogeneous vector bundle with total space
$$E=K\times_T V.$$
(I.e., we identify $(kt,v)$ with $(k,R(t)v$ for all $k\in K, t\in T, v\in V$.)
Set $E_0$ to be $E$ minus the zero section. This space is fibered over $K/T$ with fiber $V$ minus the origin.
Suppose the Euler (=top Chern) class of the vector bundle is zero. Then $H^{2d-1}(E_0,\mathbf{Z})\cong \mathbf{Z}$, because $K/T$ has only even-dimensional cells. Take a generator $a$ of $H^{2d-1}(E_0,\mathbf{Z})$. We can identify the integral cohomology of $E_0$ with $H^*(K/T,\mathbf{Z})\otimes\Lambda(a)$ where $\Lambda$ stands for exterior $\mathbf{Z}$-algebra.
We have the natural group action map $K\times E_0\to E_0$. The cohomology of $E_0$ has no torsion, so we can identify $H^*(K\times E_0,\mathbf{Z})$ with
$H^*(K,\mathbf{Z})\otimes H^*(E_0,\mathbf{Z})$ using the projections. The latter can be written as $$H^*(K,\mathbf{Z})\otimes H^*(K/T,\mathbf{Z})\otimes\Lambda(a).$$
The pullback of $a$ under the action map is $1\otimes a+x\otimes 1$ for some $x\in H^{*}(K,\mathbf{Z})\otimes H^{*}(K/T,\mathbf{Z})$. Question: find $x$. The weights of the representation are assumed to be known; for simplicity one can assume that $K=U(n)$ or $SU(n)$.
For real cohomology I know how to reduce the problem to linear algebra using Lie algebra cohomology. But the result I am able to get in this way is not very illuminating. And moreover, I have no idea how to extract the integral structure out of it.
For projectivized (and not spherized) bundles, the question becomes trivial. But this does not seem to help.

REPLY [5 votes]: I'll answer in the case of K = U(n) because I'm less familiar with the cohomology of the other compact Lie groups.  Let BK and BT be the classifying spaces.
The representation of T on V gives rise to a vector bundle on BT with total space F, and complement of the zero section F0.  There is an Euler class c in H*(BT).  As H*(BT) is a polynomial algebra in these cases, there are basically two cases: either the Euler class is zero, or it is not a zero divisor.  There is a map from K/T to BT which is, up to homotopy, the "quotient" by the action of K.
In the first case, you will actually find that your element 'a' in H*(E0) is pulled back from H*(F0), where the action of K is trivial.
In the second case, you have an exact sequence (the Gysin sequence)
$$
0 \to H^\*(BT) {\mathop\to^c} H^\*(BT) \to H^\*(F_0) \to 0
$$
of modules over H*(BK), which is actually a free resolution.
I find it easier to describe the "coaction" you want in terms of an action on homology.  The space E0 is the pullback of F0 along the map from K/T to BT.  An Eilenberg-Moore argument finds that there are isomorphisms
$$
H\_\*(K) \cong Ext_{H^\*(BK)} (\mathbb Z, \mathbb Z)
$$
and
$$
H\_*(E\_0) \cong Ext\_{H^\*(BK)} (H^\*(F\_0), \mathbb Z).
$$
(I have to be a little careful about the gradings; these Ext-groups are bigraded according to a grading on the ring and a grading on the Ext-group, and the elements in Ext^s actually have their grading shifted down by s.)
The coaction on homology you describe is actually just the Yoneda pairing on Ext.  For a specific choice of Euler class this is something that you can work out by writing down a resolution of Z over H*(BK) and using that to find generators for your Ext-algebra.
I don't know if an example would be more helpful.<|endoftext|>
TITLE: A rigid type of structure that can be put on every set?
QUESTION [16 upvotes]: Call a type of structure rigid if any automorphism of such a structure is an identity.  (This is a bit different from some other uses of the word, but hopefully I'll be forgiven.)  For example, well-orderings are rigid.  It follows that assuming the axiom of choice, there is a rigid type of structure (namely, a well-ordering) such that any set can be equipped with that type of structure (in a non-unique way, of course).
Now the axiom of choice isn't necessary for that conclusion.  Extensional well-founded relations are also rigid, and the axiom of foundation implies that any set injects into an extensional well-founded relation (its transitive closure).  Aczel's axiom of anti-foundation also suffices, since strongly extensional relations are also rigid, and anti-foundation implies that any set injects into such a relation.  But with neither foundation nor anti-foundation, the membership relation $\in$ needn't be rigid.  For instance, the set {a,b}, where a={a} and b={b} are unequal ill-founded sets with the same membership tree, has a nonidentity $\in$-automorphism which swaps a and b.
Now my question: If we don't assume choice or any sort of foundation, does there still exist a rigid type of structure with the property that any set can be equipped with that type of structure?
Edit: Of course, as Steven points out in the comments, I haven't said exactly what I mean by "type of structure."  I'm using the word "structure" in the Bourbaki-sense, not in the sense of stuff, structure, property.  Here's one way to make this question precise: does there exist a theory in higher-order logic which is rigid, in the sense that any automorphism of one of its models is the identity, and which admits a definable functor to Set which is essentially surjective?

REPLY [3 votes]: This doesn't completely answer the question, but I think it's relevant.
Theorem: Let $T$ and $W$ be theories in higher-order logic (by which I mean the internal type theory associated to elementary topoi), where $W$ has a specified underlying object $X$ (i.e. we regard a model of $W$ as an object $X$ equipped with some additional stuff).  Then it is not the case that both (1) $T$ proves that $W$ is rigid (as structure on $X$) and (2) in any topos satisfying $T$, every object can be equipped with W-structure.
Proof: Suppose that both the given conditions hold.  Let $C[T,X]$ be the free topos on the theory $T$ and an additional object $X$, and let C[T,W] be the free topos on $T$ and $W$.  (Recall that for any theory $S$ and any topos $E$, the category $\operatorname{Log}(C[S],E)$ of logical functors and natural isomorphisms from $C[S]$ to $E$ is equivalent to the category of models  of $S$ and their isomorphisms in $E$.)  Now the generic model of $W$ in $C[T,W]$ has an underlying object, giving a logical functor $C[T,X]\to C[T,W]$.  And $C[T,X]$ satisfies $T$, so by assumption, its generic object $X$ can be equipped with $W$-structure; thus we also have a logical functor $C[T,W]\to C[T,X]$, and the composite $C[T,X]\to C[T,W]\to C[T,X]$ is isomorphic to the identity.  In other words, $C[T,X]$ is a pseudo-retract of $C[T,W]$.  Now let $E$ be any other topos satisfying $T$ and $Y$ any object of $E$ admitting a nonidentity automorphism (such as $Y=1+1$); then we have a logical functor $C[T,X]\to E$ sending $X$ to $Y$, which in turn has a nonidentity automorphism.  But because $C[T,X]\to C[T,W]\to C[T,X]$ is the identity, this functor $C[T,X]\to E$ must factor through $C[T,W]$, and since $T$ proves that $W$ is rigid, any automorphism of this functor must be the identity, a contradiction. ∎
This means that if you want a rigid type of structure that can be put on every set, you need to use more about sets than the fact that they form an elementary topos, or even a Boolean topos with a NNO, or any additional property of $\mathsf{Set}$ that can be expressed as a theory $T$ in HOL.  Note that AC, although it can be phrased as a "categorical" property not referring directly to $\in$ (every surjection splits), cannot be expressed as a theory in HOL since it involves a quantification over all sets.  Likewise for the "topos-theoretic axiom of foundation" that every set injects into some well-founded extensional relation.  So the question is, what can we do with the remaining non-HOL axioms of ZF, i.e. basically replacement and (its consequence) unbounded separation.<|endoftext|>
TITLE: What's so great about blackboards?
QUESTION [40 upvotes]: Many mathematicians seem to think that the only way to give a mathematics talk is by using chalk on a blackboard.  To some, even using a whiteboard is heresy.  And we Don't Talk About Computers.
I'd like to know people's opinions on this.  In a - probably useless - attempt to forestall a flame war, let me try to narrow down the question a little.

I'm interested primarily in the effect that the medium has on the talk.  This will make comparisons difficult since saying "X gave a rubbish talk using a computer" leaves open the possibility that X would have given an even worse talk using a blackboard.  So please try to analyse why you have a preferred medium for presentations.  This isn't meant to be a collection of tips for making good presentations (though I can see the value in that).
It may be just my impression, but it seems that this attitude is peculiar to mathematicians.  Is this true, in your opinion?  Is there something special about mathematics that makes it so right for blackboard talks?
I'm interested in opinions both as speakers and listeners.  Your answers are allowed to be different from the different perspectives!
It may be that the type of talk has a bearing on the medium.  Try to take this into account as well.
I'm possibly completely out of date on this - maybe everyone has now fully Embraced Their Inner Beamer.
Please be nice.  By all means, praise good behaviour ("X gave a really nice talk using a combination of rope, blancmange, and a small aubergine; it wouldn't have worked so well with only chalk") but let bad behaviour rot in the dungeon of obscurity.

REPLY [2 votes]: Many excellent points are made in the above responses.  I suggest another point of view that may be important. It might be that at some point you will work in industry.  I've found that PowerPoint is often the preferred mode of presentation.  It is useful for everyone to learn how to use this medium if only because it may facilitate working with certain groups - industry.
For myself, I like the flexibility and simplicity of a blackboard.
Suggestion for whiteboard users - if you find a "dry" pen, throw it away!  All to often, I see someone discover that a pen doesn't write, put the cap back on, and return it to the tray.  Tossing it out would be helpful.
Final comments: 
Edward R Tufte 
Dr. Tufte, in my opinion, dislikes any PowerPoint presentation. The information desensity is just too restricted.  See his web site www.edwardtufte.com/tufte
Gapminder
There are some amazing examples of what can be done with a flexible electronic media in presentating data. See www.gapminder.org<|endoftext|>
TITLE: Beamer hints and tips
QUESTION [7 upvotes]: I deleted a rant from this question because I felt it detracted from the given answer to the specific question.  However, beamer is the "new kid on the block" in terms of giving talks (not that new!) and there are many little hints, tips, and howtos that people have found that can make the difference between a nice presentation and a smooth presentation - or make the difference between a downright annoying presentation and a nice presentation!
With the general idea that good communication is 90% of good mathematics, and also with a little bit of self-interest (I might learn something, and if not then hopefully everyone else's beamer talks will improve to the point that I can sit through them without wincing), I'd like to know what those hints and tips are.
These can be both technical (how to get the pauses right after an itemize/enumerate environment) and non-technical (the deleted rant was about not using loads of pauses just because you can).
Please be nice: say "I like it when the up-coming text is greyed out rather than invisible because I like to get a sense of where the talk's going" rather than "It's horrible when you can't read ahead because when the speaker is rambling then there's no way to work out what they're talking about".
Also, inevitably, there will be non-beamer-related stuff.  That's fine.

REPLY [15 votes]: I think using Beamer well is hard.  Even just getting the layout right is hard - I spend hours moving things around a bit so that the layout aids comprehension.  I agree that one should not insert pauses "just because one can", but I think one should insert pauses because it helps people understand what is going on, as if one were writing it gradually on a blackboard.  
My top tips would be:

Use pstricks for diagrams, because then you can increase the thickness of the arrows, and pause in the diagram so that it can be built up gradually.  There is some kind of myth out there that you can't go through postscript with beamer; this is untrue.
Keep a running frame number discreetly (eg bottom right) so that you know where you are in terms of timing.  However do not show how many slides there are in total, so that if you run out of time you can abort your talk without anyone knowing.
Keep the section heading at the top of every slide so that the audience knows where they are.  Apart from this and the page numbers I remove everything else like Tom does, but I disagree with the total blankness approach.
Read out every word on your slide. If you don't, the audience will do it in their heads anyway, and won't listen to you while they're doing it.  This is also why pauses are good (otherwise the audience will read to the end of the slide and not listen to you), and having the rest of the text in light grey before it has officially appeared is something I find extremely counter-productive.
I personally hate sans serif font for maths presentations but I seem to be the only person on earth who does.  I find it extremely difficult to read from far away.
I prefer to keep the lights on the audience. It might make your slides more legible if you dim them, but it will make your audience fall asleep, and will mean they can't write their own notes or do their own work when they're bored, which is just cruel.  Also, if you dim the lights you won't be able to see them, so you won't be able to communicate with them.
Personally I think the standard font size on beamer is too small, and I take it up by a point or two.   
I agree with Tim Gowers' point about copying material you need again instead of flicking back through the beamer.
Purchase a Kensington remote page turning thingy with laser pointer.  It's not that expensive (under 30 quid) and works extremely well.  It's better than relying on your hosts to provide one, and we've all been to talks where the speaker can't work the remote page turning thingy.  So it's better to have your own, which you know how to operate.
I have never read the manual.  I assume I can do anything I want, and look it up when I need to.  The manual is epic and I would probably reach retirement before making it through.  
I try to avoid any text that has to run over one line; I think that indicates when my sentence is too long for a presentation.  The one exception is if I'm saying something informal, like a slogan, idea, or moral, in which case I put it in a box.
My top tip for actually writing the beamer is to plan all the slides first on paper, then type in the material without any thought about layout, and then go through and sort the layout out separately.  These three stages require different types of brain function and it's hard to think about them at the same time.
I love bezier curves for pointing at things on slides.  I usually put them in purple.<|endoftext|>
TITLE: Is it best to run or walk in the rain?
QUESTION [49 upvotes]: According to the Norwegian meterological institute, the answer is that it is best to run.  According to Mythbusters (quoted in the comments to that article), the answer is that it is best to walk.
My guess would be that this is something that can be properly modelled mathematically and solved.  I suspect that I'm not alone in making a start on this, but as it's usually during a rainstorm that I think of it, I don't get far before deciding that the best strategy is to call a taxi!
I would also guess that someone has already done this and figured out the answer.  If I'm right about that, I would like to read about it and learn what the real answer is.  Thus my question is most likely: can someone provide a reference for this?
(If I'm wrong that it's already been solved, then some pointers as to how to solve it or even an explanation of what the main issues are, would be most interesting.)
As well as simply satisfying my curiosity, this is also one of those questions that non-mathematicians often ask (or at least, that non-mathematicians can easily understand the motivation for the question) so if there's a reasonable answer then it would be a good way to show off some mathematics.

Update This question has a long and curious history.  There's a discussion about it at the meta site: http://mathoverflow.tqft.net/discussion/90/question-about-walking-in-the-rain

REPLY [40 votes]: try this, the latest in a long line of recreational mathematics on the topic
"Keeping Dry: The Mathematics of Running in the Rain"
Dan Kalman and Bruce Torrence,
Mathematics Magazine, Volume 82, Number 4 (2009) 266-277<|endoftext|>
TITLE: What is a metric space?
QUESTION [42 upvotes]: According to categorical lore, objects in a category are just a way of separating morphisms.  The objects themselves are considered slightly disparagingly.  In particular, if I can't distinguish between two objects by using morphisms, then I should consider them equivalent (not equal, that would be evil).
In this view, then, metric spaces with continuous functions are just plain wrong.  The category of metric spaces is equivalent to the full subcategory of topological spaces consisting of metrisable spaces.  Choosing a metric is evil.
So what's the right view of metric spaces, that makes the metric both worth having and not (so much of) an arbitrary choice?
I put the "so much of" there because the obvious answer is having isometries as morphisms, but then the category becomes too rigid to be of any conceivable use.  So what's the best middle ground?

REPLY [2 votes]: An interesting class of morphisms that sits between isometric embeddings and bi-Lipschitz maps (both the map and its inverse are Lipschitz) is the class of almost isometric embeddings.  There are two natural definitions of this.  Let $X$ and $Y$ be metric spaces.
(1) An almost isometric embedding from $X$ into $Y$ is a sequence $(f_n)_{n\in\mathbb N}$ of maps $f_n:X\to Y$ such that for some sequence $(\varepsilon_n)_{n\in\mathbb N}$ of positive reals converging to zero, for every $n$, $f_n$ is bi-Lipschitz of constant 
$\varepsilon_n$.
The composition of two almost isometric embeddings $(f_n)_{n\in\mathbb N}$ and $(g_n)_{n\in\mathbb N}$ (of appropriate domains and ranges) is $(f_n\circ g_n)_{n\in\mathbb N}$.
(2) If you don't like the fact that an almost isometric embedding as defined above does not have a fixed range in the classical sense, change the definition by requiring that
each $f_n$ has the same range.  
$X$ and $Y$ are almost isometric if there is an almost isometric embedding as in (2) from $X$ onto $Y$ (all $f_n$ are onto $Y$).
For example, any two countable dense subsets of $\mathbb R$ are almost isometric (back and forth argument), but not necessarily isometric, simply because there are $|\mathbb R|$ many possible distances but only countably many of these are realized in a fixed countable set.<|endoftext|>
TITLE: References for equivariant K-theory
QUESTION [14 upvotes]: I want a good introduction to localization in equivariant $K$-theory. This introduction can be simple in several ways:

I only care about torus actions.
I only care about $K^0$.
I only care about very nice spaces. I would be fine if the only spaces considered were $G/P$'s and smooth projective toric varieties.

However, I want this exposition to include the following:

How to compute a $K$-class from a Hilbert series.
Given $X \to Y$ nice spaces, how describe push back and pull forward in terms of the map $X^T \to Y^T$.
Ideally, this reference would also give the generators and relations presentation of $K$-theory for the sort of examples mentioned above. I mean formulas like
$$K^0(\mathbb{P}^{n-1}) = K^0(\mathrm{pt})[t, t^{-1}]/(t-\chi\_1)(t-\chi\_2) ... (t - \chi_n)$$
where $\chi_i$ are the characters of the torus action. But I can find other references for this sort of thing.

The best reference I currently know is the appendix to Knutson-Rosu.

REPLY [11 votes]: I spent the last few days1 reading references on equivariant K-theory. I just pulled together the following bibliography for my collaborators and I can't see a reason not to make it public. This list tilts heavily towards combinatorics and classical algebraic geometry.
My general conclusion is that there are several good references on equivariant K-theory localization, but they are all too dense for someone who has never seen this stuff before. Fortunately, there are good, slow, introductions to equivariant cohomology, and the two fields are similar enough that you can get the motivation and examples from the cohomology papers and then dive into the K-theory tomes.
This post is Community Wiki, so feel free to add your own favorites.
Short papers with lots of nice examples
1:      Julianna Tymoczko's introduction to equivariant cohomology, An introduction to equivariant cohomology and homology, following Goresky, Kottwitz, and MacPherson
2:      Early parts of Knutson-Tao's paper Puzzles and (equivariant) cohomology of Grassmannians, on the equivariant cohomology of the Grassmannian
More detailed references on equivariant cohomology:
3:      Fulton's notes (since superceded by the book Equivariant Cohomology in Algebraic Geometry). See lectures 4 and 5 for localization, lectures 6 - 10 for special features of homogeneous spaces.
4:      Section 2 of Guillemin and Zara's first paper, Equivariant de Rham Theory and Graphs, explaining which parts of the story are pure combinatorics.
References specifically for K-theory
5:      Chapters 5 and 6 of Complex Geometry and Representation Theory, by Chriss and Ginzburg. Lots and lots of detail, and focuses on flag varieties in particular. One of the only places I've found that describes how to pushforward to something other than a point.
6:      Guillemin and Zara's K-theory paper G-actions on graphs. This is similar to the previous paper of theirs that I cite, but terser and for K-theory.
7: A paper of Nielsen, Diagonalizably linearized coherent sheaves, which seems to have anticipated a lot of the results in this field, and has some nice examples at the end. (Note: The MathSciNet review at the link is in French, but the paper is in English.)
8:      Knutson and Rosu's paper Equivariant K-theory and Equivariant Cohomology, establishing localization in equivariant K-theory and explaining equivariant Grothendieck-Riemann-Roch.
William Fulton also tells me that he and Graham are working on an expository article, which I expect will be excellent.
1 For those who are curious, I am not learning this material for the first time. Rather, I originally learned it by skimming a lot of papers, going to talks and asking Allen Knutson about anything that confused me. I am now playing the role of Allen to others, so I need to learn the subject better.<|endoftext|>
TITLE: Highly transitive groups (without assuming the classification of finite simple groups)
QUESTION [25 upvotes]: What is known about the classification of n-transitive group actions for n large without using the classification of finite simple groups?  With the classification of finite simple groups a complete list of all 2-transitive group actions is known, in particular there are no 6-transitive groups other than the symmetric groups and the alternating groups.  I want something like "there are no interesting n-transitive group actions for n sufficiently large" but without the classification theorem (however, I'd be happy if the n in that statement was obscenely large).  Even any partial (but unconditional) results would interest me (like any n-transitive group for n sufficiently large needs to have properties X, Y, and Z).

REPLY [15 votes]: There is a classical result of Wielandt that if you assume the Schreier conjecture (that the outer automorphism group of an finite nonabelian simple groups is solvable), then a group of degree n other that A_n or S_n is at most 7-transitive. Unfortunately, the only known proof of the Schreier conjecture uses the classification of finite simple groups.<|endoftext|>
TITLE: Number theoretic spectral properties of random graphs
QUESTION [16 upvotes]: If G is a graph then its adjacency matrix has a distinguished Peron-Frobenius eigenvalue x.  Consider the field Q(x).  I'd like a result that says that if G is a "random graph" then the Galois group of Q(x) is "large" with high probability.  (Well, what I really want is just that the Galois group is non-abelian, but I'm guessing that for a typical graph it will be the full symmetric group.)
Here's another version of this question.  Take a graph with a marked point and start adding a really long tail coming into that point.  This gives a sequence of graphs G_n.  If G is sufficiently complicated (i.e. you're not building the A or D type Dynkin diagrams as the G_n) is Gal(Q(x_n)/Q) symmetric for large enough n?
The reason behind these questions is that fusion graphs of fusion categories always have cyclotomic Peron-Frobenius eigenvalue.  So results along these lines would say things like "random graphs aren't fusion graphs" or "fusion graphs don't come in infinite families."  So the particular details of these questions aren't what's important, really any results on the number theory of the Peron-Frobenius eigenvalue of graphs would be of interest.

REPLY [19 votes]: Just in case anyone else is still thinking about this question...
The answer is the following. Either:

The eigenvalues of $G_n$ are all of the form $\zeta + \zeta^{-1}$ for roots of unity $\zeta$, and the graphs $G_n$ are subgraphs of the Dynkin diagrams
$A_n$ or $D_n$.
For sufficiently large $n$, the largest eigenvalue $\lambda$ is greater than two, and
$\mathbf{Q}(\lambda^2)$ is not abelian.

The proof is effective, but a little long to post here. The main ingredients are some basic facts about Weil height, some ideas due to Cassels, and an amplification step using Chebyshev polynomials.

The paper is now on the ArXiv.  See the last section for the proof of this result, and the second to last section for a more effective but logically weaker result.<|endoftext|>
TITLE: "Positive systems" in n * the (n-1)-simplex
QUESTION [11 upvotes]: Let S := the nonnegative integer solutions to {$a_1 + ... + a_n = n$},
and center := (1,1,1,...,1).
Call a vector v generic if v.s = v.center <=> s = center.
Then each generic v defines a positive system in S,
the subset { s in S : v.s > v.center }.
Already at n=3 it is possible for one positive system to contain another.
Up to permutation, we may as well take v strictly decreasing.
Having done so, is there a reasonable way to classify 
the maximal positive systems?

REPLY [2 votes]: If you replace S by the lattice points in the positive orthant, and forget the condition that the hyperplane passes through a particular point, and require that the hyperplane only cuts off finitely many points, then you are looking at things called "corner cuts" by Onn and Sturmfels in Cutting corners.
Anyway, I'm not sure if this will help, but maybe it'll suggest something.<|endoftext|>
TITLE: "The" random tree
QUESTION [20 upvotes]: One time I heard a talk about "the" random tree. This tree has one vertex for each natural number, and the edges are constructed probabilistically. Connect vertex $2$ to vertex $1$. Connect vertex $3$ to vertex $1$ or $2$ with probability $\frac{1}{2}$. Connect vertex $n+1$ to exactly one of vertices $1,\dots, n$ with equal probability $\frac{1}{n}$. This procedure will construct an infinite tree. The theorem is that with probability $1$, any tree constructed this way will be the same (up to permutation of the vertices). 
My question is, does anyone know of a reference for this result? What is the automorphism group of this tree? Can anyone draw a picture of it? 
I don't have any reason for knowing about this, just curiosity, and I wasn't able to turn up anything with a (not too extensive) internet/mathscinet search.

REPLY [3 votes]: The algorithm you describe is at each stage isomorphic to an algorithm for picking a particular tree from the "Uniform Spanning Tree" (which is THE uniform distribution of all possible spanning trees) when applied to any finite fully connected graph.
One standard method is to iterate choosing new points at random (uniformly from the unconnected ones) and performing a "Loop Erased Random Walk" until "hitting" the existing tree.
The isomorphism of algorithms is because such a fractal growth process is uniform so we can modify it at each LERW by removing the irrelevant intermediate points.  This is just shrinking the graph by removing nodes on a line without changing "connected-ness".
The large-scale symmetry under permutations is due to the "uniform probability" property of all finite sub-trees.
Bit of a hand-wavy argument so far I know but UST is the way to go.  A fully connected graph is a simpler starting point than the 50% connected Rado graph but ..
Corollary ..
I imagine any graph with a bounded below probability of connected-ness must almost surely have "THE random tree" as a sub-graph as n grows.  Average density of connections of "THE random tree" is $\frac2{n-1}$ and expected neighbours of $k$'th point is $\sum_{i=k+1}^n\frac1i$, so by permuting a large sub-graph of Rado to order the most connected points earliest we can comfortably exceed this everywhere and select at least one tree with probability approaching $1$ as $n\rightarrow\infty$.<|endoftext|>
TITLE: Factoring maps of handlebodies
QUESTION [9 upvotes]: Any map of finite graphs (1-dimensional CW-complexes) factors as a composition of

a finite sequence of folds;
an inclusion; and
a finite-to-one covering map.

There should be a corresponding result for handlebodies, which presumably should say that, after a homotopy, a continuous map of handlebodies factors as:

a compression (by which I mean a map of a handle into the complement of its interior);
an inclusion; and
a finite-to-one covering map.

Is my intuition correct, and does anyone have a reference?  I'm specifically interested in how well-behaved the homotopy can be taken to be.  For instance, can it be made to respect the boundary?
Notes
A fold is a map that identifies two edges with a common endpoint.  Many folds don't change the homotopy type of a graph, and one would expect not to need these in the handlebody setting.  The important folds are the ones that kill a loop.  In handlebody terms, you can think of this as gluing in a two-handle, or as cutting a one-handle - hence my use of the word "compression".  Is this word acceptable in this context?
The graph-theoretic result is due to Stallings.
By an inclusion of handlebodies, I mean that the new one should be obtained from the old by attaching 1-handles.
EDIT (prompted by Sam's comments below) I'm not quite sure what "respect the boundary" should mean, at this point.  Suggestions welcome!

REPLY [4 votes]: Suppose that we are given a PL map from a handlebody $W$ to a handlebody
$V$.  Choose a spine for $W$.  Homotope the map until the image is a
regular neighborhood of the image of the spine.  By general position,
our map is now an embedding.  Fix a pants decomposition of disks $D =
(D_i)$, for $V$.  Suppose that $P$ is a component of $V - D$ (so $P$ is a
three-ball with three distinguished disks on its boundary).  Consider a 
component $X$ of $f(W)
\cap P$.  This is essentially a knotted graph.  Via a homotopy (keeping
$X \cap D$ fixed) unknot $X$.  If the rank of $X$ is positive then another 
homotopy produces small lollipops which we shall compress a bit later.
If $X$ meets any disk $D_i \subset \partial P$ more than once 
then we may homotope a leg of
$X$ through $D_i$.  Let $Y$ be the resulting component of $f(W) \cap
P$. (Note that this reduces $f(W) \cap D$.)
Homotoping in this fashion we eventually arrive at an embedding of $W$
so that every component in every three-ball of $V - D$ is either a
tripod or an interval, possibly with lollipops attached.  The feet of
the tripod/interval lie in distinct disks in the boundary of the
containing solid pants. 
Now compress all of the lollipops to get $f'(W')$ (a new handlebody,
because we compressed and a new map because we have to extend it over
the two-handles we added).  
EDIT:
This reproduces, in our context, part of Stallings paper (eg sliding 
the leg is a fold, arriving at only tripods and intervals produces 
an immersion.)  
Since $f'$ is an immersion, it follows from Stallings paper that $f'$ 
is $\pi_1$ injective and that $W'$ embeds into a finite cover of $V$.<|endoftext|>
TITLE: What's an efficient way to calculate covariance for a large data set?
QUESTION [8 upvotes]: What is the best algorithm for computing covariance that would be accurate for a large number of values like 100,000 or more?

REPLY [2 votes]: A single pass stable algorithm has been discovered in the time since this question has been originally answered:
Bennett, Janine, et al. "Numerically stable, single-pass, parallel statistics algorithms." Cluster Computing and Workshops, 2009. CLUSTER'09. IEEE International Conference on. IEEE, 2009.
An implementation is given in Boost.<|endoftext|>
TITLE: Calculating the "Most Helpful" review
QUESTION [18 upvotes]: How would you calculate the order of a list of reviews sorted by "Most Helpful" to "Least Helpful"?
Here's an example inspired by product reviews on Amazon:
Say a product has 8 total reviews and they are sorted by "Most Helpful" to "Least Helpful" based on the part that says "x of y people found this review helpful".
Here is how the reviews are sorted starting with "Most Helpful" and ending with "Least Helpful":
7 of 7
21 of 26
9 of 10
6 of 6
8 of 9
5 of 5
7 of 8
12 of 15
What equation do I need to use to calculate this sort order correctly?  I thought I had it a few times but the "7 of 7" and "6 of 6" and "5 of 5" always throw me off.  What am I missing?

REPLY [7 votes]: David Eppstein suggests a Bayesian method in his comment.  One standard thing to do in this situation is to use a uniform prior.  That is, before assessments of a review come in, its probability $p_i$ of being helpful is assumed to be uniformly distributed on [0, 1].  Upon receiving each assessment of the review, apply Bayes' theorem.
This sounds complicated, and it would be for an arbitrary prior distribution.  But it turns out that with the uniform prior, the posterior distributions are all beta distributions.  In particular, the expected value of $p_i$ after s positive assessments and n-s negative ones is (s+1)/(n+2).  This is Laplace's rule of succession, and proofs of the facts I've mentioned can be found in that Wikipedia article.  Then one would sort on the score (s+1)/(n+2).
The constants "1" and "2" come from the use of a uniform prior, and don't actually give the same results as the sample data you provide.  But if you give a review that s out of n people have said to be helpful the score (s+3)/(n+6), then your reviews have scores
7 of 7: 10/13 = 0.769...
21 of 26: 24/32 = 0.75
9 of 10: 12/16 = 0.75
6 of 6: 9/12 = 0.75
8 of 9: 11/15 = 0.733
5 of 5: 8/11 = 0.727
7 of 8: 10/14 = 0.714
12 of 15: 15/21 = 0.714
This essentially amounts to sorting by the proportion of positive assessments of each review, except that each review starts with some "imaginary" assessments, three positive and three negative.   (I don't claim that (3,6) is the only pair of constants that reproduce the order you give; they're just the first pair I found, and in fact (3k, 4k+2) works for any $k \ge 1$.)<|endoftext|>
TITLE: Inverses in convolution algebras
QUESTION [6 upvotes]: Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant.  Let $C_c(G)$ be the space of locally constant complex functions on $G$ with compact support, which forms an algebra under convolution.  Suppose $e \in C_c(G)$ is an idempotent, so that $H = eC_c(G)e$ is an algebra with identity.  Is it now true that if $f \star g = e$ in $H$, then $g \star f = e$ as well?
This may be too general to be true, so to be more specific: suppose $K < G$ is a compact subgroup such that $K\backslash G/K$ is countable, suppose $\phi : G \to \mathbf{C}^{\times}$ is a character, and suppose the idempotent $e$ is the function with support in $K$ such that $e(x) = \phi(x^{-1})/\mu(K)$ for $x \in K$.  Now is it true that if $f \star g = e$ in $H$ then $g \star f = e$ as well?
[I am reading something that claims some $f$ is a unit but then checks it by checking the existence of $g$ such that $f \star g = e$.  So really, the question is whether it follows by some general business that $g \star f = e$, or whether one has to do another computation to check the other direction.]

REPLY [4 votes]: I don't have a solution, but here are some thoughts which might be of use or interest.
You may have seen this already, but if your group is discrete then its group von Neumann algebra $VN(G)$ is "directly finite" - that is, every left invertible element is invertible. I think this property is inherited by the algebra obtained when one compresses by an idempotent in $C_c(G)$.
The earliest reference I know of is somewhere in Kaplansky's Fields and Rings; a proof of something slightly weaker, which can in fact be boosted to prove the original result, was given in
Montgomery, M. Susan. Left and right inverses in group algebras. Bull. Amer. Math. Soc.  75  1969 539--540. MR0238967 (39 #327)
(The proof uses the existence of a faithful tracial state on $VN(G)$, plus the fact that every idempotent in a $C^*$-algebra is similar in the algebra to a self-adjoint idempotent -- something which was not all that obvious to me the first time I saw this result.)
I don't know what the state of play is for algebras of the form $H$, as described in your question. I think enough is known about $C^*$-algebras of some totally disconnected groups (work of Plymen et al.) that one might have similar results, but the arguments have to be different from the discrete case because one no longer has the faithful positive trace that is used by Kaplansky and Montgomery's arguments.
Of course in the more special case you have at hand, we might have enough structure to force left-invertibles to be right-invertible; but off the top of my head nothing comes to mind.<|endoftext|>
TITLE: Formal geometry
QUESTION [24 upvotes]: [Edit (June 20, 2010): I posted an answer to this which summarizes one that I received verbally a few weeks after posting this question.  I hope it is useful to someone.]
I am presently seeking references which introduce "formal geometry".  So far as I can tell, this idea was presented by I.M. Gel'fand at the ICM in Nice in 1970.  There is his lecture, a paper by him and Fuks, and also a paper by Bernshtein and Rozenfeld with some applications that I don't understand too well.  What I am unable to find is a thorough exposition of the foundations.  It seems like a canonical enough construction that it should have been included in some later textbook, (though apparently not called "formal geometry" since that is not turning up anything useful).
Below is what I understand, which is several main ideas, but missing many details; this will most certainly be riddled with errors, because I am only able to give what I have roughly figured out from reading incomplete (though well-written and interesting!) sources, and asking questions.  I am including it in the hopes it will be familiar to some kind reader.

I'm sorry to not ask a specific question.  Hopefully some answers will help me edit the below description to remove inaccuracies, and some others will suggest references.  Both would be very helpful.

Let $X$ be a smooth complex algebraic variety of dimension $n$ (could just as well be a complex analytic or smooth real manifold so far as I understand; probably can be algebraic over any field, at least for awhile).  There is a completely general torsor over $X$: its fiber over a point $x$ is the set of all coordinate charts on the formal neighborhood of $x$ in $X$.  This is a torsor over the infinite dimensional group G of algebra endomorphisms of $\mathbb{C}[\![x_1,...,x_n]\!]$ which preserve the augmentation ideal and are invertible modulo quadratic terms (and hence invertible over power series of endomorphisms).  It's a torsor because any two coordinate systems are related by such an endomorphism, but there isn't a canonical choice of coordinate system along the variety.
I think one can rephrase the conditions of the previous paragraph more precisely by first noting that an endomorphism of $\mathbb{C}[\![V]\!]$ preserving augmentation ideal (where we use notation $V=\operatorname{span}_\mathbb{C}(\{x_1,...,x_n\})$ is given by a linear map $V\to V\ast \mathbb{C}[\![V]\!]$, which then uniquely extends to an algebra map.  Then the condition of the last paragraph is that
$$V \to V\ast \mathbb{C}[\![V]\!]\to V\ast \mathbb{C}[\![V]\!] / V\ast V\ast \mathbb{C}[\![V]\!] = V$$ is invertible.
It's not hard to see that these in fact form a group, and that this group acts simply transitively on the set of coordinate systems.
The Lie algebra $\mathfrak{g}$ of $G$ (once one makes sense of this) is a subalgebra $W^0$ (described below) of the Lie algebra $W_n$ of derivations of $\mathbb{C}[\![x_1,...,x_n]\!]$. $W_n$ is the free $\mathbb{C}[\![x_1,...,x_n]\!]$-module generated by $\partial_1,\dots,\partial_n$, with the usual bracket.
$W=W_n$ has a subalgebra $W^0$ of vector fields which vanish at the origin (i.e. constant term in coefficients of ∂i are all zero), and another $W^{00}$ of vector fields which vanish to second order (so constant and linear terms vanish).  It's fairly clear that $W^0/W^{00}$ is isomorphic to $\mathfrak{gl}_n$. One now considers W_n modules which are locally finite for the induced $\mathfrak{gl}_n$-action.  It turns out that these can be "integrated" to the group $G$, because $G$ is built out of $\operatorname{GL}_n$ and a unipotent part consisting of those endomorphisms which are the identity modulo $V\ast V$.  So the integrability of the $\mathfrak{gl}_n$-action is all one needs to integrate to all of $G$.
Now one performs the "associated bundle construction" in this context, to produce a sheaf of vector spaces out of a W_n module of the sort above.  One could instead start with a f.d. module $V$ over $\mathfrak{gl}_n$, and there's a canonical way to turn it into a $W_n$-module (in coordinates you tensor it with $\mathbb{C}[\![x_1,...,x_n]\!]$ and take a diagonal action: W_n acts through $\mathfrak{gl}_n$ on the module $V$ and by derivations on $\mathbb{C}[\![x_1,...,x_n]\!]$).  The sheaves you get aren't a priori quasi-coherent; some can be given a quasi-coherent structure (i.e. an action of the structure sheaf on X) and some can't.  However, the sheaves you get are very interesting.  By taking the trivial $\mathfrak{gl}_n$-bundle you get the sheaf of smooth functions on the manifold (this was heuristically explained to me as saying that to give a smooth function on a manifold is to give its Taylor series at every point, together with some compatibilities under change of coordinates, which are given by the $W_n$-action).  By taking exterior powers of $\mathbb{C}^n$ you recover the sheaves of differential forms of each degree (these examples can be made into quasi-coherent sheaves in a natural way).  The $W_n$-modules associated to the exterior powers are not irreducible; they have submodules, which yield the subsheaves of closed forms (these give an example of a sheaf built this way which isn't quasi-coherent: function times closed form isn't necessarily closed).
Finally, one is supposed to see that the existence of the extra operators $\partial_i$ of W which aren't in $W^0$ further induce a flat connection on your associated bundle.  I don't yet understand the underpinnings of that, but it's very important for what I am trying to do.

Is this familiar to any readers?  Is there a good exposition, or a textbook which discusses the foundations?  Can anyone explain the last paragraph to me?

REPLY [5 votes]: I am writing to post an answer to my own question.  The answer below consists of what I was able to jot down from a seminar talk by Jacob Lurie, and a patient followup explanation by Roman Travkin, followed by a correction by David Ben-Zvi.  Of course, mistakes and naivete in the translation are solely attributed to me.  Since the answer was given to me in response to my asking on MO, it seems that karma dictates that I record it here.
The deRham stack, $X_{dR}$, of a scheme $X$ is defined as a functor on test schemes S by
$$X_{dR}(S) := \{\textrm{Maps}: S_{red} \to X \}.$$
where $S_{red}$ is the reduced scheme associated to a scheme $S$.  It is representable as a stack - I think always, but at least when $X$ is smooth - which we should assume for later purposes anyways (maybe everything should be based over $\mathbb{C}$ also?  I welcome corrections, which I'll incorporate).
Okay so $X$, viewed as the functor $X(S)=\{\textrm{Maps}:S\to X\}$ has a natural map $\pi:X\to X_{dR}$, by pre-composing with $S_{red} \to S$.
There is a scheme $Aut_X$ of infinite type over a smooth $X$ whose set of points consists of pairs $(x\in X, t_x: X_{(x)}\cong \hat{D_n})$, where $\hat{D_n}$ is the formal disc, $\mathcal{O}(\hat{D_n}):=\mathbb{C}[[x_1,\ldots,x_n]]$.  The fiber over each point $x\in X$ is the group $Aut^0(X_{(x)})$ of automorphisms of the formal neighborhood of $x$ preserving the maximal ideal, which is in turn (non-canonically) isomorphic to $Aut^0(\hat{D_n})$, the group of automorphisms of $\hat{D_n}$ preserving the origin.
Now, the scheme $Aut_X$ is actually a $\hat G$-torsor over $X_{dR}$, where (I gather from David Ben-Zvi's comments) $\hat G$ is something more like an inductive limit of the groups of automorphisms on $n$th infinitessimal neighboroods of the origin in $D_n$.
Now, suppose that $M$ is a $W_n$ module, which has the property that the operators $x_i\partial_i$ for all $i$ are diagonalizable with finite dimensional eigenspaces.  (These we should think of as being the diagonals $h_i$ for a copy of $\mathfrak{gl}_n$ sitting inside $W_n$).  In this case, one can prove some nice things.
First, in representation theory:  $M$ lies in the category generated by modules $\mathcal{F}_\lambda$ which are coinduced from $V_\lambda$, the irreducible of $\mathfrak{gl}_n$.`  That is, as vector spaces the $\mathcal{F}_\lambda$s are just $V_\lambda \otimes C[x_1,\ldots, x_n]$, and the action is given by natural formulas.  So $M$ is a (possibly infinite) extension of such things.  But there is a theory of weights which control the representation theory somewhat.  See A. N. Rudakov. Irreducible representations of infinite-dimensional Lie algebras of Cartan type. Math. USSR Izv. Vol. 8, pgs. 836-866, which has been translated to English.
Second, in geometry:  The module $M$ is a Harish-Chandra module for the pair $(W_n,Aut_0(C[[x_1,...,x_n]]))$.  This means:  $W_n$ has a Lie sub algebra $W^0$ of vector fields which vanish at the origin (so they have no constant vector field terms like $\partial_i$).  $W^0$ (or perhaps its completion w.r.t order of vanishing at origin) is the Lie algebra of $Aut_0(C[[x_1,...,x_n]])$, as it consists of derivations which preserve the maximal ideal (not sure exactly how you make precise that it is "the Lie algebra", but anyways, there is an exponential map turning an integrable $W^0$ module into an $Aut_0(C[[x_1,...,x_n]]$)-module ).  The assumptions on $M$ were precisely those that make $M$ an integrable $W^0$ module, and so we can regard $M$ as a $G$-module.  Well now we have an associated bundle construction for the $G$-torsor $Aut_X\to X$, and we can use this to produce a sheaf of vector spaces (not quasi-coherent!) over $X$ with fiber $M$.
Better $M$ has an action of the operators $\partial_i$, and we can exponentiate the action of all of $W_n$ to the group $\hat G$.  That means that we can instead construct the associated bundle for the $\hat G$ torsor $Aut_X\to X_{dR}$, which will give us a bundle over $X_{dR}$ with fiber $M$ again.  Now, we can pullback this bundle via $\pi$ to get a bundle on $X$ with fiber $M$.  This time, however, it's a pullback of a sheaf of vector spaces on the deRham stack, which by (some people's) definition is a crystal of vector spaces on $X$.<|endoftext|>
TITLE: Explicitly describing extreme points of infinite dimensional convex sets
QUESTION [8 upvotes]: I am currently trying to apply some results from Choquet theory - i.e., the generalisation of results by Minkowski and Krein-Milman for representing points in a compact, convex set C by probability measures over its extreme points, ext C = { x ∈ C : C - { x } is convex }.
My main problem is with explicitly describing the set of extreme points for a particular convex set, namely the set C of concave functions over the k-simplex that vanish at the vertices of the simplex and have sup-norm at most 1. I've convinced myself that this set of functions in compact and convex and so the Choquet's theorem applies. However, apart from the  case of the 1-simplex I am struggling to say anything about what the extreme functions might be.
In the case of the 1-simplex, the functions in ext C are "tents" with height 1, that is, functions f that are zero on the boundaries and rise linearly to a single point x where f(x)=1. I suspect that in the case of the 2-simplex the extreme functions are also piece-wise linear concave functions with height 1. I have considered a number of candidates (the functions formed by the taking the minimum of 3 affine functions, each zero on a different vertex) but am having trouble showing that the candidates are actually extreme.
Does anyone know of any techniques for identifying extreme points of convex sets? 
Pointers to applications of Choquet's theorem that explicitly construct ext C and the probability measure for a given point in C would also be much appreciated. My reading in this area has only got me as far as Phelps' monograph "Lectures on Choquet Theory" and a survey article by Nina Roy titled "Extreme Points of Convex Sets in Infinite Dimensional Spaces".

REPLY [7 votes]: There are more candidates for the extreme points. Take any compact convex subset of the simplex, then take the infimum of all affine functions that are ≥ the characteristic function of the subset. You could start with more arbitrary subsets, but the result is the same.
As far as general techniques go, I don't have any, except when the set is given by inequalities, try to make as many of them equalities as you can. In this case, make your functions equal to 1 or to 0 as much as possible, and as affine as you can make them.
By the way, don't you have a problem with compactness? Take tents on points $x_n$ and let $x_n$ converge to a vertex of the simplex.
Edit: I realized that you wanted a proof. Let S be the simplex, let K be a closed convex subset (not containing a corner of S), and let f be the inf of all affine functions ≥ 0 on S and ≥ 1 on K.
Assume $2f=g+h$. Then g and h are both equal to 1 on K, and since they are concave they are infimums of affine functions, and so both are $\ge f$. But then they must both be equal to f.

REPLY [3 votes]: The obvious candidates are the functions that have height 1 at some single point x other than a vertex, vanish on the boundary cells of the simplex that do not contain x, and are linear along any line segment from x to the boundary. They obviously belong to C, and no function with height 1 at x can have a smaller value than one of these functions at any other point without destroying concavity.<|endoftext|>
TITLE: Etale cohomology -- Why study it?
QUESTION [80 upvotes]: I know (at least I think I know) that some of the main motivating problems in the development of etale cohomology were the Weil conjectures.  I'd like to know what other problems one can solve using the machinery of etale cohomology.  I know a little bit about how etale cohomology groups appear in algebraic number theory but I'd like to know about ways that these things show up in other mathematical subjects as well.  Is there anything that an algebraic topologist should really know about etale cohomology?  What about a differential geometer?

REPLY [115 votes]: $\DeclareMathOperator{\gal}{Gal}$
Here's a comment which one can make to differential geometers which at least explains what etale cohomology "does". Given an algebraic variety over the reals, say a smooth one, its complex points are a complex manifold but with a little extra structure: the complex points admit an automorphism coming from complex conjugation. Hence the singular cohomology groups inherit an induced automorphism, which is extra information that is sometimes worth carrying around. In short: the cohomology of an algebraic variety defined over the reals inherits an action of $\gal(\mathbb{C}/\mathbb{R})$.
The great thing about etale cohomology is that a number theorist can now do the same trick with algebraic varieties defined over $\mathbb{Q}$. The etale cohomology groups of this variety will have the same dimension as the singular cohomology groups (and are indeed isomorphic to them via a comparison theorem, once the coefficient ring is big enough) but the advantage is that that they inherit a structure of the amazingly rich and complicated group $\gal(\bar{\mathbb{Q}}/\mathbb{Q})$. I've often found that this comment sees off differential geometers, with the thought "well at least I sort-of know the point of it now". A differential geometer probably doesn't want to study $\gal(\bar{\mathbb{Q}}/\mathbb{Q})$ though.
If I put my Langlands-philosophy hat on though, I can see a huge motivation for etale cohomology: Langlands says that automorphic forms should give rise to representations of Galois groups, and etale cohomology is a very powerful machine for constructing representations of Galois groups, so that's why I might be interested in it even if I'm not an algebraic geometer.
Finally, I guess a much simpler motivating good reason for etale cohomology is that geometry is definitely facilitated when you have cohomology theories around. That much is clear. But if you're doing algebraic geometry over a field that isn't $\mathbb C$ or $\mathbb R$ then classical cohomology theories aren't going to cut it, and the Zariski topology is so awful that you can't use it alone to do geometry---you're going to need some help. Hence etale cohomology, which gives the right answers: e.g. a smooth projective curve over any field has a genus, and etale cohomology is a theory which assigns to it an $H^1$ of dimension $2g$ (<pedant> at least if you use $\ell$-adic cohomology for $\ell$ not zero in the field <\pedant>).

REPLY [72 votes]: a) Conceptually an algebraic topologist should be interested in étale cohomology, because it answers a very naïve question: given an algebraic variety over $\mathbb C$, how do I calculate algebraically its singular cohomology? The obvious answer "why, I'll just take the cohomology of the constant sheaf $\mathbb Z$ " fails in a spectacular way: if the variety is irreducible ( a reasonable assumption) the cohomology will be zero in positive degree because constant sheaves are flabby, hence acyclic. This is because the Zariski topology of an algebraic variety is too coarse and doesn't allow for the innumerable singular simplices at the algebraic topologist's disposal.Of course you might say "who cares? I'll just do it my singular way " but then what is to be done in non-zero characteristic?
This is where Grothendieck's étale cohomology comes in : it allows one to compute a most reasonable cohomology for constant sheaves. And for sheaves of finite abelian groups over complex varieties, a difficult theorem ( aptly named comparison theorem) proved by Mike Artin says that étale cohomology  coincides with singular cohomology.
b)You ask : "...what other problems one can solve using the machinery of etale cohomology?"
Well, there is the proof  ( which won him a Fields medal) by Voevodski of Milnor's conjecture on quadratic forms, which had been the preeminent open problem in quadratic form theory for 30 years. He introduced $\mathbb A^1$-homotopy which opened a quite active field of research by your algebraic topology colleagues. Here is a survey article by Fabien Morel on $\mathbb A^1$- algebraic topology:
http://www.mathematik.uni-muenchen.de/~morel/ICMfinal1.pdf
Finally, talking of references, an excellent and very user-friendly set of notes by Milne on étale cohomology can be freely downloaded from
http://www.jmilne.org/math/CourseNotes/LEC.pdf

REPLY [28 votes]: This review by Bloch
 answers the first part of your question. Algebraic topology like Sullivan's "Geometric Topology" make use of etale cohomology and etale homotopy.

REPLY [8 votes]: One of the annoying aspects with sheaf cohomology in algebraic geometry is that - even for curves over the complex numbers - the cohomological dimensions are not what you want them to be, if you are use to the Betti cohomology.
Etale cohomology fixes this problem by defining a cohomology from the covers of a space. This is a standard algebro-topological phenomena: e.g. the fundamental group of a space is indeed the direct limit on the deck groups of covers of the space. 
I don't know if this helps any computations in algebraic topology though.<|endoftext|>
TITLE: What is $\overline{\text{Spec}\mathbb{Z}}$?
QUESTION [13 upvotes]: In Connes work on the Riemann Hypothesis he talks about constructing $\overline{\text{Spec}\mathbb{Z}}$ as a curve over the field with one element. I just want to know what Spec means. Is the same as spectrum from algebraic geometry?

REPLY [18 votes]: If you remove the overline, you have the affine scheme Spec Z.  It is the spectrum of a noetherian domain of Krull dimension one.  This description also holds for any affine algebraic curve over a field, so we have the basis for an analogy.  Z has many structural features in common with the ring of polynomials with coefficients in a field (e.g., basic number-theoretic machinery like a Euclidean algorithm), so there are good reasons to think of Spec Z as a curve.
However, there are several differences.  First, Z is the initial object in the category of commutative rings, so Spec Z is the final object in the category of locally ringed spaces.  In particular, Spec Z is not a curve over any base field.  Second, there are maps from the spectra of fields of many different characteristics into Spec Z.  This makes objects like zeta functions and L-functions much more transcendental (in appearance), since the logarithm doesn't behave as nicely as it does over finite fields.  You can also find a definition of genus of a number field in Neukirch's Algebraic Number Theory, and while it is zero for Q, it tends to be transcendental in general (essentially due to the presence of logs).
Despite the visible flaws in the analogy, there are good reasons for thinking that Spec Z should have a compactification, in a manner similar to the compactification of Spec F[t] into the projective line.  The most basic is that Z has a valuation that is not captured by the points of the topological space, namely the usual Archimedean absolute value.  For the ring of polynomials in a field, this translates to the degree.  By adding this extra valuation you find that if you take all of the absolute values of a rational number (normalized appropriately) and take their product, you get one.  Logarithmically, the sum of valuations is zero just like the residue theorem in the function field case.  A more sophisticated reason comes from arithmetic intersection theory.  If you take a curve defined over Z, you can compute a notion of intersection between two integral points, in an manner analogous to computing the intersection class of two curves in a surface.  In the topological setting, you need some conditions for this to be well-behaved, e.g., the surface should be compact, so you can't push the intersections off the end of the surface.  This is achieved over Z using Arakelov theory - the intersection over infinity is computed by base changing the curve to the complex numbers, making some distribution on the resulting Riemann surface using the integral points, and computing the integral of that distribution.
I don't think there is general agreement on how to make a geometric object with all of the properties we want from a compactification of Spec Z.  In particular, there doesn't seem to be a satisfactory theory of the "base field" yet.  There is a way to approximate the compactification using Berkovich spaces, which are an analytic refinement of schemes.  The Berkovich spectrum of Z already comes with an Archimedean branch, and if you weaken the triangle inequality, you can have points corresponding to arbitrary non-negative powers of the Archimedean absolute value.  Compactification means adding the point $|-|^\infty_\infty$.  The local ring at this point is the closed interval [-1,1] in R, with the multiplication operation (i.e., you have some kind of logarithmic structure, instead of an actual ring), and the global functions on this object are given by the multiplicative monoid {-1,0,1}, which is a rather boring looking candidate for the field with one element.<|endoftext|>
TITLE: Kahler differentials and Ordinary Differentials
QUESTION [68 upvotes]: What's the relationship between Kahler differentials and ordinary differential forms?

REPLY [4 votes]: The key ingredient in proofs that derivations are differential forms is the ability to evaluate coefficients: e.g. to go from $\mathrm{d}f(x) = f'(x) \mathrm{d}x$ to $\mathrm{d}f(x)|_a = f'(a) \mathrm{d}x|_a$.
Evaluation maps $\theta_P : \mathcal{C}^\infty(M) \to \mathbb{R} : f \to f(P)$ are examples of ring homomorphisms; abstractly, for any ring homomorphism $\varphi : R \to S$, we can consider the corresponding "evaluation"
$$
\varphi_* : \Omega_R \cong R \otimes_R \Omega_R \to S \otimes_R \Omega_R
$$
The $S$-module $S \otimes_R \Omega_R$ is isomorphic to the quotient of $\Omega_R$ by the relations $f \mathrm{d}g = \varphi(f) \mathrm{d}g$, and $\varphi_*(f \mathrm{d}g) = \varphi(f) \mathrm{d}g$.
In the case that we use one of the $\theta_P$, $\theta_{P*}$ is evaluating the coefficients of a Kähler differential at $P$, and so the corresponding $\mathbb{R} \otimes_{\mathcal{C}^\infty(M)} \Omega_{\mathcal{C}^\infty(M) / \mathbb{R}}$ looks like the cotangent space to $M$ at $P$.
Write $\mathrm{d}f|_P$ for $\theta_{P*}(\mathrm{d}f)$.
Now, we can apply the usual argument. For $M = \mathbb{R}^n$, to any function $f$, we can take the differential of a Taylor polynomial and get
$$ \mathrm{d}f(x)|_P = f'(P)  \mathrm{d}x|_P$$
so it's clear that we really do get the cotangent space to $M$ at $P$.
If we take all Kähler differentials and identify differentials that have the same "values" at every point of the manifold, then we get the ordinary differential forms.

Conjecture If $M$ is a compact manifold, then $\Omega_{\mathcal{C}^\infty(M) / \mathbb{R}} \cong \Omega^1(M)$.

(this conjecture has been posted to Are Kähler differentials the same as one-forms for compact manifolds?)<|endoftext|>
TITLE: How to sufficiently motivate organization of proofs in math books
QUESTION [11 upvotes]: Hello, 
I have a bit of a general question about math books. I get the feeling that in a lot of math books, the organization for the theorems and lemmas are not explained well (ex. Topics in Algebra Herstein, Linear Algebra Hoffman).
So I'm curious what you all think about the advantages/disadvantages of this. Because on the one hand, it forces one to create the connections for themselves. But on the other hand, it makes the presentation less clear and interesting.
In the classes that I have taken, I find that I have to constantly bother the Professors to motivate what is going on instead of just going along and providing the stripped down proofs.
Furthermore, I am wondering what you all do to rectify this problem. Last semester, I had a great opportunity to speak with a professor about the reading that I did each week but it makes it much harder to study things independently when there is nobody to give perspective on why we are doing what we are doing. Any thoughts would be appreciated!
--Alex

REPLY [4 votes]: I think the advantages and disadvantages depend on the purpose of the book and on the specific subject area. Some books are reference books, packed with results, and it's okay to be light on the motivation there. However, if you want or expect people to follow your proofs line-by-line, a little motivation goes a long way. Personally, when I'm reading (or listening to) a proof, I cringe whenever I see a magic number or function pulled out of thin air that "just happens" to work. Tell me what you were looking for and how you found it. You don't have to go into excruciating detail, but a little context makes it much clearer, and it helps make sure the reader knows how to find these "magic" functions in his or her own research.
I think almost any field of math would benefit from a conversational introductory book. Why are certain theorems important? Which ones are routine but tedious, and which ones rely on very novel ideas? What are some good heuristics in the field, and when do they break down? Which theorems generalize nicely, and for which is it not known? Reading such a book would be like discussing the field with a knowledgeable colleague, and I think it would encourage mathematicians to learn more math outside their field.
Come to think of it, the Princeton Companion achieved this somewhat. Now let's expand each article to a whole book!<|endoftext|>
TITLE: Algebraic Statistics textbook
QUESTION [6 upvotes]: Hey
A friend and I are thinking of having an algebraic statistics seminar next semester. Does anyone know of a good book to try learn it out of?

REPLY [8 votes]: I took a course in Algebraic Statistics from the book of Drton, Sturmfels, and Sullivant called "Lectures on Algebraic Statistics". I learned a lot from the book, and I enjoyed the material tremendously.  (My background is in algebraic geometry, by the way.)<|endoftext|>
TITLE: Applications of propositional dynamic logic
QUESTION [5 upvotes]: Propositional dynamic logic (PDL) is an example of a (multi)modal logic with a structure on the set of modalities. In particular, the set of its modalities is indexed by "programs" and one can use program constructors such as composition, choice and iteration to make new programs out of old (of course, there is a set of basic programs to begin with). So if π and ρ are programs, so are π;ρ, π∪ρ and ρ*. The intuitive interpretations are, respectively, "run π, then run ρ", "nondeterministically run either π or ρ", and "run ρ finite number of times".
The formulas in this language are formed from propositional letters using boolean connectives. Also, if π is a program and φ a formula, then <π>φ is a formula with intuitive interpretation "there is a state of computation accessible by program π in which φ is satisfied". Modality <π> is read "diamond π".
Not to dwell any deeper right now, I refer you to a very well written article about PDL.
Now to get to my question. PDL was created in a line of formal systems that were ment to be used to "talk about" programs, prove program correctness etc. Its predecessors were Hoare logic (HL, also known as Floyd-Hoare logic) and its modal version, the dynamic logic (DL). But, since PDL is propositional, it lacks the expressiveness given by first-order constructions found in HL and DL. So, somehow I haven't really been able to find any real example of its original use.
Does anyone know an example of a program and its specifications that I can express and verify for correctness in PDL? I would be happy with any academic or real-world example, or any reference of such an example.
I would even be happy with an example of use of PDL in other areas. Most books that I've seen mention its use in philosophy, artificial intelligence, linguistics etc., but never give an example of this use.

REPLY [5 votes]: The practical applications might be more obvious once you observe that these propositional "programs" are regular expressions -- which is to say, state machines. So you can expect it to have applications in the study of things like program analyses and verifying concurrent protocols.
Dexter Kozen at Cornell has done a great deal of work in this area. In fact, he's mostly focused on a subsystem of PDL, called "Kleene algebra with tests", which has an easier decision problem (PSPACE rather than EXPTIME) and tends to have nicer equational proofs.<|endoftext|>
TITLE: Anticanonical divisor of the blow up of P^2 in 9 points
QUESTION [8 upvotes]: Let $S$ the blow up of $P^2$ in nine points. Why is the anticanonical divisor $-K_S$ not semiample?

REPLY [7 votes]: Just to complement quim's answer, note that if $S$ is defined over the algebraic closure of a finite field, then the anticanonical divisor of the blow up of 9 general points in $\mathbb{P}^2$ is actually semiample!
The reason this is not in contradiction with quim's answer lies in the subtlety of the word "general".  Here the correct genericity assumption is what is sometimes called 'very general": the points need to lie in the complement of a countable union of closed subsets.  Indeed, if you read quim's answer you see that there is a degree zero line bundle that needs to be non-torsion, and this condition translates to not being of order at most $n$ for every positive integer $n$: clearly a countable union of closed conditions.  On the other hand, every degree zero line bundle on smooth curve defined over a finite field s torsion!!<|endoftext|>
TITLE: Global proof of Serre duality
QUESTION [10 upvotes]: Does anyone know of a global proof (involving no local argument) of Serre duality at the level of varieties or manifolds (as opposed to schemes).

REPLY [4 votes]: I thought I'd offer a high-tech alternative for certain varieties. If $X$ is smooth and projective over a field $k$ then Bondal and van den Bergh give a proof in Generators and representability of functors in commutative and noncommutative geometry that $D^b(\mathrm{Coh}X)$ is saturated which is a strong representability condition on cohomological/homological functors to the category of $k$ vector spaces. It follows immediately that $D^b(\mathrm{Coh}X)$ has a Serre functor by using the fact that $Hom(A,-)^*$ is representable for every bounded complex of coherent sheaves $A$.<|endoftext|>
TITLE: What are the right categories of finite-dimensional Banach spaces?
QUESTION [19 upvotes]: This is inspired partly by this question, especially Tom Leinster's answer.  
Let me start with some background.  I apologize that this will be rather long, since I'm hoping for input from people who probably don't much about the following.
Classically, there are two obvious categories whose objects are Banach spaces (let's say all Banach spaces are real for simplicity): in the first, morphisms are bounded linear maps and isomorphisms are exactly what are usually called isomorphisms of Banach spaces; in the second, morphisms are linear contractions (bounded linear maps with norm at most 1) and isomorphisms are linear isometries.  These categories distinguish between the "isomorphic" and "isometric" theories of Banach spaces.
Now if I'm interested in finite-dimensional spaces, then the "isomorphic" category is not rigid enough, because any two n-dimensional Banach spaces are isomorphic.  But the isometric category is too rigid for most purposes.  So we get more quantitative about our isomorphisms.  One way to do this is with the Banach-Mazur distance.  If X and Y are both n-dimensional Banach spaces,
$$d(X,Y) = \inf_T (\lVert T \rVert \lVert T^{-1} \rVert),$$
where the infimum is over all linear isomorphisms $T:X\to Y$.  Then $\log d$ is a metric on the class of isometry classes of n-dimensional Banach spaces.
Theorems about spaces of arbitrary dimension which include some constants independent of the dimension are characterized as "isomorphic results".  One example is Kashin's theorem: There exists a constant c>0 such that for every n, $\ell_1^n$ has a subspace X with $\dim X = m= \lfloor n/2 \rfloor$ such that $d(X,\ell_2^m) < c$.  (Here $\ell_p^n$ denotes R^n with the $\ell_p$ norm $\lVert x \rVert_p = (\sum |x_i|^p)^{1/p}$.) Thus $\ell_1^n$ contains an n/2-dimensional subspace isomorphic to Hilbert space in a dimension-independent way.
On the other hand there are "almost isometric" results, typified by Dvoretzky's theorem: There exists a function $f$ such that for every n-dimensional Banach space X and every $\varepsilon > 0$, X has a subspace Y with $\dim Y = m \ge f(\varepsilon) \log (n+1)$ such that $d(Y,\ell_2^m) < 1+\varepsilon$.  Thus any space contains subspaces, of not too small dimension, which are arbitrarily close to being isometrically Hilbert spaces.
So my question, finally, is: are there natural categories in which to interpret such results?  I suppose that the objects should not be individual spaces, but sequences of spaces with increasing dimensions.  In particular, as the two results quoted above highlight, the sequence of n-dimensional Hilbert spaces $\ell_2^n$ should play a distinguished role.  But I have no idea what the morphisms should be to accomodate quantitative control over norms in these ways.

REPLY [13 votes]: This is a rewording of Matthew's answer, but one can use nonstandard analysis to obtain a category that achieves what Mark wants.  After taking ultrapowers, one obtains non-standard integers (ultralimits of standard integers), nonstandard finite dimensional Banach spaces (ultralimits of standard finite dimensional Banach spaces), nonstandard linear transformations (ultralimits of standard linear transformations), etc.  One can then separate these non-standard objects into various classes, e.g. bounded non-standard reals (ultralimits of uniformly bounded standard reals, or equivalently non-standard reals that are bounded in magnitude by a standard real), bounded nonstandard linear transformations, poly(n)-bounded nonstandard linear transformations, polylog(n)-bounded nonstandard linear transformations, etc., where n is an unbounded nonstandard natural number (which, in practice, would be used to bound dimensions of things).  Each of these form a category.
Thus, for instance Kashin's theorem becomes the statement that every nonstandard Banach space of some nonstandard finite dimension N has an M-dimensional subspace which is isomorphic (in the category of bounded nonstandard linear transformations) to $\ell^2(M)$ whenever $M \leq N/2$ (or more generally when $M \leq (1-\epsilon)N$ for some standard $\epsilon > 0$.
Dvoretsky's theorem is trickier.  Here, I guess one needs to work with the category of almost contractions: operators whose operator norm is at most $1+o(1)$ (i.e. bounded by $1+\epsilon$ for every standard $\epsilon > 0$.  Then I think the theorem says that any nonstandard finite dimensional Banach space with some nonstandard dimension N has a M-dimensional subspace that is almost isometric to $\ell^2(M)$, whenever $M = o(\log N)$.  (I may have messed up the quantifiers slightly, but this is pretty close to what the nonstandard translation of things should be.)<|endoftext|>
TITLE: What is a cohomology theory (seriously)?
QUESTION [73 upvotes]: This question has bugged me for a long time. Is there a unifying concept behind everything that is called a "cohomology theory"? 
I know that there exist generalized cohomology theories, Weil cohomology theories and perhaps one might include delta-functors, which describe (some of) the properties of explicit cohomology theories. But is there now a concept that underlies all these concepts? Or has the term "cohomology theory" been used so inflationary that the best thing one could say is that a cohomology theory is a sequence of functors into an "algebraic category" (whatever that is)?
Moreover, what is the difference between a cohomology theory and a homology theory? Of course I am aware of the examples like singular (co-)homology and I know the difference in this situation, but in general?

REPLY [8 votes]: I'd say a cohomology theory is a misnomer.  A theory really ought to be significant, make predictions, help us think about things, help us prove theorems.  At some point mathematicians decided to start giving away the word "theory" for free.    Newtonian mechanics, evolution, calculus -- those are theories.  
Is it a lack of imagination on our part?  It seems like anything that hasn't earned a proper name gets called "X theory" nowadays, for various values of X.  I'm glad differential geometry was invented in a previous era. Our contemporaries would have saddled the subject with some glorious name like "geometry theory" or "G-theory". 
(not exactly sure if tongue is in cheek)<|endoftext|>
TITLE: Fundamental group of 3-manifold with boundary
QUESTION [24 upvotes]: Is it true that any finitely presented group can be realized as fundamental group of compact 3-manifold with boundary?

REPLY [3 votes]: If a finite group $G$ is (isomorphic to) the fundamental group of a three-manifold, then $G$ embeds in $\mathrm{SO}(4)$.
To see this, suppose that $M$ is a compact, connected three-manifold, possibly with boundary, having $G = \pi_1(M)$ finite. If $M$ has two-sphere boundary components, we can cap them off without changing the fundamental group.  Since free products of non-trivial groups are always infinite, we deduce that $G$ is freely indecomposible.  Appealing to the Poincaré conjecture [solved by Perelman], we have that $M$ is irreducible: all two-spheres in $M$ bound three-balls.
If $M$ is non-orientable, then a theorem of Livesay implies that $M$ is homeomorphic to the real projective plane crossed with a unit interval.
Thus we reduce to the case where $M$ is compact, oriented, connected, irreducible, and has finite fundamental group.  All boundary components of $M$ are now oriented, and of genus at least one.  Applying "one-half lives, one-half dies" we find that $M$ has no boundary components.  Appealing to the elliptic part of the geometrisation conjecture [solved by Perelman] we find that the universal cover of $M$ is the three-sphere, and the deck group, and thus $G$, is (conjugate to a) subgroup of $\mathrm{SO}(4)$. QED
A quick google search finds a paper of Zimmermann giving a readable introduction to the finite subgroups of $\mathrm{SO}(4)$ - see section three of that paper.  One then has to determine which of these act freely.  Finally, there is another approach to this problem via spherical Seifert fibered spaces.
As a concrete example of a finite group that is not a three-manifold group, consider $\mathbb{Z}_p \times \mathbb{Z}_q$ where $p$ and $q$ are not coprime.  This is a isomorphic to a subgroup of $\mathrm{SO}(4)$, but it cannot act freely.  [See Theorem 9.14 in Hempel's book for a proof that, in this special case, avoids geometrisation.]<|endoftext|>
TITLE: How can I learn about doing linear algebra with trace diagrams?
QUESTION [11 upvotes]: There is a wikipedia article.  There is a paper by Elisha Peterson.  I tried reading these but they don't seem to click for me.
Are there books or other resources for learning how to do linear algebra with trace diagrams?

REPLY [12 votes]: to Greg Kuperberg: Use of such diagrams probably starts in mid 19th century (Sect. 4.9 A BRIEF HISTORY OF BIRDTRACKS). If you find "spiders" or “birdtracks” too silly, I vote for the somewhat unwinged "tensor diagrams." 

So why give these diagrams a new name, and not call them “Feynman diagrams” or "tensor diagrams"? I needed a distinct name to distinguish them from the more traditional uses of diagrams. The difference is that here diagrams are not a mnemonic device, an aid in writing down a Feynman integral that is to be evaluated by other techniques. I did not call them "tensor diagrams" as that is too close to “invariant tensor operators”, the Wigner-Eckart theorem, and the 3n-j diagrams that are only a prelude to a computation. Here “birdtracks” are everything, all calculations are carried out in terms of birdtracks, from start to finish.<|endoftext|>
TITLE: Explanation for the Chern character
QUESTION [51 upvotes]: The Chern character is often seen as just being a convenient way to get a ring homomorphism from K-theory to (ordinary) cohomology.
The most usual definition in that case seems to just be to define the Chern character on a line bundle as $\mathrm{ch}(L) = \exp(c_1(L))$ and then extend this; then for example $\mathrm{ch}(L_1 \otimes L_2) = \exp(c_1(L_1 \otimes L_2)) = \exp(c_1(L_1) + c_2(L_2)) = \mathrm{ch}(L_1) \mathrm{ch}(L_2)$; then we can use this to define a Chern character on general vector bundles.
This all seems a bit ad-hoc, and it doesn't give much insight as to why such a thing exists anyway.
An explanation I like a lot better comes from even complex oriented cohomology theories. Given any complex oriented periodic cohomology theory, such as K-theory or periodic (ordinary) cohomology, we have $H(\mathbb{CP}^\infty) \cong H(P)[[t]]$ for $P$ a point. Seeing as $\mathbb{CP}^\infty$ is the classifying space for line bundles, this gives us a way of having "generalised Chern classes" for any line bundle corresponding to any cohomology theory, and even for any vector bundle.
We have a link between complex oriented periodic cohomology theories and formal group laws, corresponding to what corresponds to $c_1(L_1 \otimes L_2)$ in $\mathbb{CP}^\infty$: for ordinary cohomology, as above, we get that $c_1(L_1 \otimes L_2) = c_1(L_1) + c_1(L_2)$ which gives the additive formal group law, and for K-theory we get $c_1(L_1 \otimes L_2) = c_1(L_1) + c_1(L_2) + c_1(L_1) c_2(L_2)$ which is the multiplicative formal group law. The fact that over $\mathbb{Q}$ (but not over $\mathbb{Z}$) there is an isomorphism between the formal group laws given by the exponential map, and this reflects in the cohomology, giving the Chern character $K(X) \otimes_\mathbb{Z} \mathbb{Q} \to \prod_n H^{2n}(X,\mathbb{Q})$.
I'm not too sure what the exact formulation in that second case is, but more importantly I was wondering if there are any other, cleaner interpretations of the Chern character (I've been hearing about generalised Chern characters, and I have no idea where they would come from in this case). It seems like there should be a way to link the Chern character to things like the genus of a multiplicative sequence, and tie it in with other similar ideas for example the Todd genus or the L-genus given by similar formal power series. I guess the trouble is that I don't see how these related ideas all fit in together.

REPLY [3 votes]: I can give you my interpretation of the Chern character in terms of morphism of formal group laws. Denote with $\gamma$ the universal line bundle over $\mathbb{P}^\infty$ and denote its first Chern class with $c_1(\gamma) = x_H \in H^2(\mathbb{P}^\infty)$. Also denote with $k = \gamma-1 \in K(\mathbb{P}^\infty)$ and with $k_i : = p_i^*(\gamma)-1 \in K(\mathbb{P}^\infty \times \mathbb{P}^\infty)$. Then by definition the Chern character of $k \in K(\mathbb{P}^\infty)$ is equal to $e^{x_H}-1 \in HP^0(\mathbb{P}^\infty, \mathbb{Q}):=\prod_{i\geq0}H^{2i}(\mathbb{P^\infty},\mathbb{Q})$. The naturality of the Chern character and the isomorphism $HP^0(\mathbb{P}^\infty,\mathbb{Q}) \cong \mathbb{Q}[[x_H]], HP^0(\mathbb{P}^\infty \times \mathbb{P}^\infty,\mathbb{Q}) \cong \mathbb{Q}[[x_1,x_2]]$ gives the following commutative square
$\require{AMScd}$
\begin{CD}
    K(\mathbb{P}^\infty) @>ch>> \mathbb{Q}[[x_H]]\\
    @V \mu^* V V @VV \mu^* V\\
    K(\mathbb{P}^\infty \times \mathbb{P}^\infty) @>>ch> \mathbb{Q}[[x_1,x_2]]
\end{CD}
where $\mu$ is the multiplication in the H-space $\mathbb{P}^\infty$. If we follow the element $1+k \in K(\mathbb{P}^\infty)$ we get
$\require{AMScd}$
\begin{CD}
    1+k @>ch>> e^{x_H}\\
    @V \mu^* V V @VV \mu^* V\\
    1+k_1+k_2+k_1k_2 @>>ch> e^{x_1+x_2}
\end{CD}
Now let $m$ be te multiplicative formal group law and $a$ be te additive one. Then, by the previous diagram we get that
$
m(ch(k_1),ch(k_2)) = m(e^{x_1}-1,e^{x_2}-1) = e^{a(x_1,x_2)}-1 
$
and the Chern character is identified to a morphism (actually an isomorphism) of fgl.<|endoftext|>
TITLE: Spanier-Whitehead dual and Hopf fibration
QUESTION [5 upvotes]: Consider a map of spheres $f:S^n\to S^m$ covered by a map of trivial $\mathbb R^k$-bundles.
In other words, we take the trivial rank $k$ vector bundle over $S^m$ and pull it to $S^n$ via $f$. Consider the corresponding map of the Thom spaces, and its Spanier-Whitehead dual. How is the dual related to $f$? 
The test case I care about is when $f$ is the Hopf fibration $S^3\to S^2$. Then I think the dual can be represented by a map $F:\Sigma^{r+1}(S^2_+)\to\Sigma^r (S^3_+)$ where $X_+$ means $X$ disjoint union a point, and $\Sigma^s$ is $s$-fold suspension. Thus $F$ can be thought of as a map $S^{r+3}{\vee} S^{r+1}\to S^{r+3}{\vee} S^r$. What is this map?

REPLY [2 votes]: (I'm going to assume you only care about the homotopy class of map.)
Such maps are roughly their own duals.  The Spanier-Whitehead dual of a map $S^n \to S^m$ is a map $"S^{-m}\_+ \to S^{-n}\_+"$, which is represented by $\Sigma^{r-m} S^m\_+ \to \Sigma^{r-n} S^n\_+$ for sufficiently large r.  If r=n+m+s, this is a map $\Sigma^{s+n} S^m\_+ \to \Sigma^{s+m} S^n\_+$.  This is homotopy equivalent to a map $S^{n+m+s} \vee S^{n+s} \to S^{n+m+s} \vee S^{m+s}$.
The resulting map is the wedge of the identity map on the first factor and the (up to sign, which I will get wrong if I try) the s-fold suspension of $f$ on the second factor.
This is easier to say in the stable homotopy category, where your original map become stably the map $id \vee \Sigma^\infty f:S^0 \vee S^n \to S^0 \vee S^m$ and the dual just dualizes the maps on each factor.  It's also easier to say if you use based maps of spheres in the first place.<|endoftext|>
TITLE: Where is there a treatment of "exponential monads"?
QUESTION [13 upvotes]: I have a category $C$, which is equipped with a symmetric monoidal structure (tensor product $\otimes$, unit object $1$).  My category also has finite coproducts (I'll write them using $\oplus$, and write $0$ for the initial object), and $\otimes$ distributes over $\oplus$.
By an exponential monad, I mean a monad $(T,\eta,\mu)$ on $C$, where the functor $T:C\to C$ is equipped with some structure maps of the form
$$\nu \colon 1 \to T(0)$$
and
$$\alpha\colon T(X)\otimes T(Y) \to T(X\oplus Y).$$
The structure maps are isomorphisms, and are suitably "coherent" with respect to the two monoidal structures $\otimes$ and $\oplus$.
The simplest example is: $C$ is the category of $k$-vector spaces, and $T=\mathrm{Sym}$ is the commutative $k$-algebra monad (i.e., $\mathrm{Sym}(X)$ is the symmetric algebra $\bigoplus \mathrm{Sym}^q(X)$).
Now, I'm sure I can work out all the formalism that I need for this, if I have to.  My question is: is there a convenient place in the literature I can refer to for this?  Alternately, is there suitable categorical language which makes this concept easy to talk about?
I'd also like to have a good formalism for talking about a "grading" on $T$.  This means a decomposition of the functor $T=\bigoplus T^q$, where $T^q\colon C\to C$ are functors, which have "nice" properties (for instance, $T^m(X\oplus Y)$ is a sum of $T^p(X)\otimes T^{m-p}(Y)$).  The motivating example again comes from the symmetric algebra: $\mathrm{Sym}=\bigoplus \mathrm{Sym}^q$.

REPLY [11 votes]: I've just spent some time working out for myself the details of what an exponential monad is supposed to be, so I might as well post what I learned here.  (I don't know that I'll ever have a reason to write it up more formally.)
Here is the correct definition.  Given symmetric monoidal $(C,\otimes, 1)$, with finite coproducts $+$ and initial object $0$, an exponential structure on a monad $T$ on $C$ should be 

the structure of strong symmetric monoidal functor on $T: (C,+,0)\to (C,\otimes, 1)$, consisting of natural isomorphisms $\nu\colon 1\to T0$ and $\alpha\colon TX\otimes TY\to T(X+Y)$ satisfying a bunch of coherence properties.

There is one additional condition you need to impose.  To state this condition, let $\gamma: T(TX\otimes TY)\to TX\otimes TY$ be the composite
$$ T(TX\otimes TY) \xrightarrow{T\alpha} TT(X+Y) \xrightarrow{\mu} T(X+Y) \xrightarrow{\alpha^{-1}} TX\otimes TY,
$$
where $\mu: TT\to T$ is part of the monad structure.  The additional condition is 

for all $X$ and $Y$, we have $\gamma\circ T(\mu\otimes \mu)=(\mu\otimes \mu)\circ \gamma$ as maps $T(TTX\otimes TTY)\to TX\otimes TY$.

The map $\gamma$ defines a $T$-algebra structure on $TX\otimes TY$ (it is the free algebra structure on $T(X+Y)$, transported along the isomorphism $\alpha$); the additional property says that $\mu\otimes \mu$ is itself a map of $T$-algebras.  Note also that the map $\nu$ identifies $1$ with the initial $T$-algebra $T0$.
Given this, you can prove (with no more difficulty than you would expect) that for every pair of $T$-algebras $A$ and $B$, we can put a canonical $T$-algebra structure on $A\otimes B$, which exhibits it as the coproduct of $A$ and $B$ in the category of $T$-algebras.  In particular, the forgetful functor from $T$-algebras to $C$ becomes strong symmetric monoidal, using coproduct as the monoidal structure for $T$-algebras.  
I would have expected that you would need another condition, relating $\gamma$ to the unit map $\eta\colon I\to T$, or perhaps a condition on $\nu$, but it doesn't seem that this is necessary as far as I can tell.
You don't need the hypothesis that $\otimes$ distribute over coproduct, as I suggested in my question.  It might seem surprising, but apparently you don't even need the monoidal structure on $C$ to be symmetric or associative; being unital appears to be enough to exhibit coproducts of $T$-algebras using $\otimes$.  I suspect that for such a "unital monoidal" category $C$, you might be able to show that the existence of an exponential monad implies that $C$ is symmetric monoidal.  (This does not seem so crazy in light of the way the symmetric monoidal smash product of  EKMM spectra comes about).<|endoftext|>
TITLE: Explanation for gamma function in formula for $n$-ball volume
QUESTION [14 upvotes]: It is well-known that the volume of the unit ball in n-space is $\pi^{n/2}/\Gamma(n/2+1)$. Do you know of a proof which explains this formula? Any proof which does not treat the cases $n$ even and $n$ odd separately (like using an explicit expression for $\Gamma(n/2+1)$ in terms of factorials) should be fine.

REPLY [4 votes]: Here is an argument similar to that of S. Carnahan, but more direct (without sphere surface area.) 
Denote the volume of a unit ball in $\mathbb{R}^n$ by $c_n$, then the volume of a ball of radius $r$ equals $c_nr^n$. 
We use the (Lebesgue-Stieltjes?) formula $\int fd\mu=\int_0^\infty \mu\{x:f(x)>t\}dt$ for a positive function $f$ on a measure space $(X,\mu)$. 
Let $(X,\mu)$ be $\mathbb{R}^n$ with Lebesgue measure and $f(x)=e^{-\sum x_i^2}$. Then $$\pi^{n/2}=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^n=\int fd\mu=\int_0^1 c_n(-\log t)^{n/2}dt=\\=c_n\int_0^\infty s^{n/2}e^{-s}ds=c_n\Gamma(n/2+1),$$
where we substitute $t=e^{-s}$.<|endoftext|>
TITLE: Is there a name for the matrix equation A X B + B X A + C X C = D?
QUESTION [7 upvotes]: I happen to be working on a problem that reduces to solving the following equation:
$$\mathbf{A X B} + \mathbf{B X A} + \mathbf{C X C} = \mathbf{D}$$
where A through D are known matrices ( A, B, D are real, symmetric matrices and C is real and antisymmetric), and X is an unknown square matrix to be solved for.
Is there a name for this equation, and is there any known algorithm for solving this equation? (Without the C X C term this reduces to the continuous Lyapunov equation given either A or B is an invertible matrix. I wonder if anyone working in control theory may have seen such equations before.)

REPLY [2 votes]: Apart from very special cases (something commuting with something else), as far as I know there is no efficient algorithm for this kind of equations with more than two summands. (by "efficient" I mean "better than the Kronecker product approach"). 
May sound strange, but I would actually suggest solving the Kronecker product system with an iterative method like SYMMLQ, or CG if it's positive definite. Matrix-vector products cost "only" $O(n^3)$, and dropping a term provides a better-than-nothing preconditioner.<|endoftext|>
TITLE: Cohomology of associative algebras
QUESTION [16 upvotes]: Let $A$ be an associative algebra over a commutative ring $k$. I've read statements saying that Hochschild (co)homology is the "right" notion of (co)homology for associative algebras. When $A$ is projective over $k$, the Hochschild cohomology, say, can be written as $Ext^*_{A \otimes A^{op}}(A,A)$, where $A^{op}$ is the opposite algebra, i.e., $A$ with $xy$ redefined to be $yx$.
On the other hand, when $A$ is augmented, the ext-group $Ext^*_A(k,k)$ is also referred to as the cohomology of $A$. What is the difference between these two notions of cohomology, and why would I choose one over the other?

REPLY [5 votes]: In my view Hochschild cohomology is the most interesting cohomology on associative (and I dare say commutative algebras).  So far all that has been said is about different methods of computation.  But there are also many applications and ways of viewing it.  
Skip the following paragraph if you want, it's just a side point.

The one that sticks in my mind is the
  application to deformation theory. 
  The Hochschild cochain complex is
  actually the object of interest in
  deformation theory, its homology is
  just one invariant of it and captures
  the infinitesimal deformations.  The
  cochain complex carries a specific
  algebraic structure; it's an algebra
  for the braces operad.  And then
  there's the celebrated (and many times
  proved ;-)) Deligne conjecture which
  says that it may be viewed as a
  homotopy Gerstenhaber algebra. 
  Finally there's Kontsevich's formality
  result which says that for smooth
  commutative algebras that looking at
  homology and its Gerstenhaber algebra
  structure actually does capture all
  information about the Hochschild
  cochains and hence the deformation
  theory of the algebra.

Anyway I didn't mean the write that, but just got overexcited, my point in writing this answer was to say that there are other homology theories.  
For example there's the bar homology.  This homology is little known which is a big pity because it's actually rather special!  There's a very good reason why it's not studied though and that's because for a unital algebra its homology is always zero, but it is still interesting because it the chain complex a coalgebra and we're not interested in its homotopy type as a complex and so shouldn't be taking its homology at all!  The coalgebra actually gives generators and relations for the algebra, it's the derived functor of
$A \mapsto A/(A.A)$
from the category of associative algebras to vector spaces.
But you guys like taking homology, so I should give you a better reason for studying the bar homology.  Suppose you have an augmented algebra, so we can split the identity off and write
$A = k\oplus A'$
Then the bar homology of $A'$ is not necessarily zero and gives interesting invariants of the algebra.  In the char 0 commutative case this is well studied, you guys might know it as part of rational homotopy theory.  The commutative bar homology of the cohomology ring of a nice space is the rational homotopy of the space.<|endoftext|>
TITLE: Does there exist a meromorphic function all of whose Taylor coefficients are prime?
QUESTION [15 upvotes]: More precisely, does there exist an unbounded sequence $a_0, a_1, ... \in \mathbb{N}$ of primes such that the function
$\displaystyle O(z) = \sum_{n \ge 0} a_n z^n$
is meromorphic on $\mathbb{C}$?  
[A previous version of the question also asked about the exponential generating function of $(a_n)$.  However, such a function can trivially be entire. - GJK]

REPLY [19 votes]: Borel proved the following much stronger result: if a power series with integer coefficients
represents a function f(z) that is meromorphic in a disk of radius >1, then f(z) extends to a rational function on all of C. I found this result without a reference on page 3 of www.mathematik.uni-bielefeld.de/~anugadre/Adeles.pdf.<|endoftext|>
TITLE: What books should I read before beginning Masaki Kashiwara and Pierre Schapira's "Sheaves on Manifolds"
QUESTION [15 upvotes]: I am a beginner trying to learn about sheaves. I am trying to read Masaki Kashiwara and Pierre Schapira's book "Sheaves on Manifolds", but I find it is not easy for me to understand.
What other books should I read first, with little knowledge about abstract algebra and homological algebra?

REPLY [3 votes]: I like: Iversen, Cohomology of sheaves
http://books.google.com/books?id=JTAZAQAAIAAJ&dq=Iversen+Cohomology&hl=en&ei=gScvTO6_JpqqsQaquL21Ag&sa=X&oi=book_result&ct=result&resnum=10&ved=0CFIQ6AEwCQ
It serves as an introduction to sheaves and their cohomology without requireing 
much background. Applications to topology and algebraic geometry are
explained. Morover it has an appendix on derived categories.

REPLY [3 votes]: I think, the first volume of Harder's Lectures on Algebraic Geometry contains a nice and balanced account of sheaf theory and the cohomology of sheaves. Besides the title, it is not really a book about algebraic geometry. Instead there are many examples from algebraic topology and Riemann surfaces. One should although note that the book contains many typos.<|endoftext|>
TITLE: What is Quantization ?
QUESTION [78 upvotes]: I would like to know what quantization is, I mean I would like to have some elementary examples, some soft nontechnical definition, some explanation about what do mathematicians quantize?, can we quantize a function?, a set?, a theorem?, a definition?, a theory?

REPLY [7 votes]: A very basic answer: think about the classical Hamiltonian,
$$
a(x,\xi)=\vert \xi\vert^2-\frac{\kappa}{\vert x\vert},\quad \text{$\kappa>0$ parameter}.
$$
The classical motion is described by the integral curves of the Hamiltonian vector field of $a$,
$$
\dot x=\frac{\partial a}{\partial\xi},\quad
\dot \xi=-\frac{\partial a}{\partial x}.
$$
The attempt of describing the motion of an electron around a proton by classical mechanics leads to the study of the previous integral curves and is extremely unstable since the function $a$ is unbounded from below. If classical mechanics were governing atomic motion, matter would not exist, or would be so unstable that could not sustain its observed structure for a long time, with electrons collapsing onto the nucleus.
Now, you change the perspective and you decide, quite arbitrarily that atomic motion will be governed by the spectral theory of the quantization of $a$, i.e. by the selfadjoint operator
$$
-\Delta-\frac{\kappa}{\vert x\vert}=A.
$$
It turns out that the spectrum of that operator is bounded from below by some fixed negative constant, and this a way to explain stability of matter. Moreover the eigenvalues of $A$ are describing with an astonishing accuracy the levels of energy of an electron around a proton
(hydrogen atom).
My point is that, although quantization has many various mathematical interpretations, its success is linked to a striking physical phenomenon: matter is existing with some stability and no explanation of that fact has a classical mechanics interpretation. The atomic mechanics should be revisited, and quantization is quite surprisingly providing a rather satisfactory answer. For physicists, it remains a violence that so refined mathematical objects
(unbounded operators acting on - necessarily - infinite dimensional space) have so many things to say about nature. It's not only Einstein's "God does not play dice", but also Feynman's "Nobody understands Quantum Mechanics" or Wigner's "Unreasonable effectiveness of Mathematics."<|endoftext|>
TITLE: Are there any good nonconstructive "existential metatheorems"?
QUESTION [26 upvotes]: Are there any good examples of theorems in reasonably expressive theories (like Peano arithmetic) for which it is substantially easier to prove (in a metatheory) that a proof exists than it is actually to find the proof? When I say "substantially easier," the tediousness of formalizing an informal proof should not be considered. In other words, a rigorous but informal proof doesn't count as a nonconstructive demonstration that a formal proof exists, since there is no essential difficulty in producing a formal proof besides workload. If there is some essential barrier, then there's something wrong either with the proof or with the definition of "formal proof." (Although the concept of "informal proof" is by nature vague, it seems reasonable to say that the procedure transforming a typical informal proof to a formal proof is a primitive recursive computation, even though it's probably impossible to make that precise enough to prove.)
I don't know what such a nonconstructive meta-proof could look like, but I also don't see why one couldn't exist. The only near-example I can think of right now is in the propositional calculus: you can prove that a propositional formula is a theorem by checking its truth table, which does not explicitly provide a deduction from the axioms; however, in most interesting cases, that probably isn't much easier (if at all) than exhibiting a proof, just maybe more mechanical. (Of course, estimating the difficulty of proving or disproving a candidate for a propositional theorem is a huge open problem.) Also, I think there is actually a simple way to convert a truth-table proof into an actual deduction from the axioms of propositional calculus, although I don't remember and didn't retrace all of the details. (Anyway, I'm not even so interested in the propositional calculus for this question, since it's not very expressive.)
Another thing that seems vaguely relevant is Godel's completeness theorem, which states that a formula is a theorem of a first-order theory if and only if it is true in every model of that theory. I don't know in what cases it would be easier, though, to show that some formula were true in every model than it would be just to prove the theorem in the theory.

REPLY [5 votes]: This is almost a result of the type the OP wanted:  In his attempt to prove what is now known as the prime number theorem, Chebyshev used an elementary argument involving a cleverly chosen finite sequence of numerical weights to show that the number of primes up to $x$ is between $0.92 x/\log x$ and $1.11 x/\log x$ for large $x$.  It is natural to ask whether Chebyshev's method can be extended to get closer to the prime number theorem.  It was shown by Rosser and Erdos-Kalmar independently that for any $\varepsilon>0$, there existed a sequence of weights for which the elementary method of Chebyshev showed that the number of primes up to $x$ lay between $(1-\varepsilon) x/\log x$ and $(1+\varepsilon) x/\log x$.  Sending $\varepsilon \to 0$, this would seemingly give an elementary proof of the PNT!  But there is a very important catch: the arguments of Rosser and Erdos-Kalmar use the PNT in the proof (and at the time, there was no known elementary proof of the PNT).  So we have a non-elementary proof that elementary proofs of (arbitrary approximations to) the PNT exist!   See Chapter 9 of 
Harold G. Diamond, MR 670132 Elementary methods in the study of the distribution of prime numbers, Bull. Amer. Math. Soc. (N.S.) 7 (1982), no. 3, 553--589.
for further discussion.
As it turns out, this particular result is not quite answering the original question, because it can be made constructive: numerically established zero-free regions of zeta can be converted constructively to elementary proofs of the PNT with an epsilon loss.  But it is at least an example of a non-elementary metatheorem about the existence of elementary proofs.<|endoftext|>
TITLE: Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)
QUESTION [21 upvotes]: Let us say that a set B admits a rigid binary relation, if there is a binary relation R such that the structure (B,R) has no nontrivial automorphisms. 
Under the Axiom of Choice, every set is well-orderable, and since well-orders are rigid, it follows under AC that every set does have a rigid binary relation. 
My questions are: does the converse hold? Does one need AC to produce such rigid structures? Is this a weak choice principle? Or can one simply prove it in ZF?
(This question spins off of Question A rigid type of structure that can be put on every set?.)

REPLY [4 votes]: A variant of Justin Palumbo's answer, blackboxing the forcing construction: 
An infinite set X is called amorphous if all its subsets are finite or co-finite.  X is called strongly amorphous (or superamorphous) if every relation on X (that is, every subset of $X^n$, for any $n$)
is definable with finitely many parameters in the language of equality. (In other words, restricting to $n=2$ for simplicity: Every $R \subseteq X^2$ must be in the Boolean algebra generated by the sets $X_a:=\{(a,x): x\in X\}$, their converses and the diagonal.  Unless I have forgotten a few more generators.) 
(In particular, amorphous sets are Dedekind finite.) 
Clearly, strongly amorphous sets have no rigid relation.  Their consistency with ZF can be shown (as pointed out above) by first constructing a ZFA model with an amorphous set of atoms, then applying the Jech-Sochor theorem.  
(Strongly) amorphous sets X (and related structures, such as the powerset of X) can often be used as counterexamples showing that some choice is needed.<|endoftext|>
TITLE: Least number of non-zero coefficients to describe a degree n polynomial
QUESTION [23 upvotes]: I'd be grateful for a good reference on this, it feels like a classic subject yet I couldn't find much about it.  
Polynomials in one variable of the form $x^n+a_{n-1}x^{n-1}+\dots +a_1 x+a_0$ can be transformed into simpler expressions. For instance it is apparently well-known that the Tschirnhaus transformation allows to bring any quintic into so-called Bring-Jerrard form $x^5+ax+b$, while for degree 6 one needs at least three coefficents $x^6+ax^2+bx+c$.
Is there a name for such "generalized Bring-Jerrard form", and what is known about it?   In particular there is a cryptic footnote of Arnold (page 3 of this lecture) where he says roughly that the degrees for which more coefficients are needed occur along "a rather strange infinite sequence": could someone please describe what those degrees are (I had a look at the OEIS but I believe that sequence is different from Hamilton numbers, and couldn't find a relevant one).

REPLY [21 votes]: The modern notion of the essential dimension of a group gives a precise way to state your question (and generalizations), and there is some recent work extending the work mentioned in Scott's answer.  To get started, see the article
J. Buhler and Z. Reichstein, On the essential dimension of a group, Compositio Math. 106 (1997), 159-179.
For instance, it is proved there that for polynomials of degree $n$, at least $\lfloor n/2 \rfloor$ coefficients are required.  (This agrees with what you mentioned for $n=5$ and $n=6$.)<|endoftext|>
TITLE: When can cohomology be calculated on the coarse moduli space?
QUESTION [12 upvotes]: Suppose $\cal{X}$ is a DM-stack, and X its coarse moduli space. Let F be a sheaf on $\cal{X}$, and $\pi : \mathcal{X} \to X$ the projection. In all examples I have seen, it has been true that
$H^i(\mathcal{X},F) = H^i(X,\pi_\ast F)$.
Is there a simple example where this fails? Are there easy conditions where this is true?

REPLY [5 votes]: (This mostly just adds some references to Tyler's answer)
$\newcommand{\X}{\mathcal X}$
As Tyler said, if $F$ is a quasi-coherent $\cal O$-module, you get the Grothendieck spectral sequence
$$H^p(X,R^q\pi_*(F))\Rightarrow H^{p+q}(\X,F)$$
so it suffices to impose the condition that $R^q\pi_*=0$ for $q>0$ (i.e. that $\pi_*$ is exact). This is exactly the condition that $\X$ is tame. See Abramovich-Olsson-Vistoli's Tame stacks in positive characteristic. In particular, DM stacks in characteristic 0 are always tame.†
Note that it didn't matter that $\pi$ was a coarse space map. We just needed that $\pi_*$ is exact (that "$\pi$ is cohomologically affine"). In particular, the isomorphism holds when $F$ is quasi-coherent and $\pi:\mathcal X\to X$ is a good moduli space.
† This is asserted in the first paragraph of the paper, but I don't see how to prove it. Perhaps somebody could clarify in a comment.<|endoftext|>
TITLE: Counting points on varieties of low codimension
QUESTION [10 upvotes]: The graduate students here at MIT have been thinking about questions like the following: Over $\mathbb{F}\_q$, how many symmetric matrices are there with nonzero determinant and $0$'s on the diagonal? They are doing brute force computer searches; checking every point in $\mathbb{A}^N(\mathbb{F}\_q)$ and seeing whether it is in the variety.
Is there a better way? Their examples are, like the above, low codimension subvarieties of high dimensional affine spaces. I think they are usually looking at smooth varieties, but I wouldn't swear to it in every case. The ideal answer, of course, would be preexisting software.

REPLY [9 votes]: For a general variety and for a fixed small value of $q$, there isn't going to be a very good algorithm.  That is because you can encode a Boolean formula in a single polynomial equation with minimal overhead.  You are therefore counting solutions to a general logical expression, which is not only #P-hard, but also morally there is often nothing much better than exhaustive search.
On the other hand, if $q$ grows for a fixed formula, then you can use zeta function facts that algebraic geometers and number theorists know or conjecture to extrapolate from small values of $q$.  For instance if $q = p^k$, you can use the Weil conjectures.
Since your example variety is far from general, you can try to exploit special structure to chip away at an exponential search (or count) and make it a better exponential search or count.  I'll stick to your example problem, on the assumption that the others that they are looking at are similar.
First idea:  There is a torus action on the set of symmetric matrices with vanishing diagonal.  Thus you can assume that the first row and column is entirely 0s and 1s, and multiply by the size of the orbit.  (Refinement:  You can assume that the leading term in each row and column is 1.)
Second idea: Quadratic hypersurfaces in $\mathbb{A}^n$ are classified and you know how many points they have.  So you do not have to complete the matrix; you can instead complete all but one row and column of the matrix.  This idea combines with the first idea.
Third idea, much better than the other two: For each vector subspace $V \subseteq \mathbb{A}^n$, you can count the symmetric matrices which vanish on the diagonal and which annihilate $V$ (and maybe other vectors), because the equations for that are linear.  You can then apply Möbius inversion on the lattice of subspaces to count the matrices that do not annihilate any vectors.  You can loop over the vector spaces $V$ by assuming the unique basis in RREF form.  Moreover, many subspaces with the same RREF pivot positions have to give you the same answer, for one reason because of the torus action.<|endoftext|>
TITLE: Definition of "simplicial complex"
QUESTION [49 upvotes]: When I think of a "simplicial complex", I think of the geometric realization of a simplicial set (a simplicial object in the category of sets). I'll refer to this as "the first definition".
However, there is another definition of "simplicial complex", e.g. the one on wikipedia:  it's a collection $K$ of simplices such that any face of any simplex in $K$ is also in $K$, and the intersection of two simplices of $K$ is a face of both of the two simplices. There is also the notion of "abstract simplicial complex", which is a collection of subsets of $\{ 1, \dots, n \}$ which is closed under the operation of taking subsets. These kinds of simplicial complexes also have corresponding geometric realizations as topological spaces. I'll refer to both of these definitions as "the second definition".
The second definition looks reasonable at first sight, but then you quickly run into some horrible things, like the fact that triangulating even something simple like a torus requires some ridiculous number of simplices (more than 20?). On the other hand, you can triangulate the torus much more reasonably using the first definition (or alternatively using the definition of "Delta complex" from Hatcher's algebraic topology book, but this is not too far from the first definition anyway).
I believe you can move back and forth between the two definitions without much trouble. (I think you can go from the first to the second by doing some barycentric subdivisions, and going from the second to the first is trivial.)
Due to the fact that the second definition is the one that's listed on wikipedia, I get the impression that people still use this definition. My questions are:

Are people still using the second definition? If so, in which contexts, and why?
What are the advantages of the second definition?

REPLY [6 votes]: Abstract simplicial complexes have had quite a renaissance recently. Simplicial complexes were originally used to describe pre-existing topological spaces such as manifolds, as in the question. But now they are the key tool in constructing discrete models for topological spaces. The nerve of a covering of a set is a simplicial complex - if the set is a topological space and the subspaces are contractible (plus some technical conditions) the nerve has the same homotopy type as the space. The 
Wikipedia article http://en.wikipedia.org/wiki/Rips_complex is a good introduction.<|endoftext|>
TITLE: Why is algebraic geometry so over-represented on this site?
QUESTION [20 upvotes]: Seriously. As an undergrad my thesis was on elliptic curves and modular forms, and I've done applied industrial research that invoked toric varieties, so it's not like I'm a partisan here. But this can't be a representative cross-section of mathematical questions. How can this be fixed? (I mean the word "fixed".)

REPLY [2 votes]: I think that in some fields people are more active in the internet. There seems to be some correlation between the most popular tags here and the number of blogs in the corresponding section in the mathematics/statistics blog wiki here:
http://wiki.henryfarrell.net/wiki/index.php/Mathematics/Statistics
I do not know the relative proportion of algebraic geometers to other mathematicians but it may be in some fields the internet is more useful than others and there is more activity of the people in the field.<|endoftext|>
TITLE: The 2-sphere and $\mathbb{CP}^1$
QUESTION [6 upvotes]: As is very well known, the algebraic variety $S^2$ is isomorphic to projective variety $\mathbb{CP}^1$ as a complex manifold. As is also well known, the coordinate ring of $S^2$ is given by $< x,y,z > / < x^2 + y^2 +z^2 - 1 >$ and the function field of $\mathbb{CP}^1$ is $\mathbb{C}(\mathbb{CP}^1)$ (its coordinate ring being $\mathbb{C}$). My question is how the coordinate ring of $S^2$ and the function field of $\mathbb{CP}^1$ are related? Presumably this relation is a special case of a general variety-complex manifold relationship.

REPLY [10 votes]: Any projective variety is also a real affine variety, by using the real and imaginary parts of the coordinates $x_{jk} = z_j\overline{z_k}$.  You should first normalize the projective coordinates to have Hermitian-Euclidean length 1.  Ordinarily the projective coordinates are sections of a line bundle that is only defined up to a scalar.  But with the length normalization, the only ambiguity is the phase.  The phase cancels out in the definition of $x_{jk}$, so it is a well-defined function rather than just a section.
This realification of $\mathbb{CP}^n$ is a map to $(n+1) \times (n+1)$ Hermitian matrices.  It is important in quantum probability:  The image of the map is the set of pure states of a quantum system; its convex hull is the set of all states.  This convex region is the quantum analogue of the simplex of states (= measures = distributions) for classical probability on a finite set.  The matrices have unit trace, so the image lies in an $(n^2+2n)$-dimensional real subspace.  You can check that it is a sphere when $n=1$.
Another viewpoint that is important is that of toric varieties.  The phase part of the toric action on $\mathbb{CP}^1$ is rotation about the $z$ axis, and the moment map is the projection onto the $z$ coordinate.  This too generalizes to any projective toric variety.<|endoftext|>
TITLE: Coxeter Arrangements and an Identity
QUESTION [16 upvotes]: Let $\{A_i\}$ be a collection of $m$ hyperplanes in $\mathbb{C}^n$ which all pass through the origin (a central hyperplane arrangement).  Such an arrangement is called Coxeter if reflecting across any hyperplane in $\{A_i\}$ sends the arrangement to itself (and so the reflections automatically will generate a Coxeter group).
Now, I will define a rather random-seeming condition on an arbitrary central arrangement.  Choose a normal vector $n_i$ to each $A_i$.  Consider the function on $\gamma$ on $\mathbb{C}^n$ given by
$$ \gamma(v) = \sum_{1\leq i< j\leq m}(n_i,n_j)\left(\prod_{k\neq i,j} (n_k,v) \right) $$
The function $\gamma$ depends on the specific choice of normal vectors; however, whether $\gamma$ vanishes does not depend on the choice of normal vectors (since scaling a normal vector will scale the output).  Call a central hyperplane arrangement puzzling if $\gamma(v)=0$ for all $v$.
The 'puzzling' condition came up in studying a very specific research problem.  However, both the context and a day's worth of experimentation have lead to the following conjecture.
Conjecture: The puzzling arrangements are exactly the Coxeter arrangements.
It's worth noting that I can't show either direction of the conjecture.  Brute force computation says that for $n=2$, the conjecture is true.
Just from the form, it kind of reminds me of a Weyl character formula-type identity, but I don't really know much about those.  My hope is that this kind of identity is pretty well known to people who work with such things.

Edit: There's another way of stating this identity, that's closer to the context in which I encountered it (differential operators).  Let $n_i^*$ denote the function $(n_i,-)$, and let $d_i$ denote differentiation along the vector $n_i$.  Then
$$ \gamma = \left( \sum_i (n_i^*)^{-2}(n_i^*d_i-(n_i,n_i))\right) \prod_j n_j $$ 
Therefore, the 'puzzling' condition is then a statement about the function $\prod_jn_j$ being killed by a particular rational differential operator.

Edit 2: Fixed the definition of Coxeter arrangements... I want all the reflections, not just a generating set.

REPLY [15 votes]: Proof that Coxeter arrangements obey this identity: Group together summands according to the two-plane spanned by $n_i$ and $n_j$. For any two-plane $H$, every summand coming from that two-plane is divisible by $\prod_{n_k \not \in H} \langle n_k, v \rangle$. Factoring out this common summand, the contribution from $H$ is
$$\sum_{n_i \neq n_j,\ n_i, n_j \in H} \langle n_i, n_j \rangle \prod_{n_k \in H,\ n_k \neq n_i, n_j} \langle n_k, v \rangle.$$
This is the two dimensional example you've already done.
Proof that only Coxeter arrangements obey this identity:
Consider $H$, a two plane spanned by some $(n_i, n_j)$. Our first goal is to show that $H \cap \{ n_k \}$ is a dihedral root system.
Let $r$ be the number of hyperplanes in your arrangement. Let $S$ be the ring of polynomial functions and let $I$ be the ideal generated by the functions $\langle n, \ \rangle$, for $n \in H$. Note that every term of your sum which does not come from $(n_i, n_j)$ with $n_i$, $n_j \in H$ lies in $I^{r-1}$. So, the sum of the terms with $n_i$, $n_j \in H$ must be zero modulo $I^{r-1}$. 
As before, all of those terms are divisible by $\prod_{n_k \not \in H} \langle n_k, v \rangle$. This is not a zero divisor in $S/I^{r-1}$. So we can factor it out and deduce that
$$\sum_{n_i \neq n_j,\ n_i, n_j \in H} \langle n_i, n_j \rangle \prod_{n_k \in H,\ n_k \neq n_i, n_j} \langle n_k, v \rangle \equiv 0 \ \mathrm{mod} \ I^{r-1}$$
But the left hand side is degree $r-2$, so it must be identically zero. By the two dimensional example which you have already done, this shows that $\{ n_k: n_k \in H \}$ is a root system.
So, for any $n_i$ and $n_j$, the set of $n_k$ in the span of $(n_i, n_j)$ is a root system. In particular, the reflection of $n_i$ by $n_j$ is some $n_k$. So your whole set of vectors is a root system.
Warning: I have not, myself, checked the two dimensional case which I am relying on.<|endoftext|>
TITLE: Hermite normal form in families
QUESTION [5 upvotes]: How does Hermite normal form (over $Z$) vary in families? I.e. if I have an $n\times m$ matrix $M$ whose entries are integral polynomials in some integral variable $x$, how does the Hermite normal form of the integral matrix $M(p)$ (obtained by setting $x$ equal to $p$) vary as a function of $p$? What about the special case that the entries are (at most) linear in $x$?
The question is a bit open ended so answers could be of several kinds, eg:
(i) how certain integer programming problems associated to $M(p)$ depend on $p$; 
(ii) by explaining how the answer can be expressed in a way that generalizes Ehrhart theory;
(iii) by specializing to an important case that is well-understood;
(iv) in some other interesting way.
I would also really appreciate a pointer to any relevant literature.

REPLY [3 votes]: Hi "DC".  I think that I have worked out that the Hermite normal form is a "trichotomous quasipolynomial" in the variable $p$.  If $f:\mathbb{Z} \to \mathbb{Z}$ is a function, then my definition is that $f$ is a trichotomous quasipolynomial if it is a quasipolynomial for $x \gg 0$, possibly a different quasipolynomial for $x \ll 0$, and in between finitely many unrestricted values.
I think that if $R$ is a canonically Euclidean ring — Euclidean with canonically chosen quotients and remainders — then there is a Hermite normal form for matrices over $R$.  In particular, I think that $R$ does not have to be a Euclidean domain.
As a first try, let $A$ be the ring of all functions $f:\mathbb{Z} \to \mathbb{Z}$, using pointwise quotients and remainders.  Hermite normal form over this ring is a model of computing Hermite normal form for any $\mathbb{Z}$ family of integer matrices.   $A$ is sort-of a Euclidean ring, except it isn't Noetherian.
Let $B$ be the subring of $A$ consisting of trichotomous quasipolynomials.  Then I believe that $B$ is Noetherian and it is a Euclidean ring.  If that is correct, then you obtain a Hermite normal form that is also a trichotomous quasipolynomial.

It's not correct, at least not in any obvious way.  It is easy to check that $B$ is not only a subring, but is also closed under quotients with pointwise remainders.  In order to compute how $b(x)$ divides into $a(x)$, you can reduce to the case in which $a(x)$ and $b(x)$ are both polynomials.  Then for starters there is a quotient and a remainder in $\mathbb{Q}[x]$:
$$a(x) = q(x)b(x) + r(x).$$
Since $r(x)$ has lower degree than $b(x)$, its values are smaller than those of $b(x)$ when $x \gg 0$, so that part is okay; but it may be negative and $q(x)$ may not be integral.  We can fix all that by rounding $q(x)$, which creates quasipolynomial behavior; and by changing it by 1 to make $r(x)$ positive.  Also since $r(x)$ might have odd degree, there may be a different quasipolynomial solution when $x \ll 0$.
The part that is either not true or far from obvious is why $B$ is Euclidean.  My argument for that fell apart.  However, I can still show that the Euclidean algorithm for finitely many elements $a_1,\ldots,a_n$ of $B$ terminates in a finite number of steps, and consequently that the Hermite normal form stabilizes in a finite number of steps with trichotomous quasipolynomial entries.  The proof is a two-stage induction.  The outer stage is the sum of the degrees of $a_1,\ldots,a_n$.  If $\deg a_j < \deg a_k$ for some $j$ and $k$, then dividing $a_j$ into $a_k$ reduces the degree of $a_k$.  On the other hand, suppose that the degrees are all equal.  Then we can pass to a congruence class for the input $x$ in $\mathbb{Z}$ and apply a linear change of variables so that the leading coefficients are all integers.  Then (in the inner induction) the Euclidean algorithm on these polynomials, for $x \gg 0$, amounts to the Euclidean algorithm on their coefficients.  It is important, in this inner inductive part, to only change the variable $x$ once; don't worry if the lower-order coefficients are not integers.  Eventually a $0$ is produced and the degrees once again decrease.
Alas, this is a very informal writeup, but this time I think that it works.<|endoftext|>
TITLE: Going further on How sections of line bundles rule maps into projective spaces
QUESTION [8 upvotes]: My question is located in trying to follow the argument bellow. 
Given a normal algebraic variety $X$, and a line bundle $\mathcal{L}\rightarrow X$ which is ample, then eventually such a line bundle will have enough section to define an embedding $\phi:X\rightarrow \mathbb(H^0(X,\mathcal{L}^{\otimes d}))=\mathbb{P}^N$ (notice that $N$ depends on $d$). However, if the line bundle is NOT ample we can still say something about the existence of a certain map $\phi_d$; the so-called Iitaka fibration. I'll omit some details, but the argument of the construction goes (more or less) as follows. 
Suppose in the first place that $\mathcal{L}$ is base-point base. That is to say, that there are no points (or a set) of $X$ which all the hyperplanes of $\mathbb{P}^N$ pass through. Then such a line bundle will define a linear system $|\mathcal{L}^d|$ which gives rise to a morphism $\phi_d:X\rightarrow \phi_d(X)\subset\mathbb{P}^N$ (again $N$ depends on $d$). Such a map may not be an embedding, however $\phi_d:X\rightarrow \phi(X)$ is an algebraic fiber space. My question is the following. As I increase the value $d$ the image $\phi_d(X)\subset \mathbb{P}^N$ may change, however, as a matter of fact such an image "stabilizes" as $d$ gets larger. Meaning that if $d$ is large enough, the image if $\phi_d$ is the "same" regardless $d$. 
-What is the reason for this to happen?
-What is going on with all the sections of $\mathcal{L}^{\otimes d}$ that I am getting as I increase the value of $d$?. I'll appreciate any comment.
As a result, due to the fact that after a while we no longer care about the value of $d$, we can associate the space $X\rightarrow \phi(X)$ to the line bundle $\mathcal{L}$. Here, the variety $\phi(X)$ no longer depends on $d$.
Could someone comment further about the word "stabilizes"?.

REPLY [4 votes]: Let $R(X,\mathcal L)=\oplus_d H^0(X,\mathcal L^{\otimes d})$ as graded rings, $s_0,\dots s_m\in H^0(X,\mathcal L^{\otimes d})$ and finally $R(X,\mathcal L,s_{\cdot})$ the (graded) subring generated by the  $s_0,\dots s_m$ in $R(X,\mathcal L)$. 
Facts:

${s_0,\dots,s_m}$ define a rational map $\sigma: X\dashrightarrow \mathbb P^m$. If ${s_0,\dots,s_m}$ generate $H^0(X,\mathcal L^{\otimes d})$, then $\sigma$ agrees with $\phi_t$. 
If $\vert\{s_0,\dots,s_m\}\vert$ is a basepoint-free linear system, then $\sigma$ is a morphism.
and $\sigma(X)\simeq {\rm Proj}\, R(X,\mathcal L, s_{\cdot})$.
If $R(X,\mathcal L,s_{\cdot})$ has the property that for some $a\in\mathbb N$, $R(X,\mathcal L,s_{\cdot})_{ak}=R(X,\mathcal L)_{ak}$ for all $k\in \mathbb N, k\gg 0$, then 
${\rm Proj}\, R(X,\mathcal L, s_{\cdot})\simeq {\rm Proj}\, R(X,\mathcal L)$ (think of the $a$-uple embedding).
Since $R(X,\mathcal L)$ is finitely generated, if  $s_0,\dots s_m\in H^0(X,\mathcal  L^{\otimes d})$ are chosen so that they generate $H^0(X,\mathcal  L^{\otimes d})$, then 
for large enough $d$ the property in 3) will be satisfied, so by 1) and 2) $\phi_d(X)\simeq 
{\rm Proj}\, R(X,\mathcal L)$ which is independent of $d$, hence it is stabilized.<|endoftext|>
TITLE: Combinatorics of the Stasheff polytopes
QUESTION [12 upvotes]: First a little background for those unaware. The Stasheff polytopes (or associahedra) are certain convex polytopes that arise in the theory of $A_\infty$-algebras. There is one polytope for each $n\geq 2$ and is denoted by $K_n$. The $K_n$'s essentially encode the homotopies, higher homotopies and so on of the associativity relation. One way to describe $K_n$, which is has dimension $(n-2)$, is to take all rooted binary trees with $n$ leaves and take a suitable convex hull. For example, $K_2=\{\ast\}$, $K_3$ is an interval while $K_4$ is a pentagon. 
It is known that the number of vertices $v$ of $K_n$ is the $(n-1)^{th}$ Catalan number, i.e., $v=\frac{1}{n}{2n-2 \choose n-1}$. What can one say about the number of edges of $K_n$ and generally about counting faces of all codimension?

REPLY [11 votes]: Greg's explanation is way too complicated for a simple problem.  Here is a more direct combinatorial approach.  I know this is an old question, but just for the record.  
Fix a convex $n$-gon $Q \subset \Bbb R^2$ with set of vertices $V$. Consider a Gelfand-Kapranov-Zelevinsky realization of the associahedron in the space of functions: for every triangulation $T$ of $Q$, take $f_T: V \to \Bbb R$, where $f(v)=$area of triangles with vertex at $v \in V$.  Now consider a linear function $\phi: \Bbb R^n  \to \Bbb R$ by setting 
$$\phi(x_1,\ldots,x_n) = x_1 + \epsilon x_2 + \ldots \epsilon^{n-1} x_n,$$
where $\epsilon >0$ is very small.  Now, the triangulations $T$ of $Q$ correspond to binary trees and the edges of the associahedron are the pairs of triangulations obtained by a flip - the change of an edge in a quadrilateral. In the language of binary trees, these correspond to the edges as well. Let us compute the $h$-vector of the associahedron.  Recall that the index$(T)=$ the number of edges of $Q$ where $\phi$ is increasing, and observe that it is equal to the number of right edges in the corresponding binary tree.  Thus $h_i$ is equal to the Narayana number:
$$h_i = \frac{1}{n} \binom{n}{i}\binom{n}{i+1}$$
(you need to shift $n\gets n-2$ here).  Now the $f$-vector can be obtained via binomial identities:
$$F(t) = H(t+1), \ F(t)=\sum_k f_k t^k, \  H(t)=\sum_k h_k t^k$$
To see this example more carefully explained, see my book (chapter 8). For more on the associahedron and the ideas behind the GKZ realization, see Ziegler's Lectures on polytopes (chapter 4).<|endoftext|>
TITLE: Why forgetful functors usually have LEFT adjoint?
QUESTION [48 upvotes]: for forgetful functors, we can usually find their left adjoint as some "free objects", e.g. the forgetful functor: AbGp -> Set, its left adjoint sends a set to the "free ab. gp gen. by it". This happens even in some non-trivial cases. So my question is, why these happen? i.e. why that a functor forgets some structure (in certain cases) implies that they have a left adjoint? Thanks.

REPLY [5 votes]: the answer of reid barton is perfect and general. for categories of $\tau$-algebras, where $\tau$ is a type (thus consisting of function symbols and identities), we also have the following: let $\tau \to \sigma$ be a homomorphism of types, then there is a functor from $\sigma$-algebras to $\tau$-algebras and it has a left adjoint. this can be proved using freyd's adjoint functor theorem. this yields tons of examples:

the forgetful functor of $\tau$-algebras to sets has a left adjoint. in particular, free monoids, groups, modules, lie algebras etc. exist.
if $R \to S$ is a ring homomorphism, $S-Alg \to R-Alg$ has a left adjoint.
every ring has a free unital ring.
forgetful functors of algebra may be seen as the functors from $\sigma$-algebras to $\tau$-algebras, where $\tau \subseteq \sigma$ is a subtype; these have a left-adjoint. for example, from groups to monoids, from rings to abelian groups, from R-algebras to R-modules, and so on.

basically, it's all about representing functors, and since subobjects of Set are well-behaved in a certain sense, freyd's adjoint functor theorem tells you that you can do everything you want.<|endoftext|>
TITLE: Why does one think to Steenrod squares and powers?
QUESTION [42 upvotes]: I'm studying Steenrod operations from Hatcher's book. Like homology, one can use them only knowing the axioms, without caring for the actual construction. But while there are plenty of intuitive reasons to introduce homology, I cannot find any for the Steenrod operations. I can follow the steps in the proofs given by Hatcher, but I don't understand why one introduces all these spaces like $\Lambda X$, $\Gamma X$ and so on (in Hatcher's notation, I don't know if it's universal). Does anyone know how to get an intuitive grasp of what's going on?

REPLY [38 votes]: Here's how I explain Steenrod squares to geometers.  First, if $X$ is a manifold of dimension $d$ then one can produce classes in $H^n(X)$ by proper maps $f: V \to X$ where $V$ is a manifold of dimension $d-n$ through many possible formalisms - eg. intersection theory (the value on a transverse $n$-cycle is the count of intersection points), or using the fundamental class in locally finite homology and duality, or Thom classes, or as the pushforward $ f_*(1)$  where $1$ is the unit class in $H^0(V)$.   Taking this last approach, suppose $f$ is an immersion and thus has a normal bundle $\nu$.  If $x = f_*(1) \in H^n(X)$ then $Sq^i(x) = f_*(w_i(\nu))$.  This is essentially the Wu formula.
That is, if cohomology classes are represented by submanifolds, and for example cup product reflects intersection data, then Steenrod squares remember normal bundle data.<|endoftext|>
TITLE: What is an integrable system?
QUESTION [149 upvotes]: What is an integrable system, and what is the significance of such systems? (Maybe it is easier to explain what a non-integrable system is.) In particular, is there a dichotomy between "integrable" and "chaotic"? (There is an interesting Wikipedia article, but I don't find it completely satisfying.)
Update (Dec 2010): Thanks for the many excellent answers. I came across another quote from Nigel Hitchin:
"Integrability of a system of differential equations should manifest
itself through some generally recognizable features:

the existence of many conserved quantities

the presence of algebraic geometry

the ability to give explicit solutions.


These guidelines would be interpreted in a very broad sense."
(If there are some aspects mentioned by Hitchin not addressed by the current answers, additions are welcome...)
Closely related questions:

What does it mean for a differential equation "to be integrable"?
basic questions on quantum integrable systems

REPLY [2 votes]: "All integrable Hamiltonian systems are alike, while each nonintegrable one is nonintegrable in its own way", Valerij V. Kozlov (after L. N. Tolstoi of course), Symmetries, Topology and Resonances in Hamiltonian Mechanics, Springer, 1996<|endoftext|>
TITLE: Prime numbers $p$ not of the form $ab + bc + ac$ $(0 < a < b < c )$ (and related questions)
QUESTION [8 upvotes]: If we ask which natural numbers n are not expressible as $n = ab + bc + ca$ ($0 < a < b < c$) then this is a well known open problem. Numbers not expressible in such form are called    Euler's "numerus idoneus" and it is conjectured that they are finite.
If we omit the condition $a < b < c$ and assume $0 < a \leq b \leq c$ then it was proved (assuming Generalized Riemann Hypothesis) that there is only a finite number of such numbers $n$.
I am interested in the problem of expressing a prime number $p$ as $p = ab + ac + bc$ for $a \geq 1$ and $b,c \geq 2$.
Anybody knows if there is some known result related to expressing prime numbers in such form? This would yield (as a corollary) a very beautiful theorem related to spanning trees in graphs.

REPLY [9 votes]: Looking at the encyclopedia of integer sequnces, I've found that the set of numbers not expressible as ab + bc + ac 0 < a < b < c is finite.
http://oeis.org/A000926
Chowla showed that the list is finite and Weinberger showed that there is at most one further term.<|endoftext|>
TITLE: Lecture notes on representations of finite groups
QUESTION [31 upvotes]: Next term I am supposed to teach a course on representation of finite groups. This is a third year course for undegrads. I was thinking to use the book of Grodon James and Martin Liebeck "Representations and characters of groups", but also looking for other references. 
The question is:  could you advise some other books (or lecture notes)? Maybe you had a nice experience of teaching or listening to a course with a similar title? It would be really nice if this book (notes) has also exercises.
ADDED. I would like to thank everybody who answered the question, very helpful answers!!! The answer of John Mangual below contains a "universal" reference. 
For the moment my favourites are Serre (very clear and short introduction of main ideas), some bits from notes of Teleman and Martin, and Etingof for beautiful exposition. My last problem is to have enough of exercises, in particular to write down a good exam. So I would like to ask if there are some additional references for exercises (with or without solutions)?

REPLY [7 votes]: At the risk of tooting my own horn, I have a new book that will be published by Springer which is a course on representation theory of groups intended for undergrads and beginning grads.  It assume only linear algebra, group theory and basic ring theory.  It assumes no module theory.  Included are applications to combinatorics and probability.  The link is
http://www.springer.com/mathematics/algebra/book/978-1-4614-0775-1?detailsPage=authorsAndEditors
and it should be out by the end of the year.<|endoftext|>
TITLE: Integration in equivariant K-theory
QUESTION [9 upvotes]: Let F be a smooth classifying space for K-theory (ordinary or equivariant). If X is a smooth compact manifold and W is a real vector space of dimension n, there is an integration map from the compactly supported K-theory of X x W to that of X. It can be represented on the level of classifying maps by regarding a map f from X x W into F as a map from X into the nth loop space of F.
Does a similar picture exist in equivariant K-theory, where X is a G-space and W is a group representation? Of course it does if W is simply a trivial representation. But what if W is some other representation? Is there an integration map, and can one represent it on the level of classifying maps? Is there a good reference for this?

REPLY [7 votes]: Atiyah gave a very precise answer to this question: an equivariant vector bundle V/M is K-orientable (satisfies the Thom isomorphism for K-theory, i.e. K(B(V),S(V)) is a rank-one free module over K(M)) if and only if V has an equivariant Spin_c-structure. Atiyah's proof definitely involves analysis (the family index). The proof by Kasparov is a little more technical, since the definition of the product in Kasparov's KK theory is rather technical. (It is a generalization of the family index theorem of Atiyah and Singer.) But all of these proofs come down to the same thing: the study of families of equivariant Fredholm operators.<|endoftext|>
TITLE: Hodge theory for quasi-Kaehler manifolds: where does it break down?
QUESTION [7 upvotes]: Let $U$ be the complement of a divisor with normal crossings in a smooth compact complex manifold $X$. If $X$ is algebraic, then Deligne describes in "Th\'eorie de Hodge 2" a procedure to equip the rational cohomology of $U$ with a mixed Hodge structure. Most of this procedure can be carried out if the $\partial\bar\partial$-lemma holds for $X$, in particular, if $X$ is bimeromorphic to a K\"ahler manifold. But there are also some things that break down, for example, the functoriality and the independence of the compactification (for both the resolution of singularities is used).

Is it true that the rest of section 3 (of Th\'eorie de Hodge 2) works as soon as $X$ is, say, K\"ahler? For example, is it true that the rational cohomology of $U$ carries a mixed Hodge structure such that the weight filtration is the Leray filtration induced by the open embedding $U\to X$?
If so, are there examples of K\"ahler compactifications that give different rational mixed Hodge strutures? Non-isomorphic Hodge structures?

REPLY [4 votes]: I may be misunderstanding question 2, but there is an example (due to Serre) of a complex surface U with 2 different K\"ahler compactifications with different mixed Hodge structures.  By projectivizing a nontrivial extension of the trivial line bundle by itself over an elliptic curve E, you get a P^1 bundle over the curve, and the trivial subbundle gives a section, which you delete to get U, which is a C-bundle over E.  U is complex-analytically isomorphic to $\mathbb{C}^\times \times \mathbb{C}^\times$ and can be compactified to $\mathbb{P}^1 \times \mathbb{P}^1$.  H^1(U) in the P^1-bundle lies in W_1, while H^1(U) in the product of projective lines has pure weight 2.
(From Peters-Steenbrink Mixed Hodge Structures section 4.5)<|endoftext|>
TITLE: Morita equivalence and moduli problems
QUESTION [9 upvotes]: Two rings $A$ and $B$ are said to be Morita equivalent if the category of modules over $A$ and $B$ are equivalent as additive categories. (Here I'm considering left modules).
Ex: $M_n(R)$ (the algebra of matrices over a ring $R$) is morita equivalent to $R$.
In fact more generally whenever $A$ is a ring and $e$ is an idempotent in $A$ and $AeA = A$ then the follwing functor is a morita equivalence:
$A$  modules $\rightarrow$  $eAe$ modules
$M$ $\mapsto$ $eM$
Now under nice conditions the categories of $A$ modules (resp $B$ modules) might have a moduli description. Then can you say anything about the induced map on the moduli spaces? I'm asking for situations in which this is known.
Also is there a nice book or paper which talks about Morita equivalence and has lots of examples?

REPLY [6 votes]: In the setting you are interested in, that is, finitely generated k-algebras and GIT-quotients of closed or semistable orbits, Morita equivalence induces isomorphism on the moduli spaces provided one scales dimension(vectors) and stability structures accordingly. That is, if B=M_n(A) one should compare Mod(A,k) to Mod(B,nk) moduli. If A and B have a complete set of orthogonal idempotents e_i resp. f_i(that is, a quiver-like situation) and if the Morita equivalence induces rank(f_i) = n_i rank(e_i), then one should compare A-reps of dimension vector alpha=(alpha_i) to B-reps with dimension vector beta=(n_i alpha_i).
Geometrically, the module varieties of Morita-equivalent algebras are related via associated fibre bundle constructions. In the example above, Mod(B,kn) = GL(nk) x^GL(k) Mod(A,k). In general,one had such a description locally in the Zariski topology (coming roughly from the fact that a vectorbundle (or projective) is locally free). Anyway, this gives a natural and geometric one-to-one correspondence between GL(nk)-orbits (isoclasses) of B-reps and GL(k)-orbits of A-reps inducing the desired isos on the quotient variety level (isos of semi-simples). In quiver-like situations one should adjust the dim vectors as above.
Now as to semi-stability. Observe that semi-stable reps are just ordinary reps of a universal localization of your algebra(s), so one can reduce to the closed case by covering the variety of semi-stables by Zariski open (affine) pieces. In quiver-like situations when the Morita-data is as above and your stability structure mu for A is given by the vector (mu_i), then the corresponding stability structure for B is mu'=(1/n_i x mu_i) (or multiply it with a common factor if you want it to have integral components.<|endoftext|>
TITLE: When is a localization of a commutative ring finitely generated as an algebra?
QUESTION [7 upvotes]: Prove or counterexample: If A is a commutative ring and $A_p$ is a finitely generated algebra over A for all prime ideal p of A, then A is a product of local rings.

REPLY [11 votes]: Here is a counterexample. Let p and q be distinct prime numbers, let S denote the complement of {p,q} in the set of all primes, and let A denote Z[1/S]. Then the prime ideals of A are pA, qA, and {0}, the first two of which are maximal. Also, since A is domain (being a subring of Q) with two maximal ideals, it's not a product of local rings. On the other hand, the local rings at the various prime ideals are A[1/q], A[1/p], and A[1/p,1/q], which are clearly finitely generated as A-algebras.

REPLY [2 votes]: In the case that $A$ is noetherian and we replace finitely generated as an algebra with finitely generated as a module we can argue as follows. We have for any choice of minimal prime ideal $P$ of $\mathrm{Spec} A$ that $A_P$ is flat and finitely presented hence projective. It is also an artinian local ring. As it is projective the corresponding sheaf is locally free on $\mathrm{Spec} A$. But
$$\mathrm{Supp} A_P = \mathrm{Spec} A_P$$
so that if $\mathrm{Spec} A$ were connected it would have to agree with the spectrum of $A_P$ since a locally free sheaf has constant rank on connected components. Thus $A$ is in fact artinian and local.
$$ $$
If $A$ does not have connected spectrum then we can rewrite $A$ as $A_P \times A_1$. Where $A_P$ doesn't split up at all. We then just keep playing the same game (next with $A_1$) which terminates since $A$ is noetherian so has finitely many connected components and we in fact see that $A$ is a product of artinian local rings.
The stronger hypotheses I have are probably only somewhat necessary. I don't see a problem (except one needs to remove the word artinian) if one just assumes that the localizations are finitely presented as modules - there is the issue of whether or not this terminates but if you have an infinite product you also pick up extra prime ideals via choice and I have no idea what localizing at one of those would look like so I am not quite sure if this causes problems.<|endoftext|>
TITLE: Sets, Universes, and the small Yoneda Lemma
QUESTION [7 upvotes]: Suppose we fix a universe $U$ and a $U$-small category $C$.  The regular Yoneda lemma gives us some locally small (not necessarily locally U-small?) functor category $C'=[C^{op},Sets]$ with a fully faithful embedding $C\rightarrow C'$ and the canonical bijection between $Nat(F,Hom(-,x))$ and $F(X)$.  Suppose we consider now, the U-small Yoneda lemma, that is, we look at $[C^{op},U-Sets]$.  This is well-behaved since even though it is not U-small, $Ob([C^{op},U-Sets])$ is still a set.  
So the main question I have is: Are there any useful properties of the standard Yoneda lemma that we cannot reproduce with the $U$-small Yoneda lemma for some $U$?

REPLY [4 votes]: Well, for any universe U, the U-small sets satisfy the axioms of ZFC (or whatever your preferred set theory is).  Therefore, anything that we can prove in ZFC about the Yoneda embedding into "all" presheaves will also be true about the Yoneda embedding into U-small presheaves.  So on that score, the answer would seem to be "no."
There might be other interpretations that would make the answer "yes," since in vanilla ZFC every class is definable, whereas not every U-large set is definable in terms of U-small ones.  Thus an informal statement like "for every large thingumbob X, ..." might be true in ZFC, but no longer provable relative to a universe, since the meaning of "large" has changed (unless we change the quantifier to "for every U-small-definable large thingumbob X").  This doesn't contradict the first observation, since such a statement cannot be a single theorem of ZFC, only a meta-theorem, and each instance of the meta-theorem is about a particular definable class and therefore still true about the corresponding U-small-definable U-large set.  However, right now I can't think of any interesting or useful properties of the Yoneda embedding that would fall under this heading.
So I think that probably the answer is still "no."<|endoftext|>
TITLE: How long for a simple random walk to exceed $\sqrt{T}$?
QUESTION [13 upvotes]: Let $R_n$ be a simple random walk with $R_0 = 0$, and let $T$ be the smallest index such that $k\sqrt{T} < |R_T|$ for some positive $k$.
What is an expression for the probability distribution of $T$?

REPLY [3 votes]: Building on Yemon, who suggests that the solution is some distribution of a hitting time, if we assume the threshold for the 'hit' is $\sqrt{T}$, and that variance, $u$, of the random variable is directly proportional to the square root of time, then the mean time for that variable to exceed $\sqrt{T}$ may equal $ku\sqrt{T} \Phi(1)/2$, or approximately $ku\sqrt{T} \cdot 0.16$, distributed lognormally.<|endoftext|>
TITLE: Ramsey Theory, monochromatic subgraphs
QUESTION [10 upvotes]: If we have the complete countably infinite bipartite graph $K_{\omega,\omega}$ and we colour the edges with just two colours. Should we expect to get a monochromatic copy of $K_{\omega,\omega}$. 
Infinite Ramsey theorem gives us infinitely many edges of one colour but this is no enough.

REPLY [15 votes]: The example given by Konstantin Slutsky is, however, essentially unique, in the following sense. Let $G$ be the complete bipartite graph with both vertex sets equal to (copies of) $\mathbb{N}$ and colour its edges red or blue. For each $m< n$, colour the set $\{m,n\}$ according to whether the edge $(m,n)$ is red or blue and whether the edge $(n,m)$ is red or blue. (So this is a 4-colouring of the complete graph on $\mathbb{N}$.) By Ramsey's theorem we can find a subset $X$ of $\mathbb{N}$ such that all pairs $\{m,n\}$ with $m,n\in X$ have the same colour. If the colour is RED-RED or BLUE-BLUE then we have a monochromatic bipartite subgraph by choosing two disjoint subsets $Y$ and $Z$ of $X$ (to avoid the problem when $m=n$). If the colour is RED-BLUE then we can again pass to two disjoint subsets, which we could even make alternate, and the edge $(m,n)$ will then be RED if $m< n$ and BLUE if $m> n$, and similarly if the colour is BLUE-RED. So we either find a monochromatic subgraph or we find Konstantin Slutsky's example. This is an example of a "canonical Ramsey theorem" -- you can't find a monochromatic structure but you can find a simple "canonical" structure.<|endoftext|>
TITLE: Nimber multiplication
QUESTION [16 upvotes]: Is there a game-theoretic interpretation of nimber multiplication? There is such for addition (a single move in a+b is either a move in a or a move in b).

REPLY [4 votes]: In part 2 of Game Theory by Thomas Ferguson, example 2 'Turning Corners' on page 33, Thomas Ferguson mentions a so-called 'flipping-coin' game, where the Sprague-Grundy functions g(x, y) equals nim multiplication of x and y.

A move consists of turning over four distinct coins at the corners of a rectangle, i.e.
  (a, b), (a, y), (x, b) and (x, y), where 0 ≤ a < x and 0 ≤ b < y, the coin at (x, y) going from
  heads to tails.

This is under the (in part 2 of this book) usual assumptions that the game is for 2 players, and the last player that makes a move wins (i.e., you lose when you can't make a valid move according to the rules of the game). This might be the game that Alex Fink referred to, but I didn't find it to be such a contrived example, so it might be worth looking at (to be honest, at this point I don't understand the game completely and I don't have time for it now: this might be the same game that Lenstra mentions in the article linked to by Kevin O'Bryant: both are 'coin-turning' games).
Edit: In fact, the Sprague-Grundy value of a move in any 'product' of two coin-turning games G1 and G2 is given by the nimber product of the SG-value of the corresponding move in G1 and the corresponding move in G2. What this means exactly is explained more clearly in the document I linked to.<|endoftext|>
TITLE: Kapranov's analogies
QUESTION [14 upvotes]: I just wonder about Kapranov's "Analogies between Langlands Correspondence and topological QFT". I would like to read a more detailed exposition and how one turns that analogy into concrete conjectures. Do there exist texts from courses or seminars on that? Has this book by İkeda already been published and reviewed?

REPLY [25 votes]: There are notes from more recent (2000) lectures of Kapranov on the subject on my webpage.
Despite this, as far as I know, there isn't any convincing argument at this point that there should be a general higher dimensional Langlands theory. What we learn from physicists is that one can expect deep dualities for algebraic surfaces and threefolds, but probably not in general. And we also learn that the shape of these dualities won't be one
you are likely to guess from analogies with curves and with higher dimensional class field theory, but for which string theory is a great guide. One way to put this is that we now (post 2005) understand that geometric Langlands is an aspect of four-dimensional topological field theory (more specifically, maximally supersymmetric gauge theory) — so one ought to look for higher dimensional analogs of this, and they are very very few and very special.
Perhaps the main issue in making direct generalizations (that Kapranov addresses) is the great complexity in describing Hecke algebras in higher dimensions — in fact this is one of the most exciting areas of current research, under the name Hall Algebras (cf work of Joyce, Kontsevich-Soibelman and others). 
It turns out that Hall algebras of 3-dimensional (Calabi-Yau) varieties have a remarkably rich representation theoretic structure, which (upon dimensional reduction to surfaces or curves) encapsulates much of classical representation theory and geometric Langlands etc.
In any case the most exciting thing around as far as I'm concerned is a 6-dimensional supersymmetric field theory (which lacks a good name — it's called the (0,2) theory or the M5 brane theory) which seems to enfold everything we know in all of the above examples. It is a conformal field theory, and "explains" Langlands duality simply as the SL2(Z) conformal symmetry of tori (when one considers it on R^4 times a torus). 
One can expect this theory to lead to an interesting Langlands-like program for algebraic surfaces, which includes in particular the interesting recent work in this direction of Braverman-Finkelberg (arXiv:0711.2083 and 0908.3390). 
I should mention also that since
Kapranov's paper there has been a lot of work, by Kazhdan, Braverman, Gaitsgory and Kapranov in particular, on representation theory for higher dimensional local fields — it's just that there isn't yet a clear Langlands type picture for this.
Again one should expect something special to happen for surfaces, of which we have many glimpses — in particular the beautiful theory of Cherednik's double affine Hecke algebra and its Fourier transform, which is I think accepted as a clear hint at a Langlands picture for surfaces. But in any case I strongly believe that the current evidence indicates we should be learning from the physicists, reading papers by Gaiotto etc, to figure out what should happen for surfaces rather than just believing that we should try to extrapolate from what we know for curves.<|endoftext|>
TITLE: (Non-trivial) presentation of general linear and symplectic group over Z/mZ?
QUESTION [8 upvotes]: I know that there exists a nice presentation (generators and relations) of the general linear group over a finite field (by Steinberg, I think). Is there also a nice presentation of $GL(n,\mathbb{Z}/m\mathbb{Z})$ for an arbitrary integer $m$? And perhaps also for the symplectic group over $\mathbb{Z}/m\mathbb{Z}$?
I want to do some calculations in these groups with a computer and my first problem was to determine these groups. One solution (my current) is of course to determine both by brute force. But already for small m,n this takes too much time.

REPLY [9 votes]: Two things.  I'll talk about SL_n rather than GL_n, but that's just a technicality.

There is a short exact sequence

$1 \rightarrow SL_n(\mathbb{Z},m) \rightarrow SL_n(\mathbb{Z}) \rightarrow SL_n(\mathbb{Z}/m\mathbb{Z}) \rightarrow 1,$
where $SL_n(\mathbb{Z},m)$ is the level $m$ congruence subgroup of $SL_n(\mathbb{Z})$.  There is a nice presentation $\langle S|R \rangle$ for $SL_n(\mathbb{Z})$ due essentially to Magnus; it can be found in Milnor's book on algebraic K-theory (look for the chapter that calculates $K_2(\mathbb{Z})$).  The generating set $S$ is just the set of elementary matrices.  To get from there to $SL_n(\mathbb{Z}/m\mathbb{Z})$, you just need to add a normal generator for $SL_n(\mathbb{Z},m)$ to $R$.  By the solution to the congruence subgroup problem for $SL_n(\mathbb{Z})$ (due to Mennicke, but you are better off looking at Bass-Milnor-Serre's paper), this congruence subgroup is normally generated by the mth power of a single elementary matrix.
A similar idea also works for the symplectic group.  To find a presentation for $Sp_{2g}(\mathbb{Z})$, look at Theorem 9.2.13 of Hahn and O'Meara's book "The Classical Groups and K-Theory".

I also want to make a comment on DLS's answer.  There is a related short exact sequence

$1 \rightarrow V \rightarrow SL_n(\mathbb{Z}/p^{k+1}\mathbb{Z}) \rightarrow SL_n(\mathbb{Z}/p^k\mathbb{Z}) \rightarrow 1.$
The group $V$ has a beautiful description, due essentially to Lee and Szczarba.  Namely, it is isomorphic to the additive group underlying the special lie algebra over $\mathbb{Z} / p\mathbb{Z}$ (in particular, it is abelian).  Moreover, the action of $SL_n(\mathbb{Z}/p^k\mathbb{Z})$ on $V$ factors through the adjoint representation of $SL_n(\mathbb{Z}/p\mathbb{Z})$ on the special lie algebra.
Lee and Szczarba did not write this down in quite this form, but I wrote out a similar result for the symplectic group in Lemma 3.1 of my paper "The Picard group of the moduli space of curves with level structures".<|endoftext|>
TITLE: What is the relationship between various things called holonomic?
QUESTION [25 upvotes]: The following things are all called holonomic or holonomy:

A holonomic constraint on a physical system is one where the constraint gives a relationship between coordinates that doesn't involve the velocities at all.  (ie, $r=\ell$ for the simple pendulum of length $\ell$)
A D-module is holonomic (I might not have this one quite right) if the solutions are locally a finite dimensional vector space, rather than something more complicated depending on where on the manifold you look (this is apparently some form of being very over-determined)
A smooth function is holonomic if it satisfies a homogeneous linear ODE with polynomial coefficients.
On a smooth manifold $M$, with vector bundle $E$ with connection $\nabla$, the holonomy of the connection at a point $x$ is the subgroup of $GL(E_x)$ given by the transformations you get by parallel transporting vectors along loops via the connection.

Now, all of these clearly have something to do with differential equations, and I can see why 2 and 3 are related, but what's the common thread? Is 4 really the same type of phenomenon, or am I just looking for connections by terminological coincidence?

REPLY [2 votes]: I don't know the connection between the four points, but it may help to redefine holonomy in your first point. A constraint is generally a regular distribution (in the sense of a subbundle of the tangent bundle of the configuration manifold). 
If that distribution is integrable, then the constraint is said to be holonomic. And indeed, by restricting the system to one particular integral submanifold, one obtains the definition of holonomy in your first point.
The point is, in mechanics, "holonomic" is just another word for "integrable distribution". If the constraint distribution is not integrable, the system is called "nonholonomic".
Maybe that may help to make a connection with point 4.?<|endoftext|>
TITLE: Applications of the "other" definition of sheaves
QUESTION [24 upvotes]: In most literature, when you try to look for the definition of sheaves you will see the usual definition for presheaves as a functor from a topological space (or from a Grothendieck topology) to some category and then sheaves would require this category to be complete and you have some exactness/equalizer condition. 
But then for some categories there is another equivalent definition. You are defined a "protosheaf" (there are various names for these creatures), a sheaf space, a base space, a local homeomorphism between the sheaf space and the base space, you are even already defined a stalk.. but this definition seems not to be very abstract in the category-theoretical point of view as I only see this kind of definition for very specific categories (for instance in the category of groups or rings, you want the addition operation defined on the fiber product of the sheaf space over the base space to be continuous). What is the equivalent category theoretical way of defining a sheaf using this method? In which cases does this definition give us a more psychological advantage than the aforementioned one? I have personally found the former definition more advantageous in my practice, but there are some mathematical practices by which the latter definition might be more useful.

REPLY [6 votes]: There is an important application of the "other definition" in arithmetic geometry: It is used to give a canonical action of the frobenius on sheaf defined over a finite field:
There is an equivalence of categories between constructible sheaves ("usual sheaves") on a variety $X$ and algebraic spaces etale over $X$ ("other sheaves"). 
Now one can use the frobenius action on spaces and translate it back through the equivalence into an action on sheaves.<|endoftext|>
TITLE: Evidence for $Q^{\operatorname{solv}}$ being pseudo-algebraically-closed
QUESTION [12 upvotes]: This is a follow-up to the following answer:
Solvable class field theory
in which it is stated as a "folklore" conjecture that the maximal solvable extension of Q is pseudo algebraically closed (this means, in particular, any geometrically connected variety over Q has a point over a solvable extension).
I am curious what evidence there is to support such a conjecture.
In addition, what can be said for the analogous statement for global function fields?

REPLY [4 votes]: I'll respond to this question with another question.  Given a curve C/Q, there is a solvable extension K/Q such that C/K has local points everywhere.  (At least I think this is is the case; it requires some kind of argument on approximating local solvable extensions by global ones, which I've seen done before.) Some people believe that the Brauer-Manin obstruction is the only obstruction to the Hasse principle for curves.  If that were the case, would it say anything about your question?<|endoftext|>
TITLE: What is the algebraic closure of the field with one element?
QUESTION [28 upvotes]: If doing geometry over $\mathbb F_p$ means also using its algebraic closure, it must be interesting to talk about the algebraic closure of $\mathbb F_1$ - the field with one element.
I saw that the finite extensions of $\mathbb F_1$ are considered as $\mu_n$, but an article by Connes et al says that it is unjustified to think of the direct limit of these.
In their paper, the group ring $\mathbb Q[\mathbb Q/\mathbb Z]$ appears a lot. Maybe it's one of $\mathbb Q/\mathbb Z$, $\mathbb Q[\mathbb Q/\mathbb Z]$, $\mathbb Z[\mathbb Q/\mathbb Z]$ ?
What is the algebraic closure of the field with one element?
And then, what is $\overline{\mathbb F_1} \otimes_{\mathbb F_1}\mathbb Z$? 
This seems like a very interesting question...

REPLY [34 votes]: The algebraic closure of F_1 is the group of all roots of unity, and, tensoring it with Z gives the integral group ring Z[mu_infty], or, if you prefer Z[Q/Z].
For a readable account (and for folklore references such as Kapranov-Smirnov) see Yu. I. Manin's "Cyclotomy and analytic geometry over F_1" (http://arxiv.org/abs/0809.1564).<|endoftext|>
TITLE: Examples of mathematics motivated by technological considerations
QUESTION [33 upvotes]: I would like examples of technological advances that were made possible only by the creation of new mathematics.  I'm talking about technology that was desired in some period of history but for which the mathematics of the time were not enough.
It would be great if I could have an example in which the technology needed was associated with an industrial problem, and it would be even better if the technology developed had a significant impact in society.

REPLY [2 votes]: One of the most important tools used in Computer Science is the Fast Fourier Transform, which basically made much of image processing possible - described by Gilbert Strang as "the most important numerical algorithm of our lifetime". Note however that unknown to the inventors James Cooley and John Tukey it seems that Gauss actually had already discovered the algorithm (see this paper for more details) but not published it! 
Much of modern medical imaging is only possible due to mathematical advances - for a flavour here are a few example projects from the University of Eindhoven:

Lie Group Theory & Differential Geometry for Medical Image Analysis, Remco Duits
Riemann-Finsler Geometry for Brain Connectivity and Tractography, Luc Florack & Andrea Fuster
Differential Geometry in Complex Medical Imaging & Relativity Theory, Andrea Fuster<|endoftext|>
TITLE: Double affine Hecke algebras and mainstream mathematics
QUESTION [59 upvotes]: This is something of a followup to the question "Kapranov's analogies", where a connection between Cherednik's double affine Hecke algebras (DAHA's) and Geometric Langlands program was mentioned.
I am interested to hear about known connections of the double affine Hecke algebras, or their various degenerations called rational and trigonometric degenerate DAHA's (or simply Cherednik algebras) to other areas of mathematics.  Affine Hecke algebras answers are also interesting; I once asked a friend if he planned to attend a talk about DAHA, and was told "nah, that sounds like a little too much affine Hecke algebra for my tastes."  As prompts for answers, I'll share my limited understanding in the hopes a kind reader will elaborate:

I understand that they were introduced to solve the so-called Macdonald conjectures, though I don't know much about this and would be delighted to hear anyone's thoughts.
I understand that they can be defined in terms of certain "Dunkl-Opdam" differential operators, and can be thought of as the algebra of differential operators on a non-commutative resolution of $\mathfrak{h}/W$.  I'd be interested to hear more about this.
I understand that they have an action by automorphisms of $\widetilde{SL_2(Z)}$ coming alternately from their connection to configurations of points on a torus, or from their description via differential operators.  If anyone has another take on this action, that would be interesting to hear about.
Apparently they also relate to the geometric Langlands program, which is what really prompts my post; can anyone elaborate on this?

REPLY [52 votes]: Well the first thing to say is to look at the very enthusiastic and world-encompassing papers of Cherednik himself on DAHA as the center of the mathematical world (say his 1998 ICM). 
I'll mention a couple of more geometric aspects, but this is a huuuge area..
There are at least three distinct geometric appearances of DAHA, which you could classify by the number of loops (as in loop groups) that appear - two, one or zero.
(BTW for those in the know I will mostly intentionally ignore the difference between DAHA and its spherical subalgebra.)
Double loop picture: See e.g. Kapranov's paper arXiv:math/9812021 (notes for lectures of his on it available on my webpage) and the related arXiv:math/0012155. The intuitive idea, very hard to make precise, is that DAHA is the double loop (or 2d local field, such as F_q((s,t)) ) analog of the finite (F_q) and affine (F_q((s)) ) Hecke algebras. In other words it appears as functions on double cosets for the double loop group and its "Borel" subalgebra. (Of course you need to decide what "functions" or rather "measures" means and what "Borel" means..)  This means in particular it controls principal series type reps of double loop groups, or the geometry of moduli of G-bundles on a surface, looked at near a "flag" (meaning a point inside a curve inside the surface). The rep theory over 2d local fields that you would need to have for this to make sense is studied in a series of papers of Kazhdan with Gaitsgory (arXiv:math/0302174, 0406282, 0409543), with Braverman (0510538) and most recently with Hrushovski (0510133 and 0609115). The latter is totally awesome IMHO, using ideas from logic to define definitively what measure theory on such local fields means.
Single loop picture: Affine Hecke algebras have two presentations, the "standard" one (having to do with abstract Kac-Moody groups) and the Bernstein one (having to do specifically with loop groups). These two appear on the two sides of Langlands duality (cf eg the intro to the book of Chriss and Ginzburg). Likewise there's a picture of DAHA that's dual to the above "standard" one. This is developed first in Garland-Grojnowski (arXiv:q-alg/9508019) and more thoroughly by Vasserot arXiv:math/0207127 and several papers of Varagnolo-Vasserot. The idea here is that DAHA appears as the K-group of coherent sheaves on G(O)\G(K)/G(O) - the loop group version of the Bruhat cells in the finite flag manifold (again ignoring Borels vs parabolics). Again this is hard to make very precise. This gives in particular a geometric picture for the reps of DAHA, analogous to that for AHA due to Kazhdan-Lusztig (see again Chriss-Ginzburg). 
[EDIT: A
new survey on this topic by Varagnolo-Vasserot has just appeared.]
Here is where geometric Langlands comes in: the above interp means that DAHA is the Hecke algebra that acts on (K-groups of) coherent sheaves on T^* Bun_G X for any Riemann surface X -- it's the coherent analog of the usual Hecke operators in geometric Langlands.
Thus if you categorify DAHA (look at CATEGORIES of coherent sheaves) you get the Hecke functors for the so-called "classical limit of Langlands" (cotangent to Bun_G is the classical limit of diffops on Bun_G).
The Cherednik Fourier transform gives an identification between DAHA for G and the dual group G'. In this picture it is an isom between K-groups of coherent sheaves on Grassmannians for Langlands dual groups (the categorified version of this is conjectured in Bezrukavnikov-Finkelberg-Mirkovic arXiv:math/0306413). This is a natural part of the classical limit of Langlands: you're supposed to have an equivalence between coherent sheaves on cotangents of Langlands dual Bun_G's, and this is its local form, identifying the Hecke operators on the two sides!
In this picture DAHA appears recently in physics (since geometric Langlands in all its variants does), in the work of Kapustin (arXiv:hep-th/0612119 and with Saulina 0710.2097) as "Wilson-'t Hooft operators" --- the idea is that in SUSY gauge theory there's a full DAHA of operators (with the above names). Passing to the TFT which gives Langlands kills half of them - a different half on the two sides of Langlands duality, hence the asymmetry.. but in the classical version all the operators survive, and the SL2Z of electric-magnetic/Montonen-Olive S-duality is exactly the Cherednik SL2Z you mention..
Finally (since this is getting awfully long), the no-loop picture: this is the one you referred to in 2. via Dunkl type operators. Namely DAHA appears as difference operators on H/W (and its various degenerations, the Cherednik algebras, appear by replacing H by h and difference by differential). In this guise (and I'm not giving a million refs to papers of Etingof and many others since you know them better) DAHA is the symmetries of quantum many-body systems (Calogero-Moser and Ruijsenaars-Schneiders systems to be exact), and this is where Macdonald polynomials naturally appear as the quantum integrals of motion.
The only thing I'll say here is point to some awesome recent work of Schiffmann and Vasserot arXiv:0905.2555, where this picture too is tied to geometric Langlands..
very very roughly the idea is that H/W is itself (a degenerate version of an open piece of) a moduli of G-bundles, in the case of an elliptic curve. Thus studying DAHA is essentially studying D-modules or difference modules on Bun_G in genus one (see Nevins' paper arXiv:0804.4170 where such ideas are developed further). Schiffman-Vasserot show how to interpret Macdonald polynomials in terms of geometric Eisenstein series in genus one.. 
enough for now.<|endoftext|>
TITLE: Motivation for uniform surjectivity of mod l representations associated to elliptic curves
QUESTION [12 upvotes]: Background
Let $E$ be an elliptic curve over $\mathbb{Q}$ and let $G_{\mathbb{Q}}$ be the absolute Galois group $Aut(\overline{\mathbb{Q}})$.  For any positive integer $n$ the $n$-torsion subgroup $E[n](\overline{\mathbb{Q}})$ is stable under the $G_{\mathbb{Q}}$-action. Since $E[n](\overline{\mathbb{Q}})$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^2$ one gets a continuous (with respect to the profinite topology on the left and the discrete on the right) homomorphism
$$
\overline{\rho_{E,\, n}}\colon G_{\mathbb{Q}} \to GL_2(\mathbb{Z}/n\mathbb{Z})
$$
which one calls the mod $n$ representation associated to $E$. As $n$ varies these are compatible and taking limits gives representations $\rho_{E,\ell^{\infty}}$ and $\rho_E$ with values in $GL_2(\mathbb{Z}_l)$ and $GL_2(\widehat{\mathbb{Z}})$ which one calls respectively the $\ell$-adic and adelic representations associated to $E$.
Alternatively, $\rho_{E,n}$ is isomorphic to the representation induced by the action of $G_{\mathbb{Q}}$ on the etale cohomology $H^1_{\text{et}}(E_{\overline{\mathbb{Q}}}; \mathbb{Z}/n\mathbb{Z})$; the description of $\rho_{E,n}$ via torsion generalizes to give representations $\rho_{A,n}$ for higher dimensional abelian varieties, but for a general variety one must instead use cohomlogy.
Serre famously proved that for $E$ an elliptic curve with $End(E) = \mathbb{Z}$, $\rho_E(G_{\mathbb{Q}})$ has finite index in $GL_2(\widehat{\mathbb{Z}})$. In particular for $\ell$ large $\rho_{l^{\infty}}$ is surjective; how large one must take $\ell$ depends on $E$.
Conjecture
This last fact does not depend on $E$. I.e. there exists a constant $N$ such that for every $E/\mathbb{Q}$ with $End(E) = \mathbb{Z}$ and $\ell \geq N$, the mod $\ell$ representation $\overline{\rho_{E,\ell}}$ is surjective; equivalenty there is an upper bound (independent of $E$) on the index of $\rho_E(G_{\mathbb{Q}})$.
By a recent paper of Bilu and Parent, one knows that the image is either surjective or contained in a non-split cartan subgroup.

My  Question
Why do people expect this to be true? Does it follow from other believable conjectures? Is there some heuristic that predicts this?
One fact is that one can phrase this question as the failure of various modular curves (e.g. $X_{ns}(l)$, which have increasingly large genus) to have non-trivial rational points. Is there a geometric reason that one expects these modular curves to have no non-trivial rational points?

REPLY [5 votes]: The conjecture is known for semistable curves. There the only non-surjective $\overline{\rho_p}$ come from isogenies and they have been treated by Mazur's theorem on $X_0(N)$.
When looking at the proof of Serre's result one sees that pairs $(E,p)$ for which the representation is not surjective have to be very special. For instance if the $j$-invariant of $E$ is not integral one can exclude all but finitely many $p$ immediately. My intuition is that it is very hard for $\overline{\rho_p}$  not to be surjective when there is no isogeny of degree $p$.
For curves with integral $j$-invariant but without cm it is much harder to say anything general.<|endoftext|>
TITLE: Generalizations of Boolean posets/lattices
QUESTION [10 upvotes]: A Boolean lattice has a number of rather nice properties which give it a central role in many parts of combinatorics. For instance, it's a lattice, it can be augmented with a ring structure, it can also be augmented with an associative algebra structure, it has a complementation operation, it can be "identified" in different ways with things like the hypercube or a powerset, it satisfies the Stone representability theorem, etc.
It's kind of obvious that Boolean lattices are pretty closely related to the number 2. Like, they have a duality, finite ones all have order a power of 2, etc. One way to see this is to look at a (finite) Boolean lattice as the set of all functions from a set $S \rightarrow \{0, 1\}$, with the lattice and complementation structures acting pointwise.
To what extent can you get an analogue of Boolean lattices (or even posets) with some other natural number k taking the place of 2? You could consider the set of all functions from $S \rightarrow \{0, 1, ..., k-1\}$, which again gives you a poset with a lattice structure, but we don't get a complementation map (although I guess we do get some sort of "k-ality.")
There are a number of questions that this raises: Is this the proper generalization of Boolean lattices in this direction? Does some weird analogue of the Stone theorem hold? What's true for Boolean lattices but isn't true for these guys? What are the right notions to replace "hypercube" and "power set?" Etc.

REPLY [6 votes]: Harrison, I've been meaning to blog a bit about this somewhere, but I might as well put something down here in case it's useful to you. 
As you know, there are lots of ways of defining Boolean algebras. I'm going to focus on one (due in large part to Lawvere) that may open one's eyes to some not-too-well-known possibilities. It starts by observing that for any finitary algebraic theory $T$, the category of algebras is equivalent to the category of product-preserving functors 
$$Kl(T)^{op} \to Set$$ 
where $Kl(T)$, aka the Kleisli category, is the category of finitely generated free algebras. (The morphisms here are of course natural transformations.) This is "Lawvere Theories 101" (the Lawvere theory of an algebraic theory or monad $T$ being defined as $Kl(T)^{op}$). In the case we are concerned with, the finitely generated free Boolean algebras are finite Boolean algebras of cardinality $2^{2^n}$, and by a baby form of Stone duality, the opposite of the category of finitely generated free Boolean algebras is equivalent to the category of finite sets having cardinalities of the form $2^n$. 
Let's call this category $Fin_{2^{-}}$. Thus the category of Boolean algebras is equivalent to the category of product-preserving functors 
$$Fin_{2^{-}} \to Set$$ 
But we can say it more nicely than that. A functor $F: C \to Set$ extends uniquely (up to isomorphism) to the so-called Cauchy completion of $C$, aka the Karoubi envelope or idempotent-splitting completion of $C$, which I'll denote as $\bar{C}$. Moreover, if $C$ has finite products and $F$ preserves them, then it's an easy exercise that $\bar{C}$ acquires finite products and the extension $\bar{F}: \bar{C} \to Set$ preserves them. In the present case, it is easy to see that the idempotent-splitting completion of $C = Fin_{2^-}$ is the category $Fin$ of all finite sets, basically because any (nonempty) finite set is a retract of a finite set of cardinality $2^n$. 
Putting all this together, we obtain what I think is a pretty description of the category of Boolean algebras: it is equivalent to the category of product-preserving functors 
$$Fin \to Set$$ 
(which I believe Lawvere and Schanuel have taken to calling "distributions".) One of the beauties of this description is that it is totally unbiased: there is no special bias toward finite sets of cardinality $2^n$. 
In fact, all this shows that we could, if we want, change our bias to, say, finite sets of cardinality $3^n$. In other words, the category of such sets is also a category with finite products, and its Cauchy completion is again $Fin$, and therefore we are within our rights to describe the category of Boolean algebras as equivalent to the category of product-preserving functors of the form 
$$Fin_{3^-} \to Set$$ 
The category $Fin_{3^-}$ a perfectly legitimate single-sorted Lawvere theory $T'$; the generic object is the finite set $3^1$, of which all other objects in $T'$ are finite products. Using this to extract an alternative operations-and-equations presentation of the theory of Boolean algebras is an interesting exercise, but maybe I'll confine myself to some additional remarks that bear a bit on Stone duality. 
From Lawvere Theories 101, we know that the free $T'$-algebra on $n$-generators, $F(n)$, is in this case given by the representable (note that representable functors automatically preserve products)
$$Fin_{3^-}(3^n, -) = Fin(3^n, -): Fin_{3^-} \to Set$$ 
If $A$ is a $T'$-algebra, then the underlying set is naturally identified as 
$$U(A) = Set(1, U(A)) \cong T'\text{-Alg}(F(1), A) \cong Nat(Fin(3^1, -), A-)$$ 
(in the last step identifying $T'$-algebras with product-preserving functors $A$). For example, if $A = F(n) = Fin(3^n, -)$, then we have 
$$U(A) \cong Nat(Fin(3^1, -), Fin(3^n, -)) \cong Fin(3^n, 3)$$ 
by the Yoneda lemma, whence the underlying set of the $T'$-algebra $F(n)$ has $3^{3^n}$ elements, in perfect analogy with the standard description of the free Boolean algebra on $n$ generators having $2^{2^n}$ elements. (Obviously I shouldn't say "the" underlying set; one of the morals here is that there can be many underlying-set functors which are monadic for a given variety of algebras.) 
Next, we can analogize baby Stone duality for finite Boolean algebras, which says that homming into the free Boolean algebra on 0 generators (the Boolean algebra with two elements) induces an equivalence 
$$Bool_{fp}^{op} \to Fin$$ 
where $Bool_{fp}$ stands for the category of finitely presented Boolean algebras. Of course, notions like "finitely presented Boolean algebras" and the "free Boolean algebra on 0 generators" have perfectly invariant unbiased descriptions, but if we allow ourselves to be biased toward $T'$-algebras, where the free $T'$-algebra has $3^{3^0} = 3$ elements, then baby Stone duality reads 

The functor $T'\text{-Alg}(-, 3): T'\text{-Alg}(-, 3): T'\text{-Alg}_{fp}^{op} \to Fin$ is an equivalence. 

The other direction of the equivalence is the obvious functor $Fin(-, 3): Fin^{op} \to T'-\text{Alg}_{fp}$, which sends a finite set $X$ to $3^X$, with the pointwise-defined $T'$-algebra operations induced by the $T'$-algebra structure on $3$. 
Similarly, an ultrafilter on a set $X$ (or an ultrafilter in a $T'$-algebra $A$) may be defined as a $T'$-algebra map $3^X \to 3$ (a $T'$-algebra map $A \to 3$, resp.). Thus the general Stone duality may be lifted to the context of $T'$-algebras. 
These observations (which I discovered for myself only recently, but which are undoubtedly known to people like Lawvere and Schanuel) may help shed some light on an observation made by Lawvere which might be a bit mysterious otherwise. Namely, if we consider the subtheory of $T'$ generated by the unary operations, noticing that unary operations under composition form a monoid which is isomorphic to the monoid of endomorphisms $M = Fin(3, 3)$ (again by the Yoneda lemma), then we get a forgetful functor 
$$Bool \simeq T'\text{-Alg} \to Set^M$$ 
where $Set^M$ denotes the category of sets equipped with an $M$-action, or $M$-sets for short. It turns out that this is a full embedding; in other words, if a function between $T'$-algebras preserves just the unary operations, then it preserves all the $T'$-operations. (This is a nice exercise.) Therefore, we may identify an ultrafilter on a set $X$ as being essentially the same as a homomorphism of $M$-sets 
$$3^X \to 3$$ 
(This doesn't work if $3$ is replaced by $2$!) And similarly if $3$ is replaced by any $n \geq 3$. (Lawvere goes on to say that countably complete ultrafilters may be equivalently defined as functions $\mathbb{N}^X \to \mathbb{N}$ which preserve the evident actions by the monoid of endofunctions on $\mathbb{N}$. Thus, the canonical map 
$$prin_X: X \to Set^{End(\mathbb{N})}(\mathbb{N}^X, \mathbb{N})$$ 
which is the unit of an evident monad 
$$Set \stackrel{\mathbb{N}^-}{\to} Set^{End(\mathbb{N})} \stackrel{\hom(-, \mathbb{N})}{\to} Set$$ 
on $Set$, fails to be an isomorphism if and only if there exists a measurable cardinal.   

Thanks to a comment of Gerhard Paseman on my answer to this MO question, I recently learned that the algebras of the Lawvere theory $Fin_{3^-}$ are better known as $3$-valued Post algebras, and there is a similar notion of $n$-valued Post algebra. That gives perhaps a quicker answer to Harrison's question.<|endoftext|>
TITLE: Can different modules have the same symmetric algebra? (answered: no)
QUESTION [11 upvotes]: Algebraic geometry allows one to think of an $A$-module $M$ geometrically as a module of functions on the $A$-scheme $\mathrm{Spec}(\mathrm{Sym}(M))$.
I'm wondering if anything is lost in just replacing $M$ by this geometric object.   Since nothing is lost in taking $\mathrm{Spec}$, this amounts to asking:

(1) Can two non-isomorphic $A$-modules $M$ , $N$ have isomorphic symmetric $A$-algebras $\mathrm{Sym}(M)$ , $\mathrm{Sym}(N)$?  

(Clearly they are not isomorphic as graded $A$-algebras.)
If the answer is "No", great!  If "Yes", I would like to see a specific example.
It may be interesting to have a second interpretation, even if it doesn't help solve the problem.  Since we have the adjunction (of set-valued functors)
$hom_{A-alg}(\mathrm{Sym}(M),B) \simeq hom_{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:

(2) If the (set-valued) functors $hom_{A-\mathrm{mod}}(M,-)$ and $hom_{A-\mathrm{mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?

Edit: I emphasized "set-valued" above, thanks to a comment from Buzzard.  Also, partially in response to Mark Hovey's comment, I removed "Is it safe to think of modules geometrically" from the quesiton statement, since I don't want to assert that this is "the correct" geometric interpretation of a module.

REPLY [6 votes]: I now believe a-fortiori's argument: translations are a problem, but, as a-fortiori observed, they are the only problem. Let me spell it out.
Say $f:Sym(M)\to Sym(N)$ is an isomorphism. For $m\in M$ write $f(m)=f_0(m)+f_1(m)+f_{\geq2}(m)$ with obvious notation: $f_0(m)$ is in $A$, $f_1(m)$ is in $N$ and $f_{\geq2}(m)$ is all of the rest. Now here's another $A$-algebra map $g:Sym(M)\to Sym(N)$. To define $g$ all I have to do is to say where $m\in M$ goes so let's say $g(m)=f_1(m)+f_{\geq2}(m)$.
Claim: $g$ is an $A$-algebra isomorphism. 
The proof is that $g$ is just $f$ composed with the isomorphism $Sym(M)\to Sym(M)$ sending $m$ to $m-f_0(m)$ (one needs to check that this is an isomorphism but it is because there's an obvious inverse). 
Claim: the isomorphism of rings inverse to $g$ also has "no constant terms", i.e. it's of the form $h:Sym(N)\to Sym(M)$ where $h(n)=h_1(n)+h_{\geq2}(n)$ with no constant term.
The proof is that $g$ sends terms of degree $d$ to terms of degree $d$ or higher, so applying $g$ to $h(n)=h_0(n)+h_1(n)+h_{\geq2}(n)$ we see $n=h_0(n)+f_1(h_1(n))+$(higher order terms).
Claim: $f_1$ and $h_1$ are mutual inverses. This is easy now.
So in fact all the ideas are in a-fortiori's comments.<|endoftext|>
TITLE: What is an intuitive view of adjoints? (version 1: category theory)
QUESTION [71 upvotes]: In trying to think of an intuitive answer to a question on adjoints, I realised that I didn't have a nice conceptual understanding of what an adjoint pair actually is.
I know the definition (several of them), I've read the nlab page (and any good answers will be added there), I've worked with them, I've found examples of functors with and without adjoints, but I couldn't explain what an adjunction is to a five-year-old, the man on the Clapham omnibus, or even an advanced undergraduate.
So how should I intuitively think of adjunctions?
For more background: I'm a topologist by trade who's been learning category theory recently (and, for the most part, enjoying it) but haven't truly internalised it yet.  I'm fully convinced of the value of adjunctions, but haven't the same intuition into them as I do for, say, the uniqueness of ordinary cohomology.

REPLY [5 votes]: Adjoints can be viewed intuitively as expressing conceptual duality between mathematical notions.
Examples along these lines include $${\sf Free}\dashv {\sf Forgetful},$$ $$\varinjlim\dashv\Delta\dashv\varprojlim,$$ where $\Delta:\mathcal{C}\to\mathcal{C}^\mathcal{D}$ is the diagonal functor on a (co-)complete category $\mathcal{C}$, $$\exists\dashv*\dashv\forall,$$ where $*:{\sf Form}(\overline x)\to{\sf Form}(\overline x,y)$ is the functor adjoining a variable $y$ not appearing in $\overline x=(x_0\dots,x_n)$ which we quantify existensially or universally over, $$\lceil x\rceil\dashv i\dashv\lfloor{x}\rfloor$$ as noted by user1421 in his answer, where $i:\mathbb{Z}\to\mathbb{R}$ is the standard inclusion - there are many more that belong on this list.
In essence, if two notions are related to eachother closely enough that giving a definition of one notion (in the presence of sufficient ambient structure) defines the other notion completely, they can be expressed as a (possibly $\infty$-) adjoint pair of functors. Conversely, when we see that two notions can be expressed as an adjoint pair of functors it means we can think about one by thinking about the other plus some canonical additional structure.
In the above cases, one of the two notions is significantly simpler than the other -- forgetful functors are simpler than free object functors, diagonal functors are simpler than limit and colimit functors, adjoining a variable is simpler than existentially or universally quantifying, etc. This means we can understand complex notions by understanding simple ones and making them functorial, then considering their left or right adjoint. Although there may be another way to construct whatever notion we end up with, any two right (or left) adjoints to the same functor will be naturally isomorphic, so we've gotten at the definition safely and using tools that are very well understood and easy to manipulate.<|endoftext|>
TITLE: What is an intuitive view of adjoints? (version 2: functional analysis)
QUESTION [38 upvotes]: After realising that I don't have an intuitive understanding of adjoint functors, I then realised that I don't have an intuitive understanding of adjoint linear transformations!
Again, I can use 'em, compute 'em, and convolute 'em, but I have no real intuition as to what is going on.  My best attempts (when teaching them) were:

The adjoint allows us to shift stuff from one side of the inner product to the other, thus, in a fashion, moving it out of the way while we do something and then moving it back again.
Nice behaviour with respect to the adjoint (say, normal or unitary) translates into nice behaviour with respect to the inner product.

But neither of those is really about what the adjoint is, just what it does.

REPLY [4 votes]: I recommend this article, which gives a pretty damn good analogy between adjoints and complex conjugates.<|endoftext|>
TITLE: Terminology in category theory
QUESTION [28 upvotes]: A lot of the terminology of category theory has obvious antecedents in analysis: limits, completeness, adjunctions, continuous (functors), to name but a few.  However, analysis and category theory seem to be at opposite poles of the spectrum.
Is there anything deep here, or is it a case of "it has wings, so let's call it a duck"?
This was partly inspired by the top-rated answer to the question What is a metric space? and by the (slightly unsatisfactory answers to the) question Can adjoint linear transformations be naturally realized as adjoint functors?.  In particular, in the first case - the categorical view of metric spaces - there does seem to be an obvious route between the two worlds, do the terminologies correspond there?

REPLY [10 votes]: As for the limit example, it is rooted in analysis by way of topology.
A topology on some set X is a system of subsets, partially ordered by inclusion, that have arbitrary joins and finite meets (or possibly the other way around depending on whether you're looking for open or closed sets), with X and 0 members of the system.
This partial order gets us a category structure, consisting of open (closed?) sets. Limits of these turn out to be interpretable as limits in the classical sense; and colimits are the dual concept as usual.<|endoftext|>
TITLE: Version of Brauer-Nesbitt for summands
QUESTION [10 upvotes]: The Brauer-Nesbitt theorem (well, one of them) says that if $k$ is a field and I have two semisimple representations (on finite-dimensional $k$-vector spaces) $r_1$ and $r_2$ of a $k$-algebra $A$ with the property that the char polys of $r_1(a)$ and $r_2(a)$ coincide for all $a\in A$, then the representations are isomorphic.
Is it the case that if the char poly of $r_1(a)$ divides the char poly of $r_2(a)$ for all $a\in A$, then the smaller representation is a direct summand of the bigger?
[I came up against this recently, but fortunately in the case I was considering $A$ was commutative and $k$ had characteristic zero, and I convinced myself it was surely fine in this case (base change up to an alg closure of $k$ and convince yourself that the semisimple representations kill all the nilpotent elements, so WLOG $A$ is etale and now do it by hand). If $k$ is finite then I'm still not sure which way to bet. If $A$ were a group ring and we knew only that one char poly divided the other for all elements of the group, then my gut feeling is that this isn't enough in characteristic $p$ but maybe I'm wrong. If $k$ has characteristic zero then I am betting on yes but then again I'm no algebraist.]

REPLY [6 votes]: The result is true, regardless of characteristic. 
Lemma: Let A be a k algebra and M a semi-simple A-module which is finite dimensional as a k-algebra. Then the image of A in $\mathrm{End}_k(M)$ is a semi-simple ring. 
So, by the Artin-Wedderburn theorem, this image is a direct sum of matrix algebras over division rings.
Proof: Call this image S. Since S is finite dimensional, it is artinian. Let $M= \bigoplus V_i$ and let $t=(t_i)$ lie in the Jacobson radical of S. For each $V_i$, the condition that $V_i$ is a simple A-module implies that it is a simple S-module. So t must act trivially on $V_i$, and thus $t_i=0$. But we have proved this for all i, so $t=0$ and the Jacobson radical of S is trivial. QED.
We can now reduce to the case that A=S, and is a direct sum of division rings. Say $A = \bigoplus \mathrm{Mat}_{n_i}(\Delta_i)$. So every A-module is of the form 
$$\bigoplus (\Delta_i^{n_i})^{k_i}$$
for some $k_i$ and the corresponding characteristic polynomial is
$$\chi(\lambda, g) = \prod \chi_i(\lambda, g_i)^{k_i}$$
where, for $h \in \mathrm{Mat}_{n_i}(\Delta_i)$, the polynomial $\chi_i(\lambda, h)$ is the characteristic polynomial of $h$ acting on $\Delta_i^{n_i}$.
A much better argument, suggested by buzzard's comment below. (I am not sure whether or not the original can be fixed.) Let $M = \bigoplus (\Delta_i^{n_i})^{k_i}$ and $N = (\Delta_i^{n_i})^{\ell_i}$. Suppose M is not a summand of N, so $k_i > \ell_i$ for some $i$. Let g be 1 on the i component and 0 everywhere else.  Then the characteristic polynomials of M and N are of the form $(\lambda-1)^{n_i k_i} \lambda^{\bullet}$ and $(\lambda-1)^{n_i \ell_i} \lambda^{\bullet}$. So the former does not divide the latter.
We just need to show that the polynomials $\chi_i$, as polynomial functions on $\overline{k} \times A$, are relatively prime to each other. This is easy enough. Let $t_a$ be a basis for the $k$-linear functions on $\mathrm{Mat}_{n_i}(\Delta_i)$ and $u_b$ a basis for the $k$-linear functions on  $\mathrm{Mat}_{n_j}(\Delta_j)$. Then $\chi_i(\lambda, g_i)$ is a homogenous polynomial in $\lambda$ and $t_a$; while $\chi_j$ in homogenous in $\lambda$ and $u_b$. Their GCD must be homogenous in both ways, hence, it is of the form $\lambda^m$.
But, if $\lambda | \chi_i(\lambda)$, this means that there is no $g$ which acts invertibly on $\Delta_i^{n_i}$; contradicting that the identity does so act. So the GCD is 1, and the result is true.

REPLY [3 votes]: By a weird coincidence I've found the answer to my question. I was trying to generalise some ideas of Chenevier; I was reading his Jacquet-Langlands paper---a version I'd got from his website. For other reasons I actually went to Duke's website to get the official version---and the official version has got the lemmas proved in more generality! See Proposition 3.2 of the published version for some pertinent comments on this issue...<|endoftext|>
TITLE: Finding the new zeros of a "perturbed" polynomial
QUESTION [14 upvotes]: Given a univariate polynomial with real coefficients, p(x), with degree n, suppose we know all the zeros xj, and they are all real.  Now suppose I perturb each of the coefficients pj (for j ≤ n) by a small real perturbation εj.  What are the conditions on the perturbations (edit: for example, how large can they be, by some measure) so that the solutions remain real?  
Some thoughts: surely people have thought of this problem in terms of a differential equation valid for small ε that lets you take the known solutions to the new solutions.  Before I try to rederive that, does it have a name?  This seems like a pretty general solution technique, but perhaps it is so general as to be intractable practically, which might explain why I don't know about it.
If you ignore the smallness of the perturbation, then there is a general question here which seems like it might be related to Horn's problem: given two real polynomials p(x) and q(x) of degree n and their strictly real roots, what can you infer about the roots of p(x)+q(x)?  This question is very interesting and I would love to hear what people know about it.  But I'm also happy with the perturbed subproblem above, assuming it is indeed simpler.

REPLY [4 votes]: The following criterion might help (see for example these lecture notes): 

all roots of a polynomial are real if
  and only if the $n \times n$ Hankel
  matrix defined by $H_{ij}=s_k$ for all
  $i+j=k-2$ is positive definite,

where $s_k$ is the sum of the $k$-th powers of the $n$ roots, i.e. $s_k = \sum_{i=1}^n x_i^k$ with your notation. These sums can be computed directly from the polynomial coefficients using Newton's identities.<|endoftext|>
TITLE: Which changes of metric fix all open balls of a metric space?
QUESTION [9 upvotes]: In an earlier question, I was interested in counting the number of metric spaces on N points, where I considered two metric spaces to be the same if they had the same collection of open balls. Two questions:

What are the usual notions of metric space equivalence? Are any of them nontrivial for finite metric spaces? (For instance, the obvious one that two metric spaces are equivalent if their topology is the same is trivial for finite vector spaces).
If we say that two (labeled) metric spaces are equivalent if they have the same collection of open balls, in what ways can we operate on the metric d such that we get the same collection of open balls?

For example, any two metric spaces on 2 points are equivalent in this way, so any allowable operation on metrics yields an equivalent metric space. Clearly we can always scale d by any positive real without changing the equivalence class. Now consider the two metric spaces
x ---3--- y        x ---3--- y
 \       /          \       /
  3     4            4     5
   \   /              \   /
     z                  z

In both cases, the nontrivial open balls are {x, z} and {x, y}, so these metric spaces are equivalent. How can I describe the operation I'm performing on d in general terms? What other operations on d will yield metric spaces that are equivalent in this sense?

REPLY [4 votes]: If you take the set of distances between $n$ points, then you can define a function on the real numbers that when applied to the distances preserves the open balls on $n$ points and defines a new metric as follows. Let $m$ be the number of distinct distances. First note that zero must be fixed second the set of new distances must preserve the original open balls and also preserve the triangle inequality and be positive then we can take the sum of the polynomials $\sum_1^k x(x-d_1)(x-d_2)$...(goes through all distances but d_k)$(d_k'/d_kf_k(d_k))(f_k(x)$ summing over all such terms for all $k$ where $f_k$ are arbitrary smooth functions not zero at $d_k$ and $d_k'$ is the desired new distance we can also add a term $x(x-d_1)$...$f(x)$ where $f(x)$ is an arbitrary smooth function for all distances in this way we preserve zero and changes all the distances to new distances satisfying the triangle inequality and having positive distances and preserving the open balls on the n points and alao preserving symmetry since if $x,y$ and $y,x$ have the same distance it will be sent to a new distance which will be the distance between $x$ and $y$ as well as $y$ and $x$. Thus the new set of distances will be a metric on the $n$ points preserving the open balls on the $n$ points.
This provides a real function operating on the reals that maps one metric space to another by transforming the distances. It does not send metrics with two equal distances to metrics with two different distances so for the two equivalent metrics given it sends one to the other but that transformation is not reversible. 
For any set of inequalities on distances there is a realization as a metric space. Just
Let the distances be close enough to one to satisfy the triangle inequality and set distances from points to themselves as zero and the triangle inequality will be satisfied.
Let us look at what is happening in the case with three points mentioned in the problem. If one distance is maximum. It will determine the whole structure. Say it is the distance from $A$ to $B$ then at point $A$ it will be greater than the distance from $A$ to $C$ so at this point we will get the set containing $A$ and $C$ in the topology similarly we will get the set containing $B$ and $C$ in the topology. The set containing $A$ and $B$ will not be in the toplogy. In all cases the set of all elements and the set of individual elements will be in the topology so at this point we have determined the entire topology. Now we have done this without specifying distances $AC$ and $BC$ and the can be set arbitrarily as long as they are less than $AB$. 
In fact we can determine the entire set of topologies for three points by looking at the number of maxima. If there are two distances which are equal and larger than the third then the topology contains the entire set, single points and the two points of the third edges. If there all three distances are equal we have the topology consists of the three pionts and the entire set. 
Something similar happens in three dimensions if one distance $AB$ is greater than all the others. It forbids the three points sets containing the two points with this distance and it forces the two three point sets not containing it to be in the topology and that determines the entire structure of three points sets in the topology.<|endoftext|>
TITLE: Hypercube decomposition of perverse sheaves
QUESTION [5 upvotes]: There is a well known theorem that says that the functor associating to a perverse sheaf $F$ on $X$ the data $(F|_U,\phi_f(F),can:\psi_f(F) \to \phi_f(F),var:\phi_f(F)\to \psi_f(F)(-1))$ where $U = X \setminus (f=0)$ is an equivalence of categories. 
In dimension 1, this gives that a perverse sheaf on $(\mathbb C,0)$ is described by a quiver $\psi_z(F) = V_0 \leftrightarrows V_1 = \phi_z(F)$ . 
There is an analog statement for perverse sheaves $(\mathbb C,0)^n$. My question is how do you define directly the vector spaces in the quiver representation. For example, in dimension 2, I think we should have $V_{00} = \psi_x\psi_y(F) = \psi_x\psi_y(F)$, $V_{10} = \phi_x\psi_y(F) = \psi_y\phi_x(F)$, $V_{01} = \psi_x\phi_y(F) = \phi_y\psi_x(F)$, $V_{11} = \phi_x\phi_y(F) = \phi_y\phi_x(F)$ but I'm not sure why these functors should commute. Is the key point here that we have a normal crossing divisor? 
Also, what if we want to make things more canonical replacing nearby cycles $\psi_f(F)$ by the restriction of the Verdier specialisation $\nu_Z(F)|_{T_Z^0X}$ and $\phi_f(F)$ by the evanescent part of $\nu_Z(F)$? What would be the precise statement in this case?

REPLY [3 votes]: In some sense there is no canonical way to extract these vector spaces.    Algebraically this is because projective objects usually have automorphisms.  Geometrically because these vector spaces are the stalks of local systems on a manifold with no natural base point.  But the manifold and the local system are canonical.
Each of the 2^n coordinate subspaces of C^n has a conormal bundle in C^{2n}.  This gives an arrangement of 2^n Lagrangian subspaces of the symplectic vector space C^2n, which are in general position (as general as possible for Lagrangian subspaces).  The smooth part of this arrangement, the points that lie on exactly one of these Lagrangians, is a disjoint union of 2^n copies of (C-0)^n.  The 2^n vector spaces in your hypercube are the stalks of local systems on these.
I am not sure what Verdier specialization is, but I bet it's exactly the construction of these local systems.  In general, if Y is a smooth subvariety of X, then it is possible to extract from a perverse sheaf a local system on an open subset of the conormal variety to Y in X, by the following recipe:

Take nearby cycles for the deformation-to-the-normal-bundle family.  This family has a C^*-action so taking nearby cycles is more canonical than usual--the monodromy action (var compose can) is trivial.
Take the Fourier transform.  (As you know, I'm still pretty confused about this, but I think this has to be the topological version, or Fourier-Sato transform.  In particular even with the C^*-action on the family, I don't think the sheaf you get at the end of 1. is C^*-equivariant.)
The result is a perverse sheaf on the conormal bundle to Y.  This sheaf is locally constant on an open subset.

At one time S. Gelfand, MacPherson, and Vilonen had a project to understand the natural maps between the fibers of these local systems on different strata, i.e. the analogs of can and var.  From a certain perspective the work of Nadler and Zaslow is a (not very concrete) solution to this problem.
In general, whether \psi_f \psi_g F = \psi_g \psi_f F depends on the blowup behavior of F with respect to the map (f,g) to C^2.  For a counterexample, consider F = constant sheaf on C^2, f = x, g = xy.  But if F is constructible with respect to a stratification that satisfies Thom's condition a_{f,g}--that is, the stratification is without blowups/sans eclatement with respect to (f,g)--then you are in good shape.  (Although even here there is no canonical isomorphism between the two.  They behave like stalks of a local system on an open subset of the base C^2.)  
For sheaves without blowups I am not sure about references, but try Sabbah's "Morphismes analytiques stratifies sans eclatement et cycles evanescents," or from the etale point of view Illusie's recent note: http://www.math.u-psud.fr/~illusie/vanishing1b.pdf<|endoftext|>
TITLE: V=L and a Well-Ordering of the Reals
QUESTION [13 upvotes]: A fairly simple question:  I've read in multiple sources that Godel proved that if we accept the axiom of constructibility in ZFC, then we can create an explicit formula that well-orders the real numbers.  I tried searching for a paper or some other source that explains what this formula is, but I came up empty-handed.  Can someone explain what this formula is, or perhaps point me to a resource that does?

REPLY [28 votes]: The order is very easy. Under $V=L$, the set-theoretic universe is built according the hierarchy $(L_\alpha \mid \alpha \in \mathrm{Ord})$, where $L_0$ is empty, $L_{\alpha+1}$ consists of all definable subsets of $L_\alpha$, and $L_\lambda$ is the union of all earlier $L_\alpha$ when $\lambda$ is a limit ordinal. 
Since we can order the definitions used to go from $L_\alpha$ to $L_{\alpha+1}$, we obtain a definable well-ordering of the entire universe. Namely, $x$ is less than $y$ iff 

$x$ appears before $y$ in the hierarchy or 
they appear at the same stage, but $x$ appears with an earlier definition than $y$. 

If one analyzes the complexity of the resulting definition for real numbers, it has complexity $\Delta^1_2$ in the descriptive set theoretic hierarchy.<|endoftext|>
TITLE: de Rham cohomology and flat vector bundles
QUESTION [54 upvotes]: I was wondering whether there is some notion of "vector bundle de Rham cohomology".
To be more precise: the k-th de Rham cohomology group of a manifold $H_{dR}^{k}(M)$ is defined as the set of closed forms in $\Omega^k(M)$ modulo the set of exact forms. The coboundary operator is given by the exterior derivative.
Let now $E \rightarrow M$ be a vector bundle with connection $\nabla^E$ over $M$, and consider the $E$-valued $k$-forms on $M$: $\Omega^k(M,E)=\Gamma(\Lambda^k TM^\ast \otimes E)$.
If $E$ is a flat vector bundle, we get a coboundary operator $d^{\nabla^E}$ (since $d^{\nabla^E} \circ d^{\nabla^E} = R^{\nabla^E}=0$, with $R^{\nabla^E}$ being the curvature) and we can define 

$$H_{dR}^{k}(M,E) := \frac{ker \quad d^{\nabla^E}|_{\Omega^k(M,E)}}{im \quad d^{\nabla^E}|_{\Omega^{k-1}(M,E)}}$$

So my question: Is this somehow useful? I mean can one use this definition to make some statements about $M$ or $E$ or whatever? Or is the restriction of $E$ to be a flat vector bundle somehow disturbing? Or is this completely useless?

REPLY [5 votes]: The exposition by Eugene Xia Abelian and non-abelian cohomology explicitly describes and relates the Betti, Čech, de Rham,  and Dolbeault moduli spaces in the context of the corresponding cohomology theories (see my review here).  It is a well-written synthetic treatment with good references.<|endoftext|>
TITLE: What is the origin of this positive matrix characterization of bounded analytic functions on the unit disk?
QUESTION [6 upvotes]: Background: Let $S$ denote the so-called Schur class of complex analytic functions from the open unit disk $D$ in $\mathbb{C}$ to the closed unit disk $\overline{D}$.  Given distinct points $z_1,\ldots,z_n\in D$ and points $w_1,\ldots,w_n\in\overline{D}$, G. Pick's Theorem (1916) says that there is an $f\in S$ such that $f(z_j)=w_j$ for each $j$ if and only if the matrix $\left(\frac{1-w_j\overline{w_k}}{1-z_j\overline{z_k}}\right)_{j,k=1}^n$ is positive semidefinite.  Suppose that $g:D\to\mathbb{C}$ is a function.  By Pick's Theorem it follows immediately that a necessary condition for $g$ to be in $S$ is that for each $n$ and each $n$-tuple of distinct points $z_1,\ldots,z_n\in D$, the matrix $\left(\frac{1-g(z_j)\overline{g(z_k)}}{1-z_j\overline{z_k}}\right)_{j,k=1}^n$ is positive semidefinite.  I find it remarkable that this is also a sufficient condition (this too can be derived using Pick's Theorem, although there is a more general method, as indicated in Edit 2 and Yemon Choi's answer). However, I do not know whether Pick observed this fact, and I don't see it in I. Schur's related work from around that time (1917-1918).

Question: Who first observed that a function $g:D\to\overline{D}$ is analytic if for each $n$ and each $n$-tuple of distinct points $z_1,\ldots,z_n\in D$, the matrix $\left(\frac{1-g(z_j)\overline{g(z_k)}}{1-z_j\overline{z_k}}\right)_{j,k=1}^n$ is positive semidefinite?  
To elaborate a little, the theory of reproducing kernel Hilbert spaces and their multipliers in which this characterization can now be found was developed well after people including Pick, Nevanlinna, and Schur had developed other aspects of the theory of bounded analytic functions on the unit disk.  Part of the reason I would be interested in knowing the origin is that it would be interesting to see if completely different methods were used.  If different methods weren't used, learning of the origin of this would help me to better understand the history of the uses of Hilbert space in function theory.

Edit 1: In an attempt to increase clarity, I changed the codomain from $\mathbb{C}$ to $\overline{D}$, so that the trivial condition of being bounded by 1 is no longer emphasized.
Edit 2: I'm still curious about the history of this, but I've realized that I was confused about the logical relationship between Pick's Theorem and the positivity criterion for analyticity.  Yemon Choi's answer points out a better way to think about the latter; it is a simple criterion for being a multiplier of a reproducing kernel Hilbert space, and it works even for those RKHS for which the analogue of Pick's Theorem is false.  The general criterion appears in section 2.3 of Agler and McCarthy's book Pick Interpolation and Hilbert Function Spaces, the identification of the multipliers of Hardy space appears in section 3.4, and Pick's Theorem is proved in section 5.2.

REPLY [2 votes]: I don't know the answer to your question, but it might be of interest to observe that one can interpret the Pick condition for a function $g$ as saying that $g$ is a bounded multiplier for a certain reproducing kernel Hilbert space (see the book of Agler and McCarthy). The fact that said RKHS is the classical Hardy space, and consideration of $g\cdot 1$ where $1$ is the constant function, then tells us that $g$ is analytic. 
However, as I don't have Agler and McCarthy's book to hand, I don't know what machinery gets used in proving the above. It could well be that they deduce it from something simpler and older which proves the automatic analyticity result more directly. In fact, thinking about this a little more, that must surely be the case. You probably end up showing that $g$ must agree on any finite subset of the disc with some bounded analytic function (a priori depending on the subset), and then e.g. Morera's theorem plus approximation ought to do the job.<|endoftext|>
TITLE: Does such a subgroup exist?
QUESTION [9 upvotes]: I am looking for a certain masa in a $II_1$ factor which is singular and has nontrivial Takesaki invariant.
For this I am looking for an example of an inclusion of groups $H\subset G$ such that:

$G$ is a countable icc (infinite conjugacy class) group
$H$ is abelian
$\forall g\in G-H,\{ hgh^{-1} |h\in H \}$ is infinite
$ |H\backslash G/H| \geq 3$
there exists $g\in G-H$ and $h_1\neq h_2\in H$ with $h_1 g=gh_2$.

Does such an example exist?

REPLY [2 votes]: A more singular example:
Take an infinite index inclusion of abelian groups $K\subset H$. Let a non-trivial group $L$ act on $K$ by automorphisms. Then the amalgamated free product $G=H\underset{K}{\ast} (K\rtimes L)$ satisfies the conditions. Moreover, $L(H)\subset L(G)$ $\,$is a singular masa. On can use the results of Ioana, Peterson and Popa for this, but maybe there are more elementary ways to see this.<|endoftext|>
TITLE: The De Rham Cohomology of $\mathbb{R}^n - \mathbb{S}^k$
QUESTION [11 upvotes]: I'm reading Madsen and Tornehave's "From Calculus to Cohomology" and tried to solve this interesting problem regarding knots. 
Let $\Sigma\subset \mathbb{R}^n$ be homeomorphic to $\mathbb{S}^k$, show that $H^p(\mathbb{R}^n - \Sigma)$ equals $\mathbb{R}$ for $p=0,n-k-1, n-1$ and $0$ for all other $p$. Here $1\leq k \leq n-2$. 
Now the case $p=0$ is obvious from connectedness and the two other cases are easily solved by applying the fact that
$H^{p+1}(\mathbb{R}^{n+1} - A) \simeq H^p(\mathbb{R}^n - A),~~~~p\geq 1$
and
$H^1(\mathbb{R}^n - A) \simeq H^0(\mathbb{R}^n - A)/\mathbb{R}\cdot 1$
So what is my problem, really?
Now instead let's look at this directly from Mayer-Vietoris. If $\hat{D}^k$ is the open unit disk and $\bar{D}^k$ the closed. Then $\mathbb{R}^n - \mathbb{S}^k = (\mathbb{R}^n - \bar{D}^k)\cup (\hat{D}^k)$ and $(\mathbb{R}^n - \bar{D}^k)\cap (\hat{D}^k) = \emptyset$ 
Now
$H^p(\mathbb{R}^n - \bar{D}^k) \simeq H^p(\mathbb{R}^n - \{ 0 \})$ since $\bar{D}^k$ is contractible. And $H^p(\mathbb{R}^n - \{ 0 \})$ is $\mathbb{R}$ if $p=0,n-1$ and $0$ else. Since $\hat{D}^k$ is open star shaped we find it's cohomology to be $\mathbb{R}$ for $p=0$ and $0$ for all other $p$.
This yields and exact sequence
$\cdots\rightarrow 0\overset{I^{\ast}}\rightarrow H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \overset{J^{\ast}}\rightarrow \mathbb{R} \rightarrow 0\cdots$
So due to exactness I find that $\ker(J^*) = \text{Im}(I^*) = 0$ and that $J^*$ is surjective, hence 
$H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \simeq \mathbb{R}$. 
But ... If I apply the exact same approach to $p = n-k-1$ my answer would be $0$ for $H^{n-k-1}(\mathbb{R}^n - \mathbb{S}^k)$. 
Where does this last approach fail?

REPLY [3 votes]: Your question has been answered by Tom.  But I am also not sure if you are aware that the point of the problem was to show that the cohomology of a sphere embedded in euclidean space is independent of the embedding. You seem to think only of the standard embedding.  As Tom mentioned, duality is one way to prove this independence.
(This might have been better as a comment, but I am not yet allowed to comment.)<|endoftext|>
TITLE: What should Spec Z[\sqrt{D}] x_{F_1} Spec \bar{F_1} be?
QUESTION [6 upvotes]: What should be $\text{Spec } \mathbb{Z}[\sqrt{D}] \times_{\mathbb{F}_1} \text{Spec } \overline{\mathbb{F}}_{1}$?
Sure, there's more than one definition.
I'm looking for any answer that uses at least one definition of scheme over $\mathbb{F}_1$.
This really is more a question of opinion.
What do you think this should be?
Some monoid that has something to do with $\text{Spec }\mathbb{Z}[\sqrt{D}][\mathbb{Q}/\mathbb{Z}]$ would be my guess (where the second brackets mean group ring).
This interests me from the point of view that, say, hyperelliptic curves over a finite field come (geometrically) from the group scheme of a quadratic extension of $\overline{\mathbb{F}}_p [t]$. In this case the frobenius acts on ideal classes, and satisfies a quadratic equation.
But, from what I understand, the natural analogue of frobenius in the arithmetic case, is like taking any positive power, and taking limits to 0 (or something of the sort).
Would this satisfy some kind of equation on, say, $\text{Pic(Spec } \mathbb{Z}[\sqrt{D}] \times_{\mathbb{F}_1} \text{Spec } \overline{\mathbb{F}}_{1}\text{)}$?
(for whatever definition of Pic that should be natural here)
I've searched for information on $\mathbb{F}_1$, but most just talk about making $\text{Spec }\mathbb{Z}$ into a curve, getting zeta functions to be Riemann's, etc.
Instead, I want to ask questions that are not just about proving the Riemann hypothesis, like the one above.

REPLY [3 votes]: Sorry I didn't reply before, I somehow didn't read the question till now.
I think your question is a bit misguided. The main problem I see with it is: what is $\text{Spec} \mathbb{Z}[\sqrt{D}]$ over F1? If you think of it as the M_0 scheme given by $\mathbb{Z}[\sqrt{D}]$ as a multiplicative monoid, then it is something huge (since that monoid is not even finitely generated, for starters), so none of the current notions can effectively deal with it (so far one can mostly only control schemes of finite type). If you want to make better sense of the question, one should ask: is it possible to find an algebra $A$ over F1 such that its base extension to $\mathbb{Z}$ gives $A\otimes_{\mathbb{F}_1} \mathbb{Z} = \mathbb{Z}[\sqrt{D}]$? 
If you find an answer to this question, say in CC setting where it boils down to finding a monoid (with zero) such that the reduced semigroup ring gives you back your original ring, 
then you might consider an approximation to your problem (the tensor with the abelian closure of F1) by taking some kind of multivariable-cyclotomic completion of your monoid, as in the papers by Manin and Marcolli.<|endoftext|>
TITLE: Do subgroups have "two sided bases"?
QUESTION [13 upvotes]: Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of
$$
E(g)=\begin{cases}
g &\text{if } g\in H\\\
0 &\text{else,}
\end{cases}
$$
i.e., $E$ is the projection onto $\mathbb{C}[H]$. A finite subset $B\subset \mathbb{C}[G]$ will be called a left basis for $G$ over $H$ if
$$
x=\sum\limits_{b\in B} b E(b^\ast x)
$$
for all $x\in \mathbb{C}[G]$, where $\ast$ is the anti-linear extension of the map $g\mapsto g^{-1}$. For an example, take $B$ to be a set of left-coset representatives. Similarly, we can define a right basis to be a finite subset $B\subset \mathbb{C}[G]$ such that
$$
x=\sum\limits_{b\in B} E(x b^\ast)b
$$
for all $x\in\mathbb{C}[G]$.
Note that there exist groups for which there is a basis which is both a left and right basis, but $H$ is not a normal subgroup of $G$. One can take the subgroup of the symmetric group $S_n$ ($n\geq 3$) which fixes $1$. Then a set of left and right coset representatives is given by 
$$
\{ (1 j)|j=1,\dots,n\}.
$$
Does there always exist a basis which is both a left and right basis, or are there inclusions of groups for which there is no simultaneous left and right basis?
The motivation for this question is another question from subfactor theory: if $N\subset M$ is a finite index, extremal $II_1$-subfactor, does there always exist a Pimsner-Popa basis which is both a left and right basis? The subgroup subfactor is an example of such a subfactor, and the question posed above is a watered-down version of the subfactor question, where perhaps an answer is already known or more easily obtainable.

REPLY [3 votes]: It was recently proved  that every regular subfactor (resp., depth 2 subfactor with simple relative commutant) admits a two-sided basis - see  http://nyjm.albany.edu/j/2020/26-37.html (resp., https://arxiv.org/abs/2102.01462).<|endoftext|>
TITLE: What is known about the intersection pairing on H^{mid}?
QUESTION [6 upvotes]: When we restrict to the torsion-free part of the cohomology of a manifold, the intersection pairing is nondegenerate.  In dimension 2n, this gives a bilinear form on the free part of Hn (symmetric if n is even, skew-symmetric of n is odd).  Supposedly, this classifies simply-connected 4-manifolds pretty well; more precisely, the map from homeomorphism classes of manifolds is at most 2-to-1, and whenever a bilinear form has two preimages, at least one of them isn't smooth.
Broadly, I'm wondering what else is known about this pairing.  (Of course, if anything I've said so far is wrong, please correct me!)  For example, can we obtain the bilinear form associated to a connected-sum (or a product, or whatever other natural ways we can get manifolds) in some nice way?  It seems like in the connected-sum case, Mayer-Vietoris should tell us something -- I'd imagine the isomorphism it gives (when n>1) between Hn(connected-sum)=Hn(M1)+H(M2) is probably natural, but I don't know if we can tell anything about the cohomology ring structure from it.  As for the product of manifolds, I'm pretty sure that the Kunneth formula gives an isomorphism of algebras, so that should take care of it unless I'm missing something...  Also, is this at all related to the Kirby-Siebemann invariant?  (All I really know is that it's related to smoothness/smoothability, but perhaps, for example, we can somehow read off the possible values of the K-S invariant from the bilinear form?)

REPLY [3 votes]: Not an answer to your question, but I just want to stress that, whilst the intersection form does more-or-less classify simply-connected topological 4-manifolds, it is far from the whole story in the smooth category. There are many examples of simply-connected topological 4-manifolds which admit infinitely many different smooth structures. (This is the only dimension in which this occurs.)
These different smooth structures are usually detected via Donaldson or Seiberg-Witten invariants. These invariants also enter in the connected sum discussion. Given a connect-sum of topological 4-manifolds X=M#N, as others have said, the intersection forms add: Q(X)= Q(M)+Q(N). Freedman's work on 4-manifolds tells us that there is essentially a converse to this in the topological category: roughly, if a simply-connected topological 4-manifold X has intersection form which can be written as Q(M)+Q(N) for simply-connected 4-manifolds M,N then in fact there is connected sum decomposition X=M#N. 
On the other hand, one can show that for many smooth 4-manifolds M,N the, say, Seiberg-Witten invariants of M#N must vanish. Meanwhile, there are many X for which we know that the invariants are non-zero (e.g., symplectic 4-manifolds thanks to the work of Taubes). This means that there are many examples of smooth 4-manifolds which can be written topologically as a connected sum - by Freedman's work - but not smoothly because the SW invariant is non-zero.<|endoftext|>
TITLE: "Counter"-example for Gauss's Lemma on irreducible polynomials
QUESTION [12 upvotes]: Gauss's Lemma on irred. polynomial says,
Let R be a UFD and F its field of fractions. If a polynomial f(x) in R[x] is reducible in F[x], then it is reducible in R[x]. 
In particular, an integral coefficient polynomial is irreducible in Z iff it is irreducible in Q. For me this tells me something on how the horizontal divisors in the fibration from the arithmetic plane SpecZ[x] to SpecZ intersects the generic fiber: a prime divisor (the divisor defined by the prime ideal (f(x)) in Z[x]) intersect the generic fiber exactly at one point (i.e. the prime ideal (f(x)) in Q[x]) with multiplicity one.
Now here is my question:
Give a ring R, with Frac(R)=F, and a polynomial f(x) in R[x] such that f(x) is reducible in F[x], but is irreducible in R[x]. 
Of course, R should not be a UFD.
I'd like to see an example for number fields as well as a geometric example (where R is the affine coordinate ring of an open curve or higher dimensional stuff). Thanks

REPLY [7 votes]: For any commutative ring $B$ and subring $A$, the following are equivalent:
(1) $A$ is integrally closed in $B$
(2) If $F\in A[X]$ factorizes as $F = GH$ in $B[X]$ with $G$ and $H$ monic, then $G$ and $H$ are in $A[X]$.
These conditions imply:
(3) If $F\in A[X]$ is monic and irreducible, then $F$ remains irreducible in $B[X]$.
When $B$ is an integral domain, all three conditions are equivalent.
Proof: (2) $\Rightarrow$ (1) is trivial: if $F(b)=0$ for some $b\in B$ and $F\in A[X]$ monic, then $F = (X-b)G$ in $B[X]$ for some polynomial $G$ (remainder theorem). Then $G$ is monic, so by (2) $X-b$ and $G$ are in $A[X]$. In particular, $b$ is in $A$.
(1) $\Rightarrow$ (2) is folklore. Take a commutative ring $S$ that contains $B$, over which $G$ and $H$ can be written as products of linear factors: $G = (X-x_1)\cdots(X-x_n)$, $H = (X-y_1)\cdots(X-y_m)$. Then in $S$ the $x_i$ and $y_j$ are zeroes of the monic polynomial $F$, so they are integral over $A$. But the coefficients of $G$ and $H$ are (elementary symmetric) polynomials in the $x_i$ and $y_j$, respectively, hence they are again integral over $A$. As these coefficients are in $B$, by (1) they must lie in $A$. (Basically, one can construct such a ring $S$ in the same way as a splitting field for a polynomial over a field is found: first consider $S_1:= B[X]/(G)$; then $G = (X-x_1)G_1$ in $S_1[X]$, where we write $x_1 := X \bmod (G)$ in $S_1$. Then "adjoin another root $x_2$ of $G$" by passing to $S_2 := S_1[X]/(G_1)$, etc, until we arrive at a ring $S_n$ over which $G$ completely splits into linear factors. Then proceed to adjoin roots for $H$ in the same manner. Note that $B$ remains a subring throughout, i.e. no non-zero element of $B$ will map to $0$ in $S$.)
Condition (3) immediately follows from (2), for if $F = GH$ in $B[X]$, the leading coefficients of $G$ and $H$ are inverse units of $B$ because $F$ is monic. So we can rewrite this as $F = G_1H_1$ with $G_1$ and $H_1$ monic in $B[X]$.
When $A$ and $B$ are domains, (3) implies (1): if $F(b) = 0$ with $b\in B$ and $F\in A[X]$ monic, factor $F$ as $F_1\cdots F_r$ with the $F_i$ monic and irreducible in $A[X]$. (Factoring $F$ as a product of monic polynomials reduces the degree, so eventually we wind up with factors that are irreducible.) Since $B$ is a domain, it follows that $F_i(b) = 0$ for some $i$. By (3), $F_i$ is still irreducible in $B[X]$. But it is divisible by $X-b$ there, and so $F_i = X-b$. Hence $X-b$ is in $A[X]$ and thus $b$ belongs to $A$.
Q.e.d.
(Matthe van der Lee, Amsterdam.)<|endoftext|>
TITLE: How do I calculate the discriminant of a galois closure and its other subfields?
QUESTION [8 upvotes]: Given a number field K of dimension d over Q, and galois closure of dimension d! over Q (i.e galois group Sd), can we relate the discriminant of the galois closure to that of the discriminant of K? Assume no special ramification happens in the closure or the other subfields, for example if the discriminant of K is a prime.
Tests in sage indicate that the discriminant of the galois closure is $\Delta_K^{\frac{d!}{2}}$, and that the discriminants of the other subfields are also powers of $\Delta_K$, but the power has something to do with the corresponding subgroup of Sd. (Not just the size of the subgroup)
Is there a way to prove the first indication, and thoughts about the second?

REPLY [3 votes]: The problem of computing the relative discriminant of an extension 
of number fields can be reduced to the computation of the relative
discriminant of a local kummerian extension of degree equal to the residual
characteristic, which is easy to solve.  See for example arXiv:0711.3878v1 [math.NT], p. 51.<|endoftext|>
TITLE: Integer points of an elliptic curve
QUESTION [21 upvotes]: Where can I found some resources to learn how to determine the integer points of given elliptic curve? I would like to learn a method based on computing the rank and the torsion group of given curve. Also, how can I determine the integer points if the curve is not on its Weierstrass form?

REPLY [29 votes]: There are precisely two available "serious" implementations of the standard algorithm for computing integral points on an elliptic curve: a non-free one in Magma (http://magma.maths.usyd.edu.au/magma/) and a free one in Sage (http://sagemath.org).  The one in Sage was done by Cremona and two German masters students a few years ago, and when refereeing the Sage code, I compared the answers with Magma, and uncovered and reported numerous bugs in Magma, which were subsequently fixed.  Here's how to use Sage to find all integral (or S-integral!) points on a curve over Q:
sage: E = EllipticCurve([1,2,3,4,5])
sage: E.integral_points()
[(1 : 2 : 1)]
sage: E.S_integral_points([2])
[(-103/64 : -233/512 : 1), (1 : 2 : 1)]

and here is how to use Magma:
> E := EllipticCurve([1,2,3,4,5]);
> IntegralPoints(E);
[ (1 : 2 : 1) ]
> SIntegralPoints(E, [2]);                                                     
[ (1 : 2 : 1), (-103/64 : -233/512 : 1) ]

Note that in both cases by default the points are only returned up to sign.  In Sage you get both signs like this:
sage: E.integral_points(both_signs=True)
[(1 : -6 : 1), (1 : 2 : 1)]

Finally, you can use Magma for free online here: http://magma.maths.usyd.edu.au/calc/
and you can use Sage free here: https://sagecell.sagemath.org/.  With Sage, you can also just download it for free and install it on your computer.  With Magma, you have to pay between $100 and a few thousand dollars, depending on who you are, and deal with copy protection.
NOTE: Technically a system called SIMATH (http://tnt.math.se.tmu.ac.jp/simath/) had an implementation of computing integral points.  But it was killed by our friends at Siemens Corp.<|endoftext|>
TITLE: Weil Conjectures for Grassmannians
QUESTION [5 upvotes]: To establish the Weil conjectures for $n$-dimensional projective space over a finite field is elementary. Does there exist a simple direct proof of the conjectures for finite field Grassmannians?

REPLY [11 votes]: Yes.  Both cohomology and number of points are readily determined by looking at the Schubert cells (a cell of dimension k contributes one dimension to $H^{2k}$, and $q^k$ to the point count) and they match.
In fact, it's very easy to check Weil's conjecture directly for any smooth variety which has a decomposition into cells.

REPLY [10 votes]: The first result on the google search "zeta function of grassmannian" seems to contain quite a direct and not too long derivation of the zeta function for a grassmannian over a finite field:
http://www.math.mcgill.ca/goren/SeminarOnCohomology/GrassmannVarieties%20.pdf
From the zeta you see that it is rational, of course get the zeros (which are none), but you don't immediately get confirmation of the functional equation. Though, from the very simple combinatoric representation of the zeta function, it might be easy to prove directly, I will try with pen and paper later.
I'm glad I searched this, I didn't know the zeta was so simple in this case as well<|endoftext|>
TITLE: Weil Conjectures for nonprojective algebraic varieties
QUESTION [12 upvotes]: If we replace projective variety with algebraic variety in the statement of the Weil conjectures what happens? To me it seems the statement still makes sense. But is it still true?

REPLY [21 votes]: Correctly restated, the conjectures hold for any variety $V$ (not necessarily complete or nonsingular) over a finite field $k$.
Dwork proved that the zeta function $Z(V,t)$ of $V$ is a rational function of $t$.
Grothendieck (et al.) expressed $Z(V,t)$ as the alternating product of the characteristic
polynomials of the Frobenius map $F$ acting on the etale cohomology groups of $V$ with compact support.
Deligne showed (Weil II) that for each positive integer $i$ and each eigenvalue $a$ of $F$ acting on the $i$th etale cohomology group of $V$ with compact support, there exists an integer $j\leq i$ such that all the complex conjugates of $a$ have absolute value $q^{j/2}$ where $q=|k|$.
When $V$ is nonsingular and complete, these statements together with Poincare duality, give Weil's original conjectures.<|endoftext|>
TITLE: Completion of modules of differentials (A strange exercise in Liu's AG textbook)
QUESTION [7 upvotes]: A is a Noetherian ring, B is an f.g. algebra over A, I is an ideal of A. let $\hat B$ be B's I-adic completion. Prove that $\Omega^1_{\hat B/A}$'s I-adic completion is isomorphic to $\Omega^1_{B/A}$'s I-adic completion.
This is an exercise from Liu's "Algebraic geometry and arithmetic curves", Exercise VI.1.3. It seems strange because according to Part(a) of that problem, there is an exact sequence involving these two objects. And if this is true, we must prove the first term in that exact sequence is actually zero under only Noetherian condition! I feel a bit puzzled, can anyone help me? Thanks!

REPLY [22 votes]: Yes $K_n=0$ in this situation ($A$ noetherian and $B$ essentially of finite type over $A$). 
Hints: You need to know that $\Omega_{B/A}$ is a finitely generated $B$-module and you can use the canonical bijection 6.1.5 for both $B$ and $\hat{B}$ and with $M=\Omega_{B/A}\otimes_B \hat{B}$ to get a canonical map $\Omega_{\hat{B}/A}\to \Omega_{B/A}\otimes_B \hat{B}$...
EDIT: Let me detail a little more. Recall $\hat{B}$ is the formal completion of $B$ with respect to the $I$-adic topology for some ideal $I$ of $B$, and $B$ is essentially of finite type over $A$ (localization of a finitely generated $A$-algebra). 

Let $M$ be a finitely generated $\hat{B}$-module. Any $A$-derivation $B\to M$ induces a $A$-derivation $B/I^n \to M/I^nM$ for all $n\ge 1$. Passing to the limit, we get an $A$-derivation $\hat{B}\to \hat{M}=M$ whose restricton to $B$ is just the $A$-derviation we start with. Theorefore the natural (restriction) map ${\rm Der}_A(\hat{B}, M)\to {\rm Der}_A(B, M)$  has a section (it is therefore surjective). 
Applying the above result to $M= \Omega_{B/A}\otimes_B \hat{B}$, the canonical map
$$Hom_{\hat{B}}(\Omega_{\hat{B}/A}, M)\to Hom_A(\Omega_{B/A}, M)$$ 
has a section $\sigma$. Consider $\psi: \Omega_{B/A}\to M$, $\omega\mapsto \omega\otimes 1$ and $\psi=\sigma(\phi) : \Omega_{\hat{B}/A}\to \Omega_{B/A}\otimes_B \hat{B}$. Now tensoring by $B/I^n=\hat{B}/I^n\hat{B}$, we see that the natural map 
$$\Omega_{B/A}\otimes_B B/I^n \to \Omega_{\hat{B}/A}\otimes_{\hat{B}} \hat{B}/I^n\hat{B}$$ 
is injective. This is an isomorphism by Part (a) of the exercise.<|endoftext|>
TITLE: Definition of inner product for vector spaces over arbitrary fields
QUESTION [31 upvotes]: Is there a canonical definition of the concept of inner products for vector spaces over arbitrary fields, i.e. other fields than $\mathbb R$ or $\mathbb C$?

REPLY [5 votes]: Several approaches have been tried to conceive a well-behaved "inner product" on non-Archimedean valued fields. Let's see some examples:
Option 1:
    Let $\lambda\mapsto\lambda^*$ be a field automorphism of order $2$ defined on $K$. Let $E$ be a $K$-vector space. An inner product is a map $\langle,\rangle:E\times E\to K$ such that:

$\langle x,x\rangle\neq 0$ for all $x\neq 0$, $x\in E$.
$\langle x,y\rangle=\langle y,x\rangle^*$ for all $x,y\in E$.
$x\mapsto \langle x,y\rangle$ is linear for each $y\in E$.

Note that if axioms 1,2,3 are assumed in the complex case then either 
$\langle ,\rangle$ or $-\langle ,\rangle$ is positive definite.
It is proved in this book1 that $\langle ,\rangle$ satisfies the Cauchy-Schwartz inequality, it induces a norm, $\langle ,\rangle$ is continuous in the topology induced by the norm and more. The down side is that there is no infinite-dimensional Hilbert-like spaces over $K$.
Option 2: A similar approach was taken in the paper: A non-Archimedean inner product, L. Narici, E. Beckenstein, Contemporary Mathematics, AMS, vol.384, pp.187--202 (2005). There, the "inner product" $\langle ,\rangle$ is defined by the axioms:

$\langle x,x\rangle\neq0$ whenever $x\neq0$.
$x\mapsto \langle x,y\rangle$ is linear for each $y\in E$.
$|\langle x,y\rangle|^2\leq|\langle x,x\rangle||\langle y,y\rangle|$.

This map defines the norm $\|x\|=|\langle x,x\rangle|^{1/2}$.
Also, it is proved that in $C_0$, the sup norm is induced by $\langle x,y\rangle=\sum x_ny_n$
(for $x=(x_n)_n$ and $y=(y_n)_n$ in $C_0$) if and only if the residue class field of $K$ is real closed. 
Subsequent study of operators in $C_0$ over an ordered non-Archimedean valued field was developed in here2, here3, here4 and here5.
As in the option 1, in this case there are no infinite-dimensional Hilbert-like spaces. This is shown in this other paper6. However, when the valuations take values in ordered groups rather than the real numbers, the situation changes as we see in the following.  
Option 3: It is possible to conceive Quadratic spaces over Krull valued fields. Under certain conditions, it is even possible to have infinite-dimensional Hilbert-like spaces.
For this topic, in particular, I recommend the papers: 
a. On a class of orthomodular quadratic spaces, H. Gross, U.M. Künzi - Enseign. Math, 1985.
b. Banach spaces over fields with a infinite rank valuation - [H.Ochsenius A., W.H.Schikhof] - 1999
c. After that see: Norm Hilbert spaces over Krull valued fields - [H. Ochsenius, W.H. Schikhof] - Indagationes Mathematicae, Elsevier - 2006

1Perez-Garcia, C.; Schikhof, W. H. Locally convex spaces over non-Archimedean valued fields
2J. Aguayo and M. Nova: Non-Archimedean Hilbert like spaces, Bull. Belg. Math. Soc. Simon Stevin, Volume 14, Number 5 (2007), 787-797.
3 José Aguayo, Miguel Nova, and Khodr Shamseddine: Characterization of compact and self-adjoint operators on free Banach spaces of countable type over the complex Levi-Civita field, Journal of Mathematical Physics 54, 023503 (2013); https://doi.org/10.1063/1.4789541
4 José Aguayo, Miguel Nova, and Khodr Shamseddine: Inner product on $B^*$-algebras of operators on a free Banach space over the Levi-Civita field, Indagationes Mathematicae, Volume 26, Issue 1, January 2015, Pages 191-205; https://doi.org/10.1016/j.indag.2014.09.006
5 José Aguayo, Miguel Nova, and Khodr Shamseddine: Positive operators on a free Banach space over the complex Levi-Civita field, p-Adic Numbers, Ultrametric Analysis and Applications, April 2017, Volume 9, Issue 2, pp 122–137; https://doi.org/10.1134/S2070046617020029
6 M. P. Solèr: Characterization of hilbert spaces by orthomodular spaces, Communications in Algebra, Volume 23, 1995 - Issue 1,  Pages 219-243; https://doi.org/10.1080/00927879508825218<|endoftext|>
TITLE: How to think about CM rings?
QUESTION [40 upvotes]: There are a few questions about CM rings and depth.

Why would one consider the concept of depth? Is there any geometric meaning associated to that? The consideration of regular sequence is okay to me. (currently I'm regarding it as a generalization of not-a-zero-divisor that's needed to carry out induction argument, e.g. as in $\operatorname{dim} \frac{M}{(a_1,\cdots,a_n)M} = \operatorname{dim} M - n$ for $M$-regular sequence $a_1,\cdots,a_n$; correct me if I'm wrong!) But I don't understand why the length of a maximal regular sequence is of interest. Is it merely due to some technical consideration in cohomology that we want many $\operatorname{Ext}$ groups to vanish?
What does CM rings mean geometrically? As I read from Eisenbud's book, there doesn't seem to be an exact geometric concept that corresponds to it. Nonetheless I would still like to know about any geometric intuition of CM rings. I know that it should be locally equidimensional. Some examples of CM rings come from complete intersection (I read this from wiki). But what else?
Why do we care about CM rings? If I understand it correctly, CM rings <=> unmixedness theorem holds for every ideal for a noetherian ring, which should mean every closed subschemes have equidimensional irreducible components (and there's no embedded components). This looks quite restrictive.

Thanks!

REPLY [6 votes]: One way I think about Cohen-Macaulayness (probably not in largest generality but at least in context relevant to combinatorics) is as follows:
Think first of the ring of symmetric polynomials in $n$ variables. A remarkable fact from first year linear algebra is that this ring is a polynomial ring in some other variables, the elementary symmetric polynomials. 
Being a polynomial ring is rare (but this is a sort of role model). Being Cohen-Macaulay comes close. A Cohen Macaulay ring can be described as a direct sum where each summand $S_i$ is of the form $\eta_i R$, where $R$ is a polynomial rings (whose variables are the elements of a system of parameters) and $\eta_i$ are elements. Being a direct sum is important here. 
For graded rings such a description has remarkable combinatorial consequences.<|endoftext|>
TITLE: When and why do universal objects have extra properties?
QUESTION [21 upvotes]: I'm interested in situations where universal objects come with more structure than their definitions suggest. A classic case of this is where the free abelian group on one element has a ring structure. Proving this is a straightforward exercise using the free-underlying adjunction. So I'd like to know of other cases of this phenomenon, and if possible an explanation as to why the extra structure comes about.
Plenty of examples are given in Hazewinkel's paper Niceness Theorems, but how about very familiar examples such as the rationals? Does, say, the characterisation of $\langle \mathbb{Q}, \gt \rangle$ as the Fraïssé limit of the category of finite linearly ordered sets and order preserving injections tell us why it should support a compatible group, ring and even field structure? Do characterisations of the reals relate to each other?
My question is not completely unrelated to Theorems for nothing (and the proofs for free), as shown by the example given there of subgroups of free groups being free, which also occurs in Hazewinkel's paper.

REPLY [2 votes]: An incomplete answer on the subject of countable dense linear orders without endpoints;
I left some other thoughts at the Cafe; on further reflection, one can think of the maps $\cdot\times \frac{p}{q}:q[j,k]\rightarrow[pj,pk]$ defined on finite sets of integers as both inducing refinements of some map $[j,k]\rightarrow D$ and as rational scalings of the image.
(I don't know much about Fraïssé constructions, so this gets vague now) then from universality, we  (almost) get a $\mathbb{Q}$-module structure on the Fraïssé limit?<|endoftext|>
TITLE: Integrability of derivatives
QUESTION [45 upvotes]: Is there a (preferably simple) example of a function $f:(a,b)\to \mathbb{R}$ which is everywhere differentiable, such that $f'$ is not Riemann integrable?
I ask for pedagogical reasons.  Results in basic real analysis relating a function and its derivative can generally be proved via the mean value theorem or the fundamental theorem of calculus.  Proofs via FTC are often simpler to come up with and explain: you just integrate the hypothesis to get the conclusion. But doing this requires $f'$ (or something) to be integrable; textbooks taking such an approach typically stipulate that $f'$ is continuous.  Proofs via MVT can avoid such unnecessary assumptions but may require more creativity.  So I'd like an example to show that the extra work does actually pay off.
Note that derivatives of everywhere differentiable functions cannot be arbitrarily badly behaved.  For example, they satisfy the conclusion of the intermediate value theorem.

REPLY [6 votes]: I remember, that there was an example of such a function in the book Counterexamples in Analysis. Just wanted to mention it for the sake of completeness. It can be found in Chapter 8 (Sets and Measure on the Real Axis), Example 35 (A bounded function possessing a primitive on a closed interval but failing to be Riemann-integrable there.)<|endoftext|>
TITLE: What functor does Grassmannian represent?
QUESTION [13 upvotes]: As we know, the projective space $\mathbb{P}^n$ represents the functor sending $X$ to the set of line bundles $L$ on $X$ together with a surjection from the trivial vector bundle to $L$.
My question is, what functor does the Grassmannian $Gr(d,n)$ represent? Sending $X$ to the set of rank-$n$ vector bundles together with a sub-bundle of rank $d$?
More generally, what functors do flag varieties represent?

REPLY [8 votes]: Let me elaborate on some of the other answers. 
On the Grassmannian X = Gr(k,n) (I am using this notation to mean k-dimensional subspaces of an n-dimensional vector space), we have the trivial bundle $X \times K^n$ (here K is our field of definition), and the tautological subbundle R (naively, this is the subset {$(x,v) \in X \times K^n \mid v \in x$} which is a locally free sheaf of rank k, and its quotient Q is also a locally free sheaf of rank n-k. So we write
$0 \to R \to X \times K^n \to Q \to 0$,
which is the tautological sequence. Given a map to $f \colon Y \to X$, where Y is a k-scheme, we can pull back this sequence to get 
$0 \to f^*R \to Y \times K^n \to f^*Q \to 0$.
Conversely, given such a sequence, there is a unique map to X which gives this pullback. Naively, over a point y, the fiber of $f^*R$ is a subspace of $K^n$, so we send it to that closed point. So X represents the functor which sends Y to the set of short exact sequences as above. The quotient being locally free implies that the subsheaf is also locally free, so it really represents the functor which sends Y to the set of quotients $\mathcal{O}_Y^n \to F \to 0$ where the rank of F is n-k.
For general flag varieties, we have similar tautological sequences (but there are more subbundles to consider). This kind of analysis also makes sense for symplectic and orthogonal Grassmannians and flag varieties.
We can also do something similar if we're talking about the Grassmannian of a vector bundle instead of a vector space. Then we just work in the category of S-schemes instead of k-schemes where S is whatever the base space is.<|endoftext|>
TITLE: The inverse Galois problem, what is it good for?
QUESTION [76 upvotes]: Several years ago I attended a colloquium talk of an expert in Galois theory. He motivated some of his work on its relation with the inverse Galois problem. During the talk, a guy from the audience asked: "why should I, as a number theorist, should care about the inverse Galois problem?" 
I must say that as a young graduate student that works on Galois theory, I was amazed or even perhaps shocked from this question. But later, I realized that I should have asked myself this question long ago. 
Can you pose reasons to convince a mathematician (not just number theorist) of the importance of the inverse Galois problem? Or maybe why it's unimportant if you want to ruin the party ;)

REPLY [2 votes]: Just to illustrate that the inverse Galois problem in specific cases can be useful.
In my master thesis (2016-2017), I used the inverse Galois problem of cyclic groups over $\mathbb{Q}$ to construct Anosov $2$-step nilpotent rational Lie algebras of certain types. This can help in the classification of Anosov nilmanifolds.<|endoftext|>
TITLE: A proof of the salamander lemma without Mitchell's embedding theorem?
QUESTION [25 upvotes]: The salamander lemma is a lemma in homological algebra from which a number of theorems quickly drop out, some of the more famous ones include the snake lemma, the five lemma, the sharp 3x3 lemma (generalized nine lemma), etc.  However, the only proof I've ever seen of this lemma is by a diagram chase after reducing to R-mod by using mitchell's embedding theorem.  Is there an elementary proof of this lemma by universal properties in an abelian category (I don't know if we can weaken the requirements past an abelian category)?
If you haven't heard of the salamander lemma, here's the relevant  paper.  
And here's an article on it by our gracious administrator, Anton Geraschenko: Click!
Also, small side question, but does anyone know a good place to find some worked-out diagram-theoretic proofs that don't use mitchell and prove everything by universal property?  It's not that I have anything against doing it that way (it's certainly much faster), but I'd be interested to see some proofs done without it, just working from the axioms and universal properties.  
PLEASE NOTE THE EDIT BELOW
EDIT:  Jonathan Wise posted an edit to his answer where he provided a great proof for the original question (doesn't use any hint of elements!).  I noticed that he's only gotten four votes for the answer, so I figured I'd just bring it to everyone's attention, since I didn't know that he'd even added this answer until yesterday.  The problem is that he put his edit notice in the middle of the text without bolding it, so I missed it entirely (presumably, so did most other people).

REPLY [4 votes]: The first paper on abelian categories is "Exact Categories and Duality" D. A. Buchsbaum. It was published in 1955, two years before Grothendieck's famous Tohoku paper. You can find it on jstor. The section 5 is about "fundamental lemmas" such as the Nine Lemma (5.5), the Snake lemma (5.8) and the Five Lemma (5.9). The proofs are direct using the definition of an abelian category (called "exact category" by Buchsbaum, this term was used later by Quillen), in particular they use - of course - no elements. Unfortunately I cannot find the salamander lemma, but lots of basics of homological algebra such as homology, derived functors, satellites. Basically Buchsbaum observed that there is nothing special to module categories treated in Eilenberg-Cartan's landmark book on homological algebra.<|endoftext|>
TITLE: What functor does a Schubert variety represent?
QUESTION [10 upvotes]: I'm inspired by Yuhao's question.  The functor that takes a scheme S to the set of k-dimensional vector subbundles of C^n x S (understanding "subbundle" to mean that the quotient by it is another vector bundle) is represented by the Grassmannian G(k,n).  What functor is represented by the Schubert subvarieties of G(k,n)?
The Schubert variety associated to a sequence of numbers d = (d_0,d_1,...,d_n) is the collection of k-planes that meet the standard i-plane in a subspace of dimension $\geq$ d_i.  (The d for which the associated Schubert variety is nonempty are naturally parameterized by k-element subsets of {1,2,...,n}, or more usefully by partitions that fit in a k x (n-k) box.)
It's tempting to say that the functor represented by a Schubert variety should take a scheme S to the set of subbundles of C^n x S that, fiber-by-fiber, meet the standard i-plane in a subspace of dimension $\geq$ d_i.  But I don't know what this should mean on a non-reduced scheme.
One way to look for a moduli interpretation for the functor represented by a variety X is to find one or several "universal" or "tautological" families of things over X.  Then one hopes that families of things of the same type over another scheme S will be pulled back along maps from S to X.  The natural-looking tautological objects over a Schubert variety are the families of intersections $V \cap \mathbf{C}^i$, where V runs through k-planes belonging to the Schubert variety.  These families are not equidimensional, and in particular not flat; this might be an obstacle.
...
It's really clear what's going on from Steven's answer, but I decided to follow it to the end and write down a functor.  Beyond saying that these things are zero loci of interesting sections of vector bundles, this point of view probably doesn't much illuminate Schubert varieties.  But here it is.
Steven's answer is cleanest if we regard the Grassmannian and its subvarieties as parameterizing quotients rather than subspaces of C^n.  So a point in the Grassmannian G(k,n) is an isomorphism class of surjective maps C^n --> V, or equivalently of tuples (V,v_1,...,v_n) where V is a k-dimensional vector space and v_1,...,v_n span V.
The equivalence class of (V,v_1,...,v_n) belongs to the Schubert variety corresponding to r = (r_1,...,r_n) if for each i the first i terms on the list of vs span a subspace of dimension $\leq$ r_i.  As in the answer we can use exterior powers to make sense of these rank conditions over a scheme S.  So an S-valued point of the Schubert variety corresponding to r is an equivalence class of tuples (E,s_1,...,s_n) where

E is a vector bundle
s_1,...,s_n is a list of sections of E
these sections generate E in each fiber, and given any i and any (r_i + 1)-tuple of sections chosen from (s_1,...,s_i), we have

$$s_{j_1} \wedge s_{j_2} \wedge \cdots \wedge s_{j_{r_i + 1}} = 0$$
For example, if S = Spec(C[x]/x^2), and E is the rank 2 free sheaf on S spanned by e_1 and e_2, then (x.(ae_1 + be_2), x.(ce_1 + de_2), e_1, e_2) is an S-valued point of the Schubert variety corresponding to r = (1,1,2,2).  The complex numbers a,b,c, and d parameterize the 4-dimensional Zariski tangent space to the singular point on this 3-dimensional variety.
There's a loose end.  I think it's clear that the functor
$$S \mapsto \{(E,s_1,\ldots,s_n)\}/\sim$$
represents a subscheme of the Grassmannian with the same set-theoretic support as a Schubert variety.  Why does it represent the actual Schubert variety?  It would be enough to know that the representing object is reduced.  But I don't even know how to express "X is reduced" in terms of the functor of points Hom(-,X).

REPLY [5 votes]: Everything you said should be fine. As for the case of not necessarily reduced schemes, we have to be careful, but I think the following will work. 
Say E is our given vector bundle of rank n which is a subbundle of V which is some trivial bundle, and fix a trivial subbundle W of V. The condition $\dim(E(x) \cap W(x)) \ge k$ (here x is a closed point and E(x) denotes fiber) is equivalent to saying that rank$(E(x) \to V/W(x)) \le n - k$, or equivalently, that the map on exterior products 
$\bigwedge^{n-k+1} E(x) \to \bigwedge^{n-k+1} V/W(x)$ 
is zero. Well, this map can be thought of as a section s of 
$(\bigwedge^{n-k+1} E)^\vee \otimes \bigwedge^{n-k+1} V/W$. 
So to define the locus of points where $\dim(E(x) \cap W(x)) \ge k$ scheme-theoretically, we can just ask for the zero locus of s (is this what it's called?).
So if you're asking for several intersection conditions, we can take the scheme-theoretic intersection of all these zero loci and ask if it's equal to our scheme.
This seems to be the reasonable thing to think about since those sections can be pulled back if you have a map to the Grassmannian whose image is contained in a certain Schubert variety, since the Schubert variety can be defined by the intersection of certain zero loci in the Grassmannian.<|endoftext|>
TITLE: Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object?
QUESTION [10 upvotes]: X is a Noetherian scheme, F is an injective object in the category of quasi-coherent sheaves on X. U is an open subset of X. Why F's restriction on U is still an injective object in the category of quasi-coherent sheaves on U?

REPLY [23 votes]: The restriction-by-zero type arguments can actually be made to work, with some effort and an extra hypothesis.  Suppose $X$ is locally Noetherian, $j: U \to X$ the inclusion of an open subscheme.
Let $Mod(X)$ and $QCoh(X)$ be the categories of $O_X$-modules, and quasi-coherent $O_X$-modules, respectively.
The "some effort" is the following Lemma
Lemma If $X$ is locally Noetherian, then the injective objects in $QCoh(X)$ are precisely the injective objects of $Mod(X)$ which are quasi-coherent as sheaves of modules.
Pf: Any injective object of $Mod(X)$ which is quasi-coherent must certainly be injective in the smaller category $QCoh(X)$.  For the converse, it suffices to show that any injective object $I$ of $QCoh(X)$ injects into some $I'$ which is a quasi-coherent injective object of $Mod(X)$, for then $I$ will be a retract of $I'$ and so injective in $Mod(X)$.  This seems tricky, but is proved in Theorem 7.18 of Hartshorne's "Residues and duality".

Now, let's prove the result using the Lemma:  If $J$ is an injective object in $QCoh(X)$, then the hard direction of the Lemma implies that it is injective in $Mod(X)$.  The restriction-by-zero argument applies in this category, allowing us to conclude that $j^* J$ is injective in $Mod(U)$.  It's clearly quasi-coherent, so applying the easy direction of the Lemma we see that it is injective in $QCoh(U)$ as desired.
[Aside: On a Noetherian scheme, any quasi-coherent sheaf is a union of its coherent subsheaves and one can "extend" coherent sheaves on U to coherent sheaves on X (see e.g., Hartshorne Ex. II.5.15).  Using these facts, one should be able to give a more direct argument in the Noetherian case.]<|endoftext|>
TITLE: When is tensoring with a module representable by a scheme?
QUESTION [10 upvotes]: Consider the following: Let $A$ be a commutative ring, let $M$ be an $A$-module. When is the functor from $A$-algebras to Sets given by $R \mapsto R \otimes M$ representable by an $A$-scheme?
Unless I've made a mistake, this is always be an fpqc sheaf. When $M$ is a finitely generated free A-module, then $\mathrm{Spec}( \mathrm{Sym}^\bullet M^*)$ does the trick.

REPLY [4 votes]: When $A$ is noetherian and $M$ is finitely generated, Nitin Nitsure showed that the functor is representable if and only if $M$ is projective (see http://arxiv.org/abs/math/0308036).<|endoftext|>
TITLE: Characteristic classes of sphere bundles over spheres in terms of clutching functions
QUESTION [23 upvotes]: I'm trying to understand Milnor's proof of the existence of exotic 7-spheres.
Milnor finds his examples among $S^{3}$ bundles over $S^{4}$ (with structure group $SO(4)$ ).  Such a bundle can be described as follows:
Given $M$, an $S^{3}$ bundle over $S^{4}$, if we restrict $M$ to the northern (or southern) hemisphere of $S^{4}$, it must trivialize since each hemisphere is contractible.  Hence, we can build $M$ by specifying, for each point $p$ in $S^{3}$ = equator of $S^{4}$ = intersection of northern and southern hemispheres, an element of $SO(4)$ which glues $p\times S^{3}$ in the northern hemisphere to $p\times S^{3}$ in the southern hemisphere.
This defines a function $f:S^{3}\rightarrow SO(4)$, which is known as the clutching function for $M$.  By usual fiber bundle theory, the isomorphism type of $M$ only depends on the homotopy class of $f$.
$SO(4)$ is double covered by $S^3\times S^3$, and hence $\pi_3(SO(4)) = \mathbb{Z}\oplus\mathbb{Z}$.  Thus, $f$ is really determined (at least, up to homotopy) by an ordered pair of integers (i,j).
Now, as the bundles have structure group $SO(4)$, it makes sense to talk about the Pontryagin classes of $M$.  In Milnor's proof of the existence of exotic spheres, he needs to argue that $p_1(M) = \pm 2(i-j)$.  His first step in this argument is that "clearly $p_1(M)$ is a linear function of $i$ and $j$."
It IS clear to me that the Pontragin classes associated to $(ni, nj)$ for $n\in \mathbb{Z}$ will depend linearly on $n$.  For, if we let $N_{i,j}$ denote the principal $SO(4)$ bundle over $S^{4}$ corresponding to $(i,j)$, then $N_{ni,nj}$ is clearly obtained as the pullback of $N_{i,j}$ via a degree $n$ map from $S^{4}$ to itself.
However, it's not clear to me why $p_1(M)$ is additive in $(i,j)$.  Am I missing something simple?
And while we're talking about it, is more true?  That is, For any sphere bundle over a sphere, say, $S^{k}\rightarrow E\rightarrow S^{n}$, should any characteristic classes (Pontryagin, Stiefel-Whitney, Euler) be linear in terms of the clutching function?
For example, we can think of $p_1$ as a map from $\pi_{n-1}(SO(k+1))\rightarrow H^{4}(S^{n})$.  Is this map a homomorphism?  How about for the other characteristic classes?

REPLY [24 votes]: There is a way to explain it that's similar to what you said about multiplication by $n$.
Let $G$ be a Lie group, and let $f_1$ and $f_2$ be any two clutching functions that describe $G$-bundles $E_1$ and $E_2$ on $S^n$.  Suppose that $f_1$ and $f_2$ agree at a base point of $S^{n-1}$.  Let $c$ be some characteristic class of $G$-bundles of degree $n$; it could even be a cup product of standard classes such as Chern or Pontryagin or whatever.  Let $f_3$ be the combination of $f_1$ and $f_2$ on the one-point union $S^{n-1} \vee S^{n-1}$.  It is the clutching function of a bundle $E_3$ on the suspension $\Sigma(S^{n-1} \vee S^{n-1})$, which is the union of two $n$-spheres along an interval and thus homotopy equivalent to $S^n \vee S^n$.
Whether you define $c$ the old-fashioned way by obstructions, or the more modern way by pullbacks from classifying spaces, it is easy to argue that $c(E_3) = c(E_1) \oplus c(E_2)$. I.e., it's just the ordered pair of the characteristic classes of its parts.  Now addition in $\pi_n$ is modeled by a map $S^n \to S^n \vee S^n$, and the induced map on $H^n$ takes $a \oplus b$ to $a+b$.  (Your generalized question about $H^k(S^n)$ is of course non-trivial only when $k=n$.)

A shorter, more modern summary of the same story is as follows.  The $X$ is a space and $LX$ is its loop space, then $\pi_{n-1}(LX) \cong \pi_n(X)$.  The loop space of the classifying space $B_G$ is homotopy equivalent to $G$ itself, so $\pi_{n-1}(G) \cong \pi_n(B_G)$.  A characteristic class of degree $n$ is any cohomology class in $H^n(B_G)$.  The linearity that Milnor uses is the transposed form of the fact that the Hurewicz homomorphism $\pi_n(B_G) \to H_n(B_G)$ is linear.  (If you expand this it out more explicitly, it isn't really different from what I say above.)<|endoftext|>
TITLE: Triangle-free Lemma
QUESTION [9 upvotes]: Theorem (Triangle-free Lemma). For all $\eta>0$ there exists $c > 0$ and $n_0$ so that every graph $G$ on $n>n_0$ vertices, which contains at most $cn^3$ triangles can be made
triangle free by removing at most $\eta\binom{n}{2}$ edges.
I am trying to find some information related to this topic, I am unable to access the orignal paper by Ruzsa & Szemeredi. 
Does anyone know any useful papers/books on the triangle-free lemma?

REPLY [4 votes]: The recent survey paper 
D. Conlon and J. Fox, Graph removal lemmas, Surveys in Combinatorics, Cambridge University Press, 2013, 1-50 
is on the triangle removal lemma and its various extensions.<|endoftext|>
TITLE: Seifert surfaces of torus knots
QUESTION [8 upvotes]: Does anyone know a nice description of a Seifert surface of a torus knot? I can construct such surfaces in band projection, but what I get is ugly and unwieldy. Is there some elegant description for Seifert surfaces for such knots which I'm missing? (I'm not sure precisely what I mean by elegant...)

REPLY [12 votes]: Torus knot complements fiber over $S^1$. So the minimal Seifert surface for a $(p,q)$-torus knot is a once punctured surface of genus $\frac{(p-1)(q-1)}{2}$.  You get it as the Milnor fibre of the map from $\mathbb C^2 \to \mathbb C$ given by $f(z_1,z_2)=z_1^p-z_2^q$. 
That's pretty elegant to me.  The monodromy is an automorphism of the surface of order $pq$, it is a free action except on two orbits -- one orbit has $p$ elements, the other orbit has $q$ elements.  
These details are mostly in Milnor's "Singular points of complex hypersurfaces", also Eisenbud and Neumann's "Three-dimensional link theory and invariants of plane curve singularities".  I also have a sketch of it in my JSJ-decompositions paper, on the arXiv. I got the idea for this computation by fleshing out an example of Paul Norbury's (from Walter Neumann's canonical decompositions paper on his webpage).
I'd like to add, Eisenbud and Neumann describe the Seifert surfaces of all knots whose complements are graph manifolds in this way. Well, the ones that fibre.  They also characterise the knots with graph manifold complements that fiber over S^1. 
edit: alternatively you could construct the Seifert surface and monodromy from the Seifert-fiber data, as I sketch in this thread: Periodic mapping classes of the genus two orientable surface<|endoftext|>
TITLE: Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?
QUESTION [93 upvotes]: Is there an infinite field $k$ together with a polynomial $f \in k[x]$ such that the associated map $f \colon k \to k$ is not surjective but misses only finitely many elements in $k$ (i.e. only finitely many points $y \in k$ do not lie in the image of $f$)?
For finite fields $k$, there are such polynomials $f$. If such a poynomial $f$ exists, then $k$ cannot be algebraically closed; the field $\mathbb{R}$ doesn't work either.

REPLY [8 votes]: Recently, I have shown the following theorem with Hendrik Lenstra. 
Definition: A field $k$ is called large if every irreducible $k$-curve with a $k$-rational smooth point has infinitely many $k$-points. 
Some examples of large fields are $\mathbb{R}$, $\mathbb{Q}_p$ ($p$ prime), $l((t))$ (where $l$ is a field), infinite algebraic extensions of finite fields. Furthermore, finite extensions of large fields are large.
Theorem: Let $k$ be a perfect large field and let $f \in k[x]$. Consider the evaluation map $f_k: k \to k$. Assume that $f_k$ is not surjective. Then the set $k \setminus f_k(k)$ is infinite (in fact, it has cardinality $|k|$).
See my arXiv article for a proof.<|endoftext|>
TITLE: Difference between connected vs strongly connected vs complete graphs
QUESTION [11 upvotes]: What is the difference between
connected
strongly-connected and
complete?
My understanding is:
connected: you can get to every vertex from every other vertex.
strongly connected: every vertex has an edge connecting it to every other vertex.
complete: same as strongly connected.
Is this correct?

REPLY [3 votes]: It is also important to remember the distinction between strongly connected and unilaterally connected. A directed graph is unilaterally connected if for any two vertices a and b, there is a directed path from a to b or from b to a but not necessarily both (although there could be). Strongly connected implies that both directed paths exist. This means that strongly connected graphs are a subset of unilaterally connected graphs.
And a directed graph is weakly connected if it's underlying graph is connected.<|endoftext|>
TITLE: Kunneth formula for motivic cohomology
QUESTION [11 upvotes]: I was wondering when the Kunneth formula holds for motivic cohomology:
$$
  H^p(X,A(\alpha)) = \bigoplus_{i+j=p;\beta+\gamma = \alpha} H^j(X,A(\beta)) \otimes H^i(X,A(\gamma)) 
$$
where $H^p(X,A(\alpha))$ is defined as you wish: by higher Chow groups, Hom groups in $DM(X)$, etc... The case I'm most interested in is $A= \mathbb{Q}$ and $M(X) \in DM(\mathbb{Q})_{\mathbb{Q}}$ in the thick sub-triangagulated category generated by the $\mathbb{Q}(n)$, $n\in \mathbb{Z}$.

REPLY [3 votes]: There is a paper by Dugger & Isaksen called Motivic Cell Structures in which they establish a Künneth formula for various cohomology theories that are represented in the $\mathbb{A}^1$-homotopy category, provided the object $X$ satisfies some sort of cellularity condition which is similar to the requirement of having a Tate motive.
Of course, as Tyler suggested this Künneth formula is of the form of a spectral sequence over the motivic cohomology of the ground field, to wit $\mathrm{Tor}_{H(\mathrm{spec}\; k)} (H(X) , H(Y)) \Rightarrow H(X \times Y)$.
In general, this spectral sequence fails, as can probably be seen by all the counterexamples already given, and certainly can be seen by considering $\mathrm{spec}\; \mathbb{C} \times_\mathbb{R} \mathrm{spec} \; \mathbb{C}$ with $\mathbb{Z}/2$-coefficients, where the motivic cohomology rings are known in their entirities (These calculations also appear in papers of Dugger & Isaksen on the motivic Adams spectral sequence).<|endoftext|>
TITLE: Does a universal Frobenius map exist?
QUESTION [10 upvotes]: For any prime p, one has the Frobenius homomorphism Fp defined on rings of characteristic p.
Is there any kind of object, say U, with a "universal Frobenius map" F such that for any prime p and any ring R of characteristic p we can view the Frobenius Fp over R as "the" base change of F from U to R?
I have the following picture in mind: In some sense it should be possible to view the category of Z-algebras as a sheaf of categories over Spec Z such that the fibre over Spec Fp is just the category of F_p-algebras. A natural transformation f of the identity functor on the category of Z-algebras should restrict to a natural transformation fp of the identity functor on the category of Fp-algebras. In this naive picture one cannot expect the existence of an f such that fp is the Frobenius on Fp-algebras for all primes p. But is there way to make this picture work?
Another possible way to answer my question could be the following: Is there a classifying topos of, say, algebras with a Frobenius action? By this I mean the following: Is there a topos E with a fixed ring object R and an algebra A over it and an R-linear endomorphism f of A such that for any other topos E' with similar data R', A' there is a unique morphism of topoi E' -> E that pulls back R, A to R', A' and such that f is pulled back to the Frobenius fp of A' in case R' is of prime characteristic.
(Feel free to modify my two pictures to make them work.)

REPLY [3 votes]: I don't really get the categorical picture of what you are asking, but it feels something very similar to the relation between finite fields of characteristic $p$ and the ring (of characteristic 0) of $p$-typical Witt vectors.
You might want to have a look at Borger and Wieland work on pleythistic algebras. The "lifting of all Frobeniuses at the same time" gives you an structure of $\Lambda$-ring, so the paper The basic geometry of Witt vectors by Borger might also be useful.<|endoftext|>
TITLE: What (if anything) happened to Intersection Homology?
QUESTION [31 upvotes]: In the early 1990's, Gil Kalai introduced me to a very interesting generalization of homology theory called intersection homology, which existed for like 10 years back then I believe. Defined initially by Goresky and MacPherson, this is a version of homology which agrees with ordinary homology on manifolds, but also retains crucial properties like Poincare Duality and Hodge Theory on singular (non-)manifolds. The original definition was combinatorial, but it was later re-interpreted in sheaf-theoretic terms (perverse sheaves?). 
Back then it certainly looked like an exciting new development. So, I'm curious - where does the field stand today? Is it still thriving, or has it been merged with something else, or just faded away?

REPLY [34 votes]: Intersection homology is alive and well in a large number of guises. It's true that a lot of the work trended to algebraic geometry, representation theory, and categorical constructions, such as perverse sheaves, through the 90s, but there also continues to be work in the more topological settings by people such as me, Cappell, Shaneson, Markus Banagl, Laurentiu Maxim, and many others. At least some of this work is dedicated to extending classical manifold invariants, such as characteristic classes, in a meaningful way to stratified spaces, such as algebraic varieties, and there is a lot of recent interest (though slow) progress in figuring out how intersection homology might tie into various algebraic topology constructions. There are also analytic formulations such as L^2 cohomology (initiated by Cheeger), and much more.
Here are some good references to get started in the area:
Books:
An Introduction to Intersection Homology by Kirwan and Woolf (mostly concerned with telling the reader about the fancy early applications to algebraic geometry and representation theory, but a great overview nonetheless)
Intersection Cohomology by Borel, et.al. This is a great serious technical introduction to the area and, to my mind, the canonical source for the foundations of the subject)
Topological Invariants of Stratified Spaces by Markus Banagl (topological but mostly from the sheaf point of view)
For an overview of state-of-the-art in intersection homology and related fields, I'm co-editing a volume on Topology of Stratified Spaces that will be published in the MSRI series. Unfortunately, it's not out yet, but look for it soon.
Papers:
The original papers of Goresky and MacPherson are quite good.
Topological invariance of intersection homology without sheaves by Henry King is a good introduction to the singular version of the theory.
And for a whole pile of recent papers, I'll shamelessly plug my own web site: http://faculty.tcu.edu/gfriedman/
and Markus Banagl's: http://www.mathi.uni-heidelberg.de/~banagl/
And many further references can be found from these locations.<|endoftext|>
TITLE: Under what conditions do eigenvalues of a quadratic eigenvalue problem come in reciprocal pairs?
QUESTION [5 upvotes]: Suppose we have a quadratic eigenvalue problem $\lambda^2 M + \lambda C + K$. Under what conditions is the following statement true: If $\lambda$ is an eigenvalue, so is $1/\lambda$?
Here, $M$, $C$, and $K$ are square matrices (not necessarily full rank). This is of interest to me since I have such systems for which I know (based on physical arguments) that the eigenvalues must come in reciprocal pairs, but I don't know what this necessarily implies about the matrix properties. From a cursory look through Google, it seems that palindromic QEPs have this property, but I'm wondering if this property is more general.

REPLY [4 votes]: Consider the linearization
$$
L(\lambda)=\lambda\begin{pmatrix}M&0\cr
0&1
\end{pmatrix}+\begin{pmatrix}C&K\cr
-1&0\end{pmatrix}=\lambda A+B.
$$
The eigenvalues of the original problem coincide with those for the linearization.
Now, the eigenvalues of the linearized problem are the roots of the polynomial $\det(\lambda A+B)$, while their reciprocals are roots of the polynomial $\lambda^{2n}\det(\lambda^{-1}A+B)=\det(\lambda B+A)$. If we require that these coincide with multiplicities, then the two polynomials must be linearly dependent. (In fact, they will be $\pm 1$ times each other.) If $A$ is nondegenerate, this implies that $A^{-1}B$ is similar to its inverse matrix. The latter condition does not give us much immediately, but at least we know that $\det A=\pm\det B$; that is, $\det M=\pm\det K$. We also get that the traces of $A^{-1}B$ and $B^{-1}A$ coincide, which can be rewritten as a condition on $M$, $C$, and $K$ (hopefully).<|endoftext|>
TITLE: Computing correlation between time series with missing data.
QUESTION [5 upvotes]: Suppose you have two simple Ar[1] series of the form $y_n=y_{n-1}+e_n$ and $x_n=x_{n-1}+m_n$, where $e_n$ and $m_n$ are normal white noise processes with no auto-correlation and $Corr(e_n,m_n)=p$. Then suppose we have possibly non-overlapping data for Y and X (IE, observation 10 exists for Y but not for X), and to avoid data generating process issues, assume that the distribution of missing data is random.
Is there any way to estimate p? 
As a follow-up question, is there a way to easily generalize to a situation where $y_n$ and $x_n$ are observed with known normally distributed measurement error?

REPLY [3 votes]: The paper "Application of Two-Directional Time Series Models to Replace Missing Data" offers two methods (not necessarily under your precise model), one that minimizes "the average error associated with the missing value" (the other I can't understand from the abstract).
Edit. I have changed my (old) answer to community wiki.  Would someone please vote this up
so that the bot that reposts those questions for which there are no upvoted answers stops recycling
this one?  Thanks.<|endoftext|>
TITLE: What is the difference between a zeta function and an L-function?
QUESTION [25 upvotes]: I've been learning about Dedekind zeta functions and some basic L-functions in my introductory algebraic number theory class, and I've been wondering why some functions are called L-functions and others are called zeta functions. I know that the zeta function is a product of L-functions, so it seems like an L-function is somehow a component of a zeta function (at least in the case of Artin L-functions, they correspond to specific representations). Is this the idea behind the distinction between "zeta function" and "L-function"? How do things generalize to other kinds of zeta- and L-functions?

REPLY [11 votes]: I do like the other answers, too, but it seemed silly to append comments to all... :
To my perception, first, I tend to not feel a difference between "zeta function attached to ..." and "L-function attached to..." if only because usage is variable.
Second, in many settings (analytic/automorphic or geometric/motivic or...) a zeta function is an L-function with relatively trivial "further data", whatever that means in context. So, Dedekind zeta functions are Hecke L-functions with trivial data, for example. Analogously for schemes without or with non-trivial sheaf, in the other world. This general rule is certainly not strict... depending on usage.
Third, there are the systematic, partly proven, partly conjectural, miracles that "larger" zetas factor into "smaller" L-functions. Classfield theory and such. This does highlight the ambiguity in "usage", namely, that some "base" is necessary to understand "triviality", etc.
Edit: again significantly contingent on "usage"... If we say that an "L-function" (or "zeta function") "has an analytic continuation (provable by us)", then this would accidentally disallow Hasse-Weil zeta/L-functions of general varieties/schemes/whatever, because in this year we know "few" cases wherein we can prove this, although conjecturally it is mostly-always so (meaning that poles, if any, are finite and describable). Similarly, factorization into Artin L-functions is in one way completely fine (for decades), but, in another, unsatisfactory since we do not know their holomorphy, ... so might decide that they're not yet (in 2012) fully-legitimate "L-functions"? And/or that the "factorization" of Dedekind zetas into such things is not entirely satisfactory (as in a comment). I would not be surprised that such "technicalities" persist in things I know less about. :)<|endoftext|>
TITLE: Generalizations of the Birkhoff-von Neumann Theorem
QUESTION [31 upvotes]: The famous Birkhoff-von Neumann theorem asserts that every doubly stochastic matrix can be written as a convex combination of permutation matrices. 
The question is to point out different generalizations of this theorem, different "non-generalizations" namely cases where an expected generalization is false, and to briefly describe the context of these generalizations. 
A related MO question: Sampling from the Birkhoff polytope

REPLY [3 votes]: The theorem still holds if we ask each row and column to sum to some integer $m$; permutation matrices are replaced with zero-one matrices having the same constraints. See Watkins and Merris, Convex Sets of Doubly Stochastic Matrices and Lewandowski, Liu and Liu, An Algorithmic Proof of a Generalization of the
Birkhoff - Von Neumann Theorem.<|endoftext|>
TITLE: Gerbes for a cyclic group. (or maybe G_m too)
QUESTION [8 upvotes]: Let μn be the group scheme of n-th roots of unity. If X is a scheme and L is a line bundle on X, then I can construct a μn-gerbe Y over X by letting the S-points of Y be a S-point of X, a line bundle M on S and an isomorphism between the n-th tensor power of M and the pullback of L to S. Can anyone provide examples not of this form?
Commentary: It looks like I'm taking the image of an element of H1(X,Gm) in H2(X,μn) under the long exact sequence associated to 0-->μn-->Gm-->Gm-->0. Thus examples of gerbes for the multiplicative group Gm will likely be relevant, so people providing such examples will also be appreciated.

REPLY [8 votes]: A bit of a response to your "Commentary": 
As you point out, the failure of your construction to hit all $\mu_n$-gerbes is governed by the exact sequence
$H^1(X, \mathbb{G}_m) \to H^2(X, \mu_n) \to H^2(X,\mathbb{G}_m)[n] \to 0$ so answering your question is related to producing torsion $\mathbb{G}_m$-torsors.
The question of doing so has been studied as part of the theory of Brauer groups:
Let $Br(X)$ ("Brauer group") denote the group of Azumaya algebras, which are a generalization of the central simple algebras over a field (that is the classical Brauer group).
Let $Br'(X)$ ("Cohomological Brauer group") denote the torsion part of $H^2(X, \mathbb{G}_m)$.
In the case of $X = Spec k$, $k$ a field, the equivalence of these two groups is classical and they can be computed in various cases of number-theoretic interest (e.g., number fields/local fields/finite fields).  In this case, $H^1(X, \mathbb{G}_m)=0$ by Hilbert's Theorem 90, and yet there are plenty of examples where $Br(X)$ is very much non-trivial.  Grothendieck studied the relation between $Br(X)$ and $Br'(X)$ in general in Dix Exposes.  The upshot is that there is an injective map $Br(X) \to Br'(X)$ and it is an isomorphism in reasonable cases (e.g., I think $X$ quasi-projective over a field).  (See Dix Exp, or Ch. IV of Milne's "Etale Cohomology".)

I won't say more about the general picture, but I'll work out in detail the simple case of $X = Spec \mathbb{R}$.  In this case, $Br(Spec \mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$ generated by the class of the usual quaternions, viewed as a central simple algebra over $\mathbb{R}$.  We can give a geometric description of the resulting $\mathbb{G}_m$-torsor: 
Start with the smooth plane conic $C = Proj \mathbb{R}[x,y,z]/(x^2+y^2+z^2)$.  It's a smooth genus $0$ curve, but has no $\mathbb{R}$-points and so is not isomorphic to $\mathbb{P}^1_{\mathbb{R}}$. However, after base-change to $\mathbb{C}$ it attains a point and so becomes isomorphic to $\mathbb{P}^1_{\mathbb{C}}$; such a Galois-twisted form of $\mathbb{P}^n_{\mathbb{C}}$ is known as a Brauer-Severi variety and the elements of the Brauer-group (of a field) can also be thought of as corresponding to them (the group structure is then a bit strange).  Since $Aut(\mathbb{P}^n) = PGL_{n+1}$, these correspond to $PGL_{n+1}$-torsors and the relation to $\mathbb{G}_m$-torsors is via the exact sequence for $PGL_{n+1}=GL_{n+1}/\mathbb{G}_m$.  So, a $T$-point of the corresponding $\mathbb{G}_m$-torsor for a $\mathbb{R}$-scheme $T$ consists of the following data:
It is a rank $2$ vector bundle $V$ over $T$, together with an isomorphism of $T$-schemes $C_T \simeq \mathbb{P}(V)$ where $C_T = C \times_{Spec \mathbb{R}} T$ is the pullback of our genus $0$ curve to $T$, and $\mathbb{P}(V)$ is the associated projective space (here $\mathbb{P}^1$) bundle of our vector bundle $V$. 
Why is this a $\mathbb{G}_m$-gerbe? Well, $\mathbb{P}(V) \simeq \mathbb{P}(V')$ iff $V$ and $V'$ differ by tensoring by a line bundle.  The gerbe is non-trivial since it has no $\mathbb{R}$-points, since $C$ itself is not isomorphic to projective space.  It has $\mathbb{C}$-points because the base-change is isomorphic to projective space.<|endoftext|>
TITLE: On the proofs (and disproofs) of Riemann Hypothesis
QUESTION [14 upvotes]: As anyone who follows the arxiv, I notice every now and then "proofs" and "disproofs" of Riemann Hypothesis. I looked on several such articles, and it seemed to me quite nonsense, but I didn't make the effort to find a mistake. My question is whether someone reads these "proofs"? 
BTW, I wanted to refer to some of these papers in the arxiv, but it turned out that there are too many of them.

REPLY [44 votes]: Whenever someone claims a proof (or disproof) of a big conjecture, many people leap to the question of whether the proof is correct.  The problem then is it that it takes an enormous amount of work to confirm that a proof is correct.  Even a clear mistake in a proof could be reparable.  Moreover, attempted proofs have inferences that amount to gaps of different sizes.  Even in a naive attempt, it can take a lot of work to decide which gaps are so big that the proof has to be called incomplete.
There is a much simpler standard that experts use in practice:  "As I start to read this paper, am I learning from it?"  You would expect a proof of a big conjecture to have very interesting lemmas, and otherwise to teach you new things along the way.  This is not always obvious either; there have been a few grievous misunderstandings in which initial readers dismissed a great paper.  Even so, it's a somewhat reliable standard, and it's the most that authors can expect.
When Perelman posted the first of his three papers on geometrization, experts in differential geometry quickly embraced it as exciting and teachable, before they had even checked half of that paper or seen the other two papers.  From the beginning, this was very different from most claimed proofs of the Poincare conjecture, even most of the noble failures.  The great ideas in these papers were more important than the fact that they had a lot of gaps (by common standards) and even some inessential mistakes (or so I was told).
I know for a fact that experts sometimes do study weird-looking claims of big results, in the arXiv and elsewhere.  They have little incentive to broadcast their attention to it if they think that it's shoddy work, but sometimes they try to be fair.  For starters, the math arXiv has moderators, and they often take a look.  I think that usually (not quite always), several people have looked long enough to decide that they aren't getting anything out of the paper.  But hey, there could always be a diamond in the rough, or even a diamond in the garbage.<|endoftext|>
TITLE: Integral of the error estimate in the prime number theorem
QUESTION [8 upvotes]: This seems like something that should be in discussed in the literature, but I can't find anything. Here $\pi(x)$ is the prime counting function and $\psi(x)$ is the usual sum of the Von Mangoldt function.
Are there non-trivial estimates for the quantity $\int_{1}^{N} |\psi(x) - x| dx$? The prime number theorem asserts $|\pi(x)- x/ln(x)| = o(x/ln(x))$, or, equivalently, $|\psi(x) - x| = o(x)$. Using this we trivially have the estimate $o(N^2)$ for the expression above (which we can make a bit more quantitative using quantitative forms of the pnt) however it seems plausible that this could be improved since we are asking for average case instead of worst case information about $|\psi(x) - x|$. In fact since we know that $\psi(x) - x$ oscillates to the extremes $\pm \sqrt{x}/ln(x)$ infinitely often, it seems plausible that it might spend a fair amount of time away from these extremes.

REPLY [7 votes]: The mean square version 
$\int_X^{2X} |\psi(x) - x|^2dx$
is more natural than the one you want. This is discussed on page 423 of Montgomery and Vaughan's book, on the assumption of the Riemann Hypothesis. Briefly, you can use the truncated explicit formula for $\psi(x)$ and multiply out and integrate. It does not look like one gets a great bound only by the zero-free region of Korobov and Vinogradov. The cancellation that you want seems to be there, but depends on the behavior of the differences $\gamma_1 - \gamma_2$ where $\gamma_1$,$\gamma_2$ are imaginary parts of nontrivial zeros of the Riemann zeta function, so it is hard to get at. 
If you have a bound for the above integral, you can get one for your integral by Cauchy-Schwarz, of course. I think your integral is hopeless to attack directly; it will either have to come from a pointwise bound as you tried above, or a bound on some even power, or perhaps some combination.
The range from X to 2X is a convenience so that things do not change enormously over the range; you can bound the whole integral from 1 to X by adding integrals over dyadic ranges.<|endoftext|>
TITLE: How do we study Iwasawa theory?
QUESTION [44 upvotes]: What papers should we read to start? What basic knowledge do we need to understand the question? What is this area really about? And what are people researching on it?

REPLY [2 votes]: I collected a list of references for Iwasawa theory here, including links to various surveys which hasn't been mentioned so far.<|endoftext|>
TITLE: Maximal exotic $\mathbb{R}^4$
QUESTION [8 upvotes]: Article Exotic $\mathbb{R}^4$ on Wikipedia says that there is at least one maximal smooth structure on $\mathbb{R}^4$, that is such an atlas on $\mathbb{R}^4$ that any other smooth $\mathbb{R}^4$ can be embedded into it. Is the construction of such a maximal exotic $\mathbb{R}^4$ explicit? Can anyone give a reference to the construction? What is a good source (or sources) with examples of exotic smooth structures on $\mathbb{R}^4$? Thanks.

REPLY [12 votes]: The paper by Freedman and Taylor mentioned by Carsten Schultz (above or below) is indeed the place to find the explicit construction.
Very roughly, the idea of the construction is as follows.  Recall that a Casson handle is, among other things, a smooth 4-manifold which is homeomorphic (but not diffeomorphic) to the standard open 2-handle.  It turns out that a countable collection of diffeomorphism classes of Casson handles suffice for solving a certain 5-dimensional h-cobordism problem.  The universal $R^4$, call it $U$, is constructed by gluing together countably many copies of each of the Casson handles in the countable collection.  Given an arbitrary smooth 4-manifold homeomorphic to $R^4$, we can construct an embedding (but not a proper embedding) into $U$ using the fact that $U$ contains enough Casson handles to solve any link slice problem we might encounter along the way.
Well, that was kind of vague, but hopefully not inaccurate.<|endoftext|>
TITLE: Why is the standard definition of cocycle the one that _always_ comes up??
QUESTION [49 upvotes]: This question might not have a good answer. It was something that occurred to me yesterday when I found myself in a pub, needing to do an explicit calculation with 2-cocycles but with no references handy (!).
Review of group cohomology.
Let $G$ be a group acting (on the left) on an abelian group $M$. Then $H^0(G,M)=M^G=Hom_{\mathbf{Z}[G]}(\mathbf{Z},M)$ and hence $$H^i(G,M)=Ext^i_{\mathbf{Z}[G]}(\mathbf{Z},M).$$ Now $Ext$s can be computed using a projective resolution of the first variable, so we're going to get a formula for group cohomology in terms of "cocycles over coboundaries" if we write down a projective resolution of $\mathbf{Z}$ as a $\mathbf{Z}[G]$-module.
There's a very natural projective resolution of $\mathbf{Z}$: let $P_i=\mathbf{Z}[G^{i+1}]$ for $i\geq0$ (with $G$ acting via left multiplication on $G^{i+1}$) and let
$d:P_i\to P_{i-1}$ be the map that so often shows up in this sort of thing: $$d(g_0,\ldots,g_i)=\sum_{0\leq j\leq i}(-1)^j(g_0,\ldots,\widehat{g_j},\ldots,g_i).$$
Check: this is indeed a resolution of $\mathbf{Z}$. So now we have a "formula" for group cohomology.
But it's not the usual formula because I need to do one more trick yet. Currently, the formula looks something like this: the $i$th cohomology group is $G$-equivariant maps $G^{i+1}\to M$ which are killed by $d$ (that is, which satisfy some axiom involving an alternating sum), modulo the image under $d$ of the $G$-equivariant maps $G^i\to M$.
The standard way to proceed from this point.
The "formula" for group cohomology derived above is essentially thought-free (which was why I could get this far in a noisy pub with no sources). But we want something more useful and it was at this point I got stuck. I remembered the nature of the trick: instead of a $G$-equivariant map $G^{i+1}\to M$ we simply "dropped one of the variables", and considered arbitrary maps of sets $G^i\to M$. So we need to give a dictionary between the set-theoretic maps $G^i\to M$ and the $G$-equivariant maps $G^{i+1}\to M$. I could see several choices. For example, given $f:G^{i+1}\to M$ I could define $c:G^i\to M$ by $c(g_1,g_2,\ldots,g_i)=f(1,g_1,g_2,\ldots,g_i)$, or $f(g_1,g_2,\ldots,g_i,1)$, or pretty much anything else of this nature. The point is that given $c$ there's a unique $G$-equivariant $f$ giving rise to it. Which choice of dictionary do we use. Each one will give a definition of group cohomology as "cocycles" over "coboundaries". But which one will give the "usual" definition? Well---in some sense, who cares! But at some point maybe someone somewhere made a decision as to what the convention from moving from $f$ to $c$ was, and now we all stick with it. The standard decision was the rather clunky
$$c(g_1,g_2,\ldots,g_i)=f(1,g_1,g_1g_2,g_1g_2g_3,\ldots,g_1g_2g_3\ldots g_i).$$
Why is the standard definition ubiquitous?
Actually though, I bet that no-one really made that decision. I bet that the notion of a 1-cocycle and a 2-cocycle preceded general homological nonsense, and the dictionary between $f$s and $c$s was worked out so that it agreed with the definitions which were already standard in low degree.
But this got me thinking: if, as it seems to me, there is no "natural" way of moving from $f$s to $c$s, then why do the 1-cocycles and 2-cocycles that naturally show up in mathematics all satisfy the same axioms??. Why doesn't someone do a calculation, and end up with $c:G^2\to M$ satisfying some random axiom which happens not to be the "standard" 2-cocycle axiom but which is an axiom which, under a non-standard association of $c$s with $f$s, becomes the canonical cocycle axiom for $f$ that we derived without moving our brains? Does this ever occur in mathematics? I don't think I've ever seen a single example.
In some sense it's even a surprise to me that there is a uniform choice which specialises to the standard choices in degrees 1 and 2.
Examples of cocycles in group theory.
1) Imagine you have a 2-dimensional upper-triangular representation of a group $G$, so it sends $g$ to the $2\times 2$ matrix $(\chi_1(g),c(g);0,\chi_2(g))$. Here $\chi_1$ and $\chi_2$ are group homomorphisms. What is $c$? Well, bash it out and see that $c$ is precisely a 1-cocycle in the sense that everyone means when they're talking about 1-cocycles. So we must have used the dictionary $c(g)=f(1,g)$ when moving between $c$s and $f$s above. Why didn't we use $c(g)=f(g,1)$? 
2) Imagine you're trying to construct the boundary map $H^0(C)\to H^1(A)$. Follow your nose. Claim: your nose-following will lead you to the standard cocycle representing the cohomology class. Why did it not lead to a non-standard notion? Why do the notions (1) and (2) agree? 
3) Imagine we have a group hom $G\to P/M$ where $M$ is an abelian normal subgroup of the group $P$. Can we lift it to a group hom $G\to P$? Well, let's take a random set-theoretic lifting $L:G\to P$. What is the "error"? A completely natural thing to write down is the map $(g,h)\mapsto L(g)L(h)L(gh)^{-1}$ because this will be $M$-valued. This is a 2-cocycle in the standard sense of the word. Why did the natural thing to do come out to be the standard notion of a 2-cocycle? Aah, you say: there are other natural things that one can try. For example we could have sent $(g,h)$ to $L(gh)^{-1}L(g)L(h)$. For a start this "looks slightly less natural to me" (why does it look less natural? That's somehow the question!). Secondly, this is really just applying a canonical involution to everything: we're inverting on $G$ and inverting on $M$, which is something that we'll always be able to do. It certainly does not correspond to the "more natural" dictionary that I confess I tried down the pub, namely $c(g,h)=f(1,g,h)$, which gave much messier answers. Why does my "obvious" choice of dictionary lead me to 20 minutes of wasted calculations? Why is it "wrong"?
The real question.
Has anyone ever found themselves in a situation where a natural cocycle-like construction is staring them in the face, and they make the construction, and find themselves with a non-standard cocycle? That is, a cocycle which will induce an element of a cohomology group but only after a non-standard dictionary is applied to move from $f$s to $c$s?
Edit: Here is what I hope is a clarification of the question. To turn the cocycles from $G$-equivariant maps $G^{i+1}\to M$ to maps of sets $G^i\to M$ we need to choose a transversal for the action of $G$ on $G^{i+1}$ and identify this with $G^i$. There seems to be one, and only one, way to do this that gives rise to the "standard" definition of n-cocycle that I (possibly incorrectly) percieve to be ubiquitous in mathematics. I call this "the clunky way" because the map seems odd to me. Why is it so clunky? And why, whenever a 2-cocycle falls out of the sky, does it always seem to satisfy the axioms induced by this clunky method? Why aren't there people popping up in this thread saying "well here's a completely natural "cocycle" $c(g,h)$ coming from theory X that I study, and it doesn't satisfy the usual cocycle axioms, we have to modify it to be $c'(g,h)=c(g,gh)$ before it does, and this boils down to the fact that in theory X we would have been better off if history had chosen a non-clunky identification?"

REPLY [14 votes]: Late to the party as usual, but: the goal of this answer is to convince you that the standard convention for $2$-cocycles is so natural that you should consider it perverse to consider any other convention, modulo "applying a canonical involution to everything," as you say. To keep things simple let's only deal with trivial action on coefficients. The motivating question is the following:

What does it mean for a group $G$ to act on a category $C$?

For starters we should attach to each element of $G$ a functor $F(g) : C \to C$. Next we could require that $F(g) \circ F(h) = F(gh)$, but we should really weaken equalities of functors to natural isomorphisms whenever possible. Hence we should attach to each pair of elements of $G$ a natural isomorphism
$$\eta(g, h) : F(g) \circ F(h) \to F(gh).$$
This is the point at which we pick a convention for how we're going to represent $2$-cocycles. Instead of talking about $\eta(g, h)$ we could talk about its inverse; which we choose corresponds to whether we prefer to talk about lax monoidal or oplax monoidal functors, since what we're going to end up writing down is a lax monoidal resp. an oplax monoidal functor from $G$ (regarded as a discrete monoidal category) to $\text{Aut}(C)$ (regarded as a monoidal category under composition). 
In any case, let's stick to the above choice (the lax one). Then the isomorphisms $\eta(g, h)$ should satisfy some coherence conditions, the important one being the "associativity" condition that the two obvious ways of going from $F(g_1) \circ F(g_2) \circ F(g_3)$ to $F(g_1 g_2 g_3)$ should agree.
Now let's assume that in addition all of the functors $F(g)$ are the identity functor $\text{id}_C : C \to C$. Then the only remaining data in a group action is a collection of natural automorphisms
$$\eta(g, h) : \text{id}_C \to \text{id}_C$$
of the identity functor. For any category $C$, the natural automorphisms of the identity functor naturally form an abelian (by the Eckmann-Hilton argument) group which here I'll call its center $Z(C)$ (but this notation is also used for the commutative monoid of natural endomorphisms of the identity). So we get a function
$$\eta : G \times G \to Z(C).$$
The important coherence condition I mentioned above now reduces (again by the Eckmann-Hilton argument) to the condition that for any $g_1, g_2, g_3 \in G$ we have
$$\eta(g_1, g_2) \eta(g_1 g_2, g_3) = \eta(g_2, g_3) \eta(g_1, g_2 g_3)$$
which is precisely the standard cocycle condition. (Coboundaries come in when you ask what it means for two group actions to be equivalent; I'm going to ignore this.) 
The only reason this condition, which recall is in general just the statement that the two obvious ways of going from $F(g_1) \circ F(g_2) \circ F(g_3)$ to $F(g_1 g_2 g_3)$ should agree, could ever have looked anything other than completely natural is that it's a degenerate special case where the sources and targets of the various maps involved have been obscured because they are identical. In particular, of course I could have instead chosen to think about the natural isomorphisms
$$\eta(g, g^{-1} h) : F(g) \circ F(g^{-1} h) \to F(h)$$
(which corresponds to your $f(1, g, h)$), but now

it's no longer at all obvious how to state the associativity condition succinctly, and
this requires that I make explicit use of the fact that $G$ is a group.

The discussion up til now in fact gives a perfectly reasonable definition for what it means for a monoid to act on a category. (If I want to weaken "natural isomorphism" to "natural transformation," though, I get two genuinely different possibilities depending on whether I pick lax or oplax monoidal functors.) 
Reflecting on associativity suggests that, for a more "unbiased" point of view, we should consider families of natural isomorphisms
$$\eta(g_1, g_2, \dots g_n) : F(g_1) \circ F(g_2) \circ \dots \circ F(g_n) \to F(g_1 g_2 \dots g_n)$$
and then impose a "generalized associativity" condition that every way of composing them to get a natural isomorphism with the same source and target as $\eta(g_1, g_2, \dots g_n)$ should give $\eta(g_1, g_2, \dots g_n)$. Another way to say this is that the cocycle condition (in the $F(g) = \text{id}_C$ special case, at least) should really be written
$$\eta(g_1, g_2, g_3) = \eta(g_1, g_2) \eta(g_1 g_2, g_3) = \eta(g_2, g_3) \eta(g_1, g_2 g_3).$$
This is in the same way that we can consider a monoid operation to be a family $m(g_1, g_2,  \dots g_n) = g_1 g_2 \dots g_n$ of operations satisfying a generalized associativity condition, and in particular satisfying
$$m(g_1, g_2, g_3) = m(m(g_1, g_2), g_3) = m(g_1, m(g_2, g_3)).$$
Namely, by "associativity" we usually mean that the middle expression equals the right, but really the reason that the middle expression equals the right is that they both equal the left.<|endoftext|>
TITLE: Modular Forms - Eichler quote
QUESTION [19 upvotes]: We have the following quote from Eichler: "There are five elementary arithmetical operations: addition, subtraction, multiplication, division, and… modular forms."
Why did Eichler consider modular forms "elementary arithmetical operations"? Not to mention that subtraction and division are just addition and multiplication with inverse elements. Can anybody shed some light on his train of thought?

REPLY [2 votes]: I too was fascinated by this quote. And then I came across a very interesting set of slides by none other than Prof. Ken Ribet (herein PKR) titled The five fundamental operations of mathematics: addition, subtraction, multiplication, division, and modular forms
I'll try to summarize his slides below. As usual, any mistakes here are my own and not PKR's. 
So, to start PKR bases his around discussing the topic in terms of counting, and solving problems equations.

This talk is about counting, and it’s about solving equations

In particular, Diophantine Equations.
PKR doesn't really discuss addition, subtraction, multiplication or division. 
However he does provide simple answer to the question: what the heck is modular form:

Modular forms are special functions that are analogous to the
  trigonometric functions like $\sin$, $\cos$, $\tan$,... in that they are
  periodic in the same way that $\sin$ is periodic. (Recall the formula
  $\sin(x + 2\pi) = \sin(x)$.) Modular forms have the periodicity of the
  trigonometric functions plus enough extra symmetries that they
  are essentially unchanged under a large group of substitutions.
  Because of the symmetries, it is possible to write modular
  forms as Fourier series $\sum_{m=0}^{\infty}{}$, where the $q$ here is a
  shorthand for $e^{2\pi\iota\\z}$

Any how hope this helps<|endoftext|>
TITLE: Egalitarian measures
QUESTION [18 upvotes]: A question I got asked I while ago: 
If $T$ is a triangle in $\mathbb R^2$, is there a function $f:T\to\mathbb R$ such that the integral of $f$ over each straight segment connecting two points in the boundary of $T$ not on the same side is always $1$?
(Of course, you can change $T$ for your favorite convex set... and the problem should really be seen as asking for what sets is the answer affirmative, mostly)

REPLY [17 votes]: In general, no. For the double integral $\iint_T f(x,y)\,dx\,dy$ will be the height on any side, as is seen by turning the triangle with one side parallel to an axis and performing the integral. So at least, $T$ has to be equilateral. I don't know the answer in that case.
Edit: Wait, wait – the same trick works even if I turn the triangle at any angle, hence all heights (defined as $\sup_{RT} y-\inf_{RT} y$ where $R$ is a rotation) must be the same. That is never true for a triangle, and limits the number of convex sets seriously – but there are still non-circles that might satisfy the criterion. Once more, I don't know the answer, but for circles at least, it should in principle be straightforward to check if a radially symmetric function will do. And of course, if there is a solution, there is a radially symmetric one, as can be seen by rotating the solution and taking the average of all its rotated variants.
Edit2: For the unit disk (and radially symmetric $f$) the answer should in principle be obtainable by the Abel transform (which is really nothing but the Radon transform on radially symmetric functions). The required Abel transform $\Phi$ should be the characteristic function of the interval $[0,1]$ (we only use positive $x$ due to symmetry), and the inverse Abel transform provides the answer:
$$f(r)=\frac{-1}{\pi}\int_r^\infty \frac{\Phi'(x)\,dx}{\sqrt{x^2-r^2}}=\frac{1}{\pi\sqrt{1-r^2}}$$
when $0\le r<1$. Being lazy, I checked the answer using Maple, and it seems right.
Addendum: Anton Petrunin pointed out in a comment that the above measure is the push-forward of the surface measure on the unit sphere on the unit disk under projection. It is well-known that the surface area of the portion of a disk between two parallel planes depends only on the distance between the planes (and is proportional to said distance), which ties in nicely with the desired property of $f$ on the disk.<|endoftext|>
TITLE: What is etale descent?
QUESTION [16 upvotes]: What is etale descent?  I have a vague notion that, for example, given a variety $V$ over a number field $K$, etale descent will produce (sometimes) a variety $V'$ over $\mathbb{Q}$ of the same complex dimension which is isomorphic to $V$ over $K$ and such that $V(K)=V'(\mathbb{Q})$.  Is this at all right?  How does one do such a thing?

REPLY [2 votes]: When a functor happens to be a sheaf in étale topology you say it satisfies étale descent. So for example algebraic K-theory does not satisfy étale descent but Bott periodic algebraic K-theory modulo a suitable prime power does.<|endoftext|>
TITLE: How are graph automorphisms are affected by transformations?
QUESTION [5 upvotes]: I have a heavily symmetric regular graph whose automorphisms I know.  I remove one subgraph and insert another one in a consistent manner; for example, this could be a Delta-Y transformation (replacing a node with a complete subgraph).  I'd like to compute the automorphisms of the new graph using the automorphisms of the old graph.
I have the sense that this problem isn't too difficult, but am not sure how best to approach it and am new to algebraic graph theory.  
What are some good references or suggested starting points?

REPLY [3 votes]: I also have the sense that the problem is, in general, extremely hard. Actually, I'll give an example which hopefully will convince you that there's not a lot we can say, in general. (Unless you mean "symmetric" in the technical sense, to mean "arc-transitive," or really even "vertex-transitive," in which case my example breaks down, although it may well be possible to do something similar in those cases.) For concreteness, let's think about the delta-Y transformation.
There are certainly graphs with large automorphism group where you can perform a delta-Y tranformation and get back another graph with, um, large automorphism group. For instance, take a complete graph, or, if you're squeamish about the whole "it's not simple anymore" thing, subdivide each edge of a complete graph. But we'll take a higher level of abstraction and just say that Aut(G) is the automorphism group of our graph.
So if you're doing algebraic graph theory at all, you know about the Cayley graph of a group. The salient details here are just that it's got a vertex for each element of the group, a unique automorphism for each element corresponding to right multiplication, and no other automorphisms. The standard construction of the Cayley graph is directed and has edges assigned to color classes, but there's a combinatorial trick to make a "vanilla" Cayley graph, which is to attach little gadgets to your edges to indicate color and direction. This graph is no longer vertex-transitive, which sucks, but its automorphism group is still completely described (at least if you're careful about building your gadgets...)
Now consider an (undirected uncolored) Cayley graph $\Gamma$ of Aut(G). You can see that if we perform a delta-Y transformation in one of our gadgets, NONE of the original nontrivial automorphisms of $\Gamma$ remain, since the colored edge the gadget's standing in for couldn't stay in the same place. So depending on how your gadget is constructed, the new graph might have trivial automorphism group.<|endoftext|>
TITLE: Singular value decomposition over finite fields?
QUESTION [10 upvotes]: What is the definition of a singular value over a finite field $\mathcal{F}$ of a matrix ${\bf A}$ in $\mathcal{F}^{m\times n}$? Is there a geometric intuition in the same manner as with the real case where the eigenvalues are the radii of the ellipse $\frac{\|{\bf A}{\bf x}\|^2}{\|{\bf x}\|^2}$?

REPLY [15 votes]: There is no definition of a singular value of a matrix over a finite field.  You could
define it to be a non-zero eigenvalue of $A^TA$, but this does not really work as you might expect.
Over the reals, the eigenvalues of $A^TA$ are non-negative and the smallest singular value
is a measure of how close $A$ is to being a non-invertible matrix.  Further, there are
stable algorithms to compute it, whereas we cannot compute the rank of $A$ in a stable fashion.
Over the complex numbers, you would use the eigenvalues of ${\bar A}^TA$ as singular values
instead the eigenvalues of $A^TA$.  Over finite fields you might use $\sigma(A^T)A$,
where $\sigma$ is a field automorphism.  This is a problem, we have ore choice than we do over the reals and complexes.
There is no difficulty in computing ranks over matrices over finite fields anyway.
Over finite fields, eigenvalues are of limited use.  We can get the characteristic polynomial of a matrix, and factor it; the zeros of the factors are the eigenvalues.  These eigenvalues
would lie in some extension $E$ of your finite field $F$; if I came along with $E$ and asked
which elements of $E$ were the eigenvalues, you would have a lot of work and your final
answer would depend on exactly how I described $E$.  (Even over the reals, eigenvalues
are useful if the matrix is small, or normal, but they are much less use otherwise.
Google 'pseudospectra')<|endoftext|>
TITLE: Intuitive explanation to Probability question
QUESTION [6 upvotes]: I have \$3. I flip a coin. If I get heads, I get \$1. If I get tails, I lose \$1. The game stops when I have \$0 or \$7. What is the probability I get \$7?
I solved this by creating a system of linear equations, where $P_0 = 0$, $P_7 = 1$, and $P_x = 0.5 \cdot P_{x-1} + 0.5 \cdot P_{x+1}$. Solving them, I got $P_3 = 3/7$. Moreover, $P_x = x/7$. Why does it work out to such a simple fraction?
More generally, it seems that $P_{x,y} = x/y$, which is the probability that, starting from x dollars, I end up with y dollars. I haven't proven this, but why is this the case?
Finally, what is $P_{x,y,p}$, where I gain 1 dollar with probability $p$ instead of probability 0.5 .

REPLY [2 votes]: Ori Gurel-Gurervich's comment suggests a very simple way to use a martingale (an example of a Wald martingale) to evaluate the final question in which the probability of gaining a dollar is $p \ne \frac12$. 
If $m(t)$ is how much money you have at time $t$, then $m(t)$ is not a martingale for $p \ne \frac12$. However, for the right base $C$, $C^{m(t)}$ is a martingale: $E (C^{m(t+1)}) = E(C^{m(t)})$. That means we can use the same argument that the starting value of a martingale is the average of the stopping values to compute the probabilities of ending at $0$ or $y$, or even of escaping to $\infty$ (with a bit more technical work). 
The right value of $C$ is $(1-p)/p$. You start at $C^x$ and end at $C^y$ or $C^0 = 1$, so if you finish with probability 1 (easy to prove with another martingale) you end up at $C^y$ with probability $(1-C^x)/(1-C^y)$, and you end up at $C^0$ with the complementary probability $(C^x-C^y)/(1-C^y)$. 
When $p \sim \frac12$, $C^x \sim 1- x \epsilon$ and $C^y \sim 1-y\epsilon$, which is continuous with the case $p=\frac12$. 
If you don't stop at $y$, the probability that you escape to $\infty$ is $0$ if $p \le 1/2$ and $1-C^x$ if $p \gt 1/2$, which makes the probability of ruin $C^x$.<|endoftext|>
TITLE: Model theoretic applications to algebra and number theory(Iwasawa Theory)
QUESTION [35 upvotes]: One of my favorite results in algebraic geometry is a classical result of AX (see http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/) I'll recall the version of the theorem that I learned in an undergraduate class in model theory.
An algebraic map $F: \mathbb{C}^{n} \to \mathbb{C}^{n}$ is injective iff it is bijective. 
The model theoretic proof of this result is very simple. Without discussing details ( truth in TACF_0 is the same as truth in TACF_p for p big enough )one can replace $\mathbb{C}$ by the algebraic closure of $F_p$ and prove the result there.  Although the algebraic proof is also "simple" (Hilbert's Nullstellensatz is what one needs ) I personally think that the model theoretic proof is great by its simplicity and elegance.
So my first question is: Are there more examples like AX's theorem in which the model theoretic proof is much more simple that the ones given by other areas. (I should mention that some  algebraist I know  consider quantifier elimination for TACF non a model theoretic fact, so for them the proof that I refer above is an algebraic proof)
My second question:  One of the most famous model theoretic applications to algebra and number theory is Hrushovski's proof on Mordell-Lang of the function field Mordell-Lang conjecture. I'd like to know what are the research questions that applied model theorists are currently working on, besides continuation to Hrushovski's work . In particular, I'd like to know if there is any model theorist that work in applications to Iwasawa theory.

REPLY [11 votes]: I'd just like to expand on John Goodrick's mention of Zilber's work on exponential fields and mention that `Categoricity' of is an active area of research. In particular, model theory may be used to give some justification as to why theorems of classical mathematics should hold.
In general it's an interesting question to see what good model theoretic behaviour translates to in the world of classical mathematics. 
One way of viewing things (and this is Zilber's point of view) is that if a mathematical structure is useful, and therfore well studied by the mathematical community, then it will be complicated enough to be interesting, but nice enough to be analysed. One aspect of model theory is involved with trying to classify structures with respect to how nice (or wild) they are (e.g. a structure could be strongly minimal, O-minimal, stable, categorical etc...).
At the top of the logical hierarchy sit categorical theories. A theory is $\kappa$-categorical if it has one model up to isomorphism in cardinality $\kappa$. The stereotypical example of a categorical theory is the theory of algebraically closed fields of characteristic $0$. The unique model of cardinality continuum is $\langle \mathbb{C}, + , \cdot , 0,1 \rangle$. This mathematical structure has pretty much every nice model theoretic property that you'd want in a structure - it's strongly minimal (definable sets are very simple i.e. either finite or cofinite), $\omega$-stable (there aren't many types of elements around), homogeneous (you can extend partial automorphisms to automorphisms of the whole structure), saturated (you can realise types - i.e. solutions to polynomials are in there). This theory is also complete and `categorical in powers' i.e. $\kappa$-categorical for every uncountable cardinal. 
An amazing theorem of Morley actually says that if a first order theory is $\kappa$-categorical for one uncountable cardinal, then it is categorical for every uncountable cardinal. Morley's theorem (1965) kick started stability theory, and from there Shelah has developed an unbelievable amount of abstract model theoretic technology.
However, after initiating stability theory in the first place it seemed that the study of categorical structures had run its course (Baldwin-Lachlan theorem completely categorises theories which are categorical in powers). But recently Zilber realised that some of Shelah's abstract model theoretic technology regarding infinitary logics can be used to study concrete, well known, and very interesting mathematical structures.
For example, as John Goodrick mentions, if you try and axiomatize the interaction of the exponential function with the complex field i.e. you try and capture the theory of $\langle \mathbb{C},+ \cdot ,0,1, e^x \rangle$, and you want it to be categorical, then you need things like Schanuel's conjecture and the conjecture on the intersections of Tori (CIT) to hold. Along similar lines is the Zilber-Pink conjecture.
So model theory can give us some kind of justification as to why certain results should hold. For example, if you look the theory of the universal cover of a non CM elliptic curve over a number field, and you ask it to be $\aleph_1$-categorical, then it turns out that a famous theorem of Serre saying that the image of the Galois representation on the Tate module is open must be true.<|endoftext|>
TITLE: When can we prove constructively that a ring with unity has a maximal ideal?
QUESTION [17 upvotes]: Many commutative algebra textbooks establish that every ideal of a ring is contained in a maximal ideal by appealing to Zorn's lemma, which I dislike on grounds of non-constructivity.  For Noetherian rings I'm told one can replace Zorn's lemma with countable choice, which is nice, but still not nice enough - I'd like to do without choice entirely.
So under what additional hypotheses on a ring $R$ can we exhibit one of its maximal ideals while staying in ZF?  (I'd appreciate both hypotheses on the structure of $R$ and hypotheses on what we're given in addition to $R$ itself, e.g. if $R$ is a finitely generated algebra over a field, an explicit choice of generators.)
Edit:  I guess it's also relevant to ask whether there are decidability issues here.

REPLY [3 votes]: You should take a look at Coquand and Lombardi's "A Logical Approach to Abstract Algebra". 
They observe that commutative rings have a purely equational description, and so there are very strong metatheorems that apply to this theory: Birkhoff's completeness theorem for equational logic, of course; and also Barr's theorem, which states that if a geometric sentences is a consequences of a geometric theories with classical logic plus choice, it's also intuitionistically valid. (And all equational theories are also geometric theories.) 
They strengthen Barr's theorem a bit, by characterizing the relevant intuitionistic proofs,  and then "de-Noetherian-ize" several basic theorems which are typically proved using maximal ideals.<|endoftext|>
TITLE: Typical value of totient function
QUESTION [12 upvotes]: Can anyone tell me what the expected value of Euler's totient function φ$(n)$ is (roughly) if you choose a random integer $n$ in the range $[N,N+M]$, where $M$ is large and $N$ is larger than $M$? (I think of $M$ as being $cN$ for some small constant $c$, which, if one wanted an answer accurate to $1+o(1)$, would in reality be a slowly decreasing function of $N$.)

REPLY [4 votes]: Let me also mention the following:
You can adapt Schoenberg's result to prove that
1/M * {N <= n <= N + M : phi(n) / n <= t} --> F(t) 
uniformly in t, where F is a distribution function. The proof goes
by computing the moments sum((phi(n)/n)^k , N <= n <= N + M).
You can probably get a O(loglog N / log N) rate of convergence
(as was done by Levin ... if I recall correctly).<|endoftext|>
TITLE: What would the slice-ribbon conjecture imply?
QUESTION [26 upvotes]: What would the slice-ribbon conjecture imply for 4-dimensional topology?
I've heard people speak of the slice-ribbon conjecture as an approach to the 4-dimensional smooth Poincare conjecture, and to the classification of homology 3-spheres which bound homology 4-balls. But I've never understood what they were talking about.

REPLY [6 votes]: I have a question concerning Peter Teichner's answer. Aren't there candidate counterexamples, for instance in the following paper in `Topology and its Applications':
Some well-disguised ribbon knots
Robert E. Gompf, 1 and Katura Miyazakib, , 2
Abstract
For certain knots J in S1 × D2, the dual knot J* in S1 × D2 is defined. Let J(O) be the satellite knot of the unknot O with pattern J, and K be the satellite of J(O) with pattern J*. The knot K then bounds a smooth disk in a 4-ball, but is not obviously a ribbon knot. We show that K is, in fact, ribbon. We also show that the connected sum J(O) # J*(O) is a nonribbon knot for which all known algebraic obstructions to sliceness vanish.<|endoftext|>
TITLE: Coherent spaces
QUESTION [8 upvotes]: In Proofs and Types, Girard discusses coherent (or coherence) spaces, which is defined as a set family which is closed downward ($a\in A,b\subseteq a\Rightarrow b\in A$), and binary complete (If $M\subseteq A$ and $\forall a_1,a_2\in M (a_1\cup a_2\in A)$, then $\cup M\in A$)
It was informally related to topological spaces. Anyway, I have a couple pretty general questions:
Are they particularly useful outside of type theory? Perhaps more specifically, do coherent spaces show up in topology?
The last one raises up a philosophical question I've been pondering: Why is it that some structures seem to show up all over the place, while others that seem like they "should" be more or less equivalently useful don't seem to show up much at all? An example would be matroids versus topologies. I feel, morally, that matroids should be more useful than they seem to be.
The last question probably doesn't have any sort of solid answer, but it would be nice to hear some thought from people with a stronger background.
Cheers and thanks,
Cory
Edit:
After thinking about this some more, it has occurred to me that coherent spaces are a sort of "dual" to ultrafilters. I really don't have the background to be terribly formal, but, let me try to explain:
Let $(X,C)$ be a coherent space, and call the elements of $C$ "open" (I think the analogy is justified, because adding $X$ to $C$ makes it a topology), then the closed sets form an ultrafilter. The one problem is that the closure under intersection is a bit strong (the set of closed sets is closed under arbitrary intersections).
On the other hand, if $(X,U)$ is an ultrafilter, the set of complements of open sets almost forms a coherent space-- but the conditions on unions is just a little too weak.
So, my next question is: Has this link been explored at all? Is there even anything there to explore?
Thanks again.

REPLY [2 votes]: An analogue of your coherence spaces are used extensively in set theory, particularly with the method of forcing, the set-theoretic technique often used to prove statements independent of ZFC. But there is a variation, in that the coherence clause is weakened to cover only some M, such as M of a certain size. 
For example, in the standard forcing argument to add κ many generic Cohen reals, one considers the partial order consisting of all finite partial functions from (ω x κ) to {0,1}. This collection of partial functions satisfies your properties, if we restrict to finite M, since the union of a set of partial functions is a function if and only if these functions are coherent, in the sense that any two of them agree. If F is a maximal filter on this partial order, then the union of F is a function fully from (ω x κ) to {0,1}. If the filter is what is known as V-generic, then the slices of this function adds κ many new Cohen reals. If κ is at least ω2, then one can argue that CH fails in the resulting forcing extension. 
Many forcing notions have the form of partial functions from one set to another, restricted by size or by other features, and so they also satisfy the corresponding restricted version of your Coherence space.<|endoftext|>
TITLE: Overview of automorphic representations for $SL(2)/{\mathbf{Q}}$?
QUESTION [29 upvotes]: In short: what does Labesse-Langlands say?
Slightly more precise: what are the cuspidal automorphic representations of $SL_2(\mathbf{A}_{\mathbf{Q}})$, together with multiplicities? Let's say that I have a complete list of the cuspidal automorphic representations of $GL_2/\mathbf{Q}$ and I want to try and deduce what is happening for $SL_2$. I am looking for "concrete examples of the phenomena that occur".
Now let me show my ignorance more fully. My understanding from trying to read Labesse-Langlands is the following. The local story looks something like this: if $\pi$ is a smooth irreducible admissible representation of $GL(2,\mathbf{Q}{}_p)$ then its restriction to $SL(2,\mathbf{Q}{}_p)$ is either irreducible, or splits as a direct sum of 2 non-isomorphic representations, or, occasionally, as a direct sum of 4. One interesting case is the unramified principal series with Satake parameter $X^2+c$ for any $c$; this splits into two pieces (if I've understood correctly) (and furthermore these are the only unramified principal series which do not remain irreducible under restriction). Does precisely one of these pieces have an $SL(2,\mathbf{Z}{}_p)$-fixed vector? And does the other one have a fixed vector for the other hyperspecial max compact (more precisely, for a hyperspecial in the other conj class)? Have I got this right? 
Packets: Local $L$-packets are precisely the J-H factors for $SL(2,\mathbf{Q}{}_p)$ showing up in an irreducible $GL(2,\mathbf{Q}{}_p)$-representation. So they have size 1, 2, or 4.
Now globally. a global $L$-packet is a restricted product of local $L$-packets (all but finitely many of the components had better have an invariant vector under our fixed hyperspecial max compact coming from a global integral model). Note that global automorphic $L$-packets might be infinite (because $a_p$ can be 0 for infinitely many $p$ in the modular form case). My understanding is that it is generally the case that one element of a global $L$-packet is automorphic if and only if all of them are, and in this case, again, generally, each one shows up in the automorphic forms with the same multiplicity. What is this multiplicity? Does it depend?
Finally, my understanding is that the above principle (multiplicities all being equal) fails precisely when $\pi$ is induced from a grossencharacter on a quadratic extension of $\mathbf{Q}$. In this case there seems to be error terms in [LL]. Can someone explain an explicit example where they can say precisely which elements of the packet are automorphic, and what the multiplicities which which the automorphic representations occur in the space of cusp forms?
I find it very tough reading papers of Langlands. My instinct usually would be to press on and try and work out some examples myself (which is no doubt what I'll do anyway), but I thought I'd ask here first to see what happens (I know from experience that there's a non-zero chance that someone will point me to a website containing 10 lectures on Labesse-Langlands...)
Edit: I guess that there's no reason why I shouldn't replace "cuspidal" by "lies in the discrete series" with the above (in the sense that the questions then still seem to make sense, I still understand everything (in some sense) for $GL_2$ and I still don't know the answers for $SL_2$)

REPLY [11 votes]: Warning: not an expert, so could be major mistakes in this. 
The multiplicity is one for every element in the global packet, in the non-CM case. In the CM case, 
half of the packet has multiplicity one and the other half has multiplicity zero.  
For the general story, I'd suggest looking at Arthur's conjectures, which at least conjecturally give a very nice picture.   In general one should talk of Arthur packets, not Langlands packets.   They coincide in your case. 
For CM representations on $SL_2$, the associated global parameter has dihedral image inside $PGL_2(\mathbb{C})$. Its centralizer $S$ has size $2$. According to Arthur, the obstruction to an element of the global packet being automorphic is valued in the dual of $S$; thus, "one-half" is automorphic. More precisely, the set of irreducible representations in a global $L$-packet is a principal homogeneous space for a certain product of $Z/2Z$s (in the obvious way:  one $Z/2Z$ for each $p$ where $a_p = 0$; for simplicity, suppose that local multiplicity $4$ doesn't occur). The automorphic ones correspond exactly to one of the fibers of the summation map to Z/2Z. Thus, if you take an automorphic representation in this packet, and switch it at one place to the other element of the local packet, it won't be automorphic any more. 
Concretely: Start with $\Pi$ an automorphic representation for $GL_2$, it maps
by restriction to the space of automorphic forms for $SL_2$.  The above remarks
suggest that this restriction map must have a huge kernel when $\Pi$ is CM.  But one can almost see this by hand: $\Pi$ is isomorphic
to its twist by a certain quadratic character $\omega$. This, and multiplicity one for $GL_2$,
means that, for $f \in \Pi$, the function $g \mapsto f(g)  \omega(\det g)$ also belongs to $\Pi$. This forces extra vanishing, although I didn't work out the details: For instance, if $\omega$ is everywhere unramified and $f$ the spherical vector, then $f(g) \omega (\det g)$ must be proportional to $f$, so $f$ vanishes whenever $\omega(\det g)$ is not $1$, i.e. various translates of $f$ do vanish when restricted to $SL_2$. 
Remark/warning: If you are concerned with multiplicity one, there is a further totally distinct phenomenon that causes it to fail for $\mathrm{SL}_n, n \geq 3$:  Two non-conjugate homomorphisms of a finite group into $\mathrm{PGL}_n$ can be conjugate element by element. 
 There's a paper of Blasius about this. It has nothing to do with packets.<|endoftext|>
TITLE: k-pseudorandom measures
QUESTION [5 upvotes]: In reading the paper of Green and Tao on arithmetic progressions within the primes, I became very interested in the notion of a k-pseudorandom measure discussed in that paper. 
A measure here is a function $\nu:\mathbf{Z}_N\to\mathbf{R}$ such that $\mathbf{E}\nu=1+o(1)$, and it is k-pseudorandom if it obeys the ($k2^{k-1}$,$3k-1$,$k$) (I think) linear forms condition, which basically asserts that it behaves independently with respect to at most $k2^{k-1}$ independent linear forms in $3k-1$ variables, and if it also obeys the correlation condition, which is a weaker form controlling the linear forms $x+h_i$.
They show that a relative Szemeredi's theorem applies to functions bounded by a k-pseudorandom measure, and then construct one that (effectively) bounds the primes.
My question is where else these type of functions have been studied, whether their theory has been expanded, and whether other explicit examples have been found and applied in other situations.

REPLY [4 votes]: For a relative Szemer\'edi theorem, the correlation condition was removed and just a weak linear forms condition was shown to be sufficient in: 
D. Conlon, J. Fox, and Y. Zhao, A relative Szemerédi theorem, preprint.<|endoftext|>
TITLE: Class groups of normal domains over finite fields
QUESTION [11 upvotes]: Let R be a local, normal domain of dimension 2. Suppose that R contains a finite field. I am interested in knowing when the class group of R is torsion. In characteristic 0, this is known to be related to R being a rational singularity. 
Lipman showed that if X is a desingularization of Spec(R), then one has an exact sequence:
$0 \to Pic^{0}(X) \to Cl(R) \to H $
Here $Pic^{0}(X)$ is the numerically trivial part of the Picard group of $X$, and $H$ is a finite group. Thus the second one is torsion if and only if the first one is. I do not have much understanding of the first group, unfortunately. 
Does anyone know an answer or reference to this? Does anyone know an example in positive characteristic such that $Cl(R)$ is not torsion? Thanks a lot.

REPLY [26 votes]: As requested in the comments, here's an example of a local, normal $2$-dimensional domain R in positive characteristic such that $\mathrm{Cl}(R)$ is not torsion: choose an elliptic curve $E \subset \mathbf{P}^2$ over a field $k$ such that $E(k)$ is not torsion, and take R to be the local ring at the origin of the affine cone on $E$ (i.e., $R  = k[x,y,z]/(f)_{(x,y,z)}$ where $f$ is a homoegenous cubic defining $E$). This can be done over $k = \overline{\mathbf{F}_p(t)}$.
Proof: The normality follows from the fact that R is a hypersurface singularity (hence even Gorenstein) and isolated and $2$-dimensional (hence regular in codim 1). Blowing up at the origin defines a map $f:X \to \mathrm{Spec}(R)$. One can then show the following: $X$ is smooth, and $X$ can be identified with the Zariski localisation along the zero section of the total space of the line bundle $L = \mathcal{O}_{\mathbf{P^2}}(-1)|_E$ (these are general facts about cones). By Lipman's theorem, it suffices to show that $\mathrm{Pic}^0(X)$ contains non-torsion elements. As $X$ is fibered over $E$ with a section, the pullback $\mathrm{Pic}^0(E) \to \mathrm{Pic}^0(X)$ is a direct summand. As $\mathrm{Pic}^0(E) \simeq E(k)$ has non-torsion elements by assumption, so does $\mathrm{Pic}^0(X)$.
Also, an additional comment: In general, Lipman's theorem tells you that $\mathrm{Cl}(R)$ is torsion if and only if $\mathrm{Pic}^0(X)$ is torsion. Now $\mathrm{Pic}(X) \simeq \lim_n \mathrm{Pic}(X_n)$ where $X_n$ is the $n$-th order thickening of the exceptional fibre $E$. Because we are blowing up a point, the sheaf of ideals $I$ defining $E$ is ample on $E$. The kernel and cokernel of $\mathrm{Pic}(X_n) \to \mathrm{Pic}(X_{n-1})$ are identified with $H^1(E,I|_E^{\otimes n+1})$ and $H^2(E,I|_E^{\otimes n+1})$. As $I|_E$ is ample, it follows that the system "$\lim_n \mathrm{Pic}(X_n)$" is eventually stable. Thus, $\mathrm{Pic}(X) \simeq \mathrm{Pic}(X_n)$ for $n$ sufficiently big. As $X_n$ is a proper variety, it follows that if we are working over a finite field (resp. an algebraic closure of a finite field), then $\mathrm{Pic}^0(X)$ is finite (resp. ind-finite).<|endoftext|>
TITLE: Definition of the symmetric algebra in arbitrary characteristic for graded vector spaces
QUESTION [8 upvotes]: What is the right definition of the symmetric algebra over a graded vector space V over a field k?
More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?
Two possible definitions come to my mind:
1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.
2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).
The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.
The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.
Finally, both definitions have a shortcoming in that they don't commute well with base change.

REPLY [2 votes]: The right definition is: take the free associative (tensor) algebra generated by $V$; divide out the ideal generated by the elements $xy-(-1)^{|x||y|}yx$ for all homogeneous $x$, $y\in V$ and $z^2=0$ for all odd $z\in V$.  This should commute with the base change well (when $V$ is flat over your base).<|endoftext|>
TITLE: Higher order quandle
QUESTION [8 upvotes]: The notion of quandle is known to be closely related to knot theory. The three axioms in the definition of quandle correspond to the Reidemeister moves.
Recently I learned that there are higher analogues of Reidemeister moves. For surface knots. I've heard that seven moves of Roseman determine isotopy classes.
Is there an algebraic structure, which can be regarded as a 2-dimensinal analogue of quandle, corresponding to Roseman moves?

REPLY [7 votes]: The definition of strict 2-quandle and examples thereof has not be written down in a public forum yet. Crans, Elhamdadi, Saito, and I have a notion and examples. I think that Crans spoke about the idea in Riverside, and I will give a talk about it at Knots in Washington next weekend. 
I don't reckon that I will have slides ready before then. If I do, I will post them on my web page. I am considering blackboard talks for the up-coming conference.
It is still conjectural that a weak 2-quandle is an axiomatization of the Roseman moves. I am playing with a couple of diagrammatic schemata to make sure everything works. Also after your question, I started puzzling about the tetrahedral move. This should be a theorem, and right now it appears to be an axiom.<|endoftext|>
TITLE: A problem/conjecture related to 4-manifolds that deserves a name. What name does it deserve?
QUESTION [8 upvotes]: There's an old problem in 4-manifold theory that, as far as I know, doesn't have a name associated with it and really deserves a name.  
Let $M$ be a smooth 4-manifold with boundary.  Let $S$ be a smoothly embedded 2-dimensional sphere in $\partial M$.  Assume $S$ does not bound a ball in $\partial M$, but $S$ is null-homotopic in $M$.  Does $S$ bound a smooth 3-ball in $M$? Perhaps you need to replace $S$ by another non-trivial $S'$ in $\partial M$ before you can find a 3-ball in $M$ bounding it?
You could think of this as the co-dimension one analogue to Dehn's lemma for 4-manifolds.  Usually when people talk about a Dehn lemma for 4-manifolds they're interested in the co-dimension 2 analogue.  
Does this problem / conjecture have a name?  If not, do you have a good name for it?  Do you know of anywhere in the literature where this issue is investigated? 
Off the top of my head the only vaguely related things I know about in the literature is a 1975 paper of Swarup's.

REPLY [10 votes]: I think that the conjecture is wrong. The following leads to counterexamples in the topological category and probably also smoothly: Take a closed oriented 4-manifold N with infinite cyclic fundamental group and remove an open neighborhood of a generating circle. Then you get a 4-manifold M with boundary $S^1 \times S^2$ where only the generators can be represented by embedded 2-spheres. If S denotes $pt \times S^2$ then the following hold:

S is null homotopic in M via Poincare duality and the exact sequence of the pair.
S bounds an embedded 3-ball in M if and only if N can be decomposed into $S^1 \times S^3$, connected sum with a simply connected manifold.

Hambleton and I found topological 4-manifolds N for which the intersection form is not extended from the integers, in particular 2 doesn't hold. With Friedl and Melvin we showed later that our examples don't have a smooth structure. But now I remember a discussion with Fintushel and Stern who mentioned that they constructed a smooth counterexample to 2 (and hence to the conjecture).<|endoftext|>
TITLE: What should the definition of "Yoneda property" be?
QUESTION [12 upvotes]: Let $C$ be a category. I'd like to say that a property $P$ of objects of $C$ (or rather isomorphism
classes of objects) is a "Yoneda property" or a "maps-in property" if there is a property $P'$ of
contravariant functors $h:C\to\mathrm{Set}$ such that the functor $\mathrm{Hom}(-,X)$ has $P'$ if and
only if $X$ has $P$. We might also say a property is "co-Yoneda" or "maps-out" if the induced property
on $C^{\mathrm{op}}$ is Yoneda.
But this definition is useless because every property is a Yoneda property (and, hence, also a co-Yoneda
property) -- just take $P'$ to be the property "$h$ is isomorphic to the functor $\mathrm{Hom}(-,X)$,
for some object $X$ with property $P$". So my question is: Is there a good definition of "Yoneda
property"?
Here are some examples of what I have in mind. In the category of modules over a given ring, injectivity
should be a Yoneda property and projectivity should be a co-Yoneda property. In any category, being a
terminal object should be a Yoneda property and being an initial object should be a co-Yoneda property.
We could do the same thing with maps instead of objects, and then in the category of schemes being
proper should be a Yoneda property (by the valuative criterion), as would being separated, formally
smooth, formally unramified, locally of finite presentation, and so on.
A few more remarks:

It seems that we'd want to keep the definition I gave above but make some restriction on properties
of functors we allow. Quantifying existentially over all objects of the category (which is what
breaks the definition above) probably should not be allowed. But what exactly should be allowed?
There appear to be different kinds of Yoneda properties. For example, in the category of schemes, the
definition of formally smooth is of the form "for all diagrams of type $Y$, there exists a map $f$ such
that $Z$ holds", and the definition of formally unramified is of the form "for all diagrams of type $Y$
and all maps $f, f'$ such that $Z$ holds, we have $f=f'$". Maybe it would be better to distinguish
these different kinds of properties. So it might be more natural to define separately "Yoneda
properties of existence type" (e.g. formal smoothness), "Yoneda properties of uniqueness type" (formal
unramifiedness), and maybe others.

REPLY [4 votes]: Hi Jim, not sure if this is the sort of thing you are after, but here is one possibility. 
Let $K$ be a category and $F=(f_i:A_i\to B_i)_{i\in I}$ a family of morphisms in $K$. Say that an object $X$ is injective to $F$ if for each $i\in I$ and each $a:A_i\to X$ there exists a morphism $b:B_i\to X$ whose restriction along $f_i$ is $a$. The collection of all such objects $X$ (for given $F$) is called an injectivity class in $K$. 
Any injectivity class in $[C^{op},Set]$ defines a property of objects of $C$: those objects $c$ for which the representable functor $C(-,c)$ lies in the injectivity class. Your various examples arise in this way:

For injectivity, just take all maps of the form $C(-,i):C(-,a)\to C(-,b)$ with $i:a\to b$ mono. 
For terminal object, take all the maps $0\to C(-,a)$ and $\nabla:C(-,a+a)\to C(-,a)$ (where $\nabla$ is the codiagonal)
and similarly being formally smooth or formally unramified. 

Your properties "of uniqueness type'' can be seen as a special case of those "of existence type'', by using codiagonal maps, as in the case of terminal object. 
Of course in the first example, of injectivity, you could just as well work in the original category $C$ itself, rather than $[C^{op},Set]$. More generally, you would always be able to do this if $C$ had colimits. 
You will get better properties if the family of morphisms defining the injectivity class is, or can be taken to be, small. Not sure if this is important for your purposes.<|endoftext|>
TITLE: Irreducible polynomial over number field with roots in every completion?
QUESTION [13 upvotes]: Let K/Q be a field, probably not a finite extension. Is it possible for a polynomial to be irreducible over K but have a root in every completion of K? What about all but finitely many completions?
This question is related to the question "Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?", and should help prove that such a polynomial can not exist for any subfield of the algebraic closure of the rationals.
The idea is that we make the candidate polynomial monic and have algebraic integers for coefficients, then take any maximal ideal in the ring of integers of the candidate field and complete it using the ideal - since the polynomial must have a root in the residue field, it will have a root in the completion. I'm wondering if this forces the polynomial to have a root in the original field - hence the question.
The same question only for function fields is also interesting, in order to prove the above for subfields of the algebraic closure of Fp(t)

REPLY [24 votes]: For a finite extension $K/\mathbf{Q}$, the answer is no. Suppose that $f$ is an irreducible polynomial with coefficients in $K$ and splitting field $L$. If $G$ is the Galois group of $L/K$, then the polynomial $f$ gives rise to a faithful transitive permutation representation $G \rightarrow S_d$, where $d$ is the degree of $f$. If $P$ is a prime in $O_K$ that is unramified in $L$, then $f$ has a root over $O_{K,P}$ if and only if the corresponding Frobenius element $\sigma_P \in S_d$ has a fixed point. On the other hand, a theorem of Jordan says that every transitive subgroup of $S_d$ contains an element with no fixed points. By the Cebotarev density theorem, it follows that $f$ fails to have a point modulo $P$ for a positive density of primes $P$.
For an infinite extension $K/\mathbf{Q}$, the answer is (often) yes. Let $K$ be the compositum of all cyclotomic extensions. Then $x^5 - x - 1$ is irreducible over $K$, because its splitting field over $\mathbf{Q}$ is $S_5$. On the other hand, it has a root in every completion (easy exercise). 
Finally, your claim that "since a the polynomial must have a root in the residue field, it will have a root in the completion" is false. Hensel's lemma comes with hypotheses.

REPLY [12 votes]: Edit: I see I missed the note that $K$ might not be finite over $\mathbb{Q}$.  This answer is not correct for $K$ of infinite degree over $\mathbb{Q}$, as in FC's answer above.
No.  This is a consequence of the Chebotarev density theorem.  To see how it follows, look at exercise 6 at the end of Cassels and Frohlich's "Algebraic Number Theory".  
Briefly, the Chebotarev density theorem says that for a Galois extension of global fields $L/K$ and for a finite set $S$ of places of $K$, the proportion of primes of $K$ splitting in $L$ is $1/[L:K]$.  If $G=\text{Gal}(L/K)$ and $E$ is the fixed field of some $H\subset G$, it is possible to show that the proportion of places of $K$ with a split factor in $E$ is $|\bigcup_{\rho\in G}\rho H\rho^{-1}|/|G|$, and a lemma on finite groups says that this quotient is not $1$ unless $H=G$.  
In your case, take $E=K[x]/(f)$ and $L$ to be a normal extension of $K$ containing $E$.  Then "$v$ has a split factor" means "$f$ has a root in the completion $K_v$".  If $f$ has a root in each completion (or even a set of completions with density $1$, which includes the case "all but finitely many"), we must have $H=G$ and $E=K$.  So $f$ already had a root in $K$.<|endoftext|>
TITLE: Which is the correct ring of functions for a topological space?
QUESTION [22 upvotes]: There is a fact that I should have learned a long time ago, but never did; I was reminded that I did not know the answer by Qiaochu's excellent series of posts, the most recent of which is this one.
Let $X$ be a topological space.  I can think of at least four rings of continuous functions to attach to $X$:

$C(X)$ is the ring of continuous functions to $\mathbb R$.
$C_b(X)$ is the ring of bounded functions to $\mathbb R$.
$C_0(X)$ is the ring of continuous functions that "vanish at infinity" in the following sense: $f\in C_0(X)$ iff for every $\epsilon>0$, there is a compact subset $K \subseteq X$ such that $\left|f(x)\right| < \epsilon$ when $x \not\in K$.
The ring $C_c(X)$ of functions with compact support.

Option 2. is not very good.  For example, it cannot distinguish between a space and its Stone-Cech compactification.  My question is what kinds of spaces are distinguished by the others.
For example, I learned from this question that MaxSpec of $C(X)$ is the Stone-Cech compactification of $X$.  But it seems that $C(X)$ knows a bit more, in that $C(X)$ has residue fields that are not $\mathbb R$ if $X$ is not compact?
On the other hand, my memory from my C*-algebras class (which was a few years ago and to which I didn't attend well), is that for I think locally-compact Hausdorff spaces $C_0(X)$ knows precisely the space $X$?
So, MOers, what are the exact statements?  And if I believe that the correct notion of "space" is "algebra of observables", are there good arguments for preferring one of these algebras (or one I haven't listed) over the others?

REPLY [5 votes]: If one can distill one central fact from the answers above and elsewhere, then it is that if one wants a good duality theory for algebras of continous functions, one has to move on from Banach spaces or algebras.  This fact was recognised at least 50 years ago, and a suitable generalisation was discovered---the so-called strict topology on the space of bounded, continuous functions on a, say, completely regular space.  Despite the eminence of its discoverers and advocates (Beurling, Herz, Buck), it seems to have become forgotten lore.  A lucid argument on its behalf can be found in the 1960 Transactions paper by Herz, "The spectral theory of bounded functions" (as the title suggests, the motivation for its introduction comes from Harmonic Analysis).  The original version used weighted seminorms and was valid for locally compact spaces.  It was subsequently extended to completely regular spaces by several authors.  The topology can be defined most succinctly as the finest locally convex one which agrees with compact convergence on the unit ball of $C^b(X)$.  It is a complete locally convex spaces, its compact sets and  convergent sequences can be simply characterised (the latter are the uniformly bounded sequences which converge uniformly on compacta). Its dual is the space of bounded Radon measures on $X$, the natural version of the Stone-Weierstra\ss theorem holds for it,  its spectrum is naturally identifiable with $X$ and so one has a version of Gelfand-Naimark theory.  One can also characterise the topological properties of $X$  in terms of those of $C^b(X)$, all in principle, many in practice.  Importantly, one can characterise local compactness of $X$ in terms of the property of the algebra $C^b(X)$.  A full account can be found in the book "Saks Spaces and Applications to Functional Analysis".<|endoftext|>
TITLE: Topological Rings
QUESTION [12 upvotes]: Is it true that, if S is a subring of a separable topological Noetherian ring R, 
then S is separable, too ?

REPLY [5 votes]: While looking through old questions, I came accross this one, and decided to throw my hat into the "ring."
Partial Answers and Observations:

If the identity element has a countable neighborhood base, then trivially by the continuity of the operation of addition, every point has a countable neighborhood base (as continuous maps take a filter base at some point x, and map this onto a filter base of the image point this is a particular example of homogeneity exhibited by topologically equipped algebraic objects.). 
Provided, the topology defined on the ring is $T_0$, we have that the topology must also be $T_2$ and completely regular.
Moreover, since there is a countable dense subset $D$ by assumption, we have that given the above two assumptions, there exists a countable collection, of countable covers $U_n$, with the following property: Given any closed set $C$ from this space, and point $p\not\in C$ there exists an $n\in\omega$ such that for some $O\in U_n$, $p\in O$ and $O \cap C = \{\}$. (We get such a collection by ordering the neighborhood base of each point in $D$ by reverse inclusion, and taking fixed 'sections of fixed height' from each to form the open covers)

Putting (1), (2), and (3) together produces a topological space which is about as close to a normal Moore space as you can get without actually being one, that is to say, under these assumptions: $R$ is a separable completely regular developable Hausdorff space, (and we haven't even invoked the ACC yet) 
The interesting Part:
Ignoring for the moment the previous assumptions, intuitively, the ACC should somehow produce a covering property for this particular space. However, there is an interesting problem when it comes to the definition of subgroup/subring (which is required to get to the notion of ideal needed to apply the ACC): will they be open, closed, neither? Because of this we cannot really apply the ACC, to produce a nice covering property that might have tied everything together (like Lindelöf.) 
Edit: While poking around Wiki, I came across something I felt I needed to add
However, if you mean that the space is a Noetherian topological space, then we get some gnarlly consequences ( http://en.wikipedia.org/wiki/Noetherian_topological_space ), like the fact that the space is compact! Which is exactly the thread we would want to tie everything together, and produces a normal moore space.
The Reality of the Matter:
The question is ill-posed, in that we do not have enough information to properly deduce a valid and fully general answer. My answer to this question has tried to highlight this point by giving you a case where, you can be about as close as you might ever want to be to something genuinely interesting, and then failing to make it interesting because of the incompatibility of the algebraic assumption with the topological ones. Even if we consider the other possibility we enter one of those strange and beautiful areas in topology where things being to become independent of $ZFC$.
Final Conclusion:
Because of this freedom or lack of information, we are left with an answer of Most Likely No. (Weak answer I know) But I can make this claim, because we honestly do not completely understand the notion of hereditary separability (in fact it was only in 2006 that J T Moore was able to produce a ZFC example of an L-space Article)<|endoftext|>
TITLE: Is there a theory of differential equations for smooth correspondences?
QUESTION [6 upvotes]: This question is very closely related to another one I just asked.  The general question is to what extent there is a theory of differential equations for smooth correspondences (between a smooth manifold and $\mathbb R$, say).  But I'm wondering about a specific well-studied case.
Background
Let $N$ be a compact smooth manifold with tangent bundle $TN$ and cotangent bundle $T^\*N$.  In the usual way, pick a Lagrangian $L: TN \to \mathbb R$, and suppose that it is nondegenerate in the sense that $\frac{\partial L}{\partial v}$, thought of as a map $TN \to T^\*N$, is a fiber bundle isomorphism, where $v$ is a (vector of) fiber coordinate(s).  Then, in the usual way, we can define a Hamiltonian $H: T^\*N \to \mathbb R$ by $H = pv - L$, where $pv$ is the canonical pairing between a cotangent vector and a tangent vector, and I use $\frac{\partial L}{\partial v}$ to identify $TN$ with $T^\*N$.  In the usual way, define on $N$ the second-order ordinary differential equation $\frac{d}{dt}\bigl[ \frac{\partial L}{\partial v}\bigr] = \frac{\partial L}{\partial q}$ and $v = \frac{dq}{dt}$, or equivalently $\frac{dq}{dt} = \frac{\partial H}{\partial p}$ and $\frac{dp}{dt} = -\frac{\partial H}{\partial q}$.  (These are all written in local coordinates, where $q$ is a local coordinate on $N$ and $v = dq$ and $p = \frac{\partial}{\partial q}$ are the corresponding coordinates on $TN$ and $T^\*N$.  But the ODE is coordinate-invariant.)
Since $N$ is compact and $L$ is nondegenerate, the ODE has global solutions, and each solution is determined by its initial conditions, which are given equivalently by a point in $TN$ and a point in $T^\*N$.  In the usual way, define an action map $S: TN \times \mathbb R \to \mathbb R$ by $S(v,q,t) = \int_0^t L(\phi(v,q,s))ds$, where $\phi: TN \times \mathbb R \to TN$ is the "flow" map for the ODE.
Let $\pi$ be the projection $TN \to N$, so that $\pi(v,q) = q$, and using the flow map $\phi$, define a map $TN \times \mathbb R$ to $N \times N \times \mathbb R$ via $(v,q,t) \mapsto (q,\pi(\phi(v,q,t)),t)$.  Generically, this map is a local isomorphism, in the following sense: for generic $(v,q,t)$ (say, a dense open subset of $TN\times \mathbb R$ when $\frac{\partial^2 L}{\partial v^2}$ is positive-definite), there is a small open neighborhood such that the map to $N\times N \times \mathbb R$ takes the neighborhood diffeomorphically to its image.  Pick one such small neighborhood, and use it to push forward the action function $S$.  Abusing notation, I will call this pushforward $S$.  Then $S(q_1,q_2,t)$ satisfies the Hamilton-Jacobi equation: $\frac{\partial S}{\partial t} = - H(\frac{\partial S}{\partial q_2},q_2) = - H(-\frac{\partial S}{\partial q_1},q_1)$.
My question
Above, I defined a local function $S: U \to \mathbb R$, where $U$ is a small neighborhood of $N\times N\times \mathbb R$, and it satisfied a partial differential equation.  But really I should have talked about the correspondence
$$ N\times N\times \mathbb R \overset{(\pi, \pi\circ \phi, t)}{\longleftarrow} TN \times \mathbb R \overset{S}{\longrightarrow} \mathbb R $$
Is there language with which one can say that this correspondence satisfies the Hamilton-Jacobi equations?  For example, what happens near non-generic (sometimes called "focal") points?
Bonus questions
(If $N$ is not compact, then the flow map does not have global-time solutions.  But I can still do everything; I just have to replace the space $TN \times \mathbb R$ by an open subspace.  In fact, $\phi$ still defines on $TN \times \mathbb R$ the structure of an action groupoid, and both $(\pi, \pi\circ \phi, t)$ and $S$ are groupoid homomorphisms, so that the above correspondence is a span of groupoids.  Does this enter the discussion in any interesting way?)
If $\frac{\partial L}{\partial v}$ does not define a bundle isomorphism, then the Hamiltonian is not well-defined as a function $H: T^\*N\to \mathbb R$.  But it does make sense as a correspondence, by:
$$ T^\*N \overset{ \frac{\partial L}{\partial v}}{\longleftarrow} TN \overset{ v\frac{\partial L}{\partial v} - L}{\longrightarrow} \mathbb R$$
Imposing the condition that $\frac{\partial^2 L}{\partial v^2}(v,q)$ is an invertible matrix for each $(v,q)$, so that the ODE is still nondegenerate second-order, does the language of correspondences allow me to talk about the Hamilton-Jacobi equations when the Hamiltonian is not a function but the above correspondence?
Alternately, I could start with a function $H: T^\*N \to \mathbb R$, and construct a correspondence $L$, etc.  The Legendre transform should really be thought of as a transformation of correspondences, not of functions.
Finally, when $\frac{\partial^2 L}{\partial v^2}$ is sometimes degenerate, then the ODE degenerates, sometimes in complicated ways.  Can this be accommodated by making the flow $\phi$ into a correspondence rather than a function?

REPLY [2 votes]: In complex analysis, it happens all the time that solutions to a differential equation exist locally as functions but globally only as correspondences.  For instance z d/dz f = 1 is solved by log z, which you can think of as the correspondence
$$\mathbf{C} \stackrel{\mathrm{exp}}{\leftarrow} \mathbf{C} \stackrel{=}{\to} \mathbf{C}$$
In general for a correspondence X <-- Z --> Y to "exist locally as a function" means that the map X <-- Z is a local homeomorphism.  
I don't understand it that well but your situation is definitely more complicated, for example your map N x N x R <-- TN x R has positive-dimensional fibers when t = 0.  But that log z exists only locally as a function is usually expressed in terms of sheaves not correspondences (i.e. the sheaf of solutions to the differential equation is locally constant but not constant).  Can you formulate your Hamilton-Jacobi conditions that way?<|endoftext|>
TITLE: Monotone Lipschitz embedding ?
QUESTION [11 upvotes]: In 1974, Aharoni proved that every separable metric space (X, d) is Lipschitz isomorphic to a subset of the Banach space c_0. 
Thus, for some constant L, there is a map K: X --> c_0 that satisfies the inequality d(u,v) <= || Ku - Kv || <= Ld(u,v) for all u and v in X.
Now, suppose X = l_1 (in this case, L = 2 is best possible). I have the following
Conjecture: Let K: l_1 --> c_0 be a Lipschitz embedding. Then K cannot be monotone w.r.t. the natural duality pairing (.,.) between l_1 and c_0, 
i.e., there are some u and v in l_1 such that (u - v, Ku - Kv) < 0.

REPLY [4 votes]: To answer Bill Johnson's question, a monotone linear bi-Lipschitz embedding (actually, an isometric one) $\ell^1\to\ell^\infty$ is very easy to construct. Just take any antisymmetric matrix $A$ of $\pm 1$s with the property that for each $n$ every combination of signs in the first $n$ positions appears in some row of $A$ (you can easily build it by induction) and take $Lx=x+Ax$. Unfortunately, I do not see how to convert it into a mapping to $c_0$.<|endoftext|>
TITLE: Which is the correct universal enveloping algebra in positive characteristic?
QUESTION [15 upvotes]: This is an extension of this question about symmetric algebras in positive characteristic.  The title is also a bit tongue-in-cheek, as I am sure that there are multiple "correct" answers.
Let $\mathfrak g$ be a Lie algebra over $k$.  One can define the universal enveloping algebra $U\mathfrak g$ in terms of the adjunction: $$\text{Hom}_{\rm LieAlg}(\mathfrak g, A) = \text{Hom}_{\rm AsAlg}(U\mathfrak g, A)$$ for any associative algebra $A$.  Then it's easy enough to check that $U\mathfrak g$ is the quotient of the free tensor algebra generated by $\mathfrak g$ by the ideal generated by elements of the form $xy - yx - [x,y]$.  (At least, I'm sure of this when the characteristic is not $2$.  I don't have a good grasp in characteristic $2$, though, because I've heard that the correct notion of "Lie algebra" is different.)
But there's another good algebra, which agrees with $U\mathfrak g$ in characteristic $0$.  Namely, if $\mathfrak g$ is the Lie algebra of some algebraic group $G$, then I think that the algebra of left-invariant differential operators is some sort of "divided-power" version of $U\mathfrak g$.
So, am I correct that this notions diverge in positive characteristic?  If so, does the divided-power algebra have a nice generators-and-relations description?  More importantly, which rings are used for what?

REPLY [5 votes]: This question popped up on the feed recently, and I wanted to add another, "brave new math"-style answer. Namely, the point of Lie algebras is that, in characteristic $0$, a Lie algebra (resp., an $L_\infty$-algebra) gives necessary and sufficient information to define a formal Lie group (resp., a cogroup object in CDGA). Global sections of this group form a commutative, co-associative Hopf algebra and the (topological/filtered/etc.) dual will be the universal enveloping algebra.
There is an analogous story to the $L_\infty$ story in characteristic $p$, proved by Akhil Mathew and Lukas Brantner, https://arxiv.org/abs/1904.07352. They prove that the structure of a derived connected formal group scheme (which determines a formal moduli problem) is determined by a so-called partition Lie algebra structure on the tangent space. This is no longer an "additive" structure (not determined by a multilinear operad unlike $L_\infty$), but it is "just" some nicely combinatorial algebraic structure on the tangent space. If you modify your definition of Lie algebra using this picture then to each (partition) Lie algebra you can associate a formal moduli problem, and to every formal moduli problem you can associate a universal enveloping algebra by taking the dual to a universal Hopf algebra associated to your formal moduli problem. From what I understand, the "$\pi_0$ operations" inherent in this structure are exactly Lie algebra operations and divided powers.<|endoftext|>
TITLE: Why are planar graphs so exceptional?
QUESTION [46 upvotes]: As compared to classes of graphs embeddable in other surfaces.
Some ways in which they're exceptional:

Mac Lane's and Whitney's criteria are algebraic characterizations of planar graphs. (Well, mostly algebraic in the former case.) Before writing this question, I didn't know whether generalizations to graphs embedded in other surfaces existed, but some lucky Google-fu turned up some references -- in particular there seems to be a generalization of Whitney's criterion due to Jack Edmonds for general surfaces, although frustratingly I can't find the paper, and the main reference I found implies that there might be a small problem on the Klein bottle. Anyone know if Edmonds' result is as easy to prove as Whitney's?
Kuratowski's classic characterization of planar graphs by forbidden minors. Of course this does generalize to other surfaces, but this result is both incredibly deep and difficult (as opposed to the proof of Kuratowski, which is by no means trivial but is obtainable by a sufficiently dedicated undergraduate -- actually my working it as an exercise is largely what motivated the question) and is in some sense "essentially combinatorial" in that it applies to a wider class of families that aren't inherently topologically defined.
In the other direction of difficulty, the four-color theorem. It's apparently not difficult to show (except for the plane) that what turns out to be the tight upper bound on the chromatic number of a graph embeddable on a surface (other than, for whatever reason, the Klein bottle) is, in fact, an upper bound -- the problem is showing tightness! Whereas it's pretty much trivial to show that $K_4$ is planar (to be fair, though, tightness is easy to check for surfaces of small genus -- the problem's in the general case), but the four-color theorem requires inhuman amounts of calculation and very different, essentially ad-hoc methods.

I realize that the sphere has genus 0, which makes it unique, and has trivial fundamental group, which ditto, but while I imagine this information is related to the exceptionalness of the plane/sphere, it's not really all that satisfying as an answer. So, why is it that methods that work everywhere else tend to fail on the sphere? 
Related questions: reasons-for-the-importance-of-planarity-and-colorability

REPLY [2 votes]: Planar graphs answer many important questions in graph structure theory. 
Example 1. H-minor-free graphs have bounded treewidth if and only if H is planar. 
Example 2. The set of graphs contractible to some graph H have the Erdos-Posa property if and only if H is planar. 
Both results are due to Robertson and Seymour.<|endoftext|>
TITLE: Too old for advanced mathematics?
QUESTION [163 upvotes]: Kind of an odd question, perhaps, so I apologize in advance if it is inappropriate for this forum. I've never taken a mathematics course since high school, and didn't complete college. However, several years ago I was affected by a serious illness and ended up temporarily disabled.  I worked in the music business, and to help pass the time during my convalescence I picked up a book on musical acoustics.
That book reintroduced me to calculus with which I'd had a fleeting encounter with during high school, so to understand what I was reading I figured I needed to brush up, so I picked up a copy of Stewart's "Calculus".  Eventually I spent more time working through that book than on the original text.  I also got a copy of "Differential Equations" by Edwards and Penny after I had learned enough calculus to understand that. I've also been learning linear algebra - MIT's lectures and problem sets have really helped in this area.  I'm fascinated with the mathematics of the Fourier transform, particularly its application to music in the form of the DFT and DSP - I've enjoyed the lectures that Stanford has available on the topic immensely.  I just picked up a little book called "Introduction To Bessel Functions" by Frank Bowman that I'm looking forward to reading.
The difficulty is, I'm 30 years old, and I can tell that I'm a lot slower at this than I would have been if I had studied it at age 18.  I'm also starting to feel that I'm getting into material that is going to be very difficult to learn without structure or some kind of instruction - like I've picked all the low-hanging fruit and that I'm at the point of diminishing returns.  I am fortunate though, that after a lot of time and some great MDs my illness is mostly under control and I now have to decide what to do with "what comes after."
I feel a great deal of regret, though, that I didn't discover that I enjoyed this discipline until it was probably too late to make any difference.  I am able, however, to return to college now if I so choose.
The questions I'd like opinions on are these: is returning to school at my age for science or mathematics possible? Is it worth it? I've had a lot of difficulty finding any examples of people who have gotten their first degrees in science or mathematics at my age.  Do such people exist? Or is this avenue essentially forever closed beyond a certain point?  If anyone is familiar with older first-time students in mathematics or science - how do they fare?

REPLY [3 votes]: It's never too late, my dream when I was kid to become a doctor in math, am 42 yrs old now,
I could'nt study cause my financial situations, i was top student in math, I just enrolled to finish my last year in B.sc math, then am looking to apply directly for Phd math.
I left school at age 23.
I feel very strong and more smarter than before. good luck to all.<|endoftext|>
TITLE: determinant of a perfect complex
QUESTION [24 upvotes]: Say $K_\bullet$ is a bounded complex of vector bundles.  I seem to want the determinant of $K_\bullet$ to be the alternating tensor product of the terms of the complex:
$\det(K) = \bigotimes_n \det(K_n)^{(-1)^n}$.
Is there a reason why this is the right definition (or the wrong definition)?  Is there a better definition?

REPLY [2 votes]: As I understand the construction of the determinant of a perfect complex, this definition is quite straightforward, following from the fact that in a short exact sequence, say
$$ 0\rightarrow S\rightarrow E\rightarrow Q\rightarrow 0$$
defining the determinant of the sequence to be the alternating tensor is the canonical way to make it isomorphic to $\mathbb{1}$.
Also, I think good references to this may be the original paper by Knudsen-Mumford, a book by Kato, and also a paper by Kings which are listed below:
Finn Faye Knudsen and David Mumford, The projectivity of the moduli space of stable curves. I. Preliminaries on ''det'' and ''Div'' (pdf). The part about determinants appears in Chapter I, but note that there is a typo defining the determinant, namely in the map of the transposition of tensor product, there should be $\alpha\cdot\beta$ instead the sum of these two as a power of $-1$;
Guido Kings, An introduction to the equivariant Tamagawa number conjecture: the relation to the Birch-Swinnerton-Dyer conjecture (pdf)
There is a part about determinants in lecture 1 section 5, where there are not a lot of details but it provides a good view towards the construction of determinant.
Kazuya Kato, Lectures on the approach to Iwasawa theory for Hasse-Weil L-functions via $B_{dR}$, part I Springer LNM 1553 pp 50-163 (doi:10.1007/BFb0084729), which mentions determinant in 2.1.<|endoftext|>
TITLE: Classification of finite commutative rings
QUESTION [38 upvotes]: Is there a classification of finite commutative rings available? 
If not, what are the best structure theorem that are known at present? 
All I know is a result that every finite commutative ring is a direct product of local commutative rings (this is correct, right?) in some paper which computes the size of the general linear group over that ring.

REPLY [4 votes]: As always one should check out the OEIS for questions of this type. In this case see http://oeis.org/A027623<|endoftext|>
TITLE: Joins of simplicial sets
QUESTION [7 upvotes]: Why doesn't the join operation on the category of simplicial sets commute up to unique isomorphism?  I mean, aren't products and coproducts commutative up to isomorphism? That leads me to conclude at first glance that the join is commutative, but it's not. Recall, given two simplicial sets $S$ and $S'$, we define the join to be the simplicial set such that for all finite nonempty totally ordered sets $J$,  $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$
Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.  
Now the thing is, clearly my conclusion is stupid, because we use the fact that it doesn't commute to distinguish between over quasi-categories and under quasi-categories.  Where did I go wrong?
I hope this is up to the standards of MO, but if it's not, I'll delete the topic.

REPLY [5 votes]: As mentioned in the other answers, the join of simplicial sets is closely related to "ordered disjoint union", or "concatenation", of (totally, partially, pre-) ordered sets.  You can use this both to get simple examples of its non-commutativity, and to help reconcile that with the intuition that it should be commutative.
Any order $X$ can be seen as the simplicial set whose $n$-simplices are chains $(x_0 \leq \ldots \leq x_n)$ from $X$.  That is, there's a full and faithful "nerve" embedding $N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$.  Now if $X$ and $Y$ are  orders, seen as their nerves, $X \star Y$ is exactly (the nreve of) their ordered disjoint union.
So e.g. $1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$ is an easy and intuitive example of the non-commutativity.
On the other hand, like you say, looking at the definition, there is an immediate intuition that it should be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets $X$, $Y$,
$(X \star Y)^\mathrm{op} \cong Y^\mathrm{op} \star X^\mathrm{op}.$
So commuting $\star$ distributes over $\mathrm{op}$: so in a sense, the only asymmetry in $\star$ is an asymmetry of variance.  This is nice and intuitive for ordered sets, and easily shown for all simplicial sets.<|endoftext|>
TITLE: intuition about the "section after base-change" for flat descent and exactness of the Amitsur complex
QUESTION [10 upvotes]: Suppose $A \rightarrow B$ is a faithfully flat map of rings. Then the Amitsur complex is exact:
$0 \rightarrow A \rightarrow B \rightarrow B \otimes_A B \rightarrow \dots$
(the second map is $id \otimes 1 - 1 \otimes id$, and the subsequent maps are alternating sums of the different ways of putting in a 1.)
That this is exact makes sense! Its saying that element of $b\in B$ such that $1\otimes b=b\otimes 1$ comes from $A$. The way to prove it is exact is roughly the following (the details can be found in Milne's online notes on étale cohomology.)
First, one can check that the sequence is exact if the first map has a section. Second, it is exact iff it is exact after a faithfully flat base change. Finally, if we base change by $B$, we have a section to the first map: the map we want to find a section to is $B \rightarrow B\otimes_A B, b \mapsto b\otimes 1$, and this has a section given by multiplication.
Why do we care? This fact is important in showing schemes are sheaves on the étale site. (Again, this can be found in Milne's notes.)
Okay, so my question is: what is really going on behind the scenes? I know a section is supposed to somehow "solve an equation", but I have no idea what is really being solved here. The proof feels like magic.
(One note: Its probably instructive to look at the special case of $A \rightarrow \prod A_f$ , a cover by basic open affines. I believe this proof is showing that the sequence is exact simultaneously on each member of the cover, and because exactness is a local property we have that the sequence is exact.)

REPLY [4 votes]: The more general name this sort of argument goes by is descent theory, and chapter six of Bosch, Lutkebohmert, and Raynaud's "Neron Models" is a good introduction.  The general idea is that you have some object (say, a quasi-coherent sheaf) defined on the "open sets" in a cover (it could be a Zariski cover, or an etale cover, or an fpqc cover, or...) of $X$, and you want to know whether you can glue it in some unique way to get something on $X$ itself.
That was vague, so let's consider the case when we have a quasi-coherent sheaf of modules $M_f$ on $\text{Spec}A_f$, where $A\rightarrow \prod A_f$ is a cover by open affines.  Let's set $B=\oplus A_f$ and $M'=\oplus M_f$.  We want a quasi-coherent sheaf $M$ on $\text{Spec} A$.  What's the most obvious requirement on the $M_f$?  Well, they have to agree on overlaps.  That is, if $U=\text{Spec}B=\coprod\text{Spec}A_f$, and $p_1,p_2$ are the projections $U\times_A U\rightarrow U$, we need an isomorphism 
$p_1^* M'\rightarrow p_2^* M'$, because $U\times_A U$ is the disjoint union of all the overlaps.  Another way of saying this is that if $M'=M\otimes_A B$, then $M\rightarrow M\otimes_A B\rightrightarrows M\otimes_A B\otimes_A B$ is exact.  Note that this is not a sufficient condition for $M$ to exist!  Line bundles are locally trivial, so of course there are isomorphisms on the overlaps, but not every choice of such isomorphisms is compatible with triple overlaps ($U\times_A U\times_A U$).
As for what the assumption of a section is doing there, it's saying that locally, your map of schemes $\text{Spec}B\rightarrow \text{Spec}A$ has a section.  Not Zariski-locally, fpqc locally, and that's important.  One reason the Zariski topology isn't so great to work with is that a lot of morphisms that "look like" they should be fiber bundles (that is, locally trivial in some sense) really aren't.  But they tend to be in finer topologies, like the etale topology and the fpcq topology.  The "base change to get a section" trick you mention above is saying that fpqc-locally, $\text{Spec}B\rightarrow \text{Spec}A$ behaves like a fiber bundle, in that it has lots of sections to work with.  If the map happens to be etale, that's just saying that locally on the base, the total space is a bunch of disjoint copies of the base, which is definitely a topological property we want.
In the specific application you mention, showing that schemes are sheaves in the etale topology, what you're actually going to do is show that they're sheaves in the fpqc topology, and deduce that they're sheaves in any coarser topology, like the etale topology.<|endoftext|>
TITLE: Open affine subscheme of affine scheme which is not principal
QUESTION [49 upvotes]: I'm not sure whether this is non-trivial or not, but do there exist simple examples of an affine scheme $X$ having an open affine subscheme $U$ which is not principal in $X$? By a principal open of $X = \mathrm{Spec} \ A$, I mean anything of the form $D(f) = \{\mathfrak p \in \mathrm{Spec} \ A : f \notin \mathfrak p\}$, where $f$ is an element of $A$.

REPLY [2 votes]: Here is an example that may be considered easier than Hailong Dao's. (At least it is one dimensional smaller!)
Consider the ring $R=\mathbb{Z}[x,y,z]/\langle xy-z(z-1)\rangle$ and the ideal $I=\langle x,z\rangle$.
The ideal $I$ is not the radical of a principal ideal. So the complement of $\mathrm{Spec}(R/I)$ in $\mathrm{Spec}(R)$ is not $R_r$ for any $r$ in $R$.
In fact, this complement is given by the ring homomorphism $R\to S=\mathbb{Z}[x,t]$ where $z\mapsto 1+xt$ and $y\mapsto (1+xt)t$. In other words, the complement is $\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ via this ring homomorphism.
To see this, let $f:R\to A$ be a ring homomorphism such that $f(x)$ and $f(z)$ generate the unit ideal. This means $af(x)+bf(z)=1$ for some $a,b\in A$.
Now, let $\tilde{f}(t)=a(f(z)-1)+bf(y)$. One checks that this extends $f$ to $\tilde{f}:S\to A$.<|endoftext|>
TITLE: Famous mathematical quotes
QUESTION [179 upvotes]: Some famous quotes often give interesting insights into the vision of mathematics that certain mathematicians have. Which ones are you particularly fond of?
Standard community wiki rules apply: one quote per post.

REPLY [30 votes]: "In science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in the case of poetry, it's the exact opposite!" -- Paul Dirac (some people attribute it to Franz Kafka!?)<|endoftext|>
TITLE: Constructing Twisted K-theory 
QUESTION [10 upvotes]: There is a simple, intuitive "construction" of twisted K-theory if we are allowed to ignore that many things only hold up to homotopy. We know that maps to $K(Z,2)$ give line bundles on a space and that $K(Z,2)$ forms a group corresponding to the tensor product of line bundles. Line bundles also act as endomorphisms of K-theory given by the tensor product. Thus, there is an action of $K(Z,2)$ on $F$ (where $F$ is the classifying space for $K^0$). $K(Z,2)$ principal bundles are classified by maps to $BK(Z,2) \cong K(Z,3)$, ie, elements of $H^3$. Choosing such a map, we get a principal $K(Z,2)$ bundle, $E$, and we can form the associated bundle $E \times_{K(Z,2)} F$. Twisted K-theory is then the homotopy classes of sections of this bundle.
The usual constructions of twisted K-theory that I have seen make the above precise by choosing representatives of the relevant objects so that all the needed relations hold on the nose. My question is whether you can avoid doing that. In other words, can you define all the various notions up to homotopy and obtain a definition of twisted K-theory that way?

REPLY [12 votes]: The answer is yes if you're working on the level of $\infty$-categories (and I'm pretty sure no if you're working on the level of homotopy categories). In other words, in the $\infty$-world there's no problem talking about a principal bundle for K(Z,2)=BBZ on any space, and they're indeed classified by maps (in the $\infty$-category of spaces) to BBBZ,
as are elements of $H^3$. Moreover for any spectrum $E$ there's a well defined group object $GL_1(E)$, and we have a map $BBZ\to GL_1(E)$ in the case of E=K-theory. For any principal $GL_1(E)$ bundle (eg one induced from a BBZ bundle) we have an associated "E line bundle", and its global sections are the twisted K-theory of your space (i.e. a spectrum whose homotopy groups are the usual twisted K-groups). 
Or again in short, everything works intuitively as you think without chosing representatives or strictifying if you work in the wonderful world of $\infty$-categories.. for example, the idea of sheaves of spectra (hence twisted cohomology theories) are as simple to work with formally as ordinary sheaves. When it comes to calculating things.. well that's a whole other story.<|endoftext|>
TITLE: How can category theory help my research in set theory?
QUESTION [29 upvotes]: How can category theory help my research in set theory?  
I rarely use category theory as such in my current work, and one almost never sees any category theory in set-theoretic research papers or at conferences (except of course when the issue is to apply set theory to category theory rather than conversely). Why is this, when category theoretic language and thinking has proved so successful in other parts of mathematics?
Since there seems to be a relatively sizable community of category theorists on this site, many of whom appear to know a lot of set theory or at least have opinions about it, I hope that I might gain some insight.
Note that I am not looking for a reason to do category theory instead of set theory. I am already inspired by a collection of topics, questions and results within set theory, which I find compelling and sometimes profound. What I want to know is whether category theory can provide me with techniques to use to attack those problems.

REPLY [7 votes]: This is not a direct answer to the question but more a small note on the difference between set theory as usually done and set theory viewed through the lenses of category theory.
For me, the most striking aspect of topos theory was the unearthing of vast families of categories that appear in "ordinary" mathematics and that behave very much like the category of sets (denoted by Set).
One can reverse the point of view and ask what is special from a categorial point of view about the category of sets? Well, one partial answer is that Set is well-pointed, that is, the terminal is a separator (or generator). This means that every arrow $f:a\to b$ is determined uniquely by the family of points $fx:t\to b$, with $t$ the terminal and $x:t\to a$ a (global) point of $a$. This is a precise categorial version of our view of sets as "discrete, structureless piles of sand".
If we did set theory this way, what would be different? There is a (at least one) major philosophical difference. There is no global $\in$. Since the equality predicate of sets is defined by
$X=Y \iff (\forall x, x\in X \iff x\in Y)$
it follows that there is no global equality predicate for sets. This may be unintuitive from our conception of sets, but it is the Right Thing in category land, because equality between objects is not preserved by equivalences. The Freyd / Scedrov book Categories, allegories contains a formal treatment of a language invariant under equivalences and a considerable part of M. Makkai's work on n-categories is building this insight right into the foundation of higher categories.
Hope it helps a little, regards.<|endoftext|>
TITLE: Moduli space of flat bundles
QUESTION [7 upvotes]: Is there a good place to learn about the structure of moduli stack of flat $G$-bundles on an algebraic curve?
Of course, we're just studying representations of a group $\pi_1(X)\to G$ modulo some conjugation (that's why it should be a stack). Since this is very similar to Galois representations in number theory, I wonder if there's a reference that also explains the similarities and differences between the two cases.

REPLY [16 votes]: You have to be a bit careful here. Over $\mathbb{C}$ the stack of representations of $\pi_{1}(X)$ in $G$ and the stack of flat algebraic $G$-bundles on $X$ are isomorphic  as complex analytic stacks but are not isomorphic as algebraic stacks. In fact the algebraic structure on the stack of flat $SL_{n}(\mathbb{C})$ bundles on $X$ reconstructs the curve $X$, while the stack of representations of $\pi_{1}(X)$ into $SL_{n}(\mathbb{C})$  depends only on the genus of $X$ and not on the particular curve $X$.
As for references I suggest you take a look at Carlos Simpson's papers "Moduli of representations of the fundamental group of a smooth projective variety, I and II" which you can find here and here.<|endoftext|>
TITLE: Learning About Schubert Varieties
QUESTION [15 upvotes]: I am a combinatorist by training and I am interested in learning about the connections between combinatorics and Schubert varieties.   The theory of Schubert varieties seems to be a difficult area to break into if one has not already studied it in graduate school.  I don't have a formal course in Algebraic Geometry, but I do know some Commutative Algebra.  Does anyone have any recommendations on what Algebraic Geometry one needs to understand (and what books/papers one should read) in order to begin studying Schubert varieties?  Algebraic Geometry is such a huge subject that I am trying to figure out what is essential to this particular study and what is not.   Thanks, in advance, for any suggestions!

REPLY [6 votes]: Another reference I've heard suggested before is these notes by Brion:
Lectures on the geometry of flag varieties<|endoftext|>
TITLE: measure spaces as presheaves?
QUESTION [17 upvotes]: I recently had the idea that maybe measure spaces could be viewed as sheaves, since they attach things, specifically real numbers, to sets...
But at least as far as I can tell, it doesn't quite work - if $(X,\Sigma,\mu)$ is a measure space and $X$ is also given the topology $\Sigma$, then we do get a presheaf $M:O(X)\rightarrow\mathbb{R}_{\geq0}$, where $\mathbb{R}_{\geq0}$ is the poset of non-negative real numbers given the structure of a category, and $M(U) = \mu(U)$, and $M(U \subseteq V)$ = the unique map "$\geq$" from $\mu(V)$ to $\mu(U)$, but this is not a sheaf because for an open cover {$U_i$} of an open set $U$, $\mu(U)$ is in general not equal to sup $\mu(U_i)$ (and sup is the product in $\mathbb{R}$).
First off, is the above reasoning correct? I'm still quite a beginner in category theory, but I've seen on Wikipedia something about a sheafification functor - does applying that yield anything meaningful/interesting? Does anyone have good references for measure theory from a category theory perspective, or neat examples of how this perspective is helpful?

REPLY [2 votes]: http://etd.library.pitt.edu/ETD/available/etd-04202006-065320/unrestricted/Matthew_Jackson_Thesis_2006.pdf
http://front.math.ucdavis.edu/0912.4914
R. Borger "FUbini Theorem from a Categorical Viewpoint" in "Category Thery at work" (H.Herrlich- H-E Prost (eds) "  "Helderman Verlad Berlin" 1991<|endoftext|>
TITLE: Examples of noncommutative analogs outside operator algebras?
QUESTION [14 upvotes]: Theo's question made me wonder if there are other "noncommutative analogs" outside of operator algebras. Some noncommutative analogs from operator algebras include:

A $C^\ast$-algebra is a noncommutative topological space (cf. the Gelfand transform).
The multiplier algebra of a nonunital $C^\ast$-algebra is the noncommutative Stone-Cech compactification.
A spectral triple is a noncommutative manifold (add some extra data to the spectral triple to get a noncommutative Riemannian manifold cf. arXiv:0810.2088).
A von Neumann algebra is a noncommutative measure space.

Are there any other good examples? If you know more in operator algebras, that's great too.
EDIT: these algebras should be considered as various functions spaces for noncommutative spaces as per @Yemon's answer. I'm going to leave the above text as is unless there are requests for another edit.

REPLY [7 votes]: In a recent paper Alexei Pirkovskiy defines a class of Frechet algebras such that the subclass of commutative algebras corresponds to Stein complex manifolds (under a technical assumption of of finite embedding dimension). This may be viewed as noncommutative complex analysis. Link: http://arxiv.org/abs/1204.4936

REPLY [4 votes]: Some non-commutative analogs in lattice theory. 
von Neumann's coordinatization theorem is the 
non-commutative analogue of Stone's definitional equivalence between Boolean algebras 
(complemented distributive lattices) and Boolean rings (associative rings with 1 where all
elements are idempotent). 
The Baer - Inaba - Jonnsson - Monk coordinatization theorem gives the non-commutative 
analogue of direct products of finite chains (Łukasiewicz propositional logics); 
for this one uses not the usual formulation of the theorem 
(which coordinatizes primary lattices with modules over a artinian ring where 
one-sided ideals are two-sided and form a chain), but a reformulation that gives 
a true equivalence between the lattice (of submodules of the module) and the ring
(of endomorphisms of the module); this way one has a true analogue of the (dual) 
equivalence between commutative C^*-algebras and compact Hausdorff spaces 
(or better, their lattice of open sets). 
One has a common generalization of the two cases above: equivalence between lattices of
subobjects and rings of endomorphisms for finitely presented modules 
(of geometric dimension at least 3) over a "auxiliary" ring which is WQF 
(weakly quasi Frobenius, a.k.a. IF, injectives are flat). 
The categories of finitely presented modules over such auxiliary
rings are exactly (up to equivalence) the abelian categories with an object which is
injective, projective and finitely generates and finitely cogenerates every object.
The ultimate generalization is G.Hutchinson's coordinatization theorem, 
a correspondence between arbitrary abelian categories and 
modular lattices with 0 where each element can be doubled
and intervals are projective (in lattice theory, sense i.e. the classical projective 
geometry meaning) to initial intervals. When one looks at this theorem together 
with the Freyd - Mitchell embedding theorem, one has that three languages are fully 
adequate and equivalent ways to do linear algebra: 
(1) the usual language of sums and products (modules over associative rings); 
(2) the language of category theory (abelian categories);
(3) the old fashioned language of synthetic geometry of incidence (joins and meets in 
suitable modular lattices).
In these equivalences, the lattices are the (pointless, noncommutative) spaces,
and the rings are the rings of coordinates or functions over the space (I am not
considering the distinction between equivalences and dual equivalences because
I am more interested in structures, with their unique concept of isomorphism attached, 
rather than more general morphisms, which depend upon the particular way to define a
structure. But the complementarity between the structural and categorical views, 
where neither subsumes the other, is another long theme).
Since all these ideas have their origin in von Neumann works about continuous geometries
and rings of operators, I now explain the relation with operator algebras.
First note that the usual definition of pointless topological space as complete
Heyting algebra is not a true generalization of the "topological space" concept:
 they are a true generalization of sober spaces,
but to generalize topological spaces one must consider pairs: a complete boolean 
algebra (which in the atomic case is the same thing as a set) with a complete 
Heyting subalgebra (the lattice of open sets). To obtain the non-commutative
analogue, complete boolean algebras are generalized to meet-continuous geometries,
and algebras of measurable sets are replaced with suitable structures (projection
ortholattices of von Neumann algebras) which are embedded in the meet-continuous
geometries in the same way as a right nonsingular ring is embedded in its maximal
ring of right fractions (a regular right self-injective ring).
A meet continuous geometry is a complete lattice which is modular, complemented 
and meet distubutes over increasing joins (not arbitray joins, like Heyting algebras).
These structures were introduced by von Neumann and Halperin in 1939; they are a common
generalization of (possibly reducible) continuous geometries (the subcase where 
join distrubutes over decreasing meets) and (possibly reducible and infinite dimensional)
 projective geometries (the atomic subcase). For them one has a dimension and decomposition
theory much like the one for continuous geometries and rings of operators (the theory of
S.Maeda, in its last version of 1961, is sufficient; one does not need the 2003 theory
by Wehrung and Goodearl). One can define the components of various types, in particular
I_1 (the boolean component, i.e. classical logic), the I_2 component (the 2-distibutive 
component i.e. subdirect product of projective lines; physically these are "spin
factors" and quantum-logically it is the non-classical component which nonetheless has
non-contextual hidden variables), the I_3 nonarguesian component (subdirect product 
of projective nonarguesian planes i.e. irreducible projective geometries that cannot 
be embedded in larger irreducible projective geometries; quantum-logically this means
that interacion with other components is only possible classically, without superposition).
Once these bad low dimensional components are disregarded, von Neumann coordinatization
theorem gives a equivalence between the meet-continuos geometries and the right 
self-injective von Neumann regular rings. So meet-continuous geometries are pointeless 
quantum (i.e. non-commutative) sets in the same way as complete Boolean algebras are 
(commutative) pointless sets. Regular rings are the coordinate rings of these quantum
sets in the same way as (commutative, regular) rings of step functions (with values in 
a field) are the ring equivalent of a boolean algebra (classical propositional logic);
the important new fact is that in the "truly non-commutative case" the regular ring
is uniquely and canonically determined by the lattice (in the distributive case, on the
contrary, it is not: one can use step functions with values in any field, and one can 
change the field with the point; commutative [resp. strongly] regular rings are the
subrings of direct products of [skew] fields which are stable for the generalized inverse
operation).
The above "propositional logics" are without the negation operator; on the other hand,
the projection ortholattices of von Neumann algebras are complete orthomodular lattices
with sufficiently many completely additive probability measures and sufficiently many
internal simmetries (von Neumann said that the strict logic of orthocomplementation 
and the probability logic of the states uniquely determine each other by means of his 
symmetry axioms in his characterization of finite factors as continuos geometries with 
a transition probability. One should also note how much more physically meaningful are
von Neumann axioms when compared with the "modern" ones based on Soler's theorem,
but this is another large topic). Using Gleason's theorem (and as always in absence of 
the bad low-dimensional components) one obtains an equivalence between von Neumann's
"rings of operators" (i.e. real von Neumann algebras, or their self-adjoint part,
real JBW-algebras) and their projection ortholattices (the normal measures on the
 ortholattice give the predual of the ring of operators). One can see these logics
inside a meet-continuos geometry by equipping the geometry with a linear orthogonality
relation which has for each element a maximum orthogonal element 
(pseudo-orthocomplementation in part analogous to "external" in a stonean topological 
space, in part anti-analogous as it happens with Lowere closure when compared to
Kuratowski closure). The regular ring of the lattice is the ring generated by all 
complementary pairs in the lattice (which are the idempotents of the ring: kernel and 
image) with the 
relations corresponding to the partial operation e+f-ef which is defined on idempotents
whenever fe=0 (at the lattice level this partial operation is implemented with disjoint
 join of the images and co-disjoint meet of the kernels); in the case associated to 
a "ring of operators", this regular ring is the ring of maximal right quotients of
the von Neumann algebra, and conversely the algebra is recovered from the lattice with 
orthogonality by taking the subring generated by orthogonal projections (idempotents 
whose kernel and image are orthogonal); by a theorem of Berberian the algebra is ring
generated by its self-adjoint idempotents, and there is clearly at most one involution
on the algebra which fixes such generators.
Hence, in summary, in absence of bad low dimensional components (whose exclusion 
is physically meaningful, see their meanings above) one has equivalences
between the following concepts:
(0) right self-injective regular rings with a suitable additional structure
    (to associate a orthogonal projection onto the closure of the image to any element)
(1) real von Neumann algebras
(2) real JBW-algebras (the Jordan algebra of self-adjoint operators i.e. observables)
(3) the effect algebra (of operators with spectrum in [0,1]) i.e. unsharp quantum logic
(4) the projection ortholattice (the sharp quantum logic)
(5) the pointeless quantum set (meet-continuous geometry) with a suitable orthogonality
(6) the convex compact set of normal states
Any of the above is a adequate starting point for quantum foundation since all
the other points of view can be canonically recovered. 
All this is restricted to the level of non-commutative measure spaces; a locally compact
topological space is something more precise (like a C^* algebra when compared to a 
von Neumann algebra), and a (Riemannian) metric space is something still more precise 
(and in particular a differentiable structure, which can be seen as an equivalence class
of riemannian structures: note that a isometry for the geodesic metric between complete 
Riemannian manifolds is automatically differentiable, so the differentiable structure 
must be definable from the geodesic metric, and infact Busemann and Menger had such a 
explicit definition of the tangent spaces from the global metric. The topological 
structure is then another equivalence class of metrics, for another weaker equivalence).
Given the equivalence (0)--(6) above, one can note that all the concepts which are used
in Connes definition of spectral triples, and analogues structures, can be seen 
equivalently from each of the above points of view. In this way, all of the above points
of view are a possible starting point for non-commutative geometry.
[Sorry for the too long post and for my bad pseudo-english language]<|endoftext|>
TITLE: Asymptotics of q-Catalan numbers
QUESTION [12 upvotes]: q-Catalan numbers are defined recurrently as C0=1, $C_{N+1}=\sum_{k=0}^N q^k C_k C_{N-k}$.
What can be said about the asymptotics of Cn when 0<q<1?
P.S. In the case q>1 it is known that as n goes to infinity, $q^{-{n\choose 2}}C_n(q)$ tends to the partition function $\prod_{i=1}^\infty\frac1{1-q^{-i}}$. However, this doesn't help in the case 0<q<1.

REPLY [3 votes]: Re Leonid's comment on a previous answer.
If the ratios $C_{n+1}/C_n$ converge, their limit $c(q)$ is such that $C(q,q/c(q))=c(q)$. 
Equivalently, $1/c(q)$ is the radius of convergence of the series $z\mapsto C(q,z)$. 
Or, writing $C(q,\cdot)$ as the ratio of two $q$-hypergeometric functions, one can show that $F(q,1/c(q))=0$, where 
$$
F(q,z)=\sum_{n\ge0}(-1)^nq^{n^2-n}z^n/(q)_n.
$$ 
This implies that $c(q)$ is the sum of a series in $q$ with integer coefficients, whose signs seem to be alternating starting with the coefficient of $q$. The first terms are
$$
c(q)=1+q+q^3-q^4+2q^5-3q^6+6q^7-12q^8+25q^9-52q^{10}+111q^{11}+\ldots
$$
The function $q\mapsto c(q)$ is nondecreasing on $q\ge0$, obvious values are $c(0)=1$ and $c(1)=4$, and as a holomorphic function, $c(\cdot)$ might have a pole inside the unit disk at about $q\approx-.4$. 
But apart from that...<|endoftext|>
TITLE: Sums of injective modules, products of projective modules?
QUESTION [22 upvotes]: Under what assumptions on a noncommutative ring R does a countable direct sum of injective left R-modules necessarily have a finite injective dimension?
Analogously, under what assumptions on R does a countable product of projective left R-modules necessarily have a finite projective dimension?

These questions arise in the study of the coderived and contraderived categories of (CDG-)modules, or, if one wishes, the homotopy categories of unbounded complexes of injective or projective modules.  
There are some obvious sufficient conditions and some less-so-obvious ones.  For both #1 and #2, it clearly suffices that R have a finite left homological dimension.
More interestingly, in both cases it suffices that R be left Gorenstein, i.e., such that the classes of left R-modules of finite projective dimension and left R-modules of finite injective dimension coincide.
For #1, it also suffices that R be left Noetherian.  For #2, it suffices that R be right coherent and such that any flat left module has a finite projective dimension.
Any other sufficient conditions?

REPLY [2 votes]: For #1, it suffices that $R$ be left coherent and such that any fp-injective left $R$-module has finite injective dimension.  In particular, these conditions hold when $R$ is left coherent and every left ideal in $R$ has a set of generators of the cardinality not exceeding $\aleph_n$ for some nonnegative integer $n$ (e.g., a countable set of generators).  See Section 2 in https://arxiv.org/abs/1504.00700 .<|endoftext|>
TITLE: Topological Langlands?
QUESTION [37 upvotes]: In a workshop about the geometry of $\mathbb{F}_1$ I attended recently, it came up a question related to a mysterious but "not-so-secret-anymore" seminar about... an hypothetical Topological Langlands Correspondence!
I had never heard about this program; I have found this page via Google:
http://www.math.jhu.edu/~asalch/toplang/
I only know a bit about the Number-Theoretic Langlands Program, and I still have a hard time trying to understand what is happening in the Geometrical one, so I cannot even start to draw a global picture out of the information dispersed in that site.
So, the questions are: What do you know (or what can you infere from the web) about the Topological Langlands Correspondence? Which is the global picture? What are its analogies with the (original) Langlands Program? Is it doable, or just a "little game" for now? What has been proved until now? What implications would it have?
(Note: It is somewhat difficult to tag this one, feel free to retag it if you have a better understanding of the subject than I have!)

REPLY [34 votes]: I would cautiously venture that there is not really something we could call a topological Langlands program to outsiders at this point.  My objection is to the final word - we don't really know what we're doing.  For example, I don't think we even have a conjecture at this point relating representations of something to something else, or have the right idea of L-functions that are supposed to play a role.
The "topological Langlands program" banner is more of an idea that the scattered pieces of number-theoretic content we see in stable homotopy theory should be part of a general framework.  There is the appearance of groups whose orders are denominators of Bernoulli numbers in stable homotopy groups of spheres, Andrew Salch's calculations at higher chromatic levels that seem to be related to special values of L-functions, the relationship between K(1)-local orientation theory and measures p-adically interpolating Bernoulli numbers (see Mark Behrens' website for some material on this), the surprisingly canonical appearance of realizations of Lubin-Tate formal group laws due to work of Goerss-Hopkins-Miller in homotopy theory, et cetera, et cetera.
Perhaps at this point I'd say that we have a "topological Langlands program" program, whose goal is to figure out why on earth all this arithmetic data is entering homotopy theory and what form the overarching structure takes in a manner similar to the Langlands program itself.<|endoftext|>
TITLE: How to select a journal?
QUESTION [43 upvotes]: What are good criteria for selecting a journal to submit a paper to?
One criterion per answer, please.
It is easy to group journals by subject and prestige, but is there a thought-process that you use to determine which journal is good for your specific paper?
For example, if you have a clearly top paper, it is not hard to find a journal that will accept it (the only question is which top journal to submit it to), but do you have favorite journals for specific types of projects, papers that are not earth-shattering, or other categories of papers?

REPLY [10 votes]: (This advice isn't my own idea, I am just passing along advice from senior mathematicians', I think I heard it from Bill Fulton.)
Before submitting to journal X, go to the library and look up journal X, and browse around. Would your article fit in there? (Do they publish articles in your field? Do they publish articles that are as technical as yours? Do they publish 95-page articles, such as yours? Etc.)<|endoftext|>
TITLE: Complex vector bundles that are not holomorphic
QUESTION [41 upvotes]: Is there an example of a complex bundle on $\mathbb CP^n$ or on a Fano variety (defined over complex numbers), that does not admit a holomorphic structure? We require that the Chern classes of the bundle are $(k,k)$ Hodge classes (which is automatic for $\mathbb CP^n$ or Fanos of dimension<4). If by any chance such examples are known, what is the smallest dimension of the variety (or the bundle)?
For $\mathbb CP^1$ it is elementary to see that all bundles are holomorphic. In the book of Okonek and Schneider it is stated, that all complex bundles on $\mathbb CP^2$ and $\mathbb CP^3$ are also holomorphic. But for $\mathbb CP^n$, $n\ge 4$ this is stated as an open problem (as for 1980).

REPLY [34 votes]: Here is the answer to the question, kindly explained to me by Burt Totaro. 
EDITED. This is an OPEN PROBLEM.
0) Apparently in the case of $\mathbb CP^n$ existence of a complex bundle without holomorphic structure is still an OPEN PROBLEM. Though it is believed that there should be plenty of examples starting from $n\ge 5$, coming from topologically indecomposable rank two bundles, apparently no such bundle was proven to be non-holomorphic as for today. 
1) A topologically non-trivial rank 2 complex bundle with $c_1=0$, $c_2=0$ was constructed in 
Rees, Elmer, Some rank two bundles on ${\rm P}_{n}\mathbb C$, whose Chern classes vanish. Variétés analytiques compactes (Colloq., Nice, 1977).
It was also claimed in this article that this bundle does not admit a holomorphic structure. But this claim was deduced from an article that contained a gap. So it is yet unknown if this particular bundle has holomorphic structure or not. This is discussed in 
M. Schneider. Holomorphic vector bundles on ${\rm P}^n$. Seminaire Bourbaki
1978/79, expose 530. 
This is why Okonek and Schneider write in their book p. 137 that this is an open problem.
2) On the positive side it is proven that every complex vector bundle on a smooth projective rational 3-fold has an holomorphic structure.
C. Banica and M. Putinar. On complex vector bundles on projective
threefolds. Invent. Math. 88 (1987), 427-438.
3) If one wants to construct examples of bundles on projective manifolds that are not necessarily Fanos it is possible to use the fact that the integral Hodge conjecture fails. Namely there are elements in $H^{2p}(X,\mathbb Z)$ which are in $H^{p,p}$
but which are not represented by an algebraic cycle. Kollar gave
such examples with $dim(X)=3$. A recent reference, which refers back to earlier results, is:
C. Soule and C. Voisin. Torsion cohomology classes and algebraic cycles
on complex projective manifolds. Adv. Math. 198 (2005), 107-127
4) One reason to expect examples of such bundles in higher dimensions is 
Schwarzenberger's conjecture that every rank-2 algebraic vector bundle $E$,
on $\mathbb CP^n$ with $n\ge 5$ is a direct sum of two line bundles. So, for example,
if $c_1(E)=0$ then $c_2(E)=-d^2$ for some integer $d$, according to the
conjecture.<|endoftext|>
TITLE: What does primary decomposition of (sub) modules mean geometrically?
QUESTION [17 upvotes]: I want to know how I should visualize modules in algebraic geometry.  The way we visualize rings, via their spectra, automatically (or by the beauty of its design) depicts primary decomposition of ideals: the primary components of an ideal $I \triangleleft A$ cut out "primary subschemes" (irreducible and embedded components) whose union is $Z(I)=Spec(A/I)$.  (See, for example, Eisenbud and Harris, The Geometry of Schemes, II.3.3, pp. 66-70).  This aspect of scheme theory is essential to what makes it "geometric."
By this standard, I think however we visualize modules should allow us to depict primary decomposition of submodules; otherwise I would say it's not a very good visualization.  
If we're happy taking quotients, WLOG we can just look at primary decompositions of $0$.  So let $M$ be a finitely generated module over a Noetherian ring $A$, and  $0=N_1\cap\cdots\cap N_n$ be a primary decomposition of $0$ in $M$, with primes $P_i$ co-associated to the primary modules $N_i$, i.e. associated to the coprimary modules $M/N_i$.

How can one visualize the modules $M,N_1,\ldots,N_n$ in relation to $Spec(A)$ in a way that meaningfully depicts:
  (1) the primary decomposition of $0$ in $M$ (in particular that the $N_i$ are primary in $M$), and
  (2) the relationship of the modules $N_i$ to their co-associated primes, say { $P_i$ } $ = Ass(M/N_i) \subseteq Spec(A)$?

Some useful background results to make sense of the above (all rings and modules are Noetherian):

The primes $P_i$ co-associated to $N_i$ are precisely the associated primes of $M$ (see R. Ash, Comutative Algebra, Theorem 1.3.9)
A module $Q$ is coprimary iff it has exactly one associated prime $P$, and then $P=\sqrt{ann Q}$.  (see R. Ash, Comutative Algebra, Corollary 1.3.11)

REPLY [3 votes]: Visualizing embedded primes:
In P^2,  a one dimensional scheme cannot have embedded points unless its ideal has more than one generator, by the unmixedness theorem of Macaulay.  So imagine we have two polynomials that define a one dimensional scheme in P^2.  We will imagine this scheme as a limit of zero dimensional schemes.  First take two quadratic polynomials, one of which is a product of two linear factors, i.e. take one pair of lines meeting at p, and another irreducible conic.  In general the irreducible conic C meets each of the lines twice, away from p.  Thus the two qudratic polynomials define a zero dimensional scheme of 4 points. 
Now hold fixed the two intersections of C with one of the lines L, and let the two intersections of C with the other line M approach p, i.e. let C become tangent to M at p.  When this occurs, the conic C now contains three distinct points of L, hence C has become reducible and contains L.  Now the scheme defined by intersecting L+M with C has become one dimensional, reducible, and consists set theoretically only of the line L.  I claim the point p is an embedded point of the component L of the scheme defined by L+M and C.
This is easy algebraically, since the ideal of the given scheme is (xy,(x(x-y)) = (x^2, xy) which is the intersection of the primary ideals (x) and (x^2, xy, y^2), with associated primes (x) and (x,y).  Hence (x,y) is an embedded prime.  I.e. the origin is an embedded point on the y axis for this scheme.  This also helps explain the apparent failure of Bezout's theorem for this intersection of two conics apparently not having degree 4.
In general, in P^n, a scheme S with embedded subschemes must be defined by intersecting more hypersurfaces than the codimension of S.  Thus such an S can always be viewed as a limit of lower dimensional schemes.  It seems to me that embedded subschemes should arise when these lower dimensional schemes are reducible and some lower dimensional component comes to lie on a larger dimensional component of the limit.  I do not know if this intuition is the only possibility, and since the world is wide, probably not.<|endoftext|>
TITLE: Etale cohomology and l-adic Tate modules
QUESTION [25 upvotes]: $\newcommand{\bb}{\mathbb}\DeclareMathOperator{\gal}{Gal}$
Before stating my question I should remark that I know almost nothing about etale cohomology - all that I know, I've gleaned from hearing off hand remarks and reading encyclopedia type articles. So I'm looking for an answer that will have some meaning to an etale cohomology naif. I welcome corrections to any evident misconceptions below.
Let $E/\bb Q$ be an elliptic curve the rational numbers $\bb Q$: then to $E/\bb Q$, for each prime $\ell$, we can associate a representation $\gal(\bar{\bb Q}/\bb Q) \to GL(2n, \bb Z_\ell)$ coming from the $\ell$-adic Tate module $T_\ell(E/\bb Q)$ of $E/\bb Q$ (that is, the inverse limit of the system of $\ell^k$ torsion points on $E$ as $k\to \infty$). People say that the etale cohomology group $H^1(E/\bb Q, \bb Z_\ell)$ is  dual to $T_\ell(E/\bb Q)$  (presumably as a $\bb Z_\ell$ module) and the action of $\gal(\bar{\bb Q}/\bb Q)$ on $H^1(E/\bb Q, \bb Z_\ell)$ is is the same as the action induced by the action of $\gal(\bar{\bb Q}/\bb Q)$ induced on $T_\ell(E/\bb Q)$.
Concerning this coincidence, I could imagine two possible situations:
(a) When one takes the definition of etale cohomology and carefully unpackages it, one sees that the coincidence described is tautological, present by definition.
(b) The definition of etale cohomology (in the case of an elliptic curve variety) and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries is conceptually different from that of the dual of the $\ell$-adic Tate module and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries. The coincidence is a theorem of some substance.
Is the situation closer to (a) or to (b)? 
Aside from the action $\gal(\bar{\bb Q}/\bb Q)$ on $T_\ell(E/\bb Q)$, are there other instances where one has a similarly "concrete" description of representation of etale cohomology groups of varieties over number fields and the actions of the absolute Galois group on them?
Though I haven't seen this stated explicitly, I imagine that one has the analogy [$\gal(\bar{\bb Q}/\bb Q)$ acts on $T_\ell(E/\bb Q)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(E/\bb Q; \bb Z_\ell)$]::[$\gal(\bar{\bb Q}/\bb Q)$ acts on $T_\ell(A/K)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(A/K; \bb Z_\ell)$] where $A$ is an abelian variety of dimension $n$ and $K$ is a number field: in asking the last question I am looking for something more substantively different and/or more general than this.
I've also inferred that if one has a projective curve $C/\bb Q$, then $H^1(C/\bb Q; \bb Z_\ell)$ is the same as $H^1(J/\bb Q; \bb Z_\ell)$ where $J/\bb Q$ is the Jacobian variety of $C$ and which, by my above inference I assume to be dual to $T_\ell(J/\bb Q)$, with the Galois actions passing through functorially. If this is the case, I'm looking for something more general or substantially different from this as well.
The underlying question that I have is: where (in concrete terms, not using a reference to etale cohomology as a black box) do Galois representations come from aside from torsion points on abelian varieties?

[Edit (12/09/12): A sharper, closely related question is as follows. Let $V/\bb Q$ be a (smooth) projective algebraic variety defined over $\bb Q$, and though it may not be necessary let's take $V/\bb Q$ to have good reduction at $p = 5$. Then $V/\bb Q$ is supposed to have an attached 5-adic Galois representation to it (via etale cohomology) and therefore has an attached (mod 5) Galois representation. If $V$ is an elliptic curve, this Galois representation has a number field $K/\bb Q$ attached to it given by adjoining to $\bb Q$ the coordinates of the 5-torsion points of $V$ under the group law, and one can in fact write down a polynomial over $\bb Q$ with splitting field $K$. The field $K/\bb Q$ is Galois and the representation $\gal(\bar{\bb Q}/\bb Q)\to GL(2, \bb F_5)$ comes from a representation $\gal(K/\bb Q) \to GL(2, \bb F_5)$. (I'm aware of the possibility that knowing $K$ does not suffice to recover the representation.)
Now, remove the restriction that $V/\bb Q$ is an elliptic curve, so that $V/\bb Q$ is again an arbitrary smooth projective algebraic variety defined over $\bb Q$. Does the (mod 5) Galois representation attached to $V/\bb Q$ have an associated number field $K/\bb Q$ analogous to the (mod 5) Galois representation attached to an elliptic curve does? If so, where does this number field come from? If $V/\bb Q$ is specified by explicit polynomial equations is it possible to write down a polynomial with splitting field $K/\bb Q$ explicitly? If so, is a detailed computation of this type worked out anywhere? 
I'm posting a bounty for a good answer to the questions succeeding the "Edit" heading.

REPLY [4 votes]: Regarding the Dec 9 edit: Yes, but you probably won't like it. Take the etale cohomology of V_{Q-bar} with coefficients in Z/5Z. Then this is a finite abelian group, and Gal(Q-bar/Q) acts continuously on it. Therefore the subgroup of Gal(Q-bar/Q) acting trivially is open, and hence corresponds to a finite Galois extension K of Q.
I don't know whether, given the defining equations of V, there exists an algorithm to find a polynomial whose splitting field is K. The only idea I'd have is to use fibration by curves to try to reduce the question to the etale cohomology of curves with coefficients in local systems, which can probably be calculated by using a Tate module-style approach on Jacobians. I think it's an interesting question, but I bet no one's ever looked at it.<|endoftext|>
TITLE: Heuristic explanation of why we lose projectives in sheaves.
QUESTION [34 upvotes]: We know that presheaves of any category have enough projectives and that sheaves do not, why is this, and how does it effect our thinking?
This question was asked(and I found it very helpful) but I was hoping to get a better understanding of why.
I was thinking about the following construction(given during a course);
given an affine cover, we normally study the quasi-coherent sheaves, but in fact we could study the presheaves in the following sense:
Given an affine cover of X,
$Ker_2\left(\pi\right)\rightrightarrows^{p_1}_{p_2} U\rightarrow X$
then we can define $X_1:=Cok\left(p_1,p_2\right)$, a presheaf, to obtain refinements in presheaves where we have enough projectives and the quasi-coherent sheaves coincide. Specifically, if  $X_1\xrightarrow{\varphi}X$ for a scheme $X$, s.t. $\mathcal{S}\left(\varphi\right)\in Isom$ for $\mathcal{S}(-)$ is the sheaffication functor, then for all affine covers $U_i\xrightarrow{u_i}X$ there exists a refinement $V_{ij}\xrightarrow{u_{ij}}U_i$ which factors through $\varphi$.
This hinges on the fact that $V_{ij}$ is representable and thus projective, a result of the fact that we are working with presheaves. In sheaves, we would lose these refinements. Additionally, these presheaves do not depend on the specific topology(at the cost of gluing).
In this setting, we lose projectives because we are applying the localization functor which is not exact(only right exact). However, I don't really understand this reason, and would like a more general answer. 
A related appearance of this loss is in homological algebra. Sheaves do not have enough projectives, so we cannot always get projective resolutions. They do have injective resolutions, and this is related to the use of cohomology of sheaves rather than homology of sheaves. In paticular, in Rotman's Homological Algebra pg 314, he gives a footnote;

In The Theory of Sheaves, Swan writes "...if the base space X is not discrete, I know 
  of no examples of projective sheaves except the zero sheaf." In Bredon, Sheaf Theory: 
  on locally connected Hausdorff spaces without isolated points, the only 
  projective sheaf is 0

addressing this situation.

In essence, my question is for a
  heuristic or geometric explanation of
  why we lose projectives when we pass
  from presheaves to sheaves.

Thanks in advance!

REPLY [4 votes]: We can turn the question around to ask: why do we have projectives in module categories? One answer is that we know we have a plentiful supply of projectives because free modules are projective. Abstractly, we have a forgetful functor $U : \text{Mod}(R) \to \text{Set}$ with left adjoint $F : \text{Set} \to \text{Mod}(R)$. Then we have the following two results:

By the axiom of choice, every set is projective (in the sense that homs out of it preserve epimorphisms).
If a functor $U$ with a left adjoint preserves epimorphisms, then its left adjoint $F$ preserves projectives.

Finally, it is straightforward to verify that $U$ in fact preserves epimorphisms (that is, epimorphisms of $R$-modules are surjective on underlying sets). 
Now what is the analogous situation for sheaves? We still have a forgetful functor $U : \text{Sh}(X) \to \text{Psh}(X)$, and it still has a left adjoint, namely sheafification. However, $U$ no longer preserves epimorphisms (this is exactly Dinakar's observation that an epimorphism of sheaves need not be an epimorphism of presheaves), so the above argument doesn't go through. 
For sheaves what we can instead do is the following. There is a different forgetful functor sending a sheaf on $X$ to its stalks; it can be thought of as pullback $p^{\ast}$ along the map $p : X_d \to X$ where $X_d$ denotes $X$ with the discrete topology. As a pullback, this functor has a right adjoint, namely pushforward $p_{\ast}$. The composite $p_{\ast} p^{\ast}$ is the Godement construction. In any case, a dual argument to the above shows that because pullback $p^{\ast}$ preserves monomorphisms, its right adjoint $p_{\ast}$ preserves injectives. So now instead of a plentiful supply of projectives we have a plentiful supply of injectives.<|endoftext|>
TITLE: When is an Albanese variety principally polarized?
QUESTION [12 upvotes]: Let (X,x) be a pointed projective variety. Then there exists an abelian variety V which is universal for maps of pointed varieties $(X,x) \to (A,e_A)$, called the albanese variety. When X is a curve, the variety V is isomorphic to the Jacobian of X (in higher dimensions this fails) which is a principally polarized abelian variety. 
Question: When is an albanese variety principally polarized? If it is not always principally polarized, can one describe the degree of the polarization in terms of data intrinsic to X?

REPLY [13 votes]: In general it could happen that the Albanese variety does not admit a principal polarization at all. For instance the Albanese variety of an abelian variety is the Abelian variety itself. So choose $X$ to be some abelian variety that has no principal polarization and you will get an example. 
On the other hand it can happen that the Albanese variety is principally polarized. For instance you can take the Albanese of the $n$-th symmetric product of a curve. It is equal to the Jacobian of the curve and so admits a principal polarization. Or if you want to be fancier you can take a hyperplane section in the symmetric product of a curve. It will also have the Jacobian of the curve as its Albanese variety. 
Another useful comment is that the Albanese of $X$ is the dual of $Pic^{0}(X)$ and so $Alb(X)$ admits a principal polarization if and only if $Pic^{0}(X)$ does. If you fix an ample line bundle $L$ on an $n$-dimensional complex projective variety $X$, then $L$ induces a natural polarization on $Pic^{0}(X)$: the universal cover of $Pic^{0}(X)$ is naturally identified with $H^{1}(X,O_{X}) = H^{1,0}(X)$, the integral $(1,1)$ form $c_{1}(L)$ then induces a Hermitian pairing on $H^{1,0}(X)$ by the formula 
$$
h(\alpha,\beta) := -2i \int_{X} \alpha\wedge \bar{\beta} \wedge c_{1}(L)^{\wedge (n-1)}.
$$
This $h$ defines a polarization on $Pic^{0}(X)$. The construction of $h$  is purely cohomological and so it is straightforward to check if it defines a principal polarization by computing the divisors of this polarization.<|endoftext|>
TITLE: Is Dependent Choice all we really need?
QUESTION [21 upvotes]: http://en.wikipedia.org/wiki/Axiom_of_dependent_choice
Is DC sufficient for the understanding of objects that are countable in some suitable sense? 
For example, is DC sufficient for the full development of the theory of von Neumann algebras on a separable Hilbert space?

REPLY [23 votes]: Let me adopt an extreme interpretation of your question, in order to prove an affirmative answer. 
Yes, in the arena of the countable, DC suffices. 
To see why, let me first explain what I mean. If one wants to consider only countable sets, then the natural set-theoretic context is HC, the class of hereditarily countable sets. These are the sets that are countable, and all members are countable, and members-of-members and so on. The class HC is the land of all-countable set theory, the land of the fundamentally countable. I am not proposing that you want to remain within HC. Rather, you want to consider the properties of the objects in HC that are expressable there and what you can prove about them in ZF+DC, versus ZFC.
Since hereditarily countable sets can be coded easily by real numbers, it turns out that the structure HC is mutually interpreted in the usual structure of the reals R and vice versa.  That is, the structure HC is essentially equivalent in a highly concrete way to the usual structure of the real numbers. And so questions about the fundamentally countable are equivalent to questions in the usual projective hierarchy of descriptive set theory. Thus, in this interpretation, your question is really asking whether one needs ever needs AC as opposed to DC in order to prove a projective statement. 
Question. Are the projective consequences of ZFC the same as those of ZF+DC?
The answer is Yes, by the following theorem.
Theorem. A projective statement is provable in ZFC if and only if it is provable in ZF+DC.
Proof. It suffices to show that for every model W of ZF+DC there is a model of ZFC with the same reals. The reason this suffices is that in this case, any projective statement failing in a model of ZF+DC will also fail in a model of ZFC. 
So, suppose that W is a model of ZF+DC. Let L(R) be the inner model of W obtained by constructing relative to reals. This is also a model of ZF+DC. Consider the forcing notion P in L(R) consisting of well-ordered countable sequences of real numbers in L(R), ordered by end-extension. Let G subset P be L(R)-generic for P. Since P is countably closed in L(R), it follows using DC that the forcing extension L(R)[G] has no additional real numbers. (The usual argument that countably closed forcing adds no new $\omega$-sequences makes essential use of DC, and indeed, DC is actually equivalent to that assertion.) And since G adds a well-ordering of these reals in order type $\omega_1$, it follows that L(R)[G] has a well-ordering of the reals. From this, it follows that L(R)[G] is actually a model of ZFC, since it has the form L[A], where A is a subset of $\omega_1$ enumerating these reals in the order that G lists them. Thus, we have provided a model of ZFC, namely L(R)[G], with the same reals as W. It follows that W and L(R)[G] have the same projective truth, since this truth is obtained by quantifying only over this common set of reals. QED
This phenomenon generalizes by the same argument beyond projective statements, to statements of the form "L(R) satisfies $\phi$". The point is that ZFC and ZF+DC prove exactly the same truths for L(R), since any model W of ZF+DC has the same L(R) as the model L(R)[G] in the proof above, and this satisfies ZFC.
The conclusion is that if you wish to study mathematical objects that are essentially countable only in the weak sense that they exist in L(R) — and this includes many highly uncountable objects — then the features about them in L(R) that you can prove in ZFC are exactly the same as the features you can prove in ZF+DC. So this interpretation is not so extreme after all...<|endoftext|>
TITLE: photon propagator
QUESTION [6 upvotes]: I am reading Zee's book "QFT in a nutshell". I have a question on the photon propagator computation. For a massive photon, consider the Lagrangian
 $L = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2}m^2A_\mu A^\mu + A_\mu J^\mu$, then the path integral is $Z = \int dx ~L = \int dx ~\{ \frac{1}{2}A_\mu[(\partial^2 + m^2)g^{\mu \nu} - \partial^\mu \partial^\nu]A_\nu +  A_\mu J^\mu \}$. From this we get that the photon propagator $D_{\mu \nu}$ satisfies $[(\partial^2 + m^2)g^{\mu \nu} -\partial^\mu \partial^\nu ] D_{\nu \lambda}(x) = \delta^\mu_\lambda \delta^{(4)}(x)$, and solving this, 
$$D_{\nu \lambda}(k) = \frac{-g_{\nu \lambda} + k_\nu k_\lambda/m^2}{k^2 - m^2}.$$ 
I can not see why the numerator has a term  $ k_\nu k_\lambda/m^2$. Any ideas?

REPLY [3 votes]: Theo and David are perfectly correct. To add a bit more of a physical explanation which
might help with the why part, a massive spin one particle has 3 physical degrees of freedom
so there must be some condition on the four components $A_\mu$. The equation of motion
for $A_\mu$ is equivalent to saying that each component of $A_\mu $ satisfies the massive
Klein-Gordon equation and that in addition $\partial^\mu A_\mu=0$. This latter condition in
momentum space implies that $k^\mu D_{\mu \nu}(k)=0$. So one can understand the $1/(k^2-m^2)$ from each component obeying the massive KG equation and the factor in the numerator as ensuring that $k^\mu D_{\mu \nu}(k)=0$.<|endoftext|>
TITLE: Cyclic spaces and S^1-equivariant homotopy theory
QUESTION [10 upvotes]: I'm trying to understand the relationship between cyclic spaces and S1-equivariant homotopy theory.  More precisely, I only care about S1-spaces up to equivalence of fixed point spaces for the finite subgroups of S1.  Given a cyclic space X : ΔCop → Top, I know the geometric realization of the restriction of X to Δop is an S1-space.  Form the associated fixed point diagram Oop → Spaces where O is the full subcategory of the orbit category of S1 on the objects S1/C where C ranges over finite subgroups of S1.  I regard the category of functors Oop → Spaces as an (∞,1)-category.
My question is, what structure on X does the resulting diagram depend on?  More specifically, under what conditions does a map f : X → Y of cyclic spaces induce an equivalence of fixed point diagrams?
In O consider the full subcategory O1 on the object S1/{•}.  The restriction of this diagram to O1 is a space with S1-action in the (∞,1)-categorical sense, and I think it's just the left Kan extension of X along the functor ΔCop → BS1 induced by the fact that ΔC is the quotient of something (ΔZ) by an S1-action.  Thus it only depends on X viewed as a functor from ΔCop to the (∞,1)-category of spaces.  But to evaluate on the other objects of O, corresponding to the fixed point spaces of nontrivial finite subgroups of S1, do I need to know each X[r] as a Cr+1 space (i.e. the homotopy types of the fixed points sets for subgroups of Cr+1)?  Is there a way to encode all of that information in a functor from some (maybe (∞,1)-)category to Spaces?  Or is it possible that I need to remember even more information about X?
Edit: I guess another way to phrase the question is this: I'm looking for a model category structure on the category of functors ΔCop → Top, such that the identity functor to the injective model structure is a left Quillen functor, and such that the geometric realization to genuine S1-spaces is also a left Quillen functor.  Furthermore I would like to know whether this model category structure is Quillen equivalent to a diagram category of spaces (possibly on a topological index category) with objectwise weak equivalences.

REPLY [7 votes]: I don't know if this is exactly what you're looking for (and there's a good chance you already know what I'm going to write) but let me give it a try:
The realization functor of cyclic sets (not spaces!) to $S^1$-spaces can be made part of a Quillen equivalence for two of the three commonly desired model structures on $S^1$-spaces: The model structure that gives you "Spaces over $BS^1$" is given in a 1985 paper of Dwyer-Hopkins-Kan, while a model structure that gives you the equivalences that you want (i.e., checked on fixed sets for finite subgroups) is given in a 1995 paper "Strong homotopy theory of cyclic sets" by Jan Spalinksi.
(Irrelevant to your question, but along the same lines: A recent paper of Andrew Blumberg describes how one can throw in some extra--still combinatorial--data and obtain a combinatorial model of the third desirable model structure on $S^1$-spaces, namely where equivalences are those that induce equivalences on fixed sets for all closed subgroups.)
Spalinksi's model structure depends on the following construction of $|X_{.}|^{C_n}$ (as a space-over-$BS^1$) in terms of the subdivision construction: The simplicial set $(sd_r X)_n = X_{r(n+1)-1}$ has an action of $C_r$--since $C_r$ is a subgroup of the copy of $C_{r(n+1)}$ acting on $X_{r(n+1)-1}$; taking fixed points (in sSet) and then realizing gives  $|X_{.}|^{C_n}$.
This suggests (though I haven't checked too carefully) that remembering each $X_n$ as a $C_{n+1}$-space (in the sense you suggest, with subgroups) is enough, as you expected.  

Now begins the speculative (and probably wrong) part of this answer: I have nothing too certain to say about writing this as a functor category, but it doesn't seem too unreasonable (to me, right now, at least) based on the above simplicial subdivision construction that we might be able to construct a reasonable candidate: some sort of mix of the cyclic category and the orbit categories for the cyclic groups.  Purely combinatorially, this seems to get tricky.
But, I think we can realize this geometrically:  Let $(S^1)_r$ be the circle equipped with a $\mathbb{Z}$ action given by the rotation by $2\pi/r$.  We could try to define $Hom'([m-1]_r, [m'-1]_{r'})$ along the lines of "(htpy classes of) degree $r'/r$, increasing $\mathbb{Z}$-equivariant maps $S^1 \to S^1$ sending the $mr$-torsion points to the $m' r'$-torsion points".  This should correspond to taking all the $r$-cyclic categories and sticking them together, and in particular is bigger than what we want.  But, the $\mathbb{Z}$-action on the circles should induce one on the $Hom'$-sets and the composition should respect it. Taking the quotient, we seem to get something that looks like a reasonable candidate.  For each fixed $r$, we should be getting a copy of the cyclic category.  And, e.g. $Hom([m-1]_r, [mr-1]_1)$ should contain $Hom_{orbit}(Z/mr, Z/r)$. (Disclaimer: It's late and I haven't checked any of this too carefully!)<|endoftext|>
TITLE: What is $TC(\Sigma^\infty \Omega X)$?
QUESTION [5 upvotes]: I know that for $X$ a connected space, $THH(\Sigma^\infty \Omega X) = \Sigma^\infty \Lambda X$, the suspension spectrum of the free loop space of $X$.  The computation can be carried out in spaces and then transferred to spectra via $\Sigma^\infty$.  What is $TC(\Sigma^\infty \Omega X)$?  Can it also be computed from some kind of $TC$ on the level of spaces?
Edit: Tyler answered my question, but I want to ask a followup question: Is it fair to say that $TC(\Omega X)$ in the world of spaces, after $p$-completion, is just $X$, and is there a map (not an equivalence, because we take limits to build $TC$) $\Sigma^\infty X \to $ the thing Tyler wrote down?  (Note: all my $\Sigma^\infty$ are $\Sigma^\infty_+$.)

REPLY [5 votes]: The TC spectrum, at a prime $p$, of this is the homotopy pullback of a diagram
$S^1 \wedge (\Sigma^\infty_+ \Lambda X)_{hS^1} \to \Sigma^\infty_+ \Lambda X \leftarrow \Sigma^\infty_+ \Lambda X$
after $p$-completion.  Here the left-hand map is the $S^1$-transfer from homotopy orbits back to the spectrum and the right-hand map is the difference between the identity and the "$p$'th power" maps on the loop space.
This is in Bökstedt-Hsiang-Madsen's original paper defining topological cyclic homology, in section 5.
ADDED LATER: This doesn't really work on the space level, because they don't have all the structure necessary.  They have the $F$ maps, but not the $R$ ones which only come about from stable considerations.  Spaces with a group action really only have one notion of "fixed points," namely the honest fixed points of the group action.
However, the associated equivariant spectrum of $\Lambda X$ is built out of spaces like
$$\Omega^V \Sigma^V \Lambda X = Map(S^V, S^V \wedge \Lambda X_+)$$
where $V$ ranges over representations of $S^1$.  This has two "fixed-point" objects for any cyclic group $C$: there's the fixed points, which is the space
$$Map^C(S^V, S^V \wedge \Lambda X_+)$$
of equivariant maps.  There is also the collection of maps-on-fixed-points
$$Map((S^V)^C, (S^V \wedge \Lambda X_+)^C)$$
which is called the "geometric" fixed point object, and it accepts a map from the ordinary fixed points.  The fact that $(\Lambda X)^C \cong \Lambda X$ implies that you can interpret this as a map $(Q \Lambda X)^C \to (Q \Lambda X)$ where the latter uses an accelerated circle.  These maps give rise to the $R$ maps in the definition of $TC$, and they definitely rely on the fact that you're considering the associated spectra.<|endoftext|>
TITLE: Geometry Vs Arithmetic of schemes
QUESTION [11 upvotes]: Let's suppose we have a Scheme $X$ over the the field $k$, where such a field can be though to be either $\mathbb{C}$ or a finite field $\mathbb{F}_q$. Then having this in mind, 
Where do we find some representative examples where Geometry governs arithmetic? That is to say, examples where the geometry (or topology) of $X$ over $\mathbb{C}$ dictates the arithmetic behavior over $\mathbb{F}_q$.
Answers along with references would be highly appreciated.

REPLY [2 votes]: Look at Darmon's article on "Arithmetic of Curves".<|endoftext|>
TITLE: The Jouanolou trick
QUESTION [30 upvotes]: In Une suite exacte de Mayer-Vietoris en K-théorie algébrique (1972) Jouanolou proves that for any quasi-projective variety $X$ there is an affine variety $Y$ which maps surjectively to $X$ with fibers being affine spaces. This was used e.g. by D. Arapura to (re)prove that the Leray spectral sequence of any morphism of quasi-projective varieties is equipped from the second term on with a natural mixed Hodge structure.
Here is a proof when $X$ is $\mathbf{P}^n$ over a field $k$: take $Y$ to be the affine variety formed by all $n+1 \times n+1$ matrices which are idempotent and have rank 1. This is indeed affine since it is given by the equations $A^2=A$, the characteristic polynomial of $A$ is $x^n(x-1)$. Moreover, $Y$ is mapped to $\mathbf{P}^n(k)$ by taking a matrix to its image. The preimage of a point of $\mathbf{P}^n(k)$ is "the set of all hyperplanes not containing a given line", which is isomorphic to an affine space.
The general (quasi-projective) case follows easily from the above. However, it is not clear how to generalize Jouanolou's trick for arbitrary varieties. Nor is it clear (to me) that this is impossible.

Is there an analogue of the Jouanolou lemma for arbitrary (not necessarily quasi-projective) varieties (i.e. reduced separated schemes of finite type over say an algebraically closed field)?
(weaker version of 1 over complex numbers) Is there, given a complex algebraic variety $X$, an affine variety $Y$ that maps surjectively to $X$ and such that all fibers are contractible in the complex topology? A negative answer would be especially interesting.
(the following question is a bit vague, but if it has a reasonable answer, then it would probably imply a positive answer to 2.) Is there a quasi-projective analog of the topological join of two projective spaces? I.e., if $P_1$ and $P_2$ are two complex projective spaces, is there a quasi-projective variety $X$ which "contains the disjoint union of $P_1$ and $P_2$ and is formed by all affine lines joining a point in $P_1$ with a point in $P_2$"?

Edit 1: in 1. and 2. the varieties are required to be connected (meaning that the set of closed points is connected in the Zariski topology; in 2 one could use the complex topology instead).
Edit 2: as Vanya Cheltsov explained to me, the answer to question 3 is most likely no.

REPLY [13 votes]: Jouanolou's trick has been extended to schemes with an "ample family of line bundles" by Thomason; see Weibel: Homotopy Algebraic K-theory, Proposition 4.4. This includes all smooth varieties and more generally all varieties with torsion local class groups. However, there exist (positive dimensional) proper varieties with no non-trivial line bundles on them; it seems possible that on such varieties there are no affine bundles with affine total space.<|endoftext|>
TITLE: Explanation for the Thom-Pontryagin construction (and its generalisations)
QUESTION [38 upvotes]: In 1950, Pontryagin showed that the n-th framed cobordism group of smooth manifolds was equal to n-th stable homotopy group of spheres:
$$ \lim_{k \to \infty} \pi_{n+k}(S^k) \cong \Omega_n^{\text{framed}}.$$
Later on, in his 1954 paper, Thom generalises this with the now called Thom spaces, and shows that there is a similar correspondence for more general types of cobordism: manifolds with a $(B,f)$ structure on their normal bundle; for example, unoriented cobordism for $B = BO$, oriented cobordism for $B = BSO$, complex cobordism for $B = BU$, framed cobordism for $B = BI$ for the identity $I$ in $O$, etc. (Thom considers the cases $BO$ and $BSO$.)
The generalisation he arrived to, now called the Thom-Pontryagin construction, is the following:
$$\lim_{k \to \infty}\pi_{n+k}(TB_k) \cong \Omega_n^{(B,f)}, $$
where $TB$ is the Thom space of the universal bundle over $B$ given by the classifying map $B \to BO$; $TB$ is obtained by adding a point at infinity to each fiber and gluing all these added points to a single point in the total space of the bundle. 
In fact, the result can be generalised further by considering cobordism as a homology theory, and one arrives at the following:
$$\Omega_n^{(B,f)}(X,Y) \cong \lim_{k \to \infty} \pi_{n+k}(X/Y \wedge TB_k),$$
where, if $Y$ is empty, $X/Y$ is the disjoint union of $X$ with a point (and $\wedge$ is the smash product). Here $\Omega_n(X,\emptyset)$ is to be understood as a relative cobordism over $X$. This clearly generalises the previous result by taking $X$ to be a point (and $Y$ empty).
Now, my question is, how do you understand the Thom-Pontryagin construction? I've seen a few mentions of a particularly visual way of understanding it, but without much to actually back this up (besides from, I remember, a few mentions of blobs of ink). The standard proofs (in Stong's Notes on Cobordism Theory or in Thom's original paper for example) are quite long and I have trouble keeping hold of my geometric intuition throughout.

REPLY [13 votes]: This is an interesting question, and Greg's answer is excellent. Thinking about this was very nice!
Regarding the Pontryagin-Thom construction, as opposed to the Pontryagin construction which other comments and answers have addressed, I think Thom's original approach, whose goal was to solve Steenrod's realization problem, was quite geometric if you think of it the right way (see this question and this question, where people ask Steenrod's realization problem as an MO question).
Going one way, our initial data is a bordism class, i.e. a closed manifold M mapping into a polyhedron X, modulo cobordism. Steenrod's realization problem, by the way, is to determine which homology classes in X are realized this way for some M. Well, what Thom does first is to embed X in some $\mathbb{R}^n$ for n large enough, and take a regular neighbourhood N of X, which, as Greg points out, is diffeomorphic to the normal bundle of X. You now collapse the boundary of N to a point, and the quotient $N/\partial N$ to the Thom space of the regular neighbourhood (normal bundle) of the image of M in X. This is a map into a Thom space, so you can compose with a map into the universal Thom space. As I understand it, the point of this whole exercise is to leverage the nice properties of Euclidean space- namely the existence of the universal Thom space. So taking stock, starting with a bordism class, I have given you a map from $N/\partial N$ to the universal Thom space, whose image is a homotopy class of the right dimension. This is totally geometric- you can "see" the class as the image of $N/\partial N$ in the universal Thom space, where everything outside the image of M is mapped to the basepoint. Thom proved that this is well-defined, and that it's a surjection.
To go the other way is even easier. Starting from a class in the universal Thom space, you realize it as a map g from $N/\partial N$ to the universal Thom space, and apply the Thom isomorphism. What is the Thom isomorphism? It's the intersection of the (relative) cycle in the Thom space of $N/\partial N$ corresponding to g with the zero section (algebraically, you're multiplying by the Thom class), which gives you a closed manifold $M^\prime$ in N. Retract N to X, and you are done, modulo transversality and other technical issues.<|endoftext|>
TITLE: What are the most overloaded words in mathematics?
QUESTION [41 upvotes]: This is community wiki.  In each answer, please list one word at the top and below that list as many different meanings of that word in mathematics as you can think of, preferably with links or definitions.  ("Adjective" and "adjective noun" count as the same adjective.)  People should edit previous answers as appropriate.
(This is mostly just for fun, but I'm also curious if there have been successful attempts to rename concepts that involve overused words.)
Edit:  I may have been slightly unclear about the intent of this question.

When I say "overused" I don't mean "used too often," I mean "used in too many different ways."  So I'll change the title of the question to reflect this.
Different concepts named after the same mathematician, while potentially confusing, are understandable.  
I mostly had in mind adjectives that get recycled in different disciplines of mathematics.  Different uses of the same noun tend to be less confusing, e.g. the example of "space" below.  I think it's good to be intentionally vague about what we consider a "space."

REPLY [22 votes]: If you happen to work on del Pezzo surfaces, don't make the mistake of standing in an airport security line talking about "blowing up a plane at eight points".
(Yes, this really happened, and ended happily, or at least not in Guantanamo.)<|endoftext|>
TITLE: What is a projective space?
QUESTION [15 upvotes]: Is there a "recognition principle" for projective spaces?
What categories are there with projective spaces for objects?

Background: Although the title is a nod to What is a metric space?, this question is really built on top of my questions about Hilbert spaces (What is an intuitive view of adjoints? (version 2: functional analysis), Can one do without Riesz Representation?, Linearity of the inner product using the parallelogram law).  The answers to these questions have set me pondering about Hilbert spaces.  In short, I'm wondering whether or not it is possible to change the slogan of Hilbert spaces from "Hilbert spaces are great because they are self-dual" to "Hilbert spaces are great because you don't have to mention the dual".  So I'm looking at ways to remove any explicit mention of the dual from the basic constructions in Hilbert space theory.  My current favourite replacement is with statements about complementary subspaces.
This leads to projective spaces since they are the first non-trivial family of subspaces of some Hilbert space.  The subspace analogue of Riesz representation says that there is an isomorphism between $\mathbb{P}H$ and $\operatorname{Gr}_{\infty -1}(H)$ (the space of codimension one subspaces).  I want to be able to treat the latter space as if it were a projective space in its own right and say that this is an "isomorphism of projective spaces".  However, I don't want to explicitly construct a vector space of which $\operatorname{Gr}_{\infty - 1}(H)$ is the projective space since that vector space would be $H^*$.  So I want a recognition principle for projective spaces and a suitable notion of "isomorphism".
An example of what the recognition principle might be like would be an expression of the structure of a projective space as an algebraic theory, perhaps a many-sorted one.  That would also make it obvious what the morphisms were.
Obviously there are many versions of what the morphisms could be.  What I would like is for the obvious functor from Hilbert spaces to projective spaces to be full and faithful but not by construction.  That is, I don't simply want to take the category of Hilbert spaces and replace each Hilbert space by it's projective space because then to work out the morphisms between two projective spaces I first need to choose representing Hilbert spaces and I don't want to do that.
To give this a connection to something a little deeper than just a different way of working with Hilbert spaces, recall that in twisted K-theory the starting place is a bundle of projective spaces that cannot be globally lifted to a bundle of Hilbert spaces.  Nonetheless, some constructions related to a Hilbert space still work on such a bundle because they don't depend on the actual choice of Hilbert space so you can make a local choice and then prove it independent of that choice (for example, the space of Fredholm operators).  Is there a way to go directly to that construction without making local choices?

REPLY [2 votes]: The book "Modern Projective Geometry" has a system of axioms for projective spaces in a set $X$ using a function $f\colon X\times X\to P(X)$ (see page 30). Also there's a lot of projective geometry that can be done in the context of lattices. I thing this is related to Greg's answer through Von Neumann's "Continuous geometry". (see here). Another book to look might be Baer's classic "Linear algebra and projective geometry". 
On Greg's comment to Qiaochu's comment: let's not forget  that for every projective space $X$ we can always find a division ring $D$ such that every $X=P^n(X)$ for some n (in the the plane case you need that $X$ be a Desargue's plane).See this. (But, as far as I know,  these are all finite dimensional results so they might not be that interesting to Andrew...)<|endoftext|>
TITLE: What is a monoidal metric space?
QUESTION [14 upvotes]: At time of writing, the highest rated answer to my question What is a metric space? is Tom Leinster's account of Lawvere's description of a metric space as an enriched category.  This prompted my question on terminology in category theory.  That question was focussed on terms in category theory that were previously in use elsewhere.  It now occurs to me that there's an obvious question in the other direction: 
Given that metric spaces are enriched categories, what do the standard categorical things look like?
Obvious ones are adjoint functors (which might help me get a picture of what adjoint functors really are, answering this question), monoidal structures (and symmetric monoidal structures), products and coproducts (more generally, limits and colimits), but I'm sure that there are many more "out there" and I don't want to limit the answers.
To forestall Urs Schrieber's likely first comment, I intend to stick all of this on an n-lab page sometime soon as, if my intuition is right, I think it might be a neat case study that can help topologists like me get a picture of how categories (and enriched categories) can behave.

REPLY [5 votes]: Here are a couple of answers to your wider question.
1) Tom Leinster defined the notion of Euler characteristic for a finite categories, generalizing things like cardinality of sets, Euler characteristic of posets and Euler characteristic of finite groups.  This can be generalized to enriched categories and specialized to metric spaces, giving rise to an (occasionally undefined) invariant of metric spaces called the magnitude.  (The names cardinality and Euler characteristic were deemed to be too confusing.)  Interestingly, this was discovered in the nineties by some ecologists interested in measuring biodiversity.  See our paper [http://arxiv.org/abs/0908.1582](On the asymptotic magnitude of subsets of Euclidean space) for more details.
2) Given an endofunctor F:*C*->C of a category enriched over V there are two ways of taking the 'trace' of F that I know of, both leading to an object of V.  One is the end $\int_c C(c,F(c))$ and the other is the coend $\int^c C(c,F(c))$.  In the context of metric spaces this means that for a distance non-increasing function f:*X*->X the two traces are supxd(x,f(x))
and infxd(x,f(x)) - which can be thought of as the furthest distance that f moves points and the least distance that f moves points.<|endoftext|>
TITLE: Tropical mathematics and enriched category theory
QUESTION [16 upvotes]: Is there a connection between tropical mathematics and the Lawvere enriched category theory approach to metric spaces?  I guess I will give a partial answer to this below, but I mean can they be formally be put on the same level in some sense?
In the Lawverian point of view one does category theory with the extended non-negative real numbers, [0,∞] or R≥0∪∞, equipped with + as the 'tensor' product and max as the 'categorical' product or sum.  In tropical mathematics you work (it seems) with the the extended reals R∪∞ equipped with the 'product' + and the 'sum' max (or min depending on your point of view I think).
In the enriched category theory approach to metric spaces, one has the notion of a kernel (or bimodule or profunctor depending on your point of view) between two metric spaces X and Y which is just a distance non-increasing function K:X×Y->[0,∞].  The correct notion of function on a metric space here is a distance non-increasing function φ:X->[0,∞].  Then the transform of a function φ by a kernel K is a function on Y defined by

K(φ)(y):= infxεX  ( φ(x) + K(x,y) ).

There is similarly a dual notion which takes functions on Y to functions on X.

K^(ψ)(x):= supyεY   ( ψ(y) - K(x,y) ).

This is explained in a bit more detail in a post in at the n-Category Café.
It was pointed out to me that these look similar to the Legendre transform.  And looking on the internet I found that tropical mathematics is one way to interpret the Legendre transform as an 'integral transform'.
So has anyone ever considered any formal connections between these two points of view?

REPLY [4 votes]: Perhaps you know that over the years we've had many discussions about matrix mechanics at the Cafe. Depending on the rig (ring without negatives) used, you end up with a different form of mechanics - classical, quantum, statistical - where you multiply along paths and the sum over paths. We even took a look at homotopy theory as a truth-valued mechanics, e.g., for path connectedness between two points sum (in this case OR, so does there exist) paths for which the product (in this case AND) of truth values for points on the path being in the space. I'd hoped we'd do something interesting transforming the different mechanics using rig morphisms. So we'd know, say, that if a classical particle could move from x to y, that x and y were path connected.
In your question, you're looking beyond matrices corresponding to a single space to rig-valued kernels on two spaces X and Y. In the case of the Legendre transform the choice is for related spaces, e.g., a vector space and its dual. This transform is often a good way to form a corresponding (possibly easier to solve) dual optimisation problem. Interesting, though, to think of more general pairs of spaces.<|endoftext|>
TITLE: Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension
QUESTION [14 upvotes]: Let $(A,m_A)$ and $(B,m_B)$ be noetherian local rings and $f:A\rightarrow B$ a local homomorphism.  Let $F = B/m_AB$ be the fiber ring and assume that
$$\mathrm{dim}(B) = \mathrm{dim}(A) + \mathrm{dim}(F).$$
The following Theorem (23.1 in Matsumura's CRT) is really quite a miracle:
Theorem: If $A$ is regular and $B$ is Cohen-Macaulay then $f$ is flat.
I am wondering to what extent this theorem can be generalized.  What I have in mind is a statement of the type:
"Theorem":  If $A$ is $X$ and $B$ is $Y$ then $f$ is of finite Tor-dimension
(i.e. $\mathrm{Tor}^i_A(B,A)=0$ for all $i$ sufficiently large).
Here, $X$ and $Y$ are ring-theoretic conditions which should be strictly weaker than 
"regular" and "CM" respectively.  Is the "Theorem" above true just requiring $A$ and $B$ to be normal?
How about both CM? Or maybe CM plus finitely many $(R_i)$?  
Any thoughts/ counterexamples?

REPLY [10 votes]: I post this answer to give some intuition about what is really happening behind the scene in the theorem mentioned. If $f:A\rightarrow B$ is flat, then obviously the image of any $A$-regular sequence under $f$ is a $B$-regular sequence. This can be seen by tensoring the $A$-Koszul complex on an $A$-regular sequence, by $B$.
Now let's ask this question: Suppose a map $f:A\rightarrow B$ has the property that it maps any $A$-regular sequence to a $B$-regular sequence. Is $f$ flat then? The answer is no. As an example, you can consider the Frobenius endomorphism $F:A\rightarrow A$ of a local ring of characteristic $p>0$. Obviously it maps every regular sequence to a regular sequence, but $F$ is not flat, unless $A$ is regular, by a theorem of Kunz. Another example is any endomorphism $f$ of a local Cohen-Macaulay ring $(A,\mathfrak{m}_A)$ for which $f(\mathfrak{m}_A)A$ is $\mathfrak{m}_A$-primary. One can see quickly that the image of any regular sequence is a regular sequence, but in general $f$ need not be flat.
The reason for this failure is existence of modules of infinite projective dimension. The condition that $f$ sends any regular sequence to a regular sequence only guarantees that finite free resolutions of $A$-modules stay exact after tensoring by $B$. This quickly follows from Buhsbaum-Eisenbud exactness criterion. (cf. p. 37, Corollary 6.6 in Topics in the homological theory of modules over commutative rings, M. Hochster.)
When $A$ is regular, however, every finite $A$-module has finite projective dimension. That's why in this case the condition that every $A$-regular sequence will be mapped to a $B$ regular sequence by $f$ is equivalent to flatness! (keep in mind that flatness only needs to be checked on finite modules.) The conditions $B$ Cohen-Macaulay and  $\dim B=\dim A+\dim F$ are just meant to guarantee that any $A$-regular sequence is mapped to a $B$-regular sequence, as you can check quickly. To check this, take an $A$-regular sequence $x_1,\ldots,x_t$, extend it to a maximal regular sequence $\underline{x}:=(x_1,\ldots,x_d)$ in $A$, then use the dimension assumption and the fact that $B$ is Cohen-Macaulat to show that $f(\underline{x}):=(f(x_1),\ldots,f(x_d))$ is a regular sequence in $B$. 
(Note that on one hand, the inclusion $f(\underline{x})B\subseteq\mathfrak{m}_AB$ gives $\dim B/f(\underline{x})B\geq\dim B/\mathfrak{m}_AB$. On the other hand the map $A/\underline{x}\rightarrow B/f(\underline{x})B$ gives $\dim B/f(\underline{x})B\leq \dim A/\underline{x}+\dim B/\mathfrak{m}_AB=0+\dim B/\mathfrak{m}_AB$. Hence $\dim B/f(\underline{x})B=\dim B-\dim A$.)<|endoftext|>
TITLE: Choice of adviser
QUESTION [7 upvotes]: Not sure how to tag this one so feel free to edit and add tags. 
When I initially started graduate school my choice for an area of study was quite nebulous. I had only figured out enough to know that I wanted to do some work involving a lot of category theory. So when I applied to schools I figured I could find some interesting topic to work on since almost anything involves category theory. Now I'm a bit more mature and I have a much better perspective and a much better idea of what I would like to do. My question then is how do I go about finding an adviser working in an area I would like to specialize in and if this person is not at my current school what are my possible courses of action?

REPLY [3 votes]: Here is some advice that my department wrote some years ago for our undergraduate math majors that might be relevant. It takes a rather extreme stance, but this was done purposely to provide counterpoint to the advice usually given to Ph.D. students in math:
Choosing a Ph.D. program or advisor<|endoftext|>
TITLE: Intuition about schemes over a fixed scheme
QUESTION [15 upvotes]: I am taking a first course on Algebraic Geometry, and I am a little confused at the intuition behind looking at schemes over a fixed scheme.  Categorically, I have all the motivation in the world for looking at comma categories, but I would like to make sense of this geometrically.
Here is one piece of geometric motivation I do have: A family of deformations of schemes could be thought of as a morphism $X \rightarrow Y$, where the fibers of the morphism are the schemes which are being deformed, and these are indexed by the scheme Y.
This is all well and good, but I am really interested in Schemes over $Spec(k)$ are thought of as doing geometry "over" $k$.  I know that their is a nice "schemeification functor" taking varieties over $k$ to schemes over $k$, but this is somewhat unsatisfying. All that I see algebraically is that $k$ injects into the ring of global sections of the structure sheaf of $X$, but this does not seem like much of a geometric condition...
Any help would be appreciated. 
EDIT:  The answers I have received so far have been good, but I am looking for something more.  I will try to flesh out an example that give the same style of answer that I am asking for:
The notion of a k-valued point:  Every point $(a,b)$ of the real variety $x^2 + y^2 - 1 = 0$ has a corresponding evaluation homomorphism $\mathbb{R}[x,y] \rightarrow \mathbb{R}$ given by $x \mapsto a$ and $y \mapsto b$.  Since $a^2 + b^2 - 1 = 0$, this homomorphis factors uniquely through $\mathbb{R}[x,y]/(x^2 + y^2 -1)$. So the real valued points of the unit circle are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{R}$.  Similarly, the complex valued points are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{C}$.  Actually, points of the unit circle valued in any field $k$ are going to be in one to one correspondence with homomorphisms from $\mathbb{Z}[x,y]/(x^2 +y^2 -1)$ to $k$.
Dualizing everything in sight, we are saying that the $k$- valued points of the scheme $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$ are just given by homomorphisms from the one point scheme $Spec(k)$ into $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$.  
EDIT 2: Csar's comment comes very close to answering the question for me, and I will try and spell out the ideas in that comment a little better. I wish Csar had left his comment as an answer so I could select it.
So it seems to me that the most basic reason to think about schemes over a field $k$ is this: 
I already spelled out above why $k$-valued points of a scheme are important.  But a lot of the time, a morphism from $Spec(k)$ to $X$ will point to a generic point of $X$, not a closed point.  Different morphisms can all point to the same generic point.  For instance the dual of the any injection $\mathbb{Z}[x,y] \rightarrow \mathbb{C}$ (of which there are many), will all "point" to the generic point of $\mathbb{Z}[x,y]$.  
On the other hand if we are looking at $\mathbb{Q}[x,y]$, this is a $\mathbb{Q}$ scheme.  $Spec(\mathbb{Q})$ is also a $\mathbb{Q}$ scheme via the identity map.  A morphism of $\mathbb{Q}$ schemes from  $Spec(\mathbb{Q})$ to $\mathbb{Q}[x,y]$ will correspond to a closed $\mathbb{Q}$ -valued point!  So this is a nice geometric reason for looking at schemes over a fixed field $k$: $k$ morphisms of $Spec(k)$ to $X$ correspond to $k$-valued closed points of $X$.
It will take some work for me to internalize the vast generalization that S-schemes entail, but I think this is a good start.  Does everything I said above make sense?

REPLY [5 votes]: I tried to post this as a comment to Csar's post, but the margin was to small to contain it.
I understand what you are trying to say here, but this style of answer isn't what I am exactly looking for.  Extension of scalars should not be vague. Here is an explicit example (I am sticking with circles as per my edit.):  Say we are looking at rational points of the circle. Then naturally we are interested in $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$.  Say we want to instead look at complex valued points.  Well $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$ is  a scheme over $\mathbb{Q}$, and $\mathbb{C}$ is a scheme over $\mathbb{Q}$ (the dual of the inclusion maps give the appropriate maps here).  When you take the fiber product to get a scheme over $Spec(\mathbb{C})$, this corresponds to a pushout in the category of rings, i.e. a tensor product. So the fiber product is $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1) \otimes_\mathbb{Q} \mathbb{C})$ which is isomorphic to $Spec(\mathbb{C}[x,y]/(x^2+y^2-1))$, i.e. exactly what you would expect.
I would like something this explicit explaining the geometric meaning of a scheme over a fixed scheme.<|endoftext|>
TITLE: Is there a neat formula for the volume of a tetrahedron on $S^3$?
QUESTION [23 upvotes]: There is a nice formula for the area of a triangle on the 2-dimensional sphere;
If the triangle is the intersection of three half spheres, and has angles $\alpha$, $\beta$ and $\gamma$, and we normalize the area of the whole sphere to be $4\pi$ then the area of the triangle is 
$$
\alpha + \beta + \gamma - \pi.
$$
The proof is a cute application of inclusion-exclusion of three sets, and involves the fact
that the area we want to calculate appears on both sides of the equation, but with opposite signs.
However, when trying to copy the proof to the three dimensional sphere the parity goes the wrong way and you get 0=0.
Is there a simple formula for the volume of the intersection of four half-spheres of $S^3$ in terms of the 6 angles between the four bounding hyperplanes?

REPLY [4 votes]: Nice answer, Greg. I looked at the linked paper and was sufficiently intimidated.
I just want to point out, again, that for those (like me) who have a phobia of differential geometry, and hence don't want to use (generalized)Gauss-Bonnet, it is easy to see, using inclusion-exclusion, that the formula in even dimensions is a neat linear combination of
the formulas in lower dimensions.<|endoftext|>
TITLE: Non-integral scheme having integral local rings
QUESTION [33 upvotes]: I can show that if $X$ is a scheme such that all local rings $\mathcal{O}_{X,x}$ are integral and such that the underlying topological space is connected and Noetherian, then $X$ is itself integral. 
This doesn't seem to work without the "Noetherian" condition. But can anyone think about a nice counterexample to illustrate this? So I am looking for a non-integral scheme - with connected underlying topological space - having integral local rings.

REPLY [9 votes]: Hochster has an elegant construction which associates a commutative ring to each infinite  totally ordered set with the property that strictly between two distinct elements there is a third one. 
The spectrum of such a ring is a connected affine scheme of dimension one, all the local rings of which are domains. The ring itself, however is NOT a domain.
So, every ordered set with the property mentioned above yields a scheme with the required property.
Here is the link to Hochster's (one-page) construction
http://www.math.lsa.umich.edu/~hochster/614F08/ECdom.sol.pdf<|endoftext|>
TITLE: Differences between reflexives and projectives modules
QUESTION [6 upvotes]: Let R be a normal noetherian domain.
What is the difference between a finitely generated reflexive  module and a finitely  generated projective module?
Can anybody recommend any references or make any suggestions about this? 

Finitely generated projective modules can be identified with idempotents matrix...
Finitely generated projective modules correspond with vector bundles over topological space...
Are there similar results about reflexives modules?

REPLY [3 votes]: I asked a very similar question a few months ago and got some very good answers.<|endoftext|>
TITLE: Algebraic varieties which are topological manifolds
QUESTION [18 upvotes]: Inspired by this thread, which concludes that a non-singular variety over the complex numbers is naturally a smooth manifold, does anyone know conditions that imply that the topological space underlying a complex variety is a topological manifold without necessarily implying it is smooth?

REPLY [14 votes]: The answer from Dmitri motivates this partial answer from the topological side of the question.
It is a theorem of Mark Goresky and others that every stratified space, and in particular every complex variety $V$, has a smooth triangulation.  Moreover, I would bet (although I don't know that Goresky's paper has it) that the associated piecewise linear structure is unique.  This means that the PL homeomorphism type of the link of a singular point $p$ of $V$ is a local invariant.  I don't know how to compute this local invariant in general, but there must be some way to do it from the local ring at $p$.  There can't be a simple calculation of this invariant that is fully general.  As a special case, $V$ can be the cone of a projective variety $X$.  If so, then the link at the cone point $p$ is the total space of the tautological bundle on $X$.  $X$ and therefore the link can be all sorts of things.  If $p$ is an isolated singularity, then the type of this link is obtained by "intersecting with a small sphere", as Dmitri says. 
The variety $V$ is a PL manifold if and only if the link of every vertex is a PL sphere.  This is the case for the Brieskorn examples.
On the other hand, a theorem of Edwards (or maybe Cannon and Edwards) says that a polyhedron is a topological $n$-manifold (for $n \ge 3$) if and only if the link of every vertex is simply connected and the link of every point is a homology $(n-1)$-sphere.  In particular, the link of a simplex which is not a point does not have to be simply connected!  For example, if $\Gamma \subseteq \text{SU}(2)$ is the binary icosahedral group, then $\mathbb{C}^2/\Gamma$ is not a manifold, because the link of the singular point is the Poincaré homology sphere.  But $(\mathbb{C}^2 / \Gamma) \times \mathbb{C}$ is a topological manifold, even though it is not a PL manifold.
So for the question as stated, you would want to combine Goresky's theorem with Edwards' theorem, and with a method to compute the topology of the link of a singular point.  On the other hand, whether a variety $V$ is a PL manifold could be a more natural question than whether it is a topological manifold.

At least in the case of isolated singularities, the possible topology of the link of a singular point has been studied in the language of complex analytic geometry rather than complex algebraic geometry.  I found this paper by Xiaojun Huang on this topic.  The link of the singular point is in general a strictly pseudoconvex CR manifold.  This is a certain kind of odd-dimensional analogue of a complex manifold and you could study it with algebraic geometry tools.  (I think that strict pseudoconvexity also makes it a contact manifold?)  But the analytic style seems to be more popular, maybe because a CR manifold is not a scheme.
Sometimes, for instance in the case of a Brieskorn-Pham variety, such a CR manifold has a circle action whose quotient is a complex algebraic variety.  At a smooth point, this quotient is just the usual Hopf fibration from $S^{2n-1}$ to $\mathbb{C}P^{n-1}$.  In the famous Brieskorn examples, the link is a topological sphere with a circle action, but the circle action yields a non-trivial Seifert fibration over an orbifold-type complex variety.  On the other hand, I don't think that this circle action always exists.<|endoftext|>
TITLE: Are Jacobians principally polarized over non-algebraically closed fields?
QUESTION [18 upvotes]: How does one define the Torelli map $M_g \to A_g$ of moduli stacks? On geometric points a curve maps to its principally polarized Jacobian. 
So what I am asking is: if I have a curve $C$ over a non-algebraically closed field $k$ such that $C(k)$ is empty, is the Jacobian of C still principally polarized? After base change to $\bar{k}$ one has a theta divisor; does it descend? Also, is the relative Jacobian of a family of curves principally polarized?
The thing I am confused about is that the theta divisor naturally lives on $Pic^{g-1}$ as the image of the map from the symmetric power $C^{g-1}$; this is a torsor under $Pic_0$, but not itself an abelian variety.
Also, the classical Torelli theorem says that this map is an injection on field valued points. Is this actually a locally closed immersion of stacks?

REPLY [24 votes]: There's a more down to earth way to deal with this, which is already explained in Mumford's GIT:  make an fppf (or even etale) surjective base change to acquire a section, use that to define the principal polarization, and then show it is independent of the choice.  (Short reason: varying the choice amounts to a morphism from the smooth proper curve to a Hom or Isom scheme that is unramified over the base, hence constant.)  Thus, by descent theory one gets the polarization over the original base.
This is related to the same issue which comes up in explaining why a polarization of an abelian scheme need not arise as the "Mumford construction" $\phi_{\mathcal{L}}$ even though it automatically does so on geometric fibers (due to the special nature of $k$-simple finite commutative $k$-groups when $k = \overline{k}$).  That is, a definition of "polarization" which is better-suited to the relative case is not to mimic what one traditionally does over an algebraically closed field (the Mumford construction) but rather something which makes more effective use of the Poincar\'e bundle.  The possible lack of $\mathcal{L}$ over the base is analogous to the possible lack of a section of the curve to define the principal polarization.  See the Wikipedia page on ``abelian varieties'' for more on this. :)<|endoftext|>
TITLE: A hypersurface with many points
QUESTION [22 upvotes]: Ok, it's time for me to ask my first question on MO.
Consider the affine curve $Y+Y^q=X^{q+1}$ over the finite field $\mathbf{F}_q$.  It's interesting because it has the largest number of points over $\mathbf{F}_{q^2}$ possible relative to its genus, which is $q(q-1)/2$.  In other words, this curve realizes the Weil bound over $\mathbf{F}_{q^2}$.  This seems to be well-known in the literature. 
Now consider the following hypersurface $\mathcal{X}$ over the finite field $\mathbf{F}_q$:
$$Z+Z^q+Z^{q^2}=\det\left(\begin{matrix}
0 & X & Y \\ Y^q & 1 & X^q \\ X^{q^2} & 0 & 1\end{matrix}\right)$$
Empirical observation seems to point to the following:   The compactly supported cohomology of $\mathcal{X}$ is only nonzero in degrees 2 and 4.  In degree 2, the dimension of $H^2(\mathcal{X})$ is $q^2-1$ and the $q^3$-power frobenius acts as the scalar $q^3$.  Which is all a fancy way of saying that for all $n$,
$$\#\mathcal{X}(\mathbf{F}_{q^{3n}})=q^{6n}+q^{3n}(q^2-1).$$
Thus $\mathcal{X}$ has the largest number of $\mathbf{F}_{q^3}$-points among any hypersurface with the same compactly supported Betti numbers. 
Can anyone help me prove the above formula?  (I can do $n=1$ alright...)  Bonus points if you can also compute the automorphism group of $\mathcal{X}$.  Many more bonus points if you can formulate the generalization to hypersurfaces of higher dimension!
The above hypersurface arises in the study of the bad reduction of Shimura varieties, if anyone cares to know.
EDIT:  Admittedly, this is a narrow problem about a very particular surface.  Therefore I'm going to accept an answer to the following question:  Is there an algorithm to compute the zeta function of a hypersurface of this sort, that's quicker than counting points?  
(I've already noticed that it's enough to recur over X and Y, and to test for each pair that the expression on the right lies in the image of the linear map defined by the expression on the left.  But I can't think of anything faster than this.)

REPLY [8 votes]: Here is a complete solution to the main question when $n$ and $q$ are both odd, and a partial solution for the other parities.  The partial solution includes a reduction to the case $n=2$.
Let $\text{Tr}_k$ denote the trace map from $\mathbb{F}_{q^n}$ to $\mathbb{F}_{q^k}$, assuming that $k|n$.
The equation is 
$$z+z^q + z^{q^2} = x^{q^2+q+1} - xy^q - x^{q^2}y.$$
First, make the change of variables $y \leftarrow -yx^{q^+1},$ so that the equation becomes
$$z+z^q+z^{q^2} = x^{q^2+q+1}(y^q+y+1).$$
Second, when $q$ is odd, we can clarify matters a little with the change of variables $y \leftarrow y - \frac12$ to get rid of the constant.  The image of the $q$-linear map $z \mapsto z + z^q + z^{q^2}$, acting on $\mathbb{F}_{q^{3n}}$, consists of those elements $Z'$ such that $\text{Tr}_3(z') = \text{Tr}_1(z')$.  Whenever such a $z'$ is reached by the right side, there are $q^2$ solutions for $Z$.  On the right side, $y \mapsto y' = y^q+y$ is an $q$-linear isomorphism when $n$ is odd, while when $n$ is even its image is the locus of $\text{Tr}_2(y') = \text{Tr}_1(y')$.  Meanwhile $y' \mapsto x^{q^2+q+1}y'$ is a $q$-linear isomorphism unless $x=0$.
Thus when $n$ and $q$ are both odd, there are $(q^{3n}-q^2)(q^{3n}-1)$ solutions with $x,y \ne 0$.  There are $q^2(2q^{3n}-1)$ more solutions when one of them is zero.
When $q$ is odd and $n$ is even, then in principle different $x' = x^{q^2 + q + 1}$ could behave differently in the equation $z' = x'y'$.  For a fixed $x'$, the set of possible $x'y'$ is a certain $q$-linear hyperplane with codimension $1$, while the set of possible $z'$ is a certain $q$-linear hyperplane with codimension $2$.  In the special case that $\{x'y'\}$ contains $\{z'\}$, then there are $q^{3n+1}$ solutions for that value of $x'$.  For generic non-zero values of $x'$, there are $q^{3n}$ solutions.  Dualize the hyperplanes with respect to $\text{Tr}_1$.  The dual of $\{y'\}$ is the line $L$ of trace 0 elements in $\mathbb{F}_{q^2}$, while the dual of $\{z'\}$ is the plane $P$ of trace 0 elements in $\mathbb{F}_{q^3}$.  Meanwhile $\{x'\}$ is the subgroup $G$ of $\mathbb{F}_{q^{3n}}$ of index $q^2+q+1$.  A special value of $x'$ in this subgroup is one such that $x'L \subset P$.  A priori I am not sure that it never happens.  What I can say is that if $x'$ is special, then it must lie in $\mathbb{F}_{q^6}$ because both $L$ and $P$ do.  So you can reduce the counting problem to the case that $3n = 6$ or $n = 2$.
If $q$ is even, then the equation is 
$$z' = x'(y'+1),$$
where as before $z' = z+z^q+z^{q^2}$ and $y' = y + y^q$.  In this case $y'$ is any element with zero trace, and the dual line $L$ is just $\mathbb{F}_q$ itself.  I have not worked out exactly how it looks, but I suppose that it reduces to the case $n=1$ for similar reasons as above.
Afterthought:  I don't feel like changing all of the equations, but I'm wondering now whether  there a dual change of variables to put the right side in the form $y''(x^{q^2}+x)$.  I think that the map $x \mapsto x^{q^2}+x$ is always non-singular when $q$ is odd.

A remark about where the trace conditions come from.  If $a$ is an irreducible element of $\mathbb{F}_{q^n}$, then the map $x \mapsto x^q$ is a cyclic permutation matrix in the basis of conjugates of $a$.  A map such as $z \mapsto z+z^q+z^{q^2}$ is then a sum of disjoint permutation matrices and it easy to compute its image and cokernel.

Some remarks about Jared's second, more general question:  C.f. the answer to this other mathoverflow question about counting points on varieties.  For fixed $q$, the equation of a hypersurface is equivalent to a general Boolean expression, and there may not be much that you can do other than count one by one.  There are several strategies that work in the presence of special structure:  You can use zeta function information, if you have it, to extrapolate to large values of $q$.  You can count the points on a variety if you happen to know that it's linear, or maybe quadratic, or the coset space of a group.  And you can use standard combinatorial counting tricks, which in algebraic geometry form amount to looking at fibrations, blowups, inclusion-exclusion for constructible sets, etc.
This particular variety decomposes a lot because it can be made jointly linear in $Y$ and $Z$, and $X$ only enters in a multiplicative form.<|endoftext|>
TITLE: Reference for Deligne's construction of Galois representations attached to modular forms
QUESTION [17 upvotes]: I was wondering if anyone can suggest some good reference for learning more about Deligne's construction of Galois representations attached to modular forms. I find Deligne's original paper hard to read.

REPLY [4 votes]: These notes by Takeshi Saito also give a very helpful overview: www.ms.u-tokyo.ac.jp/~t-saito/talk/eepr.pdf<|endoftext|>
TITLE: Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group?
QUESTION [20 upvotes]: This question is closely related to this one.
Ado's theorem states that given a finite-dimensional Lie algebra $\mathfrak g$, there exists a faithful representation $\rho\colon\mathfrak g \to \mathfrak{gl}(V)$, with $V$ a finite-dimensional vector space. In the real or complex case one can take the exponent of the image and obtain a (virtual) Lie subgroup $\exp\rho(\mathfrak g)$ in $GL(V)$ having Lie algebra $\rho(\mathfrak g)$. But nothing guarantees that this subgroup will be closed in $GL(V)$. 
So the question is: is every finite-dimensional Lie algebra the Lie algebra of some closed linear Lie group? I am primarily interested in the real and complex case, but it might be interesting to ask what happens in the ultrametric case as well.

REPLY [3 votes]: The answer is yes. According to Ado's theorem,
every Lie algebra can be realized as subalgebra
of $\mathfrak{gl}(V)$ for some $V$.
According to the subalgebras-subgroups theorem,
every Lie subalgebra of $\mathfrak{gl}(V)$
is the Lie algebra of an immersed Lie subgroup of $\mathrm{GL}(V)$.
Call a Lie group linear if it admits a faithful representation.
We have shown that every Lie algebra is the Lie algebra
of a linear Lie group. It is known that every
linear Lie group $G$ admits a faithful representation
whose image is closed.
Here's a sketch of the proof of the last statement.
Let $r\colon G\rightarrow GL(V)$ be a faithful representation of $G$.
The derived group $r(G^{\prime})$ of $r(G)$ is closed $GL(V)$
and hence in $r(G)$. Hence
$G/G^{\prime}$ is an abelian Lie group. Therefore there exists a
representation $s\colon G\rightarrow GL(W)$ such that $Ker(s)=G^{\prime}$ and
$s(G)$ is closed in $GL(W)$. Consider the representation $G\rightarrow GL(V\oplus
W)$ such $a\in G$ acts on $V$ as $r(a)$ and on $W$ as $s(a)$. This is
faithful, and one can show that its image is closed --- Proposition 5 of
Djokovic, D. Ž. A closure theorem for analytic subgroups of real Lie groups. Canad. Math. Bull. 19 (1976), no. 4, 435--439. MR0442147<|endoftext|>
TITLE: Two conjectures by Gabber on Brauer and Picard groups
QUESTION [20 upvotes]: In a paper I need to make reference to two conjectures by Gabber, from

Ofer Gabber, On purity for the Brauer group, in: Arithmetic Algebraic Geometry, MFO Report No. 37/2004, doi:10.14760/OWR-2004-37

(see Conjectures 2 and 3, page 1975)

Let $R$ be a strictly henselian complete intersection noetherian
local ring of dimension at least 4. Then $Br'(U_R) = 0$ (the cohomological Brauer group of the punctured spectrum is $0$).

Let $R$ be a complete intersection noetherian local ring of dimension 3. Then $Pic(U_R)$ is torsion-free.


Does anyone know of any new developments on these conjectures beyond the Oberwolfach report above? I tried MathScinet but could not find anything. May be someone in the  Arithmetic Geometry community happen to know some news on these? Thanks a bunch.

REPLY [11 votes]: Happy news: there have been some progress over the last 10 years. The hypersurface case of Conjecture 2 was proved in

H. Dao,  Picard groups of punctured spectra of dimension three local hypersurfaces are torsion-free. Compositio Mathematica, 148(1) (2012) pp. 145-152. doi:10.1112/S0010437X11005513, arXiv:1004.0471,

Česnavičius and Scholze recently settled both Conjectures:

Kestutis Cesnavicius, Peter Scholze, Purity for flat cohomology, arXiv:1912.10932.<|endoftext|>
TITLE: Some examples of depth
QUESTION [9 upvotes]: This is related to the question I asked last time. This sounds a bit too specific, I hope this question is still acceptable on MO.
I am still not quite comfortable with the concept of depth, and there is this exercise in Matsumura's book that goes as follows:

Find an example of a noetherian local ring $A$ and a finite $A$-module $M$ such that $\operatorname{depth}M > \operatorname{depth}A$. Also find $A$, $M$ and $P \in \operatorname{Spec}A$ such that $\operatorname{depth}M_P > \operatorname{depth}_P(M)$. 

I hope I have found correct examples, but I am still quite lost about why one can find such examples, and what the generic ones are. So if someone can just give me some representative examples I would be grateful.

The examples I found myself:
For the first one, it is clear that $A$ must not be Cohen-Macaulay. Then I set $A = \frac {k[x,y,z]}{(xz,yz)}_{(x,y,z)}$, which is of depth 1, and I consider its quotient by $(z)$, which is $k[x,y]_{(x,y)}$ and should be of depth 2 (at least $x,y$ is a regular sequence I think).
For the second one, I try to fix $\operatorname{depth}_P(M) = 0$, which means $P$ should lie in some associated primes of $M$, so I consider $M = \frac {k[x,y,z]}{(x^2,xy,xz)}_{(x,y)}$, such that $(x,y)$ is not associated prime when localized.

REPLY [9 votes]: 1) Start with a regular local ring $R$. Take 2 ideals $I,J$ such that $I$ does not contain $J$, $R/I$ is CM and $\dim R/J <\dim R/I$. Then $A=R/(I\cap J)$ and $M=R/I$  work. In your example, $I=(x)$ and $J=(y,z)$. The reason is that CM means unmixed, so by having components of different dimensions one makes sure A is not CM.      
2) Take $(A,m,k)$ to be any CM rings of dimension at least 2. Let $M=A\oplus k$. Then for any non-minimal $P\in Spec(A)-{m}$, $depth M_P =depth A_P$, but $depth M=0$.
The common theme: depth is usually the minimum depth of all components, while dimension is the maximal of those.<|endoftext|>
TITLE: When is a commutative ring the limit of its local rings?
QUESTION [14 upvotes]: Let $A$ be a commutative ring. Then we get local rings $A_p$ by localizing at each prime ideal $p$. Moreover, we get $A_p \rightarrow A_q$ when $p$ contains $q$. So we get a big diagram indexed by the inclusion poset of prime ideals. When is $A$ the limit of this diagram?
When $A$ is a local ring or an integral domain it's true. I don't see any reason why it should be true for arbitrary rings. What's going on here?

REPLY [6 votes]: This is a minor variation of MH's response:
A ring R is Boolean if x^2 = x for all x in R.  (This implies R is commutative.)
In a Boolean ring R, every prime ideal is maximal,  Moreover, the only local Boolean ring is Z/2Z.  Therefore, R' := inverse limit_{p \in Spec R} R_p = (Z/2Z)^{# Spec R}.  In particular, R' is either finite or uncountably infinite.
But there are certainly countably infinite Boolean rings (a fancy justification for this is the Lowenheim-Skolem theorem in model theory): take an uncountable Boolean ring, and consider the subring generated by a countably infinite set of generators. 
For more details on Boolean rings, see e.g. Section 4.5 of 
http://math.uga.edu/~pete/integral.pdf<|endoftext|>
TITLE: finite surjective l.c.i morphism is flat
QUESTION [7 upvotes]: Let $X,Y$ be locally Noetherian schemes. Let $f:X\to Y$ be a finite, surjective, and locally complete intersection morphism, i.e., locally it can be decomposed as regular immersion followed by a smooth morphism.
Recall: an immersion $X\to Y$ is called a regular immersion at a point $x$ if $\mathcal{O}_{X,x}$ is isomorphic as $\mathcal{O}_{Y,y}$-module to $\mathcal{O}_{Y,y}$ modulo an ideal $I$ generated by a regular sequence of elements of $\mathcal{O}_{Y,y}$. 
Question: prove that $f$ is flat. In particular, $f$ will be a simultaneously open and closed morphism.

REPLY [7 votes]: Work locally, assume that $f: R\to S$ is a local homomorphism. Let $\operatorname{cmd}(R) =\operatorname{dim} R-\operatorname{depth} R$ (this is the so-called Cohen-Macaulay defect of $R$). Claim: $\operatorname{cmd}$ is preserved by l.c.i maps
(easy, essentially because both depth and dimension drop by one when you kill a regular element). 
Now since the map $\phi: \operatorname{Spec}(S) \to \operatorname{Spec} (R)$ is finite and surjective, $\operatorname{dim} R= \operatorname{dim} S$, which combines with the last claim to show that $\operatorname{depth} R = \operatorname{depth} S$. But since l.c.i  also implies finite flat dimension, we have $\operatorname{depth} R -\operatorname{depth} S = pd_RS$, so $S$ is flat over $R$.<|endoftext|>
TITLE: Naive Z/2-spectrum structure on E smash E?
QUESTION [6 upvotes]: Let $E$ be a spectrum.  Then $E \wedge E$ is a $\mathbb{Z}/2$-spectrum in the naivest possible sense, i.e., an object with $\mathbb{Z}/2$-action in the (∞,1)-category of spectra.  Can I make it a $\mathbb{Z}/2$-spectrum in the less naive, but still not genuine, sense?  (That is, a $\mathbb{Z}/2$-spectrum indexed on the trivial universe.)
I'm thinking of something like the following.  I may represent $E$ as an (reduced & continuous) excisive functor from pointed spaces to pointed spaces.  Then define
$$G(X) = \mathrm{colim}_{I \times I} \mathrm{Map}(S^{x_1} \wedge S^{x_2}, E(S^{x_1}) \wedge E(S^{x_2}) \wedge X)$$
where $I$ is the category of finite sets and inclusions.  Hopefully $G$ is a functor from spaces to $\mathbb{Z}/2$-spaces.  If I forget about the $\mathbb{Z}/2$-fixed point set, I can think of it as $E \wedge E$ with its $\mathbb{Z}/2$ action.  What spectrum does $G(X)^{\mathbb{Z}/2}$ correspond to?  Is there a more familiar name for it?  Edit: I seem to be getting $E \vee (E \wedge E)^{h\mathbb{Z}/2}$, but without much confidence.
[Leftover part of the question: If so, by my question here I can think of the resulting object as a functor from the opposite of the orbit category of $\mathbb{Z}/2$ to spectra.  Unpacking this amounts to giving some spectrum $F$ together with a map $F \to (E \wedge E)^{h\mathbb{Z}/2}$.  What is $F$?]

REPLY [3 votes]: Here's a very simple way to obtain $(E \wedge E)^{\Bbb Z_2}$ without having to resort to representations (at least if $E$ is connective). Consider the functor from spectra to spectra given by $$E \mapsto \Sigma^\infty \Omega^\infty E .$$ That is, the suspension spectrum of the zeroth space of $E$.
Let 
$$E \mapsto P_2(E)$$ be its quadratic approximation in the sense of Goodwillie's calculus of homotopy functors (that is, the second stage of the Goodwillie tower).
Then $P_2(E)$ has the homotopy type of $(E \wedge E)^{\Bbb Z_2}$ whenever $E$ is connective.
Incidentally, one can also see that this comes equipped with a fiber sequence
$$
D_2(E) \to P_2(E) \to E
$$
which amounts the tom Dieck's splitting in the case when $E$ is a suspension spectrum.
Here, $D_2(E) = (E\wedge E)_{h\Bbb Z_2}$ is the quadratic construction.
In general, this sequence needn't split.<|endoftext|>
TITLE: What are the most misleading alternate definitions in taught mathematics?
QUESTION [143 upvotes]: I suppose this question can be interpreted in two ways. It is often the case that two or more equivalent (but not necessarily semantically equivalent) definitions of the same idea/object are used in practice. Are there examples of equivalent definitions where one is more natural or intuitive? (I'm meaning so greatly more intuitive so as to not be subjective.)
Alternatively, what common examples are there in standard lecture courses where a particular symbolic definition obscures the concept being conveyed.

REPLY [7 votes]: Ok I'm joining very late but let me tell this.
I think the following definition of relation, that is often given, is misleading:

"Wrong" definition: A relation $R$ between the set $A$ and the set $B$ is an arbitrary subset of the cartesian product, $R\subseteq A\times B$.

In fact, I think it is a "wrong" definition:

it obfuscates the fact that the datum of source set and target set is important and itself part of the definition. For example, if you defined a function as a relation (in the sense of the definition above) satisfying the functional property ($\forall x\in A\exists ! y\in B: (x,y)\in R$) then the notion of codomain will not be well defined (or not explicitly defined), and one could be led to think that $x\mapsto x^2$ as a function $\mathbb R \to [0,+\infty)$ is literally equal to $x\mapsto x^2$ as a function $\mathbb R\to \mathbb R$.
It allows you to define, given $A$ and $B$ (in that order), the set $\mathsf{Rel}(A,B)$ of relations between $A$ and $B$, but not (immediately) the class of all relations (or the set of all relations within a given universe $U$).
It makes less clear that there should be a category $\mathsf{Rel}$, of sets with relations as morphisms, of which the $\mathsf{Rel}(A,B)$ are the hom-sets.

The right definition should of course be:

"Right" definition: A relation is a triple $(A,B,R)$ where $A$, $B$ are sets and $R\subseteq A\times B$.

Now the notion of codomain is well defined (or explicitly defined). And also the category $\mathsf{Rel}$ is well defined.<|endoftext|>
TITLE: Are any finitely generated reflexive module a 2nd syzygy?
QUESTION [5 upvotes]: Are any finitely generated reflexive module a second syzygy?
(I´m thinking especially in normal noetherian domains)
More general...
Are any divisorial lattice a second syzygy?
(I´m thinking especially in Krull domains)

REPLY [6 votes]: Over a normal domain (in fact, you only need Gorenstein in codimension 1, being second syzygy and reflexive are equivalent). This is Theorem 3.6 of Evans-Griffith "Syzygies" book.<|endoftext|>
TITLE: When does the zeta function take on integer values?
QUESTION [22 upvotes]: Here $\zeta(s)$ is the usual Riemann zeta function, defined as $\sum_{n=1}^\infty n^{-s}$ for $\Re(s)>1$.
Let $A_n=${$s\;:\;\zeta(s)=n$}. The behaviour of $A_0$ is basically just the Riemann hypothesis; my question concerns $A_n$ for $n\neq0$. 
1) Is determining this just as hard as the Riemann hypothesis?
2) If we know the behaviour of some $A_n$, does it help in deducing the behaviour of other $A_m$?
3) For which $n$ is $A_n$ non-empty?
Question 3 has now been answered for all strictly positive $n$ - it is non-empty, and has points on the real line to the left of $s=1$. For $n=0$, it is known to be non-empty. Any idea for negative $n$? (the same answer won't work, since $\zeta(s)$ is strictly positive on the real line to left of $s=1$.
Big Picard gives it non-empty for all but at most one $n$. How can we remove the 'at most one'?

REPLY [7 votes]: There are in fact rather precise conjectures as to the location of roots $s$ of the equation $\zeta(s) = a$ for any $a$ complex (this includes $a = n$ of course). 
For $a = 0$ this is the Riemann Hypothesis. 
For $a \neq 0$ it is conjectured that half of the points cluster slightly to the left of the half-line (precisely $\sqrt{\log\log t}/\log t$ to the left, if $t$ is the imaginary part) while the other half clusters very closely to the half-line (possibly $o(1/\log t)$). Around each of these lines the zeros spread uniformly to the left and right. In particular this led Selberg to conjecture that $3/4$ of the $a$-points lie to the right of the half-line while $1/4$ lies to the right. Selberg complemented this with the conjecture that on any given line, there are at most two $a$-points. Selberg has in addition obtained a central limit theorem for the distribution of the $a$-points around $1/2 - c \sqrt{\log\log t}/\log t$ with $c$ varying. A good reference for these questions are the paper of Selberg mentionned above, the thesis of Kai-Man Tsang (written under Selberg), which can be found on the internet, and finally for some more recent developments I would refer you to the papers: 
http://arxiv.org/abs/1402.0169
http://arxiv.org/abs/1402.6682
http://www.ams.org/journals/proc/2012-140-12/S0002-9939-2012-11275-4/home.html<|endoftext|>
TITLE: Topologically contractible algebraic varieties
QUESTION [18 upvotes]: From a post to The Jouanolou trick:

Are all topologically trivial (contractible) complex algebraic varieties necessarily affine? Are there examples of those not birationally equivalent to an affine space?

The examples  that come to my mind are similar to a singular $\mathbb P^1$ without a point given by equation $x^2 = y^3$. This particular curve is clearly birationally equivalent to affine line.
Perhaps the "affine" part follows from a comparison between Zariski cohomology and complex cohomology?

REPLY [10 votes]: About the rationality of contractible varieties: Yes for curves and surfaces and is an open question for higher dimensions. 
Any such contractible variety $X$ has $\chi_{top}(X)=1$, obviously. 
If $X$ is a curve then it must have only cusps as singularities, if any, by a simple $\chi_{top}$ calculation. Now let $Y$ be a projective model of $X$ such that it is smooth at the points in $Y-X$. Topologically, $Y$ is a real surface without boundary such that a few punctures make it contractible. The only real surface with this property is $S^2$, obviously. Hence $Y$ better be rational and so is $X$.   
If $X$ is an algebraic surface then it was a conjecture of Van de Ven that such a surface must be rational (actually his conjecture is for any homologically trivial $X$). This was proved by Gurjar & Shastri in: 

On the rationality of complex homology 2-cells
Here is the the part II of the above paper (MathSciNet review number MR0984747)<|endoftext|>
TITLE: Indexing the line bundles over a Grassmannian.
QUESTION [7 upvotes]: As is well known, the line bundles over *CP*$^1$ are indexed by the integers. My question is how are the line bundles over *CP*$^n$, $n > 1$, and *Gr*$(n,k)$ indexed? Moreover, do there exist any other interesting classifications of line bundles over spaces (I remember something about Atiyah and elliptic curves)?

REPLY [4 votes]: As a compact homogeneous Kahler manifold the Grassmannian is a coset space of the general linear group by a maximal parabolic subgroup. Homogeneous line bundles over the Grassmannian are in a one to one correspondence with the character representations of the maximal parabolic, which are indexed by one integer. According to the Bott-Borel-Weil theorem, the space of holomorphic sections of the line bundle carries an irreducible representation of the special unitary group SU(n). In the case of the Grassmannian Gr(n,k), these representations correpond to Young tableaux with k rows. The integer characterizing the line bundle is just the number of columns in this tableaux.<|endoftext|>
TITLE: When are some products of gamma functions algebraic numbers?
QUESTION [33 upvotes]: I want to know when certain expressions of the form 
$ {\Gamma(r_1/m) \Gamma(r_2/m) \ldots \Gamma(r_j/m) \over \Gamma(s_1/m) \Gamma(s_2/m) \ldots \Gamma(s_j/m)} $
are algebraic numbers.  These ratios of Γ functions occur in the asymptotic enumeration of certain classes of restricted partitions, but I don't think this is relevant.  Also, In the partition problems I'm interested in, it's natural to have $r_1 + \ldots + r_j = s_1 + \ldots + s_j$ but this isn't necessary.  This seems to happen with some frequency.  For example,a note of Albert Nijenhuis (arXiv:0907.1689) shows that $\Gamma(1/14) \Gamma(9/14) \Gamma(11/14) = 4\pi^{3/2}$; the techniques of the same paper show that $\Gamma(3/14) \Gamma(5/14) \Gamma(13/14) = 2\pi^{3/2}$, so the quotient is in fact 2!   Similarly, we can get the identity
$ {\Gamma(1/8) \Gamma(5/8) \Gamma(6/8) \over \Gamma(2/8) \Gamma(3/8) \Gamma(7/8)} = \sqrt{2}$
by applying the duplication formula
$ \Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z) $
to the first two factors in the numerator and the last two in the denominator.  In trying to prove other identities of this type, the duplication formula, its generalization to the "multiplication formula" 
$\Gamma(z) \Gamma(z+1/k) \cdots \Gamma(z+(k-1)/k) = (2\pi)^{(k-1)/2} k^{1/2-kz} \Gamma(kz)$ 
and the reflection formula 
$ \Gamma(z) \Gamma(1-z) = \pi \csc(\pi z)$
are the most obvious tools.  So this seems to be a problem in combinatorial number theory; given an expression of the form in the first displayed equation, when can we use the multiplication and reflection formulas to reduce it to a rational power of some integer times a product of trig functions of rational multiples of π?

REPLY [5 votes]: Rohrlich has conjectured that the multiplicative relations in $\mathbb{C}^\times / \overline{\mathbb{Q}}^\times$ between values of $\Gamma$ at rational numbers are generated by the multiplication formula and the reflection formula. In conceptual terms, Lang says that $\Gamma$ is an odd punctured distribution on $\mathbb{Q}/\mathbb{Z}$, and that conjecturally, it is universal, see Relations de distributions et exemples classiques, Deligne's article mentioned in Felipe Voloch's answer and this MO answer. For more details about distributions, see the book by Kubert and Lang, Modular units (Springer, 1981), Chapter 1.
On the other hand, the first Bernoulli polynomial $B_1(x)=x-\frac12$ also gives rise to a universal distribution. More precisely, the function
\begin{equation*}
h_1 \colon x \mapsto \begin{cases} B_1(\{x\}) & \textrm{if } x \not\in \mathbb{Z} \\ 0 & \textrm{if } x \in \mathbb{Z}, \end{cases}
\end{equation*}
where $\{x\}=x-\lfloor x \rfloor$ is the fractional part of $x$, is a distribution on $\mathbb{Q}/\mathbb{Z}$. Following the terminology of Kubert-Lang, the first Bernoulli distribution on $\mathbb{Q}/\mathbb{Z}$ is the Stickelberger distribution associated to $h_1$. It takes values in the direct limit of the group rings $\mathbb{Q}[(\mathbb{Z}/N\mathbb{Z})^\times]$. At each finite level $N$, it is given by
\begin{align*}
\mathbf{B}_1 \colon \bigl(\frac{1}{N}\mathbb{Z}\bigr)/\mathbb{Z} & \to \mathbb{Q}[(\mathbb{Z}/N\mathbb{Z})^\times] \\
x & \mapsto \sum_{u \in (\mathbb{Z}/N\mathbb{Z})^\times} h_1(ux) [u].
\end{align*}
The good thing is that we can show that $\mathbf{B}_1$ is universal (among odd punctured distributions on $\mathbb{Q}/\mathbb{Z}$). This follows from the non-vanishing of the Dirichlet $L$-values $L(\chi,1)$ with $\chi$ odd. Concretely, this means that there are no other linear relations than distribution and parity.
This leads to a conjectural criterion for an arbitrary product of $\Gamma$-values
\begin{equation*}
\Gamma(x_1)^{n_1} \cdots \Gamma(x_r)^{n_r}
\end{equation*}
with $x_i \in \mathbb{Q} \backslash \mathbb{Z}$ and $n_i \in \mathbb{Z}$, to be an algebraic number times a power of $\sqrt{\pi}$. Namely, just check whether the divisor $X = \sum_{i=1}^r n_i [x_i]$ on $\mathbb{Q}/\mathbb{Z}$ is in the kernel of the first Bernoulli distribution $\mathbf{B}_1$.
If it is, then using linear algebra, you will be able to write $X$ as a linear combination of the multiplication and reflection relations, because $\mathbf{B}_1$ is universal. Therefore you will be able to compute the $\Gamma$ product as an explicit algebraic number times a power of $\sqrt{\pi}$. This algebraic number will be a cyclotomic number times a product of fractional powers of prime numbers.
If one excepts the possible simplifications of this algebraic number, all this can be made into an algorithm.
In the same article, Lang also asks whether the multiplication and reflection formulas generate the ideal of polynomial relations between $\Gamma$-values over $\overline{\mathbb{Q}}(\sqrt{\pi})$. One could try, similarly as above, to give an explicit conjectural criterion for a given polynomial in $\Gamma$-values to be algebraic.<|endoftext|>
TITLE: Hypercohomology of a dg-algebra
QUESTION [9 upvotes]: Can someone give me a reference (note I am looking for a reference and not a proof) for the following:
If a complex $C$ has a dg-algebra structure, then the hypercohomology
$H^0R\pi_*C$ has an algebra structure, and if $M$ is a dg-module for $C$, 
then $H^0R\pi_*M$ is a module under $H^0R\pi_*C$. (Here I am thinking of $C$ as a complex of sheaves on some scheme $S$ with $\pi : S \rightarrow T$, but this should just be a fact of homological algebra.)

REPLY [4 votes]: Since you want references, not a proof, maybe you can look at the paper by Hinich that Leonid already mentioned, but also at the original book by Godement, "Topologie Algébrique et Théorie des Faisceaux". There he treats also the problem with multiplicative structures in sheaf cohomology (and you can adapt it to the derived functor of direct images). As for "Thom-Whitney functors", mentioned by Minhyong, you should look at the original paper by V. Navarro-Aznar, "Sur la thérorie de Hodge-Deligne", Inv. Math. 90 (1987), 11-76. But be aware that if you are not working with commutative dg algebras you don't need in fact the essential tool of this paper (the "Thom-Whitney simple/total functor"): the usual total functor of double complexes (together with the Alexander-Whitney map) will work as well and looks easier. Or you can also ask me for a preprint about that subject that I'm going to finish one of these days. Sorry for this self-advertising.  :-)  (One last hint: you don't need to restrict yourself to the H^0 cohomology: you have a dg multiplicative structure already in R\pi_*C : the multiplicative structure in the H^0 is inherited from that one.)<|endoftext|>
TITLE: Are the field norm and trace the unique "nice" maps between fields?
QUESTION [13 upvotes]: This seems like an obvious fact, but I'm not sure what the necessary meaning of "nice" is to get a result like this. I'm wondering if there is a theorem of the form: 
For any <1> field extension $K/F$, a map from $\phi:K\rightarrow F$ that satisfies <2> is the field norm (or trace).
where <1> could be something like finite, algebraic, etc., and <2> could be anything (obviously there would be different <2>'s for norm and trace).

REPLY [13 votes]: Here is a nice characterization of the norm mapping on a finite extension of fields $K/k$. 
If $K/k$ is any finite extension of fields with degree $n$, then the norm mapping from $K$ to $k$ is the unique function $f \colon K \rightarrow k$ satisfying the following three conditions:
1) $f(xy) = f(x)f(y)$ for all $x$ and $y$ in $K$.
2) $f(c) = c^n$ for all $c$ in $k$.
3) $f$ is a polynomial function over $k$ of degree at most $n$, by which I mean there is a basis $\{e_1,\dots,e_n\}$ of $K/k$ relative to which $f$ can be described by a polynomial: there's a polynomial $P(x_1,\dots,x_n)$ in $k[x_1,\dots,x_n]$ such that $f(\sum_{i=1}^n c_ie_i) = P(c_1,\dots,c_n)$ for all $c_1,\dots,c_n$ in $k$. (Being a polynomial function is independent of the choice of basis.) 
This is due to Harley Flanders. See the following two articles of his:
The Norm Function of an Algebraic Field Extension, Pacific J. Math 3 (1953), 103--113.
The Norm Function of an Algebraic Field Extension, II, Pacific J. Math 5 (1955), 519--528. 
One nice consequence of this characterization of the norm, which Flanders points out, is that it gives a slick proof of the transitivity of the norm: if $K \supset F \supset k$ then the function 
${\rm N}_{F/k} \circ {\rm N}_{K/F}$ satisfies the three conditions that characterize ${\rm N}_{K/k}$.<|endoftext|>
TITLE: Set theory and alternative foundations
QUESTION [12 upvotes]: Every foundational system for mathematics I have ever read about has been a set theory, from ETCS to ZFC to NF. Are there any proposals for a foundational system which is not, in any sense, a set theory? Is there any alternative foundation which is not a set-theory?

REPLY [3 votes]: This may not satisfy the request for something that is "not, in any sense, a set theory" but Oliver Deiser has worked out two versions of foundations, one based on lists and one on multisets.  This is in his book "Orte, Listen, Aggregate" (and his Habilitationsschrift with the same title).<|endoftext|>
TITLE: Twin Prime Conjecture Reference
QUESTION [21 upvotes]: I'm looking for a reference which has the first statement of the twin prime conjecture. According to wikipedia, nova, and several other quasi-reputable resources it is Euclid who first stated it, but according to Goldston
http://www.math.sjsu.edu/~goldston/twinprimes.pdf
it was stated nowhere until de Polignac. I'm hoping to resolve this issue by accessing either primary historical documents, or other reputable secondary sources (Goldston being one such example). I have looked at de Polignac's work, and he does indeed make a conjecture, but have been unable to find anything definitive (besides Goldston's statements) that there was no conjecture earlier. If this is too specific for MO, I'll remove the question. Thank you.

REPLY [3 votes]: Hardy & Littlewood give (in 1923) what I believe to be the first quantitative version of the twin prime conjecture (actually, generalized to all even differences) as Conjecture B in their famous "Some problems of ‘Partitio numerorum’; III: On the expression of a number as a sum of primes" on page 42.  In particular, they conjecture that the twin-prime counting function $P_2(n)$ is
$P_2(n)=2C_2\frac{n}{(\log n)^2}(1+o(1))$ where $C_2=\prod\left(1-\frac{1}{(\varpi-1)^2}\right)$ with $\varpi$ running over the odd primes.<|endoftext|>
TITLE: Generalizations of Planar Graphs
QUESTION [36 upvotes]: This is a follow up to Harrison's question: why planar graphs are so exceptional. I would like to ask about (and collect answers to) various notions, in graph theory and beyond graph theory (topology; algebra) that generalize the notion of planar graphs, and how properties of planar graphs extend in these wider contexts.

REPLY [2 votes]: Wagner-Fáry-Stein theorem states that each (finite simple) planar graph admits a straight-line crossing-free plane drawing. On the other hand, each graph (of at most $\frak c$ vertices) admits a straight-line crossing-free three-dimensional space drawing
(any placement of vertices with no four at one plane generates such a drawing). Recently we investigate the minimum number of planes that together cover a straight-line drawing of a graph, which turned out to be independent on dimension of the space in which the graph is drawn, provided this dimension is at least three. Also, since any straight-line and circular-arc drawing can be transformed into a circular-arc drawing by an inversion map, we investigate the minimum number of spheres covering circular arc graph drawings.<|endoftext|>
TITLE: Trace map attached to a finite homomorphism of noetherian rings
QUESTION [19 upvotes]: Let $f:A\rightarrow B$ be a homomorphism of noetherian rings which
makes $B$ into a finite $A$-module.  Under what conditions on $f$, $A$,
$B$ can one associate to this map a canonical "trace map"
$$\mathrm{Tr}_f:B\rightarrow A,$$ 
i.e. a homomorphism of $A$-modules $B\rightarrow A$ which is compatible with base change (perhaps with restrictions on the kind of allowable base changes), localization, and
which recovers the "usual" thing when $B$ is free over $A$ (i.e.
$\mathrm{Tr}_f(b) =$ the trace of the $A$-linear endomorphism given by
multiplication by $b$ on the finite $A$-module $B$)?
Here are my thoughts so far:
1) If $f$ is flat, one always has $\mathrm{Tr}_f$.  Just work locally,
using that finite flat over a noetherian local ring is free.
2) More generally, if $f$ is of finite Tor-dimension, I can construct
$\mathrm{Tr}_f$ by taking a finite projective resolution
$$0\rightarrow P_n\rightarrow \cdots\rightarrow P_0\rightarrow B\rightarrow 0$$
of $B$ as an $A$-module: lift multiplication by $b$ on $B$ to a map of complexes
$b:P_{\bullet}\rightarrow P_{\bullet}$ and define
$$\mathrm{Tr}_f(b) := \sum_i (-1)^i \mathrm{Tr}_i(b)$$
where $\mathrm{Tr}_i(b)$ is the trace of the (lift of the)
endomorphism $b$ of $B$ to $P_i$.
One chekcs this is independent of the choice of projective resolution.
It commutes with "tor-independent" base change (sometimes called
"cohomologically transverse" base change).
3) If $A$ and $B$ are normal, I can construct $\mathrm{Tr}_f$ as follows: the localization of $f$ at any height-1 prime ideal is automatically flat by Matsumura 23.1 as the corresponding localizations of $A$ and $B$ are regular and the dimensions work out (the fiber ring is 0-dimensional as $f$ is finite).  Thus, one has a canonical trace map on each localization, and since $A$ and $B$ are normal, they are the intersections of these localizations so we win.
4) If $A$ and $B$ are only assumed reduced, one can look at the injections $A\rightarrow A'$
and $B\rightarrow B'$ with $A'$ and $B'$ the normalizations of $A$ and $B$ in their total rings of fractions.  Let $f':A'\rightarrow B'$ be the corresponding map.  By 3), we get a trace map
for $f'$ and the whole problem of constructing 
the trace map for $f$
comes down to showing that $\mathrm{Tr}_{f'}$ carries $B$ into $A$.  Letting $C_A:=\mathrm{ann}_{A'}(A'/A)$
be the conductor ideal (with $C_B$ defined similarly), I think that a necessary condition for 
$\mathrm{Tr}_{f'}(B)$ to be contained in $A$ is
$$f'(C_A) \supseteq C_B.$$
Is this condition sufficient?  As an example of how things can go wrong if this condition is 
violated, consider the finite map between reduced $k$-algebras ($k$ a field)
$$k[x,y]/(xy) \rightarrow k[x]$$ 
given by sending $y$ to 0.  The normalization of $k[x,y]/(xy)$ is the product $k[x]\times k[y]$
and the trace map attached to $f':k[x]\times k[y]\rightarrow k[x]$ sends $b\in k[x]$ to 
$(b,0)$.  But $(b,0)\in k[x]\times k[y]$ lies in the image of $k[x,y]/(xy)\rightarrow k[x]\times k[y]$
if and only if $b(0)=0$.  It follows that the trace map on normalizations doesn't restrict to a trace map on the original rings.
I'd be happy to assume that $A$ and $B$ are flat $R$-algebras, for a regular local ring $R$, and that $f:A\rightarrow B$ is an $R$-algebra homomorphism.  I'd also be happy to assume that $A$, $B$, and $f$ are local, with $A$ and $B$ reduced complete intersections over $R$.  I'm wondering if there is a framework for trace maps in this context that is general enough to handle the different constructions given in 1) -- 4) above.

REPLY [6 votes]: This is really a response to Karl's beautiful example; I'm posting it as an "answer" only because there isn't enough room to leave it as a comment.
The condition on conductor ideals is one that I had come across by thinking about the dual picture.
Namely, let $f:Y\rightarrow X$ be a finite map of 1-dimensional proper and reduced schemes over an algebraically closed field $k$.  Then $Y$ and $X$ are Cohen-Macaulay by Serre's criterion, so the machinery of Grothendieck duality applies.  In particular, the sheaves 
$f_*O_Y$ and $f_*\omega_Y$ are dual via the duality functor $\mathcal{H}om(\cdot,\omega_X)$, as are 
$O_X$ and $\omega_X$.  Here, $\omega_X$ and $\omega_Y$ are the ralative dualizing sheaves of
$X$ and $Y$, respectively.  Thus, the existence of a trace morphism $f_*O_Y\rightarrow O_X$
is equivalent by duality to the existence of a pullback map on dualizing sheaves $\omega_X\rightarrow f_*\omega_Y$.  
In the reduced case which we are in, one has Rosenlicht's explicit description of the dualizing sheaf: for any open $V$ in $X$, the $O_X(V)$-module $\omega_X(V)$ is exactly the set of
meromorphic differentials $\eta$ on the normalization $\pi:X'\rightarrow X$ with the property that $$\sum_{x'\in \pi^{-1}(x)} res_{x'}(s\eta)=0$$
for all $x\in V(k)$ and all $s\in O_{X,x}$.  
It is not difficult to prove that if $C$ is the conductor ideal of $X'\rightarrow X$ 
(which is a coherent ideal sheaf on $X'$ supported at preimages of non-smooth points in $X$), 
then one has inclusions 
$$\pi_*\Omega^1_{X'} \subseteq \omega_X \subseteq \pi_*\Omega^1_{X'}(C).$$
Since $X'$ and $Y'$ are smooth, so one has a pullback map on $\Omega^1$'s, our question
about a pullback map on dualizing sheaves boils down the following concrete question:
 
   When does the pullback map on meromorphic differentials $\Omega^1_{k(X')}\rightarrow \pi_*\Omega^1_{k(Y')}$ carry the subsheaf $\omega_X$ into $\pi_*\omega_Y$?

By looking at the above inclusions, I was led to conjecture the necessity of conductor ideal containment as in my original post.  As Karl's example shows, this containment is not sufficient.
Here is Karl's example re-worked on the dual side:
Set $B:=k[x,y]/(xy)$ and $A:=k[u,v]/(uv)$ and let $f:A\rightarrow B$ be the $k$-algebra map
taking $u$ to $x^2$ and $v$ to $y$.  Writing $B'$ and $A'$ for the normalizations, we have
$B'$ and $A'$ as in Karl's example, and the conductor ideals are $(x,y)$ and $(u,v)$.
Now the pullback map on meromorphic differentials on $A'$ is just
$$(f(u)du,g(v)dv)\mapsto (2xf(x^2)dx,g(y)dy).$$
The condition of being a section of $\omega_A$ is exactly 
$$res_0(f(u)du)+res_0(g(v)dv)=0,$$
and similarly for being a section of $\omega_B$.  But now we notice that
$$res_0(2xf(x^2)dx)+res_0(g(y)dy) = 2 res_0(f(u)du) + res_0(g(v)dv) = res_0(f(u)du)$$
if $(f(u)du,g(v)dv)$ is a section of $\omega_A$.  Thus, as soon as $f$ is not
holomorphic (i.e. has nonzero residue) the pullback of the section 
$(f(u)du,g(v)dv)$, as a meromorphic differential on $B'$, will NOT lie in the subsheaf $\omega_B$.
Clearly what goes wrong is that the ramification indices of the map $f:A'\rightarrow B'$ 
over the two preimages of the nonsmooth point are NOT equal.  With this in mind, I propose the following addendum to my original number 4):

  In the notation of 4) above and of Karl's post, assume that $f'(C_A)=C_B^e$
for some positive integer $e$.  Then the trace map $B'\rightarrow A'$ carries $B$
into $A$.

Certainly this rules out Karl's example.  I think another way of stating the condition is that the map $f':Spec(B')\rightarrow Spec(A')$ should be "equi-ramified" over the nonsmooth locus of $Spec(A)$, i.e. that the ramification indices of $f'$ over all $x'\in Spec(A')$ which map to the same nonsmooth point in $Spec(A)$ are all equal.
Is this the right condition?<|endoftext|>
TITLE: Why does the Gamma-function complete the Riemann Zeta function?
QUESTION [69 upvotes]: Defining   $$\xi(s) := \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)$$  yields      $\xi(s) = \xi(1 - s)$ (where $\zeta$ is the Riemann Zeta function).
Is there any conceptual explanation - or intuition, even if it cannot be made into a proof - for this? Why of all functions does one have to put the Gamma-function there?
Whoever did this first probably had some reason to try out the Gamma-function. What was it?
(Best case scenario) Is there some uniform way of producing a factor out of a norm on the rationals which yields the other factors for the p-adic norms and the Gamma factor for the absolute value?

REPLY [17 votes]: Multiple answers and comments have already pointed out that the conceptual role of $\pi^{-s/2}\Gamma(s/2)$ comes from the viewpoint of Iwasawa and Tate, which for $\text{Re}(s) > 1$ creates this function as $\int_{\mathbf R^\times} e^{-\pi x^2}|x|^s\,dx/|x|$, an integral over the multiplicative group $\mathbf R^\times$ of the function $e^{-\pi x^2}$ that is self-dual for the Fourier transform on the additive group $\mathbf R$ relative to the self-duality $\langle x,y\rangle = e^{2\pi ixy}$ or $\langle x,y\rangle = e^{-2\pi ixy}$ on $\mathbf R$. (If we use another self-duality of $\mathbf R$ then $e^{-ax^2}$ would be self-dual for some $a \not= \pi$ instead.)
It's also been said elsewhere on this page that there are many self-dual Schwartz functions on $\mathbf R$, or more specifically many even self-dual Schwartz functions on $\mathbf R$: for Schwartz $f$ on $\mathbf R$ and $\text{Re}(s) > 0$, we have $\int_{\mathbf R^\times} f(x)|x|^s\,dx/|x| = \int_{0}^\infty (f(x) + f(-x))x^s\,dx/x$ and this is $0$ when $f$ is odd, so we may as well assume $f$ is even since $f(x) + f(-x)$ is even anyway and we want to avoid the silly equation $0=0$ even if it is a valid equation.
For arbitrary Schwartz $f$ on $\mathbf R$, set $\Gamma_f(s) = \int_{0}^\infty f(x)x^s\,dx/x$, which is a mild modification of the function $\Gamma(f,s)$ in Paul Garrett's answer (his $\Gamma(f,s)$ is my $\Gamma_{f(x)+f(-x)}(s)$ by a formula I wrote in the previous paragraph). This function converges absolutely and is analytic for $\text{Re}(s) > 0$, and it extends meromorphically to $\mathbf C$ by repeated integration by parts (the same way the $\Gamma$-function can be extended to $\mathbf C$ from its integral definition for $\text{Re}(s) > 0$), and Tate's thesis shows there is a general functional equation $\Gamma_f(s)\zeta(s) = \Gamma_{\hat{f}}(1-s)\zeta(1-s)$ where $\hat{f}$ is the Fourier transform of $f$ (for the self-duality on $\mathbf R$ given by $\langle x,y\rangle = e^{-2\pi ixy}$), so if $f$ is self-dual then we get $$\Gamma_f(s)\zeta(s) = \Gamma_{f}(1-s)\zeta(1-s),$$ a very nice functional equation indeed, especially if we use even $f$ to avoid $0 = 0$. 
All of what I wrote so far has appeared explicitly or implicitly in some of the other comments or answers. Since there are many self-dual even Schwartz functions $f$ on $\mathbf R$, what is it about the choice $f(x) = e^{-\pi x^2}$, leading to $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$ (an extra $1/2$ on both sides of the functional equation can be cancelled) that is so nice? I have not seen the following property pointed out yet: with this choice of $f$ and familiarity with the $\Gamma$-function we know $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (in fact for $\text{Re}(s) > 0$), so therefore $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) > 1$ from $\zeta(s)$ being nonvanishing there, and then by the functional equation $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) < 0$, which means all zeros of $\Gamma_f(s)\zeta(s)$ have $0 \leq \text{Re}(s) \leq 1$.  If you want to use a totally random even Schwartz function for $f$ in order to define a factor $\Gamma_f(s)$ that completes the Riemann zeta-function, you will get the nice-looking nontrivial functional equation displayed above but how are you going to use $\Gamma_f(s)\zeta(s)$ to analyze the location of zeros of $\zeta(s)$ (including discovering its  trivial zeros, whether or not you consider those important) if you do not know where $\Gamma_f(s)$ has its zeros and poles? 
So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually know the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)?  The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders at most $1$ and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, which by the Fundamental Theorem of Calculus is $(-1/n!)(f^{(n)}(\infty) - f^{(n)}(0)) = f^{(n)}(0)/n!$.
Therefore you could determine the poles of $\Gamma_f$ by seeing when $f^{(n)}(0)$ is 0 and not 0, but how are you going to determine where the zeros of $\Gamma_f$ are or that there are no zeros? (EDIT: for even $f$, its odd-order derivatives vanish at $0$, so the residue at $-n$ vanishes when $n$ is odd, which means the poles of $\Gamma_f(s)$ can only be at $n = 0, -2, -4, -6, \ldots$. Those are all simple poles of $\pi^{-s/2}\Gamma(s/2)$, which has no zeros, so $G(s) := \Gamma_f(s)/(\pi^{-s/2}\Gamma(s/2))$ is an entire function. Thus $\Gamma_f(s) = G(s)\pi^{-s/2}\Gamma(s/2)$ with $G$ entire, so $\pi^{-s/2}\Gamma(s/2)$ a "holomorphic gcd" of all $\Gamma_f(s)$ for even Schwartz functions $f$ on $\mathbf R$. The exponential factor $\pi^{-s/2}$ was kind of irrelevant to drag through the calculation since it has no zeros or poles, but it's traditionally seen alongside $\Gamma(s/2)$ so I used it. This addresses comments below by Will Sawin and Venkataramana.)
Example: the function $f(x) = 1/(e^{\pi x} + e^{-\pi x})$ is an even self-dual Schwartz function on $\mathbf R$. Can someone determine in a self-contained way (i.e., not using $\zeta(s)$) where $\Gamma_f(s)$ has its zeros on $\mathbf C$, or determine if it has no zeros? 
Edit: Ignoring the wacky example just above, in some comments below I work out an example with $f(x)$ being a 4th degree Hermite polynomial times a Gaussian and find that $\Gamma_f(s)$ has two zeros with positive real part, at $s = (1\pm \sqrt{-2})/2$.<|endoftext|>
TITLE: Lax Functors and Equivalence of Bicategories?
QUESTION [21 upvotes]: Lax functors of bicategories were introduced at the very inception of bicategories, and I'm trying to get a better feel for them. They are the same as ordinary 2-functors, but you only require the existence of a coherence morphism, not an isomorphism. The basic example I'm looking at are when you have a lax functor from the singleton bicategory to a bicategory B. These are just object b in B with a monad T on B. 
My Question: If I have an equivalence of bicategories A ~ A', do I a get equivalent bicategories of lax functors Fun(A, B) and Fun(A', B)? If not, is there any relation between these two categories? 
[Edit] 
So let me be more precise on the terminology I'm using. I want to look at lax functors from A to B. These are more general then strong/pseudo and much more general then just strict functors. For a lax functor we just have a map like this:
$ F(x) F(g) \to F(fg)$
for a strong or pseudo functor this map is an isomorphism, and for strict functor it is an identity. I don't care about strict functors. 
I'm guessing that these form a bicategory Fun(A,B), with the 1-morphism being some sort of lax natural transformation, etc, but I don't really know about this. Are there several reasonable possibilities?
When I said equivalence between A and A' what I meant was I had a strong functor F:A --> A' and a strong functor G the other way, and then equivalences (not isomorphisms) FG = 1, and GF = 1. This seems like the most reasonable weak notion of equivalence to me, but maybe I am naive.
I haven't thought about equivalences using lax functors. Would they automatically be strong? What I really want to understand is what sort of functoriality the lax functor bicategories Fun(A, B) have?

REPLY [10 votes]: I was surprised (and delighted) to discover that the answer is no.  Here is an example; in fact it's a special case of the example Mike referred to, but worked out in more detail.
Let S be a set and cd(S) the codiscrete category on S (Hom(x, y) = • for every x and y in S).  Let BSet denote the one-object bicategory corresponding to the monoidal category (Set, ×).  As remarked here on the nlab, a lax functor cd(S) → BSet is the same data as a category with object set S.  However, there is a catch: the morphisms (whether lax or strong natural transformations) are not just functors between categories fixing the objects.  Instead, if C and D are categories corresponding to two lax functors cd(S) → BSet, then a morphism (X, F) from C to D consists of

for each object s of S, a set X(s),
for each pair of objects s and t of S, a map F(s, t) : C(s, t) × X(t) → X(s) × D(s, t),
such that obvious identity and composition laws hold.

Depending on which notion of natural transformation you want to take, the map F(s, t) might be an isomorphism, or it might go the other way.  I'll consider all three possibilities simultaneously.
We can compose 1-morphisms: the composition (Z, H) of (X, F) and (Y, G) has Z(s) = X(s) × Y(s), and H formed from F and G in an obvious way.  The identity morphism has X(s) = • and F(s, t) = id.
We also have 2-morphisms from (X, F) to (Y, G), which are families of maps X(s) → Y(s) making some diagrams involving X, F, Y, G commute.
This doesn't yet seem to be any familiar 2-category.  Let's extract the maximal 2-groupoid by taking only invertible 1- and 2-morphisms.  (That's certainly an equivalence-invariant thing to do.)  The invertible 2-morphisms are the ones where the maps X(s) → Y(s) are isomorphisms.  For (X, F) to be invertible, we first need for every s, an object Y(s) and an isomorphism X(s) × Y(s) = •.  That can only happen if X(s) = • for every s.  (This is the only property of (Set, ×) I really care about: it has no invertible objects besides •.)  Now F is just an ordinary functor C → D which is the identity on objects, and for (X, F) to be invertible F needs to be an isomorphism.  When (X, F) and (Y, G) are invertible, so X(s) = Y(s) = •, there is at most one 2-morphism between them, exactly when F and G are equal as maps C(s, t) → D(s, t).  In conclusion, the maximal 2-groupoid contained in the 2-category of lax functors cd(S) → BSet is the 1-groupoid of categories with object set S and isomorphisms fixing the objects.  Note that it didn't matter for the argument which notion of natural transformation I chose.
Now as S varies over nonempty sets, these resulting groupoids are nonequivalent categories.  However, surely the 2-categories cd(S) are all equivalent under any reasonable definition, since they equivalent 1-categories.<|endoftext|>
TITLE: What is the geometric meaning of reconstruction of quantum group via Ringel Hall algebra
QUESTION [6 upvotes]: If I remembered correctly. There are some work done by C.M.Ringel,he defined so called Ringel-Hall algebra on abelian category and then show that Ringel-hall algebra is isomorphic to positive part of quantized enveloping algebra. Some others generalized to triangulated(derived category of coherent sheaves on projective spaces,so on so forth)
I wonder know whether there exists some explicit geometric explanation to these works. I think we can consider the quantized enveloping algebra as noncommutative affine scheme. Therefore, it seems these work are doing some sort of reconstruction of schemes in abelian category or derived category. 
I wonder whether Bondal-Orlov's work has any relationship(explicit)to these stuff. Therefore, what I am really interested in is what is the real meaning of Ringel-Hall algebra
At last, it is well known that Belinson-Bernstein's theorem established the equivalence of category of U(g)-module and category of D-modules on flag variety of Lie algebra. And some others, say Bezrukavnikov,Frenkel,gaitsgory generalized these results to Kac-Moody algebra.On the other hand, Van den Berg used generalized Ringel Hall algebra to realize quantum group of Kac-Moody algebra. I wonder whether anybody here can say something about these stuff.
All the other comments are welcomed

REPLY [2 votes]: I would not say that the Hall algebra construction is some sort of "reconstruction" (to me, reconstruction means that you find an algebraic object so that you can  reconstruct the category as the representation category of this object). 
I agree with Ben Webster that it is better related to categorification. For a more geometric approach to Hall algebras I would recommend to look at the work of Olivier Schiffmann.<|endoftext|>
TITLE: S-unit equation and small sets of places
QUESTION [7 upvotes]: Let $K$ be a number field, and let $S_x$ denote the set of primes of norm at most $x$. Is it possible to find a smaller set of places $T_x\subset S_x$ so that a lot of the solutions of the $S_x$-unit equation $a+b=1$ for $a,b\in S_x$ are solutions of the $T_x$-unit equation?
Here's a possible precise statement (although I'd be interested in other formulations as well): Does there exist a constant $0<c<1$, depending on $K$ (but not $x$), so that for each $x$, there is a $T_x\subset S_x$ with $|T_x|\le\sqrt{|S_x|}$ so that the number of solutions to the $T_x$-unit equation is at least $c$ times the number of solutions of the $S_x$-unit equation?
I'm interested in this mostly by analogy: Belabas and Gangl have a bound for the set of places of a number field one must check in order to compute $K_2$ of the ring of integers. It would be interesting to know if one could at least get a pretty good approximation for $K_2$ by looking at a much smaller set of places.

REPLY [2 votes]: This is an old question, but maybe it's worth mentioning that a conjecture of Erdös, Stewart, and Tijdeman [1] says that if $P_s=\{2,3,\ldots,p_s\}$ is the set of the first $s$ primes, then the number of solutions to the $S$-unit equation in $\mathbb Q$ should satisfy
$$ \exp(s^{2/3-\epsilon}) \le \text{\# of sols using $P_s$} \le \exp(s^{2/3+\epsilon})
\quad \text{for all $s>s_0(\epsilon)$}. $$
If this is true, and if (as Bjorn suggests) the smaller primes are the ones yielding the most solutions, this would show that if one takes the square root of the number of primes, then the number of solutions decreases by a large (and reasonably calculable) amount.
In the same article, they prove that there are sets $S$ containing $s$ primes so that the number of solutions is roughly
$$ \exp\left( \sqrt{\frac{s}{\log s}} \right). $$
Their clever construction take a lot of initial primes and throws in some carefully selected larger primes.
[1]  Erdös, P.; Stewart, C. L.; Tijdeman, R.
Some Diophantine equations with many solutions.
Compositio Math. 66 (1988), no. 1, 37–56. MR0937987<|endoftext|>
TITLE: Clifford algebra as an adjunction?
QUESTION [40 upvotes]: Background
For definiteness (even though this is a categorical question!) let's agree that a vector space is a finite-dimensional real vector space and that an associative algebra is a finite-dimensional real unital associative algebra.
Let $V$ be a vector space with a nondegenerate symmetric bilinear form $B$ and let $Q(x) = B(x,x)$ be the associated quadratic form.  Let's call the pair $(V,Q)$ a quadratic vector space.
Let $A$ be an associative algebra and let's say that a linear map $\phi:V \to A$ is Clifford if
$$\phi(x)^2 = - Q(x) 1_A,$$
where $1_A$ is the unit in $A$.
One way to define the Clifford algebra associated to $(V,Q)$ is to say that it is universal for Clifford maps from $(V,Q)$.  Categorically, one defines a category whose objects are pairs $(\phi,A)$ consisting of an associative algebra $A$ and a Clifford map $\phi: V \to A$ and whose arrows
$$h:(\phi,A)\to (\phi',A')$$
are morphisms $h: A \to A'$ of associative algebras such that the obvious triangle commutes:
$$h \circ \phi = \phi'.$$
Then the Clifford algebra of $(V,Q)$ is the universal initial object in this category.  In other words, it is a pair $(i,Cl(V,Q))$ where $Cl(V,Q)$ is an associative algebra and $i:V \to Cl(V,Q)$ is a Clifford map, such that for every Clifford map $\phi:V \to A$, there is a unique morphism
$$\Phi: Cl(V,Q) \to A$$
extending $\phi$; that is, such that $\Phi \circ i = \phi$.
(This is the usual definition one can find, say, in the nLab.)
Question
I would like to view the construction of the Clifford algebra as a functor from the category of quadratic vector spaces to the category of associative algebras.  The universal property says that if $(V,Q)$ is a quadratic vector space and $A$ is an associative algebra, then there is a bijection of hom-sets
$$\mathrm{hom}_{\mathbf{Assoc}}(Cl(V,Q), A) \cong \mathrm{cl-hom}(V,A)$$
where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms.
My question is whether I can view $Cl$ as an adjoint functor in some way.  In other words, is there some category $\mathbf{C}$ such that the right-side is
$$\mathrm{hom}_{\mathbf{C}}((V,Q), F(A))$$
for some functor $F$ from associative algebras to $\mathbf{C}$.  Naively I'd say $\mathbf{C}$ ought to be the category of quadratic vector spaces, but I cannot think of a suitable $F$.
I apologise if this question is a little vague.  I'm not a very categorical person, but I'm preparing some notes for a graduate course on spin geometry next semester and the question arose in my mind.

REPLY [14 votes]: Here is another proposal. The key will be to enlarge the category of quadratic vector spaces fairly substantially. Here are three hints leading towards the proposal:

Clifford algebras can and should be thought of as $\mathbb{Z}_2$-graded; for example, Clifford algebras over $\mathbb{R}$ give every element of the $\mathbb{Z}_2$-graded Brauer group / Brauer-Wall group of $\mathbb{R}$.
Thinking of Clifford algebras as deformations of exterior algebras, the analogous deformations of symmetric algebras are the Weyl algebras. It is possible to combine the construction of exterior algebras and symmetric algebras into a single construction, namely the construction of the symmetric algebra on a super vector space.
Weyl algebras are almost universal enveloping algebras of certain Lie algebras. More precisely, let $(V, \omega)$ be a symplectic vector space. From this data we can construct a Lie algebra $V \oplus \mathbb{R}$ such that $\mathbb{R}$ is central and $[v, w] = \omega(v, w) \in \mathbb{R}$ where $v, w \in V$. Then the Weyl algebra constructed from $(V, \omega)$ is the quotient of the universal enveloping algebra $U(V \oplus \mathbb{R})$ by the extra relation that $1 \in \mathbb{R}$ acts as the identity. 

So let's try this: 

There is a forgetful functor from super algebras to a certain category of super Lie algebras with extra structure whose left adjoint restricts to 1) the symmetric algebra functor, 2) the exterior algebra functor, 3) the Weyl algebra functor, 4) the Clifford algebra functor, and 5) the universal enveloping algebra functor on suitable subcategories. 

This left adjoint generalizes every functor discussed above. Some details: 
The first category $\text{SAlg}$, the category of super algebras, is the category of monoid objects in super vector spaces. As a category it can be thought of as the category of $\mathbb{Z}_2$-graded algebras, but as a symmetric monoidal category it has a nontrivial braiding given by the Koszul sign rule as usual. In particular, a commutative super algebra is commutative in the super sense, not the usual sense.
The second category begins from the category $\text{SLieAlg}$ of super Lie algebras, which is the category of Lie algebra objects in super vector spaces; here it's quite important to distinguish this category from the category of $\mathbb{Z}_2$-graded Lie algebras because the Koszul sign rule changes some signs in the Lie algebra axioms. In particular, the skew-symmetry axiom becomes
$$[x, y] = - (-1)^{|x| |y|} [y, x]$$
for homogeneous elements $x, y$. Hence if either $x$ or $y$ is even then this is skew-symmetry in the usual sense, but if $x$ and $y$ are both odd then we actually have symmetry. This is crucial.
The category we're actually interested is not quite this category; instead it is the category of super Lie algebras $\mathfrak{g}$ equipped with a morphism $\mathbb{R}[0] \to \mathfrak{g}$, where $\mathbb{R}[0]$ denotes the abelian Lie algebra $\mathbb{R}$ in degree $0$, with central image. This might be called the category of "centrally pointed super Lie algebras," maybe. 
The forgetful functor from $\text{SAlg}$ to the above category sends a super algebra $A$ to the underlying super vector space of $A$ equipped with the super Lie bracket
$$[x, y] = xy - (-1)^{|x| |y|} yx$$
on homogeneous elements, with the map $\mathbb{R}[0] \to A$ being given by the unit element of $A$.
The left adjoint to this forgetful functor sends $\mathbb{R}[0] \to \mathfrak{g}$ to the quotient of the universal enveloping (super) algebra $U(\mathfrak{g})$ by the extra relation that $1 \in \mathbb{R}[0]$ acts as the identity. I'll also call this functor $U$. Here are the five promised subcategories (they are not full subcategories):

Given a vector space $V$, send it to the abelian super Lie algebra $V[0] \oplus \mathbb{R}[0]$. Then $U(V[0] \oplus \mathbb{R}[0])$ is the symmetric algebra on $V$. 
Given a vector space $V$, send it to $V[1] \oplus \mathbb{R}[0]$. Then $U(V[1] \oplus \mathbb{R}[0])$ is the exterior algebra on $V$.
Given a symplectic vector space $(V, \omega)$, send it to $V[0] \oplus \mathbb{R}[0]$ with bracket given by $\omega$ as previously discussed. Then $U(V[0] \oplus \mathbb{R}[0])$ is the Weyl algebra of $(V, \omega)$.
Given a quadratic vector space $(V, Q)$, let $B(v, w) = Q(v + w) - Q(v) - Q(w)$ be twice the symmetric bilinear form determined by $Q$ and send it to $V[1] \oplus \mathbb{R}[0]$ with bracket given by $B$, as previously discussed for the Weyl algebra. Then $U(V[1] \oplus \mathbb{R}[0])$ is the Clifford algebra of $(V, Q)$.
Given a Lie algebra $\mathfrak{g}$, send it to $\mathfrak{g}[0] \oplus \mathbb{R}[0]$. Then $U(\mathfrak{g}[0] \oplus \mathbb{R}[0])$ is the usual universal enveloping algebra of $\mathfrak{g}$.<|endoftext|>
TITLE: The category of finite locally-free commutative group schemes
QUESTION [11 upvotes]: I'm trying to understand the properties of the category $\mathcal{FL}/S$ of finite locally-free commutative group schemes over an arbitrary base-scheme $S$. I know it is not in general an abelian category: Over the integers the morphism $\mathbb{Z}/2\mathbb{Z}\to\mu_2$ given by the ring homomorphism $\mathbb{Z}/(T^2-1)\to\mathbb{Z}\times\mathbb{Z}$ with $T\mapsto (1,-1)$ is a monomorphism and an epimorphism in $\mathcal{FL}/\mathbb{Z}$ but it is clearly not an isomorphism.
What I don't know: Does every morphism in $\mathcal{FL}/S$ have a kernel and/or a cokernel?
One has a notion of short exact sequences in $\mathcal{FL}/S$: If $f:G'\to G$ and $g:G\to G''$ are morphisms in $\mathcal{FL}/S$ we say that the sequence $0\to G'\to G\to G''\to 0$ is exact if its image under the embedding of $\mathcal{FL}/S$ into the category of abelian fppf-sheaves is exact. How can we detect in $\mathcal{FL}/S$ whether a sequence is exact? Specifically: If $f$ is a kernel of $g$ and if $g$ is a cokernel of $f$ (both in $\mathcal{FL}/S$), is the corresponding sequence exact?
I'd appreciate (references for) answers to any of these questions.

REPLY [2 votes]: If the base $S$ is Spec of a Dedekind domain, say $A$, and one restricts to finite flat groups schemes with etale generic fibres (which is no restriction if the fraction field of $A$ has char. 0), then one can form kernels and cokernels in the category of finite flat group schemes by taking scheme-theoretic closures of the corresponding notions on the generic fibre, and this can be useful.  (Of course, one doesnt' get an abelian category: the map
${\mathbb Z}/2 {\mathbb Z} \rightarrow \mu_2$ has trivial kernel and cokernel, but is not an isomorhpism.)  
An illustration of how this method can be used is given in e.g. in my joint paper with Calegari On the
ramification of Hecke algebras at Einstein primes.
(The point of the restriction to a Dedekind base is that in this case, torsion free, which is what you get with scheme-theoretic closure from the generic fibre, coincides with flat.)<|endoftext|>
TITLE: Why is Riemann-Roch an Index Problem?
QUESTION [17 upvotes]: I was in a lecture not long ago given by C. Teleman and at some point he said "Well, since Riemann-Roch is an index problem we can do..."
Then right after that he argued in favour of such a sentence. Could anyone tell me what did he mean exactly?. That is to say, in this case what is elliptic operator like, what is the heuristic idea which such a result relies on? ...and a little bit of more details about it.
As usual references will be appreciated.
ADD: Thanks for the comments below, but I think they do not answer the question of title : Why is RR an Index problem?. Up to this point, what I can see is that two numbers happened to be the same.

REPLY [11 votes]: Here is a self contained answer to why Riemann's original theorem as he proved it, before Roch's refinement, is indeed an index statement.
Traditionally, the “index” of a linear operator is the difference between the dimensions of its kernel and cokernel. Computing this difference is usually easier than computing either the kernel or cokernel dimensions alone, hence it is a helpful first step even where one of those dimensions is really wanted.  Moreover the index gives an estimate of the kernel dimension. Riemann’s proof of his theorem was to calculate such an index as follows.
Given a divisor $D$ of $r$ distinct points on a compact complex Riemann surface $X$ of genus g, he constructed a basis of the $g+r$ dimensional space $W \cong \mathbb{C}^{r+g}$ of corresponding differentials of “second kind”, i.e. having at worst principal part $\operatorname{const}.dz/(z-p)^2$ at each point $p$. Since the space $L(D)$ of meromorphic functions with at worst simple poles at these points maps with one dimensional kernel into the space $W$, in order to compute the dimension of $L(D)$ it suffices to compute the subspace of exact differentials in $W$, i.e. the kernel of the “period map”, obtained by integrating these differentials over a basis for the 1st homology of $X$. 
Thus he wants to compute the kernel of a linear map from $\mathbb{C}^{r+g}$ to $\mathbb{C}^{2g}$. It follows immediately from the rank/nullity theorem that the index of this period map is $(r+g)-2g = r-g = \deg(D)-g$. Hence this is a lower bound for the kernel, so $\dim L(D) – 1 \geq r-g$, i.e. $\dim L(D) \geq \deg(D) + 1-g$.
This also implies the modern index form of the theorem as follows. Riemann knew the period map is injective on holomorphic differentials, so he could, and Roch did, replace it by a normalized period map from $\mathbb{C}^r$ to $\mathbb{C}^g \cong H^1(X;\mathbb{C})/H^1(X; K) \cong H^1(X;\mathcal{O})$. The cokernel of this map is now called $H^1(X;D)$, so Riemann’s theorem can be phrased $\chi(D) = h^0(D)-h^1(D) = \deg(D) + 1-g$. Since it takes some work to compute $\chi(\mathcal{O}) = h^0(\mathcal{O})-h^1(\mathcal{O})$ as $1-g$, this is actually stronger than the modern sheaf theoretic result that $\chi(D) – \chi(\mathcal{O}) = \deg(D)$, which follows immediately from the sheaf sequence $0 \to \mathcal{O} \to \mathcal{O}(D) \to \mathcal{O}(D)|_D \to 0$.
Of course Riemann’s result should be stronger since it contains in addition to the trivial sheaf theoretic linear algebra, also the computation of both the holomorphic genus $= h^0(X; K)$, and the topological genus $= (1/2)h^0(X; \mathbb{C})$. Thus Riemann’s theorem combines the two modern results: $\chi(D) -\chi(\mathcal{O}) = \deg(D)$, and $\chi(\mathcal{O}) = 1-g$.
Roch’s refinement of the RRT is to compute the cokernel $h^1(X;D)$ of Riemann’s period map, which he does of course by making a residue calculation. He obtains that $h^1(X;D) = h^0(X;K(-D))$, hence $h^0(D)-h^0(K(-D)) = \deg(D)+1-g$, the full RRT.<|endoftext|>
TITLE: Statements in group theory which imply deep results in number theory
QUESTION [21 upvotes]: Can we name some examples of theorems in group theory which imply (in a relatively straight-forward way) interesting theorems or phenomena in number theory?
Here are two examples I thought of:
The existence of Golod-Shafarevich towers of Hilbert class fields follows from an inequality on the dimensions of the first two cohomology groups of the ground field.
Iwasawa's theorem on the size of the $p$ part of the class groups in $\mathbb{Z}_p$-extensions follows from studying the structure of $\mathbb{Z}_p[\![T]\!]$-modules.
Can you name some others?

REPLY [4 votes]: This is not a straightforward example, but the Oppenheim conjecture was proved originally by Margulis using ergodic theory and group theory.

REPLY [4 votes]: A conjecture was made by Dunfield and Calegari that certain congruence covers of an arithmetic hyperbolic 3-manifold have trivial first betti number (which corresponds to the non-existence of certain automorphic forms, conjectured based on the generalized Riemann hypothesis and the Langlands proram). This was subsequently proved by Boston and Ellenberg using methods from pro-$p$ groups.<|endoftext|>
TITLE: What is "restriction of scalars" for a torus?
QUESTION [16 upvotes]: I am starting on a Phd program and am supposed to read Colliot Thelene and Sansuc's article
on R-equivalence for tori. I find it very difficult and although I have some knowledge over schemes , I am completely baffled by this scalar restriction business of having a field extension $K/k$ , a torus over $K$ and "restricting" it to $k$.
I would be very gratefull for a reference or even better by some explanation . I found nothing in my standard books (Hartshorne, Qing Liu, Mumford etc) so I hope this question is appropriate for the site. Thank you.

REPLY [10 votes]: There is another description for tori. The category of tori over a field F is equivalent to the category of finite-dimensional $G_F$-lattices. Now, there is an operation of induction for group representations that converts a  $G_K$-lattice into a  $G_k$-lattice; this is the lattice you need.
See http://en.wikipedia.org/wiki/Algebraic_torus<|endoftext|>
TITLE: What methods exist to prove that a finitely presented group is finite?
QUESTION [22 upvotes]: Suppose I have a finitely presented group (or a family of finitely presented groups with some integer parameters), and I'd like to know if the group is finite. What methods exist to find this out? I know that there isn't a general algorithm to determine this, but I'm interested in what plans of attack do exist. 
One method that I've used with limited success is trying to identify quotients of the group I start with, hoping to find one that is known to be infinite. Sometimes, though, your finitely presented group doesn't have many normal subgroups; in that case, when you add a relation to get a quotient, you may collapse the group down to something finite.
In fact, there are two big questions here:

How do we recognize large finite simple groups? (By "large" I mean that the Todd-Coxeter algorithm takes unreasonably long on this group.) What about large groups that are the extension of finite simple groups by a small number of factors?
How do we recognize infinite groups? In particular, how do we recognize infinite simple groups?

(For those who are interested, the groups I am interested in are the symmetry groups of abstract polytopes; these groups are certain nice quotients of string Coxeter groups or their rotation subgroups.)

REPLY [6 votes]: If a discrete group is amenable and has Kazhdan's Property (T), then it is finite.  
I'm not sure it would help you, but it's a technique.
This technique was used by Margulis in his original proof of his Normal Subgroup Theorem.  It's since been used in a couple of other Normal Subgroup Theorems, which have been applied to prove simplicity of some infinite groups.  See for example "Lattices in products of trees" by Burger and Mozes, and "Simplicity and superrigidity of twin building lattices" by Caprace and Remy.<|endoftext|>
TITLE: Diameter of m-fold cover
QUESTION [30 upvotes]: Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that 
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:

This is a modification of a problem of A. Nabutovsky.  Here is yet related question about universal covers.
You can reformulate it for compact length metric space --- no difference. 
The answer is YES if the cover is regular (but that is not as easy as one might think).
The estimate $\mathop{diam}\tilde M\le 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial.
We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.

REPLY [26 votes]: I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary.
Split $\gamma$ into $m$ arcs $a_1,\dots,a_n$ of equal length: $\gamma=a_1a_2\dots a_m$, $L(a_i)=L(\gamma)/m>diam(M)$. Let $b_i$ be a shortest path in $M$ connecting the endpoints of $a_i$. Note that $L(b_i)\le diam(M)< L(a_i)$. I want to replace some of the components $a_i$ of the path $\gamma$ by their "shortcuts" $b_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction.
To switch from $a_i$ to $b_i$, you left-multiply $\gamma$ by a loop $l_i:=a_1a_2\dots a_{i-1}b_i(a_1a_2\dots a_i)^{-1}$. More precisely, if you replace the arcs $a_{i_1},a_{i_2},\dots,a_{i_k}$, where $i_1< i_2<\dots< i_k$, by their shortcuts, the resulting path is homotopic to the product $l_{i_1}l_{i_2}\dots l_{i_k}\gamma$.
So it suffices to find a product $l_{i_1}l_{i_2}\dots l_{i_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$.
Now consider the following $m+1$ elements of $\pi_1(M,p)$: $s_0=e$, $s_1=l_1$, $s_2=l_1l_2$, $s_3=l_1l_2l_3$, ..., $s_m=l_1l_2\dots l_m$. Two of them, say $s_i$ and $s_j$ where $i< j$, are in the same left coset. Then  $s_i^{-1}s_j=l_{i+1}l_{i+2}\dots l_j\in H$ and we are done.<|endoftext|>
TITLE: The work of Thurston
QUESTION [12 upvotes]: I seem to remember written or said somewhere that at some point Thurston decided to stop writing down his theorems in order not to repel mathematicians from his field (maybe this is not correct?). I am really curious if now 25-30 years later there is some nice source, book, or notes, where it is possible to learn some basic ideas about the proof of the fact that Haken manifolds admit a hyperbolic structure? Maybe some of his ideas got a more accessible explanation? Of course his beautiful notes http://www.msri.org/publications/books/gt3m/ exist, but they don't go so far.

REPLY [13 votes]: This isn't a direct answer to the actual question, but in your first sentence I think you're alluding to Thurston's article  On proof and progress in mathematics.  In section 6, entitled "Some personal experiences", he describes how his experience working on foliations influenced the way he presented his later work on geometrization.<|endoftext|>
TITLE: How to see meromorphicity of a function locally?
QUESTION [7 upvotes]: Given a germ of an analytic function on a (compact, for simplicity) Riemann surface, how can one see (locally) whether this is a "germ of meromorphic function"?  I.e. if I do analytic continuation along various paths, how can I be sure sure that I will never see an essential singularity?
Another formulation of this question is, how can one determine whether a convergent taylor series determines a meromorphic function on the universal covering space of the Riemann surface? 
The fact that there will be no essential singularity certainly implies something, e.g. when our Riemann surface is CP^1, then for a taylor series to be meromorphic, it must be rational. But how do one check this locally, in a nbhd of a point?
Thanks
P.S. I don't really know how to tag this question. Suggest a tag in comment please if possible.

REPLY [7 votes]: Checking whether a function is rational locally is straightforward, since the Taylor coefficients satisfy a very strong structure theorem.  There must exist complex numbers $\alpha_1, ... \alpha_k$ and polynomials $P_1, P_2, ... P_k$ such that the Taylor coefficients satisfy
$\displaystyle f_n = \sum_{i=1}^{k} P_i(n) \alpha_i^n$
for all but finitely many $n$.  The degree of $P_i$ is one less than the order of the pole at $\alpha_i$.  Among other things it follows that asymptotically we have $f_n \sim A \alpha^n n^k$ for some $A, \alpha \in \mathbb{C}$ and some non-negative integer $k$, and this is a very strong condition.  However, I don't know what the situation is like for other Riemann surfaces.<|endoftext|>
TITLE: Do DG-algebras have any sensible notion of integral closure?
QUESTION [23 upvotes]: Suppose R → S is a map of commutative differential graded algebras over a field of characteristic zero.  Under what conditions can we say that there is a factorization R → R' → S through an "integral closure" that extends the notion of integral closure in degree zero for connective objects, and respects quasi-isomorphism of the map R to S?
I'm willing to accept answers requiring R → S may need to have some special properties.  The motivation for this question is that I'd actually like a generalization of this construction to ring spectra; the hope is that if there is a canonical enough construction on the chain complex level, it has a sensible extension to contexts where it's usually very difficult to construct "ring" objects directly.

REPLY [17 votes]: I like this question a lot. It deserves an answer, and I really wish I had a good one. Instead, I offer the following idea. Maybe it has some merit?
Background
Let me fix some terminology. Suppose $f:R\to S$ a homomorphism of (classical, commutative) rings.

An element $s\in S$ is said to be integral over $R$ if there is a monic polynomial $p\in (R/\ker f)[x]$ such that $s$ is a root of $p$; this is equivalent to saying that the subring $(R/\ker f)[s]\subset S$ is finite over $(R/\ker f)$.
We say that $f$ is integrally closed if it is a monomorphism and if every element of $S$ that is integral over $R$ is in $R$.
At the opposite extreme, we say that $f$ is integrally surjective if every element of $S$ is integral over $R$. (This turns out to be equivalent to being a colimit of proper homomorphisms of finite presentation.)
Among the integrally surjective homomorphisms are the elementary integrally surjective homomorphisms, i.e., homomorphisms of the form $R\to (R/\mathfrak{a})[x]/(p)$, where $R$ is of finite presentation, $\mathfrak{a}\subset R$ is a finitely generated ideal, and $p$ is any monic polynomial.

The classical integral closure construction can now be described as a unique factorization of every homomorphism $f:R\to S$ into an integrally surjective homomorphism $R\to\overline{R}$ followed by an integrally closed monomorphism $\overline{R}\to S$.
In §3.6 of Mathieu Anel's (really cool!) paper, he describes this factorization system and the "proper topology" constructed from it. In particular, he observes that integrally closed monomorphisms are precisely those morphisms satisfying the unique right lifting property with respect to all elementary integrally surjective homomorphisms.
Making this work for $E_{\infty}$ ring spectra
Since you expressed interest in getting integral closure off the ground for $E_{\infty}$ ring spectra, I'll work in that context.
We can use Andre Joyal's theory of factorization systems in ∞-categories (see §5.2.8 of Jacob Lurie's Higher Topos Theory) to try to play this same game in the ∞-category $\mathcal{C}$ of connective $E_{\infty}$ ring spectra. (For reasons that will become clear, I'm worried about making this fly for nonconnective $E_{\infty}$ ring spectra.) Once a set $I$ of elementary integrally surjective morphisms is selected, integrally closed monomorphisms are determined as the class of maps that are right orthogonal to $I$, and the integral closure construction is a factorization system on $\mathcal{C}$.
So what should $I$ be? I think there might be some flexibility here, depending on your aims, but here's a proposal:

Start with the coherent connective $E_{\infty}$ ring spectra that are of finite presentation (over the sphere spectrum). (Concretely, these are the connective $E_{\infty}$ ring spectra $A$ with the following properties: (1)  $\pi_0A$ is of finite presentation, (2) for every integer $n$, $\pi_nA$ is a finitely presented module over $\pi_0A$, and (3) the absolute cotangent complex $L_A$ is a perfect $A$-module.) We'll only need these kinds of $E_{\infty}$ ring spectra in our construction of $I$.
Among these $E_{\infty}$ ring spectra, consider the set $I'$ of all morphisms $A\to B$ of finite presentation that induce a surjection on $\pi_0$. Let's call the maps of $I'$ quotients.
Now we need to enlarge $I'$ to allow ourselves morphisms that act as though they are of the form $A\to A[x]/(p)$ for $p$ monic. For any of our connective $E_{\infty}$ ring spectra $A$, we can consider any finitely generated and free $E_{\infty}$-$A$-algebra $A[X]$ (i.e., the symmetric algebra on some free and finitely generated $A$-module), and we can consider quotients (in the sense above) $A[X]\to B$ where $B$ is almost perfect as an $A$-module (equivalently, $\pi_nB$ is finitely presented as an $\pi_0A$-module for every integer $n$); let us add the resulting composites $A\to A[X]\to B$ to our set $I'$ to obtain the set $I$.

Now the morphisms that are right orthogonal to $I$ can be called integrally closed monomorphisms of $E_{\infty}$ ring spectra; call the set of them $S_R$. The morphisms that are left orthogonal to that can be called the integrally surjective morphisms of $E_{\infty}$ ring spectra. The integral closure would be a factorization system $(S_L, S_R)$. One shows the existence of a factorization (via a presentation argument), and it follows from general nonsense (more precisely, 5.2.8.17 of HTT) that it is unique.
Three observations

It's not clear that $I$ is big enough for all purposes. One might want to allow shifts of free modules to generate our finitely generated and free $E_{\infty}$-$A$-algebras in the description above. I haven't thought carefully about this.
It's not so obvious how to talk about quotients $A\to B$ when dealing with nonconnective guys. One wants to say that the fiber (in the category of $A$-modules) is "not any more nonconnective than $A$." I'm not quite sure how to formulate this. In any case, that's why I restricted attention to the connective guys above.
Predictably, this is definitely not compatible with the usual integral closure: if I take two classical rings $R$ and $S$ and a ring homomorphism $R\to S$, the integral closure of $HR$ in $HS$ is not in general an Eilenberg-Mac Lane spectrum. If $R$ is a $\mathbf{Q}$-algebra, the two notions are compatible, however.

Making any computation
... seems really hard. But maybe that's not such a big surprise. After all, integral closures are hard to compute classically as well.<|endoftext|>
TITLE: Universal property for collection of epimorphisms
QUESTION [8 upvotes]: Question Is there a nice universal property which captures the notion of "collection of all epimorphisms  out of a given object". Of course I will have to consider two epimorphisms $X \rightarrow Y$ the same if they are isomorphic over $X$. The answer to the dual question is yes, at least in a topos: The power object $P(X)=\Omega^X$ , where $\Omega$ is the subobject classifier can be thought of as "the collection of all subobjects of X".  The universal property is just the property for exponentials. 
Background (Not strictly necessary for the question): I have been reading Sheaves in Geometry and Logic by Mac Lane and Moerdijk.  Their definition of an elementary topos is this: A category with pullbacks, a terminal object (i.e. all finite limits), a subobject classifier, and a power object for every object.  They construct all other exponential objects from these axioms.  The construction they use is to basically consider the "collection" of all graphs of morphisms. This is just the standard construction in set theory suped up to toposes.
This construction agrees with the set theoretic convention that a function should be regarded as a set of ordered pairs, i.e. if $f:A \rightarrow B$, then the set theorist will define $f$ as the image of the map $A \rightarrowtail A \times B$ induced by the $1_A$ and $f$ (this may be the most convoluted sentence I have ever written).  Why not define functions dually? There is also a map $A+B \twoheadrightarrow B$ induced by $1_B$ and $f$.  Then we could define $f$ as the partition of $A$ induced by this epimorphism, which seems like a perfectly nice way to define functions.
I was wondering if this construction could be used to construct exponential objects if I was given finite colimits and some kind of epimorphism classifier, or collection of epimorphisms out of a given object. 
Comment if it turns out that there is no really nice answer to this question, do you think that has bearing on the fact that the formula for the number of subsets of a set is easy ($2^{|X|}$) but the formula for the number of partitions of a set is relatively hard (http://en.wikipedia.org/wiki/Partition_of_a_set)?

REPLY [7 votes]: I think the natural meaning of "collection of all epimorphisms out of $X$" or "epimorphism classifier" in a category $\mathbf{S}$ would be: an object $E$, an object $Y\to E$ of $\mathbf{S}/E$, and an epimorphism $p\colon E\times X \twoheadrightarrow Y$ in $\mathbf{S}/E$, such that for any object $U$ and any epimorphism $q\colon U\times X\twoheadrightarrow Z$ in $\mathbf{S}/U$, there exists a unique morphism $f\colon U\to E$ such that $(f\times 1)^*q$ is isomorphic to $p$ under $E\times X$ in $\mathbf{S}/E$.  In other words, a representing object for the presheaf on $\mathbf{S}$ which sends an object $U$ to the set of (isomorphism classes of) epimorphisms out of $U\times X$.
In a topos, such epimorphism classifiers can be constructed from power objects.  Every epimorphism $X\twoheadrightarrow Z$ in a topos is the quotient of its kernel pair, which is an internal equivalence relation on $X$, i.e. a particular element of $P(X\times X)$, and every internal equivalence relation has a quotient.  Therefore, the subobject of $P(X\times X)$ which internally "consists of all equivalence relations" can be shown to be an epimorphism classifier in the above sense.<|endoftext|>
TITLE: Are the inner automorphisms the only ones that extend to every overgroup?
QUESTION [39 upvotes]: Let $H$ be a group. Can we find an automorphism $\phi :H\rightarrow H$ which is not an inner automorphism, so that given any inclusion of groups $i:H\rightarrow G$ there is an automorphism $\Phi: G\rightarrow G$ that extends $\phi$, i.e. $\Phi\circ i=i\circ \phi$?

REPLY [17 votes]: I struggled with the same question for quite some time and solved it for finite groups, only to then discover that it had already been solved. Schupp solved it for the class of all groups. Martin Pettet later solved it for the class of finite groups, and his proof works for the classes of $p$-groups, finite $p$-groups, finite $\pi$-groups, solvable groups, etc.

Pettet, Martin R., On inner automorphisms of finite groups, Proc. Am. Math. Soc. 106, No. 1, 87-90 (1989). ZBL0675.20015.
Pettet, Martin R., Characterizing inner automorphisms of groups, Arch. Math. 55, No. 5, 422-428 (1990). ZBL0683.20025.

These proofs also show that analogous statements are true if we replace injective embeddings with quotient maps (i.e., the only automorphisms that can be pulled back over all quotient maps are the inner ones).
I also have some notes on this and similar problems here: extensible automorphisms problem.<|endoftext|>
TITLE: Intuition behind moduli space of curves
QUESTION [34 upvotes]: For a genus g compact smooth surface $M$, an algebraic structure is the same as a complex structure is the same as a conformal structure. So the moduli space of smooth curves should be the same as the moduli space of conformal structures on $M$. A conformal structure is an equivalence class of Riemmanian metrics that give the same angle measurements.
If I embed $M$ in $\mathbb{R}^3$, I get a metric on $M$. This gives me a conformal structure, hence a point in the moduli space. The moduli space is known to have complex dimension 3g-3 (except for g=0, where the moduli space is a point, and g=1, where the moduli space is 1-dimensional).
My question is: can we visualize the 6g-6 real dimensions as deformations of the embedding of $M$ in $\mathbb{R}^3$. In particular, how can we see that small deformations of a 2-sphere are conformally equivalent to the original 2-sphere.

REPLY [5 votes]: This is related to Kevin's answer but goes back to the 19th century.  A Riemann surface is planar if a simple closed curve separates it.  A Riemann surface of genus g becomes planar after cutting g handles along simple closed curves.  Then the classical uniformaization theorem states that it is equivalent to a sphere with 2g slits parallel to the "x axis".  Then the location of the slits is determine by the 2g centers (4g real coordinates) and the 2g lengths (2g real coordinates).  This give 6g real coordinates, which are then reduced by the 6 real dimensions of the conformal automorphisms of the sphere.  Thus Riemann surfaces of genus g have 6g-6 real parameters, plus the number of parameters of the conformal automorphisms of the generic surface of genus g, which adds 6 for g=0 and 2 for g=1.  (Alan Mayer explained it this way in a colloquium at Brandeis about 1967.)
As usual, induction is the easiest way to see anything, and here one degenerates a Riemann surface to one of lower genus by acquiring a node (shrinking off a loop).  Then the lower genus Riemann surface has 3(g-1)-3 complex parameters. the two identified points have 2 parameters, and we add one more for the degeneration direction.  (Such degeneration methods occur in Wirtinger's book on theta functions in 1895.)<|endoftext|>
TITLE: What are the auto-equivalences of the category of groups?
QUESTION [17 upvotes]: My question is motivated by Are the inner automorphisms the only ones that extend to every overgroup?
What are the auto-equivalences of the category of groups? What kind of structure do they form?
There are certainly some not-so-trivial examples (which are still trivial if one just focus on a single group), e.g. the functor send an abstract group G to the group G^op with the same underlying set with multiplication g,h ---> hg, where hg is the product of h and g in G. [The notation G^op will make sense if you think G as the category BG]

REPLY [13 votes]: Just to make this a little more visible: for a proof that every autoequivalence of the category of groups is naturally isomorphic to the identity, see page 31 of Peter Freyd's book "Abelian Categories".<|endoftext|>
TITLE: Feasibility of a list of prescribed distances in R^3
QUESTION [8 upvotes]: I am puzzled with the following problem:
Given $n$ real numbers it is to obtain a Yes/No answer to: "whether it is possible to arrange different points in the Euclidean $\mathbb{R}^3$ so that every of the given numbers represents a shortest distance which belongs to a distinct pair of points?"
What is an efficient algorithm to solve such problem? If I understand properly, first I have to find $m$ from $n=\frac{m(m+1)}{2}$ which is the number of such points. But what is the next step? Should I deal with checking the triangle inequality (which seems to be very inefficient) or what?
Thank you in advance!

REPLY [11 votes]: What did not seem to be mentioned in this old thread 
is that this is a well-studied problem that goes under the name 
distance geometry, which Wikipedia defines as "the characterization and study of sets of points based only on given values of the distances between member pairs."
The problem has applications especially to molecular structure.  Crippen and Havel published a
book on this in 1988: Distance geometry and molecular conformation (Research Studies Press, Taunton, Somerset, England), with an update by Crippen later ("Chemical distance geometry: Current realization and future projection").  There are many papers on the topic, published in venues like the Journal of Computational Chemistry, e.g., "Molecular conformations from distance matrices" (Volume 14, Issue 1, pages 114–120, January 1993).  Because of the many applications, there are software packages
developed to solve the problem; the Wikipedia article has a list.
The problem is computationally intractable; the proof that one version is NP-hard goes back
to a 1979 paper by Saxe, "Embeddability of weighted graphs in $k$-space is strongly NP-hard."<|endoftext|>
TITLE: Exact sequences of permutational representations?
QUESTION [7 upvotes]: Let $R$ be a commutative ring, like the ring of integers $\mathbb Z$ or the ring of $p$-adic integers $\mathbb Z_p$.  Let $G$ be a finite group; let us consider permutational representations of $G$ over $R$, i.e., $R[G]$-modules of the form $R[G/H]$, where $H\subset G$ is a subgroup, and direct sums of such modules.  These are free $R$-modules where $G$ acts so that there exists a basis of the module preserved (as a whole) by the action.
I am interested in finite exact sequences of representations of the above type, particularly in those of them that are not very long.  There are some beautiful examples, e.g., for $G=\mathbb Z/2$ and $R=\mathbb Z/2$ there is an exact sequence
$$
 0\rightarrow R\rightarrow R[G]\rightarrow R\rightarrow 0,
$$
while for $G=\mathbb Z/n$ and $R=\mathbb Z$ there is an exact sequence
$$
 0\rightarrow R\rightarrow R[G]\rightarrow R[G]\rightarrow R\rightarrow 0.
$$
For the fourth symmetric group $G=\mathbb S_4$ and $R=\mathbb Z$ there is an exact sequence
$$
 0\rightarrow R\rightarrow R[\mathbb X_4]\oplus R\rightarrow
 R[\mathbb X_6]\oplus R\rightarrow R[\mathbb X_3]\rightarrow0,
$$
where $\mathbb X_4$ is the four-element set that $\mathbb S_4$ permutes, $\mathbb X_6$ is the set of all two-element subsets of $\mathbb X_4$, and $\mathbb X_3$ is the quotient set of $\mathbb X_6$ by the obvious involution.  Dihedral groups also have some four-term exact sequences of permutational representations.
Where is one supposed to get such exact sequences?  There are some obvious ways, like e.g. one can take cones of morphisms of exact sequences of this type, or one can do restriction or induction from one group to another one.  Are there any other constructions?
For constructions to be interesting, they should of course be removed far enough from the trivial case when $|G|$ is invertible in $R$.  E.g., to have $R=\mathbb Z_p$ and $|G|$ a large $p$-group would be perhaps most highly nontrivial.
EDIT: One of the commenters asked where does the sequence for $\mathbb S_4$ come from, so let me say a few words about this.  Not that I really understand it, but there is a geometric construction using a CW complex, and not quite of the kind that Greg suggests in his second answer below.
Represent the group $\mathbb S_4$ as the group of rotations of the $3$-dimensional cube.  Consider the quotient CW complex of the cube's surface by the central symmetry involution.  The group $\mathbb S_4$ still acts on the quotient.  The set of vertices of the quotient is the $\mathbb S_4$-set $\mathbb X_4$, the set of edges is the $\mathbb S_4$-set $\mathbb X_6$, and the set of faces is the $\mathbb S_4$-set $\mathbb X_3$.
Now consider the map $R[\mathbb X_4]\rightarrow R[\mathbb X_6]$ assigning to a vertex the sum of the three edges ending in it (without any signs!) and also the map $R[\mathbb X_6]\rightarrow R[\mathbb X_3]$ assigning to an edge the sum of the two faces bordering on it (also without any signs!).  The composition of these two maps is not zero, of course; what it does is taking every vertex to twice the sum of all the three faces.  One somehow transforms this pair of arrows into a four-term exact sequence by adding the trivial $\mathbb S_4$-module direct summands $R$ in several degrees.

REPLY [6 votes]: Now that I have thought about the question some more, I can give a better answer.  I have a remark about how to search for these resolutions in general, and a construction that leads to many examples.
First, every permutation module is part of many exact sequences of permutation modules in which the subgroup is trivial:
$$\cdots \to R[G]^{n_2} \to R[G]^{n_1} \to R[G/H] \to 0.$$
The reason is very simple and standard: Any exact complex of this type is by definition a free resolution.  The way that you make a free resolution is that there is some intermediate target or kernel $K$, and you can send the generators $1 \in R[G]$ to some spanning set of $K$.  Usually the resolution is infinite, but with standard linear algebra you can search for a finite solution when there is one.
This observation generalizes to other permutation modules.  There is a part of induction-restriction reciprocity that holds over any ring.  Namely,
$$\text{Hom}_G(R[G/H],M) \cong \text{Hom}_H(I,M),$$
where $I$ is the trivial representation.  This relation is a generalization of the proof that a free module is projective.  So if there is a finite permutation resolution of a module $M$ (which could be the kernel of some incomplete sequence of permutation modules), you can search for it in the same way that you search for free resolutions.
Second (and I suspect that readers will like this answer better), you can obtain many examples from the chain homology complex of a finite CW complex $K$ with an action of $G$.  In order to make everything match, let's consider a slight generalization of a permutation module, not just $R[G/H]$ but also a module $R[G/H]_\chi$ induced from a character
$$\chi:H \to \{1,-1\}.$$
The basic idea is not hard:  Each term $C_n(K)$ is a direct sum of signed permutation modules of $G$, because $G$ acts on the cells.  If a cell $c$ has stabilizer $H$, then it makes an orbit equivalent to $G/H$, and we can define $\chi$ by examining which elements of $H$ flip over $c$.  If $K$ happens to have the same $R$-homology as a point, then you can augment its chain complex by the trivial module.  Or if it is an $R$-homology sphere, you can augment its chain complex at both ends.  If you don't like the signed permutation modules, you can subdivide the cell $c$ to get rid of them, or work in characteristic 2.
If $K$ is a line segment and $G = C_2$ acts by reflecting it, the result is Leonid's first example.
If $K$ is a polygon with $n$ sides and $G = C_n$ acts by rotation, the result is Leonid's second example.
If $K$ is a polygonal tiling of the 2-sphere and $G$ is a rotation group that acts on $K$ without reversing edges, the result is a new example.  For instance you can take a dodecahedron graph and divide each edge into two edges.  Again, the point of splitting the edges is just to get rid of the signed permutation modules.
Every finite group $G$ acts faithfully on a sphere of some dimension, because $G$ has a faithful linear representation.  So there are many sphere examples for every finite group.
At first glance, the second answer is a type of construction.  In many cases, it is also an interpretation of a chain complex $C$, because if you have $C$ you can try to build a CW complex to represent it.  In order to be an augmented chain complex, $C$ needs to end in the trivial module $I$.  If it ends in something else, you can concatenate with
$$0 \to I \to I \to 0.$$
The differentials of $C$ also need to have integer matrices.

I found several papers on a related question called the "quasi-projective dimension" of a group ring $R[G]$.  The original paper on this is Groups of finite quasi-projective dimension, by Howie and Schneebeli.  Their definition of a quasi-projective resolution is a finite resolution of a module $M$ by projective terms, and at the end a term which is a permutation module.  I assume that, certainly for finite groups, it would work just as well to use a free resolution as a projective resolution.  Among other results, Howie and Schneebeli establish that if $G$ is a finite group and $R = \mathbb{Z}$, then the quasi-projective dimension of $R[G]$ equals the period of its Tate cohomology.  But another theme of the paper is that these questions, both theirs and surely Leonid's also, are perfectly interesting for infinite groups too.
The papers that cite this initial paper use the second idea that I propose above.  They make CW complexes with an action of the group $G$, and then make chain complexes from these CW complexes.  So these CW complexes seem like a main way to understand complexes of permutation modules.  In my opinion, the CW complex picture suggests generalizing the original question to include signed permutation modules.<|endoftext|>
TITLE: Connectivity after Geometric Realization?
QUESTION [5 upvotes]: Suppose that I have a map of simplicial spaces,
$ f: X_* \to Y_*$,
and that I know that the map on zero spaces $f_0: X_0 \to Y_0$ is n-connected. Can I conclude anything about the connectivity of the map of geometric realizations?
$ |f|: |X| \to |Y|$
Are there any reasonable conditions I can place on the simplicial spaces X and Y that would allow me to conclude something along these lines? I'm especially interested in knowing when the map is 0-connected (i.e. a surjection on $\pi_0$).

REPLY [7 votes]: If you know the map on k-simplices is (n-k)-connected, you can deduce the map on realisations is n-connected. I don't think you can do better in any sort of generality.<|endoftext|>
TITLE: Normal coordinates for a manifold with volume form
QUESTION [8 upvotes]: I'm hoping that the following are true.  In fact, they are probably easy, but I'm not seeing the answers immediately.
Let $M$ be a smooth $m$-dimensional manifold with chosen positive smooth density $\mu$, i.e. a chosen (adjectives) volume form.  (A density on $M$ is a section of a certain trivial line bundle.  In local coordinates, the line bundle is given by the transition maps $\tilde\mu = \left| \det \frac{\partial \tilde x}{\partial x} \right| \mu$.  When $M$ is oriented, this bundle can be identified with the top exterior power of the cotangent bundle.)  Hope 1:  Near each point in $M$ there exist local coordinates $x: U \to \mathbb R^m$ so that $\mu$ pushes forward to the canonical volume form $dx$ on $\mathbb R^m$.
Hope 1 is certainly true for volume forms that arise as top powers of symplectic forms, for example, by always working in Darboux coordinates.  If Hope 1 is true, then $M$ has an atlas in which all transition maps are volume-preserving.  My second Hope tries to describe these coordinate-changes more carefully.
Let $U$ be a domain in $\mathbb R^m$.  Recall that a change-of-coordinates $\tilde x(x): U \to \mathbb R^m$ is oriented-volume-preserving iff $\frac{\partial \tilde x}{\partial x}$ is a section of a trivial ${\rm SL}(n)$ bundle on $U$.  An infinitesimal change-of-coordinates is a vector field $v$ on $U$, thought of as the map $x \mapsto x + \epsilon v(x)$.  An infinitesimal change-of-coordinates is necessarily orientation-preserving; it is volume preserving iff $\frac{\partial v}{\partial x}(x)$ is a section of a trivial $\mathfrak{sl}(n)$ bundle on $U$.  Hope 2:  The space of oriented-volume-preserving changes-of-coordinates is generated by the infinitesimal volume-preserving changes-of-coordinates, analogous to the way a finite-dimensional connected Lie group is generated by its Lie algebra.
Hope 2 is not particularly well-written, so Hope 2.1 is that someone will clarify the statement.  Presumably the most precise statement uses infinite-dimensional Lie groupoids.  The point is to show that a certain a priori coordinate-dependent construction in fact depends only on the volume form by showing that the infinitesimal changes of coordinates preserve the construction.
Edit: I have preciseified Hope 2 as this question.

REPLY [9 votes]: Yes for "hope" 1.  This theorem was proven by Moser using volume-preserving flows.  A manifold with a volume form is the same thing as a manifold with an atlas of charts modeled on the volume-preserving diffeomorphism pseudogroup.  He found an argument that can be adapted to either the symplectic case or the volume case.  I cited this result in my paper A volume-preserving counterexample to the Seifert conjecture (Comment. Math. Helv. 71 (1996), no. 1, 70-97), where I also established a similar result for the volume-preserving PL pseudogroup.  In the PL case, the corresponding decoration on the manifold is a piecewise constant volume form.
In my opinion, the most exciting result on this theme is the Ulam-Oxtoby theorem.  (But Moser's version is the most useful one and the most elegant.)  The theorem is that if you have a topological manifold with any Borel measure that has no atoms and no bald spots, then it is modeled on the pseudogroup of volume-preseserving homeomorphisms.  For example, you can start with Lebesgue measure in the plane and add uniform measure on a circle, and there is a homeomorphism that takes that measure to Lebesgue measure.
For a long time I have wondered about the pseudogroup of volume-preserving Lipschitz maps.  The question is whether there is a corresponding cone of measures, and if so, how to characterize it.<|endoftext|>
TITLE: Negative Gromov-Witten invariants
QUESTION [13 upvotes]: I understand the heuristic reason why Gromov-Witten invariants can be rational; roughly it's because we're doing curve counts in some stacky sense, so each curve $C$ contributes $1/|\text{Aut}(C)|$ to the count rather than $1$.
However, I don't understand why or how Gromov-Witten invariants can be negative. What is the meaning of a negative GW invariant? What are some simple examples?

REPLY [4 votes]: The simplest example that I could find is: blow-up $P^3$ at one point and consider the following correlator $\langle pt, E^2, E^2\rangle_l=-1$ , where $l$ is the pull-back of a line from $P^3$. 
Proof: use the splitting axiom for the following 4 classes $(pt, E^2|E,E)_l$. To unravel this notation take a look at page 5 in http://www.mathematik.uni-kl.de/~gathmann/pub/blowup-9804043.pdf
The splitting axiom gives:
$\langle pt, E^2, E^2\rangle_l=\langle pt, E, E\rangle_{l-E'}\cdot\langle E^2, E^2, E\rangle_{E'}=1\cdot(-1)=-1$
I hope this is corect. 
Does anybody know an explicit example in dimension 2?<|endoftext|>
TITLE: Factorization of elements vs. of ideals, and is being a UFD equivalent to any property which can be stated entirely without reference to ring elements?
QUESTION [6 upvotes]: Why exactly is the unique factorization of elements into irreducibles a natural thing to look for? Of course, it's true in $\mathbb{Z}$ and we'd like to see where else it is true; also, regardless of whether something is natural or not, studying it extends our knowledge of mathematics, which is always good. But the unique factorization of elements - being specifically a question of elements - seems completely counter to the category theory philosophy of characterizing structure via the maps between objects rather than their elements. Indeed, I feel like unique factorization of ideals into prime ideals is less a generalization of unique factorization of elements into irreducibles than the latter is a messier, unnatural special case of the former, a "purer" question (ideals, being the kernels of maps between rings, I feel meet my criteria for being a category-theoretically acceptable thing to look at). Certainly, the common theme in algebra (and most of mathematics) is to look at the decomposition of structures into simpler structures - but quite rarely at actual elements.
Now, for nice cases like rings of integers in number fields, we can characterize being a UFD in terms of the class group and other nice structures and not have to mess around with ring elements, but looking at the Wikipedia page on UFDs and the alternative characterizations they list for general rings, they all appear to depend on ring elements in some way (the link to "divisor theory" is broken, and I don't know what that is, so if someone could explain it and/or point me to some resources for it, it'd be much appreciated). 
Sorry about the rambling question, but I was wondering if anyone had any thoughts or comments? Is "being a UFD" equivalent to any property which can be stated entirely without reference to ring elements? Should we care whether it is or not?

EDIT: Here's a more straightforward way of saying what I was trying to get at: The structure theorem for f.g. modules over a PID, the Artin-Wedderburn theorem, the Jordan-Holder theorem - these are structural decompositions. Unique factorization of elements is not, because elements are not a structure. My feeling is that this makes it a fundamentally less natural question, and I ask whether being a UFD can be characterized in purely structural terms, which would redeem the concept somewhat, I think.

REPLY [3 votes]: This is sort of an anti-answer, but: my instinct is that ZC is taking the categorical perspective too far.  
To start philosophically, I think it is quite appropriate to, when given a mathematical structure like a topological space or a ring -- i.e., a set with additional structure -- refrain from inquiring as to exactly what sort of object any element of the structure is.  There is a famous essay "What numbers could not be" by Paul Benacerraf, in which he pokes fun at this idea by imagining two children who have been taught about the natural numbers by two different "militant logicists".  Their education proceeds well until one day they get into an argument as to whether 3 is an element of 17.  (The writing is very nice here and unusually witty for an essay on mathematical philosophy: the names of the children are Ernie and Johnny, an allusion to Zermelo and von Neumann, who had rival definitions of ordinal numbers.)  The point of course is that it's a silly question, and a mathematically useless one: it won't help you to understand the structure of the natural numbers any better.
On the other hand, to deny that a set is an essential part of certain (indeed, many) mathematical structures seems to be carrying things too far.  As far as I know, it is not one of the goals of category theory to eliminate sets (though one occasionally hears vague mutterings in this direction, I have never seen an explanation of this or, more critically, of the need for this).  
Coming back to rings, it seems to me that very few properties of rings can be expressed without elements.  You also seem to implicitly suggest that it is "more structural" to think about things in terms of ideals than elements.  Can you explain this?  It would seem that speaking of ideals involves more set theoretic machinery than speaking about elements: this is certainly true in model theory in the language of rings.  
It seems wrong to say that unique factorization of ideals into primes is a "generalization" of unique factorization of elements, since neither property implies the other.  
Finally a positive remark: it sounds like you might like the characterization of UFDs as Krull domains with trivial divisor class group.<|endoftext|>
TITLE: Hilbert spaces are induced by a bilinear form. How about n-linear forms?
QUESTION [9 upvotes]: A Hilbert space is a complete vector space equipped with scalar product, i.e. a symmetric positive definite bilinear form.
What if we replace 'bilinear' by 'n-linear'? One might wonder, whether the $l^3$-norm might be induced by a trilinear form in a similar fashion like the $l^2$-norm by a bilinear form is.
Is there any interesting theory on this?

REPLY [8 votes]: As long as the form is positive definite and the unit ball is convex, you get a perfectly good Banach space using any symmetric $n$-linear form on a real vector space $V$.  The degree $n$ is necessarily even.  It is equivalent to defining the norm as the $n$th root of a homogeneous degree $n$ polynomial.  $\ell^p$ is an example for any even integer $p$.   There are many other examples.  I found a paper, Banach spaces with polynomial norms, by Bruce Reznick, that studies these norms.  He obtains various results; the most appealing one to me at a glance is that these Banach spaces are all reflexive.
Off-hand I can't think of any simple way to recover positive definiteness starting with odd polynomials.  The cube of the norm on $\ell^3$ is a polynomial in the absolute values of the coordinates rather than the coordinates themselves.
Addendum: To address Darsh's comment, what you would look at in the complex case is self-conjugate polynomials of degree $(n,n)$.  Equivalently, as with all complex Banach norms, the realification is a real Banach norm which is invariant under complex scalar rotation.<|endoftext|>
TITLE: Undergraduate differential geometry texts
QUESTION [25 upvotes]: Can anyone suggest any basic undergraduate differential geometry texts on the same level as Manfredo do Carmo's Differential Geometry of Curves and Surfaces other than that particular one?
(I know a similar question was asked earlier, but most of the responses were geared towards Riemannian geometry, or some other text which defined the concept of "smooth manifold" very early on.  I am looking for something even more basic than that.)

REPLY [5 votes]: There is our book What is differential geometry: curves and surfaces.
It is written for those who either plan to work in differential geometry, or at least want to have a good reason not to do it.
(Sorry for self-advertisement.)<|endoftext|>
TITLE: Examples of finite local rings of length 2 or 3
QUESTION [9 upvotes]: What is an example of a finite local rings, that has length 2 or 3? 
I want something different from $F_{q}[x] / x^{i}$ for $i=2, 3$; I'm looking for something more interesting. If you can give me examples of higher length, yet have "simple structure" (e.g. $F_{q}[x]/x^{i}$), that would be nice too. 
I know this is related to Classification of finite commutative rings, but I didn't completely understand the answer there.

REPLY [5 votes]: Any local rings of length 2 are of the form $R/n^2$ with $(R,n)$ a regular local ring of dimension 1.
Any local rings of length 3 are of the form $R/n^2$ with $(R,n)$ a regular local ring of dimension 2 or $R/n^3$ with $(R,n)$ a regular local ring of dimension 1.
Proof for the length 3 case: Let $A,m,k$ be our ring. Then the exact sequence:
$$ 0 \to m \to A \to k  \to 0 $$
shows that $length(m)=2$. So the number of generators of $m$ is at most 2. If it is 2, then 
$length(m/m^2)=2$, thus $m^2=0$. So $A=R/n^2$ with $(R,n)$ a regular local ring of dimension 2.
If it is 1, then $A$ is a quotient of a regular local ring of dimension 1, and counting length shows that we need to kill the cube of the (principal) maximal ideal. 
Of course, $R$ may or may not contains a field, so one may have things like $k[[x,y]]/(x^2,xy,y^2)$ or $Z_p[[x]]/(p^2,xp,x^2)$.
MORE: We can assume our regular $R$ is complete, since $A$, being Artinian, is complete. Then the  Cohen Structure Theorem completely described $R$. If $R$ contains a field, then 
$R \cong k[[x,y]]$. If $R$ is mixed charateristic and unramified, then $R\cong V[[x]]$, with 
$(V,pV)$ a discrete valuation ring. If $R$ is ramified, then $R=V[[x,y]]/(f)$, with $f=p+g$ such that $g \in (x,y)^2$. From those you can get $A$ accordingly by killing the square of the maximal ideal.

REPLY [3 votes]: I'm just going to consider the local rings with residue field $\mathbf{F}_p$.
Suppose A' is an extension of $\mathbf{Z}/p$ with ideal $\mathbf{Z}/p$.  If we take the fiber product with $\mathbf{Z}$, we get an extension of $\mathbf{Z}$ with ideal $\mathbf{Z}/p$.  Up to isomorphism, there is only one such extension: $B' = \mathbf{Z}[t] / (t^2, pt)$.  The kernel of the map from B' to A' is isomorphic to $\mathbf{Z}$ and reduces modulo t to the ideal generated by p.  Therefore, the square-zero extensions of $\mathbf{Z}/p$ are all isomorphic to $\mathbf{Z}[t] / (t^2, pt, p + \lambda t)$ for some $\lambda \not= 0$.  If $\lambda$ is not a multiple of p, we get $\mathbf{Z} / p^2$; if $\lambda$ is a multiple of $p$, we get $\mathbf{Z}[t] / (p, tp, t^2) = \mathbf{F}_p[t] / t^2$.  So these are all the length 2 finite local rings with residue field $\mathbf{F}_p$.
For length 3, we'll look for extensions of $\mathbf{Z} / p^2$ by $\mathbf{Z} / p$.  The same analysis shows that these are all of the form $\mathbf{Z}[t] / (t^2, pt, p^2 + \lambda t)$.  If $\lambda$ is not divisible by $p$, we get $\mathbf{Z} / p^3$ and if $\lambda$ is divisible by $p$ we get $\mathbf{Z}[t] / (p^2, pt, t^2)$.
We also have to look for extensions of $\mathbf{F}_p[t] / t^2$.  By base change, any such extension A' gives an extension B' of $\mathbf{Z}[t]$ with ideal $\mathbf{Z} / p$.  Once again, there is only one of these up to isomorphism (since $\mathbf{Z}[t]$ is projective, for example) and it is given by $\mathbf{Z}[t,u] / (u^2, pu, tu)$.  The ideal of the map from B' to A' generates the ideal of the map from $\mathbf{Z}[t]$ to $\mathbf{F}_p[t] / t^2$.  Since this is generated by p and t^2 the ideal of A' in B' is generated by $(p + \lambda u, t^2 + \mu u)$ and A' is of the form
$\mathbf{Z}[t,u] / (u^2, pu, tu, p + \lambda u, t^2 + \mu u)$
for some polynomials $\lambda, \mu \in \mathbf{Z}[t]$.  

Edit: If $\lambda$ is not in $(p, t)$ then it is invertible in the quotient, so we get 
$\mathbf{Z}[t,u] / (p^2, tp, t^2 + \mu \lambda^{-1} p)$.
There are two possibilities up to isomorphism here, depending on whether $- \mu \lambda^{-1}$ is a quadratic residue modulo p.
If $\lambda$ is in $(p,t)$ we get
$\mathbf{Z}[t,u] / (u^2, pu, tu, p, t^2 + \mu u) = \mathbf{F}_p[t,u] / (u^2, tu, t^2 + \mu u)$.
The $\mu$ is also not in $(p, t)$ then we get $\mathbf{F}[t] / t^3$.  If $\mu$ is in $(p,t)$ we get $\mathbf{F}_p[t,u] / (u^2, tu, t^2)$.
Modulo any mistake I made above, I think a complete list is of length 3 finite local rings with residue field $\mathbf{F}_p$ is

$\mathbf{Z} / p^3$,
$\mathbf{Z}[t] / (p^2, pt, t^2)$,
$\mathbf{Z}[t] / (t^2 - p, t^3)$,
$\mathbf{Z}[t] / (t^2 - \alpha p, t^3)$ where $\alpha$ is a non-quadratic residue modulo $p$,
$\mathbf{F}_p[t] / t^3$, and
$\mathbf{F}_p[t,u] / (t,u)^2$.<|endoftext|>
TITLE: Is the space of volume-preserving maps path-connected?
QUESTION [14 upvotes]: This is a clarification of another post of mine.
Fix $n$ a positive integer.  Let $SL(n)$ have its usual matrix representation, so that it really is the codimension-one subset of $M(n) = \mathbb R^{n^2}$ cut out by the degree-$n$ condition that the determinant is $1$.  So we have $n^2$ coordinate functions $A^i_j$ on $SL(n)$, $i,j = 1,\dots,n$.
Let $U$ be a domain in $\mathbb R^n$, with coordinates $x_1,\dots,x_n$.  Consider the set $\mathcal S$ of smooth functions $f: U \to SL(n)$ satisfying the differential equation $\frac{\partial f^i_j}{\partial x^k} = \frac{\partial f^i_k}{\partial x^j}$ for each $i,j,k = 1,\dots,n$ (of course, $f^i_j = A^i_j \circ f$ is the $(i,j)$th coordinate of $f$).
(Why would you care about $\mathcal S$?  Because a smooth map $g: U \to \mathbb R^n$ is volume-preserving if and only if $\frac{\partial g^i}{\partial x^j} \in \mathcal S$, and every element of $\mathcal S$ arises this way; indeed, $\mathcal S$ is the space of volume-preserving maps up to translations.)
Let's agree that a smooth path in $\mathcal S$ is a smooth function $F: [0,1] \times U \to SL(n)$ such that for each $t\in [0,1]$, $F(t,-) \in \mathcal S$.
Question:  Is $\mathcal S$ smooth-path-connected?  I.e. given $f_0, f_1 \in \mathcal S$, does there exist a smooth path $F$ so that $F(0,-) = f_0$ and $F(1,-) = f_1$?
If the answer is "no" in general, is it "yes" for sufficiently nice domains $U$ (contractible, say, or with compact closure and require that each $f\in \mathcal S$ extend smoothly to a neighborhood of the closure, or...)?

REPLY [3 votes]: I'd like to point out a special case where there's a particularly appealing answer -- a variant of the Alexander trick. 
Consider the space of smooth volume-preserving immersions (or embeddings) $f : D^n \to \mathbb R^n$.  $D^n$ is the unit compact disc in $\mathbb R^n$. 
This space deformation-retracts to the subspace where $f(0)=0$. So we restrict to that subspace for brevity of notation. 
Given $t \in (0,1]$ define
$$F_t : D^n \to \mathbb R^n$$
by $F_t(x) = \frac{1}{t}f(tx)$ and define $F_0(x) = Df_0(x)$. 
Provided $f$ is $C^1$, this makes $F : [0,1] \times D^n \to \mathbb R^n$ continuous. 
And $F_t$ is volume preserving for all $t$ by the chain rule -- even better $D(F_t)_p = (Df)`_`{tp}$ 
for all $p \in D^n$. Moreover, if $f$ is an embedding, $F_t$ remains an embedding for all $t$. 
Since $F_1 = f$ and $F_0 \in SL_n\mathbb R$, and $SL_n\mathbb R$ is connected, we're done.<|endoftext|>
TITLE: What's the standard name for sets of a given size with maximal probability (or a given probability and minimal size)? 
QUESTION [7 upvotes]: The definition I'm going to give isn't quite the concept I really want, but it's a good approximation. I don't want to make the definition too technical and specific because if there's a standard name for a slightly different definition, then I want to know about it. 
Let $(X,\mu)$ be a measure space, and let $\rho$ be a probability measure on $X$. I call a subset $A$ of $X$ special if for all measurable $B\subseteq X$, 

$\mu(B)\leq\mu(A)$ implies $\rho(B)\leq\rho(A)$, and 
$\mu(B)=\mu(A)$ and $\rho(B)=\rho(A)$ implies $B=A$ up to measure zero (with respect to both $\mu$ and $\rho$). 

What is the standard name for my "special" sets? Equivalently, one could stipulate $\mu(A)\leq\beta$ and call $A$ "special" if it is essentially the unique maximizer of $\rho(A)$ given that constraint. 
Also equivalently, we could stipulate a particular $\rho$-measure and consider sets achieving that $\rho$-measure having the smallest possible $\mu$-measure. That's probably the most intuitive way to think about this: we're looking for sets that contain a certain (heuristically: large) fraction of of the mass of $\rho$ but are as small as possible (with respect to $\mu$). That seems like a completely natural and obvious concept, which is why I think it should have a standard name. But I have almost no training in statistics, so I don't know what the name is. 
This example might be far-fetched, but just to illustrate: suppose the FBI has knowledge that somebody is going to attempt a terrorist attack in a certain huge city at a particular hour. They might not know where, but they might have (some estimate of) a probability distribution for the location of the attack. They want to distribute agents strategically throughout the city, but they probably don't have enough agents to cover the entire city. Let's say every agent can forestall an attack if it occurs within a certain radius of his/her position (which is unrealistic, since the number of nearby agents surely also matters, but ignore that); then, to maximize the probability that the attack will be stopped, to an approximation, they should distribute their agents uniformly over a special subset of the city's area. To approach this from the other perspective, it could be the case that 99% of the mass of their probability distribution is contained in a region with very small area. (The one with the smallest area will be a special set.) Then, to save resources, if they're okay with 99:1 odds (c'est la vie), they might only distribute a relatively small number of agents to that small special region.
If $\rho$ has a density $f$ with respect to $\mu$ (when it makes sense to talk about such), then special sets are closely related to the superlevel sets of $f$, i. e., sets of the form $\{x:f(x)\geq c\}$ for $c\geq 0$. (I think they're basically the same, but specialness of $A$ is unaffected by changing $A$ by a set of measure zero, so a superlevel set actually corresponds to an equivalence class of special sets.) I mention this here because (1) the connection to superlevel sets is one of my reasons for caring about specialness, and (2) "superlevel sets of the density" is not the answer I'm looking for. 

Example 1
Here's a very simple example in which special sets can be completely characterized. Let $X=\{x\_1,\ldots,x\_n\}$ be a finite set, and let $\mu$ be counting measure on $X$. Let $\rho$ be any probability distribution on $X$, which necessarily has a density function $f:X\to\mathbb{R}\_+$, so by definition, $f(x\_1) + \ldots + f(x\_n) = 1$ and $\rho(A) = \sum\_{x\in A} f(x)$. Suppose that no two points have the same $f$-value; then, without loss of generality, $f(x\_1) > f(x\_2) > \ldots > f(x\_n)$. It's easy to see that the special sets in this setup are exactly the sets $A\_k = \{x\_1,x\_2,\ldots,x\_k\}$, i. e., which contain the largest $k$ points as measured by $\rho$, for $k=0,\ldots,n$. (Why: if you have some other candidate special set $B$, then $A\_{\\#B}$ has the same $\mu$-measure as $B$ but higher $\rho$-measure, so $B$ can't be special.) It's easy to generalize this example to the case in which $f$ isn't necessarily one-to-one: you have to treat all points with the same $f$-value as a block: either all of them are in the special set, or none of them are. (Otherwise, there's no way to satisfy the "uniqueness" part (point 2) of the definition.) 

Example 2
Here's a generalization of the first example that hopefully clarifies what I said above. Let $(X,\mu)$ be some nice measure space on which integration of functions makes sense (like a Riemannian manifold, or just $\mathbb{R}^d$). Let $f:X\to\mathbb{R}\_+$ be a nonnegative integrable function with $\int_X f(x) d\mu = 1$, and let $\rho$ be the probability measure $\rho(Y) = \int_Y f(x) d\mu$, so $f$ is the density of $\rho$ with respect to $\mu$. Fix some $c\geq 0$ and let $A=\{x:f(x)\geq c\}$. 

Claim: $A$ is a special set.

Proof: It suffices to show that if $\mu(B) = \mu(A)$, then $\rho(B)\leq \rho(A)$, with equality if and only if $B$ and $A$ differ by a set of measure zero. If $\mu(B) = \mu(A)$, then $\mu(B-A) = \mu(A-B)$. Now we write 
$\begin{align*} \rho(A) - \rho(B) &= \int\_A f(x) d\mu - \int\_B f(x) d\mu \\\\
 &= \int\_{A-B} f(x) d\mu - \int\_{B-A} f(x) d\mu \\\\
 &= \int\_{A-B} f(x) d\mu - \int\_{A-B}c\\,d\mu - \int\_{B-A} f(x) d\mu  + \int\_{B-A}c\\,d\mu\\\\
 &= \int\_{A-B} (f(x)-c) d\mu - \int\_{B-A} (f(x)-c) d\mu.
\end{align*}$
By construction, $f(x) \geq c$ on $A$ and $f(x) < c$ on $B-A$, so the first integral is nonnegative and the second integral is nonpositive, and is in fact negative unless $\mu(B-A)=0$, in which case $\mu(A-B)=0$ as well. Thus, $\rho(A)-\rho(B)\geq 0$, with strict inequality unless $A$ and $B$ differ by measure zero, QED.

REPLY [3 votes]: A similar (equivalent?) problem in statistic.
I am not sure about the name but it should be something like "critical decision area". This arises naturally in the problem of hypothesis testing of simple hypothesis where the optimization problem is: 
$$ \sup_{ A \in \mathcal{A}\;  \mu(A)\leq \beta  } \rho(A)$$ 
How this problem is solved
A more convenient set for the optimization (than the set of indicator function of set)  is
: 
$$\Psi= (\psi: X\rightarrow [0,1]\;\; measurable )$$ (it is more convenient because convex) 
The problem reformulates as:
$$ \sup_{ \psi\in \Psi\;  \int_{X} \psi d \mu \leq \beta } \int \psi d\rho=\sup_{\psi\in \Psi}\inf_{t\in [0,\infty]}\mathcal{L}(t,\psi)$$ 
where $\mathcal{L}(t,\psi)$ is the Lagrangien of the considered optimization problem:
$$
\mathcal{L}(t,\psi)=\int\psi d\rho- t\left (\int\psi d\mu-\beta \right )=t\beta-\int \psi (d\rho-td\mu).
$$
If you can use a minmax theorem (under some suitable topological assumption, not restrictive) to revert inf and sup, and if $\rho$ is absolutly continuous with respect to  $\mu$, the minimizer  of the optimization problem is given by the so famous Neymann Pearson test:
$$\psi^*= 1_{\frac{d\rho}{d\mu}\geq t(\beta)}$$
(it remains to find $t(\beta)$...)
This is called a test because it can help you to test wether the law of a random variable $X$ is $\mu$ or  $\rho$ given an observation of $X$. If $\rho$ is not absolutly continous with respect to $\mu$ you can decompose $\rho$ (by Lebesgue Theorem) and apply the same methodology.<|endoftext|>
TITLE: Sum of $n$ vectors in $(\mathbb Z/n)^k$
QUESTION [16 upvotes]: Let $n,k$ be positive integers. What is the smallest value of $N$ such that for any $N$ vectors (may be repeated) in $(\mathbb Z/(n))^k$, one can pick $n$ vectors whose sum is $0$? 
My guess is $N=2^k(n-1)+1$. It is certainly sharp: one can pick our set to be $n-1$ copies of the set $(a_1,...,a_k)$, with each $a_i=0$ or $1$. The case $k=1$ is some math competition question (I think, but can't remember the exact reference). Does anyone know of some references? Thanks. 

Thank you all! I wish I could accept all the answers, they are very helpful!

REPLY [6 votes]: There was some discussion of the case n=3 in sci.math around 1994. There is a card game called Set with an 81 card deck so that each card is naturally a point in $(\mathbb Z/3)^4$. Several cards are dealt out, and your task is to identify triples of cards called Sets which form a line, or equivalently, which add up to the 0 vector. A natural question is how many points you can deal out without the existence of a line. It's not too hard to construct 9 distinct points in affine 3-space, or 20 distinct points in affine 4-space over $\mathbb Z/3$ so that there is no line contained in the points, and these are the maximums. These correspond to $N=19$ for $(n,k) = (3,3)$ and $N=41$ for $(n,k) = (3,4)$, as in the reference Ricky Liu linked, by repeating each point twice.
The maximal configurations are highly symmetric. The 9 points in dimension 3 correspond to a nondegenerate conic, which is unique up to symmetry. The 20 points in dimension 4 actually correspond to a nondegenerate conic containing 10 points in projective 3-space viewed as lines passing through the origin in affine 4-space.
For example, there are 9 points in dimension 3 satisfying $z=x^2 + y^2:$
$\{(0,0,0),(\pm1,0,1),(0,\pm1,1),(\pm1,\pm1,-1)\}$
and this set contains no lines.<|endoftext|>
TITLE: Smooth structures on PL 4-manifolds
QUESTION [13 upvotes]: Is it known whether $O(4) \to PL(4)$, the map from the orthogonal group to the group of piecewise linear homeomorphisms of $\mathbb{R}^4$, is a homotopy equivalence? By smoothing theory for PL manifolds, this is equivalent to whether the space of smooth structures on a PL 4-manifold is contractible. (I think it's known that this map is at least 4-connected, which shows that the space of smooth structures on any PL 4-manifold is nonempty and connected.)

REPLY [10 votes]: Very little is known about that question, the same smoothing theory gives something that I'm trying to get people to call "The Cerf-Morlet Comparison Theorem"
$$ Diff(D^n) \simeq \Omega^{n+1}(PL(n)/O(n)) $$
$Diff(D^n)$ is the group of diffeomorphisms of the $n$-ball where the diffeomorphisms are pointwise fixed on the boundary.  Nobody knows if $Diff(D^4)$ is path-connected or not.  Very little is  known about the homotopy-type of $Diff(D^4)$, no seriously informative statements other than that homotopy-equivalence.  I wrote up a paper where I described in detail the iterated loop-space structure and how it arrises naturally.  Moreover, I described how that iterated loop-space structure relates to various natural maps.  That's my main relation to to topic. The paper is called "Little cubes and long knots" and is on the arXiv.  I elaborate on some of these issues in the paper "A family of embedding spaces", also on the arXiv. 
There are several natural connections here, one of the big ones being that $Diff(D^n)$ has the homotopy-type of the space of round metrics on $S^n$ -- ie the subspace of the affine-space of Riemann metrics on $S^n$, the subspace is specified by the condition that "$S^n$ with this metric is isometric to the standard $S^n$."<|endoftext|>
TITLE: Why is a variety of general type hyperbolic?
QUESTION [11 upvotes]: I heard people mentioned this in one sentence, but don't see the reason.
Why a (smooth) variety of general type, i.e. an algebraic variety X with K_X big, is hyperbolic, i.e. has no non-constant map from the complex number into it?
I don't know what are the necessary assumption on the variety, do we need properness or smoothness?
Edit: according to David Lehavi's reply, we should certainly put some more condition on it. What's the correct statement of the fact?

REPLY [5 votes]: Surely you need some assumptions on  your variety, for example for every $n$ there is a smooth surface of degree $n$ in $CP^3$ that contains a line, for $n>3$ such surfaces are minimal and of general type. So for large classes of varieties of general type you need an additional assumption that the variety should be generic.
For hypersufaces in $CP^n$ the most optimistically you can hope that a generic hypersurface of degree $2n+1$ is hyperbolic (hypersurfaces of degree $2n-1$ and less always contain lines). This is related to a conjecture of Kobayshi.
There is a very nice review of Claire Vosin on different aspects of hyperbolicity of complex projective manifold that you can find here http://people.math.jussieu.fr/~voisin/Articlesweb/harvard.pdf
(this also contains the result mentioned by Tony Pantev)
Recently there was a genuing progress in proving  of Kobayashi conjecture
http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.2346v4.pdf
though the obtained bound is very far fron optimal, worse than a triple exponent of n. At the conference due to 80 birthday of Atiyah Kirwan annonced that she can get a much more realistic bound.<|endoftext|>
TITLE: How to find all integer points on an elliptic curve?
QUESTION [10 upvotes]: How can I determine the integer points of a given elliptic curve if I know its rank and its torsion group? 
I read same basic books on elliptic curves but as a non-professional I didn't understand everything. Is it true that if rank is 0 and torsion group is isomorphic to a group of order $n$ then the number of integer points is $n-1$? And what is a good reference to learn to determine the integer points if the rank is positive? 
I tried to read the book Rational Points on Elliptic Curves but I didn't found an explicit algorithm. I just heard something like take some point and use group law to find the rest. But how can I be sure that I have found every point?
The curve I had on my mind is $2x^3 + 385x^2 + 256x - 58195 = 3y^2$. I'm not even sure if this is an elliptic curve. I mean why it is projective and why it is isomorphic to a closed subvariety of $\mathbb{P}_{\mathbb{Q}}^2$? And why it contain the priviledged rational point $(0,1,0)$?

REPLY [14 votes]: The following Sage code (which I ran with Sage 4.2.1) produces the solutions (and agrees with Magma!):
E = EllipticCurve([0, 1155,0,4608,-6285060])  
E1 = E.minimal_model()
pts1 = E1.S_integral_points([2,3])
iso = E1.isomorphism_to(E)
pts = [iso(P) for P in pts1]
solutions = [(x/6,y/18) for (x,y,z) in pts]
solutions = [(x,y) for (x,y) in solutions if x.is_integral() and y.is_integral()]
solutions

Before the first line I used pencil-and-paper to scale the equation to be in Weierstrass form (monic in both X and Y) which involves new variables 6X and 18Y;  the solutions a rescaled at the end.  In fact we found all S-integral solutions with S={2,3}, i.e. all solutions which are integral except at 2 and 3.  (There are 32 of these, of which only 15 are really integral).
John Cremona<|endoftext|>
TITLE: Non-commutative algebraic geometry
QUESTION [57 upvotes]: Suppose I tried to take Hartshorne chapter II and re-do all of it with non-commutative rings rather than commutative rings. Is this possible? Which parts work in the non-commutative setting and which parts don't? 
Edit: I also welcome any comments/references regarding any reasonable notions whatsoever of "non-commutative algebraic geometry".

REPLY [6 votes]: Rosenberg and Kontsevich will publish three volumes books soon. I have got the second draft of these books which Rosenberg has already taught us. There is a historical observations in the preface of the first volume of these books. I posted here:
Historical observations on noncommutative algebraic geometry I
Historical observations on noncommutative algebraic geometry II 
Historical observations on noncommutative algebraic geometry III

Historical observations on noncommutative algebraic geometry IV
Historical observations on noncommutative algebraic geometry V
Pseudo Geometry I II III

REPLY [3 votes]: As mentioned above, the lack of localization will be a major issue. It is interesting to note that many homological constructions will work. For example, major work was done by Van den bergh, Zhang, Yekutieli and others in constrtuing dualizing complexes over some noncommutative  rings, thus allowing some of the features of the Grothendieck duality to be recovered. See for example https://www.jstor.org/stable/2161576<|endoftext|>
TITLE: Failure of smoothing theory for topological 4-manifolds
QUESTION [27 upvotes]: Smoothing theory fails for topological 4-manifolds, in that a smooth structure on a topological 4-manifold $M$ is not equivalent to a vector bundle structure on the tangent microbundle of $M$. Is there an explicit compact counterexample, i.e., are there two compact smooth 4-manifolds which are homeomorphic, have isomorphic tangent bundles, but are not diffeomorphic? (The uncountably many smooth structures on $\mathbb{R}^4$ should give a noncompact counterexample, since $Top(4)/O(4)$ does not have uncountably many components.)
Addendum to question, added 12/11/09:
I'm also interested in the other type of counterexample, of a nonsmoothable topological 4-manifold whose tangent microbundle does admit a vector bundle structure. Does someone know such an example? Tim Perutz's answer to my first question, below, says that homeomorphic smooth 4-manifolds have isomorphic tangent bundles. If it's not true that all topological 4-manifolds have vector bundle refinements of their tangent microbundle, what is the obstruction in the homotopy of $Top(4)/O(4)$?

REPLY [13 votes]: I have always been mystified about handlebody structures on topological 4-manifolds. Already in 1970 Kirby and Siebenmann had established that topological n-manifolds have a handlebody structure for n>5 (see Essay III.2 in the 1976 K-S book), and Quinn proved this for n=5 in Ends of Maps III (1982). Finally I just sent an email to Kirby, who gave a simple argument that a topological 4-manifold has a handlebody structure if and only if it is smoothable. I have posted his email on the surgery pages of the Manifold Atlas Project.<|endoftext|>
TITLE: books well-motivated with explicit examples
QUESTION [79 upvotes]: It is ultimately a matter of personal taste, but I prefer to see a long explicit example, before jumping into the usual definition-theorem path (hopefully I am not the only one here). My problem is that a lot of math books lack the motivating examples or only provide very contrived ones stuck in between pages of definitions and theorems. Reading such books becomes a huge chore for me, even in areas in which I am interested. Besides I am certain no mathematical field was invented by someone coming up with a definition out of thin air and proving theorems with it (that is to say I know the good motivating examples are out there).
Can anyone recommend some graduate level books where the presentation is well-motivated with explicit examples. Any area will do, but the more abstract the field is, the better. I am sure there are tons of combinatorics books that match my description, but I am curious about the "heavier" fields. I don't want this to turn into discussion about the merits of this approach to math (i know Grothendieck would disapprove), just want to learn the names of some more books to take a look at them.
Please post one book per answer so other people can vote on it alone. I will start:
Fourier Analysis on Finite Groups and Applications by Terras
PS. this is a similar thread, but the main question is different.
How to sufficiently motivate organization of proofs in math books

REPLY [5 votes]: No probability book yet, so let me add a classic. 
William Feller, An Introduction to Probability Theory and Its Applications, vol I, II.
Full of examples, well motivated.<|endoftext|>
TITLE: Irreducible polynomials with constrained coefficients
QUESTION [18 upvotes]: Over at the Cafe, after reading about TWF 285, I asked more-or-less 

About how many polynomials with coefficients in $\{\pm 1\}$ and of degree $d$ are irreducible?

and that's what I want to ask here.
The first no-go analysis: since if $P$ is reducible, then it is reducible mod $3$, we get that the number of reducible such is $O(\frac{d-1}{d}3^d)$; but that's clearly too large already to help much; reducing mod $2$ we can't distinguish polynomials anymore!

REPLY [20 votes]: Your problem is hard, but here are some things that can actually be proved!
Let $S_d$ be the set of polynomials of degree $d$ with all $d+1$ coefficients in $\{\pm 1\}$.
1) $|S_d| \gg 2^d/d$ as $d \to \infty$ through even integers.
Proof: By Theorem 2 in a paper by Brillhart, Filaseta, and Odlyzko, $f$ will be irreducible if
(a) $f(2)$ is prime,
(b) $f(1) \ne 0$, and
(c) all complex zeros of $f$ have absolute value less than $3/2$.
The condition that $d$ is even guarantees (b).
To guarantee (c), restrict attention to $f$ whose first $100$ coefficients are $+1$, and use Rouch\'e's theorem to show that $f$ has the same number of zeros satisfying $|z|<3/2$ as does $x^d+x^{d-1}+\cdots+1$, i.e., all $d$ of them.  The values of these $f$ at $2$ are the odd integers in a certain interval, so the prime number theorem gives the required number of $f$ satisfying (a).  $\square$
2) The Generalized Riemann Hypothesis implies that there are infinitely many $d$ for which every $f$ in $S_d$ is irreducible.
Proof: GRH implies that there are infinitely many primes $p$ for which $2$ is a primitive root.  If $d+1$ is such a prime, then $x^d+x^{d-1}+\cdots+1$ is irreducible mod $2$, so every $f \in S_d$ will be irreducible over $\mathbf{Z}$.  $\square$
3) There exist infinitely many $d$ for which at least $50\%$ of the polynomials in $S_d$ are irreducible.
Proof: Let $d=2^n-1$ for any $n \ge 1$.  If $f \in S_d$, then $f(x+1) \equiv x^d \pmod{2}$. Thus $f(x+1)$ is Eisenstein at $2$ half of the time.  $\square$
A final remark: It is conjectured that the fraction of irreducible polynomials among those of degree $d$ with 0,1 coefficients tends to $1$ as $d \to \infty$: see this paper by Konyagin.  The same is likely true for your problem, as Greg suggests, but it also seems that the best you can hope for at the moment are partial results such as the ones above.<|endoftext|>
TITLE: Riemann surfaces: explicit algebraic equations
QUESTION [17 upvotes]: Suppose $\Gamma$ is a nice discrete subgroup of $SL(2,\mathbb{R})$ such that the genus of the Riemann surface $\mathbb{H}/\Gamma$ is larger than 1. We know that this Riemann surface is also an algebraic curve over $\mathbb{C}$ defined by a bunch of polynomials. Is there any explicit/canonical way of going back and forth between the group $\Gamma$ and a set of polynomials defining the curve? For example, if the group is given in terms of generators and relations, is there any algorithm for obtaining a set of polynomials defining the associated curve?
Somehow I couldn't add a comment so let me write it here.
Thanks for the responses. It will take me some time to digest the suggestions and look up the references provided. Please ignore the last comment about generators and relation. Let me ask a more specific question just to clarify what I wanted. What I am wondering is whether something akin to what happens for genus one curve also happens for higher genus. Recall that if $L=\mathbb{Z}+\tau \mathbb{Z}$, then we can write an equation for the elliptic curve $\mathbb{C}/L$ with the Eisenstein series $G_4(\tau)$ and $G_6(\tau)$ as coefficients. Can we do (or hope to do) something similar if we replace $\mathbb{C}$ by $\mathbb{H}$ and the lattice by a discrete group ?

REPLY [6 votes]: I'm not sure what you mean by "if the group is given by generators and relations" -- you need the group as a subgroup of $\mathrm{PSL}_2(\mathbb{R})$, not just as an abstract group.
If your group is commensurable with $\mathbb{SL}_2(\mathrm{Z})$ -- let's say for simplicity is a finite-index subgroup of the congruence subgroup $\Gamma(2)$ -- then your Riemann surface is a finite cover of the modular curve (or orbicurve, or stack) $X(2)$, which is (again, ignoring stacky issues) $\mathbb{P}^1$ with three points removed.  And your Riemann surface is an finite etale cover of $\mathbb{P}^1$ - three points, and you can easily recover the monodromy action of $\pi_1(\mathbb{P}^1 -$ three points$)$ on the sheets of this cover from the group you have in your hand.
In other words, the data you have is that of a Belyi curve.  Unfortunately it's quite hard to write down algebraic equations for Belyi curves in general, though there's a reasonably large literature working out many examples.  The best-understood cases are those where $\mathcal{H}/\Gamma$ has genus 0 ("dessins d'enfant") which is exactly what you're not interested in! 
This paper by Couveignes and Granboulan should be relevant:
http://www.di.ens.fr/~granboul/recherche/publications/abs-1994-CoGr.html
G.V. Shabat also has extensive tables of explicit equations for dessins d'enfants, but I don't think they're online.<|endoftext|>
TITLE: Elkies' supersingularity theorem in higher dimension
QUESTION [19 upvotes]: The following is a theorem of Elkies:
Let $X$ be an elliptic curve over $\mathbb{Q}$. Then there are infinitely many primes $p$ such that the action of Frobenius on $H^1(\mathcal{O}, X)$ is zero.
Allen Knutson and I would like a similar theorem for higher dimensional Calabi-Yau varieties. Unfortunately, we've been told that this is probably open. (References for the fact that it is open are appreciated.) But we don't need the full strength of Elkies' result. It would be enough for us to know the following:
Let $X$ be an $n$-dimensional, smooth, complete Calabi-Yau variety over $\mathbb{Q}$, for $n>0$. Write $X/p$ for the fiber of $X$ over $p$. Let $T(p)$ be the action of Frobenius on $H^n(\mathcal{O}, X/p)$. Are there infinitely many $p$ for which $T(p) \neq 1$? Also, in the same generality, are there infinitely many primes for which $T(p) \neq 0$?

REPLY [5 votes]: Your question is connected with Frobenius splitting. Namely, we have the following fact (Brion-Kumar "Frobenius splitting methods in geometry and representation theory", Remark 1.3.9 (ii)):

A complete smooth variety $X$ over an algebraically closed field of characteristic $p>0$ is Frobenius split if and only if the pullback map $F^*: H^n(X, \omega_X)\to H^n(X, \omega_X^p)$ is nonzero.

Since in your case $\omega_X = \mathcal{O}_X$, this is exactly the same question. We can be even more general and ask questions like "which characteristic zero varieties are (or are not) Frobenius split for $p$ sufficiently large (or infinitely many $p$)". 
One result of this kind (Brion-Kumar, exercise in paragraph 1.6) is that if $X$ is Fano, then $X/p$ is Frobenius split for $p$ large enough.
Moreover, in their recent work, Mustata and Smith propose a conjecture of this kind in the local case. I don't know enough details to state it precisely, but it says something like "$fpt(X/p) = lct(X)$ for infinitely many $p$", where $X$ is a singularity we reduce mod $p$, $fpt$ is the Frobenius pure threshold and $lct$ is the log-canonical threshold. I heard that this conjecture will imply that all characteristic zero abelian varieties become ordinary when reduced modulo infinitely many $p$.<|endoftext|>
TITLE: Endomorphisms of vector bundles
QUESTION [5 upvotes]: I'm a bit stuck, and I'm hoping someone can help me out.  I have a vector bundle $E$ on an algebraic curve (the ones I am interested in are holomorphic, but I'm sure that doesn't matter so much for the purposes of this question...), and I also have an endomorphism $\Psi\in\Gamma(\mbox{End}(E)).$
How in general does one compute the determinant of $\Psi$?  If $E$ splits as a sum of rank-1 bundles then so too does the endomorphism bundle associated with $E$, and then it's obvious to me how to compute the determinant in that case.  But, when $E$ is more general, what is an invariant way to go about computing the determinant?  (I'm looking for something similar to the invariant way of computing the trace, for which I can rewrite $\Psi$ as a map $E\otimes E^*\rightarrow\mathcal{O}$ and then take the image of the identity morphism of $E$.)
I'm sure there is a quick answer to this, that I am just not seeing.  Any help is greatly appreciated.

REPLY [4 votes]: I can't help but repeat the earlier answers but in my own way:
This is really just a question about linear algebra. It is worth remembering that a vector bundle is just a parameterized family of vector spaces (I like to call differential geometry "parameterized linear algebra"). So anything you can do naturally to a vector space, you can do to a vector bundle.
So to extend the definition of a determinant of an endomorphism of vector spaces to one for an endomorphism of vector bundles, you just need a good natural definition of determinant:
An endomorphism of a vector space $V$ naturally induces an endomorphism of $\Lambda^nV$, where $n$ is the dimension of $V$. Since $\Lambda^nV$ is 1-dimensional, an endomorphism of $\Lambda^nV$ must consist of multiplication by a fixed scalar. That scalar is the determinant of the endomorphism.
The extension to vector bundles then becomes obvious, because you just do the same thing to each fiber.<|endoftext|>
TITLE: Graphs of Tangent Spheres
QUESTION [10 upvotes]: The Koebe–Andreev–Thurston theorem gives a characterization of planar graphs in terms of disjoint circles being tangent. For every planar graph $G$ there is a circle packing whose graph is G. What happens when circles are replaced by spheres? By spheres of  higher dimension? What is known about the graphs that are formed in this case? I know that they must contain all planar graphs but that is about it. 
Let me update this. Based on the answer to this question I am beginning to think about the maximalchromatic number of this graph. If its edge density is less than 7 than for all graphs of this type there must be a point with 13 or less edges. Assume that there is such a graph with chromatic number greater than 14 then look at a graph with the minimum number of points with chromatic number greater than 14. One point must have 13 or less edges. Remove it and since the graph had minimum number of points to have chromatic number greater than 14 the new graph must have chromatic number 14 or less. Color it with 14 or less colors. Add the removed point and look at the colors of the thirteen points it is adjacent to give it the remaining color then we have a 14 coloring of the graph and hence a contradiction. So the chromatic number must be 14 or less. We have an lower bound of 4 as the graphs of tangent spheres contain the graphs of tangent circles which are the planar graphs which contain graphs with chromatic number four.
As we increase the dimensions of the tangent spheres the chromatic number goes to infinity just take $d+1$ tangent spheres. But I think we can do better than that for one thing we should be able to improve the lower bound from 4. Also I think there should be examples from lattices that give better lower bounds on the chromatic number possibly exponential.
Finally based on the existing chromatic numbers I am wondering if it is possible to answer this question. Is there a dimension where the chromatic number of the unit distance graph is different from the chromatic number of the graphs in that dimension of tangent spheres.
The unit distance graph is the set of all points in the $n$-dimensional space with two points connected if their distance is one. For dimension two the chromatic number is known to be in the range from 4 to 7. For dimension three the range is 6 to 15. For the graphs of tangent circles we have a chromatic number of 4 and for spheres a range from 4 to 15. So the possibility that the chromatic numbers of the two types of graphs are the same has not yet been eliminated. So the specific question is what is known and what can be proved about the chromatic number of the graphs of tangent spheres.

REPLY [12 votes]: The number of edges in such a graph is linear in the number of vertices, and they can be split into two equal-sized subgraphs by the removal of $O(n^{\frac{d-1}{d}})$ vertices. See e.g.
A deterministic linear time algorithm for geometric separators and its applications.
D. Eppstein, G.L. Miller, and S.-H. Teng.
Fundamenta Informaticae 22:309-330, 1995.
Unlike in the 2d case (where we know that the maximum number of edges such a graph can have is 3n-6) the precise maximum edge density is not known even in 3d. There's a lower bound of roughly $\frac{3828n}{607}\approx 6.3064n$ in another of my papers,
Fat 4-polytopes and fatter 3-spheres.
D. Eppstein, G. Kuperberg, and G. Ziegler.
arXiv:math.CO/0204007.
Discrete Geometry: In honor of W. Kuperberg's 60th birthday, Pure and Appl. Math. 253, Marcel Dekker, pp. 239-265, 2003.
and an upper bound of $(4+2\sqrt3)n \approx 6.8284n$ in
Greg Kuperberg and Oded Schramm, Average kissing numbers
for non-congruent sphere packings, Math. Res. Lett. 1 (1994),
no. 3, 339–344, arXiv:math.MG/9405218.<|endoftext|>
TITLE: Why a subvariety of a variety of general type is of general type
QUESTION [9 upvotes]: It seems a well-known fact that subvarieties of a variety of general type containing a general point are also of general type. This fact is an essential property used to prove some extension theorems of pluricanonical forms on algebraic varieties of general type, see for example the very nice survey on extension of pluricanonical forms
http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf (Internet Archive)
I want to know why this is true? Is there any thing more we can say about the relation between the canonical line bundle of a variety and that of subvarieties of codimension no less than 2.

REPLY [4 votes]: Theorem
Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have 
$$\kappa(X_{t}) \geq \kappa(X)-\dim T$$
proof:
For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism.
\begin{align*}
    \kappa(X_{t})& \geq   \dim(\phi(X_{t}))\\
    &\dim X_{t}-\dim(X/Z)\\
    &=\dim Z-\dim T\\
    &=\kappa(X)-\dim T
\end{align*}
Corollary
Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.
proof:\begin{align*}
    \kappa(X_{t})&\geq \kappa(X)-\dim T \\
    &=\dim X-\dim T\\ 
    &=\dim X_{t}\\
\end{align*}
The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.
Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.
proof:
By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have 
$$K_{Z}=g^{*}K_{X}+R$$
where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.
Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.
proof:
The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably many components. For each $P\in \mathbb{Q}[t]$, we have the universal family $$U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$$ Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our variety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. For any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form the Corollary above.<|endoftext|>
TITLE: Chebyshev-like polynomials with integral roots
QUESTION [7 upvotes]: Chebyshev polynomials have real roots and satisfy a recurrence relation. I was wondering if one can find a sequence of polynomials with integral or rational roots with similar properties. More precisely, one is looking for a sequence of polynomials $(f_n),f_n\in\mathbf{Q}[t]$ such that

$\deg f_n\to\infty$ as $n\to\infty$;
$\sum_{n=0}^\infty f_n(t) x^n$ is (the Taylor series of) a rational function $F$ in $x$ and $t$.
All roots of any $f_n$ are integer and have multiplicity 1. (A weaker version: the roots are allowed to be rational and are allowed to have multiplicity $>1$ but there should be an $a>0$ such that the number of distinct roots of $f_n$ is at least $a\deg f_n$.)

REPLY [3 votes]: Here's one thought.  For each integer k, f_n(k) satisfies a recurrence relation.  If the roots of f_n are all integers, then f_n(k) and f_n(k+1) have to be "in sync" in the sense that they never have opposite sign.  This is a strong condition!  For instance, suppose the sequences f_n(k) and f_n(k+1) each have unique largest eigenvalue:  then these eigenvalues would have to have the same argument.
Update:  Qiaochu's answer suggests that in fact working mod p would be just better than the "archimedean" picture sketched above, since it is really F_q[t], not Z[t] or Q[t], that is analogous to the integers.  Let F_n(t) be the reduction of f_n(t) to F_p[t].  If f_n(t) has all roots rational for every n, then the reduction of f_n(t) mod p has the same property.  But now we are saying something quite strong; that f_n(t) lies in a finitely generated subgroup of F_q(t)^*!  This is presumably ruled out by Mason's theorem (ABC for function fields.)  Indeed, you could probably prove in this way that not only are the roots of f_n(t) not rational, but f_n(t) has irreducible factors of arbitrarily large degree.
I don't think this approach would touch a harder question along the same lines like this one.<|endoftext|>
TITLE: What is a section?
QUESTION [21 upvotes]: This question comes out of the answers to Ho Chung Siu's question about vector bundles.  Based on my reading, it seems that the definition of the term "section" went through several phases of generality, starting with vector bundles and ending with any right inverse.  So admittedly I'm a little confused about which level of generality is the most useful.
Some specific questions:

Why can we think of sections of a bundle on a space as generalized functions on the space?  (I'm being intentionally vague about the kind of bundle and the kind of space.)
What's the relationship between sections of a bundle and sections of a sheaf?
How should I think about right inverses in general?  I essentially only have intuition for the set-theoretic right inverse.

Pointers to resources instead of answers would also be great.

REPLY [23 votes]: To your first question, "function on a space" $X$ usually means a morphism from $X$ to one of several "ground spaces" of choice, for example the reals if you work with smooth manifolds, Spec(A) if you work with schemes over a ring, etc.  (This is a fairly selective use of the word "function" which used to confuse me.)  A section $\gamma$ of a (some-kind-of) bundle $E\to X$ is thought of as a "generalized function" on $X$ by thinking of it as a funcion with "varying codomain", i.e. at each point $x\in X$, it takes value in the fibre $E_x\to x$.  If one is talking about locally free / locally trivial bundles, meaning $E$ is locally (over open sets $U\subset X$) isomorphic to some product $U\times T$, then we can locally identify the fibres with $T$.  Thus locally a section just looks like a function with codomain $T$, which is often required to be nice.
To your second question, I generally take the "right-inverse" or "pre-inverse" definition from category theory, because it relates back to others in the following precise way: 
Say $\pi: Y\to X$ is a space over $X$ (intentionaly vague).  The word "over" is used to activate the tradition of suppressing reference to the map $\pi$ and refering instead to the domain $Y$.  For $U\subseteq X$ open, the notation $\Gamma(U,Y)$ denotes sections of the map $\pi$ over $U$, i.e. maps $U\to Y$ such that the composition $U \to Y\to X$ is the identity (thus necessarily landing back in $U$).  It's not hard to see that $\Gamma(-,Y)$ actually forms a sheaf of sets on $X$.
Conversely, given any sheaf of sets $F$ on a space $X$, one can form its espace étalé, a topological space over $X$, say $\pi: \acute{E}t(F) \to X$.  Then for an open $U\subseteq X$, the elements of $F(U)$ correspond precisely to sections of the map $\pi$, which by the above notation is written $\Gamma(U,\acute{E}t(F)$.  That is to say, $F(-)\simeq\Gamma(-,\acute{E}t(F))$ as sheaves on $X$.  This explains why people often refer to sheaf elements as "sections" of the sheaf.
Moreover, what we now denote by $\acute{E}t(F)$ actually used to be the definition of a sheaf, so people tend to identify the two and write $\Gamma(-,F)$ a instead of $\Gamma(-,\acute{E}t(F))$.  This explains the otherwise bizarre tradition of writing $\Gamma(U,F)$ instead of the the more compact notation $F(U)$.
$\Big($Unfortunate linguistic warning: Many people incorrectly use the term "étale space".  However, the French word "étalé" means "spread out", whereas "étale" (without the second accent) means "calm", and they were not intended to be used interchangeably in mathematics.  This is unfortunate, because the espace étalé has very little to with with étale cohomology.  More unfortunate is the annoying coincidence that when dealing with schemes the projection map from the espace étalé happens to be an étale morphism, because it is locally on its domain an isomorphism of schemes, a much stronger condition.$\Big)$
To your third question, I think the observation that $\Gamma(-,Y)$ forms a sheaf on $X$ gives a nice context in which to think of sections $X$ to $Y$: they "live in" the sheaf $\Gamma(-,Y)$ as its globally defined elements.<|endoftext|>
TITLE: Do there exist non-PIDs in which every countably generated ideal is principal?
QUESTION [32 upvotes]: The title pretty much says it all: suppose $R$ is a commutative integral domain such that every countably generated ideal is principal.  Must $R$ be a principal ideal domain?
More generally: for which pairs of cardinals $\alpha < \beta$ is it the case that: for any commutative domain, if every ideal with a generating set of cardinality at most $\alpha$ is principal, then any ideal with a generating set of cardinality at most $\beta$ is principal?
Examples: Yes if $2 \leq \alpha < \beta < \aleph_0$; no if $\beta = \aleph_0$ and $\alpha < \beta$: 
take any non-Noetherian Bezout domain (e.g. a non-discrete valuation domain).
My guess is that valuation domains in general might be useful to answer the question, although I promise I have not yet worked out an answer on my own.

REPLY [6 votes]: Sorry to dig up an old question, but in case anybody else randomly lands here, here's a quick side note about a way that this can be generalized.
Theorem:  If every countably generated ideal of a ring $R$ is finitely generated, then $R$ is Noetherian.  Hence, if $n < \infty$ and every countably generated ideal is $n$-generated, then every ideal is $n$-generated.
Proof:  By contrapositive.  If $R$ is not Noetherian, then we can make an infinite properly ascending chain $I_1 \subsetneq I_2 \subsetneq \cdots$ of finitely generated ideals.  The union of this chain is a countably generated ideal, and it cannot be finitely generated, because that would cause the chain to terminate at some point.<|endoftext|>
TITLE: Why are spectral sequences so ubiquitous?
QUESTION [54 upvotes]: I sort of understand the definition of a spectral sequence and am aware that it is an indispensable tool in modern algebraic geometry and topology.  But why is this the case, and what can one do with it?  In other words, if one were to try to do everything without spectral sequences and only using more elementary arguments, why would it make things more difficult?

REPLY [22 votes]: Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...).  And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces.  For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences.  Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).
Long exact sequences are actually a special case of spectral sequences—those whose E1 page has only two columns.  If you have never done this before, you should check for yourself that the d1 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to.  This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds.  Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map.  These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world).  Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite.  It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.
Edit: Clark Barwick suggests that one may indeed view all "natural" spectral sequences as arising from filtered spectra.  He and I would like to know whether there are any convincing counterexamples, so please let me know if you have any!  Note however that 1 and 2 from VA's answer are not counterexamples.<|endoftext|>
TITLE: What are good non-English languages for mathematicians to know?
QUESTION [49 upvotes]: It seems that knowing French is useful if you're an algebraic geometer.  More generally, I've sometimes wished I could read German and Russian so I could read papers by great German and Russian mathematicians, but I don't know how useful this would actually be.  What non-English languages are good for a generic mathematician to know?  Are there specific languages associated to disciplines other than algebraic geometry?  
(This question is a little English-centric, but I figure it's okay because this website is run in English.)

REPLY [13 votes]: I not a native  English  speaker  and thus my intervention here is a bit odd, to say the least. Growing up behind the Iron Curtain I learned English, but I discovered that as a  mathematician  it is more useful to learn Russian,  simply because the Soviets translated  all the influential books and articles into Russian. Often, the Russian translations would be better than the original because the translators,  influential mathematicians themselves, would correct possible errors and would add illuminating footnotes and appendices.
To this day I cannot read German, but I can read Grauert's or Brieskorn'article in Russian. Today though, English is the lingua Franca of  Science, so for a youngster it is more profitable to concentrate on Math.<|endoftext|>
TITLE: Inequality of the number of integer partitions
QUESTION [8 upvotes]: I am familiar with the partition function p(k,n) where p is the number of partitions of n using only natural numbers at least as large as k.
Is there a way of determining if p(k1, n1) > p(k2, n2) that does not actually use the partition function?
To clarify, I want to know if there is a quick way to tell if the number of partitions of k1, n1 is greater than or less than the number of partitions of k2, n2 without using the partition function.
Thanks!
--Connor

REPLY [5 votes]: No, he probably means exactly what he said.  That is the way the partition function is usually defined.  But either way, the answer is no.
If $q(k,n)$ counts partitions of n into integers no bigger than k, as Jonah suggests, then note that $q(2,2m) = m+1$ for every $m$.  (A partition is determined by the number of 2's.)  So being able to compare values of $q(k,n)$ would in particular entail being able to compare $q(k,n)$ to any given integer.
As for the question as actually asked, note that $p(2k,4k-1)=k+1$ for every $k$.  Once again, knowing the relative sizes of all $p(k,n)$ is tantamount to knowing whether $p(k,n)$ is more or less than each integer, i.e. knowing the values of $p(k,n)$.<|endoftext|>
TITLE: Spin structures on 7-dimensional spherical space forms
QUESTION [15 upvotes]: Background
Let $M$ be a spin manifold and let $\Gamma$ be a finite group acting freely and isometrically on $M$ in such a way that $M/\Gamma$ is a smooth riemannian manifold.  The quotient will be spin if and only if the action of $\Gamma$ on $M$ lifts to the spin bundle.
For reasons having to do with $11 = 7 + 4$, I got interested in $M=S^7$ with the round metric.  There is a unique spin structure on $S^7$ and the spin bundle is
$$\mathrm{Spin}(7) \to \mathrm{Spin}(8) \to S^7.$$
A while back, together with one of my students, we investigated which smooth quotients $S^7/\Gamma$ are spin and how many inequivalent spin structures they admit.  This boils down to determining the isomorphic lifts of $\Gamma \subset \mathrm{SO}(8)$ to $\mathrm{Spin}(8)$.
There are lots of finite subgroups $\Gamma \subset \mathrm{SO}(8)$ acting freely on $S^7$, which are listed in Wolf's Spaces of constant curvature and to our surprise (this does not happen with $S^5$, say) we found that all quotients $S^7/\Gamma$ are spin; although they do not all have the same number of spin structures.  Our results were obtained by a case-by-case analysis, but we always remained with the sneaky suspicion that there ought to be a simple topological explanation.
Question
Is there one?  Perhaps based on the parallelizability of $S^7$?
Thanks in advance.
Edit
I'm answering Chris's questions in the first comment below.
The problem is indeed the existence of a subgroup $\Gamma' \subset \mathrm{Spin}(8)$ such that obvious square commutes:
$$\Gamma' \to \Gamma \to \mathrm{SO}(8) = \Gamma' \to \mathrm{Spin}(8) \to \mathrm{SO}(8)$$
and where the first map $\Gamma' \to \Gamma$ is an isomorphism.  This is the same as lifting $\Gamma \to \mathrm{SO}(8)$ via the spin double cover.
The simplest counterexample for $S^5$ is to take any freely acting cyclic subgroup $\Gamma \subset \mathrm{SO}(6)$ of even order.

REPLY [4 votes]: Here is a partial answer. If the order of $\Gamma$ is odd, then this is a trivial application of transfer maps. You have described your manifold as a quotient $\pi:S^7 \to M = S^7/\Gamma$, and hence $S^7$ is a covering space of $M$. The transfer map is a wrong way map in cohomology:
$ \tau^* : H^* (S^7) \to H^* (M) $
which exists for cohomology in, say, $\mathbb{Z}/2$-coefficients. The composition $\tau^* \pi^*$ is multiplication by the order of $\Gamma$, which in this case is an isomorphism when the order of $\Gamma$ is odd. But since the cohomology of $S^7$ vanishes in degrees 1 and 2, this proves that these groups also vanish for $M$ and hence $M$ has a spin structure and it is unique. 
The more interesting case is when $\Gamma$ is 2-primary. For example why does $\mathbb{R}P^7$ have a spin structure? I suspect that your intuition is spot on and that it has to do the framing of $S^7$.<|endoftext|>
TITLE: On the dimension of moduli space of pointed curves with fixed Weierstrass semigroup
QUESTION [5 upvotes]: Does anyone know any information on the question of the dimension of moduli space of pointed curves with fixed Weierstrass semigroup? Some conjecture?

REPLY [2 votes]: One good reference is "Recent progress in the study of Weierstrass points" by Eisenbud and Harris in Geometry Today, Birkhauser, 1985.<|endoftext|>
TITLE: examples of admissible representations of $GL_{n}(K)$ over p-adic field
QUESTION [6 upvotes]: I've been reading about the Langlands program (the paper by Torsten Wedhorn "Local langlands correspondence for GL(n) over p-adic fields, to be precise), and I want to get my hands dirty with examples.
What are some interesting (yet not too nasty) examples of admissible representations of $GL_{n}(K)$, where $K$ is a $p$-adic field? 
Taking Schur functors of an admissible representation, should still generally give you something admissible, right?

REPLY [7 votes]: I second L Spice's recommendation of the book by Bushnell and Henniart, called "The local Langlands conjectures for GL(2)."  
After you master the principal series representations, it's not too hard to tinker with some supercuspidals.  Easiest among these are the tamely ramified supercuspidals.  To construct these, let's start with the unramified quadratic extension $L/K$, with corresponding residue fields $\ell/k$.  Choose a character $\theta$ of $L^\times$ which has these properties:
(a) The character $\theta$ is trivial on $1+\mathfrak{p}_L$, so that $\theta\vert_{\mathcal{O}_L^\times}$ factors through a character $\chi$ of $\ell^\times$.
(b) $\chi$ is distinct from its $k$-conjugate.  (In other words, $\chi$ does not factor through the norm map to $k^\times$.)
It's a standard fact that there's a corresponding representation $\tau_\chi$ of $\text{GL}_2(k)$, characterized by the identity $\text{tr}\tau_\chi(g)=-(\chi(\alpha)+\chi(\beta))$ whenever $g\in\text{GL}_2(k)$ has eigenvalues $\alpha,\beta\in\ell\backslash k$.  (This is somewhere in Fulton and Harris, for instance.)
Inflate $\tau_\chi$ to a representation of $\text{GL}_2(\mathcal{O}_K)$, and extend this to a representation $\tau_\theta$ of $K^\times\text{GL}_2(\mathcal{O}_K)$ which agrees with $\theta$ on the center.  Finally, let $\pi_\theta$ be the induced representation of $\tau_\theta$ up to $\text{GL}_2(K)$;  then $\pi_\theta$ is an irreducible supercuspidal representation.
By local class field theory, our original character $\theta$ can be viewed as a character of the Weil group of $L$. In the local Langlands correspondence, $\pi_\theta$ lines up with the representation of the Weil group of $K$ induced from $\theta$.  All the supercuspidals of $\text{GL}_2(K)$ arise by induction from an open compact-mod-center subgroup, but the precise construction of these is a little more subtle than the above example.<|endoftext|>
TITLE: Range of the Fourier transform on $L^1$
QUESTION [27 upvotes]: It is well known that the Fourier transform $\mathcal{F}$ maps 
$L^1(\mathbb{R}^d)$ into, but not onto, $\overline{C_0^0}(\mathbb{R}^d)$, where the closure is taken in the $L^\infty$ norm. This is a consequence of the open mapping theorem, for instance. 
My question is: what's an explicit example of a function in $\overline{C_0^0}(\mathbb{R}^d)$ which is not in the image of $L^1(\mathbb{R}^d)$ under the Fourier transform? 
I would also like to know whether there is a useful characterization of $\mathcal{F}(L^1(\mathbb{R}^d))$. 
Remark: it is easy to see that the Banach space $\overline{C_0^0}(\mathbb{R}^d)$ consists of all continuous functions $f$ on $\mathbb{R}^d$ such that $f(\xi)\rightarrow 0$ as $|\xi|\rightarrow\infty$.

REPLY [4 votes]: See Page 68 (Theorem 3.2.2) in my notes on Fourier Analysis:
https://home.iitm.ac.in/mtnair/FS-2018.pdf<|endoftext|>
TITLE: Is it possible to capture a sphere in a knot?
QUESTION [94 upvotes]: You and I decide to play a game:
To start off with, I provide you with a frictionless, perfectly spherical sphere, along with a frictionless, unstretchable, infinitely thin magical rope. This rope has the magical property that if you ever touch its ends to each other, they will stick together and never come apart for all eternity. You only get one such rope, but you are allowed to specify its length.
Next, I close my eyes and plug my ears as you do something to the rope and the sphere. When you are done with whatever you have decided to do, you give me back the sphere and rope. Then I try my best to remove the rope from the sphere (i.e., make the smallest distance from a point on the rope to a point on the sphere at least 1 meter). Of course, since the rope is not stretchable, the total length of the rope cannot increase while I am trying to remove it from the sphere.
If I succeed in removing the rope from the sphere, I win. Otherwise, you win. Who has the winning strategy?
EDIT: To clarify, Zeb is looking for an answer with a finite length, piecewise smooth rope, and the sphere should be rigid.

REPLY [3 votes]: Since Anton's beautiful solution makes use of the symmetry of the sphere, I wonder how similar
results could be proven, or counterexamples given, for any other convex shape, including
2-dimensional ones - i.e. infinitely thin 3-dimensional objects.  I can't figure a way to tie a cube or a square, but it seems that an equilateral triangle could be tied by 3 connected loops starting from some knot at the center and going over each vertex (forming 3 equilateral triangles 1/3 the size of the original).
(I would have liked to just leave this post as a comment to Anton's proof, but I'm not allowed to do that. Should it perhaps be a new question?)<|endoftext|>
TITLE: Number theory textbook with an algebraic perspective
QUESTION [14 upvotes]: Most of the number theory textbooks I've dealt with take a very classical approach to the subject.  I'm looking for a textbook that's something like a first course in number theory for people who have a decent command of modern algebra (at the level of something like Lang's Algebra).  Does such a book exist, and if it does, what is it called?
Edit: As I posted in a comment below:
In the introduction to Ireland and Rosen, they note something that was bugging me for a while, "Nevertheless it is remarkable how a modicum of group and ring theory introduces unexpected order into the subject." 
This is precisely the perspective I was looking for, so if anyone passes by this topic looking for a book that approaches number theory in this way, I feel like this quote should point him (her?) in the right direction.

REPLY [5 votes]: Borevich-Shafarevich<|endoftext|>
TITLE: Higher-rank Borel sets
QUESTION [5 upvotes]: What are interesting, illustrative examples of Borel sets, situated in Borel hierarchy higher than $\Sigma^{0}_{2}$ /$\Pi^{0}_{2}$?

REPLY [4 votes]: My personal favourite is the (F-sigma-delta) set of all real numbers x for which limit as n -> infinity sin(n! pi x) exists. There is a natural way to recursively iterate this construction to get Borel additive subgroups of reals at all finite levels of Borel hierarchy.<|endoftext|>
TITLE: Existence of projective resolutions in abelian categories
QUESTION [5 upvotes]: It is a standard result of elementary homological algebra that to every R-module $A$ there exists a projective resolution.  It is often said that the category of R-modules has "enough projectives."  In which other categories is this also true? In particular is it true for abelian categories?

REPLY [17 votes]: Among the standard examples of abelian categories without enough projectives, there are

the categories of sheaves of abelian groups on a topological space (as VA said), or sheaves of modules over a ringed space, or quasi-coherent sheaves on a non-affine scheme;
the categories of comodules over a coalgebra or coring.

No abelian category where the functors of infinite product are not exact can have enough projectives.  In Grothendieck categories (i.e. abelian categories with exact functors of small filtered colimits and a set of generators) there are always enough injectives, but may be not enough projectives.
Among the standard examples of abelian categories with enough projectives, there are

the category of functors from a small category to an abelian category with enough projectives (as VA said), or the category of additive functors from any small additive category to the category of abelian groups (this class of examples includes the categories of modules over any rings);
the category of pseudo-compact modules over a pseudo-compact ring (see Gabriel's dissertation);
the category of contramodules over a coalgebra or coring (see Eilenberg-Moore, "Foundations of relative homological algebra").

REPLY [12 votes]: A simple example of an abelian category having not enough projectives is the category of finite abelian groups. In fact, it contains neither non-trivial projective objects nor non-trivial injective objects.<|endoftext|>
TITLE: Are there piecewise-linear unknots that are not metrically unknottable?
QUESTION [16 upvotes]: A stick knot is a just a piecewise linear knot. We could define "stick isotopy" as isotopy that preserves the length of each linear piece.
Are there stick knots which are topologically trival, but not trivial via a stick isotopy?

REPLY [22 votes]: Yes, there are. See "Locked and Unlocked Polygonal Chains in 3D", T. Biedl, E. Demaine, M. Demaine, S. Lazard, A. Lubiw, J. O'Rourke, M. Overmars, S. Robbins, I. Streinu, G. Toussaint, S. Whitesides, arXiv:cs.CG/9910009, figure 6.

REPLY [21 votes]: There is a survey paper on this general topic by Robert Connelly and Erik Demaine.  As David Eppstein just posted, the answer is yes in 3D.  However, it is a famous result of those two authors and Gunter Rote that the answer is no in 2D: all polygonal planar cyclic linkages can be made convex.  This second problem was a long-standing conjecture called the "Carpenter's Rule Problem".  There are at least two beautiful proofs of the conjecture: the original, and a second one using pseudo-triangulations due to Ileana Streinu.<|endoftext|>
TITLE: measurable sets not depending on even coordinates
QUESTION [11 upvotes]: Let $A\subset\{0,1\}^\omega$ be a measurable set (w.r.t. the usual borel sigma algebra) which does not depend on any even coordinate (that is, if $x\in A$ and $x$ and $y$ agree except on a finite number of even coordinates, then $y\in A$).
Is it true that $A$ belongs to the sigma-algebra generated by all the odd coordinates + the tail sigma algebra?
To clarify: the tail sigma algebra consists of all the events which do not depend on any coordinate.
It seems to me that this should be some easy/well known measure theory fact/counterexample, but perhaps I'm wrong? Suggestions on where to look for an answer would be welcome.
Note that it is well known the this statement is false if the ground space would be $[0,1]^\omega$.

REPLY [2 votes]: Write $\{0,1\}^\mathbb{N}$ as cartesian product $\{0,1\}^E \times \{0,1\}^O$, where $E$ is the evens and $O$ is the odds.  Your usual sigma-algebra is the product sigma-algebra that goes with this product.
So we need something like this:  Write $\otimes$ for product sigma-algebra.  Then is
$$\mathcal{F}\otimes\left(\bigcap_{k=1}^\infty\mathcal{G}\_k\right) = \bigcap_{k=1}^\infty\left(\mathcal{F}\otimes\mathcal{G}\_k\right)$$<|endoftext|>
TITLE: Notation/name for "Artin-Schreier roots"?
QUESTION [5 upvotes]: If x is an element of a field K and n is a positive integer, we have both a symbol and a name for a root of the polynomial t^n - x = 0: we denote it by x^{1/n} and call it an nth root of x.
Of course nth roots play a vital role in field theory, e.g. in the characterization of solvable extensions in characteristic 0.  However, in characteristic p > 0, the extraction of a p-power root is a much different business: it gives rise to purely inseparable extensions, not composition factors of solvable Galois extensions.  
To repair the characterization of solvable extensions in characteristic p as those being attainable as a tower of "radical" extensions, one needs to include the operation of taking roots of Artin-Schreier polynomials: t^q - t - x = 0, for q = p^a a power of the characteristic. 
Finally my question: do we have a name for an element t solving the equation t^q - t = x and/or a special notation for it?  I do not know one.  Similarly, whereas classically we often speak of x as being "an nth power", in this case I find myself writing "x is in the image of the Artin-Schreier isogeny \rho".  Is there something better than this?

REPLY [9 votes]: I have been using the notation $\wp^{-1}(a)$ for a root of $T^p-T-a$ (in characteristic $p$), which I borrowed from Bourbaki, but I have also used $\wp_2^{-1}(a)$ and $\wp_3^{-1}(a)$ if the characteristic had to be mentioned explicitly. 
This morning I came across an original notation for $\wp_p^{-1}(a)$ in a paper by Davenport and Hasse in the Journal für die reine und angewandte Mathematik, 172 (1935).  I cannot display it here because Knuth was not aware of it when he created TeX, so all I can do is to provide a link to their article.  The notation appears on the last line of the first page (p.151) and it is defined in the first line of the second page (p.152).
Clearly, the reason behind their notation is its similarity to $\root p\of{}$.  It is a less angular and more curvaceous version of the radical, with the loop closing upon itself.
I wish the people responsible for TeX and Unicode will give a second life to this little gem.<|endoftext|>
TITLE: Is a polynomial with 1 very large coefficient irreducible?
QUESTION [13 upvotes]: I am asking for some sort of generalization to Perron's criterion which is not dependent on the index of the "large" coefficient. (the criterion says that for a polynomial $x^n+\sum_{k=0}^{n-1} a_kx^k\in \mathbb{Z}[x]$ if the condition $|a_{n-1}|>1+|a_0|+\cdots+|a_{n-2}|$ and $a_0\neq 0$ holds then it is irreducible.) 
This would answer a second question about the existence of n+1-tuples $(a_0,\dots,a_n)$ of integers for which $\sum_{k=0}^n a_{\sigma(k)}x^k$ is always irreducible for any permutation $\sigma$.
What happens if we restrict $|a_i| \le O(n)$ ? $ |a_i| \le O(\log n)$ ?

REPLY [4 votes]: OK, about your second question. Let's consider the polynomial $x^n+2(x^{n-1}+\ldots+x^2+x)+4$. I claim that this polynomial is almost good for your purposes: if we permute all coefficients except for the leading one, it remains irreducible. Proof: if the constant term becomes 2 after the permutation, use Eisenstein, if not, look at the Newton polygon of this polynomial mod 2 - you can observe that if it is irreducible, it has to have a linear factor, which is easily impossible. 
[If we allow to permute all coefficients, I would expect that something like $9x^n+6(x^{n-1}+\ldots+x^2+x)+4$ would work for nearly the same reasons...]<|endoftext|>
TITLE: Subfields joining an algebraic element to another
QUESTION [10 upvotes]: Let $\alpha$ and $\beta$ be two algebraic numbers over $\mathbb Q$. Say that a subfield $\mathbb K$ of $\mathbb C$ joins $\alpha$ to $\beta$ iff $\beta \in {\mathbb K}[\alpha]$ but $\beta \not\in {\mathbb K}$. Now, if $\mathbb K$ joins $\alpha$ to $\beta$ and we add a completely unrelated algebraic number to $\mathbb K$, we still have a join from $\alpha$ to $\beta$. So it is natural to consider the minimal joins from $\alpha$ to $\beta$, i.e. the joins that are minimal with respect to field inclusion. Let ${\cal M}(\alpha,\beta)$ denote the set of all minimal joins from $\alpha$ to $\beta$. My guesses are that :
1) Any field in ${\cal M}(\alpha,\beta)$ is always contained in the normal (Galois) closure
of ${\mathbb Q}(\alpha,\beta)$
   2) ${\cal M}(\alpha,\beta)$ is always finite
   3) The two facts above should be provable using Galois theory.
 Note that 2) follows from 1).   
Can anyone confirm this ?
A simple example : ${\cal M}(\sqrt{2},\sqrt{3})$ consists of 
${\mathbb Q}(\sqrt{6})$. Indeed, suppose $\mathbb K$ joins
$\sqrt{2}$ to $\sqrt{3}$ and 
$x$ and $y$ are numbers in $\mathbb K$ such that $x+y\sqrt{2}=\sqrt{3}$. If $x \neq 0$ then
$\sqrt{3}=\frac{3+x^2-2y^2}{2x} \in \mathbb K$ which is absurd. So $x=0$ and $y=\frac{\sqrt{6}}{2}$.

REPLY [2 votes]: Finally my guesses seem to be correct :  let $d$ be the degree of the extension ${\mathbb K}(\alpha):{\mathbb K} $. There are exactly $d$ field homomorphisms ${\mathbb K}(\alpha)\to \mathbb C$ that coincide with the identity on $\mathbb K$. Each one of this homomorphisms may be extended to a field homomorphism from ${\mathbb K}(\alpha,\beta)$ to $\mathbb C$ (in a generally nonunique way). Let us denote by $H$ the $d$-element set of all the homomorphisms thus obtained.   
As darij said, there is a polynomial $P\in {\mathbb K}[X]$ such that $P(\alpha)=\beta$. We may take $P$ of degree smaller than $d$. If we denote the coefficients of $P$ by $p_0,p_1, \ldots , p_{d-1}$, then we have
$\sum_{k=0}^{d-1}p_k \alpha^k=\beta$.
Now, applying the elements of $H$ to this equalities yields
$\sum_{k=0}^{d-1}p_k h(\alpha)^k=h(\beta)$
for any $h\in H$. We see now that the $p_k$ may be retrieved as solutions of
a $d\times d$ system all of whose coefficients are in the closure of
${\mathbb Q}(\alpha,\beta)$. The determinant of this system is a Van der Monde determinant on the distinct conjugates of $\alpha$, so it's nonzero.
This shows statement 1) (and 2)). Although darij's proof is wrong as shown in the comments, it may still be that his statement a) is correct (so that "Galois closure" may be supressed in
the statement of 1)). Pheraps one could show it the Galois way, by showing that any field homomorphism that coincides with the identity on ${\mathbb Q}(\alpha,\beta)$ in fact preserves the coefficients of $P$.<|endoftext|>
TITLE: Elliptic Curves over F_1?
QUESTION [16 upvotes]: Is there an notion of elliptic curve over the field with one element? As I learned from a previous question,  there are several different versions of what the field with one element and what schemes over it should be (see for example this article by Javier López Peña and Oliver Lorscheid). What I want to know is whether there a good notion of elliptic curve over $F_{un}$? What about modular forms?

REPLY [11 votes]: As was mentioned by others, currently varieties over $\mathbb F_1$ look uncomfortably like toric varieties or something very close to that. But of course there is a way to think of an elliptic curve as a sort of "toric" variety, using the Tate's uniformization $E\ "="\ \mathbb G_m/\mathbb Z\ $! (Same for higher-dimensional abelian varieites, using the Mumford-Tate(-Faltings-Chai) uniformization.) So there has to be a way...
Put it another way, an elliptic curve is very similar to $\mathbb P^1$ with two points, $0$ and $\infty$, identified. Now that is an object that is really defined over $\mathbb F_1$.<|endoftext|>
TITLE: How can I really motivate the Zariski topology on a scheme?
QUESTION [28 upvotes]: First of all, I am aware of the questions about the Zariski topology asked here and I am also aware of the discussion at the Secret Blogging Seminar. But I could not find an answer to a question that bugged me  right from my first steps in algebraic geometry: how can I really motivate the Zariski topology on a scheme?
For example in classical algebraic geometry over an algebraically closed field I can define the Zariski topology as the coarsest $T_1$-topology such that all polynomial functions are continuous. I think that this is a great definition when I say that I am working with polynomials and want to make my algebraic set into a local ringed space. But what can I say in the general case of an affine scheme? 
Of course I can say that I want to have a fully faithful functor from rings into local ringed spaces and this construction works, but this is not a motivation. 
For example for the prime spectrum itself, all motivations I came across so far are as follows: well, over an algebraically closed field we can identify the points with maximal ideals, but in general inverse images of maximal ideals are not maximal ideals, so let's just take prime ideals and...wow, it works. But now that I know that one gets the prime spectrum from the corresponding functor (one can of course also start with a functor) by imposing an equivalence relation on geometric points (which I find very geometric!), I finally found a great motivation for this. What is left is the Zariski topology, and so far I just came across similar strange motivations as above...

REPLY [3 votes]: think there are two questions here: (1) why study the prime spectrum, and (2) why think of it in terms of the axioms for a topology. (1) has been pretty well handled by other commenters. And a number of them point out that (2) isn't really especially useful.
Part of the problem, according to a very interesting article I read by Grothendieck (maybe where he introduces dessins d'enfants?), is that the axioms for topological spaces are "wrong". Alas, he doesn't know what the right axioms are; he's just sure that the field of general topology should never have existed. From that point of view, discovering that the prime spectrum has a topology automatically isn't that interesting. (This guy has a contrary viewpoint on the field of general topology, but unsurprisingly I find Grothendieck more convincing.)<|endoftext|>
TITLE: less elementary group theory
QUESTION [12 upvotes]: Most of the group theory that is taught in introductory graduate classes is of the form $$(\mbox{number theory} + \mbox{ group actions} + \mbox{ orbit-stabilizer thm}) + \mbox{group axioms} \Rightarrow \mbox{theorems}$$ So what is the equivalent of "(number theory + group actions + orbit-stabilizer thm)" in the more advanced parts of group theory?
To clarify based on some comments: The techniques I learned in a graduate group theory class were just the orbit-stabilizer thm + some number theory + Lagrange's thm. Adding in some more constructions like semi-direct products allows one to make some inroads into some less elementary parts of group theory, i.e. we get some classification theorems for groups of small order with the help of Sylow theorems which is really just clever number theory + orbit-stabilizer theorem. So I would like to expand my toolbox a little bit by seeing what other tools are used in more advanced group theory but are still applicable to the elementary parts of group theory like classifying groups of small order.
Tidbits collected from comments: Kevin McGerty makes some excellent points about the extension of the theory from actions on sets which allow number theoretic arguments to actions on vector spaces which increase the sophistication and depth of the theory. The move from mere sets to vector spaces allows the use of linear algebra as another tool which in turn allows some tools from homological algebra to enter into the game.

REPLY [2 votes]: The impression I get is that a large chunk finite group theory can be built up from the beginner's toolset: orbit-stabiliser, the isomorphism theorems, and a lot of fiddling around with conjugation, normalisers and centralisers, and induction on the order of the group.  You can achieve a lot with surprisingly little.
Character theory (over the complex numbers) is probably the non-'elementary' tool that sees the heaviest use.  For instance, one often wants to solve the equation $x y = z$, where $z$ is given and $x$ and $y$ must come from specified conjugacy classes.  It turns out that there is a formula for the number of solutions in terms of characters.  So instead of trying to find an explicit $(x,y)$, one can try to estimate the value of the formula and prove that the answer is non-zero.  (Typically, the trivial character makes a large positive contribution, and the aim is to show that all the other characters make small contributions.)<|endoftext|>
TITLE: Elementary theory of finite fields
QUESTION [5 upvotes]: I read on Ax's article that the elementary theory of finite fields is decidable if one assumes the continuum hypothesis to be true. What about if one assumes the hypothesis to be false?

REPLY [10 votes]: Let me explain how Ax eliminates his use of CH in the proof
of decidability. 
The point is that if a given proof of
decidability of any first order theory in a countable
language uses CH, then it may be uniformly omitted. Thus,
the decidability of a theory like this can never depend on
CH or on the Axiom of Choice.
Theorem. The following are equivalent, for any first
order theory T in a countable language.

ZFC+CH proves that T is decidable.
ZFC proves that T is decidable.
ZF proves that T is decidable

In particular, if a given proof of decidability uses CH+AC,
then these uses can be omitted.
Proof. Clearly, each statement implies the previous, since
the theories are progressively weaker. Suppose now that we
have a proof that t is decidable, but the proof uses
ZFC+CH. The assertion that T is decidable is a statement of
arithmetic, since it says "there is a computer program,
which accepts all and only the theorems of T and rejects
all other strings." This statement has complexity
Sigma^0_3(T), which is to say, that it is fairly low in the
arithmetic hierarchy (relative to the set of Goedel codes
of T).
Let us now establish that T is decidable by arguing merely
in ZF. Assume that ZF holds in the set-theoretic universe
V. Goedel proved that ZFC+CH holds in the constructible
universe L. This argument relativizes to show that ZFC+CH
also holds in L[T], the constructible universe relative to
T. (Note, the theory T of the question above is actually in
L, so in this case, L[T]=L.) Thus, T is decidable in L[T].
Since this assertion is arithmetic, it has the same truth
value in L[T] as in V, since these two set-theoretic models
have the same arithmetic. So T is decidable. QED
One can push the argument much lower than ZF, down to
something like second order number theory. Of course, what
we would really like is to prove the decidability of T in
PA, but I don't think that this is always possible.
(Although PA proves all true existential statements, the
complexity of the assertion that T is decidable rises above
this.)
The more general fact is that arithmetic statements can never depend on AC or CH, because if one proves them in ZFC+CH, then they would be true in L, and hence true in V, where one might have only ZF. In fact, the same argument shows via
the Shoenfield Absoluteness theorem that a Sigma^1_2
statement is provable in ZFC+CH if and only it is provable
in ZF.
An instance of this would be: Fermat's Last Theorem is
provable in ZFC+CH if and only if it is provable in ZF.
In particular, any use of the Axiom of Choice in Wiles' proof can be
uniformly eliminated.
Edit: I changed the presentation to be clearer.<|endoftext|>
TITLE: Encoding fuzzy logic with the topos of set-valued sheaves
QUESTION [7 upvotes]: One of the canonical examples used by Barr & Wells in order to motivate the use of topoi is that we can construct a theory for fuzzy logic and fuzzy set theory as set-valued sheaves on a poset (Heyting algebra) of confidence values for the fuzziness. Doing this constructs a fuzzy theory where both the membership and equality relations have more truth values than just true and false.
How would one construct a ternary approach using this mindset? In other words, is there an easy way to see a sheaf on a poset with three values as a fuzzy set theory for the truth values (true, maybe, false)? Or is this formulation even the wrong approach? Do I want a different Heyting algebra, so that the subobject classifier ends up having three elements?

REPLY [7 votes]: First I must warn you that there is a difference between fuzzy logics and topos theory.  There are some categories of fuzzy sets which are almost toposes, but not quite - they form a quasitopos, which is like a topos, but epi + mono need not imply iso.  There is a construction of such a quasitopos in Johnstone's Sketches of an Elephant - Vol 1 A2.6.4(e).
Now for the three valued logic I think a good example is a time-like logic.  Suppose you have a fixed point T in time.  This gives you two regions of time - before T and after T.  Our logic will have three truth values - always true, true after T but not before, and never true.  Note that we don't have a case "true before T, but not after", since once something is true, it is always true from that time on.  Like knowledge of mathematical theorems (assuming there are no mistakes!).
The topos with this logic is the arrow category of set: Set$^\to$.  Objects consist of Set functions $A \to B$, and morphism consist of pairs of set functions forming a commutative square.
For other three valued logics look at different Heyting algebras, but pay close attention to the implication operation, as it is a vital part of topos logic.  For the true, false, maybe case I am not sure on how to construct a Heyting algebra which reflects this logic.<|endoftext|>
TITLE: Chromatic number of graphs of tangent closed balls
QUESTION [16 upvotes]: The Koebe–Andreev–Thurston theorem gives a characterization of planar graphs in terms of disjoint circles being tangent. For every planar graph $G$ there is a disk packing whose graph is $G$. What happens when disks are replaced by closed balls? By closed balls of higher dimension? I have already asked one question about this here:
Graphs of Tangent Spheres
The question I want to ask here is what is known about the chromatic numbers of these graphs? I have updated the numbers and changed the arguments in the following based on some of the answers.
Assume the chromatic number is 14 or more and we have the smallest such graph that is colorable with 14 or more colors. Take one of the smallest closed balls then since the kissing number for three dimensions is 12 there are at most 12 closed balls tangent to this closed ball. Remove this closed ball then the remaining graph can be colored in 13 or less colors. Color it with 13 colors. Then add the closed ball back in since it is tangent to only 12 closed balls it can be given one of the thirteen colors so we have the entire graph can be colored with thirteen colors which gives a contradiction so the chromatic number must be 13 or less. We have an lower bound of 6 from a spindle constructed according to David Eppstein's answer. Can we improve on the 6 to 13 range? 
We have the lower bound is a quadratic function and we have an upper bound that is exponential. Which of these two is right?  
Is there a case where closed balls of different sizes raise the chromatic number from closed balls the same size?
Finally based on the existing chromatic numbers I am wondering if it is possible to answer this question. Is there a dimension where the chromatic number of the unit distance graph is different from the chromatic number of the graphs in that dimension of tangent closed balls. The unit distance graph is the set of all points in the $n$-dimensional space with two points connected if their distance is one. For dimension two the chromatic number is known to be in the range from 4 to 7. For dimension three the range is 6 to 15. For the graphs of tangent disks we have a chromatic number of 4 and for closed balls a range from 6 to 13. So the possibility that the chromatic numbers of the two types of graphs are the same has not yet been eliminated. So the specific question is what is known and what can be proved about the chromatic number of the graphs of tangent closed balls?

REPLY [12 votes]: Update May 2016
I removed the updates in Oct 2015.  I was trying to combine two copies of strongly regular ball packings to double the chromatic number.  But it has been point out that my construction was buggy.
Previous examples obtained in January 2015
Inspired by Bodarenko's counter-example to the Borsuk conjecture, I recently find many ball packings whose chromatic number is significantly higher than the dimension.  There tangency graphs are all strongly regular.  A note is available on arXiv and here (up to date).  
Here I list their parameters, dimensions and lower bounds (Hoffman) for the chromatic numbers.  Many of these parameters are for the complement to the more famous graph, e.g. Higman-Sims graph.

$(100, 77, 60, 56)$ (Higman-Sims graph), dimension 22, $\chi\ge 80/3$.
$(105, 72, 51, 45)$, dimension 20, $\chi\ge 25$.
$(120, 77, 52, 44)$, dimension 20, $\chi\ge 80/3$.
$(126, 75, 48, 39)$, dimension 20, $\chi\ge 26$.
$(162, 105, 72, 60)$ (Local McLaughlin), dimension 21, $\chi\ge 36$.
$(175, 102, 65, 51)$, dimension 21, $\chi\ge 35$.
$(176, 105, 68, 54)$, dimension 21, $\chi\ge 36$.
$(176, 85, 48, 34)$, dimension 22, $\chi\ge 88/3$.
$(243, 132, 81, 60)$ (Delsarte graph), dimension 22 $\chi\ge 45$.
$(253, 140, 87, 65)$, dimension 22, $\chi\ge 143/3$.
$(275, 162, 105, 81)$ (McLaughlin graph), dimension 22, $\chi\ge 55$.
$(276, 135, 78, 54)$, dimension 23, $\chi\ge 46$.
$(729, 520, 379, 350)$, dimension 112, $\chi\ge 621/5$.

Furthermore, there are two infinite families with high chromatic number (here $q$ is a prime power).

$(q^3, (q+1)(q^2-1)/2, (q+3)(q^2-3)/4+1, (q+1)(q^2-1)/4)$ (complements to Hubaut's C20), dimension $q^2-q$, $\chi\ge q^2$.
$((q^3+1)(q+1), q^4, (q^2+1)(q-1)q, q^3(q-1))$ (complement to the point graph of the generalized quadrangle $(q,q^2)$), dimension $q^3-q^2+q$, $\chi=q^3+1$.

Note that the last case is an equality.  This is the first non-constant lower bound for $\chi-d$.
Hope this helps future improvement.<|endoftext|>
TITLE: Four Dimensional Origami Axioms
QUESTION [19 upvotes]: What are the axioms of four dimensional Origami.
If standard Origami is considered three dimensional, it has points, lines, surfaces and folds to create a three dimensional form from the folded surface. This standard origami has seven axioms which have been proved complete.
The question I have is whether these same axioms apply to four dimensional origami which has lines, surfaces, shapes, and produces shapes folded along surfaces in the fourth dimension. Or are there more axioms in four dimensional Origami.
I cannot find this question approached in the literature. Four Dimensional Origami does not seem to be a field of study yet.

REPLY [14 votes]: In the $n$-dimensional origami question, you start with a generic set of hyperplanes and their intersections, which can then be some collection of $k$-dimensional planes.  An "axiom" is a set of incidence constraints that determines a unique reflection hyperplane, or conceivably a reflection hyperplane that is an isolated solution even if it is not unique.  The space of available hyperplanes is also $n$-dimensional.  Each incidence constraint has a codimension.  A set of independent constraints makes an axiom when their codimensions add to $n$.
It is easy to write down a simple incidence constraint on a reflection $R$ and compute its codimension.  If $n=2$, then the simple constraints are as follows:

$R(L) = L$ for a line $L$,
$R(x) = x$ for a point $x$,
$R(x) \in L$,
$R(L_1) = L_2$, and
$R(x_1) = x_2$

The first three constraints have codimension 1 and the last two have codimension 2.  Formally there are 8 ways that to combine these constraints to make axioms.  However, 1 cannot be used twice in Euclidean geometry, so that leaves 7 others.  These are the seven axioms listed on Robert Lang's page.  As it happens, Huzita missed combining 1 and 3.  If you could have hyperbolic origami, you would have 8 axioms.
You can go through the same reasoning in $n=3$ dimensions.  The simple constraints can again be written down:

$R(x) = x$ for a point $x$ (1)
$R$ fixes a line $L$ pointwise (2)
$R(L) = L$ by reflecting it (2)
$R(P) = P$ for a plane $P$ (1)
$R(x) \in L$ (2)
$R(x) \in P$ (1)
$R(L) \subset P$ (2)
$R(x_1) = x_2$ (3)
$R(L_1) \cap L_2 \ne \emptyset$ (1)
$R(P_1) = P_2$ (3)

I've written the codimensions of these constraints in parentheses.  As before, you can combine these constraints.  Certain pairs, such as 3 and 4, can't combine.  You also can't use 4 three times.  If you go through all of this properly, it is not all that hard to make a list of axioms that resembles the ones in 2 dimensions.  (Possibly I made a mistake in this list or missed something, but it is not hard to go through this properly.)
However, there is a possible subtlety that I don't know how to address.  Namely, suppose that you do make some complicated configuration using combinations of these constraints, at first.  Can you create a configuration with the property that the codimensions don't simply add?  For instance, ordinarily you can't use condition 7 twice to define the reflection $R$, because the total codimension is 4, which is too large.  However, if the lines and planes in this condition are related to each other, is the true codimension sometimes 3?  I would guess that you can make this happen.  If so, then potentially you'd have to add "unstable" construction axioms to the list.  But then it is not clear whether an unstable construction axiom is actually needed, or whether an unstable axiom can always be replaced by a sequence of stable axioms.<|endoftext|>
TITLE: What is the name of the following categorical property?
QUESTION [6 upvotes]: Is there a name for those categories where objects posses a given structure and every bijective morphism determines an isomorphism between the corresponding objects?
Examples of categories of that type abound: Gr, Set, ...
An specific example of a category where the constraint doesn't hold is given by Top: a morphism there is a continuous function between topological spaces. Now, it is easy to give here a concrete example of a bijective morphism between [0,1) and $\mathbb{S}^{1}$ that fails to be an isomorphism of topological spaces (in point of fact, much more is known in this case, right?).

REPLY [7 votes]: "every bijective morphism determines an isomorphism"
I think you mean that the forgetful functor reflects invertibility.
Let $\bf A$ be the "category of objects with structure" and their structure-preserving maps (homomorphisms), $\bf S$ the category of carriers (maybe sets and functions) and $U:{\bf A}\to{\bf S}$ the "forgetful" functor between them.  In fact, just let $U:{\bf A}\to{\bf S}$ be any functor you like.
Now let $f:X\to Y$ be any morphism of $\bf A$.  You are saying that, whenever $U f:U X\to U Y$ is an isomorphism (such as a bijection) then $f$ was already an isomorphism.
The forgetful functor from any category of algebras has this property, as more generally does the right adjoint of any monadic adjunction.   However, the underlying set functor from the usual category of topological spaces does not, because there are many different topologies that can be put on a set.<|endoftext|>
TITLE: Is there a meaningful difference between biased and unbiased composition?
QUESTION [13 upvotes]: In higher category theory, there are notions of biased and unbiased definitions of composition of $n$-morphisms (or, as a special case, tensor products of objects). In the biased framework, we define what it means to have a $2$-fold composition of $n$-morphisms as well as the $0$-fold composition (i.e., the identity $n$-morphism), and then we add in some sort of coherent associativity relating the various ways of composing $k$ $n$-morphisms for all $k$. In the unbiased framework, we define, for all $k$, the notion of a $k$-fold composition of morphisms, and then give compatible isomorphisms between a $k$-fold composition and the various composites of smaller-fold compositions.
The traditional definitions of mathematics lean strongly towards the biased framework. Perhaps the simplest example is a group (or monoid), which is usally defined as having a binary operation ($2$-fold product) and identity ($0$-fold product), with the condition that $(xy)z = x(yz)$. Of course, this definition is easier to write out than an unbiased one. In the unbiased world, we would define a group to have a $k$-fold product for all $k$ and then say that for any $k_1, k_2, \ldots, k_r \in \mathbb{N}$ such that $\sum_{i = 1}^r k_i = k$, and any $x_{11}, \ldots, x_{1k_1}, x_{21}, \ldots, x_{2k_2}, \ldots, x_{r1}, \ldots, x_{rk_r}$, the $k$-fold product of the $x_{ij}$ is equal to the $r$-fold product of the $k_i$-fold products of the $x_{ij}$. However, the relative simplicity of biased definitions seems to vanish as one moves up the $n$-categorical ladder, as more and more complicated coherence axioms are required for fully weak notions of composition of $n$-morphisms.
It seems that in many cases, an unbiased definition feels more natural. For instance, when defining tensor products of modules, we use a universal property, and this universal property definition works just as well for any number of modules as it does for two. Moreover, the universal property immediately provides the relevant maps from the $k$-fold tensor product to composites of smaller-fold tensor products, whereas I'm not sure of an obvious direct way to give an isomorphism $(X \otimes Y) \otimes Z \to X \otimes (Y \otimes Z)$ without using elements or at least implicitly going through the three-fold tensor product.
Ross Street, in his review of Leinster's book Higher Operads, Higher Categories (arXiv:math/0305049, doi:10.1017/CBO9780511525896) (which is a good reference for biased and unbiased definitions), seems to imply that the difference between unbiased and biased notions is more technical than foundational. Is this the case, or are some concepts of tensor product or morphism really more suited to a biased or an unbiased interpretation? It seems, for instance, that the theory of Lie algebras of Lie groups is rather firmly planted in a biased definition of group, as it relates to the failure of $2$-fold products to commute. Is there a formulation of Lie algebras that is unbiased? Do biased or unbiased definitions better lend themselves to categorification?
EDIT: I should probably clarify what I mean by technical vs. foundational. I imagine any biased gadget should be equivalent to an unbiased one and vice versa, so I'm not envisaging the creation of new types of objects by taking an unbiased viewpoint. Rather, I'm asking if the unbiased viewpoint can provide fundamental insights that the biased viewpoint cannot (or vice versa).
An example of such a foundational reformulation would be the use of the functor of points in algebraic geometry. That you can view schemes in terms of their functors of points is a triviality, but this change of reference frame is more than a technical convenience; in fact, it buys us a great deal. For instance, it leads to the theory of algebraic stacks. It might have been possible for algebraic stacks to have been developed in the absence of the functor of points, but I imagine that it would have taken much longer and would be less elegant and more inaccessible.
So my overarching question is, might viewing composition in one way or the other be more than just a technical convenience?

REPLY [7 votes]: Certainly unbiased definitions are the norm in modern homotopy theory.  I guess an example of a biased definition is the (original?) definition by Stasheff of an $A_\infty$ space—the homotopy theorist's monoid.  (For simplicity, I'll ignore matters related to the unit.)  Informally, it consists of a space $X$, a multiplication $\mu : X \times X \to X$, a homotopy $\alpha$ between the maps $\mu(\mu(x,y),z)$ and $\mu(x,\mu(y,z))$ from $X \times X \times X$ to $X$, a "pentagonator" $\pi$ which extends a map $S^1 \times X^4 \to X$ built from $\mu$ and $\alpha$ to $D^2 \times X^4$, and so on.  This seems to be the appropriate generalization of your biased definition of monoids to the situation where we need infinitely many higher coherence homotopies.  More precisely, an $A_\infty$ space under this definition consists of a space $X$ together with maps $\mu_n : A_n \times X^n \to X$, where $A_n$ is the $n$th associahedron—a certain convex polyhedron in $\mathbb{R}^{n-2}$—and on the boundary of $A_n$, $\mu_n$ is required to be equal to a certain expression built out of the $\mu_k$ for $k$ smaller than $n$.  Thus, we may think of it as a null-homotopy of a prescribed map $\partial A_n \times X^n \to X$.
This definition is quite complicated combinatorially, since it depends on the polyhedra $A_n$.  Peter May realized that certain features of the $A_n$ could be abstracted into the notion of a (non-symmetric) operad.  I don't want to define exactly what an operad is, but instead just observe the change in perspective: rather than thinking of $A_n$ as a space which encodes a null-homotopy of a map defined on its boundary, we think of the points of $A_n$ as describing various $n$-ary multiplications on $X$ via the map $\mu_n : A_n \to \mathrm{Map}(X^n, X)$.  The contractibility of $A_n$ guarantees that there is essentially just one way to multiply $n$-tuples.  So we see that the same definition can be viewed as either biased or unbiased.  However, in the world of operads, we are free to consider algebras over any operad whose $n$th space is contractible for all $n$, and any such space is an $A_\infty$ algebra.  This definition no longer has the combinatorial complexity of the associahedra, although operads require a certain amount of machinery.  I would call it an unbiased definition.  From this point of view, the biased definition consists of giving a certain kind of homotopical presentation of the terminal operad, which is a useful thing to have, but quite unnecessary from the point of view of giving a definition!
Since homotopy theory is "just the easy part of ∞-category theory", I expect that as higher-categorical structures become more commonplace, we will see a similar phenomenon: the easy definitions to write down and work with abstractly are the unbiased ones, and biased definitions are certain presentations of the unbiased notions, which might become quite complicated as the categorical level increases.<|endoftext|>
TITLE: What's a nice argument that shows the volume of the unit ball in $\mathbb R^n$ approaches 0?
QUESTION [81 upvotes]: Before you close for "homework problem", please note the tags.
Last week, I gave my calculus 1 class the assignment to calculate the $n$-volume of the $n$-ball. They had finished up talking about finding volume by integrating the area of the cross-sections. I asked them to calculate a formula for $4$ and $5$, and take the limit of the general formula to get 0.
Tomorrow I would like to give them a more geometric idea of why the volume goes to zero. Anyone have any ideas? :)
Comm wiki in case people want to add/modify this a bit.

REPLY [5 votes]: Permit me to suggest another "intuitive" approach, which hardly uses any calculations, just some basic Arithmetic (that means: Combinatorics)
Well, it is a fact that even for big numbers of n (dimension) the unit n-sphere inscribed in the unit n-cube is still the LARGEST possible. Its diameter is always equal to the "side length" of the cube, and the sphere's surface touches every "face" of the cube (as "Face" for an n-cube, is to be considered of course an (n-1)-cube).
So, why the "shrinkage" of the sphere for higher and higher dimensions? It is simple a matter of Arithmetic. The sphere always occupies the "central area"/center of the n-cube, but there is not much "center" left, as n goes to infinity. Most of the cube's volume "escapes" (centrifugally, sort of) towards the corners/"vertices". For example a 100-cube has 200 "faces" but $2^{100}$ vertices. Let's consider that we produce the "diameters" of the cube, namely all the straights that pass through the Center of the cube. 
There are two main types, as far as "length" is concerned. The smallest, let's call them "good" or "short" which start from the center of a face, pass through center C of the cube and end on the center of the opposite face. For the usual 3-dimensional cube that prescribes a unit sphere (r=1), this length of the short diameter is 2 (since 2 is both the diameter of the sphere and the edge length of the cube). That shows that a "good/short" diameter is spreading entirely (its whole length) INSIDE the sphere. A "bad/long" one from the other side, are these diameters of course which start from a vertex---trough center C---to the opposite vertex. For an n-cube with side length=2 , the long "bad" diameter has length = $2\sqrt n$. Thus, for example for a 100-cube the long/bad diameter has length equal to 2*10=20. We notice that just 1/10th of this length lies in this case inside the sphere!
Moreover, there are 100 "good/short"/full" diameters, but $2^{99}$ long ones! ("empty/bad"). The above argument "by induction", may be perhaps not so mathematically strict, but it is logically very powerful, I think.<|endoftext|>
TITLE: Proof assistants for mathematics
QUESTION [44 upvotes]: This question is related to (maybe even the same in intent as) Intro to automatic theorem proving / logical foundations?, but none of the answers seem to address what I'm looking for.
There are a lot of resources available for people who want to use proof assistants like Coq, Isabelle, …, to prove properties about programs—and that's no surprise, since a lot of the development of these programs is done by computer scientists.  However, I am interested in resources, and especially in course materials (because I'm trying to put together an independent study for a CS student), involving the use of proof assistants to prove mathematical statements—see the work of Hales and Weedijk for examples.  Does anyone know of any such?

REPLY [6 votes]: I would like to mention Mizar, proof verification system based on language which is human readable and very near to natural mathematical language used in mathematical practice. It is one of the longest working proof checkers, and it is one of the most successful one. 
Here You may find some about it: http://www.cs.ru.nl/~freek/mizar/
There is the whole library congaing presently more that 40Mb zipped proofs ( in pure ascii files!).<|endoftext|>
TITLE: Is there a natural family of languages whose generating functions are holonomic (i.e. D-finite)?
QUESTION [17 upvotes]: Let $L$ be a language on a finite alphabet and let $L_n$ be the number of words of length $n$.  Let $f_L(x) = \sum_{n \ge 0} L_n x^n$.  The following are well-known:

If $L$ is regular, then $f_L$ is rational.  
If $L$ is unambiguous and context-free, then $f_L$ is algebraic.

Does there exist a natural family of languages $\mathcal{L}$ containing the context-free languages such that if $L \in \mathcal{L}$, then $f_L$ is holonomic?  Is that class of languages also associated to a natural class of automata?  
This question is prompted by a remark in Flajolet and Sedgewick where they assert that there is no meaningful generating function formalism associated to context-sensitive languages because of the significant undecidability issues.  However, holonomic functions have proven a robust and incredibly useful framework in combinatorics, so I think this is a natural question to ask.

REPLY [2 votes]: There are such natural examples, such as the very recent ICALP'20 contribution on "unambiguous finite and Pushdown Parikh automata" (paper not yet available),
or the related RCM class by Massazza and Castiglione.
In both cases the generating function of these language classes are holonomic and moreover contain examples which are not algebraic.
EDIT: The ICALP'20 paper is now available.<|endoftext|>
TITLE: Just starting with [combinatorial] game theory
QUESTION [10 upvotes]: I have recently become interested in game theory by way of John Conway's on Numbers and Games.  Having virtually no prior knowledge of game theory, what is the best place to start?

REPLY [4 votes]: If you have played Go, then you may enjoy Mathematical Go: Chilling Gets the Last Point by Berlekamp and Wolfe. Some go players, especially Bill Spight, have developed combinatorial game theory within go further.<|endoftext|>
TITLE: Adding a random real makes the set of ground model reals meager
QUESTION [10 upvotes]: This is a question about forcing. I have seen the following fact mentioned in multiple places, but have not been able to find a proof: if a random real is added to a transitive model of ZFC, then in the generic extension the set of reals in the ground model becomes meager. 
My guess is that one should be able to, in some natural way, directly construct from a random real a countable sequence of nowhere dense sets covering the ground model reals, but I am not sure.

REPLY [7 votes]: The proof is based on the fact that there is a decomposition ${\bf R}=A\cup B$ of the reals such that $A$, $B$ are (very simple) Borel sets, $A$ is meager, $B$ is of measure zero, and ${\bf R}=A\cup B$ even holds if after forcing we reinterpret the sets. Nos let $s$ be a random real. If $r\in {\bf R}$ is an old real, then $s\notin r+B$, so $s\in r+A$, that is, the meager 
$s-A$ contains all old reals.<|endoftext|>
TITLE: Holomorphic and antiholomorphic forms of projective space
QUESTION [7 upvotes]: For $\mathbb{CP}^1$ the bundles of holomorphic and antiholomorphic forms are equal to the $\mathcal{O}(-2)$ and $\mathcal{O}(2)$ respectively. Do the holomorphic and antiholomorphic bundles of $\mathbb{CP}^2$ (or indeed) $\mathbb{CP}^n$) have a description in terms of line bundles. What happens in the Grassmannian setting?

REPLY [7 votes]: Greg's otherwise excellent answer gives the impression that computing Chern classes on projective space requires a computer algebra system. I'm writing to repell this impression. The cohomology ring of $\mathbb{P}^{n-1}$ is $\mathbb{Z}[h]/h^n$ where $h$ is Poincare dual to the class of a hyperplane. We have the short exact sequence
$$0 \to S \to \mathbb{C}^n \to Q \to 0$$,
where $S$ is the tautological line bundle, whose fiber over a point of $\mathbb{P}^{n-1}$ is the corresponding line in $\mathbb{C}^n$. The line bundle $S$ is also called $\mathcal{O}(-1)$, and has Chern class $1-h$. So $$c(Q) = 1/(1-h) = 1+h+h^2 + \cdots h^{n-1}.$$
As Greg explained, the tangent bundle is $\mathrm{Hom}(S, Q) = S^{\star} \otimes Q$. The formula for the Chern class of a general tensor product is  painful to use in practice, but we can circumvent that here by tensoring the above exact sequence by $S^{\star}$.
$$0 \to \mathbb{C} \to (S^{\star})^{\oplus n} \to T_{\mathbb{P}^{n-1}} \to 0$$,
So the Chern class of the tangent bundle to $\mathbb{P}^{n-1}$ is
$$(1+h)^n = 1 + n h + \binom{n}{2} h + \cdots + n h^{n-1}.$$
Yes, I deliberately ended that sum one term early. Remember that $h^n$ is $0$.
As Greg says, if this is to be expressible as a direct sum1 of line bundles, then this should be a product of $n-1$ linear forms, $c(T) = \prod_{i=1}^{n-1} (1+ a_i h)$. Since this is an equality of polynomials of degree $n-1$, we can forget that we are working modulo $h^n$ and just check whether the honest polynomial $f(x) := 1 + n x + \binom{n}{2} x + \cdots + n x^{n-1}$ factors in this way.
The answer is it does not, except when $n=2$ (the case of $\mathbb{P}^1$.) 
The unique factorization of $f(x)$ is 
$$\prod_{\omega^n=1,\  \omega \neq 1} (1+x (1-\omega)).$$
Thus does raise an interesting question. When $n$ is even, $(1+2x)$ divides $f(x)$. This suggests that $\mathcal{O}(2)$ might be a subquotient of $T_{\mathbb{P}^{n-1}}$. I can't figure out whether or not this happens.

The case of Grassmannians is going to be worse for three reasons. The two minor ones are that (1) we may honestly have to use the formula for the chern class of a tensor product. and (2) the polynomials in question will be multivariate polynomials. The big problem will be that $H^{\star}(G(k,n))$ has relations in degree lower than $\dim G(k,n)+1$, so we can't pull the trick of forgetting that $h^n=0$ and working with honest polynomials.
I suspect that your question was more "give a nice description of the tangent bundle to the Grassmannian" than "can that tangent bundle be expressed as a direct sum of line bundles?". If you seriously care about the latter, I'll give it more thought.
1 Direct sum is the natural thing to ask for in the categories of smooth, or of topological, complex vector bundles. If you like the algebraic or holomorphic categories, as I do, it is more natural to ask for the weaker property that there is a filtration of the vector bundle, all of whose quotients are line bundles.<|endoftext|>
TITLE: Badiou and Mathematics
QUESTION [7 upvotes]: Does anyone have an opinion on Alain Badiou's use of set theory? Is there anything interesting mathematically there? Also could anyone shed any light on the comment in the Wikipedia article link text that says: 

This effort leads him, in Being and Event, to combine rigorous mathematical formulae with his readings of poets such as Mallarmé and Hölderlin and religious thinkers such as Pascal.

REPLY [10 votes]: Badiou's got some mathematical training; reading back and forth between the relevant sections of Goldblatt's Topoi and Badiou's account of Ω-sets in Logics of Worlds, for example, you can see that the one tracks the other closely. It's not just blind quotation, followed by hand-wavy inference-drawing either: you could actually learn about Ω-sets from Badiou's presentation of them alone, and not be too horribly surprised or confused when you came to read the technical presentation in Goldblatt (this was, in fact, the order in which I did it).
On the axiom of choice and "infinite liberty":
The AoC says that given a set {A, B, C...} none of whose members are the empty set, there exists a set {x ε A, y ε B, z ε C...} which takes one element from each of the first set's members. The point here is that the AoC "freely" chooses an element from each set rather than (for example) identifying a "least" element and choosing that: even when there's no rule that can tell you which element should be chosen, the AoC says that a set exists representing some choice.
The AoC only has any work to do in situations where no rule can be found (for example, no-one knows of a rule that will well-order the reals, but the AoC entails that a real can be chosen, then another from the remaining reals, then another etc. - so "axiomatically" a well-ordering of the reals exists, provided one accepts AoC) - hence it represents, in this sense, the possibility of a predicatively undetermined choice. That's the "infinite liberty" he's on about. It is nowhere asserted that AoC "proves" that such a liberty exists, but rather that introducing AoC into ZF makes such a liberty thinkable within the confines of its axiomatic system (this is in line with Badiou's general program of treating mathematics as "ontology", as a means for systematically demarcating what is thinkable of "being as such").
In terms of "interest to mathematicians": Badiou's early text The Concept of Model is a good philosophical introduction to model theory, and his Number and Numbers is an interesting and accessible guide to the philosophy of number, covering Frege, Peano, Cantor, Dedekind and Conway (surreal numbers).

REPLY [6 votes]: There's an interesting review of Badiou's "Number and Numbers" at the Notre Dame Philosophical Review by John Kadvany.<|endoftext|>
TITLE: What does "linearly disjoint" mean for abstract field extensions?
QUESTION [42 upvotes]: All definitions I've seen for the statement "$E,F$ are linearly disjoint extensions of $k$" are only meaningful when $E,F$ are given as subfields of a larger field, say $K$.  I am happy with the equivalence of the various definitions I've seen in this case.  Lang's Algebra VIII.3-4 and (thanks to Pete) Zariski & Samuel's Commutative Algebra 1 II.15-16 have good coverage of this.
"Ambient" definitions of linear disjointness:
Wikipedia says it means the map $E\otimes_k F\to E.F$ is injective, where $E.F$ denotes their compositum in $K$, the smallest subfield of $K$ containing them both.  
An equivalent (and asymmetric) condition is that any subset of $E$ which is linearly independent over $k$ is also linearly independent over $F$ (hence the name); this all happens inside $K$.
However, I often see the term used for field extensions which are NOT subfields of a larger one, even when the field extensions are not algebraic (so there is no tacit assumption that they live in the algebraic closure).  Some examples of these situations are given below.

Question: What is the definition of "linearly disjoint" for field extensions which are not specified inside a larger field?


ANSWER: (After reading the helpful responses of Pete L. Clark, Hagen Knaf, Greg Kuperberg, and JS Milne -- thanks guys! -- I now have a satisfying and fairly exhaustive analysis of the situation.)
There are two possible notions of abstract linear disjointness for two field extensions $E,F$ of $k$ (proofs below):
(1) "Somewhere linearly disjoint", meaning 
"There exists an extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$."
This is equivalent to the tensor product $E\otimes_k F$ being a domain.
(2) "Everywhere linearly disjoint", meaning 
"For any extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$."
This is equivalent to the tensor product $E\otimes_k F$ being a field.
Results:
(A) If either $E$ or $F$ is algebraic, then (1) and (2) are equivalent.
(B) If neither $E$ nor $F$ is algebraic, then (2) is impossible.
Depending on when theorems would read correctly, I'm not sure which of these should be the "right" definition...  (1) applies in more situations, but (2) is a good hypothesis for implicitly ruling out pairs of transcendental extensions. So I'm just going to remember both of them :)

PROOFS: (for future frustratees of linear disjointness!)
(1) Linear disjointness in some field $K$, by the Wikipedia defintion above, means  the tensor product injects to $K$, making it a domain.  Conversely, if the tensor product is a domain, then $E,F$ are linearly disjoint in its field of fractions.
(2) If the tensor product $E\otimes_k F$ is a field, since any map from a field is injective, by the Wikipedia definition above, $E,F$ are linearly disjoint in any $K$.  Conversely, if $E \otimes_k F$ is not a field, then it has a non-trivial maximal ideal $m$, with quotient field say $K$, and then since $E\otimes_k F\to K$ has non-trivial kernel $m$, by definition $E,F$ are not linearly disjoint in $K$.
(A) Any two field extensions have some common extension (take a quotient of their tensor product by any maximal ideal), so (2) always implies (1).  
Now let us first show that (1) implies (2) supposing $E/k$ is a finite extension.  By hypothesis the tensor product $E\otimes_k F$ is a domain, and finite-dimensional as a $F$-vector space, and a finite dimensional domain over a field is automatically a field: multiplication by an element is injective, hence surjective by finite dimensionality over $F$, so it has an inverse map, and the image of $1$ under this map is an inverse for the element.  Hence (1) implies (2) when $E/k$ is finite.
Finally, supposing (1) and only that $E/k$ is algebraic, we can write $E$ as a union of its finite sub-extensions $E_\lambda/k$.  Since tensoring with fields is exact, $E_\lambda\otimes_k F$ naturally includes in $E\otimes_k F$, making it a domain and hence a field by the previous argument.  Then $E\otimes_k F$ is a union of fields, making it a field, proving (1) implies (2).
(B) Now this is easy.  Let $t_1\in E$, $t_2\in F$ be transcendental elements.  Identify $k(t)=k(t_1)=k(t_2)$ by $t\mapsto t_1 \mapsto t_2$, making $E$,$F$ extensions of $k(t)$.  Let $K$ be a common extension of $E,F$ over $k(s)$ (any quotient of $E\otimes_{k(s)} F$ by a maximal ideal will do).  Then $E,F$ are not linearly disjoint in $K$ because their intersection is not $k$: for example the set { $1,t$ }$\subseteq E$ is linearly independent over $k$ but not over $F$, so they are not linearly disjoint by the equivalent definition at the top.

Examples in literature  of linear disjointness referring to abstract field extensions:

Eisenbud, Commutative Algebra, Theorem A.13 (p.564 in my edition) says, in characteristic $p$,


"$K$ is separable over $k$ iff $k^{1/p^{\infty}}$ is linearly disjoint from $K$."


Liu, Algebraic Geometry and Arithmetic Curves, Corollary 2.3 (c) (p. 91) says, for an integral algebraic variety $X$ over a field $k$ with function field $K(X)$, 


"$X$ is geometrically integral iff $K(X)$ and $\overline{k}$ are linearly disjoint over $k$.

(Follow-up: Since in both these situations, one extension is algebraic, the two definitions summarized in the answer above are equivalent, so everything is fantastic.)
Old edit: My first guess was (and still is) to say that the tensor product is a domain...

REPLY [8 votes]: Linear disjointness and its relation to tensor products is explained in
detail in Zariski+Samuel, Commutative Algebra - I forgot which volume
of the two.
There linear disjointness over $k$ of two $k$-algebras $A,B$ is defined
only for algebras that are contained in some larger ring $C$.
The tensor product is introduced as follows: a product of $A$ and $B$ is a
$k$-algebra $C$ and $k$-algebra morphisms $f:A\rightarrow C$, $g:B\rightarrow C$
such that the smallest subalgebra of $C$ containing $f(A),g(B)$ is $C$ itself.
A product $C$ of $A$ and $B$ is called tensor product, if $f(A)$ and $g(B)$ are
linearly disjoint over $k$.
As for the case of two field extensions $E,F$ of $k$ one of which is algebraic over $k$ 
($F$ say) one has a surjective map $E\otimes_k F\rightarrow E.F$, where $E.F$
denotes the smallest subring of the algebraic closure of $E$ that contains $E$ and $F$.
Since $F/k$ is algebraic, $E.F$ also is the smallest subfield that contains $E$ and $F$.
We get the equivalent statements:
(1) $E$ and $F$ are linearly disjoint over $k$ within the algebraic closure of $E$.
(2) the tensor product $E\otimes_k F$ is a field.<|endoftext|>
TITLE: Line bundles trivial after extension of the base-field
QUESTION [10 upvotes]: Let k be a field and let X be scheme over k. Let K be a field extension of k and denote by $X_K$ the base-change of X to Spec K. Under what conditions is the canonical map of Picard groups $Pic(X)\to Pic(X_K)$, induced by the projection, injective? I know that this is true if X is geometrically integral and proper over k, but what if X is only of finite type, separated and geometrically integral over k?

REPLY [8 votes]: Perhaps it is worth recording a simple example: let $X_k= \text{Spec}(k[x,y]/(x^2+y^2-1))$. Then $\text{Pic}(X_{\mathbb R}) = \mathbb Z/(2)$ while $\text{Pic}(X_{\mathbb C}) = \mathbb 0$, see Fossum's book "Divisor class groups of Krull domains", Prop 11.8.<|endoftext|>
TITLE: Reference for Learning Global Class Field Theory Using the Original Analytic Proofs?
QUESTION [13 upvotes]: Hi Everyone!
I'm wondering if anyone knows of a reference for learning global class field theory using the original analytic proofs developed in the 1920s and 1930s. Almost every book I can find either does local class field theory first or uses ideles/cohomology to prove global class field theory. This is not how it was historically done - the ideal-theoretical formulation of class field theory was proven first, using more elementary analytic methods. So I'm wondering if anyone knows of any resources which would teach these proofs.
I'm currently about to take a class which follows global class field theory in this way, and our teacher says he does not know of any textbook for this, so I'm wondering if anyone here would know.

REPLY [5 votes]: "The" classical approach to class field theory can be found in Hasse's Marburg lectures from the early 1930s (in German, as of now). The key arguments are contained in Artin's three lectures on class field theory from 1932, given (in English) in the appendix of Harvey Cohn's book "A classical invitation of algebraic numbers and class fields".
Another nice introduction along the classical lines is provided by Iyanaga's "Class-field theory notes". Perhaps this is the book you're looking for. It's definitely more classical than Janusz. Recently, Nancy Childress has published a book on class field theory, which is
at least "semi-classical".<|endoftext|>
TITLE: The monodromy-weight-, Ramanujan-, Langlands-landscape
QUESTION [12 upvotes]: The drawing on the last page of Yoshida's notes make me puzzle, perhaps you can help? It shows a "landscape" featuring the monodromy-weight conj., the general Ramanujan-conj., the Langlands correspondence and some connections between them. However, I guess the Ramanujan-conj. "is" the monodromy-weight-conj. expressed via Langlands correpondence and the Rapoport-Zink spectral sequence is a tool for showing the MWC in some cases and reducing the computation of L-functions to semi-simple L-functions. That seems not to fit to the drawing. Can you can clearify that? 
Edit: A fascinating new survey on the "philosophy of weights" with many conjectures by Jannses. (Posted here because I don't know an entry where it would fit better)

REPLY [23 votes]: Dear Thomas,
This "landscape" is, I think, a sketch of the proof of the following theorem of Taylor and Yoshida:  if $\Pi$ is a self-dual cuspdidal automorphic form on $GL_n$ over $E$ (a CM field)
(satisfying some further technical conditions) and $\rho$ is the associated $n$-dimensional
Galois representation (constructed by Harris and Taylor in their book "The geometry and cohomology of some simple Shimura varieties"), then the local factors of $\Pi$ at any prime
$p$ of $E$ matches, via local Langlands, with the restriction of $\rho$ to a decomposition group at $p$.
What was proved in Harris--Taylor was that this matching is correct, up to the question
of matching the $N$ on each side.  What Taylor and Yoshida verified is that the $N$ on each side
also matches.  
The top part of the landscape represents the reduction to the unipotent situation (i.e. the context in which $\Pi$, locally at $p$, has an Iwahori fixed vector --- the analogue
of classical level $\Gamma_0(p)$), by
base-changing to an extension of $E$.  On the automorphic side, this is the process
labelled by the names Arthur--Clozel--Selberg; on the Galois side, this is just restricting
from $Gal(\overline{E}/E)$ to $Gal(\overline{E}/E')$ for some appropriately chosen
extension $E'$ of $E$.  If one thinks of $\rho$ as appearing in the cohomology of a Shimura
variety (which is, after all, how it is constructed), then this just corresponds to
base-changing the variety from $E$ to $E'$, which is why this process is labelled as
geometric base change in the "landscape".
On the lower right of the diagram, one has the generalized Ramanujan conjecture, stating that the local factor of $\Pi$ at $p$ should be tempered,
which in turn implies that the local (at $p$) Galois representation (more precisely,
Weil--Deligne representation) attached to this local factor satisfies the monodromy
weight conjecture.   This temperedness was proved by Harris--Taylor.
Now we already know that $\rho$ locally at $p$ matches with the local factor of $\Pi$ at
$p$ up to $N$, and it is not hard to show that there is a unique way to add $N$ so as
to obtain a local Weil--Deligne representation satisfying the monodromy weight conjecture.
So to prove their theorem, Taylor and Yoshida are reduced to proving that $\rho$, locally
at $p$, satisfies the monodromy weight conjecture.
They do this via an application of the Rapoport--Zink spectral sequence and a careful analysis of the bad reduction of the Shimura variety in whose cohomology $\rho$ lives.  This is represented in the lower left part of the diagram. 
The very bottom part of the diagram on the left represents the fact that a priori one knows
that $\rho$ is mixed, i.e. its semi-simplification has Frobenius eigenvalues that are
Weil numbers of various weights.    But the main part of the landscape here is the monodromy weight conjecture, which describes the precise relationship between the Frobenius eigenvalues and the $N$ operator.
To prove the MWC in their context, Taylor and Yoshida
use the interplay between the geometry of the
special fibre of the Shimura variety and the representation theory of $\Pi$ that is the main subject of Harris and Taylor's book, as well as Section 2 of Taylor and Yoshida's article.  In particular, the fact that $\Pi$, locally at $p$, is tempered unitary, and is assumed to have an Iwahori fixed vector, puts strong restrictions on its structure (using the known classification of unitary reps. of $GL_n$ over local fields; this is why Tadic's name appears on the right hand side of the landscape), which when fed over to
the geometric side implies that the RZ spectral sequence degenerates at $E_1$, giving the desired monodromy weight conjecture.
My suggestion is that if you want to understand this in more detail, you should read Taylor and Yoshida's article.  Start at the very end of the paper, where the main theorem is proved (i.e. the paragraph beginning "We can now conclude ... ").  The lemmas referred to in section 1 are more or less elementary.  Theorem 3.2 is the heart of the argument; it is where degeneration of the RZ spectral sequence is proved.  One can try to read it more or less formally, taking various assertions as a black box,
at least to see what role the temperedness of $\Pi$ locally at $p$ plays.  The earlier parts of Section 3 are just establishing notation, and making contact with the book of Harris--Taylor.   Section 2 is devoted to establishing the basic properties of the semi-stable models of the Shimura varieties in whose cohomology $\rho$ lives; I recommend treating it as a black box on first reading.  
Added:  First, I should note that the references I make are to the version of Taylor--Yoshida that is currently posted on Taylor's web-page, which I am told may differ quite substantially in its organization from the published version of the paper.
Also, let me add something about the role of the classification of unitary representations
of $GL_n$, which will help illuminate the structure of the lower part of the landscape:
First, one uses the general result that geometrically obtained Galois representations are mixed (i.e. have Frobenius eigenvalues that are Weil numbers), coupled with the local-global compatibility without the $N$, to deduce that the Frobenius eigenvalues 
of the Weil--Deligne representation attached to the local factor of $\Pi$ at $p$ are Weil numbers. Second, results of Tadic on the classification of unitary representations of $GL_n$ of $p$-adic fields greatly restrict the structure of this
local factor (since it is both unitary and generic, being the local factor of a cuspidal
automorphic representation for $GL_n$).  Finally, when this is combined with the fact that 
the Weil--Deligne rep'n associated to it via local Langlands is mixed, ones sees that this
local factor is forced to be tempered.   This is how Harris--Taylor deduce temperedness;
it is an analogue at $p$ of an argument at the archimedean prime made by Clozel in his Ann Arbor paper.  (See Lemma 4.9 on p.144 of volume I of the Ann Arbor conference.  A scan
is available on Jim Milne's web-page here,
but note that it is quite a big file.)
Finally, let me now explain how to correctly read the landscape: one starts in the lower left, following the arrow.  When you come to a bridge, you cross the river from the motivic/Galois side to the automorphic side, or back again, continuing to follow the arrows.  The areas marked with
x's are impenetrable (or, at least, you don't try to cross them directly --- e.g. you don't prove GRC directly, you don't prove WMC directly, you don't study the general semi-stable reduction problem directly); the only way to proceed is by crossing the river.  This then gives
the structure of the Taylor--Yoshida argument.<|endoftext|>
TITLE: N=(2,2) supersymmetry in two-dimensional Euclidean space
QUESTION [7 upvotes]: Can anyone describe (or give a reference for) the 2d superspace formulation of N=(2,2) SUSY in Euclidean signature?
I'm reading Hori's excellent introduction to QFT in the book 'Mirror symmetry', and my question is basically Ex. 12.1.1. page 273. What I imagine the answer is is a super version of the usual story of differential forms on complex manifolds, i.e. we complexify, find square roots of the $\partial_z$ and $\partial_{\bar{z}}$ operators, then find a subalgebra of 'chiral' fields analogous to the subalgebra of holomorphic forms.

REPLY [5 votes]: You imagine well.  Hori is talking about $\mathbb{R}^{2|2}$, which is arguably the simplest super Riemann surface.
There is lots on this subject, mostly in the Physics literature, which I'm hesitant to recommend.  In the Mathematics literature, you might wish to read Deligne and Freed's Supersolutions and in particular section 2.6.<|endoftext|>
TITLE: Battleship Permutations
QUESTION [14 upvotes]: Using the game of Battleship as an example, is there a general solution for determining the number of arrangements of a given set of 1xN rectangles on a X by Y grid?
Example: In Battleship, each player has a 10x10 grid on which they must place each of the following rectangular 1xN ships which may not overlap or be placed diagonally:

1x5-square "carrier"
1x4-square "battleship"
1x3-square "submarine"
1x3-square "cruiser"
1x2-square "destroyer"

A) How would one calculate the number of arrangements for this example that use each "ship" exactly once?
B) What is the general solution for a grid of height Y and width X, and a given set of 1xN "ships"?

REPLY [3 votes]: I've recently done some coding to try and solve this problem, and I've arrived at the answer
30,093,975,536
where we consider the two length 3 ships as distinct. I can't say that I am certain of the result [edit: having received confirmation by a second party, I believe that I can say that I am certain now], and you can find my (python) code at
http://forums.xkcd.com/viewtopic.php?f=17&t=97981&p=3186720#p3186509
If someone can find a clear logical error in it, or has comments on the algorithm I use, I'd be keen to talk specifics.
@Lemming: Regarding the table found at
http://www.haskell.org/haskellwiki/Battleship_game_combinatorics
the second result there is wrong. If we consider placing 2 ships of length 2, there are 8 ways to put the first in a corner which blocks 4 moves for the second ship, 28 ways to place it along a side which blocks 5, 32 ways to place it perpendicular to a side which blocks 6 moves, and 112 ways to place the first ship in the middle of the board, which block 7 moves for the second ship ie there are (8*176 + 28 * 175 + 32 * 174 + 112 * 173)/2 = 15626 ways to place 2 indistinguishable 2s, and the table at the haskell wiki says there are 13952.
Edit: I believe I've determined where the disagreement in numbers comes from. If one assumes that ships can be placed in any way so long as they don't overlap anywhere, then you get the figures that I reported. But, if you assume that ships cannot be adjacent, neither horizontally, vertically or diagonally, then the numbers that arise are those reported by Lemmings.
Nobody is wrong, people are just using different sets of rules.<|endoftext|>
TITLE: Are two probability distributions uniquely constrained by the sum of their p-norms?
QUESTION [14 upvotes]: Let A, B and C be finitely supported probability distributions with at most d nonzero probabilities each.  Now consider the following simultaneous equations using p-norms, for each value of p≥1, given by
||A||p + ||B||p = ||C||p
where A, B and C are still non-negative, but we relax normalization on A and B.  Imagine that C is fixed and, without loss of generality, normalized.  We want to solve for A and B.
First, note that one obvious family of solutions is 
A = (1-x) C , B = x C , 0≤x≤1 .
Question: Ignoring the obvious permutation symmetries, are these the only solutions?
Edit: By p-norm, I mean the vector p-norm: ||A||p = (∑j |aj|p )1/p.  Although we don't really need the absolute values, since the aj are all non-negative.

REPLY [12 votes]: Here is a proof that Steve's rescaling gives you all solutions, together with the trivial operation of permuting the components of $A$, $B$, and $C$ if you view them as vectors with positive coeifficients.  (If you view them this way, then Steve's notation $||A||_p$ is just the usual $p$-norm.)
I first tried what Alekk tried:  You can take the limit as $p \to \infty$ and eventually obtain certain power series expansions in $1/p$.  Or you can take the limit $p \to 0$ and obtain certain power series expansions in $p$.  The problem with both approaches is that the information in the terms of these expansions is complicated.  To help understand the second limit, I observed that the two sides of Steve's equation are analytic in $p$, but it only helped so much.
Then I realized that when you have a complex analytic function of one variable, you can get a lot of information from looking at singularities.  So let's look at that.  Let
$\alpha_k = \ln a_k$, so that
$$||A||_p = \exp\left( \frac{\ln \bigl[\exp(\alpha_1 p) + \exp(\alpha_2 p) + \cdots + \exp(\alpha_d p) \bigr]}{p} \right).$$
The expression inside the logarithm has been called an exponential polynomial in the literature, which I'll call $a(p)$.  As indicated, $||A||_p$ has a logarithmic singularity when $a(p) = 0$.  $||A||_p$ has another kind of singularity when $p = 0$, but won't matter for anything.  Also $a(p)$ is an entire function, which means in particular that it is univalent and has isolated zeroes.  Also, none of the zeroes of $a(p)$ are on the real axis.  Let $b(p)$ and $c(p)$ be the corresponding exponential polynomials for $B$ and $C$.
Suppose that you follow a loop that starts on the positive real axis, encircles an $m$-fold zero of $a(p)$ at $p_0$, and then retraces to its starting point.  Then the value of $||A||_p$, which is non-zero for $p > 0$, gains a factor of $\exp(2m\pi i/p_0)$.  Thus Steve's equation is not consistent unless all three of $a(p)$, $b(p)$ and, $c(p)$ have the same zeroes with the same multiplicity.  (Since $\exp(2m\pi i/p_0)$ cannot have norm 1, geometric sequences with this ratio but with different values of $m$ are linearly independent.)
At this point, the problem is solved by a very interesting paper of Ritt, On the zeros of exponential polynomials.  Ritt reviews certain results of Tamarkin, Polya, and Schwengler, which imply in particular that if an exponential polynomial $f(z)$ does not have any zeroes, then it is a monomial $f_\alpha \exp(\alpha z)$.  Ritt's own theorem is that if $f(z)$ and $g(z)$ are exponential polynomials, and if the roots of $f(z)$ are all roots of $g(z)$ (with multiplicity), then their ratio is another exponential polynomial.  Thus in our situation $a(p)$, $b(p)$, and $c(p)$ are all proportional up to a constant and an exponential factor.  Thus, $A$, $B$, and $C$ must be the same vectors up to permutation, repetition, and rescaling of the coordinates.  Repetition is an operation that hasn't yet been analyzed.  If $A^{\oplus n}$ denotes the $n$-fold repetition of $A$, then $||A^{\oplus n}||_p = n^{1/p}||A||_p$.  Again, since geometric sequences with distinct ratios are linearly independent, Steve's equation is not consistent if $A$, $B$, and $C$ are repetitions of the same vector by different amounts.
The same argument works for the generalized equation
$$x_1||A_1||_p + x_2||A_2||_p + \cdots + x_n||A_n||_p = 0.$$
The result is that any such linear dependence trivializes, after rescaling the vectors and permuting their coordinates.
Update (by J.O'Rourke): Greg's paper based on this solution was just published: 

"Norms as a function of $p$ are linearly
  independent in finite dimensions," Amer. Math. Monthly, Vol. 119, No. 7, Aug-Sep 2012, pp. 601-3
  (JSTOR link).<|endoftext|>
TITLE: Projective closure of affine curve
QUESTION [6 upvotes]: Is there a generalized method to find the projective closure of an affine curve? For example, I read that the projective closure of $y^2 = x^3−x+1$ in $\mathbb{P}^2$ is $y^2z = x^3−xz^2+z^3$.
If I want to find the the closure of another affine curve, what method should I employ? I can't seem to find an adequate description in any text.
Thanks

REPLY [4 votes]: Let me add a comment with some geometric flavor to Charles's clean answer. Once the field is algebraic closed, then the affine curve $C\subset \mathbb{A}^2$ is never compact. However it admits an embedding into the projective space (remember that we know that $\mathbb{A}^2\subset \mathbb{P}^2$) where we can look for a compactification. The construction above, geometrically speaking, takes a hyperplane in $\mathbb{P}^2$ and compactifies with it. In other words, there is a hyperplane (divisor) $H=[z=0]\subset \mathbb{P}^2$ which tells you which points to add to your curve $C$, in order to get it compact: let's denote it by $\tilde{C}$. At the end of the day, you will have that $\tilde{C}\cap H$ are the points that you have added to $C$.
In a more general setting (related to the Groebner basis), if you have a manifold $M$, you may ask yourself that it would be desirable for a compactification of it to be algebraic. Now, as far as I now, there are two main obstructions for a compactification to be algebraic. There may not be a "nice" compactification and it may not have enough meromorphic functions. By "nice" here, I mean a compactification given by divisors (with normal crossings usually is asked), and by "enough meromorphic functions", there may not be ample line bundles. Questions on -When such obstructions do not cause troubles?- are big theorems in algebraic geometry. Not referring to self-confidence issues... "here, positivity helps a lot!"<|endoftext|>
TITLE: Combinatorial Techniques for Counting Conjugacy Classes
QUESTION [24 upvotes]: The number of conjugacy classes in $S_n$ is given by the number of partitions of $n$. Do other families of finite groups have a highly combinatorial structure to their number of conjugacy classes? For example, how much is known about conjugacy classes in $A_n$?

REPLY [14 votes]: For an explicit formula for the size of the conjugacy classes of GL$(n,q)$ (going back to Frobenius and Philip Hall), see equation (1.107) on page 92 of http://math.mit.edu/~rstan/ec/ec1.pdf. (There are formulas for the quantities appearing in (1.107) earlier in the text and in (1.108).)
As for the number $\omega^\ast(n,q)$ of conjugacy classes, see Exercise
1.190 on page 156 of the above reference. In particular, for fixed $n$,
$\omega^\ast(n,q)$ is a polynomial in $q$ satisfying 
  $$ \sum_{n\geq 0}\omega^\ast(n,q)x^n = \prod_{j\geq 1}
    \frac{1-x^j}{1-qx^j} $$
  $$ \omega^\ast(n,q) = q^n-q^m- q^{m-1}-q^{m-2}-\cdots
    -q^{\lfloor n/3\rfloor}+O(q^{\lfloor n/3\rfloor-1}), $$
where $m=\lfloor (n-1)/2\rfloor$.<|endoftext|>
TITLE: Freeing a sphere from within a sphere
QUESTION [5 upvotes]: We can embed $S^2\times I$ into $\mathbb{R}^3$ by taking a compact 3-ball and removing an open 3-ball from its interior.  Taking the boundary gives an embedding $i: S^2\sqcup S^2\hookrightarrow\mathbb{R}^3$, as "a sphere contained inside another sphere". Now it's intuitively clear that this embedding is not ambient-isotopic to the embedding $j$ given by putting these two spheres "side-by-side" in $\mathbb{R}^3$.  That is, there is no isotopy $F_t:\mathbb{R}^3\to \mathbb{R}^3$ with $F_0=1_{\mathbb{R}^3}$ and $F_1\circ i=j$.  At least, this looks visually obvious.  Assuming my intuition isn't betraying me and this isn't false, what's an elegant way to prove this?  
Also, (why) does there (not) exist an embedding $S^2\times I\hookrightarrow\mathbb{R}^3$ whose boundary is ambient-isotopic to the "side-by-side" embedding?  How about for $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?

REPLY [2 votes]: Edit: This is meant to answer the question of why we can't have an embedding $\mathbb S^{n-1}\times I\hookrightarrow\mathbb R^n$ such that the boundary is two side-by-side spheres rather than two nested spheres. I said "the second question" but it changed.
It seems to me that if you have an embedding of $S^{n-1}\times I$ into $\mathbb R^n$, this is the same as an ambient isotopy of one copy of $S^{n-1}$ to another.  (I found this easier to visualize when thinking of the fact that you can't embed a cylinder in $\mathbb R^2$ except as an annulus of concentric circles.)
In particular, you're trying to get one sphere to bound the same disk as the other sphere does, by a homotopy whose image at any time never intersects the image at another time.  You've already noted that two concentric spheres bound the same disk: so you're trying to get the first sphere $A$ to surround the other sphere $B$. Since $B$ cuts off a component of $\mathbb R^n$, we can remove that component and know the isotopy will not pass through it without first passing through $B$, which is not allowed.  Therefore remove a ball $C$ or a point from the interior of $B$.  $A$ is contractible in $\mathbb R^n-C$, and $B$ is not: $B$ generates the $n-1$st homology group of the resulting manifold.  This also is a nice way to see why it will work in $\mathbb R^{n+1}$, though of course in that case you really do have a "cylinder" (take $I$ to run in the $n+1$st dimension with each $S^{n-1}$ in an $n$-hyperplane).<|endoftext|>
TITLE: Barrelled, bornological, ultrabornological, semi-reflexive, ... how are these used?
QUESTION [70 upvotes]: I'm not a functional analyst (though I like to pretend that I am from time to time) but I use it and I think it's a great subject.  But whenever I read about locally convex topological vector spaces, I often get bamboozled by all the different types that there are.  At one point, I made a little summary of the properties of my favourite space (smooth functions from the circle to Euclidean space) and found that it was:

metrisable, barrelled, bornological, Mackey, infrabarrelled, Montel, reflexive, separable, Schwartz, convenient, semi-reflexive, reflexive, $c^\infty$-top is LCS top, quasi-complete, complete, Baire, nuclear

Its dual space (with the strong topology) is

reflexive, semi-reflexive, barrelled, infrabarrelled, quasi-complete, complete, bornological, nuclear, Mackey, convenient, $c^\infty$ top = LCS top, Schwartz, Montel, separable, DF space

I get the impression that many of these properties (and there are more!) are not "front line" properties but rather are conditions that guarantee that certain Big Theorems (like uniform boundedness, or open mapping theorem) hold.  But as an outsider of functional analysis, it's not always clear to me which are "front line" and which are "supporters".
So that's my question: which of these properties (and others that I haven't specified) are main properties and which have more of a supporting role?
I realise that there's a little vagueness there as to exactly where the division lies - but that's part of the point of the question!  If pressed, I would refine it to "Which of these properties would you expect to find used outside functional analysis, and which are more part of the internal machine?".

REPLY [18 votes]: Apology of one maker of a slight mess:
Whether properties are front line or supporting depends very much on which war you are fighting. If you are interested in the linear locally convex theory then many properties
are front line. But if you are mainly interested in smooth mappings, then quite few properties are front line; especially those which allow for uniform boundedness theorems (Greg should have added webbed in the diagram).
In fact, the space of smooth curves in a lcs (= locally convex space) does not change if you change the topology, as long as the bornology stays the same. In fact you can start from a dual pair of spaces  $(E, E')$ (separating points on each other), check, whether Mackey Cauchy sequences for the weak topology converge (then $E$ is convenient), and then you can choose any lcs topology which is compatible with the duality. The finest such topology is bornological. This was part of the approach in the book of Froelicher and Kriegl, who started from scratch and reconstructed everything. 
The most natural topology on $E$ from the point of view of Calculus is the final topology with respect to all smooth curves (equiv: all Mackey convergent sequences, equiv: all locally Lipschitz curves, ...); this is denoted $c^\infty E$. The finest lcs topology coarser than $c^\infty E$ is the bornologification of any lcs toplogy which is compatible with dual pair.  
Or you can try to allow users to make use of knowledge in the theory of lcs. This was the approach in the book: Convenient setting ... 
There you are allowed to use any lcs topology on $E$ that you know well or can describe well and which still has the given bornology. Spaces $E$ are identified if they are 
biboundedly linearly isomorphic (equiv.: diffeomorphic). To be convenient is a property of such equivalence classes of spaces. 
[Added in edit:] Or, to be convienient is a property shared by all spaces in such an equivalence class (or a space with all lcs topologies with the same system of bounded sets). The relation to the Frölicher-Kriegl notion is: take the bornologification of the space in question, as in their book only bornological spaces are considered. 
So, in Greg's nice diagram above, the place of convenient should be:
Sequentially complete $\implies$ convenient. 
In fact, each "naturally described" lcs is convenient. You have to force it to be not convenient by choosing a not Mackey complete subspace with the induced lcs topology.<|endoftext|>
TITLE: Learning Topology
QUESTION [23 upvotes]: EDIT (Harry):  Since this question in its original form was poorly stated (asked about topology rather than graph theory), but we have a list of Topology books in the answers, I guess you should go ahead and post with regard to that topic, rather than graph theory, which the questioner can ask again in another topic.
EDIT (David): The original question was asking for places to learn topology with an eye towards applying it to computer science (artificial neural networks in particular)

REPLY [2 votes]: This question originally asked about places to learn topology with an eye towards applying it to programming. The OP then clarified that he was interested in Artificial Neural Networks. Most answers have been about which textbooks are good for topology, but none address connections to computer science. One mentions computational topology, which seeks to develop efficient algorithms for solving topological problems, e.g. computing homology and homotopy. This doesn't seem to be what the OP was after, since he wanted to apply topology to computer science and not the other way around.
As an algebraic topology PhD student who's simultaneously getting a masters in CS, I have thought quite a bit about ways to combine the two. I've seen talks in which algorithms are developed to apply homology to detect information about graph properties, but these are mostly useless in practice because they're so slow. I haven't seen any algorithms using more than $H_1$ except in the work of Robert Ghrist, who uses sheaf cohomology and similarly high-brow mathematics to do pretty impressive things with sensor networks and engineering (e.g. flows). He seems to be the primary pioneer in this new field and calls it Applied Topology. He recently updated his website to include lots of powerpoints and pdfs to help a beginner get into this field. Here is the link. I've also seen some conferences about Applied Topology in the works, so keep your eyes open for them if you're interested in this field.<|endoftext|>
TITLE: Definition of étale for rings
QUESTION [11 upvotes]: Let $A \to B$ be a ring extension.
What is the definition of $B/A$ étale ?
When $A$ is a field, do we get a nice characterization ?

REPLY [3 votes]: If f is a map of local rings
$$f:A\rightarrow B$$ is étale iff it is flat and unramified (check out Bhargav Bhatt's notes at the stacks project link text). If A is a field and B is finite over A, then f is étale iff B is isomorphic to a finite product of separable field extensions of A (see proposition I.3.1 of Milne's book "Étale cohomology"). More generally, for f any ring homomorphism, check out definition II.1.1 of SGA 4.5 (B is a finitely presented A-algebra and satisfies a Jacobian criterion is a possible definition. Or B is a finitely presented A-algebra and B is flat and the relative differentials are trivial). The definition comes down to "smooth of relative dimension 0".<|endoftext|>
TITLE: Chevalley Eilenberg complex definitions?
QUESTION [21 upvotes]: In Weibel's An Introduction to Homological Algebra, the Chevalley-Eilenberg complex of a Lie algebra $g$ is defined as $\Lambda^*(g) \otimes Ug$ where $Ug$ is the universal enveloping algebra of $g$.  The differential here has degree -1.
I have been told that the Chevalley-Eilenberg complex for $g$ is 
    $C^*(g) = \text{Sym}(g^*[-1])$,
the free graded commutative algebra on the vector space dual of $g$ placed in degree 1.  The bracket $[,]$ is a map $\Lambda^2 g \to g$ so its dual $d : = [,]* \colon g^* \to \Lambda^2 g^*$ is a map from $C^1(g) \to C^2(g)$.  Since $C^*(g)$ is free, this defines a derivation, also called $d$, from $C^*(g)$ to itself.  This derivation satisfies $d^2 = 0$ precisely because $[,]$ satisfies the Jacobi identity.
Finally, Kontsevich and Soibelman in Deformation Theory I leave it as an exercise to construct the Chevalley-Eilenberg complex in analogy to the way that the Hochschild complex is constructed for an associative algebra by considering formal deformations.
The first is a chain complex, the second a cochain complex, and what do either have to do with formal deformations of $g$?

REPLY [10 votes]: Just to add to what Mariano has written and focussing only on the deformations aspect, the exercise in Kontsevich-Soibelman hints at the "principle" that deformations of algebraic structures are always governed by a cohomology theory and that if you didn't know which, you would discover it by analysing the conditions defining infinitesimal deformations, trivial infinitesimal deformations and obstructions to integrating infinitesimal deformations.
I think that this is a very good exercise and it helps motivate the classical formulas for the differentials, at least in the case of cohomology with values in the adjoint module.
I would add another reference to the one of Gerstenhaber et al, and that is the older paper by Nijenhuis and Richardson Deformations of Lie algebra structures.  In general there are a number of classic papers of Nijenhuis and Richardson on this topic.  In particular they define a graded Lie algebra structure on the deformation complex which makes very clear the nature of the obstructions.

REPLY [10 votes]: The other comment revealed most of the story. Let me add some points. First, the Chevalley-Eilenberg complex is defined for the most general case of $H^\bullet(g,M)$ --- cohomology with coefficients in a module M. In the case M=k the trivial module, you get your second complex. In the case $M=g$ the adjoint representation, you get the comples Kontsevich and Soibelman ask to construct. 
The first complex is some version of Koszul complex $(A^!)^* \otimes A$ for quadratic algebras: one can think of the Koszul dual of the dg-commutative algebra $\Lambda^\bullet( g^* )$ as the quadratic-linear algebra $U(\mathfrak{g})$. It is acyclic, and gives a resolution of the trivial module by free $U(g)$-modules, so you can use is to compute Ext groups $Ext_{U(g)}^\bullet(k,M)=H^\bullet(g,M)$.
Finally, another way to think of your second complex is as follows. If $g$ is the Lie algebra of a Lie group $G$, one can consider the subcomplex of the de Rham complex consisting of left-invariant differential forms. A left-invariant form is defined by its behaviour at the unit of the group, 1-forms are dual to the tangent space and form $ g^* $ , 2-forms give $\Lambda^2(g^*)$ etc., and you get precisely the CE complex, where the differential is what the de Rham differential induces. If $G$ is compact, it is easy to show that this complex has the same cohomology as the de Rham complex, so we compute the de Rham cohomology of the group in this case!<|endoftext|>
TITLE: Models with SLE scaling limit
QUESTION [10 upvotes]: What discrete processes/models have been proven to have scaling limits to $\text{SLE}(\kappa)$, for various $\kappa$?
I know about loop-erased random walk and uniform spanning trees.
What about conjectures in this direction? (Such as the double-dimer-cover cycles, which I read are conjectured to be $\text{SLE}(4)$)
I'm very new to this, so if you please could, together with the answer refer to a paper/article/survey accompanying the result, that would be greatly appreciated!

REPLY [7 votes]: From Cardy's article http://arxiv.org/abs/cond-mat/0503313

Some important special cases are therefore: 
$\kappa = 2$: loop-erased random walks (proven in [24]); 
$\kappa = 8/3$: self-avoiding walks, as already suggested by the restriction property, Sec. 3.5.2; 
unproven, but see [22] for many consequences; 
$\kappa = 3$: cluster boundaries in the Ising model, however as yet unproven; 
$\kappa = 4$: BCSOS model of roughening transition (equivalent to the 4-state Potts 
model and the double dimer model), as yet unproven; also certain level lines of a 
gaussian random field and the ‘harmonic explorer’ (proven in [23]); also believed to 
be dual to the Kosterlitz-Thouless transition in the XY model; 
$\kappa = 6$: cluster boundaries in percolation (proven in [7]); 
$\kappa = 8$: dense phase of self-avoiding walks; boundaries of uniform spanning trees 
(proven in [24]). 
It should be noted that no lattice candidates for κ > 8, or for the dual values κ < 2, have 
been proposed.

REPLY [5 votes]: There are several other models proved to converge to SLE: critical percolation on the triangular lattice, Gaussian Free Field, Harmonic Explorer, and recently also the critical Ising model. You can check the paper Kevin linked or Schramm's slides from ICM2006 for some highlights. Just keep in mind that this is a fast changing field at the moment so there has been some progress since 2006.

REPLY [3 votes]: There are several examples in this readable survey article by Schramm.<|endoftext|>
TITLE: Example of non-closed convex hull in a CAT(0) space
QUESTION [7 upvotes]: this is related to this question but is simpler, and hopefully is well-known. There are a number of references that say that the convex hull of a collection of points in a CAT(0) space need not be closed. I was wondering if anyone was aware of an explicit example ?

REPLY [4 votes]: There is a natural example if you don't ask the collection of point to be finite (but where it is closed): take the subset $A_1$ of the real Hilbert space $X=L^2(\mathbb{R})$ consisting of function taking at most one value beside $0$. Then it is easily seen that the geodesic segments between pairs of points in $A_1$ cover the set $A_3$ of function taking at most $3$ values beside $0$. The convex hull of $A_1$ is then the set of functions taking a finite number of values, and is dense in $X$.
For the story, this is a variation on an example I came across in optimal transportation: $X$ was the Wasserstein space of the real line with quadratic cost, which is isometric to the subset of $L^2([0,1])$ consisting of non-decreasing functions.<|endoftext|>
TITLE: Weierstrass points on rigid-analytic surfaces
QUESTION [8 upvotes]: Does a rigid-analytic surface defined over a nonarchimedean complete field have Weierstrass points (if its genus is big enough let's say)? Is there a good reference that (ideally) lists theorems for rigid-analytic spaces that are the analog of commonly known theorems about complex analytic spaces?

REPLY [3 votes]: About references I think Brian Conrad's lecture notes (I'd check mainly his references) from his course in AWS07 might be a good place to look at. http://swc.math.arizona.edu/aws/07/ConradNotes11Mar.pdf
From Conrad's notes I remember that two references were helpful,
http://wwwmath.uni-muenster.de/sfb/about/publ/heft378.pdf
and "Non-Archimedean analysis" by S. Bosch, U. Güntzer, R. Remmert<|endoftext|>
TITLE: Reps of $U(n)$ for the bundles of holomorphic and antiholomorphic forms of projective space
QUESTION [5 upvotes]: What are the representations of $U(n)$ that induce (see link text) the bundles of holomorphic $\Omega ^{(1,0)}$ and antiholomorphic $\Omega ^{(0,1)}$ forms of $\mathbb{CP}^n$ (recalling the well-known fact that $\mathbb{CP}^n = SU(n+1)/U(n)$). 
Also, what are the representations that induce the line bundles?

REPLY [4 votes]: In the comments above, I've been trying to figure out what the vector bundle of antiholomorphic forms is. José Figueroa-O'Farrill writes

I think that holomorphic and antiholomorphic refer, respectively, to (1,0) and (0,1) forms.  

I write 

It ... seems that "antiholomorphic forms" is the dual vector bundle to "holomorphic forms", that is to say, it is the holomorphic tangent bundle.

The point of this answer is to explain that we are both right.
First of all, I am used to distinguishing between two notions. For me, a "$(1,0)$-form" is a $\mathbb{C}$-valued $1$-form which locally looks like $\sum f_i(z) d z_i$, for some $C^{\infty}$ functions $f_i$. In a "holomorphic $1$-form" we also require the $f_i$ to be holomorphic functions. Apparently, not everyone makes this distinction. That's fine, but in this post I am going to use my language because it appears to be more precise.
Let $X$ be a complex manifold. The sheaf of $(1,0)$-forms on $X$ is a sheaf of modules for the sheaf of $C^{\infty}$ functions. Similarly, the sheaf of holomorphic $1$-forms is a sheaf of modules for the sheaf of holmorphic functions. Using the appropriate version of the Serre-Swan theorem in each case; we get a complex vector bundle over $X$, in the $C^{\infty}$ and holomorphic categories respectively. Taking the forgetful functor from holomorphic vector bundles to smooth vector bundles, we get naturally isomorphic vector bundles in either case, and this bundle is the cotangent bundle $T^*(X)$. So, no matter how you think about it, the "bundle of holomorphic $1$-forms" is the cotangent bundle.
That's what happens with $(1,0)$-forms. What happens with $(0,1)$-forms? In this case, I think it is best to generalize the $C^{\infty}$ approach above. That is to say, we consider the sheaf of all $1$-forms that locally look like $\sum f_i(z) d \overline{z}_i$, for some $C^{\infty}$ functions $f_i$. Again, Serre-Swan gives us a $C^{\infty}$ complex vector bundle which I'll call $A$. From this perspective, $A$ does not have an obvious holomorphic structure.
However, I claim that $A$ is noncanonically isomorphic to the tangent bundle to $X$. Here is the isomorphism. Choose a positive definite Hermitian structure on tangent bundle of $X$. This is always possible by a partition of unity argument. (My convention is that Hermitian forms are linear in the second argument and conjugate linear in the first.) For any $(1,0)$-vector field $v$, the $1$-form $\langle \ , v \rangle$ is a $(0,1)$-form.
The tangent bundle to $X$, of course, does have a natural holomorphic structure; the holomorphic sections are of the form $\sum f_i(z) \partial/\partial z_i$ where the $f_i$ are holomorphic. But this structure is very hidden in the presentation of $A$ as the bundle of $(0,1)$-forms, and that was the point that was confusing me.<|endoftext|>
TITLE: Cotangent bundle of a submanifold
QUESTION [8 upvotes]: Maybe this is a silly question (or not even a question), but I was wondering whether the cotangent bundle of a submanifold is somehow canonically related to the cotangent bundle of the ambient space.
To be more precise:
Let $N$ be a manifold and $\iota:M \hookrightarrow  N$ be an embedded (immersed) submanifold. Is the cotangent bundle $T^\ast M$ somehow canonical related to the cotangent bundle $T^\ast N$. Canonical means, without choosing a metric on $N$. The choice of a metric gives an isomorphism of $TN$ and $T^\ast N$ and therefore a "relation", since the tangent bundle of the submanifold $M$ can be viewed in a natural way as a subspace of the tangent bundle of the ambient space $N$ ($\iota$ induces an injective linear map at each point $\iota_\ast : T_pM \rightarrow T_pN$). I think this is not true for the cotangent space (without a metric)
Moreover, the cotangent bundle $T^\ast N$ of a manifold $N$ is a kind of "prototype" of a symplectic manifold. The symplectic structure on $T^\ast N$ is given by $\omega_{T^\ast N} = -d\lambda$, where $\lambda$ is the Liouville form on the cotangent bundle. (tautological one-form, canonical one-form, symplectic potential or however you want). The cotangent bundle of the submanifold $T^\ast M$ inherits in the same way a canonical symplectic structure. So, is there a relation between $T^\ast N$ and $T^\ast M$ respecting the canonical symplectic structures. (I think the isomorphism given by a metric is respecting (relating) these structures, or am I wrong?) As I said, this question is perhaps strange, but the canonical existence of the symplectic structure on the cotangent bundle is "quite strong". For example:
A given diffeomorphism $f:X \rightarrow Y$ induces a canonical symplectomorphism $T^\ast f : T^\ast Y \rightarrow T^\ast X$ (this can be proved by the special "pullback cancellation" property of the Liouville form).
So in the case of a diffeomorphism the symplectic structures are "the same". 
Ok, a diffeomorphism has more structure than an embedding, but perhaps there is a similar relation between $T^\ast M$ and $T^\ast N$?

EDIT: Sorry fot the confusion, but Kevins post is exactly a reformulation of the problem, I'm interested in. To clarify things: with the notation of Kevin's post:

When (or whether) are the pulled back symplectic structures the same ? Under what circumstances holds $a^\ast \omega_{T^\ast N} = b^\ast \omega_{T^\ast N}$

I think this isn't true for any submanifold $M \subset N$, but what is a nice counterexample? Is it true for more restricted submanifolds as for example embedded submanifolds which are not just homoeomorphisms onto its image, but diffeomorphisms (perhaps here the answer is yes, using the diffeomorphism remark above?)?

REPLY [5 votes]: My take on this: I will repeat something I've said elsewhere: Differential geometry is just parameterized linear algebra.
So first just understand the underlying linear algebra. If you have a submanifold $S \subset M$, then for each $x \in S$, you know that $T_xS$ is a linear subspace of $T_xM$. So what does this tell you about the respective dual spaces?
Well, if you have a subspace $L$ of a vector space $V$, then corresponding natural subspace of  $V^\star$ is the annihilator $L^\perp$ of all $\xi \in V^\star$ such that $\langle\xi,v\rangle = 0$, for every $v \in L$. Moreover, the dual space $L^\star$ is naturally isomorphic to $V/L^\perp$.
So returning to the submanifold $S$, there is a natural subbundle $N^*S \subset T^\star M$, where $N^\star_xS = (T_xS)^\perp$. This is commonly called the conormal bundle. The cotangent bundle of $S$ is the quotient bundle $T^\star M|_{S}/N^\star S$.
Just remember that the dual of "is a subspace of" is "is a quotient of".<|endoftext|>
TITLE: Number of spanning trees in a grid
QUESTION [7 upvotes]: Given a $\sqrt{n}\times\sqrt{n}$ piece of the integer $\mathbb{Z}^2$ grid, define a graph by joining any two of these points at unit distance apart. How many spanning trees does this graph have (asymptotically as $n\to\infty$)?
Can you also say something about the triangular grid generated by $(1,0)$ and $(1/2,\sqrt{3}/2)$?

REPLY [4 votes]: Expanding on Steve Huntsman's answer, call the product which appears in A007341 f(n).  That is, 
$$f(n) = \prod_{k=0}^{n-1} {\prod_{l=0}^{n-1}}^\prime \left(2 - \cos {\pi k \over n} - \cos {\pi l \over n } \right)$$
where the $\prime$ on the second product indicates that we start at $l=1$ in the case $k = 0$.  The number of interest here is $a(n) = 2^{n^2-1} f(n)/n^2$ .
The product is the exponential of a sum, so
$$\log f(n) = \sum_{k=0}^{n-1} {\sum_{l=0}^{n-1}}^\prime \log \left(2 - \cos {\pi k \over n} - \cos {\pi l \over n } \right).$$
This sum is, in turn, $n^2$ times a Riemann sum for the integral
$$  C = \int_0^1 \int_0^1 \log(2-\cos x\pi - \cos y\pi) \: dx \: dy $$
which I believe converges, although actually evaluating it numerically is tricky.  If you believe that, then $\log f(n) \sim Cn^2$ as $n \to \infty$, and $\log a(n) \sim (C+\log 2) n^2$ as $n \to \infty$.  From evaluating $f(n)$ for various $n$, it appears that $C$ is near $0.473$, $e^C$ is near $1.605$ and so we have
$$ a(n) \approx 3.21^{n^2} $$
where I write $p(n) \approx q(n)$ for $\log p(n)/\log  q(n) \to 1 $ as $n \to \infty$, i. e. $\log p(n) \sim \log q(n)$.<|endoftext|>
TITLE: Characterization of Riemannian metrics
QUESTION [11 upvotes]: This is probably an insanely hard question, but given an abstract metric space, is there some way to determine whether it's a manifold with a Riemannian, or more generally a Finslerian, metric? If that's too hard, one could start off by assuming that the underlying space is a manifold. The example that got me thinking about this was the induced metric on the 2-sphere embedded in $R^3$...the underlying space is obviously a smooth manifold, and the metric should be smooth(the geodesics would even be great circles, as they are for the standard metric on $S^2$,) but I don't see how you could prove the triangle inequality pointwise, as you'd have to to show that it's Finslerian, let alone Riemannian. This example is already incredibly simple, since it's a homogeneous space with a the metric induced by being a subspace of another homogeneous space.

REPLY [18 votes]: If $X$ is a metric space and $x$, $y\in X$, a segment from $x$ to $y$ is a subset $S\subseteq X$ such that $x$, $y\in S$ and $S$ is isometric to $[0,d(x,y)]$. 
Let now $n\in\{1,2,3\}$ and let $X$ be a metric space which is locally compact, $n$-dimensional and such that (i) every two points are the endpoints of a unique segment, (ii) if two segments have an endpoint and one other point in common, then one is contained in the other, and (iii) every segment can be extended, at either end, to a larger segment. Then $X$ is homeomorphic to $\mathbb R^n$.
This is a metric characterization of $\mathbb R^n$ for small $n$. Using it locally, you get a characterization of topological manifolds of small dimension. I imagine higher dimensions or smoothness are harder to come by...
The theorem above, along with quite a few other nice results, is proved in [Berg, Gordon O. Metric characterizations of Euclidean spaces. Pacific J. Math.  48  (1973), 11--28.]

LATER...
Of course, the paper [Nikolaev, I. G. A metric characterization of Riemannian spaces. Siberian Adv. Math.  9  (1999), no. 4, 1--58.] is relevant here! The paper gives purely metric characterizations of Riemannian manifolds of all smoothnesses up to $C^\infty$.
The MR review gives a small history of the problem and references to earlier work.

REPLY [6 votes]: I think that if the underlying space is a smooth manifold $M$, a metric $d$ (in the "metric space" sense) on it will give rise to a Riemannian metric if and only if it satisfies a smoothness condition (I assume we want the Riemannian metric to be smooth). Fix a point $p \in M$. Since our manifold is smooth, there is a natural identification between a neighborhood of the origin of the vector space $TM_p$ and a neighborhood of $p$ in $M$. So, we can assume that our metric $d$ is defined on this neghborhood of origin in $TM_p$. For a vector $v \in TM_p$, we define its length to be $||\vec v||=\lim_{t\to 0} (d(t\vec v,0)/t)$. If this limit converges, then $\vec v \mapsto ||\vec v||^2$ should be a quadratic form on $TM_p$ and the polarization identity should give us a positive definite bilinear form $g$ on $TM_p$ such that $g(\vec v,\vec v)=||\vec v||^2$ for all $\vec v$. The smoothness condition on $d$ should, I guess, be that the function $d(t\vec v,0)$ is always smooth (so the limit always exists), and that the resulting $g$ should be smooth as a function of $p$. In this case, we have our Riemannian metric.
So, the problem seems to reduce to the question of when does the topological manifold $M$ have a smooth structure such that the metric $d$ satisfies the above smoothness condition. It would also be good to find a simpler way to state the smoothness condition.
I'm also thinking that people might know a way to define Riemannian metrics in categories other than that of smooth manifolds. This might be more appropriate for this question, but I have no idea how to deal with that. 
(I would've put this in a comment, but I don't have enough karma for that. I'm a bit intimidated that nobody has posted something like this yet -- I hope I haven't missed something stupid.)<|endoftext|>
TITLE: Diameter of universal cover
QUESTION [15 upvotes]: Let $M$ be Riemannian manifold and $\tilde M$ be its universal cover (with induced metric).
What is the upper bound for $k=\mathop{diam}\tilde M/\mathop{diam} M$ in terms of $m=|\pi_1(M)|$ (or $\pi_1(M)$)?
Comments:

There is a similar answered question here, but there  cover is NOT universal. So we get $k\leqslant m$, but $k\ll m$ is expected.
The question is open even in case $\pi_1=\mathbb Z_m$ (even asymptotics is not known).   
Clearly, $\sup k$ for given finite group $\Gamma=\pi_1(M)$ is an invariant of $\Gamma$. Is it an interesting invariant?

Examples:

For $\pi_1=\mathbb Z_{3\cdot 2^n}$ one can make $k\sim n$ or $k=O(\log m)$  (see my answer below).
For $\pi_1=S_n$, one can make $k$ of order $n^2$ or (see Greg's answer and also this question). It is much more than $\log m$, but still $k=o(\log^2 m)$.

REPLY [5 votes]: As both Anton and Greg pointed out, it is enough to look at the Cayley graph of a presented finite group $G$ with only quadratic and cubic relators and study how large $diam(G)$ can be in terms of $\vert G \vert$. Note that the property that all relators are quadratic or cubic implies that $G$ is the $1$-skeleton of a simply connected $2$-dimensional simplicial complex. 
It can be shown that $diam(G) \leq \sqrt{ 4 \vert G \vert +1}-2$, which implies the effective bound $diam(\tilde{M} )/ diam(M) \leq 4 \sqrt{\vert \pi_1(M) \vert } $. For the strictly asymptotic behavior, the main result of this paper implies that $diam (G) = o (\vert G \vert ^p) $ for any $p>0$, implying that $diam(\tilde{M} ) / diam(M) = o ( \vert \pi_1 (M) \vert ^p)$ for any $p>0$.
An sketch of the proof of the first inequality follows. Take $h \in G$ with $d(h,e)= diam(G)$ and a path $e=g_0, g_1 , \ldots, g_{diam(G)}=h$. Set $S_i$ to be the induced subgraph by the set $ \{  g \in G  \mid d(g,e)= i \}$ for each $i \in \mathbb{N}$ and $T_i  $ the connected component of $g_i$ in $S_i$. Fix $1 \leq i < diam(G)$. Removing $T_i$ disconnects $G$, otherwise there would be a loop in $G$ which cannot be contracted in any simplicial complex with $G$ as its $1$-dimensional skeleton. 
Denote by $C_1$ and $C_2$ the connected components of $e$ and $h$, respectively in $G \backslash T_i$. Since the left multiplications are graph isomorphisms, removing $ g_i^{-1}T_i$ disconnects $G$ in two components $C_1^{\prime}$ and $C_2^{\prime}$ isomorphic to $C_1$ and $C_2$ respectively. If $g_i^{-1} T_i$ does not intersect $T_i$, then one of $C_1^{\prime}$ or $C_2^{\prime}$ properly contains $C_2$, but since $\vert C_2^{\prime} \vert = \vert C_2 \vert$, we have $C_1^{\prime } \subsetneq C_2$ and $\vert C_1 \vert > \vert C_2 \vert$. The same way, if $hg^{-1}_iT_i$ does not intersect $T_i$, then $\vert C_2 \vert > \vert C_1 \vert$.
Therefore $T_i$ has to intersect either $g_i^{-1}T_i$ or $hg_i^{-1}T_i$, so $\vert T_i \vert \geq diam(T_i) \geq \min \{   i, diam(G) -i  \} +1$. Since the $T_i$'s are disjoint, summing over all $i$'s yield
$$\vert G \vert \geq \frac{(diam(G)+2)^2 +1}{4}.$$
Which is the desired inequality.<|endoftext|>
TITLE: synthetic differential geometry and other alternative theories
QUESTION [18 upvotes]: There are models of differential geometry in which the intermediate value theorem is not true but every function is smooth. In fact I have a book sitting on my desk called "Models for Smooth Infinitesimal Analysis" by Ieke Moerdijk and Gonzalo E. Reyes in which the actual construction of such models is carried out. I'm quite new to this entire subject and I only stumbled upon it because I was trying to find something like non-standard analysis for differential geometry. 
Already I'm liking the more natural formulations for differentials and tangent vectors in the new setting although I can see that true mastery of all the intricacies will require more background in category theory like Grothendieck topologies. So my questions are a bit philosophical. Suppose some big conjecture is refuted in one of these models but proven to be true in the classical setting then what exactly would that mean for classical differential geometry? Is such a state of affairs possible or am I missing something that rules out such a possibility like a metatheorem that says anything that can be proven in the new models can be proven in the usual classical model? More specifically what is the exact relationship between the new models and the classical one? Could one even make any non-trivial comparisons? References for such discussions are welcome. I'm asking the question here because I suspect there might be some experts familiar with synthetic differential geometry that will be able to illuminate the connection to the classical theory.
Edit: Found a very lively and interesting discussion by John Baez, Andrew Stacy, Urs Schreiber, Tom Leinster and many others on n-category cafe called Comparative Smootheology although I couldn't make out the exact relation to SDG.

REPLY [14 votes]: Perhaps I can make the implications of what Harry said a bit more explicit. A well-adapted model of SDG embeds smooths manifolds fully and faithfully. This in particualar means that the SDG model and the smooth manifolds "believe" in the same smooths maps between smooth manifolds (but SDG model contains generalized spaces which do no correspond to any manifold), and moreover precisely the same equations hold in the SDG model and in smooth manifolds. In this sense SDG is conservative: the model will never validate an invalid equation involving smooth maps between smooth manifolds.
The situation is really quite similar to other situations where we have to distinguish between truth and meaning inside a model and truth and meaning outside the model. For example, there are models of set theory which violate the axiom of choice, but these models are built in a setting where the axiom of choice holds. This is no mystery or magic, as long as we remember that a statement can have a different meanings inside the model and outside. The same applies to SDG: when the internal meaning of statements in the model is appropriately interpreted on the outside, nothing can go wrong (that's what a model is, after all).<|endoftext|>
TITLE: Is the category of Banach spaces with contractions an algebraic theory?
QUESTION [12 upvotes]: Consider the category of Banach spaces with contractions as morphisms (weak, so $\|T\| \le 1$).  Is this an algebraic theory?
I suspect that this is true.  The "operations" will be weighted sums, where the sum of the weights is at most $1$.  The "free Banach space" on a set $X$ should be $\ell^1(X)$.  (Note that the "underlying set" functor sends a Banach space to its unit ball.)  So, part one:

Is this correct?  If so, can anyone supply a reference (unfortunately, searching for "algebraic theory" and "Banach" doesn't turn up anything obvious).
Has anything useful/unusual come out of this point of view?
If this is correct, then the algebraic theory seems to be commutative, in which case it's a symmetric closed category.  Has this angle been used?
The norm can't be encoded as an operation, can it be categorically recovered?

Part two says: can we do this for Hilbert spaces?
Edit: Anyone even vaguely intrigued by this question should read the paper linked in Yemon's answer.  In particular, it also answers a question that I was going to ask as a follow-up: what's the nearest algebraic theory to Banach spaces (the answer being totally convex spaces).

REPLY [7 votes]: One of the major-league experts on this topic is Michael Barr. In a message to the categories list (dated October 22, 2003), he writes:

"For Banach spaces, if you take as underlying functor the closed unit ball, it has an adjoint. It is not tripleable, however, but $C^*$-algebras are (with the unit ball underlying functor)."

"Tripleable" means the same as monadic. See the discussion in the book by Barr-Wells, Toposes, Theories, and Triples, section 4.4. I think that I saw somewhere though that the forgetful functor from Banach spaces and linear contractions to metric spaces with basepoint and contractions is monadic.<|endoftext|>
TITLE: Coupling of Wiener processes
QUESTION [8 upvotes]: Is it possible to find a coupling of two Wiener processes $W^0, W^x$ (i.e. two Wiener processes defined on a common probability space). One starting from $0$ and the other from $x$ such that 
$W_t^0 - W_t^x \rightarrow_t 0$ almost surely and in $L^1$.
Using some random-walks considerations I suspect that it is not possible to have convergence in $L^1$ but I do not know how to prove it.

The answer is: $W_t^0 - W_t^x$ cannot converge to $0$ in $L^1$
The proof is as follow (it is a slightly extended version of the proof by smalldeviations). Obviously
$|W_t^0 - W_t^x|_1\geq  \inf _{\gamma\in \Gamma } \{\int _{\mathbf{R}^d\times \mathbf{R}^d} |x-y| \text{d}\gamma(x,y)\},$
where $\Gamma$ is the set of all couplings of $\mathcal{L} (W_t^0)$ and $\mathcal{L} (W_t^x)$ ($\mathcal{L}$ is the law of given variable). By the duality formula (see http://en.wikipedia.org/wiki/Transportation_theory) the right hand side is equal to
$\sup \{ \int_{\mathbf{R}^d} \phi (x)  \mathrm{d} \mu (x) + \int_{\mathbf{R}^d} \psi (y)  \mathrm{d} \nu (y) \},$
where the supremum runs over all pairs of bounded and continuous functions such that $\phi(x)+\psi(y)\leq |x-y|$ and $\mu =\mathcal{L}(W_t^x)$ and $\nu = \mathcal{L} (W_t^0)$. In our case it is sufficent to take $\phi(x) = x, \psi(y)=y$. Then the expression under the sup is equal to:
$\mathbf{E}(W^x_t) - \mathbf{E}(W^0_t) = x-0 =x$. Therefore $|W_t^0 - W_t^x|_1\geq x$.

REPLY [2 votes]: Here is another solution to this problem: $W^0$ and $W^x$ are both martingales and therefore so is their difference $W^0-W^x$. Then $|W^0-W^x|$ is a submartingale ($|\cdot|$ is convex). It follows that $E[|W^0_t - W^x_t|] \ge E[|W^0_0 -W^x_0|] =|x|$ for all $t$.<|endoftext|>
TITLE: How much choice is needed to show that formally real fields can be ordered?
QUESTION [26 upvotes]: Background: a field is formally real if -1 is not a sum of squares of elements in that field.  An ordering on a field is a linear ordering which is (in exactly the sense that you would guess if you haven't seen this before) compatible with the field operations.  
It is immediate to see that a field which can be ordered is formally real.  The converse is a famous result of Artin-Schreier.  (For a graceful exposition, see Jacobson's Basic Algebra.  For a not particularly graceful exposition which is freely available online, see http://math.uga.edu/~pete/realspectrum.pdf.)
The proof is neither long nor difficult, but it appeals to Zorn's Lemma.  One suspects that the reliance on the Axiom of Choice is crucial, because a field which is formally real can have many different orderings (loc. cit. gives a brief introduction to the real spectrum of a field, the set of all orderings endowed with a certain topology making it a compact, totally disconnected topological space).  
Can someone give a reference or an argument that AC is required in the technical sense (i.e., there are models of ZF in which it is false)?  Does assuming that formally real fields can be ordered recover some weak version of AC, e.g. that any Boolean algebra has a prime ideal?  (Or, what seems less likely to me, is it equivalent to AC?)

REPLY [26 votes]: This is equivalent (in ZF) to the Boolean Prime Ideal Theorem (which is equivalent to the Ultrafilter Lemma).
Reference:
R. Berr, F. Delon, J. Schmid, Ordered fields and the ultrafilter theorem, Fund Math 159 (1999), 231-241. online<|endoftext|>
TITLE: Are all polynomial inequalities deducible from the trivial inequality?
QUESTION [37 upvotes]: I remember learning some years ago of a theorem to the effect that if a polynomial $p(x_1, ... x_n)$ with real coefficients is non-negative on $\mathbb{R}^n$, then it is a sum of squares of polynomials in the variables $x_i$.  Unfortunately, I'm not sure if I'm remembering correctly.  (The context in which I saw this theorem was someone asking whether there was a sum-of-squares proof of the AM-GM inequality in $n$ variables, so I'm not 100% certain if the quoted theorem was specific to that case.)
So: does anyone know a reference for the correct statement of this theorem, if in fact something like it is true?  (Feel free to retag if you don't think it's appropriate, by the way.)

REPLY [19 votes]: Let me draw your attention to the paper "There are significantly more nonegative polynomials than sums of squares" by Grigoriy Blekherman. Here are slides of a related talk by Grigoriy.
The abstract reads: "We study the quantitative relationship between the cones of nonnegative polynomials, cones of sums of squares and cones of sums of even powers of linear forms. We derive bounds on the volumes (raised to the power reciprocal to the ambient dimension) of compact sections of the three cones. We show that the bounds are asymptotically exact if the degree is fixed and number of variables tends to infinity. When the degree is larger than two, it follows that there are significantly more nonnegative polynomials than sums of squares and there are significantly more sums of squares than sums of even powers of linear forms. Moreover, we quantify the exact discrepancy between the cones; from our bounds it follows that the discrepancy grows as the number of variables increases."<|endoftext|>
TITLE: What is the meaning of symplectic structure?
QUESTION [13 upvotes]: Answers can come in mathematical, physical, and philosophical flavors.
Edit: There seems to be a consensus that this question is not formulated well. I must respectfully disagree. My interest in the question is immaterial to the question itself. It is manifestly not a "what is" question. I see no reason to write more than is necessary for the formulation of the question, and I invite nothing more than a sentence giving me something to think about or an idea of where to look.

REPLY [16 votes]: Here's how I understand it. Classical mechanics is done on a phase space M. If we are trying to describe a mechanical system with n particles, the phase space will be 6*n*-dimensional: 3*n* dimensions to describe the coordinates of particles, and 3*n* dimensions to describe the momenta. The most important property of all this is that given a Hamiltonian function $H:M\to \mathbb R$, and a point of the phase space (the initial condition of the system), we get a differential equation that will predict the future behavior of the system. In  other words, the function $H$ gives you a flow $\gamma:M\times \mathbb R \to M$ that maps a point p and a time t to a point $\gamma_t (p)$ which is the state of the system if it started at p after time t passes.
Now, if we do classical mechanics, we are very interested in changes of coordinates of our phase space. In other words, we want to describe M in a coordinate-free fashion (so that, for a particular problem, we can pick whatever coordinates are most convenient at the moment). Now, if $M$ is an abstract manifold, and $H:M\to\mathbb R$ is a function, you cannot write down the differential equation you want. In a sense, the problem is that you don't know which directions are "coordinates" and which are "momenta", and the distinction matters. 
However, if you have a symplectic form $\omega$ on M, then every Hamiltonian will indeed give you the differential equation and a flow $M\times\mathbb R \to M$, and moreover the symplectic form is precisely the necessary and sufficient additional structure.
I must say, that although this may be related to why this subject was invented a hundred years ago, this seems to have little to do to why people are studying it now. It seems that the main reason for current work is, first of all, that new tools appeared which can solve problems in this subject that couldn't be solved before, and secondly that there is a very non-intuitive, but very powerful, connection that allows people to understand 3- and 4-manifolds using symplectic tools.

 EDIT: I just realized that I'm not quite happy with the above. The issue is that the phase spaces that come up in classical mechanics are always a very specific kind of symplectic manifold: namely, the contangent bundlde of some base space. In fact, in physics it is usually very clear which directions are "coordinates of particles" and which are "momenta": the base space is precisely the space of "coordinates", and momenta naturally correspond to covectors. Moreover, the changes of coordinates we'd be interested in are always just changes of coordinates of the base space (which of course induce a change of coordinates of the cotangent bundle).
So, a symplectic manifold is an attempts to generalize the above to spaces more general than the cotangent bundle, or to changes of coordinates that mess up the cotangent-bundle structure. I have no idea how to motivate this. 
UPDATE: I just sat in a talk by Sam Lisi where he gave one good reason to study symplectic manifolds other than the cotangent bundle. Namely, suppose you are studying the physical system of just two particles on a plane. Then, their positions can be described as a point in $P = \mathbb R^2 \times \mathbb R^2$, and the phase space is the cotangent bundle $M=T^* P$. 
Notice, however, that this problem has a lot of symmetry. We can translate both points, rotate them, look at them from a moving frame, etc., all without changing the problem. So, it is natural to want to study not the space M itself, but to quotient it out by the action of some Lie group G. 
Apparently, $M/G$ (or something closely related; see Ben's comment below) will still be a symplectic manifold, but it is not usually the cotangent bundle of anything. The difference is significant: in particular, the canonical one-form on $M=T^*P$ will not necessarily descend to $M/G$.<|endoftext|>
TITLE: Tips on cohomology for number theory
QUESTION [34 upvotes]: I am curious about what is a good approach to the machinery of cohomology, especially in number-theoretic settings, but also in algebraic-geometric settings.
Do people just remember all the rules and go through the formal manipulations of the cohomology groups of  class field theory mechanically, or are people actually "feeling" what is going on here. If it is the latter, could you give a list of mnemonics, or cheat-sheet, or a little fairy tale involving all the characters, so that it is easy to associate with the abstraction. Is there some strong intuition coming from algebraic-topology that would help here? Do people think of Cech covers when they do diagram chasing on crazy grids of exact sequences?
I personally was extremely happy with the central simple algebra approach to class field theory, because it gave me a whole new kind of creature, a CSA, which I could learn to adapt to and love. On the other hand, I don't see myself ever making friends with a 2-cocycle.

REPLY [10 votes]: "Do people just remember all the rules and go through the formal manipulations of the cohomology groups of class field theory mechanically, or are people actually "feeling" what is going on here. If it is the latter"
Let me be one to, if not advocate, at least defend the formal manipulation point of view.  Perhaps ironically, given this, this will be an extremely hand-wavy response:  Cohomology problems emanate from failure of exactness of a functor (i.e., "something you want to do to something else"), e.g., $A\rightarrow A^G$ for a $G$-module $A$.  So you start with a short exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$, apply the "fixed by G" functor, and lo and behold, instead of a 0 on the right, you need a $H^1(G,A)$.  Forgetting completely about what this object means/represents/is, it is the thing that is standing in the way of exactness.  Then you go and notice that Hilbert's Theorem 90, or some other amazing result, tells you that the $H^1$ that you just ran into is zero.  Voila!  Exactness.
Now.  Sometimes you find that $H^1$'s aren't always zero -- sad but true.  So you start studying $H^1$'s in their own right.  Some times these are manageable, or have been previously calculated, and some times not.  Then you find, by looking at the long exact sequences in cohomology, that you could figure out an $H^1$ you need to know by looking at an $H^2$ (maybe you want an $H^1(G,C)$ given an $H^1(G,B)$ and an $H^2(G,A)$).  And then lo and behold -- $H^2(G,A)$ happens to be a Brauer group or something else well-studied.  Knowing this $H^2(G,A)$ trickles down to give you newfound knowledge of $H^1(G,C)$, which in turn gives you information about some failure of exactness on $H^0$'s (say, via some new short exact sequence $0\rightarrow C\rightarrow D\rightarrow E\rightarrow 0$ for which one would want $H^1(G,C)$), which is what you were trying to understand in the first place ("The house that Jack built" comes to mind).  All this not having any idea what $H^2$ is!  
And the process doesn't stop there -- $H^3$'s help you control $H^2$, which in turn control $H^1$'s, etc.  I've never run into an $H^4$ in the wild, but they're not that scary for exactly this reason.  They're just the thing standing in the way of an $H^3$ computation, and fit into the same exact sequences everything else does -- one leg at a time.
In any case, my point is not that you shouldn't try to understand cohomology groups on an intuitive level.  It's that you shouldn't wait for a complete understanding of cohomology groups before you play around with the theorems to see what these groups are good for.  The two should be learned in tandem, maybe even with a preference going toward being able to use them over being able to intuitively understand them.<|endoftext|>
TITLE: What does "quantization is not a functor" really mean?
QUESTION [41 upvotes]: The answers to this question do a good job of exploring, at a heuristic level, what "quantization" should be.  From my perspective, quantization involves replacing a (commutative) Poisson algebra by some related noncommutative associative algebra.  Poisson algebras arise naturally especially as algebras of functions in geometry and physics.  Noncommutative algebras arise naturally as  algebras of operators on linear spaces.
I've often heard it said that "quantization is not a functor".  I'm wondering what a precise statement of this is.
For example, I could imagine statements of the following form.

There is no functor from the category of Poisson manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.
There is no functor from the category of symplectic manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.
Recall that for any smooth manifold, its cotangent bundle is naturally symplectic.  There is no functor from the category of smooth manifolds to the category of associative algebras that quantizes the cotangent bundle.
Recall that the dual to the universal enveloping algebra of a Lie bialgebra is naturally Poisson Hopf.  There is no functor from the category of Lie bialgebras to the category of Hopf algebras satisfying some nice property.

Actually, 4. is false.  Indeed, Etingof and Khazdan constructed a functor from bialgebras to Hopf algebras satisfying a host of properties, and Enriquez classified all the ones with nice properties.  Note that Kontsevich does give a quantization of any Poisson manifold, but perhaps his isn't functorial?

REPLY [23 votes]: This is meant to explain a bit more some of the things that were already mentioned before. 
Quantization of Lie bialgebras is indeed a functor, as was shown in my work with Kazhdan. However, the Konstevich or Fedosov deformation quantization are not functors in this sense. In fact, there is not even a functorial deformation quantization of the symplectic plane, which is sometimes referred to as "Groenewald-van Hove theorem", mentioned in one of the previous answers. The essence of this theorem is that there is no way to quantize the symplectic plane preserving all its symmetries, i.e. symplectic diffeomorphisms. Speaking algebraically, the Lie algebra of infinitesimal symmetries of the symplectic plane is $L_0=C^\infty(\Bbb R^2)/\Bbb R$, and the Lie algebra of symmetries of the (say, Moyal) quantization is the quotient of the quantized algebra by its center, $L=C^\infty(\Bbb R^2)[[h]]_\ast/\Bbb R[[h]]$, with bracket $[a,b]:=(a\ast b-b\ast a)/h$. Now, $L$ is a flat deformation of $L_0$, but the key point is that it is ${\bf nontrivial}$. This nontriviality (which is the gist of the Groenewald-van Hove theorem) is the obstruction to existence of a quantization functor. 
However, as was mentioned in a previous answer, a Drinfeld associator does, according to Tamarkin, give a map between sets of isomorphism classes of Poisson structures and star products (which is not a functor, however, since we have modded out by automorphisms).<|endoftext|>
TITLE: Simplicial homotopy book suggestion for HTT computations
QUESTION [19 upvotes]: I'm struggling through Lurie's Higher Topos Theory, since it appears that someone reading through the book is expected to be somewhat comfortable with simplicial homotopy theory.  The main trouble I've had is computing things like the join, product, coproduct, pullback, pushout, and so forth.  I understand them as far as their universal properties, and maybe have a little intuition because the category of simplicial sets is a presheaf category over the simplex category, but Lurie uses geometrical language, so I can't even compute like when working with presheaves.  So, could you lot recommend some books or lecture notes, preferably with suggested sections, that won't go too deep into simplicial homotopy theory, but deep enough for me to learn how to compute?
Note: Most of the time I waste is sitting around with that book is trying to make sense of the computations, while I understand the arguments just fine.  So please, don't suggest an entire book detailing all of simplicial homotopy theory from start to finish.  I have a goal in mind here, and I'm only trying to learn as much as is necessary for me to continue reading HTT.

REPLY [5 votes]: Echoing what Urs said, "But you should all be using the nLab more: if you want literature on quasi-categories, look up the references section on quasi-categories! :-) " and in regard to your question in particular, "See the nLab page on simplicial sets for links."
By looking at this nLab page on simplicial sets, in the references section you'll have found (as of this date, July 21, 2011 10:19 pm, EST):
Greg Friedman: An elementary illustrated introduction to simplicial sets 
which is chock full of pictures illustrating the geometric ideas underlying the combinatorics of simplicial sets. As Friedman discusses in the introduction on the second and third pages of  this paper, 

"Here, for the most part, you won't
  find many complete proofs of theorems,
  and so these notes will not be
  completely self-contained. Rather, I
  try  primarily to show by example how
  the very basic combinatorics,
  including the definitions, arise out
  of geometric ideas and to show the
  geometric ideas underlying the most
  elementary proofs and properties."

Suffice it to say, those notes may be the end of your search in finding a concise introduction to simplicial sets that also helps develop your geometric intuition and the computation of products. You also might find the references I gave in answer to the question here helpful.<|endoftext|>
TITLE: Favorite popular math book
QUESTION [86 upvotes]: Christmas is almost here, so imagine you want to buy a good popular math book for your aunt (or whoever you want). Which book would you buy or recommend?
It would be nice if you could answer in the following way:
Title: The Poincaré Conjecture: In Search of the Shape of the Universe
Author: Donal O'Shea 
Short description: The history of the Poincaré Conjecture.
(Perhaps something like "difficulty level": + (no prior knowledge of math, as the book mentioned above), ++ (some prior knowledge of math is helpful), +++ (Roger Penrose: Road to Reality (?))
I hope this is appropriate for MO, since I think is of interest to mathematicians (at least for those who want to buy a popular math book for some aunt :-) ).

REPLY [2 votes]: Title: Prisoner's Dilemma 
Author: William Poundstone
Short description (from New York Times Book Review): The real originality of PRISONER'S DILEMMA lies in its colorful synthesis of logical material and historical and biographical narration [which] takes us in parallel lines through cold war history, strategic games of the nuclear age and the life of von Neumann . . . Lively, open and multifaceted.
Indeed, the book can be read as a whole or just by following one of those "parallel lines". Depending of the line you choose, the level of difficulty would be + or ++!<|endoftext|>
TITLE: When is a Riemannian manifold an open subset of a complete one?
QUESTION [18 upvotes]: Under what conditions can a Riemannian manifold be embedded isometrically as a submanifold of a complete one of the same dimension?    There should some kinds of necessary conditions.    For instance, any ball in $M$ (considered as a metric space) must be totally bounded.  Is this sufficient? 
I am curious because it seems that many theorems are stated and proved only for the complete case, and I was wondering how to what extent they could be generalized using a completion tool (if it existed).
Also, is there any kind of uniqueness (there is for $C^{\omega}$ manifolds--implied by the Myers-Rinow theorem)?

REPLY [3 votes]: I doubt it's possible to give necessary and sufficient conditions for an incomplete Riemannian manifold to be embeddable in a complete Riemannian manifold of the same dimension. It's too easy to construct incomplete Riemannian manifolds that do terrible things at its metric boundary.
You need to have some control over the topology and metric near the metric boundary.
Some possible sufficient conditions that come to mind:
a) The metric completion of the incomplete Riemannian manifold is a smooth Riemannian manifold with boundary.
b) The metric completion of the incomplete Riemannian manifold is smooth Riemannian manifold without boundary.
I don't even know how to resolve the following simple case (a point singularity): an incomplete Riemannian manifold with bounded sectional curvature whose metric completion is the manifold plus one additional point. It's easy enough to give examples where completion is not a complete Riemannian manifold. But what I don't know how to do, even in this example, is how to give necessary and sufficient conditions for the completion to be a complete Riemannian manifold.<|endoftext|>
TITLE: cut elimination
QUESTION [11 upvotes]: What is the cut rule? I don't mean the rule itself but an explanation of what it means and why are proof theorists always trying to eliminate it? Why is a cut-free system more special than one with cut?

REPLY [11 votes]: Cut elimination is indispensable for studying fragments of arithmetic. Consider for example the classical Parsons–Mints–Takeuti theorem:
Theorem If $I\Sigma_1\vdash\forall x\,\exists y\,\phi(x,y)$ with $\phi\in\Sigma^0_1$, then there exists a primitive recursive function $f$ such that $\mathrm{PRA}\vdash\forall x\,\phi(x,f(x))$.
The proof goes roughly as follows. We formulate $\Sigma^0_1$-induction as a sequent rule
$$\frac{\Gamma,\phi(x)\longrightarrow\phi(x+1),\Delta}{\Gamma,\phi(0)\longrightarrow\phi(t),\Delta},$$
include axioms of Q as extra initial sequents, and apply cut elimination to a proof of the sequent $\longrightarrow\exists y\,\phi(x,y)$ so that the only remaining cut formulas appear as principal formulas in the induction rule or in some axiom of Q. Since other rules have the subformula property, all formulas in the proof are now $\Sigma^0_1$, and we can prove by induction on the length of the derivation that existential quantifiers in the succedent are (provably in PRA) witnessed by a primitive recursive function given witnesses to existential quantifiers in the antecedent.
Now, why did we need to eliminate cuts here? Because even if the sequent $\phi\longrightarrow\psi$ consists of formulas of low complexity (here: $\Sigma^0_1$), we could have derived it by a cut
$$\frac{\phi\longrightarrow\chi\qquad\chi\longrightarrow\psi}{\phi\longrightarrow\psi}$$
where $\chi$ is an arbitrarily complex formula, and then the witnessing argument above breaks.
To give an example from a completely different area: cut elimination is often used to prove decidability of (usually propositional) non-classical logics. If you show that the logic has a complete calculus enjoying cut elimination and therefore subformula property, there are only finitely many possible sequents that can appear in a proof of a given formula. One can thus systematically list all possible proofs, either producing a proof of the formula, or showing that it is unprovable. Again, cut elimination is needed here to have a bound on the complexity of formulas appearing in the proof.
Sigfpe wrote above in his answer that cut elimination makes proofs more complex, but that’s not actually true: cut elimination makes proofs longer, but more elementary, it eliminates complex concepts (formulas) from the proof. The latter is often useful, and it is the primary reason why so much time and energy is devoted to cut elimination in proof theory. In most applications of cut elimination one does not really care about having no cuts in the proof, but about having control of which formulas can appear in the proof.<|endoftext|>
TITLE: Easiest way to determine the singular locus of projective variety & resolution of singularities
QUESTION [7 upvotes]: For an affine variety, I know how to compute the set of singular points by simply looking at the points where the Jacobian matrix for the set of defining equations has too small a rank. 
But what is the corresponding method for a variety that is a projective variety,and also a variety is a subset of a product of some projective space and affine space? The way I can think of is covering it by sets that are affine, and doing it for each affine set in this open cover - but that seems tedious for practical purposes (but fine for theoretical definitions & theoretical properties). 
Also for resolution of singularities, what is a simple method that is guaranteed to work? The way suggested in the definitions in Hartshorne and other books, is to blow up along the singular locus, then look at the singular locus of the blow-up, and blow up again, and so on - is that guaranteed to terminate? What are some more efficient methods? I have looked at the reference "Resolution of Singularities", a book by someone - that's what he also seems to suggest (though his proof is very general, and I didn't read all of it).

REPLY [7 votes]: The other answers are already very good. A few additional notes:
(1) As others have explained, the Jacobian ideal method works for projective space. It also works for other toric varieties. Any smooth toric variety can be written as $(\mathbb{C}^n \setminus \Sigma)/(\mathbb{C}^*)^k$ where $\Sigma$ is an arrangement of linear spaces and $(\mathbb{C}^*)^k$ acts on $\mathbb{C}^n$ by some linear representation. For example, $\mathbb{P}^n = (\mathbb{C}^{n+1} \setminus \{ 0 \} )/\mathbb{C}^*$. So you can use Greg's trick of unquotienting, using the ordinary Jacobi criterion, and ignoring singularities on $\Sigma$. See David Cox's notes (Erratum to "The Homogeneous Coordinate Ring of a Toric Variety", along with the original paper) for how to write a toric variety in this manner.
(2) For varieties of dimension greater than $1$, resolution of singularities is very computationally intensive. It has been implemented in Macaulay 2, but it tends to tax the memory resources of the system.<|endoftext|>
TITLE: Is the tangent space functor from PD formal groups to Lie algebras an equivalence?
QUESTION [16 upvotes]: The previous version of this question was rather badly broken, and I hope this version makes some sense.
There have been a few questions on MathOverflow about how much representation-theoretic information is lost when passing from a Lie group to its Lie algebra, e.g., away from the semisimple case, Lie algebras have many more representations.  In the algebraic setting, there is an intermediate construction between an algebraic group and its Lie algebra, given by the formal group.  One completes the algebraic group along the identity to get a formal scheme equipped with a group law, and one can pass from there to the tangent space to get the Lie algebra.  In characteristic zero, the tangent space functor is an equivalence of categories from formal groups to Lie algebras, but in positive characteristic, formal groups form an honest intermediate category since the tangent space can lose a lot of information.  For example, there is only one isomorphism class of one-dimensional Lie algebra, but one-dimensional formal groups have a rich arithmetic theory, with a moduli space stratified by positive integer heights.  The completions at the identity of the additive group and the multiplicative group have very distinct formal group structures, and one way to explain the lack of isomorphism is by the presence of denominators in the usual logarithm and exponential power series.
It seems to me that in positive characteristic, there could be an intermediate construction between formal groups and Lie algebras, given by passing to PD rings and replacing the coordinate ring of the formal group with the divided power envelope of the identity section.  If I'm not mistaken, this construction yields a group object in PD formal schemes.
Here is a bit of explanation for the uninitiated (see Berthelot-Ogus for more): PD rings are triples $(A,I,\gamma)$, where $A$ is a commutative ring, $I$ is an ideal, and $\gamma = \{ \gamma_n: I \to A \}_{n \geq 0}$ is a system of divided power operations.  I think they arose when Grothendieck tried to get De Rham cohomology to give the expected answers for proper varieties in characteristic $p$, since the naïve definition tended to yield infinite dimensional spaces.  There is a forgetful functor $(A,I,\gamma) \mapsto (A,I)$ from PD rings to ring-ideal pairs, and it has a left adjoint, called the divided power envelope.  In characteristic zero, $\gamma$ is canonically given as $\gamma_n(x) = x^n/n!$, so both functors are equivalences in that case.  The notion of PD ring can be sheafified and localizations have canonical PD structures, so one has notions of PD scheme and PD formal scheme.

Question: Do PD formal groups contain any more information than the underlying Lie algebra?

I have a suspicion that the answer is "no" and the answer to the title question is "yes".  Vague word-association suggests that the divided power structure is exactly what one needs to get a formal logarithm, but maybe there is a more fundamental obstruction.
I was originally motivated by the question of how Gelfand-Kazhdan formal geometry would differ in charateristic $p$ if I switched between ordinary and PD structures (cf. David Jordan's question).  Unfortunately, I was laboring under some misconceptions about formal completions, and I'm still a bit confused about the precise structure of the automorphism group of the completion (PD or ordinary) of a smooth variety at a point in characteristic $p$.

REPLY [6 votes]: I think that this MR0277590 (43 #3323)
André, M.
Hopf algebras with divided powers.
J. Algebra 18 1971 19--50 may be relevant. It says that a graded commutative divided power Hopf algebra is the co-enveloping algebra of a graded Lie algebra. I think that the grading could be replaced by a condition of completion instead which would give your correspondence.<|endoftext|>
TITLE: (infinity,1)-categories directly from model categories
QUESTION [9 upvotes]: Edit & Note: I'm declaring a convention here because I don't feel like trying to fix this in a bunch of spots:  If I said model category and it doesn't make sense, I meant a model-category "model" of an (infinity,1)-category.  Also, "model" in quotes means the English word model, whereas without quotes it has do do with model categories.
At the very beginning of Lurie's higher topos theory, he mentions that there is a theory of $(\infty,1)$-categories that can be directly constructed by using model categories.  
What I'd like to know is: 
Where can I find related papers (Lurie mentions two books that are not available for download)?
How dependent on quasicategories is the theory developed in HTT?  Can the important results be proven for these $(\infty,1)$-model-categories by proving some sort of equivalence (not equivalence of categories, but some weaker kind of equivalence) to the theory of quasicategories?
When would we want to use quasicategories rather than these more abstract model categories?  
And also, conversely, when would we want to look at model categories rather than quasicategories?  
Does one subsume the other?  Are there disadvantages to the model category construction just because it requires you to have all of the machinery of model categories? Are quasicategories better in every way?
The only "models" of infinity categories that I'm familiar with are the ones presented in HTT.

REPLY [6 votes]: Here is an extended comment regarding Charles answer, including some more references.
In this paper, Theorem 2.5.9, it is shown that every model category (not necessarily a combinatorial one) has all limits and colimits. However, it is not hard to find examples of model categories whose underlying $\infty$-categories are neither presentable nor co-presentable. For instance, Isaksen's strict model structure on pro-simplicial sets. It is shown in the paper mentioned above that the underlying $\infty$-category of this model category is the pro category of spaces considered in Lurie's "Higher Topos Theory" Definition 7.1.6.1. The pro category of a large cocomplete and finitely complete $\infty$-category is complete and cocomplete but neither presentable nor copresentable. 
In this paper, Proposition 1.5.1, it is shown that any Quillen pair between model categories (not necessarily combinatorial ones) gives rise to an adjoint pair of $\infty$-categories. 
It seems plausible to me that the (underlying $\infty$-category of the) relative category of model categories and left Quillen functors between them, with weak equivalences taken to be the Quillen equivalences is equivalent to the $\infty$-category of complete and cocomplete $\infty$-categories and left adjoints between them. Restricting to combinatorial model categories would correspond to restricting to presentable $\infty$-categories in the image. The latter statement is probably already proven, whereas I am pretty sure the former is not.<|endoftext|>
TITLE: Smallest k-term AP of primes
QUESTION [6 upvotes]: Let $S(k)$ denote the smallest integer such that there exists a k-term arithmetic progression of primes among the integers $[1,S(k)]$. Green and Tao have an unpublished note that gives a very large upper bound for $S(k)$. Conversely, an elementary argument (sketched below) gives the lower bound $S(k) > (k-1)e^{k}$. My question is, can this lower bound be improved at all?
By considering how the AP reduces mod $p$, it is easy to see that if $q$ is the step of a $k$-term AP of primes then every prime less than $k$ must divide $q$.  Thus we have that $S(k) > (k-1) \prod_{p_{i} \leq k } p_{i}$. By taking logarithms and applying the prime number theorem we have that $\prod_{p_{i} \leq k } p_{i} ~ e^{k}$. This gives $S(k) > (k-1)e^{k}$.
Update: As Thomas Bloom points out, the inequality I record isn't true for small $k$. I should have written something more along the lines of $S(k) >(k-1)e^{k-o(k)}$ or $S(k) = \Omega((k-1)e^{k})$ since we know that $psi(x) > x$ infinitely often. The spirit of my question is, are there any arguments (presumably using more sophisticated machinery) that produces a better lower bound (here, I'll accept either bounds that hold for all k, or just infinitely many k).

REPLY [3 votes]: I doubt that you can do substantially better than $\text{log}~S(k) > k$; actually, Granville believes that $\text{log}~S(k) = O(k~\text{log}~k)$.
"Combinatorially speaking," the given lower bound seems asymptotically tight (i.e., if we only use local information and PNT). But heuristically the combinatorial arguments seem to work really well when it comes to questions about additive patterns -- I'm sure there are counterexamples, but I don't know them off the top of my head.
Of course it may not be hard to improve on the $(k-1)$ factor.<|endoftext|>
TITLE: Embeddings of $S^2$ in $\mathbb{CP}^2$
QUESTION [29 upvotes]: Suppose we are given an embedding of $S^2$ in $\mathbb{CP}^2$ with self-intersection 1. Is there a diffeomorphism of $\mathbb{CP}^2$ which takes the given sphere to a complex line?
Note: I suspect that either it is known that there is such a diffeomorphism, or the problem is open. This is because if there was an embedding for which no such diffemorphism existed, you could use it to produce an exotic 4-sphere. To see this, reverse the orientation on $\mathbb{CP}^2$ then blow down the sphere.
EDIT: for a counter-example, it is tempting to look for the connect-sum of a line and a knotted $S^2$. The problem is to prove that the result cannot be taken to a complex line. For example, the fundamental group of the complement $C$ is no help, since it must be simply connected. This is because the boundary of a small neighbourhood $N$ of the sphere is $S^3$ and so $\mathbb{CP}^2$ is the sum of $N$ and $C$ across $S^3$ and so in particuar $C$ must be simply-connected.

REPLY [4 votes]: Take a self-intersection one $S^2$ in $\mathbb CP^2$ -- it appears to me that several people have made the observation in this thread that the complement is simply-connected.  If you don't see this right away then a nice way to see this is that the unit normal bundle is $S^3$ and the Hopf fibration $S^3 \to S^2$ then gives a relator that kills the $S^2$-linking elements of $\pi_1$ but these generate $\pi_1 (\mathbb CP^2 \setminus S^2)$ (generalized Wirthinger presentation). So Poincare/Alexander duality tells you $\mathbb CP^2 \setminus S^2$ is contractible.  So "blowing-down" on this embedded $S^2$ basically amounts to a twisted sphere construction -- gluing two discs $D^4$ together along their common boundary.   There's the issue of whether or not one of the discs has an exotic smooth structure or not, but that's the only issue. 
So I guess there's a question lurking in this -- is there a "natural" way to unknot a self-intersection one embedded $S^2$ in $\mathbb CP^2$?   If there were, you couldn't produce any interesting $S^4$'s via this construction.
I've seen this kind of "unnatural unknotting" phenomena with other constructions -- deform-spinning / twist-spinning knots is another example.<|endoftext|>
TITLE: On limits and Colimits
QUESTION [8 upvotes]: I want to ask a stupid question. Let $I$ be an infinite set and suppose $i$ belongs to $I$. I wonder whether following morphisms exist in general:
Hom($A$,colim $B_i) \to$ lim Hom($A,B_i$) and 
Colim Hom($A,B_i) \to$ Hom($A$,colim $B_i$)
What I know is: if we replace lim by infinite product and colim by infinite coproduct, it exists. But I am not sure in this general case above.

REPLY [21 votes]: For any diagram $B_i$ and an object $A$ in a category, there are natural maps of sets:

colim Hom($A,B_i) \to$ Hom($A$, colim $B_i$)
colim Hom($B_i,A) \to$ Hom(lim $B_i, A$)

These maps need not be isomorphisms, in general (neither even when the diagram is filtered, nor when it is finite).  Nor are they isomorphisms for infinite products and coproducts, in general (for finite products and coproducts in an additive category they are isomorphisms, though).
Besides, for any diagram $B_i$ and an object $A$ there are natural isomorphisms of sets:

Hom($A$, lim $B_i$) = lim Hom($A,B_i$)
Hom(colim $B_i, A$) = lim Hom($B_i,A$)

These isomorphisms hold for any diagram (it does not have to be filtered, nor does it have to be finite).  Actually, they hold by the definition of lim and colim.

REPLY [3 votes]: In general, your category has to admit small limits for that to even start to begin to make any sense at all.  Also, as I noted in my question, I'm fairly sure that you've got it backwards.  It should be:
Hom(colim(F(-)),X) is isomorpic to lim Hom(F(-),X), and Hom(Y,lim(F(-))) is isomorphic to lim Hom(Y,F(-)), where we're limiting and colimiting over the domain of F, where F is a functor into our category from some other category (Diagrams for example.)
I don't know if this is what you actually wanted, but if I read your question the way you typed it out, the first one doesn't make sense, since covariant hom is covariant.  The second one might be true provided that the limit has certain restrictions on it or if covariant hom has an appropriate adjoint.  There might be other cases, but it's not true in general.  If you're just looking for the existence of a map in the second one, then it's trivial.<|endoftext|>
TITLE: Graded commutativity of cup in Hochschild cohomology
QUESTION [15 upvotes]: I am trying to get used to Hochschild cohomology of algebras by proving its properties. I am currently trying to show that the cup product is graded-commutative (because I heard this somewhere); however, my trouble is that I have no idea what the exact conditions are for this to hold.
As always in Hochschild cohomology, we start with an algebra A and an A-bimodule M. Is A supposed to be unital? In the original article by Hochschild, it is not, and the proofs would even become harder if we suppose it to be. On the other hand, I have some troubles working with non-unital algebras - they seem just so uncommon to me. Is it still standard to use non-unital algebras 60 years after Hochschild's articles?
Anyway, this is not the main problem. The main problem is that I have no idea what we require from A and M. Of course, M should be an A-algebra, not just a module, for the cup product to make sense. Now:

Must A be commutative?
Must M be commutative?
Must M be a symmetric A-bimodule? (That is, am = ma for all a in A and m in M.)

I have some doubts that if we require this all, then we still get something useful (in fact, the most common particular case of Hochschild cohomology is group cohomology, and it is as far from the "symmetric A-bimodule" case as it can be), but I may be completely mistaken.
As there seem to be different definitions of Hochschild cohomology in literature, let me record mine:
For a k-algebra A and an A-bimodule M, we define the n-th chain group $C^n \left(A, M\right)$ as $\mathrm{Hom}\left(A^{\otimes n}, M\right)$, with differential map
$\delta : C^n \left(A, M\right) \to C^{n+1} \left(A, M\right)$,
$\left(\delta f\right) \left(a_1 \otimes ... \otimes a_{n+1}\right) = a_1 f\left(a_2 \otimes ... \otimes a_{n+1}\right)$
$ + \sum\limits_{i=1}^{n} \left(-1\right)^if\left(a_1 \otimes ... \otimes a_{i-1} \otimes a_{i}a_{i+1} \otimes a_{i+2} \otimes ... \otimes a_{n+1}\right) + \left(-1\right)^{n+1} f\left(a_1 \otimes ... \otimes a_n\right) a_{n+1}$.
The cohomology is then the homology of the resulting complex, and if M is an A-algebra, the cup product is given by
$\left(f\cup g\right)\left(a_1 \otimes ... \otimes a_{n+m}\right) = f\left(a_1 \otimes ... \otimes a_n\right) g\left(a_{n+1} \otimes ... \otimes a_{n+m}\right)$.
I am aware of this article by Arne B. Sletsjøe, but it defines Hochschild cohomology differently (by using $\left(-1\right)^{n+1} a_{n+1} f\left(a_1 \otimes ... \otimes a_n\right)$ in lieu of $\left(-1\right)^{n+1} f\left(a_1 \otimes ... \otimes a_n\right) a_{n+1}$); this definition is only equivalent to mine if M is a symmetric A-module, so it won't help me find out whether this is necessary to assume.
Thanks for any help, and sorry if the counterexamples are so obvious that I am an idiot not to find them on my own...

REPLY [11 votes]: The cup product in Hochschild cohomology$H^\bullet(A,A)$ is graded commutative for all unitary algebras. If $M$ is an $A$-bimodule, then the cohomology $H^\bullet(A,M)$ with values in $M$ is a symmetric graded bimodule over $H^\bullet(A,A)$.
(If $M$ itself is also an algebra such that its multiplication map $M\otimes M\to M$ is a map of $A$-bimodules, then in general $H^\bullet(A,M)$ is not commutative (for example, take $A=k$ to be the ground field, and $M$ to be an arbitrary non-commutative algebra! I do not know of a criterion for commutatitivity in this case)
These results originally appeared in [M. Gerstenhaber, The cohomology structure of an associative ring, Ann. of Math. (2) 78 (1963), 267–288.], and they are discussed at length in [Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory.  Deformation theory of algebras and structures and applications (Il Ciocco, 1986),  11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.] Both references give proofs of a rather computational nature.
You can find an element-free proof of the graded commutativity in this paper, which moreover applies to the cup-products of many other cohomologies.
As for what happens with non-unital algebras, I do not know. But they are very much used today as they were before. One particular context in which they show up constantly is in the intersection of K-theory and functional analysis, where people study `algebras' which are really just ideals in rings of operators of functional spaces---one egregius example is the algebra of compact operators in a Hilbert space.
By the way, you say that group cohomology is a special case of Hochschild cohomology: it is only in a sense... There is a close relationship between the group cohomology $H^\bullet(G,\mathord-)$ of a group $G$, and the Hochschild chomology $H^\bullet(kG,\mathord-)$ of the group algebra, but they are not the same. You can use the second to compute the first (because more generally if $M$ and $N$ are $G$-modules, then $\mathord{Ext}_{kG}^\bullet(M,N)=H^\bullet(kG, hom(M,N))$, where on the right we have Hochschild cohomology of the group algebra $kG$ with values in the $kG$-bimodule $\hom(M,N)$), but the «principal» group cohomology $H^\bullet(G,k)$ is only a little part of the «principal» Hochschild cohomology $H^\bullet(kG,kG)$.
Finally, in the paper by Sletsjøe the definition for the boundary is given as you say because he only considers commutative algebras and only principal coefficients, that is $H^\bullet(A,A)$.<|endoftext|>
TITLE: Higher genus closed string B-model
QUESTION [15 upvotes]: The closed string A-model is mathematically described by Gromov-Witten invariants of a compact symplectic manifold $X$. The genus 0 GW invariants give the structure of quantum cohomology of $X$, which is then an example of a so-called Frobenius manifold. The mirror genus 0 B-model theory on the mirror manifold (or Landau-Ginzburg model) is usually described, mathematically, in terms of variation of Hodge structure data, or some generalization thereof.
Since the higher genus A-model has a nice mathematical description as higher genus GW invariants, I am wondering whether the higher genus B-model has a nice mathematical description as well. Costello's paper on TCFTs and Calabi-Yau categories gives a partial answer to this. One can say that GW theory is the study of algebras over the (homology) operad of compactified (Deligne-Mumford) moduli space; I say that Costello gives a partial answer because he only gives an algebra over the operad of uncompactified moduli space. Though, according to Kontsevich (see Kontsevich-Soibelman "Notes on A-infinity..." and Katzarkov-Kontsevich-Pantev), we can extend this to the operad of compactified moduli space given some assumptions (a version of Hodge-de Rham degeneration). There are also various results (e.g. Katzarkov-Kontsevich-Pantev, Teleman/Givental) which say that the higher genus theory is uniquely determined by the genus 0 theory. But --- despite these sorts of results, I still have not seen any nice mathematical description of the higher genus B-model which "stands on its own", as the higher genus GW invariants do. I have only seen the higher genus B-model described as some structure which is obtained formally from genus 0 data, or, as in the situation of Costello's paper, a Calabi-Yau category, e.g. derived category of coherent sheaves of a Calabi-Yau manifold, matrix factorizations category of a Landau-Ginzburg model, etc.
So, my questions are: 

Are there any mathematical descriptions of the higher genus closed string B-model which "stand on their own"? What I mean is something that can be defined without any reference to other structure, just as genus 47 GW invariants can be defined without any reference to genus 0 GW invariants or the Fukaya category.
Genus 0 theory is essentially equivalent to the theory of Frobenius manifolds. What about higher genus theory? Is there any nice geometric structure behind, say, the genus 1 theory, analogous to the Frobenius manifold structure that we get out of genus 0 theory?

REPLY [6 votes]: One thing missing from this discussion is the even-more-mysterious
holomorphic ambiguity (not "anomaly").  BCOV is not deterministic,
and should probably be thought of as part of a general schema for a
B-model-type topological theory.  What I mean is that there are
other solutions to the BCOV equation which do not yield GW
invariants.  The reason is that BCOV determines the partition
function only up to a holomorphic function at each genus.  The choice of this
function -- analogous to initial conditions -- is usually set
by matching to GW invariants or by imposing some structure at
the singularities of CY moduli space.  
I don't think there is a comprehensive understanding of how
this ambiguity is resolved, though there are cases where it
is fixed by other symmetries or properties of the full
partition function.<|endoftext|>
TITLE: The inverse limit of locally free module
QUESTION [6 upvotes]: A is an I-adic complete Noetherian ring. M is a finitely generated A module. For any n>0, $M/I^nM$ is a finitely generated locally free A/I^n-module. Is M necessarily a locally free A-module?

REPLY [4 votes]: The answer is yes.
1) $A$ is $I$-adic complete implies that $I \subset rad(A)$, the intersection of all maximal primes. Indeed, pick any $a \in I$. Look at $1-a + a^2 -a^3 ... \in A $ (this is where we use completeness). The inverse of this is $1+a$, so $1+a$ is an unit. Since this is true for any $a\in I$, we have $I\subset rad(A)$ (see Section 1 Matsumura). 
2) It suffices to prove that $M$ is free at any maximal ideal $m$ of $A$. Since $I$ is inside $m$, we may as well replace $A$ by $A_m$ and assume $A$ is local. By assumption then $M/I \cong (A/I)^l$, Nakayama Lemma shows that $l$ is the mimimum number of generators of $M$.
Look at the beginning  of a minimal resolution of M:  $ N \to F=A^l \to M $.The last map become isomorphism when tensoring with $A/I^n$, so the first map has to become $0$. This means that  $N \subset I^nF$ for all $n>0$. This forces $N=0$ (use Artin-Rees lemma), thus $M \cong F$.<|endoftext|>
TITLE: Ramified cover of 4-sphere
QUESTION [13 upvotes]: Is it true that any closed oriented $4$-dimensional manifold can be obtained as a result of the following construction: 
Take $S^4$ with a finite collection of immersed closed 2-manifolds (with transversal intersections and self-intersections) and construct ramified cover of $S^4$ with a ramification of order at most 2 only at these submanifolds.
Comments:

Two related questions: Ramified covers of 3-torus, Ramified covers of $S^n$
According to Feighn's Branched covers according to J.W. Alexander any closed oriented 4-manifold is a branched cover of $S^4$ with a ramification along 2-skeleton of 4-tetrahedron embedded in $S^4$ (which is not at all a 2-manifold).

REPLY [11 votes]: The answer is yes, at least if we interpret your phrase "ramification of order 2" to mean "simple branched covering". See Piergallini, R., Four-manifolds as $4$-fold branched covers of $S^4$. Topology 34 (1995), no. 3, 497--508. Any closed, orientable PL 4-manifold can be expressed as a 4-fold simple branched covering of S4 branched along an immersed surface with only transverse double points. It is apparently still an open question whether the branch set can be chosen to be nonsingular. A simple branched covering of degree d is a branched covering in which each branch point is covered by d-1 points, only one of which is singular, of local degree 2.<|endoftext|>
TITLE: Finding cocycles that square to zero
QUESTION [14 upvotes]: Suppose $x$ is a chosen class in the singular cohomology (integer coefficients) of a space $X$. I'm thinking primarily of classes of odd degree on a simply connected  space. What are necessary conditions (besides $x^2=0$) for the existence of a cocycle representing $x$ whose cup-square equals zero as a cocycle? Sufficient conditions?
Take your pick of the precise form of the question: you can fix a cochain model for cup products before or after choosing $x$, or even allow a DGA quasi-isomorphic to the singular cochains on $X$.
You may feel inclined to mutter "Steenrod square" or "Massey product" - but which, and why?

REPLY [5 votes]: I don't know what your application is, so the below may not be sufficient, but it was fun to think about.

The issue we run into here is that being (respresentable by) a square-zero cocycle is not a homotopical condition, and is certainly not invariant under quasi-isomorphisms. For instance, $C^*(S^3)$ contains no square-zero 3-cocycles representing top cohomology: if $\sigma:\Delta^3 \to S^3$ is a singular 3-simplex  with $\sigma$ constant on $\partial \Delta^3$, on which $\phi(\sigma) = 1$, then we can always find a 6-simplex $\overline \sigma$ with $\overline \sigma_{[0, 3]} = \sigma$ and $\overline \sigma_{[3, 6]} = \sigma$. (Choose a projection $\Delta^6 \to \Delta^3 \vee \Delta^3$ and compose with the map $\sigma \vee \sigma$, which makes sense because $\sigma$ is constant on the boundary.) Then $(\phi \cup \phi)(\overline \sigma) = 1$. Of course, if you instead work with the quasi-isomorphic dg-algebra of simplicial cochains, every representative is square-zero because $C^6_\Delta(S^3) = 0$.
A more homotopical condition is being coherently square zero (in the literature, being a 'twisting element', though this is a very special case of a general theory).  If $[x_1]$ is a cohomology class of odd degree $2n+1$, then a coherently square-zero extension is a sequence of cochains $x_2, x_3, \cdots$ with $|x_i| = 2ni + 1$, so that $$dx_k + \sum_{\substack{i+j = k \\ i,j \geq 1}} x_i x_j = 0.$$
There are two natural ways to come up with this formula.
First: consider the dg-algebra $\Bbb Z[2n+1]$, which is a copy of $\Bbb Z$ in degree 3 with zero differential and zero product. Then a dg-algebra homomorphism $\Bbb Z[2n+1] \to C^*(X;\Bbb Z)$ is precisely a square-zero cocycle. When we want more "homotopical" conditions, we learn to generalize from dg-algebra maps to $A_\infty$ maps. And the above is (maybe up to sign, I didn't check) the same thing as an $A_\infty$-homomorphism $\Bbb Z[2n+1] \to C^*(X;\Bbb Z)$.
Second: you might be looking for a "twisted differential" on the cochain complex $C^*(X;\Bbb Z) \otimes \Bbb Z[T, T^{-1}]$ which incorporates cup-product with $x_1$. The first formula you would guess is $d_{tw} = d + L_1 T$, where $L_1(y) = x_1 y$. The conditon $d_{tw}^2 = 0$ is equivalent to the demand that $dx_1 = 0$ and $x_1^2 = 0$.
When you give up on finding $x_1^2 = 0$, you might try to correct for this by adding higher terms. You could then guess that the right differential is $d_{tw} = d + L_1 T + L_2 T^2 + \cdots$ where $L_m(y) = x_m y$. The claim that $d_{tw}^2 = 0$ is now precisely the formulae I wrote above.

(There is also a notion of homotopy of these. A non-trivial but not too difficult fact is that if $A$ and $B$ are algebras equipped with a cup-1 product satisfying the Hirsch formula, and if $f: A \to B$ is a quasi-isomorphism of dg-algebras which preserves the cup-1 product, then $f$ induces a bijection on the sets of twisting sequences as above up to homotopy.)


Theorem. Suppose $X$ is a topological space with $H^*(X;\Bbb Z)$ is torsion-free. Then if $c \in H^{2n+1}(X;\Bbb Z)$ is an odd class, there is always a coherently square-zero extension $(x_1, x_2, \cdots) \in \prod_{i \geq 1} C^{2ni+1}(X;\Bbb Z)$ with $[x_1] = c$.

More generally --- as in Mariano's nice answer --- all we need to demand is that all Massey powers $\langle c, \cdots, c\rangle$ is well-defined and equal to zero.
Proof: Construct this by induction.
Choose $x_1$ to be any cocycle with $[x_1] = c$. Inductively suppose that we have constructed $x_1, \cdots, x_m$ with $$dx_k + \sum_{\substack{i+j = k \\ i,j \geq 1}} x_i x_j = 0$$ for all $k \leq m$. Then by explicit computation the term $$\sum_{\substack{i+j = m + 1 \\ i,j \geq 1}} x_i x_j$$ is a cocycle, and referring to Kraines, "Massey higher products", Section 3, you see that this is a representative for the so-called Massey power $\langle c\rangle^{m+1}$.
By Kraines, "Massey higher products", Theorem 15, these vanish in rational cohomology; by naturality of Massey products under change-of-coefficients and injectivity of the map $H^*(X;\Bbb Z) \to H^*(X;\Bbb Q)$, it follows that these vanish integrally. Thus there exists an $x_{m+1}$ so that $$dx_{m+1} + \sum_{\substack{i+j = m + 1 \\ i,j \geq 1}} x_i x_j = 0,$$ completing the induction.

Notice that the simplest case of this is that $c^2 = 0$ in rational cohomology because $c^2$ is 2-torsion. The Massey powers are a sequence of higher torsion obstructions to such an extension.
One can further analyze the obstructions to such sequences being homotopic, but I will not do so here. A special case is that if $x_1^2 = y_1^2 = 0$ in your dga with cup-1 products satisfying the Hirsch formula, then if you have torsion-free cohomology then in fact $(x_1, 0, \cdots)$ and $(y_1, 0, \cdots)$ are homotopic as twisting sequences if and only if $[x_1] = [y_1]$. To manage the higher cases one has to also ask that certain higher cohomology classes are equal, where the higher cohomology classes are made out of iterated cup-1 products of lower ones.

In particular you can extend every odd cohomology class on $SU(n)$ to a twisting sequence. (If you're trying to extend the 3-dimensional generator and you work with simplicial cochains, I can prove that you can find a representative with $x_k = 0$ for $k \geq n$, but this is a little tedious.)
The argument I outlined above that you will never be able to construct an honest square-zero cocycle in the singular chains on $S^3$ works just as well to show you won't be able to do that in the singular chains on $SU(n)$. I don't think you can do it in simplicial chains on $SU(n)$ for $n > 2$, either, but I haven't thought about that so much.<|endoftext|>
TITLE: Knots that unknot in a manifold
QUESTION [15 upvotes]: Take a closed $n$-manifold $M$ and fix a nice $n$-ball $B$ in $M$. How much information about $M$ does the set of knotted $(n-2)$-spheres of $B$ which are unknotted in $M$ give?

REPLY [4 votes]: Ryan gives a very nice answer in dimension 3, leaving the higher-dimensional case of the question open.  I can't discuss the question with as much authority as I would like, but I can start to piece together an answer.  My impression is that in any dimension $n$, an $(n-2)$-knot $K$ that is unknotted in $M$ is already unknotted in $B$ in the continuous category.  In the smooth and PL categories, it should be the same except in $n=4$ dimensions, where the question runs into a lack of understanding of smooth (or equivalently PL) structures on 4-manifolds.
First, my impression from skimming some geometric group theory is that a finitely presented group cannot be an infinite free product.  Moreover, that it if it is a finite free product, the factors have unique isomorphism types.  If this impression is correct, then you can identify the knot group $\pi_1(S^n \setminus K)$ from the embedded fundamental group $\pi_1(M \setminus K)$.  In order to be trivial in $M$, the knot group of $K$ would have to be $\mathbb{Z}$.
Then, Aspherical manifolds and higher-dimensional knots, by Beno Eckmann, begins:

E. Dyer and R. Vasquez proved that the complement of a higher-dimensional
knot $S^{n-2} \subset S^n$, $n \ge 4$, is never aspherical unless the knot group is infinite cyclic (and hence, for $n \ge 5$, the imbedding is unknotted). [Footnote referring to papers of Levine, Stallings, Wall, Shaneson.] In the present note we give a simple proof of this fact based on some remarks concerning compact $\partial$-manifolds.

My impression is that the surgery methods in dimension $n \ge 5$ used to prove the unknottedness assertion are valid in the smooth, PL, and continuous categories.  Dimension $n=3$ is of course a special case where you do not need to consider algebraic topology, but instead prove things with direct cut-and-paste arguments as Ryan describes.  In dimension $n=4$, the surgery theory is (a) much more difficult in the continuous category, and (b) non-existent in the smooth/PL category.
The paper The algebraic characterization of the exteriors of certain 2-knots, by Jonathan Hillman, credits Freedman with the result that a 2-knot with knot group $\mathbb{Z}$ is unknotted.  If this requires Freedman's work, then I would think that it is open in the smooth/PL case.  What would be open is whether the 4-manifold with boundary $B^3 \times S^1$ has more than one smooth structure.  If it does, I have seen a principle that smooth structures on a 4-manifold can merge together when you take a connected sum with manifolds such as $\pm \mathbb{C}P^2$ an $S^2 \times S^2$.  I do not know if this principle is established for manifolds with a fundamental group.  But at the very least, I have heard that the semigroup of smooth structures on $S^4$ is completely unknown, and its action on the smooth structures on another 4-manifold is completely unknown.  So if the smooth Poincare conjecture is sufficiently false, presumably you could have a non-trivial 2-knot in $B$ that becomes trivial in $M$.<|endoftext|>
TITLE: What are natural transformations in 1-categories?
QUESTION [5 upvotes]: It's well-known that, for lots of concrete categories (but by no means all), we can think of the objects as themselves being small categories, and morphisms are the functors between these categories. Examples include Grp, Ab, Top... When we apply such a construction, we turn a 1-category into a (strict, I think?) 2-category. But 2-categories carry some extra structure, namely the notion of a natural transformation. When we "decategorify" back down, where does this extra structure go?
I can work it out in some specific cases; for instance, if we categorify Ab in the obvious way, there are no nontrivial natural transformations. I don't have a characterization for when two morphisms of groups are naturally isomorphic as functors between the underlying categories, although I have a feel for how the question behaves.
Are there any sort of general results on what natural transformations between morphisms look like if we categorify thusly? Is it at least independent of how we realize the objects as small categories? (I suspect the answer to the second question is "no," but don't have the skills to construct a counterexample. I hope I'm wrong, though.)
More generally, if we can categorify an n-category into an (n+k)-category forgetting the higher morphisms, do the higher morphisms go downstairs to the n-category in any nice way?

REPLY [5 votes]: Here is a counterexample for your next-to-last question.  Let S be a set with more than one element and consider the two full subcategories of Cat on, respectively, the single category which is the discrete category on S, and the single category which is the codiscrete category on S.  In each case, when viewing Cat as a 1-category, the resulting full subcategory has a single object with endomorphisms Hom(S, S).  However, if we view Cat as a 2-category, the former subcategory has no nontrivial natural transformations and thus really is BHom(S, S), while the latter has a unique natural transformation between any two functors and thus is actually • up to 2-equivalence.
Cat-the-1-category and Cat-the-2-category are very different constructs which unfortunately usually go by the same name.  Even though they have "the same" objects, I suggest thinking of their objects as being different kinds of things.  An object of Cat-the-1-category has more information than an object of Cat-the-2-category; we may talk about the cardinality of its set of objects, not just the cardinality of its set of isomorphism classes of objects.  (This shouldn't seem too strange, since an object of Cat-the-0-category is a "specific" category, of which we may talk about the actual set of objects.)  Put differently, an object of Cat-the-1-category is a "monoid with many objects", while an object of Cat-the-2-category is what we more often think of when thinking about categories (especially large ones).
In your example, you expressed Ab as a full subcategory of Cat-the-1-category.  The full subcategory of Cat-the-2-category on the same objects is not Ab, since it has nontrivial natural automorphisms, as others have pointed out.  It only becomes Ab after truncation—replacing each Hom-category by its set of isomorphism classes of objects.  For Grp, the situation is worse, since distinct group homomorphisms may be naturally isomorphic as functors.  The usual way to repair this is to work with "pointed categories", as described at this nlab page.  But of course this is a kind of extra structure on a category, and if I'm allowed to introduce arbitrary extra structure then the question is too easy.  Anyways, I'm not sure that one should expect various concrete categories to naturally be full subcategories of either Cat-the-1-category or Cat-the-2-category.<|endoftext|>
TITLE: Local-globalism for similar matrices?
QUESTION [11 upvotes]: My background on number theory is very weak, so please bear with me...
Given two matrices $A$ and $B$ in $\mathbb{Z}^{n\times n}$. Assume that for every prime $p$, the images of $A$ and $B$ in $\mathbb{F}_p^{n\times n}$ are similar to each other. Does this yield the existence of a matrix $X\in\mathrm{SL}_n\left(\mathbb{Z}\right)$ satisfying $AX=XB$ ? What if we additionally assume $A$ and $B$ to be similar in $\mathbb{Q}^{n\times n}$ ?

REPLY [11 votes]: The answer is no. Here is a counter-example
$$\left( \begin{matrix} 0 & -5 \\\\ 1 & 0 \end{matrix} \right) \quad \mbox{and} \quad \left( \begin{matrix} -1 & -3 \\\\ 2 & 1 \end{matrix} \right).$$
Both of these matrices have characteristic polynomial $x^2+5$. For $p \neq 2$, $5$, this polynomial has no repeated factors so any matrices with this polynomial are similar. By the same argument, they are similar over $\mathbb{Q}$. By brute force computation, they are also similar at $2$ and $5$. 
However, they are not similar over $\mathbb{Z}$. Consider $\mathbb{Z}^2$ as a module for $\mathbb{Z}[t]$ where $t$ acts by one of the two matrices above. Both of these matrices square to $-5$, so these are in fact $\mathbb{Z}[\sqrt{-5}]$-modules. If the matrices were similar, the similarity would give an isomorphism of $\mathbb{Z}[\sqrt{-5}]$-modules. But these are not isomorphic: the former is free on one generator while the latter is isomorphic to the ideal $\langle 2, 1+ \sqrt{-5} \rangle$.
In general, the way to classify similarity of matrices over $\mathbb{Z}$ is the following: If the matrices do not have the same characteristic polynomial over $\mathbb{Q}$, they are not similar. If they do, let $f$ be the characteristic polynomial and let $R=\mathbb{Z}[t]/f(t)$. Then your matrices give $R$-modules, and the matrices are similar if and only if the $R$-modules are isomorphic. If $R$ is the ring of integers of a number field, then $R$-modules are classified by the ideal class group. In general, they are related to the ideal class group, but there are various correction factors related to how $R$ fails to be the ring of integers of its fraction field (or how it fails to be a domain at all). I don't know the details here.<|endoftext|>
TITLE: Generating ribbon diagrams for knots known to be ribbon knots
QUESTION [6 upvotes]: Is there a source in the literature for ribbon diagrams for the knot-table knots known to be ribbon knots?
For example, I'm interested in doing a computation which needs as input a ribbon diagram for the knot $8_{20}$ (Rolfsen knot table notation).  This knot is known to be ribbon, but I don't know a ribbon diagram for the knot.  
Usually when I encounter a claim of the sort "knot X is ribbon" either the author supplies the ribbon diagram, or nothing.  Citations to information of this sort seem kind of sparse. Or am I just unaware of a standard source for this type of information?

REPLY [3 votes]: See also 'A refined Jones polynomial for symmetric unions', Michael Eisermann and Christoph Lamm, Osaka J. Math. (2011), https://projecteuclid.org/euclid.ojm/1315318344
All prime ribbon knots up to 10 crossings are given as symmetric diagrams (examples 1.14, 6.6, 6.7 and 6.8, see especially the table on page 363) which are simpler than the diagrams in Kawauchi's book.
Update 2022: My paper "The search for nonsymmetric ribbon knots" contains ribbon diagrams for all 11 and 12 crossing (prime, ribbon) knots.<|endoftext|>
TITLE: Topological version of Bogomolov’s question
QUESTION [16 upvotes]: I'm quoting a question from p. 753 of Gromov's recent paper Singularities, Expanders and Topology of Maps:

Does there exist, for every closed oriented $n$-manifold $X_0$, a closed oriented $n$-manifold $X$ that admits a map
$X \to X_0$ of positive degree and, at the same time, can be smoothly fibered over some
$Y$ with $dim(Y) = n − 2$ ? (Bogomolov’s original question concerns parametrization
of complex algebraic manifolds $X_0$ by algebraic manifolds fibred by surfaces.)

I'm wondering where Bogomolov's question came from, and for what cases it is known and why it may be of interest?
Gromov doesn't give a reference, and searching for "Bogomolov conjecture" seems to bring up a different conjecture. Maybe an expert can point me quickly to a reference.
Also, if anyone has partial answers to Gromov's question, I would be interested. I think I can prove it for 3-manifolds, and there may be a 4-manifold topologist who might know something about the 4-dimensional case.
Addendum: I'll add my argument in the 3-dimensional case, then ask a question which would answer Gromov's question in dimension 4.
Any orientable 3-manifold is a cover of $S^3$ branched over the figure 8 knot $K$. The 3-fold cyclic cover branched over $K$ is a Euclidean manifold, which has a finite-sheeted cover which is a 3-torus. The preimage of the branched locus is geodesic, so we may find a product fibering of the 3-torus by tori transverse to the preimage of the branched locus. Thus, any covering of this cover will also fiber (by taking the preimage of the fibering). For a 3-manifold, we may take the common branched cover over $K$ with the 3-torus branched cover. This fibers since it covers the 3-torus, and has a non-zero degree (in fact bounded degree) map to the 3-manifold.
As mentioned in an answer to Ramified cover of 4-sphere, any closed orientable PL 4-manifold is a branched cover over $S^4$ with branched locus a surface. One could try to generalize my 3-dimensional argument to this context. The key fact is that there is a cover of $S^3$
branched over the figure 8 knot which has a fibration transverse to the branched locus. So I ask the question: for any surface in $S^4$, is there a cover of $S^4$ branched over the surface which has a fibration by surfaces which are transverse to the preimage of the branched locus? I have no evidence for or against this question, it just seems like a natural generalization of the 3-D case.
One could also ask the analogous question in higher dimensions and in the algebraic category. But one would need to know the corresponding branched covering result first. For projective algebraic varieties, maybe it would be natural to consider branched covers over projective space, analogous to Belyi's theorem.

REPLY [6 votes]: There is strong circumstantial evidence that the question came from a personal conversation with Bogomolov.  First, Gromov's paper has more than 50 references, but he doesn't cite Bogomolov at all.  Second, he does later in the paper thank Bogomolov for a helpful conversation, referring to him in nickname form as Fedya.  They are about the same age, they both have appointments at NYU, and Gromov has thanked Bogomolov in papers before.  Third, when I skimmed over Bogomolov's papers (okay, the abstracts and some introductions), I couldn't find anything directly bearing on the question.
On the other hand, I did find a paper by Carlson and Toledo that said that 30 years ago, Gromov introduced the partial ordering on $n$-manifolds given by the existence of a map $M \to N$ with non-zero degree.  The broad question is what this partial ordering looks like.  Obviously Gromov norm is an important invariant for this question.
Unless you ask Gromov or Bogomolov directly, it may make more sense to make an educated guess based on Gromov's summary.  I would say, if $X$ is a closed, smooth, $n$-dimensional algebraic variety over $\mathbb{C}$, is there a closed, smooth $n$-fold $Y$ with a non-constant morphism $Y \to X$, such that $Y$ also fibers smoothly over a complex curve? Unless there is some clear reason that this question is misstated, it surely works as motivation for Gromov's question.<|endoftext|>
TITLE: Categorical foundations without set theory
QUESTION [47 upvotes]: Can there be a foundations of mathematics using only category theory, i.e. no set theory? More precisely, the definition of a category is a class/set of objects and a class/set of arrows, satisfying some axioms that make commuting diagrams possible. So although in question 7627, where psihodelia asked for alternative foundations for mathematics without set theory, there Steven Gubkin said that Lawvere and McCarthy did some work in reformulation the set theoretic axioms ZFC as the axioms of elementary topoi, this manner of foundations is still not complete since a category is still ultimately a set!
J Williams in his answer below noted that via metacategories, we can have a first order axiomatization of categories. However, this does not provide a foundations of mathematics using only category theory, since set theory permeates the formulation of first order logic. In first order logic, structures are sets together with constants, functions and relations. Here constants, functions and relations are also sets. So even if we say that categories are first order axiomatizable, at the very end, categories are still defined in terms of sets.
I admit in wanting foundations totally in terms of categories, then there will be some kind of recursiveness. However, this recursiveness should not be seen as a problem since as described above, first order axiomatization of sets like ZFC, are written in the language of first order logic which (at least in a meta-level) are sets themselves. In fact, this recursiveness is very much a feature of symbolic logic and is partially responsible for the successful that a single primitve concept of set/set-membership can describe so much (or all?) of mathematics.
I'm aware also in certain proofs of equivalence of categories in mainstream math, like GAGA theorems by Serre, there is a need to use categories where the objects are of classes of different levels, like the NBG set theory. In the end, the reasons provided for why the argument of using classes can be pushed down to essentially small category, this in the end invokes NBG set theory.

REPLY [26 votes]: Since Tom Leinster queries my reference to actual/completed versus
potential/incomplete infinities, maybe we should ask a philosopher
whether I am using these terms in the standard way.
In any case, I am not doing metaphysics. I am just describing the way in
which it appears to me that mathematicians actually work,
in contrast to the way they say they work because they have
been trained to say such things.
When you compare my remarks with the others on this page,
please note that they are based on thinking about these things for
myself over 25 years, originally from a categorical perspective
but increasingly influenced by symbolic logic,
and not on reciting bits of textbooks.
To do ordinary arithmetic, you may need very (arbitrarily) large
numbers, but you don't all of them together.
So far as I can gather from history, mathematicians up to the mid-19th
century managed very well to deal with things in this way,
for example defining functions as expressions.
Post-Cantor, 20th century mathematicians got into the habit of
introducing the completed infinity before the structure.
For example, we say "a group is a set with...",
relegating the essence of symmetry to second place.
This is like saying that humanity is a collection of pieces of flesh,
onto which faces are painted as an afterthought.
Categorists, being part of the pure mathematical culture,
did the same thing, in the vast majority of cases with great profit.
However, when it comes to foundations, treating the universe
first as a completed infinity (and only afterwards containing
products, function-spaces, powersets or whatever other structure
you require) inevitably leads into the set-theoretic trap.
By contrast, type theoretic methods build up the universe
by means of the actual operations that you actually want to 
consider, just as the symmetry group of the Rubik cube is built
up from individual rotations.  Moreover, despite the fact that
type theory looks completely different from category theory or
algebra, it is an accurate underpinning of the actual methods
of reasoning of mathematics.  See, for example, my discussion
of the idiom "there exists" in
my book.
This is not dogmatic Finitism or Logicism and is readily 
adaptable to considering the object $\bf N$ along with individual
natural numbers, an internal category $\bf Set$ along with 
individual types, and so on.
Now let me consider the other approaches to this question.
First order logic.
This was the first usable general technique
in mathematical logic. Like other disciplines, it starts with the
completed infinity and adds properties to it.
Does it presuppose a set theory? Well, yes, in the same sense
that a boot-loader presupposes a primitive operating system.
I would be more convinced that first-order logic is independent
of set theory if there were a branch of model theory that had
examples of structures whose carriers were topological spaces
or algebraic varieties.
In fact, first order logic can be set up in the type-theoretic
way that I have described above.
But if you're going to do that, you may as well set up the type
theory that you actually want to use.
If we're looking for a metalanguage specificaly for
categorical logic
(say, in which to construct toposes) then first order logic is
not the right structure.  It is easy to describe an internal
category in a category with all finite finits, and, by adding
more diagrams, we can talk about internal toposes too.
However, it's much more interesting to consider free
internal structures, for which we need an arithmetic universe,
although unfortunately there is next to zero literature on this topic.
Fibrations, 2-categories, etc.
None of what I have said 
contradicts the use of these categorical techniques.
I personally consider that fibrations,
and especially hyperdoctrines,
are obfuscation, but other people find them useful.
However, they organise the world,
but they do not bring it into existence,
which was the thrust of the original question.<|endoftext|>
TITLE: Justifying a theory by a seemingly unrelated example
QUESTION [15 upvotes]: Here is a topic in the vein of  Describe a topic in one sentence and  Fundamental examples  : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but  which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model.   
A classical example is Galois theory for solving polynomial equations. 
Any examples for homological algebra ? For Fourier analysis ? For category theory ?

REPLY [4 votes]: Proving the termination of Goodstein sequences (a problem in natural numbers) via arithmetic on infinite ordinals.<|endoftext|>
TITLE: The density hex
QUESTION [21 upvotes]: Gale famously showed that the determinacy of n-player, n-dimensional Hex is equivalent to the Brouwer fixed point theorem in n dimensions.
We can (and Gale does) view this as saying that if you d-color the vertices of a certain graph  specifically, the graph with vertex set $[n]^d$ and two vertices $v, w$ adjacent iff the max norm of $v - w$ is 1 and all the nonzero components of $v - w$ have the same sign -- then there's a certain monochromatic path. Alternatively, you can think of d-coloring a d-dimensional $n \times \ldots \times n$ cube, and the determinacy of Hex/Brouwer fixed-point says that a certain "twisted path" must exist.
Here's what I want to know: 

Is there a topological proof of the density version of the determinacy of Hex? 

The density version ends up following from density Hales-Jewett, since combinatorial lines are paths in the underlying graph. But density Hales-Jewett is hard, and this seems like it should admit a proof along the lines of Gale's.
What I mean by the "density version" is: for any $\delta > 0$, and fixed n, for sufficiently large dimension d any choice of $\delta n^d$ moves must connect two opposite sides of the hypercube/d-dimensional Hex board. (I'm fairly sure this is the correct statement, but it's possible I'm wrong. Let me know if this is for some reason utterly trivial or false.)

REPLY [6 votes]: For a closely related question when you do not insist that all non zero components of v-w has the same sign, then the answer is known: See the following paper: B. Bollobas, G. Kindler, I. Leader, and R. O'Donnell, Eliminating cycles in the discrete torus, LATIN 2006: Theoretical informatics, 202{210, Lecture Notes in Comput. Sci., 3887, Springer, Berlin, 2006. Also: Algorithmica 50 (2008), no. 4, 446-454. This Graph is referred to as G_\inf and there is a beautiful new proof via the Brunn Minkowski's theorem by Alon and Feldheim. For this graph a rather strong form of a density result follows, and the results are completely sharp.
The paper by Alon and Klartag http://www.math.tau.ac.il/~nogaa/PDFS/torus3.pdf is a good source and it also studies the case where we allow only a single non zero coordinate in v-u. An even sharper result is given in another paper by Noga Alon. There, there is a log n gap which can be problematic if we are interested in the case that n is fixed and d large. See also this post.
As Harrison points out, the graph he proposes (that we can call the Gale-Brown graph) is in-between the two graphs. So the unswer is not known but we can hope that some discrete isoperimetric methods can be helpful. 
The statement is an isoperimetric-type result so this can be regarded as a quantitative version of the topological notion of connectivity. 
Two more remarks: 1) The Gale result seems to give an example of a graph where there might be a large gap between coloring number and fractional coloring. This is rare and an important other example is the Kneser graph where analyzing its chromatic number is a famous use (of Lovasz) of a topological method.
2) Hex is closely related to planar percolation and the topological property based on planar duality is very important in the study of planar percolation and 1/2 being the critical probability. (See eg this paper) It seems that we might have here an interesting high dimensional extension with some special significance to chosing each vertex with probability 1/d.<|endoftext|>
TITLE: Examples of algebraic closures of finite index
QUESTION [29 upvotes]: So there are easy examples for algebraic closures that have index two and infinite index: $\mathbb{C}$ over $\mathbb{R}$ and the algebraic numbers over $\mathbb{Q}$. What about the other indices?
EDIT: Of course $\overline{\mathbb{Q}} \neq \mathbb{C}$. I don't know what I was thinking.

REPLY [51 votes]: Theorem (Artin-Schreier, 1927): Let K be an algebraically closed field and F a proper subfield of K with $[K:F] < \infty$.  Then F is real-closed and $K = F(\sqrt{-1})$.
See e.g. Jacobson, Basic Algebra II, Theorem 11.14.

REPLY [30 votes]: The Artin-Schreier theorem says that every algebraic closure of finite index has index 2, and it's the algebraic closure of a real-closed field.
Page 299 of Algebra by Serge Lang.  Google Books Link  to the page.<|endoftext|>
TITLE: Cohomology rings and 2D TQFTs
QUESTION [19 upvotes]: There is a "folk theorem" (alternatively, a fun and easy exercise) which asserts that a 2D TQFT is the same as a commutative Frobenius algebra. Now, to every compact oriented manifold $X$ we can associate a natural Frobenius algebra, namely the cohomology ring $H^\ast(X)$ with the Poincare duality pairing. Thus to every compact oriented manifold $X$ we can associate a 2D TQFT.
Is this a coincidence? Is there any reason we might have expected this TQFT to pop up?
When $X$ is a compact symplectic manifold, perhaps the appearance of the Frobenius algebra can be explained by the fact that the quantum cohomology of $X$, which comes from the A-twisted sigma-model with target $X$, becomes the ordinary cohomology of $X$ upon passing to the "large volume limit".
But for a general compact oriented $X$? I don't see how we might interpret the appearance of the Frobenius algebra in some quantum-field-theoretic way. Maybe there is an explanation via Morse homology?

REPLY [14 votes]: These 2D TQFTs do not come from extended theories (unless X is discrete). I interpret this as saying that these theories are non-local (in the 2D bordism) and so you will have trouble interpreting them in a traditional QFT framework. You will have to do something funny and non-local, like squashing your circles to points and surfaces to graphs, as in the Cohen work mentioned by Tim.<|endoftext|>
TITLE: Statistics of irreps of S_n that can be read off the Young diagram, and consequences of Kerov–Vershik
QUESTION [25 upvotes]: Alexei Oblomkov recently told me about the beautiful theorem of Kerov and Vershik, which says that "almost all Young diagrams look the same."  More precisely: take a random irreducible representation of $S_n$ (in Plancherel measure, which assigns a probability of $(\dim \chi)^2 / (n!)$ to an irrep $\chi$) and draw its Young diagram, normalized to have fixed area.  Rotate the diagram 45 degrees and place the vertex at the origin of the Cartesian plane, so that it lies above the graph of $y = \lvert x\rvert$.  Then there is a fixed curve, with equation
$$ y = \begin{cases} \lvert x\rvert, &  \lvert x\rvert > 2 \\ \frac 2\pi \left(x \arcsin\frac x2 + \sqrt{4-x^2}\right), & \lvert x\rvert \leq 2 \end{cases}$$
such that the normalized Young diagram is very close to the curve with probability very close to 1.
My question:  what asymptotic statements about irreducible characters of $S_n$ can be "read off" the Kerov–Vershik theorem?  In a sense, this isn't a question about Kerov–Virshik at all, but a question about which interesting statistics of irreducible characters can be read off the shape of the Young diagram.
There are some tautological answers:  for instance, if $f(\chi)$ is the height of the first column of $\chi$, then I guess Kerov-Virshik shows that
$$\frac{f(\chi)}{\sqrt n}$$
is very probably very close to 2 as $n$ gets large (if I did this computation right — in any event it concentrates around a fixed value). But I don't really have in mind any representation-theoretic meaning for $f(\chi)$.

REPLY [6 votes]: A good paper related to this question is P. Biane,  Representations of symmetric groups and free probability , Advances in Mathematics 138, 126-181 (1998); doi: 10.1006/aima.1998.1745, item [4] at http://www.dma.ens.fr/~biane/publi.html (Wayback Machine).<|endoftext|>
TITLE: Why is Lie's Third Theorem difficult?
QUESTION [29 upvotes]: Recall the following classical theorem of Cartan (!):
Theorem (Lie III): Any finite-dimensional Lie algebra over $\mathbb R$ is the Lie algebra of some analytic Lie group.
Similarly, one can propose "Lie III" statements for Lie algebras over other fields, for super Lie algebras, for Lie algebroids, etc.
The proof I know of the classical Lie III is very difficult: it requires most of the structure theory of Lie algebras.
But why should it be difficult?  For example, for finite-dimensional Lie algebra $\mathfrak g$ over $\mathbb R$, the Baker-Campbell-Hausdorff formula (the power series given by $B(x,y) = \log(\exp x \exp y)$ in noncommuting variables $x,y$; it can be written with only the Lie bracket, no multiplication) converges in an open neighborhood of the origin, and so defines a unital associative partial group operation on (an open neighborhood in) $\mathfrak g$.  What happens if one were to try to simply glue together copies of this open neighborhood?
Alternately, are there natural variations of Lie III that are so badly false that any easy proof of Lie III is bound to fail?

REPLY [3 votes]: You can find below the proof due to G.M; Tuynman, of the third Lie theorem.
The proof is similar to using Ado's theorem, but requires an 'advanced' result:
the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.
http://ifile.it/hy0q139
I posted a related question in math.stackexchange.com
https://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem<|endoftext|>
TITLE: Can every manifold be given an analytic structure?
QUESTION [82 upvotes]: Let $M$ be a (real) manifold.  Recall that an analytic structure on $M$ is an atlas such that all transition maps are real-analytic (and maximal with respect to this property).  (There's also a sheafy definition.)  So in particular being analytic is a structure, not a property.
Q1: Is it true that any topological manifold can be equipped with an analytic structure?
Q2: Can any $C^\infty$ manifold (replace "analytic" by "$C^\infty$" in the first paragraph) be equipped with an analytic structure (consistent with the smooth structure)?

REPLY [5 votes]: Let me correct the misconception appearing in Greg's answer, which also made its way into this nLab article:

"Anyway, the result is much harder than what Whitney did, which is a very good but expected application of the Weierstrass approximation theorem."

What one can do with the Weierstrass approximation theorem alone, in combination with Whitney's smooth embedding theorem, is to prove:
If $M$ is a compact stably parallelizable manifold then it can be embedded in some ${\mathbb R}^N$ as a smooth algebraic subvariety.
If one looks at Whitney's proof (incidentally, I am unaware of any textbook treatment of this proof) in
Whitney, H., Differentiable manifolds., Ann. Math., Princeton, (2) 37, 645-680 (1936). ZBL62.1454.01.
it clear that the proof is not an application of Weierstrass' theorem but needs much more delicate approximation arguments which are based on Whitney's earlier work. Whitney is quite clear on this point in the introduction to his paper:

That said, Whitney indeed proves that every smooth $m$-dimensional manifold admits an embedding in ${\mathbb R}^{2m+1}$ such that the image is a real-analytic submanifold.<|endoftext|>
TITLE: Proofs of Bott periodicity
QUESTION [164 upvotes]: K-theory sits in an intersection of a whole bunch of different fields, which has resulted in a huge variety of proof techniques for its basic results.  For instance, here's a scattering of proofs of the Bott periodicity theorem for topological complex K-theory that I've found in the literature:

Bott's original proof used Morse theory, which reappeared in Milnor's book Morse Theory in a much less condensed form.
Pressley and Segal managed to produce the homotopy inverse of the usual Bott map as a corollary in their book Loop Groups.
Behrens recently produced a novel proof based on Aguilar and Prieto, which shows that various relevant maps are quasifibrations, therefore inducing the right maps on homotopy and resulting in Bott periodicity.
Snaith showed that $BU$ is homotopy equivalent to $CP^\infty$ once you adjoin an invertible element.  (He and Gepner also recently showed that this works in the motivic setting too, though this other proof relies on the reader having already seen Bott periodicity for motivic complex K-theory.)
Atiyah, Bott, and Shapiro in their seminal paper titled Clifford Modules produced an algebraic proof of the periodicity theorem. EDIT: Whoops x2!  They proved the periodicity of the Grothendieck group of Clifford modules, as cdouglas points out, then used topological periodicity to connect back up with $BU$.  Wood later gave a more general discussion of this in Banach algebras and Bott periodicity.
Atiyah and Bott produced a proof using elementary methods, which boils down to thinking hard about matrix arithmetic and clutching functions.  Variations on this have been reproduced in lots of books, e.g., Switzer's Algebraic Topology: Homotopy and Homology.
A proof of the periodicity theorem also appears in Atiyah's book K-Theory, which makes use of some basic facts about Fredholm operators.  A differently flavored proof that also rests on Fredholm operators appears in Atiyah's paper Algebraic topology and operations on Hilbert space.
Atiyah wrote a paper titled Bott Periodicity and the Index of Elliptic Operators that uses his index theorem; this one is particularly nice, since it additionally specifies a fairly minimal set of conditions for a map to be the inverse of the Bott map.
Seminaire Cartan in the winter of '59-'60 produced a proof of the periodicity theorem using "only standard techniques from homotopy theory," which I haven't looked into too deeply, but I know it's around.

Now, for my question: the proofs of the periodicity theorem that make use of index theory are in some vague sense appealing to the existence of various Thom isomorphisms.  It seems reasonable to expect that one could produce a proof of Bott periodicity that explicitly makes use of the facts that:

The Thom space of the tautological line bundle over $CP^n$ is homeomorphic to $CP^{n+1}$.
Taking a colimit, the Thom space of the tautological line bundle over $CP^\infty$ (call it $L$) is homeomorphic to $CP^\infty$.
The Thom space of the difference bundle $(L - 1)$ over $CP^\infty$ is, stably, $\Sigma^{-2} CP^\infty$.  This seems to me like a route to producing a representative of the Bott map.  Ideally, it would even have good enough properties to produce another proof of the periodicity theorem.

But I can't find anything about this in the literature.  Any ideas on how to squeeze a proof out of this -- or, better yet, any ideas about where I can find someone who's already done the squeezing?
Hope this isn't less of this is nonsense!
-- edit --
Given the positive response but lack of answers, I thought I ought to broaden the question a bit to start discussion.  What I was originally looking for was a moral proof of the periodicity theorem -- something short that I could show to someone with a little knowledge of stable homotopy as why we should expect the whole thing to be true.  The proofs labeled as elementary contained too much matrix algebra to fit into parlor talk, while the proofs with Fredholm operators didn't seem -- uh -- homotopy-y enough.  While this business with Thom spaces over $CP^\infty$ seemed like a good place to look, I knew it probably wasn't the only place.  In light of Lawson's response, now I'm sure it isn't the only place!
So: does anyone have a good Bott periodicity punchline, aimed at a homotopy theorist?
(Note: I'll probably reserve the accepted answer flag for something addressing the original question.)

REPLY [6 votes]: I like very much the idea of Dusa McDuff to strip the Atiyah-Singer approach of any analysis. In a sense one might say that it is an elaboration on $e^{2\pi i}=1$ :)
She considers restriction of the exponential map $\textrm{Exp}:\mathbb H\to\mathbb U$ from Hermitian operators to unitary operators to the subspace $\mathbb H_{[0,1]}\subset\mathbb H$ of weakly contracting operators (that is, operators with eigenvalues in [0,1]). Fibre over $U\in\mathbb U$ turns out to be the set of those $H\in\mathbb H_{[0,1]}$ for which the space  $\textrm{Fix}(H)$ of fixed points of $H$ is a subspace of $\textrm{Fix}(U)$. (Very roughly, "$e^{2\pi iH}$ is that unitary thing which is like 1 wherever $H$ is like 1".)
Moreover any subspace of $\textrm{Fix}(U)$ can be realized as $\textrm{Fix}(H)$ for a unique weakly contracting $H$ with $\textrm{Exp}(H)=U$. This establishes a homeomorphism between the fibre over $U$ and the set of subspaces of $\textrm{Fix}(U)$, i. e. the Grassmanian of $\textrm{Fix}(U)$.
After some massaging one obtains a (quasi)fibration with base the unitary group, fibre the Grassmanian, and contractible total space.
This is in the last section of McDuff's 1977 paper "Configuration spaces"; a detailed proof with the extension to real Bott periodicity by Behrens is here, although there are some flaws later corrected in an addendum. That proof is itself a cleanup of a 1999 version by Aguilar and Prieto.
PS I later realized that some additional care is needed for the eigenvalue 0 (of $H$)...<|endoftext|>
TITLE: Fukaya categories of hyperkahler reductions: general request for information
QUESTION [10 upvotes]: I'd really like to hear any references or information people have about the Fukaya categories of hyperkahler reductions of vector spaces (for more informations on the varieties, see Nick Proudfoot's thesis Hyperkahler analogues of Kahler quotients).  These are very nice symplectic manifolds (hyperkahler and exact, in particular), so I feel like their Fukaya categories should themselves be nice, but I've never found a good reference on them.
The marquee example of this is Nakajima quiver varieties.  I would be very interested to hear anything about the Fukaya categories of these.
A question of particular interest to me the Floer homology of images of invariant Lagrangian subspaces in the quotient.

REPLY [5 votes]: At the risk of writing things that are obvious to those listening in: this is Nadler-land, no?  
If $X$ is a smooth complex variety with reductive group $G$ acting, and $\mu_{\mathbb C}: T^*X\rightarrow {\mathfrak g}^*$ is the complex moment map, then $\mu_{\mathbb C}^{-1}(0)/G = T^*(X/G)$ provided one interprets all quotients as stacks.  
If $T^*X$ is hyperkahler and we do the hyperkahler quotient for the maximal compact of $G$, picking a nontrivial real moment value $\mu_{\mathbb R}^{-1}(\zeta)$ at which to reduce amounts (by Kirwan) to imposing a GIT stability on $\mu_{\mathbb C}^{-1}(0)$---i.e. to picking a nice open subset of the cotangent stack $T^*(X/G)$ that is actually a variety.  A stack version of Nadler's "microlocal branes" theorem would describe the (suitable, undoubtedly homotopical/derived) exact Fukaya category as the constructible derived category of $X/G$.  
Since I'm completely ignorant of how the Nadler-Zaslow/Nadler story actually works, I'd like to then imagine that such an equivalence microlocalizes properly to give an equivalence over the hyperkahler reduction (i.e. the nice open set) as well?  Admittedly, by microlocalizing to the stable locus one should avoid all the derived unpleasantness (this should be analogous to what happens in Bezrukavnikov-Braverman's proof of "generic" geometric Langlands for $GL_n$ in characteristic $p$, where by localizing to the generic locus, ${\mathcal D}$-module really means ${\mathcal D}$-module, not "module over the enveloping algebroid of the tangent complex" or something like that).
Admittedly, I don't have a clue how to deal with the issue that the base $X$ in the important examples is typically affine...maybe if one forces some kind of boundary conditions also in the $X$-direction one could make the Fukaya category nontrivial in Tim's example of the Hilbert scheme??<|endoftext|>
TITLE: Why do automorphism groups of algebraic varieties have natural algebraic group structure?
QUESTION [17 upvotes]: I am not sure that all automorphism groups of algebraic varieties have natrual algebraic group structure. 
But if the automorphism group of a variety has algebraic group structure, how do I know the automorphism group is an algebraic group.
 For example, the automorphism group of an elliptic curve $A$ is an extension of the group $G$ of automorphisms which preserve the structure of the elliptic curve, by the group $A(k)$  of translations in the points of $A$, i.e. the sequence of groups
$0\to A(k)\to \text{Aut}(A)\to G \to 0$ is exact, see  Springer online ref - automorphism group of algebraic variaties. 
In this example, how do I know $\text{Aut}(A)$ is an algebraic group.

REPLY [25 votes]: This is really a comment on Pete's comment for Mikhail's answer, but I am making it an answer because it raises a question which I think should be more widely known.
The construction of Aut-scheme uses the entire Hilbert scheme, which has countably many components (due to varying Hilbert polynomials), and it is not obvious how many of these intervene in the construction. Mikhail's answer illustrates the basic example showing it can be infinite.  But this leads to the following problem (suggested by what is known about Neron-Severi groups, whose construction encounters the same issue via construction of Pic schemes in terms of Hilbert schemes, at least in the original Grothendieck construction):  
Q. Does the Aut scheme of a proper scheme over an alg. closed field have finitely generated component group? (This is a more basic question than finiteness, which is really Pete's comment to Mikhail's answer.) 
Incredibly, even for smooth projective varieties over  C this appears to be a wide open problem!! I have mentioned this to several experts in alg. geom. (including Oort), and nobody has an idea. If anyone has an idea or a solution, please let me know right away.<|endoftext|>
TITLE: What manifolds are bounded by RP^odd?
QUESTION [38 upvotes]: Real projective spaces $\mathbb{R}P^n$ have $\mathbb{Z}/2$ cohomology rings $\mathbb{Z}/2[x]/(x^{n+1})$ and total Stiefel-Whitney class $(1+x)^{n+1}$ which is $1$ when $n$ is odd, so it follows that odd dimensional ones are boundaries of compact $(n+1)$-manifolds. My question is: are there any especially nice constructions of these $(n+1)$-manifolds? 
I'm especially interested in the case $n=3$. I believe we can get an explicit example of a 4-manifold bound by $\mathbb{R}P^3$ using Rokhlin/Lickorish-Wallace but it doesn't look like that would generalize to higher dimensions at all easily. Are there a lot of different 4-manifolds with this property?

REPLY [7 votes]: Here is a simple general fact: if a manifold $M$ admits a free involution $T$, then $M$ is a boundary.  Proof: $M$ will be the boundary of $W =(M \times [-1,1])/((Tx,t) \sim (x,-t))$.  In your particular case: when $n$ is odd, the standard free action of $C_4$ on $S^n$ induces a free action of $C_2$ on $\mathbb RP^n$.  (Note that this $W$ will tend to be nonorientable, even if $M$ is.  Exercise: let $M = S^1$.)<|endoftext|>
TITLE: Elementary questions in arithmetic geometry
QUESTION [7 upvotes]: In many theories there is a rough divide between elementary problems that can be solved with "one's hands", and "deep results that require powerful tools". For example, I am told that Hodge theory is a such a tool in the topology of complex algebraic varieties.
As somebody that is just beginning to learn about arithmetic questions, it is not clear to me what these divides are (if they exist) in all the fields under the heading of arithmetic geometry. So, I have a few naive questions.

Roughly, what are all of the subfields?
Generally, what type of problem is elementary? 
What are some examples of elementary results?
What are the powerful tools, and how are they broadly used?
What are the overarching principles (e.g. Hasse principle; now calculate obstructions)?

If, like me, you know very little about this field, I can recommend this fantastic article by Jordan Ellenberg and these notes by Bjorn Poonen. Please edit this question if it is nonsensical. Thanks to everybody who responds.

REPLY [9 votes]: Lots of the answers are going to be of the form "read book X" so I'll start:  Serre's Course in Arithmetic is the best conceivable answer to the question "where could I find a concise, clearly written compilation of results in arithmetic geometry that can be proved without advanced machinery?"  It's really a beautiful book.
The kind words about my article are much appreciated.<|endoftext|>
TITLE: Proofs without words
QUESTION [380 upvotes]: Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results?
(One could ask if this is of interest to mathematicians, and I would say yes, in so far as the kind of little gems that usually fall under the title of 'proofs without words' is quite capable of providing the aesthetic rush we all so professionally appreciate. That is why we will sometimes stubbornly stare at one of these mathematical autostereograms with determination until we joyously see it.)
(I'll provide an answer as an example of what I have in mind in a second)

REPLY [8 votes]: Gluing two Mobius strips along their edges is a Klein bottle.<|endoftext|>
TITLE: The "ultimate" indefinite inner product space
QUESTION [6 upvotes]: This can be considered as a relative of Splitting a space into positive and negative parts.
Is there a real (non-trivial) vector space $V$, endowed with a nondegenerate symmetric bilinear pairing $\langle-,-\rangle : V^2 \to \mathbb{R}$, satisfying the following property:
for each function $f \in \mathbb{R}^\mathbb{R}$ with $f(0) = 0$ there is some $F \in V^V$ such that $\langle Fx - Fy, Fx - Fy\rangle = f( \langle x - y, x - y\rangle )$ ($x,y \in V$)?
I'll bet no Krein space can do this.
Somewhat vaguely, the (generally nonlinear) mapping $F$ may be viewed as an "internal realization" of the function $f$.
P.S. By "non-trivial" I mean $V \ne {0}$.
P.P.S. $A^B$ means $\lbrace u | u: B \to A \rbrace$, so that $V^2$ is the Cartesian square (of $V$). Just to avoid any confusion (I hope).

REPLY [5 votes]: I think that it is possible with a large enough vector space $V$.  I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not.  The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.
Version 1
I'll construct $F$ and $V$ together, and then construct the bilinear pairing last.  Let $V_0 = \mathbb{R}$ with its basis vector $1$.  Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$.  If $x \in V_n \setminus V_{n-1}$, let $[x]$ denote the corresponding element in $W_n \subset V_{n+1}$.  Let $V$ be the union of all $V_n$, and let $F(x) = [x]$.  Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V_n$.
To construct the pairing, let $\langle 1,1 \rangle = 1$.  We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$.  I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction.  Also the diagonal values $\langle e, e \rangle$ are unrestricted.  To see this, consider your equation
$$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$
with $x \ne y$.  By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$.  Then $d(x-y) \le n$.  In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.
Version 2
Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$.  In this case, let $W_n$ be the vector space of formal linear combinations of elements of $(V_n \setminus V_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V_{n+1} = V_n \oplus W_n$.  In this case, $W_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$.  For any fixed $f$, define $F(x) = [x,f]$.
As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted.  Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$.  Finally, as before,
$$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$
with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.
Version 3
Ady reminds me that the second version still misses the condition that the bilinear form on $V$ should be non-degenerate.  I think that the same trick works a third time:  We can just enlarge $V$ to also guarantee this condition.  This time let $W_n$ be as in the second version, and let
$$V_{n+1} = V_n \oplus W_n \oplus V_n^*,$$
where $V_n^*$ is the (algebraic) dual vector space to $V_n$.
Define the bilinear form on $V_n \oplus W_n$ as in version 2, and define $F$ as in version 2.  The bilinear form on $V_n^*$ is unrestricted, and so is the bilinear pairing between $V_n^*$ and $W_n$.  Finally the bilinear pairing between $V_n^*$ and $V_n$ should be the canonical pairing $\langle \phi, x \rangle = \phi(x)$.  This guarantees that for every vector $x \in V_n$, there exists $y \in V_{n+1}$ such that $\langle y,x \rangle = 1$.
Every version of the construction is cheap in the sense that the image of $F$ is a linearly independent set. Moreover, in the second and third versions, the image of $F$ is far from a basis.  My feeling is that it is difficult to ask for much better than that in a universal construction.<|endoftext|>
TITLE: Is there a way to check if a relative Hilbert Scheme is reduced?
QUESTION [7 upvotes]: More specifically, suppose I have a rational curve on a complete intersection, and I know that the relative Hilbert Scheme is not smooth at the point corresponding to my rational curve. Is there any algorithm that will eventually tell me whether the Hilbert Scheme is reduced or not there?
Just to make it harder, this curve happens to pass through the singular locus of the complete intersection.

REPLY [7 votes]: Showing nonreducedness of a Hilbert scheme is a hard question in general.  The most direct algorithm would involve producing a Grobner basis for the defining ideal of the component in question, and then computing its radical and seeing if the two are equal.  But with the exception of rather simple cases, this computation is not going to be feasible.  Here are two shortcut answers that I've seen in the past.

You could show that the entire component of the Hilbert scheme in question is nonreduced.  Mumford produces a famous such example, in his "Further Pathologies" paper from 1962.  I have also seen works by J.O. Kleppe and some by Scott Nollet which provide many further examples and techniques.
You could produce a global description of component of the Hilbert scheme in question.  For instance, perhaps the component is parametrized by something that you understand (a Grassmanian, say).  The best example of this that I've ever read is Vainsencher-Avritzer "Compactifying the space of ellpitic quartic curves".<|endoftext|>
TITLE: What are some slogans that express mathematical tricks?
QUESTION [40 upvotes]: Many "tricks" that we use to solve mathematical problems don't correspond nicely to theorems or lemmas, or anything close to that rigorous. Instead they take the form of analogies, or general methods of proof more specialized than "induction" or "reductio ad absurdum" but applicable to a number of problems. These can often be summed up in a "slogan" of a couple of sentences or less, that's not fully precise but still manages to convey information. What are some of your favorite such tricks, expressed as slogans?
(Note: By "slogan" I don't necessarily mean that it has to be a well-known statement, like Hadamard's "the shortest path..." quote. Just that it's fairly short and reasonably catchy.)
Justifying blather: Yes, I'm aware of the Tricki, but I still think this is a useful question for the following reasons:

Right now, MO is considerably more active than the Tricki, which still posts new articles occasionally but not at anything like the rate at which people contribute to MO.
Perhaps causally related to (1), writing a Tricki article requires a fairly solid investment of time and effort. The point of slogans is that they can be communicated without much of either. If you want, you can think of this question as "Possible titles for Tricki articles," although that's by no means its only or even main purpose.

REPLY [5 votes]: "It is better to study a nice category of ugly objects than an ugly category of nice objects"
Examples:

Say you want to study projective varieties, perhaps smooth, you
better study them into the category of schemes which is nicer (you
are allowed to perform more constructions) but some objects look weird  at first glance.

You want to study (finite dimensional) vector bundles over a variety but is is convenient to consider the category of coherent sheaves that it is abelian. Moreover, if you want to take infinite limits, then you generalizer further to the category of quasi-coherent sheaves.

The previous example can be read algebraically: the category of projective finitely generated modules is not abelian but finite presentation modules is.

REPLY [3 votes]: One of my favourite such slogans is "Just do it". This is very well covered on the Tricki:
http://www.tricki.org/article/Just-do-it_proofs
but as you say it is not as active as MO. I first discovered this trick when trying to solve a problem which is now a favourite of mine: Does there exist an enumeration $q_1,q_2,\dots$ of $\mathbb{Q}$ such that the series
$$
\sum_{n=1}^{\infty} (q_{n+1} - q_n)^2 
$$
converges?
Writing the problem here already gives away that the answer is 'yes'
I won't write out a solution but there's a certain kind of solution which, once you've thought of it is very much: Well...just...do it. Just enumerate them so it converges!<|endoftext|>
TITLE: Contracting  maps of hyperbolic manifolds
QUESTION [5 upvotes]: Is it possible for SOME positive $c$, $c<1$ to find a pair of COMPACT hyperbolic manfiolds $M^3$ and $N^3$
with a positive degree map $$f: M^3 \to N^3,$$ such that $f$ is contacting with constant $c$? Are there may examples like this?
One can ask the same question of Riemann surfaces, and it seems to me that this should be possible. For example we can take a double cover of Riemann surface with many points or ramification. Though I don't know a proof even in this case. Of course for non-ramified cover the best possible constant $c$ is $1$. 
ADDED. Following the answer of Sam Need, let me give an approximative "proof" of the fact that this works in dimesnion 2. 
Let us triangulate a hyperbolic surface $N^2$ in triangles of very small size, that have acute angles (this is always possible). We want to show that a double cover of $N^2$ with ramifications at vertices of the triangulation will do the job. For this we need a lemma (without a proof).
Lemma. Suppose we have two hyperbolic trianlges, one very small and acute with angles $a$, $b$, $c$, and the over with angles a/2, b/2, c/2. Then there is a contacting map from the second triangle to the first one. 
The lemma is true, since the second trianlge will be large.
Now on the double cover we can take a trangulation that comes from $N^2$ and glue it from these triangles with half angles. Half angles come from  doble cover. Then we just need to "adjust" the map.
Of course this is not a real proof, but I am 100% it can be made real.

REPLY [5 votes]: In general, for any non-zero degree map from one closed negatively curved manifold to another, there is a canonical map (due to Besson-Courtois-Gallot) called the "natural map". However, it's only known to be pointwise volume decreasing, not necessarily contracting. They call this the "real Schwarz-Lemma". Applying the Schwarz lemma for Riemann surfaces I think gives the contracting map in this case for branched covers. Think of the induced map on the universal cover, which is the unit disk, or $\mathbb{H}^2$. The Schwarz lemma says that any conformal map from the disk to the disk is contracting, unless it's an isometry. 
I thought of one (not very explicit) example in 3-D. Take two simplices in hyperbolic space. There is a canonical affine map (say in the Lorentzian model) taking one simplex to the other. This will be a contracting map for the hyperbolic metric if one simplex sits inside the other [Edit: actually I'm not sure about this now, but in the example below there exists a contracting map]. There are finitely many tetrahedra in $\mathbb{H}^3$ which give rise to fundamental domains for discrete reflection groups (see Ratcliffe). Two of these have one dihedral angle $\pi/5$, with opposite edge angle $\pi/2$ and $\pi/4$, respectively, and all other angles the same. There is a 1-parameter family of polyhedra interpolating between these (basically, just "push" the two faces closer together along the dihedral angle $\pi/5$ edge) which decreases distances. Also, the orbifold fundamental group (i.e. reflection group) from the $\pi/4$ one maps to that of the $\pi/2$ one. So there's a distance decreasing map from one orbifold to the other. Using Selberg's lemma, one may find finite-sheeted manifold covers with the same property.<|endoftext|>
TITLE: Building elliptic curves into a family
QUESTION [11 upvotes]: Suppose $E/ \mathbb{Q}$ is an elliptic curve whose Mordell-Weil group $E(\mathbb{Q})$  has rank r.  When can we realize E as a fiber of an elliptic surface $S\to C$ fibered over some curve, with everything defined over $\mathbb{Q}$, such that the group of $\mathbb{Q}$-rational sections of $S$ has rank at least r?
Edit: Let me also demand that the resulting family is not isotrivial, i.e. the j-invariants of the fibers are not all equal.

REPLY [4 votes]: Evidence that it is not known to always be possible over $\mathbb{QP}^1$: If you look at the records of the elliptic curves with high rank, they are broken down into three categories 

Elliptic curves over $\mathbb{Q}$.
Nonisotrivial curves over $E$, where $E$ is an elliptic curve over $\mathbb{Q}$ with $|E(\mathbb{Q})|$ infinite.
Nonisotrivial curves over $\mathbb{QP}^1$.

If we could always deform, these records would be the same. In fact, according to the tables here and here, the highest known rank of type 1 is 28, of type 2 is 19 and of type 3 is 18.
I do not know whether there are examples where it is known that such a deformation is impossible.<|endoftext|>
TITLE: Legitimacy of reducing mod p a complex multiplication action of an elliptic curve?
QUESTION [10 upvotes]: I scoured Silverman's two books on arithmetic of elliptic curves to find an answer to the following question, and did not find an answer:
Given an elliptic curve E defined over H, a number field, with complex multiplication by R, and P is a prime ideal in the maximal order of H and E has good reduction at P. Is it legitimate to reduce an endomorphism of E mod P?
In the chapter "Complex Multiplication" of the advanced arithmetic topics book by Silverman, a few propositions and theorems mention reducing an endomorphism mod P.
A priori, this doesn't seem trivial to me. Sure, the endomorphism is comprised of two polynomials with coefficients in H. But I still don't see why if a point Q is in the kernel of reduction mod P, why is phi(Q) also there. When I put Q inside the two polynomials, how can I be sure that P is still in the "denominator" of phi(Q)?
(*) I looked at the curves with CM by sqrt(-1), sqrt(-2) and sqrt(-3), and it seems convincing that one can reduce the CM action mod every prime, except maybe in the case of sqrt(-2) at the ramified prime.

REPLY [13 votes]: Let $\mathcal{E}$ denote the Neron model of $E$ over $R$ and $k=R/P$.   Thus $\mathcal{E}$ is the unique (up to isomorphism) smooth commutative group scheme over $R$ with generic fiber $E$ such that for any smooth $X/R$ the natural map $\mathcal{E}(X)\to E(X_H)$ is an isomorphism.  Then "$E$ mod $P$" is the special fiber $\mathcal{E}_k = \mathcal{E}\times_R k$.   (This makes sense and is canonical even if $E$ has bad reduction at $P$.)     Now take $X=\mathcal{E}$ in the definition of Neron model.  Your endomorphism $\varphi$ is just an element of $E(E) = \mathcal{E}(\mathcal{E})$, so extends uniquely to a morphism $\tilde{\varphi}:\mathcal{E} \to \mathcal{E}$.   Base extending $\tilde{\varphi}$ by $k$ yields the reduction $\overline{\varphi}: \mathcal{E}_k \to \mathcal{E}_k$.     In particular, $\overline{\varphi}$ is defined (it is "legitimate").  It even makes sense when the reduction is bad at $P$, but of course then it is a map on special fibers of Neron models.    
The above all makes sense for abelian varieties as well.  If you want to understand any question about reduction of abelian varieties, learning Neron models is a very good idea.   Some references are Silverman's "Advanced topics in the arithmetic of elliptic curves" which you mention above (only for elliptic curves), and Bosch-Lutkebohmert-Raynaud "Neron Models" in general.<|endoftext|>
TITLE: What is a logic?
QUESTION [24 upvotes]: I am not interested in the philosophical part of this question :-)
When I look at mathematics, I see that lots of different logics are used : classical, intuitionistic, linear, modal ones and weirder ones ...
For someone new to the field, it is not easy to really see what they have in common for justifying the use of the word "logic". Is it just because
of a filiation with classical logic ?
I have attempted to find an answer in the literature. Some papers are telling me that a logic is a pre-order. It is not a satisfactory answer to me.
I imagined that it may be related to the use of some specific connectors : but linear logic is telling me it is not so simple.
I imagined that it may be related to some symmetry properties of the rules of the system : but it is dependent on how the logic is formalized.
Then, I had the crazy idea (after discovering the Curry-Howard isomorphism) that it may be related to the computational content of the system. But, it is obviously wrong. 
So, I have not progressed and I am still wondering if there may be a point of view allowing to see what all these systems have in common.
I have avoided the use of the word "truth" in this question. I am expecting a mathematical answer if there is one. There are too many philosophical problems related to the notion of truth.
But, perhaps my question is a naive one ...

REPLY [4 votes]: Here are three research traditions that both illustrate how the problem can be approached, and give rather different perspectives on what counts as a logic.
(this really is just to complement Dan Piponi's answer).

Tarski's consequence relations and abstract algebraic logic

Tarski's basic idea was to define a logic as an abstract pair of the form $\langle \mathcal{F},C\rangle$ where $\mathcal{F}$ is a free algebra of formulas and $C$ is an operator on $\mathcal{P}(\mathcal{F})$ [I write $\mathcal{F}$ for the domain of the free algebra]. For any set of formulas $A$, the set $C(A)\subseteq\mathcal{F}$ is meant to represent the 'consequences' of $A$ -- so that $C$ generates a consequence relation $\vdash_{C}$ defined as $A\vdash_{C} B$ iff $B\subseteq C(A)$. 
Next, Tarski gave several structural conditions that the operator $C$ ought to satisfy in order for the resulting consequence relation to count as well-behaved, or logical (roughly, those conditions consist of reflexivity, monotonicity, compactness, as well as invariance under uniform substitution of variables). See here for more detail.
This view really treats logic as a (very special) branch of abstract algebra. One idea is to try to differentiate between logics, and classify them, by looking at their different algebraic properties. It was one of the earliest systematic attempts at answering the question of 'what a logic is' in such general terms.
On the other hand, this framework is generally too weak to account for quantification of any sort; the attention is almost exclusively restricted to propositional logics.
(NB. Tarski's approach eventually gave rise to some very interesting work on the general process of algebraization of a logic, under the guise of Abstract Algebraic Logic -- see, e.g. here. Interesting monographs on the topic include Rasiowa and Sikorski's An Algebraic Approach to Non-Classical Logics as well as Blok and Pigozzi's Algebraizable Logics.)

Model-theoretic logics, generalized quantifiers

For an introduction see this book. The model-theoretic approach studies various extensions of first-order logic: predominantly infinitary logics of the form $\mathcal{L}_{\alpha\kappa}$, where $\alpha, \kappa$ are ordinals (the logic $\mathcal{L}_{\alpha\kappa}$ allows conjunctions/disjunctions of less than $\alpha$-many formulas, and quantification over less than $\kappa$-many variables). It also covers topics like abstract characterisations of first-order logic (cf. Lindstrom's theorem mentioned in John Goodrick's answer) as well as connections with probabilistic logics. 
There is related research on what makes a quantifier 'logical'. The idea is to characterise the logical quantifiers as operations of a certain type that are invariant under certain groups of transformations -- there appears to be some controversy about what exact transformations truly define the 'logical' operations (see here). 

Applied logics

Another approach that gives a slightly different perspective on what counts as a logic is work in `applied' logic: this is a broad field of study which has at its root a dynamic view of logic (see here). Here, one uses so-called 'dynamic' modal logics to model processes that change over time, such as transitions between the states of a program (see Propositional Dynamic Logic) or informational states of agents (see Dynamic Epistemic Logic). Those logics are studied either for their intrinsic mathematical interest, or can also be applied to the study of information exchange protocols in game theory, cryptography, and various topics in formal philosophy.
The approach here is less algebraic, but focused more on model-theoretic and computational aspects. This research often bears close links to computer science and philosophy.<|endoftext|>
TITLE: Flat SU(2) bundles over hyperbolic 3-manifolds
QUESTION [13 upvotes]: Can someone give me a non-trivial example of a flat SU(2)-connection over a compact orientable hyperbolic 3-manifold?
The literature on such bundles over 3-manifolds is huge and my naive searches don't seem to turn up specific examples. 
Roughly speaking, the Casson invariant counts flat bundles over 3-manifolds, so in principal I suppose I would be happy with an example of a hyperbolic 3-manifold with non-zero SU(2) Casson invariant (and surely such things are known). In practice, I would really like to see the non-trivial bundle (or corresponding representation) more-or-less explicitly.  
Finally, I would also be happy with just being told precisely where I should go and look in the literature!

REPLY [2 votes]: There is a huge literature on this. I second starting with Klassen's article. You should also go back to Riley's old article and then look at Burde for 2-bridge knots. But I suspect that noone knows an "explicit" description in the sense that "this loop goes to this matrix" in general the representations are given by the real points of a variety defined over Z.<|endoftext|>
TITLE: Analogue of the Chebyshev polynomials over C?
QUESTION [21 upvotes]: I've been driven up a wall by the following question: let p be a complex polynomial of degree d.  Suppose that |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ (for some small δ>0).  Then what's the best upper bound one can prove on |p(1)|?  (I only care about the asymptotic dependence on d and δ, not the constants.)
For the analogous question where p is a degree-d real polynomial such that |p(x)|≤1 for all 0≤x≤1-δ, I know that the right upper bound on |p(1)| is |p(1)|≤exp(d√δ).  The extremal example here is p(x)=Td((1+δ)x), where Td is the dth Chebyshev polynomial.
Indeed, by using the Chebyshev polynomial, it's not hard to construct a polynomial p in z as well as its complex conjugate z*, such that
(i) |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ, and
(ii) p(1) ~ exp(dδ).
One can also show that this is optimal, for polynomials in both z and its complex conjugate.
The question is whether one can get a better upper bound on |p(1)| by exploiting the fact that p is really a polynomial in z.  The fastest-growing example I could find has the form p(z)=Cd,δ(1+z)d.  Here, if we choose the constant Cd,δ so that |p(z)|≤1 whenever |z|=1 and |z-1|≥δ, we find that
p(1) ~ exp(dδ2)
For my application, the difference between exp(dδ) and exp(dδ2) is all the difference in the world!
I searched about 6 approximation theory books---and as often the case, they answer every conceivable question except the one I want.  If anyone versed in approximation theory can give me a pointer, I'd be incredibly grateful.
Thanks so much!
--Scott Aaronson
PS. The question is answered below by David Speyer.  For anyone who wants to see explicitly the polynomial implied by David's argument, here it is:
pd,δ(z) = zd Td((z+z-1)(1+δ)/2+δ),
where Td is the dth Chebyshev polynomial.

REPLY [31 votes]: I may be missing something obvious here. Let $f(z, z^{*})$ be the polynomial in $z$ and $z^{*}$ of degree $d$ which achieves $\exp(d \delta)$. Let $g(z)$ be the Laurent polynomial obtained from $f$ by replacing $z^{*}$ by $z^{-1}$. On the unit circle, we have $f=g$.
Now, let $h$ be the polynomial $z^d g$. This is an honest polynomial, because we multiplied by a high enough power of $z$ to clear out all the denominators and, for $z$ on the unit circle, we have $|h|=|f|$.
Doesn't this mean that $h$ is a polynomial of degree $2d$, obeying your conditions, with $|h(1)| \sim \exp(d \delta)$?<|endoftext|>
TITLE: A regularity property of transition matrices for the cat map
QUESTION [8 upvotes]: I've noticed a rather strange phenomenon (not important for my particular research, but interesting) and wouldn't be surprised if someone more versed in symbolic dynamics (i.e., just about anyone who knows what those words mean together) could easily explain it. 
Consider the cat map $A$ and the Markov partition $\mathcal{R} =$ {$R_1,\dots,R_5$} shown below: 

The rectangles in the partition are numbered from 1 (darkest) to 5 (lightest). 
Now for a given initial point $x$ with rational coordinates (so that the period $t(x)$ of the sequence $A^\ell x$ is finite) consider the matrix $T(x)$ with entries $T_{jk}(x)$ equal to the cardinality of {$\ell < t(x): A^\ell x \in R_j \land A^{\ell + 1}x \in R_k$}, i.e., the number of times per period that the trajectory goes from the $j$th rectangle to the $k$th rectangle. Clearly the sparsity pattern of $T(x)$ is inherited from the matrix defining the corresponding subshift of finite type. 
Let $L_q$ denote the set of rational points in $[0,1)^2$ with denominator $q$. When I compute the sum $T_{(q)} := \sum_{x \in L_q} T(x)$ I get some surprising near-equalities. For instance, with $q = 240$ I get 
  301468           0      301310      186567           0
  186567           0      186407      114903           0
  301310           0      301251      186407           0
       0      301470           0           0      186407
       0      186407           0           0      115060

and when $q = 322$ I get
  262625           0      262624      162291           0
  162291           0      162312      100312           0
  262624           0      262632      162312           0
       0      262603           0           0      162312
       0      162312           0           0      100312

The entries of each matrix are bunched around 3 values. What's more, the stochastic matrices obtained by adding unity to each entry and then row-normalizing agree to one part in a thousand.
Is there a (simple) explanation for this?

REPLY [9 votes]: The reason that the matrices are alike is that, up to a multiplicative factor, each is approximately encoding $Area(T(R_i) \cap R_j))$. This is because the rational points with denominator $q$ are close being distributed across the $R_i$ in proportion to their areas. (I think you're counting these points with multiplicity equal to the period, but I don't think this makes a difference.) In particular, notice that the ratio of the three values that the entries of each matrix bunch around is about $\phi^2 : \phi : 1$, where $\phi$ is the golden ratio.<|endoftext|>
TITLE: Is an algebraic space group always a scheme?
QUESTION [37 upvotes]: Suppose G is a group object in the category of algebraic spaces (over a field, if you like, or even over ℂ if you really want). Is G necessarily a scheme?
My feeling is that the answer is "yes" because an algebraic space group which is not a scheme would be too awesome. Any group homomorphism from such a G to an algebraic group (a scheme group) would have to have infinite kernel since an algebraic space which is quasi-finite over a scheme is itself a scheme. In particular, G would have no faithful representations or faithful actions on projective varieties (probably). There can't be a surjective group homomorphism from an algebraic group H to G, since that would identify G with H/K, which is a scheme. In particular, you cannot put a group structure on any étale cover of G.

REPLY [4 votes]: In Raynaud's paper "Specialisation du foncteur de Picard", Theorem 3.3.1 says that algebraic group spaces, which are locally of finite type and separated over any locally noetherian normal 1-dim scheme, are group schemes. 
On the other hand, relative Picard functors for proper flat finite presentation schemes $f:X\rightarrow S$ over DVR such that $f$ is cohomologically flat are represented by algebraic sapces, but are often not schemes, see also that paper or the book Neron Model.<|endoftext|>
TITLE: Diffeomorphism of 3-manifolds
QUESTION [19 upvotes]: Surgery theory aims to measure the difference between simple homotopy types and diffeomorphism types. In 3 dimensions, geometrization achieves something much more nuanced than that. Still, I wonder whether the surgeons' key problem has been solved. Is every simple homotopy equivalence between smooth, closed 3-manifolds homotopic to a diffeomorphism?
In related vein, it follows from J.H.C. Whitehead's theorem that a map of closed, connected smooth 3-manifolds is a homotopy equivalence if it has degree $\pm 1$ and induces an isomorphism on $\pi_1$. Is there a reasonable criterion for such a homotopy equivalence to be simple? One could, for instance, ask about maps that preserve abelian torsion invariants (e.g. Turaev's).

REPLY [20 votes]: Turaev defined a simple-homotopy invariant which is a complete invariant of homeomorphism type (originally assuming geometrization).
Here is the Springer link if you have a subscription:
Towards the topological classification of geometric 3-manifolds
He claims in the paper that a map between closed 3-manifolds is a
homotopy equivalence if and only if it is a simple homotopy equivalence,
but he says that the proof of this result will appear in a later paper. I'm
not sure if this has appeared though (I haven't searched through his
later papers on torsion, and there's no MathScinet link).<|endoftext|>
TITLE: Probability vertices are adjacent in a polygon
QUESTION [7 upvotes]: With regard to my original question:

A subset of k vertices is chosen from the vertices of a regular N-gon. What is the probability that two vertices are adjacent?

I suppose that the responses that were elicited to my question were to be expected. You see, I have been in your position, and thought what you did, on many occasions. But mostly not in the area of mathematics.
By way of providing background, I am not a student at all. In fact, I am a biochemist and part-time university instructor. I have often provided answers to students who post chemistry/biochemistry questions (samples will be provided upon request). And, like you, I hope that I have not (or had not!) become a vehicle for students to avoid thinking through THEIR chemistry homework assignments.
The above question, believe it or not, comes from the local, Boston-based television program "extrahelp", hosted by a somewhat sarcastic character who went by the name "Mr. Math". It was intended for the K-12 demographic, but, according to the story behind the video, an M.I.T. student was listening, and asked the question, ostensibly to give this man his comeuppance. The video is no longer available online, but I can send you a copy, if you don't mind that it is 25.1 MB, and that I don't have access to any FTP server at the university where I teach. It is hilarious.
When I saw this video (and after I stopped laughing), I was interested by the question itself. And since at least one of you asked for any work that I have done on my own, here is as far as I got before I made my original post:
1) N must, of course, be a positive integer >= 3. k must be a positive integer <= N.
2) When k = 1, the solution is trivial (p = 0). In fact, the non-trivial values of k are: 2 <= k <= (N \ 2), where "\" is integer division. For other values of k, p = 1.
3) For non-trivial values of k, the denominator is C(N,k).
4) For k = 2, the numerator is N.
5) For k = 3, the numerator is N(N-k).
6) For N even, and k = N / 2, the numerator is N – 2. For N odd, and k = N \ 2, the numerator is C(N,k) – N.
7) For k = (N \ 2) -1, the numerator may be C(N,k) – N(N-k).
Where I am having trouble is, obviously, getting from here to a general solution. It has been suggested to me that I take the approach of finding the general expression for the probability of NOT selecting adjacent vertices, but the general answer of p = 1 - [C(N-k,k) / C(N,k)] is not confirmed by the examples that I am sure about (that is, where N is so small that the answers can be confirmed by enumeration)
I am sure that you would agree that, by now, enough time has passed that, if this really were a homework question, that the due date for such homework would have passed. Further, I hope that I have convinced you beyond a reasonable suspicion that a) this is definitely not a homework question, and b) that I have worked on the problem myself to such a degree that I haven't just passed this off on the respondents without making some effort.

REPLY [5 votes]: It's not so hard to calculate the probability that no two points are adjacent:
We may as well assume that the first vertex is chosen for us. So let's ignore it and unroll the rest of the polygon into a line. Now imagine that I write down a sequence of $0$s and $1$s along this line to indicate whether the corresponding vertices are chosen or not. The condition on this sequence is that: the first and last numbers in this sequence must not be $1$s, and any $1$ must be followed by a $0$. So, let's replace any subsequence of the form $01$ with an $x$ and ignore the last $0$ in the sequence, so for instance we'd have the transformation $01001010 \rightarrow x0xx$. Now there is no condition on the sequence of $0$s and $x$s, other than the length and number of $x$s.
The length of the new sequence will be $(n-1)-(k-1)-1 = n-k-1$, and will contain $k-1$ $x$s, so the number of such sequences is just $\binom{n-k-1}{k-1}$, while the number of all sequences of $0$s and $1$s (including ones violating our condition) is $\binom{n-1}{k-1}$. So the probability you want is just $1-\frac{\binom{n-k-1}{k-1}}{\binom{n-1}{k-1}}=1-\frac{(n-k-1)!(n-k)!}{(n-1)!(n-2k)!}$.
Just to be sure, let's check that with $n = 6, k = 3$. Our formula gives us a probability of $1-\frac{2!3!}{5!0!} = 1-\frac{12}{120} = \frac{9}{10}$. We can check by hand that there are two ways for the vertices to not all be adjacent - the two triangles in the hexagon, and there are a total of $\binom{6}{3} = 20$ ways to choose three vertices, and this gives us a probability of $1-\frac{2}{20} = \frac{9}{10}$. Hooray! No off by one errors!<|endoftext|>
TITLE: Is there a good differential calculus for quantum SU(3)?
QUESTION [7 upvotes]: For quantum $\operatorname{SU}(2)$, Woronowicz gave a well differential calculus. If we denote the generators of quantum $\operatorname{SU}(2)$ by $a$, $b$, $c$, $d$, then the ideal of $\ker(\epsilon)$ corresponding to this calculus is
$$
\langle a+ q^2d - (1+q^2),b^2,c^2,bc,(a-1)b,(d-1)c\rangle.
$$
This calculus can be shown to generalise the classical calculus on $\operatorname{SU}(2)$ when $q=1$. Does anyone know of a (good) calculus (and its ideal) for quantum $\operatorname{SU}(3)$?

REPLY [3 votes]: I do not know whether it fits all of your requirements, but at least going by the abstract, some version of Woronowicz' result was generalized to all of the quantum groups of classical type in Differential calculus on quantized simple Lie groups, by Branislav Jurčo.<|endoftext|>
TITLE: Is the Torelli map an immersion?
QUESTION [16 upvotes]: The Torelli map $\tau\colon M_g \to A_g$ sends a curve C to its Jacobian (along with the canonical principal polarization associated to C); see this question for a description which works for families. 

Theorem (Torelli): If $\tau(C) \cong \tau(C')$, then $C \cong C'$.

If one prefers to work with coarse spaces (instead of stacks) it is okay to just say that $\tau$ is injective.

Question: Is $\tau$ an immersion?

(One remark: $\tau$ isn't a closed immersion -- the closure of its image consists of products of Jacobians!)

REPLY [23 votes]: Respectfully, I disagree with Tony's answer. The infinitesimal Torelli problem fails for $g>2$ at the points of $M_g$ corresponding to the hyperelliptic curves. And in general the situation is trickier than one would expect.
The tangent space to the deformation space of a curve $C$ is $H^1(T_C)$, and the tangent space to the deformation space of its Jacobian is $Sym^2(H^1(\mathcal O_C))$. The infinitesimal Torelli map is an immersion iff the map of these tangent spaces 
$$ H^1(T_C) \to Sym^2( H^1(\mathcal O_C) )$$
is an injection. Dually, the following map should be a surjection:
$$ Sym^2 ( H^0(K_C) ) \to H^0( 2K_C ), $$
where $K_C$ denotes the canonical class of the curve $C$. This is a surjection iff $g=1,2$ or $g=3$ and $C$ is not hyperelliptic; by a result of Max Noether.
Therefore, for $g\ge 3$ the Torelli map OF STACKS $\tau:M_g\to A_g$ is not an immersion. It is an immersion outside of the hyperelliptic locus $H_g$. Also, the restriction $\tau_{H_g}:H_g\to A_g$ is an immersion.
On the other hand, the Torelli map between the coarse moduli spaces IS an immersion in char 0. This is a result of Oort and Steenbrink "The local Torelli problem for algebraic curves" (1979). 
F. Catanese gave a nice overview of the various flavors of Torelli maps (infinitesimal, local, global, generic) in "Infinitesimal Torelli problems and counterexamples to Torelli problems" (chapter 8 in "Topics in transcendental algebraic geometry" edited by Griffiths).
P.S. "Stacks" can be replaced everywhere by the "moduli spaces with level structure of level $l\ge3$ (which are fine moduli spaces).
P.P.S. The space of the first-order deformations of an abelian variety $A$ is $H^1(T_A)$. Since $T_A$ is a trivial vector bundle of rank $g$, and the cotangent space at the origin is $H^0(\Omega^1_A)$, this space equals $H^1(\mathcal O_A) \otimes H^0(\Omega^1_A)^{\vee}$ and has dimension $g^2$. 
A polarization is a homomorphism $\lambda:A\to A^t$ from $A$ to the dual abelian variety $A^t$. It induces an isomorphism (in char 0, or for a principal polarization) from the tangent space at the origin $T_{A,0}=H^0(\Omega_A^1)^{\vee}$ to the tangent space at the origin $T_{A^t,0}=H^1(\mathcal O_A)$.  This gives an isomorphism 
$$ H^1(\mathcal O_A) \otimes H^0(\Omega^1_A)^{\vee} \to 
H^1(\mathcal O_A) \otimes H^1(\mathcal O_A). $$
The subspace of first-order deformations which preserve the polarization $\lambda$ can be identified with the tensors mapping to zero in $\wedge^2 H^1(\mathcal O_A)$, and so is isomorphic to $Sym^2 H^1(\mathcal O_A)$, outside of characteristic 2.<|endoftext|>
TITLE: Homotopy groups of Lie groups
QUESTION [139 upvotes]: Several times I've heard the claim that any Lie group $G$ has trivial second fundamental group $\pi_2(G)$, but I have never actually come across a proof of this fact.  Is there a nice argument, perhaps like a more clever version of the proof that $\pi_1(G)$ must be abelian?

REPLY [2 votes]: There is a sketch of the proof using the maximal torus as an exercise in Greub-Halperin-Vanstone's book.<|endoftext|>
TITLE: Number of valid topologies on a finite set of n elements
QUESTION [41 upvotes]: I've heard that the problem of counting topologies is hard, but I couldn't really find anything about it on the rest of the internet.  Has this problem been solved?  If not, is there some feature that makes it pretty much intractable?

REPLY [3 votes]: You can find the latest results here: https://arxiv.org/pdf/1503.08359.pdf<|endoftext|>
TITLE: Do sets with positive Lebesgue measure have same cardinality as R?
QUESTION [38 upvotes]: I have been thinking about which kind of wild non-measurable functions you can define. This led me to the question:
Is it possible to prove in ZFC, that if a (Edit: measurabel) set $A\subset \mathbb{R}$ has positive Lebesgue-measure, it has the same cardinality as $\mathbb{R}$? It is obvious if you assume CH, but can you prove it without CH?

REPLY [10 votes]: A set of positive measure contains a closed subset of positive measure. It is known that closed subset of reals may have cardinality either continuum or at most countable
(http://en.wikipedia.org/wiki/Perfect_set_property)<|endoftext|>
TITLE: Chain homotopy: Why du+ud and not du+vd?
QUESTION [36 upvotes]: When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is zero in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.
However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.
My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?
I am sorry if this makes no sense...
EDIT: Thanks David, it seems I can't write a single absatz without a stupid mistake.

REPLY [2 votes]: I have just found another article on this: Absolute Homology by Michael Barr.<|endoftext|>
TITLE: Ordinals that are not sets
QUESTION [11 upvotes]: The class of all ordinal numbers $\mathbf{Ord}$, aside being a proper class, can be thought of an ordinal number (of course it contains all ordinal numbers that are sets, not itself). Then one could consider $\mathbf{Ord}+1$, $\mathbf{Ord}+\mathbf{Ord}$, $\mathbf{Ord} \cdot \mathbf{Ord}$ and so on. Does this extension of ordinals make sense/is interesting? Maybe it was described by someone? Could it go deeper - to create a "superclass" of all ordinals that are classes?

REPLY [6 votes]: According to Kanamori, Reinhardt worked on extending the ordinals beyond the height of the universe. Kanamori talks about this in The Higher Infinite, page 313, where he cites the following reference from his bibliography.
Reinhardt, William N. Remarks on reflection principles, large cardinals, and elementary embeddings, pages 189-205
Jech, Thomas J. (ed.), Axiomatic Set Theory. Proceedings of Symposia in Pure Mathematics vol. 13, part 2. Providence, American Mathematical Society 1974.<|endoftext|>
TITLE: Closest point on Bezier spline
QUESTION [11 upvotes]: Given a two-dimensional cubic Bezier spline defined by 4 control-points as described here, is there a way to solve analytically for the parameter along the curve (ranging from 0 to 1) which yields the point closest to an arbitrary point in space?
$$
\mathbf{B}(t) = (1-t)^3 \,\mathbf{P}_0 + 3(1-t)^2 t\,\mathbf{P}_1 + 3(1-t) t^2\,\mathbf{P}_2 + t^3\,\mathbf{P}_3, ~~~~~  t \in [0,1]
$$
where P0, P1, P2 and P3 are the four control-points of the curve.
I can solve it pretty reliably and quickly with a divide-and-conquer algorithm, but it makes me feel dirty...

REPLY [12 votes]: If you have a Bezier curve $(x(t),y(t))$, the closest point to the origin (say) is given by the minimum of $f(t) = x(t)^2 + y(t)^2$.  By calculus, this minimum is either at the endpoints or when the derivative vanishes, $f'(t) = 0$.  This latter condition is evidently a quintic polynomial.  Now, there is no exact formula in radicals for solving the quintic.  However, there is a really nifty new iterative algorithm based on the symmetry group of the icosahedron due to Doyle and McMullen.  They make the point that you use a dynamical iteration anyway to find radicals via Newton's method; if you think of a quintic equation as a generalized radical, then it has an iteration that it just as robust numerically as finding radicals with Newton's method.
Contrary to what lhf said, Cardano's formula for the cubic polynomial is perfectly stable numerically.  You just need arithmetic with complex numbers even if, indeed exactly when, all three roots are real.
There is also a more ordinary approach to finding real roots of a quintic polynomial.  (Like Cardano's formula, the Doyle-McMullen solution requires complex numbers and finds the complex roots equally easily.)  Namely, you can use a cutoff procedure to switch from divide-and-conquer to Newton's method.  For example, if your quintic $q(x)$ on a unit interval $[0,1]$ is $40-100x+x^5$, then it is clearly close enough to linear that Newton's method will work; you don't need divide-and-conquer.  So if you have cut down the solution space to any interval, you can change the interval to $[0,1]$ (or maybe better $[-1,1]$), and then in the new variable decide whether the norms of the coefficients guarantee that Newton's method will converge.  This method should only make you feel "a little dirty", because for general high-degree polynomials it's a competitive numerical algorithm.  (Higher than quintic, maybe; Doyle-McMullen is really pretty good.)
See also this related MO question on the multivariate situation, which you would encounter for bicubic patches in 3D.  The multivariate situation is pretty much the same:  You have a choice between polynomial algebra and divide-and-conquer plus Newton's method.  The higher the dimension, the more justification there is for the latter over the former.<|endoftext|>
TITLE: Truncated exact sequence of homotopy groups
QUESTION [5 upvotes]: This is a question about a name of a very useful lemma, 
that permits one in particular to show that smooth birational complex projective
varieties have isomorphic fundamental groups. 
If this lemma has no name, I would like at least to have a reference (if it exits).
The lemma can be seen as a
truncated version of the basic fact, that if we have a locally trivial 
fibration (say of finite dimensional CW complexes) $F\to E\to B$ then 
we get a long exact sequence 
$\to \pi_i(F)\to \pi_i(E)\to \pi_i(B)\to \pi_{i-1}(F)\to$
Lemma. Let $E\to B$ be a surjective map of finite dimensional $CW$ complexes, 
such that every fiber is connected, simply connected and is a
deformation retract of a small neighbourhood. 
Then $\pi_1(E)=\pi_1(B)$. 
Question. 
Do you know the name of such a lemma, or of some of its generalizations? Is there a reference for this?
The result about $\pi_1$ of birationaly equivalent varieties follows
since any birational transformation can be decomposed in blow-ups 
and blow downs along smooth submanifolds. And it is not hard 
to check that the conditions of lemma are satisfied for such 
elementary blow ups.

REPLY [2 votes]: Check out the paper "A Vietoris Mapping Theorem for Homotopy," by S. Smale, Proc. Amer. Math. Soc. 8 (1957), 604-610, available at http://www.jstor.org/stable/2033527 . 
Paraphrase of the main theorem: If $f:X\to Y$ is a proper, onto map of 0-connected, locally compact, separable metric spaces, X is $LC^n$, and each point inverse is $LC^{n-1}$ and $(n-1$-connected, then the induced homomorphism $\pi_r(X)\to \pi_r(Y)$ is an isomorphism for $r\le n-1$ and surjective for $r=n$. 
$LC^n$ is a local connectedness condition surely satisfied by CW complexes, which are locally contractible.<|endoftext|>
TITLE: Discrete harmonic function on a planar graph
QUESTION [15 upvotes]: Given a graph $G$ we will call a function $f:V(G)\to \mathbb{R}$ discrete harmonic if for all $v\in V(G)$ , the value of $f(v)$ is equal to the average of the values of $f$ at all the neighbors of $v$. This is equivalent to saying the discrete Laplacian vanishes.
Discrete harmonic functions are sometimes used to approximate harmonic functions and most of the time they have similar properties. For the plane we have Liouville's theorem which says that a bounded harmonic function has to be constant. If we take a discrete harmonic function on $\mathbb{Z}^2$ it satisfies the same property (either constant or unbounded).
Now my question is: If we take a planar graph $G$ so that every point in the plane is contained in an edge of $G$ or is inside a face of $G$ that has less than $n\in \mathbb{N}$ edges, does a discrete harmonic function necessarily have to be either constant or unbounded?
I know the answer is positive if $G$ is $\mathbb{Z}^2$, the hexagonal lattice and triangular lattice, I suspect the answer to my question is positive, but I have no idea how to prove it.
Edited the condition of the graph to "contain enough cycles". (So trees are ruled out for example)

REPLY [3 votes]: For instance, any regular $H_{p,q}$ tessellation of the hyperbolic space $\mathbb{H}_2$ with $\frac{1}{q}+\frac{1}{q}<4$ does the job.<|endoftext|>
TITLE: Intermediate value theorem on computable reals
QUESTION [22 upvotes]: Wikipedia says that the intermediate value theorem “depends on (and is actually equivalent to) the completeness of the real numbers.” It then offers a simple counterexample to the analogous proposition on ℚ and a proof of the theorem in terms of the completeness property.
Does an analogous result hold for the computable reals (perhaps assuming that the function in question is computable)? If not, is there a nice counterexample?

REPLY [14 votes]: Thanks first to Andrej for drawing attention to
my paper on the IVT,
and indeed for his contributions to the work itself.
This paper is the introduction to Abstract Stone Duality
(my theory of computable general topology) for the general mathematician,
but Sections 1 and 2 discuss the IVT in traditional language first.
The following are hints at the ideas that you will find there and
at the end of Section 14.
I think it's worth starting with a warning about the computable
situation in ${\bf R}^2$, where it is customary to talk about fixed
points instead of zeroes.
Gunter Baigger

described
a computable endofunction of the square.
The classical Brouwer theorem says that it has a fixed point,
but no such fixed point can be defined by a program.
This is in contrast to the classical response to the constructive
IVT, that either there is a computable zero, or the function
hovers at zero over an interval.
(I have not yet managed to incorporate Baigger's counterexample
into my thinking.)
Returning to ${\bf R}^1$, we have a lamentable failure of classical
and constructive mathematicians to engage in a meaningful debate.
The former claim that the result in full generality is "obvious",
and argue by

quoting random fragments of what their opponents have said in
order to make them look stupid.
On the other hand, to say that
"constructively, the intermediate value theorem fails"
by showing that it implies excluded middle
is equally unconstructive.
Even amongst mainstream mathematicians several arguments are conflated,
so I would like to sort them out on the basis of
the generality of the functions to which they apply.
On the cone hand we have the classical IVT, and the approximate
constructive one that Neel mentions.  These apply to any
continuous function with $f(0) < 0 < f(1)$.
There are several other results that impose other pre-conditions:

the exact constructive IVT, for non-hovering functions,
described by Reid;

using Newton's algorithm,
for continuously differentiable functions
such that $f(x)$ and $f'(x)$ are never simultaneously zero; and

the Brouwer degree,
with an analogous condition in higher dimensions.


These conditions are all weaker forms of saying that the function is
an open map.
Any continuous function $f:X\to Y$ between compact Hausdorff spaces
is proper: the inverse image $Z=f^{-1}(0)\subset X$
of $0\in Y$ is compact (albeit possibly empty).
If $f:X\to Y$ is also an open map then $Z$ is overt too.
I'll come back to that word in a moment.
When $f$ is an open map between compact Hausdorff spaces and $Z$
is nonempty, there is a compact subspace $K\subset X$ and an open
one $V\subset Y$ with $0\in V$ and $V\subset f(K)$.
So for real manifolds we might think of $K$ is a (filled-in) ball
and $f(K)\setminus V$ as the non-zero values that $f$
takes on the enclosing sphere.
Could I have forgotten that the original question was about
computability?
No, that's exactly what I'm getting at.
In ${\bf R}^1$ an enclosing sphere is a straddling interval,
$[d,u]$ such that $f(d) < 0 < f(u)$ or $f(d) > 0 > f(u)$.
The interval-halving (or, I suspect, any computational) algorithm
generates a convergent sequence of straddling intervals.
More abstractly, write $\lozenge U$ if the open subset $U$ contains
a straddling interval.
The interval-halving algorithm (known historically as the
Bolzano--Weierstrass theorem or
lion hunting)
depends exactly on the property that $\lozenge$ takes unions to
disjunctions, and in particular
$$ \lozenge(U\cup V) \Longrightarrow \lozenge U \lor \lozenge V. $$
(Compare this with the Brouwer degree, which takes disjoint unions
to sums of integers.)
I claim, therefore, that the formulation of the constructive IVT
should be the identification of suitable conditions (more than
continuity but less than openness) on $f$ in order to prove the
above property of $\lozenge$.
Alternatively, instead of restricting the function $f$,
we could restrict the open subsets $U$ and $V$.
This is what the argument at the end of
Section 14
of my paper does.
This gives a factorisation $f=g\cdot p$ of any continuous
function $f:{\bf R}\to{\bf R}$ into a proper surjection $p$
with compact connected fibres and a non-hovering map $g$.
To a classical mathematician, $p$ is obviously surjective
in the pointwise sense, whereas this is precisely the situation
that a constructivist finds unacceptable.
Meanwhile, they agree on finding zeroes of $g$.
In fact, this process finds interval-valued zeroes of
any continuous function that takes opposite signs, which was
the common sense answer to the question in the first place.
The operator $\lozenge$ defines an overt subspace,
but I'll leave you to read the paper to find out what that means.<|endoftext|>
TITLE: Inverting a covariance matrix numerically stable
QUESTION [5 upvotes]: Given an $n\times n$ covariance matrix $C$ where $n$ around $250$, I need to calculate $x\cdot C^{-1}\cdot x^t$ for many vectors $x \in \mathbb{R}^n$ (the problem comes from approximating noise by an $n$-dimensional Gaussian distribution).
What is the best way (in the sense of numerically stable) to solve these equations?
One option is the Cholesky Decomposition, because it is numerically quite stable and fast. Is the higher computational complexity of the Singular Value Decomposition worth it? Or is there another better possibility?

REPLY [7 votes]: Cholesky sounds like a good option for the following reason: once you've done the Cholesky factorization to get $C=LL^T$, where $L$ is triangular, $x^TC^{-1}x = ||L^{-1}x||^2$, and $L^{-1}x$ is easy to compute because it's a triangular system. The downsides to this are that even if $C$ is sparse, $L$ is probably dense, and also that you do the same amount of work for all $C$ and $x$ while other methods may allow you to exploit some special structure and get good approximations to the solution with less work. For those reasons, you might also consider Krylov subspace methods for computing $C^{-1}x$, like conjugate gradients (since $C$ is symmetric and positive definite), especially if $C$ is sparse. $n=250$ isn't terribly large, but still large enough that Krylov subspace methods could pay off if $C$ is sufficiently sparse. (There might actually be special methods for computing $x^TC^{-1}x$ itself as opposed to $C^{-1}x$, but I don't know of any.) 
Edit: Since you care about stability, let me address that: Cholesky is pretty stable, as you note. Conjugate gradients is notoriously *un*stable, but it tends to work anyway, apparently.<|endoftext|>
TITLE: Representations of surface groups via holomorphic connections
QUESTION [15 upvotes]: EDIT: Tony Pantev has pointed out that the answer to this question will appear in forthcoming work of Bogomolov-Soloviev-Yotov. I look forward to reading it! 
Background
Let $E \to X$ be a holomorphic vector bundle over a complex manifold. A connection $A$ in $E$ is called holomorphic if in local holomorphic trivialisations of $E$, $A$ is given by a holomorphic 1-form with values in End(E).
Notice that the curvature of $A$ is necessarily a (2,0)-form. In particluar, holomorphic connections over Riemann surfaces are flat. This will be important for my question.
The Question
I am interested in the following situation. Let $E \to S$ be a rank 2 holomorphic vector bundle over a Riemann surface of genus $g \geq 2$. I suppose that $E$ admits a global holomorphic trivialisation (which I do not fix) and that we choose a nowhere vanishing section $v$ of $\Lambda^2 E$. (So I do fix a trivialisation of the determinant bundle.) I want to consider holomorphic connections in $E$ which make $v$ parallel. The holonomy of such a connection takes values in $\mathrm{SL}(2,\mathbb{C})$ (modulo conjugation). 
My question: if I allow you to change the complex structure on $S$, which conjugacy classes of representations of $\pi_1(S)$ in $\mathrm{SL}(2,\mathbb C)$ arise as the holonomy of such holomorphic connections?
EDIT: As jvp points out, some reducible representations never arise this way. I actually had in mind irreducible representations, moreover with discrete image in $\mathrm{SL}(2,\mathbb{C})$. Sorry for not mentioning that in the beginning!
Motivation
A naive dimension count shows that in fact the two spaces have the same dimension: 
For the holomorphic connections, if you choose a holomorphic trivialisation of $E\to S$, then the connection is given by a holomorphic 1-form with values in $sl(2, \mathbb C)$. This is a $3g$ dimensional space. Changing the trivialisation corresponds to an action of $\mathrm{SL(2,\mathbb C)}$ and so there are in fact $3g-3$ inequivalent holomorphic connections for a fixed complex structure. Combined with the $3g -3$ dimensional space of complex structures on $S$ we see a moduli space of dimension $6g-6$.
For the representations, the group $\pi_1(S)$ has a standard presentation with $2g$-generators and 1 relation. Hence the space of representations in $\mathrm{SL}(2,\mathbb{C})$ has dimension $6g-3$. Considering representations up to conjugation we subtract another 3 to arrive at the same number $6g-6$.
A curious remark
Notice that if we play this game with another group besides $\mathrm{SL}(2,\mathbb{C})$ which doesn't have dimension 3, then the two moduli spaces do not have the same dimension. So it seems that $\mathrm{SL}(2,\mathbb{C})$ should be important in the answer somehow.

REPLY [4 votes]: I'm too new to add this to my previous comment so apologies. 
Dmitri, the trivial bundle can be part of a stable Hitchin Pair (specifically if A and B are two matrices without common eigenspaces tensored with independent sections of the canonical bundle). 
The construction above generates maps from the Hitchin moduli space to itself if we start with the flat $SU(2)$ connections and use the Higgs field to define a holomorphic connection and then map the flat connection induced by the holomorphic connection to the one defined by the self-duality equations. Is this not likely to be holomorphic or well behaved under any of the complex structures? 
Likewise for a fixed $SU(2)$ representation there will be a map from the space constructed as suggested above to the Hitchin moduli space for any family of curves (or $SL(2,\mathbb{R})$ representation) and holomorphic connections.<|endoftext|>
TITLE: Cech to derived spectral sequence and sheafification
QUESTION [16 upvotes]: Let X be a topological space, let $\mathcal{U} = \{U_i\}$ be a cover of X, and let $\mathcal{F}$ be a sheaf of abelian groups on X. If X is separated, each $U_i$ is affine, and $\mathcal{F}$ is quasi-coherent, then Cech cohomology computes derived functor cohomology; in general one only gets a spectral sequence
$$
H^p(\mathcal{U},\underline{H}^q(\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F})
$$
where $\underline{H}^q(\mathcal{F})$ is the presheaf $U \mapsto H^q(U,\mathcal{F}|_U)$.

Question: For q > 0, $\underline{H}^q(\mathcal{F})$ sheafifies to 0.

For a quasi-coherent sheaf $\mathcal{F}$ this is clear because cohomology vanishes on affines. Is this really true in general? Brian Conrad states this in the introduction to his notes on cohomological descent.

REPLY [13 votes]: Here's another short proof: denote by $I,J$ the inclusion of sheaves on X into presheaves and sheafification respectively, then
$$
\underline{H}^p(\mathcal{F})\cong R^pI(\mathcal{F}).
$$
Since $J$ is exact,
$$
J\circ R^pI\cong R^p(J\circ I)
$$
and the later vanishes for $p>0$ as $J\circ I=id_{\mathfrak{Ab}(X)}$. So $\underline{H}^p(\mathcal{F})$ sheafifies to 0 for $p>0$.<|endoftext|>
TITLE: Approximating with translated Gaussians and low-frequency trig functions
QUESTION [8 upvotes]: Defining the translated Gaussians by $f_t(x)=\exp(-(x-t)^2)$  for $t,x\in\Bbb{R}$, we showed that the linear span of $\{f_t \mid 0 \le t < \epsilon\}$ is dense in $L^2(\Bbb{R})$, for any $\epsilon>0$. As a consequence, low-frequency trigonometric functions are dense in $L^2([a,b])$. The proof of the first result uses Hermite functions, and the second one follows by taking the Fourier transform of the first. (Arxiv link.)
Two reviewers told us that these results "must be known" but didn't provide references. 
Two questions:

Are these results folklore?
What's a good place to look for similar/related results?

REPLY [2 votes]: Wiener's approximation theorem says that
Given a function $h: \mathbb{R} \to  \mathbb{R}$, the set $\{\sum a_i h(\cdot - x_i): a_i, x_i \in \mathbb{R}\}$ is dense in $L^2(\mathbb{R})$ if and only if zeros of the Fourier transform of $h$ has zero Lebesgue measure.
See Wiener's book "The Fourier Integral and Certain of Its Applications" or Chandrasekharan's "Classical Fourier Transforms".
Further question when the translating parameters are restricted to a smaller set are considered by a series of authors. See this paper for instance and the reference therein.<|endoftext|>
TITLE: Field structure for R^n
QUESTION [5 upvotes]: Hi!
Is it possible to define a product on R^n for n>2 such that R^n can be made into a field?
R is a field in its own right with the standard operations and R^2 can be made into a field by introducing the product (a,b)*(c,d) = (ac-bd,ad+bc) i.e. the product of the complex numbers.

REPLY [3 votes]: No if you use the usual additive structure on $\mathbb{R}^{n}$ for the field addition; but if you give up commutativity of multiplication, you have the skew-field of hamiltonians, or quaternions, on $\mathbb{R}^{4}$, and if you then give up associativity of multiplication, you have the non-associative Cayley algebra, or octonians , on $\mathbb{R}^{8}$.  The Cayley-Dickson process builds the complex field from the real field, the skew-field of hamiltonians from the complex field, the non-associative Cayley algebra from the hamiltonians, and in general a $2^{n+1}$-dimensional involutive Cayley-Dickson algebra from the $2^{n}$-dimensional involutive Cayley-Dickson algebra.  A.A. Albert did much to articulate the state of affairs in the early part of the twentieth century, if memory serves.<|endoftext|>
TITLE: Quotient of a Hausdorff topological group by a closed subgroup 
QUESTION [8 upvotes]: Sorry if this question is below the level of this site: I've read that the quotient of a Hausdorff topological group by a closed subgroup is again Hausdorff. I've thought about it but can't seem to figure out why. Is it obvious? A simple yes or no (with reference is possible) is all I need.

REPLY [14 votes]: In fact, an even stronger statement holds: If $G$ is a topological group and $H$ is an (abstract) subgroup, then $G/H$ is Hausdorff if and only if $H$ is closed (cf Bourbaki, General Topology, III.2.5, prop 13). It's not hard to prove.<|endoftext|>
TITLE: Residual finiteness for graph manifold groups
QUESTION [8 upvotes]: Is there a simple proof that 3-dimensional graph manifolds have residually finite fundamental groups? 
By "simple" I mean the proof that does not use any hard 3d topology. I care because I wish to generalize this to higher-dimensional analogs of graph manifolds.

REPLY [9 votes]: As far as I'm aware, every proof of this fact is essentially the same as Hempel's original proof.  I don't know whether it's "simple" enough for you!  The key point is that the fundamental group G of a Seifert-fibred piece has the following property.
Property. There exists an integer K such that for any positive integer n there is a finite-index normal subgroup Gn of G such that any peripheral subgroup P intersects Gn in KnP.
It's not too hard to prove.  There's a nice account in a paper by Emily Hamilton (which generalizes Hempel's result).
The other important fact is that peripheral subgroups in Seifert-fibred manifold groups are separable (ie closed in the profinite topology, for any non-experts out there).
Using these two pieces of information, you can piece together finite quotients of Seifert-fibred pieces into a virtually free quotient of π1 of the graph manifold in which your favourite element doesn't die.
Note on separability of peripheral subgroups.  Of course, Scott proved that Seifert-fibred manifold groups are LERF.  But, by a pretty argument of Long and Niblo, a  subgroup is separable if and only if the double along it is residually finite.  In particular, you can deduce peripheral separability from the easier fact that Seifert-fibred manifold groups are residually finite.<|endoftext|>
TITLE: Why is this not an algebraic space?
QUESTION [27 upvotes]: This question is related to the question Is an algebraic space group always a scheme? which I've just seen which was posted by Anton. His question is whether an algebraic space which is a group object is necessarily a group scheme, and the answer appears to be YES. Now my naive idea of what an algebraic space is is that it is the quotient of a scheme by an étale equivalence relation, but I seem to be confusing myself. I'm hoping someone will help lead me out of my confusion.
Let me first consider an analogous topological situation, where the answer is no. We can consider the category of smooth spaces, by which I mean the category of sheaves on the site of smooth manifolds which are quotients of manifolds by smooth equivalence relations (with discrete fibers).
Here is an example: If we have a discrete group $G$ acting (smoothly) on a space $X$, we can form the equivalence relation $R \subset X \times X$, where $R$ consists of all pairs of points $(x,y) \in X \times X$ where $y= gx$ for some $g \in G$. If R is a manifold, then the sheaf $[X/R]$ (which is defined as a coequalizer of sheaves) is a smooth space.
Here is an example of a group object in smooth spaces: We start with the commutative Lie group $S^1 = U(1)$. Now pick an irrational number $r \in \mathbb{R}$ which we think of as the point $w = e^{2 \pi i r}$. We let $\mathbb{Z}$ act on $S^1$ by "rotation by $r$" i.e.
\begin{align*}
\mathbb{Z} \times S^1 & {}\to S^1\\
(n, z) & {}\mapsto w^n z.
\end{align*}
This gives us an equivalence relation $R = \mathbb{Z} \times S^1 \rightrightarrows S^1$, where one map is the action and the other projection. The fibers are discrete and the quotient sheaf is thus a smooth space, which is not a manifold. However the groupoid $R \rightrightarrows S^1$ has extra structure. It is a group object in groupoids, and this gives the quotient sheaf a group structure.
The group structure on the objects $S^1$ and morphisms $R$ are just given by the obvious group structures. Incidentally, this group object in groupoids serves as a sort of model for the "quantum torus": Blohmann–Tang–Weinstein, Hopfish structure and modules over irrational rotation algebras, arXiv:math/0604405.
Now what happens when we try to copy this example in the setting of algebraic spaces and schemes?
Let's make it easy and work over the complex numbers. An analog of $S^1$ is the group scheme
$$
\mathbb{G}_m / \mathbb{C} = \operatorname{Spec} \mathbb{C}[t,t^{-1}].
$$
Any discrete group gives rise to a group scheme over $\operatorname{Spec} \mathbb{C}$ by viewing the set $G$ as the scheme
$$
\bigsqcup_G \operatorname{Spec} \mathbb{C}.
$$
So for example we can view the integers $\mathbb{Z}$ as a group scheme. This (commutative) group scheme should have the property that a homomorphism from it to any other group scheme is the same as specifying a single $\operatorname{Spec} \mathbb{C}$-point of the target (commutative) group scheme.
A $\operatorname{Spec} \mathbb{C}$-point of $\operatorname{Spec} \mathbb{C}[t,t^{-1}]$ is specified by an invertible element of $\mathbb{C}$. Let's fix one, namely the one given by the element $w \in S^1 \subset \mathbb{C}^\times$. So this gives rise to a homomorphism $\mathbb{Z} \to \mathbb{G}_m$ and hence to an action of $\mathbb{Z}$ on $\mathbb{G}_m$.
Naïvely, the same construction seems to work to produce a group object in algebraic spaces which is not a scheme. So my question is: where does this break down?
There are a few possibilities I thought of, but haven't been able to check:

Does $R = \mathbb{Z} \times \mathbb{G}_m$ fail to be an equivalence relation for some technical reason?
Do the maps $R \rightrightarrows \mathbb{G}_m$ fail to be étale?
Is there something else that I am missing?

REPLY [12 votes]: Though Knutson requires algebraic spaces to be quasi-separated, I don't think it's a reasonable requirement. After all, not all schemes are quasi-separated, and we certainly want all schemes to be algebraic spaces.
Perhaps $\mathbb{G}_m/\mathbb{Z}$ feels like a strange algebraic space because it is not locally separated (i.e. its diagonal is not an immersion). You can see this because the morphism $R\to \mathbb{G}_m\times \mathbb{G}_m$ is not an immersion (this map is a base change of the diagonal, and immersions are stable under base change). The other classic example of a non-locally-separated (but quasi-separated) algebraic space is the line with a doubled tangent direction (Example 1 on page 9 of Knutson's book).<|endoftext|>
TITLE: Hopf algebras arising as Group Algebras
QUESTION [12 upvotes]: Every commutative $C^*$-algebra is isomorphic to the set of continuous functions, that vanish at infinity, of a locally compact Hausdorff space. Every commutative finite dimensional Hopf algebra is the group algebra of some finite group. Does there exist a characterisation of the finitely generated commutative Hopf algebras that arise as group algebras?

REPLY [14 votes]: Since the questioner starts off asking about $C^*$ algebras, I am going to assume that he only cares about Hopf algebras over $\mathbb{C}$. Every finitely generated, commutative $\mathbb{C}$-Hopf-algebra is the polynomial functions on an algebraic group $G$. As Ben says, we can just take $G$ to be Spec of the Hopf algebra, and then the comultiplication gives a group structure on this Spec.
There are several ways that working over $\mathbb{C}$ makes things nicer than working over arbitrary fields. Other answers have pointed them out but, IMO, have done so in a way that makes things sound more confusing. Let me instead point out how nice algebraic groups over $\mathbb{C}$ are:

You might worry that $G$ would not be reduced. This happens over fields of characteristic $p$, but not characteristic zero. See my earlier question. Also, you might worry that $G$ has singularities, but it doesn't. In short, $G$ is a complex Lie group.
Over non-algebraically closed fields $k$, the behavior of the $k$-points of $G$ may not determine the behavior of $G$. For example, let $X = \operatorname{Spec} \mathbb{R}[x]/(x^n-1)$. Then $X$ can be equipped with the structure of an algebraic group of order $n$; the coproduct is $x \mapsto x \otimes x$. Intuitively, you should think $x_1 \times x_2 \mapsto x_1 x_2$, so this group is the $n$-th roots of $1$ under multiplication. By the Nullstellensatz, this can't happen over $\mathbb{C}$. The $\mathbb{C}$-points will be dense (in both the ordinary and Zariski topologies) and any map will be determined by what it does to the $\mathbb{C}$-points.<|endoftext|>
TITLE: Two-dimensional quotient singularities are rational: why?
QUESTION [15 upvotes]: I've read that quotient singularities (that is, spectra of invariant subrings of finite groups acting linearly on polynomial rings) have rational singularities.  Is there an elementary proof of this fact in dimension two?  I have one proof, but it uses big guns like Grothendieck spectral sequences etc.

REPLY [10 votes]: Let me add a quick proof, also working in characteristic zero, that doesn't rely on classification (this proof is essentially originally due to S\'andor, see his Duke paper on rational singularities).  It does rely on canonical modules and a certain functoriality for them.  I will use the following characterization of rational singularities though (which I believe is originally due to Kempf), $Z$ has rational singularities if and only if it is Cohen-Macaulay and for a resolution $\pi : Z' \to Z$, we have $$\pi_* \omega_{Z'} \cong \omega_Z.$$ 
Suppose that $R \subseteq S$ is the invariant subring of some finite group action.  We know $f : R \to S$ splits as a map of $R$-module (trace map splits everything).  
Take a resolution $\pi : X \to \text{Spec} R$ and another proper birational map $\pi' : Y \to \text{Spec} S$ such that $f \circ \pi'$ factors through $\pi$ (we can always do this, we can even assume $Y$ is also smooth if we feel like it).  So $f \circ \pi' = \pi \circ \eta$ and $\eta : Y \to X$ is some proper map.  
Now, we have the following map
$$\pi_* \eta_* \omega_Y \to \omega_R$$
which can either be factored as 
$$\pi_*  \eta_* \omega_Y \to \pi_* \omega_X \to \omega_R$$
or
$$f_* \pi'_* \omega_Y \to f_* \omega_S \to \omega_R.$$
In the second factorization, the first map is surjective because $S$ is regular and the second map $f_* \omega_S \to \omega_R$ is surjective because $R \to S$ splits (that map can be obtained by applying $Hom_R( \cdot , \omega_R)$ for instance).  Thus the original map is surjective and so is $\pi_* \omega_X \to \omega_R$.
If we want to prove rational singularities, we now just need to prove Cohen-Macaulayness.  But that is obvious for normal surfaces.  Of course, the Cohen-Macaulayness of a summand of a regular ring is also pretty easy ($H^i_{\mathfrak{m}}(R) \to H^i_{\mathfrak{m}}(S)$ injects).<|endoftext|>
TITLE: Maximum degree in maximal triangle free graphs
QUESTION [10 upvotes]: It's easy to see that in bipartite maximal triangle free graphs (n vertices), the maximum degree is at least $\lceil n/2 \rceil$.  What about mtf graphs in general?  Must there always be some vertex of high degree?  Before I jump in with both feet, is there an obvious reason why there cannot be an absolute constant $c$ such that for every mtf $G$, $\Delta(G) > cn$?  A handy counterexample to c = 1/4?

REPLY [7 votes]: A maximal triangle-free graph has diameter two, so its breadth-first spanning tree has a vertex with a number of children that is at least proportional to $\sqrt n$.
I have the feeling that it should be possible to cook up a matching $O(\sqrt n)$ upper bound on the maximum degree, by using incidence graphs of the points and lines in a projective plane, but that doesn't quite work: it is triangle-free (and bipartite), has maximum degree $O(\sqrt n)$, and prevents adding any more point-point or line-line edges, but there is no obstacle to adding more point-line edges.
(Edit to add, after Greg left a comment on an earlier version of this answer): See the Kim 1995 reference from http://en.wikipedia.org/wiki/Triangle-free_graph:
Kim, J. H. (1995), "The Ramsey number $R(3,t)$ has order of magnitude $\frac{t^2}{\log t}$", Random Structures and Algorithms 7: 173–207 .
He describes a construction for a triangle-free graph in which every independent set has size $O(\sqrt{n\log n})$. If one makes the graph maximal, this doesn't increase the size of its independent sets. But in a triangle-free graph, the neighbors of every vertex form an independent set, so the maximalized version of Kim's graph has degree $O(\sqrt{n\log n})$.<|endoftext|>
TITLE: Is there any rational curve on an Abelian variety?
QUESTION [12 upvotes]: Is that true that there is no rational curves contained in an Abelian variety? If it's true,  is that because abelian varieties are not uniruled? How do I know whether an abelian variety is not uniruled?

REPLY [2 votes]: Assume there is a rational curve $C$ in an Abelian variety $A$. Let $p\in C$ be a point and $q\in A$ another point. There exists an automorphism $\sigma:A\rightarrow A$ such that $\sigma(p) = q$. Then $\Gamma = \sigma(C)$ is a rational curve through $q$. In this way we see that $A$ is covered by rational curves and its Kodaira dimension is negative. A contradiction because $k(A) = 0$. We conclude that there are not rational curves on an Abelian variety.<|endoftext|>
TITLE: How many L-values determine a modular form?
QUESTION [9 upvotes]: Suppose $f$ and $g$ are two newforms of certain levels, weights etc. If we know that
L(f,n)=L(g,n) for all sufficiently large $n$, can we conclude that $f=g$? 
Same question when the forms have the same weight and $n$ runs over critical points.

REPLY [5 votes]: The answer to the first question is "yes".  The standard proof of the uniqueness of a Dirichlet series expansion actually generalizes to show the following.
Theorem.  Suppose that $A(s) = \sum_n a_n n^{-s}$ and $B(s) = \sum_n b_n n^{-s}$ are Dirichlet series with coefficients $a_n, b_n$ bounded by a polynomial.  If there exists a sequence of complex numbers $s_k$ with real part approaching infinity such that $A(s_k) = B(s_k)$ for all $k$, then $a_n = b_n$ for all $n$.
Proof (sketch).  Proceed by induction.  For $k$ big we have $A(s_k) = a_1 + O(2^{-\sigma_k})$ where $\sigma_k$ is the real part of $s_k$.  Similarly, $B(s_k) = b_1 + O(2^{-\sigma_k})$.  Since $A(s_k) = B(s_k)$, we conclude that $a_1 = b_1$.  A similar argument shows $a_2 = b_2$, $a_3 = b_3$, etc.<|endoftext|>
TITLE: what was Hilbert's geometric construction in his 17th problem?
QUESTION [10 upvotes]: Hilbert's 17th problem asked if a nonnegative real polynomial is the sum of squares of rational functions. It was answered affirmative by Artin in around 1920. However, in his speech, he also asked if the rational functions could have coefficients over Q rather than over R. Here is the relavant part of his speech
"At the same time it is desirable, for certain questions as to the possibility of certain geometrical constructions, to know whether the coefficients of the forms to be used in the expression may always be taken from the realm of rationality given by the coefficients of the form represented."
Does anyone know what these "certain geometrical constructions" are?
It seems maybe to me that Hilbert was attempting to embed rational projective space into higher dimensional rational projective space via these polynomials. Briefly, given a nonnegative homogeneous function $f(x_0,\ldots, x_n)$ with rational coefficients induces a metric on $QP^n$. Suppose Hilbert's dream holds that $f=p_0^2+\cdots + p_N^2$ where $p_i$'s are polynomials with rational coefficients. Then the map $p: QP^n\to QP^N$ where $p(x)=(p_0(x),\ldots, p_N(x))$ is an isometric embedding (almost!) where the metric induced by $f$ is the pullback back of the Euclidean metric on $QP^N$.
The above is just my hazard. But I would be delighted if anyone is aware of what exactly Hilbert's intended "geometrical constructions" are.

REPLY [10 votes]: Actually the answer is in the sections 36 to 39 of Hilbert's "Foundations of geometry", which can be found on the web.
The constructions are construction with "straightedge" (ruler) and "transferrer of segments". 
I quote a result from Hilbert's book :
Theorem 41. A problem in geometrical construction is, then, possible of solution
by the drawing of straight lines and the laying off of segments, that is to say, by
the use of the straight-edge and a transferrer of segments, when and only when, by
the analytical solution of the problem, the co-ordinates of the desired points are
such functions of the co-ordinates of the given points as may be determined by the
rational operations and, in addition, the extraction of the square root of the sum of
two squares.
This result explains relatively clearly why this kind of geometrical constructions leads to the question of the determination of those functions of $x_1,\ldots,x_n$ which can be written as sums of squares of rational functions with rational coefficients.<|endoftext|>
TITLE: A good example of a curve for geometric Langlands
QUESTION [21 upvotes]: I'm currently working through Frenkel's beautiful paper: 
http://arxiv.org/PS_cache/hep-th/pdf/0512/0512172v1.pdf. 
I'm looking for a good example of a projective curve to get my hands dirty, and go through the general constructions that Frenkel shows there and try to do them manually for this example of a curve. Are there any good instructive examples for doing this? (Or does it always get out of hand very quickly?)

REPLY [4 votes]: Mosuli space of SL_2 bundles on any genus 2 curve - is P^3 by Narashimhan and Ramanan. So working with P^3 should be accessible, but may be not so easy as might seems...
Classical Hitchin system has been described by van Geemen, Previato, Gawedzki, et. al.  
http://arxiv.org/abs/alg-geom/9410015 
http://arxiv.org/abs/solv-int/9710025
Quantum hamiltonians should be contained in Geemen de Jong paper:
http://arxiv.org/abs/alg-geom/9701007 
So one needs to 1) Quantize (?may be done) it 2) Make Hecke transform and see that result is as predicted by Beilison and Drinfeld - product of initial Hitchin D-module on D-module give by SL_2-oper.
If I would think on this  I would try to do the following:
1) find Lax matrix L(z) - it might be done by Gawedzki, et.al.
2) try to guess what is the "qauntum spectral curve" "det"(d/dz - L(z)) - it gives both quantum Hitchin hamiltonians and SL_2-oper (this done by Talalaev for Gaudin-Hitchin for P^1 http://arxiv.org/abs/hep-th/0404153 )
3) try to prove that Hecke transformation agrees with "qauntum spectal curve"
As in previous post I would strongly suggest to use down-to-earth point of view on Hecke transformation.
So in such a way one may prove that the Hitchin's D-module is Hecke eigen-sheave with "eigenvalue" given by the SL_2-oper="quantum spectral curve" (D-module on basic curve).
Considerations of other Hecke eigensheaves (not Hitchin's) is another story...

In our paper http://arxiv.org/abs/hep-th/0303069 we described classical Hitchin system for  the degenerate genus 2 curve
y^2 = (x-a)^3 (x-b)^3. However we did not check that our "analogs" of Narasimhan-Ramanan coordinates are indeed limits of true Narashimhan-Ramanan coordinates, so it might not be very helpful. Also Talalaev's formula cannot be applied directly for the Lax matrix in these coordinates. It is coordinate dependent, it is solvable problem but requires some work.<|endoftext|>
TITLE: Detecting etale maps on reduced points
QUESTION [6 upvotes]: Suppose I have a morphism of schemes for which I know the relative cotangent complex is trivial, and the map on reduced subschemes is an isomorphism. Is the map an isomorphism? 
More generally, given a morphism of schemes with zero relative cotangent complex, which is of finite presentation on the reduced points. Is the map of finite presentation, and thus etale?
(Maybe a better way to phrase this is - what's the reference for these statements?
are they in SGA or in Illusie somewhere?)

REPLY [4 votes]: Illusie, Complexe cotangent et deformations I, Prop. 3.1.1 (p. 203) is essentially the second thing you asked.  Just a technical point: I don't think people would use the term "etale" unless the morphism is locally finitely presented or something like that (you seem to be wanting to assume that only at the level of reduced schemes or something?).  Without thinking too hard, though, Prop. 3.1.2 (same page) says that L_{X/Y} of perfect amplitude in [0,0] implies f is formally smooth...surely also your condition implies it's formally etale, which is what you're asking in general (without finiteness assumption), no?<|endoftext|>
TITLE: Algebraic geometry for cocommutative corings with counit.
QUESTION [8 upvotes]: Is there a notion of algebraic geometry for these objects? If we take the dual category of the category of cocommutative corings with counit, is there geometry in it in a sense dual to affine schemes? Can we look at the set of coideals of a coring, put a space structure on it and sheaves (maybe cosheaves) of sections?

REPLY [5 votes]: One can do (relative) algebraic geometry with respect to any symmetric monoidal category, with affine schemes corresponding to the opposite category of (commutative) monoid objects. This viewpoint is developed in the paper Au-dessous de Spec Z (in French) by Toen and Vaquie. By taking the category of $\mathbb{Z}$--modules with tensor products over $\mathbb{Z}$ one recovers usual algebraic geometry.
For (cocommutative) corings, you just have to do the same in the opposite category $(\mathbb{Z}-Mod)^{op}$. The big question is: what do you take as your monoidal structure? If you just dualize diagrams, your new coproduct would be the old product, i.e. direct product of the underlying abelian groups. Or instead of this you take usual tensor products. Each of these approaches will produce a different geometry.
I remember a course given by Lieven Le Bruyn on Kontsevich-Soibelman (the noncommutative coalgebra thing) and he mentioning that by the "discrete" nature of coalgebras one could never expect them to describe anything but the etale topology of a scheme, which I think is precisely what Leonid observed in his answer.

REPLY [4 votes]: Leonid speaks with great authority.  I'd like to point out that you get a different answer in the category of graded coalgebras.  If $A$ is a graded vector space, it has a graded dual $A'$ which is smaller than its full vector space dual $A^*$.  Indeed, if the grading is locally finite, then this $A'$ has the same dimension sequence as $A$ and $A'' = A$.  If $A$ is a locally finite, graded coalgebra, then its ring structure is given by an infinite sequence of finite tensors, and $A'$ is equivalently a graded algebra.  Graded coalgebra homomorphisms also transpose to graded algebra homomorphisms.  Recall that if $A'$ is also finitely generated (and scalar in degree $0$), then it corresponds to a projective variety with a choice of an ample line bundle.  The morphisms between these are a perfectly good non-full subcategory of the projective schemes.
The full dual $A^*$ is also the graded completion of $A'$.  So what Leonid is saying in this case is that if you take the graded completion of a finitely generated, graded algera, it localizes the projective variety to the apex of its affine cone.  This is similar to what Leonid describes for $\text{Spec}(A^*)$ in general.
I said in a previous version of this answer that $\text{Spec}(A^*)$ atomizes the variety $\text{Proj}(A')$ to a "Cantor-set-like" structure.  As Leonid points out in a comment, this is totally wrong in context.  However, it is true that if $A$ is a finitely generated algebra, it has a completion $\hat{A}$ related to coalgebras such that $\text{Spec}(\hat{A})$ is an atomized form of $\text{Spec}(A)$.  Namely, let $A'$ be the vector space of those dual vectors on $A$ that factor through finite-dimensional algebra quotients of $A$.  Then $A'$ is a coalgebra, and $\hat{A} = (A')^*$ is an atomization in the sense that $\text{Spec}(\hat{A})$ is a 0-dimensional scheme whose points are the closed points of $\text{Spec}(A)$.  (It isn't a Cantor set though.)<|endoftext|>
TITLE: Recursive presentations
QUESTION [7 upvotes]: A recursive presentation of a group is a one in which there is a finite number of generators and the set of relations is recursively enumerable. I found the following quote in Lyndon-Schupp, chapter II.1:
"This usage may seem a bit strange, but we shall see that if G has a presentation with the set of relations recursively enumerable, then it has another presentation with the set of relations recursive."
It is not clear to me however how one proves it. Does it go through Higman theorem? I.e. one first proves that group with a recursive presentation embeds in a finitely presented group and then one proves that every subgroup of a finitely presented group has a presentation with a recursive set of relations?
And in any case, can one see it somehow directly that having a presentation with a recursively enumerable set of relations (i.e. being recursive) implies having a presentation with a recursive set of relations?

REPLY [19 votes]: The answer is a simple trick. Essentially no group theory is involved.
Suppose that we are given a group presentation with a set of generators, and relations R_0, R_1, etc. that have been given by a computably enumerable procedure. Let us view each relation as a word in the generators that is to become trivial in the group. 
Now, the trick. Introduce a new generator x. Also, add the relation x, which means that x will be the identity. Now, let S_i be the relation (R_i)x^(t_i), where t_i is the time it takes for the word R_i to be enumerated in the enumeration algorithm. That is, we simply pad R_i with an enormous number of x's, depending on how long it takes for R_i to be enumerated into the set of relations. Clearly, S_i and R_i are going to give the same group, once we have said that x is trivial, since the enormous number of copies of x in S_i will all cancel out. But the point is that the presentation of the group with this new presentation becomes computably decidable (rather than merely enumerable), because given a relation, we look at it to see if it has the form Sx^t for some t, then we run the enumeration algorithm for t steps, and see if S has been added. If not, then we reject; if so, then we accept. 
One can get rid of the extra generator x simply by using the relation R_0 from the original presentation. This gives a computable set of relations in the same generating set that generates the same group.
The essence of this trick is that every relation is equivalent to a ridiculously long relation, and you make the length long enough so that one can check that it really should be there.<|endoftext|>
TITLE: What is the origin of the term "spectrum" in mathematics?
QUESTION [18 upvotes]: The use of the term "spectrum" to denote the prime ideals of a ring originates from the case that the ring is, say, $\mathbb{C}[T]$ where $T$ is a linear operator on a finite-dimensional vector space; then the prime spectrum (which is equal to the maximal spectrum) is precisely the set of eigenvalues of $T$.  The use of the term "spectrum" in the operator sense, in turn, seems to have originated with Hilbert, and was apparently not inspired by the connection to atomic spectra.  (This appears to have been a coincidence.)
A cursory Google search indicates that Hilbert may have been inspired by the significance of the eigenvalues of Laplacians, but I don't understand what this has to do with non-mathematical uses of the word "spectrum."  Does anyone know the full story here?

REPLY [6 votes]: Here is a url for "Beiträge zu Riemann’s Integrationsmethode für hyperbolische Differentialgleichungen, und deren Anwendungen auf Schwingungsprobleme"
by Wilhelm Wirtinger<|endoftext|>
TITLE: Inequality in Gaussian space -- possibly provable by rearrangement?
QUESTION [11 upvotes]: The following problem arose for my collaborators and me when studying the computational complexity of the Maximum-Cut problem.
Let $f : \mathbb{R} \to \mathbb{R}$ be an odd function.  Let $\rho \in [0,1]$.  Let $X$ and $Y$ be standard Gaussians with covariance $\rho$.  Prove that $\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[f(X)^2 \mathrm{sgn}(X) \mathrm{sgn}(Y)]$.
The quantity on the left-hand side arises naturally in many contexts; e.g., it is the integral of $f(x)f(y)$ against the Mehler kernel (with parameter $\rho$).  
I have some reason to believe this inequality is true.  For one piece of evidence, suppose $f$'s range is $\pm 1$.  Then the inequality reduces to 
$\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[\mathrm{sgn}(X) \mathrm{sgn}(Y)]$;
i.e., it's saying that $\mathrm{sgn}$ is the $\pm 1$-valued odd function maximizing the LHS.  This is indeed true; it follows from a result of Christer Borell ("Geometric bounds on the Ornstein-Uhlenbeck velocity process"), proved by Ehrhard symmetrization.  It was also given a different proof by Beckner, deducing it from a rearrangment inequality on the sphere.  
The second inequality generalizes to the case of functions $f : \mathbb{R}^n \to$ {$-1,1$}, but I believe the first inequality, which I would like to prove, is inherently $1$-dimensional.
Any ideas, or pointers to literature that might help?
Thanks!

REPLY [10 votes]: That is just Cuchy-Schwartz (though pretty well hidden). Writing everything down in terms of the joint density, dropping irrelevant factors, and taking into account that $f(-t)=-f(t)$, we see that the inequality in question is equivalent to
$$
\iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}(e^{\rho xy}-e^{-\rho xy})f(x)f(y)dxdy\le 
\iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}(e^{\rho xy}-e^{-\rho xy})f(x)^2 dxdy 
$$
Now, 
$$
e^{\rho xy}-e^{-\rho xy}=\sum_n c_n(\rho) x^ny^n
$$
with $c_n(\rho)\ge 0$. Thus, it suffices to show that
$$
\iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}x^ny^nf(x)f(y)dxdy\le 
\iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}x^ny^n f(x)^2 dxdy
$$
But, since the integrands are pure products now, this rewrites as
$$
\left[\int_{(0,+\infty)}e^{-\frac 12 x^2}x^n f(x)dx\right]^2\le 
\left[\int_{(0,+\infty)}e^{-\frac 12 x^2}x^n f(x)^2dx\right]\cdot
\left[\int_{(0,+\infty)}e^{-\frac 12 x^2}x^n dx\right],
$$
which is pure Cauchy-Schwartz.

REPLY [5 votes]: Using the antisymmetry of $f$ and $\mathrm{sgn}$ to bring the expectations to integral expressions
over $[0, \infty) \times [0, \infty)$, the first expectation takes the form:
$const\times \int f(x) f(y) \exp\left(-\frac{x^2+y^2}{2(1-\rho^2)}\right)\sinh\left(\frac{2\rho xy}{2(1-\rho^2)}\right) dx dy$
while for the second case $f(y)$ is replaced by $f(x)$. 
The proof becomes an application of the Cauchy–Schwarz inequality.<|endoftext|>
TITLE: Analytic density of the set of primes starting with 1
QUESTION [15 upvotes]: In 'Cours d'arithmetique', Serre mentions in passing the following fact (communicated to him by Bombieri): Let P be the set of primes whose first (most significant) digit in decimal notation is 1. Then P possesses an analytic density, defined as
$\lim_{s \to 1^+} \frac{\sum_{p \in P} p^{-s}}{\log(\frac{1}{s-1})}$.
This is an interesting example since it's easy to see that this set does not have a 'natural' density, defined simply as the limit of the proportion of elements in P to the # of all primes up to $x$, as $x$ tends to infinity. Therefore the notion of analytic density is a genuine extension of the naive notion (they do coincide when both exist).
How would one go about proving that P has an analytic density?
EDIT: This question has been asked again a few years ago, and it is my fault. I did accept Ben Weiss' answer but I couldn't back then check the papers, and it turns out that they don't actually answer my question! So, please refer to the newer version of the question for additional information.

REPLY [13 votes]: I think instead of posting my own explanation (which will only lose something in the translation) I'll instead refer you to two very interesting papers (thanks for posting this question, I haven't thought about this stuff in a couple years, and these papers were interesting reads to solve your problem.)
The first (among other things) proves that the density of primes with leading coefficient $k$ is $\log_{10}\left(\frac{k +1}{k}\right).$
Prime numbers and the first digit phenomenon
by
Daniel I. A. Cohen* and Talbot M. Katz
in Journal of number theory 18, 261-268 (1984)
The second is a more general statement about first digits. It is
The first digit problem
by
Ralph Raimi
in American Math Monthly vol 83 No 7
Hope this all helps.<|endoftext|>
TITLE: Noncommutative rational homotopy type
QUESTION [32 upvotes]: Ok, this question is much less ambitious than it might sound, but still:
Two commutative differential graded algebras (cdga's) are quasi-isomorphic if they can be connected by a chain of cdga quasi-isomorphisms. There is a similar definition for not necessarily commutative differential graded algebras (dga's).

If two $\mathbf{Q}$-cdga's, $A$ and $B$, are quasi-isomorphic as dga's, are they necessarily quasi-isomorphic as cdga's? I suspect that the answer is no, but don't know any counter-examples, nor can prove that such counter-examples exist.
Same question as 1 when $A$ and $B$ are the Sullivan $\mathbf{Q}$-polynomial cochain algebras of simply connected compact polyhedra. In other words, is the "rational noncommutative homotopy type" of compact simply-connected polyhedra the same as the usual rational homotopy type?

REPLY [11 votes]: An affirmative answer (to both questions) appeared today on the arxiv, due to Campos, Petersen, Robert-Nicoud, Wierstra: https://arxiv.org/abs/1904.03585<|endoftext|>
TITLE: How to think about parabolic induction.
QUESTION [38 upvotes]: In studying representations of a reductive group G, a standard technique is to use parabolic induction. The idea is that one studies such groups as a family (or perhaps in smaller families, like say just taking (products of) GL_n's) and then trying to understand representations of larger groups via representations of their smaller subgroups in the same family. Induction is the natural functor for doing this, but inducing directly from a Levi subgroup L (a natural class of reductive subgroups) gives you very large representations -- the homogenous space G/L is not projective and so bundles on it will have lots of sections. The remedy is to use parabolic induction: take the bigger subgroup P which contains L as it's reductive part, and extend representations of L to representations of P by letting the unipotent radical U of P act trivially. Then you induce the result from P to G and (since G/P is projective) you get a much smaller representation, which you can feasibly study via Hecke algebras etc.
Now in all of the above, it seems to me to make more sense to think of L not as a subgroup of G at all, but rather the reductive quotient of the parabolic group P -- that is, as a subquotient of G only. Now however, comes my question. One of the theorems you prove when studying parabolic induction (I'm begin sloppy about the context deliberately, but if you like, when studying finite group of Lie type say) is that it does not depend of the choice of a parabolic containing L. Since the same Levi can lie in two non-conjugate parabolic subgroups, this draws you back to the conclusion that it was better to think of parabolic induction as an operation on representations of subgroups of G after all. 
How do people think about this? "Why" is parabolic induction independent of the choice of parabolic? The proof of this in the context of finite groups of Lie type goes as follows: prove a formula for the composition of a parabolic induction followed by a parabolic restriction (using two arbitrary pairs of Levi inside parabolic) which expresses the composition as a sum of (parabolic) inductions and restrictions for smaller groups (as for the normal Mackey identity in finite group). Then consider the Hom space between two parabolic induction functors for the same Levi, and notice that Frobenius reciprocity lets you write this in terms of the Mackey identity you have, and then induction on rank finishes you off. 
This argument doesn't feel very enlightening me (though that might be because I don't really understand it), and I know another slick geometric proof in the context of character sheaves on a Lie algebra, which I'm not sure I understand the representation-theoretic content of. The result also is not just a curiosity -- the Harish-Chandra strategy of classifying representations of G via cuspidal data wants to associate to an irreducible representation of G a unique "cuspidal datum" $(L,\rho)$ consisting of a Levi subgroup L and a cuspidal representation of L (up to conjugation in G), and this doesn't make sense without the independence of parabolic. 
Also I don't remember the situation for p-adic groups: the "Mackey formula" argument I sketched should show that the parabolic induction functors you get don't depend on the parabolic at least in at the level of the K-group (which would be all you need to get a cuspidal theory running) but the functors themselves are perhaps not isomorphic (because the identity becomes some sort of filtration on the composition of the induction and restrictions, which is probably already is really in the finite groups type, so perhaps the same thing happens already for finite groups of Lie type when you study modular representations?

REPLY [7 votes]: Let me contribute some confusion.
In the situation I am familiar with (algebraic groups), one may first restrict to
a Borel subgroup $B$ with the property that $B\cap L$ is a Borel subgroup of $L$.
Recall that if $P$ contains both $L$ and $B$, then restricting from $P$ to $B$ and
next inducing back up to $P$ does nothing to $P$-modules. 
So instead of inducing up from $P$ one might as well first restrict to $B$ and
then induce up from $B$ to $G$.
And all Borel subgroups are conjugate.<|endoftext|>
TITLE: What's the nearest algebraic theory to inner product spaces?
QUESTION [10 upvotes]: Following the references to the accepted answer to Is the category of Banach spaces with contractions an algebraic theory? one discovers that there is an algebraic theory (infinitary) which is closely related to Banach-spaces-with-contractions but which contains some extra stuff (this wider category is called the category of totally convex spaces).  This category has a remarkably nice presentation, and the relevant paper (MR1223636) also discusses Banach algebras (not algebraic) and $C^*$-algebras (algebraic).
Inspired by this, my question is:

What's the nearest algebraic theory for inner product spaces?

There's also a connection to the question In what sense are fields an algebraic theory?.  Fields aren't an algebraic theory, but there are "nearby" algebraic theories that can be used instead.  Most often used is that of commutative, unital rings but there's also the theory of meadows which is "nearer" in some vague sense.
Also, in my answer to Clifford algebra as an adjunction?, I argued that vector+spaces+with+bilinear+form was sort of algebraic over pointed sets (I'd like an expert to check that if one's around!).  But in this question, I don't want to change the underlying category (i.e. that of sets) but rather change (as little as possible) the category of inner product spaces.
Let me give a little more detail on the kind of answer I'd like to see.  If one thinks about defining an inner product space, then it feels almost as though it is algebraic.  The vector space structure is certainly algebraic and then on top of that there is a bilinear form with certain properties.  All but one of these properties is expressed as an equality so these look like the usual sort of identities that one gets in an algebraic theory.  However, there are two stumbling blocks:

The positive definiteness of the inner product (and nondegeneracy, depending on whether you interpret "positive definite" to imply this or not)
The codomain of the inner product.

The first is easy to deal with: simply allow arbitrary symmetric bilinear (or sesquilinear if over $\mathbb{C}$) forms.  That's not a huge problem.
The second is more problematic.  "Operations" in an algebraic theory go from products of the "base" object to the base object itself, $X^n \to X$, not to some other object no matter how special.  One way around this is to start in the monoidal category $\operatorname{Vect}$ and use PROPs but that's not what I want.  So we need to "beef up" the binary form to a genuine operation.  One possibility would be to define a ternary operation $X^3 \to X$ which "morally" is the operation $(u,v,w) \mapsto \langle u, v \rangle w$.
Then, of course, there should be a variety (ha ha) of identities that this new operation should satisfy.  So one set of possible answers to this question will be simply listing these identities.  However, there may be other ways of making a bilinear form into an actual operation so other answers could give suggestions for these.  It would also be interesting to see, for any particular answer, an example of a model of the theory that isn't a vector space with a symmetric bilinear form.

REPLY [3 votes]: Certainly you can get fairly close with your ternary operation $T(x,y,z) = \langle x,y \rangle z$.  You can impose conditions on $T$ so that it comes from a bilinear form that takes values in a commutative ring of extended scalars acting on the vector space $V$.  This is not entirely a bad thing; you could instead start with an abelian group $A$ and let $T$ induce both the bilinear form and the ground ring.  To stick close to your original question, let's suppose that $V$ is a vector space over a fixed field $k$, and that the elements of $k$ are written into the algebraic theory.  Then still the scalars could extend further, but you can write axioms to make sure that that is all that happens.
In detail, you can first suppose that $T(x,y,z)$ is trilinear and that $T(x,y,z) = T(y,x,z)$.  Then $T$ can already be read as a bilinear form that takes values in operators acting on $V$.  We can use the shorthand $U(z) = T(a,b,z)$, with $U$ an implicit function of $a$ and $b$, and see what conditions can be further imposed.  You can impose the axiom:
$$T(a,b,T(c,d,x)) = T(c,d,T(a,b,x)),$$
which says that the different values of $U$ all commute, and thus generate a commutative algebra $R$.  You can also impose the relation:
$$T(T(a,b,x),y,z) = T(a,b,T(x,y,z)),$$
in other words $T(U(x),y,z) = U(T(x,y,z))$.  This relation says that $T$, as an operator-valued inner product, is $U$-linear.
With these relations, every word $W$ in $T$ collapses like this, after permuting inputs:
$$W(x_0,x_1,\ldots,x_{2n}) = \langle x_1, x_2 \rangle \cdots \langle x_{2n-1}, x_{2n} \rangle x_0,$$
where the product is interpreted over $R$ rather than over $k$.  (The number of inputs must be even because $T$ is ternary.)  This sort of collapse is the most that you can expect from any $(n,1)$-ary operation formed from a bilinear form.  I think that that proves that with this approach, you can't do better than inner products with extension of scalars.
For all I know, it is possible that you could hard-code Euclidean geometry in some more subtle way using inequalities and names of elements in $\mathbb{C}$ or $\mathbb{R}$ in addition to using multilinear algebra.  I do not know how to do that, though.<|endoftext|>
TITLE: Is there an existing name for "piecewise vector multiplication"
QUESTION [7 upvotes]: Given two vectors of size $n$
$$u = [u_1, u_2, u_3, ..., u_n ] $$
and 
$$v = [v_1, v_2, v_3, ..., v_n ] $$
What is the name of the operation "$u ? v$" such that the result is a vector of size $n$ of the form
$$u ? v = [v_1 \times u_1, v_2 \times u_2, v_3\times u_3, ..., v_n \times u_n ]$$
For want of a better name, I have termed it "piecewise vector multiplication".
What is this operation normally known as in the literature?

REPLY [19 votes]: That is the Hadamard product---which usually, though, is only used with matrices of a more matrixy shape.

REPLY [13 votes]: If $k$ is a field, the vector space $k^n$ endowed with the componentwise multiplication is called a diagonal algebra ( and so are isomorphic algebras). The terminology is due to Bourbaki and is justified by the following result.
If $M$ is a square matrix over $k$, it is diagonalizable over $k$ if and only if the algebra $k[M]$ is diagonal.The proof results from  the diagonalization criterion (the minimal polynomial of $M$ should be split over $k$  and have distinct roots) , the isomorphism of $k$-algebras   $ \frac {k[X]} {polmin_M (X)}\to k[M]$ and the Chinese remainder theorem.
These algebras are important to algebraic geometers because they are a model for étale algebras over a field. Indeed, a $k$ -algebra $A$ is étale if and only it becomes diagonal after some extension of the base field . More explicitly $A$ is étale if and only if for some field extension $K/k$ we have an isomorphism  $A\otimes_k K \simeq K^n$  of K-algebras.

REPLY [7 votes]: It's pointwise product. See Wikipedia articles here and here

REPLY [2 votes]: If you look at u and v as functions on the set S={1,2,3,...,n}, then they are elements of a function space on S, which is an algebra under pointwise addition and multiplication.<|endoftext|>
TITLE: Request for reference: Banach-type spaces as algebraic theories.
QUESTION [5 upvotes]: Sparked by Yemon Choi's answer to Is the category of Banach spaces with contractions an algebraic theory? I've just spent a merry time reading and doing a bit of reference chasing.  Imagine my delight at finding that one of my old favourites (functional analysis) and one of my new fads (category theory, and in particular algebraic theories) are actually very closely connected!
I was going to ask about the state of play of these things as it's a little unclear exactly what stage has been achieved.  Reading the paper On the equational theory of $C^\ast$-algebras and its review on MathSciNet then it appears that although it's known that $C^\ast$-algebras do form an algebraic theory, an exact presentation in terms of operations and identities is still missing (at least at the time of that paper being written), though I may be misreading things there.  It's possible to do a little reference chasing through the MathSciNet database, but the trail does seem to go a little cold and it's very hard to search for "$C^\ast$ algebra"!
But now I've decided that I don't want to just know about the current state of play, I'd like to learn what's going on here in a lot more detail since, as I said, it brings together two seemingly disparate areas of mathematics both of which I quite like.
So my real question is

Where should I start reading?

Obviously, the paper Yemon pointed me to is one place to start but there may be a good summary out there that I wouldn't reach (in finite) time by a reference chase starting with that paper.  So, any other suggestions?  I'm reasonably well acquainted with algebraic theories in general so I'm looking for specifics to this particular instance.
Also, I'll write up my findings as I find them on the n-lab so anyone who wants to join me is welcome to follow along there.  I probably won't actually start until the new year though.

REPLY [3 votes]: Do an emath search for Waelbroeck, L*; note especially his paper "The Taylor spectrum and quotient Banach spaces".  For more recent things, search for Castillo, J*.  Also, Mariusz Wodzicki at Berkeley has unpublished notes that contain many things.  I don't know if they are in a form for distribution.<|endoftext|>
TITLE: Disjoint stable sets in tournaments
QUESTION [24 upvotes]: Let $(V,A)$ be a tournament. A subset of vertices $V'\subseteq V$ is stable if 
there exists no $v\in V\setminus V'$ such that $V'\cup${$v$} contains an inclusion-maximal transitive subtournament with source $v$.
(In other words, $V'$ is stable if for every transitive subtournament $T\subseteq V'\cup${$v$} with $v\in T$ and $(v,x)\in A$ for all $x\in T\setminus${$v$}, there is a $w\in V'$ such that $(w, x)\in A$ for all $x\in T$.)
Is it true that no tournament contains two disjoint stable sets?
The statement would imply that every tournament contains a unique minimal stable set, which would have several appealing consequences in the social sciences.  The statement is a weak version of a conjecture by Schwartz (see this paper and the references therein).  Computer analysis has shown that there exists no counter-example with less than 13 vertices.

REPLY [8 votes]: The statement is false. The existence of a counter-example using a probabilistic argument was shown by Chudnovsky, Kim, Liu, Norin, Scott, Seymour, and Thomasse.<|endoftext|>
TITLE: Pascal Triangle and Prime Numbers
QUESTION [18 upvotes]: Back in the days when I was in high-school I developed a big interest about number theory specifically prime numbers and prefect numbers, I used to stay awake all night long with a bunch of sketch papers trying to come up with a formula to generate / test prime numbers. I discovered a lot of things by my own like $p(p + 1)/2$ is a perfect number when p is a Mersenne prime.
I was so obsessed back then that I used to make mental calculations when I was asleep, I remember one day waking up really excited because I had discovered that $2^p - 2  = 0 \pmod p$ when $p$ is a prime, only to discover a few weeks later that Pierre de Fermat had a similar idea but, unfortunately it didn't work for pseudoprimes. I was very disappointed back then and I started playing with the Pascal triangle.
Blaise Pascal, Marin Mersenne and Pierre de Fermat were contemporaneous and shared thoughts with letters, in fact if you think a bit both the Mersenne prime formula and the Fermat primality test seem to be closely related with the rows of the Pascal triangle (the sum of all numbers in row $n$ is $2^n$ where the first and last numbers are $1$, hence the $-1$ in the Mersenne formula and $-1$ or $-2$ in the primality tests).
I coded a Pascal Triangle generator with PHP and HTML that highlighted all the numbers that were multiples of a specific number and the results amazed me, and until this day I don't know why this happens and I would very much like to know why. Instead of trying to explain, I'll post here the images.
Composite Example:

Prime Example:

I think the difference [between the prime and composite cases] is obvious, but if you're confused just say so and I'll try to go into it a bit more...
Can anyone explain me why does this happens?

REPLY [4 votes]: I also wanted to give a very short answer. Let $p$ be a prime number. It is easy to see that the binomial coefficient $\left(p\atop n\right)$ is divisible by $p$ for $1\le n\le p-1$. So the $p$-th line looks like $1,0,0,\ldots,0,1$ mod $p$. Then by the recursive definition of the Pascal triangle a new triangle starts at the left and at the right (until they meet in the mid somewhere). And this process goes on and on. Probably the line $\left(p^2\atop *\right)$ is also a line with this property, etc. 
This explains the recursive nature of this phenomenon.<|endoftext|>
TITLE: How to generate random points in $\ell_p$ balls?
QUESTION [11 upvotes]: How do I feasibly generate a random sample from an $n$-dimensional $\ell_p$ ball? Specifically, I'm interested in $p=1$ and large $n$. I'm looking for descriptions analogous to the statement for $p=2$: Take $n$ standard gaussian random variables and normalize.

REPLY [26 votes]: For arbitrary p, this paper does exactly what you want.  Specifically, pick $X_1,\ldots,X_n$ independently with density proportional to $\exp(-|x|^p)$, and $Y$ an independent exponential random variable with mean 1. Then the random vector
$$\frac{(X_1,\ldots,X_n)}{(Y+\sum |X_i|^p)^{1/p}}$$
is uniformly distributed in the unit ball of $\ell_p^n$.
The paper also shows how to generate certain other distributions on the $\ell_p^n$ ball by modifying the distribution of $Y$.<|endoftext|>
TITLE: Classifying Algebra Extensions over a fixed extension?
QUESTION [5 upvotes]: There are lots of "Ext groups" in homological algebra which measure extensions of various things. I'm sure there must be a homological algebra machine for computing the following, and I'm hoping that someone out there knows about it. 
I'm interested in the following situation. Let R and S be commutative rings and fix a ring homomorphism $f:R \to S$.  Also fix a commutative S-algebra A. I'm interested in understanding/classifying those R-algebras B, together with a (surjective?) ring homomorphism $g: B \to A$ which intertwines the algebra structures in the following sense:
$ g(rb) = f(r) g(b)$
for all $r \in R$, and  $ b \in B$. Is there a homological algebra way to do this? 
A particular example that I am interested in is when we have the equality $B \otimes_R S = A$, but I am also interested in other cases as well.

REPLY [3 votes]: I think your problem is not constrained enough to have an interesting answer. Notice that your intertwining condition can be rephrased by saying that $g: B \to A$ is a homomorphism of $R$-algebras, where $A$ is given the structure of $R$-algebra given by $f$. In these terms, what you are looking for is the comma category $\mathcal{C} = (\mathbf{Alg}_R \downarrow A)$, whose objects are precisely the pairs $(B, g)$ as above, and whose morphisms $\operatorname{Hom}_{\mathcal{C}}((B, g), (B', g'))$ are the $R$-algebra homomorphisms $h: B \to B'$ such that $g = g' \circ h$. I am not sure it is possible to capture this beast with a cohomology group of any sort.
What you can do is restrict the class of objects that you are looking at. For example, you can classify square-zero extensions: fixing an $A$-module $I$, you can look at $R$-algebras $B$ such that you have a short exact sequence $0 \to I \to B \to A \to 0$; the name square-zero comes from the fact that $I$ is an ideal of $B$ with $I^2 = 0$. You can read about them in the first chapter of Sernesi's Deformations of Algebraic Schemes.<|endoftext|>
TITLE: Geometry of the multilagrangian Grassmannian 
QUESTION [11 upvotes]: Let's introduce the following variety $MG(3,6)$, which is a "multisymplectic" analog of a Lagrangian Grassmannian $LG(3,6)$.
Consider a 3-form $\omega = dx1 \wedge dx2 \wedge dx^3 - dx4 \wedge dx5 \wedge dx^6$ on a 6-dimensional vector space $V$ over an algebraically closed field $K$, $char(K) = 0$ (I feel that it's safe to assume $K = \mathbb C$).
We can consider a subvariety $MG(3,6)$ of Grassmannian $Gr(3,6)$ consisting of 3-planes $E$ satisfying $\omega(E)=0$.
$MG(3,6)$ turns out to be an 8-dimensional smooth hyperplane section in $Gr(3,6)$ w.r.t the Plucker embedding. I believe that $MG(3,6)$ is NOT a homogenous space for a group action.
Now the question is: What can we say about geometry of $MG(3,6)$? More precisely, I'd like to compute the cohomology $H^*(MG(3,6))$ in terms of some "canonical" cycles, like Chern classes of something etc. 
According to the Weak Lefschetz theorem, there's no problem in small codimensions: the map $H^{2k}(Gr(3,6)) \to H^{2k}(MG(3,6))$ is an isomorphism for $k < 4$,
and therefore lower codimensional cohomology groups are generated by Chern classes of canonical bundle coming from Grassmannian.
So the question really is: how do I find a nice description of cycles in the middle dimension?
I know for sure, that there is one cycle which doesn't pull-back from Grassmannian.
Remark. The reason I'm looking for a "canonical" representation of cycles is that in fact I have some twisted form $X/F$ of $MG(3,6)/\bar F$.
What I'm really looking at is cohomology (Chow groups, actually) of $X$, so I hope to find a basis of cycles which would descend to the field of definition.
Thanks.

REPLY [4 votes]: Evgeny, here are some remarks on your variety. First, I think it should be possible indeed to prove that the whole group of symmetries of this variety is $G=SL(3,C)xSL(3,C)$ (plus a finite discrete bit, as Evgeny noted below).
One first needs to show that the stabiliser of your 3-form is G,
this is worked out in details in a very nice paper of Hitchin 
"The geometry of three-forms in six and seven dimensions"
http://arxiv.org/PS_cache/math/pdf/0010/0010054v1.pdf 
You can check pages 3-5. Then it is sufficient to show that all symmetries of your variety extend to symmetries of  $G(3,6)$ an my feeling this will be the case...
As for your idea of constructing vanishing cycles, you can try to deform your variety, which just corresponds to deforming the 3-form. Now, you should take the less degenerate 3-form, maybe this will be $dx_1\wedge (dx_2\wedge dx_3+dx_4\wedge dx_5)$? (If not, Hitchin's article should help I guess). You can study the singularites of the deformed variety and if by a miracle the singularity will be just a double point, that can hint you tovards what is the vanishing cycle...
ADDED. Let me try to give an argument that could prove potentially that all automorphism of $M(3,6)=X$ may well come from $GL(6,C)$.
The Grassmanian $G(3,6)=G$ has a natural (Plucker) embedding to $CP^{19}$, the projectivisation of the third exterior power $\Lamda^3(C^6)$. The subvariety $X$ is given by a hyperplane section (indeed, every exerior 3-form of $C^6$ difines a hyperplane $CP^{18}$ in it.
Let us argue first, that Aut(X) extends to $Aut(CP^{18})$. $X$ is a Fano variety with $Pic(X)=Z$. So the bundle $O(1)$ on $CP^{18}$ should be a positive power of the anti-canonical bundle of X. If $O(1)=-nK(X)$  with n positive integer, then, since the action of $Aut(X)$ naturally lifts to the action on $K(X)$, it also acts on the sections of $-nK(X)=O(1)$, and so on $C^{19}$ as well as on its projectivisation $CP^{19}$. If $n$ is just a positive rational nubmer we don't get an action on $C^{19}$, but we still get an action on its projectivisation, I guess. Now we need to show, that in reality this action of $Aut(X)$ on $CP^{18}$ extends to the action on $CP^{19}$ that moreover preserves $Gr(3,6)$. I don't see how to do it for the moment. Maybe it is worth to ask a specialist on Fano varieties...<|endoftext|>
TITLE: What do Weierstrass points look like?
QUESTION [55 upvotes]: As somebody who mostly works with smooth, real manifolds, I've always been a little uncomfortable with Weierstrass points.  Smooth manifolds are totally homogeneous, but in the complex category you are just walking around, minding your own business, when suddenly you say to yourself: "this point feels different from all the others!"
I think I'd be happier if I could see a geometric feature which corresponds to the Weierstrass points.  So here is a more precise sub-question:  take a compact complex curve $X$ of genus $g \geq 2$, equipped with the hyperbolic metric.  Is there something special happening for this metric at the Weierstrass points?  Some kind of geodesic focusing, maybe?

REPLY [2 votes]: Associated to a dessin d'enfant (or hypermap) there is an associated complex analytic structure.
(See Grothendieck's esquisse d'un programme or 
 D. Singerman Automorphisms of maps, permutation groups and Riemann surfaces,
Bull. London Math. Soc. 8 (1976), 65-68.) 
Grothendieck observed that Belyi's theorem implies that the corresponding algebraic curve is defined over the field of algebraic numbers.
The most symmetric dessins are the regular maps, long studied by Coxeter and many others. For example, the famous Klein quartic corresponds to the well-known  map of type {3,7} of 24 heptagons on a genus 3 surface. Here the 24 Weierstrass points are the face-centres.
More generally we can ask whether vertices, face-centres or edge-centres are Weierstrass points. 
The answer is yes for genus 2, 3, 4 and with one possible exception for genus 5 too.
See D. Singerman and P. Watson, Weierstrass points on regular maps, Geom. Dedicata, 66
(1997), 69-88.<|endoftext|>
TITLE: Where can I find the text of Weyl's Fields Medal speech for Serre?
QUESTION [7 upvotes]: I thought about asking this question a while ago, but decided against it. But now I see a question about Eichler's "modular forms" quote, so while I guess it's probably still, um, questionable, what the hey.
So when Serre won the Fields Medal in 1954, Hermann Weyl (I guess) presented the award and described Serre's work. The Wikipedia article on Serre describes it thus: "...Weyl praised Serre in seemingly extravagant terms, and also made the point that the award was for the first time awarded to an algebraist."
If you're still not sure what I'm talking about, this is the speech where Weyl says something like "Never before have I seen such a rapid or bright ascension of a star in the mathematical sky as yours," if you've heard that quote.
Anyway, I've been trying to find a full version of Weyl's remarks (just out of curiosity), but to no avail. I would guess it would probably be in the congress proceedings somewhere, but I don't really have a copy on me. Anyone know where else I could find it?

REPLY [12 votes]: The IMU seems to have a great deal (all?) of the proceedings of older ICM conferences online now on the IMU website.  The written version of Weyl's address is on page 161 of Volume 1 of the 1954 proceedings.<|endoftext|>
TITLE: What techniques exist to show that a problem is not NP-complete?
QUESTION [16 upvotes]: The standard way to show that a problem is NP-complete is to show that another problem known to be NP-complete reduces to it.  That much is clear.  Given a problem in NP, what's known about how to show that it is not NP-complete?  
(My real question is likely to be inappropriate for this site for one or more reasons; I'm curious about what a proof that factoring isn't NP-complete would look like.)

REPLY [19 votes]: Here's one more pointer. 
Ladner's theorem states that if $P \neq NP$ then there is a problem in $NP-P$ which is not $NP$-complete. 

Richard E. Ladner: On the Structure of Polynomial Time Reducibility. J. ACM 22(1): 155-171 (1975)

Unfortunately the known proofs of this result produce a fairly unnatural problem. The zombie-metaphorical explanation is you take the head and torso of an $NP$-complete set and surgically stick on arms and legs of a polynomial time computable set. (If that ain't unnatural, I don't know what is.)
See this writeup for two proofs.<|endoftext|>
TITLE: Is the theory of incidence geometry complete?
QUESTION [6 upvotes]: Consider the basic axioms of planar incidence geometry, which allow us to speak of in-betweeness, collinearity and concurrency. These axioms per se are not complete, since for example, Desargues theorem may not always hold. in fact, Desargues theorem holds if and only if the model of incidence geometry can be coordinatized by a field, i.e. KP^2 serves as a model for some field K.
My question, then, is whether the theory of planar incidence geometry together with Desargues theorem is complete? (Call this theory IG + D)
If it is not, then what time is true in RP^2 (resp CP^2) that is indepedent of the theory IG + D?

REPLY [9 votes]: Pappus theorem implies Desargues'. The theory is far from being complete, not in the logical but in the philosophical and even aesthetic sense. Why do these incidence identities look so beautiful? :) Also what about combinatorics of, say, free or projective or some other submodules of the free module over the ring? noncommutative? (There was some old activity on this subject.) What about combinatorics of geodesic surfaces in nice Riemannian manifolds, and so on?<|endoftext|>
TITLE: Can you determine whether a graph is the 1-skeleton of a polytope?
QUESTION [21 upvotes]: How do I test whether a given undirected graph is the 1-skeleton of a polytope?
How can I tell the dimension of a given 1-skeleton?

REPLY [3 votes]: Not an answer, but potentially useful:

A matrix whose rows form an orthogonal basis of an eigenspace of a graph's adjacency matrix has columns that serve as coordinate vectors of an harmonious geometric realization of the graph.

("harmonious" == automorphisms of the graph induce rigid isometries the realization)
When the graph has a high degree of symmetry, these realizations --which I call "spectral"[*]-- have a great visual appeal; in general, though, these realizations are jumbles of points in one-dimensional space. In most cases, multiple vertices (and edges) are collapsed to single points, so that the realizations aren't faithful.
If a graph happens to admit a faithful spectral realization, you might be able to tease out a cell structure (which may not be unique), though I've not investigated this. I'll note that, even for polyhedra, there's no guarantee that the "faces" of a spectral realization are bounded by planar cycles of edges.
[*] More precisely, the realizations as described here are orthogonal projections of realizations I call "spectral". (See my still-drafty note, "Spectral Realizations of Graphs", the bulk of which is dedicated to a gallery of spectral realizations of the uniform polyhedra.)<|endoftext|>
TITLE: Why is 3 a bad constant in the Vitali covering lemma?
QUESTION [7 upvotes]: Hi,
recently I had to do with the Hardy-Littlewood maximal function and we used there the Vitali covering lemma with constant 5. Then, given an advice, I proved it with constant k>3. But I cannot find an counterexapmle why 3 is not enough (it is enough in the finite version of the lemma).
Has anyone seen such an example?

REPLY [5 votes]: An example in $\mathbb{R}$ was posted by Oded Schramm in May 2008 on the discussion page of the Wikipedia article on the Vitali covering lemma.  Namely, consider $\{B(x,r):|x|<1/2\text{ and }|x|<r<(|x|+1)/3\}$.<|endoftext|>
TITLE: Category of categories as a foundation of mathematics
QUESTION [30 upvotes]: In

Lawvere, F. W., 1966, “The Category of
  Categories as a Foundation for
  Mathematics”, Proceedings of the
  Conference on Categorical Algebra, La
  Jolla, New York: Springer-Verlag,
  1–21.

Lawvere proposed an elementary theory of the category of categories which can serve as a foundation for mathematics.
So far I have heard from several sources that there are some flaws with this theory so that it does not completely work as proposed.
So my question is whether there is currently any (accepted) elementary theory of the category of categories that is rich enough so that one can formulate, say, the following things in the theory:

The category of sets.
Basic notions of category theory (functor categories, adjoints, Kan extensions, etc.).
Other important categories (like the category of rings or the category of schemes).

The elementary theory I am looking for should allow me to identify what should be called a category of commutative rings (at best I would like to see this category defined by a universal 2-categorical property) or how to work with this category. I am not interested in defining groups, rings, etc. as special categories as this seems to be better done in an elementary theory of sets.
P.S.: The same question has an analogue one level higher. Assume that we have constructed an object in the category of categories (=: CAT) which can serve as a, say, category C of spaces. Classically, we can associate to each space X in C the sheaf topos over it. In the picture I have in mind, one should ask whether there is a similar elementary theory of the category of 2-categories (=: 2-CAT). Then one should be able to lift the object C from CAT to 2-CAT (as one is able to form the discrete category from a set), define an object T in 2-CAT that serves as the 2-category of toposes, and a functor C -> T in 2-CAT.

REPLY [22 votes]: I've just run into your question now. I realized it's over a year since it was asked, but since the question is not closed and I happen to have at hand some references (as I'd been interested in such developments for a while), perhaps you should take a look at the following proposals.
Lawvere's original paper was indeed flawed, as reviewed by Prof. Isbell. Something later known as "category description theorem", a way to generating objects in the theory (i.e. categories) from some description of its intuitive structure, happened to be non provable in the original theory. A thorough review of the paper including ways to fix this and save a considerable portion can be found in: Blanc G., Preller A., Lawvere’s basic theory of the category of categories. Journal of Symbolic Logic 40 (1975, no. 1), 14–18 (doi:10.2307/2272263, JSTOR).
Subsequent proposals also try to address the point of the "category description theorem" by taking a variant of it as an axiom. This is done in Blanc G., Donnadieu M. R., Axiomatisation de la catégorie des catégories. Cahiers de topologie et géométrie différentielle 17 (1976, no. 2), 1–35 (Numdam).
Finally, McLarty proposed a clean, elegant and well presented axiomatization where proofs of independence and relative consistency are given. This can be found in McLarty, C., Axiomatizing a category of categories. Journal of Symbolic Logic 56 (1991, no. 4), 1243–1260 (doi:10.2307/2275472, JSTOR). McLarty theory was meant to be taken in conjunction with certain axioms for specific categories or functors needed for specific purposes.
But I believe that all these three proposals are able to formulate the three bullets you mentioned in the first part of your question. For some other purposes one should take a closer look at them.<|endoftext|>
TITLE: Are the “identity object axioms” in the definition of a braided monoidal category needed? (Answered: No)
QUESTION [11 upvotes]: I am asking because the literature seems to contain some inconsistencies as to the definition of a braided monoidal category, and I'd like to get it straight.  According Chari and Pressley's book ``A guide to quantum groups," a braided monoidal category is a monoidal category $\mathcal{C}$ along with a natural system of isomorphisms $\sigma_{U,V}: U \otimes V \rightarrow V \otimes U$ for all pairs of objects $U$ and $V$, such that
(i) The ``Hexagon" axioms (two commutative diagrams) hold. 
(ii) The ``identity object" axioms: $\rho_V= \lambda_V \circ \sigma_{{\bf 1},V}: {\bf 1} \otimes V \rightarrow V$
and
$\lambda_V= \rho_V \circ \sigma_{V, {\bf 1}}: {V} \otimes {\bf 1} \rightarrow V$,
where $\lambda_V$ and $\rho_V$ are the isomorphisms of $V \otimes {\bf 1}$ and ${\bf 1} \otimes V$ with $V$ that are part of the definition of monoidal category. See Chari-Pressley Definitions 5.2.1 and 5.2.4. They use the term "quasitensor category," but note on p153 that the term "braided monoidal category" is equivalent.
However, in some references (ii) seems to have been dropped. I am thinking in particular of Definition 3.1 is this
expository paper, 
and the
wikipedia article. 
The wikipedia article goes further, and suggests that (ii) somehow follows from (i) and the axioms of a monoidal category. So, my questions are.
1) Is (ii) needed? That is if we do not impose (ii), does it follow from (i) and the axioms of a monoidal category?
2) If (ii) is needed, can someone provide an example demonstrating why? That is, provide an example of a monoidal category $\mathcal{C}$ along with maps $\sigma_{U,V}$ such that (i) holds but (ii) fails. Alternatively, if (ii) is not needed, I'd like a proof (or reference to a proof) that it follows from other axioms.

REPLY [7 votes]: This is Proposition 1 in the seminal paper "Braided Monoidal Categories" by Joyal and Street. Relation (ii) is implied by the others.<|endoftext|>
TITLE: In model theory, does compactness easily imply completeness?
QUESTION [22 upvotes]: Recall the two following fundamental theorems of mathematical logic:
Completeness Theorem: A theory T is syntactically consistent -- i.e., for no statement P can the statement "P and (not P)" be formally deduced from T -- if and only if it is semantically consistent: i.e., there exists a model of T.
Compactness Theorem: A theory T is semantically consistent iff every finite subset of T is semantically consistent.  
It is well-known that the Compactness Theorem is an almost immediate consequence of the 
Completeness Theorem: assuming completeness, if T is inconsistent, then one can deduce 
"P and (not P)" in a finite number of steps, hence using only finitely many sentences of T.
The traditional proof of the completeness theorem is rather long and tedious: for instance, the book Models and Ultraproducts by Bell and Slomson takes two chapters to establish it, and Marker's Model Theory: An Introduction omits the proof entirely.  There is a quicker proof due to Henkin (it appears e.g. on Terry Tao's blog), but it is still relatively involved.  
On the other hand, there is a short and elegant proof of the compactness theorem using ultraproducts (again given in Bell and Slomson).  
So I wonder: can one deduce completeness from compactness by some argument which is easier than Henkin's proof of completeness?  
As a remark, I believe that these two theorems are equivalent in a formal sense: i.e., they are each equivalent in ZF to the Boolean Prime Ideal Theorem.  I am asking about a more informal notion of equivalence.  

UPDATE: I accepted Joel David Hamkins' answer because it was interesting and informative.  Nevertheless, I remain open to the possibility that (some reasonable particular version of) the completeness theorem can be easily deduced from compactness.

REPLY [7 votes]: About the equivalence of the compactness, completeness, prime ideal theorems over ZF: what really matters here is the case when the language $L$ over which the theory $T$ is defined is not well-ordered. Otherwise, the Henkin proof gives a model of $T$ without using any form of axiom of choice. In particular, it is OK when the considered language is countable.
Now, the implication compactness $\Rightarrow$ completeness in the general case goes as follows (although it still uses completeness for well-ordered theories, which is a theorem of ZF).
Fix a first-order language $L$. Let $T$ be a syntactically consistent theory. Then any finite $F\subseteq T$ is syntactically consistent. Define $L_F$ to be the language whose operational and relational symbols are the ones occurring in $F$. Since $F$ is finite, $L_F$ is finite. Then $F$ is a syntactically consistent theory in the language $L_F$. We have completeness for countable languages, so we have a model $M_F$ of $F$ treated as a theory over $L_F$. The model $M_F$ can easily be extended to a model $M_F'$ of $F$ treated as a theory over $L$ (just give the unused symbols trivial interpretations: empty relations and constant operations). By compactness we now have a model of $T$.<|endoftext|>
TITLE: Maps between K-groups induced by rings homomorphism
QUESTION [12 upvotes]: Let $f: R\to S$ be a map between two commutative Noetherian rings. Let $G_0(R)=K_0(mod R)$ be the Grothendieck group of finite generated modules over $R$. It means $G_0(R)$ is the quotient of the free abelian group on all isomorphism class of finitely generated modules over $R$ by the subgroup generated by relations coming from short exact sequences.
If $fd_RS<\infty$, one can define a map $f^*:G_0(R)\to G_0(S)$ by:
$$f^*([M]) = \sum_{i\geq0} (-1)^i [Tor_R^i(M,S)] $$
(for reference, see Section 7,  Chapter 2 
of Weibel's book on K-theory.
Now, if $R$ is not regular or $S$ is not a complete intersection in $R$, then having finite flat dimension is somewhat a miraculous condition. So my question is: Can a map $f^*$ be defined in a more general situation than for finite flat dimension maps?
EDIT: Let me elaborate a little bit because of some interesting answers and comments below (especially Clark's answer). The main motivation I have in mind is the case of $R$ being a hypersurface. Then most $R$- modules have infinite resolutions, but it is well known that their resolution is eventually periodic. So, even though they are not homologically finite, the modules can be homologically described with finite data (i.e. finite number of matrices).
The fact above is crucial in many results I know about hypersurfaces. For a random recent example, see here. In particular, in this situation one can define( at least when $S$ is finite $R$-module):
$$f^*([M])= [Tor^{2n}(M,S)] - [Tor^{2n-1}(M,S)]$$
for sufficiently big $n$.
So that's one of the reasons I wonder if there is more systematic map  one can define.

REPLY [7 votes]: Probably not for $G$-theory, unfortunately. That $G_0$ is a contravariant for morphisms of schemes that are globally of finite Tor-dimension is SGA VI, Exp. IV, 2.12. The corresponding statement for the $G$-theory spectra is in Thomason-Trobaugh, 3.14.1.
In effect, given a morphism of schemes $f:X\to Y$, one wishes to show that the induced functor $f^{\star}$ on categories of modules gives rise to a functor between the ∞-categories of cohomologically bounded psuedocoherent complexes of $\mathcal{O}$-modules. Preservation of pseudocoherence is automatic, but the statement that $f^{\star}E$ is cohomologially bounded when $E$ is so is equivalent to the assertion that $f$ is globally of finite Tor-dimension [SGA VI, Exp. III, Pr. 3.3].
This doesn't quite show that there is no way to produce a pullback map on $G$-theory for more general kinds of morphisms, but it does make it clear that it cannot be induced by the functoriality of the ∞-categories of complexes of modules.<|endoftext|>
TITLE: (Co-) Homology associated to Waldhausen K-Theory
QUESTION [7 upvotes]: Waldhausen K-Theory takes as input a Waldhausen category C and produces a spectrum K(C). I would like to know what is known about generalized (co-) homology theories that can be realized by this spectra. I guess that stable homotopy groups are known to be representable this way? Can you help me to get an overview?

REPLY [6 votes]: Thomason's paper "Symmetric monoidal categories model all connective spectra" claims to show exactly what the title claims - namely, you can model all generalized homology theories E with En(*) = 0 for n < 0 by taking the spectrum associated to a symmetric monoidal category.  So in principle, these are what you might feel like you should get.
However, Waldhausen categories are more restrictive - they require that the symmetric monoidal structure is actually the underlying categorical coproduct.  I don't know of any results along this line.
It appears that Thomason's proof is something like the category of weakly contractible spaces over X, but - I will be blunt - I have never managed to sort through Thomason's paper.  It seems conceivable that the object he constructs might be equivalent to something coming from a Waldhausen category or its opposite, but this might be optimistic.

REPLY [3 votes]: The Waldhausen category C which is the category of finite pointed sets, with cofibrations the monomorphisms and weak equivalences the isomorphisms, has K(C) = the sphere spectrum—or so I am told.<|endoftext|>
TITLE: Does Milnor K-Theory arise from Waldhausen K-Theory
QUESTION [24 upvotes]: Quillens higher K-groups of rings can be realized as πnK(C) - the Waldhausen K-Theory of a suitable Waldhausen category C. Is this also true for Milnor K-Theory of Rings? Is there a functor F from rings to waldhausen categories s.t. $K^M_n(R)\cong \pi_n(K(F(R))$?

REPLY [21 votes]: The original question seems not to have been answered yet.  One answer might be that it would be unnatural to expect all the Milnor K-groups of a field R to arise as the homotopy groups of a single space $K(F(R))$, because the natural way they currently arise is as homotopy groups of separate spaces, or better, of separate spectra.  The spectra are the Eilenberg-MacLane spectra $\mathbb Z(n)$ associated to the chain complexes that compute motivic cohomology of $R$, namely, $K_n^M R = \pi_{-n} \mathbb Z(n)$.<|endoftext|>
TITLE: Definable collections of non measurable sets of reals
QUESTION [7 upvotes]: Is there a definable (in Zermelo Fraenkel set theory with choice) collection of non measurable sets of reals of size continuum? More verbosely: Is there a class A = {x: \phi(x)} such that ZFC proves "A is a collection, of size continuum, consisting of non Lebesgue measurable subsets of reals"?

REPLY [4 votes]: The following appears in On definability of non measurable sets, Harvey Friedman, Canadian Journal of Mathematics, Vol. 32, No. 3, 1980.
Let $M$ be the Solovay's model for $ZF + DC + V = L(R) +$ every set of reals is Lebesgue measurable etc. Let $\kappa$ be a regular cardinal of cofinality bigger than $\omega_1$ in $M$. Then forcing with countable partial functions from $\kappa$ to $2$ gives a model $N$ which satisfies choice and the statement: "Every definable, with ordinal and real parameters, set of sets of reals of size less than continuum has only Lebesgue measurable sets".<|endoftext|>
TITLE: linear recurrence relations with random coefficients
QUESTION [8 upvotes]: Are there such things as recurrence equations with random variable coefficients. For example, $$W_n=W_{n-1}+F\cdot W_{n-1}$$ where $F$ is a random variable. I tried to see if I could make sense of it using the simplest possible case of $F$ being a uniform discrete random variable on 2 points but I didn't get far because even if the initial data is not random the succeeding terms in the sequence are and each term seems to live on a different space. I couldn't figure out what space $W_n$ and $W_{n-1}$ should live on. A google search turned up nothing for the obvious keywords "random recurrence equation".
Edit in response to Alekk's answer: More specifically suppose I wanted to find the probability $P(W_{200}>3000)$. Is there a way to compute the distribution of $W_{200}$ explicitly given the distribution of $F$ and some non-random initial data $W_0$?
Edit: $F$ does not depend on $n$ and to make things even more explicit lets say $F$ has the distribution $P(F=2f)=\dfrac{1}{2}, P(F=-f)=\dfrac{1}{2}$.

REPLY [2 votes]: There are such things as probabilistic recurrence relations that come up in the analysis of randomized algorithms. The recurrence form is slightly different to the way you phrase it: rather than the coefficients of the recurrence being random, it's the "jump" itself that can be random. For example, a protoptypical example would be 
$T(n) = T(H(n)) + f(n)$, 
where H(n) is a random function of n (i.e a random variable that takes n as input and returns some random number less than n), and f(n) is some (deterministic) function. 
Richard Karp first studied these recurrences in a classic paper, and there was later followup work by Chaudhari and Dubhashi.<|endoftext|>
TITLE: Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?
QUESTION [15 upvotes]: That is, for any symplectomorphism $\psi: D^2 \to D^2$, there should be a time-dependent Hamiltonian Ht on D2 such that the corresponding flow at time 1 is equal to $\psi$.
I found this in claim a paper, and I think it should be easy, but nothing comes to mind. I'd be happy with a reference to a page in McDuff-Salomon, but I couldn't find this there immediately.
Thanks!

REPLY [4 votes]: I just wanted to expand two points of Greg's answer. Both are rather trivial additions, but took me a short while to understand, so I'm putting them here for completeness's sake and my own future reference.
First, here's a picture of Greg's idea of fixing the problem with Moser's trick.  In his notation, we have volume forms $\mu_{\alpha,t} = \mu^t \mu_\alpha^{1-t}$. They are defined because the two volume form must have the same sign everywhere (since everything is orientation-preserving; note that because of this I doubt that this argument can be generalized to symplectic forms in higher dimensions).
The obvious, but wrong, solution (described in my comments to his answer) would be to apply Moser's theorem to $\mu_{\alpha,0}$ in this notation. This would correspond to flowing along the horizontal axes of the picture below. However, Greg's idea is to flow in a different direction: we fix $\alpha$ and vary $t$. Strictly speaking, we should also show that the resulting flow $\phi_{\alpha,1}$ will be smooth, but this should follow from the proof of Moser's theorem quite easily.
                                         Moser flow,
                                t        defines phi_alpha,t
        (mu_{alpha,1} = mu)   1 ^          ^
                                |          .
                                |          .
                                |          .
                                |          .
   (mu_{alpha,0} = mu_alpha)  0 ----------------> alpha
 (phi_{alpha,0} = phi_alpha)    

(Note that, for all t, $\mu_{0,t}=\mu_{1,t}=\mu$) 

Secondly, Greg's answer implies that there is an isotopy $\psi_t: D^2 \to D^2$ such that each $\psi_t$ is volume preserving (or, equivalently, a symplectomorphism), $\psi_0$ is the identity, and $\psi_1=\psi$. Here's how we find the time-dependent Hamiltonian such that this is the Hamiltonian flow. Let Xt be the time-dependent vector field that is the derivative of the flow $\psi_t$. The fact our flow is volume-preserving is equivalent to the fact that the 1-form $\iota_{X_t} \omega$ is be closed for all $t$. Since we are on a disk, this form will also be exact. So, let $H_t$ be the function such that $dH_t = \iota_{X_t} \omega$. The Hamiltonian flow of the function $H_t$ is precisely $\psi_t$; this is immediate from the definitions.<|endoftext|>
TITLE: the affine coordinate ring of orbit closures in the ordinary nilpotent cone
QUESTION [5 upvotes]: Given a partition $\lambda$ of $n$, consider the orbit closure $\overline{ \mathcal{O}_{\lambda}}$ of the nilpotent orbit corresponding to that partition. My question, is how to explicitly construct the affine coordinate ring of the (singular) variety that is the closure of this orbit?
My second question, is the same but for the orbit closure of an orbit in the enhanced nilpotent cone (see, for instance, Achar-Henderson's ``Orbit closures in the enhanced nilpotent cone"). How would you explicitly find equations generating the ideal killing that variety in this case? 
At least the first question is probably well-known and an exercise in combinatorics/linear algebra, but I am having trouble finding a reference - the closest I could find is a paper describing the Springer correspondence for types other than A via the definition of the Weyl group acting on the weight $0$ space, but I couldn't find the answer there.

REPLY [6 votes]: Weyman has calculated generators of the radical ideal of $\overline{\mathcal{O}_{\lambda}}$ for every $\lambda$. See "The equations of conjugacy classes of nilpotent matrices", 
Invent. Math. 98 (1989), no. 2, 229–245. In particular, see Theorem 4.6 and the definitions which precede it.<|endoftext|>
TITLE: Depth Zero Ideals in the Homogenized Weyl Algebra
QUESTION [7 upvotes]: Let $\mathcal{D}$ be the $n$th Weyl algebra $ \mathcal{D} :=k[x_1,...,x_n,\partial_1,...,\partial_n] $, where $\partial_ix_i-x_i\partial_i=1$.  
Let $\widetilde{\mathcal{D}}$ be its Rees algebra, which is $ \mathcal{D} :=k[t, x_1,...,x_n,\partial_1,...,\partial_n] $, where $\partial_ix_i-x_i\partial_i=t$, and $t$ is central.  
Let $\mathcal{O}_X$ denote the polynomial algebra $k[x_1,...,x_n]$, which is a left $\widetilde{\mathcal{D}}$-module, where $t$ and all the $\partial_i$ act by zero.  NOTE: this is different than the homogenization of the standard $\mathcal{D}$-module structure on $\mathcal{O}_X$.

The question I am interested in is, how many generators does a left ideal $M$ in $\widetilde{\mathcal{D}}$ need before $Hom_{\widetilde{\mathcal{D}}}(\mathcal{O}_X,\widetilde{\mathcal{D}}/M)$ can be non-zero?  My conjecture is that $M$ needs at least $n+1$ generators.  NOTE: Savvy Weyl algebra veterans will know every left ideal in $\mathcal{D}$ can be generated by two elements; however, this is not true in $\widetilde{\mathcal{D}}$. There can be ideals generated by $n+1$ elements and no fewer.
The functor $Hom_{\widetilde{\mathcal{D}}}(\mathcal{O}_X,-)$ acts as a relative analog of the more familiar functor $Hom_R(k,-)$ (where $R=k[t,y_1,...y_n]$).  Therefore, the above question is analogous to asking "how many generators must an ideal $I\subseteq R$ have before $R/I$ can have depth zero?"  The answer here is $n+1$, which follows from Thm 13.4, pg 98 of Matsumura (essentially a souped up version of the Hauptidealsatz).
In the noncommutative case, if you try to make this work with the $\widetilde{\mathcal{D}}$-module $k$ (where $t$, $x_i$ and $\partial_i$ all act by zero), it doesn't work.  The natural conjecture would be that $Hom_{\widetilde{D}}(k,\widetilde{\mathcal{D}}/M)\neq 0$ implies $M$ had at least $2n+1$ generators, except this fails even for the first Weyl algebra and $M=\widetilde{\mathcal{D}}x_1+\widetilde{\mathcal{D}}\partial_1$ (since $\widetilde{\mathcal{D}}/M=k$).
However, it seems that things might work right for the relative module $O_X$, based on a fair amount of experimentation.  It is easily true in the first Weyl algebra.  Oh, and equivalent condition is to ask when $Hom_{\overline{\mathcal{D}}}(\mathcal{O}_X,\overline{M})\neq 0 $, where $\overline{\mathcal{D}}=\widetilde{\mathcal{D}}/t$, and $\overline{M}$ is $M/Mt$.

REPLY [5 votes]: I am not sure I understand the analogue correctly, but in the commutative case, one can get to depth zero with 3 generators. That is because any second syzygy of a module of depth at least $1$ is isomorphic to a second syzygy of a 3-generated ideal by a result of Bruns. It is even implemented here (be warned that the statement misses the at least depth $1$ part):
http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Bruns/html/
You can take the N to be second syzygy of $m=(t,y_1,...,y_n)$, so $N$ has depth 3. Produce a three generated ideal $I$ such that $syz^2(I)\cong N$. So $depth I =3-2=1$, and $depth R/I=0$.    
I think Theorem 13.4 shows that $dim R/I=0$ implies $I$ is at least $n+1$-generated.<|endoftext|>
TITLE: randomness in nature
QUESTION [8 upvotes]: What is the explanation of the apparent randomness of high-level phenomena in nature?
For example the distribution of females vs. males in a population (I am referring to randomness in terms of the unpredictability and not in the sense of it necessarily having to be evenly distributed).
1. Is it accepted that these phenomena are not really random, meaning that given enough information one could predict it? If so isn't that the case for all random phenomena?
2. If there is true randomness and the outcome cannot be predicted - what is the origin of that randomness? (is it a result of the randomness in the micro world - quantum phenomena etc...)
where can i find resources about the subject?

REPLY [2 votes]: The field of statistical physics exists for this question. Basically when you have a nonequilibrium state that is complicated (e.g., has high entropy, Kolmogorov complexity, or whatever you like) and some kind of hyperbolic dynamics, the process of averaging leads to effective parabolicity. Thus you have things like the heat equation emerging from the effectively deterministic but complex Newtonian (quantum effects really aren't responsible for anything but perhaps the averaging scale, which is extremely small) microdynamics of particle collisions.<|endoftext|>
TITLE: Number Theory and Geometry/Several Complex Variables
QUESTION [10 upvotes]: This is a question for all you number theorists out there...based on my skimming of number theory textbooks and survey articles, it seems like most of the applications of geometry and complex variables to number theory are restricted to surfaces and the theory of a single complex variable. My questions are
1) Is this impression indeed accurate?
and, if so
2) Why is this? Is it because the theories of surfaces and of a single complex variable are "easier"(in the sense that they're simple enough to have a single unified theory,) or is there some deeper reason? And if the former is the case, are there some deep conjectures in number theory that could be solved using higher-dimensional geometry/complex variables?

REPLY [14 votes]: No, I wouldn't say that "most" applications of algebraic and complex geometry to number theory are limited to the case of spaces of complex dimension 1.  A more detailed answer follows:
1) Classical algebraic number theory and classical algebraic geometry both fit under the aegis of scheme theory.  In this regard, there is an analogy between the ring of integers Z_K of a number field K and an affine algebraic curve C over a field k: both are one-dimensional, normal (implies regular, here) integral affine schemes of finite type.  (The analogy is especially close if the field k is finite.)  My colleague Dino Lorenzini has written a very nice textbook An Invitation to Arithmetic Geometry, which focuses on this analogy.  I might argue that it could be pushed even further, e.g. that students and researchers should be as familiar with non-maximal orders in K as they are with singular curves...
2) Algebraic number theory is closely related to arithmetic geometry: the latter studies rational points on geometrically connected varieties.  To do so it is essential to understand the "underlying" complex analytic space, and it is undeniable that by far the best understood case thus far is when this space has dimension one: then the theorems of Mordell-Weil and Faltings are available.  Greg Kuperberg's remark about mixing two things which are in themselves nontrivial is apt here: it is certainly advantageous in the arithmetic study of curves that the complex picture is so well understood: by now the algebraic geometers / Riemann surface theorists understand a single complex Riemann surface (as opposed to moduli spaces of Riemann surfaces) rather well, and this firm knowledge is very useful in the arithmetic study.
3) In considering a scheme X over a number field K, one often "gains a dimension" in thinking about its geometry because key questions require one to understand models of X over the ring of integers Z_K of K.  For instance, the study of algebraic number fields as fields is the study of zero-dimensional objects, but algebraic number theory proper (e.g. ramification, splitting of primes) begins when one looks at properties not primarily of the field K but of its Dedekind ring of integers Z_K.  
A consequence of this is that in the modern study of curves over a number field, one makes critical use of the theory of algebraic surfaces, or rather of arithmetic surfaces, but the latter is certainly modeled on the former and would be hopeless if we didn't know, e.g. the classical theory of complex surfaces.
4) On the automorphic side of number theory we are very concerned with a large class of Hermitian symmetric domains and their quotients by discrete subgroups.  For instance, Hilbert and Siegel modular forms come up naturally when studying quadratic forms over a general number field.  More generally the theory of Shimura varieties is playing an increasingly important role in modern number theory.  
5) Also classical Hodge theory (a certain additional structure on the complex cohomology groups of a projective complex variety) is important to number theorists via Galois representations, Mumford-Tate groups of abelian varieties, etc.
And so forth!

Addendum: An (only a few years) older and (ever so much) wiser colleague of mine who does not yet MO has contacted me and asked me to mention the following paper of Bombieri:
MR0306201 (46 #5328)
Bombieri, Enrico
Algebraic values of meromorphic maps.
Invent. Math. 10 (1970), 267--287.
32A20 (10F35 14E99 32F05)
He says it is "an extreme counterexample to the premise of the question."  Because my august institution does not give me electronic access to this volume of Inventiones, I'm afraid I haven't even looked at the paper myself, but I believe my colleague that it's relevant and well worth reading.
Edit: He is now a MO regular: Emerton.<|endoftext|>
TITLE: Pedagogical question about linear algebra
QUESTION [23 upvotes]: Last semester I taught a linear algebra class that is intended to introduce young students (at a sophmore-junior level) to "abstract mathematics".  It seems that a major conceptual hurdle for many of the students is understanding the definition of a vector space.  More specifically, a vector space is some set of things to which we can perform the operations of addition and scalar multiplication.  Despite an enormous amount of effort on my part, many of the students insisted that it makes sense to do things like "take the real/imaginary part" of a vector or look at the components of a vector.
What strategies have you found useful for getting students to understand this type of definition?
I made this community wiki -- please edit it if the question seems badly phrased.

REPLY [5 votes]: Maybe I overlooked it, but I didn't see, in the previous answers, anything about a really geometric view of vectors.  When introducing vector spaces, I like to use 2-dimensional vectors (arrows drawn on the blackboard, with the understanding that only length and direction matter, not the location on the board), with geometric definitions of addition and scalar multiplication.  It is, of course, easy to explain that these geometric vectors are "really the same" as 2-component algebraic vectors (i.e., elements of $\mathbb R^2$), and also that the sameness depends on the choice of a coordinate system.  This approach provides me with a lot of analogies for more complicated things that come up later in the course.<|endoftext|>
TITLE: Presentation for the double cover of A_n
QUESTION [9 upvotes]: The wikipedia page Covering groups of the alternating and symmetric groups gives explicit presentations for the double covers of the symmetric group Sn (n ≥ 4).  Can someone provide a similar presentation, or better yet an explicit combinatorial description, of the double cover of the alternating group An?  (I really only care about the case of large n, in case it matters.)

REPLY [12 votes]: Yeah, Schur did this a long time ago.   Let $\tilde \Sigma_n \to \Sigma_n$ be a double cover (there are two) -- lets denote them $\tilde \Sigma_n = \Sigma_n^\epsilon$ where $\epsilon \in \{+1, -1\}$. 
Schur uses the notation $[a_1 a_2 \cdots a_k]$ for a specific lift of the cycle $(a_1 a_2 \cdots a_k) \in \Sigma_n$ to $\Sigma_n^\epsilon$ -- might as well call these $k$-cycles.  Then his presentation goes like this:
$$[a_1 a_2 \cdots a_k] = [a_1 a_2 \cdots a_i][a_i a_{i+1} \cdots a_k] \ \ \forall 1 < i< k$$
and all $k$-cycles, $k>1$.
$$[a_1 a_2 \cdots a_k]^{[b_1 b_2 \cdots b_j]} = (-1)^{j-1}[\phi(a_1) \phi(a_2) \cdots \phi(a_k)]$$ 
where $\phi$ is the cycle $(b_1 b_2 \cdots b_j)$
$$[a_1 a_2 \cdots a_k]^k = \epsilon$$
for all $k$-cycles -- ie this is always $+1$ or $-1$ depending on which extension of $\Sigma_n$ you're interested in. And:
$$[a_1 a_2 \cdots a_k][b_1 b_2 \cdots b_j] = (-1)^{(k-1)(j-1)}[b_1b_2 \cdots b_j][a_1 a_2 \cdots a_k]$$
provided the cycles $(a_1 a_2 \cdots a_k)$ and $(b_1 b_2 \cdots b_j)$ are disjoint. 
The map $\tilde \Sigma_n \to \Sigma_n$ sends $[a_1 \cdots a_k]$ to $(a_1 \cdots a_k)$. So this gives you a corresponding presentation of the double of $A_n$ -- take your favourite presentation of $A_n$, lift the relators and see what happens using the above relations. 
A small extra tidbit -- think of $\Sigma_n$ as being the group of orientation-preserving isometries of $\mathbb R^{n}$ that preserves a regular (n-1)-simplex.   Then if you lift this group to $Spin(n)$, the extension you want is the one where $[a_1 a_2 \cdots a_k]^k = -1$.<|endoftext|>
TITLE: Why do finite homotopy groups imply finite homology groups?
QUESTION [36 upvotes]: Why does a space with finite homotopy groups [for every n] have finite homology groups? How can I proof this [not only for connected spaces with trivial fundamental group]? The converse is false. $\mathbb{R}P^2$ is a counterexample.
Do finitely generated homotopy groups imply finitely generated homology groups? I can proof this only for connected spaces with trivial fundamental group. The converse is false. $S^1\vee S^2$ is a counterexample.

REPLY [9 votes]: Using transfer map the question "why finite homotopy groups implies finite homology groups" can be reduced to the simply-connected case, where Serre's mod finite groups theory applies.
Recall that for a finitely-sheeted covering map $p:\tilde X\to X$ the homology transfer 
$t: H_*(X)\to H_*(\tilde X)$ is the homomorphism induced by associating to an oriented cell in $X$ the sum of its (finitely many) lifts in $\tilde X$. So the composition 
$p_* t$ is the self-map
of $H_*(X)$ that simply multiplies by the order of the cover; this map is an isomorphism when we use rational coefficients. Thus the tranfer is injective on rational homology.
In particular, if $X$ is a space with finite homotopy groups and some rational homology group nonzero, then the same is true for its universal (finite) cover.<|endoftext|>
TITLE: When does a real polynomial have a pair of complex conjugate roots?
QUESTION [11 upvotes]: Suppose we have a polynomial function $f(z)=a_0+a_1z+a_2z^2+...+z^n$ with each $a_i$ between 0 and 1. Is there a method to determine if $f$ has a pair of complex conjugate roots?
There are many results on radius of roots, but I never see similar facts concerning the question here.

REPLY [23 votes]: We can assume $f$ has no multiple root (if the gcd of $f$ and $f'$ is not constant, divide by this gcd). Let $n$ be the degree of $f$. Compute
$$\frac{f(X)f'(Y)-f(Y)f'(X)}{X-Y} = \sum_{i,j=i}^{n}a_{i,j}\; X^{i-1}\; Y^{j-1}\;.$$
Then $f$ has all roots real iff the symmetric matrix $(a_{i,j})_{i,j=1,\ldots,n}$ is positive definite. This can be checked for instance by computing the principal minors of this matrix and verifying whether they are all positive.
There are several methods for computing the number of real roots using signature of quadratic form : see for instance this note (in french).<|endoftext|>
TITLE: reduced ⊗ reduced = reduced; what about connected?
QUESTION [12 upvotes]: Several questions actually.
All rings and algebras are supposed to be commutative and with $1$ here.
(1) Let $k$ be a field, and let $A$ and $B$ be two $k$-algebras. I need a proof that if $A$ and $B$ are reduced (i. e., the only nilpotents are $0$) and $\operatorname{char}k=0$, then $A\otimes_k B$ is reduced as well.
The condition $\operatorname{char}k=0$ can be replaced by "$k$ is perfect", but I already know a proof for the $\operatorname{char}k>0$ case (the main idea is that every nilpotent $x$ satisfies $x^{p^n}=0$ for some $n$, where $p=\operatorname{char}k$), so I am only interested in the $\operatorname{char}k=0$ case.
Please don't use too much algebraic geometry - what I am looking for is a constructive proof, and while most ZFC proofs can be made constructive using Coquand's dynamic techniques, the more complicated and geometric the proof, the more work this will mean.
BTW the reason why I am so sure the above holds is that some algebraist I have spoken with has told me that he has a proof using minimal prime ideals, but I haven't ever seen him afterwards.
Ah, and I know that this is proven in J. S. Milne's Algebraic Geometry (version 6.01, Proposition 5.17 (a)) for the case $k$ algebraically closed.
(2) What if $k$ is not a field anymore, but a ring with certain properties? $\mathbb{Z}$, for instance? Can we still say something? (Probably only to be thought about once (1) is solved.)
(3) Now assume that $k$ is algebraically closed. Can we replace reduced by connected (which means that the only idempotents are $0$ and $1$, or, equivalently, that the spectre of the ring is connected)? In fact, this even seems easier due to the geometric definition of connectedness, but I don't know the relation between $\operatorname{Spec}\left(A\otimes_k B\right)$ and $\operatorname{Spec}A$ and $\operatorname{Spec}B$. (I know that $\operatorname{Spm}\left(A\otimes_k B\right)=\operatorname{Spm}A\times\operatorname{Spm}B$ however, but this doesn't help me.)
PS. All algebras are finitely generated if necessary.

REPLY [6 votes]: The answer to (2) is no. For example, for every prime $p$,
$$\mathbb{Z}[\sqrt{p}] \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{Z}[x]/(x^2-p) \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{F}_p[x]/(x^2)$$
is not reduced.<|endoftext|>
TITLE: Poincaré quasi-isomorphism
QUESTION [8 upvotes]: Suppose we have a simplicial combinatorial manifold (just a triangulated manifold) and its Poincaré dual cell complex. 
Corresponding homology simplicial and homology cell complexes are quasi-isomorphic of cause. But this quasi-isomormohism as it usually quoted from Solomon Lefschetz "Alg Topology" book is transcendental. Can we have a reasonable formula for such a quasi-isomorphism (for homology over a good field at least)? 
Update
1) sorry for Russian math-slang use of "transcendental" here "transcendental"="non-constructive" 
2) among the others one motivation is to see really the Poincare duality for simplicial chains and cochains of the given triangulation of a manifold. 
UUpdate
The problem has a nice very canonical solution, with Laplas operator Green function heat kerenel etc. It allows to solve some problems. The preprint(s) are in preparation.

REPLY [4 votes]: Both the cell complex, $C$, and the dual cell complex $C'$ are refined by the first barycentric subdivision $BC$. There are maps $C \to BC$ and $C' \to BC$, sending a cell $\sigma$ to the sum of all cells of the same dimension contained in $\sigma$; these maps are both quasi-isomorphisms. 
So, if you allow me to formally invert quasi-isomorphisms, I'm done. 
Is the question whether there is an honest map of chain complexes between $C$ and $C'$, without subdividing?

UPDATE Here is something you can do, and something you can't do. 
With $C$ and $BC$ as above, and $r : C \to BC$ the refinement map, there is a homotopy inverse $s: BC \to C$. (More precisely, $C \to BC \to C$ is the identity, and $BC \to C \to BC$ is homotopic to the identity.) Working the same trick with $r' : C' \to BC$, we get quasi-isomorphisms between $C$ and $C'$ which are homotopy inverse to each other. As you will see, however, this construction is very nongeometric and inelegant.
Construction: Let $q:BC \to Q$ be the cokernel of $C \to BC$. An easy computation checks that each $Q_i$ is free. Since $C \to BC$ is a quasi-isomorphism, $Q$ is exact. An exact complex of free $\mathbb{Z}$ modules must be isomorphic to a direct sum of complexes of the form $\cdots \to 0 \to \mathbb{Z} \to \mathbb{Z} \to 0 \to \cdots$. Choose such a decomposition of $Q$, so $Q_i = A_{i+1} \oplus A_{i}$ and the map $Q_i \to Q_{i-1}$ is the projection onto $A_{i}$. 
Now, consider the map $q_i^{-1}(A_i) \to A_i$ in degree $i$. This is surjective, and  $A_i$ is free, so choose a section $p^1_i$. We also define a map $p^2_i$ from the $A_{i+1}$ summand of $Q_{i}$ to $BC_i$ by $p^2_i = d p^1_{i+1} d^{-1}$. In this way, we get maps $p_i = p^1_i \oplus p^2_i: Q_{i} \to BC_i$ which give a map of chain complexes.
We note that $qp: Q \to Q$ is the identity. Therefore, $1-pq$, a map from $BC \to BC$, lands in the subcomplex $C$ and gives a section $s:BC \to C$. Proof of the claim about homotopies will be provided on request.

On the other hand, here is something you can't do: Get the quasi-isomorphism to respect the symmetries of your original space. For example, let $C$ be the chain complex of the cube, and $C'$ the chain complex of the octahedron. I claim that there is no quasi-isomorphism $C \to C'$ which commutes with the group $S_4$ of orientation preserving symmetries. 
Consider what would happen in degree $0$. A vertex of the cube must be sent to some linear combination of the vertices of the octahedron. By symmetry, it must be set to 
$$a (\mbox{sum of the "near" vertices}) + b (\mbox{sum of the "far" vertices})$$
for some integers $a$ and $b$. But then the map on $H_0$ is multiplication by $3(a+b)$, and cannot be $1$. 

I imagine you want something stronger then my first answer, but weaker than my second. I am not sure what it it, though.<|endoftext|>
TITLE: A conjecture of Montesinos
QUESTION [8 upvotes]: Not every orientable 3-manifold is a double cover of $S^3$ branched over a link. For example, the 3-torus isn't. However, in 1975 Montesinos conjectured (Surjery on links and double branched covers of $S^3$, in: "Knots, groups and 3-manifolds", papers dedicated to the memory of R. Fox) that every orientable 3-manifold is a double branched cover of a sphere with handles i.e. the connected sum of a certain number of copies of $S^1\times S^2$ (this number can be zero, in which case we get $S^3$). Notice that this time $T^3$ does not provide a counter-example since if we take the quotient of $T^3$ by the involution $(x,y,z)\mapsto (x^{-1}, y^{-1},z)$ we get $S^2\times S^1$.
I was wondering what the status of this conjecture is.

REPLY [11 votes]: It is false. For example, there are closed, orientable, aspherical 3-manifolds that admit no nontrivial action of a finite group whatsoever. The first examples were due to F. Raymond and J. Tollefson in the 1970s, I believe.<|endoftext|>
TITLE: Do chains and cochains know the same thing about the manifold?
QUESTION [18 upvotes]: This question was inspired by Poincaré quasi-isomorphism
Let $M$ be a closed oriented $n$-manifold. The cap product with the fundamental class of $M$ induces an isomorphism $H^i(M,\mathbf{Z})\to H_{n-i}(M,\mathbf{Z})$. Both the source and the target of this are rings. (For the definition of the homology intersection product see e.g. McClure http://arxiv.org/abs/math/0410450 or M. Goresky and R. MacPherson's first paper on the intersection homology.) It is not too difficult to show that the Poincar\'e isomorphism respects the ring structure.
The question is: to which extent is this true on the chain level?
More precisely, Goresky and MacPherson's PL chains of a manifold form a partial commutative dga (see McClure's paper mentioned above). Singular cochains form a non-commmutative dga that can be completed to an $E_{\infty}$-algebra, which is a different kind of structure. So one way to make the above question precise would be as follows:

Is there a natural way to turn the PL-chains on a PL-manifold into an $E_\infty$ algebra? (In the above-mentioned paper McClure promises to do this in another paper, but I don't know if the details are available.)
If the answer to 1. is positive, then can one complete the chain level cap product with the fundamental cycle into an $E_{\infty}$ morphism?

REPLY [8 votes]: You may also want to look up David Chataur's work on the subject. I've heard that he proves that for any Poincare Duality space, each "Poincare Duality isomorphism" from cohomology to homology gives rise to a "unique" $E_{\infty}$ quasi-isomorphism from the $E_{\infty}$ structure on cochains to Wilson's $E_{\infty}$ structure on chains.<|endoftext|>
TITLE: Is there a "finitary" solution to the Basel problem?
QUESTION [28 upvotes]: Gabor Toth's Glimpses of Algebra and Geometry contains the following beautiful proof (perhaps I should say "interpretation") of the formula $\displaystyle \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} \mp ...$, which I don't think I've ever seen before.  Given a non-negative integer $r$, let $N(r)$ be the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 \le r^2$, i.e. the number of lattice points in the ball of radius $r$.  Then if $r_2(n)$ is the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 = n$, it follows that $N(r^2) = 1 + r_2(1) + ... + r_2(r^2)$.
On the other hand, once one has characterized the primes which are a sums of squares, it's not hard to show that $r_2(n) = 4(d_1(n) - d_3(n))$ where $d_i(n)$ is the number of divisors of $n$ congruent to $i \bmod 4$.  So we want to count the number of divisors of numbers less than or equal to $r^2$ congruent to $i \bmod 4$ for $i = 1, 3$ and take the difference.  This gives
$\displaystyle \frac{N(r^2) - 1}{4} = \left\lfloor r^2 \right\rfloor - \left\lfloor \frac{r^2}{3} \right\rfloor + \left\lfloor \frac{r^2}{5} \right\rfloor \mp ...$
and now the desired result follows by dividing by $r^2$ and taking the limit.
Question:  Does a similar proof exist of the formula $\displaystyle \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ...$?
By "similar" I mean one first establishes a finitary result with a clear number-theoretic or combinatorial meaning and then takes a limit.

REPLY [5 votes]: A somewhat different perspective to the Basel problem relates $\zeta(2)$ to the volume of $SL_2(\mathbb{R})/SL_2(\mathbb{Z})=\zeta(2)/2$. They compute this volume via a count of lattice points. One can also compute this via Gauss-Bonnet as a circle bundle over the modular curve $\mathbb{H}^2/PSL_2(\mathbb{Z})$ and deduce the Basel identity. There is some subtlety here about how the volume forms are defined in the comparison, but I think that this can be made into a proof.<|endoftext|>
TITLE: Why these particular numerical factors in the definition of Gaussian curvature?
QUESTION [6 upvotes]: Wikipedia tells me that:
Gaussian curvature is the limiting difference between the circumference of a geodesic circle and a circle in the plane:
$K = \lim_{r \rightarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}$
Gaussian curvature is the limiting difference between the area of a geodesic circle and a circle in the plane:
$K = \lim_{r \rightarrow 0} (\pi r^2 - \mbox{A}(r)) \cdot \frac{12}{\pi r^4}$
Can anyone explain to me why we are dividing by the factors $\frac{3}{\pi r^3}$ and $\frac{12}{\pi r^4}$ respectively? I don't understand why we are dividing by these particular factors?

REPLY [6 votes]: Joel is right that it is partly just a convention to scale Gaussian curvature so that the curvature of a unit sphere is $1$.  However, there are three natural motivations for this scale besides matching 1 to 1 in the case of a sphere.
First, Gauss defined his curvature as the product of the extrinsic curvatures of a surface in $\mathbb{R}^3$.  So there is a coefficient of 1 in this natural formula.
Second, the unit sphere has the property that the deviation from Euclid's parallel postulate has a factor of 1.  In other words, the area $A$ of a triangle with angles $\alpha, \beta, \gamma$ is $\alpha + \beta + \gamma - \pi$.  In general, if you have a very small triangle with area $A$ at a point of local curvature $K$, its angle deviation is $KA$ to first order.  This factor of 1 leads to a factor of $2\pi$ in the Gauss-Bonnet theorem, that the integral of Gaussian curvature is $2\pi \chi$.
Third, Gaussian curvature is the ratio of the Ricci curvature tensor to the metric, and it is also half of the scalar curvature.
In comparing these formulas, the most reasonable scales for Gaussian curvature are the standard choice, the standard choice times 2 to match scalar curvature, and the standard choice divided by $2\pi$ to match the Gauss-Bonnet theorem.  The volume and area formulas are some justification for a 1/3 or a 1/12 or similar, but these are taken to be less fundamental scales.
(One irony of the discussion is that $\pi$ itself is half of the most important value in trigonometry.)
Also the volume and surface area ratios are given in Wikipedia in $n$ dimensions.  It is also worth looking at the generalized Gauss-Bonnet theorem in $2n$ dimensions.<|endoftext|>
TITLE: Riemannian Geometry 
QUESTION [14 upvotes]: I come from a background of having done undergraduate and graduate courses in General Relativity and elementary course in riemannian geometry. 
Jurgen Jost's book does give somewhat of an argument for the the statements below but I would like to know if there is a reference where the following two things are proven explicitly, 

That the sectional curvature of a 2-dimensional subspace of a tangent space at a point on the Riemannian manifold is independent of the choice of basis.  That is the definition of the sectional curvature depends only on the choice of the 2-dim subspace. 
That the sectional curvature determines the Riemannian curvature fully. 

Secondly can one give me a reference where I can see how in practice is sectional curvature computed. To a first timer to this subject it is not obvious how one does a calculation on "all" 2-dimensional subspaces of a high-dimensional space. Especially when people talk of manifolds with "constant sectional curvature". How are they realized?  
I would like to see some explicit examples to understand this point.  
Further some studies about homogeneous spaces (needed to understand some issues in Quantum Field Theory) got me to the following 4 very non-trivial ideas in Riemannian manifolds which I am stating in my own way here , 

That the isometry group of a Riemannian manifold is always a lie group.
The isotropy subgroup of any point on a Riemannian manifold under the
smooth transitive action of its own isometry group on itself is a compact
subgroup. (The context being what is called a "Riemannian Homogeneous Space") 
{This point was earlier framed in a way which made the bi-implication false as pointed out by some people} The formulation should be as follows. 
A Riemannian Homogeneous Space is a riemannian manifold on which the isometry group acts transitively. Now the theorem is that such a space is compact IFF its isometry group is compact. 
Thats the statement whose intuition I am looking for. 
Apologies for the confusion caused.  
{ This question too was not framed properly. Basically I could not figure out how to write the nabla for the connection! It should be as Jose has pointed out.} 
A riemannian manifold is locally symmetric if and only if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection.  

Can one give me the intuition behind these or give me specific references where these are proven in explicit details?

REPLY [3 votes]: Two answers (by Deane Yang and Matt Noonan) have addressed the question about sectional curvatures determining the full curvature tensor, but they seem incomplete to me. The proofs I know use the additional symmetry $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$.
Of course there's a basis for $\Lambda^2 T_xM$ consisting of decomposable two-vectors, as Deane says, but knowing a quadratic form on a basis doesn't determine it – you need to know the bilinear form on pairs of basis elements.  And in Matt's argument, why should the eigenvectors of $S$ be decomposable two-vectors?<|endoftext|>
TITLE: Do there exist modern expositions of Klein's Icosahedron?
QUESTION [31 upvotes]: Reading Serre's letter to Gray
, I wonder if now modern expositions of the themes in Klein's book
 exist. Do you know any?

REPLY [2 votes]: You could also take a look at Section 1.6 of Finite Mobius Groups, Immersion of Spheres, and Moduli, by Gabor Toth.<|endoftext|>
TITLE: Searching the symmetric group
QUESTION [8 upvotes]: You want to design a set of yes/no questions for quickly searching the symmetric group. The questions have to be of the form "Does your permutation move $a_1$ to $b_1$ or $a_2$ to $b_2$ or ... or $a_k$ to $b_k$?" Given a random permutation, you will ask all of your questions about that permutation, and your goal is to know what the permutation is once you know the answers to your questions.
In particular, note that you aren't allowed to have later questions depend on the answers to earlier questions. You always ask the same questions of whichever random permutation you get.
Here's a slightly different way to phrase the type of question you're allowed to ask. If we represent the elements of the symmetric group by permutation matrices, then the questions you can ask involve picking a (0,1)-matrix and asking if the random permutation matrix has any 1 where your chosen matrix has a 1 (i.e., if I apply the componentwise AND function to the random matrix and your chosen matrix, do I get 0 or 1?)
Two questions:

How many such questions are needed in order to search $S_n$? How much bigger than $log_2 n!$ is this?
Are there "good" strategies for designing sets of questions that can be used for any n? Can we achieve the minimum number of questions for each n with a single strategy?

REPLY [6 votes]: It certainly isn't difficult to obtain an upper bound of $n\lceil \log_2 n \rceil = O(\log_2 n!)$.  All you need to do is to separately learn each bit of each value of the permutation.  This looks like it is only roughly optimal, but actually it is already $(1+o(1))(\log_2 n!)$.  (Thanks to t3suji for noticing this.)  However, the $o(1)$ term only decays logarithmically in this algorithm, so you could ask whether you could make it $(1+O(n^{-\alpha}))(\log_2 n!)$.
I agree with Michael that it seems similar to the asymptotic question of sorting with comparisons.  However, in the usual version of that problem, you are allowed to move elements based on an incomplete list of comparisons, which then amounts to adaptive comparisons.  You could call this new question "non-adaptive Mastermind for permutations with matrix guesses".  Michael also suggested the restricted problem of permutation guesses, which seems like a more difficult problem to me.<|endoftext|>
TITLE: What are the fibrant objects in the injective model structure?
QUESTION [18 upvotes]: If C is a small category, we can consider the category of simplicial presheaves on C. This is a model category in two natural ways which are compatible with the usual model structure on simplicial sets. These are called the injective and projective model structures, and in both the weak equivalences are the levelwise weak equivalences, i.e. those natural transformations $X \to Y$ such that $X(c) \to Y(c)$ is a weak equivalence for all $c \in C$. 
The cofibrations in the injective model structure are the levelwise cofibrations. The injective fibrations are then defined as those maps which have the left-lifting property with respect to all cofibrations which are also weak equivalences. The injective fibrant objects are those simplicial presheaves in which the map to the terminal presheaf is an injective fibration. 
My Question: Is there a better characterization of the fibrant objects? If I have an object are there any shortcuts which I can use to test if it is fibrant? An obvious necessary condition is that each simplicial set $X(c)$ must be fibrant (i.e. a Kan complex), but I want sufficient conditions. I'm willing to put restrictions on C. 
Dually, we can look at the projective model structure where the fibrations are the levelwise fibrations, and  cofibrations have the extension property with respect to these. I'd also be interested in knowing about the cofibrant objects in this model structure, although for now I'm more interested in the injective model structure.

REPLY [8 votes]: In the introduction to his paper "Flasque Model Structures for Presheaves" (in fact simplicial presheaves) Isaksen states on the top of page 2 that his model structure has a nice characterisation of fibrant objects and that "This is entirely unlike the injective model structures, where there is no explicit description of the fibrant objects". This would answer your question. It might be my ignorance, but I think there is no justification for Isaksen quoted statement except that no characterisation is known as yet.<|endoftext|>
TITLE: Why is Top_4 a reflective subcategory of Top_3?
QUESTION [10 upvotes]: Hi, 
I’m studying some category theory by reading Mac Lane linearly and solving exercises.  
In question 5.9.4 of the second edition, the reader is asked to construct left adjoints for each of the inclusion functors Top_{n+1} in Top_n, for n=0, 1, 2, 3, where Top_n is the full subcategory of all T_n-spaces in Top, with T_4=Normal, T_3=Regular, etc.
For n=0, 1, 2, it seems to me that I can use the AFT, with the solution set constructed similarly to the one constructed for proving that Haus (=Top_2) is a reflective subcategory of Top (Proposition 5.9.2, p. 135 of Mac Lane).
But I can’t figure out what should I do with the case of n=3, that is, with the inclusion functor Top_4 in Top_3:  Top_4 doesn’t even have products, so it seems that I cannot use the AFT.  
Is there some direct construction of this left adjoint (by universal arrows, perhaps)?  Answers including a reference would be especially helpful.

REPLY [12 votes]: I think that MacLane made a mistake.  I think that he just forgot that the category of $T_4$ spaces lacks closure properties.
Claim: If $\mathcal{A} \subseteq \mathcal{C}$ is a (full) reflective subcategory and $\mathcal{C}$ has finite products, then $\mathcal{A}$ is closed under $\mathcal{C}$'s finite products, up to isomorphism.
If $\mathcal{A}$ is reflective, it means that an object $X \in \mathcal{C}$ has an "$\mathcal{A}$-ification" $X'$, e.g., an abelianization in the case where $\mathcal{A}$ is abelian groups and $\mathcal{C}$ is groups.  If $A, B \in \mathcal{A}$ are objects, then they have a product $A \times B$ in $\mathcal{C}$, and then that has an $\mathcal{A}$-ification $(A \times B)'$.  There is an $\mathcal{A}$-ification morphism $A \times B \to (A \times B)'$, and there are also projection morphisms $A \times B \to A, B$.  Since $A, B \in \mathcal{A}$, the projection morphisms factor through $(A \times B)'$, and then the universal property gives you a morphism $(A \times B)' \to A \times B$.  So you get canonical morphisms in both directions between $A \times B$ and $(A \times B)'$, and I think that some routine bookkeeping shows that they are inverses.
So the Sorgenfrey plane (which is the non-normal Cartesian square of the Sorgenfrey line) does not have a "$T_4$-ification".
I got onto this track after I found the paper, Reflective subcategories and generalized covering spaces, by Kennison.  Kennison lists closure under products as a necessary condition for reflectiveness.<|endoftext|>
TITLE: Satisfiability of general Boolean formulas with at most two occurrences per variable
QUESTION [20 upvotes]: (If you know basics in theoretical computer science, you may skip immediately to the dark box below. I thought I would try to explain my question very carefully, to maximize the number of people that understand it.)
We say that a Boolean formula is a propositional formula over some 0-1 variables $x_1,\ldots,x_n$ involving AND and OR connectives, where the atoms are literals which are instances of either $x_i$ or $\neg x_i$ for some $i$. That is, a Boolean formula is a "monotone" formula over the $2n$ atoms $x_1,\ldots, x_n, \neg x_1,\ldots, \neg x_n$. For example, $$((\neg x_1 ~OR~ x_2) ~AND~ (\neg x_2 ~OR~ x_1)) ~OR~ x_3$$ is a Boolean formula expressing that either $x_1$ and $x_2$ take on the same value, or $x_3$ must be $1$. We say that a 0-1 assignment to the variables of a formula is satisfying if the formula evaluates to $1$ on the assignment.
The Cook-Levin theorem says that the general satisfiability of Boolean formulas problem is $NP$-complete: given an arbitrary formula, it is $NP$-hard to find a satisfying assignment for it. In fact, even satisfying Boolean formulas where each variable $x_i$ appears at most three times in the formula is $NP$-complete. (Here is a reduction from the general case to this case: Suppose a variable $x$ appears $k > 3$ times. Replace each of its occurrences with fresh new variables $x^1, x^2, \ldots, x^k$, and constrain these $k$ variables to all take on the same value, by ANDing the formula with $$(\neg x^1 ~OR~ x^2) ~AND~ (\neg x^2 ~OR~ x^3) ~AND~ \cdots ~AND~ (\neg x^{k-1} ~OR~ x^k) ~AND~ (\neg x^k ~OR~ x^1).$$ The total number of occurrences of each variable $x^j$ is now three.) On the other hand, if each variable appears only once in the formula, then the satisfiability algorithm is very easy: since we have a monotone formula in $x_1,\ldots, x_n, \neg x_1,\ldots, \neg x_n$, we set $x_i$ to $0$ if $\neg x_i$ appears, otherwise we set $x_i$ to $1$. If this assignment does not get the formula to output $1$, then no assignment will.

My question is, suppose every variable appears at most twice in a general Boolean formula. Is the satisfiability problem for this class of formulas $NP$-complete, or solvable in polynomial time? 

EDIT: To clarify further, here is an example instance of the problem:
$$((x_1 ~AND~ x_3) ~OR~ (x_2 ~AND~ x_4 ~AND~ x_5)) ~AND~ (\neg x_1 ~OR~ \neg x_4) ~AND~ (\neg x_2 ~OR~ (\neg x_3 ~AND~ \neg x_5))$$
Note that when we restrict the class of formulas further to conjunctive normal form (i.e. a depth-2 circuit, an AND of ORs of literals) then this "at most twice" problem is known to be solvable in polynomial time. In fact, applying the "resolution rule" repeatedly will work. But it is not clear (at least, not to me) how to extend resolution for the class of general formulas to get a polytime algorithm. Note when we reduce a formula to conjunctive normal form in the usual way, this reduction introduces variables with three occurrences. So it seems plausible that perhaps one might be able to encode an $NP$-complete problem in the additional structure provided by a formula, even one with only two occurrences per variable.
My guess is that the problem is polynomial time solvable. I'm very surprised that I could not find a reference to this problem in the literature. Perhaps I'm just not looking in the right places.
UPDATE: Please think about the problem before looking below. The answer is surprisingly simple.

REPLY [20 votes]: A theorem in a paper of Peter Heusch, "The Complexity of the Falsifiability Problem for Pure Implicational Formulas" (MFCS'95), seems to suggest the problem is NP-hard.  I repeat the first part of its proof here:
By reduction from the restricted version of 3SAT where every variable occurs at most 3 times.
Given such a CNF, WOLOG every variable with 3 occurrences occurs once positively and twice negatively.  If $x$ is such a variable, let $C_1$, $C_2$ be the clauses such that $C_1 = \neg x \vee C_1'$ and $C_2 = \neg x \vee C_2'$.  Introduce new variables $x'$ and $x''$ and replace $C_1$ and $C_2$ by $\neg x \vee (x' \wedge x'')$, $\neg x' \vee C_1'$, $\neg x'' \vee C_2'$.  Repeat.<|endoftext|>
TITLE: Recursively dependent types?
QUESTION [6 upvotes]: Is there such a thing as "recursively dependent types"?  Specifically, I would like a dependent type theory containing a type $A(x)$ which depends on a variable $x: A(z)$, where $z$ is a particular constant of type $A(z)$.
This may be more "impredicative" than some type-theorists would like, but from the perspective of semantics in locally cartesian closed categories, I can't see any reason it would be a problem: the type $A$ comes with a display map to $A_z$, while $A_z$ itself is the pullback of this display map along a particular morphism $z \colon 1\to A_z$.  But I want to know whether a corresponding syntax exists.

REPLY [3 votes]: Mike, if you consider that locally cartesian closed categories
provide the canonical semantics for dependent type theories then
you may as well just use sets, for which any function
$p:A\to X$ provides a dependent type $A[x]=\{a|p((a)=x\}$.
Not only is this a very dull notion of dependent type,
but it gives no account of the way in which $A{x]$ might depend
"continuously" on $x$,
something that we probably need to understand in order to
give a meaning to the word "recursive".
(Local, relative or ordinary) cartesian closure is needed to
interpret function- or Pi-types, which do not feature in your
question.  The appropriate arena is a category with some finite
limits and something infinitary to capture the recursion.
A class of display maps is a class of morphisms that is
closed under (composition with isomorphisms and) pullback against
arbitrary maps in the category.
This categorical notion is equivalent to that of a dependent type
theory in the basic algebraic sense, ie with types, terms, equations
and structural rules.
As I believe you are more comfortable with a categorical language,
you can solve your problem in that setting and then use the equivalence
to reformulate it symbolically.
In particular, the class of display maps includes
- all isomorphisms iff the type theory includes singleton dependent types;
- composites iff the type theory has Sigma types;
- inclusions of diagonals and hence all maps iff the type theory has
  equality types;
- relative cartesian closure corresponds to Pi types.
I had originally interpreted your "recursively dependent" types
to mean an infinite chain of dependencies, and hence of display maps.
For that you would want the class of displays to be closed under
cofiltered limits.
Neel, on the other hand, read it as a fixed point equation,
which we can interpret categorically as the fixed point of a functor.
Unsurprisingly, domain theory would be a useful setting in which
to look for models of these situations.
Indeed
my PhD thesis
introduced classes of display maps in order to study dependent types
in domain theory, 
and you might like to look at the last chapter for investigations
of appropriate notions of displays of domains.
For the theory of display maps and their equivalence with dependent
types, my thesis was completely superseded by 
Chapter VIII
of my book, "Practical Foundations of Mathematics" (CUP 1999).
For the interpretation of dependent types in domain theory,
Martin Hyland and Andrew Pitts gave a comprehensive account in
their paper
The Theory of Constructions: Categorical Semantics and
Topos-Theoretic Models in
Categories in Computer Science and Logic
edited by John Gray and Andre Scedrov, 
AMS Contemporary Mathematics 92 (1989).<|endoftext|>
TITLE: Mapping Class Groups of Punctured Surfaces (and maybe Billiards)
QUESTION [14 upvotes]: Where can I find a concrete description of mapping class group of surfaces?  I know the mapping class group of the torus is $SL(2, \mathbb{Z})$.  Perhaps, there is a simple description for the sphere with punctures or the torus with punctures.  Also, I would appreciate any literature reference for an arbitrary surface of genus g with n punctures.  
Mapping class groups come up in my reading about billiards and the geodesic flow on flat surfaces.  I wonder: the moduli space of complex structures on the torus is $\mathbb{H}/SL(2, \mathbb{Z})$, is it a coincidence the mapping class group appears here?

REPLY [4 votes]: Here is an arcticle of Luo, 
A Presentation of the Mapping Class Groups
http://arxiv.org/PS_cache/math/pdf/9801/9801025v1.pdf
I remember seen somewhere a neat presentation of the hyperelliptic mapping class group, 
but I can't remeber now the reference, but you can check page 12 in the article 
Hyperelliptic Szpiro inequality 
http://arxiv.org/PS_cache/math/pdf/0106/0106212v1.pdf<|endoftext|>
TITLE: On Quadratic Integer Programming
QUESTION [5 upvotes]: I have the quadratic integer program over $\mathbb{Z}^n$
$\displaystyle\min_{z \in \mathbb{Z}^n} \Phi (z) = \frac{1}{2} z^T Q z - r^T z + s$
subject to $G z = h$, and $z_i \in \{0,1,2,\dots, b_i\}$ for all $i \in \{1,2,\dots,n\}$, where $Q$ is symmetric positive-definite. Moreover, $G, h$ are integer-valued and, thus, I suppose that $\{z \in \mathbb{Z}^n : G z = h\}$ defines a sublattice of $\mathbb{Z}^n$. 
Let us suppose we solve the relaxed quadratic program over $\mathbb{R}^n$
$\displaystyle\min_{x \in \mathbb{R}^n} \Psi (x) = \frac{1}{2} x^T Q x - r^T x + s$
subject to $G x = h$, and $0 \leq x_i \leq b_i$ for all $i \in \{1,2,\dots,n\}$. Let $x_{opt} \in \mathbb{R}^n$ be the minimizer of $\Psi$. We can define an $n$-cube (in $\mathbb{Z}^n$) containing $x_{opt}$ by taking the floor/ceil of each component of $x_{opt}$. We intersect this $n$-cube with the integer sublattice defined by $G z = h$, and evaluate $\Phi$ at all points of this intersection. Let $z^{\ast}$ be the point in such intersection that minimizes $\Phi$. Let $z_{opt}$ be the minimizer of the original quadratic integer program. Can we say that $z_{opt} = z^*$? I know nothing of integer programming, so I preemptively apologize if my question is silly / elementary...
In other words, can one solve a quadratic integer program by solving a relaxed quadratic real program, and then searching in the neighborhood of the real solution? This seems to work for $n = 1$ and $n = 2$... but it also seems too good to be true in general. If in general $z_{opt} \neq z^{\ast}$, can we (at least) quantify how sub-optimal $z^{\ast}$ is?
Any feedback will be most welcome!

REPLY [9 votes]: The relaxed quadratic programming problem is a red herring.  It is true that quadratic programming over $\mathbb{R}$ with linear inequalities can be solved in practice, for one reason because it is a special case of convex programming.  But in the stated question, the inequality $0 \le x_i \le b_i$ came from nowhere.  The correct relaxation is even simpler:  You should just minimize $\Phi(z)$ over all of $\mathbb{R}^n$, and the minimum is directly at $z_0 = Q^{-1}r$.
After that, Mitch is roughly correct.  The question as stated is exactly the closest vector problem, which is related to the shortest vector problem that Mitch mentions.  The constant $s$ is not important.  The question is to find the integer lattice point $z$ which is the closest to $z_0$ in the metric defined by $Q$.  If you like, you can change distance to Euclidean distance, and change the lattice from the standard integer lattice to something else, by applying the operator $Q^{-1/2}$.  $L = Q^{-1/2}(\mathbb{Z}^n)$ is a certain lattice, and you are looking for the point which is the closest in Euclidean distance to $Q^{-1/2}(z_0)$.
In any fixed dimension $n$, the closest and shortest vector problems can be solved in polynomial time.  There are various lattice reduction algorithms that only search polynomially many points.
If the dimension $n$ is a parameter, then the situation is very different.  For many purposes, people are happy with just a close vector or a short vector, not necessarily the closest or shortest one.  The problem varies greatly in difficulty depending on how close is good enough, or equivalently whether there are few lattice points that stick out as much closer than all of the others.  Close vector is intuitively harder than short vector, but there is a theoretical result that they are roughly equivalent in difficulty.
Taking the strictest possible requirements, finding a close vector is NP-hard.  There are intermediate levels of closeness, given by some tolerance that grows with $n$, that seem hard but are probably not NP-hard.  Other levels of closeness can be done in polynomial time.  There are lots of papers on the these two problems, and the Wikipedia page that Mitch mentions is a pretty good review:  The GapCVP section addresses the approximate versions of the question that I mention briefly here.
One weakness of the Wikipedia page is that it has more to say about hardness than practical algorithms.  But it does mention two important algorithms:  Lenstra-Lenstra-Lovasz and Ajtai-Kumar-Sivakumar.<|endoftext|>
TITLE: Simultaneous diagonalization
QUESTION [12 upvotes]: I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let $k$ be a field, let $V$ be a finite-dimensional $k$-vector space and let $S \subseteq \mathrm{End}_k(V)$ be a subset of pairwise commuting (i.e. $\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds:

If all $f \in S$ are diagonalizable, then there exist maps $\chi_i:S \rightarrow k$, $i=1,\ldots,r$, such that $V = \bigoplus_{i=1}^r E_{\chi_i}(S)$, where $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. 
The maps $\chi_i$ in 1 are unique.
1 is equivalent to the existence of a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is diagonal. (I believe that this might not be true)
If all $f \in S$ are trigonalizable, then there exists a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is upper triangular and for each diagonalizable $f \in S$ the matrix $M_{\mathcal{B}}(f)$ is diagonal.

I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the $\chi$-maps when I try to prove this by induction on $\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think.
Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :)
One additional question: Suppose that $k$ is algebraically closed and that $G$ is an affine commutative algebraic group over $k$ which coincides with its semisimple part, embedded as a closed subgroup in some $GL(V)$. Are the maps $\chi_i:G \rightarrow \mathbb{G}_{m}$ morphisms of algebraic groups?

REPLY [6 votes]: And the answer to the additional question (which Ben skipped) is also positive.
Indeed, there are two parts here:

$\chi_i$ is a group morphism (this is obvious for any subgroup $G\subset GL(V)$
$\chi_i$ is algebraic (this is obvious because matrix elements of matrices in $GL(V)$
in any basis are algebraic functions for tautological reasons).<|endoftext|>
TITLE: References for Donaldson-Thomas theory and Pandharipande-Thomas theory?
QUESTION [11 upvotes]: I'm looking for good introductory references for Donaldson-Thomas theory and Pandharipande-Thomas theory. I know a bit about Gromov-Witten theory, but almost nothing about Donaldson-Thomas and Pandharipande-Thomas. Are there some canonical (or good non-canonical) references for Donaldson-Thomas theory and Pandharipande-Thomas theory? References that assume knowledge of Gromov-Witten theory are fine.

REPLY [9 votes]: This survey by Pandharipande and Thomas appeared today on arXiv:
13/2 ways to count curves
It should give a partial answer to your question. 
They give an elementary review of 6 ways to count curves and the relations between them.
These approaches are based on Gromov-Witten theory, Copakumar-Vafa / BPS invariants,
Donaldson-Thomas theory, Stable pairs, Stable unramified maps and Stable quotients. 
In the appendix of the paper they give a nice review of Virtual classes.
As they say at the beginning of the paper their goal is to provide a guide for graduate students looking for an elementary route into the subject.
I wrote this since I am very excited seeing the paper and wanted to post it here for those who have not yet looked at it.<|endoftext|>
TITLE: The shortest path in first passage percolation
QUESTION [39 upvotes]: Update (January 17): The problem has now been solved by Daniel Ahlberg and Christopher Hoffman.  (Thanks to Matt Kahle for informing us.)

Consider a square planar grid. (The vertices are pair of points in the plane with integer coordinates and two vertices are adjacent if they agree in one coordinate and differ by one in the other.)
Give every edge a length one with probability a half and length two with probability a half.
Consider a shortest path between the origin and the vertex $(n,0)$.
Show that with probability that tends to one as $n$ tends to infinity the shortest path will not contain the "middle edge" on the x-axis inbetween the orgin and $(n,0)$. (Namely, the edge between the vertices $(\lfloor\frac{n}{2}\rfloor,0)$ and $(\lfloor\frac{n}{2}\rfloor+1,0)$.)

This question is in the category of "a missing lemma". It is not really a full fledged open problem but rather a statement which looks correct that was needed in some paper and resisted proof. Of course, some such "lemmas" turn out to be very difficult, but sometimes maybe a simple argument was simply missed. The relevant paper is

Itai Benjamini, Gil Kalai, and Oded Schramm, First Passage Percolation Has Sublinear Distance Variance,  Ann. Probab. 31(4) (2003) pp 1970-1978, doi:10.1214/aop/1068646373, arXiv:math/0203262.


While MO have chosen to accept one answer, and there were some nice suggestions, the problem is still wide open.

REPLY [7 votes]: It seems that this problem was recently solved by Ahlberg and Hoffman.
https://arxiv.org/abs/1609.02447<|endoftext|>
TITLE: Canonical topology on the category of schemes?
QUESTION [21 upvotes]: Every category admits a Grothendieck topology, called canonical, which is the finest topology which makes representable functor into sheaves.
Is there a concrete description of the canonical topology on the category of schemes? By Grothendieck's results on descent this is at least as fine as the fpqc topology, but I don't even know if the two actually coincide. If not, what is known about it?

REPLY [5 votes]: following the answers given by Pantev, I will give you more example of canonical topology on some category(universally strict epimorphism)
if C is an abelian category or a topos, then canonical topology consists of all epimorphism
if C is a quasi abelian category,then canonical topology consists of all strict epimorphism(Note that strict epimorphism is subcanonical in general)
if C is a category of associative unital k-algebras(opposite category of affine schemes,not necessarily commutative). Canonical topology consists of all strict epimorphism which are precisely surjective morphism of algebras. In this case, univerally strict epimorphism coincides with strict epimorphism. 
Rosenberg has a very detailed treatment for a category and 2-category(taken as category of spaces)with canonical topology(he called right exact structure).It is in MPIM preprint series," Homological algebra of noncommutative 'space' I"
What pantev mentioned is related to my answers in another question: 
Does sheafification preserve sheaves for a different topology?
The effective descent topology is finer that fpqc topology, fppf topology,smooth topology(in Kontsevich-Rosenberg sense)
The descent topology on category of affine schemes(not necessarily commutative) coincides with subcanonical topology
The reference is Orlov's paper and Kontsevich-Rosenberg MPIM preprint series. Noncommutative stack and Noncommutative grassmannian and related construction<|endoftext|>
TITLE: Smooth proper scheme over Z
QUESTION [62 upvotes]: Does every smooth proper morphism $X \to \operatorname{Spec} \mathbf{Z}$ with $X$ nonempty have a section?
EDIT [Bjorn gave additional information in a comment below, which I am recopying here.  -- Pete L. Clark]
Here are some special cases, according to the relative dimension $d$. If $d=0$, a positive answer follows from Minkowski's theorem that every nontrivial finite extension of $\mathbf{Q}$  ramifies at at least one prime. If $d=1$, it is a consequence (via taking the Jacobian) of the theorem of Abrashkin and Fontaine that there is no nonzero abelian scheme over $\mathbf{Z}$, together with (for the genus $0$ case) the fact that a quaternion algebra over $\mathbf{Q}$ split at every finite place is trivial.

REPLY [68 votes]: Hey Bjorn. Let me try for a counterexample. Consider a hypersurface in projective $N$-space, defined by one degree 2 equation with integral coefficients. When is such a gadget smooth? Well the partial derivatives are all linear and we have $N+1$ of them, so we want some $(N+1)$ times $(N+1)$ matrix to have non-zero determinant mod $p$ for all $p$, so we want the determinant to be +-1. The determinant we're taking is that of a symmetric matrix with even entries down the diagonal (because the derivative of $X^2$ is $2X$) and conversely every symmetric integer matrix with even entries down the diagonal comes from a projective quadric hypersurface. So aren't we now looking for a positive-definite (to stop there being any Q-points or R-points) even unimodular lattice?
So in conclusion I think that the hypersurface cut out by the quadratic form associated in this way to e.g. the $E_8$ lattice or the Leech lattice gives a counterexample!<|endoftext|>
TITLE: Equivariant singular cohomology
QUESTION [10 upvotes]: One can define the $G$-equivariant cohomology of a space $X$ as being the ordinary singular cohomology of $X \times_G EG$ --- I think this is due to Borel? (See e.g. section 2 of these notes)
Alternatively if $X$ is a manifold, we also have $G$-equivariant de Rham cohomology, defined in terms of $G$-equivariant differential forms --- I think this is due to Cartan? (See e.g. section 3 of loc. cit.)
I suspect this is extremely standard or obvious, but if it is, I don't know where it's written down: Is it possible to define equivariant cohomology of a topological space in terms of some notion of "equivariant singular cochains", that is, without using the Borel construction?

REPLY [10 votes]: In 1965 or so, Glen Bredon defined ordinary equivariant cohomology, ordinary meaning 
that it satisfies the dimension axiom:  For each coefficient system $M$ (contravariant 
functor from the orbit category of G to the category of Abelian groups), there is a
unique cohomology theory $H^*_G(-;M)$ such that, when restricted to the orbit category, 
it spits out the functor $M$.  Just as in the nonequivariant world, it can be defined
using either singular or cellular cochains, the latter defined using $G$-CW complexes.
This works as stated for any topological group $G$.
For an abelian group $A$, Borel cohomology with coefficients in $A$, $H^*(EG\times_G X;A)$
is the extremely special case in which one takes $M$ to be the constant coefficient system
$\underline{A}$ at the group $A$ and replaces $X$ by $EG\times X$.  That is,
$$ H^*_G(EG\times X; \underline{A}) = H^*(EG\times_G X;A) $$<|endoftext|>
TITLE: Generalizations of "standard" calculus
QUESTION [19 upvotes]: We have the usual analogy between infinitesimal calculus (integrals and derivatives) and finite calculus (sums and forward differences), and also the generalization of infinitesimal calculus to fractional calculus (which allows for real and even complex powers of the differential operator). Have people worked on a "fractional finite" calculus, where instead of a differintegral we had some sort of "differsum"? I don't know much about it, but I was thinking maybe the answer might come from umbral calculus?
To give a motivating example/special case for this question: the Wikipedia article on fractional calculus uses the example of the $\frac{1}{2}$th derivative, which when applied twice gives the standard derivative. What is the operator $D$ on sequences such that, when applied twice, it gives the forward difference of the original sequence?
Also, I have perhaps a related question: The solution to $\frac{d}{dx}f=f$ is $f=e^x$, while the solution to $\Delta f = f$ is $f=2^x$. Is the fact that $e$ is close to 2 a coincidence, or is there something connecting these results? Is there more generally some sort of spectrum of calculi lying between "finite" and "infinitesimal" each with its own "$e$"?
EDIT: After looking around some more I found time scales, which are pretty much what I was thinking of in the second part of my question (though many of the answers people have provided are along the same general lines). I'm surprised I don't hear more about this in analysis - unifying discrete and continuous should make it a pretty fundamental concept!

REPLY [2 votes]: Just to draw the attention of the curious reader to take a look at the book titled "Discrete
Calculus by Analogy" witten by Izadi, Aliev, and Bagirev to see the differences as well as 
similirities between Discrete Calculs and Continous one.<|endoftext|>
TITLE: Eichler-Shimura isomorphism and mixed Hodge theory
QUESTION [18 upvotes]: Let $Y(N),N>2$ be the quotient of the upper half-plane by $\Gamma(N)$ (which is formed by the elements of $SL(2,\mathbf{Z})$ congruent to $I$ mod $N$). Let $V_k$ be the $k$-th symmetric power of the Hodge local system on $X(N)$ tensored by $\mathbf{Q}$ (the Hodge local system corresponds to the standard action of $\Gamma(N)$ on $\mathbf{Z}^2$).
$V_k$ is a part of a variation of polarized Hodge structure of weight $k$. So the cohomology $H^1(Y(N),V_k)$ is equipped with a mixed Hodge structure (the structure will be mixed despite the fact that $V_k$ is pure because $Y(N)$ is not complete). The complexification $H^1(Y(N),V_k\otimes\mathbf{C})$ splits
$$H^1(Y(N),V_k\otimes\mathbf{C})=H^{k+1,0}\oplus H^{0,k+1}\oplus H^{k+1,k+1}.$$
There is a natural way to get cohomology classes $\in H^1(Y(N),V_k\otimes\mathbf{C})$ from modular forms for $\Gamma(N)$. Namely, to a modular form $f$ of weight $k+2$ one associates the secion
$$z\mapsto f(z)(ze_1+e_2)^k dz$$
of $$Sym^k(\mathbf{C}^2)\otimes \Omega^1_{\mathbf{H}}.$$
Here $\mathbf{H}$ is the upper half plane and $(e_1,e_2)$ is a basis of $\mathbf{C}^2$ coming from a basis of $\mathbf{Z}^2$. This pushes down to a holomorphic section of 
$V_k\otimes \mathbf{C}$.
Deligne had conjectured (Formes modulaires et repr\'esentations l-adiques, Bourbaki talk, 1968/69) that the above correspondence gives a bijection between the cusp forms of weight $k+2$ and $H^{k+1,0}$. (This was before he had even constructed the Hodge theory, so strictly speaking this can't be called a conjecture, but anyway.) Subsequently this was proved by Zucker (Hodge theory with degenerating coefficients, Anns of Maths 109, no 3, 1979). See also Bayer, Neukirch, On automorphic forms and Hodge theory, (Math Ann, 257, no 2, 1981).
The above results concern cusp forms and it is natural to ask what all modular forms correspond to in terms of Hodge theory. It turns out that all weight $k+2$ modular forms give precisely the $k+1$-st term of the Hodge filtration on $H^1(Y(N),V_k\otimes\mathbf{C})$ i.e. $H^{k+1,0}\oplus H^{k+1,k+1}$.
The proof of this is not too difficult but a bit tedious. So I would like to ask: is there a reference for this?
upd: The original posting contained non-standard notation; this has been fixed.

REPLY [7 votes]: The argument given in Elkik, Le théorème de Manin-Drinfelʹd, seems to generalize also to nontrivial coefficient systems. The "usual" Manin-Drinfel'd theorem claims that the difference of two cusps has finite order in the Jacobian of the modular curve. Elkik reformulates this as asserting that the mixed Hodge structure on $H^1$ of the open modular curve is the direct sum of pure Hodge structures. It is this fact that she then proves by realizing the weight spaces as eigenspaces for the action of the Hecke operators.<|endoftext|>
TITLE: Subspaces of $L^{2}$ 
QUESTION [6 upvotes]: [In what follows $0^{0}$= 1 by convention.]
Is there some closed infinite dimensional linear subspace $F$ of $L^{2}(0,1)$
such that $\left|f\right|^{\left|f\right|}$ belongs to $L^{2}(0,1)$
for all $f$ in $F$ ?
This problem is related to the Erdos - Shapiro - Shields paper [ESS].
From this paper it follows that the answer is negative if $\left|f\right|^{\left|f\right|}$ is replaced by $\left|f\right|^{\left|f\right|^{2}}$.
Some thoughts. Suppose that such an $F$ exists, and take some $p > 2$.
Let $f$ be in $F$.  
Then clearly $g:=(p/2)\cdot f$  is in $F$, too, hence $h :=\left|g\right|^{\left|g\right|}$
belongs to $L^{2}(0,1)$.
Next, it is easy to see that $\left\Vert f\right\Vert _{p}^{p}\leq1+\left\Vert h\right\Vert _{2}^{2}<+\infty$. 
Therefore, $F$ is contained in $L^{p}(0,1)$ as a linear subspace
(i.e., algebraically). 
Now, applying the Closed Graph Theorem to the natural linear embedding
$j:(F, ||.||_{2})\rightarrow L^{p}(0,1)$, it follows that $j$ is
continuous. Consequently, the Hilbertian 2-norm and the $p$-norm
are equivalent on $F$. Moreover, it follows that $F$ is complete
w.r.t. the $p$-norm, and, in turn, it is a closed subspace of $L^{p}(0,1)$.
And this is true for all $p > 2$.
[ESS] http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.mmj/1028999306

REPLY [4 votes]: The classical lacunary series example allows you to integrate anything that is $e^{O(|f|^2)}$, so it works for your question. It seems that for reasonable (say, positive, increasing, and convex) functions $\Phi$, the complete answer is the following: A closed infinite-dimensional subspace $F$ such that $\Phi(|f|)$ is integrable for all $f\in F$ exists if and only if $\log\Phi(x)=O(x^2)$. The reason is that you can always create a random function in the unit ball that is pointwise almost Gaussian by taking a random linear combination of large number $n$ of the elements in the orthonormal basis of $F$ with random coefficients of size $n^{-1/2}$.<|endoftext|>
TITLE: Finitely presented sub-groups of GL(n,C)
QUESTION [19 upvotes]: Here are two questions about finitely generated and finitely presented groups (FP):
1) Is there an example of an FP group that does not admit a homomorphism to $GL(n,C)$ with trivial kernel for any n?
The second question is modified according to the sujestion of Greg below.
2) For which $n$ given two subgroups of $GL(n,C)$ generated by explicit lists of matrices, together with finite lists of relations and the promise that they are sufficient, is there an algorithm to determine if they are isomorphic as groups?" 
In both cases we don't impose any conidtion on the group (apart from been FP), in particular it need not be discrete in $GL(n,C)$.

REPLY [10 votes]: I just wanted to make a comment on Mal'cev's theorem (if I could leave this as a comment, I would).
Mal'cev's paper is a great exposition of the theorem, as well as a lot of other related material, all written in a basic yet enlightening style.
Also, if you know a little commutative algebra (as in the Nullstellensatz, the one given in Eisenbud pg. 132), there is quick and easy proof of Mal'cev's theorem. I could sketch it if necessary, but I am right now in the process of LaTeX-ing it, so I'll probably just come back and post a link.
Steve
EDIT - a sketch of the argument:
Mal'cev's theorem says a finitely generated linear group is residually finite. So let $X\subset GL(n,F)$ be a finite subset of the general linear group over some field $F$, and $G=\langle X \rangle$.  First, make $X$ symmetric, so that if $x\in X$ then also $x^{-1}\in X$.  Each $x\in X$ is an $n\times n$ matrix, so we can assemble all entries from all elements of $X$, getting a finite subset of $F$.  Let $R$ denote the subring of $F$ generated by this subset (along with $1$).  Then $R$ is a Jacobson ring, and since it's a subring of $F$, it's Jacobson radical is $0$.  Now $G$ is  a subgroup of $GL(n,R)$;  let $g\in G$ be a non-identity element, so that $g-I_n\neq 0$, where $I_n$ is the identity matrix.  Thus $g-I_n$ has a non-zero element, and thus there is some maximal ideal $m\subset R$ not containing this non-zero element.  The matrix ring homomorphism $M_n(R)\rightarrow M_n(R/m)$ (reducing everything mod $m$) induces a group homomorphism $G\rightarrow GL(n,R/m)$, where $g$ is not in the kernel. But $R/m$ is finite (by the Nullstellensatz), so $GL(n,R/m)$ is a finite group.<|endoftext|>
TITLE: Sites which are stacks over themselves
QUESTION [8 upvotes]: A site C with pullbacks is subcanonical (all representable presheaves are sheaves) if and only if its codomain fibration $Arr(C) \to C$ is a prestack (all hom-presheaves are sheaves).  Is there a common name for a site whose codomain fibration is a stack?  The canonical topology on a Grothendieck topos has this property, as does the coherent topology on a pretopos, the regular topology on a Barr-exact category, the extensive topology on a lextensive category, etc.

REPLY [10 votes]: I don't have an answer to your question, but I'm going to post whatever thoughts I had about it. Maybe something here will help someone answer the question, or at least help more people understand what's involved. I'm sorry that it's come out so long.

Definitions(skip this unless you suspect we mean different things by "(pre)stack")
A functor $F\to C$ is a fibered category if for every arrow $f:U\to X$ in $C$ and every object $Y$ in $F$ lying over $X$, there is a cartesian arrow $V\to Y$ in $F$ lying over $f$ (see Definition 3.1 of Vistoli's notes). This arrow is determined up to unique isomorphism (by the cartesian property), so I'll call $V$ "the" pullback of $Y$ along $f$ and maybe denote it $f^*Y$. A fibered category is roughly a "category-valued presheaf (contravariant functor) on $C$".
Given an object $X$ in $C$, let $F(X)$ be the subcategory of objects in $F$ lying over $X$, with morphisms being those morphisms in $F$ which lie over the identity morphism of $X$. I'll call $F(X)$ the "fiber over $X$." Given a morphism $f:U\to X$ in $C$, let $F(U\to X)$ be "the category of descent data along $f$," whose objects consist of an element $Z$ of $F(U)$ and an isomorphism $\sigma:p_2^*Z\to p_1^*Z$ (where $p_1,p_2:U\times_XU\to U$ are the projections) satisfying the usual cocycle condition over $U\times_XU\times_XU$ (see Definition 4.2 of Vistoli's notes). A morphism in $F(U\to X)$ is a morphism $Z\to Z'$ in $F(U)$ such that the following square commutes:
$\begin{matrix}
p_2^*Z & \xrightarrow{\sigma} & p_1^*Z \\
\downarrow & & \downarrow\\
p_2^*Z' & \xrightarrow{\sigma'} & p_1^*Z'
\end{matrix}$
Suppose $C$ has the structure of a site. Then we say that $F$ is a prestack (resp. stack) over $C$ if for any cover $U\to X$ in $C$, the functor $F(X)\to F(U\to X)$ given by pullback is fully faithful (resp. an equivalence). Roughly, a prestack is a "separated presheaf of categories" and a stack is a "sheaf of categories" over $C$.

The domain fibration (not your question, but related)
Consider the domain functor $Arr(C)\to C$ given by $(X\to Y)\mapsto X$. You can check that a cartesian arrow over $f:U\to X$ is a commutative square
$\begin{matrix}
U & \xrightarrow{f} & X \\
\downarrow & & \downarrow\\
Y & = & Y
\end{matrix}$
If I haven't made a mistake,

This fibered category is a prestack iff every cover $U\to X$ is an epimorphism.
It is a stack if furthermore every cover $U\to X$ is the coequalizer of the projection maps $p_1,p_2:U\times_XU\to U$. This last condition is equivalent to saying that every object $Y$ of $C$ satisfies the sheaf axiom with respect to the morphism $U\to X$. In particular, the domain fibration is a stack if and only if the topology is subcanonical.


The codomain fibration (your question)
Consider the codomain functor $Arr(C)\to C$ given by $(U\to X)\mapsto X$. You can check that a cartesian arrow over a morphism $f:U\to X$ is a cartesian square
$\begin{matrix}
V & \to & U \\
\downarrow & & \downarrow\\
Y & \xrightarrow{f} & X
\end{matrix}$
There is a general result that says that the fibered category of sheaves on a site is itself a stack (I usually call this result "descent for sheaves on a site"). If you're working with the canonical topology on a topos (where every sheaf is representable), it follows that the codomain fibration is a stack. If the topology is subcanonical, then objects are sheaves, so descent for sheaves tells you that the pullback functor is fully faithful (i.e. the codomain fibration is a prestack), but when you "descend" a representable sheaf, it may no longer be representable, so the codomain fibration may not be a stack. In your question you say that being a prestack is actually equivalent to the topology being subcanonical, but I can't see the other implication (prestack⇒subcanonical).
Supposing the codomain fibration is a prestack, saying that it is a stack roughly says that when you glue representable sheaves along a "cover relation," you get a representable sheaf, but with the strange condition that the "cover relation" you started with came from a relation where you could glue to get a representable sheaf. That is, given this diagram, where the squares on the left are cartesian ($\Rightarrow$ is meant to be two right arrows), can you fill in the "?" so that the square on the right is cartesian?
$\begin{matrix}
Z' & \Rightarrow & Z & \to & ?\\
\downarrow & & \downarrow & & \downarrow\\
U\times_XU & \Rightarrow & U & \to & X
\end{matrix}$
A more natural (to me) condition is to ask that the only sheaves you can glue together from representable sheaves are already representable. That is, if $R\Rightarrow U$ is a "covering relation" (i.e. each of the maps $R\to U$ is a covering and $R\to U\times U$ is an equivalence relation), then the quotient sheaf $U/R$ is representable. I would call such a site "closed under gluing."
For example, the category of schemes with the Zariski topology is closed under gluing (it's the "Zariski gluing closure" of the category affine schemes). The category of algebraic spaces with the etale topology is closed under gluing (it's the "etale gluing closure" of the category of affine schemes). In fact, I think that a standard structure theorem for smooth morphisms and a theorem of Artin (∃ fppf cover ⇒ ∃ smooth cover) imply that the category of algebraic spaces with the fppf topology is closed under gluing.<|endoftext|>
TITLE: How far is  Lindelöf  from compactness?
QUESTION [36 upvotes]: A while ago I heard of  a nice characterization of compactness but I have never seen a written source  of it,  so I'm starting to doubt  it.  I'm looking  for a reference, or counterexample,  for the following .  Let $X$ be a Hausdorff topological space.  Then, $X$ is compact if and only if $X^{\kappa}$ is Lindelöf  for any cardinal $\kappa$.  
If the above is indeed a fact, can one restrict the class of $\kappa$'s for which the characterization is still valid? 
Note:  Here I'm thinking under ZFC.

REPLY [6 votes]: Here is a very surprising fact I was completely unaware of until yesterday, when I found it in Herrlich´s book "Axiom of Choice". It shows that the answer to the question in the title could be "very very close":

There are models of $ZF$ in which for every $T_1$ space $X$, $X$ is Lindelöf if and only if $X$ is compact.

For instance this holds in what is called Cohen first model. Also in this model, Tychonoff´s theorem holds for Hausdorff spaces, so arbitrary products of Lindelöf Hausdorff spaces are Lindelöf.<|endoftext|>
TITLE: Is every curve birational to a smooth affine plane curve?
QUESTION [25 upvotes]: Is every curve over $\mathbf{C}$ birational to a smooth affine plane curve?
Bonnie Huggins asked me this question back in 2003, but neither I nor the few people I passed it on to were able to answer it.  It is true at least up to genus 5.

REPLY [22 votes]: Yes. Here is a proof. 
It is classical that every curve is birational to a smooth one which in turn is birational to a closed curve $X$ in $\mathbb{C}^2$ with atmost double points. Now my strategy is to choose coordinates such that by an automorphism of $\mathbb{C}^2$ all the singular points lie on the $y$-axis avoiding the origin. Now the map $(x,y)\rightarrow(x,xy)$ from $\mathbb{C}^2$ to itself will do the trick of embedding the smooth part of $X$ in a closed manner. Below are the details. 
The only thing we need to show is that the smooth locus of a closed curve $X\in\mathbb{C}^2$ with only double points can again be embedded in the plane as a closed curve. 
Step 1. Let $S$ be the set of singular points of $X$. Choose coordinates on $\mathbb{C}^2$ such that the projection of $X$ onto both the axes gives embeddings of $S$. Call the projection of $S$ on the $y$-axis as $S'$. By sliding the $x$-axis a little bit we can make sure that $S'$ doesn't contain the origin of the plane. Now I claim that there is an automorphism of $\mathbb{C}^2$ which takes $S$ to $S'$. This is easy to construct by a Chinese remainder kind of argument: There is an isomorphism of the coordinate rings of $S$ and $S'$ and we need to lift this to an isomorphism of $\mathbb{C}[x,y]$. I will illustrate with an example where #$\{S\}=3$. Let $(a_i,b_i)$ be the points in $S$. Then there exists a function $h(y)$ such that $h|S'=x|S$ as functions restricted to the sets $S$ and $S'$. Here is one recipe: $h(y)=c_1(\frac{y}{b_2}-b_2)(\frac{y}{b_3}-b_3)(y-b_1+1)+\dots$ where $c_1=a_1(\frac{b_1}{b_2}-b_2)^{-1}(\frac{b_1}{b_3}-b_3)^{-1}$ etc.
Look at the map $\phi:(x,y)\rightarrow(x-h(y),y)$ on $\mathbb{C}^2$. It is clearly an automorphism and takes the set $S$ to $S'$. 
Step 2. Now consider the map $\psi:(x,y)\rightarrow(x,xy)$ from the affine plane to itself. It is an easy check that $\psi^{-1}\circ\phi:X-S\rightarrow\mathbb{C}^2$ is a closed embedding.<|endoftext|>
TITLE: Sequence of Diophantine Equations 
QUESTION [6 upvotes]: Is there some (huge) positive integer $M$ with the following property:
for any $z>M$, there exist positive integers $x, y_{1}, y_{2},..., y_{z}$
such that $x^x$ $=$ $y_{1}^{y_{1}}$+ $y_{2}^{y_{2}}$+ ... +$y_{z}^{y_{z}}$
? 
[Please remark that the $y$'s are $\geq$ $1$ and need not
to be necessarily distinct.]
As a [rather naive] way to attack this problem [which may  (perhaps) be
related to some works of Robinson, Matiasevich, M. Davis, and Chao-Ko],
I'm thinking about lots of $1$'s, lots of $2$'s, and lots of 
$(x-1)$'s.
Also, let us observe that, if $z$ has this property and $y_{i}$ $=$ $2$
for some $i$, then $z+3$ has the same property, too...

REPLY [2 votes]: Have you tried looking at the density of possible $z$'s?
I think the answer might be "no". Here's my heuristics (hoping it's not bogus): the smallest $z$ we can achieve using $x$ is at least $\frac{x^x}{(x-1)^{(x-1)}}\approx ex$, the second smallest would be at least $\frac{x^x-(ex-1)(x-1)^{(x-1)}}{(x-2)^{(x-2)}}\approx e^2x$ and so on. 
Let $N$ be a very large integer. we look at the interval $[1,N]$ and see how many such $z$'s are in it. From the above argument there are at most $\approx \frac{N}{e}+\frac{N}{e^2}+\cdots=\frac{N}{e-1}< N$ solutions. This contradicts your conjecture that there are asymptotically $\approx N$ solutions (in terms of values of $z$)<|endoftext|>
TITLE: Is "semisimple" a dense condition among Lie algebras?
QUESTION [56 upvotes]: The "Motivation" section is a cute story, and may be skipped; the "Definitions" section establishes notation and background results; my question is in "My Question", and in brief in the title.  Some of my statements go wrong in non-zero characteristic, but I don't know that story well enough, so you are welcome to point them out, but this is my characteristic-zero disclaimer.
Motivation
In his 1972 talk "Missed Opportunities" (MR0522147), F. Dyson tells the following story of how mathematicians could have invented special and much of general relativity long before the physicists did.  Following the physicists, I will talk about Lie groups, but I really mean Lie algebras, or maybe connected Lie groups, ...
The physics of Galileo and Newton is invariant under the action of the Galileo Group (indeed, this is the largest group leaving classical physics invariant and fixing a point), which is the group $G_\infty = \text{SO}(3) \ltimes \mathbb R^3 \subseteq \text{GL}(4)$, where $\text{SO}(3)$ acts as rotations of space, fixing the time axis, and $\mathbb R^3$ are the nonrelativistic boosts $\vec x \mapsto t\vec u$, $t\mapsto t$.  This group is not semisimple.  But it is the limit as $c\to \infty$ of the Lorentz Group $G_c = \text{SO}(3,1)$, generated by the same $\text{SO}(3)$ part but the boosts are now
$$ t \mapsto \frac{t + c^{-2}\vec u \cdot \vec x}{\sqrt{1 - c^{-2}u^2}} \quad \quad \vec x \mapsto \frac{\vec x + t\vec u}{\sqrt{1 - c^{-2}u^2}} $$
Since semisimple groups are easier to deal with than nonsemisimple ones, by Occam's Razor we should prefer the Lorentz group.  Fortunately, Maxwell's equations are not invariant under $G_\infty$, but rather under $G_c$, where $c^{-2}$ is the product of the electric permititivity and magnetic permeability of free space (each of which is directly measurable, giving the first accurate measurement of the speed of light).  Actually, from this perspective, $c^{-2}$ is the fundamental number, and we should really think of the Galileo Group as a limit to $0$, not  $\infty$, of something.
But of course from this perspective we should go further.  Actual physics is invariant under more than just the Lorentz group, which is the group that physics physics and a point.  So Special Relativity is invariant under the Poincare Group $P = G_c \ltimes \mathbb R^{3+1}$.  Again this is not semisimple.  It is the limit as $r \to \infty$ of the DeSitter Group $D_r$, which in modern language "is the invariance group of an empty expanding universe whose radius of curvature $R$ is a linear function of time" (Dyson), so that $R = rt$ in absolute time units.  Hubble measured the expansion of the universe in the first half of the twentieth century.
Anyway, I'm curious to know if it's true that every Lie group is a limit of a semisimple one: how typical are these physics examples?  To make this more precise, I'll switch to Lie algebras.
Definitions
Let $V$ be an $n$-dimensional vector space.  If you like, pick a basis $e_1,\dots,e_n$ of $V$, and adopt Einstein's repeated index notation, so that given $n$-tuples $a^1,\dots,a^n$ and $b_1,\dots,b_n$, then $a^ib_i = \sum_{i=1}^n a^ib_i$, and if $v \in V$, we define the numbers $v^i$ by $v = v^ie_i$; better, work in Penrose's index notation.  Anyway, a Lie algebra structure on $V$ is a map $\Gamma: V \otimes V \to V$ satisfying two conditions, one homogeneous linear in (the matrix coefficients) of $\Gamma$ and the other homogeneous quadratic:
$$ \Gamma^k_{ij} + \Gamma^k_{ji} = 0 \quad\text{and}\quad \Gamma^l_{im}\Gamma^m_{jk} + \Gamma^l_{jm}\Gamma^m_{ki} + \Gamma^l_{km}\Gamma^m_{ij} = 0 $$
Thus the space of Lie algebra structures on $V$ is an algebraic variety in $V \otimes V^{*} \otimes V^{*}$, where $V^{*}$ is the dual space to $V$.
If $\Gamma$ is a Lie algebra structure on $V$, the corresponding Killing form $\beta$ is the symmetric bilinear pairing $\beta_{ij} = \Gamma^k_{im}\Gamma^m_{jk}$.  Then $\Gamma$ is semisimple if and only if $\beta$ is nondegenerate.  Nondegeneracy is a Zariski-open condition on bilinear forms, since $\beta$ is degenerate if and only if a certain homogeneous-of-degree-$n$ expression in $\beta$ vanishes (namely $\sum_{\sigma \in S_n} (-1)^\sigma \prod_{k=1}^n \beta_{i_k,j_{\sigma(k)}} = 0$ as a map out of $V^{\otimes n} \otimes V^{\otimes n}$, where $S_n$ is the symmetric group on $n$ objects and $(-1)^\sigma$ is the "sign" character of $S_n$).  Since $\beta$ is expressed algebraically in terms of $\Gamma$, semisimplicity is a Zariski-open condition on the variety of Lie algebra structures on a given vector space.
Incidentally, Cartan classified all semisimple Lie algebras (at least over $\mathbb C$ and $\mathbb R$) up to isomorphism, and the classification is discrete.  So any two semisimple Lie algebras in the same connected component of the space of semisimple structures are isomorphic.
My Question
Is the space of semisimple Lie algebra structures on $V$ dense in the space of all Lie algebra structures on $V$?  (I.e. if $\Gamma$ is a Lie algebra structure on $V$ and $U \ni \Gamma$ is an open set of Lie algebra structures, does it necessarily contain a semisimple one?)  This is really two questions.  One is whether it is Zariski-dense.  But we can also work over other fields, e.g. $\mathbb R$ or $\mathbb C$, which have topologies of their own.  Is the space of semisimple Lie algebra structures on a real vector space $V$ dense with respect to the usual real topology?
(The answer is no when $\dim V = 1$, as then the only Lie algebra structure is the abelian one $\Gamma = 0$, which is not semisimple, and it is no when $\dim V = 2$, as there are nontrivial two-dimensional Lie algebras but no semisimple ones.  So I should ask my question for higher-dimensional things.)
Edit: I have posted the rest of these as this follow-up question.
If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part?
A related question is whether given a nonsemisimple Lie algebra structure, are all its nearby semisimple neighbors isomorphic?  Of course the answer is no: the abelian Lie algebra structure $\Gamma = 0$ is near every Lie algebra but in general there are nonisomoprhic semisimple Lie algebras of the same dimension, and more generally we could always split $V = V_1 \oplus V_2$, and put a semisimple Lie algebra structure on $G_1$ and a trivial one on $V_2$.  So the converse question: are there any nonsemisimple Lie algebras so that all their semisimple deformations are isomorphic?  Yes, e.g. anyone on the three-dimensional vector space over $\mathbb C$.  So: is there a (computationally useful) characterization of those that are?
If the answer to all my questions are "yes", then it's probably been done somewhere, so a complete response could consist of a good link.  The further-further question is to what extent one can deform representations, but that's probably pushing it.

REPLY [8 votes]: I studied deformations of Lie algebras a lot and was able to construct a moduli space and a universal family for subgroups of an algebraic group:
http://www.mpim-bonn.mpg.de/preblob/4183
(I am very sorry to seemingly promote my work, please be sure I am not aware of any similar reference.)
I think you are interested by Lie algebras at large, while I can only give you informations on algebraic subalgebras of the Lie algebra of an algebraic group—
i.e. subalgebras that are Lie subalgebras of a subgroup, the terminology is of Chevalley.  You might however find some these results interessant or useful.
In front of all, this simple but striking consequence of a deformation theorem of Richardson—and the classification of semi-simple algebras:
3.18 COROLLARY Let $\mathfrak{g}$ be a Lie algebra and $k \le \dim\mathfrak{g}$. The map from the variety of $k$-dimensional subalgebras to the set of isomorphism class of semi-simple Lie algebras sending a Lie algebra to the isomorphism class of its semi-simple factors is lower semi-continuous, in the Zariski topology.
(Isomorphism classes of semi-simple algebras are ordred by injections.)
I cannot totally answer your question (for deformations of subalgebras):
“If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part?”
However it is worth noting, that all what you need to get the description you are looking for, is to understand degenerations on nilpotent algebras.  Indeed, consider a group $G$ and two subgroups $H$ and $M$ such that $M$ is in the closure of the conjugacy class of $H$. (The closure is taken in the moduli space, but replacing connected groups by their Lie algebras, you can embed the closure of the orbit of $H$ live in an appropriate Grassmannain variety.)  So you have a curve $C$ in $G$ such that $M$ is in the closure of the translated of $H$ under elements of $C$.  A consequence of the corollary is that if $M$ contains a semi-simple group $S$ then you can replace $H$ by a conjugate containing $S$ and assume that $C$ is in the centralizer of $S$.
So there is a natural way to enumerate possible degenerations, by looking at possible degenerations of the semi simple part (in the Levi-Malcev decomposition) and then study what can happen when you fix the smallest semi-simple part.  In practice, small rank semi-simple examples should be very tractable, because we know very well the centralizers in the large group of maximal tori of the small semi-simple part.
A few words about the moduli space.
The moduli space is not an algebraic variety but rather a countable union of (projective) algebraic varieties.  This unpleasant phenomenon is indebted to tori: let me explain what happens for tori and for one dimensional subgroups of $SL_3$.
Tori.  Let $T$ be a torus of dimension $r$, its Lie algebra ${\mathfrak t}$ contains a lattice $M$, the kernel of the exponential.  As we know, the only $k$-dimensional subspaces of ${\mathfrak t}$ that are algebraic algebras are spanned by points of $M$.  Hence the set $\mathfrak{P}_k(T)$ of $k$ dimensional subgroups of $T$ is the set of rational points if the Grasmann variety of $k$. Its topology is not exactly the topology induced by the Grassmann variety: it is on each irreducible component—which in this special case just means that $\mathfrak{P}_k(T)$ is discrete.
$SL_3$. Let me describe the set of $\mathfrak{P}_1(SL_3)$ of all one-dimensional subgroups of $SL_3$.  We are then looking for $1$-dimensional algebraic subalgebras of the $\mathfrak{sl}_3$.  There is the two nilpotent orbits, and the countably many orbits of $SL_3$ through $\mathfrak{P}_1(T)$, once we choosed a torus $T$ (these orbits almost never meet, by the normalizer theorem).  Each of these orbits contains a nilpotent orbit in its closure, so from the smallest nilpotent orbit, you can generalise to any semi-simple orbit: subgroups in two distinct semi-simple orbits are abstractly isomorphic but not isomorphic as subgroups of $SL_3$.
Closing remark. I do not know enough scheme theory to write my work in a suitable language, so you may be puzzled by some terminology I use. I am using algebraic varieties as in Shafarevic I or in the book of Hanspeter Kraft (Geometrische Methoden der Invarianten Theorie).  It is probably very clumsy!  I am quite confident that these results would survive in the world of schemes, there is actually a scheme-theoretic version of Richardson's theorem, that Brian Conrad once showed me.<|endoftext|>
TITLE: What are some results in mathematics that have snappy proofs using model theory?
QUESTION [69 upvotes]: I am preparing to teach a short course on "applied model theory" at UGA this summer.  To draw people in, I am looking to create a BIG LIST of results in mathematics that have nice proofs using model theory.  (I do not require that model theory be the first or only proof of the result in question.)
I will begin with some examples of my own (the attribution is for the model-theoretic proof, not the result itself).

An injective regular map from a complex variety to itself is surjective (Ax).

The projection of a constructible set is constructible (Tarski).

Solution of Hilbert's 17th problem (Tarski?).

p-adic fields are "almost C_2" (Ax-Kochen).

"Almost" every rationally connected variety over Q_p^{unr} has a rational point (Duesler-Knecht).

Mordell-Lang in positive characteristic (Hrushovski).

Nonstandard analysis (Robinson).


[But better would be: a particular result in analysis which has a snappy nonstandard proof.]
Added: The course was given in July of 2010.  So far as I am concerned, it went well.  If you are interested, the notes are available at
http://alpha.math.uga.edu/~pete/MATH8900.html
Thanks to everyone who answered the question.  I enjoyed and learned from all of the answers, even though (unsurprisingly) many of them could not be included in this introductory half-course.  I am still interested in hearing about snappy applications of model theory, so further answers are most welcome.

REPLY [5 votes]: Sorry to dig up such an old question, but I wanted to mention some more applications of Model theory that has perhaps been not covered in the other answers:
1) Using the basic algebraic fact that given a field $\mathbb F$ and finitely many polynomials $f_1,...,f_n \in \mathbb F[X]$, there is a field extension $\mathbb K_{|\mathbb F}$ in which all the $f_i$ s has a root; one can show by using the Compactness theorem  , that given a field $\mathbb F$, there is a field extension $\mathbb L_{|\mathbb F}$ in which every polynomial $f \in \mathbb F[X]$ has a root. This in turn can be used iteratively to build an algebraically closed field extension $\mathbb K_{|\mathbb F}$ . Then taking the field of all elements of $\mathbb K$ which are algebraic over $\mathbb F$, we get an algebraic extension of $\mathbb F$ which is algebraically closed i.e. we get an algebraic closure of $\mathbb F$. The Compactness theorem can also be used to prove the uniqueness of algebraic closure. 
2) The first order theory of Real Closed fields in the language of ordered fields has quantifier elimination. This can be easily used to prove certain facts about " semi-algebraic sets" (i.e. sets which are Boolean combination i.e. finite union , intersection and compliment of sets of the form $\{\bar x : p(\bar x) =0\}$ and $\{\bar x : p(\bar x) <0\}$, where $<$ is the canonical order on a real closed field and $p$ is a multivariate polynomial with coefficients in the field)  
(i) If $\mathbb F$ is a real closed field and $f: \mathbb F^m \to \mathbb F^n$ is a semi-algebraic function (the graph of the function is a semi-algebraic set) , then $f(A)$ and $f^{-1}(B)$ are semi-algebraic for any semi-algebraic sets $A \subseteq \mathbb F^m ,  B\subseteq \mathbb F^n$. This is often called the generalized Tarski-Seidenberg theorem.
(ii)  Every non-negative element in a real closed field is the square of a unique non-negative element; this defines the square root of a non-negative element. For a real closed field $\mathbb F$, using the $\mathbb F$-valued norm $|(x_1,...,x_n)|:=\sqrt {x_1^2+...+x_n^2}$ on $\mathbb F^n$ , we get a topology on $\mathbb F^n$. Using quantifier elimination, it can be easily shown that the closure (under the above mentioned topology) of any semi-algebraic set is again semi-algebraic. 
(iii) the theory of RCF has quantifier elimination and has a prime model, hence it is a complete theory. Using this it can be shown that If $\mathbb F$ is a Real closed field and $f : \mathbb F^m \to \mathbb F^n  $ is a continuous (w.r.t. the topology ) semi-algebraic function, then $f$ carries closed and bounded sets to closed and bounded sets (bounded in the sense of the $\mathbb F$-valued norm). 
(iv) Using the completeness of RCF and quantifier elimination , it can also be shown that if $f: \mathbb F \to \mathbb F$ is semi-algebraic , then for any open-interval $U\subseteq \mathbb F$, there is $x \in U$ such that $f$ is continuous at $x$. 
And there are many more similar results.<|endoftext|>
TITLE: Rings of integers of function fields
QUESTION [10 upvotes]: This might be a somewhat silly and inconsequential question, but it's aroused my curiosity. One has the theorem in commutative algebra that the integral closure of a domain $A$ in its field of fractions $Q(A)$ is the intersection of all the valuation subrings of $Q(A)$ containing $A$. This naturally leads to the impulse to define the ring of integers of an arbitrary field $k$, independent of any particular subring, to be the intersection of all valuation subrings of $k$. Really what one is doing here is taking the integral closure of the prime ring, which is to say the minimal integrally closed subring. This gives the answer one would expect for $\mathbb{Q}$ and for number fields. I believe that the ring of integers under this definition of $\mathbb{C}$ is $\overline{\mathbb{Z}}$, and the ring of integers of $\mathbb{R}$ is $\overline{\mathbb{Z}} \cap \mathbb{R}$. So far, so good.
Now here's the question: what happens in algebraic geometry? We could ask about the ring of integers $\mathcal{O}$ of the function field $Q(A)$ of an irreducible affine variety $\textbf{Spec
 }A$ over an algebraically closed field $k$, and naïvely hope that it coincides with $A$; but there may be many subalgebras of $Q(A)$ whose field of fractions is $Q(A)$, and at most one of these will be its ring of integers. Can we characterise geometrically those affine varieties whose coordinate ring is the ring of integers of its function field?
And what happens for function fields of more general varieties? For projective varieties, whose global ring of functions isn't much to write home about, we might hope for a more interesting answer.
Edit: It occurs to me that as it stands, the (boring) answer to my question is that the ring of integers is always inside the ground field. Let me instead ask the following, related question: for affine or general function fields $k(X)$, is it always true that there is a minimal (but not necessarily unique minimal) subring $R$ such that
$R$ is integrally closed in $k(X)$, and
$k(X)$ is algebraic over the field of fractions $Q(R)$?
If so, what is the geometric meaning of $R$, and when must we have $Q(R) = k(X)$?

REPLY [3 votes]: I understand this answer might be coming a little late, but the stack exchange "Related Questions" algorithm brought me here to a question that does not yet have a complete answer, so I'll answer it.
1) For such an $X,R$, $k(X)$ is indeed equal to $Q(R)$. The reason is that every element $\alpha$ of $k(X)$ is algebraic over $Q(R)$, and hence satisfies a monic polynomial equation with coefficients in $Q(R)$. If $f$ is the lcm of the denominators of the fractions appearing in the coefficients, then $f\alpha$ satisfies a monic polynomial equation with coefficients in $R$, and lies in $k(X)$, so by assumption lies in $R$, and $\alpha=\frac{f\alpha}{f}$ lies in $\mathbb Q(R)$.
2) As Felipe pointed out 7 years ago, if $X$ is a curve, then $R$ satisfies this condition if and only if it is the ring of functions on the complement of a closed point of $R$.
3) This cannot happen for a higher-dimensional variety. Let $Y = \operatorname{Spec} R$ and let $f$ be a nonconstant function on $Y$. Choose a projective compactification $\overline{Y}$ such that $f$ is well-defined: $\overline{Y} \to \mathbb P^1$. Then the divisor $\overline{Y}-Y$ is ample, and the zero locus of $f$ is effective, so because the dimension is at least $2$, they must have some nontrivial intersection. Blow this up, producing a compactification $\overline{Y}^*$ of $Y$ containing an exceptional divisor $E$ on which $f$ vanishes (because it is the blowup of a closed subscheme where $f$ vanishes).
Now consider the subring of $R$ consisting of all functions which do not have a pole at the generic point of $E$. Because it is the intersection of $R$ with a valuation ring, it is is integrally closed. Because $f$ vanishes at the generic point of $E$, we can for any element $g\in R$ there is some power $n$ such that $g f^n$ does not have a pole at the generic point of $E$, and so $g = \frac{ gf^n}{f^n}$ lies in $K(R)$. So the quotient ring of this subring is the quotient ring of $R$, which is the function field of $X$, as desired.
For the simplest example of this phenomenon, take $R= k[x,y]$, which is integrally closed in its field of fractions. Then the subring $k[x,xy]$ is integrally closed in the same field of fractions, and so on, ad infinitum, so none of these are minimal. By the geometric argument, we see that there is no minimal subring.<|endoftext|>
TITLE: Characteristic classes in generalized cohomology theories?
QUESTION [17 upvotes]: Hello,
'ordinary' Stiefel-Whitney classes are elements of the singular cohomology ring and are constructed using the Thom isomorphism and Steenrod squares. So I think they should exist for any (generalized) multiplicative cohomology theory for which  the Thom isomorphism and cohomology operations like the Steenrod squares exist. If I am not wrong, I would be really happy about some references on this.
Thanks in advance (and merry christmas)
Jonas

REPLY [6 votes]: Maybe this isn't the "right" way to think about them, but I have often found the following characterization much more appealing: the characteristic classes are just elements of $E^*(BG)$ for some cohomology theory and some group $G$, if $E$ is ordinary cohomology with integer coefficients and $G$ is the infinite unitary group you get Chern classes etc. The universal property of $BG$ gives you a unique map that classifies each $G$-bundle, now look at the map in cohomology that you get from this and you get the characteristic class of the desired bundle. This probably requires some sort of orientability for things to be "nice" but you can certainly define characteristic classes for any cohomology theory this way, they just might not have nice relations.<|endoftext|>
TITLE: Why are non-singleton covering families often ignored?
QUESTION [12 upvotes]: It seems to me that frequently when discussing stack conditions and descent, people consider only singleton covering families, i.e. there is some single covering map $U\to X$, for which one constructs a category of descent data out of $U$ and $U\times_X U$, and so on.  But many sites have non-singleton covering families $(U_i \to X)_{i\in I}$; why are those ignored?  Is there some deep reason why it suffices to consider singleton families?
My efforts to understand this so far have led me to think about "superextensive sites", which are sites whose covering families are generated by singleton covers together with inclusions into coproducts (the "extensive topology").  In particular, a fibered category is a stack on a superextensive site iff it is a stack for the singleton covers and also for the extensive topology.  But while stack conditions for the extensive topology have an exceptionally simple form (the compatibility conditions being mostly vacuous), they are still not automatic.  So why are they often not mentioned?

REPLY [6 votes]: In applications I've seen, what matters is the topos, not the site. If this is true for your applications, you should feel free to replace your site by any site that produces the same topos. I think you can always make the following conventions without changing the topos (edit: not true, you need some hypothesis on the site; see comments):

If $\{U_i\to X\}_{i\in I}$ is a covering, then so is the singleton $\bigsqcup_I U_i\to X$.
$\{U_i\to \bigsqcup_{j\in I} U_j\}_{i\in I}$ is always a covering.

After that, you can use convention 1 to replace any non-singleton covering by a singleton covering, which is easier to think about (at least easier to symbolically manipulate). But you do have to keep 2 around to be able to prove, for example, that $F(\bigsqcup_{I} U_i)=\prod_{I} F(U_i)$ for any sheaf $F$. As far as I can tell, everybody I know adopts these two conventions and then just talks about singleton covers.
An alternative (probably better) explanation comes from the sieve-theoretic formulation of Grothendieck topologies. All that matters about a covering is the sieve that it generates, and $\bigsqcup_I U_i\to X$ generates the same sieve as $\{U_i\to X\}_{i\in I}$ (edit: also not true; perhaps someone who understands the sieve approach could say what the right statement is). I think the canonical reference for this approach is SGA 4 ("sieve"="crible").<|endoftext|>
TITLE: What's the status of the following relationship between Ramanujan's $\tau$ function and the simple Lie algebras?
QUESTION [26 upvotes]: Qiaochu asked this in the comments to this question.  Since this is really his question, not mine, I will make this one Community Wiki.  In MR0522147, Dyson mentions the generating function $\tau(n)$ given by
$$ \sum_{n=1}^\infty \tau(n)\,x^n = x\prod_{m=1}^\infty (1 - x^m)^{24} = \eta(x)^{24}, $$
which is apparently of interest to the number theorists ($\eta$ is Dedekind's function).  He mentions the following formula for $\tau$:
$$\tau(n) = \frac{1}{1!\,2!\,3!\,4!} \sum \prod_{1 \leq i < j \leq 5} (x_i - x_j)$$
where the sum ranges over $5$-tuples $(x_1,\dots,x_5)$ of integers satisfying $x_i \equiv i \mod 5$,  $\sum x_i = 0$, and $\sum x_i^2 = 10n$.  Apparently, the $5$ and $10$ are because this formula comes from some identity of $\eta(x)^{10}$.  Dyson mentions that there are similar formulas coming from identities with $\eta(x)^d$ when $d$ is on the list $d = 3, 8, 10, 14,15, 21, 24, 26, 28, 35, 36, \dots$.  The list is exactly the dimensions of the simple Lie algebras, except for the number $26$, which doesn't have a good explanation.  The explanation of the others is in I. G. Macdonald, Affine root systems and Dedekind's $\eta$-function, Invent. Math. 15 (1972), 91--143, MR0357528, and the reviewer at MathSciNet also mentions that the explanation for $d=26$ is lacking.
So: in the last almost-40 years, has the $d=26$ case explained?

REPLY [25 votes]: The case of $d=26$ is related to the exceptional Lie algebra $F_4$.  Let me quote from the 1980 paper by Monastyrsky which was originally published as a supplement to the Russian translation of the Dyson's paper:
A more careful study of Macdonald's article reveals that the identity for the
26th power of $\eta(x)$ is not really such a mystery. It is related to the exceptional
group $F_4$ of dimension 52, where the space of dual roots $F_4^V$ and the space of roots $F_4$ are not the same. Thus, there are two distinct identities associated
with $F_4$, one for $\eta^{52} (x)$ and the other for $\eta^{26} (x)$. A similar situation prevails in
the case of the algebra $G_2$ of dimension 14, which yields identities for $\eta^{14} (x)$
and $\eta^{7} (x)$. The identities for $\eta^{26} (x)$ and $\eta^{7} (x)$ are considerably more complicated.<|endoftext|>
TITLE: What is the Zariski closure of the space of semisimple Lie algebras?
QUESTION [18 upvotes]: Given Leonid Positselski's excellent answer and comments to this question, I expect that the present one is a hard question.  Recall that the Lie algebra structures on a (finite-dimensional over $\mathbb C$, say) vector space $V$ are the maps $\Gamma: V^{\otimes 2} \to V$ satisfying $\Gamma^k_{ij} = -\Gamma^k_{ji}$ and $\Gamma^m_{il}\Gamma^l_{jk} + \Gamma^m_{jl}\Gamma^l_{ki} + \Gamma^m_{kl}\Gamma^l_{ij} = 0$; thus the space of Lie algbera structures is an algebraic variety in $(V^\*)^{\otimes 2} \otimes V$.  A Lie algebra structure $\Gamma$ is semisimple if the bilinear pairing $\beta_{ij} = \Gamma^m_{il}\Gamma^l_{jm}$ is nondegenerate; thus the semisimples are a Zariski-open subset of the space of all Lie algebra structures.  Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes.  The semisimples are not dense among all Lie algebra structures: if $\Gamma$ is semisimple, then $\Gamma_{il}^l = 0$, whereas this is not true for the product of the two-dimensional nonabelian with an abelian.

Is there a (computationally useful) characterization of the Zariski closure of the space  of semisimple Lie algebra structures?  (LP gives more equations any semisimple satisfies.)
Suppose that $\Gamma$ is not semisimple but is in the closure of the semisimples.  How can I tell for which isomorphism classes of semisimples is $\Gamma$ in the closure of the isomorphism class?  (I.e. $\Gamma$ is a limit of what algebras?)
To what extent can I understand the representation theory of algebras in the closure of the semisimples based on understanding their neighboring semisimple algebras?

For 3., I could imagine the following situation.  There is some natural "blow up" of the closure of the semisimples, with at the very least each element of the boundary is a limit of only one isomorphism class in the Cartan classification.  Then any representation of the blown-down boundary is some combination of representations of the blown-up parts.

REPLY [5 votes]: As a warmup to this question you might want to think about the closure of semisimple associative algebras inside all finite dimensional asociative algebras of a given dimension.  As a warmup to that question you might want to think about the commutative case.  As a warmup to that question you might look at a beautiful paper by Bjorn Poonen (in particular, Section 6 and Remark 6.11).

REPLY [3 votes]: A (hopefully helpful) comment. I've thought about the problem of finding the closure of one isomorphism class - I haven't got an answer but I had an idea that I hope is helpful towards a solution. 
Consider the closure of the set S of Lie algebras isomorphic to a fixed semisimple lie algebra $L$ of dimension $n$, and fix a basis of $L$ which gives you the structure constants. Then there is a surjection from invertible matrices $GL_n(\mathbb{C})$  to $S$ , namely by acting on a fixed standard basis of $V$  with $x \in GL_{n}(\mathbb{C})$ to give you another basis, and now you force this other basis to have the properties of the basis of $L$ fixed above, i.e. the structure constants - then trace this back to get the values of $\Gamma^k_{ij}$ defining this particular Lie algebra.
To be precise with the above, I think it is best described as a transitive action of the algebraic group $GL_{n}(\mathbb{C})$ on the variety $S$. I think there are some matrices which act trivially however, and that these matrices correspond to automorphisms of the Lie algebra (which leave the structure constants invariant) – i.e. the point stabilizers correspond to automorphisms of the Lie algebras, so the homogenous space has the structure of the quotient of $GL_{n}(\mathbb{C})$ by this point stabilizer, which is the Lie group of automorphisms of $L$. 
I think this could help getting the closure of a single isomorphism class of Lie algebras (and since there are only finitely many isomorphism classes of semisimple Lie algebras of fixed dimension, should help with that problem too). But I’m not sure how – I tried naively by saying that perhaps this closure consists of the union of isomorphism classes which you get, in an intuitive sense, by replacing the invertible matrix $x$, by allowing singular matrices as well; but what I get from that seems to be some rubbish so I’m sure that path is mistaken.

REPLY [2 votes]: This is a really a comment to rajamanikkam's answer, but it does not fit in the comment box.
What rajamanikkam is describing in the last paragraph is essentially a Lie algebra contraction of $L$.  It is well known that this is how one obtains Lie algebras which are close in some sense to the original Lie algebra.  (I'm afraid that I do not speak the right language, so I am not sure whether this is in the Zariski closure.)
One way to define a contraction of a given Lie algebra $L$, say complex of dimension $n$, is to consider a continuous curve $A(t)$ in $\mathfrak{gl}_n(\mathbb{C})$ which for $t$ in some interval, say $[1,\infty)$, lies in $\mathrm{GL}_n(\mathbb{C})$.  For $t$ in that interval the Lie algebras $L(t)$ related to $L$ via $A(t)$ will be be isomorphic to $L$, but if $\lim_{t\to\infty} L(t)$ exists -- which is by no means the case for all curves $A(t)$ -- then it will give rise to a Lie algebra which may or may not be isomorphic to $L$.
I'm not sure if contractions are sufficient to generate the full closure, or indeed whether this is the sort of closure that the originally question intended.<|endoftext|>
TITLE: Highbrow interpretations of Stirling number reciprocity
QUESTION [11 upvotes]: The number ${n \choose k}$ of $k$-element subsets of an $n$-element set and the number $\left( {n \choose k} \right)$ of $k$-element multisets of an $n$-element set satisfy the reciprocity formula
$\displaystyle {-n \choose k} = (-1)^k \left( {n \choose k} \right)$
when extended to negative integer indices, for example by applying the usual recurrence relations to all integers.  There's an interesting way to think about the "negative cardinalities" involved here using Euler characteristic, which is due to Schanuel; see, for example, this paper of Jim Propp.  Another (related?) way to think about this relationship is in terms of the symmetric and exterior algebras; see, for example, this blog post.
The number $S(n, k)$ of $k$-block partitions of a set with $n$ elements and the number $c(n, k)$ of permutations of a set with $n$ elements with $k$ cycles satisfy a well-known inverse matrix relationship, but they also satisfy the reciprocity formula
$c(n, k) = S(-k, -n)$
when extended to negative integer indices, again by applying the usual recurrence relations.  
Question:  Are there any known highbrow interpretations of this reciprocity formula?

REPLY [22 votes]: Supplementary Exercise 3.2(d,e) on page 313 of my book Enumerative Combinatorics, vol. 1, second printing, shows that this Stirling number reciprocity is a special case of the reciprocity theorem for order polynomials (Exercise 3.61(a)). Thus it is related to a lot of "highbrow" math, such as the reciprocity between a Cohen-Macaulay ring and its canonical module. (For the basic properties of canonical modules, see Section I.12 of my other book Combinatorics and Commutative Algebra, second ed.)

REPLY [7 votes]: Another interpretation (which is essentially the same as Richard's example of order polynomials if uses order polytopes) is via Ehrhart reciprocity (Stanley Enumerative Combinatorics 1, Section 4.6). This says that if P is an integral polytope, the number of integer points in tP for positive integers t is a polynomial $L_P(t)$, and its we have the identity $L_P(-t) = (-1)^d L_{P^\circ}(t)$ where $d = \dim P$ and $P^\circ$ denotes the interior of P. Now apply this result to the standard k-simplex $\Delta$ to get the identity 
$\left( \binom{-t+1}{k} \right) = L_\Delta(-t) = (-1)^d L_{\Delta^\circ}(t) = (-1)^k \binom{t-1}{k}$,
and replace t by 1-n.
Thinking back to the exact sequence you have in your blog, my comment mentioned it as a special case of a "Schur complex." It's also a linear strand (pulling out some linear part) of the Koszul complex for a polynomial ring, which is a special case of "Priddy complexes" (for example, Eisenbud, Commutative Algebra, exercise 17.22, and the references therein). From a different point of view, it's a linear strand of the Chevalley-Eilenberg complex for an Abelian Lie algebra (Weibel, An Introduction to Homological Algebra, Section 7.7).<|endoftext|>
TITLE: Polynomial representing all nonnegative integers
QUESTION [288 upvotes]: Lagrange proved that every nonnegative integer is a sum of 4 squares.
Gauss proved that every nonnegative integer is a sum of 3 triangular numbers.
Is there a 2-variable polynomial $f(x,y) \in \mathbf{Q}[x,y]$ such that $f(\mathbf{Z} \times \mathbf{Z})=\mathbf{N}$?

REPLY [25 votes]: After thinking about this problem for a bit using rather a naive approach, looking at regions where f grows faster than quadratically (as mentioned in Qiaochu's attempt), it certainly appears that obtaining a negative answer to this problem is very difficult. To obtain a positive answer might be easier since we only need to exhibit a single polynomial with $f(\mathbb{Z}\times\mathbb{Z})=\mathbb{N}$ although, in all likelihood, there do not exist such examples. However, even using the naive approach, it quickly becomes clear what kind of behaviour could lead to a polynomial f having the required properties. Polynomials of degree less than four can be quickly dismissed. Then, assuming f has degree 2n, look at the leading order terms $f_{2n}$, which is a homogeneous polynomial of degree 2n. Away from the zeros of $f_{2n}$, it grows at rate $R^{2n}$ ($R=\sqrt{x^2+y^2}$) and dominates the lower order terms, so f cannot cover a strictly positive density of the integers. The difficulties occur when we look close to certain curves which are asymptotic to the zeros of $f_{2n}$ (being homogeneous, the zeros of $f_{2n}$ lie on a finite set of lines radiating out from the origin). On such curves, $f_{2n}$ will grow at a rate less than $O(R^{2n})$ and cancellation with lower order terms can occur. Looking at how much cancellation can occur seems to lead to difficult problems of Diophantine approximation.
The kind of polynomial which are plausible candidates for a polynomial mapping onto the positive integers are as follows.
$$
f(x,y)=a\prod_{i=1}^d\left(x-\alpha_iy\right)^{2n} - q(x,y)\qquad\qquad{\rm(1)}
$$
where $\alpha$ is a real algebraic integer with minimal polynomial of degree $d > 2$ over the rationals and conjugates $\alpha_1,\cdots,\alpha_d$, $a$ is a positive integer, and $q(x,y)$ is a polynomial of degree $2n(d-2)$. By Dirichlet's theorem, we know that there are infinitely many integer x,y such that $\vert x/y-\alpha_i\vert < y^{-2}$, so the leading order term of (1) is less than some fixed multiple of $R^{2n(d-2)}$ infinitely often. So there will be some cancellation with q. On the other hand, by the Thue-Siegel-Roth theorem, we know that $\vert x/y-\alpha\vert > y^{-2-\epsilon}$ for all large x,y, so the leading order terms of (1) grow at least as fast as $O(R^{2n(d-2)-\epsilon})$ which, at least, means that (1) cannot go negative very quickly. The question then, is there an algebraic number $\alpha$ such that $\vert x/y-\alpha\vert\ge cy^{-2}$ for some positive constant c and all integer x,y? In that case, the leading order term of (1) would be bounded below by a multiple $R^{2n(d-2)}$ and q could be chosen such that $f(\mathbb{Z}\times\mathbb{Z})\subseteq\mathbb{N}$. It is then possible, but still unlikely, that (1) gives a polynomial mapping onto the positive integers. Looking in my copy of Hindry & Silverman (Diophantine Geometry, An Introduction) it mentions that it is an open problem whether there exist such algebraic numbers (and there are no known examples or counterexamples with $d > 2$) but it is conjectured that there aren't any. So, polynomials such as (1) appear unlikely to do what we want, but proving this seems to be very difficult. Of course, that f actually covers $\mathbb{N}$ would be a much stronger statement than $\vert x/y-\alpha\vert \ge cy^{-2}$ so, maybe an expert in this area could actually rule out such examples, but it still looks like a very tricky problem.
We can also try polynomials such as
$$
f(x,y)=a\prod_{i=1}^d\left((x-\alpha_iy)^r-p(\alpha_i)y^s\right)^{2n} - q(x,y)\qquad\qquad{\rm(2)}
$$
where, now, p is a polynomial with integer coefficients,r,s are positive integers and q has degree less than $2n$. This is even more difficult than (1) to deal with and whether or not such polynomials can provide what the question is asking for depends on how small $(x/y-\alpha)-p(\alpha)^{1/r}y^{s/r-1}$ can be. This also looks like a very difficult problem in Diophantine approximation.
So, although I expect that the answer to this is no, there are no such $f$, any method of proving this has to cope with possibilities such as (1) and (2). Just these two cases look extremely difficult to handle. Maybe it is possible though, and an expert on such areas would be able to say something more about them than I can.<|endoftext|>
TITLE: Matrix factorizations and physics
QUESTION [34 upvotes]: I have heard during some seminar talks that there are applications of the theory of 
 matrix factorizations  in string theory. A quick search shows mostly papers written by physicists. Are there any survey type papers aimed at an algebraic audience on this topic, especially  with current state/open questions motivated by physics? 
Background: A matrix factorization of an element $x$ in a ring $R$ is a pair of square matrices  $A,B$ of size $n$ such that $AB=BA=xId_n$. For more see, for example, Section 3 of 
 this paper .

REPLY [28 votes]: Indeed matrix factorizations come up in string theory. I don't know if there are good survey articles on this stuff, but here is what I can say about it. There might be an outline in the big Mirror Symmetry book by Hori-Katz-Klemm-etc., but I am not sure.
When we are considering the B-model of a manifold, for example a compact Calabi-Yau, the D-branes (boundary states of open strings) are given by coherent sheaves on the manifold (or to be more precise, objects of the derived category of coherent sheaves). Matrix factorizations come up in a different situation, namely, they are the D-branes in the B-model of a Landau-Ginzburg model. Mathematically, a Landau-Ginzburg model is just a manifold (or variety) $X$, typically non-compact, plus the data of a holomorphic function $W: X \to \mathbb{C}$ called the superpotential. In this general situation, a matrix factorization is defined to be a pair of coherent sheaves $P_0, P_1$ with maps $d : P_0 \to P_1$, $d : P_1 \to P_0$ such that $d^2 = W$. I guess you could call this a "twisted (or maybe it's 'curved'? I forget the terminology) 2-periodic complex of coherent sheaves". When $X = \text{Spec}R$ is affine, and when the coherent sheaves are free $R$-modules, this is the same as the definition that you gave.
The relationship between matrix factorization categories and derived categories of coherent sheaves was worked out by Orlov: http://arxiv.org/abs/math/0503630 http://arxiv.org/abs/math/0503632 http://arxiv.org/abs/math/0302304
I believe that the suggestion to look at matrix factorizations was first proposed by Kontsevich. I think the first paper that explained Kontsevich's proposal was this paper by Kapustin-Li: http://arXiv.org/abs/hep-th/0210296v2
There are some interesting recent papers regarding the relationship between the open-string B-model of a Landau Ginzburg model (which is, again, mathematically given by the matrix factorizations category) and the closed-string B-model, which I haven't described, but an important ingredient is the Hochschild (co)homology of the matrix factorizations category. Take a look at Katzarkov-Kontsevich-Pantev http://arxiv.org/abs/0806.0107 section 3.2. There is a paper of Tobias Dyckerhoff http://arxiv.org/abs/0904.4713 and a paper of Ed Segal http://arxiv.org/abs/0904.1339 which work out in particular the Hochschild (co)homology of some matrix factorization categories. The answer is it's the Jacobian ring of the superpotential. This is the correct answer in terms of physics: the Jacobian ring is the closed state space of the theory.
Katzarkov-Kontsevich-Pantev also has some interesting stuff about viewing matrix factorization categories as "non-commutative spaces" or "non-commutative schemes".
Edit 1: I forgot to mention: Kontsevich's original homological mirror symmetry conjecture stated that the Fukaya category of a Calabi-Yau is equivalent to the derived category of coherent sheaves of the mirror Calabi-Yau. Homological mirror symmetry has since been generalized to non-Calabi-Yaus. The rough expectation is that given any compact symplectic manifold, there is a mirror Landau-Ginzburg model such that, among other things, the Fukaya category of the symplectic manifold should be equivalent to the matrix factorizations category of the Landau-Ginzburg model. For example, if your symplectic manifold is $\mathbb{CP}^n$, the mirror Landau-Ginzburg model is given by the function $x_1+\cdots+x_n + \frac{1}{x_1\cdots x_n}$ on $(\mathbb{C}^\ast)^n$. This is sometimes referred to as the Hori-Vafa mirror http://arxiv.org/abs/hep-th/0002222
I think that various experts probably know how to prove this form of homological mirror symmetry, at least when the symplectic manifold is, for example, a toric manifold or toric Fano manifold, but it seems that very little of this has been published. There may be some hints in this direction in Fukaya-Ohta-Oh-Ono http://arxiv.org/abs/0802.1703 http://arxiv.org/abs/0810.5654, but I'm not sure. There is an exposition of the case of $\mathbb{CP}^1$ in this paper of Matthew Ballard http://arxiv.org/abs/0801.2014 -- this case is already non-trivial and very interesting, and the answer is very nice: the categories in this case are equivalent to the derived category of modules over a Clifford algebra. I quite like Ballard's paper; you might be interested in taking a look at it anyway.
Edit 2: Seidel also has a proof of this form of homological mirror symmetry for the case of the genus two curve. Here is the paper http://arxiv.org/abs/0812.1171 and here is a video http://www.maths.ed.ac.uk/~aar/atiyah80.htm of a talk he gave on this stuff at the Atiyah 80 conference.<|endoftext|>
TITLE: Global dimenson of quivers with relations
QUESTION [8 upvotes]: Let Q be a finite quiver without loops. Then its global dimension is 1 if it contains at least one arrow.
I'm trying to get some intuition about how much the global dimension can grow when we quotient by some homogeneous ideal of relations I. In general, if Q is acyclic (is this necessary?), then the global dimension is bounded by the number of vertices, but I want something that uses information from I. For simplicity, let's assume that there is at most 1 edge between any two vertices. If I add the relation that a single path of length 2 is 0, then the global dimension goes up to 2, and the same is true if 2 is replaced by any r>2 (right?). I can get higher global dimensions by the following: take some consecutive arrows $a_1, a_2, \dots, a_n$, and require that each path of length 2 $a_{i+1} a_i$ is 0, then the global dimension goes up to n-1. 
The way I am trying to picture this is by thinking of projective modules as flowing water which gets stopped by some rock placed where the relations are, and seeing how many times the flow needs to restart before it can reach the end (I don't know if this is a useful comment.)
Anyway, here is my question: is there some simple way to bound the global dimension of Q/I assuming that Q is acyclic and no multiple edges between any two vertices? In this case, we're only allowed to say that certain paths are 0, so I am suspecting this has something to do with "number of overlaps." My guess would be something like, define an overlap to be when an initial segment of one path coincides with an ending segment of another path, and then the global dimension should be less than or equal to number of overlaps + 2.

REPLY [15 votes]: If you are only considering monomial algebras (that is, if you are generating the ideal I by paths) then your intuition about overlaps is correct, once you see which overlaps you need to consider. There is a paper by Bardzell (The alternating behaviour of monomial algebras) where he constructs explicitely a projective resolution of the quotient algebra as a bimodule over itself (whose length bounds the gldim of the algebra) which is constructed precisely by considering overlaps.
By the way, if the graph is not acyclic, then the global dimension can very well be infinite. The simplest example is a quiver with one vertex and a loop, and the ideal geberated by the square of the loop.
Later: Let me be more explicit about what I meant by "once you see which overlaps you need to consider"... Consider the quiver $Q$ which is an oriented path with 15 arrows, and let $I$ be ideal generated by all paths in $Q$ of length 8. There are then 8 minimal relations, they all overlap, but if you work through the construction of minimal projective resolutions of the simple modules of $kQ/I$ you'll see that most of those overlaps do not matter, and that the global dimension is $3$ in this case. You can play this game with longer paths, as long as you divide by not too short relations. 
It is not too hard to single out precisely which are the overlaps that do matter when the quiver is a path. The general case is not horribly more complicated, yet it always manages to confuse me.<|endoftext|>
TITLE: The Yoneda Lemma for $(\infty,1)$-categories?
QUESTION [13 upvotes]: According to this page on the nLab, it is currently unclear whether or not the entire Yoneda lemma generalizes to $(\infty ,1)$-categories rather than just the Yoneda embedding.  Have there been counterexamples to the stronger statement?  If not, what complications are there in generalizing the entire Yoneda lemma?  The whole lemma is a very powerful tool in ordinary category theory, and if there is no definitive answer, are there any reasons to believe that the generalization should or shouldn't be true?
By the "entire" Yoneda lemma, I mean the isomorphism $F(X)\cong Nat(h_X,F)$

REPLY [6 votes]: I realize it is a bit late to respond to the question... but this is precisely Lemma 5.1.5.2 of HTT (and the proof is very much along the same lines as the accepted one).<|endoftext|>
TITLE: Non-smooth algebra with smooth representation variety
QUESTION [5 upvotes]: A not necessarily commutative algebra A (over C, say) is called formally smooth (or quasi-free) if, given any map $f:A \to B/I$, where $I \subset B$ is a nilpotent ideal, there is a lifting $F:A \to B$ that commutes with the projection. (The reason for the terminology is that if we restrict to the category of finitely generated commutative algebras, this condition is equivalent to Spec(A) being smooth. For more info see the paper "algebra extensions and nonsingularity" by Cuntz and Quillen, 1995.) It isn't hard to see that if A is formally smooth, then the representation varieties $Rep_\mathbb{C}(A,V)$ are smooth (V is finite dimensional). Does anyone know of an example of an algebra that is not formally smooth, but whose representation varieties are smooth? 
One almost-answer is the Weyl algebra $A = \mathbb{C}\langle x,y\rangle/(xy - yx = 1)$. This isn't formally smooth, but its representation varieties are all empty. (To see this, take the trace of $xy - yx = 1$ to get $0 = n$.) This doesn't seem like it should count as answer, does anyone know a better one?

REPLY [11 votes]: Take any semi-simple lie algebra g and consider its enveloping algebra U(g). As all finite dimensional representations are semi-simple, every representation variety rep_n U(g) is a finite union of orbits, whence smooth. No such U(g) is formally smooth.<|endoftext|>
TITLE: Who invented the gamma function?
QUESTION [20 upvotes]: Who was the first person who solved the problem of extending the factorial to non-integer arguments? 
Detlef Gronau writes [1]: "As a matter of fact, it was Daniel Bernoulli who gave in 1729 the first representation of an interpolating function of the factorials in form of an infinite product, later known as gamma function."
On the other hand many other places say it was Leonhard Euler. Will the real inventor please stand up? 
[1] "Why is the gamma function so as it is?" by Detlef Gronau, Teaching Mathematics and Computer Science, 1/1 (2003), 43-53.

REPLY [8 votes]: In his first letter to Goldbach (already linked to in Jonas' answer) Euler writes that he communicated his interpolation of the sequence of factorials (or Wallis's hypergeometric series, as it was called back then) to Daniel Bernoulli:
"I communicated this to Mr. Bernoulli, who by his own method arrived at nearly the same final expression" (this is the sentence starting with "communicavi haec . . . " on p. 4).<|endoftext|>
TITLE: Characterising extendable automorphisms
QUESTION [9 upvotes]: Suppose H is a subgroup of a finite group G. Can the group of all automorphisms of H that extend to G can be characterized somehow? What condition on the group extension would guarantee that any automorphism of H can be extended to G?

REPLY [5 votes]: The papers in the accepted answer by @SixWingedSeraph all refer to a somewhat more specific problem, that of extending an automorphism of a normal subgroup to a larger group.  Although the questioner says something about a group extension (which would usually imply normality of $H$), he did not specify that $H$ should be normal.
Extending automorphisms of normal subgroups
The Wells paper "Automorphisms of group extensions" is the first paper I know about on this topic.  In the situation where $N$ is a normal subgroup of $G$, such that $\alpha : G/N \rightarrow \operatorname{Out}(N)$ is the action of the quotient on $N$, Wells finds an exact sequence
$$1 \rightarrow Z^1_\alpha(G/N, Z(N)) \rightarrow \operatorname{Aut}(G; N) \rightarrow \operatorname{Compat}(G/N; N) \rightarrow H^2_\alpha(G/N, Z(N)). $$
Here $\operatorname{Aut}(G; N)$ is the group of all automorphisms of $G$ fixing $N$, and $\operatorname{Compat}(G/N; N)$ is all pairs of automorphisms of the quotient $G/N$ and subgroup $N$ satisfying a certain compatibility condition.  $Z^1$ and $H^2$ are the 1st cocycle space and the 2nd group cohomology, respectively.  It should be noticed that the last map is not a group homomorphism.
I haven't managed to get my hands on the Robinson paper "Applications of cohomology to the theory of groups", but I understand it has some explanation and applications of Wells' result, as well as a new proof.  The Jin paper "Automorphisms of groups" restates the result of Wells in somewhat different language, and applies it to the case where one wants to find an automorphism of $N$ which acts trivially on $G/N$.
I'll remark that one interesting case is where $N$ is the direct product of isomorphic finite simple groups (e.g., a nonabelian chief factor).  Here $Z(N)$ is trivial, which makes the long exact sequence above collapse rather nicely.  But in this case one can obtain similar results by using the fact that if $N$ is center free, then any homomorphism $\alpha : Q \rightarrow \operatorname{Out}(N)$ uniquely determines an extension of $Q$ by $N$.  This is however only mildly more elementary, as proving this fact also requires group cohomology techniques -- see chapter IV.6 of Ken Brown's book "Cohomology of groups".
Extending automorphisms of arbitrary subgroups
Skupp shows in "A characterization of inner automorphisms" that an automorphism $\varphi$ of a group $H$ extends to an automorphism in every group $G$ with $H$ embedded in $G$ if and only if $\varphi$ is inner (i.e., obtained via conjugation by some element).  Pettet showed that the same holds if we restrict $G$ to be finite in "On inner automorphisms of finite groups" and "Characterizing inner automorphisms of groups".
So I guess the overall answer to the original question is "sometimes not".  If you have a specific situation which is still of interest, you might try looking at the second Pettet paper to see if it helps you show the automorphism extends or does not extend.  It looks like it would be a hard problem to find a general characterization.<|endoftext|>
TITLE: Is every smooth affine curve isomorphic to a smooth affine plane curve?
QUESTION [22 upvotes]: As suggested by Poonen in a comment to an answer of his question about the birationality of any curve with a smooth affine plane curve we ask the following questions: 
Q) Is it true that every smooth affine curve is isomorphic to a smooth affine plane curve?
(a) In particular, given a smooth affine plane curve $X$ with an arbitrary Zariski open set $U$ in it, can one give a closed embedding of $U$ in the plane again?  
(b) An extremely interesting special case of (Q) above: Suppose $X$ is a singular plane algebraic curve with $X_{sm}$ the smooth locus. Can one give a closed embedding of $X_{sm}$ in the plane? 
All varieties in question are over $\mathbb{C}$. 

UPDATE:
Bloch, Murthy and Szpiro have already proven in their paper "Zero cycles and the number of generators of an ideal" , a much more general result (see Theorem 5.7, op.cit), that every reduced and irreducible prjective variety has an affine open set which is a hypersurface. This settles the above question birationally, in particular.
The authors give a very short and beautiful alternate proof of their result by M.V. Nori which I include here for its brevity and for anyone who may not have access to the paper:
Proof:  Suppose $X$ is an integral projective variety of dimension $d$. By a generic projection, easily reduce to the case of a (possibly singular) integral hypersurface $X$ of $\mathbb{A}^{d+1}$. Suppose the coordinate ring of $X$ is $A=\mathbb{C}[x_1,\dots,x_{d+1}]$ and its defining equation is $F=\Sigma_0^m{f_i}x_{d+1}^{i}=0$ with $f_0\neq{0}$. For some element $h$ in $J\cap\mathbb{C}[x_1,\dots,x_d]$, where $J$ defines the singular locus of $X$, put $x_{d+1}'=x_{d+1}/(hf_0^2)$ in $F=0$ to observe that $1/(hf_0)\in\mathbb{C}[x_1,\dots,x_{d+1}']$ and $A_{hf_0}=\mathbb{C}[x_1,\dots,x_{d+1}']$. Clearly $\rm{Spec}\ {A_{hf_0}}$ admits a closed immersion in $\mathbb{A}^{d+1}$.  

However, the above authors also prove in their Theorem 5.8 that there exist affine varieties of any dimension, which are not hypersurfaces. This answers our question in negative. This was also known to Sathaye for curves, see On planar curves. He gives a nice example of a double cover of a punctured elliptic curve, ramified at 9 points and also at the point at infinity. This curve cannot be embedded in $\mathbb{A}^2$. Sathaye uses the value semigroup at the only point at infinity to prove this. His example has trivial canonical divisor. So it answers Poonen's question in the comments below, negatively.
In short, $K=0$ for an affine curve is necessary but not sufficient for the curve to be planar, however one should note that $K=0$ is necessary and sufficient for an affine curve to be a complete intersection.

REPLY [14 votes]: You can try this:
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.52.6348<|endoftext|>
TITLE: Is $\varphi(n)/n$ the maximal portion of $n$-cycles in a degree $n$ group?
QUESTION [26 upvotes]: Let $G$ be a degree $n$ group, i.e., a subgroup of the symmetric group $S_n$. Let $p(G)$ be the number of $n$-cycles in $G$ divided by the size of $G$. 
Examples: 

If $G$ is a cyclic transitive group, then $p=\varphi(n)/n$.
If $G=S_n$, then $p=1/n$.
(If $G$ is not transitive, then $p=0$)

The question is whether $p(G)\leq \varphi(n)/n$ for every degree $n$ group? 
Note: 

One can see that $p(G)=k/n$, where $k$ is the number of conjugacy classes of $n$-cycles, so the answer is YES if $n$ is prime.
Numerical testing shows the answer is YES for $n\leq 30$ and for primitive groups for $n\leq 1000$.
There are non-cyclic groups achieving the bound $\varphi(n)/n$, e.g., the wreath product of cyclic groups.

Edit: Recently Joachim König solved this using the classification both in the induction basis as Michael Giudici mentioned and also in the induction step. I guess we should wait for the paper which is now in refereeing process.

REPLY [10 votes]: It is true for all primitive groups: The primitive groups of degree n containing an n-cycle were independently classified in
Li, Cai Heng The finite primitive permutation groups containing an abelian regular subgroup.
Proc. London Math. Soc. (3) 87 (2003), no. 3, 725--747. ) 
and 
Jones, Gareth A.
Cyclic regular subgroups of primitive permutation groups.
J. Group Theory 5 (2002), no. 4, 403--407. 
They are the groups G such that 
-$C_p\leqslant  G\leqslant AGL(1,p)$ for p a prime
-$A_n$ for n odd, or $S_n$
-$PGL(d,q)\leqslant G \leqslant P\Gamma L(d,q)$: here there is a unique class of cyclic subgroups generated by an n-cycle except for $G=P\Gamma L(2,8)$ in which case there are two.
-$(G,n)=(PSL(2,11),11), (M_{11},11), (M_{23},23)$
All these groups satisfy the bound.
Gordon Royle has pointed out to me that the bound does not hold for elements of order n. The smallest examples which do not meet the bound are of degree 12 and are the groups numbered 263 and 298 in the catalogue of groups of degree 12 in Magma.<|endoftext|>
TITLE: Magic trick based on deep mathematics
QUESTION [181 upvotes]: I am interested in magic tricks whose explanation requires deep mathematics. The trick should be one that would actually appeal to a layman. An example is the following: the magician asks Alice to choose two integers between 1 and 50 and add them. Then add the largest two of the three integers at hand. Then add the largest two again. Repeat this around ten times. Alice tells the magician her final number $n$. The magician then tells Alice the next number. This is done by computing $(1.61803398\cdots) n$ and rounding to the nearest integer. The explanation is beyond the comprehension of a random mathematical layperson, but for a mathematician it is not very deep. Can anyone do better?

REPLY [5 votes]: A recent episode of Penn and Teller's Fool Us had a trick by Hans Petter Secker which exploits the parity of a permutation in a lovely way.  It may be useful the next time one has to teach the sign of a permutation!
Brief description of the trick in the video:  The magician has sent a box over.  Penn picks up a Rock from the box; Teller then picks Scissors from the box; and Alysson gets the remaining crumpled sheet of paper.  The magician (on video) says that he will predict what they'll do (in the context of the Rock-Paper-Scissors game).  He writes down a prediction, and invites Alysson to pick any two of them and swap their objects. The magician reveals his correct prediction.  Next he writes down two predictions, and asks Teller to perform a swap, and then Penn to perform a swap (I might be forgetting who selects the swaps, but two swaps are made).  Again the magician reveals his correct predictions.  Finally he gets them to make three swaps.  This time he has not written down any predictions, but when the crumpled paper is smoothed out, there are three correct predictions written on it. Enjoy!
https://www.youtube.com/watch?app=desktop&v=hf7Sy7FPal0<|endoftext|>
TITLE: Explicit computation of induced modules of semidirect products with the symmetric group
QUESTION [7 upvotes]: I've gotten stuck in a project I have been working on, essentially on the following combinatorial question about the symmetric group.
One can obtain a 1-dimensional representation $M^n_c$ of the algebra $T_n := S_n \rtimes \mathbb{C}[y_1, \dots, y_n] $ by letting each $y_i$ act by $c$ and $S_n$ act trivially.
Given a partition $\pi$ of $n  =n_1 + \dots + n_k$ and $c_1, \dots, c_k \in \mathbb{C}$, we can consider the standard induced module
$$ M^{\pi}_c = \mathrm{Ind} ( M^{n_1}_{c_1} \otimes \dots \otimes M^{n_k}_{c_k}) $$
where the induction is from the subalgebra $T_{n_1} \otimes \dots \otimes T_{n_k} \subset T_n$ to $T_n$.
As far as I know, the simple quotients of these standard modules form a complete class for the simple representations of $T_n$.  (I think this is a general fact about semidirect products of a finite group with a commutative algebra, together with the fact that representations of the symmetric group $S_n$ can be obtained by taking quotients induction of the trivial representation of a Young subgroup $S_{n_1} \times \dots \times S_{n_k}$.)
My question is how to represent this explicitly in terms of the simple objects in the semisimple category $Rep(S_n)$.  First of all, the $S_n$-module $M^{\pi}_c$ can be described in a combinatorial manner in terms of the simple objects in $Rep(S_n)$ using the Kostka numbers.  The action of the $y_i$'s on $M^{\pi}_c$ can be given in terms of a suitable morphism $y: \mathfrak{h} \otimes M^{\pi}_c \to M^{\pi}_c$, where $\mathfrak{h} \in Rep(S_n)$ is the regular representation.
The Pieri rule allows one to compute the decomposition in irreducibles (as parametrized by Young diagrams, of course) of $\mathfrak{h} \otimes M^{\pi}_c $, so we can view $y$ as a bunch of matrices based upon this decomposition (matrices w.r.t. the simple objects in $Rep(S_n)$, not as vector spaces).  
Is there an approach  to compute these matrices?  
I am interested in this because it may allow a way to directly interpolate the construction of these modules to complex rank, via Pavel Etingof's program.  (I believe one can interpolate the construction in another way, by reasoning more directly on the definition of the category $Rep(S_t)$ given by Deligne, but this seems to be useless as far as explicit computations--which might be helpful to study the "degeneracy phenomena" that Etingof has suggested might exist--are concerned.) In this case we have tensor categories $Rep(S_t)$ for $t$ not necessarily an integer, and while the interpretation in terms of vector spaces fails, the one in terms of Young diagrams does not.  
Edit (12/27) I added a bounty today and here is some additional information that may be useful:
It should come out that the matrices representing the $y$-morphism $\mathfrak{h} \otimes M^{\pi}_c \to M^{\pi}_c$ are polynomials in the dimension $n$.  When increasing $n$, we change the partition $\pi$ by adding to the first (largest) element and leaving the rows below fixed.  Since a simple object in $Rep(S_t)$ for $t$ not an integer can be represented as a normal Young diagram (of size, say, $N$) with a "very long line" of "size" $t-N$ at the top, this kind of a polynomial interpolation will allow for an interpolation of the $M^{\pi}_c$ to complex rank.  My claim that it should come out as a polynomial was based upon studying induction directly on these categories and finding it was interpolable.  However, I don't know how to compute the $y$'s directly as matrices via the simple object decomposition.
My hope was that there is a clean not-too-computationally-intensive way to do this, but unfortunately I'm not yet sufficiently comfortable with the theory of the symmetric group to have any ideas as to how to proceed.
I am also interested in the degenerate affine Hecke algebra of type A, where these kinds of standard induced modules can be defined similarly.  Their simple quotients form a complete collection of irreducible modules for the Hecke algebra according to a theorem of Zelevinsky, and I know that these, too, can be interpolated by reasoning on the definitions in Deligne's paper (so one gets objects in the interpolated category $Rep(H_t)$, which is defined in Etingof's talk).  But I am curious here too how it is possible to compute the $y$-morphisms as matrices using the decomposition into irreducibles in $Rep(S_t)$.

REPLY [2 votes]: I suspect I know the answer, but I don't yet have a proof (not because I think it would be hard to prove, but because I didn't try really; when you see my guess, you'll likely want to believe it).  The answer is stated not in the basis of simples, because I didn't compute the decomposition of $\mathbb{C}[S_n/S_{pi}]$. However, it is stated in the tensor category S_n-mod, so that given that decomposition, you can easily adjust what I write here.

Fact: Let H be a finite dimensional semi-simple Hopf algebra (e.g. H=\mathbb{C}[S_n]), and let $V\in H$-mod be an irrep.  Let us regard H as an H-module via the left action.  Then $V\otimes H\cong H^{\oplus dim(V)}$.

The proof of this fact is given as follows (see http://www-math.mit.edu/~etingof/tenscat.pdf, or Akhil's comments below):
$Hom_H(V\otimes H,W)=Hom_H(H,^\ast V\otimes W) = \widetilde{^\ast V\otimes W}$
On the other hand, $Hom_H(\tilde{V}\otimes H,W) =  \tilde{V}\otimes Hom_H(H,W) = \widetilde{V\otimes W}$,
where $\tilde{M}$ means we forget the module $M$ down to a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted originally isn't canonically given, you just know that there's this multiplicity space)
(above we took right duals since I didn't assume $H$ is commutative or co-commutative; for $C[G]$ there is no need to distinguish.)  One could (and should) be uncomfortable that we got duals on the one hand and not on the other.  However, the standard representation for $S_n$ is special in that it is isomorphic to its own dual, by sending $e_i$ to $e^i$ (the point is that the standard rep for $S_n$ has a basis build into its definition).
The general fact above about Hopf algebras is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space; indeed the regular representation is the unique eigenvector which realizes the Frobenius Perron dimension as an eigenvalue.
Okay so now we are considering $\mathfrak{h}\otimes \mathbb{C}[S_n/S_{\pi}]\to \mathbb{C}[S_n/S_{\pi}]$.  This is then isomorphic to $(\mathfrak{h}\otimes \mathbb{C}[S_n])\otimes_{S_\pi}\mathbf{1}$, where we tensor the trivial $S_\pi$-module on the right.  This is because $C[S_n]$ is a $S_n-S_\pi$ bi-module, so that the map
$\mathfrak{h}\otimes \mathbb{C}[S_n] \to \mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$ given by right multiplying with the symmetrizer $a_\pi=\sum_{g\in S_\pi} g$ is an $S_n$-morphism, and allows us to identify $\mathfrak{h}\otimes \mathbb{C}[S_n]\otimes_{S_\pi}\mathbf{1}$ with $\mathfrak{h}\otimes \mathbb{C}[S_\pi]$.
Morally, this is just because $S_\pi$ acts on the right, while the other action is on the left.
[edited an error from preceding paragraph]
Together with the Fact, this implies that $\mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$ is in fact just isomorphic to $\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})},$ which I should really write as 
$\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}^\ast}$.
Well, now we have this function $c: \mathfrak{h}\to \mathbb{C}$.  We will project $\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})}$ (or rather $\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$) to $\mathbb{C}[S_n/S_\pi]$ by just applying $c$ to the multiplicity space.
I haven't really proved that this last paragraph is what happens, but once one has applied "Fact" above, this seems like the only natural guess.  I imagine verifying it would be pretty straightforward.
Note that it doesn't seem to matter how $\mathbb{C}[S_n/S_{\pi}]$ decomposes into simples, since they all get lumped together.<|endoftext|>
TITLE: Sheaves and Differential Equations
QUESTION [23 upvotes]: How do sheaves arise in studying solutions to ordinary differential equations?
EDIT: Is it possible to construct non-isomorphic sheaves on a domain $D \subset \mathbb{R}^n$ using solution sets to differential equations?
EDIT: Is the sheaf of vector spaces arising from the solution set of a linear ODE necessarily a vector bundle?

REPLY [11 votes]: I will start commenting on Mariano's answer. I believe it is a perfect answer for the question 

How do sheaves arise in studying
  solutions of differential equations ?

but not for the question 

How do sheaves arise in studying
  solutions to ordinary differential
  equations ? 

According to the current terminology a function $f$ satisfying $X(f)=0$ is not a solution of the vector field $X$ but a first integral. Moreover, if $X = a(x,y) \partial_x + b(x,y) \partial_y$ then 
$$
X(f) = a \partial_x f + b \partial_y f  .
$$
Thus $X(f)=0$ is a PDE and not an ODE. Indeed t3suji made the same point at a  comment on Mariano's answer.  I understand the solutions of (the ODE determined by)  $X$ as  functions $\gamma : V \subset \mathbb R \to U$ satisfying $X(\gamma(t))=\gamma'(t)$ for every $t \in V$. Notice that here indeed we have a system of ODEs. 
A vector field can be thought as autonomous differential equation and I do not see clearly how to consider the sheaf of its solutions. 
On the other hand when we have a non-autonomous ordinary differential equation  then there is its sheaf of solutions. This sheaf is a sheaf over the time variable 
only and not the whole space. ( At this point it is natural  to talk about connections and/or jet bundles but I will try to keep things as elementary as possible. ) 
Note that in general the sheaf of solutions will not be a sheaf of vector spaces: the sum of two solutions, or the multiplication of a solution by a constant need not to be a solution. This will  occur only when the differential equation is linear.
The differential equations $y'(t) = y$ and $y'(t) = y^2$, both defined over the whole real line, are examples of differential equations with non-isomorphic sheaves of solutions. The solutions of the first ODE are the multiples of $\exp t $ and define a sheaf of $\mathbb R$-modules. The solutions of the second ODE are zero and $\frac{1}{\lambda - t}$ with $ \lambda \in \mathbb R$. They do define a sheaf of sets, but  not a sheaf of $\mathbb R$-modules.
To obtain examples of linear differential equations with non-isomorphic sheaves, one has to have nontrivial  fundamental group on the time-variable of the differential equation. Thus it is natural to consider complex differential equations over $\mathbb C^{\ast}$. 
The equations  $y'(z) = \frac{ \lambda y(z)}{z}$ parametrized by $\lambda \in \mathbb C$ have non-isomorphic sheaves of solutions. More precisely,

if $\lambda \in \mathbb Z$ then the solution sheaf is the free $\mathbb C$-sheaf of rank one (solutions of the ODE are complex multiples of $z^{ \lambda }$); 
if $\lambda \in \mathbb Q - \mathbb Z $ then the solution sheaf has no global sections but some tensor power of it does;
if $\lambda \in \mathbb C - \mathbb Q$ then the solution sheaf has no global sections nor any of its powers does.<|endoftext|>
TITLE: Complete discrete valuation rings with residue field ℤ/p
QUESTION [40 upvotes]: There are two great first examples of complete discrete valuation ring with residue field $\mathbb{F}_p = \mathbb{Z}/p$:  The $p$-adic integers $\mathbb{Z}_p$, and the ring of formal power series $(\mathbb{Z}/p)[[x]]$.  Any complete DVR over $\mathbb{Z}/p$ is a ring structure on left-infinite strings of digits in $\mathbb{Z}/p$.  The difference between these two examples is that in the $p$-adic integers, you add the strings of digits with carries.  (In any such ring, you can say that a $1$ in the $j$th place times a 1 in the $k$th place is a 1 in the $(j+k)$th place.)  At one point I realized that these two examples are not everything:  You can also add with carries but move the carry $k$ places to the left instead of one place to the left.  The ring that you get can be described as $\mathbb{Z}_p[p^{1/k}]$, or as the $x$-adic completion of $\mathbb{Z}[x]/(x^k - p)$.  This sequence has the interesting feature that the terms are made from $\mathbb{Z}_p$ and have characteristic $0$, but the ring structure converges topologically to $(\mathbb{Z}/p)[[x]]$, which has characteristic $p$.
(Edit: Per Mariano's answer, I have in mind a discrete valuation in the old-fashioned sense of taking values in $\mathbb{Z}$, not $\mathbb{Z}^n$.)
I learned from Jonathan Wise in a question on mathoverflow that these examples are still not everything.  If $p$ is odd and $\lambda$ is a non-quadratic residue, then the $x$-adic completion of $\mathbb{Z}[x]/(x^2-\lambda p)$ is a different example.  You can also call it $\mathbb{Z}_p[\sqrt{\lambda p}]$.
So my question is, is there is a classification or a reasonable moduli space of complete DVRs with residue field $\mathbb{Z}/p$?  Or whose residue field is any given finite field? Or if not a classification, an indexed family that includes every example at least once?  
I suppose that the question must be related to the Galois theory of $\mathbb{Q}_p$; maybe the best answer would be a relevant sketch of that theory.  But part of my interest is in continuous families of DVR structures on the Cantor set of strings of digits in base $p$.

As question authors often say in mathoverflow, I learned stuff from many of the answers and it felt incomplete to only accept one of them.  I upvoted several others, though.  Following Kevin, the set of pairs $(R,\pi)$, where $R$ is a CDVR and $\pi$ is a uniformizer, is an explicit indexed family that includes every example (of $R$) at least once.  The mixed characteristic cases are parametrized by Eisenstein polynomials, and there is only one same-characteristic choice of $R$.
One side issue that was not addressed as much is continuous families of CDVRs.  To explain that concern a little better, all CDVRs with finite residue field are homeomorphic (to a Cantor set).  For any given finite residue field, there are many homeomorphisms that commute with the valuation.  So you could ask what a continuous family of CDVRs can look like, where both the addition and multiplication laws can vary, but the topology and the valuation stay the same.  Now that I have been told what the CDVRs are, I suppose that not all that much can happen.  It seems key to look at $\nu(p)$, the valuation of the integer element $p$, in a sequence of such structures (say).  If $\nu(p) \to \infty$, then the CDVRs have to converge to $k[[x]]$.  Otherwise $\nu(p)$ is eventually constant; it eventually equals some $n$.  Then it looks like you can convert the convergent sequence of CDVRs to a convergent sequence of Eisenstein polynomials of degree $n$.

REPLY [4 votes]: Let $k$ be a finite field of characteristic $p$.  There is only one
characteristic-$p$ local field with residue field $k$, namely
$k((\pi))$, where $\pi$ is transcendental, and there is a "smallest"
characteristic-$0$ local field with residue field $k$, namely the
degree-$a$ unramified extension $K$ of $\mathbb{Q}_p$, where $q=p^a$ is
the cardinality of $k$.
Every characteristic-$0$ local field with residue field $k$ is a
finite totally ramified extension of $K$, and every finite totally
ramified extension of $K$ has residue field $k$.
As $K$ has only finitely many extensions of any given degree, we might
ask for the number of totally ramified extensions $L|K$ of given
degree $[L:K]=n$.  Serre's "mass formula" (Comptes rendus 1978),
says that there are exactly $n$ such extensions, when counted
properly.  This formula was also proved by Krasner by his methods.
Let Serre explain:
"Let $K$ be a local field, with finite residue
field with $q$ elements. Let $n$ be a positive integer, and let
$\Sigma_n$ be the set of all totally ramified extensions of $K$ of
degree $n$ contained in a given separable closure of $K$. If
$\operatorname{gcd}(n,p)=1$, it is easy to show that $\text{Card}(\Sigma_n)=n$. If $L$
belongs to $\Sigma_n$, put $c(L)=d(L)-n+1$, where $d(L)$ is the
valuation of the discriminant of $L/K$. Our 'mass formula' is
$\sum_{L\in \Sigma_n} q^{-c(L)}=n$. 
We give two proofs. The first one
uses Eisenstein polynomials, while the second one applies the H. Weyl
integration formula to the multiplicative group of a division
algebra."  MR0500361 (80a:12018)
(Note that $k((\pi))$ has many totally ramified extensions $L$, but they are all of the form $L=k((\varpi))$ for some uniformiser $\varpi$ of $L$.)<|endoftext|>
TITLE: Does torsion-freeness of class group localize? 
QUESTION [13 upvotes]: Let $R$ be a local normal domain, and let $P \in Spec (R)$. It is well known that $Cl(R) \to Cl(R_P)$ is surjective. However, I do not know any example where $Cl(R)$ is torsion-free, but $Cl(R_P)$ is not (we consider $(0)$ to be torsion-free). So:
If the class group of $R$ is torsion-free, must all the class groups of local rings of $R$ torsion-free? What if $R$ is not local?   
I do not have a motivation for this, but it just seems an intriguing question.

REPLY [2 votes]: Perhaps something like the following works (I have not checked all the details):
Let $C$ be a smooth plane conic and let $Y$ be the projective cone over $C$. Then $Cl(Y) = \mathbb{Z}$ but the class group of the local ring of the vertex of $Y$ is $\mathbb{Z}/2$. Let $X$ be affine cone over $Y$. Then the class group of the local ring
$R$ of the vertex of $X$ is $\mathbb{Z}$ but it seems that the class group of $R$ localised at the prime ideal corresponding to the cone over the vertex of $Y$ is $\mathbb{Z}/2$.
EDIT
The above is wrong as pointed out by Hailong Dao in his comment. I try to fix it below:
Let Y be as above i.e. the singular quadric in $\mathbb{P}^3$ given by the equation $x^2 + y^2 +z^2 = 0$. It may be viewed as the toric surface given by the complete fan with rays passing through $(1,0)$, $(0,1)$ and $(-1,-2)$. Then $Cl(Y) = \mathbb{Z}$ and $Pic(Y)$ is of index $2$ in $Cl(Y)$. Let $Y'$ be the blowup of $Y$ at a non-singular torus fixed point. We may view
Y as the surface obtained from the fan for $Y$ by adding the ray through the point $(1,1)$. Let $\pi:Y' \to Y$ be the blowup map and let $E$ be the exceptional divisor.
Then $Cl(Y') \cong \mathbb{Z} \oplus \mathbb{Z}$ and $Cl(Y')/\mathbb{Z}E = Cl(Y)$.
Let $H$ be an ample divisor on $Y$. Then for $n >> 0$, $H':= n\pi^*(H) - E$ is an ample divisor on $Y'$ (this is true for the blowup of a point on any surface). Note that $Cl(Y')/\mathbb{Z}H' \cong \mathbb{Z}$, so it is torsion free. SInce $Y'$ is a projective toric surface and $H'$ is an ample divisor, it follows that $H'$ is very ample and gives a projectively normal embedding of $Y'$ in $\mathbb{P}(H^0(Y',\mathcal{O}(H')))$. 
As before, we now let $X$ be the cone over $Y'$ and let $R$ be the local ring of the vertex. We let $P$ be the prime ideal corresponding to the cone over the singular point of $Y'$.
($Cl(Y)$ and $Cl(Y')$  can be computed by hand or using the toric description I gave and the results in Fulton, Toric Varieties, Sections 3.3, 3.4; the fact that an ample divisor on a projective toric surface is very ample is an Exercise at the bottom of p.70.)
Note that by letting $R$ be the coordinate ring of $Y' \backslash D$ , where $D$ is a general divisor linearly equivalent to $H'$  (so not containing the singular point) one gets a normal 2-dimensional (non-local) ring with $Cl(R)= \mathbb{Z}$  and with a prime ideal $P$ such that $Cl(R_P)=\mathbb{Z}/2$ . In all of the above one can replace $2$ by any integer $n>1$ (by considering the projective cone over the rational normal curve on degree $n$, or, in the toric description, replacing $(-2,-1)$  by $(-n,-1)$.<|endoftext|>
TITLE: Models for, and motivation for, (oo,n)-categories for general n
QUESTION [5 upvotes]: First: Is there a precise meaning to the term "model for (oo,n)-categories"? A related question, which might actually be the same question, is: what exactly do we want to get out of (oo,n)-category theory for general n? Whatever definition of (oo,n)-categories we use, what are the desired things it should satisfy? What should the main examples be, for general n? I know that bordism categories should be examples. What else? Actually, aside from the cobordism hypothesis and other TQFT-y things I don't really know what the motivations are for (oo,n)-category theory for general n (or at least n bigger than 1), so I hope that people can say some words about that as well. (For n=1, there seems to be a lot of motivation, see for example this or this or this or this or ...)
Second: Presently, what are the models that we have for (oo,n)-categories? Which models have been proven to be equivalent? Of course, there's already a lot about this on the nLab:
(oo,1)-categories
(oo,2)-categories
(oo,n)-categories
I'm mainly just curious about the current status on (oo,n)-categories for general n. Aside from n-fold complete Segal spaces, are there other definitions/"models"?

REPLY [4 votes]: The nlab page on n-categories includes a list of known definitions and comparisons, which should in particular include all definitions of (∞,n)-categories (since any definition of (∞,n)-category can be specialized to a definition of n-category = (n,n)-category by requiring all k-cells to be trivial for k>n).  This list is almost certainly incomplete, but I humbly propose that anyone with knowledge of something it omits should rectify the situation!  This sort of question is so common, and the comparisons are so important to the subject, that it would be useful to have a definitive list of what is known.<|endoftext|>
TITLE: What to do with antique math books?
QUESTION [19 upvotes]: My grandfather had a PhD in math.  When he died, he left a lot of math textbooks, which I took.  These include things like Van der Waerden's 2-volume algebra set from the 1970s, 
"Studies in Global Geometry and Analysis" by Shiing-Shen Chern, a series called "Mathematics: it's content, methods, and meaning," and many more.
I'm keeping about 20 of them, but there are 103 which I don't want to keep, but which I don't know what to do with.  I obviously don't want to throw them away, and I don't really know what will happen to them if I donate them to the giant used-books depository in downtown Baltimore (called "the book thing," where people drop off and pick up used books for free).  I'd like to donate them to some math collector or math library.  But maybe there are just too many used antique math books floating around. 
RECAP:  I have 103 antique used math books which I cannot keep.  Do you have a suggestion for what to do with them?
Thanks,
David

REPLY [11 votes]: I would suggest scanning them all and donate them to "the" "internet" book library (for example, a thepiratebay.se); many people would be grateful and the legacy of your father shall be preserved. And one hopes that the laws shall eventually change so that it becomes legal (and maybe is already legal in some countries)..<|endoftext|>
TITLE: How Much Work Does it Take to be a Successful Mathematician?
QUESTION [95 upvotes]: Hi Everyone,
Famous anecdotes of G.H. Hardy relay that his work habits consisted of working no more than four hours a day in the morning and then reserving the rest of the day for cricket and tennis. Apparently his best ideas came to him when he wasn't "doing work." Poincare also said that he solved problems after working on them intensely, getting stuck and then letting his subconscious digest the problem. This is communicated in another anecdote where right as he stepped on a bus he had a profound insight in hyperbolic geometry.
I am less interested in hearing more of these anecdotes, but rather I am interested in what people consider an appropriate amount of time to spend on doing mathematics in a given day if one has career ambitions of eventually being a tenured mathematician at a university. 
I imagine everyone has different work habits, but I'd like to hear them and in particular I'd like to hear how the number of hours per day spent doing mathematics changes during different times in a person's career: undergrad, grad school, post doc and finally while climbing the faculty ladder. "Work" is meant to include working on problems, reading papers, math books, etcetera (I'll leave the question of whether or not answering questions on MO counts as work to you). Also, since teaching is considered an integral part of most mathematicians' careers, it might be good to track, but I am interested in primarily hours spent on learning the preliminaries for and directly doing research.
I ask this question in part because I have many colleagues and friends in computer science and physics, where pulling late nights or all-nighters is commonplace among grad students and even faculty. I wonder if the nature of mathematics is such that putting in such long hours is neither necessary nor sufficient for being "successful" or getting a post-doc/faculty job at a good university. In particular, does Malcom Gladwell's 10,000 hour rule apply to mathematicians?
Happy Holidays!

REPLY [24 votes]: Every excellent mathematician that I know works extremely hard and for very long hours; however there are many others who also work extremely hard who are just average or "journeyman" mathematicians.
In other words, hard work is necessary but not sufficient - the existence of prodigies such as the Terry Taos and Akshay Venkateshs (just to name two Aussies that I've at least met) seems to me to be sufficient evidence that some natural talent/creativity/imagination/genius is necessary beyond mere hard work. 
The working habits of mathematicians can also be endlessly amusing - we currently have a visitor who feels that he is full of creative energy immediately on waking, and so rather than getting out of bed and wasting that energy showering and having breakfast etc., he puts in an hour or two of maths and only starts the daily routine when he reaches his first dip in energy.<|endoftext|>
TITLE: Does an inverse polynomial map on the taylor coefficients of a rational function preserve rationality?
QUESTION [5 upvotes]: Supppose there are integers $a_1,a_2,\dots$ and a polynomial $p$ so that the integers $p(a_1),p(a_2)...$ satisfy some linear recurrence, i.e. $\sum p(a_i)x^i$ is a rational function of $x$. Must integers $b_i\in p^{-1}(p(a_i))$ so that $\sum b_ix^i$ is a rational function, necessarily exist?
(The answer is no if we ask for the function $\sum a_i x^i$ to be rational, as can be seen when $p(t)=t^2$ and $a_i$ being a random sequence of $\pm1$)

REPLY [3 votes]: If $p(x)=x^d$ then this question coincides with Pisot's d'th root conjecture. A proof is given in this paper of Zannier, I'm not sure if it's the one that Qiaochu was referring to in the comments.
Edit: A more general question was answered in the subsequent article Equations in the Hadamard ring of rational functions by Andrea Ferretti and Umberto Zannier.<|endoftext|>
TITLE: Is the volume form on an oriented Riemannian manifold parallel?
QUESTION [5 upvotes]: This question actually has two parts:

Am I correct that the Riemannian connection induces a connection on the k-th exterior power of the cotangent bundle, for any k?  What I had in mind was: view the Riemannian connection as a connection on the frame bundle and then think of the k-th exterior power of the cotangent bundle as the associated bundle to the frame bundle under the natural action of GL(n) on the k-th exterior algebra of (R^n)*.   
If you view the volume form of an oriented Riemannian manifold as a section of the n-th exterior power of the cotangent bundle, is it always parallel with respect to this connection?

REPLY [2 votes]: These are important questions.
When one starts to define connections, one usually imposes the Leibnitz condition on it. But I am proceeding too fast. First, let's recall what a connection is.
Here we come to one very confusing point, at least, it was for me, when I started learning differential geometry: There are various ways of defining connections.
Just to name some of them:

Define the Levi-Civita connection on a (Riemannian) manifold.
The Levi-Civita connection is an affine connection that preserves the metric structure, i.e.,
$\nabla g = 0$ and that is torsion-free, i.e., $\nabla_{X}Y-\nabla{Y}X = [Y, X]$, where $[. , .]$ denotes the Lie-bracket of two elements of $\Gamma(TM)$. Then one usually starts by showing that, by this definition, the Levi-Civita connection (hereafter called LC connection) is uniquely determined.
However, one can even generalize this approach by at first considering linear connections, then metric ones and then affine ones. After all that one can turn to the derivation of the Levi-Civita connection. Personally, I like this approach pretty much because it shows very well why the LC connection is so important. Namely, it eases computations for the Lie-derivative which normally does not require the notion of the LC connection at all. Also there are many fascinating mathematical objects, such as Killing fields which can be calculated (or at least defined) effectively, if one introduces the notion of the LC connection.
One can even start more generally, by considering fiber bundles. This approach was taken by the French mathematician Charles Ehresmann (1905-1979). This approach has one fundamental advantage: Generality, however, it lacks the easy-to-grasp-effect (I like to assign a definition this value if it has a very intuitive meaning). Interestingly, this approach does yield the LC connection as a special case when one reduces fiber bundles to vector bundles and considers the bundles $\pi : TM \longrightarrow M$ as a special case and requires the base manifold $M$ to be equipped with a symmetric, positive definite 2-times covariant tensor field, i.e., the metric tensor $(g_{ij})_{i,j}$.

Why am I writing all this? Simply to sum up what we already know and how beautifully, at least, in my opinion, all these definitions harmonize.
To come to your question (I apologize for relying so much on your patience):
If one has finally obtained a manifold that is equipped with a Riemannian metric, then one goes further and asks what the term orientation means for general Riemannian manifolds. Although for submanifolds of $\mathbb{R}^{n}$ one has a very intuitive picture of what orientability means. We could expand this picture by making use of the Nash Embedding theorem, but it turns that this is not necessary.
Example: Consider for instance the so-called Möbius strip. It is a typical example of a non-orientable two-dimensional submanifold of the Euclidean 3-space.
Then, one can ask further what it means for a manifold to be orientable. Consider now the tangent and the cotangent bundle. It is now possible to associate to each vector space a certain orientation if one introduces one starting point for that. Consider a basis that is called "unit basis". Define the orientation of this basis to be positive. Then, one imposes further conditions on the unit basis, namely orthonormality w.r.t the Riemannian metric. This way, one can define an orientation on the cotangent bundle $TM^{*}$ as well (Duality principle).
Now we simply ask what the Riemannian volume form is: It is certainly $SO(n)$-invariant, because we have chosen an orientation and require the 1-forms that constitute a basis of $TM^{*}$ to be positively oriented (c.f. what I said above).
OK, now to question one:
Via the Leibnitz rule and the duality principle it is possible to define an induced connection on the cotangent bundle and, by that way, further expanding this definition to the k-th exterior power of the cotangent bundle.
Namely, from the tensor product spaces can be obtained the k-th exterior cotangent space. I would like to refer you to the book Introduction to Kähler Manifolds which contains and excellnt treatment of exactly this. Via the equivalence relation one then breaks down the Leibnitz rule to the k-h exterior power cotangent bundle. The rules is exactly what Matt has written above. 
Let's come to question two: We know that the volume form is $SO(n)$ invariant, where $n=dim(M)$ Since we define the volume form $\omega := dx^{1} \wedge \dots \wedge dx^{n}$, with the $dx^{i}$ for $i=1, \dots, n$ to be the product of the basis of the cotangent bundle, we can make the following thought.
We know how to "differentiate" the basis w.r.t to the LC connection. When we calculate all that, we finally obtain that the volume form is parallel. This was how I learned it. Of course one can use representation theory here which is, in my eyes, very interesting and also more elegant.
If there are questions, don't hesitate to contact me.<|endoftext|>
TITLE: Classification of symplectic surfaces
QUESTION [17 upvotes]: Is there a classification of symplectic surfaces, i.e. of surfaces equipped with an area form?  Symplectic topology references like McDuff-Salamon seem to start their discussion of open questions with dimension four.

A surface admits a symplectic form iff it is orientable.
The Moser trick seems to show that on a compact orientable surface $M$, the unique invariant of a symplectic form is its total area?  So the set of symplectic forms on $M$ (up to symplectomorphism) is parametrized by $\mathbb{R}^+$?
And, for non-compact orientable surfaces, ...?

REPLY [7 votes]: This is a bit late answer to an old MO question, but Moser's theorem was generalized to open manifolds in 
R. Greene and K. Shiohama, Diffeomorphisms and volume-preserving embeddings of noncompact manifolds, Trans. Amer. Math. Soc. 255 (1979), p. 403-414.
http://www.ams.org/journals/tran/1979-255-00/S0002-9947-1979-0542888-3/S0002-9947-1979-0542888-3.pdf
They gave a necessary and sufficient condition for existence of volume-preserving diffeomorphisms between open manifolds in terms of volumes of ends and overall volumes. This takes care of surfaces as a very special case.<|endoftext|>
TITLE: Heuristic behind the Fourier-Mukai transform
QUESTION [38 upvotes]: What is the heuristic idea behind the Fourier-Mukai transform? What is the connection to the classical Fourier transform?
Moreover, could someone recommend a concise introduction to the subject?

REPLY [4 votes]: This question is a little old but I want to complement the multiple great answers given above with some more similarities between the ``classical'' Fourier transform and the Fourier-Mukai transform.
Compositions
Let us denote by $\phi$ the ``classical'' Fourier transform. It has already been mentioned above that $\phi^4$ is the identity, but it has not been mentioned yet that $\phi^2(f) = f(-x)$. 
Let us denote by $\Phi_{\mathcal P}$ the Fourier-Mukai transform on the derived category of coherent sheaves on an abelian variety $X$ and its dual $\hat X$. So $\Phi_{\mathcal P}$ sends a complex of coherent sheaves on $X$ to a complex of coherent sheaves on $\hat X$. It can be proved that $\Phi_{\mathcal P}^4$ is the identity and moreover $\Phi_{\mathcal P}^2 = -1_X^*[-g]$. Here $-1_X : X \to X $ denotes the inverse morphism of $X$ and $[-g]$ denotes a shift by $-g = -\dim X$.
So the above shows that compositions of $\Phi_{\mathcal P}$ behave roughly ``the same'' as compositions of $\phi$.
Product and Convolution
Another striking similarity is how $\phi$ and $\Phi_{\mathcal P}$ handle products and convolutions. Recall that for functions $f$ and $g$ satisfying certain requirements we can define their convolution 
$$ f * g (x) := \int_{-\infty}^{\infty} f(t)g(t - x) \text{d}t. $$
A well known theorem then states that $\phi(f * g) = \phi(f) \cdot \phi(g)$ and $\phi(f \cdot g) = \phi(f) * \phi(g)$.
We can also define a sort of convolution for coherent sheaves on abelian varieties. Suppose $\mathcal F, \mathcal G$ are coherent sheaves on $X$. We define the convolution of these sheaves as
$$ \mathcal F * \mathcal G := \mu_*(p_1^*\mathcal F \otimes p^*_2\mathcal G ). $$
Here $\mu : X \times_k X \to X$ denotes the group law of the abelian variety $X$. 
Now it can be shown that $\Phi_{\mathcal P}(\mathcal F * \mathcal G) = \Phi_{\mathcal P}(\mathcal F) \otimes \Phi_{\mathcal P}(\mathcal G)$ and $\Phi_{\mathcal P}(\mathcal F \otimes \mathcal G) = \Phi_{\mathcal P}(\mathcal F) * \Phi_{\mathcal P}(\mathcal G)[g]$<|endoftext|>
TITLE: Polynomial representing prime numbers
QUESTION [6 upvotes]: Along the lines of Polynomial representing all nonnegative integers, but likely well-known question:

is there a polynomial $f \in \mathbb Q[x_1, \dots, x_n]$ such that $f(\mathbb Z\times\mathbb Z\times\dots\times\mathbb Z) = P$, the set of primes?

REPLY [19 votes]: No.  Any such polynomial would have the property that any of its restrictions $f(x)$ to one variable consist only of primes, but this is easily seen to be impossible, since if $p(a)$ is prime then $p(k p(a) + a)$ is divisible by $p(a)$.  (Even accounting for the coefficients in $\mathbb{Q}$ is straightforward by multiplying by the common denominator and using CRT; in fact, we can show that given an integer polynomial $q(x)$ and a positive integer $n$ there exists $x_n$ such that $q(x_n)$ is divisible by $n$ distinct primes.)
However, there do exist multivariate polynomials with the property that their positive integer outputs consist of the set of primes.  See the Wikipedia article.<|endoftext|>
TITLE: Why is the Alexander polynomial a quantum invariant?
QUESTION [27 upvotes]: When we think of quantum invariants, we usually think of the Jones polynomial or of the coloured HOMFLYPT. But (arguably) the simplest example of a quantum invariant of a knot or link is its Alexander polynomial. From the beginning, the central problem in the study of quantum invariants has been what do they mean topologically? The Alexander polynomial has clear algebraic topological meaning as the order of the Alexander module (first homology of the infinite cyclic cover as a module over the group of deck transformations). Can people conceptually explain (in terms of both physics and mathematics) why the representation theory of certain small quantum groups naturally gives rise to this quantity? Computationally I can understand it, but not conceptually.
A somewhat related question was already asked here.
Update: I posted on this question here and here. See also this question.

REPLY [2 votes]: Cimasoni and Turaev, A Lagrangian representation of tangles (Topology 44 (2005), 747-767, doi:10.1016/j.top.2005.01.001) have a very natural generalization of the Alexander module that's hovering around your concerns.
I found out about this paper for completely different reasons — Paolo Salvatore pointed it out when we were trying to come up with an argument that the group completion of the monoid of string links can't be abelian.<|endoftext|>
TITLE: Iterated adjoint functors
QUESTION [15 upvotes]: Let $F_0 : C \to D$ be a functor.  If it exists, let $G_0 : D \to C$ be its left adjoint.  If it exists, let $F_1 : C \to D$ be its left adjoint.  And so forth.  In situations where the infinite sequence $(F_0, G_0, F_1, G_1, ...)$ exists, when is it periodic?  Aperiodic?  (Feel free to replace all "lefts" by "rights," of course.)

REPLY [5 votes]: Let $C$ be an $\infty$-category.

If $C$ has a zero object, then the unique functor $C \to pt$ is an ambidextrous adjoint (i.e. fits into an adjoint string periodic of order 2).

(If $C$ is semiadditive, then the diagonal functor $C \to C^2$ likewise is an ambidextrous adjunction.)

Let $C^{[1]}$ be the category of arrows of $C$. There is always an adjunction $cod \dashv id \dashv dom$ between the codomain functor, the functor which takes the identity arrow, and the domain functor. If $C$ has an initial object, this chain extends one step further to the left, and if $C$ has a terminal object, it extends one further to the right. If $C$ is pointed and has cokernels it extends an additional step to the left, and if $C$ has kernels it extends an additional step to the right.
If $C$ is stable, this adjoint string extends infinitely in both directions. It is periodic of order 6 up to a twist by the suspension functor $\Sigma$.

Likewise, let $C^{[2]}$ be the category of composable pairs of arrows of $C$. The composition map $C^{[2]} \to C^{[1]}$ always fits into the middle of an adjoint string of length 5, which extends an additional step with intitial / terminal objects and again with kernels / cokernels.
If $C$ is stable, this adjoint string extends infinitely in both directions. It is periodic of order 24 up to a twist by $\Sigma^3$.


Examples $(1,3,4)$ clearly suggest that if $C$ is stable, then between $C^{[n-1]}$ and $C^{[n]}$ we have a bi-infinite adjoint string, perhaps periodic (up to a shift) of order $(n+1)!$. But I'm not sure.
Various reindexing functors for functor categories into a stable $\infty$-category from various other finite categories also induce infinite adjoint strings. For instance,

the constant functor $C \to C^{\bullet \leftarrow \bullet \to \bullet}$ fits in a bi-infinite adjoint string which is periodic of order 4 up to a shift by $\Sigma$.

If $C$ is ambidextrous in the sense of (Hopkins and) Lurie, even more functor categories do this.
Another place this happens is in Grothendieck duality -- it turns out that an exact symmetric monoidal functor between nice tensor-triangulated categories either fits in an adjoint string of length 1, 3, 5, or bi-infinity. In the infinite case, it is periodic of order 6 up to a twist by the so-called "dualizing object".<|endoftext|>
TITLE: Uniformly Sampling from Convex Polytopes
QUESTION [20 upvotes]: How to choose a point uniformly from a convex polytope $P \subset [0,1]^n$ defined by some inequalities,  $Ax < b$? (Here $A$ is an $m \times n$ matrix, $x \in \mathbb{R}^n$, and $b \in \mathbb{R}^m$.)  I imagine that you could start with a uniformly chosen point in the cube and do some process to get a point with $Ax < b$.

REPLY [16 votes]: Hit and run sampling will perform much better than rejection sampling for higher dimensions.
It converges to a uniform distribution in polynomial time. The basic idea is to start at any point $x_0$ inside the polytope, then follow the following procedure iteratively:

pick a direction $\alpha$ uniformly at random.
find the minimum and maximum value of $\theta$ such that $x_n+\theta\cdot\alpha$ is contained in the polytope
pick a $\theta^*$ uniformly at random from $[\theta_{min},\theta_{max}]$
your new point is $x_{n+1}=x_n+\theta^*\cdot\alpha$<|endoftext|>
TITLE: When f(x)-a and f(x)-b yield the same field extension
QUESTION [17 upvotes]: An interesting mathoverflow question was one due to Philipp Lampe that asked whether a non-surjective polynomial function on an infinite field can miss only finitely many values.  In my interpretation of the question, if $k$ is a starting field and $f$ is a polynomial, you could ask what happens if you repeatedly adjoin a root of $f(x)-a$, except for a finite set of values $a \in S \subset k$ for which you hope a root never appears.  You have to adjoin a root for all $a \in \tilde{k} \setminus S$, where $\tilde{k}$ is the growing field.  Either a root of $f(x)-a$ for some $a \in S$ will eventually appear by accident, or $f$ as a polynomial over the limiting field $\tilde{k}$ is an example.
(Edit: I call this an interpretation rather than a construction, because in generality it is equivalent to Philipp's original question.  I also don't mean to claim credit for  the idea; it was already under discussion when I posted my answer then.  Maybe an answer to the question below was already implied in the previous discussion, but if so, I didn't follow it.)
For some choices of $f$ and a non-value $a$, you can know that you are sunk at the first stage.  For instance, suppose that $f(x) = x^n$.  When you adjoin a root of $x^n-a$, you also adjoin a root of $x^n-b^na$ for every $b \in k$.  You cannot miss $a$ without also missing every $b^na$, which is then infinitely many values when $k$ is infinite.
So let $k$ be an infinite field, and let $f \in k[x]$ be a polynomial.  Define an equivalence relation on those elements $a \in k$ such that $f(x)-a$ is irreducible.  The relation is that $a \sim b$ if adjoining one root of $f(x)-a$ and $f(x)-b$ yield isomorphic field extensions of $k$.  Is any such equivalence class finite?  What if $k$ is $\mathbb{Q}$ or a number field?
In my partial answer to the original MO question, I calculated that if $f$ is cubic and the characteristic of $k$ is not 2 or 3, then the equivalence classes are all infinite.

REPLY [24 votes]: The answer is that over a number field $k$, an equivalence class can be finite, and in fact it is usually so for $f$ of moderately large degree.  Consider $f(x):=x^7+x$ over $\mathbf{Q}$, for example.  If the equations $f(x)=1$ and $f(x)=t$ for some other $t \in \mathbf{Q}$ yield the same degree $7$ extension, then in particular the discriminant $D(t)$ of the polynomial $f(x)-t$ in $x$ must equal $D(1)$ times a square.  In fact, $D(t) = -823543 t^6 - 46656$, so the necessary condition is $-823543 t^6 - 46656 = -870199 u^2$.
This defines a genus $2$ curve, so Faltings' theorem implies that this equation has only finitely many rational solutions.
Moreover, for a typical $f$ of slightly higher degree, it is reasonable to expect that all the equivalence classes are singletons, although proving such a statement would seem to require understanding the rational points on surfaces of general type, which is probably beyond the current state of knowledge.  
Reasoning along these lines suggests to me that the iterative procedure you and others have proposed should succeed in constructing a field with a nonsurjective polynomial as in Philipp Lampe's question, even if we can't prove this.<|endoftext|>
TITLE: Presburger Arithmetic
QUESTION [23 upvotes]: Presburger arithmetic apparently proves its own consistency. Does anyone have a reference to an exposition of this? It's not clear to me how to encode the statement "Presburger arithmetic is consistent" in Presburger arithmetic. 
In Peano arithmetic this is possible since recursive functions are representable, so a recursive method of assigning Godel numbers to formulas and proofs means that Peano arithmetic can represent its own provability relation (of course, showing all that requires a lot of work). In particular, we can write a Peano arithmetic sentence which says "there is no natural number which encodes a proof of $\bot$". 
On the other hand, Presburger arithmetic can't represent all recursive functions. It can't even represent all the primitive recursive ones, so this same trick doesn't work. If it did, the first incompleteness theorem would apply.

REPLY [36 votes]: Presburger arithmetic does NOT prove its own consistency. Its only function symbols are addition and successor, which are not sufficient to represent Godel encodings of propositions.
However, consistent self-verifying axiom systems do exist -- see the work of Dan Willard ("Self-Verifying Axiom Systems, the Incompleteness Theorem and Related Reflection Principles"). The basic idea is to  include enough arithmetic to make Godel codings work, but not enough to make the incompleteness theorem go through. In particular, you remove the addition and multiplication function symbols, and replace them with subtraction and division. This is enough to permit representing the theory arithmetically, but the totality of multiplication (which is essential for the proof of the incompleteness theorem) is not provable, which lets you consistently add a reflection principle to the logic.<|endoftext|>
TITLE: Can topologies induce a metric? (revised)
QUESTION [5 upvotes]: This is a revised version of a question I already posted, but which patently was ill posed. Please give me another try.

For comparison's sake, the axioms of a metric:
Axiom A1: $(\forall x)\ d(x,x) = 0$
Axiom A2: $(\forall x,y)\ d(x,y) = 0 \rightarrow x = y$
Axiom A3: $(\forall x,y)\ d(x,y) = d(y,x)$
Axiom A4: $(\forall x,y,z)\ d(x,y) + d(y,z) \geq d(x,z)$

Let $T$ = {X,T} be a topology, $B$ a base of $T$, $x, y, z$ $\in$ X
Definition D0: $x$ is nearer to $y$ than to $z$ with respect to $B$ ($N_Bxyz$) iff $(\exists b \in B)\  x, y \in b \ \& \ z \notin b \ \&\ (\nexists b \in B)\  x, z \in b \ \& \ y \notin b$
Definition D1: $B$ is pre-metric1 iff $(\forall x,y)\ x \neq y \rightarrow N_Bxxy$
Definition D2: $B$ is pre-metric2 iff $(\forall x,y,z)\ ((z \neq x\ \&\ z \neq y) \rightarrow N_Bxyz) \rightarrow x = y$
Definition D3: $B$ is pre-metric3 iff $(\forall x,y,z)\ N_Bzyx \rightarrow (N_Byxz \rightarrow N_Bxyz)$

Definition: $T$ is pre-metrici iff $(\exists B)\ B$ is pre-metrici (i = 1,2,3).
Definition: $B$ is pre-metric iff $B$ is pre-metric1, pre-metric2 and pre-metric3.
Definition: $T$ is pre-metric iff $(\exists B)\ B$ is pre-metric.
Remark: D1 is an analogue of axiom A1, D2 of axiom A2, D3 of axiom A3.
Remark: $T$ is pre-metric1 iff $T$ is T1 [not quite sure].
Remark: If $T$ is induced by a metric, then $T$ is pre-metric.

Question: Can a property pre-metric4 be defined such that $T$ induces a metric iff $T$ is induced by a metric with
Definition: $B$ is metric iff $B$ is pre-metric and pre-metric4.
Definition: $T$ induces a metric iff $(\exists B)\ B$ is metric.
Remark: Property pre-metric4 should be an analogue of A4 (the triangle inequality).
If provably no such property can be defined does this shed a light on the difference (an asymmetry) between topologies and metric spaces? ("It's the triangle inequality, that cannot be captured topologically.")

REPLY [10 votes]: I shall prove that there can be no characterization of metrizability along the lines that you seek. (This argument fleshes out and fulfills the expectation of Mariano in the comments.)
Your axioms are all stated in the language with two types of objects: points and basis elements. So let us be generous here, and entertain the possibility of any statement of a similar type, expressed in the same language. You are considering structures consisting of a set X and a collection B of subsets of X, to be used to generate a topology on X. Your language allows you to quantify over points and over basis elements, and to say that a given point is an element of a given basis element, and so on. This is a perfectly fine formal language to work in. One can express that B is indeed a basis for a topology in this language, by the assertion first that every point is in some basis element and second that for every two basis element b and c, and every point x in both b in c, there is a basis element d such that x is in d and for all points z, if z is in d, then it is both b and c. Similarly, one can state the Hausdorff condition and other topological properties in this language.
But I claim that metrizability is simply not expressible in this language. To see why, suppose toward contradiction that there were an axiom Phi expressible in the language of points and basic open sets in the way we have described, which held exactly of the bases that generate metric topologies. Consider now the real line (R,<). This is a linear order, and the natural basis for the order topology, the collection of open intervals, is first-order definable in the language consisting only of the order. Thus, the collection of basic open sets is describable in the language of the linear order. For example, to quantify over basic open sets, one says, "there are two points a<b such that..." and then refers to the interval (a,b). Since the real line is indeed a metric topology, the collection of intervals in the real line (R,<) will satisfy your axiom Phi. 
Now comes the problem. By the methods of nonstandard analysis, there is an extension of the real line to the nonstandard real line, denoted (R*,<), which contains infinitesimal elements and infinite integers, but which has the amazing property that any statement expressible in the language of < which is true in the real line (R,<) is also true in the nonstandard line (R*,<). In fact, nonstandard analysis is rather powerful, for one can add the field structure and indeed, any structure whatsover to R and get the nonstandard version in R*. In particular, your axiom Phi will remain true for the nonstandard reals (R*,<). That is, your axiom Phi, whatever it is, will remain true for the order topology on the nonstandard real line. But this is a serious problem, because there are nonstandard extensions R* whose order topology is not metrizable. For example, there are instances of R* where the cofinality of the positive infinitesimals is uncountable, and in these instances, there are no nontrivial convergent sequences in the order topology of R*, but the topology is not trivial. Thus, the axiom Phi has made a mistake with R*, contradiction.
Thus, there can be no axiomatization of metrizability along the lines you suggest.
Edit: In case you suggest that the standard basis for the order should not be the witness for Phi in the real line, but instead some other weird basis B is the one satisfying Phi, 
then we simply consider the nonstandard (R*,<) together with B*. By the transfer principle, B* is a basis for the order topology on (R*,<) and continues to satisfy Phi, but R* is not metrizable, for the same contradiction.<|endoftext|>
TITLE: Using TikZ in papers
QUESTION [17 upvotes]: I've started using TikZ for a paper I'm writing and am worrying that a journal might not accept the paper with inline TikZ. I had to update my PGF installation for all the examples to work and I'm not even sure that my collaborator will have an updated version...
So here are the questions.
-Do people know if journals (math and physics) accept inline TikZ? 
-Is there a nice way to create a pdf figure from a tikzpicture ? Perhaps I could have a latex file for each figure and only one \begin{tikpicture} \end{tikzpicture} pair and somehow compile the result to get a pdf or eps containing just the figure (and not a whole page with a figure in it...)
Cheers,
Yossi.

REPLY [11 votes]: It is not exactly answering either of your two questions, but here is another work-around.  I had problems putting papers on the arxiv which used pgf/tikz because the version of pgf/tikz they used at the arxiv was not as up to date as my version.  The admin at the arxiv told me to do the following.  LaTeX your file with the option -recorder.  This will create a .fls file containing a list of all of the FiLeS used by LaTeX when typesetting your document.  Choose all of the files in the list containing "pgf" or "tikz" and move them into the directory containing your document.  You can then send that directory to your collaborator/the arxiv/the journal without worrying about how up to date their set up is. 
[Unfortunately, I then had the problem that the tikz graphics I produced required an enormous amount of working memory which was greater than the allocation on the arxiv server, so I resorted to using 'grab' on my mac to take a high resolution snap-shot of the graphic, which I then incorporated into the LaTeX file :-(.  However, I have used this including-all-the-files technique for subsequent uploads to the arxiv.]<|endoftext|>
TITLE: Notation for the all-ones vector
QUESTION [27 upvotes]: What's the most common way of writing the all-ones vector, that is, the vector, when projected onto each standard basis vector of a given vector space, having length one? The zero vector is frequently written $\vec{0}$, so I'm partial to writing the all-ones vector as $\vec{1}$, but I don't know how popular this is, and I don't know if a reader might confuse it with the identity matrix.
I'm writing for a graph theory audience, if that helps pick a notation.

REPLY [5 votes]: Let $I \subset \{ 1,2,3,\ldots, n \} $. Let $e_I = \sum_{i\in I} e_i.$ Let $[n]=\{ 1,2,3, \ldots, n \} $. 
Then $\vec{1}=e_{[n]}$. Also $e_{\{i\}} = e_i$.
This is not satisfactory to your context, but may have the advantage of alternative usages in subsequent contexts.<|endoftext|>
TITLE: Which are the rigid suborders of the real line?
QUESTION [23 upvotes]: Which are the rigid suborders of the real line? 
If A is any set of reals, then it can be viewed as an order structure itself under the induced order (A,<). The question is, when is this structure rigid? That is, for which sets A does the structure (A,<) have no nontrivial order automorphisms? 
For example, the positive integers are rigid under the usual order. More generally, any well-ordered subset of R is rigid. Similarly, any anti-well-ordered set, such as the negative integers, is also rigid. 
It is also true that the order sum of any well-order plus an anti-well-order is rigid. For example, a sequence converging upward to 0 plus a sequence converging downward to 0 will have order type ω+ω*, which is rigid.  (Whereas it is easy to see that the sum of an infinite anti-well-order and an infinite  well-order will not be rigid, since it has a copy of Z in the center.) A more elaborate example will be a well-ordered sum of anti-well-orders, such as the set consisting of k+1/n for any positive integers k and n. 
All these examples are countable; are there uncountable examples? 
Perhaps there will be some ZFC independence for certain types of examples?  I am primarily interested in the situation under ZFC. In ZF without the Axiom of Choice, there can be weird anomalies of uncoutable sets that are Dedekind finite. All such sets are rigid, as I explained in this question. But if someone can provide a ZF characterization, that would also be interesting.

REPLY [25 votes]: Here is a simple example of size continuum. Do the ordinary middle-third construction of the Cantor set, except that whenever you delete the $n$-th (numbered by level and then left to right, say) middle-third interval leave in exactly $n$ points from that interval. Let's call the resulting set $X$. Any automorphism of $X$ must map (resp. left-, right-) isolated points to (resp. left-, right-) isolated points. Since we left a different number of points in each middle-third interval, the automorphism must fix every inner point of each middle-third interval as well as its two endpoints. By density, it follows that the automorphism fixes every point of $X$.
All of your examples are scattered, I checked the Rosenstein's Linear Orderings to see if he had anything good to say about which scattered linear orders are rigid. To my dismay, this is what I found: "These considerations seem to make impossible an inductive argument (on $F$-rank or $VD$-rank) to determine which scattered types are rigid." (p. 133) However, he does cite a result of Anne Morel (Ordering relations admitting automorphisms, Fund. Math. 54 (1964), 279-284.) which says that a linear order $A$ is not rigid if and only if $A \cong A_1 + A_2\times\mathbb{Z} + A_3$ for some linear orderings $A_1,A_2,A_3$, with $A_2$ nonempty.
Dense examples of rigid subsets of $\mathbb{R}$ would be interesting to see. This is probably not too difficult to construct under CH. But a ZFC example might have to deal with important barriers such as Baumgartner's result that All ${\aleph_1}$-dense subsets of $\mathbb{R}$ can be isomorphic, Fund. Math. 79 (1973), 101-106. Maybe there are some examples in this classic paper of Sierpinski Sur les types d'ordre des ensembles linéaires, Fund. Math. 37 (1950), 253-264.
All three papers can be found here.

Addendum (after sdcvvc's comment): For the sake of completeness, I'm including a simplification of the Dushnik-Miller argument that produces a dense subset $X$ of $\mathbb{R}$ which is rigid (though not the stronger result that $X$ has no self-embeddings).
To ensure density, the set $X$ will contain all rational numbers. Note that an automorphism $f$ of $X$ is then completely determined by its restriction to $\mathbb{Q}$. Indeed, since $f[\mathbb{Q}]$ must be dense (in $X$ and) in $\mathbb{R}$, we always have
$f(x) = \sup\{f(q):q \in (-\infty,x)\cap\mathbb{Q}\} = \inf\{f(q):q \in (x,\infty)\cap\mathbb{Q}\}.$
There are only $c = 2^{\aleph_0}$ increasing maps $f:\mathbb{Q}\to\mathbb{R}$ with dense range. Let $\langle f_\alpha:\alpha<c \rangle$ enumerate all such maps, except for the identity on $\mathbb{Q}$.
We will define by induction a sequence $\langle (x_\alpha,y_\alpha) : \alpha<c \rangle$ of pairs of irrational numbers. The $x_\alpha$ will be points of $X$ while the $y_\alpha$ will be in the complement of $X$. For each $\alpha$, we will have $f_\alpha(x_\alpha) = y_\alpha$ (in the sense of the inf/sup definition above).
Suppose we have defined $(x_\beta,y_\beta)$ for $\beta<\alpha$. Since $f_\alpha$ is not the identity, there is a rational $q$ such that $f_\alpha(q) \neq q$. Let's suppose that $f_\alpha(q) > q$ (the case $f_\alpha(q) < q$ is symmetric). Since the real interval $(q,f_\alpha(q))$ has size $c$ and the extension of $f_\alpha$ to all of $\mathbb{R}$ is injective, we can always pick
$x_\alpha \in (q,f_\alpha(q)) \setminus(\mathbb{Q}\cup\{y_\beta:\beta<\alpha\})$
such that $y_\alpha = f_\alpha(x_\alpha) \notin \mathbb{Q}\cup\{x_\beta:\beta<\alpha\}$. Note that $x_\alpha < f_\alpha(q) < y_\alpha$ so $x_\alpha \neq y_\alpha$.
In the end, we will have 
$\{x_\alpha: \alpha < c\} \cap \{y_\alpha : \alpha<c\} = \varnothing$
and any set $X$ such that
$\mathbb{Q}\cup\{x_\alpha:\alpha<c\} \subseteq X \subseteq \mathbb{R}\setminus\{y_\alpha:\alpha<c\}$
is necessarily rigid since $f_\alpha(x_\alpha) = y_\alpha \notin X$ for each $\alpha<c$.<|endoftext|>
TITLE: Are Dynkin diagrams of some universal construction?
QUESTION [14 upvotes]: This is a general question. The classification of semisimple Lie algebras by using Dynkin diagrams has always amazed me. And these A,B,C,D,E,F,G diagrams seem to appear quite often in the realm of representation theory (of all kinds of things! Lie algebra, Lie group, quivers,etc). 
My question is (vaguely put), WHY are these diagrams useful? Are they of some more universal (thus more imaginable, more trivial) constructions? They always seem very mysterious for me.

REPLY [4 votes]: I'm not an expert and would gladly learn about any updates but, as far as I know, your question has no good answer yet. The problem of deeper (universal) origin of the Dynkin diagrams, i.e., of explaining why they show up in many apparently unrelated areas of mathematics (the A-D-E problem) was posed by V.I. Arnold more than 30 years ago, see here and here, and apparently is still open; at least the 2004 book Arnold's problems lists no solution.
Update: you may wish to look also at this question, in particular at the paper mentioned by Thomas Riepe.<|endoftext|>
TITLE: Affine Weyl groups as Coxeter groups
QUESTION [12 upvotes]: If G is a reductive algebraic group (say over ℂ), T a maximal torus, then we can consider its Weyl group W which acts on the abelian group Y of one parameter subgroups of T. Thus we may form the semidirect product, which I will call the affine Weyl group.
In the semisimple simply connected case, this affine Weyl group is a Coxeter group. In the general setup I have here, the affine Weyl group is no longer a Coxeter group, but shares some properties of Coxeter groups. For example one can still define a length function, and considering an Iwahori subgroup of a loop group, we get something that looks like it wants to be a BN-pair/Tits system (but isn't, since we don't have a Coxeter group).
Since the axiomatic setup of Coxeter groups and BN-pairs is so convenient, I ask if there has been developed a generalisation of this setup to include the affine Weyl groups I mention above. I have some vague feeling that this should include some sort of 'disconnected' (I am thinking in the algebraic group sense when I use this word) Coxeter group, where the connected component is a genuine Coxeter group.

REPLY [12 votes]: In the abstract Bourbaki set-up, the affine Weyl group is defined to be a
semidirect product of an irreducible Weyl group with its coroot lattice.
This is naturally a Coxeter group, characterized in terms of its
positive semidefinite Coxeter matrix.  The basic theory is developed
independently of applications in Lie theory, but is directly usable if you
start with a connected semisimple algebraic group (over an algebraically
closed field) and require its root system to be irreducible of type A, B,
etc.  Most of the time this causes no trouble.  While it is natural to work
with a connected reductive group, people often use the expression "affine
Weyl group" too loosely in this general context.   For example, the standard
features of alcove geometry require irreducibility.   Otherwise you have
to deal with products of simplexes, etc.  In any case, the difference between
reductive and semisimple groups such as general linear and special linear
is sometimes significant.
In the Iwahori-Matsumoto (or Bruhat-Tits) setting over local fields, a
more intrinsic affine Weyl group occurs directly within the structure of
the group itself.  Here one has to be cautious about applying
abstract Coxeter group theory or BN-pair theory, as I believe most authors
are.  Already in the proceedings of the 1965 Boulder AMS summer institute,
Iwahori had to formulate a more complicated "generalized BN-pair" formalism
for this situation.   I'm not sure what has become standard by now in the
literature.
In other situations (the classical study of compact Lie groups, or
the later application of affine Weyl groups in modular representation
theory starting with Verma) there is usually no difficulty about specializing
to the irreducible case.   Here the affine Weyl group lives outside the
actual group under study.  This is the situation I'm most comfortable with.
You need to make precise the setting in which you really want to study reductive groups, in order to adapt the Bourbaki language and results. 
There are several distinct issues here: 1) Extra care is needed in treating
disconnected algebraic groups such as orthogonal groups, or in treating 
reductive rather than semisimple groups.   2) Adjoint groups, simply
connected groups, and the occasional intermediate type: not all details of
structure are exactly the same.    3) Most important for working over
local fields is the natural use of an "extended affine Weyl group" (as in
much of Lusztig's work involving Hecke algebras, cells, etc.).  Here you
start with the Bourbaki version of the affine Weyl group (a Coxeter group)
and form a semidirect product with a finite group $\Omega$ isomorphic to the weight lattice mod root lattice (universal center).   This amounts to 
working with a semidirect product of the Weyl group and the full (co)weight
lattice rather than the (co)root lattice.    Fortunately it's easy to extend
notions like length function to this extended group. 
EDIT: Besides Bourbaki's treatment of Coxeter groups and root systems (1968),
foundational papers from that era include Iwahori-Matsumoto (IHES Publ. Math. 25, 1965), at http://www.numdam.org,  and Iwahori's 1965 article in the AMS Boulder proceedings
http://www.ams.org/books/pspum/009/0215858/pspum0215858.pdf, followed by much more technical work by Bruhat-Tits.
Lusztig has written many technical papers on extended affine Weyl groups and corresponding affine Hecke algebras, including his four part series on cells in affine Weyl groups and later work on multiparameter cases.    Much of this work is motivated by reductive groups over local fields, as well as the modular representation theory of reductive groups and their Lie algebras (where "linkage" of weights appears at first relative to an extended affine Weyl group).<|endoftext|>
TITLE: yoneda-embedding vs. dual vector space
QUESTION [11 upvotes]: I've made the following observation: let V be a vector space over $\mathbb{R}$ with a inner product $\langle , \rangle$. then there is a "natural contravariant" injective map $V \to \hom(V,\mathbb{R})$. if we apply this twice, we get a "natural covariant" injective map $V \to \hom(\hom(V,\mathbb{R}),\mathbb{R}), v \mapsto (\phi \mapsto \phi(v))$. but the same things happen in category theory: let $C$ be a category (which I assume to be locally-small), then $C \to \hom(C,Set), x \mapsto \hom(x,-)$ is a natural contravariant fully-faithful functor and applying this twice yields (up to natural isomorphism) the natural covariant functor $C \to \hom(\hom(C,Set),Set), x \mapsto (F \mapsto F(x))$ (yoneda-lemma).
so how can we unify these two phenomena? perhaps we can make $V$ to a enriched category over $\mathbb{R}$ and hope that the yoneda-lemma for symmetric closed monoidal categories is the common generalization? what is the right structure on $\mathbb{R}$?

REPLY [4 votes]: There are many different things that Martin may have had in mind in asking this question, but the obvious answer requires less category theory and not more.
The basic operation is   ${-}\to S$   for some fixed "dualising" object in some cartesian closed (or more generally monoidal closed) category.
This operation is part of a monad, of which Martin's natural map is the unit.<|endoftext|>
TITLE: projections of finitely presented groups
QUESTION [5 upvotes]: let's call an object $x$ of a cocomplete category (categorical) finitely generated if $\hom(x,-)$ commutes with filtered colimits of monomorphisms, and finitely presented if $\hom(x,-)$ even commutes with arbitrary filtered colimits. I know that in the literature there are some other definitions for these notions. at least in categories of algebraic structures, it would be nice that these notations coincide with the usual ones. let's concentrate on the category of groups, for these make problems ;).
it's easy to see that categorical finitely generated means the usual finitely generated. but I've figured out that a group $G$ is categorical finitely presented if and only if there is a finitely presented group $H$, over which $id_G : G \to G$ factors through. is this the same as beeing finitely presented?
the question is equivalent to: let $H$ be a finitely presented group and $p : H \to H$ a projection, i.e. a homomorphism satisfying $p^2 = p$. is then $im(p)$ also finitely presented?

REPLY [5 votes]: Here's a geometric proof that a retract of a finitely presented group is finitely presented.
This uses the fact that a f.g. group is coarsely simply connected iff it is finitely presented (Gromov calls this a triviality; it takes a little time to realize he's right).
Coarsely simply connected passes to retracts: it's clear because if you have a loop in the retract, coarse fill it in the all space and apply the retraction to the homotopy to get a homotopy inside the retract.
[The same argument shows that the Dehn function decreases when passing to retracts.]<|endoftext|>
TITLE: Order of the Tate-Shafarevich group
QUESTION [11 upvotes]: I thought that the order of the Tate-Shafarevich group should always be a square (it's also supposed to be finite, but for the purposes of this question let's assume we know this) but I don't seem to find a good explanation; Wikipedia is silent on the matter.
While I know it may be an open problem, is there a good argument pro or contra this?

REPLY [35 votes]: The first example of an abelian variety with nonsquare Sha was discovered in a computation by Michael Stoll in 1996.  He emailed it to me and Ed Schaefer, because his calculation depended on a paper that Ed and I had written.  At first none of us believed that it was what it was: instead we thought it must be due to either an error in Stoll's calculations or an error in the Poonen-Schaefer paper.  Stoll and I worked together over the next few weeks to develop a theory that explained the phenomenon, and this led to the paper http://math.mit.edu/~poonen/papers/sha.ps - that paper contains a detailed answer to your question.
To summarize a few of the key points: If the abelian variety over a global field $k$ has a principal polarization coming from a $k$-rational divisor (as is the case for every elliptic curve), then the order of Sha is a square (if finite), because it carries an alternating pairing - this is what Tate proved, generalizing Cassels' result for elliptic curves.  For principally polarized abelian varieties in general, the pairing satisfies the skew-symmetry condition $\langle x,y \rangle = - \langle y,x \rangle$ but not necessarily the stronger, alternating condition $\langle x,x \rangle=0$, so all one can say is that the order of Sha is either a square or twice a square (if finite).  Stoll and I gave an explicit example of a genus 2 curve over $\mathbf{Q}$ whose Jacobian had Sha isomorphic to $\mathbf{Z}/2\mathbf{Z}$ unconditionally (in particular, finiteness could be proved in this example).
If the polarization on the abelian variety is not a principal polarization, then the corresponding pairing need not be even skew-symmetric, so there is no reason to expect Sha to be even within a factor of $2$ of a square.  And indeed, William Stein eventually found explicit examples and published them in the 2004 paper cited by Simon.
A final remark: Ironically, my result with Stoll quantifying the failure of Sha to be a square is used by Liu-Lorenzini-Raynaud to prove that the Brauer group $\operatorname{Br}(X)$ of a surface over a finite field is a square (if finite)!<|endoftext|>
TITLE: Algorithm or theory of diagram chasing
QUESTION [84 upvotes]: One of the standard parts of homological algebra is "diagram chasing", or equivalent arguments with universal properties in abelian categories.  Is there a rigorous theory of diagram chasing, and ideally also an algorithm?
To be precise about what I mean, a diagram is a directed graph $D$ whose vertices are labeled by objects in an abelian category, and whose arrows are labeled by morphisms.  The diagram might have various triangles, and we can require that certain triangles commute or anticommute.   We can require that certain arrows vanish, which can be used to ask that certain compositions vanish.  We can require that certain compositions are exact.  Maybe some of the arrows are sums or direct sums of other arrows, and maybe some of the vertices are projective or injective objects.  Then a diagram "lemma" is a construction of another diagram $D'$, with some new objects and arrows constructed from those of $D$, or at least some new restrictions.
As described so far, the diagram $D$ can express a functor from any category $\mathcal{C}$ to the abelian category $\mathcal{A}$.  This looks too general for a reasonable algorithm.  So let's take the case that $D$ is acyclic and finite.  This is still too general to yield a complete classification of diagram structures, since acyclic diagrams include all acyclic quivers, and some of these have a "wild" representation theory.  (For example, three arrows from $A$ to $B$ are a wild quiver.  The representations of this quiver are not tractable, even working over a field.)  In this case, I'm not asking for a full classification, only in a restricted algebraic theory that captures what is taught as diagram chasing.
Maybe the properties of a diagram that I listed in the second paragraph already yield a wild theory.  It's fine to ditch some of them as necessary to have a tractable answer.  Or to restrict to the category $\textbf{Vect}(k)$ if necessary, although I am interested in greater generality than that.
To make an analogy, there is a theory of Lie bracket words.  There is an algorithm related to Lyndon words that tells you when two sums of Lie bracket words are formally equal via the Jacobi identity.  This is a satisfactory answer, even though it is not a classification of actual Lie algebras.  In the case of commutative diagrams, I don't know a reasonable set of axioms — maybe they are related to triangulated categories — much less an algorithm to characterize their formal implications.
(This question was inspired by a mathoverflow question about George Bergman's salamander lemma.)

David's reference is interesting and it could be a part of what I had in mind with my question, but it is not the main part.  My thinking is that diagram chasing is boring, and that ideally there would be an algorithm to obtain all finite diagram chasing arguments, at least in the acyclic case.  Here is a simplification of the question that is entirely rigorous.
Suppose that the diagram $D$ is finite and acyclic and that all pairs of paths commute, so that it is equivalent to a functor from a finite poset category $\mathcal{P}$ to the abelian category $\mathcal{A}$.  Suppose that the only other decorations of $D$ are that: (1) certain arrows are the zero morphism, (2) certain vertices are the zero object, and (3) certain composable pairs of arrows are exact.  (Actually condition 2 can be forced by conditions 1 and 3.)  Then is there an algorithm to determine all pairs of arrows that are forced to be exact?  Can it be done in polynomial time?
This rigorous simplification does not consider many of the possible features of lemmas in homological algebra.  Nothing is said about projective or injective objects, taking kernels and cokernels, taking direct sums of objects and morphisms (or more generally finite limits and colimits), or making connecting morphisms.  For example, it does not include the snake lemma.  It also does not include diagrams in which only some pairs of paths commute.  But it is enough to express the monomorphism and epimorphism conditions, so it includes for instance the five lemma.

REPLY [2 votes]: This is an old question, but I think I have a reasonably complete solution to this problem.  It gives you a criterion for when you can answer a question by a diagram chase, and a procedure for actually doing it.  I will sketch the answer here, and if anyone is still interested in this question, I will make a more formal write-up.
Diagram chase arguments make sense for pointed sets, which I will think of as vector spaces over $F_1$ (the field of one element).  Maps between pointed sets are required to have well-defined kernels and cokernels, i.e. they are partial bijections for nonzero elements.  In this setting, diagram chases are even simpler because when you chase an element backward or forward, there's only one possibility.
So what good does this do us?  It turns out that when the diagram is simple enough,
theorems over $F_1$ can be lifted to arbitrary abelian categories.  I will call these diagrams “chaseable”.  You can associate with each object in the diagram the lattice of subobjects containing the zero object, the whole object, and closed under taking images and inverse images.  If that lattice is finite and distributive for each object, then the diagram is chaseable.  The main reason it would fail to be finite is if the diagram has loops.  It is easy to find cases where it fails to be distributive (the $D_3$ quiver is a ready example), but double complexes give you lattices that are both finite and distributive.  (To me this ``explains'' the prominence of double complexes -- they are much simpler than arbitrary diagrams.)
Once you associate a finite distributive lattice to each object, you can use the Birkhoff representation theorem to understand the lattice.  You can associate to each join-irreducible element a one-element generator of a $F_1$-vector space.  If you follow that element around, you get an indecomposable representation of the diagram over $F_1$.  If a condition such as being exact holds for indecomposable $F_1$ representations, then it holds for the diagram.
This is sufficient to prove results such as the five or nine lemmas.  Results that imply the existence of maps, such as the snake lemma or the connecting homomorphism for long exact sequences require slightly more work -- the join-irreducible elements have an order, and you need to keep track of the order -- but it is doable.
You can also use the same technique to reprove the characterization of finite-dimensional indecomposable modules over string algebras -- the indecomposable representations over a field correspond to the indecomposable representations over $F_1$.<|endoftext|>
TITLE: "Kummerian" fields?
QUESTION [12 upvotes]: This is sort of a random, spur of the moment question, but here goes:
We define [with apologies to Conan the Barbarian] a field K to be $\textbf{Kummerian}$ if there exists 
an index set I, and functions $x: I \rightarrow K, n: I \rightarrow \mathbb{Z}^+$ such that 
the algebraic closure of K is equal to $K[(x(i)^{\frac{1}{n(i)}})_{i \in I}]$.  More plainly, the algebraic closure is obtained by adjoining roots of elements of the ground field, not iteratively, but all at once. 
Questions:
QI) Is there a classification of Kummerian fields?
QII) What about a classification of "Kummerian (topological) groups", i.e., the absolute Galois groups of Kummerian fields? 
Here are some easy observations:
1) An algebraically closed or real-closed field is Kummerian.  In particular, the groups of order 1 and 2 are Galois groups of Kummerian fields.  By Artin-Schreier, these are the only finite absolute Galois groups, Kummerian or otherwise.
2) A finite field is Kummerian: the algebraic closure is obtained by adjoining roots of unity.  Thus $\hat{\mathbb{Z}}$ is a Kummerian group.  
3) An algebraic extension of a Kummerian field is Kummerian.  Thus the class of Kummerian groups is closed under passage to closed subgroup.   Combining with 2), this shows that any torsionfree procyclic group is Kummerian.  On the other hand, the class of Kummerian groups is certainly not closed under passage to the quotient, since $\mathbb{Z}/3\mathbb{Z}$ is not a Kummerian group.
4) A Kummerian group is metabelian: i.e., is an extension of one abelian group by another.  This follows from Kummer theory, using the tower $\overline{K} \supset K^{\operatorname{cyc}} \supset K$, where $K^{\operatorname{cyc}}$ is the extension obtained by adjoining all roots of unity.  
In particular no local or global field (except $\mathbb{R}$ and $\mathbb{C}$) is Kummerian.
5) The field $\mathbb{R}((t))$ is Kummerian.  Its absolute Galois group is the profinite completion of the infinite dihedral group $\langle x,y \ | \ x^2 = 1, \ xyx^{-1} = y^{-1} \rangle$.  In particular a Kummerian group need not be abelian.  
Can anyone give a more interesting example?
ADDENDUM: In particular, it would be interesting to see a Kummerian group that does not have a finite index abelian subgroup or know that no such exists.

REPLY [4 votes]: Recall that a field $K$ is pseudo finite if (1) $K$ is perfect (2) the absolute Galois group of $K$ is $\hat{\mathbb{Z}}$, and (3) every absolutely irreducible non-void variety defined over $K$ has a $K$-rational point. 
Ax proved that large finite fields have the same elementary theory as a pseudo finite field.  
I guess that the statement: the unique extension of degree $n$ is generated by a root of an element of the field is elementary, hence from since finite fields are Kummerian, so are pseudo finite fields are Kummerian.
If I got it right so far, this gives an abundance of examples using Jarden's theorem (non of them explicit): Choose a Galois automorphism $\sigma$ in the absolute Galois group of $\mathbb{Q}$. Then with probability one, the fixed field of $\sigma$ in the algebraic closure is pseudo finite. (Recall that Galois groups are profinite, hence compact, hence equipped with Haar probability measure.)
Jarden's theorem holds in more generality, namely when we replace $\mathbb{Q}$ with any countable Hilbertian field (for some examples see here).
Anyway, I guess it doesn't answer any of your questions, in particular, it doesn't give any new absolute Galois group. But I thought some more examples might be interesting...<|endoftext|>
TITLE: When is Aut(G) abelian
QUESTION [33 upvotes]: let $G$ be a group such that $\mathrm{Aut}(G)$ is abelian. is then $G$ abelian?
This is a sort of generalization of the well-known exercise, that $G$ is abelian when $\mathrm{Aut}(G)$ is cyclic, but I have no idea how to answer it in general. At least, the finitely generated abelian groups $G$ such that $\mathrm{Aut}(G)$ is abelian can be classified.

REPLY [7 votes]: A detailed survey on this problem for finite groups is now published 
 and is available at https://link.springer.com/chapter/10.1007/978-981-13-2047-7_7. A pre-published version is  available at https://arxiv.org/abs/1708.00615.
There have been some activities on this topic recently. No example of a non-special finite $p$-group having abelian automorphism group was know until quite recently. A class of such groups is constructed in "V. K. Jain, M. K. Yadav, On finite p-groups whose automorphisms are all central, Israel J. Math. 189 (2012), 225 - 236." This paper also contains a quick  survey of results on the topic and a big bibliography. Some more, different kind of examples are available at http://arxiv.org/pdf/1304.1974.pdf<|endoftext|>
TITLE: Limits in category theory and analysis
QUESTION [35 upvotes]: Is it possible to regard limits in analysis (say, of real sequences or more generally nets in topological spaces) as limits in category theory? Is there some formal connection?
Edit ('13): Perhaps it is more interesting to ask whether limits in category theory can be seen as special limits of ultrafilters or nets.

REPLY [2 votes]: I am not completely satisfied by the accepted answer because the functor which characterizes the convergence of a filter depends on the limit. I therefore add another quite simple answer (written for sequences but this easily generalizes to filters and nets) to this old post.

The definition of a limit as a universal cone of a functor resembles the infimum (greatest lower bound) of a set in a very transparent way: Considering a partially ordered set $(X,\le)$ as a category with only one morphism from $x$ to $y$ if $x\le y$ and none otherwise, a subset $A$ of $X$ has an infimum if and only if the inclusion functor $A\hookrightarrow X$ has a limit. In particular, if the power set $\mathscr P(X)$ is ordered by inclusion the intersection of any subfamily $\mathscr A$ is a limit.

Let now $(x_n)_{n\in\mathbb N}$ be a sequence in some topological space $X$. Then the limit of the contravariant functor $F:\mathbb N\to \mathscr P(X)$ assigning to $n$ the set $F(n)=\overline{\{x_k:k\ge n\}}$ is the set of all limit points of the sequence.
I think that this is a strong relation between analytical and functorial limits although it does not yet characterize convergence of sequences. At least, if either $X$ is a compact Hausdorff space or $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence in a Hausdorff uniform space, the sequence converges if and only if the set of limit points is a singleton.<|endoftext|>
TITLE: Is this a sleight of hand or a video edit?
QUESTION [9 upvotes]: In another question someone linked to this video where the lady ties a knot in a seemingly impossible manner.
What I don't understand is how the end sticking out beyond her right hand gets longer as she pulls through (seems that if it's to change it should get shorter). This suggests to me that there is some footage cut out.

REPLY [14 votes]: Sleight of hand!  I learned to do this by watching the video - there is a moment when you let go of the end of the rope and grab another part.  Your observation is telling you which end to let go of, when to let go, and where to grab.<|endoftext|>
TITLE: distance regular metric spaces
QUESTION [12 upvotes]: A metric space (V,d) will be called distance regular if for every distances a>0, b, c a nonnegative integer p(a,b,c) is defined, so that whenever d(B,C)=a, there are precisely p(a,b,c) points A such that d(A,B)=c, d(A,C)=b. 
The Euclidean plane is an example: p(a,b,c)=0,1, or 2 when the triangle inequality for a,b,c, correspondingly, fails, turns into equality, or is strict.
If we also require that p(a,b,c)>0 whenever the triangle inequality does not fail, then I conjecture that this is the only possibility for the parameters p(a,b,c). That is, there may be many non-isomorphic examples, but the parameters will be the same for all of them. (Thanks to Heather for this clarification.)
Has anybody formulated/proved/refuted this conjecture before? It looks very natural.
UPD. I should have mentioned this: "for every distances a>0, b, c" means all nonnegative reals, and the same for "whenever the triangle inequality does not fail". In particular this means that all positive real distances actually occur.
UPD2. After a week trial, the question seems to be new, open, and interesting. Anton suugested a line of attack, and I believe I can write down the proof of the first step: that p(a,b,c)=1 when the triangle inequality turns into equality. Fedja produced examples showing that this first step is indeed essential.
I'm adding the "open problem" tag to the question.

REPLY [4 votes]: It all would be nice if not for the following. Using the transfinite induction, you can easily construct a set in $\mathbb R^3$ that intersects each circle by any number of points greater than 2 you want depending on the radius of the circle only. Indeed, the cardinality of the set of all circles is continuum. Now take any complete ordering of this set so that each beginning interval has cardinality less than continuum and start adding points to the least circle that still has too few points on it. Clearly, the points that can overload other circles should lie on some circles with at least 3 points already, but those are fewer than continuum at each step of induction, so they intersect our circle by less than continuum points and there are continuum points on our circle to choose from. So, at each step we can make a good choice.
Needless to say, this can be generalized quite a lot, so I'm not sure if there are any restrictions on $p$ whatsoever as long as we talk about arbitrary metric spaces. Now, if we impose additional restrictions, the situation changes a lot. The minimal restriction to impose is that the space be complete. Can we make nontrivial examples now?<|endoftext|>
TITLE: Colimits of schemes
QUESTION [50 upvotes]: This is related to another question.
I've found many remarks that the category of schemes is not cocomplete. The category of locally ringed spaces is cocomplete, and in some special cases this turns out to be the colimit of schemes, but in other cases not (which is, of course, no evidence that the colimit does not exist). However, I want to understand in detail a counterexample where the colimit does not exist, but I hardly found one. In FGA explained I've found the reference, that Example 3.4.1 in Hartshorne, Appendix B is a smooth proper scheme over $\mathbb{C}$ with a free $\mathbb{Z}/2$-action, but the quotient does not exist (without proof). To be honest, this is too complicated to me. Are there easy examples? You won't help me just giving the example, because there are lots of them, but the hard part is to prove that the colimit really does not exist.

REPLY [2 votes]: This answer is mainly in response to Matt Emerton's question above about whether a collection of maps $\operatorname{Spec} A/I^n \to X$ algebraizes to a map $\operatorname{Spec} \widehat{A} \to X$. The answer is indeed yes if $X$ is qcqs and is a result of Bhargav from 2014. Note that $A$ and $X$ need not be Noetherian!<|endoftext|>
TITLE: Coarse moduli spaces over Z and F_p
QUESTION [16 upvotes]: I would like to know to what extent it is possible to compare fibers over $\mathbb{F}_p$ of coarse moduli spaces over $\mathbb{Z}$, and coarse moduli spaces over $\mathbb{F}_p$. I ask a more precise question below.
Let $\mathcal{M}_g^{\mathbb{Z}}$ be the moduli stack of smooth genus $g$ curves over $\mathbb{Z}$. Let $M_g^{\mathbb{Z}}$ be its coarse moduli space, and $(M_g^{\mathbb{Z}})_p$ the fiber of this coarse moduli space over $\mathbb{F}_p$.
Let $\mathcal{M}_g^{\mathbb{F}_p}$ be the moduli stack of smooth genus $g$ curves over $\mathbb{F}_p$ and $M_g^{\mathbb{F}_p}$ its coarse moduli space.
The universal property gives a map $\phi:M_g^{\mathbb{F}_p}\rightarrow(M_g^{\mathbb{Z}})_p$. My question is : is $\phi$ an isomorphism ?
In fact, since $\phi$ is a bijection between geometric points, and $M_g^{\mathbb{F}_p}$ is normal, the question can be reformulated as : is $(M_g^{\mathbb{Z}})_p$ normal ? This shows that when $g$ is fixed, the answer is "yes" except for a finite number of primes $p$.

REPLY [9 votes]: So you're asking if formation of coarse spaces commutes with (certain types of) base change. In general the answer is no; one needs the notion of a tame moduli space. A good starting point for this is Jarod Alper's paper "Good Moduli Spaces for Artin Stacks", available on his web page; he explains the notion and cites the relevant papers for tame moduli spaces. This should help you to work out your particular example (I don't know the answer off the top of my head).<|endoftext|>
TITLE: What does Faltings' theorem look like over function fields?
QUESTION [18 upvotes]: Minhyong Kim's reply to a question John Baez once asked about the analogy between $\text{Spec } \mathbb{Z}$ and 3-manifolds contains the following snippet:
Finally, regarding the field with one element.  I'm all for general theory building, but I think this is one area where having some definite problems in mind might help to focus ideas better.  From this perspective, there are two things to look for in the theory of $\mathbb{F}_1$.

1)  A theory of differentiation with respect to the ground field.  A well-known consequence of such a theory could include an array of effective theorems in Diophantine geometry, like an effective Mordell conjecture or the ABC conjecture.  Over function fields, the ability to differentiate with respect to the field of constants is responsible for the considerably stronger theorems of Mordell conjecture type [emphasis mine], and makes the ABC conjecture trivial.
What is the strongest such theorem?  Does anyone have a reference?  (A preliminary search led me to results that are too general for me to understand them.  I'd prefer to just see effective bounds on the number of rational points on a curve of genus greater than 1 in the function field case.)

REPLY [14 votes]: The "function field analogues" of Faltings' theorem were proved by Manin, Grauert and Samuel: see
29/PMIHES_1966_29_55_0/PMIHES_1966_29_55_0.pdf">http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1966_29/PMIHES_1966_29_55_0/PMIHES_1966_29_55_0.pdf
especially Theorem 4.  (The quotation marks above are because all of this function field work came first: the above link is to Samuel's 1966 paper, whereas Faltings' theorem was proved circa 1982.)
The statement is the same as the Mordell Conjecture, except that there is an extra hypothesis on "nonisotriviality", i.e., one does not want the curve have constant moduli.  For some discussion on why this hypothesis is necessary, see e.g. p. 7 of 
http://math.uga.edu/~pete/hassebjornv2.pdf
An effective height bound in the function field case is given in Corollaire 2, Section 8 of

Szpiro, L.(F-PARIS6-G)
Discriminant et conducteur des courbes elliptiques. (French) [Discriminant and conductor of elliptic curves]
Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988).
Astérisque No. 183 (1990), 7--18. 

Note that effectivity on the height is much better than effectivity on the number of rational points (Faltings' proof does give the latter).  This is not to be confused with uniform bounds on the number of rational points, for which I believe there are only conditional results known in any case.

REPLY [2 votes]: I don't know enough arithmetic geometry to understand more than, like, half the words that I had to think about to find this reference, but if I'm understanding everything correctly, I think that there's an effective bound on the number of rational points on (nonisotrivial, genus greater than one) varieties over function fields given in this paper (p. 16). There's also a paper of Miyaoka, apparently, but I haven't found an open-access version.<|endoftext|>
TITLE: Homotopies of triangulations
QUESTION [18 upvotes]: I imagine this is pretty much standard, but surely someone here will be able to provide useful references...
Suppose $X$ is a topological space. Let me say that two triangulations $T$ and $T'$ of $X$, are homotopic if there is a triangulation on $X\times [0,1]$ which induces $T$ and $T'$ on $X\times\{0\}$ and on $X\times\{1\}$, respectively.

This rather natural equivalence relation has probably been studied. Does it have a standard name? Are there standard references?
In particular, can this equivalence be redefined by using local moves, ie, as the transitive symmetric closure of a (small) set of local changes in triangulations?

REPLY [24 votes]: The standard name for this type of relation between two structures on $X$ is concordance rather than homotopy.  If two structures on $X$ are isotopic (with the respect to the appropriate homeomorphism group), then they are concordant, but not necessarily vice versa.  In some cases you can also assign a separate meaning to homotopy, but I don't think that it means the same thing as concordance.
There are then two levels to your question.  A triangulation $T$ of $X$ induces a piecewise linear structure $\mathcal{P}$.  You could ask whether the PL structures $\mathcal{P}$ and $\mathcal{P}'$ are isotopic or concordant, without worrying about the original triangulations.  For simplicity suppose that $X$ is a closed manifold.  Then at least in dimension $n \ge 5$, Kirby-Siebenmann theory says that the set of PL structures on $X$ up to isotopy are an affine space (or torsor) of $H^3(X,\mathbb{Z}/2)$.  I think that the concordance answer is the same, because the Kirby-Siebenmann invariant comes from the stable germ-theoretic tangent bundle of $X$.  In other words, two PL structures give you two different sections of a bundle on $X$ whose fiber is $\text{TOP}(n)/\text{PL}(n)$, the group of germs of homeomorphisms of $\mathbb{R}^n$ divided by the subgroup of germs of PL homeomorphisms.  Stabilization in this case means replacing $n$ by $\infty$ by adding extra factors of $\mathbb{R}$.  If $n = 4$, then up to isotopy there are lots of PL structures on many 4-manifolds, as established by gauge theory.  But I think that the concordance answer is once again Kirby-Siebenmann.  (I learned about this stuff in a seminar given by Rob Kirby  — I hope that I remembered it correctly!  You can also try the reference by Kirby and Siebenmann, although it is not very easy to read.)
There is a coarser answer than the one that I just gave.  I tacitly assumed that the triangulations not only give you a PL structure (which always happens), but that they specifically give you a PL manifold structure, with the restriction that the link of every vertex is a PL sphere.  These are called "combinatorial triangulations".  It is a theorem of Edwards and Cannon that $S^5$ and other manifolds also have non-combinatorial triangulations.  If your question is about these, then it is known that they are described by some quotient of Kirby-Siebenmann theory, but it is not known how much you should quotient.  It is possible that every manifold of dimension $n \ge 5$ has a non-combinatorial triangulation, and that PL structures are always concordant in this weaker sense.  It is known that you should quotient more than trivially, that there are manifolds that have a non-combinatorial triangulation but no PL structure.  (I think.)
The other half of the question is to give $X$ a distinguished PL structure $\mathcal{P}$, and to look at triangulations $T$ and $T'$ that are both PL with respect to $\mathcal{P}$.  In this case there are two good sets of moves to convert $T$ to $T'$.  First, you can use stellar moves and their inverses.  A stellar move consists of adding a vertex $v$ to the interior of a simplex $\Delta$ (of some dimension) and supporting structure to turn the star of $\Delta$ into the star of $v$.  The theorem that these moves suffice is called the stellar subdivision theorem.  (The theorem is due to Alexander and Newman and it is explained pretty well in the book by Rourke and Sanderson.)  The other set of moves are specific to manifolds and they are the bistellar moves or Pachner moves.  One definition is that a bistellar move is a stellar move that adds a vertex $v$ plus a different inverse stellar move that removes $v$ (hence the name).  But a clearer definition is that in dimension $n$, a bistellar move replaces $k$ simplices by $n-k+1$ simplices in the minimal way, given by a local $n+1$ concordance that consists of attaching a single $n+1$-dimensional simplex.  The theorem that these moves work is due to Pachner.  Pachner's moves in particular give you a shellable triangulation of $X \times [0,1]$.
Even though the bistellar moves are motivated by concordance and the stellar moves are not, it is not hard to make concordance triangulations for stellar moves as well.<|endoftext|>
TITLE: Is every left fibration of simplicial sets with nonempty fibers a trivial kan fibration?
QUESTION [7 upvotes]: In Lemma 2.1.3.4 of Higher Topos Theory, the statement of the lemma requires that the fibers are not only nonempty but contractible.  However, in the proof, I don't see where contractibility is directly used, only the fact that the fibers are nonempty.  There is one other place where contractibility is mentioned, "Since the boundary of this simplex maps entirely into the contractible kan complex $S_t$, it is possible to extend $f'$ to $X(n+1)$."  However, I don't see how contractibility directly factors in, since that would only attest to the uniqueness of the extension.  The existence of the extension comes from the fact that the inclusion $\partial \Delta^n \times \Delta^1 \subseteq X(n+1)$ is left anodyne and $S_t$ is a nonempty Kan complex and the fact that the map f' factors through the inclusion of $S_t$.  
Please correct me if I'm wrong.  Also, there is a relevant post on meta where I first asked if this question is appropriate, and I was greenlighted by Anton.

REPLY [5 votes]: Whenever I get confused about quasicategories I think back to ordinary categories.  A left fibration between (nerves of) ordinary categories is the same as an opfibration with groupoid fibers, and a trivial fibration between ordinary categories is the same as an equivalence that is surjective on objects.  Clearly an opfibration in groupoids with nonempty fibers need not be an equivalence unless the fibers are contractible.<|endoftext|>
TITLE: is localization of category of categories equivalent to |Cat|
QUESTION [9 upvotes]: It might be a stupid question. 
Suppose There is a category of categories,denoted by CAT,where objects are categories, morpshims are functors between categories
Take multiplicative system S={category equivalences}. Then we take localization at S. Then we get localized category S^(-1)CAT 
Another Category, denoted by |CAT|,where objects are categories, morphism are isomorphism classes of functors between categories. 
I want to prove this two categories are equivalent. I want to prove this problem in two ways:

Naive prove
Because either |CAT| or S^(-1)CAT has same objects as CAT. So one can prove the morphism class of these two category has same equivalent relationship. For the S^(-1)Cat, suppose F and G are two functors in the same equivalent class, we have sF=sG,for s belongs to S. Then I can prove F and G belongs to the same equivalent class in |CAT|. But on the other hand, if F and G are isomorphic functor. I can not prove sF=sG,for some s belongs to S. What I can only prove is sF is isomorphic to sG.
Using adjoint functors
Because I want to prove the projection functor CAT--->|CAT| is a localization functor at S. So I want to construct an adjoint functor |CAT|--->CAT which is fully faithful. But what is this functor. It seems I can not make it well defined.

This is an exercise in Toen's lecture: Lectures on DG-categories
I attach the lecture notes on DG-categories here Toen notes
page 5,exercise 2

REPLY [5 votes]: Bertrand Toën is using a different definition of localization than what you describe in your question. Let $\mathcal{C}$ be a category and let $\Sigma$ be a class of morphisms in $\mathcal{C}$. A category $\mathcal{D}$ with a functor $L\colon \mathcal{C} \rightarrow \mathcal{D}$ is called a localization of $\mathcal{C}$ at $\Sigma$ if $L(f)$ is an isomorphism for every $f \in \Sigma$, and $L$ is universal with this property, that is, whenever $F \colon \mathcal{C} \rightarrow \mathcal{E}$ is a functor such that $F(f)$ is an isomorphism for all $f\in \Sigma$, then there exists a unique functor $G \colon \mathcal{C} \rightarrow \mathcal{E}$ with $F=G \circ L$. If a localization of $\mathcal{C}$ at $\Sigma$ exists, then it is unique up to isomorphism and ususally denoted by $\Sigma^{-1}\mathcal{C}$ or $\mathcal{C}[\Sigma^{-1}]$.
Note that we do not make any assumptions about the class $\Sigma$. As with all universal properties, such an object need not exist in the most general case: usually the resulting category might be "too big", that is, there would be a proper class of morphisms between certain objects in the localization. If the class $\Sigma$ admits a right calculus of fractions in the sense of (e.g.) Borceux, Handbook of Categorical Algebra, Volume I, Chapter 5, then the localization of $\mathcal{C}$ at $\Sigma$ does exist. However, there are lots of examples of categories with classes $\Sigma$ which do have a localization but don't admit a right calculus of fractions. In many of these cases, the functor $L \colon \mathcal{C} \rightarrow \mathcal{D}$ will then not have a right adjoint.
One of the main sources of such categories are Quillen model categories $\mathcal{M}$, with $\Sigma$ the class of weak equivalences. The localization at $\Sigma$ is then called the homotopy category of $\mathcal{M}$. In fact, the notion of Quillen categories was invented mainly to deal with the difficulties of formally inverting a class of arrows in a category. There is a model structure on the category of small categories (sometimes called the folk model structure) whose weak equivalences are the equivalences of categories, so the desired localization does exist. Moreover, there is a notion of left and right homotopy in a model category. For nice objects (objects which are both fibrant and cofibrant), these notions agree and define an equivalence relation on maps. In the model category of categories, all objects are fibrant and cofibrant. Two functors are homotopic if and only if they are naturally isomorphic.
The usual construction of the homotopy category of a model category (see e.g. Mark Hovey's book "Model Categories") shows that the localization of $\mathbf{Cat}$ at the equivalences is given by the category whose objects are again categories and whose morphisms are homotopy classes of morphisms. Thus we do indeed find that the localization of $\mathbf{Cat}$ at the class of equivalences is the category $\vert \mathbf{Cat} \vert$ of categories and isomorphism classes of functors, as Toën claimed.
Charles Rezk has a nice writeup of the folk model structure here.<|endoftext|>
TITLE: What do you call this ring?
QUESTION [10 upvotes]: I want a ring $R$ of "numbers" such that:
For any sequence of congruences $x\equiv a_1 \pmod{n_1}, x\equiv a_2 \pmod{n_2},\dots$ with $a_i\in \mathbb{Z}$ and $n_i\in \mathbb{N}$ such than any finite set of these congruences has a solution $x\in\mathbb{Z}$, there is a $r\in R$ such that $r\equiv a_1 \pmod{n_1}, r\equiv a_2 \pmod{n_2},\dots$ 
and
For any $r\in R$ and $n\in\mathbb{N}$ there is a $a, 0\leq a< n $ such that $r\equiv a \pmod{n}$.
I think that $R$ has to be the product set of the p-adic integers over all primes p, but what do you call this ring?
(Perhaps there should be a "terminology" tag? Edit: It already exists but it is called "names")

REPLY [3 votes]: Another (fancy) name is "the free profinite group of rank 1"<|endoftext|>
TITLE: A local-to-global principle for being a rational surface
QUESTION [14 upvotes]: Let $k$ be a number field and $F$ a $1$-variable function field over $k$ (a finitely generated extension of $k$, of transcendence degree $1$, in which $k$ is algebraically closed).  If $F$ becomes the rational function field over every completion $k_v$ of $k$, then $F$ is the rational function field over $k$.  This is a restatement of the local-to-global principle for the existence of rational points on $k$-conics.
Now suppose that $F$ is a $2$-variable function field over $k$, and that $F$ becomes the rational functional field over every completion $k_v$ of $k$.  Does it follow that $F$ is the rational function field over $k$ ?  (In other words, if a $k$-surface is birational of ${\mathbb P}_2$ over every $k_v$, is it $k$-birational to ${\mathbb P}_2$ ?)
I don't think the answer is known, but perhaps someone can put together what is known about the group of $\bar k$-automorphisms of  the rational function field ${\bar k}(x,y)$ (the Cremona group) to decide one way or the other.

REPLY [5 votes]: It seems to me that there are irrational surfaces over $\mathbb Q$ that are $\mathbb Q_v$-rational for all $v$. (I couldn't find them in the literature, but didn't look very hard. Almost certainly they are to be found there, in papers by either Iskovskikh or Colliot-Thelene.) 
Take the affine surface $S$ given by $y^2+byz+cz^2=f(x)$, where $f$ is an irreducible cubic and $b^2-4c$ equals the discriminant $D(f)$ of $f$, up to a square in $\mathbb Q^*$, and $D(f)$ is not a square. According to Beauville-Colliot--Thelene--Sansuc--Swinnerton-Dyer $S$ is not $\mathbb Q$-rational, but is stably rational. (Irrationality is Iskovskikh I think, in fact.) Via projection to the $x$-line a projective model $V$ of $S$ is a conic bundle over $\mathbb P^1$ with $4$ singular fibers (one is at infinity). There is an embedding of $V$ into a weighted projective space $\mathbb P(2,2,1,1)$; the defining equation is $Y^2+bYZ+cZ^2=F(X,T)T$, where $F$ is the homogeneous version of $f$. By construction the Galois action on the $8$ lines that comprise the singular fibers is via the symmetric group $S_3$: the two lines in the fiber at infinity are conjugated, and the other six are permuted transitively. 
Claim: Assume that $D(f)$ is square-free and prime to $6$. Then $S$ is $\mathbb Q_v$-rational for all $v$. 
Proof: Suppose that the decomposition group $G_v$ at $v$ is cyclic. Whatever its order ($1,2$ or $3$) there are at least $2$ disjoint lines among the $8$ that are $G_v$-conjugate, so they can be blown down to give a conic bundle over $\mathbb P^1$ with at most $2$ singular fibres and a $\mathbb Q_v$-point; it is well known that such a surface is $\mathbb Q_v$-rational.
Now suppose that $G_v= S_3$. Then $v$ is non-archimedean and $V$ has bad reduction there. In fact, exactly two of the singular fibers are equal modulo $v$; it follows that $G_v=S_3$ is impossible, and we are done.
E.g., $f=x^3+x+1$, of discriminant $-31$, $c=8$, $b=1$.
(This doesn't use stable rationality, but rather the fact that these surfaces, although irrational, are very close to being rational, in the sense that the action of $Gal_{\mathbb Q}$ on the lines is as small as possible subject to the surface being irrational, and the action of the decomposition groups is even smaller.)<|endoftext|>
TITLE: Everywhere locally isomorphic abelian varieties
QUESTION [19 upvotes]: Is there a standard example of two abelian varieties $A$, $B$ over some number field $k$ which are $k_v$-isomorphic for every place $v$ of $k$ but not $k$-isomorphic ?

REPLY [26 votes]: (If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.)
The smallest examples are in dimension $2$.  Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X_0(11)$.  We will construct two twists of $E^2$ that are isomorphic over $\mathbf{Q}_p$ for all $p \le \infty$ but not isomorphic over $\mathbf{Q}$.
Let $K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$.  Let $G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$.  Let $\alpha \colon G \to \operatorname{GL}_2(\mathbf{Z}) = \operatorname{Aut}(E^2)$ be a homomorphism sending the two generators to the reflections in the coordinate axes of $\mathbf{Z}^2$, and let $A$ be the $K/\mathbf{Q}$-twist of $E^2$ given by $\alpha$.  Define $\beta$ and $B$ similarly, but with the lines $y=x$ and $y=-x$ in place of the coordinate axes.  The representations $\alpha$ and $\beta$ of $G$ on $\mathbf{Z}^2$ are not conjugate: only the former is such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$.  Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$.
On the other hand, every decomposition group $D_p$ in $G$ is smaller than $G$ since $-1$ is a square in $\mathbf{Q}_{17}$ and $17$ is a square in $\mathbf{Q}_2$.  Also, the restrictions of $\alpha$ and $\beta$ to any proper subgroup of $G$ are conjugate: any single line spanned by a primitive vector in $\mathbf{Z}^2$ can be mapped to any other by an element of $\operatorname{GL}_2(\mathbf{Z})$.  Thus $A$ and $B$ become isomorphic after base extension to $\mathbf{Q}_p$ for any $p \le \infty$. $\square$
Remark: The abelian surfaces $A$ and $B$ constructed above are isogenous even over $\mathbf{Q}$, because the $\mathbf{Z}^2$ with one Galois action can be embedded into the $\mathbf{Z}^2$ with the other Galois action: rotate $45^\circ$ and dilate.
Remark: The nonexistence of examples in dimension $1$ follows from these two well-known facts:
1) Twists of an elliptic curve over a field $k$ of characteristic $0$ 
are classified by
$H^1(k,\mu_n)=k^\times/k^{\times n}$ where $n$ is 2, 4, or 6.  
2) If $n<8$, the map $k^\times/k^{\times n} \to \prod_v k_v^\times/k_v^{\times n}$ is injective.  
[Edit: This answer was edited to simplify the construction and to add those remarks at the end.]<|endoftext|>
TITLE: Which is the correct version of a quantum group at a root of unity?
QUESTION [29 upvotes]: By this I mean the specialisation of the quantum group Uq(g) with q a root of unity, and the 'correct' meaning of 'correct' (enclosed in quotations since there isn't necessarily a correct answer) is likely to mean that its category of representations is the 'correct' one.
My suspicion is that I want to take an integral form of Uq(g) defined over ℤ[q±1/2] and base change to the appropriate ring of integers in a cyclotomic field. Having heard of the 'small quantum group' and Lusztigs algebra U dot (notation in his quantum groups book), I suspect the existence of multiple approaches, which diverge at least when an integral form is desired, and hence turn to mathoverflow for clarification and enlightenment.
Afficionados of this type of question can consider it as a continuation in a sequence initiated by this question on the universal enveloping algebra in positive characteristic.

REPLY [57 votes]: There are (at least) five interesting versions of the quantum group at a root of unity.
The Kac-De Concini form:
This is what you get if you just take the obvious integral form and specialize q to a root of unity (you may want to clear the denominators first, but that only affects a few small roots of unity).  This is best thought of as a quantized version of jets of functions on the Poisson dual group.  It's most important characteristic is that it has a very large central Hopf subalgebre (generated by the lth powers of the standard generators).  In particular, its representation theory is sits over Spec of the large center, which is necessarily a group and turns out to be the Poisson dual group.  It also has a small quotient Hopf algebra when you kill the large center.
The main sources for the structure of the finite dimensional representations are papers by subsets of Kac-DeConcini-Procesi (the structure of the representations depends on the symplectic leaf in G*, in particular there are "generic" ones coming from the big cell) as well as some more recent work by Kremnitzer (proving some stronger results about the dimensions of the non-generic representations) and by DeConcini-Procesi-Reshetikhin-Rosso (giving the tensor product rules for generic reps).  The main application that I know of this integral form is to invariants of knots together with a hyperbolic structure on the compliment and to invariants of hyperbolic 3-manifolds due to Kashaev, Baseilhac-Bennedetti, and Kashaev-Reshetikhin.  The hope is that these invariants will shed some light on the volume conjecture.
The Lusztig form:
Here you start with the integral form that has divided powers.  Structurally this has a small subalgebra generated by the usual generators (E_i, F_i, K_i)  since E^l = 0.  The quotient by this subalgebra gives the usual universal enveloping algebra via something called the quantum Frobenius map.  The main representation that people look at are the "tilting modules."  Tilting modules have a technical description, but the important point is that the indecomposable tilting modules are exactly the summands of the tensor products of the fundamental representations.  Indecomposable tilting modules are indexed by weights in the Weyl chamber.  The "linkage principle" tells you that inside the decomposition series of a given indecomposable tilting module you only need to look at the Weyl modules with highest weights given by smaller elements in a certain affine Weyl group orbit.
It is the Lusztig integral form (not specialized) that is important for categorification.  The Lusztig form at a root of unity is important for relationships between quantum groups and representations of algebraic groups and for relationships to affine lie algebras.  The main sources are Lusztig and HH Andersen (and his colaborators).  I'm also fond of a paper of Sawin's that does a very nice job cleaning up the literature.
The Lusztig integral form is also the natural one from a quantum topology point of view.  For example, if you start with the Temperley-Lieb algebra (or equivalently, tangles modulo the Kauffman bracket relations) and specialize q to a root of unity what you end up with is the planar algebra for the tilting modules for the Lusztig form at that root of unity.
The small quantum group:
This is a finite dimensional Hopf algebra, it appears as a quotient of the K-DC form (quotienting by the large central subalgebra) and as a subalgebra of the Lusztig form (generated by the standard generators).  I gather that the representation theory is not very well understood.  But there has been some work recently by Roman Bezrukavnikov and others.  I also wrote a blog post on what the representation theory looks like here for one of the smallest examples.
The semisimplified category:
Unlike the other examples, this is not the category of representations of a Hopf algebra!  (Although like all fusion categories it is the category of representations of a weak Hopf algebra.)  You start with either the category of tilting modules for the Lusztig form or the category of finite dimensional representations of the small quantum group and then you "semisimplify" by killing all "negligible morphisms."  A morphism is negligible if it gives you 0 no matter how you "close it off." Alternately the negligible morphisms are the kernel of a certain inner product on the Hom spaces.  The resulting category is semisimple, its representation theory is a "truncated" version of the usual representation theory.  In particular the only surviving representations are those in the "Weyl alcove" which is like the Weyl chamber except its been cut off by a line perpendicular to l times a certain fundamental weight (see Sawin's paper for the correct line which depends subtly on the kind of root of unity).
This example is the main source of modular categories and of interesting fusion categories.  Its main application is the 3-manifold invariants of Reshetikhin-Turaev (where this quotient first appears, I think) and Turaev-Viro.  For those invariants its very important that your braided tensor category only have finitely many different simple objects.
The half-divided powers integral form:
This appears in the work of Habiro on universal versions of the Reshetikhin-Turaev invariants and on integrality results concerning these invariants.  This integral form looks like the Lusztig form on the upper Borel and like the K-DC form on the lower Borel.  The key advantage is that in the construction of the R-matrix via the Drinfeld double you should be looking at something like U_q(B+) \otimes U_q(B+)* and it turns out that the dual of the Borel without divided powers is the Borel with divided powers and vice-versa.  There's been very little work done on this case beyond the work of Habiro.<|endoftext|>
TITLE: Characterizing faces of 3-dimensional polyhedra. (Related to Victor Eberhard's Theorem [1890]:)
QUESTION [11 upvotes]: Originally asked by Ali Dino Jumani;  EXTENSIVELY EDITED by David Speyer. The previous version was a bit confused, but Steven Sivek and Graham, in the comments, figured out what was going on.

G. C. Shephard, in his paper "Twenty Problems on Convex Polyhedra: Part I", associates to a three dimensional polyhedron the sequence $(p_3, p_4, p_5, p_6, \ldots)$, with $p_k$ being the number of facets that are $k$-gons. He poses the problem of characterizing all sequences of integers which arise in this way.
Are there any developments and references on this problem?
the original question as appeared in Shephard's paper "Given any finite sequence (f3, f4, ...., fm) of non-negative integers, find a necessary and sufficient condition for it to be assocoated with some convex polyhedron". Following Grunbaum's Convex Polytope 2e notation, section 13.3 under Eberhard's Therorem heading, f3 is triangle renamed as p3 and f4 as p4 is square and fm as pk is n-gon. A convex polyhedron containing these faces satisfies the Eberhard's criterion.
Revisions and corrections are highly appreciated, Thanks.

REPLY [20 votes]: Eberhard Theorem
Consider a simple 3-polytope P, (so every vertex has 3 neighbors). If $p_k$ is the number of faces of P which are k-gonal, Euler's theorem implies that
$$(*) ~~\sum_{k \ge 3} (6-k)p_k=12.$$
Note that 6-gonal faces do not contribute to the LHS.
One way to think about it is that polygonal faces with 7 and more sides contribute "negative curvature", small faces  namely triangles quadrangles and pentagons, contribute positive "curvature" and hexagons are "flat".
Eberhard's theorem asserts that if you have a sequence of numbers $p_k, k \ne 6$ such that $\sum_{k \ge 3} (6-k)p_k=12$ then you can find a simple 3-polytope with $p_k$ k-gonal faces. (But you have no control on $p_6$).
Extensions of Eberhard theorem
There are various results extending Eberhard's theorem in various directions. Chapter 13 in Grunbaum's book "Convex Polytopes" and especially the supplemantary material at the end of the chapter in the new 2nd edition is a good source. Another general source is my chapter from the "Handbook of Discrete and Computational geometry" on garphs and skeleta of polytops.
A relatively recent paper on the subject is by Stanislav Jendrol "On the face vectors of trivalent convex polyhedra". Another paper by Jendrol which deals with general 3-polytopes from the same year is "On face vectors and vertex vectors of convex polyhedra" Discrete Math 118 (1993)119-144. There are analogs of Eberhard theorem for 4-regular planar graphs, for toroidal graphs and in other directions.
A far as I know, there is no good answer known for the question posed by Shephard of characterizing all sequences $(p_3,p_4,\dots)$, and no such characterization is known even for the simple case.
High dimensions
In higher dimensions and even in four dimensions there are various different ways to extend these problems, these problems become very difficult and very little is known.
2-dimensional faces
You can ask again about the numbers of k-gonal 2-dimensional faces. While the formula above implies that in dimension 3 and more $p_3+p_4+p_5>0$ it is known that in dimensions 5 and more $p_3+p_4>0$.
Types of facets
Perhaps an even more natural extension is to consider the type of facets a given d-simensional polytope have. You can ask for a simple 4 polytope (a 4-polytope whose graph is 4-regular) what are the number of facets $p_Q$ isomorphic to a given 3-polytope Q. This gives you a vector indexed by combinatorial types of simple 3-polytopes, but I am not aware of any Eberhard type theorem and I do not know even which 3-polytopes should be considered as the analogs of the hexagons in the above formula. Dually stated and extented to triangulations of 3-spheres the question is to associate to a triangulated 3-simensional sphere (or just simplicial 3-polytope) the list of links of vertices (with multiplicities) it has. A related MO question is this one.
Type of facets according to their number of their facets
Rather than classifying the facets according to their full combinatorial type we can classify them according to their own number of facets. Here, for dimension greater than 4 there is no analog for (*). It is possible that under wide circumstences the numbers of facets with k facets can be prescribed, but I am not aware of results in this direction.
Type of facets according to their f-vectors
Precscribing the entire vector of face numbers of the individual facets, may well be the most interesting extension from 3 to higher dimensions.
There are some reasons to jump from dimension 3 directly to dimension 5. The nature of the problem is different in even and odd dimensions. Let me try to elaborate on that.
Suppose you have a simple 5-polytope P and denote by $p_{a,b}$ the number of facets F so that $f_3(F)=a$ and $f_2(F)=b$. (For a polytope or some other cell complex X, $f_i(X)$ denotes the number of $i$-dimensional faces of $X$.) Dually, we can consider a triangulation $K$ of the 4-simensional sphere $S^4$ and let $p_{a,b}$ be the number of vertices whose link has $a$ vertices and $b$ edges.
Now the Dehn-Sommervill relations imply that
$$(**) \sum p_{a,b} (b-6a+30) = 60 .$$
So now we can consider 4 dimensional facets as "small" if b-6a+30 is negative, as "large" if $b-6a+30$ is positive, and as neutral if $b=6a-30$. An analog of Eberhard theorem will say that you can prescribe the types of small and large facets and realize the polytope (or the triangulated sphere) by allowing to add many "neutral" facets.
Facets and non facets
A related question that was studied in several papers is: for which d-polytopes P there is a (d+1) polytopes Q such that all faces of Q are isomorphic to P. Such polytopes P are called "facets". An intriguing open problem is if the icosahedron is a facet.
Cubical complexes
Simple polytopes are dual to simplicial polytopes, and there is some special interest in these problems for cubical complexes (or their duals). Indeed there is an analog of Eberhard's theorem for 3-polytopes with all vertex degrees 4. There should be an analog for (**) and this may be related to interesting questions about curvature of cubical complexes.
Stacked polytopes
A stack polytope is a polytope obtained by gluing simplices along facets. (They are related to Appolonian circle packing.) Stacked polytopes are simplicial. I do not know if there is an analog of Eberhard's  theorem when we consider only simple 3-polytope which are dual to stacked 3-polytopes. This is also interesting for other dimensions. All facets of stacked 5 polytopes are stacked and their number of edges is four times the number of vertices minus 10. Therefore, for a dual-to-stacked 5 polytope formula (**) reduces to $$\sum p_{a,b}(10-a)=30.$$ (Here, $b=4a-10$ so we can omit the subscript $b$.) So in this case, the distinction is between facets with $a<10$, those with $a>10$ and those with $a=10$. Again we can look for some analogs for Eberhard's theorem, which in this case, might be easier.<|endoftext|>
TITLE: A parametrization of Heronian triangles
QUESTION [8 upvotes]: Let $a,b,c$ be integers which are the sides of a triangle with integral area, a so called Heronian triangle. This website attributes to Gauss the result that there must then exist integers $m,n,p,q$ such that
$a = mn(p^2+q^2)$
$b = (mp)^2+(nq)^2$
$c = (m+n)(mp^2-nq^2)$
(where I left out a $4pq$ factor designed to make the radius of the circumscribed circle integral as well). It's not hard to see that the triangle defined by these formulas is indeed Heronian, however I could neither prove nor find a reference for the fact that this parametrization is exhaustive. 
Can someone do one of these two things?
Thanks!
(Note: I'm communicating this question on behalf of my dad, who is really the person who looked into that but is not easily capable of asking it himself over here. I may be slow to respond on his behalf if questions come up).

REPLY [5 votes]: Let your triangle $\triangle{ABC}$ have side lengths $a,b,c \in \mathbb{Q}$ and rational area. Assume WLOG that $c$ is the longest side and drop the altitude from $C$ with length $h\in Q$. The triangle is divided into two right triangles one with hypotenuse $a$ and legs $d,h$, and one with hypotenuse $b$ and legs $e,h$. We have $d+e=c\in \mathbb{Q}$. Also notice that $$d-e=\frac{d^2-e^2}{d+e}=\frac{a^2-b^2}{c}\in \mathbb{Q}$$ so we conclude that $d,e$ are rational. From the pythagorean triples we have relations $$a-d=r,\quad a+d=\frac{h^2}{r},\quad b-e=s,\quad b+e=\frac{h^2}{s}$$ and therefore $$a=\frac{1}{2r}(h^2+r^2),\quad b=\frac{1}{2s}(h^2+s^2),\quad c=\frac{r+s}{2rs}(rs-h^2)$$
This is exactly your parametrization up to scaling.<|endoftext|>
TITLE: substitute for Serre's twisting when the "twisting" is outer
QUESTION [8 upvotes]: Does anyone know if there is something that can be said (ideally under at most very mild hypotheses) in group cohomology (let's even restrict to degree 1) that is similar to Serre's twisting, but in the case where the thing you want to twist by is not inner? I'm not expecting the statement to hold word-for-word, just that there is some general and reasonably sharp relationship between the original cohomology set and the twisted one. Thanks!
Edit in response to Chris Schommer-Pries: 
A complete discussion of Serre's twisting can be found in section 5.3 of his book "Galois Cohomology". It's both a "process" and a proposition I guess. It goes as follows. Start with a group $G$ and a (not necessarily abelian) group $A$ on which $G$ acts, and form the pointed set $H^1(G,A)$. Now pick a $1$-cocycle $c \in H^1(G,A)$. Let $A_c$ denote the same abstract group $A$, but now it has the "$c$-twisted" action defined by $g*a = c(g) \cdot (g \cdot a) \cdot c(g)^{-1} $ (the 1st and 3rd $\cdot$ are operations in $A$ and the 2nd $\cdot$ is the original action of $G$). That's the "process". The proposition (cf. Prop 35 bis in the book) is that $d \mapsto d \cdot c $ (product of functions) induces a bijection of pointed sets $ H^1(G,A_c) \cong H^1(G,A)$ (note that the first set has a different basepoint than the second). It is frequently used to show that some "monomorphism of pointed sets" is actually injective as a function.
Edit in response to Mariano Suárez-Alvarez:
By "inner" above I am referring to the fact that, in Serre twisting, the action of $G$ on $A$ was modified by conjugating by an element of $A$. By "outer", I mean the case where you instead replaced the action by $g*a = f(g)( g \cdot a ) $ where $f : G \rightarrow Aut( A )$ is a prescribed cocycle ($Aut( A )$ has the conjgation action using the original action of $G$ on $A$).

REPLY [6 votes]: The short answer is that there is not much to say about the relationship between $H^1(G, B)$ and a twist $H^1(G, B_c)$ where $c$ is a cocycle taking values in $Aut(B)$. (I am going to write $B_c$ for the twist instead of Serre's notation $_cB$ for the sake of easy typesetting.)  You can get a good feel for what is possible by soaking in sections I.5.7 and III.1.4 of Serre's Galois Cohomology.
Section I.5.7
One thing you can do -- as exhibited in I.5.7 -- is twist all three terms in a short exact sequence of $G$-modules and get a new short exact sequence, assuming the obvious compatibility conditions hold.  Serre starts with an exact sequence

$1 \to A \to B \to C \to 1$

where $A$ is assumed central in $B$.  Then he fixes a 1-cocycle $c$ with values in $C$ and twists to get an exact sequence

$1 \to A \to B_c \to C_c \to 1$.

Note that this twist $B_c$ is not an inner twist of $B$, because $c$ need not be in the image of $H^1(G, B) \to H^1(G, C)$.  
This may look like a lame example, in that the twist of $B$ is "pretty close" to being inner.  But already here you don't have any results regarding a connection between $H^1(G, B)$ and $H^1(G, B_c)$.  That's a pretty fuzzy statement; Serre says as much as you can say with precision in Remark 1: "it is, in general, false that $H^1(G, B_c)$ is in bijective correspondence with $H^1(G, B)$."
Section III.1.4
This section discusses your question for the specific case where $G$ is the absolute Galois group of a field $k$ and $B$ is the group of $n$-by-$n$ matrices of determinant 1 with entries in a separable closure of $k$.  Serre explains what you get as $B_c$ when you twist $B$ by a cocycle with values in $Aut(B)$. You can get, for example, a special unitary group.  
You can find explicit descriptions of $H^1(G, B_c)$ for some $B_c$'s in The Book of Involutions, pages 393 (Cor. 29.4) and 404 (box in middle of page).  Note that for $B_c$ as in Section I.5.7, $H^1(G, B_c)$ is a group (a nice coincidence) but in the case where you get a true special unitary group, $H^1(G, B_c)$ does not have a reasonable group structure--it is just a pointed set like you expect.<|endoftext|>
TITLE: Definition of Category of Locales
QUESTION [5 upvotes]: In the wikipedia entry for 'frames and locales', pains are taken to distinguish between the category of locales - defined to be the opposite of the category of frames - and the category whose objects are the complete Heyting algebras but whose arrows are the adjoints of the frame arrows.  The two are clearly isomorphic as categories.  How far are they from being identical, though?  I became acquainted with locales through Borceux's excellent handbook and he defines arrow in locales to be the adjoints. So this is a little worrisome.  Ok, I know this isn't precise...  Let me put it this way:  What are concrete examples of how these two differently defined categories actually differ?  
Thank you in advance

REPLY [6 votes]: The article on Heyting algebras and frames is one of many that are truly awful in Wikipedia.
Frames and complete Heyting algebras are completely different things.  They are algebras for different theories and (so) their homomophisms are different.
Johnstone's convention, which Tom has described and to which there are now few dissenters, allows one to use the algebraic machinery to speak in topological language, but without mentioning points.  For example, in Johnstone's book you will find definitions of locally compact locales and of open (continous) maps.<|endoftext|>
TITLE: Equivalence of ordered and unordered cech cohomology. 
QUESTION [20 upvotes]: Given a topological space X and a finite cover X = $\cup X_i$, one can define Cech cohomology of a sheaf of abelian groups F with respect to the cover $\{X_i\}$  in two different ways: 

(Ordered): The kth term of the Cech
    complex is $\bigoplus_{i_1 < \ldots
        < i_k} \Gamma(X_{i_1} \cap \ldots
        \cap X_{i_k}, F)$.
(Unordered): The kth term of the Cech
    complex is $\bigoplus_{i_1, \ldots
        , i_k} \Gamma(X_{i_1} \cap \ldots
        \cap X_{i_k}, F)$.

In particular, the second description involves repetition and is non-zero in every degree. These two descriptions give isomorphic cohomology (the first maps you try to write down will likely be homotopy equivalences). 

Question: Is there a canonical reference for this fact?

REPLY [5 votes]: I'll make use the category $\mathcal{P}=\mathrm{Fun}(\mathrm{Open}_X^{\mathrm{op}}, \mathrm{Ab})$ of presheaves of abelian groups on $X$.  This is an abelian category with enough projectives: for any open set $U$, the free presheaf $\mathbb{Z}[U]$ defined by $\mathbb{Z}[U](V)=\mathbb{Z}$ if $V\subseteq U$ and $=0$ if not, is a projective object in presheaves.
We can define two chain complexes in $\mathcal{P}$:
$$
A_k = \bigoplus_{i_0<\cdots<i_k} \mathbb{Z}[X_{i_0}\cap \cdots \cap X_{i_k}],
\qquad 
B_k = \bigoplus_{i_0,\dots,i_k} \mathbb{Z}[X_{i_0}\cap \cdots \cap X_{i_k}], 
$$
with an evident chain map $\gamma\colon A_\bullet\to B_\bullet$.  It is clear that for a sheaf $F$, the complexes $\mathrm{Hom}_{\mathcal{P}}(A_\bullet, F)$ and $\mathrm{Hom}_{\mathcal{P}}(B_\bullet, F)$ are respectively the ordered and unordered Cech complexes of $F$.  So it suffices to show that $\gamma$ is a chain homotopy equivalence in $\mathrm{Ch}(\mathcal{P})$, as such data will thereby induce a chain homotopy equivalence between the two versions of the Cech complex.
Both $A_\bullet$ and $B_\bullet$ are bounded-below complexes of projectives in $\mathcal{P}$, so it suffices to show that $\gamma$ induces an isomorphism on homology.  That is, we need to show that $H_k A_\bullet(U)\to H_k B_\bullet(U)$ is an isomorphism for all open sets $U$.  
If $U$ is not contained in any $X_i$, then it is immediate that $A_\bullet(U)\equiv 0\equiv B_\bullet(U)$.  If $U$ is contained in some $X_i$, then you can show by explicit calculation (e.g., an explicit chain homotopy, see below) that
$$
H_0A_\bullet(U)\approx \mathbb{Z}\approx H_0B_\bullet(U),\qquad
H_kA_\bullet(U)\approx 0 \approx H_kB_\bullet(U),\quad k>0.
$$
To see how this works, write $I_U=\{i\in I\;\mid\; U\subseteq X_i\}$ where $I$ is the (well-ordered) indexing set of the cover.  We see that each degree of $A_\bullet(U)$ and $B_\bullet(U)$ are free abelian groups:
$$
A_k(U)=\mathbb{Z}\bigl\{ (i_0<\cdots<i_k),\; i_j\in I_U\bigr\},\qquad
B_k(U)=\mathbb{Z}\bigl\{ (i_0,\dots,i_k),\; i_j\in I_U\bigr\}.
$$
In fact, $A_\bullet(U)$ is the complex of normalized chains on the nerve of the poset $(I_U, \leq)$ with ordering inherited from $I$, which can be contracted to a point corresponding to the minimal element of $I_U$.  Likewise, $B_\bullet(U)$ is the complex of unnormalized chains on the nerve of the category $(I_U, I_U\times I_U)$ with objects $I_U$ and a unique morphism between any pair of objects, and this nerve can also be contracted to a point.
(Explicit contractions for the augmented chain complexes $A_\bullet(U)\to \mathbb{Z}$ and $B_\bullet(U)\to \mathbb{Z}$ are given by 
$$
h_A(i_0<\cdots<i_k)=\text{$(m<i_0<\cdots<i_k)$ if $m\neq i_0$, or $0$ if $m=i_0$},
$$
and
$$
h_B(i_0,\dots,i_k)=(m,i_0,\dots,i_k),
$$
where $m\in I_U$ is the minimal element.)<|endoftext|>
TITLE: Conformal maps in higher dimensions
QUESTION [19 upvotes]: In dimension 2 we know by the Riemann mapping theorem that any simply connected domain ( $\neq \mathbb{R}^{2}$) can be mapped bijectively to the unit disk with a function that preserves angles between curves, ie is conformal. 
I have read the claim that conformal maps in higher dimensions are pretty boring but does anyone know a proof or even a intuitive argument that conformal maps in higher dimensions are trivial?

REPLY [6 votes]: I think this is a good reference for it.
Iwaniec, Tadeusz; Martin, Gaven Geometric function theory and non-linear analysis. Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2001. xvi+552.<|endoftext|>
TITLE: The Category of Representations of a Group
QUESTION [7 upvotes]: Do people study the category of representations of a compact finite group (not just irreducible ones)?  I'm more interested in small cases like S_3 and SU(2) but I'd be curious about general cases like $S_n, SU(n)$.  These must be tensor categories since - well... they admit tensor products and direct sums.  Can these representations be considered a ring?
** What does this category look like in the case of S_3?

REPLY [11 votes]: The category Rep(G) is a symmetric tensor category, and it is a theorem that this structure determines G (Tannaka-Krein duality, but I'm not familiar with it).  Each object is dualizable because there is a dual representation, from which appropriate evaluation and coevaluation maps can be constructed.  The unital object is the trivial representation.
(Incidentally, a symmetric tensor category is a kind of categorification of a commutative ring.)
In the finite case, It is a fusion category because there are finitely many simple objects, there is rigidity as stated above, and because $\mathbb{C}[G]$ (or more generally $k[G]$ for $k$ a field of characteristic prime to the order of $G$) is a semisimple algebra (Maschke's theorem).  Semisimplicity is also true for continuous finite-dimensional representations of compact groups by the same "averaging" argument used in Maschke's theorem, though the group algebra is not necessarily semisimple.    In the finite group case, the number of simple objects is equal to the number of conjugacy classes in G.  In the infinite group case, for instance, the rotation group $SO(n)$ has infinitely many irreducible finite-dimensional representations obtained by the action on the spherical harmonics of various degrees (i.e. harmonic polynomials restricted to the sphere).  
In the case of S_n, generators of this ring can be indexed by the Young diagrams of size n.  The relations are given by the tensor product rules, and while the Pieri rule gives a special case of this, as far as I know, there is no a simple general way to express the tensor product of two representations associated to Young diagrams as a sum of Young diagrams.
However, there are apparently algorithms to do this.<|endoftext|>
TITLE: Definition of homotopy limits
QUESTION [7 upvotes]: Here's a definition for homotopy limits that isn't quite right, but seems salvageable. Does anyone know how to fix it?
Suppose the category $C$ is some reasonable setting for homotopy theory, say it's enriched over some kind of category of spaces (e.g. chain complexes, simplicial sets, ...).
Def: Let F: Dop $\to$ C be a diagram (functor). An object X together with a map η from X to the diagram is a LIMIT for the diagram iff the induced natural transformation of functors HomC(-, X) $\to \lim$ HomC(-,F) is an isomorphism. A pair (X,η) is a homotopy limit for the diagram F iff the induced transformation of functors HomC(-,X) $\to \lim$ HomC(-, F) is a weak equivalence.
This definition doesn't quite cut it since, in most of the motivating examples I know, though the homotopy limit object X does come equipped with a morphism to each object in the diagram, these do not commute with the morphisms in the diagram---they only commute up to homotopy. So a homotopy limit won't even come with a map to the diagram, so it doesn't come with an induced natural transformation. How then can I characterize the object X by a similar universal property as the (strict) limit?

REPLY [15 votes]: Reid's answer is quite right, but long before "quasicategories" became fashionable, algebraic topologists were doing exactly the same thing using the "simplicial bar construction" and plain old topological or simplicial enriched categories.
For a fixed x, the limit $\lim \hom_C(x,F)$ is equivalent to the set of natural transformations from the constant functor $\Delta_1\colon D\to Set$ at $1$ to the functor $\hom_C(x,F(-))$.  So to replace it by something "coherent" we need a notion of "homotopy coherent transformation."  Now the set of natural transformations from a functor $G\colon X\to Top$ to a functor $H\colon Y\to Top$ can be defined as an "end," and computed as an equalizer of the two maps $\prod_x \hom(G x, H x) \rightrightarrows \prod_{x \to y} \hom(G x, H y)$.  But these two maps are the first two coface maps of a cosimplicial object that continues with $\prod_{x\to y\to z} \hom(G x, H z)$ and so on, so we can define the space of "homotopy coherent transformations" to be its totalization (the dual of geometric realization).
Now a homotopy limit can be defined as a representing object for the space of homotopy coherent transformations from $\Delta_1$ to $\hom_C(x,F(-))$ (where now $C$ is topologically enriched, so that these functors take values in spaces).  Moreover, if $C$ admits "totalizations" as a topologically enriched category (a sort of "weighted limit"---it suffices to have ordinary limits and "cotensors"), then the homotopy limit can be constructed by "internalizing" the above construction.<|endoftext|>
TITLE: A GAGA question
QUESTION [11 upvotes]: A GAGA question.
Say I have a ``quasi-projective''  (*) subvariety X over the complex
numbers within a smooth complex  algebraic variety Z.   
True or False:   The analytic and   algebraic closure  of X (within Z) coincide.  
I guess the answer must be  `True' and is contained somewhere within Serre's GAGA,
or some elucidation thereof.  If I'm right
could someone point me to a   precise reference, either within GAGA or
elsewhere? If I'm wrong I'd love to hear about it. 
Elucidation:   
(*) By `quasi-projective'' I mean  X is  defined by 
a finite number ofalgebraic equations'  $f_i = 0$ 
and inequalities $g_a \ne 0$, As is the case when Z is projective
 space,   the   $f_i$ may  not be globally defined; 
 same for the $g_a$.   In  my situation,
the Zariski open set defined by    intersecting the sets   $g_a \ne 0$
is an  affine set  (in the usual schemy sense) and the $f_i$ are polynomials
on this affine set.
My Z is probably projective -- I'm not positive here, just pretty sure.
(My Z is obtained by  iterating the construction of taking the
bundle over a smooth projective variety whose
fiber is the  Grassmannian  of d-planes within  said variety's tangent space.  )

REPLY [14 votes]: Yes, it is true: the analytic and algebraic closures of $X$ in $Z$ coincide (and you don't need at all to assume that $Z$ is smooth).
You may suppose that X is open in $Z$: if it isn't, just replace $Z$ by the Zariski (=algebraic) closure of $X$ in $Z$.  Then Serre's Proposition 5, page 11 of his famous GAGA article "Géométrie Algébrique et Géométrie Analytique" says exactly that the analytic closure of $X$
in $Z$ is $Z$, i.e. coincides with its Zariski closure. [ Caution:  his "X" takes the place of your "Z" , the big ambient variety]
You can also find the result (in English !) in the fine book by Joseph L. Taylor  "Several Complex Variables with Connections to Algebraic Geometry and Lie Groups", published by the AMS in its series Graduate Studies in Mathematics ( Volume 46).There the result you need is Proposition 13.4.6, page 344.<|endoftext|>
TITLE: What are interesting families of subsets of a given set?
QUESTION [35 upvotes]: Motivation
The usual starting point of both Topology and Measure Theory is the definition of a family of subsets of a set $S$.
Indeed, one defines a topology on $S$ to be a family of subsets including the empty set $\emptyset$ and $S$ itself and which is closed under arbitrary unions and finite intersections.  These are the open sets.  (One can of course also define a topology by stipulating which are the closed sets, which are now closed under finite unions and arbitrary intersections.)
In Measure Theory one starts by defining on $S$ the notion of a $\sigma$-algebra, which is a collection of subsets again including $S$ and which is closed under complementation and countable unions, so in particular it also includes $\emptyset$ and is closed under countable intersections.  The subsets in the $\sigma$-algebra are the measurable sets.
When I learnt these subjects I was always intrigued by the similarity of both definitions.  This suggests other family of subsets of a set $S$ defined by demanding that both $\emptyset$ and $S$ belong to the family and that the family be closed under some operations.

Question
Are there any interesting families of subsets, other than topologies and $\sigma$-algebras, which can be defined in this way?  And if so, to which areas of mathematics are they germane?

REPLY [7 votes]: Another ultrafilter cousin is the concept of a majority
space. This is a family $M$ of nonempty subsets of $X$,
called the majorities, such that any superset of a
majority is a majority, every subset of $X$ or its
complement is a majority, and if disjoint sets are
majorities, then they are complements. A strict majority
space has $Y\in M\to Y^c\notin M$, and otherwise they are
called weak majorities. A vast majority space is closed
under finite differences in majorities. There are other
various overwhelming majority concepts.
The main point is that the majority space concept
generalizes the ultrafilter concept by omitting the
intersection rule. Every ultrafilter on $X$ is a majority
space. But there are others. For example, on a finite set,
one may take the the subsets with at least half the size, and of course this situation motivates the voting theory terminology.
On an infinite set $X$, one can divide it into a finite odd
number of disjoint pieces $X_i$, each carrying an
ultrafilter $\mu_i$ on $X_i$, and then saying that
$Y\subset X$ is a majority if for most $i$ one has
$Y\cap X_i\in\mu_i$. This produces a vast majority on
$X$ that is not an ultrafilter.
Eric Pacuit has investigated majority logic, and I recall that Andreas Blass has some very interesting work showing that it is consistent with ZFC that every majority space derives from ultrafilters in a simple way.<|endoftext|>
TITLE: Great mathematicians born 1850-1920 (ET Bell's book ≲ x ≲ Fields Medalists)
QUESTION [36 upvotes]: When I was a teenager, I was given the book Men of Mathematics by E. T. Bell, and I rather enjoyed it.  I know that this book has been criticized for various reasons and I might even agree with some of the criticism, but let's not digress onto that. E. T. Bell made a reasonable list of 34 of the greatest mathematicians from the ancient Greek period to the end of the 19th century.  His list isn't perfect — maybe he should have included Klein or skipped Poncelet — but no such list can be perfect anyway.  I think that his selection was good.  He also made the careers of these mathematicians exciting and he addressed why mathematicians care about them today.  It helped me learn what achievements and topics in mathematics are important.
(Just to head off discussion, the standard complaints include that the title is sexist and so is the book, that Bell was loose with biographical facts, and that he "chewed the scenery".)
After the period covered by Bell, there is a 50-70 year gap, followed by the the Fields Medals.  The list of Fields Medalists has its own limitations, but it is another interesting, comparably long list of great mathematicians.  Somewhat accidentally, this list also orients and motivates advanced mathematics students today.
But who are the great mathematicians in the gap itself, that is, those born between 1850 and 1900, or say between 1860 and 1910?  (Or 1920 at the latest; that is what I had before.)  There is a good expanded list of mathematicians with biographies at the St. Andrews site.  However, it is too long to work as a sequel to Bell's book.  (Not that I plan to write one; I'm just asking.)  If you were to make a list of 20 to 50 great mathematicians in this period, how would you do it or who would they be?  Presumably it would include Hilbert, but who else?  (Poincaré was born in 1854 and is the latest born in Bell's list.)
For example, I know that there was the IBM poster "Men of Modern Mathematics" that also earned some criticism.  But I don't remember who was listed, and my recollection is that the 1850-1900 period was somewhat cramped.
I decided based on the earliest responses to convert this list to community wiki.  But I am not just asking people to throw out names one by one and then vote them up.  If a good reference for this question already exists, or if there is some kind of science to make a list, then that would be ideal.  If you would like to post a full list, great.  If you would like to list one person who surely should be included and hasn't yet been mentioned, then that's also reasonable.  It may be better to add years of birth and death in parentheses, for instance "Hilbert (1862-1943)".
Note: To make lists, you should either add two spaces to the end of each line, or " - " (space dash space) to the beginning of each line.

REPLY [9 votes]: I feel some of the Russian/Soviet mathematicians are missing in the lists, like:

Aleksandr Mikhailovich Lyapunov (1857-1918)
Nikolai Nikolaevich Luzin (1883-1950)
Aleksandr Yakovlevich Khinchin (1894-1959)
Pavel Sergeyevich  Alexandrov (1896-1982)  
Andrey Nikolayevich Tikhonov (1906-1993)

though I think one could include many others.<|endoftext|>
TITLE: Explicit computations of small Deligne-Lusztig varieties (e.g. Drinfeld curve)
QUESTION [7 upvotes]: Background:
I am focusing on $G=GL_{2}(\overline{\mathbb{F_q}})$ here. If you wonder why I am interested in this, I am trying a problem relating to the Deligne-Lusztig varieties defined over local rings by Stasinksi, and this background theory is relevant there. The definition I am using is this: 

The first definition I have of Deligne-Lusztig varieties is this: Consider the Lang map $L(g) = g^{-1} F(g)$. Let $T$ be an $F$-stable maximal torus of $G$, and $B$ a Borel subgroup containing $T$ (not necessarily $F$-stable), and $U$ the unipotent radical of this Borel subgroup. Then the Deligne-Lusztig variety is defined as $X = L^{-1}(U)$

Question
Roughly: Using these definitions (I am not sure exactly to what extent these two definitions are "compatible"), how do I explicitly compute the Deligne-Lusztig variety for $GL_{2}(\mathbb{F}_{q})$ to be the Drinfeldt curve $xy^{q} - yx^{q} = 1$ in the non-split case? More precisely:

If I pick a torus that is not maximally split, $T$, and a Borel subgroup $B$ containing $T$, then apply the definition $1$ above - how do we relate this variety to the Drinfeldt curve? Do we explicitly get the Drinfeldt curve, and if not what do we get? If it is not the Drinfeldt curve that we get using this definition, how do we express this variety "nicely" (with a view towards counting $F_{q}$ points on it). 
Roughly speaking (not explicitly!) how would I go about doing this for $GL_{3}(\mathbb{F_q})$? Is this computationally feasible? Are there any tricks that would help significantly with the computation? Even better, are there any references you know which do this? 
I understand Deligne-Lusztig varieties, and these tori, correspond in some sense to Weyl group elements. Are there any specific Weyl group elements in $S_n$ for which the Deligne-Lusztig variety always has a "nice" / "tractable" description, or do they get out of hand very quickly? 

My Attempts at (1)
(I'm not entirely sure how to write matrices in Latex here, so I did it crudely by writing it as $4$ numbers, Row 1 followed by Row 2).
Pick $\alpha, \beta \in F_{q^2}$, so that $(x- \alpha)(x- \beta)$ is irreducible over $F_{q}$. Then a unipotent subgroup of a Borel subgroup for a non-split torus by conjugating the ordinary unipotent subgroup of strictly upper triangular matrices, by the matrix $M$ with entries $(1, 1, \alpha, \beta)$. This is because we can obtain the matrix with entries $(0, 1, -\alpha \beta, \alpha + \beta)$ (lying inside $GL_{2}(\mathbb{F_q})$ as $M^{-1} X M$, where $X$ is the matrix with entries $( \alpha, 0, 0, \beta)$. Since when we conjugate the nilpotent matrix with entries $(0, 1, 0, 0)$  by $M$ we get a scalar multiple of the matrix with entries $ ( - \alpha, 1, - \alpha^2, \alpha)$, the end result of this calculation is that our non-split maximal torus consists of matrices of the form $(1-s \alpha, s, - s \alpha^2, 1 + s \alpha)$.
Now if $g = (a,b,c,d)$ (i.e. the matrix with those 4 entries), then $g^{-1} = \frac{1}{ad-bc}( d, -b, -c, a)$, $F(g) = (a^q, b^q, c^q, d^q)$, and $g^{-1} F(g) = \frac{1}{ad-bc} (da^q - bc^q, db^q - bd^q, -ca^q + ac^q, -cb^q + ad^q)$, so equating that $g^{-1} F(g)$ lies in the subgroup calculated in the last paragraph gives the following description of the Deligne-Lusztig variety (let $D = ad-bc$), by comparing entry by entry:

$ \frac{1}{D} ( da^q - bc^q -cb^q + ad^q ) = 2 $
$ -ca^q + ac^q = - \alpha^2 (db^q - bd^q) $

This might be easy but I cannot see how to finish off from here: why is this related to the Drinfeldt curve? All I can see from a first glimpse is that the terms $db^q - bd^q$ appears, but I don't know how to get rid of everything else. How can I simplify the defining equations of this variety (hopefully in a suitably simple version, so that counting $F_{q^{k}}$ points is straightforward, which is what I really need to do).

REPLY [5 votes]: Hi Vinoth, you might be interested in the following recent book:
"Representations of $SL_2(\mathbb{F_q})$" by Cédric Bonnafé
"Deligne-Lusztig theory aims to study representations of finite reductive groups by means of geometric methods, and particularly l-adic cohomology. Many excellent texts present, with different goals and perspectives, this theory in the general setting. This book focuses on the smallest non-trivial example, namely the group $SL_2(\mathbb{F_q})$, which not only provide the simplicity required for a complete description of the theory, but also the richness needed for illustrating the most delicate aspects.
The development of Deligne-Lusztig theory was inspired by Drinfeld's example in 1974, and Representations of $SL_2(\mathbb{F_q})$ is based upon this example, and extends it to modular representation theory. To this end, the author makes use of fundamental results of l-adic cohomology. In order to efficiently use this machinery, a precise study of the geometric properties of the action of SL2(Fq) on the Drinfeld curve is conducted, with particular attention to the construction of quotients by various finite groups.
At the end of the text, a succinct overview (without proof) of Deligne-Lusztig theory is given, as well as links to examples demonstrated in the text. With the provision of both a gentle introduction and several recent materials (for instance, Rouquier's theorem on derived equivalences of geometric nature), this book will be of use to graduate and postgraduate students, as well as researchers and lecturers with an interest in Deligne-Lusztig theory."
See http://www.springer.com/mathematics/algebra/book/978-0-85729-156-1
Best, Daniel.<|endoftext|>
TITLE: on chern classes and Riemann Roch theorem for torsion-free sheaves on singular (possibly multiple) curve
QUESTION [10 upvotes]: I'm looking for a definition of Chern class (at least the first one) for a torsion-free sheaf  $F$ (not necessarily locally free) on a singular curve (for simplicity can assume all the singularities are planar).

The Chern class can be, of course, extracted from an exact sequence relating $F$ to some  locally free sheaves. But I would like some more direct definition, like the one given by Hartshorne (Generalized divisors on Gorenstein curves and a theorem of Noether. J. Math. Kyoto Univ. 26 (1986), no. 3, 375--386). 
At least for the first Chern class. 
1.5 Even if one wants to define $c_1(F)$ from some resolution: torsion free sheaves on singular curves sometimes have no finite locally free resolutions.  What would you do in this case?

I'm looking for the Riemann-Roch for torsion-free sheaves on a singular curve (can assume the singularities to be planar). For example Hartshorne in the paper above does it for rank one.

Of course, if the only definition of the first Chern class is from the exact sequence, then Riemann-Roch is tautological (an alternative way to define $c_1(F)$).  So this question is meaningful modulo the first question.
Somehow I do not find all this in classical textbooks.
Thanks to everybody!!!!

REPLY [4 votes]: I must be missing something here. Why not use the standard Baum-Fulton-MacPherson $\tau$ map? This is the map one needs for Riemann-Roch to work anyway as in sections 18.2 and 18.3 of Fulton's "Intersection theory" book. The definition is very natural -  embed the curve in a smooth variety (e.g. in projective space) and resolve the push-forward by vector bundles on the target. You only need to be careful and correct your computation by intersecting with the Todd class of the target. Baum-Fulton-MacPherson check that the definition is independent of the embedding and that Riemann-Roch holds for l.c.i. proper morphisms, e.g. for projective curves with planar singularities.
If we want to cheat we can also define the degree of a torsion free sheaf $F$ on a curve $C$ from the Hilbert polynomial, or simply  as $\deg(F) = \chi(F) - rk(F)\chi(\mathcal{O}_C)$.<|endoftext|>
TITLE: Reference for this theorem in representation theory?
QUESTION [16 upvotes]: Let $G$ be a finite group and $\chi$ be an irreducible character of
$G$ (characteristic zero algebraically closed base field). If $H$ is
the kernel of $\chi$ then the irreducible representations of $G/H$
are exactly all the irreducible constituents of all tensor powers
$\chi^n$.

Do you know any reference for this theorem?
Is it also working in positive characteristic?
Is it also working for some infinite groups? (maybe some special
classes: reductive, Lie type, etc)

Thank you very much!

REPLY [3 votes]: Explicit references: 

BURNSIDE, W. Theory of Groups of Finite Order, 2nd ed. Cambridge University
Press, Cambridge, 1911; Dover, New York, 1955.  (§226 Theorem IV)   
BRAUER, R. A note on theorems of Burnside and Blichfeldt. Proc. Am. Math.
soc. 15 (1964), 31-34.
(Theorem 1)  

Generalizations: 

STEINBERG, R. Complete sets of representations of algebras. Proc. Am. Math.
Soc. 13 (1962), 746-747.
RIEFFEL, M.A. Burnside’s Theorem for Representations of Hopf Algebras. J. Alg. 6 (1967), 123-130.<|endoftext|>
TITLE: Sum of radical ideals
QUESTION [9 upvotes]: Let $A$ be a commutative ring and endow the closed subsets of $\operatorname{Spec}(A)$ with the Grothendieck topology of finite covers. One may ask if the presheaf $V \mapsto A/I(V)$ is a sheaf. This is not true in general and is related (but not equivalent) to the following pure algebraic question:

In which commutative rings $A$ are the radical ideals closed under sum?

The property can be checked locally. It holds in dimension $0$, and also for integral domains of dimension 1. It doesn't hold for the $2$-dimensional ring $k[x,y]$ (consider $(x^2 + y)+(y) = (x^2,y)$), nor for the $1$-dimensional ring $\bigl(k[x,y]/(x^2 y + y^2)\bigr)_{(x,y)}$.
Are there other interesting examples/counterexamples or approaches for a general classification? I think the property has some algebro-geometric interpretation: All intersections of closed subschemes are transversal. See also Conditions for $\sqrt{\mathfrak{a + b}} = \sqrt{\mathfrak{a}} + \sqrt{\mathfrak{b}}$.

REPLY [4 votes]: With regards to the original question — I don't have any good ideas, probably valuation rings satisfy this (but they are not generally Noetherian except in the cases already outlined).
With regards to other classes of ideals that satisfy this (now in characteristic zero), the ideals of “unions of log canonical centers” satisfy this property in characteristic zero (this is mostly due to a result of Florin Ambro I think).  See for example Ambro - Basic properties of log canonical centers.<|endoftext|>
TITLE: When is $A$ isomorphic to $A^3$?
QUESTION [81 upvotes]: This is totally elementary, but I have no idea how to solve it: let $A$ be an abelian group such that $A$ is isomorphic to $A^3$. is then $A$ isomorphic to $A^2$? probably no, but how construct a counterexample? you can also ask this in other categories as well, for example rings. if you restrict to Boolean rings, the question becomes a topological one which makes you think about fractals: let $X$ be stone space such that $X \cong X + X + X$, does it follow that $X \cong X + X$ (here + means disjoint union)?
edit: In the answers there are already counterexamples. but you may add others in other categories (with products/coproducts), especially if they are easy to understand :).

REPLY [44 votes]: Given a class of structures equipped with a product $(K, \times)$, the question of whether $X^3 \cong X \implies X^2 \cong X$ holds for every $X \in K$ is sometimes called the cube problem for $K$, and if it has a positive answer then $K$ is said to have the cube property. For the question to be nontrivial there need to be infinite structures $X \in K$ that are isomorphic to $X^3$. If there are such structures, it is usually possible to find one that witnesses the failure of the cube property for $K$, that is, an $X \in K$ such that $X \cong X^3$ but $X \not\cong X^2$. On the other hand, in rare cases the cube property does hold nontrivially. 
I worked on the cube problem for the class of linear orders under the lexicographical product, and while doing so had a chance to look into the history of the problem for other classes of structures. The following list contains most of the results that I am aware of. 
When the cube property fails
-- As far as I know, the first result concerning the failure of the cube property is due to Hanf, who showed in [1] that there exists a Boolean algebra $B$ isomorphic to $B^3$ but not $B^2$. Hanf's example is of size $2^{\aleph_0}$. 
-- Tarski [2] and Jónsson [3] adapted Hanf's result to get examples showing the failure of the cube property for the class of semigroups, the class of groups, the class of rings, as well as other classes of algebraic structures. Most of their examples are also of size continuum. 
It was unknown for some time after these results were published whether there exist countable examples witnessing the failure of the cube property for these various classes. Especially famous was the so-called "Tarski cube problem," which asked whether there exists a countable Boolean algebra isomorphic to its cube but not its square. 
-- As Tom Leinster answered, Corner [4] showed, by a very different route, that indeed there exists a countable abelian group isomorphic to its cube but not its square. Later, Jones [5] constructed a finitely generated (necessarily non-abelian) group isomorphic to its cube but not its square.
-- Around the same time as Corner's result, several authors [6, 7] showed that there exist modules over certain rings isomorphic to their cubes but not their squares.
-- As Asher Kach answered, Tarski's cube problem was eventually solved by Ketonen, who showed in [8] that there does exist a countable Boolean algebra isomorphic to its cube but not its square.   
Ketonen's result is actually far more general. Let $(BA, \times)$ denote the class of countable Boolean algebras under the direct product. If $(S, \cdot)$ is a semigroup, then $S$ is said to be represented in $(BA, \times)$ if there exists a map $i: S \rightarrow BA$ such that $i(a \cdot b) \cong i(a) \times i(b)$ and $a \neq b$ implies $i(a) \not\cong i(b)$. The statement that there exists a countable Boolean algebra isomorphic to its cube but not its square is equivalent to the statement that $\mathbb{Z}_2$ can be represented in $(BA, \times)$. Ketonen showed that every countable commutative semigroup can be represented in $(BA, \times)$. 
-- Beginning in the 1970s, examples began to appear showing the failure of the cube property for various classes of relational structures. For example, Koubek, Nešetril, and Rödl showed that the cube property fails for the class of partial orders, as well as many other classes of relational structures in their paper [9].
-- Throughout the 70s and 80s, Trnková and her collaborators showed the failure of the cube property for a vast array of topological and relational classes of structures. Like Ketonen's result, her results are typically much more general. 
Her topological results are summarized in [10], and references are given there. Some highlights:

There exists a compact metric space $X$ homeomorphic to $X^3$ but not
$X^2$. More generally, every finite abelian group can be represented
in the class of compact metric spaces.
Every finite abelian group can be represented in the class of
separable, compact, Hausdorff, 0-dimensional spaces.
Every countable commutative semigroup can be represented in the class
of countable paracompact spaces.
Every countable commutative semigroup can be represented in the class
of countable Hausdorff spaces.

Her relational results mostly concern showing the failure of the cube property for the class of graphs. For example:

Every commutative semigroup can be represented in $(K, \times)$,
where $K$ is the class of graphs and $\times$ can be taken to be the
tensor (categorical) product, the cartesian product, or the strong
product [11].
There exists a connected graph $G$ isomorphic to $G \times G \times
   G$ but not $G \times G$, where $\times$ can be taken to be the tensor
product, or strong product. As of 1984, it was unknown whether
$\times$ could be the cartesian product [12].

--Answering a question of Trnková, Orsatti and Rodino showed that there is even a connected topological space homeomorphic to its cube but not its square [13]. 
--More recently, as Bill Johnson answered, Gowers showed that there exists a Banach space linearly homeomorphic to its cube but not its square [14]. 
--Eklof and Shelah constructed in [15] an $\aleph_1$-separable group $G$ isomorphic to $G^3$ but not $G^2$, answering a question in ZFC that had previously only been answered under extra set theoretic hypotheses. 
--Eklof revisited the cube problem for modules in [16].
When the cube property holds 
There are rare instances when the cube property holds nontrivially.
-- It holds for the class of sets under the cartesian product: any set in bijection with its cube is either infinite, empty, or a singleton, and hence in bijection with its square. This can be proved easily using the Schroeder-Bernstein theorem, and thus holds even in the absence of choice.
-- Also easily, it also holds for the class of vector spaces over a given field.
-- Less trivially, it holds for the class of $\sigma$-complete Boolean algebras, since there is a Schroeder-Bernstein theorem for such algebras.
-- Trnková  showed in [17] that the cube property holds for the class of countable metrizable spaces (where isomorphism means homeomorphism), and in [18] that it holds for the class of closed subspaces of Cantor space. The cube property fails for the class of $F_{\sigma \delta}$ subspaces of Cantor space. It is unknown if it holds or fails for $F_{\sigma}$ subspaces of Cantor space. See [10].
-- Koubek, Nešetril, and Rödl showed in [9] that the cube property holds for the class of equivalence relations. 
-- I recently showed that the cube property holds for the class of linear orders under the lexicographical product. (My paper is here. See also this MO answer.)
A theme that comes out of the proofs of these results is that when the cube property holds nontrivially, usually some version of the Schroeder-Bernstein theorem is in play. 
References:

William Hanf, MR 108451 On some fundamental problems concerning isomorphism of Boolean algebras, Math. Scand. 5 (1957), 205--217.
Alfred Tarski, MR 108452 Remarks on direct products of commutative semigroups, Math. Scand. 5 (1957), 218--223.
Bjarni Jónsson, MR 108453 On isomorphism types of groups and other algebraic systems, Math. Scand. 5 (1957), 224--229.
Corner, A. L. S., "On a conjecture of Pierce concerning direct decompositions of Abelian groups." Proc. Colloq. Abelian Groups. 1964.
Jones, JM Tyrer, "On isomorphisms of direct powers." Studies in Logic and the Foundations of Mathematics 95 (1980): 215-245.
P. M. Cohn, MR 197511 Some remarks on the invariant basis property, Topology 5 (1966), 215--228.
W. G. Leavitt, MR 132764 The module type of a ring, Trans. Amer. Math. Soc. 103 (1962), 113--130.
Jussi Ketonen, MR 491391 The structure of countable Boolean algebras, Ann. of Math. (2) 108 (1978), no. 1, 41--89.
V. Koubek, J. Nešetril, and V. Rödl, MR 357669 Representing of groups and semigroups by products in categories of relations, Algebra Universalis 4 (1974), 336--341.
Vera Trnková, MR 2380275 Categorical aspects are useful for topology—after 30 years, Topology Appl. 155 (2008), no. 4, 362--373.
Trnková, Věra, and Václav Koubek, "Isomorphisms of products of infinite graphs." Commentationes Mathematicae Universitatis Carolinae 19.4 (1978): 639-652.
Trnková, Věra, "Isomorphisms of products of infinite connected graphs." Commentationes Mathematicae Universitatis Carolinae 25.2 (1984): 303-317.
A. Orsatti and N. Rodinò, MR 858335 Homeomorphisms between finite powers of topological spaces, Topology Appl. 23 (1986), no. 3, 271--277.
W. T. Gowers, MR 1374409 A solution to the Schroeder-Bernstein problem for Banach spaces, Bull. London Math. Soc. 28 (1996), no. 3, 297--304.
Paul C. Eklof and Saharon Shelah, MR 1485469 The Kaplansky test problems for $\aleph_1$-separable groups, Proc. Amer. Math. Soc. 126 (1998), no. 7, 1901--1907.
Eklof, Paul C., "Modules with strange decomposition properties." Infinite Length Modules. Birkhäuser Basel, 2000. 75-87.
Trnková, Věra, "Homeomorphisms of powers of metric spaces." Commentationes Mathematicae Universitatis Carolinae 21.1 (1980): 41-53.
Vera Trnková, MR 580990 Isomorphisms of sums of countable Boolean algebras, Proc. Amer. Math. Soc. 80 (1980), no. 3, 389--392.<|endoftext|>
TITLE: When does local invertibility imply invertibility?
QUESTION [11 upvotes]: Generally, local invertibility does not imply invertibility.  However, for differentiable functions from $\mathbb{R}$ to $\mathbb{R}$ then surjectivity and local invertibility do imply invertibility.
As well as being the most obvious, it's also the only (non-contrived) case that I can think of.  Are there any more?
Specifically, I'm looking for examples of spaces $X$ which are at least topological spaces (but may be more structured) and subsets of endomorphisms on $X$ for which local invertibility implies invertibility.

REPLY [5 votes]: Following Deane Yang, the answer is a definite yes: the map 
in question is a global diffeo,
provided that 
(a) it is `locally invertible': i.e. its   derivative is everywhere invertible,  and
(b) the domain and range are compact, simply connected, without boundary.
Proof: the map must be a covering map (``stack of records theorem'' -- see
for example Guillemin and Pollack). But a covering map between simply connected spaces
is an isomorphism -- here a diffeo. 
To make this `non-contrived' take domain and range to be the two-sphere.<|endoftext|>
TITLE: Casson's invariant and the trivial connection contribution to witten's 3-manifold invariant
QUESTION [9 upvotes]: This question might turn out not to make any sense, but here it is: Witten's (and Reshetikhin and Turaev's) 3-manifold invariant can be "defined" as an integral over the space of connections on the trivial SU(2) bundle over the 3-manifold M, modulo gauge transformation. In the stationary phase approximation as the level k-->infinity, we write this integral as the sum of functions f_i(k), where the sum is over the set {c_i} of critical points of the chern-simons functional (which is basically the log of the integrand). The critical points are the flat connections. In his analysis of the function f_tr(k) corresponding to the trivial flat connection (product connection), Rozansky (as well as previous authors, I believe) found the Casson invariant (as the second coefficient if you write f_i as a power series in 1/k, or something like that). 
Now, at first glance it strikes me as odd that the casson invariant shows up here, because the casson invariant can, by definition, be thought of as a (signed) sum over ALL flat connections, EXCEPT the trivial connection. So my question is: what gives? Why does the contribution from the trivial connection give the casson invariant? Why isn't it instead, say, the sum of the 2nd coefficients of the contributions from all other flat connections? is there some explanation for this? Have contributions from other connections, say reducible connections for a homology 3-sphere, been analyzed? Does the casson invariant show up there as well?
If you've gotten this far, here's another question: right now, i can write |H_1(M)|*(Casson-Walker invariant of M), say for lens spaces M=L(p,q) as the sum over certain INTEGERS, one per conjugacy class of reducible flat connection on SU(2)XM, PLUS the signature of the 2-bridge knot b(p,q). I'm trying to identify this decomposition of the  casson invariant with others in the literature, of which there are many, but I'm having the problem that none of these are (always) integral decompositions, even though they ADD UP to an integer. Well this is a long shot, but any ideas would be appreciated!

REPLY [6 votes]: The Casson invariant is not the same sum or integral over connections that you would derive from the perturbative expansion Cherns-Simons quantum field theory at all flat connections.  There is more than one way to rigorously interpret that expansion; one method uses Kontsevich's configuration space integrals.  Dylan Thurston and I proved (in Perturbative 3-manifold invariants by cut-and-paste topology) that the configuration space invariant using only the flat connection is proportional to the Casson invariant, for any simple Lie group as the gauge group.  (The configuration space integral is known as the theta invariant, because the Feynman-Jacobi diagram is a theta.)  By contrast, Casson's invariant has been interpreted by Witten as a gauge theory with a certain Lie supergroup, whose underlying Lie group is SU(2).
So what gives?  One answer is that the Casson invariant is the unique finite-type invariant of homology 3-spheres of degree 1, up to a scalar factor.   (The Wikipedia link discusses finite-type invariants of knots, but as the page mentions briefly, there is an extension to 3-manifolds due to Garoufalidis, Habiro, Levine, and Ohtsuki.)  As such, you should expect it to show up many times.  The corresponding phenomenon at the level of link invariants is that all of the standard quantum link invariants that are polynomials in $q$ have the same second derivatives at $q=1$, again up to a scalar factor.  In fact, Dylan and I didn't do anything directly with the Casson invariant; instead we showed that the theta invariant has finite-type degree 1.<|endoftext|>
TITLE: Categorifications of the Fibonacci Fusion Ring arising from Conformal Field Theory
QUESTION [6 upvotes]: I was reading about realizations of the "Fibonacci" fusion ring $X \otimes X = X \oplus 1$ in Fusion Categories of Rank 2 by Victor Ostrik.  Apparently, there are two of them and they arise in various ways:

integer-spin representations of integrable $\widehat{sl}_2$-modules of level 3
the minimal model $\mathcal{M}(2,5)$ of the Virasoro algebra (central charge c = -22/5)
representations of $U_q(sl_2)$ with $q = e^{\pi i /5}, e^{3\pi i / 5}$.

In any of these cases, how is the Fibonacci category realized?
I would like to understand each of these specific categorifications of the Fibonacci fusion ring.  Can someone here explain the basics of integrable $\widehat{sl}_2$-modules or about the $\mathcal{M}(2,5)$ minimal model from Conformal Field Theory?  I would also like to learn about $U_q(sl_2)$, though it's probably written in many texts.

REPLY [5 votes]: Unfortunately all three of those realizations are the sort of thing you need to read a book about not a MO post.  I agree with Greg that Kassel's book is a great place to start for the quantum group construction (I don't know the other two constructions well, presumably for the affine algebra construction you'd want to start with Kac's book?).
On the other hand there is an easier to explain elementary diagrammatic description.  As usual with diagram categories you only construct a full subcategory and then you'll need to take the additive and idempotent completions to get an abelian category.
Consider the Temperley-Lieb subcategory, whose objects are indexed by integers and whose morphisms m->n are given by linear combinations of planar diagrams of nonintersecting arcs with m boundary points at the bottom and n boundary points at the top modulo a single relation that a circle can be removed for a multiplicative factor of either the golden ratio or its conjugate.  Composition is stacking, tensor product is disjoint union.  There's an explicit 4-strand projection (called a Jones-Wenzl idempotent) here that has the property that any way you close it off you get zero.  Kill that idempotent.  Now look at the "even part" i.e. the full subcategory whose objects are even integers.  This is your category.  Its simple objects are the 0 and 2-strand Jones-Wenzl idempotents.
There's another way to think of this example.  First checkboard shade the regions of all your even diagrams so that they're unshaded on the outside.  Then collapse all the dark regions to lines.  What you end up with now has half as many boundary points and is allowed to have internal 3-valent and 1-valent vertices.  It's easy to see that they satisfy an I=H relation and a relation allowing absorbing vertices.  This gives a construction of the Fibonacci category using the Yamada polynomial relations (I think to get the usual Yamada polynomial on the nose here you want to actually throw in a bunch of JW2s everywhere but its six of one half dozen the other).
Finally there's a slightly different diagram description given in the appendix of one of my papers with Emily Peters and Scott Morrison.  In our notation there the Fibonacci category is (the additive and idempotent completion of) the tadpole planar algebra T_2.<|endoftext|>
TITLE: Does a regular neighborhood always exist for a properly embedded surface in a 3-manifold? 
QUESTION [6 upvotes]: Can someone please clarify if there always exist regular neighborhoods for a properly embedded surface in a 3-manifold? More precisely, if $F$ is a properly embedded surface in a 3-manifold $M$ and I give a simplicial complex structure to $M$, then will $F$ automatically recieve a subcomplex structure (after probably finitely many barycentric subdivisions of the triangulation of $M$) ? If this is so, then we can obviously construct a regular neighborhood of $F$. But I am feeling unsure about it now due to the following example where $F$ might be too big to allow such a compatible subdivision of the triangulation of $M$:  
Example: Consider a mobius band without a contractible neighborhood in a genus 1 handlebody. One can construct one as follows: First take a solid cylinder $D^2\times{[-1,1]}$ and look at the central strip. Now glue the ends of the cylinder with $180$ degrees twist to make it a genus 1 handlebody. This will glue the central strip with a twist to make it a mobius band and it has no small neighborhood in the ambient genus 1 handlebody. It seems to me that this mobius band will not have a subcomplex structure for any given simplicial complex structure on the genus 1 handlebody.   
In case the answer to my question is yes, I would also like to ask if there are no issues regarding the orientability of $F$ and $M$ while considering regular neighborhoods. That is whether it matters for constructing such a neighborhood if $F$ is non-orientable but $M$ is orientable or vice-versa or other combinations. 
I am unable to clarify this by looking at Hempel's book on 3-manifolds. It would be great if someone could elucidate this.

REPLY [4 votes]: The answer is yes, with the correct technical hypothesis of "local flatness". (Local flatness rules out, for example, the sort of behavior shown by the Alexander horned sphere.)  You are correct to think that this is a foundational issue in three-manifolds.  The reference you mention (Hempel) is also the correct one.  You could perhaps look at Bing's book.
In your example the Mobius band does have a regular neighborhood.  Note that the open regular neighborhood is homeomorphic to the normal bundle of the surface F in the three-manifold M.  The regular neighborhood need not (and in your example, is not) be homeomorphic to the product FxI.<|endoftext|>
TITLE: Schemes of Representations of Groups
QUESTION [12 upvotes]: Let $G$ be a group, say finitely presented as $\langle x_1,\ldots,x_k|r_1,\ldots,r_\ell\rangle$.  Fix $n\geq 1$ a natural number.  Then there exists a scheme $V_G(n)$ contained in $GL(n)^k$ given by the relations.  This scheme parameterizes $n$ dimensional representations of $G$.
Now, I've known this scheme since I first started learning algebraic geometry (one of the first examples shown to me of an algebraic set was $V_{S_3}(2)$) but I've never found a good reference for this.  So my first question is:
Is there a good reference for the geometry of schemes of representations?

Now, I have some much more specific questions.  The main one being a point I'd been wondering about idly and tangentially since reading about some open problems related to the Calogero-Moser Integrable System:
Are there natural conditions on $G$ that will guarantee that $V_G(n)$ be smooth? Reduced? Now, this is on the affine variety, I know that the projective closure will generally be singular, but in the case of $V_{S_3}(2)$, I know that the affine variety defined above is actually smooth, of four irreducible components.
Finally, for any $G$ and $n$, we have $V_G(n)\subset V_G(n+1)$ (By taking the subscheme where the extra row and column are zeros, except on the diagonal, where it is 1).  We can take the limit and get an ind-scheme, $V_G$.  What is the relationship between $V_G$ and the category $Rep(G)$? Can the latter be realized as a category of sheaves on the former? I know nothing here, and as I said, most of these questions are the result of idle speculation while reading about something else.
Edit: It occurs to me that as defined, $V_G(n)$ and $V_G$ may not be invariants of $G$, but really of the presentation.  So two things to add: one, $V_G(n)$ is intended to really be the scheme $Hom(G,GL(n))$ (there's some issues I want to sweep under the rug with finitely generated infinite groups here, which is part of why I was thinking in presentations), and second, the situations that I'm thinking of are often the data of group with a presentation, so for that situation, $V_G(n)$ as defined should be good enough.

REPLY [9 votes]: Charlie, as Dmitri pointed out there is a big difference between compact Kaehler and non-Kaehler manifolds as far as the structure of the representation varieties of their fundamental groups are concerned. 
By the way, by a theorem of Taubes,  every finitely presentable group is the fundamental of a compact complex three dimensional complex manifold so you don't really get any restriction by saying that you want your group to be the fundamental group of a complex manifold. Taubes' manifolds however are constructed as twistor spaces of anti-self dual real 4-manifolds and are never  Kaehler. The condition that your group can be realized as the fundamental group of a compact Kaehler manifold is a serious condition and puts many constraints on the representation variety. 
Another comment is that the representation scheme does not depend on the presentation of your group. You only need to know that the group is finitely generated.
There is a huge literature on this subject. I will list just a few of the landmarks:

A very nice classical source is the Lubotzky-Magid book "Varieties of representations of finitely generated groups"
The paper of Goldman-Milson that Dmitri mentioned is a must-read. 
There are two fundamental papers of Simpson that I mentioned in this MO post.
On the topic of Kaehler fundamental groups you can start with this book by Amoros et al. and with this article by Arapura.<|endoftext|>
TITLE: Generalization of the shakehands/condom puzzle?
QUESTION [18 upvotes]: The classic handshake puzzle goes something like this: 

"Given that everyone has a different skin disease, how can you safely shake hands with 3 people when you have only 2 gloves?"

Its common variations are:

"How can a man engage in safe sex with 3 women using 2 condoms?"
"How can a doctor operate on 3 patients with only 2 gloves while avoiding skin-blood contact between any two people"

Let's say N is the number of other people (patients/women...etc) and K is the number of gloves (or condoms).  The above case of N=3 and K=2 is not hard (and its solution readily available on the net).
QUESTION 1: In general, what can we say about the feasible N's and K's?  It seems like (2K >= N+1) is a necessary condition (K gloves has 2K sides and there are a total of N+1 people involved).  Is this also sufficient?
While researching on Google, I came across a posting that claimed the generalization of this similar puzzle is an open problem:

"Two couples get together for an evening of hetero swinging. What is the minimum number of condoms necessary for safe sex in all of the male-female pairings?"
http://mathematicsontribe.tribe.net/thread/d0f5c284-762d-4045-be74-21e6ede7e31e

QUESTION 2: I assume the general form of the question would study the feasibility of N couples and K condoms.  What is known about the general problem? Is it still open?

(Qiaochu Yuan:)  Based on the downvotes, which I would guess are directed at the way in which the problem is stated rather than its content, here is a "cleaned up" version appropriate for mathematicians:
You have a collection of $K$ tokens which have two sides, each of which can be marked.  There are two families of marks, $N$ of which are of the first family and $M$ of which are of the second family.  For each pair $(i, j)$ of a mark of the first family and the second family, attempt the following:

Stack a collection of tokens from left to right.  (Tokens may be rotated.)
If two tokens are adjacent, the adjacent sides share marks.  
Mark the left side of the leftmost token with mark $i$ and the right side of the rightmost token with mark $j$.  This move is only possible if each of the sides to be marked is either initially unmarked or is marked only with the mark you are trying to mark it with and with no other marks.

For which values of $K, N, M$ is this possible?

REPLY [38 votes]: Thank you for posting my solution. I am still amazed that this problem remains controversial after 20 years. Indeed, before publication, my book editors communicated statements to the effect that the chapter of my book entitled "The Condom Problem" was sexist. This is echoed by some of the above comments. The editors decided to go ahead with my proposed formulation when it was pointed out by (female) reviewers that my formulation (as opposed to what is given here) was gender neutral in the sense that the names "men" and "women" were interchangeable whereas in the euphemistic formulation involving gloves all the doctors were male and all the nurses were all female, so the politically correct version was the truly sexist one. The above comments that the condom problem is offensive to women imply that sexual content should naturally offend females, which is obviously incorrect and is more a reflection of the persistence of certain puritanical United States values. In any case, Addison Wesley decided to publish it, and the rest is history. If the decision of the most reputed technical publisher is not good enough for you, then I wonder what is. Moreover, I solved the problem, so if you respect my efforts which were fairly involved since the solution is quite tricky, then please formulate it as I have. As for the mathematics, it is always true that an explicit solution is best.
-Ilan Vardi<|endoftext|>
TITLE: Important results that use infinite-dimensional manifolds?
QUESTION [29 upvotes]: Are Banach manifolds (or other types of infinite-dimensional manifolds) just curiosities, or have they been utilized to prove some interesting/important results? Where do they turn up? Important examples?

REPLY [5 votes]: David Ebin and Jerry Marsden proved in a 1970 paper in the Annals that the Euler and Navier-Stokes equations are well-posed for short amounts of time.  This result uses the geometry of infinite-dimensional manifolds in a very fundamental way: the configuration space of an incompressible fluid in a container $M$ is the group of volume-preserving diffeomorphisms $\mathrm{Diff}_{\mathrm{vol}}(M)$ of $M$ and this is of course an infinite-dimensional manifold.  The proof then relies on the fact that the Euler equations define a Hamiltonian vector field on this space, and by carefully unwinding the definitions and putting everything in the right Sobolev space you can essentially use existence and uniqueness theorems for ODEs to prove the desired result.
I'm of course glossing over many details.  One of the problems, for instance, when trying to make this idea rigorous is that the convective term $\mathbf{u} \cdot \nabla \mathbf{u}$ in the Euler equations leads to derivative loss, but this can be surmounted by rewriting the equations in Lagrangian form.  If you want to do all this going back and forth in a proper way, you need a good understanding of the underlying geometry of $\mathrm{Diff}_{\mathrm{vol}}(M)$.
In my opinion, this is quite an amazing result, since it is both powerful and conceptually very clear.  There are essentially no hard estimates in the paper, just applications of the Sobolev embedding theorems, Hodge decompositions, properties of vector fields, ....  Global analysis at its best!<|endoftext|>
TITLE: A problem of Shimura and its relation to class field theory
QUESTION [27 upvotes]: In Chapter II.10 of The Map of My Life, Goro Shimura mentions a certain problem:

The second topic concerns a polynomial $F(x)$ with integer coefficients. Take
  $$
F(x) = x^3 + x^2 - 2x - 1,
$$
  for example. For an integer $n$, we consider the decomposition of $F(n)$ into the product of prime numbers. We can allow $n$ to be negative, but let us assume $n$ to be positive here. Thus
  $$
F(1) = -1, F(2) = 7, F(3) = 29, F(4) = 71, F(5) = 139,
$$
  $$
F(6) = 239, F(7) = 13 \cdot 29, F(8) = 13 \cdot 43, F(9) = 7 \cdot 113, \ldots
$$
  The prime numbers appearing as factors of $F(n)$ form a sequence
  $$
7, 13, 29, 43, 71, 113, 139, 239, \ldots
$$
  Now the question is: What are these prime numbers? In fact, we can prove that every such prime number $p$, excluding 7, has the property that $p+1$ or $p-1$ is divisible by 7. Conversely, every such prime number appears as a factor $F(n)$ for some positive integer $n$.
While learning class field theory on my own, I realized that the main theorem in easier cases can be formulated in terms of prime factors of $F(n)$ as above, and at that moment I was very happy. The polynomial $F$ cannot be taken arbitrarily. Actually, the equation $F(x) = 0$ has $2 \cos (2\pi/7)$ as a root, and that fact is essential. If $F(x) = x^2 - a$ with an integer $a$, the problem can be solved by the quadratic reciprocity law. In fact, my later work on the so-called complex multiplication is closely connected with this question of finding $F$ for which the sequence corresponding to [the sequence of integers above] can be determined. 

My question concerns the arguments in the final excerpted paragraph. Namely, how exactly are easier cases of the Main Theorem of Class Field Theory related to this problem? Additionally, how does complex multiplication help in finding a polynomial $F$ given a sequence of primes as above?

REPLY [3 votes]: I just saw this and would like to add that Shimura's example is the smallest case ($a=-1$) of Shank's simplest cubic $$F(x)=x^3-ax^2-(a+3)x-1$$ with discriminat $P^2$, $P=a^2+3a+9$. When $P$ is prime say eg. $P=13$ ($a=1$) also noted above, $19$, $37$, $79$, $97$, $139$, $163$,...,then the primes dividing $F(m)$ for some $m$ are again $P$ and primes $q$ which are cubic residues mod $P$. The Galois group is again $\mathbf Z/3$ so the proof above still works -- An explicit way to see $\mathbf Z/3$ is to verify the relation  $F(x)=-(1+x)^3 F(g(x))$ where $g(x)=-1/(x+1)$ is of order $3$ in $\mathrm{PSL2}(\mathbf Z)$, so the roots are $\alpha$, $g(\alpha)$, $g^2(\alpha)$ and hence must be all real.<|endoftext|>
TITLE: Moduli spaces of coherent sheaves on K3s
QUESTION [5 upvotes]: Reading 2007 paper A tour of theta dualities on moduli spaces of sheaves by Alina Marian and Dragos Oprea.

Why is any moduli space of coherent sheaves on a K3 surface deformation equivalent to a moduli space of sheaves on an elliptic K3?

(The authors consider a space of "Gieseker H-semistable sheaves", if that is important)

REPLY [9 votes]: This follows from a result of Yoshioka. In Theorem 8.1 of this paper Yoshioka showed that every moduli space of coherent sheaves on a K3 surface $X$ is deformation equivalent to an appropriate Hilbert scheme of points of $X$. Since every K3 is deformation equivalent to an elliptic K3 it follows that their Hilbert schemes are deformation equivalent and so you get the statement that you wanted.<|endoftext|>
TITLE: Linear equation with primes
QUESTION [10 upvotes]: Is there an integer $n$ with an infinite number of representations of the form
$n=2q-p$, where $p$ and $q$ are both primes?
Given a positive integer $k>1$, I would like to know for which (if any) integers $n$ the linear equation $q-kp=n$ admits an infinite number of solutions, where $p$ and $q$ are primes.
(I'm not including $k=1$ because it reduces to well know open problems, $k=1$ and $n=2$ would be the twin primes conjecture)
The density of the prime numbers implies that at least there are integers $n$ with an arbitrarily large number of representations.

REPLY [15 votes]: Assuming the Hardy-Littlewood prime tuples conjecture, any n which is coprime to k will have infinitely many representations of the form q-kp.
Assuming the Elliot-Halberstam conjecture, the work of Goldston-Pintz-Yildirim on prime gaps (which, among other things, shows infinitely many solutions to 0 < q-p <= 16) should also imply the existence of some n with infinitely many representations of the form q-kp for each k (and with a reasonable upper bound on n).  [UPDATE, much later: Now that I understand the Goldston-Pintz-Yildirim argument much better, I retract this claim; the GPY argument (combined with the more recent methods of Zhang) would be able to produce infinitely many $m$ such that at least two of $m + h_i$ and $km + h'_i$ are prime for some suitably admissible $h_i$ and $h'_i$, but this does not quite show that $q-kp$ is bounded for infinitely many $p,q$, because the two primes produced by GPY could both be of the form $m+h_i$ or both of the form $km+h'_i$.  So it is actually quite an interesting open question as to whether some modification of the GPY+Zhang methods could give a result of this form.]
Unconditionally, I doubt one can say very much with current technology.  For any N, one can use the circle method to show that almost all numbers of size o(N) coprime to k have roughly the expected number of representations of the form q-kp with q,p = O(N).  However we cannot yet rule out the (very unlikely) possibility that as N increases, the small set of exceptional integers with no representations covers all the small numbers, and eventually grows to encompass all numbers as N goes to infinity.<|endoftext|>
TITLE: In set theories where Continuum Hypothesis is false, what are the new sets?
QUESTION [27 upvotes]: So, say we are working with non-CH mathematics. This means, AFAIK, that there is at least one set $S$ in our non-CH mathematics, whose cardinality is intermediate between $|\mathbb{N}|$ (card. of naturals) and $|\mathbb{R}|=2^\mathbb{N}$, the continuum.
Question: what kind of objects would we find in this set $S$?
Also: is this mathematics radically different from the one where CH holds?
Specifically, are there results that are used in everyday math , at a relatively introductory level, which do not hold on our non-CH math.?. What results that we find in everyday math would not hold in our new math?  Would there, e.g., still exist non-measurable sets? Maybe more specifically: what results depend on the CH?

REPLY [2 votes]: A very natural question for algebraists is the Whitehead problem: is there an abelian group A which is not free but for which Ext^1(A, Z) = 0? Shelah showed that this natural problem was independent of ZFC. If you assume CH is false and Martin's axiom is true, then such a group exists. On the other hand, V=L, which implies CH, also implies such a group does not exist.<|endoftext|>
TITLE: When are epimorphisms of algebraic objects surjective?
QUESTION [13 upvotes]: Let $C$ be the category of $\tau$-algebras for some type $\tau$. Consider the statements:

Every monomorphism is regular.
Every epimorphism in $C$ is surjective.

It is easy to see that 1. implies 2. what about the converse?

REPLY [13 votes]: Update: the following exchange appeared on the categories mailing list several years ago: http://article.gmane.org/gmane.science.mathematics.categories/3094. Walter Tholen's response strongly suggests that the answer to Martin's question is that the converse does not hold, although I don't have access to the four-author article he cites as reference. It's probably worth a look though, and if I learn anything more I'll post another update.
Second update: My surmise was correct. Walter Tholen kindly emailed to me the relevant two pages (pp. 88-89) of the four-author paper

E.W. Kiss, L. Marki, P. Prohle, W. Tholen: Studia Sci. Math. Hungaricum 18
(1983) 79-141.

where the following example is given on page 89: in the category of semigroups with zero such that all 4-fold products are zero, all epimorphisms are surjective but not all monos are regular [a specific nonregular mono is described]. (I can forward this email if you write to me at topological dot musings at gmail dot com.)
Assuming amalgamated products exist (as they do in categories of algebras of a Lawvere theory), a mono $i: A \rightarrowtail B$ is regular if it is the equalizer of the pair of canonical maps from $B$ to the amalgamated product $B *_A B$ (i.e., the coprojections of the pushout of $i$ with itself, aka the cokernel pair of $i$). The equalizer of the cokernel pair defines a closure operator on the lattice of subalgebras $\mathsf{Sub}(B)$, called the dominion operator $\mathsf{Dom}_B$. So to prove a subalgebra is not regular is to show that it is not $\mathsf{Dom}$-closed. The key technical result needed to prove the claim above is Isbell's Zig-Zag theorem (given in his paper Epimorphisms and Dominions in the 1965 La Jolla conference proceedings on categorical algebra), as recalled in Peter M. Higgins. "A short proof of Isbell's Zigzag Theorem." Pacific J. Math. 144(1):47–50 (1990), which gives a precise and useful criterion for an element to belong to the dominion (i.e., the $\mathsf{Dom}$-closure) of a subalgebra.
Hope this helps. I am voting up your question, Martin, since it's rather nontrivial!<|endoftext|>
TITLE: Choosing a fast computer algebra system that works in characteristic p?
QUESTION [8 upvotes]: Hi all,
I want to compute in $\mathbb{F}_q (x)((y))$ i.e. a Laurent series ring over the rational functions over $\mathbb{F}_q$. The computations are fairly basic, but they involve raising to the qth power a lot. I thought that this would be easy (I thought that it will merely shirt powers around), so I tried it in SAGE. I have to say that I am highly impressed with the ease of programming in SAGE, but I think it is too big (and slow) for the calculation I need (I know that SAGE has lots of components (PARI, GAP, etc.) some of them may be what I need).
So I wanted to ask the people who have more experience then me for a recommendation. Which algebra system is good at Laurent series over rational function fields in char p if you need to do a lot of raising to the qth power. 
~AP

REPLY [3 votes]: My personal experience is a few years old, but I don't think things have changed much. Sage is (or actually, was) more about ease of use then about performance. The only three CAS's you want to consider are

Singular (Macaulay 2 uses Singular's engine)
Cocoa.
Magma.

Back then the fastest of the bunch was Magma, but not by much. Regarding ease of use, it was a tie between Macaulay 2 and Magma. 
And now to some criticism: I never looked at Magma's code (proprietary), but I did look at both Singular and Cocoa. None of them uses SSE/GPGPU, which could probably give you an acceleration factor of 10-100.<|endoftext|>
TITLE: Help me understand boundary terms in actions over nontrivial manifolds
QUESTION [5 upvotes]: So I have this manifold $M$, along with a metric $g_{\mu\nu}(x)$ and metric-compatible covariant derivative $\nabla_\mu$ (which is not necessarily the one corresponding to the Levi--Civita connection).
When dealing with the action principle, we can ignore boundary terms. My question is, which of the following is a boundary term? Let $g(x) = \mathrm{det}(g_{\mu\nu}(x))$.
$$\int_M \partial_\mu A^\mu \sqrt{|g(x)|} \, \mathrm{d} x$$
or
$$\int_M \nabla_\mu A^\mu \sqrt{|g(x)|} \, \mathrm{d} x \ ?$$
(Or both, or neither? Maybe the latter, but only when $\nabla$ is Levi--Civita?)
I guess my question reduces to, how does Stokes's theorem work for general covariant derivatives, and with the natural volume element $\sqrt{|g(x)|} \, \mathrm{d} x$?

REPLY [3 votes]: It is worth reiterating that there is only one Stokes theorem, so an integral is a "boundary term" if and only if it can be written in local co-ordinates as $\int \partial_\mu(B^\mu)\,dx$. If there is any factor outside the partial derivative, then the Stokes theorem cannot be applied.
So the determinant of the metric must be inside the partial derivative. When you expand out the derivative, Christoffel symbols appear, allowing the integrand to be written in terms of the connection instead.<|endoftext|>
TITLE: does the "convolution theorem" apply to weaker algebraic structures?
QUESTION [23 upvotes]: The Convolution Theorem is often exploited to compute the convolution of two sequences efficiently: take the (discrete) Fourier transform of each sequence, multiply them, and then perform the inverse transform on the result.
The same thing can be done for convolutions in the quotient ring Z/pZ via the analogous Number Theoretic Transform.
Does this procedure generalize to other algebraic structures?  Arbitrary rings?  Semirings?  Fast convolutions over the semiring (min,+) would be particularly useful.
I'm led to suspect this because both sorting and FFT can be computed using the same butterfly-like network with simple operations like "min", "max", "+", and "*" at the nodes of the network, and sorting can be thought of as a kind of convolution.

REPLY [3 votes]: I stumbled onto this question after encountering the problem in Bayesian inference (specifically, performing max-product inference on random variables where $M = L + R$ where $L$ and $R$ (or, an an alternate form of the problem, $M$ and $L$) have known discrete distributions and the distribution on $M$ (or $R$, in the alternative problem) is sought. The probability mass function (PMF) on $M$ that is achieved by maxing out $L$ and $R$ will be the result of the max-convolution between the PMFs of $L$ and $R$.
After playing around with it a bit, I noticed that you can rewrite the problem by generating $u^{(m)}$ for each index $m$ of the result (where $u^{(m)}[\ell] =
L[\ell] R[{m-\ell}]$). When the vectors are nonnegative (in my case, this was true because they are PMFs), then you can perform the $\max_\ell u^{(m)}_\ell$ with the Chebyshev norm:
$$
M[m] = \max_\ell u^{(m)}_\ell \\
= \lim_{p \to \infty} \| u^{(m)} \|_p \\
 = \lim_{p \to \infty} {\left( \sum_\ell {\left( u^{(m)}[\ell] \right)}^{p} \right)}^{\frac{1}{p}}
$$
This in turn can be approximated by substituting the $p^*$-norm, where $p^*$ is a sufficiently large constant (essentially, this exploits the fact that the $p^*$-norm does have inverse operations, whereas the $\max$ does not, but the $p^*$-norm can still approximate the $\max$):
$$
  \approx {\left( \sum_\ell {\left( u^{(m)}[\ell] \right)}^{p^*} \right)}^{\frac{1}{p^*}}\\
  = {\left( \sum_\ell {L[\ell]}^{p^*} ~ {R[{m-\ell}]}^{p^*} \right)}^{\frac{1}{p^*}}\\
  = {\left( \sum_\ell {\left(L^{p^*}\right)}[\ell] ~ {\left(R^{p^*}\right)}[{m-\ell}] \right)}^{\frac{1}{p^*}}\\
  = {\left( L^{p^*} ~*~ R^{p^*} \right)}^{\frac{1}{p^*}}[m]
$$
where $*$ is standard convolution, which can be performed in $n \log(n)$ via FFT. A short paper illustrating the approximation for large-scale probabilistic inference problems is in press at the Journal of Computational Biology (Serang 2015 arXiv preprint). 
Afterward a Ph.D. student, Julianus Pfeuffer, and I hacked out a preliminary bound on the absolute error ($p^*$-norm approximations of the Chebyshev norm are poor when $p^*$ is small, but on indices where the normalized result $\frac{M[m]}{\max_{m'} M[m']}$ is very small, large $p^*$ can be numerically unstable). Julianus and I worked out a modified method for cases when the dynamic range of the result $M$ is very large (if not, then the simple method from the first paper works fine). The modified method operates piecewise over $\log(\log(n))$ different $p^*$ values (including runtime constants, this amounts to $\leq 18$ FFTs even when $n$ is on the scale of particles in the observable universe, and those 18 FFTs can be done in parallel). (Pfeuffer and Serang arXiv link which has a link to a Python demonstration of the approach) The approach (and the Python demo code) generalize to tensors (i.e., numpy.arrays) by essentially combining the element-wise Frobenius norm with multidimensional FFT. 
Oliver Serang<|endoftext|>
TITLE: Is it true that, as $\Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other?
QUESTION [50 upvotes]: Is it true that, in the category of $\mathbb{Z}$-modules, $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$,  but it doesn't quite work since it is not clear how to map $S$.

REPLY [23 votes]: I give this problem each year in a problem-solving seminar. Here is the solution that I wrote up. I am using $f$ instead of $\varphi$ and $e_n$ instead of $x^n$.
Proof that if $f(e_k) = 0$ for each $k$ then $f$ is identically zero.
Let $x=(x_1,x_2,\dots)$. Since $2^n$
and $3^n$ are relatively prime, there are integers $a_n$ and $b_n$
for which $x_n=a_n2^n+b_n3^n$. Hence $f(x)=f(y)+f(z)$, where $y =                   
(2a_1, 4a_2, 8a_3,\dots)$ and $z=(3b_1,9b_2,27b_3,\dots)$. Now for
any $k\geq 1$ we have
$$ f(y)  =  f(2a_1,4a_2,\dots,2^{k-1}a_{k-1},0,0,                    
\dots) $$
$$ \qquad + f(0,0,\dots,0,2^ka_k,2^{k+1}a_{k+1},\dots) $$
$$ \qquad=  0+2^kf(0,0,\dots,0,a_k,2a_{k+1},4a_{k+2},\dots). $$
Hence $f(y)$ is divisible by $2^k$ for all $k\geq 1$, so
$f(y)=0$. Similarly $f(z)$ is divisible by $3^k$ for all $k\geq                     
1$, so $f(z)=0$. Hence $f(x)=0$.
Proof that $f(e_k) = 0$ for $k \gg 1$.
Now let $a_i=f(e_i)$. Define integers $0< n_1 <                                 
  n_2 <\cdots$ such that for all $k\geq 1$,
$$ \sum_{i=1}^k|a_i|2^{n_i} < \frac 12 2^{n_{k+1}}. $$
(Clearly this is possible --- once $n_1,\dots,n_k$ have been chosen,
simply choose $n_{k+1}$ sufficiently large.) Consider $x=(2^{n_1},                     
 2^{n_2}, \dots)$. Then
$$ f(x)  =  f(2^{n_1}e_1 + \cdots + 2^{n_k} e_k +2^{n_{k+1}}
 (e_{k+1}+2^{n_{k+2}-n_{k+1}}e_{k+2}+\cdots))$$ $$ \qquad= 
\sum_{i=1}^ka_i 2^{n_i}+2^{n_{k+1}}b_k, $$
where $b_k=f(e_{k+1}+2^{n_{k+2}-n_{k+1}}e_{k+2}+\cdots)$. Thus by the
triangle inequality,
$$\left| 2^{n_{k+1}}b_k\right| < \left| \sum_{i=1}^k a_i
   2^{n_i}\right| + |f(x)| $$ $$ \qquad < 
  \frac 12 2^{n_{k+1}} + |f(x)|. $$
Thus for sufficiently large $k$ we have $b_k=0$ [why?]. Since
$$ b_j - 2^{n_{j+2}-n_{j+1}}b_{j+1}=f(e_{j+1})\ \ \mbox{[why?]},                     
   $$
we have $f(e_k)=0$ for $k$ sufficiently large.<|endoftext|>
TITLE: Model category structure on Set without axiom of choice
QUESTION [15 upvotes]: There is a model category structure on Set in which the cofibrations are the monomorphisms, the fibrations are maps which are either epimorphisms or have empty domain, and the weak equivalences are the maps $f : X \rightarrow Y$ such that $X$ and $Y$ are both empty or both nonempty.
In order for the lifting axioms to hold we need the axiom of choice.  Suppose we want to avoid the axiom of choice.  One option seems to be to replace "epimorphism" with "map which has a section" everywhere.  Can we instead leave the definition of fibration unchanged and change the definition of cofibration?
Note that if $A$ is a cofibrant set in this hypothetical model structure then any surjection $X \rightarrow A$ has a section.  So it would be necessary that every set admits a surjection from a set $A$ of this type, which seems rather implausible to me.  Perhaps the notion of model category needs to be modified in a setting without the axiom of choice.
(Apologies if this question turns out to be meaningless or trivial; I have not thought about it much nor do I often try to avoid using the axiom of choice.)

Let me explain the motivation behind this question.  I am trying to get a better picture of what category theory without the axiom of choice looks like.
My rough understanding is that in the absence of the axiom of choice we should use anafunctors in place of functors in some if not all contexts.  An anafunctor from $C$ to $D$ is a span of functors $C \leftarrow E \rightarrow D$ in which the left leg $E \leftarrow C$ is a surjection on objects and fully faithful.  The point is that even under these conditions $E \rightarrow C$ may not have a section.  For instance, suppose $C$ is a category with all binary products.  There may not be a product functor $- \times -: C \times C \rightarrow C$, because we cannot simply choose a distinguished product of each pair of objects.  However, if we define $Prod(C)$ to be the category whose objects are diagrams of the shape $\bullet \leftarrow \bullet \rightarrow \bullet$ in $C$ which express the center object as the product of the outer two, there is a forgetful functor $Prod(C) \rightarrow C \times C$ remembering only the outer objects, and it is surjective on objects (because $C$ has binary products) and fully faithful.  Furthermore there is another forgetful functor $Prod(C) \rightarrow C$ which remembers only the center object.  Together these define a product anafunctor $C \times C \leftarrow Prod(C) \rightarrow C$.
"Classically" (:= under $AC$) the following paragraph holds:  There is a model category structure on $Cat$ in which the cofibrations are functors which are injective on objects, fibrations are the functors with the right lifting property w.r.t. the inclusion of an object into the contractible groupoid on two objects, and weak equivalences are equivalences of categories.  In particular the acyclic fibrations are the functors which are surjective on objects and fully faithful, exactly the functors we allow as left legs of anafunctors.  Since every category is fibrant in this model structure, we can view an anafunctor from $C$ to $D$ as a representative of an element of $RHom(C, D)$, i.e., a functor from a cofibrant replacement for $C$ to $D$.  Of course, every category is cofibrant too so for our cofibrant "replacement" we can just take $C$, and we learn that anafunctors from $C$ to $D$ are the same as functors when we consider both up to natural isomorphism (homotopy).
I would like to understand anafunctors as a kind of $RHom$ also without the axiom of choice.  But I cannot use the same definition of the model category structure, because the lifting axioms require $AC$. I would like to keep the same acyclic fibrations, since they appear in the definition of anafunctor, and I would like every object to be fibrant.  I cannot really imagine what a cofibrant replacement could look like in this model structure, but then I am not accustomed to working without $AC$.
My original question is related to this one via the functor which assigns to a set $S$ the codiscrete category on $S$, and it seems to contain the same kinds of difficulties.

REPLY [2 votes]: For any elementary topos $T$, there is a model category structure on $T$, whose cofibrations are the monomorphisms, and whose weak equivalences are the maps $X\to Y$ such that, either $Y$ is empty, either $X$ is non-empty (the fibrations are the split epimorphisms). In a topos, there exists an internal Hom (written here $\underline{Hom}$), as well as a subobject classifier $\Omega$: it comes with a map $t:1\to \Omega$ which is the universal subobject, in the sense that, for any object $X$ in $T$, pulling back along $t$ induces a bijection
$$\text{ { subobjects of $X$} }\simeq \ Hom(X,\Omega)$$
Using the universal property of $\Omega$, the diagonal $X\to X\times X$ defines a map $X\times X\to \Omega$, whence an embedding of $X$ into its power object: $X\to P(X)=\underline{Hom}(X,\Omega)$. This gives you injective resolutions. We thus have a functorial fibrant replacement $X\mapsto I(X)$, where $I(X)=X$ if $X$ is empty, and $I(X)=P(X)$ otherwise (note that an object of $T$ is fibrant if and only if it is either empty, either injective). It remains to factor maps into a trivial cofibration (resp. cofibration) followed by a fibration (resp. trivial fibration). Let $f:X\to Y$ be a map in $T$. Such a factorization for $f$ is given by the map $X\to F\times Y$ followed by the projection $F\times Y\to Y$, for $F=I(X)$ (resp. $F=P(X)$). Hence we get a model structure on $T$, while we never used the axiom of choice.
Edit: and now, some genuine non sense (not abstract):
This model category structure on $Set$ is rather degenerate though. It seems that, to get model categories in general (on $Cat$ or on simplicial sets), we should change the definition of a model category by asking for lifting properties only internally: given maps $i:A\to B$ and $p:X\to Y$, $i$ has the (internal) left lifting property with respect to $p$ is the map
$$Hom(B,X)\to Hom(X,A)\times_{Hom(Y,A)}Hom(Y,B)$$
is an epimorphism. In this way $Cat$ and $SSet$ remain model categories without the axiom of choice.
Edit: here, it seems I have been rather optimistic (see Mike comment below).
I was thinking about working above an arbitrary topos (instead of sets), but,
if we work externally in a setting without axiom of choice, then the lifting property suggested above is just the usual one, so that my remark seems to be (and is) silly.
But what follows still makes sense.
Note that it is easy to get a structure of category of fibrant objects (in the sense of Brown, Abstract homotopy theory and generalized sheaf cohomology, Trans. Amer. Math. Soc. 186  (1974), 419–458) on $Cat$, which is sufficient to have calculus of fractions (up to homotopy), and gives you the abstract tools to explain the good behaviour of anafunctors. As for the size problems when we avoid the axiom of choice, maybe it would be worth looking for an "internal notion of size" to get that the Hom's of any homotopy category of a model category are small in some sense (but I had never thought seriously about this before).<|endoftext|>
TITLE: A local-to-global principle for isogeny
QUESTION [10 upvotes]: If two elliptic curves over $\mathbb{Q}$ are $\mathbb{Q}_p$-isogneous for almost all primes $p$, then they are $\mathbb{Q}$-isogenous.  
This follows from the fact that they have the same number of $\mathbb{F}_p$-points for almost all $p$, hence their $L$-functions have the same local factors at all these $p$, therefore a combination of "multiplicity one" and Faltings' isogeny theorem implies that they are $\mathbb{Q}$-isogenous.  Correct me if I'm wrong.
Here $\mathbb{Q}$ can be replaced by any number field $k$.
Question :  Does the same argument work for any two abelian varieties $A$, $B$ over $k$ ?  It should, since $H^i$ is $\wedge^i H^1$ for $i>0$.
If so, this explains why Poonen's abelian surfaces $A,B$  (everywhere locally isomorphic but not isomorphic) are $\mathbb{Q}$-isogenous.

REPLY [7 votes]: A propos Dimitrov's answer, one should also mention two beautiful papers of Masser and Wustholz. 
MR1037140 Masser, D. W.(1-MI); Wüstholz, G.(CH-ETHZ); Estimating isogenies on elliptic curves. Invent. Math. 100 (1990), no. 1, 1–24.
MR1217345 Masser, David(1-MI); Wüstholz, Gisbert(CH-ETHZ); Isogeny estimates for abelian varieties, and finiteness theorems. Ann. of Math. (2) 137 (1993), no. 3, 459–472. 
Given a number field $K$ and abelian varieies $A/K$ and $A'/K$ that are isogenous over $K$, they give an explicit value $M(K,A,A')$ such that there is a $K$-isogeny $A\to A'$ of degree less then $M$. (The first article treats the case of elliptic curves.) Such estimates have many applications, including a proof of the hardest implication of the theorem stated in Keller's answer (by an argument appearing in Tate's original paper). As in the articles by the Chudnovskys, the tools used in the proof are those of Diophantine approxmation and transcendence theory.<|endoftext|>
TITLE: effective teaching
QUESTION [46 upvotes]: Eric Mazur has a wonderful video describing how physics is taught at many universities and his description applies word for word to the way I learned mathematics and the way it is still being taught, i.e. professors lecture to students and sketch some proofs. Suffice it to say I'm not a fan of the current methods and I don't think it would be too far from the truth to say that I do all the actual learning outside the classroom. Has anyone tried anything different and seen any difference in student understanding and comprehension in graduate or undergraduate courses?
Some background motivation: I'm a TA and my current method of doing things is to just write some problems on the board and then go through their solutions. This is fine and it's what the students expect but sometimes I feel guilty because I'm just teaching them problem/solution patterns and reinforcing all the bad stereotypes about what mathematics is instead of showing them the underlying conceptual tapestry and helping them rethink their attitudes toward mathematics. It's kinda like the old saying “Give a man a fish; you have fed him for today.  Teach a man to fish; and you have fed him for a lifetime”. So basically I throw a bunch of fish at the students hoping it will feed them for the semester.

REPLY [3 votes]: This very stimulating presentation on teaching math effectively came out in May 2010 and addresses some of the issues presented in the other answers:
Dan Meyer: Math class needs a makeover
Although applied to high school math, at least some aspects of the technique could be incorporated, if only in a few sessions, into advanced classes. Comments to the video by educators and students provide some feedback on the technique.
Another potential method for revamping math classes for the 21'st century (maybe start viewing at time stamp 6:50):
Salman Khan: Let's use video to reinvent education
And always a good read and reminder: V.I. Arnold, On teaching mathematics<|endoftext|>
TITLE: Counting lattice points on an n-simplex
QUESTION [5 upvotes]: Imagine an n-simplex, the solution set for the expression: $a_1$*$x_1$ + $a_2$*$x_2$ + ... + $a_n$*$x_n$ = S, where: 

$a_1$ through $a_n$ are positive bounded integers
$x_1$ through $x_n$ are positive bounded real numbers  
'S' is the sum of the expression 

This n-simplex therefore has a single vertex on the origin, as well as a single vertex on each axis at some arbitrary (strictly positive) distance from the origin. 
What is the lattice integer-point count?  
Can one use Ehrhardt polynomials to compute the integer point count for the n-simplex, perhaps under the restriction that we have vertices strictly at integer coordinates?  

From "Geometry for N-Dimensional Graphics" (by Andrew J. Hanson, CS Dept., Indiana University) we know that the oriented volume for the n-simplex with vertices {$v_1$, ..., $v_n$}, or {$a_1$*$x_1$, ..., $a_n$*$x_n$} is: 

$V_n$ = $\dfrac{1}{n!}$ * det([($v_1$-$v_0$), ..., ($v_n$-$v_0$)])
(Problems writing LaTeX for matrices here, please see terms as column vectors to obtain square matrix.)

Previous formulation of question: Imagine an expression of the form: $a_1$*$x_1$ + $a_2$*$x_2$ + ... + $a_n$*$x_n$ = S, where: 

$a_1$ through $a_n$ are positive bounded integers
$x_1$ through $x_n$ are positive bounded real numbers  
'S' is the sum of the expression 

Can we say anything about the maximum value of 'S' (for a given $x_1$ through $x_n$) below which there is only one solution for positive integer coefficients $a_1$ through $a_n$?  For example, given the expression $a_1$*98 + $a_2$*99 = 'S', where coefficients $a_1$ and $a_2$ = [1 through 100], one finds that you can always exactly recover the original $a_1$ and $a_2$ if 'S' < 9899.  Is there an analytical or more elegant method for obtaining that bound?   
[Above such a bound, is there an efficient way to obtain all possible sets of integers $a_1$ through $a_n$ that satisfy the relation for a given 'S'? Can the LLL or PSLQ algorithms be used?]  <--  This seems to be a restricted/special case of the subset-sum problem, so existing dynamic programming algorithms would work.  Can one do better here?

REPLY [4 votes]: For a polynomial-time method of counting integer lattice points for the n-simplex (with fixed dimension):
Review article - Crites, A., Goff, M., Korson, M., Patrolia, L., Wolcott, L. "Counting Lattice Points in Polyhedra." 
Available here with references for Barvinok's 1994 & 1999 algorithms - http://www.math.washington.edu/~thomas/teaching/m583_s2008_web/Barvinok.pdf
For an implementation of Barvinok's algorithm, see J.A. De Loera's LattE program (hosted at UC Davis):
http://www.math.ucdavis.edu/~latte/group.htm<|endoftext|>
TITLE: distribution of non-solvable group orders
QUESTION [6 upvotes]: let $M$ be the set of natural numbers such that there is a group of this order, which is not solvable. what is the minimal distance $D$ of two numbers in $M$?
the examples $660$ and $672$ show $D \leq 12$. the famous theorem of feit-thompson implies $D>1$.

REPLY [8 votes]: The encyclopedia of integer sequences gives the following criteria for a number being a non-solvable number:
A positive integer n is a non-solvable number if and only if it is a multiple of any of the following numbers: 
a) $2^p (2^{2p}-1)$, p any prime. 
b) $3^p (3^{2p}-1)/2$, p odd prime. 
c) $p(p^2-1)/2$, p prime greater than 3 and congruent to 2 or 3 mod 5
d) $2^4 3^3 13$
e) $2^{2p}(2^{2p}+1)(2^p-1)$, p odd prime.
It's not hard to check that all these orders are divisible by 4, so there will never be two non-solvable numbers differing by less than 4.
In fact, they're all divisible by 12 except those generated by (e), which are all divisible by 20.
So, for example, all numbers of the form 29120n are nonsolvable, since $29120 = 2^6 (2^6+1) (2^3-1)$.  And all numbers of the form 25308n are nonsolvable, since $25308 = 37(37^2-1)/2$.  We have the prime factorizations $25308 = 2^2 3^2 19^1 37^1$ and $29120 = 2^6 5^1 7^1 13^1$.  
So we just need to find multiples of 29120 and 25308 which differ by 4.  From the Euclidean algorithm, $29120 \cdot 2483 = 72304960$ and $25308 \cdot 2857 = 72304956$.
I haven't searched exhaustively, so it's possible that there's a smaller pair of non-solvable numbers that differ by 4; I chose 25308 and 29120 by just looking at the prime factorizations of the numbers generated by (a) through (e) until I found two that had gcd 4.<|endoftext|>
TITLE: p-adic noninvariance of dimension
QUESTION [5 upvotes]: Let $p$ be a prime number. Let $n,m \geq 1$ be such that the topological spaces $\mathbb{Q}_p^n$ and $\mathbb{Q}_p^m$ are homeomorphic. Can we conclude $n=m$?
For $\mathbb{Z}_p$ it's false: In fact, Brouwer's theorem implies that $\mathbb{Z}_p$ is homeomorphic to the Cantor set $C$, which of course satisfies $C^n \cong C^m$ for all $n,m$.

REPLY [10 votes]: $\mathbb{Q}_p$ is homeomorphic to a countable direct sum of copies of the Cantor set $C$.  Indeed, because the valuation is discrete, for each $n \geq 1$ the "annulus"
$A_n =$ {$x \in \mathbb{Q}_p \ | \ p^{n-1} < ||x|| \leq p^{n}$}
is closed and homeomorphic to the Cantor set $C$.  (Take of course $A_0 = \mathbb{Z}_p$.)
Since as you observed above, $C \times C \cong C$, it follows that $\mathbb{Q}_p^n$ and 
$\mathbb{Q}_p^m$ are homeomorphic for all $m, \ n \in \mathbb{Z}^+$ (and the homeomorphism type is independent of $p$).<|endoftext|>
TITLE: What does the typical non-solvable group look like?
QUESTION [8 upvotes]: According to a result of Higman and Sims (which I learned about in this paper of Poonen's) the typical p-group is 3-step nilpotent of a particular form.  In particular the typical group is a 3-step nilpotent 2-group of a particular form.  By typical here I mean that eventually the number of these groups dominate.
Is anything known about what the typical non-solvable group looks like?  Probably some sort of modification of PSL_2(F_p)?

REPLY [3 votes]: Even if you are asking about a "typical" group of order $\leq n$, the answer is nobody knows, so this $A_5$ idea for a much more delicate question is little more than a pure speculation.  If I understand the state of art correctly, it is completely open whether solvable groups are a majority.  The data, of course, supports a stronger conjecture: that asymptotically almost all groups are $2$-groups.  What is known now, is a number of good asymptotic estimates, notably a Pyber's paper in the Annals (1993), getting a tight asymptotic for the log of the number of groups of order $\le n$ (he gets an upper bound matching the Higman-Sims lower bound). For more on this, read up 
this terrific recent monograph.<|endoftext|>
TITLE: How unhelpful is graph minors theorem?
QUESTION [7 upvotes]: A very interesting Robertson-Seymour (graphs minors) theorem says:

Any infinite collection of graphs $C$ with the property that if $G\in C $ then its minors also are has the form $\{$graphs $G$ that don't contain any $E_i\}$ for some finite collection $E = \{E_i\}$.

So, the theorem says that you could create a list of forbidden minors to find out if the graph is torically embeddable, but this doesn't help much, since the list is both not fully known and large.
I wonder whether the above difficulty is because 

it is indeed hard to test this property of a graph
the theorem does turn easily testable properties into long lists
it is not known how to effectively turn easily testable properties into lists

Here's the formal question:

Consider a polynomial algorithm $P$ that returns a yes/no question given a graph as an input and which always returns yes for minors of any graph for which it returns yes. There exists $E$, the exceptional list of a collection defined by $P$. What is known about the computability of the map $P\mapsto E$?

REPLY [3 votes]: To answer some of your questions:
1. Yes, testing such properties is usually hard. For instance, before the graph minors theorem we had properties for which no algorithm was known at all! (Maybe a recursively enumerable algorithm was known, I don't remember.) After the graph-minors theorem, such properties became testable in polynomial time! Now that's a big jump from no algorithm known to polynomial time. (If I remember correctly, the polynomial is like O(n log n), which is almost linear time.)
As for 2. and 3., I don't know any property which we can easily test, for which we don't know the forbidden minors. I'd like to know such properties, if they are known. It seems to me that if we have an algorithm for easily testing a property, we really understand the property, and therefore should be able to come up with a list of forbidden minors somehow. Of course, this is just a feeling.
EDIT: I've been corrected in the comments. Please read Gil Kalai and David Eppstein's comments.<|endoftext|>
TITLE: Can a topos ever be an abelian category?
QUESTION [19 upvotes]: Can a topos ever be a nontrivial abelian category?  If not, where does the contradiction lie?  If a topos can be an abelian category, can you give a (notrivial!) example?

REPLY [41 votes]: No.  In fact no nontrivial cartesian closed category can have a zero object 0 (one which is both initial and final), as then for any X, 0 = 0 × X = X.  (The first equality uses the fact that – × X commutes with colimits and in particular the empty colimit, and the second holds because 0 is also the final object.)<|endoftext|>
TITLE: Infinity de Rham quasi-isomorphism
QUESTION [14 upvotes]: This question is similar to Do chains and cochains know the same thing about the manifold? in the sence that both deal with a natural "comparison" quasi-isomorphism that does not preserve the ring structure.
Let $M$ be a smooth manifold. There is a natural comparison map $Comp$ from the differential forms on $M$ to the smooth singular cochains of $M$ (i.e. the linear dual of the vector space spanned by smooth singular simplices). It is defined as follows: take a form $\omega$ of degree $p$ and set $Comp(\omega)$ to be the cochain $\sigma\mapsto \int_\triangle \sigma^*\omega$ where $\triangle$ is the standard $p$-dimensional simplex and $\sigma:\triangle\to M$ is a smooth singular simplex.
$Comp$ is a map of complexes (Stokes' theorem) and moreover, a quasi-isomorphism (the de Rham theorem). But as simple examples show, it does not preserve the ring structure. However it is natural to ask whether the ring structures, up to quasi-isomorphism, of the differential forms and of the cochains contain the same information about $M$. This translates into the following questions.

Can $Comp$ be completed to a morphism of $A_\infty$-algebras?
If the answer to 1. is positive (it presumably is), what about the $E_\infty$ case?

These questions also have natural rational versions. Namely, we can take an arbitrary polyhedron $X$ instead of $M$ and consider Sullivan's $\mathbf{Q}$-polynomial forms. There is a comparison quasi-isomorphism similar to the one above that will go from the $\mathbf{Q}$-polynomial forms of $X$ to the piecewise linear $\mathbf{Q}$-cochains. Can it be completed to a map of $A_\infty$ or $E_\infty$ algebras?

REPLY [7 votes]: This theorem was proven in 1977 in "V. K. A. M. Gugenheim, On Chen’s iterated integrals, Illinois J. Math. Volume 21, Issue 3 (1977), 703–715."<|endoftext|>
TITLE: Using ends to construct categorical fixed points
QUESTION [13 upvotes]: Under advice from Toby Bartels, I am posting this question here; it falls under the general heading of constructing data types categorically as fixed points of functors.
The first question I have is a warm-up. There's a way to interpret a natural number n in any cartesian closed category $C$, as a dinatural transformation of the form
$$c^c \to c^c$$
which intuitively takes an element $f\colon c \to c$ to its $n$-th iterate $f^{(n)}\colon c \to c$. One may hope that if $C$ is "nice", then every such dinatural transformation will be of this form, or better still, that the end
$$\int_{c: C} (c^c)^{c^c}$$
(assuming it exists) behaves as a natural numbers object in $C$. So first, I am interested in what "nice" might mean here: what are some general conditions on $C$ that ensure we can construct a natural numbers object as an end in this way?
Second, this end can be rewritten as
$$\int_{c: C} c^{c^{1 + c}}$$
provided that $C$ has coproducts, and the assertion that this behaves as a natural numbers object is equivalent to saying it is an initial algebra for the endofunctor $F(c) = 1 + c$ (that's algebra for an endofunctor, not for a monad), making it a "fixed point" of $F$ by a famous old result of Lambek.
This suggests a second more general question: given an endofunctor $F: C \to C$ on cartesian closed $C$ with a strength (essentially, a structure of $C$-enrichment), I want to know what "nice" conditions on $C$ and/or $F$ guarantee that the end
$$e = \int_{c: C} c^{c^{F(c)}}$$
if it exists, is an initial $F$-algebra. It's not hard to write down an an $F$-algebra structure for this end $e$, and show that it is weakly initial, i.e., show that if $x$ is any $F$-algebra, then there at least exists an $F$-algebra map $e \to x$. The issue then is over the uniqueness of this map, or rather what nice conditions would guarantee that.
Discussion of specific cases like PER models would be alright, but I'd probably be a lot more excited if it led to consideration of more general abstract conditions on $C$ or $F$.

REPLY [2 votes]: I asked Marcelo Fiore about this last week, and he instantly recognized the type of problem --- right down to the fact that one typically has existence of a map from the supposedly-initial algebra to an arbitrary given algebra, but not necessarily uniqueness.  This answer consists of my best recollections of what Marcelo said.
The key phrase is "parametric polymorphism".  In the general setting of polymorphism, one has the existence property just mentioned, but not necessarily uniqueness.  In order to get uniqueness, one adds the extra ingredient of parametricity.  If I understand correctly (and I'm not confident that I do), adding in parametricity is something like passing from a category of objects and maps to the resulting category of objects and relations.  
One paper that Marcelo found was 
Categorical data types in parametric polymorphism by Ryu Hasegawa.  I think it wasn't quite what he was looking for, in that it doesn't contain any ends.  There are also relevant papers by both Freyd and Rosolini, apparently.
Sorry this is a bit garbled.<|endoftext|>
TITLE: "Every scheme as a sheaf" references?
QUESTION [16 upvotes]: I have sometimes hard time reading papers that are written in the language of schemes being replaced by the functors they represent (I have especially homotopy scheme theory in mind). 
I think the topic is connected to topoi and Grothendieck topologies, but for now I'm looking for something simple, just the working overview of the language of representable functors, what is a scheme, etc.

REPLY [3 votes]: You may also find the notes "Introduction to Functorial Algebraic Geometry" useful.  They are based on a course given by Groethendieck - and can be found here.  Unfortunately they're a little rough around the edges sometimes and slightly dizzying to read because of the scanning.<|endoftext|>
TITLE: Formulas for vector fields on Grassmannians?
QUESTION [8 upvotes]: The Wikipedia article on (real) Grassmannians gives a simple argument that the Euler characteristic satisfies a recurrence relation $$\chi G_{n,r} = \chi G_{n-1,r-1} + (-1)^r \chi G_{n-1,r}$$.  This implies that the euler characteristic is zero if and only if $n$ even and $r$ odd.  So $\chi G_{6,3} = 0$, for example.  Basic obstruction theory on manifolds tells us that if $\chi M=0$ then there is a non-vanishing vector field on $M$.  

Does anyone have any simple, explicit examples of such vector fields?  Let's say $G_{n,1}$ for $n$ even does not count.  Not to say those aren't interesting examples -- I'd love solutions that simple. But it's not immediately clear how they can be generalized.

REPLY [11 votes]: Identify $\mathbb R^2$ with $\mathbb C$ and consider the $S^1$ action on $\mathbb R^{2n} \simeq \mathbb C^n$ induced by cordinatewise complex multiplication. These of course lead to the trivial examples on $G_{2n,1}$. For $n$ even and $r$ odd the very same examples do the trick. One has just to observe that these $S^1$ actions have no invariant odd dimensional subspaces, and therefore induce $S^1$-actions without fixed points on $G_{n,r}$.

REPLY [10 votes]: Pick an even dimensional real vector space $V$ of dimension $n$ and fix a symplectic form $\omega$ on $V$. Look at it as a map $\omega:V\otimes V\to\mathbb R$ by extending it from $\Lambda^2V$ to $V\otimes V$ as zero on the symmetric part. Fix also an inner product $\langle\mathord-,\mathord-\rangle$ in $V$.
Let $k$ be odd and such that $1\leq k\leq n$.
Let $W\subseteq V$ be an $k$-dimensional subspace, and let $W^\perp$ and $W^{\perp\omega}$ be the subspaces orthogonal to $W$ with respect to $\langle\mathord-,\mathord-\rangle$ and to $\omega$, respectively. We know that $\dim W^\perp=\dim W^{\perp\omega}=\dim V-\dim W$. 
The restriction $\omega|_{W\otimes W^\perp}:W\otimes W^\perp\to\mathbb R$, which I will write for simplicity just $\omega_W$, is not zero. Indeed, if it were zero, we would have that $W^\perp$ is contained in $W^{\perp\omega}$, so in fact these two orthogonal subspaces would be equal, and in consequence we would have that $W\cap W^{\perp\omega}=W\cap W^\perp=0$. This would tell us that $W$ is in fact a symplectic subspace of $V$, which is absurd because it is odd dimensional.
Now $\omega_W$ is an element of $\hom(W\otimes W^\perp,\mathbb R)$, which identifies canonically with $\hom(W,\hom(W^\perp,\mathbb R))$. The inner product $\langle\mathord-,\mathord-\rangle$ restricts to an inner product on $W^\perp$ which allows us to identify canonically (because the inner product is fixed!) $\hom(W^\perp,\mathbb R)$ with $W^\perp$. After all these identifications, we have a non zero vector $\omega_W$ in $\hom(W,W^\perp)$.
Now, as explained in an answer to a MO question,  $\hom(W,W^\perp)$ parametrizes a neighborhood of $W$ in $G(n,k)$, so it also can be identified with the tangent space to $G(n,k)$ at $W$. 
We thus see that the rule $W\mapsto \omega_W$ gives a non-zero tangent vector field.<|endoftext|>
TITLE: What practical applications does set theory have?
QUESTION [76 upvotes]: I am a non-mathematician. I'm reading up on set theory. It's fascinating, but I wonder if it's found any 'real-world' applications yet. For instance, in high school when we were learning the properties of i, a lot of the kids wondered what it was used for. The teacher responded that it was used to describe the properties of electricity in circuits. So is there a similar practical app of set theory? Something we wouldn't be able to do or build without set theory?

Edit: Actually, I'm asking about the practicality of the knowledge of the properties of infinite sets, and their cardinality. I'm reading Peter Suber's [A Crash Course in the Mathematics Of Infinite Sets][1] ([Wayback Machine](https://web.archive.org/web/20110703003113/https://earlham.edu/~peters/writing/infapp.htm)). The properties of infinite sets seem unintuitive, but of course, the proofs show that they are true. 
My guess is that whoever came up with the square root of -1 did so many years before it 'escaped' from mathematics and found a practical use. Before then perhaps people thought it was clever, but not necessarily useful or even 'true'. So then, if you need to understand electricity, and you can do it best by using i, then even someone who thinks it's silly to have a square root of negative -1 would have to grudgingly admit that there's some 'reality' to it, despite its unintuitiveness, because electricity behaves as if it 'exists'.
Seeing as how there was so much resistance to infinite sets at the beginning, even among mathematicians, I wonder: has the math of infinite sets been 'proven worthwhile' by having a practical application outside of mathematics, so that no one can say it's just some imaginative games?

REPLY [6 votes]: I recently gave a talk about how rank-into-rank cardinals and the cardinals around the $n$-huge level could be used to construct new public key cryptosystems.
The basic idea is to select some finite subalgebra $(X,*)$ of the quotient algebra of rank-into-rank embeddings $\mathcal{E}_{\lambda}/\equiv^{\gamma}$. Since $(X,*)$ is finite, the algebra $(X,*)$ is always computable. One then extrapolates from $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ a ternary self-distributive operation $t^{\bullet}$ on the set $X$. The operation $t^{\bullet}$ is typically of the form $t^{\bullet}(x,y)=T(x,y)*z$ for some operation $T$ that satisfies $x*T(y,z)=T(x*y,x*z)$. Then from the algebra $(X,t^{\bullet})$, one extrapolates a new algebra $(\Diamond(X,t^{\bullet}),t^{\sharp})$ ($t^{\sharp}$ is ternary) which I call a functional endomorphic Laver table. Functional endomorphic Laver tables may be used as platforms for several public key cryptosystems such as the Ko-Lee key exchange and the Kalka-Teicher key exchange.
Since such cryptosystems are very new, nobody knows anything about the security of these cryptosystems or even about the efficiency of such cryptosystems (set theorists, get to work). However, if these cryptosystems are secure against classical computation, then these cryptosystems will likely remain secure against attacks from adversaries with quantum computers as well (quantum algorithms currently use very few specialized techniques that have nothing to do with large cardinals). Depending on how you count, there are about five other different kinds of public key cryptosystems which are not known to be vulnerable to attacks from quantum computers.<|endoftext|>
TITLE: about Hilbert and Siegel modular varieties (forms)
QUESTION [5 upvotes]: It seems to me that Hilbert modular varieties (forms) are generalization from Q to totally real fields. While Siegel modular varieties (forms) are generalization from 1 dimensional to higher dimensional abelian varieties. But they should both be some kind of Shimura variety (automorphic forms), right?
According to Milne's note of Shimura varieties, Siegel modular varieties are Shimura varieties coming from the Shimura datum (G, X) where G is the symplectic similitude group of a symplectic space (V, \phi). So what is the corresponding Shimura datum for the Hilbert modular variety? Or am I asking a wrong question?
Also, in the definition of Shimura varieties G(Q)\G(A_f)X/K, why we only consider Q and its adele group? why not general number fields and their adeles? (again, maybe a wrong question)
Thank you.

REPLY [5 votes]: The Shimura datum for a Hilbert modular variety is the pair (G,X) where G is the group over Q obtained from GL(2) over a totally real number field F by (Weil) restriction of scalars and X is a product of copies of C minus the real axis indexed by the real embeddings of F.
In the definition of Shimura varieties we work only over Q by convention: we could work over number fields, but this gives nothing new (one can always take Weil restriction). However, in some cases, e.g., Hilbert modular varieties, it seems (to me to be) more natural to work over a number field other than Q.<|endoftext|>
TITLE: The orthodrome of n-spheres.
QUESTION [6 upvotes]: I am a Computer Science undergraduate who does a lot of other tinkering in his free time. Right now, I'm tinkering with n-spheres. Specifically, I'm looking at the distances between a collection of points on n-sphere surfaces. Euclidean distances are trivial (but in this particular application still interesting). I would like to look at "great-circle" distances between points on an n-sphere, but unfortunately I am not familiar with Riemannian Geometry or anything of the sort.
How can one go about calculating the distance between two points on an n-sphere? Can you make this digestible for an undergraduate student who is unfamiliar with the literature?

REPLY [9 votes]: The two-dimensional formula applies (why?): the great-circle distance is $\cos^{-1}(\vec u\cdot \vec v)$ where $\vec u$ and $\vec v$ are position vectors of the points.<|endoftext|>
TITLE: Level set of a harmonic function
QUESTION [8 upvotes]: Let $u$ be a nonconstant real-valued harmonic function defined in the open unit disk $D$. Suppose that $\Gamma\subset D$ is a smooth connected curve such that $u=0$ on $\Gamma$. Is there a universal upper bound for the length of $\Gamma$? 
Remark: by the Hayman-Wu theorem, the answer is yes if $u$ is the real part of an injective holomorphic function; in fact, in this case there is a universal upper bound for the length of the entire level set in $D$. For general harmonic functions, level sets can have arbitrarily large length, e.g. $\Re z^n$.

REPLY [12 votes]: It can get arbitrarily ugly. Indeed, approximate $1/z$ by a polynomial $p$ in the domain $K\subset\mathbb D$ whose complement is connected but goes from $0$ to the boundary along a long winding narrow path. Then each connected component of the set $\mbox{Re}p=A$ with large $A$ will have to escape the circle along essentially the same path and there are only finitely many $A$ for which we have branching points in these sets ($p'$ has finitely many roots).<|endoftext|>
TITLE: Smoothness of Symmetric Powers
QUESTION [11 upvotes]: Here's something that's been bothering me, and that's come up again for me recently while reading some stuff about Hilbert schemes of points (Nakajima's lectures, specifically):
Let $C$ be an algebraic curve.  Define $S^nC$ to be $C\times\ldots\times C/S_n$, the symmetric power.
Now, over $\mathbb{C}$, I can show that $C$ a complex manifold implies that $S^nC$ is, and that if $X$ is a variety with $S^nX$ smooth, then $X$ is one dimensional, but the argument I have involves looking in analytic open sets and reducing to the case of $\mathbb{C}^n$, and additionally is fairly unhelpful for identifying the total space (ie, that $S^n\mathbb{P}^1\cong\mathbb{P}^n$)
So here's my question: how can we, in a fairly quick and natural way, show that

If $C$ is a smooth, 1-dimensional variety over an algebraically closed field $k$,, then $S^nC$ is smooth.
If $X$ is a smooth variety over an algebraically closed field $k$, and $S^nX$ is smooth, then $X$ is one dimensional.

Now, I don't want any projectivity hypotheses here, and I'm curious, with the more arithmetically inclined, is this still true over an arbitrary field?

REPLY [11 votes]: I also wanted to mention a `high-technology' answer to (1). Namely, if $C$ is a smooth algebraic curve, its $n$-th symmetric power coincides with the variety of all degree $n$ effective divisors on $C$ (that is, we consider $n$-element subsets of $C$, and then allow points to merge). In a fancier language, we are looking at the Hilbert scheme of length $n$
subschemes of $C$.
Note that this agrees with the explicit choice of coordinates if $C={\mathbb A}^1$ (and also $C={\mathbb P}^1$): namely, if you want to parametrize $n$-element subsets of ${\mathbb A}^1$ with multiplicities, you do so (essentially by unique factorization) by saying that such a subset is the zero locus of a degree $n$ monic polynomial $p$, so it is uniquely determined 
by coefficients of $p$ (which of course are the elementary symmetric polynomials in terms of
original $n$ points).
OK, so why do finite subschemes of $C$ form a smooth variety? Because deformation theory
says that at a particular finite subscheme $D\subset C$, the obstructions to smoothness belong to $Ext^2_C(O_D,O_D)$. This vanishes because $C$ is a smooth curve, and so there are no $Ext$'s beyond first (OK, no local $Ext$'s, but $D$ is finite, so it's the same thing).
I think this is a useful point of view. For instance, if $S$ is a smooth surface, $Sym^n S$
is singular, but one can check that the Hilbert scheme $Hilb_n S$ of degree $n$ finite subschemes is still smooth; it provides a resolution of singularities of $Sym^n S$. One can thus argue that the Hilbert scheme is a 'better behaved' object compared to the symmetric power...

REPLY [3 votes]: Edit: It seems appropriate to recall here Chevalley-Shephard-Todd's Theorem.
It says  the quotient of $\mathbb A^n_k$ by a finite linear group $G$ with order prime to the characteristic of $k$ is smooth (i.e. the algebra of invariants is polynomial) if and only if $G$ is generated by pseudo-reflections (codimension one fixed point set).
Once one localizes the problem at a point, as VA and Ben Webster did, this settles both 1 and 2 over arbitrary fields of characteristic zero. Of course VA argument is preferable as it is more direct/elementary.

Original answer. Below is my original answer commented by David Speyer below:
Over the category of smooth real manifolds the symmetric power of smooth curves is not smooth in general. The second symmetric power is a smooth surface with boundary and starting from the third symmetric power what we get are varieties with corners at the boundary.
Symmetric powers of smooth surfaces are still smooth as they are locally diffeomorphic to complex curves.
If nothing else these examples show that smoothness might mean different things for algebraic geometers and differential geometers in algebraic geometry over $\mathbb R$ and in differential geometry.<|endoftext|>
TITLE: number of irreducible representations over general fields
QUESTION [8 upvotes]: For a finite group, there are finitely many irreducible representations of complex numbers.
What if the field is changed to some other fields? Like real numbers, p-adic field, finite field?
In particular, for a (finite) Galois group over a p-adic field, and consider p-adic Galois representation. Are there finitely many irreducible representations? If there are, can we actually construct some kind of varieteis s.t. the geometric representations (etale cohomology) coming from these varieties are exactly the irreducible ones?
And what if we replace the finite groups to other groups? Say, profinite groups, or even Lie groups, algebraic groups with non-discrete topologies?

REPLY [5 votes]: If a group is finite, it has finitely many simple representations over any field. Indeed, in this case the group algebra over the field is artinian, and this is true for all artinian algebras.
When the field characteristic divides the order of the group, the so called "modular situation", these finitely many representations are very far away from the whole story. For example, if the characteristic is $p$ and the group is a $p$-group, there is exactly one simple module, yet as long as the group is not cyclic, the category of representations is wild so there are extraordinarily many other indecomposable representations.<|endoftext|>
TITLE: Topological "Interpolation" ?
QUESTION [9 upvotes]: Let E be a normed space, and let $T$:E *  $\rightarrow$ E * be
a nonlinear operator.
Suppose that :
1) $T$ is continuous from (E *, ||.||) to itself (i.e., it is norm-continuous).
and
2) $T$ is continuous from (E *, w *) to itself (i.e., it is  weakly-star continuous).
Then does it follow that
3) $T$ is continuous from (E *, w) to itself (i.e., it is  weakly continuous) ? 

I guess the case E = ℓ∞ would be particularly interesting.

REPLY [9 votes]: I think Ady's question has a negative answer.
Recall that  the Mazur map, $T$,  from the unit ball of $\ell_2$ onto the unit ball of $\ell_1$, is defined by 
$T(\sum a_i e_i)= \sum a_i^2 b_i   e_i$, where $b_i$ is the sign of $a_i$. It is a uniform homeomorphism in the norm topologies and obviously is coordinatewise continuous, which means that it is weak ($=$ weak$*$ in $\ell_2$) to weak$^*$ continuous (since on the unit ball of $\ell_p$ with $p$ finite  the 
weak* topology is the topology of coordinatewise convergence).
$T$  is not weak to weak continuous since the unit vector basis converges weakly to zero in $\ell_2$ but not in $\ell_1$.
The problem is to extend $T$ to all of $\ell_2$.  I think that the easiest way to do this is to show that there is a retraction $R$ from $\ell_2$ onto its unit ball which is both norm to norm continuous and weak to weak continuous.  Define $R$ on the complement of the unit ball by
$$
R(\sum a_i e_i) = \sum_{i=1}^n a_i e_i + t e_{n+1},
$$
where $\sum_{i=1}^n a_i^2 \le 1 < \sum_{i=1}^{n+1} a_i^2$ and $t$ is chosen to have the same sign as $a_{n+1}$ and to make the image vector have norm one.  $R$ is obviously continuous (even Lipschitz)  in the norm topology and is continuous in the topology of coordinatewise convergence, hence is weak to weak continuous (since in all of $\ell_2$, the weak topology is stronger than the topology of coordinatewise convergence).
To get a counterexample that maps one dual space to itself, work with the space $\ell_2 \oplus \ell_1$.<|endoftext|>
TITLE: categorical homotopy colimits
QUESTION [6 upvotes]: let $hTop_*$ denote the homotopy category of pointed spaces. I believe that it has no pushouts, in general. the reason is that you can't expect the involved homotopies to be compatible. can anyone give an explicit example, with proof? I know that homotopy colimits are related to this, but they don't seem to be categorical colimits, so I don't think that they fit here.
especially I'm interested in the following special case: let $G= \langle X | R \rangle$ a presentation of a group and consider the resulting map $\omega : \vee_{r \in R} S^1 \to \vee_{x \in X} S^1$. does the cokernel of $\omega$ exist in $hTop_*$? in $Top_*$, the cokernel is just  consider the 2-dimensional CW-complex $Q$, which is optained from $\vee_{x \in X} S^1$ via the attaching map $\omega$. now if $f : \vee_{x \in X} S^1 \to T$ is a pointed map such that $f \omega$ is nullhomotopic, it is easy to see that it extends to a map $\overline{f} : Q \to T$. but I think that we cannot expect that $\overline{f}, \overline{g}$ are homotopic, when $f,g$ are homotopic: the homotopies between $f$ and $g$ don't have to be compatible. can you give an example for that? probably it already works for $\omega : S^1 \to S^1, z \mapsto z^2$, thus $Q = \mathbb{R} P^2$.
anyway, this only would show that $Q$ is not the cokernel in the category $hTop_*$. the proof, that the cokernel does not exist at all, will be even more difficult and I don't know how to approach it. 
you may also replace the category by $hCW_*$ (CW-complexes), $hCG_*$ (compacty generated spaces) etc., if it's useful.

REPLY [12 votes]: Your example (the "cokernel" of the multiplication by 2 map) also works.
Consider the diagram $S^1 \leftarrow S^1 \rightarrow D^2$ in the based homotopy category of CW-complexes,  where the left-hand map is multiplication by 2.  Suppose it had a pushout $X$ in the homotopy category.  Then for any $Y$, $[X,Y]$ is isomorphic to the set of 2-torsion elements in $\pi_1(Y)$.
Taking $Y = S^0$, we find $X$ is connected.
Taking $Y = K(\pi,1)$, we find that $\pi_1(X)$ must be isomorphic to $\mathbb{Z}/2$.  This means that there is a map from ${\mathbb{RP}^2}$ to $X$ inducing an isomorphism on $\pi_1$, and that there is a map $X \to K(\mathbb{Z}/2,1)$ that also induces an isomorphism on $\pi_1$.
Net result, we get a composite sequence of maps $\mathbb{RP}^2 \to X \to \mathbb{RP}^\infty \to \mathbb{CP}^\infty$.  The final space is simply connected, so the map from $X$ would be nullhomotopic and hence so would the map from $\mathbb{RP}^2$.
However, the composite of the first two maps is an isomorphism on $\pi_1$, hence on $H_1$.  Looking at induced maps on the second cohomology group $H^2$, we get the sequence of maps:
$$\mathbb{Z}/2 \leftarrow H^2(X) \leftarrow \mathbb{Z}/2 \leftarrow \mathbb{Z}$$
The rightmost map is surjective, the composite of the two leftmost maps is an isomorphism by the universal coefficient theorem, and the composite of the two rightmost maps is supposed to be nullhomotopic and hence zero.  Contradiction.<|endoftext|>
TITLE: Upper bound for number of k-term arithmetic progressions in the primes
QUESTION [8 upvotes]: Normal heuristics give that number of k-term arithmetic progressions in [1,N] should be about
$$c_k\frac{N^2}{\log^kN}$$
for some constant $c_k$ dependent on k. The paper of Green and Tao gives a similar lower bound for all k (with a much worse constant, but still), and recent work by Green, Tao and Ziegler have established the correct asymptotic for k=3 and k=4.
I am looking for a reference which establishes an upper bound for all k - I'm sure I've heard of one, but I can't find mention of the relevant paper anywhere. Of course, if there is a simple proof, that would appreciated as well.
That is, I am looking for a reference and/or proof which establishes that the number of k-term arithmetic progressions of primes in [1,N] is at most
$c_k'\frac{N^2}{\log^kN}$.
for some constant $c_k'$.

REPLY [15 votes]: Well, any standard upper bound sieve (e.g. Selberg sieve, combinatorial sieve, beta sieve, etc.) will give this type of result.  I'm not sure where you can find an easily citeable formulation, though.  One can get this bound from Theorem D.3 of this paper of Ben and myself on page 67 (see in particular the remark at the bottom of that page).  But this is certainly not the first place where such a bound appears.  (The Goldston-Yildirim papers will give this result too, but this is also not the first place either.)<|endoftext|>
TITLE: Yoneda embedding target
QUESTION [6 upvotes]: I'm learning about representable functors from Vistoli notes thanks to Charles Siegel's answer.
I see that any category $\mathcal C$ can be embedded into $\text{Hom}\\,(\mathcal C^{op}, \mathcal Set)$ by means of Yoneda embedding. I wonder if there are examples where the latter category would be interesting in itself, other then for these purposes?

REPLY [2 votes]: Just thought of another one: Kripke models of intuitionistic logic are (very closely related to) presheaves on the poset of possible worlds. The interpretation of intuitionistic logic in a Kripke model coincides with the internal logic of the corresponding presheaf topos.<|endoftext|>
TITLE: Zeta function for curves in a manifold
QUESTION [7 upvotes]: Motivation
In the analogy between prime numbers and knots, the prime number is thought sometimes as the circle of length $l([p]) = \text{log}\,p$. This is so you can express the zeta function as 
$$ \zeta(s) = \sum_{D\ge0} e^{-l(D)s}$$
where the sum goes over effective divisors on $\text{Spec}\,\mathbb Z$ and length is extended there by additivity. Similarly, you can do it to rewrite Dedekind zeta function for other number fields.
Question
I wonder, what is the right analogue of above formula for a manifold with metric? Perhaps:

integration over all closed curves of the expression $e^{-l(D)s}$
summation over positive sums of classes of closed geodesics.

I think I've heard something about definition 2, but I suspect if the two above are defined correctly they will be the same. Is it possible to formalize this definition? Do different formalizations lead to the same zeta-function?
Updates
Yes, I think this should be related to Laplacians, Selberg trace formula and dynamical system zetas. What I said I've heard about definition 2 was probably the Selberg zeta, but I can't say it clearly, hence questions.

REPLY [4 votes]: There's been a lot of work since Smale's idea of a dynamical zeta function for general flows (in particular geodesic flows).  A good starting point would be this 12 year old review by Baladi. There is a large and more recent literature but I'm no expert, although this other review by Liverani and Tsuji is probably not far from current knowledge.
There's also a whole branch of physics around those ideas, indeed related to the spectrum of the Laplacian and applications to quantum physics and statistical physics. This nice physics book is a good start (in particular if you read the quote of Smale at page 3 of this chapter, and then remark 19.2 at the very end of that chapter you'll get a quick sense of the stuff you've aked for).<|endoftext|>
TITLE: Stable ∞-categories as spectral categories
QUESTION [13 upvotes]: Let C be a stable ∞-category in the sense of Lurie's DAG I.  (In particular I do not assume that C has all colimits.)  Then C does have all finite colimits, the suspension functor on C is an equivalence, and C is enriched in Spectra in a way I don't want to make too precise (basically the Hom functor Cop × C → Spaces factors through Spectra and there are composition maps on the level of spectra).
Now suppose instead that C is an ∞-category which has all finite colimits and comes equipped with an enrichment in Spectra in the above sense.  One can show easily that C then has a zero object which allows us to define a suspension on C.  Suppose it is an equivalence.  Is C then a stable ∞-category?  Moreover, is the enrichment on C the one which comes from the fact that it is a stable ∞-category?

REPLY [8 votes]: According to Corollary 8.28 in DAG I a pointed $\infty$-category is stable iff it has finite colimits and the suspension functor is an equivalence.<|endoftext|>
TITLE: Proof of Bloch-Kato conjecture of K-theory?
QUESTION [5 upvotes]: Wikipedia says:

this circle of ideas is distinct from the Bloch–Kato conjecture of K-theory, extending the Milnor conjecture, a proof of which was announced in 2009

What exactly is the K-theory conjecture of Bloch-Kato and has it been proven?

REPLY [3 votes]: On 22 June 2013, Joël Riou is going to give a Bourbaki talk on
La conjecture de Bloch-Kato [d'après M. Rost et V. Voevodsky].

La conjecture de Bloch-Kato énonce que pour tout corps $k$ et 
  tout nombre premier $l$ différent de la caractéristique 
  de $k$, l'algèbre de K-théorie de Milnor de $k$ modulo $l$ 
  (qui est définie par générateurs et relations) s'identifie 
  à une algèbre de cohomologie galoisienne associée à $k$. La 
  démonstration de cet énoncé, qui admet de nombreuses applications, 
  utilise de façon essentielle d'une part les théories motiviques 
  (cohomologie, homotopie, opérations de Steenrod) et d'autre 
  part des constructions géometriques de variétés algébriques 
  ayant des propriétés remarquables par rapport à des symboles 
  en K-théorie de Milnor.  

Usually the notes are put up on http://www.bourbaki.ens.fr/ some time after the actual talk.  You can also write to Joël Riou directly at http://www.math.u-psud.fr/~riou/.<|endoftext|>
TITLE: Rolle's theorem in n dimensions
QUESTION [27 upvotes]: This looks like a statement from a calculus textbook, which perhaps it should be. 
"Rolle's theorem". Let $F\colon [a,b]\to\mathbb R^n$ be a continuous function such that $F(a)=F(b)$ and $F'(t)$ exists for all $a<t<b$. Then there exist numbers $a < t_1 < t_2 < \dots < t_n < b$ such that the vectors $F'(t_1),\dots,F'(t_{n})$ are linearly dependent. 
We are all familiar with the case $n=1$. The case $n=2$ is not hard either: pick any $a<t^\ast<b$ and find, using the mean value theorem, numbers $a<t_1<t^\ast$ and $t^*<t_2<b$ such that $F'(t_j)$ is collinear with $F(t^\ast)-F(a)$. Note that we avoided using the parameter value $t^\ast$, which will be important in a moment. When $n=3$, we pick $a<t^\ast<b$ and project F onto the orthogonal complement of $F'(t^\ast)$, then apply the case $n=2$ to the projection ($t^\ast$ will become the third chosen parameter value). So far so good.
But I get stuck at $n=4$. If the above process is followed, then after $F$ is projected down to two dimensions, we must avoid two particular parameter values. Which is not possible in general: if in two dimensions $F$ parametrizes a triangle and $F(a)$ is a vertex, then one of points $F(t_j)$ must be one of two other vertices. Presumably this problem can avoided by a generic choice of points of projection, but how?

REPLY [7 votes]: Let's add a bit of Anton Petrunin's answer into algori's: namely, I'm going to spell out the 'little  bit of work' from Anton's answer, which is a version of algori's Claim. I am not proving the stronger statement made by Anton though, only the original statement. (This was supposed to be a comment: my attempt to write algori's proof without taking limits, but it turned out to be too long for comments.)
Claim 1 (Petrunin): $0\in Conv(f'(a,b))$.
Proof: Otherwise there is a hyperplane separating $0$ from the convex hull.
By Caratheodory's Theorem, we get 
$$a < t_1 < \dots < t_{n+1} < b$$ such that 
$$0\in Conv(f'(t_1),\dots,f'(t_{n+1}).$$
Claim 2 (algori): There is $s\in[t_n,t_{n+1}]$ such that 
$$0\in Span(f'(t_1),\dots,f'(t_{n-1}),f'(s)).$$
Proof: Set $$l(x)=\det(f'(t_1),\dots,f'(t_n),f(x)).$$ The condition on $s$ is that $l'(s)=0$. By Claim 1, either $$0\in Conv(f'(t_1),\dots,f'(t_{n-1}))$$ (in which case the claim is obvious) or $0$ lies between $l'(t_n)$ and $l'(t_{n-1})$. In the latter case, such $s$ exists by the usual Rolle's theorem.<|endoftext|>
TITLE: Introduction to W-Algebras/Why W-algebras?
QUESTION [22 upvotes]: Does anyone know of an introduction and motivation for W-algebras?
Edit: Okay, sorry I try to add some more background. W algebras occur, for example when you study nilpotent orbits: Take a nice algebraic/Lie group. It acts on its Lie-algebra by the adjoint action. Fix a nilpotent element e and make a sl_2 triple out of it. A W algebra is some modification of the universal enveloping, based on this data.
A precise definition is for example given here: https://arxiv.org/pdf/0707.3108.
But this definition looks quite complicated and not very natural to me. I don't see what's going on. Therefor I wonder if there exists an easier introduction.

REPLY [2 votes]: The motivation did not change. The theory of finite W-algebras advanced quite a bit, imho, mostly thanks to the works of Losev.
There is a good introductory text by Arakawa now.<|endoftext|>
TITLE: Definable collections without definable members (in ZF)
QUESTION [26 upvotes]: Denote Zermelo Fraenkel set theory without choice by ZF. Is the following true: In ZF, every definable non empty class A has a definable member; i.e. for every class $A = \lbrace x : \phi(x)\rbrace$ for which ZF proves "A is non empty", there is a class $a = \lbrace x : \psi(x)\rbrace$ such that ZF proves "a belongs to A"?

REPLY [24 votes]: There is a nice model-theoretic interpretation of the property you state, which I will call property (A). 
By a theory of sets, I mean any theory $T$ with an extensional relation ${\in}$.
Fact 1. If $T$ is a theory of sets with property (A) then so does every extension of $T$.
Fact 2. If $T$ is a complete theory of sets with property (A) then $T$ has a unique prime model.
Fact 3. If $T$ is a complete theory of sets with a wellfounded prime model then $T$ has property (A).
The fact that ZF does not have property (A) follows from Fact 1 and the (simpler) observation that ZFC does not have property (A). However, Facts 2 and 3 give nice ways of finding extensions of ZF which do have property (A).
Without loss of generality, $T$ has enough closed terms so that if $T$ proves that the formula $\phi(x)$ has a unique solution then $T \vdash \phi(c)$ for some closed term $c$. (Otherwise, add a new constant $c$ for each such formula $\phi(x)$ together with the defining axiom $\phi(c)$. This is a conservative extension of $T$ with the desired property.) If $M$ is any model of $T$, then let $K_M$ be the substructure of $M$ whose ground set consists of all interpretations of the closed terms. Note that $K_M$ is completely determined by the complete theory of $M$.
Observe that if $\theta(z)$ is any formula, then 
$\psi(x) \equiv \forall z(z \in x \leftrightarrow \theta(z))$ 
has at most one solution by extensionality. Therefore, $T$ proves that 
$\phi(x) \equiv \psi(x) \lor (\forall z\lnot\psi(z) \land x = c_0)$
has exactly one solution (where $c_0$ is any fixed closed term). So there is a constant $c$ such that, in any model of $T$, if $\{z : \theta(z)\}$ is a set then $c = \{z:\theta(z)\}$. Of course, we always have $c = \{z:z\in c\}$. So in any model $M$ of $T$ the constants $c$ are precisely all the (parameter-free) instances of comprehension which exist in $M$.
In view of this, property (A) can be formulated as:
(A) For every formula $\phi(x)$, if $T \vdash \exists x\phi(x)$ then $T \vdash \phi(c)$ for some constant $c$.
Replacing $\phi(x)$ with $\phi(x) \lor \forall z\lnot\phi(z)$, we see that property (A) implies the seemingly stronger statement:
(A') For every formula $\phi(x)$ there is a constant $c$ such that $T \vdash \exists x \phi(x) \leftrightarrow \phi(c)$.
This reformulation of property (A) immediately implies Fact 1. By the Tarski-Vaught Test, property (A') says that if $M$ is any model of $T$ then the substructure $K_M$ of $M$ is an elementary submodel of $M$. Therefore, $K_M$ is the necessarily unique prime model of the complete theory of $M$, which establishes Fact 2.
For Fact 3, suppose that $T$ is complete and that $M$ is a wellfounded prime model of $T$. I will show that $M = K_M$, which immediately implies that $T$ has property (A). For the sake of contradiction, suppose that $M \neq K_M$. Since $M$ is wellfounded, there is an $a \in M \setminus K_M$ such that $a \subseteq K_M$. Since $M$ is prime, $a$ realizes a principal type $\phi(x)$. If $b$ also realizes $\phi(x)$, then $b \cap K_M = a$ since all elements of $K_M$ are definable. Therefore, $\phi(x)$ must imply $\forall z(\phi(z) \rightarrow x \subseteq z)$ over $T$. This means that $T$ proves that $\phi(x)$ has a unique solution, which contradicts the fact that $a \notin K_M$.

To partly answer Joel's comment whether wellfoundedness can be eliminated from Fact 3, Ali Enayat [Models of set theory with definable ordinals
Arch. Math. Logic 44 (2005), 363-385. MR2140616] has shown that for a completion $T$ of $ZF$ the following three properties are equivalent:

$T \vdash V = OD$.
$T$ has a unique Paris model up to isomorphism.
$T$ has a prime model.

A Paris model is a model where every ordinal is first-order definable. This is slightly weaker than property (A), but it is equivalent when $T \vdash V = OD$.<|endoftext|>
TITLE: Depressed graduate student.
QUESTION [80 upvotes]: How does a depressed graduate student go about recovering his enthusiasm for the subject and the question at hand?
Edit: I am not that grad student; it is a very talented friend of mine.
Moderator's update:
The discussion about this question should happen at this link on meta. This page is reserved for answers to the question as stated.

REPLY [3 votes]: We're all approaching this from the perspective that the correct thing for your friend is to stay in mathematics. Perhaps not. I think of three very talented friends who eventually left mathematics. One left to become a successful orthodontist, another actually got his Ph.D., but decided to become a doctor instead. The third sold all his worldly possessions and took up with the Worker's Party. I have two children in theatre, and I remember the advice they received from faculty at the schools they were considering:"You've got to want this more than anything else." I don 't think mathematics is much different.  Once the desire is gone, one can try to get it back, or consider if it might be time to move on. As a mathematician, I hope your friend finds some way to get back in the groove. Time will tell.<|endoftext|>
TITLE: Are all coproducts of 1 in a topos distinct ?
QUESTION [5 upvotes]: Inspired by the two solutions to Harry's question
Can a topos ever be an abelian category?
I was wondering whether all coproducts of 1 in a topos are distinct up to isomorphism?  That is $1 + 1 + \dots + 1 \cong 1 + 1 + \dots + 1$ iff there are an equal number of 1s on each side?
Edit: In order to make the question (possibly) non-trivial, let's assume that the topos is not equivalent to the terminal category.

REPLY [7 votes]: At least if you're talking about finite coproducts, then the answer is yes.  If $n\le m$, then we have a canonical inclusion $\sum_{i=1}^n 1 \hookrightarrow \sum_{j=1}^m 1$, which is in fact a complemented subobject with complement $\sum_{k=1}^{m-n} 1$.  If this inclusion is an isomorphism, then its complement is initial, and hence (assuming the topos is nontrivial) $n=m$.  Now if we have an arbitrary isomorphism $\sum_{i=1}^n 1 \cong \sum_{j=1}^m 1$, then composing with the above inclusion we get a monic $\sum_{i=1}^m 1 \hookrightarrow \sum_{j=1}^m 1$.  However, one can show by induction that any finite coproduct of copies of $1$ in a topos is Dedekind-finite, i.e. any monic from it to itself is an isomorphism.  (See D5.2.9 in "Sketches of an Elephant" vol 2.)  Thus, the standard inclusion is also an isomorphism, so again $n=m$.<|endoftext|>
TITLE: Newton Puiseux expansions for singular surfaces? 
QUESTION [5 upvotes]: Is there a theory of Newton-Puiseux type  expansions
which works to parameterize  singular surface germs $F(x,y,z) =0$?
Ideally, each branch would  be the image of map of  the form  the 
$x = u^m, y = v^n,  z =  \Sigma_{i + j > n} a_{i j} u^i v^j$
(after a linear change of the variables $x, y, z$).

REPLY [6 votes]: I don't believe there is anything as general as that, but when your polynomial is over the complex numbers one can do the following. First shift and rotate coordinates so that you're working on a neighborhood of the origin where $F(0,0,0) = 0$ and $\partial_z^n F(0,0,0) = 0$ for some $n$. Then you can use the Weierstrass preparation theorem and (ignoring a nonvanishing factor) write $F(x,y,z)$ as $z^n + a_{n-1}(x,y)z^{n-1} + ... + a_0(x,y)$. If the discriminant of this polynomial, viewed as a function of $x$ and $y$ has normal crossings (i.e. is a monomial times a nonvanishing factor), one can use the Jung-Abhyankar theorem to get a factorization of the form $(z - b_1(x,y))...(z - b_n(x,y))$ where the $b_i$ are analytic in fractional powers of $x$ and $y$ as you want. 
Even if the discriminant is not normal crossings, you can first resolve the singularities of the discriminant to make it normal crossings and there are a few ways to do this. One way involves subdividing the $x-y$ space into wedges, and on each wedge there is a way of first doing a coordinate change of the form $(x,y) --> (x,y - g(x))$ then a "blowing up" coordinate change $(x,y) = (x',(x')^m y')$ and get the discriminant as you want. ($m$ may not be an integer, and also you might have to reverse the roles of the $x$ and $y$ variables). Then one proceeds as above. Of course the final result isn't as nice as you wanted, but this is in the nature of things.
One can iterate the above to deal with analogues in higher dimensions, but the formulas get a lot more elaborate as one might guess. There's a paper by Parusinski "On the preparation theorem for subanalytic functions" that discusses a lot of these issues.<|endoftext|>
TITLE: Torsion of an abelian variety under reduction.
QUESTION [11 upvotes]: Let $p$ be a prime. Suppose you have an Abelian scheme $A$ over $Spec\ \mathbb{Z}_p$. How do you prove that if $q$ is another prime, then the $q$-torsion of $A$ injects into the torsion of $A_p$, under the reduction map?

REPLY [5 votes]: To complete Pete and Milne's answers when A is not an abelian scheme (for example, when it is the Néron model of an abelian variety over ${\mathbb Q}_p$ with not necessary good reduction), then for any $n$ prime to $p$, the kernel $A[n]$ is still étale over ${\mathbb Z}_p$ (because the tangent map at $0$ of the multiplication by $n$ is just multiplication by $n$ for any commutative algebraic group), but not necessarily finite. There is a biggest closed subscheme $H$ of $A[n]$ which is étale and finite over ${\mathbb Z}_p$. The reduction map on $H$ is injective (see Pete's proof). The generic fiber of $H$ corresponds to the points of the generic fiber of $A[n]$ having specialization mod $p$. You may read Bosch, Lütkebohmert and Raynaud ''Néron Models'', § 7.3.<|endoftext|>
TITLE: A nice variety without a smooth model
QUESTION [8 upvotes]: Is there a simple example of a smooth proper variety $X$ over $K=\mathbb{Q}_p$ ($p$ prime) such that
--- $X(K)\neq\emptyset$,
--- the $l$-adic étale cohomology $H^i(X\times_K\bar K,\mathbb{Q}_l)$ is unramified for every prime $l\neq p$ and every $i\in\mathbb{N}$,
--- the $p$-adic étale cohomology $H^i(X\times_K\bar K,\mathbb{Q}_p)$ is crystalline for every $i\in\mathbb{N}$,
and yet,
--- $X$ is not the generic fibre of any smooth proper $\mathbb{Z}_p$-scheme ?
My example of a Châtelet surface with these properties is simple enough, but can one do better ?

REPLY [3 votes]: As Minhyong suggests, a curve $C$ (with $C(\mathbb{Q}_p)\neq0$) which has bad reduction but whose jacobian $J$ has good reduction would do the affair.  This works because the cohomology of $C$ is essentially the same as that of $J$, and because an abelian $\mathbb{Q}_p$-variety has good reduction if and only if its $l$-adic étale cohomology is unramified for some (and hence for every) prime $l\neq p$ (Néron-Ogg-Shafarevich) or its $p$-adic étale cohomology is crystalline (Fontaine-Coleman-Iovita).
I asked Qing Liu for explicit examples.  He suggested the curve
$$
y^2=(x^3+1)(x^3+ap^6)\qquad  (a\in\mathbb{Z}_p^\times)
$$
when $p\neq2,3$, and $y^2=(x^3+x+1)(x^3+a3^4x+b3^6)$, with $a,b\in\mathbb{Z}_3^\times$, for $p=3$.
He refers to Proposition 10.3.44 in his book for computing the stable reduction of these $C$, and to Bosch-Lütkebohmert-Raynaud,
Néron models, Chapter 9, for showing that $J$ has good reduction.
I "accept" this answer as coming from Minhyong Kim and Qing Liu.<|endoftext|>
TITLE: Killing Chern classes
QUESTION [5 upvotes]: Let $G$ be a compact connected Lie group and let $E\to B$ be a principal $G$-bundle. Suppose $a$ is a rational cohomology class of $E$ such that its pullback $b$ under an orbit inclusion map $G\to E$ is one of the standard multiplicative generators of $H^{{\bullet}}(G,\mathbf{Q})$ . Let $E'=E\times EG/G$ be the Borel construction (corresponding to the action of $G$ on $E$) and let $(E^{pq}_r,d_r)$ be the Leray spectral sequence corresponding to the fiber bundle $E'\to BG$.
The class $a$ gives an element $a'\in E^{0,2k-1}_2$ for some $k$. Assume that $d_i(a')=0,i< 2k$. Is it true that $d_{2k}(a')$ is what has remained in $E_{2k}$ of the multiplicative generator of $H^{{\bullet}}(BG,\mathbf{Q})$ corresponding to $b$?
For simplicity one can assume $G=U(n)$, in which case what remains in $E_\infty$ of the generator of $H^{{{\bullet}}}(BG,\mathbf{Q})\cong E^{{\bullet},0}_2$ corresponding to $b$ is precisely the $k$-th Chern class of $E$, under the natural isomorphism $H^{{\bullet}}(E',\mathbf{Q})\cong H^{{\bullet}}(B,\mathbf{Q})$.
This is probably standard, but for some reason I don't see how to prove it nor can construct a counter-example off hand.
upd: here is a weaker version, which would be easier to (dis)prove: take $G=U(n)\times H$ where $H$ is another Lie group and suppose that the pullback of $a$ to $G$ is the canonical generator of $H^{\bullet}(U(n),\mathbf{Q})\subset H^{\bullet}(G,\mathbf{Q})$ in degree $2k-1$. Is it true that $d_{2k}(a')$ is mapped to zero under the mapping of the spectral sequences induced by the pullback of $E'$ to $BH$ i.e. by the map
$$(E\times EG)/H\to (E\times EG)/G$$
To prove this it would suffice, of course, to show that $d_{2k}(a')$ is represented in $E_2$ by a class in $$H^{\bullet}(BG,\mathbf{Q})\cong H^{\bullet}(BU(n),\mathbf{Q})\otimes H^{\bullet}(H,\mathbf{Q})$$ that is mapped to zero under $H^{\bullet}(BG,\mathbf{Q})\to H^{\bullet}(BH,\mathbf{Q})$.

REPLY [3 votes]: The inclusion $G \to E$ induces a map $EG \to E'$ of spaces over $BG$, where the map of fibers is $G \to E$, and there is a map backwards of Serre spectral sequences.  Because $a$ lifts a standard generator of the cohomology of $G$, and $a'$ is a cycle up to the $E_{2k}$-page, the differential of this element maps to the differential of the standard generator in the spectral sequence $H^p(BG;H^q(G)) \Rightarrow H^{p+q}(*)$.  The $d_{2k}$-differential on this class in $H^{2k-1}(G)$ is the "corresponding generator" in $H^{2k}(BG)$.<|endoftext|>
TITLE: Gelfand duality in NCG
QUESTION [8 upvotes]: In non-commutative geometry, Gelfand duality is the construction of multiplicative linear functionals of a commutative C*-algebra, which can be viewed as the space of all its irreducible complex representations. 
When encountered with a non-commutative C*-algebra, we can speak of the space of pure states, which is the generalization of multiplicative linear functionals in commutative cases. The GNS construction can then be viewed as a map from the space of states to the space of all representations, and also from the space of pure states to the space of all irreducible representations.
My question is, is the "GNS map" surjective, or injective? What should be viewed as the non-commutative topological space, the space of all irreducible representations, or the space of pure states? And why?

REPLY [3 votes]: @Meyer has provided a very good answer, so I just want to add something from my notes from Professor Rieffel's $C^\ast$-algebra class about the primitive ideal space, which is a canonical topological space associated to a $C^\ast$-algebra (in fact, it is associated to any normed $*$-algebra, but let's keep it "easy"...).
Let $A$ be a $C^\ast$-algebra.

Definition: An ideal $I\subset A$ is called primitive if it is the kernel of an irreducible representation. Denote the set of all primitive ideals by $Prim(A)$.

Note that $Prim(A)$ is always a set as it is a subset of $P(A)$, the power set of $A$. An important theorem:

If $A$ is GCR, then the map $\widehat{A}\to Prim(A)$ by $(\pi, H)\mapsto \ker(\pi)$ is a bijection (where $\widehat{A}$ is the "set" of equivalence classes of irreducible representations). 

This is far from the case if $A$ is NCR (there is some set theory here, such as the term "unclassifiable," that I don't want to get into as I am not a set theorist), but $Prim(A)$ is still a set, so we still get a topological space using the following fact:

All primitive ideals are prime. 

The space of prime ideals of $A$ comes with the Jacobson, or "hull-kernel" topology, so we get the relative topology on $Prim(A)$.
A few facts:

In general, $Prim(A)$ is not Hausdorff, but it is $T_0$ (see @Meyer's comment for a counterexample).
$Prim(A)$ is 
locally compact.
If $A$ is separable, $Prim(A)$ has the Baire Category Property.

Once again, this is all from a course I took from Professor Rieffel. I hope it helps!<|endoftext|>
TITLE: When is a classification problem "wild"?
QUESTION [28 upvotes]: I hope someone can point me to a quick definition of the following terminology.
I keep coming across wild and tame in the context of classification problems, often adorned with quotes, leading me to believe that the terms are perhaps not being used in a formal sense.  Yet I am sure that there is some formal definition.
For example, the classification problem for nilpotent Lie algebras is said to be wild in dimension $\geq 7$.  All that happens is that in dimension $7$ and above there are moduli.  In what sense is this wild?
Thanks in advance!

REPLY [22 votes]: Although your specific request for terminology in the case of Lie algebras has now been answered, there is a very interesting broader question underlying your inquiry. Namely, how can we understand in a precise general way the idea that a given classification problem is complicated? How are we to compare the relative difficulty of two classification problems? 
These questions form the central motivation for the emerging subject known as Borel equivalence relation theory (see Greg Hjorth's survey article). The main idea is that many of the most natural equivalence relations arising in many parts of mathematics turn out to be Borel relations on a standard Borel space. To give one example, the isomorphism problem on finitely generated groups, but of course, there are hundreds of other examples. A classification problem for an equivalence relation E is really the problem of finding a way to describe the E-equivalence classes, of finding an E-invariant function that distinguishes the classes. 
Harvey Friedman defined that one equivalence relation E is Borel-reducible to another relation F if there is a Borel function f such that x E y if and only if f(x) F f(y). That is, the function f maps E classes to F classes in such a way that different E classes get mapped to different F classes. This provides a classification of the E classes by using the F classes. The concept of reducibility provides a precise, robust way to say that one relation F is at least as complex as another E. Two relations are Borel equivalent if they reduce to each other, and we are led to the hierarchy of equivalence relations under Borel reducibility. By placing an equivalence relation into this hierarchy, we come to understand how complex it is in comparision with other equivalence relations. In particular, we say that one equivalence relation E is strictly simpler than F, if E reduces to F but not conversely.
It sometimes happens that one has a classification problem E and is able to provide a classification by assigning to each structure a countable list of data, such that two structures are equivalent iff they have the same data. This amounts to a reduction of E to the equality relation =, for two structures are E equivalent iff their data is equal. Such relations that reduce to equality are called smooth, and lay near the bottom of the hierarchy of Borel equivalence relations. These are the simplest equivalence relations. Thus, one way of showing that a relation is comparatively simple, is to show that it is smooth, and to show it is comparatively hard, show that it is not smooth.
The subject of Borel equivalence relation theory, as now developed by A. Kechris, G. Hjorth, S. Thomas and many others, is focused on placing many of the natural classification problems of mathematics into this hierarchy. Some of the main early results are the following interesting dichotomies:
Theorem.(Silver dichotomy) Every Borel equivalence relation E either has only countably many equivalence classes or = reduces to E.
The relation E0 says that two binary sequences are equivalent iff they agree from some point onward. It is easy to see that = reduces to E0, and an elementary argument shows that E0 does not reduce to =. Thus, E0 is strictly harder than equality. Moreover, it is a kind of next-step up in the hiearchy, in light of the following.
Theorem.(Glimm-Effros dichotomy) Every Borel equivalence relation E either reduces to = or E0 reduces to E. 
The subject continues with many interesting results that gradually illuminate more and more of the hierarchy of Borel equivalence relations. For example, the Feldman-Moore theorem shows that every Borel equivalence relation E having every equivalence class countable is the orbit equivalence of a countable group of Borel bijections of the space. The relation Eoo is the orbit equivalence of the left-translation action of the free group F2 on its power set. This relation is complete for the countable Borel equivalence relations, in the sense that every countable Borel equivalence relation reduces to it. It's great stuff!<|endoftext|>
TITLE: Faithful characters of finite groups 
QUESTION [8 upvotes]: Related to a previous question I am asking furthermore a proof
for the following:
Question 1: If $\chi$ is a faithful irreducible character of a
finite group $G$ then the regular character of $G$ is a polynomial
with integer coefficients in $\chi$?
I know this fact is true since there is a generalization of it for
Hopf algebras in Corollary 19 of the paper FSU96-08 from here.
The proof from that paper is a little complicated using some
(although elementary) results on norms and inner products.
I was wondering if  anyone knows a different proof of this.
Using the Stone - Weierstrass method mentioned in the previous
question, I am asking further if the following is true:
Question 2: If $\chi$ is a faithful irreducible character of a
finite group $G$ does any character of $G$ is a complex polynomial
in $\chi$?

REPLY [5 votes]: I put an answer (due to Blichfeldt, not me) to essentially this question
at your earlier question. To address the problem raised by Richard Stanley,
one result I know in this direction is by John Thompson: if $\chi$ is an irreducible
character of a finite group $G$, then there are more than $|G|/3$ elements at which 
the value taken by $\chi$ is either zero or a root of unity.<|endoftext|>
TITLE: The matrix tree theorem for weighted graphs
QUESTION [21 upvotes]: I am interested in the general form of the Kirchoff Matrix Tree Theorem for weighted graphs, and in particular what interesting weightings one can choose.
Let $G = (V,E, \omega)$ be a weighted graph where $\omega: E \rightarrow K$, for a given field $K$; I assume that the graph is without loops.
For any spanning tree $T \subseteq G$ the weight of the tree is given to be,
$$m(T) = \prod_{e \in T}\omega(e)$$
and the tree polynomial (or statistical sum) of the graph is given to be the sum over all spanning trees in G,
$$P(G) = \sum_{T \subseteq G}m(T)$$
The combinatorial laplacian of the graph G is given by $L_G$, where:
$$L_G = \begin{pmatrix} \sum_{k = 1}^n\omega(e_{1k}) & -\omega(e_{12}) & \cdots & -\omega(e_{1n}) \\\ -\omega(e_{12}) & \sum_{k = 1}^n\omega(e_{2k}) & \cdots & -\omega(e_{2n})\\\
\vdots & \vdots & \ddots & \vdots \\\
-\omega(e_{1n}) & -\omega(e_{2n}) & \cdots & \sum_{k = 1}^n\omega(e_{nk}) \end{pmatrix} $$
where $e_{ik}$ is the edge between vertices $i$ and $k$, if no edge exists then the entry is 0 (this is the same as considering the complete graph on n vertices with an extended weighting function that gives weight 0 to any edge not in G). The matrix tree theorem then says that the tree polynomial is equal to the absolute value of any cofactor of the matrix. That is,
$$P(G) = \det(L_G(s|t))$$
where $A(s|t)$ denotes the matrix obtained by deleting row $s$ and column $t$ from a matrix $A$.
By choosing different weightings one would expect to find interesting properties of a graph G. Two simple applications are to give the weighting of all 1's. Then the theorem allows us to count the number of spanning trees with ease (this yields the standard statement of the Matrix Tree Theorem for graphs). Alternatively, by giving every edge a distinct formal symbol as its label then by computing the relevant determinant, the sum obtained can be read as a list of all the spanning trees.
My question is whether there are other interesting weightings that can be used to derive other interesting properties of graphs, or for applied problems.

REPLY [2 votes]: Another interesting weighting comes from stochastic matrices, and was used by Tuncel and me to find a completely new invariant for Markov shifts (Douglas Lind and Selim Tuncel, A Spanning Tree Invariant for Markov Shifts, Codes, Systems, and Graphical Models, IMA Volumes in Mathematics and its Applications 123 (2001), 487--497).
Let $P=[p_{ij}]$ be an $r \times r$ stochastic matrix, so that $p_{ij}\ge0$ and row sums are all 1. Let $G_P$ denote the directed graph with $r$ vertices and with one edge from $i$ to $j$ if and only if $p_{ij}>0$ and no other edges. Assume that $G_P$ is irreducible. The shift of finite type $X_P$ determined by $G_P$ is the space of all bi-infinite trips on $G_P$, which is compact. Also, $P$ determines a unique probability measure $\mu_P$ (called the Parry measure) that is invariant under the left shift $\sigma_P $ on $X_P$. These measure-preserving transformations are key examples in ergodic theory and have been extensively studied.
Given $P$, assign an edge in $G_P$ from $i$ to $j$ the weight $p_{ij}>0$. If $T$ is a spanning tree in $G_P$, define the weight of $T$ to be the product of the weights of its edges. Finally, define $\tau(P)$ to be the sum of the weights of all spanning trees in $G_P$.
If $Q$ is another such matrix, of possibly a different size, we say that the dynamical systems $(X_P,\sigma_P,\mu_P)$ and $(X_Q,\sigma_Q,\mu_Q)$ are block isomorphic if there is a shift commuting measure-preserving homeomorphism between them. The main result is that $\tau$ is an invariant of block isomorphism.
We initially observed this experimentally using Mathematica. The $\tau$ invariant is defined using a simple finite sum, and differs from more standard dynamical invariants like entropy that use asymptotic limiting behavior.<|endoftext|>
TITLE: Flat cohomology and Picard groups
QUESTION [11 upvotes]: Let $(R,m)$ be a local complete intersection of dimension $3$. Let $X=Spec(R)$ and $U=Spec(R) -\{m\}$ be the punctured spectrum of $R$. I am trying to understand the following comment by Gabber (see it  here , page 1975-1976):
$Pic(U)$ is torsion-free is equivalent to the flat local cohomology $H_{\{m\}}^2(X, \mu_n)=0$ for $n>0$
So let's try to go through a possible argument (this is from my very limited understanding, so feel free to correct me here). The sequence: $  0 \to \mu_n \to {\mathbb G_m}   \xrightarrow{t\mapsto t^n} {\mathbb G_m} \to 0 $
is exact on the flat site over $U$,  $U_{fl}$. Thus the long exact sequence of flat cohomology shows:
$  0 \to H^1(U_{fl},\mu_n) \to H^1(U_{fl},{\mathbb G_m})   \xrightarrow{\times n} H^1(U_{fl},{\mathbb G_m})$
We know that $  H^1(U_{fl},{\mathbb G_m})  \cong Pic(U)$. Now excision gives:
$H^1(X_{fl},\mu_n)  \to H^1(U_{fl},\mu_n) \to H^2_{\{m\}}(X_{fl},\mu_n) \to H^2(X_{fl},\mu_n)$
Obviously $H^1(X_{fl},\mu_n)=0$ (note that $H^1(X_{fl},{\mathbb G_m})\cong Pic(X)=0$, as $R$ is local). EDIT: obviously, it is not true, the argument becomes quite subtle, please see Emerton's answer below. 
But why is $H^2(X_{fl},\mu_n)  =0$? It is true for curves, but Milne warned against using flat cohomology for higher dimensions in his note  on duality theorems (see the first page of Chapter III). In fact, I could not find any reference about flat cohomology in higher dimensions. So:
Can some one help finish/fix the argument above? Is there a reference from which I can quote the ``facts" I used above?

REPLY [11 votes]: This is not a complete answer by any means, but is intended to get the ball rolling.
First of all, it need not be the case that $H^1(X_{fl},\mu_n) = 0.$  Rather, what follows
from the vanishing of $H^1(X_{fl},{\mathbb G}_m)$ is that 
$H^1(X_{fl},\mu_n) = R^{\times}/(R^{\times})^n.$
(This is not always trivial; imagine 
e.g. that $R$ is a non-algebraically closed field.  You might have been thinking of the case
when $X$ is projective and smooth over an algebraically closed field, when the $H^0$-part of the exact sequence is itself exact, and so can be omitted from consideration.  That is not the case here.)
Secondly, this doesn't hurt your arguments, because the same consideration of $H^0$-terms has to be made for the cohomology of $U$. Since $X$ is three dimensional and a complete intersection, restriction induces an isomorphism 
$H^0(X,\mathcal O) \cong H^0(U,\mathcal O)$, and so also an isomorphism 
$H^0(X,\mathcal O^{\times})\cong H^0(U,\mathcal O^{\times}),$ and so also isomorphisms
$H^0(X_{fl},\mu_n) \cong H^0(U_{fl},\mu_n)$ and
$H^0(X_{fl},{\mathbb G}_m)\cong H^0(U_{fl},{\mathbb G}_m).$  Thus in fact one finds that
the $n$-torsion in Pic$(U)$ is equal to the cokernel of the injection
$H^1(X_{fl},\mu_n) \hookrightarrow H^1(U_{fl},\mu_n).$   And as your analysis shows, this cokernel embeds into $H^2_{\{m\}}(X_{fl},\mu_n)$, with the cokernel of that embedding itself embedding into $H^2(X_{fl},\mu_n).$
So what can be said about this latter cohomology group?
Since $H^1(X_{fl},{\mathbb G}_m)$ vanishes, as you observed, one finds that $H^2(X_{fl}, \mu_n)$ coincides with the $n$-torsion in the cohomological Brauer group $H^2(X_{fl},{\mathbb G}_m).$  (Here I am using the fact that since ${\mathbb G}_m$ is smooth, flat and etale cohomology coincide, so $H^2(X_{fl},{\mathbb G}_m) = H^2(X_{et}, {\mathbb G}_m).$)   So it seems that one wants to kill off the torsion in this Brauer group.
I don't see why this need be true, but what one actually needs is that $H^2(X_{fl},\mu_n)
\rightarrow H^2(U_{fl},\mu_n)$ is injective.  Since $H^2(X_{fl},\mu_n)$ embeds into
$H^2(X_{fl},{\mathbb G}_m)$, it would be enough to show that the restriction
$H^2(X_{fl},{\mathbb G}_m) \to H^2(U_{fl},{\mathbb G}_m)$ induces an injection on torsion.   Might this be some kind of purity result on Brauer groups of the kind Gabber discusses in his abstract?  It would be related to a vanishing of (torsion  in) $H^2_{\{m\}}(X_{fl}, {\mathbb G}_m)$.  Somewhere (maybe here?) one presumably has to make use of the dimension and lci hypotheses.
P.S.  You may well just want to email Gabber to ask him about this.  If you do, and you get an answer, please share it!
EDIT:  This is an excerpt from the email referred to in Hai Long's comment below:
To learn about these kinds of arguments, my advice is to
do just what you are doing.  One works with the exact sequence
linking $\mu_n$ and ${\mathbb G}_m$, as you did.
Number theorists (at least of a certain stripe) have some advantages
with this, because the case $X =$ Spec $K$ ($K$ a field) comes up a lot
under the name of Kummer theory, and also Mazur in one of his
famous papers uses a lot of flat cohomology.  But in the end, the
formalism is just the one you used in your question.
Then, typically, one has to inject something additional that
is less formal.  My suggestion would be to look at de Jong's proof
of Gabber's result showing $Br'(X) = Br(X)$ discussed in his abstract.
(There is a write-up on de Jong's web-page.)
Reading the proof of a result like this might give some insight
into how to work with Brauer groups in a less formal way.<|endoftext|>
TITLE: Examples of the varying strengths of topological invariants
QUESTION [8 upvotes]: In my first algebraic topology class, I remember being told that the simplest reason for homology was to distinguish spaces. For example, if is X=circle and a Y= wedge of a circle and a 2-sphere then X and Y have the same fundamental group, so the fundamental group isn't strong enough to distinguish them. We need to look at the other homotopy groups or homology to tell them apart. I'm looking for a variety of other examples of this nature. The examples I'm wondering about are 

Same homology groups
Same cohomology groups, but different cohomology rings
Same cohomology rings (but maybe different Steenrod operations?)

If I put more thought into it, I could come up with others questions like these. Any other examples/thoughts along these lines would be very welcome! (I have examples for the first one, but I'm wondering what others will say.)

REPLY [6 votes]: To change up the nature of the responses some, IMO a good theorem to think about is the Kan-Thurston theorem. It states that given any space $X$ you can find a $K(\pi, 1)$ space $Y$ and a map $f : Y \to X$ inducing isomorphisms $f_* : H_i Y \to H_i X$, $f^* : H^i X \to H^i Y$ for all coefficients (it can be souped-up to allow local coefficients) and all $i$.  The map $\pi_1 Y \to \pi_1 X$ is onto. 
So from the point of view of cohomology algebras with Steenrod operations, these spaces are the same.  One way to "spin" this would be to say the fundamental group is a far stronger invariant than anything (co)homological.<|endoftext|>
TITLE: Simplicial and cubical decompositions of low valence
QUESTION [7 upvotes]: Every surface can be triangulated in such a way that at most 7 trianlges meet at one vertex. Every surface can be decomposed in squares such that at every vertex at most 5 suqares meet. For surfaces of genus more than 1 this is the low bound. 
What happen in higher dimensions, for example for 3 and 4-manifolds, ect...? It should be easy to show that for every dimension $n$ there are numbers $S(n)$ and and $C(n)$ such that every manifold $M^n$ admits a simplicial decomposition with at most $S(n)$ 
simplexes at every vertex and a cubical decomposition with at most $C(n)$ cubes at every vertex. The refference of Gil below confirms this for $n=3$.
Here are three questions (I suspect they are hard).
1) Can it be proven that $C(n)>2^n$?
2) Can it be proven that $S(n)>\frac{Vol(S^n)}{Vol(\Delta^n)}$, where $\Delta^n$ is the spherical tetrahedron with edge of length $\frac{\pi}{3}$ in the unit sphere $S^n$.
3) Is there any reasonable estimation for $C(n)$ and $S(n)$ from above?

REPLY [6 votes]: For n = 3, Cooper and Thurston show (see Gil's answer for citation) that any 3-manifold can be paved with cubes using only 3 vertex link types.  These three types have 6, 8 and 10 cubes at each vertex.  Cooper and Thurston then subdivide the cubes into simplices to show that any 3-manifold can be triangulated with 5 vertex link types.  A slightly more clever subdivision strategy shows that in fact one only needs 3 vertex link types in a triangulation.  The largest of these types has 30 simplices around a vertex.
C&T use a result of Montesinos which says that any 3-manifold branch covers the 3-sphere with branch locus the borromean rings.  This implies that any 3-manifold has a Euclidean orbifold-ish structure where the non-smooth points have neighborhoods isomorphic to an interval cross a 2-dimensional cone with cone angle 3/2 pi or 5/2 pi.  If an analogous result were true in higher dimensions then one could show that any n-manifold could be paved with hypercubes using only 3 vertex link types.  The largest vertex link would use 5/4 * 2^n cubes.<|endoftext|>
TITLE: Goldbach-type theorems from dense models?
QUESTION [15 upvotes]: I'm not a number theorist, so apologies if this is trivial or obvious.
From what I understand of the results of Green-Tao-Ziegler on additive combinatorics in the primes, the main new technical tool is the "dense model theorem," which -- informally speaking -- is as follows:

If a set of integers $S \subset N$ is a dense subset of another "pseudorandom" set of integers, then there's another set of integers $S' \subset N$ such that $S'$ has positive density in the integers and $S, S'$ are "indistinguishable" by a certain class of test functions.

They then use some work of Goldston and Yildirim to show that the primes satisfy the given hypothesis, and note that if the primes failed to contain long arithmetic/polynomial progressions and $S'$ did, they'd be distinguishable by the class of functions. Applying Szemeredi's theorem, the proof is complete.
Obviously I'm skimming over a great deal of technical detail, but I'm led to believe that this is a reasonably accurate high-level view of the basic approach.
My question(s), then: Can one use a similar approach to obtain "Goldbach-type" results, stating that every sufficiently large integer is the sum of at most k primes? Is this already implicit in the Goldston-Yildirim "black box?" If we can't get Goldbach-type theorems by using dense models, what's the central obstacle to doing so?

REPLY [26 votes]: Yes, this can be done, provided that k is at least 3.  A typical example is given in this paper: http://arxiv.org/abs/math/0701240 .  (This uses a slightly older Fourier-based method of Ben that predates his work with me, but is definitely in the same spirit - Ben's paper was very inspirational for our joint work.)  
For k=2, unfortunately, the type of indistinguishability offered by the dense model theorem is  not strong enough to say anything.  The k=2 problem is very similar to the twin prime conjecture (in both cases, one is trying to seek an additive pattern in the primes with only one degree of freedom).  If one deletes all the twin primes from the set of primes, one gets a new set which is indistinguishable from the set of primes in the sense of the dense model theorem (because the twin primes have density zero inside the primes, by Brun's theorem), but of course the latter set has no more twins.  One can pull off a similar trick with representations of N as the sum of two primes.  But one can't do it with sums of three or more primes - there are too many representations to delete them all just by removing a few primes.<|endoftext|>
TITLE: Theories of Noncommutative Geometry
QUESTION [44 upvotes]: [I have rewritten this post in a way which I hope will remain faithful to the questioner and make it seem more acceptable to the community.  I have also voted to reopen it. -- PLC]
There are many ways to approach noncommutative geometry. 
What are some of the most important currently known approaches?  Who are the principal creators of each of these approaches, and where are they coming from?  E.g., what more established mathematical fields are they using as jumping off points?  What problems are they trying to solve?
Two examples: 
1) The Connes school, with an approach from C$^*$-algebras/mathematical physics.
2) The Kontsevich school, with an approach from algebraic geometry.

REPLY [9 votes]: There is a beautiful three-part series of lectures by Jonathan Block that introduces both the Connes and Kontsevich schools of non-commutative geometry:
http://www.math.upenn.edu/~tpantev/rtg09bc/lecnotes/block-ncg.pdf
One of the main motivating examples throughout the lectures is Lusztig's proof proof of the Novikov conjecture.<|endoftext|>
TITLE: Teichmuller Theory introduction
QUESTION [23 upvotes]: What is a good introduction to Teichmuller theory, mapping class groups etc., and relation to moduli space of curves or Riemann surfaces?

REPLY [2 votes]: McMullen's notes (http://www.math.harvard.edu/~ctm/home/text/class/harvard/275/09/html/base/rs/rs.pdf)<|endoftext|>
TITLE: Hasse principle for a group
QUESTION [7 upvotes]: In the paper "Hasse principle" for $PSL_2 (\mathbb Z)$ and $PSL_2(\mathbb F)$ there's a definition of a Hasse principle for a group $G$, but I don't completely get it. Is there a more motivated reformulation of this definition?
Why I am interested: local-global principles are often very interesting in arithmetic geometry, so when I noticed a paper with this title I looked at it to see whether this proves something geometric. 
As said below, one can formulate a problem of computing a group $Sha$ defined by a $g$-module $G$ and a family of subgroups $h_i\in G$, but the actual computation in the paper is for a specific choice of $h_i$, and I can't parse if there is an application of interest. Is it so?
(I suspect this problem arises when you try to prove Hasse principle for equations, like $x^n = a$ but with different Galois groups, see their first paper, though the results there have a mistake, corrected in the next one)

REPLY [6 votes]: Given a group $g$, a possibly nonabelian $g$-module $G$, and a family of subgroups $(h_i)_{i \in I}$ of $g$, Ono defines a pointed Shafarevich-Tate set, say $(S,0)$.  He says the Hasse Principle holds for $G$ (with respect to the data of the $g$-action and the set of subgroups ${h_i}$) if the $S = {0}$.
Namely, for each subgroup $h_i$ of $g$ there is a restriction map in Galois cohomology 
$r_i: H^1(g,G) \rightarrow H^1(h_i,G)$.
(The restriction map is defined on one-cocycles merely by pulling back by the inclusion map $h_i \hookrightarrow g$.)  Restriction carries the distinguished (trivial) class to the trivial class.  
Then $S$ is defined as the intersection of the kernels of all the $r_i$'s.  Evidently the 
trivial class $0$ lies in $S$, so $(S,0)$ is a pointed set.
All of the above was just a detailed review of the beginning of Ono's paper.  Now let me explain why this generalizes the notion of the Shafarevich-Tate group of an abelian variety A over $\mathbb{Q}$ [or take a more general global field, if you like].
The Shafarevich-Tate group $Sha(A,\mathbb{Q})$ is the set of all principal homogeneous spaces (henceforth phs) $X$ under $A$ which have $\mathbb{Q}_p$-points for every prime $p$ and also $\mathbb{R}$-points.  Because the automorphism group of a phs under a group $A$ is just $A$ itself, by [what I call] the first principle of Galois cohomology, the pointed set of all phs under A is isomorphic to the Galois cohomology set
$H^1(\mathbb{Q},A) = H^1(\mathfrak{g}_{\mathbb{Q}},A(\overline{\mathbb{Q}}))$,
where $\mathfrak{g}_{\mathbb{Q}} = Aut(\overline{\mathbb{Q}}/\mathbb{Q})$ is the absolute Galois group of $\mathbb{Q}$.  [Since $A$ is commutative, the $H^1$ is itself a commutative group, whereas for nonabelian $A$ it would in general be only a pointed set.]
Thus here we have $G = A(\overline{\mathbb{Q}})$ and $g = \mathfrak{g}_{\mathbb{Q}}$. What are the $h_i$'s?  For each prime $p$, $h_p$ is the Galois group of $\mathbb{Q}_p$, 
viewed as a subgroup of $\mathbb{Q}$ (i.e., as a decomposition group at $p$) via choosing an embedding of the algebraic closure of $\mathbb{Q}$ into the algebraic closure of $Q_p$; also we define $h_{\infty}$ to be the restriction to the subgroup generated by a complex conjugation, i.e., a group isomorphic to $Aut(\mathbb{C}/\mathbb{R})$.  A cohomology class lies in the kernel of $h_p$ iff the corresponding phs acquires a point after base extension to $Q_p$ (and similarly for $h_{\infty}$.  
Thus the Shafarevich-Tate pointed set of $A$ is indeed a special case of Ono's construction.
That's the motivation I can give you.  As to exactly why Ono's particular choice of Shafarevich-Tate set for an arbitrary group $G$ -- namely take $g = G$ with the conjugation action, and let $(h_i)_{i \in I}$ be the family of cyclic subgroups of $G$ -- is interesting and natural...I can't help you there, and I'd like to know myself.
Why are you interested in this paper?<|endoftext|>
TITLE: When is a coarse moduli space also a fine moduli space?
QUESTION [9 upvotes]: Given a moduli problem, it appears that nonexistence of automorphisms is a necessary condition for existence of a fine moduli space(is this strictly true?). 
In any case, assuming the above, what additional condition on a moduli problem in algebraic geometry will make sure that a coarse moduli space is in fact a fine moduli space?
In the n-lab page on Deligne-Mumford, the following appears.
Deligne-Mumford stacks correspond to moduli problems in which the objects being parametrized have finite automorphism groups.
Also, a few problematic examples I have heard of, had infinite automorphism groups. 
Therefore, is it true that for a moduli problem in which the stack is Deligne-Mumford, and where there are no automorphisms, existence of a coarse moduli space would imply the existence of a fine moduli space?

REPLY [14 votes]: Since nobody gave a reference yet, in my paper "Artithmetic moduli of generalized elliptic curves" I included a proof that an Artin stack whose geometric points have trivial automorphism schemes is necessarily an algebraic space. See Theorem 2.2.5(1) there; I am sure this is a folklore fact (which I inserted there because I didn't know a reference, and to my surprise seems to not be stated in the L-MB book).  So that answers the original question: if the moduli problem is an Artin stack and a coarse moduli space exists then it is a fine moduli space (meaning that the moduli problem is an algebraic space) if and only if objects over algebraically closed fields have trivial automorphism schemes (stronger than just trivial automorphism groups, except in the DM case when equivalent since then such groups are etale).
I wrote that paper in the days before I realized that non-qs algebraic spaces made sense, so I had the convention throughout (following the L-MB book) that diagonals are separated and especially quasi-compact.  I have not revisited the proof to see the effect of weakening these assumptions (especially the q-c assumption) on the diagonal.  I should do that some day.<|endoftext|>
TITLE: Ways to prove the fundamental theorem of algebra
QUESTION [202 upvotes]: This seems to be a favorite question everywhere, including Princeton quals. How many ways are there?
Please give a new way in each answer, and if possible give reference. I start by giving two:

Ahlfors, Complex Analysis, using Liouville's theorem.
Courant and Robbins, What is Mathematics?, using elementary topological considerations.

I won't be choosing a best answer, because that is not the point.

REPLY [11 votes]: Here is a variant of d'Alembert's argument using the minimum of $|p(z)|$. It has the advantage that it proves more generally the Gelfand-Mazur theorem (usually proved by complex analysis): Any Banach field $K$ over $\mathbb C$ is $\mathbb C$ itself. Indeed, this gives the fundamental theorem of algebra by applying it to any putative finite extension of $\mathbb C$.
Let $x\in K$. The map
$$
\mathbb C\to \mathbb R_{\geq 0}: z\mapsto |x-z|
$$
is continuous and gets large for $|z|$ large, so attains its minimum at some $z\in \mathbb C$. Replacing $x$ by $x-z$, we get $(\ast)$: for all $w\in \mathbb C$, one has $|x-w|\geq |x|$. We claim that then $x=0$. If not, we can rescale $x$ by a real number so that $|x|=2$. Now take some large $n$ and consider the identity
$$ \prod_{i=0}^{n-1} (x-\zeta_n^i) = x^n-1.$$
Each factor on the left is of norm at least $|x|=2$ (by $(\ast)$), so the left-hand side is of norm at least $|x-1|2^{n-1}\geq 2^n$. The right-hand side is bounded in norm by $2^n+1$. Taking the limit as $n\to \infty$, we see that $|x-1|=2$. Applying this argument to $x-1$ in place of $x$, this gives the absurd $2=|x|=|x-1|=|x-2|=\ldots=|x-5|\geq 5-|x|=3$.
Edit: This proof is due to Ostrowski, see Section 7 in "Über einige Lösungen der Funktionalgleichung $\varphi(x) \varphi(y) = \varphi(xy)$", 1916. (h/t Mohan Ramachandran!)
The proof uses the existence of infinitely many roots of unity; but $\zeta_4=i$ and if $\zeta_{2^n} = a_n + b_n i$ then $\zeta_{2^{n+1}} = a_{n+1} + b_{n+1} i$ where $a_{n+1}=\sqrt{\frac{1+a_n}2}$ and $b_{n+1}=\sqrt{\frac{1-a_n}2}$. (But one can also rewrite the proof so as to only use the existence of $\zeta_8=\frac{1+i}{\sqrt{2}}$.)<|endoftext|>
TITLE: D-modules, deRham spaces and microlocalization
QUESTION [47 upvotes]: Given a variety (or scheme, or stack, or presheaf on the category of rings), some geometers, myself included, like to study D-modules.  The usual definition of a D-module is as sheaves of modules over a sheaf of differential operators, but for spaces that aren't smooth in some sense, this definition doesn't work that well, and you want to use a different definition.  My overall question is how to reinterpret microlocalization in this alternative definition.
deRham spaces
This definition is that a D-module on $X$ is a quasi-coherent sheaf on a new space $X_{dR}$, the deRham space of $X$.  It's easiest to define this is in terms of its functor of points:  a map of Spec R to $X_{dR}$ is by definition a map of Spec $R/J_R$ to $X$ where $J_R$ is the nilpotent radical of $R$.  So this is not a topological space, but it is a sheaf on the big Zariski site, and I can make sense of a quasi-0-coherent sheaf on one of those.  For more details, you can see the notes of Jacob Lurie on these.
More informally $X_{dR}$ is $X$ "with all infinitesimally close points identified."  A sheaf on this space is like a D-module in that a D-module is a sheaf with a connection, i.e. where the fibers of infinitesimally close points are identified.  You'll note, I say "space" here, since I want to be vague about what this object is.  It's very hard from being a scheme, but I believe it is a (EDIT: not actually algebraic!) stack.
microlocalization
Now, one of the lovely things about D-modules is that they have a secret life on the cotangent bundle of X.  You might think a D-module is a sheaf on X, but this is not the whole picture: there is also a microlocal version of things.  
The sheaf of functions on $T^*X$ has a quantization $\mathcal{O}^h$; this is a non-commutative algebra over $\mathbb{C}[[h]]$ such that $\mathcal{O}^h/h\mathcal{O}^h\cong \mathcal{O}_{T^*X}$, defined using Moyal quantization.
There's a ring map $p^{-1}\mathcal{D}\to \mathcal{O}^h[h^{-1}]$, and thus a functor from D-modules to sheaves of $\mathcal{O}^h[h^{-1}]$-modules on $T^*X$ given by $\mathcal{O}^h[h^{-1}]\otimes_{p^{-1}\mathcal{D}}\mathcal{M}$, called microlocalization, because it makes D-modules even more local than they were before.  This is an equivalence between D-modules and $\mathbb{C}^*$-equivariant $\mathcal{O}^h[h^{-1}]$-modules. 
Given an $\mathbb{C}^*$ invariant open subset $U$ of $T^*X$, one can look at $\mathcal{O}^h[h^{-1}]$-modules on $U$, and obtain a microlocalized category of D-modules, which has all kinds of interesting geometry one couldn't see before.  I'm particularly interested in the semi-stable points for the action of some group $G$ on $X$ (extended to $T^*X$).
my question:
Now, I'm something of a convert to derived algebraic geometry, so it feels intuitive to me that anything one has to say about D-modules should be sayable using deRham spaces.  On the other hand, I have no idea how microlocalization can be phrased in this way.  Do any of you out in MathOverflowLand?

REPLY [39 votes]: The first thing to say is that the abelian category of 
sheaves on the de Rham space is only a good model 
for D-modules if you're in the smooth setting, or very close to it (see for example arXiv:math/0212094 for a setting where all the different notions agree).. so unless you're fully derived you need to be careful with this identification.
In any case, rather than talk about the de Rham space you can talk about crystals,
which give the right notion in general -- see for example chapter 7 (section 10 or 11) of Beilinson-Drinfeld on Quantization of Hitchin for an excellent discussion.
Another picture you might like better is as modules over the enveloping algebroid of the tangent complex -- ie everything is fine if you replace naive differential operators by their correct derived analog. Yet another picture is as dg modules for the de Rham complex. My favorite is as S^1-equivariant sheaves on the derived loop space of X.
(I assume for all of these we're in characteristic zero, otherwise there are many different analogs of D-modules..)
In any case micrlocalization can be said in terms of the deformation to the normal cone of the de Rham groupoid - or the Hodge filtration on nonabelian cohomology, after Simpson. (A good reference for this is Simpson's paper with that title and the awesome preprint on D-modules on stacks by Simpson and Teleman available on the latter's webpage.) Namely there's a canonical deformation (the Hodge stack) from the de Rham space to a stack (not a space this time!) which is the quotient of X by the tangent bundle (acting trivially) -- sheaves on which are the same as quasicoherent sheaves on the cotangent bundle. On the level of sheaves this is just the deformation quantization from sheaves on the cotangent bundle to D-modules with a parameter h - ie we consider modules over the Rees algebra of D rather than over D itself. The category of Rees modules sheafifies over the projectivized cotangent bundle, and so you can define microlocal categories by restricting to your favorite open subsets.
This is not so special to the de Rham stack: we're just saying jets of sections of a formal deformation of the category of sheaves on a variety sheafify over this variety. Once you know in general that D-modules degenerate to sheaves on the cotangent bundle, and this is true in arbitrary generality once you define both sides correctly, you can microlocalize (if you take into account correctly the C^* equivariance of the deformation).<|endoftext|>
TITLE: Universality of zeta- and L-functions
QUESTION [24 upvotes]: Voronin´s Universality Theorem (for the Riemann zeta-Function) according to Wikipedia: Let $U$ be a compact subset of the "critical half-strip" $\{s\in\mathbb{C}:\frac{1}{2}<Re(s)<1\}$ with connected complement. Let $f:U \rightarrow\mathbb{C}$ be continuous and non-vanishing on $U$ and holomorphic on $U^{int}$. Then $\forall\varepsilon >0$ $\exists t=t(\varepsilon)$ $\forall s\in U: |\zeta(s+it)-f(s)|<\varepsilon $. 
(Q1) Is this the accurate statement of Voronin´s Universality Theorem? If so, are there any (recent) generalisations of this statement with respect to, say, shape of $U$ or conditions on $f$ ? (If I am not mistaken, the theorem dates back to 1975.)
(Q2) Historically, were the Riemann zeta-function and Dirichlet L-functions the first examples for functions on the complex plane with such "universality"? Are there any examples for functions (on the complex plane) with such properties beyond the theory of zeta- and L-functions?
(Q3) Is there any known general argument why such functions (on $\mathbb{C}$) "must" exist, i.e. in the sense of a non-constructive proof of existence? (with Riemann zeta-function being considered as a proof of existence by construction). 
(Q4) Is anything known about the structure of the class of functions with such universality property, say, on some given strip in the complex plane? 
(Q5) Are there similar examples when dealing with $C^r$-functions from some open subset of $\mathbb{R}^n$ into $\mathbb{R}^m$ ?
Thanks in advance and Happy New Year!

REPLY [4 votes]: To (Q4): I am not a specialist in the field of universal functions, but I believe what you are loooking for are rather manifolds of universal functions. Try googling for manifolds of Birkhoff-universal functions. There are definitely plenty of interesting results in that topic.<|endoftext|>
TITLE: Number of uniform rvs needed to cross a threshold
QUESTION [14 upvotes]: Let $N(x)$ be the number of uniform random variables (distributed in $[0,1]$) that one needs to add for the sum to cross $x$ ($x > 0$). The expected value of $N(x)$ can be calculated and it is a very cool result that $E(N(1)) = e$. The expression for general $x$ is 
$E(N(x)) = \sum_{k=0}^{[x]} (-1)^k \frac{(x-k)^k}{k!} e^{x-k}$
where $[x]$ is the largest integer less than $x$. When $x$ is large, this function $E(N(x))$ grows linearly (as one would expect). Computer simulations suggest that $E(N(x)) \approx 2x + 2/3$. My question is as follows: is it possible to guess this form for $E(N(x))$ without actually computing it? The $2x$ part is intuitive but is there a good intuition for why there is an additive constant and why that value is $2/3$?
My motivation is as follows: When I computed $E(N(x))$ using the formula above for large values of $x$, I came across this asymptotic and found it surprising that a polynomial equation in powers of $e$ gives values that are very close to a rational number. I am very interested in knowing the reason (if any) behind it. Thanks in advance for any help.

REPLY [7 votes]: This problem has delighted me since I first encountered it before college. I wrote up a generalization for a less mathematical audience in a poker forum.
One way to look at the original is that if f(x) = E(N(X)), then $f(x) = 1 + \int_{x-1}^x f(t) dt$, satisfying the initial condition that $f(x) = 0$ on $(-1,0)$. $f$ is 1 more than the average of $f$ on the previous interval.
We can get rid of the constant by using $g(x) = f(x) - 2x$ which satisfies $g(x) = \int_{x-1}^x g(t) dt$. $g$ is equal to its average on the previous interval. Now the initial condition becomes nontrivial: $g(x) = -2x$ on $(-1,0)$, and it is from this $-2x$ that the asymptotic value of $2/3$ comes.
$g$ is asymptotically constant, and we can guess from geometry and verify that the weighted average 
$\int_0^1 2t~g(x+t)dt$ is independent of $x$, hence equal to the asymptotic value of $g$. The value at $x=-1$ is 
$\int_0^1 4t(1-t)dt = \frac23$, so that is the asymptotic value of $g$. 
The same technique works for more general continuous distributions supported on $\mathbb R^+$ than the uniform distribution, and the answer turns out to be remarkably simple. Let's find an analogous conserved quantity. 
Let $\alpha$ be the density function. Let the $n$th moment be $\alpha_n$. We'll want $\alpha_2$ to exist.
Let $f(x)$ be the average index of the first partial sum which is at least $x$ of IID random variables with density $\alpha$. $f(x) = 1 + \int_{-\infty}^x f(t) \alpha(x-t) dt$, and $f(x) = 0$ for $x\lt0$. (We could let the lower limit be 0, too.)
Let $g(x) = f(x) - x/\alpha_1$. Then $g(x) = \int_{-\infty}^xg(t)\alpha(x-t)dt$, with $g(x) = -x/\alpha_1$ on $\mathbb R^-$.
We'll choose $h$ so that $H(x) = \int_{-\infty}^x g(t) h(x-t) dt$ is constant. 
$0 = H'(x) = g(x)h(0) + \int_{\infty}^x g(t) h'(x-t)dt.$ 
This integral looks like the integral equation for $g$ if we choose $h'(x) = c\alpha(x)$. $c=-1$ satisfies the equation. So, if $h(x) = \int_x^\infty \alpha(t)dt$ then $H(x)$ is constant. 
Let the asymptotic value of $g$ be $v$. Then the value of $H(x)$ is both $H(\infty) = v~\alpha_1$ and 
$H(0) = \int_{-\infty}^0 g(t) h(0-t)dt$
$H(0) = 1/\alpha_1 \int_0^\infty y~h(y) dy$
$H(0) = 1/(2\alpha_1) \int_0^\infty y^2 ~\alpha(y)dy~~$ (by parts)
$H(0) = \alpha_2 / (2 \alpha_1)$
$v~ \alpha_1 = \alpha_2 / (2 \alpha_1)$
$v = \alpha_2 / (2 \alpha_1^2)$.
So, $f(x)$ is asymptotic to $x/\alpha_1 + \alpha_2/(2 \alpha_1^2)$.
For the original uniform distribution on [0,1], $\alpha_1 = \frac12$ and $\alpha_2 = \frac13$, so $f(x) = 2x + \frac23 + o(1)$. 
As a check, an exponential distribution with mean 1 has second moment 2, and we get that $f(x)$ is asymptotic to $x+1$. In fact, in that case, $f(x) = x+1$ on $\mathbb R^+$. If you have memoryless light bulbs with average life 1, then at time $x$, an average of $x$ bulbs have burned out, and you are on the $x+1$st bulb.<|endoftext|>
TITLE: Smooth proper schemes over rings of integers with points everywhere locally
QUESTION [14 upvotes]: [Edit: Question 1 has been moved elsewhere so that an answer to Question 2 can be accepted.]
Question 2. Is there a number field $K$, and a smooth proper scheme $X\to\operatorname{Spec}(\mathfrak{o})$ over its ring of integers, such that $X(K_v)\neq\emptyset$ for every place $v$ of $K$, and yet $X(K)=\emptyset$ ?
I believe the answer is Yes.
Remark.  Let $K$ be a real quadratic field, $\mathfrak{o}$ the ring of integers of $K$, and $A$ the quaternion algebra over $K$ which is ramified exactly at the two real places.  Then the conic $C$ corresponding to $A$ is a smooth projective $\mathfrak{o}$-scheme such that $C(\mathfrak{o})=\emptyset$ (because $C(K_v)=\emptyset$ for each of the real places $v$).  But if we insist that $C(K_v)\neq\emptyset$ at these two real places $v$, then $A$ would have to split at these $v$ (in addition to all the finite places), and we would have $C=\mathbb{P}_{1,\mathfrak{o}}$.
More generally, let $K$ be a number field, $\mathfrak{o}$ its ring of integers, and let $C$ be a smooth proper $\mathfrak{o}$-scheme whose generic fibre $C_{K}$ is a twisted $K$-form of the projective space of some dimension $n>0$.  If $C$ has points everywhere locally, then  $C=\mathbb{P}_{n,\mathfrak{o}}$.  This remark shows that $X$ cannot be a twisted form of a projective space.

REPLY [12 votes]: Chandan asked Vladimir and me for an example of an elliptic curve over a real quadratic field that has everywhere good reduction and non-trivial sha, with an explicit genus $1$ curve representing some element of sha. Here's one we found:
The elliptic curve $y^2+xy+y = x^3+x^2-23x-44$ over $\mathbb Q$ (Cremona's reference 4225m1) has reduction type III at 5 and 13. These become I0* over $K=\mathbb Q(\sqrt{65})$, and I0* can be killed by a quadratic twist. Specifically, the original curve can also be written as $y^2 = x^3+5x^2-360x-2800$ over $\mathbb Q$, and its quadratic twist over $K$.
$E: \sqrt{65}Uy^2 = x^3+5x^2-360x-2800$
has everywhere good reduction over $K$; here $U = 8+\sqrt{65}$ is the fundamental unit of $K$ of norm $-1$. 
Now 2-descent in Magma says that the 2-Selmer group of $E/K$ is $(\mathbb Z/2\mathbb Z)^4$, of which $(\mathbb Z/2\mathbb Z)^2$ is accounted by torsion. So it has either has rank over K or non-trivial Sha[2], and according to BSD its rank is 0 as L(E/K,1)<>0 (again in Magma). Actually, because $K$ is totally real, I think results like those of Bertolini and Darmon might prove that E has Mordell-Weil rank $0$ over $K$ unconditionally. So it has non-trivial Sha[2]. After some slightly painful minimisation, one of its non-trivial elements corresponds to a homogeneous space  
$C: y^2 = (23562U+1462)x^4 + (4960U+240)x^3 + (1124U-291)x^2 + (141U-833)x + (50U-733)$
with $U$ as above. So here is a curve such that $J(C)$ has everywhere good reduction and the Hasse principle fails for C.
Hope this helps!
Tim<|endoftext|>
TITLE: Is the category of representations of a finite W-algebra monoidal?
QUESTION [10 upvotes]: My question is prompted by Ben Webster's answer to this question.
Is there a notion of tensor product for representations of a finite W-algebra?
I thought about this question years ago in the context of the infinite-dimensional W-algebras in conformal field theory, but failed to reach a satisfactory conclusion.  I was hoping that now that the professional representation theorists have started to study finite W-algebras an answer might be forthcoming.

Edit (Added in response to Ben's answer below.)
My (basic) intuition about finite W-algebras is that they are more akin to universal enveloping algebras than to Lie algebras, so one way to rephrase this question is whether perhaps there is some additional structure (coproduct or what have you) on finite W-algebras which would allow one to "fuse" representations in some way.
I think I understand the remark about the Slodowy slice not possessing enough of an algebraic structure, but there is a different description of finite W-algebras besides the one coming from the Slodowy slice and which perhaps suggests that there may be some more structure.  This is the interpretation of finite W-algebras as the quantisation of the Poisson algebra of casimirs.  Interpreting the symmetric algebra $\mathfrak{S} := \mathrm{Sym}(\mathfrak{g})$ as the polynomial functions on $\mathfrak{g}^*$, it becomes a Poisson algebra à la Kirillov-Kostant.  The adjoint Lie group $G$ acts on $\mathfrak{S}$ via automorphisms and the invariant subalgebra is a Poisson subalgebra, which is generated by (the image in the symmetric algebra of) the centre of the universal enveloping algebra: the so-called casimirs.
Another version of the question is whether anything of the Hopf algebra structure present in the universal enveloping algebra survives this procedure.

Edit (in response to comments below)
Here's one more stab at motivating why one would expect some "fusing"  of representations to exist.  The homological construction of the W-algebras actually gives you more: it gives you a machine (I have not checked if it's a functor) to which you feed it a representation of the Lie algebra $\mathfrak{g}$ and you get a representation of the W-algebra.  Let us call this machine $H$.  If $R,S$ are representations of $\mathfrak{g}$ and $H(R)$ and $H(S)$ the corresponding representations of the W-algebra, one could define a sort of product $\boxtimes$ by
$$H(R) \boxtimes H(S) := H(R \otimes S)$$
Is this an interesting notion of tensor product?
I guess I should have mentioned that the reason I was interested in this question was as a first step in the definition of W-covariant operator product expansion.  This would allow one to determine the correlators of any fields in a conformal field theory with chiral algebra some W-algebra, in terms of the correlators of the W-primaries.  This is all by analogy with the case of superconformal field theories, where one can work in a manifestly supersymmetric formalism.

REPLY [5 votes]: This is just to add another point of view to Ben's and David's comments and mainly for your Edit regarding superconformal field theories, as this is more about the affine case than the finite one. 
1) There is a formal relation between the finite W-algebra and the affine one: The finite is the Zhu algebra of the affine one (Kac-De Sole '05). 
This in particular implies that irreducible modules for the affine algebra are in 1-1 correspondence with irreducible modules for the finite one. From here perhaps your suspicion that if you expect tensor products for modules of the affine ones then you should expect them for the finite one. That article also answer your question regarding a BRST construction of the finite W-algebra (see the appendix, and also the very nice article by Gan-Ginzburg in the finite case) 
2) There is a fusion category structure for the affine W-algebra at certain levels, namely, it is proved in some cases when these W-algebras are rational, and it is still conjectured in many others. At any rate, we know of at least a few examples when we indeed have the tensor structure in the categories of modules for the affine W-algebra (See articles by Arakawa and Kac-Wakimoto starting in 2005 up to late '10). 
One observation is that the affine W-algebra that is rational is the simple quotient of the affine W-algebra mentioned in 1) above.
3) The functor that you mention is just the "top component" of the homological reduction functor in the affine situation, where you feed a representation for the affine algebra g and you get a representation for the affine W-algebra (the simple one). This functor has been studied by many, there was a conjecture of Frenkel-Kac-Wakimoto regarding the exactness and behaviour of this functor. Arakawa proved (for the principal nilpotent first in Invent. '05 and then in more generality starting in '08, see also Kac-Wakimoto '07) that this functor is exact and it sends irreducible modules to either zero or irreducible modules. He used this to prove existence of modular invariant representations of the W-algebra. In particular, as you mention this gives a way of fusing W-representations by using the fusion of the affine ones. 
Actually, Arakawa proved the irreducibility part in the principal nilpotent case, and as far as I know the almost-irreducibility in general (0802.1564). In some cases he can use a previous result in the finite case: this property of sending irreducibles to irreducibles or zero was proved by Brundan-Kleschev in type A (they proved more: that every simple module over W-finite arises in this way). 
4) In the cases when you can prove that a) the affine W-algebra is rational, and b) that every module over the affine W-algebra is the Hamiltonian reduction of a module over the affine lie algebra g, then you get fusion for the W-algebra from fusion on g, and I do not see anything wrong with passing to the finite W-algebra by taking Zhu's algebras everywhere, so in this setting I agree with you, and looking at the list in Kac-Wakimoto '07 of possible rational W-algebras, you should get several examples of finite ones. I don't see anything wrong with this but I may be missing something, perhaps some subtlety between the W-algebra and its simple quotient?<|endoftext|>
TITLE: What is the current status of the function fields Langlands conjectures?
QUESTION [20 upvotes]: My question, roughly speaking is, what happened to the function fields Langlands conjecture? I understand around 2000 (or slightly earlier perhaps), Lafforgue proved the function fields Langlands correspondence for $GL(n)$ in full generality (proving all aspects of the conjectures). Since then, what's happened to the function fields Langlands conjectures, and what work have people been doing in this direction? (Or has this field died out after Lafforgue's monumental achievement?)
I have been trying but haven't found a good reference for this, but since the function fields Langlands conjectures can be defined for all reductive groups (though I understand in some sense, it is not as "strong" as it is for $GL$ in the general case) - what is the status of these conjectures? Have partial results been obtained? 
I understand that geometric Langlands is very intensely researched today, but I consider geometric Langlands as being slightly different (even though it is an analogue of the function fields Langlands correspondence - from reading Frenkel's article on geometric Langlands, my impression was not that geometric Langlands encapsulates all of the information in function fields Langlands conjectures), and I'm asking what work has been done since specifically on function fields Langlands. Or am I misunderstanding things and do the geometric Langlands conjectures actually encapsulate all the information from function fields Langlands as well? 
I understand that the Fundamental Lemma has been proven recently, and that Lafforgue is doing some things relating to Langlands functoriality currently. Here's a related thread about Langlands functoriality: Where stands functoriality in 2009?..

REPLY [22 votes]: (1) Regarding the relationship between geometric Langlands and function field Langlands:
typically research in geometric Langlands takes place in the context of rather restricted ramification (everywhere unramified, or perhaps Iwahori level structure at a finite number of points).  There are investigations in some circumstances involving wild ramification (which is roughly the same thing as higher than Iwahori level), but I believe that there is not a definitive program in this direction at this stage.  
Also, Lafforgue's result was about constructing Galois reps. attached to automorphic forms.  Given this, the other direction (from Galois reps. to automorphic forms), follows immediatly, via
converse theorems, the theory of local constants, and Grothendieck's theory of $L$-functions in the function field setting.  
On the other hand, much work in the geometric Langlands setting is about going from local systems (the geometric incarnation of an everywhere unramified Galois rep.) to automorphic sheaves (the geometric incarnation of an automorphic Hecke eigenform) --- e.g. the work of Gaitsgory, Mirkovic, and Vilonen in the $GL_n$ setting does this.  I don't know how much is
known in the geometric setting about going backwards, from automorphic sheaves to local systems.
(2) Regarding the status of function field Langlands in general: it is important, and open, other than in the $GL_n$ case of Lafforgue, and various other special cases.  (As in the number field setting, there are many special cases known, but these are far from the general problem of functoriality.  Langlands writes in the notes on his  collected works that "I do not believe that much has yet been done beyond the group $GL(n)$''.)   Langlands has initiated a program called ``Beyond endoscopy'' to approach the general question of functoriality.  In the number field case, it seems to rely on unknown (and seemingly out of reach) problems of analytic number theory, but in the function field case there is some chance to approach these questions geometrically instead.  This is a subject of ongoing research.<|endoftext|>
TITLE: Comprehensive reference for synthetic euclidean geometry
QUESTION [11 upvotes]: Euclidean geometry is a special case of the theory of Hilbert spaces; but in order to convince small children of basic facts, e.g. that the line segments from each of the vertices of a triangle to the midpoint of the opposite side are concurrent, I've found I need to resort to synthetic arguments.
Do you know of a comprehensive reference for synthetic euclidean geometry?

REPLY [3 votes]: Try Continuous Symmetry by Howe and Barker! 
Howe and Barker take the Erlangen program very seriously and prove Euclidean geometry via a synthetic fashion guided by symmetries. 
A very important book that concretely fleshes the Erlangen Programme which at times appears to be high-brow philosophy.<|endoftext|>
TITLE: Is there any value in studying divisors with coefficients in a ring R?
QUESTION [9 upvotes]: As a rule, the various groups and quotients of the divisor group on a variety have coefficients in $\mathbb{Z}$.  That is, you take $\mathbb{Z}$-linear combinations of Weil divisors or Cartier divisors, and then to construct other groups you take quotients.
However, in some cases, people tensor with $\mathbb{Q}$ and $\mathbb{R}$.  So my question is:

Are these the only rings that people use as coefficients for divisors on a variety?

My vague intuition is that it probably is, because $\mathbb{Z}$ is initial in commutative rings with identity, $\mathbb{Q}$ is a field of characteristic zero, so we can use it to kill torsion, and $\mathbb{R}$ is complete, so we can guarantee that there is an $\mathbb{R}$-divisor, plus with orbifolds, rational coefficients seem to show up naturally.  But is this it? More generally, what about for cycles and cocycles?  There's an analogy with cohomology and the Chow ring, and we do sometimes take cohomology with coefficients either in an arbitrary ring or in some other rings (finite fields, for instance, when studying things like nonorientable manifolds), which is why I started wondering about this.

REPLY [3 votes]: I think that line bundles over gerbes have a notion of associated divisor where the coefficients are not integers.<|endoftext|>
TITLE: Does p-adic $L$- function determine the $L$ function
QUESTION [20 upvotes]: Let $E_1$  and $E_2$ be two Elliptic curve defined over $\mathbb Q$ . Let $p$ be an fixed given  odd prime of $\mathbb Q$ at which both the curves have good ordinary reduction. Moreover  p-adic $L$-function  of $E_1$ and $E_2$ are same . Does it mean that  the complex $L$-function of $E_1$ and $E_2$  are also same ?
Is there some sufficient criteria on p-adic $L$-functions such that such that the $L$ function is determined?

REPLY [4 votes]: Yes, it is true that the $p$-adic $L$-function of an elliptic curve $E$ over $\mathbf{Q}$ determines the isogeny class of $E$. As mentioned in the Luo-Ramakrishnan article pointed out by Idoneal, this follows from analytical results of Rohrlich. I thought I would sketch the argument of Rohrlich because it's interesting.
So, let us assume $L_p(E_1,\cdot)=L_p(E_2,\cdot)$ for two elliptic curves $E_1,E_2$ with good ordinary reduction at $p$. By the interpolation property of the $p$-adic $L$-function, this means that
\begin{equation}
\frac{L(E_1,\chi,1)}{\Omega_{E_1}^\pm} = \frac{L(E_2,\chi,1)}{\Omega_{E_2}^\pm}
\end{equation}
for all Dirichlet characters $\chi$ of $p$-power conductor. Here $\pm = \chi(-1)$ and $\Omega^+$ (resp. $\Omega^{-}$) is the real (resp. imaginary) period of a Néron differential.
Now there is this nice analytical result by Rohrlich (On $L$-functions of elliptic curves and cyclotomic towers, Invent. Mat. 75, 1984) : if $\chi$ is a Dirichlet character of conductor $p^m$, then the average value of $L(E,\chi^{\sigma},1)$, where $\chi^\sigma$ runs through the conjugate characters of $\chi$, tends to $1$ as $m \to +\infty$. This is proved by writing $L(E,\chi,1)$ as a quickly converging series and using clever analytical arguments.
From this we deduce that $\Omega_{E_1}^{\pm}=\Omega_{E_2}^{\pm}$. This implies that there is an isogeny $\varphi : E_1 \to E_2$ (which is a priori defined over $\mathbf{C}$) such that $\varphi^* \omega_{E_2}$ is a non-zero rational multiple of $\omega_{E_1}$. By taking the sums of the conjugates of $\varphi$, we deduce that $E_1$ and $E_2$ are isogenous over $\mathbf{Q}$ (and thus their complex $L$-functions coincide).
Remark : we cannot always deduce that $E_1$ and $E_2$ are isomorphic over $\mathbf{Q}$. For example $E_1=15a1$ and $E_2=15a4$ are $2$-isogenous and have the same Néron periods $\Omega_E^{\pm}$, so they have the same $p$-adic $L$-function, but they are not isomorphic. To me this rather indicates that the definition of $p$-adic $L$-function is not the right one (as pointed out by Olivier in his comment). If we want a theory of $p$-adic $L$-functions which looks like as much as possible as the theory of complex $L$-functions, then the $p$-adic $L$-function should be invariant under isogeny.<|endoftext|>
TITLE: Writing down minimal Weierstrass equations
QUESTION [7 upvotes]: Let $E$ be an elliptic curve over $\mathbb Q_p$. It is possible that $E$ has bad reduction but then when you see $E$ as a curve over a finite extension $K$ of $\mathbb Q_p$, it obtains good reduction. Let $v$ be the valuation defined on $K$ and $R$ its valuation ring. I was interested in checking $E$ has good reduction over $K$ by hand, using the Weierstrass equation. What that amounts to then is writing down the Weierstrass equation $y^2+a_1xy + a_3y = x^3 + a_2x^2+a_4x + a_6$ with the $a_i \in R$ and considering changes of coordinates $x=u^2x' + r$ and $y=u^3y' + u^2sx' + t$ for $u,r,s,t \in R$ in hopes of finding an equation with $v(\Delta')$ minimized, subject to each $a_i'$ being in $R$. There are certain congruence conditions that guarantee minimality of the new equation, e.g. $v(\Delta') < 12$, which only depend on the choice of $u$. However, guaranteeing the new equation has coefficients in $R$ requires solving other congruence relations depending on $r,s$ and $t$, e.g. you need $v(a_1+2s)\geq v(u)$ (because $a_1' = u^{-1}(a_1+2s)$). The few times I have done this by hand, I have just had to look at the equations and make some choices until something worked out.
My question is whether or not there exists a general method for obtaining a good change of coordinates $u,r,s,t$ and if not, then how do people go about writing down minimal Weierstrass models. I can't imagine there should be general methods for solving the system of non-linear congruences (higher powers of $u,r,s$ and $t$ appear in the other congruences) in the ring $R$, but if there is then I would also be interested in understanding that as well.

REPLY [9 votes]: If you want a very fast algorithm that computes a minimal Weierstrass equation over $\mathbf{Z}$ (or more generally over any PID), you can use a simple algorithm due to Laska. ("An algorithm for finding a minimal Weierstrass equation for an elliptic curve," Math. Comp. 38 (1982), 257-260.) The basic idea is that it's easy to find a minimal equation modulo all primes $p \ge 5$. Further, there's always a minimal model with $a_1,a_2,a_3$ equal to $-1$, $0$, or $1$. So you just have to check a small number of possibilities.<|endoftext|>
TITLE: Graphs preserved under the Hamiltonian path operator
QUESTION [13 upvotes]: Given a graph $G$ with vertex set $V$, let $HP(G)$ be the graph on $V$ where there's an edge from $u$ to $v$ if and only if there's a Hamiltonian path in $G$ from $u$ to $v$. (I believe this is called the Hamiltonian path operator, but all references I can find to it are from a computational perspective.)
Anyway, here's my question: for which graphs $G$ is $HP(G) \simeq G$? There are some obvious examples (complete graphs, their complements, n-cycles, and $K_{n,n}$s), and at least one less-obvious example ($K_{2,1,1}$, or a square with one diagonal). The latter is certainly more interesting, because the isomorphism between $HP(G)$ and $G$ doesn't just map each vertex to itself. Are there others like it, or any obvious ones I missed? Even better, can we completely classify these graphs?

REPLY [3 votes]: Two more large families (one of which includes both David's example and $K_{2,1,1}$) are:
• A 4n-cycle plus n of the 2n long diagonals (every other one).
• A 2n-cycle with an n-cycle inscribed in it along the odd vertices.<|endoftext|>
TITLE: A Galois Theory Computation
QUESTION [18 upvotes]: Excuse me for the specificity of this question, but this is a silly computation that's been giving me trouble for some time.
I want to explicitly realize the order 21 Frobenius group over ℂ(x), as ℂ(x,y,z) where y3=g(x) and z7=h(x,y). The order 21 Frobenius group is C7⋊C3, where the generator of C3 acts by taking the generator of C7 to its square. Or in other words < a,b|a3=1, b7=1, ba=a(b2) >. Furthermore, I want it to branch at exactly three points, two of which will have 7 preimages, each with ramification 3, and the third will have 3 points over it with ramification 7 each.
This can easily be shown to exist: Take ℙ1 minus three points (say x=6,5,2); look at its algebraic fundamental group (=the profinite completion of < c, d, t| cdt = 1 >), and map it to the Frobenius group surjectively by c goes to a, d goes to b, and t goes to b-1a-1. This gives you a Galois cover of ℙ1, with said ramification (because order(a)=3, order(b)=7, and order(b-1a-1)=3), and group the 21 order Frobenius group.
Of course, this construction is extremely difficult to track because of the topology involved. It would be much easier to deduce the field extension from the ramification behavior. 
So: this can be broken down to two cyclic Galois extensions. The first, a ℂ(x,y), of the form y3=(x-2)2(x-6) is pretty easy to deduce (I need it to ramify at x=2 and x=6 and nowhere else -- this must be the equation up to change of variables). The second, a z7=h(x,y) is tricky. I want it to ramify only above x=5. There's some Abhyankar's lemma things going on here, and that makes the guesswork difficult, and my life much harder.
I should note that the distinction of the Frobenius group of order 21, and the reason that I'm at all interested in this example, is that it is the only order 21 group which isn't cyclic. Geometrically, it means that h is a function of both x and y, and not just x.
Thanks in advance.

REPLY [8 votes]: I'd like to change your numbers slightly.  (EDIT: Slight adjustment to make the formula nicer and address the correction in the comments, nothing to see here)
One solution is to set $y^3 = x$, triply ramified only over $0$ and $\infty$, and if we want the 7-fold ramification over $x=1$ (which has solutions $y=1,\omega,\omega^2$) we set $z^7 = (1-y)(1 - \omega y)^2(1-\omega^2 y)^4$, which only ramifies over the three preimages.  To show this is Galois, it suffices to show that the automorphism $y \mapsto \omega^2 y$ lifts to an automorphism of the whole field.  This automorphism maps $(1-y)(1-\omega y)^2(1-\omega^2 y)^4$ to $(1-\omega^2 y)(1- y)^2(1- \omega y)^4$, which has seventh root $z^2/(1-\omega^2 y)$.<|endoftext|>
TITLE: Why do Todd classes appear in Grothendieck-Riemann-Roch formula?
QUESTION [57 upvotes]: Suppose for some reason one would be expecting a formula of the kind 
$$\mathop{\text{ch}}(f_!\mathcal F)\ =\ f_*(\mathop{\text{ch}}(\mathcal F)\cdot t_f)$$
valid in $H^*(Y)$ where

$f:X\to Y$ is a proper morphism with $X$ and $Y$ smooth and quasiprojective, 
$\mathcal F\in D^b(X)$ is a bounded complex of coherent sheaves on $X$,  
$f_!: D^b(X)\to D^b(Y)$ is the derived pushforward, 
$\text{ch}:D^b(-)\to H^*(-)$ denotes the Chern character,
and $t_f$ is some cohomology class that depends only on $f$ but not $\mathcal F$.

According to the Grothendieck–Hirzebruch–Riemann–Roch theorem (did I get it right?) this formula is true with $t_f$ being the relative Todd class of $f$, defined as the Todd class of relative tangent bundle $T_f$.
So, let's play at "guessing" the $t_f$ pretending we didn't know GHRR ($t_f$ are not uniquely defined, so add conditions on $t_f$ in necessary). 

Question. Expecting the formula of the above kind, how to find out that $t_f = \text{td}\, T_f$?

You don't have to show this choice works (that is, prove GHRR), but you have to show no other choice works.
Also, let's not use Hirzebruch–Riemann–Roch: I'm curious exactly how and where Todd classes will appear.

REPLY [3 votes]: A conceptual answer to "exactly how and where Todd classes will appear:"
You don't have to use Todd classes if you stick to Hochschild homology. This might not be the best reference, but I learned this from 1.2.1 in
https://arxiv.org/pdf/1804.00879.pdf
You can stick to just the Chern character and factor through HH(Perf(X)); the Todd class corrects the fact that the HKR isomorphisms $HH(Perf(X)) \simeq H^*(X, \bigoplus \Omega_X^p)$ are NOT natural in $X$ until correction by Todd.
My apologies -- I'm sure there's a more down-to-earth reference of which I'm not aware.<|endoftext|>
TITLE: Why are the characters of the symmetric group integer-valued?
QUESTION [46 upvotes]: I remember one of my professors mentioning this fact during a class I took a while back, but when I searched my notes (and my textbook) I couldn't find any mention of it, let alone the proof.
My best guess is that it has something to do with Galois theory, since it's enough to prove that the characters are rational - maybe we have to find some way to have the symmetric group act on the Galois group of a representation or something. It would be nice if an idea along these lines worked, because then we could probably generalize to draw conclusions about the field generated by the characters of any group. Is this the case?

REPLY [2 votes]: Galois theory is probably the best and simplest explanation, and the most amenable to generalization for other groups (and more general number fields). This is already well-expounded in earlier answers. I just want to point out that it was already known to Frobenius that all irreducible characters of $S_{n}$ are $\mathbb{Z}$-combinations of permutation characters ( ie characters induced from trivial characters of subgroups). More precise information became available later (Young tableaux, etc, as mentioned in other answers), but what Frobenius knew already covered the question.<|endoftext|>
TITLE: Isomorphism types or structure theory for nonstandard analysis
QUESTION [19 upvotes]: My question is about nonstandard analysis, and the diverse possibilities for the choice of the nonstandard model R*. Although one hears talk of the nonstandard reals R*, there are of course many non-isomorphic possibilities for R*. My question is, what kind of structure theorems are there for the isomorphism types of these models? 
Background. In nonstandard analysis, one considers the real numbers R, together with whatever structure on the reals is deemed relevant, and constructs a nonstandard version R*, which will have infinitesimal and infinite elements useful for many purposes. In addition, there will be a nonstandard version of whatever structure was placed on the original model. The amazing thing is that there is a Transfer Principle, which states that any first order property about the original structure true in the reals, is also true of the nonstandard reals R* with its structure. In ordinary model-theoretic language, the Transfer Principle is just the assertion that the structure (R,...) is an elementary substructure of the nonstandard reals (R*,...). Let us be generous here, and consider as the standard reals the structure with the reals as the underlying set, and having all possible functions and predicates on R, of every finite arity. (I guess it is also common to consider higher type analogues, where one iterates the power set ω many times, or even ORD many times, but let us leave that alone for now.) 
The collection I am interested in is the collection of all possible nontrivial elementary extensions of this structure. Any such extension R* will have the useful infinitesimal and infinite elements that motivate nonstandard analysis. It is an exercise in elementary mathematical logic to find such models R* as ultrapowers or as a consequence of the Compactness theorem in model theory. 
Since there will be extensions of any desired cardinality above the continuum, there are many non-isomorphic versions of R*. Even when we consider R* of size continuum, the models arising via ultrapowers will presumably exhibit some saturation properties, whereas it seems we could also construct non-saturated examples. 
So my question is: what kind of structure theorems are there for the class of all nonstandard models R*?  How many isomorphism types are there for models of size continuum? How much or little of the isomorphism type of a structure is determined by the isomorphism type of the ordered field structure of R*, or even by the order structure of R*?

REPLY [2 votes]: The following was useful in a recent paper on asymptotic cones with Kramer, Shelah and Tent. How many ultraproducts $\prod_{\mathcal{U}} \mathbb{N}$ exist up to isomorphism, where $\mathcal{U}$ is a non-principal ultrafilter over $\mathbb{N}$? If $CH$ holds, then obviously just one ... if $CH$ fails, then $2^{2^{\aleph_{0}}}$.
In the case when $CH$ fails, the ultraproducts are already nonisomorphic as linearly ordered sets. The proof uses the techniques of Chapter VI of Shelah's
book "Classification Theory and the Number of Non-isomorphic Models".<|endoftext|>
TITLE: Relating Euler characteristic, intersection product, Morse theory (plus SU(2) and 3-manifolds)
QUESTION [5 upvotes]: Suppose we have a (closed, oriented) 3-manifold M with a Heegard surface F of genus g. Let F* denote F with a puncture. Then the space H of representations of pi_1(F*) on SU(2) is just SU(2)^2g, and the representation spaces of the two handlbodies sit inside H. Call these spaces Q_1 and Q_2 -- we will always think of them as subspaces of H. Finally, the intersection R = Q_1 \cap Q_2 is the representation space for M (Note: we haven't quotiented out by conjugation or anything).
Question 1: In the paper http://www.jstor.org/pss/2001712, Boyer and Nicas claim that if M is a \Q-homology sphere, the homological intersection [Q_1 . Q_2 ] is equal to |H_1(M)|, and they say it's easy to prove. I can't seem to figure out how to do it though, and I've tried for a bit... it seems like there's some bit of theory I must be missing. Can anyone see how to prove it?
Question 2: Is the Euler characteristic of R (that is, Q_1 \cap Q_2) also |H_1(M)|? If so, how could we prove this? In particular, is there a general relationship between the intersection pairing between two complementary submanifolds, and the Euler characteristic of their intersection (even when the intersection is not a finite number of points)?
The above is reminiscent of Morse-Bott theory, where the differential forms on the critical set of your morse function give a basis for the chain groups of your homology, and therefore the Euler characteristic of the critical set is the Euler characteristic of the manifold (or something like that... do I have this right?) This requires Morse-Bott non degeneracy of the critical set.
Final Question: What's an explicit relationship between the morse theory and the inersection theory? And when we just care about Euler characteristic, can we relax the Morse-Bott non-degeneracy? It seems that Q_1 and Q_2 don't always intersect "non-degenerately" (not only non-transversely, but the intersection might not even be smooth, for example) but Boyer and Nicas still claim that the intersection number is something nice (and computable on general grounds). Under what conditions could the same thing happen with a non-Morse-Bott morse function?
Thanks! I hope there aren't too many questions here...

REPLY [2 votes]: It might be helpful to look at the book of Akbulut and McCarthy on Casson's Invariant.  I think the answer to Question 1 is fairly clearly explained in Proposition 1.1b of of Chapter III.<|endoftext|>
TITLE: What are important examples of filtered/graded rings in physics?
QUESTION [11 upvotes]: Hi,
what comes to the mind of a physicist, when he hears words like filtered ring and associated graded? What do these guys describe? What are basic/typical/illuminating examples in physics? 
Of course also mathematicians may answer from their perspective :)
Edit: Information on my background: My "primeval" motivation to understand filtered rings/graded rings, comes from mathematics: They are the basics for D-Module theory. 
Though I would also like to see these rings from a different perspective, therefor I asked the question.
 My knowledge in physics is a bit limited, I only attended some undergraduate courses (quantum mechanics, electro dynamics, classical mechanics). Though more sophisticated answers are fine, I don't need to understand every detail, I just want to get the flavor.

REPLY [4 votes]: The deformation quantization construction, famously carried into full generality by Kontsevich, is the construction, from the datum of a manifold $M$ and a Poisson structure $\{\mathord-,\mathord-\}$ on it, of a generally non-commutative associative algebra $\mathcal A$ which is endowed with a filtration such that the associated graded algebra (which is naturally a Poisson algebra) $\mathrm{gr}\mathcal A$ is isomorphic to the algebra of smooth functions on $M$ and its Poisson structure. This should be an example of interest to physicists! (Also, this includes José's example 1, as the Weyl algebra is a deformation quantization of the ring of (polynomial) functions on the plane, and more or less example 2)
In general, a filtered object $X$ is an objects with a chosen aproximation, given by its associated graded object $\mathrm{gr}\\,X$, which ideally reflects some of the interesing propoerties of $X$ but which is, at the same time, simpler than $X$. Indeed, one can in most cases think of elements of $\mathrm{gr}\\,X$ as «elements of $X$ up to details».<|endoftext|>
TITLE: Sperner's theorem and "pushing shadows around"
QUESTION [8 upvotes]: To head off any confusion: I'm talking about the extremal-combinatorics Sperner's theorem, bounding the sizes of antichains in a Boolean lattice.
So the "canonical proof" of this theorem seems to be essentially Lubell's -- it's formulated in several different ways, but my favorite is this. Let $A$ be an antichain in $2^{[n]}$. Pick an arbitrary saturated chain and consider a random automorphism $\phi$ of the poset. Since $A$ is an antichain, so is the image $\phi(A)$, and the expected number of elements of $\phi(A)$ that lie in the distinguished chain is at most 1. Now if $S \subset [n]$ has k elements, then $\phi$ maps $S$ into our distinguished chain with probability $1/\binom{n}{k}$; linearity of expectation gives us the LYM inequality and Sperner's theorem follows.
This is a lovely proof, but reading over the DHJ polymath threads I heard about another approach -- apparently the approach Sperner himself used -- which one commenter described by the wonderful phrase "pushing shadows around." Evocative as this phrase is, I'm not actually sure what it means: presumably the idea is to replace an antichain by another antichain, at least as large and whose elements are "closer to the center?" This idea seems like it should work, but I can't quite get it to go through, which makes me think I'm missing something. Does anyone know the details of this proof?

REPLY [4 votes]: Gjergji already provided the answer, but want to add a reference to an excellent book which contains more about Sperner's theorem than any other source: "Sperner Theory" by Konrad Engel. As the name suggests the book contains many different proofs of the the theorem, but a lot of other information that is hard to find in anywhere else: various generalizations, lots references to results that one would never even hear about otherwise, etc. It is from this book that I learned the argument in the question.<|endoftext|>
TITLE: Self-similar matrices?
QUESTION [5 upvotes]: Does anyone know anything about self-similar (infinite) matrices, with more or less fractal(-like) structure and admitting meaningful matrix-algebra operations?

REPLY [2 votes]: You should learn about the hyperfinite II_1 factor, which is the limit of the inclusions
M_1 --> M_2 --> M_4 --> M_8 --> ....

(here M_k is the k by k matrices over $\mathbb{C}$) where each inclusion is given by tensoring with the identity matrix in M_2. Every 'finite' element is "self-similar" in a sense.<|endoftext|>
TITLE: obstruction to smooth lifting of smooth schemes
QUESTION [8 upvotes]: According to general theory, for a square zero thickening defined by an ideal I: SpecA -> SpecA', there is an obstruction of lifting a smooth scheme X over A to a smooth scheme over A' living in H^2(X,T_X \otimes I).
Can anyone give an example of being obstructed?
e.g. a smooth scheme over F_p which does not lift smoothly to Z/(p^2)?

REPLY [12 votes]: Ravi Vakil's paper Murphy's Law in Algebraic Geometry … gives many references of such things: see Section 2 of
http://arxiv.org/abs/math/0411469
The first example is due to Serre:

Serre, Jean-Pierre
Exemples de variétés projectives en caractéristique $p$ non relevables en caractéristique zéro. (French)
Proc. Nat. Acad. Sci. U.S.A. 47 1961 108--109. 
Here is the MathReview by I. Barsotti:
An example of a non-singular projective variety $X_0$, over an algebraically closed field $k$ of characteristic $p$, which is not the image, $\text{mod}\,p$, of any variety $X$ over a complete local ring of characteristic 0 with $k$ as residue field. The variety $X_0$ is obtained by selecting, in a 5-dimensional projective space $S$, and for $p>5$, a non-singular variety $Y_0$ which has no fixed point for an abelian finite subgroup $G$ with at least 5 generators of period $p$, of the group $\Pi(k)$ of projective transformations of $S$, but which is transformed into itself by $G$; then $X_0=Y_0/G$. The reason for the impossibility is that $\Pi(K)$, for a $K$ of characteristic 0, does not contain a subgroup isomorphic to $G$. {Misprint: on the last line on p. 108 one should read $s(\sigma)=\exp(h(\sigma)N)$.} 

ADDENDUM: 
After looking back at the question, my reference to Serre's paper is inappropriate: this is an example of a variety which does not lift to characteristic zero, whereas the poster asked for one which didn't lift (even) to $\mathbb{Z}/p^2 \mathbb{Z}$.  It is still true that this sort of thing is discussed in Vakil's paper, but now the canonical primary source seems to be

Deligne, Pierre; Illusie, Luc
Relèvements modulo $p^2$ et décomposition du complexe de de Rham. (French) [Liftings modulo $p^2$ and decomposition of the de Rham complex]
Invent. Math. 89 (1987), no. 2, 247--270. 
MathReview by Thomas Zink:
The degeneration of the Hodge spectral sequence $H^q(X,\Omega^p_{X/k}) \Rightarrow H^n_{\text{DR}}(X/k)$ for a smooth and projective algebraic variety over a field of characteristic zero is a basic fact in algebraic geometry. Nevertheless, only recently has one found an algebraic proof in connection with the comparison of étale and crystalline cohomology (Faltings, Fontaine, Messing).
In this beautiful paper the authors give a short, elementary, algebraic proof for the degeneration by methods in characteristic $p>0$.
Let $k$ be a perfect field of characteristic $p>0$, and let $X$ be a smooth variety over $k$, which lifts to the Witt ring $W_2(k)$. Denote by $X'$ the variety obtained by base change via the Frobenius automorphism, and let $F:X\to X'$ be the relative Frobenius morphism ….  More precisely, such splittings correspond to liftings of $X'$ to $W_2(k)$. This theorem implies the degeneration of the Hodge spectral sequence for $X$ in dimension $<p$, and, by Raynaud, a Kodaira vanishing theorem for $X$. The corresponding facts in characteristic zero may be deduced by the usual reduction process.
The authors show that their argument extends to the case where $k$ is replaced by an arbitrary base $S$ of characteristic $p$ and $\Omega^\cdot_{X/S}$ is replaced by differentials with logarithmic poles.<|endoftext|>
TITLE: classification of irreducible admissible representations of GL(n)
QUESTION [9 upvotes]: Does anyone know the classification of irreducible admissible representations of GL(n) (over real,complex and p-adic fields), or some references?
Sorry if this question is not appropriate here.

REPLY [8 votes]: Let me focus on the case when $K$ is non-archimedean; the archimedean case is somewhat easier.
There is a coarse classification, valid for any reductive group, into supercuspidals, 
and all the others --- the others are ones that can be parabolically induced from supercuspidals of proper Levi's, and so in principle are understood by induction,
while the supercuspidals are the basic building blocks.  There is a subtley that
certain parablic inductions of irreducibles are not themselves irreducible, which leads
to so-called special representations (EDIT: and in the general, ie. non-$GL_n$ case,
so-called packets), but at least in the $GL_n$ case these are well-understood too. 
(EDIT: In particular the packets are actually just singletons).
So everything comes down to the supercuspidals.  (This is explained in the introduction
to Harris and Taylor's book, among many other places.)  This coarse classification is also
compatible in a natural way with the local Langlands correspondence.
For $GL_n(K)$ the supercuspidals are completely classified.  (This is the difference between
$GL_n$ and most other groups.)  In fact there are two forms of the classification.
(1) Via the local Langlands correspondence (a theorem of Harris and Taylor).
(2) Via the theory of types (a theorem of Bushnell and Kutzko).
The first classification relates them to local $n$-dimensional Galois reps.  The second
relates more directly to the internal group theoretic structure of the representations.
As far as I know, the two classifications are not reconciled in general (say for large $n$,
where large might be $n > 3,$ or something of that magnitude), and this is an ongoing topic of investigation by experts in the area.  (Any updates/corrections to this statement would be welcome!)
The difference in the archimedean case is that there are no supercuspidals, so everything comes down to inducing characters of tori, and understanding the reducibility of these parabolic inductions.<|endoftext|>
TITLE: Relationship between algebraic and holomorphic differential forms
QUESTION [8 upvotes]: I'm a little confused and in need of some clarification about the
relationship between algebraic and holomorphic differential forms:
(1) What is the exact definition of the module of differential
forms of a complex projective variety?
(2) What is the definition of its differential?
(3) Am I right in assuming that the algebraic forms are a
submodule of the holomorphic forms with the two differentials
coinciding in some obvious sense?

REPLY [9 votes]: Yes, every algebraic differential form is holomorphic and yes, the differential preserves the algebraic differential forms. If you are interested in projective smooth varieties then every holomorphic differential form is automatically algebraic thanks to Serre's GAGA. This answers (3).
Concerning (1) and (2) I suggest that you consult some standard reference as Hartshorne's Algebraic Geometry.
Edit: As pointed out by Mariano in the comments below there are subtle points when one compares Kahler differentials and holomorphic differentials. I have to confess that I have not thought about them when I first posted my answer above.
Algebraic differential $1$-forms
over a Zariski open set $U$ are  elements of the module generated by $adb$ with $a$ and $b$
regular functions over $U$ (hence algebraic) by the relations $d(ab) = adb + b da$, $d (a + b)  = da + db$ and $d \lambda =0$ for any complex number $\lambda$. Since these are the rules of calculus there is a natural map to the module of holomorphic $1$-forms over $U$. This map is injective, since the regular functions on $U$ are not very different from quotients of polynomials.
If instead of considering the ring $B$ of regular functions over $U$ one considers the ring $A$ of holomorphic functions over $U$ then one can still consider its $A$-module of Kahler differentials. If $U$ has sufficiently many holomorphic functions, for instance if $U$ is Stein, then one now has a surjective map to the holomorphic $1$-forms on $U$ which is no longer injective as pointed out by Georges Elencwajg in this other MO question.<|endoftext|>
TITLE: References for theorem about unipotent algebraic groups in char=0?
QUESTION [7 upvotes]: There is a textbook theorem that the categories of unipotent algebraic groups and nilpotent finite-dimensional Lie algebras are equivalent in characteristic zero.  Indeed, the exponential map is an algebraic isomorphism in this case and the group structure can be defined in terms of the Lie algebra structure and vice versa via the Campbell-Hausdorff series, which is finite due to nilpotency. 
My problem is that I am unable to locate any textbook where this textbook theorem is stated.  The books by Borel, Humphreys, Springer, Serre do not seem to mention this theorem.
The only reference I was able to locate is this original paper by Hochschild (which refers to his earlier papers), but he does it in a heavy Hopf-algebra language that is good, too, but still leaves one desiring to find also a simple textbook-style exposition.  Later Hochschild wrote a book "Basic Theory of Algebraic Groups and Lie Algebras" on the subject, to which I have presently no access, but judging by Parshall's review, it is certainly not textbook-style.
Could anyone suggest a simple reference for this textbook theorem?

REPLY [6 votes]: Demazure-Gabriel, Groupes algebriques, Tome I (published in 1970) is a more explicit source, if available.   Chapitre IV treats "groupes affines,
nilpotents, resolubles", while Chapitre V specializes to commutative affine groups.
Typically they work over an (almost) arbitrary field $k$, but IV.2.4 is
devoted to "groupes unipotents en caracteristique 0".   This seems to be as
full an account as you will find in a textbook; see especially Corollaire
4.5 for the category equivalence you want.   
Later textbooks in English
including Hochschild focus mainly on the structure/classification of reductive
rather than arbitrary affine groups.  Even this much of the story told without scheme language is fairly long for graduate courses.  It's regrettable from the reference viewpoint that Demazure-Gabriel gave up after one volume.<|endoftext|>
TITLE: Possible formal smoothness mistake in EGA
QUESTION [42 upvotes]: EGA IV 17.1.6(i) states that formal smoothness is a source-local property. In other words, a map $X\to Y$ of schemes is formally smooth if and only if there is an open cover $U_i$ ($i\in I$) of $X$ such that each restriction $U_i\to Y$ is formally smooth.
It seems however that there is a gap in the proof. The problem is in the third paragraph on page 59 (Pub IHES v 32). The reference to (16.5.17) does not give the conclusion they need. Corollary (16.5.18) does give this conclusion, but it requires a finite presentation assumption. (So, everything is OK for smoothness instead of formal smoothness.)
Quesiton 1: Can someone give a counterexample or a complete proof of 17.1.6(i)? (My bet would be that there's a counterexample.)
I think the right way of fixing this is to change the definition of formal smoothness. Recall that a map $X\to Y$ is said to be formally smooth if for any closed immersion $X'\to Y'$ of affine schemes defined by a square-zero ideal and for map $X'\to X$ and $Y'\to Y$ making the induced square commute, there is a map $Y'\to X$ commuting with all the other maps in the diagram. I think a better definition would to require only that the map $Y'\to X$ exists locally on $Y'$. 
If I'm not mistaken, this definition has the following advantages over the old one: a) The definition of smoothness (=formally smooth and locally of finite presentation) would remain unchanged. b) It would make formal smoothness a source-local property. (Or if there is no counterexample to 17.1.16(i), then the argument with the new definition would be much easier than an argument like the one in EGA, in that it would not depend on the facts in scheme theory that the sheaf of lifts $Y'\to X$ is a torsor for a sheaf derivations and that therefore, since $Y'$ is affine, there is always a global section.) In particular, it would probably be better suited for maps of general sheaves of sets on the big Zariski topology, rather than just schemes. d) It's a general rule of thumb that, in sheaf theory, it's easier to work with local existential quantifiers than global ones.
Question 2: Does anyone know of any reason why this new definition would be bad?

REPLY [42 votes]: Let me point out the following remark made by Grothendieck in his book "Catégories Cofibreés Additives et Complexe Cotangent Relatif", 9.5.8
Please excuse my translation:
"Let $f:X \rightarrow Y$ be a morphism of schemes.  We say that $f$ is "locally formally smooth" if X can be covered by opens $X_i$ which are formally smooth over $Y$.  Evidently, if $f$ is formally smooth, then it is locally formally smooth; I don't know if the converse is true in general.  It was this which was affirmed hastily in EGA IV 17.1.6 but the proof is only valid when we assume that the relative $\Omega^1$ is of finite presentation, for example if $f$ is locally of finite type.  The Lemma 9.5.7 [loc. cit., not reproduced] implies the following criterion : the map $f$ is locally formally smooth if and only if one has $N_{X/Y} = 0$ and $\Omega^1_{X/Y}$ is "locally projective" in the following sense : one can cover $X$ by open affines $X_i$ with rings $B_i$ such that over $X_i$ the quasi-coherent module $\Omega^1_{X/Y}$ is given by a projective $B_i$ module.  We will know that this condition implies the formal smoothness of $f$ if we can show that for a commutative ring $B$, every $B$ module $N$ which is locally projective is also projective - or what amounts to the same - it satisfies $H^1(X, \operatorname{Hom}(\tilde{N}, J)) = 0$ for each quasi-coherent module $J$ on $\operatorname{Spec}(B)$."
Apparently, this local nature of projectivity was shown soon thereafter by Raynaud and Gruson ("Critéres de platitude et projectivité").  In fact I think they show that it is an fpqc local condition.  I think this implies that formal smoothness is even an étale local property.<|endoftext|>
TITLE: Reference request: The stable Kronecker ring is isomorphic to the ring of symmetric polynomials
QUESTION [13 upvotes]: Background
For $\lambda$ any partition and $n$ a positive integer, write $\lambda[n]$ for the sequence $(n - |\lambda|, \lambda_1, \lambda_2, \ldots, \lambda_r)$. For $n$ large enough, this is a partition of $n$.
The irreducible representations of $S_n$ are indexed by partitions of $n$; we denote them by $S_{\lambda}$. The Kronecker coefficients $g_{\lambda \mu}^{\nu}$ are defined by the equality
$$S_{\lambda} \otimes S_{\mu} \cong \bigoplus g_{\lambda \mu}^{\nu} S_{\nu}$$
of $S_n$ representations.
It is a theorem of Murnaghan that $g_{\lambda[n] \mu[n]}^{\nu[n]}$ becomes constant as $n \to \infty$. This constant value is called the stable Kronecker coefficient, and denoted $\overline{g}_{\lambda \mu}^{\nu}$. It is also a result of Murnaghan that, for given $\lambda$ and $\mu$, there are only finitely many $\nu$ for which $\overline{g}_{\lambda \mu}^{\nu} \neq 0$.
Therefore, we can define a commutative, associative ring to be spanned by the generators $\overline{S}_{\lambda}$, with relations
$$\overline{S}_{\lambda} \overline{S}_{\mu} = \sum \overline{g}_{\lambda \mu}^{\nu} \overline{S}_{\nu}.$$
I'll call this the stable Kronecker ring.
 Question
I can prove that the stable Kronecker ring is isomorphic to the ring of symmetric functions. Is this fact already in the literature?

REPLY [5 votes]: I've been looking into this and here are two references that I think answer this question.

In "Products and Plethysms of Characters with Orthogonal, Symplectic and Symmetric Groups" (Canad. J. Math., 10, 1958, 17--32), Littlewood writes p. 25:


The characters of the symmetric group can be obtained from those of the full linear group in a similar manner to that used for the orthogonal group, namely by considering a tensor corresponding to any partition $(\lambda)$ of any integer n, and removing all possible contractions with the fundamental forms (2, p. 392). The remainder when all contractions are removed is an irreducible character, provided that $n-p > \lambda_1$, and it is not difficult to see that it is in fact the character of the symmetric group corresponding to the partition
  $(n-p, \lambda_1,..., \lambda_i)$. It is convenient to represent by $[\lambda]$ not this character, but the corresponding S-function $$[\lambda] = \{n-p,\lambda_1,...,\lambda_i\}$$


In "The symmetric group: Characters, products and plethysms", J. Math Phys, 14, no. 9, 1973, 1176-1183, P.H. Butler and R.C. King write:


The symmetric groups are thus treated quite differently from the linear and other continuous groups: the orthogonal, rotation and symplectic groups.  The characters of the groups are know in terms of S functions and the usual method of calculating such things as Kronecker products of the representations of these groups is to use S-functional expressions for their characters and the powerful algebra of S functions associated with the n-independent outer product rule.  The labels that arise from this approach are the same as those that arise from tensorial arguments.  The aim of this paper is to show that the symmetric groups, $\Sigma_n$, may be treated in an n-independent manner similar to that used for the restriced groups $O_n$ and $Sp_n$, rather than in the usual n-dependent manner requiring a development of the somewhat complicated algebra of inner products of S functions.

After reading these references (and perhaps a few others, but I found these to be the most explicit) I would say the answer to your question "Is this fact already in the literature?" is a qualified 'yes.'  It seems some mathematicians knew that characters of the symmetric group could be realized similar to S-functions and orthogonal/symplectic characters, but I don't think they are particularly explicit (and even Littlewood seems to equate the character with the S function).
Rosa Orellana and I posted a paper in 2015 (Symmetric group characters as symmetric functions, https://arxiv.org/abs/1605.06672) that found (exactly as you and Sami Assaf did) the symmetric group realized as permutation matrices has characters that are symmetric functions in the eigenvalues of the permutation matrices.  King and Butler provide tables of these characters in terms of Schur functions up to n=5 for the embedding of $S_{n}$ in $Gl_{n-1}$.  This is equivalent to providing the Schur expansion of the characters we were calling ${\tilde s}_\lambda[X+1]$.<|endoftext|>
TITLE: Families of number fields of prime discriminant
QUESTION [15 upvotes]: When I am testing conjectures I have about number fields, I usually want to control the ramification, especially minimize to a single prime with tame ramification. Hence, I usually look for fields of prime discriminant (sometimes positive, sometimes negative).
I get the feeling that I cannot be the only one who does this...
And so, are there families of number fields of prime discriminant for each degree? Or at least degree 3 and 4? (They are the coolest. Except quadratics. Of course.) What about: given a prime - can I find a polynomial of degree d with the prime as its discriminant?

REPLY [5 votes]: I am going to annoy all the people who answered above, but I am pretty sure the answer to Dror's question is basically no. In particular, is it known that there are infinitely many cubic fields of prime discriminant? I have not heard of such a result -- if one is out there then I would be extremely grateful if someone would share the appropriate references with me.
As is pointed out above, there is a classical correspondence between such fields and subgroups of $Cl(\mathbb{Q}(\sqrt{p}))$ of index 3. However, I'm not aware that this makes the question easier to answer.
There is also the work of Bhargava and Ghate, Delone-Faddeev, Davenport-Heilbronn, etc. which says that cubic (and quartic and quintic) fields are parameterized by integral orbits on nice prehomogeneous vector spaces which meet certain local conditions. For example, in the cubic case, cubic rings are parameterized by integral binary cubic forms up to $GL_2(\mathbb{Z})$ equivalence, and maximal cubic orders are those cubic rings which meet a certain local condition at each prime.
This allows you to prove formulas for the number of cubic fields with $Disc(K) < X$ with good error terms, and this works if you ask for the condition $d | Disc(K)$. This allows you to run a sieve.
However sieves are notoriously bad at finding primes! The information above is also essentially available in the twin prime problem, but all we can prove there is that there are infinitely many primes $p$ so that $p + 2$ has at most two prime factors.
You can use this argument to find cubic fields with three (I think) prime factors -- there is a paper of Belabas and Fouvry that does this. Maybe you could push their arguments a little bit better. But one cannot hope to find primes this way.
Of course there are excellent computational results, and I don't want to take anything away from these. But I feel like the question is asking if there are infinite families, and I'm pretty sure this is widely expected but not at all known.<|endoftext|>
TITLE: Projective to Affine?
QUESTION [5 upvotes]: Perhaps a basic question... how does one study affine algebraic geometry via projective geometry? For example, suppose I have two affine varieties which I want to prove are isomorphic, would it help to look at the projective closures (assuming I don't know of any other method for proving they are isomorphic)? How does one go back to the affine case from the projective closure (or "projectivization")? Sorry if this sounds confusing.
Thanks,
Morton
Edit: Thanks for the replies. Being new to AG let me try and rephrase my quandary: It seems the projective setting is the most convenient to study AG but if I want to study properties of affine varieties, how does one use results of projective varieties in the affine case? I know this sounds vague but it is a fundamental doubt I have.

REPLY [3 votes]: Here is a simple example.  Someone asked me if the unit disc is an affine variety. the answer is no.  To see why not, assume yes, and take the projective closure.  then one gets a projective curve which is a compact 2 dimensional surface with some points identified, and which differs from the original surface by at most adding a finite number of points.  But this is impossible.  No compact surface can be reduced to a disc by removing a finite number of points, even topologically, except for removing one point from P^1.  But that does not give the disc by Liouville's theorem.
A more significant and pervasive example is the fact that at every singular point of an affine variety, the tangent cone determines a projective variety.  Thus projective geometry is the local aspect of affine geometry.  Put another way, blowing up an affine variety, at a point say, introduces projective geometry into it as a picture of its infinitesimal structure.
One can sometimes use this trick to compute the degree of a proper map of affine or other varieties, by restricting to the behavior over the projective normal bundle of a single fiber.  See e.g. Friedman - Smith (Inventiones, 67, (1982)), who compute the degree of the prym map by showing that a single projective fiber of the proper prym map is embedded in projective space by the derivative of the prym map acting on the normal bundle to the fiber.<|endoftext|>
TITLE: Are automorphism groups of hypersurfaces reduced ?
QUESTION [10 upvotes]: In the following article : "H. Matsumura, P. Monsky, On the automorphisms of hypersurfaces, J. Math. Kyoto Univ. 3 (1964) 347-361", it is shown that in finite characteristic, automorphism groups of smooth hypersurfaces of $\mathbb{P}^N$ are finite (with known exceptions such as quadrics, elliptic curves, K3 surfaces). However, the question of their reducedness is left open. Does anyone know something about it ?
In fact, we need to show that $H^0(X,T_X)=0$. In characteristic $0$, you can use Bott's theorem to do that. What can you do in finite characteristic ?

REPLY [17 votes]: If $X$ is a smooth hypersurface in $\mathbf{P}^{n+1}$ of degree $d$, where $n \ge 1$, $d \ge 3$, and $(n,d) \ne (1,3)$, then $H^0(X,T_X)=0$ by Theorem 11.5.2 in Katz and Sarnak, Random matrices, Frobenius eigenvalues, and monodromy, AMS Colloquium Publications, vol. 45, 1999.  So the connected component of the identity of the automorphism group scheme is trivial in these cases.  See also Theorem 11.1 in http://www-math.mit.edu/~poonen/papers/projaut.pdf ,
which is Poonen, Varieties without extra automorphisms III: hypersurfaces, Finite fields and their applications 11 (2005), 230-268.<|endoftext|>
TITLE: Localizing at the primitive polynomials?
QUESTION [11 upvotes]: For any UFD $R$, the concept of a primitive polynomial (gcd of the coefficients is 1) makes sense in $R[x]$. The product of two primitive polynomials is primitive (Gauss's Lemma), and certainly 1 is a primitive polynomial, so the primitive polynomials form a multiplicative subset $S$ of $R[x]$ - hence we can form the ring $S^{-1}R[x]$. What can we say about it? What does this look like geometrically?

REPLY [3 votes]: Google "Kronecker function ring". This is the germ of the idea behind Kronecker's divisor theory. An elementary historically-minded introduction can be found in Harold Edward's book "Divisor Theory".<|endoftext|>
TITLE: Categorical duals in Banach spaces
QUESTION [6 upvotes]: Near the bottom of the nlab page for Banach space I see "To be described: duals (p+q=pq)".
Are $(\mathbb{R}^n)_p$ and $(\mathbb{R}^n)_q$ dual objects in the closed symmetric monoidal category of Banach spaces and linear contractions (with the tensor product described on that page)?
Edit: take n=2, p=1, q=∞.  Then the question becomes whether $V \times V$ (which is $V^2$ with the $l_\infty$ norm) is isomorphic to $(\mathbb{R}^2)_\infty \otimes V$.  But it seems to me that the functor $V \mapsto V \times V$ does not even commute with coproducts... is that right?

REPLY [2 votes]: My suspicion is "no", because if I recall correctly the map $I \to V \otimes V^*$ naturally lands in the injective tensor product, not the projective tensor product, and it is the latter which appears as the ``correct'' tensor product for the SMC category of Banach spaces and linear contractions.
In the toy example given, $V\oplus V$ with the sup norm is the same as continuous maps from a 2-point set to $V$, equipped with sup-norm, and I'm pretty sure that this is indeed isometrically linearly isomorphic to ${\mathbb R}^2 \check{\otimes} V$ i.e. the injective tensor product.
EDIT: as Reid points out my remarks above assume without justification that the inj. t.p. does differ from the proj t.p. in the specific case being considered. I think this is indeed the case. Take $V$ to be ${\mathbb R}^2$ with usual Euclidean norm. The projective tensor product of $V$ with $V^\*$ can be identified with $M_2({\mathbb R})$ equipped with the trace class norm; the injective tensor product would lead to the `same' underlying vector space, equipped with the operator norm. The 2 x 2 identity matrix has trace class norm 2 and operator norm 1, so the two norms are genuinely different.
My answer is still not as clear as it should be, because due to a sluggish and temperamental internet connection I'm having trouble looking up just what the axioms for categorical duals in a SMC are. But if I recall correctly the natural map from $I \to V \otimes V^\*$ should be given by multiplying a scalar by the vector $e_1\otimes e_1 + e_2\otimes e_2$ where $e_1,e_2$ is an o.n. basis of ${\mathbb R}^2$ -- and that vector does not have norm 1 in the proj t.p. althought it does have norm 1 in the inj t.p.<|endoftext|>
TITLE: Quantitatively speaking, which subject area in mathematics is currently the most research active?
QUESTION [22 upvotes]: I was wondering if there is a list of the most active branches of mathematics?
If MathOverflow is a representative sample, then algebraic geometry is by far the most popular.  Is this the case?

REPLY [22 votes]: Sorry to add to the noise, but here it goes.  With a little script-fu (and emacs, of course!) I retrieved the data from MSC corresponding to the last ten years in each of the Primary Classifications.  Annoyingly the AMS changed their subject classification scheme recently, so that the numbers I queried were interpreted as MSC2010, whereas the papers are published from the year 2000.

43465     35 Partial differential equations
38151     62 Statistics
35994     81 Quantum theory
35633     68 Computer science
34474     65 Numerical analysis
28593     05 Combinatorics
28296     90 Operations research, mathematical programming
26406     34 Ordinary differential equations
26192     60 Probability theory and stochastic processes
23879     93 Systems theory; control
22361     11 Number theory
21689     76 Fluid mechanics
20787     91 Game theory, economics, social and behavioral sciences
19440     37 Dynamical systems and ergodic theory
18425     83 Relativity and gravitational theory
17323     94 Information and communication, circuits
17247     53 Differential geometry
16465     47 Operator theory
16134     03 Mathematical logic and foundations
15408     20 Group theory and generalizations
14225     92 Biology and other natural sciences
14051     82 Statistical mechanics, structure of matter
13663     46 Functional analysis
12894     74 Mechanics of deformable solids
11241     14 Algebraic geometry
10237     49 Calculus of variations and optimal control; optimization
10215     30 Functions of a complex variable
10154     16 Associative rings and algebras
 9801     01 History and biography
 9781     54 General Topology
 8014     42 Fourier analysis
 7103     58 Global analysis, analysis on manifolds
 6780     15 Linear and multilinear algebra; matrix theory
 6410     70 Mechanics of particles and systems
 6359     32 Several complex variables and analytic spaces
 6348     57 Manifolds and cell complexes
 6185     41 Approximations and expansions
 5935     39 Difference and functional equations
 5684     26 Real functions
 5349     17 Nonassociative rings and algebras
 5226     13 Commutative rings and algebras
 4840     78 Optics, electromagnetic theory
 4439     52 Convex and discrete geometry
 4418     33 Special functions
 4350     00 General
 3818     06 Order, lattices, ordered algebraic structures
 3511     28 Measure and integration
 3295     51 Geometry
 2948     22 Topological groups, Lie groups
 2944     55 Algebraic topology
 2538     86 Geophysics
 2089     45 Integral equations
 2052     18 Category theory; homological algebra
 1679     80 Classical thermodynamics, heat transfer
 1523     31 Potential theory
 1444     43 Abstract harmonic analysis
 1343     12 Field theory and polynomials
 1161     40 Sequences, series, summability
 1108     08 General algebraic systems
  898     44 Integral transforms, operational calculus
  775     19 K-theory
  534     85 Astronomy and astrophysics

Usual disclaimers apply.  In particular, before concluding that nobody works in astrophysics, go and check the submission statistics for astro-ph: more than 11,000 submissions in 2009 alone!  Clearly the AMS does not index very widely in this area.
Let me reiterate that I do not believe for a second that this data allows one to conclude anything of value about mathematics, just perhaps about mathematicians :)
Added (incorporating Gerald Edgar's summary in the comment below)
This is the summary of "pure maths" defined as classifications 00-60, with a total of 411902 articles reviewed in the decade that has just finished.  That, in case you are wondering is 55.38% of all papers reviewed.

00--08      Logic and Combinatorics         63804    15.49%
11--20      Algebra and Number Theory       80689    19.59%
22--49,60   Analysis and Probability       216252    52.50%
51--58      Geometry and Topology           51157    12.42%<|endoftext|>
TITLE: If associated-graded of a filtered bialgebra is Hopf, does it follow that the original bialgebra was Hopf?
QUESTION [8 upvotes]: Warning: older texts use the word "Hopf algebra" for what's now commonly called "bialgebra", whereas now "Hopf" is an extra condition.  So as to avoid any confusion, I'll give my definitions before concluding with my question.
Definitions
Let $C$ be a category with symmetric monoidal structure $\otimes$ and unit $1$ (and either strictify, or decorate all the following equations with associators and unitators and so on).  An (associative, unital) algebra in $(C,\otimes)$ is an object $V$ along with maps $e: 1\to V$ and $m: V\otimes V \to V$ satisfying associativity and unit axioms: $m\circ(m\otimes \text{id}) = m\circ (\text{id}\otimes m)$ and $m\circ (\text{id}\otimes e) = \text{id} = m\circ (e\otimes \text{id})$.  A (coassociative, counital) coalgebra is an object $V$ along with maps $\epsilon: V\to 1$ and $\Delta: V \to V\otimes V$ satisfying coassociativity and counit axioms.  A bialgebra is any of the following equivalent things:

A coalgebra in the category of algebras and algebra-homomorphisms ($1$ has its canonical algebra structure coming from the $\otimes$ axioms that $1\otimes 1 = 1$; in the tensor product of algebras, elements in the different multiplicands commute)
An algebra in the category of coalgebras and coalgebra-homomorphisms
An object $V$ with maps $e,m,\epsilon,\Delta$ satisfying the axioms above and a compatibility axiom:
$$ \Delta \circ m = (m\otimes m) \circ (\text{id} \otimes \text{flip} \otimes \text{id}) \circ (\Delta \otimes \Delta) $$

A bialgebra can have the property of being Hopf (it is a property, not extra data): a bialgebra $V$ is Hopf if there exists an antipode map $s: V\to V$ satisfying
$$ m \circ (s\otimes \text{id}) \circ \Delta = e\circ \epsilon = m \circ (\text{id} \otimes s) \circ \Delta $$
Naturally, it's better to see these definitions than read them; check e.g. the Wikipedia article.  If an antipode exists for a bialgebra, it is unique (justifying considering Hopfness a property rather than a structure) and it is an antihomomorphism for both the algebra and coalgebra structures.
Let VECT be the category of vector spaces (over your favorite field), with $\otimes$ the usual tensor product and $1$ the ground field.  A ($\mathbb N$-)filtered vector space is a sequence $V = \{V_0 \hookrightarrow V_1 \hookrightarrow V_2 \hookrightarrow \dots\}$ in VECT.  A morphism of filtered vector spaces $V \to W$ is a sequence of morphisms $V_n \to W_n$ so that every square commutes: $\{V_n \hookrightarrow V_{n+1} \to W_{n+1}\} = \{V_n \to W_n \hookrightarrow W_{n+1}\}$.  Equivalently, a filtered vector space is a space $V \in $VECT along with an increasing sequence of subspaces $V_0 \subseteq V_1 \subseteq \dots \subseteq V$ such that $V = \bigcup V_n$, and a linear map of filtered vector spaces $V \to W$ is filtered if the image of $V_n$ lies in $W_n$ for each $n$.
Because $\otimes$ is exact in VECT (because every monomorphism splits), to a pair $V,W$ of filtered vector spaces we can define an $\mathbb N^2$-filtered space with $(p,q)$-part $V_p\otimes W_q$, and then we can define the $\mathbb N$-filtered space $V\otimes W$ by setting $(V\otimes W)_n$ to be the colimit of the diagram given by all $V_p\otimes W_q$ with $p+q \leq n$.  Equivalently, we can take the tensor product in VECT of the unions $V = \bigcup V_n$ and $W = \bigcup W_n$, and then filter it by declaring that the $n$th part is the union of the $(p\otimes q)$th parts for $p+q = n$.
A ($\mathbb N$-)graded vector space is a sequence $\{V_0,V_1,V_2,\dots\}$ in VECT, or equivalently a space $V$ along with a direct sum decomposition $V = \bigoplus V_n$.  A morphism of graded vector spaces preserves the grading.
Let $V$ be a filtered vector space.  Its associated graded space $\text{gr}V$ is given by $(\text{gr}V)_n = V_n / V_{n-1}$, where $V_{-1} = 0$, of course.  Then $\text{gr}$ is a symmetric monoidal functor, and so takes filtered bialgebras to graded bialgebras.
Question
Let $V$ be a filtered bialgebra, i.e. a bialgebra in the category of filtered vector spaces.  Then $\text{gr}V$ is a graded bialgebra.  Suppose that $\text{gr}V$ is Hopf.  Does it follow that $V$ is Hopf?  I.e. suppose that $\text{gr}V$ has an antipode map.  Must $V$ have an antipode map?
(Or perhaps it requires additional hypotheses, e.g. that we be in characteristic 0, or that $V$ is locally finite in the sense that each $V_n$ is finite-dimensional?)

REPLY [7 votes]: There is a theorem by Takeuchi [Takeuchi, Mitsuhiro. Free Hopf algebras generated by coalgebras. J. Math. Soc. Japan  23  (1971), 561--582. MR0292876] that states that if $A$ is an algebra and $C$ is a coalgebra, then a map $f\in\hom(C,A)$ is convolution invertible iff its restriction $f|_{C_0}\in\hom(C_0,A)$ to the coradical $C_0$ of $C$ is convolution invertible. In particular, if $A=C$ is a bialgebra, to check that it is a Hopf algebra you need only check that $\mathrm{id}_{H_0}:H_0\to H$ is convolution invertible. If your filtration has $H_0\subseteq F_0$, and $F_0$ is a Hopf algebra, then this is automatic. Another simple instance of this is the classical case where $H_0=K$.<|endoftext|>
TITLE: Example of continuous function that is analytic on the interior but cannot be analytically continued?
QUESTION [18 upvotes]: I am looking for an example of a function $f$ that is 1) continuous on the closed unit disk, 2) analytic in the interior and 3) cannot be extended analytically to any larger set. A concrete example would be the best but just a proof that some exist would also be nice. (In fact I am not sure they do.)
I know of examples of analytic functions that cannot be extended from the unit disk. Take a lacuanary power series for example with radius of convergence 1. But I am not sure if any of them define a continuous function on the closed unit disk.

REPLY [9 votes]: I found some neat stuff in Remmert's Classical topics in complex function theory.
Fabry's gap theorem gives a way to construct many examples including some already mentioned.  Stated for the unit disk, it says:

If $m_1,m_2,\ldots$ is a sequence of positive integers such that $\displaystyle{\lim_{n\to\infty}}\frac{m_n}{n}=\infty$ and if $\displaystyle{f(z)=\sum_{n=1}^\infty a_nz^{m_n}}$ has radius of convergence 1, then the unit disk is the domain of holomorphy of $f$.

For example, if $p_n$ is the $n^{th}$ prime, then $$f(z)=\sum_{n=1}^\infty \frac{z^{p_n}}{n^2}$$ converges uniformly on the closed disk and is therefore continuous.  It is not analytically extendable to any larger set because it satisfies the hypotheses of Fabry's theorem.  

An interesting result that yields many such functions in a nonconstructive way is a theorem of Fatou-Hurwitz-Pólya:

If $\displaystyle{f(z)=\sum_{n=0}^\infty a_n z^n}$ has radius of convergence 1, then the set of functions $$f_\epsilon(z)=\sum_{n=0}^\infty \epsilon_na_nz^n$$ for $\epsilon_n\in\{\pm1\}$ whose domain of holomorphy is the unit disk has cardinality $2^{\aleph_0}$.

Hausdorff showed further that if $\displaystyle{\lim_{n\to\infty} |a_n|^{1/n}}$ exists (and equals 1) then the set of such functions whose domain of holomorphy is not the unit disk is at most countable.  This applies in particular to the function $\displaystyle{f(z)=\sum_{n=1}^\infty \frac{z^n}{n^2}}$, which therefore yields examples by changing the signs of the coefficients in all but countably many ways.

One more, this time an explicit example from Remmert: The series $$f(z)=1+2z+\sum_{n=1}^\infty\frac{z^{2^n}}{2^{n^2}}$$ is one-to-one and has real derivatives of all orders on the closed disk, and has the open disk as domain of holomorphy.  
Reference: Remmert's Classical topics in complex function theory, pages 252-258.  (Fatou-Hurwitz-Pólya is stated on a page without preview.)<|endoftext|>
TITLE: Are all Hawaiian Earrings homeomorphic?
QUESTION [31 upvotes]: The Hawaiian Earring is usually constructed as the union of circles of radius 1/n centered at (0,1/n): $\bigcup_1^\infty \left[   (0, \frac{1}{n}) + \frac{1}{n}S^1 \right]$.  However, nothing stops us from using the sequence of radii $1/n^2$ or any other sequence of numbers $a_n$.
I will call a Hawaiian Earring for a sequence (of distinct real numbers), $A = \lbrace a_n \rbrace$, the union of circles of radius $a_n$ centered at $(0, a_n )$.  Let the union inherit its topological structure from $\mathbb{R}^2$.  Are all of these spaces homeomorphic?
If $a_n$ is a monotone decreasing sequence converging to $0$, is its Hawaiian Earring homeomorphic to that of the sequence $\lbrace 1/n \rbrace$?

REPLY [4 votes]: Consider $\mathbb{R}^2 \subset \mathbb{C}\cup i \infty$. The inversion $z \mapsto \frac{1}{z}$ sends the circles $(a_n,0)+a_nS^1$ to the vertical lines $\{\frac{1}{2a_n}+i\mathbb{R}\}\cup i\infty$. Let $b_i=\frac{1}{a_i}$ for $i\geq 1$ and $b_0=0$. The function $f(x)=b_{\lfloor x \rfloor} + \{x\}(b_{\lfloor x + 1\rfloor} - b_{\lfloor x \rfloor})$ is a homeomorphism of $\mathbb{R}^+$ that sends the positive integers to $b_1, b_2,...$. Letting $\phi(x+iy)=\frac{1}{2}f(2x)+iy$, one gets an automorphism of the right-hand plane that sends the vertical lines $\frac{n}{2}+i\mathbb{R}$ to $\frac{1}{2a_n}+i\mathbb{R}$. Extending $\phi$ so that it sends $i\infty$ to $i\infty$, the function $z\mapsto \frac{1}{\phi(1/z)}$ then is a homeomorphism between the earring defined with $a_i$ and the standard one.<|endoftext|>
TITLE: Octonionic Unitary Group?
QUESTION [8 upvotes]: Hi all.
I was wondering if anyone has any references on work related to the Octonionic Unitary group.  I would imagine that such a group would be generated by Octonionic skew-Hermitian matrices (at the lie-algebra level) in analogy to the Complex Unitary and the Quaternion Unitary (aka Compact Symplectic) groups.  Some hints of the construction appear on page 28 of the paper "The Octonions" by Baez.
http://arxiv.org/abs/math/0105155
Thanks

REPLY [5 votes]: As others have pointed out, the nonassociativity of the octonions prevents one from constructing a group.  For example, any subgroup of the octonions lives inside of a quaternion subalgebra.  Having said that, the Clifford algebra $Cl(\mathbb{R}^7)$ has two inequivalent irreducible representations which are each as real vector spaces isomorphic to the octonions (i.e., they are eight-dimensional) and provided that we identify $\mathbb{R}^7$ with the imaginary octonions, the action of $Cl(\mathbb{R}^7)$ is given by left and right octonionic multiplications.  This is analogous to what happens with $Cl(\mathbb{R}^3)$ substituting octonion for quaternion in what I said above.  Now the Spin group $Spin(3)$ is the one-dimensional quaternionic unitary group and lives naturally inside $Cl(\mathbb{R}^3)$, so one could think of the group $Spin(7)$ as being the analogue of the one-dimensional octonionic unitary group.
By the same token, and given the low-dimensional isomorphisms
$$Spin(2,1) \cong SL(2,\mathbb{R})$$
$$Spin(3,1) \cong SL(2,\mathbb{C})$$
$$Spin(5,1) \cong SL(2,\mathbb{H})$$
one would be tempted to think of $Spin(9,1)$ as $SL(2,\mathbb{O})$, even though such a group as written does not exist.<|endoftext|>
TITLE: Field of Definition of a Meromorphic Function
QUESTION [5 upvotes]: Question
Let X be a smooth, projective curve over the algebraic closure of ℚ. Let f:X->ℙ1 be a meromorphic function. Assume that the zeros and the poles are defined over some number field, K. Then does this imply that cf is defined over K, for some c?
If so, do we have to assume that the zeros and poles are individually defined over K, or would this work if they are collectively defined over K as well?
Clarification
By being collectively defined I mean that there's some K-model of X, XK, and a closed subscheme YK of XK, such that after base change YK becomes the ramification locus of f.
Thoughts
Let D:=(f). Obviously, H0(O(D),X) is 1 dimensional. DK:=(fK) will also be degree 0, so all that's left to show is that H0(O(DK),XK) is also nonzero. This smacks of some invariance of cohomology theorem. I couldn't quite find the right one to use.
If this  line of argument works, this seems to imply that the ramification locus may be defined merely collectively.

REPLY [4 votes]: It is not sufficient that the subscheme of poles and zeroes is defined together over $K$, as the example of the function $(z+i)/(z-i):\mathbb P^1 \to \mathbb P^1$, defined only over $\mathbb Q(i)$, illustrates.
If poles of $f$, which form together a subscheme $D_\infty$, are defined over $K$, then the line bundle $\mathcal O(D_\infty)$ is defined over $K$. $f$ can be defined as the section of this line bundle that has the most poles, so it should be defined over $K$ as well.
More formally, suppose $f$ is not defined over $K$ (where $K$ is a separable extension of base field, which is $\mathbb Q$ here anyway). Then for some element $\sigma$ of Galois group of base field $f$ and $\sigma f$ would be sections of $\mathcal O(D_\infty)$ that have the same poles and zeroes, so their ratio would be a holomorphic function on $X$, thus constant. 
Therefore every $\sigma$ must act as a multiplication by constant. It remains to select any source point $x$ and divide the function by $f(x)$. After this, action of $\sigma$ will have to be multiplication by 1, thus a multiple of $f$ is defined over $K$ indeed.<|endoftext|>
TITLE: Is tensor product exact in abelian tensor categories with duals?
QUESTION [8 upvotes]: Suppose we are in an abelian tensor category with duals, where all objects have finite length.  Let $0 \to A \to B \to C \to 0$ be a short exact sequence and $Z$ an object of the category. Is 
$$0 \to Z \otimes A \to Z \otimes B \to Z \otimes C \to 0$$
exact?
Motivation: I am reading the proof of Proposition 5.7 in this paper of Deligne and trying to figure out why the lower sequence at the bottom of page 23 is exact. I believe $\mathcal{H}om(X,Y)$ here is $X^{\vee} \otimes Y$, although I have not actually found the point in the paper where he defines it. What he is trying to prove is that the corresponding sequence of external Hom's is exact, so he can't be using that fact.
There are, of course, tons of abelian tensor categories where $\otimes$ is not exact. For example, modules for any commutative ring $A$ which is not a field, with tensor product $\otimes_A$. But these don't usually have duals for all of their objects.

REPLY [11 votes]: Yes, because if Z has a dual, then in particular Z ⊗ – has a left adjoint (Z* ⊗ –) and hence commutes with limits (and similarly with colimits, but that's automatic if the category is closed monoidal).<|endoftext|>
TITLE: Why no abelian varieties over Z?
QUESTION [31 upvotes]: Motivation
I learned about this question from a wonderful article Rational points on curves by Henri Darmon. He gives a list of statements (some are theorems, some conjectures) of the form 

the set $\{$ objects $\dots$ over field $K$ with good reduction everywhere except set $S$ $\}$ is finite/empty

One interesting thing he mentions is about abelian schemes in the most natural case $K = \mathbb Q$, $S$ empty. I think according to the definition we have a trivial example of relative dimension 0.
Question

Why is the set of non-trivial abelian schemes over $\mathop{\text{Spec}}\mathbb Z$ empty?

Reference
This is proven in Il n'y a pas de variété abélienne sur Z by Fontaine, but I'm asking because: (1) Springer requires subscription, (2) there could be new ideas after 25 years, (3) the text is French and could be hard to read (4) this knowledge is worth disseminating.

REPLY [12 votes]: Here's another motivation for believing that there is no abelian variety over Z: the existence of such a beast would imply that the "motivic" expectations concerning L-functions are false. More precisely, it is not very difficult to show (Mestre does this in a paper in Compositio in 1986 using explicit formulas) that if an abelian variety over the rationals of dimension g at least 1 has the property that its L-function is entire with the expected functional equation, then its conductor is at least 10^{g}, and in particular is >1.  
(There's a very short proof of a weaker fact, sufficient to "imply" the theorem of Fontaine and Abrashkin, in Th. 5.51 of my book with Iwaniec, though the printed version has an unfortunate mistake, and -- I must apologize here for not checking the history at the time -- we did not mention Abrashkin...)<|endoftext|>
TITLE: Regularizing graphs
QUESTION [12 upvotes]: Let $G$ be a simple graph (undirected, no loops or parallel edges), with maximum degree $\Delta(G)$. I would like to add edges to the graph to make it regular, without increasing the maximum degree.
In general this is not possible. (For example, take the 5-vertex graph formed by taking a triangle ($K_3$) and adding two pendant edges to different vertices.)
However, what if we are also allowed to add vertices? I think I can see how to do it by creating many copies of the graph - so my question is: what is the least number of vertices we need to add?

REPLY [6 votes]: It is always enough to add k+2 more vertices where k denotes the maximum degree. This is sharp as shown by the graph which is a cycle of length 5 plus two independent edges.
The proof is the following. Add edges between non adjacent vertices whose degree is smaller than k until we can. After we cannot, we have some vertices with degree k and a clique of size l where l is at most k. This means we have l vertices whose degree is at least l-1 and at most k-1. Let us add k+1 or k+2 new vertices to our graph depending on the parity to be specified later. Connect the new vertices to the ones forming a clique such that the degrees of the new vertices differ by at most one from each other. Now we only have to add some edges among the new vertices, thus we have reduced the problem to degree sequences and it is an easy corollary of the Havel-Hakimi theorem that if the parity is correct, then a sequence consisting of only d-1 and d where d is less than the number of vertices is always a valid degree sequence.<|endoftext|>
TITLE: Which topological spaces admit a nonstandard metric?
QUESTION [42 upvotes]: My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R* in place of the usual R-valued metric.
That is, let us define that a topological space X is a nonstandard metric space, if there is a distance function, not into the reals R, but into some nonstandard R* in the sense of nonstandard analysis. That is, there should be a distance function d from X2 into R*, such that d(x,y)=0 iff x=y,   d(x,y)=d(y,x)
and  d(x,z) <= d(x,y)+d(y,z). Such a nonstandard metric would give rise to the nonstandard open balls, which would generate a metric-like
topology on X.
There are numerous examples of such spaces, beginning with R* itself. Indeed, every metric space Y has a nonstandard analogue Y*, which is a
nonstandard metric space. In addition, there are nonstandard metric spaces that do not arise as Y* for any metric space Y. Most of these examples
will not be metrizable, since we may assume that R* has uncountable cofinality (every countable set is bounded), and this will usually prevent
the existence of a countable local basis. That is, the nested sequence of balls around a given point will include the balls of infinitesimal
radius, and the intersection of any countably many will still be bounded away from 0. For example, R* itself will not be metrizable. The space R* is not connected, since it is the union of the infinitesimal neighborhoods of each point. In fact, one can show it is totally disconnected.
Nevertheless, it appears to me that these nonstandard metric spaces are as useful in many ways as their standard counterparts. After all, one can still
reason about open balls, with the triangle inequality and whatnot. It's just that the distances might be nonstandard. What's more, the
nonstandard reals offer some opportunities for new topological constructions: in a nonstandard metric space, one has the standard-part operation,
which would be a kind of open-closure of a set---For any set Y, let Y+ be all points infinitesimally close to a point in Y. This is something
like the closure of Y, except that Y+ is open! But we have Y subset Y+, and Y++=Y+, and + respects unions, etc.
In classical topology, we have various metrization theorems, such as Urysohn's theorem that any second-countable regular Hausdorff space is metrizable.
Question. Which topological spaces admit a nonstandard metric? Do any of the classical metrization theorems admit an analogue for nonstandard metrizability? How can we tell if a given topological space admits a nonstandard metric?
I would also be keen to learn of other interesting aspects of nonstandard metrizability, or to hear of interesting applications.
I have many related questions, such as when is a nonstandard metric space actually metrizable? Is there any version of R* itself which is metrizable? (e.g. dropping the uncountable cofinality hypothesis)

REPLY [3 votes]: I once considered the following situation:
$[0;1]^X$ is metrizable (with values in $\mathbb{R}$), if $X$ is countable. So I wondered whether $[0;1]^\mathbb{R}$ is nonstandard-metrizable for some totally ordered field $F$.
However it turns out that it does not admit a totally ordered local basis, so it can't be nonstandard-metrizable. I could give more details, if somebody is interested in them.<|endoftext|>
TITLE: Intuition for Group Cohomology
QUESTION [68 upvotes]: I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how these are used in the proofs of class field theory.
So far, all I can tell is that the cohomology groups are given by some ad hoc modding out process, then we derive some random properties (like the long exact sequence), and then we compute things like $H^2(\mathrm{Gal}(L/K),I_{L})$, where $I_L$ denotes the group of fractional ideals of a number field $L$, and it just happens to be something interesting for the study of class field theory such as $I_K/\mathrm{N}(I_L)$, where $L/K$ is cyclic and $\mathrm{N}$ denotes the ideal norm. We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups.
What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel modulo the image is interesting), but I don't know why we take these particular kernels modulo these particular images. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture.
For context, if $A$ is a $G$-module where $G$ is cyclic and $\sigma$ is a generator of $G$, then we define the endomorphisms $D=1+\sigma+\sigma^2+\cdots+\sigma^{|G|-1}$ and $N=1-\sigma$ of $A$, and then $H^0(G,A)=\mathrm{ker}(N)/\mathrm{im}(D)$ and $H^1(G,A)=\mathrm{ker}(D)/\mathrm{im}(N)$. Note that this is a slight modification of group cohomology, i.e. Tate cohomology, which the cohomology theory primarily used for Class Field Theory. Group cohomology is the same but with $H^0(G,A) = \mathrm{ker}(N)$. The advantage of Tate cohomology is that it is $2$-periodic for $G$ cyclic.

REPLY [2 votes]: Group cohomology is sheaf cohomology on a certain site, see e.g. Tamme's book on étale cohomology. The Hochschild-Serre spectral sequence is a Leray spectral sequence. Galois cohomology is étale cohomology of fields.
Using étale cohomology, it is trivial to compute the Brauer group of a local field: $\mathrm{Br}(K) = H^2(K,\mathbf{G}_m)$ and one has an exact sequence $0 \to H^2(\mathfrak{O}_K,\mathbf{G_m}) \to H^2(K,\mathbf{G}_m) \to H^1(k,\mathbf{Q}/\mathbf{Z}) \to 0$.  But $H^1(k,\mathbf{Q}/\mathbf{Z}) = \mathbf{Q}/\mathbf{Z}$ and $H^2(\mathfrak{O}_K,\mathbf{G_m}) \hookrightarrow H^2(k,\mathbf{G}_m) = 0$ since $G_k = \hat{\mathbf{Z}}$ and $H^1(k,\mathbf{G}_m) = 0$ (Hilbert 90) and the Herbrand quotient is trivial.<|endoftext|>
TITLE: English reference for a result of Kronecker?
QUESTION [26 upvotes]: Kronecker's paper Zwei Sätze über Gleichungen mit ganzzahligen Coefficienten apparently proves the following result that I'd like to reference:

Let $f$ be a monic polynomial with integer coefficients in $x$. If all roots of $f$ have absolute value at most 1, then $f$ is a product of cyclotomic polynomials and/or a power of $x$ (that is, all nonzero roots are roots of unity).

However, I don't have access to this article, and even if I did my 19th century German skills are lacking; does anyone know a reference in English I could check for details of the proof?

REPLY [7 votes]: In case anyone is interested, here's a run-down of the history of this result and a comparison of available proofs, as far as I could uncover in a rainy evening.

To be clear, Kronecker's original article, cited in P.A. Damianou's article, actually proves the following: given a monic polynomial with integer coefficients all of whose roots have norm 1, the roots are all roots of unity.
Kronecker's argument (in German) is almost identical to David Speyer's above. Both are almost identical to P.A. Damianou's---Damianou seems to have largely translated Kronecker, I think.
G. Gretier's Monthly article (cited by Damianou), J. Spencer's Fibonacci Quarterly article (cited by Greiter), and Kevin Buzzard's argument in brackets all prove this statement in related but different ways. Greiter uses companion matrices and diagonalization. Spencer's argument is the longest and builds a Fibonacci-type linear recurrence out of such a polynomial, using the corresponding Binet-type formula to show the recurrence must repeat.
Bombieri and Gluber's version is essentially a repackaging of Kronecker's argument in more number-theoretic language.
Polya and Szego's question 200 seems to be mostly just the theorem statement, though questions 198 and 199 are similar to the first part of Kevin Buzzard's argument.<|endoftext|>
TITLE: Lifting the p-torsion of a supersingular elliptic curve.
QUESTION [14 upvotes]: Let $K$ be a finite extension of $\mathbf{Q}_p$, with integer ring $R$ and residue field $k$. Say $G/R$ is a finite flat (commutative) group scheme of order $p^2$, killed by $p$. Say the special fibre of $G$ is isomorphic to the $p$-torsion in a supersingular elliptic curve over $k$. Is there some finite extension $L/K$ and an elliptic curve $E$ over $R_L$, the integers of $L$, such that $G$ becomes isomorphic to $E[p]$ over $R_L$?
The reason I ask is that in a lemma in Vincent Pilloni's thesis (which I unfortunately cannot find online) he proves that if the special fibre of $G$ is as above then the "degree" of such a $G$ is 1 (see Fargues' work on Harder-Narasimhan filtrations, for example, for the definition of "degree") via a brute force calculation. Pilloni's assertion would however also be a consequence of an affirmative answer to my question (and provided the motivation for my question). People are so good at deformation theory of finite flat group schemes nowadays that I thought this might be well-known to some people nowadays.

REPLY [5 votes]: This is a rough translation of Laurent F.'s answer (see this meta.MO thread). It's community wiki so you can improve the translation if you have 100 rep.

The answer is yes. The reason is as follows. If $S$ is a scheme where $p$ is locally nilpotent, then by definition (cf. Messing's thesis, Chapter I), a truncated Barsotti-Tate group of level of $1$ over $S$ is a finite locally free group scheme over $S$ annihilated by $p$ such that if $G_0$ denotes the reduction of $G$ modulo $p$ we have
$$
Im (F_{G_0}) = \ker (V_{G_0})
$$
as fppf sheaves on $S_0$, $S_0$ denoting the reduction modulo $p$ of $S$. However, it follows from the fiberwise criterion for flatness of EGA IV [Ed: he means EGA IV cor 11.3.11] that the morphism of $S_0$-group schemes of finite presentation
$$
F:G_0 \rightarrow \ker (V_{G_0})
$$
is faithfully flat if and only if it is fiberwise over $S_0$. From this we deduce that in the definition of a $BT_1$ one can replace $S_0$ by $S_{red}$! 
In the above considerations I took a scheme $S$ where $p$ is locally nilpotent, it follows that we have the same type of definition-result over a base which is a $p$-adic formal scheme.
We now return to the subject, namely the question of Kevin. We have that $G$ is a finite flat group scheme over the ring of integers of $K$ whose special fiber over the residue field of $K$ is a $BT_1$. The group scheme $G$ is hence a $BT_1$. [Ed: in fact this statement already implies Pilloni's result, by other results in the above cited article of Illusie.]
We will now use the following theorem (cf. the article of Illusie in Journees Arithmetiques de Rennes: "Déformations de groupes de Barsotti-Tate (d'après A. Grothendieck)", Astérisque 127). If $\mathcal{BT}_1$, resp. $\mathcal{BT}$ denotes the stack of truncated Barsotti-Tate groups of level $1$, resp. of Barsotti-Tate groups, over bases that are schemes in which $p$ is locally nilpotent, then the morphism "points of $p$-torsion"
$$
\mathcal{BT} \longrightarrow \mathcal{BT}_1
$$
is formally smooth. From this we deduce the following. Let $k$ be a perfect field of characteristic $p$ and $H$ a $p$-divisible group over $k$. Let $\mathfrak{X}$ be the space of deformations of $H$, a $spf (W(k))$-formal scheme non-canonically isomorphic to $spf \big(W(k)[[x_1,\dots,x_{d(h-d)}]]\big)$ where $h$ denotes the height of $H$ and $d$ dimension. Let $\mathcal{H}$ be the universal deformation of $\mathfrak{X}$. Then $\mathcal{H}[p]$ is a versal deformation of $H[p]$.
To conclude and answer the question of Kevin it suffices to invoke the theorem of Serre-Tate, which shows that if $H=E[p^\infty]$ where $E$ is a supersingular elliptic curve over $k$ then $\mathfrak{X}$ is the space of deformations of $E$. Applying the algebraization theorem of Grothendieck (GAGF) we deduce that if $\mathfrak{X}=spf(R)$, the universal deformation of the elliptic curve $E$ over $\mathfrak{X}$ actually becomes an elliptic curve over $spec(R)$. The result follows by specialization to $\mathcal{O}_K$.<|endoftext|>
TITLE: Notation for eventually less than
QUESTION [8 upvotes]: Is there some existing notation for

\[f(n)\leq g(n)\] for sufficiently large n

Apart from just writing that itself?
I'm thinking of something compact like the Landau notation $f\ll g$.
(Apologies if this is too specific for MathOverflow - just close it if so. I was also unsure what tags to add, so just edit it accordingly).

REPLY [2 votes]: I agree with Joel Hamkins's answer, but I don't entirely agree with his comment on that answer.  I generally use asterisks to mean "with finitely many exceptions" or "modulo finite sets", so I'd use $f\leq^*g$ and $A\subseteq^*B$ as Joel says.  But when working modulo some ideal $I$ other than the ideal of finite sets, I'd ordinarily avoid asterisks and instead write $f\leq_Ig$ and $A\subseteq_IB$.
I'd like to protest vigorously against the use of $\ll$ in this situation.  To me, $f\ll g$ means that $f$ is a lot smaller than $g$ (at least eventually), whereas here you might have $f(n)=g(n)-1$ for all $n$.<|endoftext|>
TITLE: elementary equivalence of infinitary symmetric groups
QUESTION [7 upvotes]: Two questions:

Suppose a and b are two uncountable cardinals. Consider the symmetric groups on sets of sizes a and b respectively (the symmetric group on a set is the group of all bijections from the set to itself, under composition). Consider the first-order theories of these as "pure groups" (i.e., just the group structure, no additional information). Are these elementarily equivalent? (does the answer change if we allow one of a or b to be the countable cardinal?)
Suppose $a_1$, $a_2$, $b_1$ and $b_2$ are uncountable cardinals with $a_1 < a_2$ and $b_1 < b_2$. Consider the symmetric group on a set of size $a_2$ and the subgroup of those bijections that have support of size at most $a_1$. Consider the pure theory of this group-subgroup pair (i.e., the pure theory of the group along with a membership predicate for the subgroup). Similarly, for $b_1$ and $b_2$. Are these two pure theories elementarily equivalent? Does the answer change if, instead, we look at the subgroup of those bijections that have support of size strictly less than $a_1$? Does the answer change if we allow the countable cardinal? (The support of a bijection is the set of elements that are moved). 

By the Baer-Schreier-Ulam theorem, the only normal subgroups of symmetric groups on infinite sets are the subgroups comprising bijections with support of size strictly less than a for some infinite cardinal a or the subgroups comprising bijections with support of size less than or equal to a for some infinite cardinal a, plus the trivial subgroup and the finitary alternating group. All these are also characteristic subgroups. 
If (2) is true, this would give examples of distinct characteristic subgroups of the same group that are elementarily equivalently embedded as subgroups (i.e., there is no first-order sentence true for one subgroup that is not true for the other).

REPLY [12 votes]: For each ordinal $\alpha < \omega^\omega$ the symmetric group on $\aleph_{\alpha}$ is first order definable in the class of all symmetric groups; i.e. there is a sentence in the first order language of groups that is true in $Sym(A)$ iff $|A| = \aleph_\alpha$. See Mckenzie - On elementary types of symmetric groups, Algebra Universalis  1 (1971), 13--20.
Shelah provides an interesting necessary and sufficient condition for the elementary equivalence of $Sym(\aleph_\alpha)$ and $Sym(\aleph_\beta)$ - First order theory of permutation groups, Israel J. Math. 14 (1973), 149–162; corrections: ibid. 15 (1973), 437–44.<|endoftext|>
TITLE: Class number measuring the failure of unique factorization
QUESTION [27 upvotes]: The statement that the class number measures the failure of the ring of integers to be a ufd is very common in books. ufd iff class number is 1. This inspires the following question:
Is there a quantitative statement relating the class number of a number field to the failure of unique factorization in the maximal order - other than $h = 1$ iff $R$ is a ufd?
In what sense does a maximal order of class number 3 "fail more" to be a ufd than a maximal order of class number 2?
Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?

REPLY [3 votes]: I know I'm late to the party, but it seems to me the other answers don't directly answer the last part of the question:

Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?

No.  Of course it's not necessarily true that a given integer in two fields $K_1 \cap K_2$ will have more distinct factorizations in $K_1$ than $K_2$ if $h(K_1) \ge h(K_2)$.  But even in simple situations where you could hope for this to be true, the opposite behavior can happen. 
In this paper, I gave a combinatorial expression for the number $\eta(x)$ of distinct factorizations of $x \in \mathcal O_K$ ($K$ a number field) in terms of the ideal classes of $\mathfrak p_i$ and the exponents $e_i$, where $x \mathcal O_K$ has prime ideal factorization $\prod \mathfrak p_i^{e_i}$.  There is also a precise description of the structure of distinct factorizations, which makes it clear that the structure of the distinct factorizations can get more complicated with more complicated class groups.  (In particular, you can use this description to get Carlitz's theorem and related results.)
But to return to your question, here is an example where you can describe $\eta(x)$ simply.  Suppose $K$ is quadratic, $x=p$ is a rational prime and $\mathfrak p$ is the prime ideal of $\mathcal O_K$ above $p$, then for any $n \in \mathbb N$,
$$ \eta(p^n) = \begin{cases}
1 & \mathfrak p \text{ principal or $p$ ramifies} \\
\lfloor \frac nm \rfloor + 1 & \text{else},
\end{cases} $$
where $m$ is the order of $\mathfrak p$ in the class group.  
In particular, suppose $K_1$ and $K_2$ are quadratic fields with class numbers 3 and 2 respectively, and $p$ is rational prime unramified in both $K_1$ and $K_2$.  Let $\mathfrak p_1$ and $\mathfrak p_2$ be primes of $K_1$ and $K_2$ above $p$.  Then $p^n$ (for any $n > 3$) has more distinct factorizations in $K_2$ (a class number 2 field) than $K_1$ (a class number 3 field) if and only if $\mathfrak p_2$ is not principal.<|endoftext|>
TITLE: How probable is it that a rational prime will split into principal factors in a Galois number field?
QUESTION [17 upvotes]: Let $L$ be a Galois number field over $\mathbb{Q}$.  Based on classical algebraic number theory (specifically, the Chebotarev density theorem), I can answer lots of numerical questions about how primes $p\in\mathbb{Z}$ split in $\mathfrak{O}_L$. For example the probability that $p$ splits completely is $1/[L:\mathbb{Q}]$, and the average number of factors is the sum over the Galois group of the reciprocals of the orders (so, for a quadratic extension, we get $1+1/2=3/2$ since half the primes split, and half don't.
My question: Another question one can ask is if the factors of $p$ are principal (since $L$ is Galois, they are all conjugate, and thus all principal or all not).  Are there any quantitative results about this?
One might optimistically hope that the probability of this is the reciprocal of the class number, but I have no idea if this is true.

REPLY [5 votes]: I believe that Felipe's answer is not correct.  [Edit: rather, it is correct according to a different interpretation of the question.  But my interpretation is also natural, I think.]
Say a prime $p$ is inert in $K$ if $p \mathbb{Z}_K$ remains prime.  In particular, inert primes have all of their factors principal.
If $L/\mathbb{Q}$ is not cyclic, then there are no inert primes.  However, if it is cyclic, say of degree $d$, then the density of inert primes is $\frac{\varphi(d)}{d}$, which gives a lower bound on the answer to Ben's question.  This lower bound can certainly be greater than
$\frac{1}{h(L)}$: take for instance imaginary quadratic fields with sufficiently large discriminant.<|endoftext|>
TITLE: What's the analogue of the Hilbert class field in the following analogy?
QUESTION [21 upvotes]: There's a wonderful analogy I've been trying to understand which asserts that field extensions are analogous to covering spaces, Galois groups are analogous to deck transformation groups, and algebraic closures are analogous to universal covering spaces, hence the absolute Galois group is analogous to the fundamental group.  (My vague understanding is that the machinery around etale cohomology makes this analogy precise.)  
Does the Hilbert class field (of a number field) fit anywhere into this analogy, and how?
Phrased another way, what does the Hilbert class field of the function field of a nonsinguar curve defined over $\mathbb{C}$ (say) look like geometrically?

REPLY [7 votes]: The absolute Galois group is $\pi_1( Spec K)$ and the Galois group of the maximal unramified extension of $K$ is $\pi_1( Spec R)$, where $R$ is the ring of integers of $K$ and the class group of $K$ is the abelianization of $\pi_1( Spec R)$ and the Hilbert class field is the field corresponding to this quotient.
A curve has a function field and therefore an absolute Galois group, which is much bigger than the fundamental group and describes all branched covers of the curve. The fundamental group corresponds of course to unramified covers. Working over C gives you infinite groups even for the abelianization. If you work over a finite field, then the analogy is more precise.<|endoftext|>
TITLE: What is $\sum (x+\mathbb{Z})^{-2}$?
QUESTION [14 upvotes]: This is a simple question, but its been bugging me.  Define the function $\gamma$ on $\mathbb{R}\backslash \mathbb{Z}$ by
$$\gamma(x):=\sum_{i\in \mathbb{Z}}\frac{1}{(x+i)^2}$$
The sum converges absolutely because it behaves roughly like $\sum_{i>0}i^{-2}$.
Some quick facts:

Pretty much by construction, $\gamma$ is periodic with period $1$.  
As it approaches any integer from the left or right, it goes to positive infinity.
It is symmetric at every half integer; that is, $\gamma(n+1/2+x)=\gamma(n+1/2-x)$ for all $n\in \mathbb{Z}$ and $x\in \mathbb{R}$.


Can $\gamma$ be expressed in terms of more familiar (presumably trigonometric) functions?

My best guess is $\gamma(x)=\sin^{-2}(\pi x)$, but this is based more on what I hope it would be, rather than what it is.

REPLY [3 votes]: For the record, one can prove the product formula for the sine without complex analysis (and without the Gamma function), from which (as David Speyer noted) one can recover $\sum_{i \in \bf Z} (x+i)^{-2}$ as the second logarithmic derivative.  See this Stackexchange answer.
[Later I learned that the case $x=1/2$ recovers Wallis's original proof
of his product formula for $\pi$; perhaps this generalization too was known before the complex-analytic proof.]<|endoftext|>
TITLE: What are the points of Spec(Vassiliev Invariants)?
QUESTION [18 upvotes]: Background
Recall that a (oriented) knot is a smoothly embedded circle $S^1$ in $\mathbb R^3$, up to some natural equivalence relation (which is not quite trivial to write down).  The collection of oriented knots has a binary operation called connected sum: if $K_1,K_2$ are knots, then $K_1 \# K_2$ is formed by spatially separating the knots, then connecting them by a very thin rectangle, which is glued on so that all the orientations are correct.  Connect sum is commutative and associative, making the space of knots into a commutative monoid.  In fact, by a theorem of Schubert, this is the free commutative monoid on countably many generators.  A ($\mathbb C$-valued) knot invariant is a $\mathbb C$-valued function on this monoid; under "pointwise" multiplication, the space of knot invariants is a commutative algebra $I$, and $\#$ makes $I$ into a cocommutative bialgebra.  I.e. $I$ is a commutative monoid object in $(\text{CAlg})^{\rm{op}}$, where $\text{CAlg}$ is the category of commutative algebras.

Warm-up question: Any knot $K$ defines an algebra morphism $I \to \mathbb C$, i.e. a global point of $I \in (\text{CAlg})^{\rm{op}}$.  Are there any other global points?

Edit: In response to Ilya N's comment below, I've made this into its own question.
Finite type invariants
Recall that a singular knot is a smooth map $S^1 \to \mathbb R^3$ with finitely many transverse self-intersections (and otherwise it is an embedding), again up to a natural equivalence.  Any knot invariant extends to an invariant of singular knots, as follows: in a singular knot $K_0$, there are two ways to blow up any singularity, and the orientation determines one as the "right-handed" blow-up $K_+$ and the other as the "left-handed" blow-up $K_0$.  Evaluate your knot invariant $i$ on each blow-up, and then define $i(K_0) = i(K_+) - i(K_-)$.  Note that although the connect-sum of singular knots is not well-defined as a singular knot, if $i\in I$ is a knot-invariant, then it cannot distinguish different connect-sums of singular knots.  Note also that the product of knot invariants (i.e. the product in the algebra $I$) is not the point-wise product on singular knots.
A Vassiliev (or finite type) invariant of type $\leq n$ is any knot invariant that vanishes on singular knots with $> n$ self-intersections.  The space of all Vassiliev invariants is a filtered bialgebra $V$ (filtered by type).  The corresponding associated-graded bialgebra $W$ (of "weight systems") has been well-studied (some names: Kontsevich, Bar-Natan, Vaintrob, and I'm sure there are others I haven't read yet) and in fact is more-or-less completely understood (e.g. Hinich and Vaintrob, 2002, "Cyclic operads and algebra of chord diagrams", MR1913297, where its graded dual $A$ of "chord diagrams" is described as a sort of universal enveloping algebra).  In fact, this algebra $W$ is Hopf.  I learned from this question that this implies that the bialgebra $V$ of Vassiliev invariants is also Hopf.  Thus it is a Hopf sub-bialgebra of the algebra $I$ of knot invariants.
I believe that it is an open question whether Vassiliev invariants separate knots (i.e. whether two knots all of whose Vassiliev invariants agree are necessarily the same).  But perhaps this has been answered — I feel reasonably caught-up with the state of knowledge in the mid- to late-90s, but I don't know the literature from the 00s.
Geometrically, then, $V \in (\text{CAlg})^{\rm{op}}$ is a commutative group object, and is a quotient (or something) of the monoid-object $I \in (\text{CAlg})^{\rm{op}}$ of knot invariants.  The global points of $V$ (i.e. the algebra maps $V \to \mathbb C$ in $\text{CAlg}$) are a group.
Main Questions
Supposing that Vassiliev invariants separate knots, there must be global points of $V$ that do not correspond to knots, as by Mazur's swindle there are no "negative knots" among the monoid $I$. Thus my question.

Main question. What do the global points of $V$ look like?  

If Vassiliev invariants do separate knots, are there still more global points of $V$ than just the free abelian group on countably many generators (i.e. the group generated by the free monoid of knots)?  Yes: the singular knots.  (Edit: The rule for being a global point is that you can evaluate any knot invariant at it, and that the value of the invariant given by pointwise multiplication on knots is the multiplication of the values at the global point.  Let $K_0$ be a singular knot with one crossing and with non-singular blow-ups $K_+$ and $K_-$, and let $f,g$ be two knot invariants.  Then $$\begin{aligned} (f\cdot g)(K_0) & = (f\cdot g)(K_+) - (f\cdot g)(K_-) = f(K_+)\cdot g(K_+) - f(K_-)\cdot g(K_-) \neq \\\\ f(K_0) \cdot g(K_0) & = f(K_+)\cdot g(K_+) - f(K_+)\cdot g(K_-) - f(K_-)\cdot g(K_+) + f(K_-)\cdot g(K_-)\end{aligned}$$.)  What else is there?  
What can be said without knowing whether Vassiliev invariants separate knots?

REPLY [8 votes]: The points of Spec $V$ live n a somewhat complicated completion of the space of knots (not formal linear combinations), but I think it's illuminating to think about the case of braids and their "Vassiliev" invariants.  In that context the global points of Vassiliev invariants are elements of the completion of the braid group with respect to the lower central series.  The braid group embeds in this completion, and so Vassiliev invariants do separate braids.  (This is not hard.)
By the way, the knot invariants $I$ also form a bi-algebra, where the coproduct is dual to connect sum of knots, although that coproduct plays no role for this question.<|endoftext|>
TITLE: Blow up along codimension one closed subscheme
QUESTION [9 upvotes]: Suppose X is dimension two locally Noetherian scheme. Y is a closed subscheme of X, with codimension 1. Denote X' to be the blow up of X along Y. Prove that the structure morphism f:X'-->X is a finite morphism. 
It suffices to show it's quasi-finite according to Zariski's main theorem. But I can't exclude the possibility that an irreducible component of $f^{-1}(Y)$ maps to a closed point of Y.

REPLY [14 votes]: I think it's not true : 
Let $X=Spec(A)$ with $A=k[x,y,z]/(x^2-y^2-z^2)$ be a quadratic cone. Let $Y$ be a line through the origin of the cone : its ideal is $I=(z,x-y)$. We calculate : 
$$X'=Proj_{A}A[t,u]/(zt-(x+y)u,(x-y)t-zu),$$ [EDIT : THE FORMULA HAS BEEN CORRECTED]
where, in the graded $A$-algebra $A+I+I^2+....$ we denoted $t$ and $u$ the degree one generators corresponding to $z$ and $x-y$.
Now, quotienting by $x$, $y$, and $z$, we calculate the fiber over the origin of this blow-up It is Proj(k[t,u]), which is a positive-dimensional projective line !<|endoftext|>
TITLE: Two kinds of orientability/orientation for a differentiable manifold
QUESTION [38 upvotes]: Let $M$ be a differentiable manifold of dimension $n$. First I give two definitions of Orientability.
The first definition should coincide with what is given in most differential topology text books, for instance Warner's book. 

Orientability using differential forms: There exists a nowhere vanishing differential form $\omega$ of degree $n$ on $M$.

The second one is from Greenberg and Harper, "Algebraic Topology". This is the "fundamental class" approach. Let $x$ be a point on $X$, and let $R$ be a commutative ring and in the following the homologies are with coefficients in $R$.

Local orientability: A local $R$ orientation of $X$ at $x$ is a choice of a generator of the $R$-module $H_n(X, X-x)$.

By a simple application of Excision, it is seen that the above homology module is indeed isomorphic to $R$. We can also so arrange a neighborhood around every point that this local orientation can be "continued to a neighborhood" and is "coherent". Forgive me for being imprecise here; the detailed lemmas are in the reference given above. With this background in mind, we define:

A Global $R$-orientation of $X$ consists of: 1. A family $U_i$ of open sets covering of $X$, 2. For each $i$, a local orientation $\alpha_i \in H_n(X, X -U_i)$ of along $X$, such that a "compatibility condition" holds.

Here again I am imprecise about the compatibility condition; please check in the reference given above for details. I mean this basically as a question for those who already know both the definitions, as fully writing down the second definition would take 2-3 pages with all the necessary lemmas.
Also we define "orientation" to be a such a global choice.
Now the question:

How do the two definitions, the first one using differential forms, and the second one using homology, match?

Of course, to match we have to take $\mathbb{Z}$ to be the base ring for homology. A related question is about the meaning of orientability and orientation when we take a base ring other than $\mathbb{Z}$. It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring to be $\mathbb{Z}/5\mathbb{Z}$?
Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?

REPLY [7 votes]: Speaking personally, I was not really comfortable with the notion of orientation until I understood the notion for vector bundles, so I will tell you about that.
Given a vector bundle $\pi: E \to B$, first select an orientation for each fiber $\pi^{-1}(b)$.  The bundle will be oriented is you made these choices in a coherent manner and the following two are equivalent notions of 'coherent'.
1) For every point $b$ in $B$ has a neighborhood $N$ such that there are sections $s_1, \ldots, s_r: N \to E$ such that for all $n \in N$: {$s_1(n), \ldots, s_r(n)$} is an oriented basis for the fiber $\pi^{-1}(n)$.
2) Every point $b$ in $B$ is in a vector bundle chart $\phi:N \times \mathbb{R}^r \to \pi^{-1}(N)$ such that $\phi(n,\cdot): n \times \mathbb{R}^r \to \pi^{-1}(n)$ is orientation preserving.
Forgetting about picking an orientation for each fiber ahead of time, being orientable is also equivalent to:
3) You can cover $B$ with vector bundle charts $\phi:N \times \mathbb{R}^r \to \pi^{-1}(N)$ such that for any two $\phi$ and $\psi$ the linear isomorphism $n \times \mathbb{R}^r \stackrel{\phi(n,\cdot)}{\to} \pi^{-1}(n) \stackrel{\psi(n,\cdot)^{-1}}{\to} n \times \mathbb{R}^r$ is orientation preserving.
4) There is a nonzero section of the line bundle $\wedge^rE \to B$.
Now a manifold $M$ being orientable is equivalent to its tangent bundle being orientable.  Given what has been said, the quickest way to see this is to note that a nonzero n-form on $M$ is by definition a nonzero section of the bundle $\wedge^n (T^\*M)$.  (Note: $T^\*M$ is orientable iff $TM$ is orientable since they are isomorphic as bundles by picking a Riemannian metric.)
The canonical example of a nonorientable bundle is the Mobius bundle which is the line bundle over the circle whose total space looks like a Mobius band.  In terms of 1) this bundle is not orientable since if you pick a nonzero section (vector) at a point and try to extend to the whole circle, by the time you get back to where you started your vector is now pointing the other way.<|endoftext|>
TITLE: Why isn't Likelihood a Probability Density Function?
QUESTION [16 upvotes]: I've been trying to get my head around why a likelihood isn't a probability density function. My understanding says that for an event $X$ and a model parameter $m$:
$P(X|m)$ is a probability density function
$P(m|X)$ is not
It feels like it should be, and I can't find a clear explanation of why it's not. Does it also mean that a Likelihood can take a value greater than 1?

REPLY [13 votes]: The accepted answer is wrong. The likelihood is the function $\theta \mapsto L(\theta \mid x)=f(x \mid \theta)$ for a given $x$ in the observations space and $f(\cdot \mid \theta)$ is a Radon-Nikodym derivative of $P_\theta$ when the statistical model is given by a family of probabilities ${(P_\theta)}_{\theta\in\Theta}$ on the observations space.
In general there's not even a $\sigma$-field in the parameter space $\Theta$, hence the question "is the likelihood a pdf ?" is not even meaningful!
For more information see https://stats.stackexchange.com/questions/31238/what-is-the-reason-that-a-likelihood-function-is-not-a-pdf and https://stats.stackexchange.com/questions/29682/how-to-rigorously-define-the-likelihood<|endoftext|>
TITLE: Does homology detect chain homotopy equivalence?
QUESTION [22 upvotes]: Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.

REPLY [14 votes]: Yes, this is true, and it does not matter whether the complexes are bounded from any side (nor of course does it matter whether the homology is finitely generated).  This is so because:

The homotopy category of free abelian groups is equivalent to the derived category of abelian groups.  This holds even for unbounded complexes, since the category of abelian groups has a finite homological dimension.
Any complex of abelian groups is quasi-isomorphic to its homology, since the category of abelian groups has homological dimension 1.<|endoftext|>
TITLE: Why is edge-coloring less interesting than vertex-coloring?
QUESTION [21 upvotes]: I was wondering why there is (apparently) much more research directed towards vertex-coloring than edge-coloring? Prima facie, it seems that edge-coloring is just as "natural" a thing to investigate.
I can think of a few reasons:

Vertex coloring is well behaved under deletion and contraction of edges.
Vertex colorability is closely linked to the cycle matroid.
Edge-coloring can be regarded as vertex-coloring restricted to line graphs.
Since Vizing's theorem (that the chromatic index of $G$ is either $\Delta(G)$ or $\Delta(G)+1$) edge-coloring has been solved (asymptotically).

But is it really true that edge-coloring is less interesting than vertex-coloring?

REPLY [3 votes]: If you're going to be completely honest about the question, you need to consider edge colourings of multigraphs.  Even in this case there are not a lot of open problems.  Vizing's theorem already tells us that the chromatic index differs from the maximum degree by at most the maximum multiplicity.  However, more is known:

We can compute the fractional chromatic index in polynomial time; even approximating the fractional chromatic number of a graph is hard.
The fractional chromatic index and the chromatic index agree asymptotically, as first proven by Kahn.  However, the chromatic number is not even bounded by any function of the fractional chromatic number.
The Goldberg-Seymour conjecture implies that $\chi'_f$ and $\chi'$ differ by at most one.  This is a long-standing open problem on the chromatic index.  So what does that leave?  Well, proving that the difference is bounded by some universal constant, but not too much else.<|endoftext|>
TITLE: Founding of homological without quite involving derived categories
QUESTION [14 upvotes]: I am looking at the foundations of homological algebra, e.g. the introduction
of Ext and Tor, and am unsatisfied. The references I look at start with
"this is called a projective module, this is called a projective resolution,
now pick one and use it to define the right derived functors of your
left exact functor". I would like to see a presentation more along the
following lines:

The functor Hom(A,*), applied to a short exact sequence of modules,
doesn't produce another such. An oracle tells us that it does produce
a long exact sequence; what could it be?
We already know (from antiquity) that a short exact sequence of
complexes induces a long exact sequence on cohomology.
But in #1 we put in modules, not complexes. So let's fix that by hoping 
that Hom(A,*) can be extended in a natural way to the category of complexes
(and really, to descend to the derived category).
Such an extension might be required to have the following properties: ???
Now I'd like it to be easy to see that the extension is unique if it
exists. When is it easy to compute? At this point I'd like the definition
of "projective module" to suggest itself.
Finally, the usual boring checks that using projective resolutions
to define it, the extension does indeed exist.

One way to answer this is to say "In part 4, define the derived category,
and its t-structure, then ask that the extension be exact in the 
appropriate sense". I'm hoping to avoid going quite that far, or at least,
doing it in a way that doesn't involve introducing too many more definitions.

REPLY [2 votes]: On rereading your question, Jacobson is not what you want after all.
Anton gave a very nice answer along these lines here. In comments to that post, Tyler Lawson recommends Cartan and Eilenberg.<|endoftext|>
TITLE: Do pushouts along universal homeomorphisms exist?
QUESTION [12 upvotes]: References and background on universal homeomorphisms
Definition [EGA I (2d ed.) 3.8.1]. A morphism $f:V\to U$ is a universal homeomorphism if for any morphism $U'\to U$, the pullback $V\times_UU'\to U'$ is a homeomorphism.
Theorem [EGA IV 18.12.11]. A morphism is a universal homeomorphism if and only if it is surjective, integral, and radicial.
Theorem ["Topological invariance of the étale topos," SGA I Exp IX, 4.10 and SGA IV Exp. VIII, 1.1] If $f:V\to U$ is a universal homeomorphism, then the induced morphism $f:V_{\textrm{ét}}\to U_{\textrm{ét}}$ of the small étale topoi is an equivalence.
General examples. Any nilimmersion, any purely inseparable field extension (or any base change thereby), the geometric Frobenius of an $\mathbf{F}_p$-scheme [SGA V Exp. XIV=XV, § 1, No. 2, Pr. 2(a)].
Theorem. Suppose $X$ a reduced scheme with finitely many irreducible components. Denote by $X'$ its normalization. Then the natural morphism $X'\to X$ is a universal homeomorphism if and only if $X$ is geometrically unibranch.
Specific example. Suppose $k$ an algebraically closed field of characteristic $2$. Consider the subring $k[x^2,xy,y]\subset k[x,y]$. The induced morphism
$$\mathrm{Spec}k[x,y]\to\mathrm{Spec}k[x^2,xy,y]$$
is a universal homeomorphism.
Question

Do pushouts along universal homeomorphisms exist in the category of schemes?

In more detail. Suppose $f:V\to U$ a universal homeomorphism, and suppose $p:V\to W$ a morphism. Everything here is a scheme; I can assume $W$ quasicompact and quasiseparated, but I have no control over the map $V\to W$. Now of course I can construct the pushout $P$ of $V\to U$ along $V\to W$ as a locally ringed space with no trouble (just take the underlying space of $W$ along with the fiber product $O_W \times_{p_{\star}O_V}p_{\star}O_U$), but I can't show that $P$ is a scheme. Is it?
Thoughts
Of course the key point here is that $f$ is a universal homeomorphism, not just some run-of-the-mill morphism. So one can try to treat the cases where $f$ is schematically dominant or a nilimmersion separately.
Update
If $f$ is a nilimmersion, then I now see how to prove this completely. I still have no idea how to proceed the schematically dominant case.
[EDIT: I removed the additional question.]

REPLY [7 votes]: The answer to your question is yes in positive characteristic. Check Kollár's paper "Quotient Spaces Modulo Algebraic Groups" (Ann. of Math. 1997), Lem 8.4. The reason is that universal homeomorphisms behave as nil-immersions in positive characteristic (due to Frobenius). I think the answer is no in characteristic zero.
A general comment is that without assuming that one of the two morphisms in a push-out is a monomorphism the "correct" push-out should be a stack. This partly explains why most results on push-outs are when one of the maps is a monomorphism (e.g. push-outs of a finite map and a closed immersion, push-outs of an etale morphism and an open immersion).<|endoftext|>
TITLE: Tensor product is to flat as Hom is to ?
QUESTION [6 upvotes]: Sorry if I'm missing something here, but what do we call $M$ if the functor $H_M:N\mapsto Hom(M,N)$ is exact? Is this in fact equivalent to being flat through some adjointness properties?

REPLY [14 votes]: We call such modules projective.  If you take $N\mapsto Hom(N,M)$ then you get injective modules.  This is fairly basic, and covered in any homological algebra book, and mentioned on wikipedia.

REPLY [12 votes]: It might also be helpful to know that projective is equivalent to being a summand of a free module (apply $Hom(M,\text{--})$ to a presentation of $M$), and hence projectives are flat.
The converse is not true in general (e.g. $\mathbb Q$ is flat as a $\mathbb Z$-module, but not projective), but for finitely presented modules over commutative rings, flat and projective are equivalent.

REPLY [6 votes]: I'm pretty sure that $M$ is called projective in this case, and if $N \rightarrow Hom(N,M)$ is exact then $M$ is called injective. I might have it backwards, though.<|endoftext|>
TITLE: Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set?
QUESTION [11 upvotes]: There are two version of the singular simplicial space of a topological space $X$, one discrete and one internal. At least if X is nice, both of them have homotopy equivalent geometric realizations (and are both equivalent to X itself). I want to know why?
Background/Motivation
Let $\Delta$ be a skeleton of the category of finite non-empty ordered sets. The objects of $\Delta$ are the ordered sets [n]. A simplicial object of a category C is a functor $X: \Delta^{op} \to C$. The category $\Delta$ can be realized as a sub-category of topological spaces (the category of n-simplices, $\Delta^n$) and (via left Kan extension) this gives rise to a geometric realization functor from simplicial sets to topological spaces. It lands in the nice category of CW-complexes, and is denoted $|X|$.
The same geometric realization formula works for simplicial spaces and defines a functor from simplicial spaces to topological spaces. It doesn't always land in CW-complexes.
By general non-sense there is a right adjoint to the realization which is the singular functor. It associates to a topological space the simplicial set given by $Sing(X):[n] \mapsto map(\Delta^n, X)$. The realization from simplicial spaces to topological space also has an adjoint which is given by the simplicial space $\underline{Sing}(X): [n] \mapsto \underline{map}(\Delta^n, X)$, where $\underline{map}$ denotes the mapping space with the compactly generated compact open topology (I'm assuming all spaces are compactly generated).
A simplcial set can be viewed as a discrete simplicial space and so we have two different singular functors and a natural map of simplicial spaces between them:
$Sing(X) \to \underline{Sing}(X)$
This gives, on geometric realization a map of spaces: $|Sing(X)| \to |\underline{Sing}(X)|$. When X is sufficiently nice this map is known to be a homotopy equivalence, and both spaces are homotopy equivalent to X.

Why is this the map $|Sing(X)| \to |\underline{Sing}(X)|$ a homotopy equivalence? Can this be deduced from some sort of connectivity estimate between the constituent spaces of  these simplicial spaces?

I know there is an indirect way to prove this equivalence by comparing both spaces to X, but I'm interested in generalizing this result to some related constructions and so I would like a direct argument which uses the map between them. I'm willing to replaces "simplicial space" by bisimplicial set or to assume that the simplicial space is sufficiently "good". I'm hoping for something related to this question.

REPLY [7 votes]: There are maps $|Sing(X)| \to |\underline{Sing}(X)| \to X$ which realize to weak homotopy equivalences.  The inclusion of the n-skeleton $|Sing(X)|^{(n)}| \to |Sing(X)|$ is n-connected, because this is always true for CW-complexes, and so the map $|Sing(X)|^{(n)} \to X$ is n-connected.  You can't really do any better than this estimate because the n-skeleton has zero homology groups in degrees above n.
The simplicial space $\underline{Sing}(X)$ contains the sub-simplicial space of constant simplices $\Delta^n \to X$.  This is homeomorphic to $X$ itself and so, if we write $cX$ for the constant simplicial space with value $X$, we get a map $cX \to \underline{Sing}(X)$.  This inclusion $X \to \underline{map}(\Delta^n,X)$ is a homotopy equivalence because the simplex is contractible, so this map of simplicial spaces is levelwise a weak equivalence.  The geometric realization of $cX$ is $X$ itself, and so is its n-skeleton for all n.  An excision argument (which takes some work) will show that the same is true for the simplicial space (at least under good conditions), and so each of the skeleta of $|\underline{Sing}(X)|$ is homotopy equivalent to $X$.
Therefore the map on n-skeleta is n-connected.  I realize that this is the "compare with $X$" game that you mentioned, but my point is that because the simplicial space $\underline{Sing}(X)$ is homotopically constant the comparison $|Sing(X)| \to |\underline{Sing}(X)|$ really is comparing with $X$.  From the point of view of homotopy theory it's not arising from good levelwise structure of the map at all, but comes from the simplicial assemblage.
I don't know whether this helps your generalization.<|endoftext|>
TITLE: Automatic groups - recent progress
QUESTION [27 upvotes]: Epstein's (et al.) "Word Processing in Groups" is a quite comprehensive monograph on automatic groups, finite automata in geometric group theory, specific examples like braid groups, fundamental groups of 3-dim manifolds etc. However, the book is from 1992, so much of the material summarizes research done by Cannon, Thurston, Holt etc. back in the '80s. I'm interested in how the theory of automatic groups (and, more generally, applications of formal languages in group theory) has progressed since then - have there been any significant new results, open problems, novel ideas, examples?

REPLY [14 votes]: I'm not an expert in the area, but here's a few highlights:
Bridson distinguished automatic and combable groups.
Burger and Mozes found examples of biautomatic simple groups.
Mapping class groups were originally shown to be automatic by Mosher. Recently, Hamenstadt has shown that they are biautomatic.
See part 3 of McCammond's survey for an update on open problems.
A well-known open problem is whether automatic groups have solvable conjugacy problem.<|endoftext|>
TITLE: Are non-empty finite sets a Grothendieck test category?
QUESTION [18 upvotes]: A "test category" is a certain kind of small category $A$ which turns out to have the following property: the category $\widehat{A}$ of presheaves of sets on $A$ admits a model category structure, which is Quillen equivalent to the usual model category structure on spaces.  
The notion of test category was proposed by Grothendieck, and the above result was proved by Cisinski (Les préfaisceaux comme modèles des types d'homotopie.  Astérisque No. 308 (2006)).
Examples of test categories include the category $\Delta$ of non-empty finite ordered sets (i.e., the indexing category for simplicial sets), and $\square$, the indexing category for cubical sets.
It's hard for me to give the precise definition of test category here: it involves the counit of an adjunction $i_A: \widehat{A} \rightleftarrows \mathrm{Cat} :i_A^*$, where the left adjoint $i_A$ sends a presheaf $X$ to the comma category $A/X$ (where we think of $A\subset \widehat{A}$ by yoneda).  An online introduction to test categories, which includes the full definition and an account of Cisinski's results, is given in Jardine, "Categorical homotopy theory".
I don't really understand how one should try to prove that a particular category is a test category.  The example I have in mind is $G$, the (skeleton of) the category of non-empty finite sets, and all maps between them.  I believe this should be a test category; is this true?  
Note that there is a "forgetful functor" $\Delta\rightarrow G$, which induces some pairs of adjoint functors between $\widehat{\Delta}$ and $\widehat{G}$.  If $G$ is really a test category, I would expect one of these adjoint pairs to be a Quillen equivalence.
Another note: $G$ is equivalent to the category of finite, contractible groupoids, which is how I am thinking about it.

REPLY [23 votes]: The fact that G is a test category falls in large class of examples. Grothendieck proved that a small category $A$ is a local test category if and only if there exists a presheaf $I$ on $A$ which is an interval (i.e. which has two disjoint global sections) such that, for any representable presheaf $a$, the cartesian product $a\times I$ is aspherical (a presheaf $X$ is aspherical if the classifying space of its category of elements is contractible; for instance, any representable presheaf is aspherical because any category with a terminal object is contractible). A small category $A$ is a test category if and only if it is a local test category with contractible classifying space. For instance, any small category with a terminal object, with finite products, and with a representable interval is a test category. This is the case for the category of non-empty finite sets; see Theorem 1.5.6 and Corollary 1.5.7 in Maltsiniotis book.
I just would like to point out two nice things about Reid's last remark: the category of $1$-groupoids is canonically equivalent to the full subcategory of the category of presheaves on $G$ spanned by the presheaves which satisfy the (strict) Segal condition. This imply that the "classical model structure" and the "Joyal model structure" coincide on $\widehat{G}$, and define the homotopy theory of $(\infty,0)$-categories (aka $\infty$-groupoids). If we consider the Joyal model structure on $\widehat{\Delta}$, the Quillen adjunction with $\widehat{G}$ then really extends the adjunction between categories and groupoids.
I cannot resist to assert that such a picture can be pushed higher: we can replace $\Delta$ by Joyal's categories $\Theta_n$ (or even by some category of operators $\Theta_A$, corresponding to a contractible $n$-operad à la Batanin), and then produce an analog of $G$ with respect to $\Theta_n$, in such a way that we shall get the same picture, but relating $(\infty,n)$-categories to $(\infty,0)$-categories. The construction of such symmetrization of (weakenings) of Joyal's categories and the proof that they lead to test categories is the subject of the PhD thesis (in preparation) of Dimitri Ara (Paris 7). Such symmetrizations have been constructed explicitely by Grothendieck at the very begining of Pursuing stacks, and lead to a definition of weak $\infty$-groupoids, very close to Batanin's notion of weak higher categories; see these two preprints (1 2) of Maltsiniotis (in French). Grothendieck's conjecture that weak $\infty$-groupoids model homotopy types is stated very explicitely using this very definition of higher groupoids. I guess this is one of the starting points/motivations of his theory of test categories.<|endoftext|>
TITLE: Model Structure/Homotopy Pushouts in topological monoids?
QUESTION [12 upvotes]: Let C be the category of topological monoids, that is, the category of monoids in (Top, $\times$). 

Can the model category structure on Top (Serre fibrations, cofibrations, weak homotopy equivalence) be transferred to C along the free and forgetful pair of functors ? 
What are the functorial factorizations in C? Is there a cylinder object in C?

I'm mostly interested in computing a homotopy pushout in the category C, so any ideas how to do that would be helpful too.

REPLY [7 votes]: Clark Barwick's answer is excellent and you should accept it. This is more of an addendum. The category Top is cofibrantly generated, so $\mathcal{C} =$ Mon(Top) is also cofibrantly generated. The key paper is by Schwede and Shipley, and gives conditions on a model category $\mathcal{M}$ such that Mon$(\mathcal{M})$ is a model category. In the special case of $\mathcal{M}$ cofibrantly generated it explains how to get your hands on the cofibrations of Mon$(\mathcal{M})$. See Theorem 4.1 on page 8. Of course, now that you have your hands on the fibrations, trivial fibrations, cofibrations, and trivial cofibrations question (2) is also answered. A nice reference for relating the cylinder object to the functorial factorizations is Hovey page 9
Furthermore, every element in Top is fibrant, so the paper above gives you even stronger results, which may help you with your computations. See remark 4.5 on page 10.
The authors also wrote a second paper giving further results. It's here.<|endoftext|>
TITLE: Non-simply-connected smooth proper scheme over Z?
QUESTION [14 upvotes]: Source
This question came up in the discussion between Kevin Buzzard and Minhyong Kim in the comments to Smooth proper scheme over Z. It was 2 weeks ago, so I took the liberty of posting it as community wiki.
Question

Is there an example of smooth proper variety $X \to \mathop{\text{Spec}}\mathbb Z$ such that $\pi_1(X) \ne 0$?

About tags
We recently had other questions of the form "Example of ... with everywhere good reduction at $\mathbb Z$" (local-global, abelian varieties). I think it would be interesting to create a tag to group these. Thoughts?

REPLY [3 votes]: Here is a proof that if $X$ is smooth and proper over $\mathbb Z$ and of (relative) dimension $\leqslant 3$, then it is simply connected. The dimensional restriction is isolated to a particular step and I believe that theorem is conjectured to generalize to all dimensions.
Fontaine's letter to Messing proves that if $Y$ is smooth and proper over $\mathbb Z$, the Dolbeault cohomology $H^q(Y_{\mathbb Q};\Omega^p)$ vanishes off of the diagonal $p\ne q$ in low degree $p+q\leqslant 3$. I believe the low degree restriction is conjectured not to be necessary. By the Atiyah-Bott fixed-point formula, the Lefschetz number of an element of a finite group acting on a complex variety is the same as the Lefschetz number acting on its cohomology of the structure sheaf. Thus if $H^q(Y;\mathcal O)$ vanishes for $q>0$, Fontaine's theorem with $p=0$, then the Lefschetz number is $1$ and the action cannot be free. If $X$ were smooth and proper over $\mathbb Z$ with non-trivial pro-finite fundamental group*, then some finite cover $Y$ of $X$ would be canonical, thus defined over $\mathbb Z$ (eg, the composite of all covers of degree $\leqslant N$). Then $Y$ would be smooth and proper over $\mathbb Z$ with a free action by the finite covering group, a contradiction.
* If I recall correctly, there are varieties whose complex points have a nontrivial fundamental group, but that group has no finite quotients, and thus the étale fundamental group is trivial.<|endoftext|>
TITLE: Induced Grothendieck topology on a presheaf or sheaf category of a site?
QUESTION [6 upvotes]: While reading Demazure-Gabriel's construction of $\mathcal{S}ch$ as a full subcategory of $\mathcal{P}sh(CRing^{op})$, I've been trying to translate their exposition into the language of covering sieves and Grothendieck topologies.  The requirement that a scheme be a sheaf in the Zariski topology on $CRing^{op}$ is essentially already in the language of sieves, but the requirement that there exists a cover of affines is giving me some trouble.
Question, then:
In general, is there a natural induced Grothendieck topology on $\mathcal{P}sh(\mathcal{C})$ or $\mathcal{S}h(\mathcal{C})$, where $\mathcal{C}$ is a site (not a pre-site!)?
Edit: I removed the other parts of the question regarding t-Schemes and algebraic spaces as functors of points to ask them at some other time.

REPLY [7 votes]: The answer to your first question is yes. Suppose $C$ a site.
The category $C^{\sim}$ of sheaves on $C$ simply has the canonical topology (SGA 4, Vol. 1, Exp. II, 2.5). The sheaves here are precisely the representable sheaves. (I'm ignoring questions about universes.)
The category $C^{\wedge}$ of presheaves on $C$ inherits a topology in the following manner (SGA 4, Vol. 1, Exp. II, §5). Declare a morphism $F\to G$ of presheaves on $C$ to be a covering morphism if the induced morphism $aF\to aG$ of associated sheaves is an epimorphism. Now say that a collection of morphisms $F_i\to G$ is a covering family if the morphism $\amalg F_i\to G$ is a covering morphism. This gives $C^{\wedge}$ the finest subcanonical topology such that covering families in $C$ give rise to covering families in $C^{\wedge}$. This topology on $C^{\wedge}$ can be seen as the lift of the topology on  $C^{\sim}$ described above along the sheafification functor $a$.<|endoftext|>
TITLE: Partial sums of multiplicative functions
QUESTION [28 upvotes]: It is well known that some statements about partial sums of multiplicative functions are extremely hard. For example, the Riemann hypothesis is equivalent to the assertion that $|\mu(1)+\mu(2)+\dots+\mu(n)|$ is bounded above by $n^{1/2+\epsilon}$ for all $\epsilon > 0$. However, I want to know about lower bounds for such partial sums, valid for infinitely many n. For instance, is it known that the sums of the Möbius function must infinitely often be somewhere near $n^{1/2}$ in magnitude? Can one at least prove that they are unbounded? And what about the Liouville function $\lambda$? Are its partial sums unbounded? If so, can anyone give me a good reference or quick proof? And how about general completely multiplicative functions that take values of modulus 1?  Is anything known about those?

REPLY [7 votes]: This paper shows that $L(n) > .061867\sqrt{n}$ for infinitely many $n$.
As for somewhat elementary methods (in the sense of avoiding the Riemann zeta function) to show that $L(n)$ is "usually" of order $\sqrt{n}$, one can use the Lambert series
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)q^n}{1-q^n}} = \sum_{n=1}^{\infty}{q^{n^2}}.$$
As
$$\frac{q^n}{1+q^n} = \frac{q^n}{1-q^n} - 2\frac{q^{2n}}{1-q^{2n}},$$
we have
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{q^{-n}+1}} = \sum_{n=1}^{\infty}{q^{n^2}} - 2\sum_{n=1}^{\infty}{q^{2n^2}}$$
or equivalently, letting $q = e^{-\pi/x}$ and $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}}$, where $x$ is large,
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \psi(1/x) - 2\psi(2/x)$$
Now $\psi(x)$ satisfies the functional equation
$$\frac{1+2\psi(x)}{1+2\psi(1/x)} = \frac{1}{\sqrt{x}},$$
and so we can rewrite this as
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x}.$$
For large $x$, the left-hand side "looks like" $L(x)$, whereas the right-hand side is dominated by the term $\frac{1-\sqrt{2}}{2}\sqrt{x}$. This also explains why $L(n)$ is predominantly negative, as $\frac{1-\sqrt{2}}{2}$ is negative.<|endoftext|>
TITLE: What does ! above = mean
QUESTION [18 upvotes]: Can someone please explain what the symbol $\stackrel{!}{=}$, consisting of an exclamation mark (!) above an equals sign (=) means? Below is the example I'm trying to decipher:

The normalization factor is chosen such that in average, Dynamic Θ Time passes as fast as physical time. In practice it is determiend by the condition that the interval in Dynamic Θ Time corresponding to a 4-year reference period [$T_0$, $T_1$] should be of exactly the same length:
$T_1 - T_0 \stackrel{!}{=} \int_{T_0}^{T_1} a(t) dt$

REPLY [19 votes]: I propose that we adopt the usage of algebraic chess annotations, where the exclamation point ! indicates a particularly good move, which is also suprising or unexpected. 
In mathematics, $\stackrel{!}{=}$ should denote a useful, important, but unexpected equality.<|endoftext|>
TITLE: Hochschild (co)homology of Fukaya categories and (quantum) (co)homology
QUESTION [19 upvotes]: There is a conjecture of Kontsevich which states that Hochschild (co)homology of the Fukaya category of a compact symplectic manifold $X$ is the (co)homology of the manifold. (See page 18 of Kontsevich's "Homological algebra of mirror symmetry" paper and page 16 of Costello's paper "TCFTs and CY categories".) 
Moreover: 

The Hochschild cohomology of the Fukaya category of $X$ should be the Lagrangian Floer cohomology $HF^\ast(X,X)$ of the diagonal $X \to X \times X$.
$HF^\ast(X,X)$ is, at least according to Costello's paper, known to coincide with the quantum cohomology of $X$. But I don't know a reference for this?

What is the current status of these conjectures? Are there any cases where any of this is known to be true, or known to be false? Costello's paper states "I really don't know of much evidence" --- perhaps our state of knowledge is better by now?

REPLY [23 votes]: The statement that $HF^{\ast}(X,X)$ is isomorphic to $QH^\ast(X)$ is a version of the Piunikhin-Salamon-Schwarz (PSS) isomorphism (proved, under certain assumptions, in McDuff-Salamon's book "J-holomorphic curves in symplectic topology"). PSS is a canonical ring isomorphism from $QH^{\ast}(X)$ to the Hamiltonian Floer cohomology of $X$, and the latter can be compared straightforwardly to the Lagrangian Floer cohomology of the diagonal.
Now to Hochschild cohomology of the Fukaya category $F(X)$. There's a geometrically-defined map $QH^{\ast}(X) \to HH^{\ast}(F(X))$, due to Seidel in a slightly different setting (see his "Fukaya categories and deformations"), inspired by the slightly vague but prescient remarks of Kontsevich from 1994. One could define this map without too much trouble, say, for monotone manifolds. It's constructed via moduli spaces of pseudo-holomorphic polygons subject to Lagrangian boundary conditions, with an incidence condition of an interior marked point with chosen cycles in $X$. The question is whether this is an isomorphism.
This statement is open, and will probably not be proven true in the near future, for a simple reason: $QH^*(X)$ is non-trivial, while we have no general construction of Floer-theoretically essential Lagrangians. 
There are two positive things I can say. One is that Kontsevich's heuristics, which involve interpreting $HH^{\ast}$ as deformations of the identity functor, now have a natural setting in the quilted Floer theory of Mau-Wehrheim-Woodward (in progress). This says that the Fukaya category $F(X\times X)$ naturally embeds into the $A_\infty$-category of $A_\infty$-endofunctors of $F(X)$. 
The other is that for Weinstein manifolds (a class of exact symplectic manifolds with contact type boundary), there seems to be an analogous map from the symplectic cohomology $SH^{\ast}(X)$ (a version of Hamiltonian Floer cohomology on the conical completion of $X$) to $HH^{\ast}$ of the wrapped Fukaya category, which involves non-compact Lagrangians. (Edit August 2010: I was careless about homology versus cohomology. I should have said that $HH_{\ast}$ maps to $SH^{\ast}$.) Proving that this is an isomorphism is more feasible because one may be able to prove that Weinstein manifolds admit Lefschetz fibrations. The Lefschetz thimbles are then objects in the wrapped Fukaya category. 
One might then proceed as follows. The thimbles for a Lefschetz fibration should generate the triangulated envelope of the wrapped category (maybe I should split-close here; not sure) - this would be an enhancement of results from Seidel's book. Consequently, one should be able to compute $HH_{\ast}$ just in terms of $HH_{\ast}$ for the full subcategory generated by the thimbles. The latter should be related to $SH^{\ast}$ by ideas closely related to those in Seidel's paper "Symplectic homology as Hochschild homology". 
What could be simpler?
ADDED: Kevin asks for evidence for or against $QH^{\ast}\to HH^{\ast}$ being an isomorphism. I don't know any evidence contra. Verifying it for a given $X$ would presumably go in two steps: (i) identify generators for the (triangulated envelope of) $F(X)$, and (ii) show that the map from $QH^{\ast}$ to $HH^{\ast}$ for the full subcategory that they generate is an isomorphism. There's been lots of progress on (i), less on (ii), though the case of toric Fanos has been studied by Fukaya-Oh-Ohta-Ono, and in this case mirror symmetry makes predictions for (i) which I expect will soon be proved. In simply connected disc-cotangent bundles, the zero-section generates, and both $HH_{\ast}$ for the compact Fukaya category and $SH^{\ast}$ are isomorphic to loop-space homology, but I don't think it's known that the resulting isomorphism is Seidel's.
Added August 2010: Abouzaid (1001.4593) has made major progress in this area.<|endoftext|>
TITLE: The other classical limit of a quantum enveloping algebra?
QUESTION [9 upvotes]: Let $\mathbb K$ be a field (of characteristic 0, say), $\mathfrak g$ a Lie bialgebra over $\mathbb K$, and $\mathcal U \mathfrak g$ its usual universal enveloping algebra.  Then the coalgebra structure on $\mathfrak g$ is equivalent to a co-Poisson structure on $\mathcal U \mathfrak g$, i.e. a map $\hat\delta : \mathcal U \mathfrak g \to (\mathcal U \mathfrak g)^{\otimes 2}$ satisfying some axioms.  A formal quantization of $g$ is a Hopf algebra $\mathcal U_\hbar \mathfrak g$ over $\mathbb K[[\hbar]]$ (topologically free as a $\mathbb K[[\hbar]]$-module) that deforms $\mathcal U \mathfrak g$, in the sense that it comes with an isomorphism $\mathcal U_\hbar \mathfrak g / \hbar \mathcal U_\hbar \mathfrak g \cong \mathcal U \mathfrak g$, and moreover that deforms the comultiplication in the direction of $\hat\delta$: $$\Delta = \Delta_0 + \hbar \hat\delta + O(\hbar^2),$$ where $\Delta$ is the comultiplication on $\mathcal U_\hbar \mathfrak g$ and $\Delta_0$ is the (trivial, i.e. which $\mathfrak g$ is primitive) comultiplication on $\mathcal U\mathfrak g$.  This makes precise the "classical limit" criterion: "$\lim_{\hbar \to 0} \mathcal U_\hbar \mathfrak g = \mathcal U \mathfrak g$"
I am wondering about "the other" classical limit of $\mathcal U_\hbar \mathfrak g$.  Recall that $\mathcal U\mathfrak g$ is filtered by declaring that $\mathbb K \hookrightarrow \mathcal U\mathfrak g$ has degree $0$ and that $\mathfrak g \hookrightarrow \mathcal U\mathfrak g$ has degree $\leq 1$ (this generates $\mathcal U\mathfrak g$, and so defines the filtration on everything).  Then the associated graded algebra of $\mathcal U\mathfrak g$ is the symmetric (i.e. polynomial) algebra $\mathcal S\mathfrak g$.  On the other hand, the Lie structure on $\mathfrak g$ induces a Poisson structure on $\mathcal S\mathfrak g$, one should understand $\mathcal U \mathfrak g$ as a "quantization" of $\mathcal S\mathfrak g$ in the direction of the Poisson structure.  Alternately, let $k$ range over non-zero elements of $\mathbb K$, and consider the endomorphism of $\mathfrak g$ given by multiplication by $k$.  Then for $x,y \in \mathfrak g$, we have $[kx,ky] = k(k[x,y])$.  Let $\mathfrak g_k$ be $\mathfrak g$ with $[,]\_k = k[,]$.  Then $\lim\_{k\to 0} \mathcal U\mathfrak g_k = \mathcal S\mathfrak g$ with the desired Poisson structure.
I know that there are functorial quantizations of Lie bialgebras, and these quantizations give rise to the Drinfeld-Jimbo quantum groups.  So presumably I can just stick $\mathfrak g_k$ into one of these, and watch what happens, but these functors are hard to compute with, in the sense that I don't know any of them explicitly.  So:

How should I understand the "other" classical limit of $\mathcal U_\hbar \mathfrak g$, the one that gives a commutative (but not cocommutative) algebra?

If there is any order to the world, in the finite-dimensional case it should give the dual to $\mathcal U(\mathfrak g^\*)$, where $\mathfrak g^\*$ is the Lie algebra with bracket given by the Lie cobracket on $\mathfrak g$.  Indeed, B. Enriquez has a series of papers (which I'm in the process of reading) with abstracts like "functorial quantization that respects duals and doubles".
On answer that does not work: there is no non-trivial filtered $\hbar$-formal deformation of $\mathcal U\mathfrak g$.  If you demand that the comultiplication $\Delta$ respect the filtration on $\mathcal U\mathfrak g \otimes \mathbb K[[\hbar]]$ and that $\Delta = \Delta_0 + O(\hbar)$, then the coassociativity constraints imply that $\Delta = \Delta_0$.
This makes it hard to do the $\mathfrak g \mapsto \mathfrak g_k$ trick, as well.  The most naive thing gives terms of degree $k^{-1}$ in the description of the comultiplication.

REPLY [4 votes]: The answer is already in Drinfeld's "quantum groups" ICM report. To a 
QUE algebra $U =U_\hbar g$ with classical limit the Lie bialgebra $g$, he associates 
a QFSH algebra $U^\vee$ which turns out to be a formal deformation of the formal series Hopf 
algebra $\hat S(g)$; when $g$ is finite dimensional, this is the formal function ring 
over the formal group $G^\ast$ with Lie algebra $g^\ast$ and indeed the dual to $Ug^*$. 
All this was later developed in a paper by Gavarini (Ann Inst Fourier).  
For example, if the deformation is trivial, so $U=Ug[[\hbar]]$, one finds 
$U^\vee$ to be the complete subalgebra generated by $\hbar g[[\hbar]]$
which is roughly speaking $U(\hbar g[[\hbar]])$ i.e. $U(g_\hbar)$ where 
$g_\hbar$ is $g[[\hbar]]$ but with Lie bracket multiplied by $\hbar$. 
So $U^\vee$ is a quasi-commutative algebra (a flat deformation of $\hat S(g)$
actually).<|endoftext|>
TITLE: What programming languages do mathematicians use?
QUESTION [68 upvotes]: I understand this might be a slightly subjective question, but I am honestly curious what programming languages are used by the mathematics community.
I would imagine that there is a group of mathematicians out there that use haskell because it might be more consistent with ideas from mathematics. Is this true? 
What about APL? Do mathematicians today use APL or is that just a relic of the past?

REPLY [2 votes]: I'm in an MS in stats program and we don't use the programming languages as much as software packages like R, SAS, STATA, MATLAB and MiniTab. The closest to low a low level programming language is C-Sim which is used for simulating systems.<|endoftext|>
TITLE: А generalization of Gromov's theorem on polynomial growth
QUESTION [20 upvotes]: I was sure it is known, but it appears to be an open problem (see the answer of Terry Tao).

Assume for a group $G$ there is a
  polynomial $P$ such that given
  $n\in\mathbb N$ there is  set of
  generators $S=S^{-1}$ such that 
  $$|S^n|\leqslant P(n)\cdot|S|\ \ (\text{or stronger condition}\ |S^k|\leqslant P(k)\cdot|S|\ \text{for all}\ k\le n) .$$
  Then $G$ is virtually nilpotent (or equivalently it has polynomial growth).

Comments:

Note that typically, $|S|\to \infty$ as $n \to\infty$ (otherwise it follows easily from the original Gromov's theorem).
If it is known, then it would give a group-theoretical proof that manifolds with almost non-negative Ricci curvature have virtually nilpotent fundamental group (see Kapovitch--Wilking, "Structure of fundamental groups..."). This proof would use only one result in diff-geometry: Bishop--Gromov inequality. 

Edit: The actual answer is "Done now: arxiv.org/abs/1110.5008" --- it is a comment to the accepted answer. (Published reference: E. Breuillard, B. Green, T. Tao. The structure of approximate groups. Publ. Math. Inst. Hautes Études Sci. 116 (2012), 115–221. Link under Springer paywall)

REPLY [17 votes]: My paper with Shalom does settle the question when $S = S_n$ is known to have size polynomial in n (and maybe is allowed to grow just a little bit faster than this, something like $n^{(\log \log n)^c}$ or so), but I doubt that the result is known yet if S is allowed to be arbitrarily large.
Note that even the bounded case is nontrivial - it's not obvious why having $|S_n^n| \leq n^{O(1)} |S_n|$ implies polynomial growth.  (There is no reason why growth has to be uniform for fixed cardinality of generators; for instance, I believe it is a major open problem (due to Gromov?) as to whether  exponential growth is the same as uniform exponential growth for finitely generated groups.)
If we had a good non-commutative Freiman theorem, then one may possibly be able to settle your question affirmatively (note from the pigeonhole principle that if $|S_n^n| \leq n^C |S_n|$ for some large n, then there exists an intermediate $m=m_n$ between 1 and n such that the set $B := S_n^m$ has small doubling, in the sense that $|B^2| = O(|B|)$).  The best result in this direction for general groups currently is due to Hrushovski, which does show that sets of small doubling contain some vaguely "virtually nilpotent" structure, but it is not yet enough to give Gromov's theorem, let alone the generalisation you mention above.  More is known if one already has some additional structure on the group (e.g. it has a faithful linear representation of bounded dimension, or if it is already known to be virtually solvable).<|endoftext|>
TITLE: An example of two elements without a greatest common divisor
QUESTION [10 upvotes]: Is there an easy example of an integral domain and two elements on it which do not have a greatest common divisor?  It will have to be a non-UFD, obviously.
"Easy" means that I can explain it to my undergrad students, although I will be happy with any example.

REPLY [6 votes]: It deserves to be much better known that nonexistant GCDs (and, similarly, nonprincipal ideals)
arise immediately from any failure of Euclid's Lemma, and this provides an illuminating way to view many of the standard examples. Below is a detailed explanation extracted from one of my sci.math.research posts [2]. The results below hold true in any domain D.
LEMMA:  (a,b) = (ac,bc)/c  if  (ac,bc)  exists
Proof: d|a,b <=> dc|ac,bc <=> dc|(ac,bc) <=> d|(ac,bc)/c. QED
EUCLID'S LEMMA:  a|bc and (a,b)=1 => a|c, if (ac,bc) exists
Proof:   a|ac,bc => a|(ac,bc) = (a,b)c = c  via Lemma. QED
Therefore if a,b,c fail to satisfy the implication in Euclid's Lemma,
namely  if  (a,b) = 1  and  a|bc, not a|c, then one
immediately deduces that the gcd (ac,bc) fails to exist in D.
E.g. David Speyer's example above, and Khurana's example in [1] (= Theorem 41 in Pete L. Clark's [0]) are simply specializations where  a,b,c = p,1+w,1-w  in a quadratic number (sub)ring Z[w], ww = -d.
[0] Clark, Pete. L. Factorization in integral domains. 29pp. 2010.
http://alpha.math.uga.edu/~pete/factorization2010.pdf
[1] D. Khurana, On GCD and LCM in domains: A Conjecture of Gauss.
Resonance 8 (2003), 72-79.
https://www.ias.ac.in/article/fulltext/reso/008/06/0072-0079
[2] sci.math.research, 3/12/09, seeking comments on expository article on factorization
http://groups.google.com/group/sci.math.research/msg/88343de90a4cf6b7
http://google.com/groups?selm=gparte%24si4%241%40dizzy.math.ohio-state.edu<|endoftext|>
TITLE: Classical Enumerative Geometry References
QUESTION [11 upvotes]: I want to start out by making this clear: I'm NOT looking for the modern proofs and rigorous statements of things.
What I am looking for are references for classical enumerative geometry, back before Hilbert's 15th Problem asked people to actually make it work as rigorous mathematics.  Are there good references for the original (flawed!) arguments? I'd prefer perhaps something more recent than the original papers and books (many are hard to find, and even when I can, I tend to be a bit uncomfortable just handling 150 year old books if there's another option.)
More specifically, are there modern expositions of the original arguments by Schubert, Zeuthen and their contemporaries? And if not, are there translations or modern (20th century, say...) reprints of their work available, or are scanned copies available online (I couldn't find much, though I admit my German is awful enough that I might have missed them by not having the right search terms, so I'm hoping for English review papers or the like, though I'll deal with it if I need to.)

REPLY [2 votes]: Have you looked at Semple and Roth, Introduction to Algebraic Geometry? It was published in 1949 and contains a wealth of classical results (there is a chapter devoted to enumerative geometry). Going back a bit further, both German and French Encyclopaedias of Mathematical Sciences published in the early 20th century had surveys of algebraic geometry. Moving in the opposite direction, Fulton's "Intersection theory" discusses applications of his theory to classical enumerative geometry problems where excessive intersections play crucial role (such as finding the number of conics touching 5 given ones).
I know you said you've decided to move away from GW theory, but I thought I'd just throw it in here: Sheldon Katz's book "Enumerative geometry and string theory" (Student Mathematical Library, vol 32) is actually very readable.<|endoftext|>
TITLE: A terminology issue with the Killing Form
QUESTION [7 upvotes]: I understand the definition of a Killing Form $B$ as $B(X,Y)=Tr(ad(X)ad(Y))$
And when the Lie group is semi-simple the negative of the Killing Form can serve as a Riemannian metric.
{Wonder if thats why some people say that a semi-simple Lie Group has only one metric!}
I would like to know what is the difference between a "Killing Metric" and "Cartan-Killing metric" and the "Killing Form" ?
Or are they just different names for the same concept?
{Like the books by Knapp or Fulton and Harris have nothing called "Cartan-Killing metric"}

REPLY [15 votes]: Let me add a few comments to the answers by Mariano and Theo.
There is a one-to-one correspondence between bi-invariant metrics (of any signature) in a Lie group and ad-invariant nondegenerate symmetric bilinear forms on its Lie algebra.  
In a simple Lie algebra every nondegenerate symmetric bilinear form is proportional to the Killing form which you wrote down in your question, hence a simple Lie group has precisely one conformal class of bi-invariant metrics.  If (and only if) the group is compact, are these metrics positive-definite.  (Some people call this riemannian, reserving the word pseudo-riemannian (or sometimes also semi-riemannian) for indefinite signature metrics.  Personally I prefer to use riemannian for general metrics.)
One can ask the question: which Lie groups admit bi-invariant metrics of any signature? which is the same thing as asking which Lie algebras admint ad-invariant non-degenerate symmetric bilinear forms.  Such Lie algebras are called metric (or also sometimes quadratic, orthogonal,...) and although there is no classification except in small index (index 0 = positive-definite, index 1 = lorentzian, etc...) there is a structure theorem proved by Alberto Medina and Philippe Revoy in this paper (in French).  Their theorem says that the class of such Lie algebras is generated by the simple and the one-dimensional Lie algebras under two operations: orthogonal direct sum and double extension, a construction explained in that paper.
Double extension always results in indefinite signature, so if you are only interested in the positive-definite case, you get back to the well-known result that every positive-definite metric Lie algebra $\mathfrak{g}$ is isomorphic to the orthogonal direct sum of a compact semisimple Lie algebra and an abelian Lie algebra, or in other words,
$$\mathfrak{g} \cong \mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_N \oplus \mathfrak{a}$$
where the $\mathfrak{s}_i$ are the simple factors and $\mathfrak{a}$ is abelian.
Up to automorphisms, the most general positive-definite inner product on such a Lie algebra is given by choosing for each simple factor $\mathfrak{s}_i$ a positive multiple $\lambda_i > 0$ of the Killing form.
These Lie algebras are precisely the Lie algebras of compact Lie groups.  Their metricity can also be understood as follows: take any positive-definite inner product on $\mathfrak{g}$ and averageng it over the adjoint representation.
So in summary, although there are metric Lie algebras which are not semisimple (or even reductive), their inner product is always an additional structure, unlike the Killing form which comes for free with the Lie algebra.
As for question concerning the difference between Killing form and Cartan(-Killing) metric it depends on who says this.  In much of the Physics literature people refer to an inner product on a vector space as a "metric".  But assuming that this is not the case, then the Killing form is a bilinear form on the Lie algebra, whereas the metric is a metric (in the sense of riemannian geometry) on the Lie group.  If $G$ is a Lie group whose Lie algebra is semisimple, then the Killing form on its Lie algebra defines a bi-invariant metric on the Lie group, which I suppose you could call the Cartan-Killing metric.<|endoftext|>
TITLE: How important are publications for undergrads?
QUESTION [15 upvotes]: I have heard vastly conflicting statements about whether undergrads applying for PhD programs should have published already, or what level of research will be expected of them. Looking at CVs of some of my school's professors, almost none of them seem to have publications from earlier than the 2nd half of their graduate studies, meaning they spent most of their time before getting their PhD without any publications or those that they had weren't worth listing, in their eyes.
Obviously, I'm going to try to get the best experience I can as an undergrad, and I hope that means getting published research, but in every area I've dipped my toe in, from probability to dynamical systems to complexity theory, the sheer amount of additional knowledge I'd need to understand even a upper-level graduate text seems intimidating.
When did you first publish, and what sort of research experience (if it's something other than publishing an article) should an undergraduate aiming for a PhD have?
Disclaimer: I'm an undergrad in CS, pretty average or maybe above average in my progress so far, and I'd like to make a career in researching some of the theoretical (and obviously math-heavy) parts of computer science, rather than software or interface.

REPLY [17 votes]: My advice (as a pure mathematician!) is that the best way to get going in research is to write mathematics, and rewrite it till you have got it clear to yourself. The composer Ravel once commented: "Do copy! If you have some originality, it will show itself. If not,  do not worry!"  In fact the originality may show itself at say the 5th rewrite, when the creaking wheels of the brain have been finally knocked into a little motion. 
I did an essay on set theory for a College prize, which helped in an exam question. As a PhD student, I most enjoyed working with a visiting researcher (Dick Swan) on notes of his lectures on sheaves. Much later, I stumbled on the area of groupoids by trying to write a new proof that the fundamental group of the circle was the integers, and writing and rewriting my proof made me see that the arguments should generalise to higher dimensions.  
Daniel Quillen was said to have written thousands of pages of notes as he tried to understand various areas, and put them in his own terms. 
The question of publication of what you have written is more problematic.<|endoftext|>
TITLE: H-space structure on infinite projective spaces
QUESTION [15 upvotes]: Any Eilenberg-MacLane space $K(A,n)$ for abelian $A$ can be given the structure of an $H$-space by lifting the addition on $A$ to a continuous map $K(A\times A,n)=K(A,n)\times K(A,n)\to K(A,n)$.
Does somebody know an explicit way to describe this structure in the cases $K({\mathbb Z}/2{\mathbb Z},1)={\mathbb R}P^\infty$ and $K({\mathbb Z},2)={\mathbb C}P^{\infty}$?

REPLY [2 votes]: John Baez has a nice expository page about this called  Classifying spaces made easy. About two thirds down the page he talks about multiplicative structure on $\mathbb{C}P^\infty$<|endoftext|>
TITLE: Space whose product with paracompact space is paracompact
QUESTION [9 upvotes]: Is there a nice characterization of topological spaces with the property that the product with any paracompact space is paracompact?
All compact spaces have this property (this can be shown from the tube lemma). But somebody once gave me an example (that I cannot locate) of a non-compact space with the property. I didn't check the example carefully, so I cannot vouch for its accuracy.
If a characterization is too hard, an example of a non-compact space would also be great.
[NOTE: I don't assume Hausdorffness in my definitions of compact and paracompact, though it would be nice if the example were a Hausdorff space.]
ADDED LATER: I forgot to mention this, but a product of paracompact spaces need not be paracompact. The standard example is the Sorgenfrey line (the real line with the lower limit topology), which is paracompact, whose product with itself, the Sorgenfrey plane, is not paracompact.

REPLY [3 votes]: The product of a CW-complex with a paracompact-Hausdorff space is again paracompact Hausdorff. I'm not sure what happens if you remove the word Hausdorff.<|endoftext|>
TITLE: Why would I want to know (equivariant) quantum cohomology?
QUESTION [12 upvotes]: Let's say that I have a variety I think is interesting, and based on some papers I don't fully understand, I can compute quite explicitly its equivariant quantum cohomology in terms of explicit formulae for multiplying by a degree 2 class.
Being something of a newcomer to quantum cohomology, I'm genuinely a bit unsure of how interesting a result this is, and have doubts about writing a paper whose content is "The quantum cohomology of variety X is blah."
Does it tell me anything particularly interesting?   Might it have cool implications in integrable systems or something like that?

REPLY [12 votes]: You might have already found an answer by now but still let me add my motivation for the
subject. The reason I am interested in quantum cohomology is that I am thinking
about it as the correct version of semi-infinite cohomology of the loop space of the variety
in question (this point of view by the way is sometimes very useful for guessing the answer).
Personally I find the definition of quantum cohomology rather ugly, but in geometric
representation theory this kind of phenomenon is sort of well-known: one way to do 
"semi-infinite geometry" which formally is not clear how to tackle, is by working with
the global curve ${\mathbb P}^1$ instead of the formal disc. For example, one can develop
rather non-trivial theory of "semi-infinite flag varieties" using spaces of quasi-maps,
which is defined using global  ${\mathbb P}^1$ as well. I sort of regard this as a similar
phenomenon to why quantum cohomology (which is supposed to be a local object) is defined
via a global curve.
A more "physical" explanation of the above phenomenon is this: in 2-dimensional conformal
field theory it happens very often that some genus 0 correlators actually coincide
with some local objects (like coefficients of the OPE and such things). One example of this
is the Kazhdan-Lusztig tensor product for representations of affine Lie algebras.
It is actually a local thing, but the Kazhdan-Lusztig definition goes through coinvariants
on global ${\mathbb P}^1$.
Let me conclude by saying that while the "global" approach is often useful, because
that's the only way to define things regorously, the price you have to pay for this is the
fact that calculations become rather difficult and mysterious. Quantum cohomology is probably a good example of this -- many answers there become much more understandable when you think in terms of loop spaces.<|endoftext|>
TITLE: What does the nilpotent cone represent?
QUESTION [14 upvotes]: Notation
Let $\mathfrak g$ be a the Lie algebra of an algebraic group $G\subseteq GL(V)$ over a(n algebraically closed) field $k$ (I'm actually thinking $G=GL_n$, so $\mathfrak g=\mathfrak{gl}_n$). Then any element $X$ of $\mathfrak g$ can be uniquely written as the sum of a semi-simple (diagonalizable) element $X_s$ and a nilpotent element $X_n$ of $\mathfrak g$, where $X_s$ and $X_n$ are polynomials in $X$. The nilpotent cone $\mathcal N$ is the subset of nilpotent elements of $\mathfrak g$ (elements $X$ such that $X=X_n$).

People often talk about the nilpotent cone as having the structure of a subvariety of $\mathfrak g$, regarded as an affine space, but usually don't say what the scheme structure really is. To really understand a scheme, I'd like to know what its functor of points is. That is, I don't just want to know what a nilpotent matrix is, I want to know what a family of nilpotent matrices is (i.e. what a map from an arbitrary scheme $T$ to $\mathcal N$ is). Since any scheme is covered by affine schemes, it's enough to understand what an $A$-valued point (a map $\mathrm{Spec}(A)\to \mathcal N$) is for any $k$-algebra $A$. So my question is

What functor should $\mathcal N$ represent?

A guess
Well, an $A$-point of $\mathfrak g$ is "an element of $\mathfrak g$ with entries in $A$" (again, I'm really thinking $\mathfrak g=\mathfrak{gl}_n$, so just think "a matrix with entries in $A$"), so I would expect that such an $A$-point happens to be in $\mathcal N$ exactly when the given matrix is nilpotent. That is, $\mathcal N(\mathrm{Spec}(A))=\{X\in \mathfrak{g}(\mathrm{Spec}(A))| X^N=0$ for some $N\}$.
However, this is wrong. That functor isn't even an algebraic space, even for the nilpotent cone of $\mathfrak{gl}_1$. If it were, the identity map on it would correspond to a nilpotent regular function $f$ (a nilpotent $1\times 1$ matrix), and this would be the universal nilpotent regular function; every other nilpotent regular function anywhere else would be a pullback of this one. But whatever the degree of nilpotence of this function (say $f^{17}=0$), there are some nilpotent regular functions which cannot be a pullback of it (something with nilpotence degree bigger than 17). If this version of the nilpotent cone were representable, you can show that the $\mathfrak{gl}_1$ version would be too.
Another guess
I think the answer might be that an $A$ point of $\mathcal N$ is a matrix ($A$ point of $\mathfrak g$) so that all the coefficients of the characteristic polynomial vanish. This is a scheme and it has the right field-valued points, but why should this be the nilpotent cone? What is the meaning of having all coefficients of the characteristic polynomial vanish for a matrix with entries in $A$?

REPLY [13 votes]: The ideal which defines the nilpotent cone is generated by the homogeneous elements of positive degree in $\mathbb{C}[\mathfrak g]^G$. In the case of $GL_n$, this ideal is generated by the functions $\mathrm{tr}\;X^k$ for $k$ up to the dimension, and also by the coefficients of the characteristic polynomial. See, for example, [Springer, T. A. Invariant theory. Lecture Notes in Mathematics, Vol. 585. Springer-Verlag, Berlin-New York, 1977. iv+112 pp. MR0447428]<|endoftext|>
TITLE: Smooth maps considered as locally ringed space morphisms?
QUESTION [8 upvotes]: I have heard that for a locally ringed space $X$ whose topology is second countable and Hausdorff, $X$ is a smooth manifold if and only if it is locally ringed space which is locally isomorphic to the sheaf of differentiable functions of some open sets in $\mathbb{R}^n$.
Question: What about the maps between smooth manifolds? Let $M, N$ smooth manifolds and $\phi: M \rightarrow N$ be a continuous map. Does every locally ringed space morphism  $\mathcal{O}_N \rightarrow \phi_* \mathcal{O}_M$ come from $\phi$?
I'm asking this because I thought it was true - so that I agree that locally ringed spaces are useful generalizations of (whatever) manifolds - until I tried to prove it tonight. In the process it seems to me that this is NOT true. But if this is not true, why is locally ringed space a good generalization of manifolds if the maps don't behave well? (As opposed to rings $A \rightarrow B$ and spectrum maps $SpecB \rightarrow SpecA$?)
Thanks!

REPLY [9 votes]: For a morphism $f: (X,{\mathcal O}_X)\to (Y,{\mathcal O}_Y)$ of locally ringed spaces the locality of the maps $f^{\sharp}_x: {\mathcal O} _{Y,f(x)}\to {\mathcal O} _{X,x}$ implies that for each function $\lambda\in{\mathcal O}_Y(U)$ we have $v(f^{\sharp}_U(\lambda)) = f^{-1}(v(\lambda))$ (here $v(\lambda)$ denotes the zero set of $\lambda$, i.e. the set of all points $y\in U$ such that $\lambda_y\in{\mathfrak m}_y$.
(*) Therefore the zero set of $f^{\sharp}_U (\lambda)$ is determined entirely by the underlying continuous map of $f$, and it equals $f^{-1}(v(\lambda))$
From now on, let $X,Y$ be topological spaces and ${\mathcal O} _X$, ${\mathcal O} _Y$ denote the sheaves of continuous real-valued functions on $X$ and $Y$, respectively. If you know that $f^{\sharp}$ preserves constant functions, then by translating a function $\lambda\in{\mathcal O}_Y (U)$ by a real number $\alpha\in{\mathbb R}$, you see by (*) that the niveau set $v_\alpha(f^{\sharp}_U(\lambda)) := \{x\in f^{-1}(U)\ |\ f^{\sharp}_U(\lambda)(x) = \alpha\}$ equals $f^{-1}(v_\alpha(\lambda))$, so $f^{\sharp}_U(\lambda) = \lambda\circ f$ as suspected.
Now $f^{\sharp}$ has to preserve constant functions, because each map $\overline{f^{\sharp}_x}: {\mathbb R} = {\mathcal O}_{Y,f(x)}/{\mathfrak m}_{Y,f(x)}\to{\mathcal O}_{X,x}/{\mathfrak m}_{X,x} = {\mathbb R}$ is an automorphism of ${\mathbb R}$, thus equals the identity. (Ok, one could use this to argue directly that $f^{\sharp}_U(\lambda)(x)=\lambda(f(x))$ - didn't see this when starting to write)
For a field different from ${\mathbb R}$, this argument doesn't work any more, because there may be nontrivial automorphims. For example, if you take ${\mathbb C}$-valued continuous functions, you could set $f^{\sharp}_U(\lambda) := \text{conj}\circ \lambda\circ f$ for the algebraic component of $f$.<|endoftext|>
TITLE: Homotopy Pushouts via Model Structure in Top
QUESTION [5 upvotes]: As far as I know, one way to take a homotopy colimit in a model category is to replace (up to acyclic fibration) all arrows in the diagram with cofibrations, and take the strict colimit of the resulting diagram.
In Top with the model structure given by Serre fibrations, cofibrations, and weak equivalences, if one wants to obtain a homotopy pushout of the diagram $X \leftarrow A \rightarrow Y$, it is "enough" to replace only one of these arrows with a cofibration: that is, there is a natural map (by the universal property of the pushout) $Cyl(X)\cup_A Cyl(Y) \to Cyl(X) \cup_A Y$ that is a homotopy equivalence of spaces.

Question 1: What conditions on the model category $\mathcal{C}$ (or objects $X,Y,A$) will guarantee that the natural map $Cyl(X) \cup_A Cyl(Y) \to Cyl(X) \cup_A Y$ is a weak equivalence?
Question 2: This question is less precise, but if the map above is a weak equivalence, does that mean $Cyl(X) \cup_A Y$ is a good model for the homotopy pushout?

REPLY [10 votes]: (I'll assume that in a general model category $\mathcal{C}$, $\mathrm{Cyl}(X)$ really means: a factorization of $A\to X$ into a cofibration $A\to \mathrm{Cyl}(X)$ followed by a trivial fibration $\mathrm{Cyl}(X)\to X$.)
A sufficient condition on objects for the map in question 1 to be a weak equivalence, is that the objects $X,Y,A$ be cofibrant.  (The fact you want to use is the statement due to Reedy (which you can find at the start of the chapter on Proper Model Categories in Hirschhorn's book), that a pushout of a weak equivalence between cofibrant objects along a cofibration is a weak equivalence.)
A sufficient condition on $\mathcal{C}$ for the map in question 1 to be a weak equivalence, is that the model category be left proper.
It's an interesting fact that in Top, the this also works if $\mathrm{Cyl}(X)$ denotes the "classical" mapping cylinder construction ($X$ union a cylinder on $A$), which isn't necessarily a cofibration in the Quillen model structure.  The slickest proof is to use the "excisive triad theorem", as proved by May in A Consise Course in Algebraic Topology, p.79.  This also leads to a proof that Top is left proper.

REPLY [9 votes]: Question 1: The model category $\mathcal{C}$ should be left proper, i.e. the pushout of a weak equivalence along a cofibration is again a weak equivalence.  (Dually, there is a notion of right proper.)  Top is left proper, as is any model category in which every object is cofibrant, such as SSet.  There is some information on this notion of properness at the nlab, and I think it's also discussed more thoroughly in Hirschhorn's book Model Categories and their Localizations (and probably many other places).
Question 2: Yes.  People often say that a square in a model category is a homotopy pushout square if the induced map from the (strict) pushout of a cofibrant replacement (meaning cofibrant objects and maps) of the "initial" three objects to the last object is a weak equivalence, and that is the case here.<|endoftext|>
TITLE: Riemann mapping for doubly connected regions
QUESTION [7 upvotes]: Remove the closure of simply connected region from the interior of a simply connected region. Is it true that the resulting domain can be mapped conformally to some annulus?

REPLY [7 votes]: The answer is yes. This is a special case of theorem 10 in Ahlfors' Complex Analysis, section 5, chapter 6. (Special in that the theorem more generally says that if the complement of the domain has $n$ connected components not reduced to points in the extended plane, then the domain is equivalent to an annulus from which $n-2$ concentric slits have been removed. In your case $n=2$.)

REPLY [5 votes]: See Wikipedia.  The third entry of the list gives an affirmative answer.

REPLY [5 votes]: No. The resulting set need not even be connected. And if it is, it need not be doubly connected, as the interior region may have boundary points in common with the original region. Aside from these crude objections, however, the answers of Mariano and Scott are OK.<|endoftext|>
TITLE: nonhausdorff dimension
QUESTION [12 upvotes]: if $X$ is a topological space, a first step in making $X$ hausdorff is taking the quotient $H(X)=X/\sim$, where $\sim$ is the equivalence relation generated by: if $x,y$ cannot be seperated by disjoint open sets, then $x \sim y$. observe that $X$ is hausdorff, when $X \to H(X)$ is an isomorphism, and that for every hausdorff space $K$ the map $Hom(H(X),K) \to Hom(X,K)$ induced by the projection $X \to H(X)$ is bijective.
by a fairly general categorical argument, we can construct from this the free functor from topological spaces to hausdorff spaces (i.e. it's left adjoint to the forgetful functor): for ordinal numbers $\alpha$, define the functor $H^\alpha$ (together with natural transformations $H^{\alpha} \to H^{\beta}, \alpha < \beta$) by $H^0 = id, H^{\alpha+1} = H \circ H^\alpha$ and $H^\alpha = colim_{\delta < \alpha} H^\delta$. for every topological space $X$ there is an ordinal number $\alpha$ such that $H^\alpha(X) = H^{\alpha+1}(X)$, then $H^\alpha(X)$ is the free hausdorff space associated to $X$. define $h(X)$, the "nonhausdorff dimension" to be the smallest such ordinal number $\alpha$. every ordinal number arises as a nonhausdorff dimension(!).
I've came up with this with a friend and we don't know of any literature about it. perhaps someone of you has already seen it elsewhere? there are some further questions: every $H^\alpha(X)$ is a quotient of $X$, but how can we describe the equivalence relation explicitely? what is the intuition for a space $X$ to have nonhausdorff dimension $\alpha$? are there known classes of topological spaces whose nonhausdorff dimension can be bounded? and of course: is there some use for the nonhausdorff dimension? ;-)

REPLY [5 votes]: An unpublished, but quite well distributed, reference is "Appendix A, Compactly Generated Spaces" to the 1978 Ph.D. thesis of L. Gaunce Lewis, Jr.  Proposition 4.1 discusses the existence of the left adjoint (Hausdorffification) by way of Freyd's adjoint functor theorem, and Construction 4.3 explains the transfinite induction by iterating $X \to JX$ where JX is essentially the quotient of X discussed by Martin Brandenburg.  (Lewis works with weak Hausdorff spaces, but the story is the same.)  A published version of something much like this appears in Lewis, May and Steinberger's book "Equivariant Stable Homotopy Theory" (Springer LNM 1213), where the spectrification functor L is constructed in Section 1
of the Appendix (pages 475-481) by this kind of transfinite induction.<|endoftext|>
TITLE: When is a Banach space a Hilbert space?
QUESTION [32 upvotes]: Let $\mathcal{X}$ be a real or complex Banach space.
It is a well known fact that $\mathcal{X}$ is a Hilbert space (i.e. the norm comes from an inner product) if the parallelogram identity holds.
Question: Are there other (simple) characterizations for a Banach space to be a Hilbert space?

REPLY [7 votes]: From the point of view of manifolds and curvature the following result is valid:
A Banach space is a Hilbert space if and only if it is a NPC (non-positive curvature) space.
 http://www.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/papers/paper41.pdf<|endoftext|>
TITLE: How does the order of a pole of a zeta function indicate any geometric information?
QUESTION [8 upvotes]: Here, I'm primarily concerced about zeta functions of hypersurfaces over fields of finite characteristic.  
Assume $F_q$ to be a finite field with q elements. Consider the zeta function of the hypersurface defined by $-y_0^2+y_1^2+y_2^2+y_3^2=0$ in $\mathbb{P}^3$.
If $-1$ is a square in $F_q$, the zeta function is 
$$Z(u)=\frac{1}{(1-uq^2)(1-uq)^2(1-u)}.$$
It has a pole of order $2$ at $1/q$. If not, it's 
$$Z(u)=\frac{1}{(1-uq^2)(1-uq)(1+uq)(1-u)}.$$ 
It has a pole of order $1$ at $1/q$.

How does orders of poles indicate any geometric information?

REPLY [9 votes]: This is an expository note filling in the background between Steven Sam's comment and Felipe Voloch's answer.
If $X$ is a smooth projective variety, then the Weil conjectures (now theorems) describe the zeroes and poles of the zeta function in terms of the cohomology of $X$, and the action of Frobenius on it. In particular, the poles on the circle $|u|=1/q$ are the reciprocals of the eigenvalues of Frobenius acting on $H^2(X, \mathbb{Q}_{\ell})$. 
In your example, $H^2$ is two dimensional. Over the algebraic closure $\overline{F_q}$, your variety is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$. $H^2$ is spanned by the two classes $\mathbb{P}^1 \times \{ \mbox{point} \}$ and $ \{ \mbox{point} \}  \times \mathbb{P}^1$.
If $-1$ is a square, Frobenius acts on this two dimensional vector space by multiplication by $q$, so you get a double pole at $1/q$. If $-1$ is not a square, then Frobenius multiplies by $q$ and switches the two generators. So the eigenvalues are $q$ and $-q$.<|endoftext|>
TITLE: Is there a simple relationship between K-theory and Galois theory?
QUESTION [11 upvotes]: I can (barely) understand the definition of the higher algebraic K-groups a la the plus construction right now (I have some past familiarity with K-theory for C*-algebras and can recall the rudiments of the situation for vector bundles). So by "simple", I mean to a mathematical layman. If you have a complicated answer, feel free to answer as well, but I probably won't be able to understand much.

REPLY [15 votes]: A simpler statement may be that if $F \to E$ is a $G$-Galois extension, then
there is a map $K(F) \to K(E)^{hG}$ from the algebraic $K$-theory space of $F$
to the $G$-homotopy fixed points of the algebraic $K$-theory space of $E$.
The Lichtenbaum--Quillen conjecture identifies situations where this map
is ``almost'' a weak equivalence, in the sense that it induces an isomorphism
on homotopy groups with finite coefficients and in sufficiently high degrees.
There is a spectral sequence $E^2_{s,t} = H^{-s}(G; K_t(E))$
(group cohomology) converging to $\pi_{s+t}(K(E)^{hG})$, and similarly with finite coefficients $Z/p$,
so when this is equivalent to $\pi_{s+t} K(F) = K_{s+t}(F)$ you can
recover the algebraic $K$-theory of $F$ from the algebraic $K$-theory of $E$.
This is a form of Galois descent.
Letting $E$ grow to the separable closure of $F$, you can use Suslin's
theorem about the algebraic $K$-theory of separably closed fields to get
at the algebraic $K$-theory of other fields $F$ by way of the cohomology of
their absolute Galois groups, i.e., their Galois cohomology.
If you work with more general commutative rings, or schemes, and replace Galois
extensions by etale covers, your answer will involve etale cohomology
instead of Galois cohomology, but the idea is much the same.
The Lichtenbaum--Quillen conjecture (e.g. for number fields) is known to
follow from the Milnor conjecture (at p=2) and the Bloch--Kato conjecture (at odd
primes p).  The argument compares etale cohomology to motivic cohomology,
and uses the Atiyah--Hirzebruch type spectral sequence from motivic cohomology
to algebraic $K$-theory.  Voevodsky proved the Milnor conjecture in 1996.
The (presumed confirmed) status of the full Bloch--Kato conjecture is discussed in other posts.
Given these results, the algebraic $K$-groups of (rings of integers in)
number fields or local fields of characteristic zero are computed in terms of the corresponding etale cohomology groups.
If you work with brave new rings, or structured ring spectra, the Galois descent analogue of the Lichtenbaum--Quillen conjectures is open, the etale descent version is still
in need of an optimal formulation, and I think there is no known definition of motivic cohomology.  In these cases the best computational results are obtained using
the cyclotomic trace map from algebraic $K$-theory to topological cyclic homology, not by descent.

John<|endoftext|>
TITLE: What is a good roadmap for learning Shimura curves?
QUESTION [56 upvotes]: I am interested in learning about Shimura curves. Unlike most of the people who post reference requests however (see this question for example), my problem is not sorting through an abundance of books but rather dealing with what appears to be an extreme paucity of sources.
Anyway, I'm a graduate student and have spent the last year or so thinking about the arithmetic of orders in quaternion algebras (and more generally in central simple algebras). The study of orders in quaternion algebras seems to play an important role in Shimura curves, and I'd like to study these connections more carefully.
Unfortunately, it has been very difficult for me to find a good place to start. I only really know of two books that explicitly deal with Shimura curves:

Shimura's Introduction to the arithmetic theory of automorphic functions
Alsina and Bayer's Quaternion Orders, Quadratic Forms, and Shimura Curves

Neither book has been particularly helpful however; the first only mentions them briefly in the final section, and the second has much more of a computational focus then I'd like.
Question 1: Is there a book along the lines of Silverman's The Arithmetic of Elliptic Curves for Shimura curves? 
I kind of doubt that such a book exists. Thus I've tried to read the introductory sections of a few papers & theses, but have run into a problem. There seem to be various ways of thinking about a Shimura curve, and it has been the case that every time I look at an article I'm confronted with a different one. For example, this talk by Voight and this paper by Milne. By analogy, it seems to be a lot like trying to learn class field theory by switching between articles with ideal-theoretic statements and articles taking an adelic slant without having a definitive source which tells you that both are describing the same theorems.
My second question is therefore:
Question 2: Can anyone suggest a 'roadmap' to Shimura curves? Which theses or papers have especially good expository accounts of the basic properties that one needs in order to understand the literature.
Clearly I need to say something about my background. As I mentioned above, I'm an algebraic number theorist with a particular interest in quaternion algebras. I don't have the best algebraic geometry background in the world, but have read Mumford's Red Book, the first few chapters of Hartshorne and Qing Liu's Algebraic Geometry and Arithmetic Curves. I've also read Silverman's book The Arithmetic of Elliptic Curves and Diamond and Shurman's A First Course in Modular Forms.
Thanks.

REPLY [14 votes]: [I decided my previous answer was long enough, so I'm adding this one separately and making it Community Wiki.  Feel free to add to it!]
Other people's PhD theses that have nice expositions on Shimura curves include:
David Helm (Berkeley 2003)
Bruce Jordan (Harvard 1981)
David Roberts (Harvard 1989)
Victor Rotger Cerda (Universitat de Barcelona 2002)
John Voight (Berkeley 2005) 
[Edit (Emerton):] Ken Ribet's Inventiones 100 article describes certain instances of
Shimura curves over $\mathbb Q$, including some relations with orders in quaternion algebras, and some information about $p$-adic uniformization and their bad reduction at primes describing the discriminant.  Like all of Ribet's papers, it is a masterpiece of exposition.<|endoftext|>
TITLE: Brownian Approximation of Downswings of Walks with Positive Drift
QUESTION [7 upvotes]: I'm interested in the downswings of discrete walks w(t) whose steps are IID, bounded, and have positive mean. A simple example might have steps which are +1 with probability 2/3, and -1 with probability 1/3. A downswing of size at least D on [0,t] means 0<=a<b<=t and w(a)-w(b)>D. 
Natural questions include the expected size of the largest downswing within [0,t] and the expected minimum b so that there is a downswing of size D ending at b. 
One possible approach is to use a Brownian approximation with the same mean and standard deviation. This has the advantage that the distribution of the largest downswing on [0,t] has been studied. The expected time before a downswing of size D is computable and has a simple formula. Asymptotic expressions for the average size of the largest downswing on [0,t] have been computed. See Amrit Pratap's MS thesis. 
However, the Brownian approximation has the disadvantage that it is wrong, and sometimes it is wrong by a lot. For example, a walk with only positive steps has no downswings at all. 
I'd like to know how bad I should expect the Brownian approximation should be for steps which can be negative, with relatively small positive mean relative to the standard deviation. For example, -1 with probability 4/5, +5 with probability 1/5. I'd like to know if a skew in the positive direction means that large downswings are less common in the discrete walk than in the Brownian approximation. 
Any help would be appreciated.

REPLY [3 votes]: I just accidentally stumbled upon this nice question. I suspect that by now you know the answer yourself, but still, let me do one simple computation. If you like it, I'll think more of the question.
The process in question is just the following. A particle departs from the origin and does i.i.d. random steps with positive mean. If it ever ends up to the right of the origin, it is put back there. The question is what is the leftmost position it visited after $t$ steps.
I want to argue as follows. Let's look at the probability that it'll reach $-D$ before it comes back to the origin. It is about $Ce^{-\lambda D}$ where $\lambda>0$ is the only positive solution of the equation $\int p(x)e^{-\lambda x}dx=1$ and $p$ is the density of the step distribution. The value of $C$ is also possible to compute. Now, each attempt to depart from the origin lasts for some time with exponentially decaying tails. Let $T$ be the average time of travel. Then, by the time $t$, the number of attempted departures is about $t/T$. Thus, the probability of success is about $(1-Ce^{-\lambda D})^{t/T}$ meaning that $ED_{\text{min}}\approx \lambda^{-1}(\log t+\log(C/T)+U)$ where $U$ is some universal constant ($U=\int_0^\infty (e^{-e^{-x}}+e^{-e^x}-1)dx$, if I haven't mistaken). This would mean that, for large times,  you are always a constant number of times off with the Brownian approximation.
Of course, this is just a back of envelope computation, but, since I don't even know if you are still interested, I'd rather stop here.<|endoftext|>
TITLE: Commutative rings to algebraic spaces in one jump?
QUESTION [40 upvotes]: Typically, in the functor of points approach, one constructs the category of algebraic spaces by first constructing the category of locally representable sheaves for the global Zariski topology (Schemes) on $CRing^{op}$. That is, taking the full subcategory of $Psh(CRing^{op})$ which consists of objects $S$ such that $S$ is a sheaf in the global Zariski topology and $S$ has a cover by representables in the induced topology on $Psh(CRing^{op})$.  This is the category of schemes.  Then, one takes this category and equips it with the etale topology and repeats the construction of locally representable sheaves on this site (Sch with the etale topology) to get the category of algebraic spaces.  
Can we "skip" the category of schemes entirely by putting a different topology on $CRing^{op}$? 
My intuition is that since every scheme can be covered by affines, and every algebraic space can be covered by schemes, we can cut out the middle-man and just define algebraic spaces as locally representable sheaves for the global etale topology on $CRing^{op}$.  If this ends up being the case, is there any sort of interesting further generalization before stacks, perhaps taking locally representable sheaves in a flat Zariski-friendly topology like fppf or fpqc?  
Some motivation: In algebraic geometry, all of our data comes from commutative rings in a functorial way (intentionally vague).  All of the grothendieck topologies with nice notions of descent used in Algebraic geometry can be expressed in terms of commutative rings, e.g., the algebraic and geometric forms of Zariski's Main theorem are equivalent, we can describe etale morphisms in terms of etale ring maps, et cetera. What I'm trying to see is whether or not we can really express all of algebraic geometry as "left-handed commutative algebra + sheaves (including higher sheaves like stacks)".  The functor of points approach for schemes validates this intuition in the simplest case, but does it actually generalize further?
The main question is italicized, but feel free to tell me if I've incorrectly characterized something in the motivation or the background.

REPLY [11 votes]: A Deligne--Mumford stack is an étale-locally ringed topos that is locally equivalent to the étale-locally ringed topos of an affine scheme.  A Deligne--Mumford stack is an algebraic space if its diagonal is an embedding.<|endoftext|>
TITLE: von Staudt-Clausen over a totally real field
QUESTION [17 upvotes]: Before I ask the question, I need to recall what Bernoulli numbers
$(B_k)_{k\in\mathbb{N}}$ are, and what von Staudt and Clausen discovered about
them in 1840.  The numbers $B_k\in\mathbb{Q}$ are the coefficients in the formal power series
$$
{T\over e^T-1}=\sum_{k\in\mathbb{N}}B_k{T^k\over k!}
$$
so that $B_0=1$, $B_1=-1/2$, and it is easily seen that $B_k=0$ for $k>1$
odd. 
Theorem (von Staudt-Clausen, 1840)  Let $k>0$ be an even integer, and let $p$ run through the primes.  Then
$$
B_k+\sum_{p-1|k}{1\over p}\in\mathbb{Z}.
$$
John Coates remarked at a recent workshop that the analogue of this theorem
for a totally real number field $F$ (other than $\mathbb{Q}$) is an open problem;
even a weak analogue would imply Leopoldt's conjecture for $F$.  I missed the
opportunity of pressing him for details.
Question :  What is the analogous statement over a totally real number field ?

REPLY [8 votes]: I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number $B_k$: it is precisely, the product of primes p for which $p-1\mid k$ (when $p-1\nmid k$, a result of Kummer says that $B_k/k$ is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of p-adic L-functions, specifically, for k a positive integer
$$\zeta_p(1-k)=(1-p^{k-1})(-B_k/k),$$
where $\zeta_p$ is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example).
For a totally real field F, a generalization of the p-adic Riemann zeta function exists, namely the p-adic Dedekind zeta function $\zeta_{F,p}$ (as proved independently by Deligne–Ribet (Inv Math 59), Cassou-Noguès (Inv Math 51), and Barsky (1978)). One link between these and the Leopoldt conjecture is through the p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s = 1 des fonctions zêta p-adiques" (Inv Math 91):
$$\lim_{s\rightarrow1}(s-1)\zeta_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R_phE_p}{w\sqrt{D}}$$
where h is the class number,
$$E_p=\prod_{\mathfrak{p}\mid p}\left(1-\mathcal{N}(\mathfrak{p})^{-1}\right)$$ is a product of Euler-like factors, w = 2 is the number of roots of unity, D is the discriminant and $R_p$ is the interesting part here: the p-adic regulator (as Colmez notes, $\sqrt{D}$ and $R_p$ both depend on a choice of sign, but their ratio does not).
Theorem: The Leopoldt conjecture is equivalent to the non-vanishing of the p-adic regulator.
(For this, see, for example, chapter X of Neukirch-Schmidt-Wingberg's "Cohomology of number fields").
A clear consequence of this is that if $\zeta_{F,p}$ does not have a pole at s = 1, then the Leopoldt conjecture is false for (F, p). Perhaps an understanding of the denominators of values of $\zeta_{F,p}$ could lead to an understanding of the pole at s = 1 of $\zeta_{F,p}$.
Added (2010/04/09): So here's how you can use von Staudt–Clausen to see that the $p$-adic zeta function (of Q) has a pole at s = 1. It is clear from your statment of vS–C that it is saying that for $k\equiv0\text{ (mod }p-1)$, $B_k\equiv -1/p\text{ (mod }\mathbf{Z}_p)$ (i.e. it is not $p$-integral). Let $k_i=(p-1)p^i$, the $k_i$ is $p$-adically converging to 0, so $\zeta_p(1-k_i)$ is approaching $\zeta_p(1)$ (since $\zeta_p(s)$ is $p$-adically continuous, at least for $s\neq1$). By the aforementioned interpolation property of $\zeta_p(1-k)$, we have
$$v_p(\zeta_p(1-k_i))=v_p(B_{k_i}/k_i)=-1-i\rightarrow -\infty$$
hence $1/\zeta_p(1-k_i)$ is approaching 0.<|endoftext|>
TITLE: Conformal maps of doubly connected regions to annuli.
QUESTION [11 upvotes]: In another question here on MO, Anweshi asks if any doubly connected region in the complex plane can be conformally mapped to some annulus. The answer to this is yes. But the fact is that two annuli are conformally equivalent iff the ratio of the outer radius and the inner radius is the same for the two. Thus each is conformally equivalent to a unique "standard" annulus 
$r < |z| < 1$ with $ 0 < r < 1$. Now my question is the following:
Is there any way to see what the radius of a "standard" annulus conformally equivalent to a doubly connected region is, just by looking at the region? That is without constructing an explicit map.

REPLY [16 votes]: By "see" I will assume you mean in a geometric sense. Then your question falls within a standard topic in geometric complex analysis. First some terminology: A doubly connected domain $R$ on the Riemann sphere is called a ring domain, and if you map it onto $r < |z| < s$ as a canonical domain, then $\mathrm{mod}(R) = (2\pi)^{-1}\log(s/r)$ is called the conformal modulus or just modulus of the ring domain. By the way I have defined it, it is nearly trivially a conformal invariant, but it need not be defined this way. There is a geometric theory, the Ahlfors-Beurling theory of extremal length of curve families, within which the modulus of a ring domain can be defined directly and geometrically, without any preliminary conformal mapping onto some canonical domain. Extremal length can be proved to be a conformal invariant, and then one quickly sees that the two definitions coincide. There is an exposition of the theory of extremal length in Conformal Invariants by Ahlfors.
It would be unreasonable to expect to "see" the exact value of the modulus of a ring domain. The boundary of a ring domain can be extremely complicated geometrically, and every tiny wiggle impacts on the modulus. But extremal length yields inequalities for the modulus from geometric data. I will quote a single, rather striking, result of this kind:
If a ring domain $R$ contains no circle on the Riemann sphere separating its two boundary components, then $\mathrm{mod}(R) \leq 1/4$. The constant $1/4$ is sharp. The result is due to D. A. Herron, X. Y. Liu and D. Minda.
Small modulus means "thin" ring domain; if the modulus is large enough, the ring domain is so "fat" that it has to contain a separating circle.<|endoftext|>
TITLE: Existence of fine moduli space for curves and elliptic curves
QUESTION [13 upvotes]: For the moduli problem of a curve of genus $g$ with $n$ marked points, how large an $n$ is needed to ensure the existence of a fine moduli space? For this question, terminology is that of Mumford's GIT.
For the following three moduli problems, how big an $N$ is required for existence of a fine moduli space? The terminology is from the exposes of Deligne-Rapoport and Katz-Mazur, or Shimura. The first is in French, the second is too big, and the third is using old language and never mentions the modern terminology of universal elliptic curve, etc.. Therefore it is not possible for me to dig up the information myself.

i) Elliptic curves equipped with a cyclic subgroup of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_0(N)$.
ii) Elliptic curves equipped with a point of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_1(N)$.
ii) Elliptic curves equipped with a symplectic pairing on $N$-torsion points -- this moduli problem corresponds to the modular group $\Gamma (N)$.
References other than the above, will be appreciated.

REPLY [15 votes]: If you want to work over a base ring such as $\mathbf{Z}[1/n]$ rather than over $\mathbf{Q}$ or $\mathbf{C}$ then the relevant numerical condition is that the part of $N$ coprime to $n$ not be "too small" in the $\Gamma_1$ and full level cases.  For an extreme example, if $N$ is a $p$-power and you work over $\mathbf{Z}_{(p)}$ then you'll always have problems in characteristic $p$ at the supersingular points.
On the other hand, if you're willing to go beyond schemes and work with algebraic spaces or Deligne-Mumford or Artin stacks then these issues go away (at the expense of more technical background) in the sense that one has a reasonable "moduli space" over $\mathbf{Z}$ with nice regularity properties for all $N$ (even incorporating degenerations in the sense of generalized elliptic curves with level structure). It has better properties than a coarse moduli space (aside from perhaps not being a scheme...).<|endoftext|>
TITLE: Asymptotics of a Bernoulli-number-like function
QUESTION [13 upvotes]: Tony Lezard asked me the following question which seemed like it should not be too hard but which I did not immediately see how to answer.  Define $f(n,k)$ recursively by $f(1,k) = 1$ and
$$f(n,k) = \frac{1}{1-(1-1/k)^n} \sum_{r=1}^{n-1} \binom{n}{r} \left(\frac{1}{k}\right)^{n-r} \left(1 - \frac{1}{k}\right)^r f(r,k).$$
What can we say about $\lim_{n\to\infty} f(n,k)$?  Even the special case $k=2$ would be interesting:
$$f(n,2) = \frac{1}{2^n-1} \sum_{r=1}^{n-1} \binom{n}{r} f(r,2).$$

Addendum. As explained in Tony Lezard’s answer below, $f(n,k)$ arises as the termination probability of a certain recursive elimination process. The process in the case $n=2$ was studied by Helmut Prodinger, “How to select a loser,” Discrete Math. 120 (1993), 149–159, although Prodinger does not seem to have analyzed $f(n,2)$ specifically.  For other references, see the MO question on The Dance Marathon Problem.

REPLY [17 votes]: [Revised and expanded to give the answer for all $k>1$ and incorporate
further terms of an asymptotic expansion as $n \rightarrow \infty$]
Fix $k>1$, and write $a_1=f(1,k)=1$ and
$$
a_n = f(n,k) =
\frac1{1-q^{-n}} \sum_{r=1}^{n-1} {n \choose r} (1/k)^{n-r} (1/q)^r a_r
\phantom{for}(n>1),
$$
where $q := k/(k-1)$, so $(1/k) + (1/q) = 1$.  Set
$$
a_\infty := \frac1{k \log q}.
$$
For example, if $k=2$ then $a_\infty = 1 / \log 4 = 0.72134752\ldots$,
which $a_n$ seems to approach for large $n$, and likewise for
$k=6$ (the dice-throwing case) with $a_\infty = 1/(6 \log 1.2) =
0.9141358\ldots$.  Indeed as $n \rightarrow \infty$ we have
"$a_n \rightarrow a_\infty$ on average", in the sense that (for instance)
$\sum_{n=1}^N (a_n/n) \sim a_\infty \phantom. \sum_{n=1}^N (1/n)$
as $N \rightarrow \infty$.  But, as suggested by earlier
posted answers to Tim Chow's question, $a_n$ does not converge,
though it stays quite close to $a_\infty$: we have
$$
a_n = a_\infty + \epsilon^{\phantom.}_0(\log_q n) + O(1/n)
$$
as $n \rightarrow \infty$, where $\epsilon^{\phantom.}_0$ is a smooth
function of period $1$ whose average over ${\bf R} / {\bf Z}$ vanishes
but is not identically zero; for large $k$ (already $k=2$ is large
enough), $\epsilon^{\phantom.}_0$ is a nearly perfect sine wave
with a tiny amplitude $\exp(-\pi^2 k + O(\log k))$, namely
$$
\frac2{k\log q}\left|\phantom.\Gamma\bigl(1 + \frac{2\pi i}{\log q}\bigr)\right|
\phantom.=\phantom.
\frac2{k \log q}
 \left[\frac{(2\pi^2/ \log q)}{\sinh(2\pi^2/ \log q)}\right]^{1/2}.
$$
For example, for $k=2$ the amplitude is $7.130117\ldots \cdot 10^{-6}$,
in accordance with numerical observation (see previously posted answers
and the plot below).  For $k=6$ the amplitude is only
$8.3206735\ldots \cdot 10^{-23}$ so one must compute well beyond
the usual "double precision" to see the oscillations.
More precisely, there is an asymptotic expansion
$$
a_n \sim a_\infty + \epsilon^{\phantom.}_0(\log_q n)
 + n^{-1} \epsilon^{\phantom.}_1(\log_q n)
 + n^{-2} \epsilon^{\phantom.}_2(\log_q n)
 + n^{-3} \epsilon^{\phantom.}_3(\log_q n)
 + \cdots,
$$
where each $\epsilon^{\phantom.}_j$ is smooth function of period $1$
whose average over ${\bf R} / {\bf Z}$ vanishes, and
— while the series need not converge —
truncating it before the term $n^{-j} \epsilon^{\phantom.}_j(\log_q n)$
yields an approximation good to within $O(n^{-j})$.  The first few
$\epsilon^{\phantom.}_j$ still have exponentially small amplitudes,
but larger that of $\epsilon^{\phantom.}_0$ by a factor
$\sim C_j k^{2j}$ for some $C_j > 0$; for instance, the amplitude of
$\epsilon^{\phantom.}_1$ exceeds that of $\epsilon^{\phantom.}_0$
by about $2(\pi / \log q)^2 \sim 2 \pi^2 k^2$.  So $a_n$ must be
computed up to a somewhat large multiple of $k^2$ before it becomes
experimentally plausible that the residual oscillation $a_n - a_\infty$
won't tend to zero in the limit as $n \rightarrow \infty$.
Here's a plot that shows $a_n$ for $k=2$ (so also $q=2$) and
$2^6 \leq n \leq 2^{13}$, and compares with the periodic approximation
$a_\infty + \epsilon^{\phantom.}_0(\log_q n)$ and the refined approximation
$a_\infty + \sum_{j=0}^2 n^{-j} \epsilon^{\phantom.}_j(\log_q n)$.
(See http://math.harvard.edu/~elkies/mo11255+.pdf for
the original PDF plot, which can be "zoomed in" to view details.)  The
horizontal coordinate is $\log_2 n$; the vertical coordinate is centered at
$a_\infty = 1/(2 \log 2)$, showing also the lines $a_\infty \pm 2|a_1|$;
black cross-hairs, eventually merging visually into a continuous curve,
show the numerical values of $a_n$; and the red and green contours 
show the smooth approximations.

To obtain this asymptotic expansion, we start by generalizing
R.Barton's formula from $k=2$ to arbitrary $k>1$:
$$
a_n = \frac1k \sum_{r=0}^\infty \phantom. n q^{-r} (1-q^{-r})^{n-1}.
$$
[The proof is the same, but note the exponent $n$ has been corrected
to $n-1$ since we want $n-1$ players eliminated at the $r$-th step,
not all $n$; this does not affect the limiting behavior
$a_\infty+\epsilon^{\phantom.}_0(\log_q n)$, but is needed to get
$\epsilon^{\phantom.}_m$ right for $m>1$.]  We would like to approximate
the sum by an integral, which can be evaluated by the change of variable
$q^{-r} = z$:
$$
\frac1k \int_{r=0}^\infty \phantom. n q^{-r} (1-q^{-r})^{n-1}
= \frac1{k \log q} \int_0^1 \phantom. n (1-z)^{n-1} dz
= \left[-a_\infty(1-z)^n\right]_{z=0}^1 = a_\infty.
$$
But it takes some effort to get at the error in this approximation.
We start by comparing $(1-q^{-r})^{n-1}$ with $\exp(-nq^{-r})$:
$$
\begin{eqnarray}
(1-q^{-r})^{n-1}
&=&
\exp(-nq^{-r}) \cdot \exp \phantom. [nq^{-r} + (n-1) \log(1-q^{-r})]
\cr
&=&
\exp(-nq^{-r}) \cdot
 \exp \left[q^{-r} - (n-1) \left(
    \frac{q^{-2r}}2 + \frac{q^{-3r}}3 + \frac{q^{-4r}}4 + \cdots
    \right)
  \right].
\end{eqnarray}
$$
The next two steps require justification (as R.Barton noted for
the corresponding steps at the end of his analysis), but the
justification should be straightforward.  Expand the second factor
in powers of $u := nq^{-r}$, and collect like powers of $n$, obtaining
$$
\exp(-nq^{-r}) \cdot \left(
1 - \frac{u^2-2u}{2n} + \frac{3u^4-20u^3+24u^2}{24n^2}
- \frac{u^6-14u^5+52u^4-48u^3}{48n^3} + - \cdots \right).
$$
Each term $n^{-j} \epsilon_j(\log_q(n))$ ($j=0,1,2,3,\ldots$)
will arise from the $n^{-j}$ term in this expansion.
We start with the main term, for $j=0$,
which is the only one that does not decay with $n$.  Define
$$
\varphi_0(x) := q^x \exp(-q^x),
$$
which as Reid observed decays rapidly both as $x \rightarrow \infty$
and as $x \rightarrow -\infty$.  Our zeroth-order approximation to $a_n$ is
$$
\frac1k \sum_{r=0}^\infty \phantom. \varphi_0(\log_q(n)-r),
$$
which as $n \rightarrow \infty$ rapidly nears
$$
\frac1k \sum_{r=-\infty}^\infty \varphi_0(\log_q(n)-r).
$$
For $k=q=2$, this is equivalent with Reid's formula for $R(n)$,
even though he wrote it in terms of the fractional part of $\log_2(n)$,
because the sum is clearly invariant under translation of $\log_q(n)$
by integers.
We next apply Poisson summation.  Since
$\sum_{r \in {\bf Z}} \phantom. \varphi_0(t+r)$
is a smooth ${\bf Z}$-periodic function of $t$, it has a Fourier expansion
$$
\sum_{m\in{\bf Z}} \phantom. c_m e^{2\pi i m t}
$$
where
$$
c_m = \int_0^1 \left[ \sum_{r \in {\bf Z}} \phantom. \varphi_0(t+r) \right]
 \phantom. e^{-2\pi i m t} dt
= \int_{-\infty}^\infty \varphi_0(t) \phantom. e^{-2\pi i m t} dt
= \hat\varphi_0(-m).
$$
Changing the variable of integration from $t$ to $q^t$ lets us recognize the
Fourier transform $\hat\varphi$ as $1/\log(q)$ times a Gamma integral:
$$
\hat\varphi_0(y)
= \frac1{\log q} \Gamma\Bigl(1 + \frac{2 \pi i y} {\log q}\Bigr).
$$
This gives us the coefficients $a_m$ in closed form.  The constant coefficient
$a_0 = 1 / (\log q)$ can again be interpreted as the approximation of the
Riemann sum $\sum_{r \in {\bf Z}} \phantom. \varphi_0(t+r)$ by an integral;
the oscillating terms $a_m e^{2\pi i m t}$ for $m \neq 0$
are the corrections to this approximation, and are small due to the
exponential decay of the Gamma function on vertical strips — indeed
we can compute the magnitude $|a_m|$ in elementary closed form using the formula
$|\Gamma(1+i\tau)| = (\pi\tau / \sinh(\pi\tau))^{1/2}$.  So we have
$$
\frac1k \sum_{r \in \bf Z} \phantom. \varphi_0(\log_q(n)-r)
= \frac1k \sum_{m \in \bf Z} \phantom. \hat\varphi_0(-m) e^{2\pi i \log_q(n)}
= a_\infty + \epsilon_0(\log_q(n))
$$
where $a_\infty = a_0 / k = 1 / (k \log q)$ as above, and
$\epsilon^{\phantom.}_0$, defined by
$$
\epsilon^{\phantom.}_0(t) =
\left[ \sum_{r\in\bf Z} \phantom. \varphi_0(t+r) \right] - a_\infty,
$$
has the Fourier series
$$
\epsilon^{\phantom.}_0(t)
= \frac1k \sum_{m \neq 0} \hat\varphi_0(-m) e^{2\pi i m t}.
$$
Taking $m = \pm 1$ recovers the amplitude $2|a_1|/k$ exhibited above;
the $m = \pm 2$ and further terms yield faster but tinier oscillations,
e.g. for $k=2$ the $m=\pm 2$ terms oscillate twice as fast but with
amplitude only $6.6033857\ldots \cdot 10^{-12}$.
The functions $\epsilon^{\phantom.}_j$ appearing in the further terms
$n^{-j} \epsilon^{\phantom.}_j(\log_q(n))$
of the asymptotic expansion of $a_n$ are defined similarly by
$$
\epsilon^{\phantom.}_j(t) =
\frac1k \sum_{r\in\bf Z} \phantom. \varphi_j(t+r),
$$
where
$$
\varphi_j(x) = P_j(q^x) \varphi_0(x) = P_j(q^x) q^x \exp(-q^x)
$$
and $P_j$ is the coefficient of $n^{-j}$ in our power series
$$
(1-q^r)^{n-1} = \exp(-nq^{-r}) \phantom.
\sum_{j=0}^\infty \frac{P_j(nq^{-r})}{n^j}.
$$
Thus
$
P_0(u)=1, \phantom+
P_1(u) = -(u^2-2u)/2, \phantom+
P_2(u) = (3u^4-20u^3+24u^2)/24
$, etc.
Again we apply Poisson to expand $\epsilon^{\phantom.}_j(\log_q(n))$
in a Fourier series:
$$
\epsilon^{\phantom.}_j(t)
= \frac1k \sum_{m \in \bf Z} \hat\varphi_j(-m) e^{2\pi i m t},
$$
and evaluate the Fourier transform $\hat\varphi_j$ by integrating
with respect to $q^t$.  This yields a linear combination of Gamma
integrals evaluated at $1 + (2\pi i y / \log q) + j'$ for
integers $j' \in [j,2j]$, giving $\hat\varphi_j$ as a
degree-$2j$ polynomial multiple of $\hat\varphi_0$.  The first case is
$$
\begin{eqnarray*}
\hat\varphi_1(y)
&=& \frac1{\log q} \left[
  \Gamma\Bigl(2 + \frac{2 \pi i y} {\log q}\Bigr)
  - \frac12 \Gamma\Bigl(3 + \frac{2 \pi i y} {\log q}\Bigr)
  \right]
\\
&=& \frac1{\log q} \frac{\pi y}{\log q} \left(\frac{2 \pi y}{\log q} - i\right)
  \phantom.
  \Gamma\Bigl(1 + \frac{2 \pi i y} {\log q}\Bigr)
\\
&=& \frac{\pi y}{\log q} \left(\frac{2 \pi y}{\log q} - i\right)
 \phantom. \hat\varphi_0(y).
\end{eqnarray*}
$$
Note that $\varphi_1(0) = 0$, so the constant coefficient of the
Fourier series for $\epsilon^{\phantom.}_1(t)$ vanishes;
this is indeed true of $\epsilon^{\phantom.}_j(t)$ for each $j>0$,
because $\hat\varphi_j(0) = \int_{-\infty}^\infty \phi_j(x) \phantom. dx$
is the $n^{-j}$ coefficient of a power series in $n^{-1}$ that we've
already identified with the constant $a_\infty$.  Hence (as can also
be seen in the plot above) none of the decaying corrections
$n^{-j} \epsilon^{\phantom.}_j(\log_q n)$ biases the average of $a_n$
away from $a_\infty$, even when $n$ is small enough that
those corrections are a substantial fraction of the
residual oscillation $\epsilon_0(\log_q n)$.  This leaves
$\hat\varphi_j(\mp1) e^{\pm 2 \pi i t} / k$ as the leading terms
in the expansion of each $\epsilon^{\phantom.}_j(t)$, so we see as promised
that $\epsilon^{\phantom.}_j$ is still exponentially small but with
an extra factor whose leading term is a multiple of $(2\pi / \log q)^{2j}$.<|endoftext|>
TITLE: Is the ultraproduct concept fundamentally category-theoretic?
QUESTION [50 upvotes]: Once again, I would like to take advantage of the large number of knowledgable category theorists on this site for a question I have about category-theoretic aspects of a fundamental logic concept. 
My question is whether the ultraproduct construction is fundamentally a category-theoretic concept. 
The ultraproduct/ultrapower construction of Łos is used pervasively in logic, particularly in model theory and also in set theory, where nearly all of the larger large cardinal axioms can be formulated in terms of the existence of certain kinds of ultrapowers of the universe.
My question is, is the ultraproduct fundamentally a category-theoretic construction, in the sense that it is characterized by some natural category-theoretic universal property? How about the special case of ultrapowers? 
I would be very interested, if there were a natural universal characterization in terms of the usual Hom sets for these first order structures, namely, first order elementary embeddings and/or homomorphisms. (Needless to say, I would be much less interested in a characterization that amounted merely to a translation of the Łos construction or of Łos's theorem into category-theoretic language.)
Background. Suppose we have a collection of structures Mi for i in J, all of the same first order type (e.g. groups, partial orders, graphs, fields, whatever), and U is an ultrafilter on the index set J. This means that U is a nonempty collection of nonempty subsets of J, containing every set or its complement, and closed under intersection and superset. The ultraproduct ΠMi /U consists of equivalence classes [f]U, where f is a function with domain J, with f(i) in Mi, and f ∼Ug iff {i in J | f(i)=g(i)} in U.  One imposes structure on the ultraproduct by saying that a relation holds in the product, if it holds on a set in U, and similarly for functions. Łos's theorem then states that the ultraproduct satisfies a first order formula φ([f]u) if and only if {i in J | Mi satisfies φ(f(i))} is in U. That is, truth in the ultraproduct amounts to truth on a U-large set of coordinates. The special case when all Mi are the same model M, we arrive at the ultrapower MJ /U. In this case, there is a natural map from M into MJ /U, defined by x maps to [cx]U, where cx is the constant function with value x. It is easy to see that this map is an elementary embedding from M into the ultrapower. 
This question is a more focused instance of a probably-too-general question I asked here, and I may have several more in the future.

REPLY [5 votes]: The notion of an ultrapower and more generally reduced powers and their generalizations are essentially category theoretic. More specifically, reduced powers are essentially pro-sets. This answer is a part of my own research, but these results are not ready to publish yet. Although these results relate the ultrapower construction to categories, I do not see how these results could generalize to possibly relate ultraproducts to categories.
In the other answers to this question, people have explained how ultraproducts are direct limits. It turns out that reduced powers are directed limits as well. Furthermore, I will show that as categories, the category of generalized reduced powers is equivalent to the category of inverse systems of sets. By the category of generalized reduced powers, I mean the category of most model-theoretic generalizations of the ultrapower and reduced power construction such as extenders, iterated ultrapowers, limit ultrapowers 3, Boolean ultrapowers 4, and their reduced power analogues such as limit reduced powers. 
If $\mathcal{C}$ is a category, then the category $\mathbf{Pro}(\mathcal{C})$ is essentially the category of inverse systems over the category $\mathcal{C}$ and it should be thought of as the category of inverse limits from the category $\mathcal{C}$. I claim that the category $\mathbf{Pro}(\mathbf{Set})$ of pro-sets is equivalent to a full subcategory of pro-filters. Furthermore, these pro-filters are the things that we want to construct reduced powers. In fact, we obtain a three way duality between the category of pro-sets, the full subcategory of pro-filters where the transitional mappings are epimorphisms, and categories of reduced powers of structures.
$\large\mathbf{Categories}$
In this section, I will first define the categories and I will state the equivalences between these categories.
Let $A$ be a fixed infinite set. Let $\Omega(A)$ be the algebra with universe $A$ and where every operation is a fundamental operation. Then every object in $V(\Omega(A))$ is isomorphic to a limit reduced power of $\Omega(A)$ and any elementary extension of $\Omega(A)$ is isomorphic to a limit ultrapower of $\Omega(A)$ (see 3 for information on limit ultrapowers). Also, one may represent algebras in $V(\Omega(A))$ as limit reduced powers in such a way so that the homomorphisms between algebras in $V(\Omega(A))$ are induced by mappings between the underlying sets of the limit reduced powers. Since every algebra in $V(\Omega(A))$ is isomorphic to a limit reduced power of of $\Omega(A)$, one should think of the algebras in $V(\Omega(A))$ as limit reduced powers.
A small nonempty category $D$ is said to be a cofiltrant category if whenever $e_{1},e_{2}\in D$, then there is some $d\in D$ and morphisms $\ell_{1}:d\rightarrow e_{1},\ell_{2}:d\rightarrow e_{2}$, and if $\ell_{1},\ell_{2}:d\rightarrow e$ are morphisms, then there is some object $c$ and a morphism $m:c\rightarrow d$ with $\ell_{1}m=\ell_{2}m$. Every downward directed set is a cofiltrant category, and the notion of a cofiltrant category is the categorization of the notion of a downward directed set. 
If $\mathcal{C}$ is a category, then $\mathbf{Pro}(\mathcal{C})$ is the category of all functors $F:\mathcal{D}\rightarrow\mathcal{C}$ for some cofiltrant category $\mathcal{D}$. One should think of the category $\mathbf{Pro}(\mathcal{C})$ as the category of all inverse limits in $\mathcal{C}$ since the notion of a cofiltrant category is the categorization of the notion of a downward directed set.
If $F:\mathcal{D}\rightarrow\mathcal{C},G:\mathcal{E}\rightarrow\mathcal{C}$ are objects in $\mathbf{Pro}(\mathcal{C})$, then define the set of morphisms by
$$\mathrm{Hom}(F,G)=\varprojlim_{e\in\mathcal{E}}\ \varinjlim_{d\in\mathcal{D}}\ \mathrm{Hom}(F(d),G(e)).$$
The transitional mappings in the direct and inverse limits are the canonical ones, and the definition of composition of morphisms is defined in the natural way.
The following categories are the object of study in the papers 1 and 2.
We shall now construct a category $\mathfrak{F}$. The objects in $\mathfrak{F}$ are pairs 
$(X,\mathcal{F})$ where $X$ is a set and $\mathcal{F}$ is a filter on $X$. If $(X,\mathcal{F}),(Y,\mathcal{G})\in\mathfrak{F}$, then function $f:X\rightarrow Y$ is a morphism from $(X,\mathcal{F}),(Y,\mathcal{G})$ if $f^{-1}[R]\in\mathcal{F}$ whenever $R\in\mathcal{G}$. It is easy to show that $f$ is a morphism from $(X,\mathcal{F})$ to $(Y,\mathcal{G})$ if and only if whenever $R\subseteq X$ is non-negligible (i.e. $R^{c}\not\in\mathcal{F}$), then the image $f[R]$ is non-negligible. The intuition behind defining the category $\mathfrak{F}$ this way is that we do not want to map non-negligible sets to negligible sets since that is like a function mapping a non-empty set to an empty set.
Let $\mathfrak{G}$ be the quotient category of $\mathfrak{F}$ where we relate two morphisms if they differ by a negligible set. In other words, the objects in $\mathfrak{G}$ and $\mathfrak{F}$ are the same. If $(X,\mathcal{F}),(Y,\mathcal{G})$ are objects in $\mathfrak{F}$, and $f,g\in\mathrm{Hom}_{\mathfrak{F}}[(X,\mathcal{F}),(Y,\mathcal{G})]$, then $f\simeq g$ iff $\{x\in X|f(x)=g(x)\}\in\mathcal{F}$. Then $\mathrm{Hom}_{\mathcal{G}}[(X,\mathcal{F}),(Y,\mathcal{G})]=\mathrm{Hom}_{\mathfrak{F}}[(X,\mathcal{F}),(Y,\mathcal{G})]/\simeq$. The composition in $\mathfrak{G}$ is defined in the obvious manner.
While the categories $\mathfrak{F}$ and $\mathfrak{G}$ were both studied in 1 and 2, the category $\mathfrak{G}$ is more fundamental than $\mathfrak{F}$ and the category $\mathfrak{G}$ deserves to be called the category of filters while $\mathfrak{F}$ does not seem to be very useful.
Morphisms between algebras in $\mathfrak{G}$ induce homomorphisms between reduced powers. If $\mathcal{A}$ is an algebraic structure and $[f]:(X,\mathcal{F})\rightarrow(X,\mathcal{G})$ is a morphism in $\mathfrak{G}$, then we define a morphism $[f]^{\bullet}:\mathcal{A}^{X}/\mathcal{G}\rightarrow\mathcal{A}^{Y}/\mathcal{F}$ by letting $[f]^{\bullet}[\ell]=[\ell\circ f]$.

$\mathbf{Proposition}$ Let $[f]\in \mathrm{Hom}_{\mathfrak{G}}[(X,\mathcal{F}),(Y,\mathcal{G})]$. The
  following are equivalent.

If $[f]$ is an epimorphism in $\mathfrak{G}$.
$\mathcal{G}=\{S\subseteq Y|f^{-1}[S]\in\mathcal{F}\}$
If $R\in\mathcal{F}$, then $f[R]\in\mathcal{G}$.
The canonical mapping $[f]^{\bullet}:A^{Y}/\mathcal{G}\rightarrow A^{X}/\mathcal{F}$ is injective for each set $A$.


Let $\mathbf{PF}$ be the full subcategory of $\mathbf{Pro}(\mathfrak{G})$ whose objects consist of inverse systems $((X_{d},\mathcal{F}_{d})_{d\in D},(\ell_{d_{1},d_{2}})_{d_{1}\leq d_{2}})$ over directed sets $D$ such that each $\ell_{d_{1},d_{2}}$ is an epimorphism in $\mathfrak{G}$.
$\mathbf{Humor}:$ I call the objects in $\mathbf{PF}$ pro-filters. That is, professional filters. The pro-filters play football in the National Filter League (NFL).
If $\kappa$ is a cardinal, then let $\mathbf{PF}_{\kappa}$ denote the full subcategory of $\mathbf{PF}$ consisting of inverse systems $(X_{d},\mathcal{F}_{d})_{d\in D}$ such that $|X_{d}|<\kappa$ for each $d\in D$. Let $\mathbf{Set}_{\kappa}$ denote the category of sets of cardinality less than $\kappa$. Let $\mathbf{U}_{\kappa}$ be the subcategory of $\mathbf{PF}_{\kappa}$ consisting of all inverse systems $(X_{d},\mathcal{F}_{d})_{d\in D}$ where each $\mathcal{F}_{d}$ is an ultrafilter.

$\mathbf{Theorem}$ Let $A$ be an infinite set.

The category $\mathbf{PF}$ is covariantly equivalent to the category $\mathbf{Pro}(\mathbf{Set})$.
The functor defined by $(X_{d},\mathcal{F}_{d})_{d\in D}\mapsto \Omega(A)^{X_{d}}/\mathcal{F}_{d}$ is a contravariant equivalence
  between the category $\mathbf{PF}_{|A|^{+}}$ and the category
  $V(\Omega(A))$. Furthermore, this functor restricts to an equivalence
  between the category $\mathbf{Pro}(\mathbf{U}_{|A|^{+}})$ and the
  elementary extensions of $\Omega(A)$.
The category $\mathbf{Pro}(\mathbf{Set}_{|A|^{+}})$ is contravariantly equivalent to the category $V(\Omega(A))$.


The equivalences between the category $\mathbf{PF}$ and $\mathbf{Pro}(\mathbf{Set})$ can be described explicitly. If $((X_{d},\mathcal{F}_{d})_{d\in D},(\ell_{d_{1},d_{2}})_{d_{1}\leq d_{2}})\in\mathbf{PF}$, then add a transitional mapping $f:X_{d}\rightarrow X_{e}$ whenever $f\in\textrm{Hom}_{\mathfrak{F}}((X_{d},\mathcal{F}_{d}),(X_{e},\mathcal{F}_{e}))$ and $[f]=\ell_{d,e}$.
If $F:\mathcal{D}\rightarrow\mathbf{Set}$ is a pro-set, then for each $d\in\mathcal{D}$, then $\{\mathrm{Im}(F(f))|f:e\rightarrow d\,\textrm{for some}\,e\in\mathcal{D}\}$ is a filterbase that generates a filter on $F(d)$ which we shall denote by $\mathcal{F}_{d}$.
Partial order $\mathcal{D}$ where $d\leq e$ iff there is a morphism from $d$ to $e$.  Then if $d\leq e$, then let $\ell_{d,e}:(X_{d},\mathcal{F}_{d})\rightarrow(X_{e},\mathcal{G}_{e})$ be the morphism where
$\ell_{d,e}=[F(f)]$ for some $f:d\rightarrow e$. It is easy to show that the morphism $\ell_{d,e}$ does not depend on $[F(f)]$ and $((X_{d},\mathcal{F}_{d}),(\ell_{d,e})_{d\leq e})\in\mathbf{PF}$.

$\mathbf{Corollary}$ If $|A|<|B|$, then the category $V(\Omega(A))$ is
  equivalent to a full subcategory of $V(\Omega(B))$.
$\mathbf{Corollary}$ If $(X_{d},\mathcal{F}_{d})_{d\in
D},(Y_{e},\mathcal{G}_{e})_{e\in E}\in\mathbf{PF}_{|A|^{+}}$, and the algebras
  $^{\lim}_{\longrightarrow}\Omega(A)^{X_{d}}/\mathcal{F}_{d}$ and
  $^{\lim}_{\longrightarrow}\Omega(A)^{Y_{e}}/\mathcal{G}_{e}$ are
  isomorphic, then 
  $^{\lim}_{\longrightarrow}\mathcal{A}^{X_{d}}/\mathcal{F}_{d}$ and
  $^{\lim}_{\longrightarrow}\mathcal{A}^{Y_{e}}/\mathcal{G}_{e}$ are
  isomorphic for each structure $\mathcal{A}$.

The above result still holds when one replaces the direct limit of reduced powers with other reduced power and ultrapower constructions.
$\large\textbf{An Application}$
We shall now give an application that shows that going between algebras and different categories may be useful. The following result can be proved using the duality between pro-filters and algebras (the proof also uses a version of the Feferman-Vaught theorem, and the Keisler-Shelah isomorphism theorem or Frayne's theorem).

$\mathbf{Theorem}$ Let $\mathbf{2}$ denote the two element Boolean
  algebra. Let
  $\mathcal{A}\mapsto\mathbf{R}(\mathcal{A}),\mathcal{A}\mapsto\mathbf{S}(\mathcal{A})$ 
  be two distinct reduced power constructions(such as limit reduced
  powers, Boolean powers etc.). If $\mathbf{R}(\mathbf{2})$ and
  $\mathbf{S}(\mathbf{2})$ are elementarily equivalent, then there is a
  sequence of ultrafilters $\mathcal{U}_{n}$ such that for every
  structure $\mathcal{A}$, we have
  $$^{\lim}_{n\rightarrow\infty}(\mathbf{R}(\mathcal{A}))^{\mathcal{U}_{0}\cdots\mathcal{U}_{n}}\simeq^{\lim}_{n\rightarrow\infty}(\mathbf{S}(\mathcal{A}))^{\mathcal{U}_{0}\cdots\mathcal{U}_{n}}.$$

We take note that in the above result we cannot replace the direct limit of ultrapowers with a single ultrapower. For example, if $\mathcal{U}$ is a non-principal ultrafilter, and
$\mathbf{R}(\mathcal{A})=\mathcal{A}$ and $\mathbf{S}(\mathcal{A})=\mathcal{A}^{\mathcal{U}}$ for all structures $\mathcal{A}$, then for each non-principal ultrafilter $\mathcal{V}$, we have $\mathcal{V}<_{RK}\mathcal{U}\cdot\mathcal{V}$. In particular, the ultrafilters
$\mathcal{V}$ and $\mathcal{U}\cdot\mathcal{V}$ are not Rudin-Kiesler equivalent, so there is some structure $\mathcal{A}$ where the structures $\mathcal{A}^{\mathcal{V}}$ and $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}\simeq(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ are not isomorphic.
I must also mention that the elementary classes of Boolean algebras have a particularly nice and simple classification. There is a countable set of Boolean algebra invariants called the elementary invariants, and two Boolean algebras are elementarily equivalent if and only if they satisfy the same elementary invariants. In particular, to check that $\mathbf{R}(2)$ and $\mathbf{S}(2)$ are elementarily equivalent, it suffices to show that these Boolean algebras have the same elementary invariants. The reader is referred to [5] or [6][Vol. 1] for more information on these elementary invariants.
Oh the joy of cats!!!
1 Blass, Andreas
Two closed categories of filters. 
Fund. Math. 94 (1977), no. 2, 129–143.
2 Koubek, Vaclav; Reiterman, Jan
On the category of filters.
Comment. Math. Univ. Carolinae 11 1970 19–29
3 Keisler, H. Jerome. Limit ultrapowers.
Trans. Amer. Math. Soc. 107 1963 382–408
4 Mansfield, Richard. The theory of Boolean ultrapowers. 
Ann. Math. Logic 2 (1970/71), no. 3, 297–323. 
[5] Chang, Chen Chung, and H. Jerome Keisler. Model Theory. Amsterdam: North-Holland Pub., 1973. 
[6] Monk, J. Donald, Robert Bonnet, and Sabine Koppelberg. Handbook of Boolean Algebras. Amsterdam: North-Holland, 1989.<|endoftext|>
TITLE: Has anything precise been written about the Fukaya category and Lagrangian skeletons?
QUESTION [33 upvotes]: At some point in this past year, some Fukaya people I know got very
excited about the Fukaya categories of symplectic manifolds with "Lagrangian skeletons."  As I understand it, a
Lagrangian skeleton is a union of Lagrangian submanifolds which a
symplectic manifold retracts to.  One good example would be the
zero-section of a cotangent bundle, but there are others; for example,
the exceptional fiber of the crepant resolution of $\mathbb
C^2/\Gamma$ for $\Gamma$ a finite subgroup of $SL(2,\mathbb C)$.  From the rumors I've heard, apparently there's some connection between the geometry of the skeleton and the Fukaya category of the symplectic manifold; this is understood well in the case of a cotangent bundle from work of Nadler and Nadler-Zaslow
I'm very interested in the Fukaya categories of some manifolds like
this, but the only thing I've actually seen written on the subject is
Paul Seidel's moderately famous picture of Kontsevich carpet-bombing
his research program

which may be amusing, but isn't very
mathematically rigorous.  Google searching hasn't turned up much, so I
was wondering if any of you have anything to suggest.

REPLY [13 votes]: A comment re. Jonny's nice answer: there was indeed a time when that was the envisioned strategy of proof.  However our present approach does not require the arborealization.  Because: now we know that the Fukaya and microlocal categories associated to any (singular) Legendrian in a cosphere bundle agree.
--
In GPS2, we prove descent for general sectorial covers.  In particular, one could deduce the existence of said cosheaf by an (unwritten) purely geometric argument that an open cover of the skeleton lifts to a sectorial cover of the symplectic manifold.  However, the notion of sectorial cover is more flexible, as it is not tied to a particular skeleton; in a certain sense it captures all skeleta at once.
The real virtue of the cosheaf on the skeleton is that it is not some arbitrary cosheaf of categories, but in fact is the Kashiwara-Schapira stack of microlocal sheaf theory. (Historical note: this assertion goes beyond the original conjecture of Kontsevich, and is probably best attributed to Nadler.) As mentioned above, GPS3 is the required local
calculation.  In principle, one could use GPS2 to glue, but actually there is a short-cut for most Weinstein manifolds of interest in mirror symmetry and geometric representation theory.  The reason is that these manifolds typically have trivial stable symplectic normal bundle, so one can use the embedding trick without messing about with descent from a Lagrangian Grassmannian bundle. This reduces the problem to the cotangent comparison of GPS3, provided one has certain full faithfulness of embedding results in both partially wrapped Floer theory and in microlocal sheaf theory.
Full faithful embeddings come from a certain doubling construction, which was originally introduced in sheaf theory by Guillermou, it is described in e.g. section 11.4 of his omnibus.  Guillermou assumes there that the relevant skeleton is smooth, but similar ideas give a construction in general and appear in Sec. 6 of my paper with Nadler.  On the wrapped Floer theory side the construction is Example 8.6 in GPS2.  Combining the above one learns that for a Weinstein manifold with stably trivial symplectic normal bundle, the wrapped Fukaya category is equivalent to the category of microlocal sheaves on any Lagrangian skeleton.
(Most likely the `embedding trick' also works to establish the result in the general case, but this requires a bit more complicated argument.)<|endoftext|>
TITLE: Is there an analogue Beilinson-Bernstein localization for quantized enveloping algebra
QUESTION [6 upvotes]: I am completely a beginner in this field. I wonder know whether there is appropriate notion for quantum flag variety of finite dimensional Lie algebra. If so, what is the correspondent notion for "quantum differential operator on this quantum flag variety". If such notions exist. I think there should be some kind of quantum analogue of Beilinson-Bernstein localization for quantum group (or quantized enveloping algebra)?
The second question is related to the question D-module theory Scott Carnahan mentioned that the category of D-module can be taken as category of quasi coherent sheaves on DeRham stack. So, if there exits "quantum D-module". Then in this case, the category of D-module can be taken as category of quasi coherent sheaves on "quantized" DeRham stack? How do we define the quantized DeRham stack? Or more generally, is there an appropriate notion for "quantized" stack?

REPLY [5 votes]: Yes, there exists the result for quantized enveloping algebra. It is developed by Lunts-Rosenberg and proved by them and Tanisaki.
There are several notions:

quantized flag variety
quantum D-module.

In the framework of noncommutative algebraic geometry. quantized flag variety is defined as a noncommutative projective scheme. It is a proj-category of quantized enveloping algebra of Lie algebra g. It is a noncommutative separated scheme with affine covers discovered by A.Joseph.
Lunts and Rosenberg defined differential calculus in noncommutative algebraic geometry. They introduced the noncommutative version of Grothendieck differential operators in differential operator on noncommutative ring and then applied this construction to define quantized D-modules in localization for quantum group. In this paper, they formulated the quantized Beilinson Bernstein localization for quantized enveloping algebra in generic case. They proved the global section functor is exact and  conjectured it is indeed the correct quantized version, which means it is an equivalent.
Later, under this framework. Tanisaki proved in his paper The Beilinson-Bernstein correspondence for quantized enveloping algebras that the conjecture of Lunts-Rosenberg is indeed true. Moreover, Tanisaki proved this result in root of unity case,see D-modules on quantized flag manifolds at roots of 1.
More comments: In the paper of Lunts-Rosenberg, they pointed out the localization for the quantum group sl2 was constructed by "hand" by T.J.Hodges.<|endoftext|>
TITLE: Determination of a symmetric convex region by parallel sections
QUESTION [11 upvotes]: This question is partly inspired by a problem in Stewart's Calculus: "Find the area of the region enclosed by $y=x^2$ and $x=y^2$."
Suppose $f\colon [0,1]\to [0,1]$ is a convex increasing function that fixes 0 and 1. The graphs $y=f(x)$ and $x=f(y)$ enclose a convex region. To find its area, students will likely integrate the length of vertical sections: $f^{-1}(x)-f(x)$. They are unlikely to ask if two different functions can yield the same integrand. But I'll ask: 

Does $f^{-1}-f$ determine $f$? (Among all convex increasing functions that fix 0 and 1).

REPLY [9 votes]: Yes.
Assume we have two distinct functions $f$ and $g$ such that $f^{-1}-f\equiv g^{-1}-g$.
Take a sequence $x_n=f(x_{n-1})$.
Clearly  $f(x_n)-g(x_n)=0$ or $(-1)^n[f(x_n)-g(x_n)]$ has the same sign for all $n$. 
Sinse $\int_0^1f=\int_0^1g$, there are two sequences $x_n$ and $y_n$ as above such that $f(x_n)=g(x_n)$, $f(y_n)=g(y_n)$ and say $(-1)^n[f(x)-g(x)]>0$ for any $x\in(x_n,y_n)$.
Note that $x_n,y_n\to 0$ and $\int_{x_n}^{y_n}|f-g|=const>0$.
It follows that $\limsup_{x\to0} |f(x)-g(x)|\to\infty$, a contradiction.<|endoftext|>
TITLE: Geometry meaning of higher cohomology of sheaves?
QUESTION [27 upvotes]: Let $X$ be a projective algebraic variety over a algebraic closed field $k$. Let $\mathcal{F}$ be a coherent sheaf on $X$. We know that $H^0(X, \mathcal{F})$ is the vector space of global sections of $\mathcal{F}$. This gives us a geometric illustration of $H^0$. For example, let $I_D$ be the ideal sheaf of a hypersurface $D$ of degree $>1$ in a projective space $\mathbb{P}^n$, then it is easy to see that
$$H^0(\mathbb{P}^n,I_D\otimes\mathcal{O}_{\mathbb{P}^n}(1))=0.$$ 
In fact, there is no hyperplane containing $D$, which means that there is no global section of $\mathcal{O}_{\mathbb{P}^n}(1)$, which are hyperplanes, containing $D$. Hence  $H^0(\mathbb{P}^n, I_D\otimes\mathcal{O}_{\mathbb{P}^n}(1))=0$. However, my first question is how to understand higher cohomologies of sheaves in geometric ways. The following questions then come out:
1) How to understand Serre's vanishing theorem, i.e., is there a geometric way to think about the vanishing of $H^q(X, \mathcal{F}\otimes A^n)$ for $n\gg1$, where $\mathcal{F}$ is coherent and $A$ is ample.
2) How to understan Kodaira's vanishing theorem geometrically.
Maybe a concrete question will help, say $D$ a subvariety of $\mathbb{P}^n$, how to  determine geometrically whether $H^0(\mathbb{P}^n, I_D\otimes\mathcal{O}_{\mathbb{P}^n}(1))$ vanishes or not.

REPLY [19 votes]: Let's start way back. The invention of schemes moved algebraic geometry away from thinking about varieties as embedded objects. However, embedding an abstract scheme into projective space has a lot of advantages, so if we can do that, it's useful. And even if we cannot embed our scheme into projective space, but we can find a non-trivial map, that gives us some way to understand our abstract scheme. In order to find a non-trivial map we need a line bundle with sections.
So we are interested in finding sections of various sheaves, but primarily line bundles (and of course for this sometimes we need to deal with other kind of sheaves). Thus we are interested in $H^0$. 
On the other hand, computing $H^0$ is non-trivial. There are no good general methods. One reason for this is that, for instance, $H^0$ is not constant in families, or put it another way it is not deformation invariant. On the other hand, $\chi(X,\mathscr F)$ behaves much better. It is constant in flat families and if $\mathscr F$ is a line bundle, then it is computable using Riemann-Roch. 
Then, if we know that $H^i=0$ for $i>0$, then $H^0=\chi$ and we're good.

Here is an explicit example for a typical use of Serre vanishing:
Example 1 Suppose $X$ is a smooth projective variety and $\mathscr L$ is an ample line bundle on $X$. 
Then we know that $\mathscr L^{\otimes n}$ is very ample and $H^i(X, \mathscr L^{\otimes n})=0$ for $i>0$ and $n\gg 0$. Then $\mathscr L^{\otimes n}$ induces an embedding $X\hookrightarrow \mathbb P^N$
where $N=\dim H^0(X,\mathscr L^{\otimes n})-1=\chi(X,\mathscr L^{\otimes n})-1$ by Serre's vanishing and hence $N$ is now computable by Riemann-Roch. 
The only shortcoming of the above is that in general there is no way to tell what $n\gg0$ really means and so it is hard to get any explicit numerical estimates out of this. This is where Kodaira vanishing can help.
Example 2 In addition to the above assume that $\mathscr L=\omega_X$, or in other words assume that $X$ is a smooth canonically polarized projective variety. There are many of these, for instance all smooth projective curves of genus at least $2$ or all hypersurfaces satisfying $\deg > \dim +2$. 
In particular, these are those of which we like to have a moduli space.
Anyway, the way Kodaira vanishing changes the above computation is that now we know that already $H^i(X,\omega_X^{\otimes n})=0$ for $i>0$ and $n>1$! In other words, as soon as we know that $\omega_X^{\otimes n}$ is very ample and $n>1$, then we can compute the dimension of the projective space into which we can embed our canonically polarized varieties. In fact, perhaps more importantly than that we can compute it, we know (from the above) without computation that this value is constant in families.
So, once we have a boundedness result that says that this happens for any $n\geq n_0$ for a given $n_0$, and Matsusaka's Big Theorem says exactly that, then we know that all such canonically polarized smooth projective varieties (with a fixed Hilbert polynomial) can be embedded into $\mathbb P^N$, that is, into the same space. 
This implies that then all of these varieties show up in the appropriate Hilbert scheme of $\mathbb P^N$ and we're on our way to construct our moduli space. 
Of course, there is a lot more to do to finish the whole construction and also this method works in other situations, so this is just an example.

There is one more thing one might think regarding your question, that is, ask the more abstract question:
"What does higher cohomology of sheaves mean (e.g., geometrically)?"
This is arguable, but I think that the essence of higher cohomology is that it measures the failure of something we wish were true all the time, but isn't. More specifically, if you're given a short exact sequence of sheaves on $X$
$$
0\to \mathscr F' \to \mathscr F \to \mathscr F'' \to 0
$$
then we know that even though $\mathscr F \to \mathscr F''$ is surjective, the induced map on the global sections $H^0(X,\mathscr F) \to H^0(X,\mathscr F'')$ is not. However, the vanishing of $H^1(X,\mathscr F')$ implies that for any surjective map of sheaves with kernel $\mathscr F'$ as above the induced map on global sections is also surjective. Since you already have a geometric interpretation of $H^0$, this gives one for $H^1$: it measures (or more precisel
For a more detailed explanation of the same idea see this MO answer. 
In my opinion the best way to understand higher ($>1$) cohomology is that it is the lower cohomology of syzygies. In other words, consider a sheaf $\mathscr F$ and embed it into an acyclic (e.g., flasque or injective or flabby or soft) sheaf. So you get a short exact sequence:
$$
0\to \mathscr F\to \mathscr A\to \mathscr G \to 0 
$$
Since $\mathscr A$ is acyclic, we have that for $i>0$
$$
H^{i+1}(X,\mathscr F)\simeq H^i(X,\mathscr G),
$$
so if you understand what $H^1$ means, then $H^2$ of $\mathscr F$ is just $H^1$ of $\mathscr G$, 
$H^3$ of $\mathscr F$ is just $H^2$ of $\mathscr G$ and so on.<|endoftext|>
TITLE: Were Bourbaki committed to set-theoretical reductionism?
QUESTION [38 upvotes]: A set-theoretical reductionist holds that sets are the only abstract objects, and that (e.g.) numbers are identical to sets.  (Which sets? A reductionist is a relativist if she is (e.g.) indifferent among von Neumann, Zermelo, etc. ordinals, an absolutist if she makes an argument for a priviledged reduction, such as identifying cardinal numbers with equivalence classes under equipotence).  Contrasting views: classical platonism, which holds that (e.g.) numbers exist independently of sets; and nominalism, which holds that there are no abstract particulars.
I'm interested in the relationship between "structuralism" as it is understood by philosophers of science and mathematics and the structuralist methodology in mathematics for which Bourbaki is well known.  A small point that I'm hung up on is the place of set theory in Bourbaki structuralism.  I'm weighing two readings.  

(1) conventionalism: Bourbaki used set theory as a convenient "foundation", a setting in which models of structures may be freely constructed, but "structure" as understood in later chapters is not essentially dependent on the formal theory of structure developed in Theory of Sets,
(2) reductionism: sets provide a ground floor ontology for mathematics; mathematicians study structures in the realm of sets.

In favor of conventionalism: 

(a) Leo Corry's arguments in “Nicolas Bourbaki and the Concept of Mathematical Structure” that the formal structures of Theory of Sets are to be distinguished from and play only a marginal role in the subsequent investigation of mathematical structure,
(b) ordered pairs: definitions reducing pairs to sets like Kuratowski's bring "baggage" (i.e., extra structure) and Bourbaki used primitive ordered pairs in the first edition of Theory of Sets, showing no excess concern for complete reduction,
(c) statements of Dieudonne to the Romanian Institute indicating chs. 1 and 2 are mostly to satisfy bothersome philosophers (like me I suppose) before getting on to topics of greater interest,
(d) the discussion of axiomatics and structure in "The Architecture of Mathematics", placing no special emphasis on sets,
(e) this interpretation serves my selfish philosophical agenda.

In favor of reductionism: 

(a) linear ordering of texts suggests perceived logical dependence on Theory of Sets, 
(b) reductionism makes sense of unity of mathematics,
(c) 1970 edition includes Kuratowski pairs,
(d) makes sense of controversies over category theory,
(e) makes sense of some outsider criticisms (e.g., Mac Lane in "Mathematical Models" that Bourbaki was dogmatic and stifling),
(f) I fear that in leaning towards conventionalism I'm self-deceiving to serve my selfish philosophical agenda.

Apologies: not sure this is MO appropriate, any answers may be anachronistic, probably no univocality of opinion among Bourbaki members, my views are based on popular expositions, interviews, and secondary literature and not close study of the primary texts.  
Discussion related to this question has recently occurred at n-category cafe, occasioned by Manin's recent claim that Bourbaki provided "pragmatic foundations".  The conventionalist interpretation, I think, helps make sense of Manin's claim and would show some criticisms levelled toward Bourbakism to misapprehend their intention (if not their impact).  I have Borel's "Twenty-Five Years With Bourbaki" which discusses Grothendieck and the controversy over the direction following the first six books.  Corry makes the claim that the Theory of Sets approach had limitations in dealing with category theory.  I would especially appreciate references or answers that help me better understand these issues in particular, which are accessible to a philosopher with some grad coursework in mathematics and with only a self taught rudimentary understanding of categories.

REPLY [5 votes]: Geoffrey Hellman has written something on structuralism that compares Bourbaki structuralism with category theoretic structuralism. Here. His take seems to be that they were being reductionist.<|endoftext|>
TITLE: Resolution of singularities, nature of
QUESTION [6 upvotes]: Hironaka's theorem guarantees an existence of resolution of singularities in characteristic 0. If I am not wrong, it also guarantees (or at least some other result does), that if the resolution is a singular point, one can get the "Exceptional Fiber" to be a simple normal crossing divisor. Very likely, if the singular locus is of higher dimension, then too one can get the "Exceptional Fiber" to be a simple normal crossing divisor.
However, if the nature of singularity varies along the singular locus, (perhaps) one cannot expect the dimensions of the fibers at each point to be constant in the given resolution.
What should be the most general result known in this direction? Can one expect, for example, a stratification such that inverse image of each strata, is "like simple normal crossing" (eg smooth irreducible components, as well as all k-fold intersections being smooth)?

REPLY [8 votes]: Hironaka in fact says that you can resolve singularities by a sequence of blow ups, and the universal property of blowing up is that the exceptional locus is a Cartier divisor.  So in fact, the exceptional locus of the whole thing will be a Cartier divisor.  Making sure that the exceptional locus is a snc divisor is called "embedded resolution" and is also known to be true.  This is covered by Kollár in his book Lectures on Resolution of Singularities, which I believe is an expanded version of his notes Resolution of Singularities – Seattle Lecture, arXiv:math/0508332, but also pretty much everywhere else that proves resolution of singularities.<|endoftext|>
TITLE: Mumford conjecture: Heuristic reasons? Generalizations? ... Algebraic geometry approaches?
QUESTION [29 upvotes]: The Mumford conjecture states that for each integer $n$, we have: the map $\mathbb{Q}[x_1,x_2,\dots] \to H^\ast(M_g ; \mathbb{Q})$ sending $x_i$ to the kappa class $\kappa_i$, is an isomorphism in degrees less than $n$, for sufficiently large $g$. Here $M_g$ denotes the moduli of genus $g$ curves, and the degree of $x_i$ is the degree of the kappa class $\kappa_i$. This conjecture was proved by Madsen-Weiss a few years ago.

What are the heuristic or moral reasons for the conjecture? (EDIT: I am particularly interested in algebraic geometric reasons, if there are any. Though algebraic topologial reasons are very welcome too.) What lead Mumford to formulating the conjecture in the first place?
I know very little about the Madsen-Weiss proof, but I know that it mainly uses algebraic topology methods. Are there any approaches to the conjecture which are more algebraic-geometric?
Is there any analogous theorem or conjecture regarding the (topological) $K$-theory of $M_g$? Or the Chow ring of $M_g$? etc.

REPLY [8 votes]: Here is a belated and perhaps too naive answer for your first question, i.e. a heuristic reason for why to believe in the Mumford conjecture.
One can think of cohomology classes on $M_g$ as characteristic classes for families of curves. That is, a class is the same thing  as a rule such that whenever you are given a smooth family of curves $X \to B$ of genus $g$, this rule associates a cohomology class in $H^\bullet(B)$ in a functorial manner. (Topologically, you can also think of it as a characteristic class of oriented surface bundles of genus $g$, since $\mathrm{MCG}(\Sigma_g) \simeq \mathrm{Diff}^+(\Sigma_g)$.)
In a similar way, one might then think of a cohomology class on $M_\infty$ as a rule that assigns a cohomology class to a family of curves of arbitrary genus, or as a characteristic class of arbitrary oriented surface bundles. Of course this should be made precise since "functoriality" of such a characteristic class seems meaningless without any way of comparing surface bundles of different genus, and so one needs to define comparison maps between different moduli spaces by working with boundary components, gluing on tori/pants, etc. Nevertheless the intuitive picture is clear: a class on $M_\infty$ should be a rule that assigns in a uniform manner, to any family of curves $X \to B$ whatsoever, a cohomology class in $H^\bullet(B)$.
Now the $\kappa$ classes seem to fit the bill for being classes on $M_\infty$: given any $\pi \colon X \to B$ whatsoever, we may form the vertical tangent bundle, take its Euler class, multiply, push forward. This seems as canonical as one could hope for. Are there any others? Well, there are the $\lambda$-classes, i.e. the Chern classes of $\pi_\ast \Omega_{X/B}^1$, but Mumford showed via Grothendieck-Riemann-Roch that these are polynomials in the $\kappa$'s. It's hard to think of anything else.
Now surjectivity of $\mathbf Q[\kappa_1,\kappa_2,\ldots] \to H^\bullet(M_\infty)$ asserts that these obvious classes are really the only ones that you can write down in a uniform way, and injectivity says that there are no uniform relations between the $\kappa$'s (i.e. all relations are "low-genus accidents"). Now one might believe in the Mumford conjecture simply because people thought hard about these things and could not find any other genus-invariant characteristic classes, nor any genus-invariant relations between the $\kappa$'s. Mumford's conjecture is the simplest possible explanation for this failure.<|endoftext|>
TITLE: Constructing the Hecke-Algebra from the Burau representation
QUESTION [8 upvotes]: I'm currently learning about knot theory, so please correct me if I'm saying something senseless. I'll try to describe the things just as I think they are.
First, suppose we have constructed the reduced Burau representation 
$\psi_n^r: B_n\to\text{GL}_{n-1}({\mathbb Z}[t^{\pm 1}])$
of the Artin Braid Group on $n$ strands. 
(1) From this representation, one can construct the Alexander-Polynomial $\nabla$ as the knot invariant corresponding to the Markov function 
(X) $\beta\mapsto (-1)^{n+1}\frac{s^{-\langle\beta\rangle} (s-s^{-1})}{s^n-s^{-n}} \text{det}(\psi^r_n(\beta) - I_{n-1}))$
(here $\langle\beta\rangle\in{\mathbb Z}$ is the image of $\beta$ under $B_n\to B_n/[B_n,B_n]={\mathbb Z}$ and $s=t^{1/2}$.) 
Now the Alexander polynomial satisfies the Skein relation $\nabla(L_+) - \nabla(L_-) = (s^{-1} - s)\nabla(L_0)$, and this suffests to look at the quotient of ${\mathbb Z}[s^{\pm 1}][B_n]$ by the relation $\sigma_i - \sigma_i^{-1} = (s^{-1}-s)\cdot 1$, because the Markov function above factors through this quotient. This was the first motivation for me to study the Hecke algebra - just take some knot invariant and mod out every relation in the group algebra of $B_n$ which is satisfied by the invariant; in fact, viewing it in this way, I'd rather say "Hecke-Algebra of the Alexander Polynomial".
(2) On the other hand, one could start from a more representation theoretic viewpoint and define the Hecke-algebra ${\mathcal H}_n^s$ to be the quotient of ${\mathbb Z}[t^{\pm 1}][B_n]$ by the relation $T_i^2 = (t-1) T_i + t\cdot 1$ in order to study those representations of $B_n$ where the representing matrices of the $T_i$ satisfy one fixed quadratic relation. The representing matrices in the reduced Burau representation do satisfy the above quadratic equation, and so one gets a representation of the Hecke algebra ${\mathcal H}_n^t$.
These are two quite different ways which lead to the study of Hecke algebras -- can somebody tell me what the relation between these two constructions is? I'd also like to get some geometric intuition for (X), if there is one (the homological construction of the Bureau representation is very natural to me, but in the definition of the Markov function (X) I'm struggling to see the motivation - I'd like to "see" that this definition is the right one in order to get a Markov function, without just doing a huge calculation).
I know that this is not a very precise question, but I'd just like to hear about what do you think is the "right" way to think about and motivate the study of the Hecke algebra.
Thank you.

REPLY [6 votes]: There's a more old-fashioned way to see the connection between the Burau representation and the Alexander polynomial. The Burau representation of a braid is the action on the first homology of the punctured disk with local coefficients. The Alexander polynomial of a knot is an invariant of the first homology of the knot complement with local coefficients. When you close up the braid, each element of homology of the punctured disk on the bottom becomes identified with its image in the punctured disk at the top. Thus you're killing all the columns in the matrix $\psi^r_n(\beta) - I$. The actual proof uses some long exact sequences, and I remember it as being a bit fiddly. It's probably in Birman and/or Rolfsen's books.
This doesn't (yet?) generalize to other quantum invariants, so it might not be the "right" way to think about it.<|endoftext|>
TITLE: Character table does not determine group Vs Tannaka duality
QUESTION [42 upvotes]: From the example $D_4$, $Q$, we see that the character table of a group doesn't determine the group up to isomorphism. On the other hand, Tannaka duality says that a group $G$ is determined by its representation ring $R(G)$.
What is the additional information contained in $R(G)$ as opposed to the character table?

REPLY [4 votes]: An  RWTH Aachen thesis by Helena Skrzipczyk [1992]  is referred to by Eick and J. Mu"ller
J. ALg 304 (2006) On Brauer pairs. She gives several examples of non-isomorphic groups with
bijections between the character tables and power maps. They are the smallest order Brauer
pairs assuming such to be p-groups.<|endoftext|>
TITLE: any linear algebraic group rational?
QUESTION [9 upvotes]: Somewhere in Mumford's GIT, he seems to imply that any linear algebraic group is rational? This seems strange to me. Is it true?

REPLY [7 votes]: This is really just Pete Clark's answer -- the new bit is to note that
the Levi decomposition isn't needed.
Let G be a (reduced, connected) linear algebraic group over an
alg. closed k, and let R be the unipotent radical of G. Choose a Borel
group B of G with unipotent radical U < B (so R < U).
There is a dense B-orbit ("big cell") V in G/B which is a rational
variety.
Since U and R are (split) unipotent, [Springer, LAG 14.2.6] shows that there
is a section $s:U/R \to U$ to the natural projection $U \to U/R$.
If $f:G \to G/B$ is the quotient mapping, using $s$ you can find a "local
section" of $f$ over the big cell V. 
This show that $f$ is a locally trivial B-bundle (in the Zariski
topology) and in particular $f^{-1}(V)$ is an open subvariety of G
isomorphic to the rational variety V x B.<|endoftext|>
TITLE: Preschemes and schemes
QUESTION [8 upvotes]: This is a very minor point, but one which had been grating me for a while. I apologize for asking a relatively trivial question, but nevertheless hope that it is suitable for MO since it should have a definite answer.
In Mumford's books, for instance Curves on Surfaces or Red Book, there is thing called "prescheme" which looks like a scheme, and scheme is something else.
But this terminology does not seem to be used elsewhere, and if at all is the case, prescheme seems to be something cruder than scheme.
I will be grateful for clarifications regarding this terminology. "Curves on surfaces" is a nice book, but whenever I pick it up I find myself wondering about this without any avail.

REPLY [9 votes]: In the 1971 edition of EGA (this is a revised version of the original 1960 EGA) you can find the following remark in the introduction:
Signalons enfin, par rapport à la première édition, un changement
important de terminologie: le mot «schéma» désigne maintenant ce qui
était appelé «préschéma» dans la première édition, et les mots «schéma
séparé» ce qui était appelé «schéma».
The 1971 terminology should be standard today.<|endoftext|>
TITLE: What proof of quadratic reciprocity is Hilbert referring to in this quote?
QUESTION [18 upvotes]: Let $(a, b)_v$ denote the Hilbert symbol on the completion $K_v$ of a global field $K$ at a place $v$.  The Hilbert reciprocity law $\prod_v (a, b)_v = 1$ is a strict generalization of quadratic reciprocity, to which it reduces in the case $K = \mathbb{Q}, a = p, b = q$.  Hilbert had this to say about his law:
The reciprocity law... reminds [sic] the Cauchy integral theorem, according to which the integral of a function over a path enclosing all of its singularities always yields the value $0$.  One of the known proofs of the ordinary quadratic reciprocity law suggests an intrinsic connection between this number-theoretic law and Cauchy's fundamental function-theoretic theorem.
(I am working off of a translation here.)  Does anyone have any idea what proof Hilbert could be referring to?

REPLY [10 votes]: $\def\FF{\mathbb{F}}$I'm just guessing, but I would have thought it was the following: Hilbert reciprocity for function fields can be deduced from Weil reciprocity. Weil reciprocity is the following statement: Let $X$ be a complete curve over an algebraically closed field $k$. For any point $x \in X$ and nonzero meromorphic functions $f$ and $g$, define $(f,g)_x = (-1)^{(\mathrm{ord}_x f)(\mathrm{ord}_x g)}(f^{\mathrm{ord}_x g}/g^{\mathrm{ord}_x f})(x)$. Then $\prod_{x \in X} (f,g)_x=1$.
See here and here for the connection.
Now, over $\mathbb{C}$, we can prove Weil reciprocity as follows: Choose a path $\delta$ connecting $0$ to $\infty$ in $\mathbb{CP}^1$ and avoiding the critical values of $f$. For simplicity, let us assume $f$ has simple zeroes and poles $\zeta^{\pm}_1$, $\zeta^{\pm}_2$, ..., $\zeta^{\pm}_n$. Set $\gamma = f^{-1}(\delta)$. Then $\gamma$ is the union of $\deg(f)$ closed line segments. After reordering, we may assume  $\zeta^+_i$ is joined to $\zeta^-_i$, say by $\gamma_i$. 
We can define $\log(f)$ on $X \setminus \gamma$, by composing $f$ with a branch of $\log$ on $\mathbb{CP}^1 \setminus \delta$. The differential form $\omega:= \tfrac{1}{2 \pi i} \log(f) \tfrac{dg}{g}$ therefore makes sense on $X \setminus (\gamma \cup g^{-1}(\{ 0,\infty \}))$. If we integrate $\omega$ on little contours around the zeroes and poles of $g$, we get $\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x))$. 
On the other hand, if we integrate around a tubular neighborhood of $\gamma_i$, we pick up $\int_{\gamma_i} \tfrac{dg}{g} = \log(g(\zeta^{+}_i) - \log(g(\zeta^-_i))$ for some branch of $\log$. Summing on $i$, this is
$\sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$
The sum of the contours around the zeroes of $f$ is homologous to the sum over the neighborhoods of the $\gamma_i$, so we deduce
$$\sum_{x \in X} \mathrm{ord}_x(g) \log(f(x)) = \sum_{x \in X} \mathrm{ord}_x(f) \log(g(x))$$
and exponentiating gives the result.<|endoftext|>
TITLE: A binomial sum is divisible by p^2
QUESTION [18 upvotes]: This is a question I have since longer time, but I have absolutely no idea how to proceed on it.
Let $p>3$ be a prime. Prove that $\displaystyle\sum\limits_{k=1}^{p-1}\frac{1}{k}\binom{2k}{k}\equiv 0\mod p^2$. Here, we work in $\mathbb{Z}_{\mathbb{Z}\setminus p\mathbb{Z}}$ (that is, $\mathbb{Z}$ localized at all numbers not divisible by $p$).
I know that it is $0\mod p$ (though I can't find the reference at the moment; it was some hard olympiad problem on MathLinks). The $0\mod p^2$ assertion is backed up by computation for all $p<100$. I am sorry if this is trivial or known. I would be delighted to see a combinatorial proof (= finding a binomial identity which reduces to the above when computed $\mod p^2$). Some number-theoretical arguments would be nice, too. However, I fear that if you use analytic number theory, I will not understand a single word.
EDIT: Epic fail at question title fixed.

REPLY [4 votes]: For the asymptotics: begin with the well-known generating function
$$ \sum_{k \ge 0} {2k \choose k} z^k = (1-4z)^{-1/2}. $$
We can remove the first term to get
$$ \sum_{k \ge 1} {2k \choose k} z^k = (1-4z)^{-1/2} - 1. $$
Dividing by $z$, and then integrating with respect to $z$, turns $z^k$ into $z^k/k$.
$$ \sum_{k \ge 1} {2k \choose k} {1 \over k} z^k = -2 \log (1 + \sqrt{1-4z}) + 2 \log 2. $$
(The $2 \log 2$ is a constant of integration.)  
Finally, let $f(n) = \sum_{k=1}^n {1 \over k} {2k \choose k}$.  These are the partial sums of the previous sequence, so we have
$$ F(z) = \sum_{n \ge 1}  f(n) z^n = {-2 \over 1-z} \log( 1 + \sqrt{1-4z} ) $$
We can find the asymptotics of the coefficients of this generating function by singularity analysis.  The singularity closest to the origin is at $z = 1/4$.  Near $z = 1/4$,
$$ F(z) \sim {-8 \over 3} \sqrt{1-4z} $$
and by a transfer theorem (probably originally due to Flajolet and Odlyzko and in the Flajolet-Sedgewick book Analytic Combinatorics, although it's too late to look up the details right now) we get
$$ f(n) = [z^n] F(z) \sim {-8 \over 3} [z^n] \sqrt{1-4z} $$
Finally, $[z^n] \sqrt{1-4z} = -{2 \over n} {2n-2 \choose n-1}$, and so by Stirling's formula, as $n \to \infty$
$$ f(n) \sim {-8 \over 3} {-2 \over n} {4^{n-1} \over \sqrt{\pi n}} = {4 \over 3} {4^n \over \sqrt{\pi n^3}}. $$
Finally, because I was somewhat awkward in my indexing, we have to take $p = n-1$, and so
$$ f(p) \sim {1 \over 3} {4^p \over \sqrt{\pi p^3}} $$.
As has been observed by John Mangual and Mariano Suárez-Alvarez, this is about one-third of the $p$th Catalan number.  Examining the singularity near $z = 1/4$ more closely would lead to higher-order terms.<|endoftext|>
TITLE: algorithm for calculating the Chow groups of a variety over a finite field
QUESTION [9 upvotes]: Is there an algorithm for calculating the Chow groups of a variety over a finite field?
It is know that $H^{2i,i}_\mathrm{mot}(X,\mathbf{Z}) = CH^i(X)$. In how many cases does this help us?

REPLY [6 votes]: I am not an expert, but let me point out that computing $CH^0(X)$ (which is freely generated by the irreducible components) is already quite hard. Algorithms do exist in this case, see page 206 of "Ideals, varieties and algorithms" by Cox, Little, O'Shea for references. I know of no way to compute the class groups (which can be identified with $CH^1(X)$ for smooth $X$) in general, but I will be very interested in what other people have to say about this.      
Of course, in special situations, more is known. For example, the total Chow group of  quadric hypersurfaces  (at least up to tensoring with $\mathbb Q$).<|endoftext|>
TITLE: Are most cubic plane curves over the rationals elliptic?
QUESTION [29 upvotes]: %This is a new version of the original question modified in the light of the answers and comments.
The word 'most' in the title is ambiguous. The following is one way of making it precise.
Question1: (This seems to be open. See Poonen's answer below)
A cubic projective curve over $\mathbb{Q}$ is given by ten relatively prime  integers (the coefficients of its equation after clearing the denominators). Suppose we take a ten dimensional box $[-N,N]^{10}$ and choose points with integer coordinates with respect to the uniform measure and form the equation of the associated cubic curve. Suppose the number of points which give rise to a curve with a rational point is $E(N)$. Then what can we say about $E(N)/(2N+1)^{10}$ as $N\rightarrow \infty$?
Should the limit exist and if it does, should it be one, zero, or some other number?
Another question of interest is:
Question 2: (There is a satisfactory answer to this. See Voloch's response below.)
Are either of the sets {cubics with no rational point} and {cubics with at least one rational point} Zariski dense?

REPLY [12 votes]: Manjul Bhargava has answered this question yesterday : 
A positive proportion of plane cubics fail the Hasse principle
Manjul Bhargava
(Submitted on 5 Feb 2014)
When all ternary cubic forms over $\mathbf{Z}$ are ordered by the heights of their coefficients, we show that a positive proportion of them fail the Hasse principle, i.e., they have a zero over every completion of $\mathbf{Q}$ but no zero over $\mathbf{Q}$. We also show that a positive proportion of all ternary cubic forms over $\mathbf{Z}$ nontrivially satisfy the Hasse principle, i.e., they possess a zero over every completion of $\mathbf{Q}$ and also possess a zero over $\mathbf{Q}$. Analogous results are proven for other genus one models, namely, for equations of the form $z^2=f(x,y)$ where $f$ is a binary quartic form over $\mathbf{Z}$, and for intersections of pairs of quadrics in $\mathbf{P}_3$.<|endoftext|>
TITLE: Extension of induced reps over Z: is it a sum of induced reps?
QUESTION [11 upvotes]: Let $G$ be a finite group. If $L$ is a finite free $\mathbf{Z}$-module with an action of $G$, say $L$ is induced if it's isomorphic as a $G$-module to $Ind_H^G(\mathbf{Z})$ with $H$ a subgroup of $G$ and $H$ acting trivially on $\mathbf{Z}$. And, for want of better terminology, let's say $L$ is a sum-of-induceds if it's isomorphic to a direct sum of induced modules in the sense above. [EDIT: Ben Webster points out that "permutation representation" is a rather better name for this notion! It's just the $\mathbf{Z}$-module coming from the action of $G$ on a finite set.]
The question: Is a finite free $\mathbf{Z}$-module which is an extension of one sum-of-induceds by another, also a sum-of-induceds?
Over $\mathbf{Q}$ this is trivial because every short exact sequence splits. But this is not true over $\mathbf{Z}$. For example there are two actions of $\mathbf{Z}/2\mathbf{Z}$ on $\mathbf{Z}^2$ which become the group ring over $\mathbf{Q}$: one has the non-trivial element acting as $(1,0;0,-1)$ and the other has it acting as $(0,1;1,0)$. The latter is a non-split extension of the trivial 1-d representation by the non-trivial one (and also a non-split extension of the non-trivial one by the trivial one). Note that this non-split extension is induced.
There are mod $p$ extensions that don't split either. For example over $\mathbf{Z}/p\mathbf{Z}$ there is a non-trivial extension of the trivial representation by itself. But this extension does not lift to an extension of the trivial $\mathbf{Z}$-module by itself.
Why am I interested? For those that know what a $z$-extension of a connected reductive group over a number field is, my "real" question is: is a $z$-extension of a $z$-extension still a $z$-extension? I've checked the geometric issues here but the arithmetic one above is the one I haven't resolved. $G$ is a Galois group and the $\mathbf{Z}$-modules are the character groups of the central tori in question. If I've understood things correctly, a $z$-extension of a $z$-extension is a $z$-extension iff the question I ask above has a positive answer.
Note finally that applying the long exact sequence of cohomology, and using the fact that induced representations have no cohomology by Shapiro's Lemma, we see that the extension I'm interested in also has no cohomology (and furthermore its restriction to any subgroup has no cohomology either). Is this enough to show it's induced?

REPLY [12 votes]: In my interpretation of your description, a sum-of-induceds is just the permutation representation of some (not necessarily connected) G-set. The representation $\mathrm{Hom}(V_1,V_2)$ for two permutation representations is itself permutation: it's the permutation action on product of the G-sets.
So, $\mathrm{Ext}^1(V_1,V_2)=H^1(G,\mathrm{Hom}(V_1,V_2))$ which you claim vanishes.  So it sounds to me like every extension between permutation representations splits. [EDIT: I'm no longer nervous] I'm still a little nervous about this cohomology vanishing, but I don't know integral representation theory so well, so it's hard for me to say.<|endoftext|>
TITLE: Cohomology of fibrations over the circle: how to compute the ring structure?
QUESTION [20 upvotes]: This question is inspired by Cohomology of fibrations over the circle Moreover, it can be considered a subquestion of the above, but somehow it seems to me that some of the more interesting points were not addressed there. So I decided to ask a bit more specific question to emphasize some of those points, but I would not mind at all if someone merges this question with the above.
Let $E\to S^1$ be a fiber bundle with fiber $F$, and assume we know $H^{\bullet}(F,\mathbf{Q})$ as a ring and the monodromy action on it. Notice that since the base is a circle, the Leray spectral sequence degenerates in the second term for dimension reasons. So we have an exact sequence $0\to I\to H^{\bullet}(E,\mathbf{Q})\to Q\to 0$ where $I$ is the kernel and $Q$ is the image of $H^*(E,\mathbf{Q})\to H^{\bullet}(F,\mathbf{Q})$. In this sequence we know $Q,I$ and the action of $Q$ on $I$ from the Leray spectral sequence.

Does this suffice to determine the rational cohomology of $E$ as a ring, up to isomorphism? My guess is that probably not, but I can't find a counter-example off hand.
If not, would the higher Massey products on $H^*(F,\mathbf{Q})$ allow one to compute the cup product on $H^{\bullet}(E,\mathbf{Q})$?
If not, would a rational homotopy model $A$ of the fiber suffice, together with an automorphism $A\to A$ that covers up to homotopy the "monodromy" automorphism of the differential forms on $F$? My guess is that probably yes, but notice that computing models of fibrations with non-simply connected bases can be tricky in general: take for example the space $X$ of all (ordered) couples of distinct points of the real projective plane and the projection on the first factor: the fiber is a M\"obius band, which contracts to the circle and the monodromy action changes the sign of the generator in degree 1; but we have $H^i(X,\mathbf{Q})=\mathbf{Q}$ if $i=0,3$ and zero otherwise.

REPLY [15 votes]: Here's a high dimensional "no" to 1) and 2).   Let $M = S^2 \times S^3$. There is an orientation preserving diffeomorphism of $f \colon M \cong M$ which induces the identity on all integral homology groups.  However the map on $\pi_3(M) \cong \mathbb{Z} \oplus \mathbb{Z}$ is non-trivial - it is "triangluar" with the Hopf map in a corner and ones on the diagonal.  If you take the mapping torus of this diffeomorphism, $T_f$, it is an oriented $6$-manifold:  however there is a class $z \in H^2(T_f; \mathbb{Z})$ such that $z^3 \in H^6(T_f, \mathbb{Z})$ is a non-zero multiple of the fundamental class of $T_f$.<|endoftext|>
TITLE: CW-structures and Morse functions: a reference request
QUESTION [19 upvotes]: The following is probably well known, but I wasn't able to locate a reference in the literature.

Let $f$ be a Morse function on a smooth compact manifold $M$ without boundary and let $\rho$ be a Riemannian metric on $M$. As explained in Milnor's Morse theory and many other sources, starting from $f$ and $\rho$ we can construct a CW-complex $M'$ homotopy equivalent to $M$. However, it seems natural to ask whether $f$ gives a CW-structure on $M$ itself, say, such that the corresponding cellular chain complex is isomorphic to the cellular chain complex of $M'$. Is there a reference for that (preferably, one that contains detailed proofs)?
For a generic choice of the couple $(\rho,f)$ one can construct a chain complex (which I believe is called the Morse complex and) which computes the homology of $M$. What is the standard reference for that? This is implicitly done in Milnor's h-cobordism book, chapter 7. Is it true that the Morse complex is isomorphic to the cellular chain complex of $M'$ from question 1?

upd: the original version of the posting contained some very wrong claims and had to be rewritten.
upd1: restored part of question 2 from the original posting. I deleted it thinking it would be trivial, but it seems that it isn't.

REPLY [3 votes]: I have not read it (It is on my ever-growing todo list), but the paper
Qin, Lizhen(1-WYNS)
On moduli spaces and CW structures arising from Morse theory on Hilbert manifolds.
J. Topol. Anal. 2 (2010), no. 4, 469–526.
58E05 (37D15 57R19)

should contain the proofs of what you want. From the Mathscinet review of D. Hurtubise

This paper contains precise statements and careful proofs of several
  essential results that are fundamental to the moduli space approach to
  Morse theory. Most of the results in this paper have appeared and/or
  been used in other papers, but this is the first self-contained
  reference that provides clear and complete proofs of all of the
  following: (1) the smooth structures on the compactified spaces that
  arise from the gradient flow of a Morse-Smale function, (2)
  orientation formulas for the strata of the compactified spaces, and
  (3) the CW structure determined by the unstable manifolds of a
  Morse-Smale function. The results are proved for a Morse function on a
  complete Hilbert manifold that satisfies the Palais-Smale Condition
  (C) and has finite index at each critical point (the CF case).

Of course this also proves the CW structure in the finite dimensional case.<|endoftext|>
TITLE: Why the similarity between Hodge theory for compact Riemannian and complex manifolds?
QUESTION [16 upvotes]: I'm aware to varying extents of the existence of certain decompositions of the space of $k$-forms on a compact complex or compact Riemannian manifold that split into closed, co-closed, and harmonic forms, and that the space of harmonic forms becomes isomorphic to the de Rham cohomology groups.  However, these are defined differently, but there seems to be analogy here in the theorems and to some extent in the proofs; this is also suggested by the common name.  Is there one?  Or is there a more general way to encapsulate all this?

REPLY [14 votes]: For any Riemannian (in particular, for any Hermitean) manifold there is the decomposition: all forms are the direct sum of the harmonic ones, exact ones and those in the image of the conjugate operator of the de Rham differential. Every de Rham cohomology class is represented by a unique $d$-harmonic form.
For any complex (Hermitean) manifold there is a similar theory for the $\bar\partial$ operator, and we get a similar decomposition for each complex $({\cal E}^{p,\bullet},\bar\partial)$ where ${\cal E}^{p,q}$ stands for complex valued smooth $(p,q)$-forms. See e.g. Chern, Complex manifolds. Every Dolbeault cohomology class is represented by a unique $\bar\partial$-harmonic form.
For general complex hermitean manifolds the above decompositions have nothing to do with one another. However, if the metric is Kaehler, some miracles happen:

the Laplacians of $d$ and of $\bar\partial$ coincide (more precisely, one is twice the other), so the spaces of harmonic forms coincide as well. 
the $(p,q)$-projection of a $d$-harmonic form is again $d$-harmonic. Indeed, a local check shows that the $(p,q)$ projection operator commutes with the $d$-Laplacian. So each of the $(p,q)$ components of a $d$-harmonic form is closed and none is exact (if non-zero). So the $(p,q)$-decomposition of forms gives a $(p,q)$-decomposition of the cohomology classes (this does not exist for general complex manifolds). This is the Hodge decomposition.
since the conjugate of a $d$-harmonic form is again $d$-harmonic, the $(p,q)$ and the $(q,p)$ parts of the Hodge decomposition are conjugate to one another.

As a consequence of the above, the Hodge-to-de Rham spectral sequence degenerates in the first term. This implies that the $dd^c$-lemma holds for Kaehler manifolds, and hence, they are formal. See Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kaehler manifolds.
The cohomological Hodge decomposition holds also for arbitrary bimeromorphically Kaehler compact complex manifolds (in particular, for smooth, complete but not necessarily projective complex algebraic varieties). See e.g. Peters, Steenbrink, Mixed Hodge structures, p. 49.<|endoftext|>
TITLE: Galois cohomology of linear groups over local fields
QUESTION [5 upvotes]: Let $F$ be a local field of characteristic zero (for simplicity), $\overline{F}$ an algebraic closure of $F$ and $L/F$ a fixed finite Galois extension. If $G$ is a linear algebraic group defined over $F$, then the Galois cohomology group $H^1(F,G)$ can be defined as a direct limit of $H^1(K/F,G)$, where $K$ runs through finite Galois subextensions of $\overline{F}$.
Now the question is: under what conditions is this direct limit just $H^1(L/F,G)$?
I guess there might be restrictions on both $L$ and $G$.

REPLY [4 votes]: @Brian Conrad: The sequence $$1 \rightarrow \mu \rightarrow Z' \times \mathcal{G} \rightarrow G' \rightarrow 1$$ is not exact in general. Namely, the kernel $\nu$ of the map
$Z' \times \mathcal{G} \rightarrow G'$ can be bigger than $\mu$. For example, if $G=G'=GL_n$, then $\mathcal{G}= \mathcal{D}(G')=SL_n$, hence $\mu=1$, while $\nu=\mu_n$, the group of roots of unity of order dividing $n$.
Excuse me for posting this as an answer. As a new user, I am not permitted to post a comment.<|endoftext|>
TITLE: what is an Euler system and the motivation for it?
QUESTION [22 upvotes]: I tried to read the definition of it on Rubin's book "Euler systems" but it looks highly technical. Can anyone shed some light on it? In particular, is there some starting examples?
The wiki entry is too short and does not contain much useful information.
And what is the motivation of such objects? In particular, why is it called "Euler"?
And if possible, can someone say something about how it is used in number theory? such as Iwasawa theory?

REPLY [5 votes]: These are good answers.  I would just like to add that while the applications of Euler systems have been mainly p-adic, they are actually motivic (ie units if the motive is h0(Spec F)).  One might hope that if one is able to attach an L-function to a geometric object, there is also an Euler system living in an appropriate motivic cohomology.  A sort of cohomological Euler product if you will.  At least that is motivation for the sweeping special values conjectures such as the Tamagawa Number Conjecture.<|endoftext|>
TITLE: Looking for book with good general overview of math and its various branches
QUESTION [5 upvotes]: Is there something in the order of a Goedel Escher Bach type book? If you've read it you know what I mean. Something compelling that you have to read a couple of times in order to start to get it, but it's so interesting you can't put it down!

REPLY [6 votes]: Mathematics: Its Content, Methods and Meaning is an excellent overview of the full body of mathematics.  It is large (3 volumes), but comes in a paperback edition that includes all three.
The draw is that it is edited by three well-known Russian mathematicians (Aleksandrov, Kolmogorov, Lavrentev) who wrote some of the articles and solicited the rest from many other Russian luminaries.  It was developed as a compendium able to communicate both the vibrancy as well as the importance of each of the areas of the mathematics so that science ministers in Russia could better understand mathematics as mathematicians do. 
The translation into English is excellent.
The first article, a General View of Mathematics, is highly recommended from a philosophical, historical, and phenomenological point of view.<|endoftext|>
TITLE: What notions of universe does predicative type theory admit?
QUESTION [6 upvotes]: Palmgren (1997), On universes in type theory, discusses work of several theorists that provide what we might call a family of Large Universe Axioms (LUAs) for predicative type theory, culminating in Rathjen's MLF_w that corresponds to Kripke-Platek set theory with a weakened epsilon-induction principle.  He also provides a criterion for recognising predicative universe-forming operations: does it come equipped with an induction rule?
That work and work since has been suggestive of a partial analogy between LCAs in set theory (specifically KP set theory) and LUAs in type theory.  This is provocative and interesting, but what I have read has left me unclear about how productive the analogy is. How well-founded is the analogy?  What are its limits?

REPLY [4 votes]: The analogy between universes in type theory and the Mahlo hierarchy in set theory has been analyzed in many different ways by Michael Rathjen. (This builds on his analysis of KPM, but ML type theories with universes came in later in the game.)
I don't have the Palmgren paper you refer to, but I think the following paper is closely related to your questions:
Rathjen, Griffor, Palmgren, Inaccessibility in constructive set theory and type theory, Ann. Pure Appl. Logic 94 (1998), 181-200.
This is not the only way of relating universes in type theory and inaccessibles in set theory, another one is presented by Anton Setzer in Extending Martin-Löf type theory by one Mahlo-universe and further investigated by Rathjen in Realizing Mahlo set theory in type theory.<|endoftext|>
TITLE: Is Higher K-functor the derived functor of K0?
QUESTION [15 upvotes]: It might be a stupid question. I wonder whether the derived functor of functor K0 is Quillen Higher K-functor?
If not, is there any relationship between derived functor of K0(or satellites of K0-functor) and Quillen Higher K-functor?
The motivation to ask this question is if this statement holds, then Quillen higher K-functor is universal.

REPLY [4 votes]: The algebraic $K$-groups of a commutative unital ring can indeed be defined as derived functors, but one needs to work in the context of non-abelian homological algebra in the sense of A. Dold, D. Puppe, Homologie nicht-additiver Funktoren, Ann. Inst. Fourier 11 (1961), and
M. Tierney, W. Vogel: Simplicial resolutions and derived functors, Math. Zeit. 111 (1969).
A useful introduction is given in the book `Non-abelian homological algebra and its applications' by Hvedri Inassaridze (Kluwer, 1997). 
For a group $G$, define $\displaystyle Z_\infty (G)=\lim_{\leftarrow} G/\Gamma_i(G)$, 
where $\{\Gamma_i(G)\}$ is the lower central series of $G$. This defines a functor $Z_\infty: Gr\rightarrow Gr$.
Now Theorem 5.1 in the cited book roughly reads as follows:
Let $L_*Z_\infty$ be the left derived functors of the functor $Z_\infty$. Then $L_i Z_\infty(GL(R))$ is isomorphic to Quillen's $K_i(R)$.<|endoftext|>
TITLE: What is the Hirzebruch-Riemann-Roch formula for the flag variety of a Lie algebra?
QUESTION [27 upvotes]: If we have a finite dimensional Lie algebra g, then the flag variety of g is a projective scheme.
My question is what is Hirzebruch-Riemann-Roch formula for this projective scheme? Are there any interesting results? 
I wonder know whether Riemann-Roch in this setting have some beautiful representation theory explanations.
Thanks

REPLY [52 votes]: Riemann-Roch for the flag variety is the Weyl Character formula!
More specifically, let $L$ be an ample line bundle on $G/B$, corresponding to the weight $\lambda$. According to Borel-Weil-Bott, $H^0(G/B,L)$ is $V_{\lambda}$, the irreducible representation of $G$ with highest weight $\lambda$, and $H^i(G/B,L)=0$ for $i>0$. So the holomorphic Euler characteristic of $L$ is $\mathrm{dim} \ V_{\lambda}$.
As we will see, computing  the holomorphic Euler characteristic of $L$ by Hirzebruch-Riemann-Roch gives the Weyl character formula for $\mathrm{dim} \ V_{\lambda}$.
 Notation:
$G$ is a simply-connected semi-simple algebraic group, $B$ a Borel and $T$ the maximal torus in $B$. The corresponding Lie algebras are $\mathfrak{g}$, $\mathfrak{b}$, $\mathfrak{t}$. The Weyl group is $W$, the length function on $W$ is $\ell$ and the positive roots are $\Phi^{+}$. It will simplify many signs later to take $B$ to be a lower Borel,  so the weights of $T$ acting on $\mathfrak{b}$ are $\Phi^{-}$.
We will need notations for the following objects:
$$\rho = (1/2) \sum_{\alpha \in \Phi^{+}} \alpha.$$
$$\Delta = \prod_{\alpha \in \Phi^{+}} \alpha.$$
$$\delta = \prod_{\alpha \in \Phi^{+}} (e^{\alpha/2}-e^{-\alpha/2}).$$
They respectively live in $\mathfrak{t}^*$, in the polynomial ring $\mathbb{C}[\mathfrak{t}^*]$ and in the power series ring $\mathbb{C}[[\mathfrak{t}^*]]$.
Geometry of flag varieties
Every line bundle $L$ on $G/B$ can be made $G$-equivariant in a unique way. Writing $x$ for the point $B/B$, the Borel $B$ acts on the fiber $L_x$ by some character of $T$. This is a bijection between line bundles on $G/B$ and characters of $T$. Taking chern classes of line bundles gives classes in $H^2(G/B)$. This extends to an isomorphism $\mathfrak{t}^* \to H^2(G/B, \mathbb{C})$ and a surjection $\mathbb{C}[[\mathfrak{t}^*]] \to H^*(G/B, \mathbb{C})$.  We will often abuse notation by identifiying a power series in $\mathbb{C}[[\mathfrak{t}^*]]$ with its image in $H^*(G/B)$.
We will need to know the Chern roots of the cotangent bundle to $G/B$.
Again writing $x$ for the point $B/B$, the Borel $B$ acts on the tangent space $T_x(G/B)$ by the adjoint action of $B$ on $\mathfrak{g}/\mathfrak{b}$. As a $T$-representation, $\mathfrak{g}/\mathfrak{b}$ breaks into a sum of one dimensional representations, with characters the positive roots. We can order these summands to give a $B$-equivariant filtration of $\mathfrak{g}/\mathfrak{b}$ whose quotients are the corresponding characters of $B$. Translating this filtration around $G/B$, we get a filtration on the tangent bundle whose associated graded is the direct sum of line bundles indexed by the positive roots. So the Chern roots of the tangent bundle are $\Phi^{+}$. (The signs in this paragraph would be reversed if $B$ were an upper Borel.)
The Weyl group $W$ acts on $\mathfrak{t}^*$. This extends to an action of $W$ on $H^*(G/B)$. The easiest way to see this is to use the diffeomorphism between $G/B$ and $K/(K \cap T)$, where $K$ is a maximal compact subgroup of $G$; the Weyl group normalizes $K$ and $T$ so it gives an action on $K/(K \cap T)$.
We need the following formula, valid for any $h \in \mathbb{C}[[\mathfrak{t}^*]]$:
$$\int h = \ \mbox{constant term of}\left( (\sum_{w \in W} (-1)^{\ell(w)} w^*h)/\Delta \right). \quad (*)$$
Two comments: on the left hand side,  we are considering $h \in H^*(G/B)$ and using the standard notation that $\int$ means "discard all components not in top degree and integrate." On the right hand side, we are working in $\mathbb{C}[[\mathfrak{t}^*]]$, as $\Delta$ is a zero divisor in $H^*(G/B)$.
Sketch of proof of (*): The action of $w$ is orientation reversing or preserving according to the sign of $\ell(w)$. So $\int h = \int (\sum_{w \in W} (-1)^{\ell(w)} w^*h) / |W|$. Since the power series $\sum_{w \in W} (-1)^{\ell(w)} w^*h$ is alternating, it is divisible by $\Delta$ and must be of the form $\Delta(k + (\mbox{higher order terms}))$ for some constant $k$. The higher order terms, multiplied by $\Delta$, all vanish in $H^*(G/B)$, so we have $\int h = k \int \Delta/|W|$. The right hand side of $(*)$ is just $k$.
By the Chern root computation above, the top chern class of the tangent bundle is $\Delta$. So $\int \Delta$ is the (topological) Euler characteristic of $G/B$. The Bruhat decomposition of $G/B$ has one even-dimensional cell for every element of $W$, and no odd cells, so $\int \Delta = |W|$ and we have proved formula $(*)$.
The computation
We now have all the ingredients. Consider an ample line bundle $L$  on $G/B$, corresponding to the weight $\lambda$ of $T$. The Chern character is $e^{\lambda}$. HRR tells us that the holomorphic Euler characteristic of $L$ is
$$\int e^{\lambda} \prod_{\alpha \in \Phi^{+}} \frac{\alpha}{1 - e^{- \alpha}}.$$
Elementary manipulations show that this is
$$\int \frac{ e^{\lambda + \rho} \Delta}{\delta}.$$
Applying $(*)$, and noticing that $\Delta/\delta$ is fixed by $W$, this is
$$\mbox{Constant term of} \left( \frac{1}{\Delta} \frac{\Delta}{\delta} \sum_{w \in W} (-1)^{\ell(w)} w^* e^{\lambda + \rho} \right)= $$
$$\mbox{Constant term of} \left( \frac{\sum_{w \in W} (-1)^{\ell(w)} e^{w(\lambda + \rho)}}{\delta} \right).$$
Let $s_{\lambda}$ be the character of the $G$-irrep with highest weight $\lambda$. By the Weyl character formula, the term in parentheses is $s_{\lambda}$ as an element of $\mathbb{C}[[\mathfrak{t}^*]]$. More precisely, a character is a function on $G$. Restrict to $T$, and pull back by the exponential to get an analytic function on $\mathfrak{t}$. The power series of this function is the expression in parentheses. Taking the constant term means evaluating this character at the origin, so we get $\dim V_{\lambda}$, as desired.<|endoftext|>
TITLE: Collection of subsets closed under union and intersection
QUESTION [12 upvotes]: Suppose A is a set and S is a collection of subsets closed under arbitrary unions and intersections. Can we find a collection F of functions from A to itself such that a subset B of A is in S if and only if $f(B) \subseteq B$ for all $f \in F$ (in other words, is S precisely the collection of invariant subsets under a collection of functions)?
P.S.: I don't really know what subject tag to give this, so I'm giving it "combinatorics", which seems the closest, though it is more like a question from lattice theory.

REPLY [15 votes]: The answer is Yes.  Furthermore, such a family can be found of size at most the cardinality of A, even when S is much larger.
The key to the solution is to realize that every such family S arises as the collection of downward-closed sets for a certain partial pre-order on A, which I shall define.  (Conversely, every such order also leads to such a family.)
An interesting special case occurs when the family S is linearly ordered by inclusion. For example, one might consider the family of cuts in the rational line, that is, downward-closed subsets of Q. (I had thought briefly at first that this might be a counterexample, but after solving it, I realized a general solution was possible by moving to partial orders.)
Suppose that S is such a collection of subsets of A. Define the induced partial pre-order on A by 

a <= b     if whenever B in S and b in B, then also a in B. 

It is easy to see that this relation is transitive and reflexive. 
I claim, first, that S consists of exactly the subsets of A that are downward closed in this order. It is clear that every set in S is downward closed in this order. Conversely, suppose that X is downward closed with respect to <=. For any b in X, consider the set Xb, which the intersection of all sets in S containing b as an element. This is in S. Also, Xb consists of precisely of the predecessors of b with respect to <=. So Xb subset X. Thus, X is the union of the Xb for b in X. So X is in S. 
Next, define fa(b) = a if a <= b, and otherwise fa(b) = b. Let F be the family of all such functions fa for a in A. 
Clearly, every B in S is closed under every fa, by the definition of <=. Conversely, suppose that X is closed under all fa. Thus, whenever b is in X and a <= b, then a is in X also. So X is downward closed, and hence by the claim above, X is in S.  
Incidently, the sets S are exactly the open sets in the topology on A induced by the lower cones of <=.<|endoftext|>
TITLE: Good algorithm for finding the diameter of a (sparse) graph?
QUESTION [15 upvotes]: My question on Stack Overflow was recently tagged "math".  Despite a bounty, it never received a satisfactory answer, so I thought I would ask it here:
I have a large, connected, sparse graph in adjacency-list form.  I would like to find the diameter of the graph and two vertices achieving it.
Is there a better approach than computing all-pairs shortest paths?
I am interested in this problem in both the undirected and directed cases, for different applications. In the directed case, I of course care about directed distance (the maximum over pairs of vertices of the length of the shortest directed path from the first vertex to the second).

REPLY [2 votes]: For road networks:
http://arxiv.org/abs/1209.4761<|endoftext|>
TITLE: Applications of Artin's holomorphy conjecture
QUESTION [18 upvotes]: I wonder why the Artin conjecture is so important. The only reason I could figure out is that one could use the holomorphy of Artin L-series and Weil's converse theorem to show modularity of two-dimensional Galois representations.
Are there other reasons?

REPLY [21 votes]: I am waking up an old and already well-answered question, to offer another point of view,
The Artin's conjecture appears very naturally in the context of Chebotarev's density theorem.
In fact, we can see Cheobtarev's contribution as a clever trick to circumvent the Artin conjecture by reducing the proof to cases where it is known (by works of Dirichlet and Hecke).
But the proof of Chebotarev will be much simpler and more natural if we had the Artin's conjecture, which moreover would give better results as far as the error term is concerned.
This is, I think, a good justification of the importance of Artin's conjecture.
To explain the role of Artin's conjecture, let us also assume for simplicity GRH.
Then, for $G=Gal(K/\mathbb Q)$ a finite Galois group, and $\rho$ an irreducible Artin representation of $G$, the $L$-function $L(\rho,s)$ has no zero on Re $s>1/2$ (by GRH)
and no pole either (by Artin), except for a simple pole at $s=1$ if $\rho$ is trivial.
Thus the logarithmic derivative, $L'/L(\rho,s)$ has no pole on Re $s>1/2$ (except perhaps...): this illustrates clearly the symmetric and complementary role played by Artin and Riemann's conjectures; both poles and zeros of $L$ contribute to simple poles of $L'/L$, and Artin eliminates some of them, Riemann the others. Now standard techniques of analytic number theory allows us, by integrating $L'/L$ on a vertical line $2+i \mathbb R$ and moving it near the critical line, to $1/2+\epsilon + i \mathbb R$, to get an estimate of the quantity:
$$\pi(\rho,x) = \sum_{p^n < x} \log(p) tr \rho(frob_p)^n$$
where the sum is on prime power less than $x$.
This estimate is $O(x^{1/2+\epsilon})$ if $\rho$ is non trivial, and $x + O(x^{1/2+\epsilon})$ if $\rho$ is trivial, because of the pole at $s=1$.
Now let $C$ be subset of $G$ stable by conjugaison, $1_C$ its characteristic function.
Since $1_C$ is a central function, it is a linear combinaison of character of irreducible representation of $G$, say
$$1_C = \sum_\rho a_\rho tr \rho.$$
Hence
$\pi(C,x) := \sum_{p^n< x} \log p 1_C(frob_p) = \sum_\rho a_\rho \pi_\rho(x)$.
Since $a_1$ is easily computed as $|C|/|G|$, we get $\pi(C,x)=|C|x/|G| + O(x^{1/2+\epsilon})$, which is up to standard manipulation Chebotarev's density theorem. 
Hence we can say that Artin's conjecture play to Chebotarev's density theorem a role analog to the role played by the standard conjectures for the Weil's conjecture proved by Deligne. In both cases, a clever and beautiful trick was used (by Chebotarev and Deligne, respectively) to prove a theorem
(Chebotarev's theorem, aka Frobenius' conjecture, and Deligne's theorem, aka the last Weil's conjecture) without proving the conjectural statement that makes the theorem limpid (Artin's conjecture, resp. standard conjectures). This is great, but doesn't make the conjectures any less interesting. 
Important addendum A friend of mine made me notice something that kind of weakens significantly the point I was trying to make above. Indeed, for the argument outlined above, one doesn't need GRH + Artin's conjecture: GRH is enough. Or, more precisely, 
the part of Artin's conjecture that is needed is already known under GRH. Indeed, it is clear that in the argument the absence of poles of $L(\rho,s)$ is only used in the region Re $s>1/2$. But by Brauer's theorem, we know that $L(\rho,s)$ is meromorphic on $\mathbb C$,
and by elementary reasoning that $\prod_\rho L(\rho,s) = \zeta_K(s)$, where $\rho$ runs amongst Artin's representations of Gal$(K/\mathbb Q)$. Moreover, by Hecke $\zeta_K$ has no poles (except a simple one at $s=1$). Hence if all the $L(\rho,s)$ have no zero on any given region, then they don't have a polo either on that region -- expected for $L(1,s)$ with its simple pole at $s=1$. 
I leave the argument above, because it shows that, the absence of poles for $L(\rho,s)$ is important for questions on the distribution of primes, even if it is a theorem rather than a conjecture. Moreover, this argument has an historical interest, as Artin's conjecture was made in the 1920's, and Brauer's theorem proved in 1946.
I ignore if the question of absence of poles on the critical line for the $L(\rho,s)$ (the only part of Artin's conjecture that is still open) has any direct application on the
distribution of primes.<|endoftext|>
TITLE: Which lattices have more than one minimal periodic coloring?
QUESTION [10 upvotes]: The lattice $\mathbb{Z}^n$ has an essentially unique (up to permutation) minimal periodic coloring for all $n$, namely the "checkerboard" 2-coloring. Here a coloring of a lattice $L$ is a coloring of the graph $G = (V,E)$ with $V = L$ and $(x,y) \in E$ if $x$ and $y$ differ by a reduced basis element. (NB. I am not quite sure that this graph is the proper one to consider in general, so comments on this would also be nice.)
The root lattice $A_n$ has many minimal periodic colorings if $n+1$ is not prime (I have sketched this here, and some motivation is in the last post in that series); if $n+1$ is prime, then it has essentially one $n+1$-coloring. Two minimal periodic colorings for $A_3$ are shown below (for convenience, compare the tops of the figures):
 
The generic ("cyclic") coloring.
 
A nontrivial example.
The lattices $D_n$ are also trivially 2-colored. 
So: are there other lattices that admit more than one minimal periodic coloring? I'd be especially interested to know if $E_8$ or the Leech lattice do.
(A related question: does every minimal periodic coloring of $A_n$ arise from a group of order $n+1$?)

REPLY [4 votes]: When you say "reduced basis", I assume you mean that two lattice points are connected in your graph if the distance between them is the minimum distance of the lattice (i.e. the shortest distance between any two lattice points).
There is a simple way to generate a $24$ colouring of $E_8$ using the $8$ colouring you have for $A_8$.  It so happens that $E_8$ is isomorphic to the union of 3 translations of $A_8$, 
$E_8 = A_8 + (A_8 + g) + (A_8 + 2g)$
where $g = \left( \tfrac{8}{3}, -\left(\tfrac{1}{3}\right)^8 \right)$.  That is, $g$ is a vector with one $\tfrac{8}{3}$ and eight $-\tfrac{1}{3}$'s. See Martinet, Perfect Lattices in Euclidean Spaces. So you just need 3 independently coloured $A_8$'s.  The resultant colouring will be periodic if the colourings for $A_8$ are periodic.  It's likely that this is not the best colouring possible for $E_8$.<|endoftext|>
TITLE: Unpointed Brown representability theorem
QUESTION [13 upvotes]: The classical Brown Representability Theorem states: Denote $hCW_*$ the homotopy category of pointed CW-complexes. Let $F : hCW_* \to Set_*$ be a contravariant functor. Then $F$ is representable if and only if

$F$ respects coproducts, i.e. $F(\vee_{i \in I} X_i) = \prod_{i \in I} F(X_i)$ for all families $X_i$ of pointed CW-complexes.
$F$ satisfies a sort of Mayer-Vietoris axiom: If $X$ is a pointed CW-complex which is the union of two pointed subcomplexes $A,B$, then the canonical map $F(X) \to F(A) \times_{F(A \cap B)} F(B)$ is surjective1.

What about omitting the base points? So let $F :hCW \to Set$ be a contravariant functor that satisfies the analogous properties as above (replace the wedge-sum by the disjoint union). Is then $F$ representable? I'm not sure if we just can copy the proof of the pointed case (which can be found, e.g., in Switzer's book "Algebraic Topology - Homology and Homotopy", Representability Theorems). For example, $F(pt)$ can be anything (in contrast to the pointed case), it will be the set of path components in the classifying space. Besides, the proof uses homotopy groups and in particular the famous theorem of Whitehead, which deal with pointed CW-complexes. Nevertheless, I hope that $F$ is representable ... what do you think?
As a first step, we may define for every $i \in F(pt)$ the subfunctor $F_i$ of $F$ by $F_i(Y) = \{f \in F(Y) : \forall y : pt \to Y : f|_{y} = i \in F(pt)\}$, which should be thought as the connected component associated to $i$. Then it's not hard to show that $F_i$ satisfies the same properties as $F$ and that $F_i = [-,X_i]$ implies $F = [-,\coprod_i X_i]$. In other words, we may assume that $F(pt)=pt$ (so that the classifying space will be connected).
1 You can't expect it to be bijective, cf. question about categorical homotopy colimits

REPLY [14 votes]: This is a copy of my answer to Brown representability for non-connected spaces which I repost here per request in the comment.
A negative answer to the question can be concluded from this paper:

Peter Freyd and Alex Heller, Splitting homotopy idempotents. II. J. Pure Appl. Algebra 89 no. 1-2 (1993) pp 93–106, doi:10.1016/0022-4049(93)90088-B.

This paper introduces a notion of conjugacy idempotent. It is a triple $(G, g, b)$ consisting of a group $G$, an endomorphism $g \colon G \to G$ and an element $b \in G$ such that for all $x \in G$ we have $g^2(x) = b^{-1} g(x) b$. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent $(F, f, a)$. The Main Theorem of the paper says (among other things) that $f$ does not split in the quotient of the category of groups by the conjugacy congruence.
Now $f$ induces an endomorphism $B f \colon B F \to B F$ which is an idempotent in $\mathrm{Ho} \mathrm{Top}$ and it follows (by the Main Lemma of the paper) that it doesn't split. It is then easily concluded that $(B f)_+ \colon (B F)_+ \to (B F)_+$ doesn't split in $\mathrm{Ho} \mathrm{Top}_*$.
The map $(B f)_+$ induces an idempotent of the representable functor $[-, (B F)_+]_*$ which does split since this is a $\mathrm{Set}$ valued functor. Let $H \colon \mathrm{Ho} \mathrm{Top}_*^\mathrm{op} \to \mathrm{Set}$ be the resulting retract of $[-, (B F)_+]_*$. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is not representable since a representation would provide a splitting for $(B f)_+$.
The same argument with $B f$ in place of $(B f)_+$ shows the failure of Brown's Representability in the unbased case.<|endoftext|>
TITLE: Strict Class Numbers of Totally Real Fields
QUESTION [9 upvotes]: In their paper Computing Systems of Hecke Eigenvalues Associated to Hilbert Modular Forms, Greenberg and Voight remark that
...it is a folklore conjecture that if one orders totally real fields by their discriminant, then a (substantial) positive proportion of fields will have strict class number 1.
I've tried searching for more details about this, but haven't found anything. 
Is this conjecture based solely on calculations, or are there heuristics which explain why this should be true?

REPLY [9 votes]: Maybe it's worth a word about why Cohen-Lenstra predicts this behavior.  Suppose K is a field with r archimedean places.  Then Spec O_K can be thought of as analogous to a curve over a finite field k with r punctures, which is an affine scheme Spec R.  Write C for the (unpunctured) curve.  Then the class group of R is the quotient of Pic(C)(k) by the subgroup generated by the classes of the punctures -- or, what is the same, the quotient of Jac(C)(k) by the subgroup generated by degree-0 divisors supported on the punctures.  (This last subgroup is just the image of a natural homomorphim from Z^{r-1} to Jac(C)(k).)
The Cohen-Lenstra philosophy is that these groups and the puncture data are "random" -- that is, you should expect that the p-part of the class group of R looks just like what you would get if you chose a random finite abelian p-group (where a group A is weighted by 1/|Aut(A)|) and mod out by the image of a random homomorphism from Z^{r-1}.  (There are various ways in which this description is slightly off the mark but this gives the general point.)
It turns out that when r > 1 the chance is quite good that a random homomorphism from Z^{r-1} to A is surjective.  In fact, the probability is close enough to 1 that when you take a product over all p you still get a positive number.  In other words, when r > 1 Cohen-Lenstra predicts a positive probability that the class group will have trivial p-part for all p; in other words, it is trivial.  (In fact, it predicts a precise probability, which fits experimental data quite well.)
When r = 1, on the other hand, the class group is just A itself, and the probability its p-part is trivial is on order 1-1/p.  Now the product over all p is 0, so one does NOT expect to see a positive proportion of trivial class groups.  And in fact, when there is just one archimedean place -- i.e. when K is imaginary quadratic -- this is just what happens!<|endoftext|>
TITLE: Applications of the Brown Representability Theorem
QUESTION [16 upvotes]: Probably you can "google" this question, but I can't find anything relevant. The classical Brown Representability Theorem states: Denote $hCW_*$ the homotopy category of pointed CW-complexes. Let $F : hCW_* \to Set_*$ be a contravariant functor. Then $F$ is representable if and only if

$F$ respects coproducts, i.e. $F(\vee_{i \in I} X_i) = \prod_{i \in I} F(X_i)$ for all families $X_i$ of pointed CW-complexes.
$F$ satisfies a sort of mayer-vietoris-axiom: If $X$ is a pointed CW-complex which is the union of two pointed subcomplexes $A,B$, then the canonical map $F(X) \to F(A) \times_{F(A \cap B)} F(B)$ is surjective.

I just know two applications: classifying spaces for $G$-principal bundles (in particular, vector bundles) for a locally compact topological group $G$ and for generalized cohomology theories on $CW_*$; the yoneda-lemma also yields functorial relations (cf. Switzer, Algebraic Topology). I'm interested in other explicit applications. I've read that there are categorical generalizations, but in this question I'm just asking whether there are explicit functors defined on CW-complexes, whose representabilty is of interest and can be shown with the theorem above. Also, these examples should really differ from the two ones mentioned above. :-)

REPLY [6 votes]: Brown Representability combined with the Landweber Exact Functor theorem allows one to construct homotopy types out of purely algebro-geometric data, and is in particular the starting point for theories such as that of topological modular forms.  Thus, this old theorem underlies one of the key themes which modern algebraic topology is fleshing out.<|endoftext|>
TITLE: Balancing problem
QUESTION [17 upvotes]: There was a problem in an Olympiad selection test, which went as follows: Consider the set $\{1,2,\dots,3n \}$ and partition it into three sets A, B and C of size n each. Then, show that there exist x, y and z, one in each of the three sets, such that x + y = z.
This has a tricky-to-get but otherwise straightforward solution, that starts by assuming 1 to be in A, finding the smallest k not in A, assuming that to be in B, and then arguing that no two consecutive elements can be present in C (for that would give an infinite descent). Finally, cardinality considerations solve the problem.
I managed to prove a corresponding statement for 4n, namely: for $\{ 1,2,3, \dots, 4n \}$, partitioned into four sets of size n each, there exist x, y, z, and w, one in each set, such that x + y = z + w.
The question here is whether analogues of this hold for all m, with $m \ge 3$ and $n \ge 2$. In other words, if $\{ 1,2, \dots, mn \}$ is divided into $m$ sets of size $n$ each, can we always make a choice of one element in each set such that the sum of floor $m/2$ of the elements equals the sum of the remaining ceiling $m/2$ elements ($(m-1)/2$ and $(m + 1)/2$ for $m$ odd, $m/2$ each for $m$ even). Note we need $n \ge 2$ due to parity considerations when $m$ is congruent to $1$ or $2$ modulo $4$.

REPLY [16 votes]: This article is a nice survey of "Rainbow Ramsey theory". In this jargon what you are trying to prove is that the vector $(1,1,\dots,1,-1,-1,\dots,-1)$ is rainbow partition $m$-regular.
The case of $(1,1,-1,-1)$ being rainbow partition 4-regular, was proved in "Rainbow solutions for the sidon equation x+y=z+w", J. Fox, M. Mahdian, and R. Radoicic. They actually proved that as long as each of the four parts of $[n]$ has at least $(n+1)/6$ members then one can always find rainbow solutions to $x+y=z+w$ (i.e. each variable coming from a different partition.)
Though these results were mostly inspired from their monochromatic version (the 3 variable case dates back to Schur, and then R.Rado classified all linear equations that are partition regular), the analogy hasn't proven very faithful. The rainbow Hales-Jewett theorem is false, and so is the rainbow Van der Waerden theorem! 
Another thing worth mentioning is that if we color $\mathbb{Z}/p\mathbb{Z}$ in $k$ colors with each color having at least $k$ elements, then the equation $\sum_{i=1}^k a_i x_i\equiv b\pmod{p}$ always has a rainbow solution given that not all $a_i$ are the same. A proof is in this article by D.Conlon.<|endoftext|>
TITLE: How seriously should a graduate student take teaching evaluations?
QUESTION [17 upvotes]: Pretty much the question in the title.  If a grad student gets bad reviews as a TA, how much does that hurt them later? How much do good reviews help? What if the situation is more complex? (For instance, bad reviews when TAing, but good reviews when actually teaching/lecturing a summer course).
Edit: I asked this question with the situation of a student hoping for a career at research universities in mind, however, I am also interested in other cases.
Edit: In your answer, please mention what your background is: have you served on hiring committees?  Are you reporting just what you've heard?  Were you successful/unsuccessful in a job search and were told that your teaching evals did/did not make a difference?

REPLY [4 votes]: You should keep all your evaluations and keep track of them.
Are you improving over time ?
Are you improving in some areas over time ?
Are you better/worse in some topics than others ?
Take every comment with a grain of salt.
Some students will hate you and/or your teaching style no matter how you teach
either because they hate the topic, hate the class time, hate you, or hate your style
of teaching/grading.
[It is hard to give someone the highest possible evaluation when you get a 'D' in
the class.]
Look for patterns in the comments and accept that you cannot teach in such a way
to please all possible students.
Strive for clarity and helpful pacing.
Even if you end up at a Research University or Lab you will have to either teach
Lectures or Grad Seminars or give talks and the clearer and better paced they are
the better for everyone involved.<|endoftext|>
TITLE: On statements independent of ZFC + V=L
QUESTION [16 upvotes]: Let $V=L$ denote the axiom of constructibility. Are there any interesting  examples of set theoretic statements which are independent of $ZFC + V=L$? And how do we construct such independence proofs? The (apparent) difficulty is as follows: Let $\phi$ be independent of $ZFC + V=L$. We want models of $ZFC + V=L + \phi$ and $ZFC + V=L + \neg\phi$. An inner model doesn't work for either one of these since the only inner model of $ZFC + V = L$ is $L$ and whatever $ZFC$ can prove to hold in $L$ is a consequence of $ZFC + V=L$. Forcing models are of no use either, since all of them satisfy $V \neq L$.

REPLY [8 votes]: What I would call the standard source of examples is the series of very nice "finitary" combinatorial statements that Harvey Friedman has been working on. You can see plenty of such statements in his numbered series of FOM posts, and for details of some of the arguments, see for example his nice paper "Finite Functions and the Necessary Use of Large Cardinals", Annals of Math., Vol. 148, No. 3, 1998, pp. 803-893.
The examples Harvey examines are arithmetic. In the "true" model (which is well-founded) they are decided one way, and there are ill-founded (in fact, not $\omega$-) models where they are decided the other way.
It would be highly desirable to have examples of nice arithmetic statements that are not just independent but for which we can produce differing answers in different well-founded models. I don't think anybody has any clue at the moment on how to do such thing. To achieve this would be akin to the invention of forcing.<|endoftext|>
TITLE: Varieties cut by quadrics
QUESTION [25 upvotes]: Is there a characterization of the class of varieties which can be described as an intersection of quadrics, apart from the taulogical one?
Lots of varieties arise in this way (my favorite examples are the Grassmanianns and Schubert varieties and some toric varieties) and I wonder how far can one go.

REPLY [4 votes]: Let $X\subset\mathbb{P}^N$ be a quadratic smooth variety of dimension $n$ and condimension $c$. Then:

if $n\geq c$ then $X$ is Fano.
If $n\geq c+1$ the $X$ is covered by lines. Let $x$ be a general point of $X$ and let $\mathcal{L}_x\subset\mathbb{P}^{n-1} = \mathbb{P}(T_xX)$ be the variety parametrizing lines in $X$ passing through $x$. Then $\mathcal{L}_x\subset\mathbb{P}^{n-1}$ is scheme theoretically defined by $c$ idependent quadratic equations.
If $n\geq c+2$ then: $X\subset\mathbb{P}^N$ is a complete intersection $\Leftrightarrow$ $\mathcal{L}_x\subset\mathbb{P}^{n-1}$ is a complete intersection $\Leftrightarrow$ $dim(\mathcal{L}_x)=n-c-1$.
(Hartshorne's conjecture) If $n\geq 2c+1$ then $X$ is a scheme theoretical complete intersection.
If $X$ is prime Fano (i.e. $X$ is Fano and $Pic(X) = \mathbb{Z}\left\langle H\right\rangle$) and the Fano index satifies $i(X)\geq\frac{2n+5}{3}$, then $X$ is a complete intersection.

This has been a breakthrough in the direction of Hartshorne's conjecture on complete intersections. Before it was known just for very special varieties and for Fano varieties of codimension two. You can find all of this in this paper: http://arxiv.org/abs/1209.2047.<|endoftext|>
TITLE: Reference for elementary and "cool" statistics or financial math
QUESTION [5 upvotes]: I signed up for a Math Mentorship Program (for high school students) this term, but one of the students assigned to me is more interested in Statistics and Finance - something that would help him to do business :-)
The closest I could come up with (using ideas of a friend) is some game theoretic stuff like  price of anarchy  (a related and possibly with much simpler mathematics is  The Economics of Caste and of the Rat Race and Other Woeful Tales  by Akerlof),  Arrow's Impossibility Theorem, and possibly some  Prospect theory.  
Does anyone have any idea about elementary interesting math related to Statistics or Finance?

REPLY [2 votes]: One topic that I would cover is the St. Petersburg Game (or paradox). Daniel Bernoulli's solution is the basis for expected utility theory and the rivalry between himself and his cousin Nicolas just makes it more fun. (The problem is well-suited to a visual description as well.)
Good luck.<|endoftext|>
TITLE: The current status of the Birch & Swinnerton-Dyer Conjecture
QUESTION [35 upvotes]: [Une traduction française suit la version anglaise.]
The question is only about elliptic curves $E$ over $\mathbb{Q}$ and concerns only the aspect 
(order of vanishing of $L(E,s)$ at $s=1$)$\ =\ $(rank of $E(\mathbb{Q})$).
Let $r$ be the LHS and $d$ the RHS, so that (a special case of ) the Birch and Swinnerton-Dyer Conjecture is
BSD?. $r=d$.
By the end of the last millenium, we knew
Theorem (1977--2000). If $\ r=0,1$, then $d=r$ (and $\ \operatorname{Sha}(E)$ is finite).
Some years ago, I heard that there was some progress in proving $(r>0)\Longrightarrow (d>0)$ under the assumption of the finiteness of $\operatorname{Sha}(E)$.  What is the current status of the
Statement. Suppose that $\operatorname{Sha}(E)$ is finite.  If $r>1$, then $d>0$ ?

L'état actuel de la conjecture de Birch et Swinnerton-Dyer
On s'interesse uniquement aux courbes abéliennes $A$ sur $\mathbf{Q}$ et à
l'aspect 
(ordre d'annulation de $L(A,s)$ en $s=1$)$\ =\ $(rang de $A(\mathbf{Q})$).
Désignons par $r$ le membre de gauche et par $d$ le membre de droite, de sorte
que la conjecture de Birch et Swinnerton-Dyer prédit (en particulier)
BSD? $r=d$.
Vers la fin du millénaire dernier, on avait démontré le
Théorème (1977--2000).  Si $r=0,1$, alors $d=r$ (et $\ \operatorname{Cha}(A)$ est fini). 
Il y a quelques années, j'avais entendu dire qu'on a fait des progrès
concernant l'implication $(r>0)\Longrightarrow(d>0)$ sous l'hypothèse de la
finitude de Cha$(A)$. Quel est l'état actuel de l'
Énoncé. Supposons que $\operatorname{Cha}(A)$ est fini.  Si $r>1$, alors
$d>0$ ?

REPLY [4 votes]: Alice Silverberg has a "cheat sheet" on all things currently known related to the rank of an elliptic curve. The link is below.
http://math.uci.edu/~asilverb/connectionstalk.pdf<|endoftext|>
TITLE: Does War have infinite expected length? 
QUESTION [71 upvotes]: My question concerns the (completely deterministic) card game known as War, played by seven-year-olds everywhere, such as my son Horatio, and sometimes also by others, such as their fathers.
The question is: Is the expected length of the game infinite? 


The Rules. (from http://en.wikipedia.org/wiki/War_(card_game)) The deck is divided evenly among the two players, giving each a face-down stack. In unison, each player reveals the top card on his stack (a "battle"), and the player with the higher card takes both the cards played and moves them to the bottom of his stack. If the two cards played are of equal value, each player lays down three face-down cards and a fourth card face-up (a "war"), and the higher-valued card wins all of the cards on the table, which are then added to the bottom of the player's stack. In the case of another tie, the war process is repeated until there is no tie. A player wins by collecting all the cards. If a player runs out of cards while dealing the face-down cards of a war, he may play the last card in his deck as his face-up card and still have a chance to stay in the game.


Let us assume that the cards are returned to the deck in a well-defined manner. For example, in the order that the cards are played, with the previous round's winner's cards going first (and a first player selected for the opening battle). 
On the Wikipedia page, they tabulate the results of 1 million simulated random games, reporting an average length game of 248 battles. But this does not actually answer the question, because it could be that there is a devious initial arrangement of the cards leading to a periodic game lasting forever. Since there are only finitely many shuffles, this devious shuffle will contribute infinitely to the Expected Value. Thus, the question really amounts to: 
Question. Is there a devious shuffle in War, which leads to an infinitely long game? 
Of course, the game described above is merely a special case of the more general game that might be called Universal War, played with N players using a deck of cards representing elements of a finite partial pre-order. Any strictly dominating card wins the trick; otherwise, there is war amongst the players whose cards were not strictly dominated. Does any instance of Universal War have infinite expected length?

REPLY [3 votes]: [I can't comment (no reputation) so I am adding this as an answer.]
I just happened to write a War program on my own, for practicing my (beginning) Python.  I had always been adding the cards to the bottom of the winner's stack in the same way, and I frequently had games where, after several wars (probably between 5 and 20) there were no more wars and the game never ended (I cut it off at 10,000 turns, and later at 100,000 turns).
   Based on some of the above, I started shuffling the "won" cards and stopped seeing these "looping" games. However, I had also fixed some other bugs.  So, just now, as an experiment, I removed the shuffle of the "won" cards.  Once again, I frequently see games which don't end after 100,000 turns.
   By the way, I'm shuffling the deck five times (and with the system-time based seed). So, as far as I know the initial order should be different every time.
I'm curious.  I've just had 28 games, out of thirty, go to 100,000 turns. It's obviously possible that I still have some bug in my shuffle, or in my game play.  Any guesses as to whether placing the won cards on the bottom of the winners stack, always in the same order, could really lead to this many games looping?
I'm fine with giving my code, if anyone is interested.  Or the starting piles, after shuffling.  
thanks.
     Paul<|endoftext|>
TITLE: closure of separative quotients
QUESTION [7 upvotes]: Does there exist a partial order, nontrivial for forcing, that is countably closed, but whose separative quotient is not countably closed? Supposing the answer is yes, then is there a partial order, nontrivial for forcing, that is countably closed, but is not forcing equivalent to any countably closed separative partial order? 
For those of you unfamiliar with the separative quotient of a partial order, it is defined as follows. Two elements of a partial order are compatible iff there is some element below both of them. We form the separative quotient of a partial order by taking equivalence classes: x is equivalent to y when x and y are compatible with the exact same things. We then define a new partial order for the separative quotient -- $x \leq y$ iff everything compatible with x is compatible with y.
A partial order is said to be separative if whenever $x \nleq y$, there is $z \leq x$ such that z is incompatible with y. The separative quotient of any partial order is separative. 
Some of the ways, order-theoretically speaking, that two partial orders can be forcing equivalent are
(1) They are isomorphic, or more generally,
(2) A dense subset of one of them is isomorphic to a dense subset of the other.

REPLY [4 votes]: Stevo Todorcevic answered this question for me at the MAMLS conference in honor of Richard Laver last weekend in Boulder, CO. Apparently, the answer is that examples of forcings that are closed, whose separative quotients are not closed, come up frequently, with one particular example being forcings involving semi-selective coideals studied by Ilija and Farah.<|endoftext|>
TITLE: Deciding membership in a convex hull
QUESTION [18 upvotes]: Given points $u, v_1, \dots,v_n \in \mathbb{R}^m$, decide if $u$ is contained in the convex hull of $v_1, \dots, v_n$.
This can be done efficiently by linear programming (time polynomial in $n,m$) in the obvious way. I have two questions:

Is there a different (or more efficient algorithm) for this? If not, there ought to be a simple reduction from linear programming to the above problem as well. What is it?

Are there interesting families of instances for which the problem can be solved significantly faster than by means of linear programming, e.g., nearly linear time in $m+n$.

REPLY [4 votes]: "Are there interesting families of instances...nearly linear time..."

For the record it should be mentioned that if the points $v_1,\ldots,v_n \in \mathbb{R}^d$
are chosen randomly, under a variety of distributions and within a
variety of regions (e.g., balls, hypercubes),
then the number of vertices of the hull is $O(n^{1/3})$ in the plane
and $O(\log^{d-1}n)$ in dimension $d$.
A recent reference, which cites many of the original papers is:

Sariel Har-Peled, "On the Expected Complexity of Random Convex Hulls." 2011.
  (arXiv abstract link).<|endoftext|>
TITLE: Computer algebra errors
QUESTION [136 upvotes]: In the course of doing mathematics, I make extensive use of computer-based calculations. There's one CAS that I use mostly, even though I occasionally come across out-and-out wrong answers.
After googling around a bit, I am unable to find a list of such bugs. Having such a list would help us remain skeptical and help our students become skeptical. So here's the question: 
What are some mathematical bugs in computer algebra systems? 
Please include a specific version of the software that has the bug. Please note that I'm not asking for bad design decisions, and I'm not asking for a discussion of the relative merits of different CAS's.

REPLY [5 votes]: http://oeis.org/A110375

A110375   Numbers n such that Maple 9.5, Maple 10, Maple 11 and Maple 12 give the wrong answers for the number of partitions of n.<|endoftext|>
TITLE: Yet another graph invariant: the similarity matrix
QUESTION [7 upvotes]: Preliminaries
Let $n \in \mathbb{N}$ and $v$ be a vertex of a graph $G$. Let the $n$-neighbourhood of $v$, $N_n(v)$, be the induced subgraph of $G$ containing $v$ and all vertices at most $n$ edges away from $v$.
With $\epsilon(v)$ the eccentricity of $v$, $N_{\epsilon(v)}(v)$ is obviously nothing but the connected component of $G$ containing $v$, so it is natural to restrict $n$ for a given $v$ to the values $0,1, ..., \epsilon(v)$.
Consider for any two vertices $v$, $w$ the greatest $s = s(v,w)$ such that $N_s(v)$ and $N_s(w)$ are isomorphic [added:] with the isomorphism sending $v$ to $w$ (for short: $N_s(v) \cong_{vw} N_s(w)$). If $s(v,w) = \epsilon(v) = \epsilon(w)$, $v$ and $w$ are conjugate. For non-conjugate vertices $v$, $w$ the number $s= s(v,w)$ reflects the size of the smallest neighbourhood that is needed to distinguish $v$ and $w$, since $N_{s+1}(v) \ncong_{vw} N_{s+1}(w)$ by definition.
Let's call the positive number
$$\sigma(v,w) = \frac{2 \cdot s(v,w)}{\epsilon(v) + \epsilon(w)}$$
the similarity index of $v$ and $w$.
$\sigma(v,w) = 0$ indicates that $v$, $w$ have different $1$-neighbourhoods (and are maximally dissimilar), $\sigma(v,w) = 1$ indicates that $v$, $w$ are conjugate (i.e. maximally similar = indistinguishable by their neighbourhoods).
The matrix $\Sigma(G) = \lbrace \sigma(v,w) \rbrace_{v,w \in V(G)}$ reflects the symmetry of the graph $G$:

If $\sigma(v,w) = 1$ only if $v = w$, the graph is asymmetric.
If the $1$'s of the matrix come in square blocks along the diagonal, these blocks indicate the orbits of the graph.

$\Sigma(G)$ is a graph invariant up to matrix equivalence. (Is this the right wording?)

Definition: An $n \times n$-matrix $S$
is a similarity matrix iff there is
a graph $G$ such that $S = \Sigma(G)$.

Questions

Is the notion of $n$-neighbourhood treated in other contexts, maybe under another name?



Is there already research on this concept of similarity (or a related one)?



How might similarity matrices be
characterized (sufficient/necessary conditions)? ("A matrix is a similarity matrix if ...") Any idea?



How will graphs with the same similarity matrix (plus same eccentricity vector) be related? (They will probably not be don't have to be isomorphic, but maybe something weaker?)

REPLY [3 votes]: This is nice. What is the motivation for dividing by $\epsilon(u)+\epsilon(v)$?
To partly answer your first question, I suspect the notion of $n$-neighborhood is relatively common, but one place where it is very common, if not a staple, in Finite Model Theory (read also Database Theory). It is a key component of various locality properties, notably Gaifman locality. The quantity $s(u,v)$ is closely related to the complexity of distinguishing $u$ from $v$ by a first-order query in the language of graphs. Specifically, a query of quantifier depth $k$ cannot distinguish between vertices such that $s(u,v) > (3^{k+1}-1)/2$. This has many uses, a typical example is to measure the complexity of distinguishing database entries. A good intro can be found in the first few chapters of Libkin's Elements of Finite Model Theory.<|endoftext|>
TITLE: Quantum equivariant $K$-theory and DAHA.
QUESTION [9 upvotes]: Theorem 3.2 of the paper  "Quantum cohomology of the Springer resolution" by Braverman, Maulik and Okounkov relates equivariant quantum cohomology of the cotangent bundle of $G/B$ to the trigonometric Dunkl operators.  A naive guess (or maybe WAG?) is that replacing cohomology with $K$-theory gives the Dunkl representation of the DAHA.  An explanation (or reference to an explanation) of what (if any) significant obstacles remain for doing the $K$-theoretic analog of BMO will be greatly appreciated!  Is it the sort of thing that should be straightforward (though perhaps technical) at this point?  I'd prefer an answer that addresses these particular varieties, rather than general foundational problems.

REPLY [2 votes]: My best guess is that either

this is true for $\mathbb{CP}^1$, and it's pretty easy to generalize that given what's already in that paper, or
this is false for $\mathbb{CP}^1$, and you're hosed.

The bit one has to understand is the map from the 2 point genus 0 moduli space to the Steinberg variety.  BMO get away with just noting that the two spaces have the same dimension, so the pushforward of the fundamental class of the moduli space has to be a sum of fundamental classes of components of Steinberg, whose coefficients they work out by deforming to an almost generic situation and doing the calculation for $\mathbb{CP}^1$.
I think by looking at the pushforward of the structure sheaf on the 2-point moduli space, you'll find that quantum correction is some K-class on the Steinberg variety and thus something in the affine Hecke algebra, and I think it should be the sum of SL(2) contributions for each root by the same deformation arguments that BMO use.

I just spoke to Davesh Maulik about this, and it seems my intuition has failed me: he claims it is just hard, and the techniques of that paper will not work.<|endoftext|>
TITLE: Why is this generality in Vitali's Lemma useful?
QUESTION [8 upvotes]: In Vitali's Lemma it uses outer measure rather than measure. What are some results that depend on it this theorem applying to sets with only outer measure rather than measurable sets?
Vitali's Lemma:
Let $E$ be a set of finite outer measure and $G$ a collection of intervals that cover $E$ in the sense of Vitali. Then given $\varepsilon> 0$ there is a finite disjoint collection of intervals in $G$ such that $m^*(E - \bigcup_{n=1}^N I_n) < \varepsilon$.
I'm trying to learn this theorem and I keep replacing "outer measure" with "measure" and I want a reason to stop doing that.

REPLY [8 votes]: The short answer (to the question in the title) is so that you don't need to worry about whether E is measurable.  If you happen to know that E is measurable then you can drop "outer" everywhere.
There may be a longer answer (better addressing the question as stated in the page) involving specific applications where E is in fact nonmeasurable, but I personally don't know of such applications offhand.<|endoftext|>
TITLE: What's the use of a complete measure?
QUESTION [46 upvotes]: A complete measure space is one in which any subset of a measure-zero set is measurable.
For what reasons would I want a complete measure space?  The only reason I can think of is in the context of probability theory: using complete probability spaces forces almost-everywhere equal random variables to generate the same sigma-sub-algebra.
Am I missing some other important technical reasons?

REPLY [4 votes]: Hi Tom E, 
Here is another reason: Let $E$ be a Borel set in Euclidean space. Then its image under a continuous map is always Lebesgue-measurable but in general not Borel measurable. Results like this make the completion useful; The theory behind this is the theory of analytic sets or the Souslin operation.<|endoftext|>
TITLE: a proof that L_min is not in coRE?
QUESTION [8 upvotes]: Define $L_{min}$ to be the language of all minimal Turing machines, in some standard encoding. (A Turing Maching is minimal if it has the shortest encoding among all the TMs recognizing the same language.) Sipser gives a slick proof that $L_{min}$ is not a Recursively Enumerable (RE) language. The argument goes like this: suppose to the contrary that $L_{min}$ is in RE, with some enumerator $E$. Define the Turing machine $B$, which obtains its own description via the recursion theorem, waits until $E$ generates a program $C$ that is longer than $B$, and then simulates the behavior of $C$. The contradiction results from the assumption that $E$ only generates minimal programs and the construction of $B$ as a program that's shorter than some "minimal" program.
Now I want to show that $L_{min}$ is not in coRE (meaning that its complement is not in RE). Any ideas?

REPLY [7 votes]: This is a problem with a surprisingly unique history. See the survey

Marcus Schaefer: A guided tour of minimal indices and shortest descriptions. Arch. Math. Log. 37(8): 521-548 (1998)

for more discussion. A proof that the problem is not coRE can be extracted from page 6 of that survey. For convenience I will give a self-contained argument here. Unfortunately it is not as simple as the "not-RE" proof!
We will show a Turing reduction from $L_{ALL}$ = { $ M~|~L(M) = \Sigma^{\star} $} to $L_{min}$, which suffices since $L_{ALL}$ is complete for the second level of the arithmetic hierarchy (implying it is not coRE). Intuitively, this means that given an oracle for $L_{min}$ we could decide the halting problem for machines which have oracles to the halting problem. A coRE language cannot have this property. 
Let $\hat{M}$ be the lexicographically first machine that accepts no inputs, under some encoding of machines. 
First, we observe that given an oracle for $L_{min}$, we can effectively determine whether $M(M)$ halts or not, for a given $M$. (Recall this is enough to determine whether $M(x)$ halts or not, for given $M$, $x$.)
Define a machine $N_M$ which on input $t$, simulates $M(M)$ for up to $t$ steps and halts iff the simulation halts. To determine if $M(M)$ halts, make a list ${\cal L}$ of all machines $M' \neq \hat{M}$ with $M' \leq N_M$ such that $M' \in L_{min}$. This list can be computed using an oracle for $L_{min}$. Observe that all such machines accept at least one input, since we excluded $\hat{M}$ and they are all minimal. Via dovetailing, we can effectively find integers $t'$ such that each $M'(M')$ halts in $t'$ steps. Let $t''$ be the largest such $t'$. Then $M(M)$ halts iff $M(M)$ halts in at most $t''$ steps. 
Now that we can decide the halting problem, we turn to computing a function version of $L_{min}$: given an oracle for $L_{min}$, we can output the minimum equivalent machine $M'$ to the input $M$. If $M \in L_{min}$ this is easy. Otherwise we make a list ${\cal L}$ of all machines that are in $L_{min}$ and are smaller than $M$. Then we begin to compute, for increasing inputs $x$, bit tables indicating the accept/reject behavior of all machines in ${\cal L}$, on the inputs seen so far. (We can do this because we can solve the halting problem with the oracle!) When we find that a machine $M''$ differs on an input from $M$, we remove it from ${\cal L}$. If $M$ is not minimal, there will eventually be only one machine $M'$ left which has not yet differed from $M$, since all machines in ${\cal L}$ are minimal. If $M$ is not minimal then this $M'$ must be the minimum machine.
Finally, we define a machine $N$ that recognizes $L_{ALL}$ with an oracle for $L_{min}$. Let $M_{ALL}$ be the minimum Turing machine that accepts everything. If the input $M$ is less than $M_{ALL}$, output NO. ($M$ must reject something.) Otherwise compute the minimum machine equivalent to $M$. If $M = M_{ALL}$ then output YES otherwise output NO. Note $L(N) = L_{ALL}$.<|endoftext|>
TITLE: What is the right version of "partitions of unity implies vanishing sheaf cohomology"
QUESTION [32 upvotes]: There are several theorems I know of the form "Let $X$ be a locally ringed space obeying some condition like existence of partitions of unity. Let $E$ be a sheaf of $\mathcal{O}_X$ modules obeying some nice condition. Then $H^i(X, E)=0$ for $i>0$."
What is the best way to formulate this result? I ask because I'm sure I'll wind up teaching this material one day, and I'd like to get this right. 
I asked a similar question over at nLab. Anyone who really understands this material might want to write something over there. If I come to be such a person, I'll do the writing!

Two versions I know: 
(1) Suppose that, for any open cover $U_i$ of $X$, there are functions $f_i$ and open sets $V_i$ such that $\sum f_i=1$ and $\mathrm{Supp}(f_i) \subseteq U_i$. Then, for $E$ any sheaf of $\mathcal{O}_X$ modules, $H^i(X,E)=0$. Unravelling the definition of support, $\mathrm{Supp}(f_i) \subseteq U_i$ means that there exist open sets $V_i$ such that $X = U_i \cup V_i$ and $f_i|_{V_i}=0$.
Notice that the existence of partitions of unity is sometimes stated as the weaker condition that $f_i$ is zero on the closed set $X \setminus U_i$. If $X$ is regular, I believe the existence of partitions of unity in one sense implies the other. However, I care about algebraic geometry, and affine schemes have partitions of unity in the weak sense but not the strong.
(2) Any quasi-coherent sheaf on an affine scheme has no higher sheaf cohomology. (Hartshorne III.3.5 in the noetherian case; he cites EGA III.1.3.1 for the general case.) There is a similar result for the sheaf of analytic functions: see Cartan's Theorems.
I have some ideas about how this might generalize to locally ringed spaces other than schemes, but I am holding off because someone probably knows a better answer.

It looks like the answer I'm getting is "no one knows a criterion better than fine/soft sheaves." Thanks for all the help. I've written a blog post explaining why I think that fine sheaves aren't such a great answer on non-Hausdorff spaces like schemes.

REPLY [13 votes]: Although we clearly all have more or less the same answers, here is how I like to organize things.
I) Let $\mathcal F$ be a sheaf  of abelian groups on the topological space $X$. It is said to be soft if every section $s \in \Gamma (A,\mathcal F)$ over a closed subset $A\subset X$ can be extended to $X$. 
Notice carefully that the definition of $s$ is NOT that it is the restriction to $A$ of some section of $\mathcal F$ on an open neighbourhood of $A$  [but that it is an element  $ s\in \prod \limits_{x\in X} \mathcal F_x$  satisfying some more or less obvious conditions] 
II) Consider the following condition on the [not necessarily locally] ringed space $(X, \mathcal O)$ :
The space $X$ is metrizable and given an inclusion $A\subset U \subset X$ with $A$ closed and $U$ open there exists a global section $s\in \Gamma (X,\mathcal O)$ such that 
$s|A=1$ and $ s|X \setminus U=0 \quad \quad (SOFT)$.
We then have the 
$\textbf{Theorem }$ : If the ringed space satisfies (SOFT), then every sheaf of $\mathcal O_ X -Modules$ is soft.
III) A metrizable space endowed with its sheaf of continuous functions satisfies $(SOFT)$.
A metrizable differential manifold endowed with its sheaf of smooth functions satisfies $(SOFT)$.
IV) On a metrizable space every soft sheaf is acyclic
Put together these results yield all standard acyclicity results on functions,vector bundles, distributions,etc.
It is interesting to notice that you use partitions of unity only once: in the proof of III). But never more afterwards; you just check that your sheaves are $\mathcal O -Modules$. I like this approach (which I learned from Grauert-Remmert)  more than the usual one, where a proof of acyclicity is  given for the sheaf of smooth functions, followed by the ( correct!) assertion that you have to repeat it with minor changes for, say, vector bundles. Moreover fine sheaves needn't even be mentioned if you follow this route.<|endoftext|>
TITLE: Characterization of Boolean-valued functions on the discrete cube based on its Fourier coefficients.
QUESTION [12 upvotes]: Consider functions on the discrete cube $\{-1,1\}^n$. 
We consider the Discrete Fourier Transform of such functions. Suppose we denote the parity function on a subset $S \subseteq [n]$ of co-ordinates by $\Pi_S(x)=\pi_{i \in S}(x_i)$, then the Fourier coefficient $\hat{f}_S$ is simply the expectation: $\hat{f}_S=\mathbb{E}_x[f(x)\Pi_S(x)]$; and by orthonormality of the parity functions, any $f$ may be represented as $f(x)=\sum_{S \subseteq [n]}\hat{f}_S \Pi_S(x)$ (the summation being in $\mathbb{R}$.)
I am interested in knowing the difference between boolean-valued ($\{-1,1\}^n \rightarrow \{-1,1\}$) and real-valued functions ($\{-1,1\}^n \rightarrow \mathbb{R}$). More specifically, I would like to know the difference between the Fourier spectra of either class of functions.
What properties of the Fourier spectra hold for one class but not for the other?
(As an example: It can be proved that for Boolean valued functions, if all the weight is concentrated on Fourier coefficients of size at most 1 then the function is either a constant or a dictator. This is not true for real-valued functions.)

REPLY [8 votes]: This is a good question which is the subject of intensive research in mathematics and theoretical computer science. The blog (which is the serialization of a book in progress) "Analysis of Boolean Functions" by Ryan O'Donnell is a good source, and so is the Book: Lectures on noise sensitivity and percolation by Garban and Steif. 
Here is some information
1) Of course, the Fourier coefficients of real functions over the discrete cube can be arbitrary. The question is therefore what restrictions apply for Boolean functions. 
Boolean functions are characterized by $f^2(x)=1$ and since product translates to convolution for the Fourier transform, being Boolean is characterized by a property of the Fourier transform convolved with itself. However, this characterization by itself is not very useful.
Parseval formula asserts that for Boolean functions the sum of square of the Fourier 
coefficients is 1. It also give the following formula for the variance of $f$, 
$$\operatorname{var}(f) = \sum \{ \hat f^2(S): S \ne \emptyset \} $$
2) An important tool which gives much information is Bonami (or Bonami–Gross–Beckner) inequality. It asserts that for every function $f$, $$\|T_\epsilon (f) \|_2 \le \|f\| _{1+\epsilon^2}.$$ 
This implies that if $f$ has values $0$, $1$, and $-1$ and the the support of $f$ has measure $t$ then most of the Fourier spectrum of $f$ is for $S$ with $|A|> \log n/10$ (say). 
3) A similar reasoning gives the so called KKL's theorem. It asserts that for a Boolean function $f$ there is an index $k$ so that $$\sum \{ \hat f^2(S): S \subset [n], i \in S \} \ge \operatorname{var}(f) \log n/n.$$ 
4) A theorem of Friedgut asserts that for a Boolean function $f$ if $\sum \hat f^2(S) |S|$ is bounded above by a constant $c$ then $f$ is "$\epsilon$-close" to a "Junta. " A Junta is a Boolean function depending on a bounded number $C$ of variables. ($C$ is a function of $c$ and $\epsilon$.)
5) A theorem by Green and Sanders from the paper Boolean functions with small spectral norm,  asserts that a Boolean function all whose Fourier coefficients are bounded by $M$ is a linear combination of bounded number $(\le 2^{2^{O(M^4)}}$) of characteristic functions of subspaces. 
6) A result by Talagrand asserts that for a Boolean functions $f_n$ if $\sum_i^n\hat f_n^2(\{i\})$ is o(1) then so is $\sum_i^n\hat f_n^2(\{i\})$. An extension of this result to higher levels was given by Benjamini, Kalai and Schramm, and a sharp quantitative version by Keller and Kindler. 
7) A theorem of Bourgain asserts that for a Boolean function if the decay of the Fourier coefficients squared  is larger than quadratic in $|S|$, then again $f$ is approximately a Junta.
8) There is a conjecture called the Entropy influence conjecture that asserts that $\sum \hat f^2 (S)|S|$ is bounded from below by an absolute constant times $\sum \hat f^2(S) \log (\hat f^2(S))$.<|endoftext|>
TITLE: A quadratic form
QUESTION [7 upvotes]: Let $q$ be a power of 2.  Let $P$ be the set of  polynomials in 
$F_q [x]$ of degree d or less.
Let $\mathbb{Z}$ be the ring of integers. For any $f \in P$, let $\psi(f)$ be the
number of distinct roots of $f$ in $F_q$. Note that $\psi(0) = q$.
For any map  $A$ from $P$ to $\mathbb{Z}$, one can  compute the summation
  $$\sum_{f, g \in P} A(f)A(g) \psi (f+g).$$
My question is:
what is the minimum positive value of the summation?
For $d=0,1$, the minimum is $q$. What happens if $d$ is bigger?
I am especially interested in the case when d=q/2-1.
Thanks a lot,
Qi

REPLY [2 votes]: (Inspired from Charles Siegel's answer) what you can do for the $d=\frac{q}{2}$ case is:
Divide the $q$ elements of $F_q/\{0\}$ into two parts $\{\alpha_{1},\dots, \alpha_{\frac{q}{2}}\}$ and $\{\beta_1,\dots,\beta_{\frac{q}{2}-1}\}$ so that $\prod \alpha_i\neq \prod \beta_j$. Denote $S(x)=\prod (x-\alpha_i)$ and $T(x)=\prod(x-\beta_j)$. Let $r_1(x)+r_2(x)=S(x)+T(x)$ be arbitrary and $f_1=S+r_1,f_2=S+r_2$. Now we let $A(f_1)=A(f_2)=1,A(r_1)=A(r_2)=-1$ and all other $A(g)=0$. The quadratic form has the value
$$-2\psi(f_1+r_1)-2\psi(f_1+r_2)-2\psi(f_2+r_1)-2\psi(f_2+r_2)+4\psi(r_1+r_2)+4q=4$$
I don't know if this can be lowered to 2...<|endoftext|>
TITLE: Are there analogues of Desargues and Pappus for block designs?
QUESTION [18 upvotes]: Finite projective planes are fascinating objects from many perspectives. In addition to the geometric view, they can be viewed as combinatorial block designs.
From the geometric perspective, there are two very important structural properties for projective planes: the Theorem of Desargues, which holds exactly when the plane can be coordinatized by a division ring, and the Theorem of Pappus, which holds precisely when the plane can be coordinatized by a field. It is a famous theorem of Wedderburn that every finite division ring is a field, so the two properties are equivalent for finite projective planes.
Although they are both very combinatorial statements, I don't recall having seen anything similar to Desargues and Pappus for other classes of block designs.

Are there interesting analogues or generalizations of the properties of Desargues and Pappus for other classes of block designs? Of particular interest would be analogues and generalizations that correspond (not necessarily exactly) to some form of coordinatization of designs.

Comment. This question has generated a fair amount of interest. I've been considering accepting the answer by John Conway and Charles Roque even though it is not quite satisfactory. (It only answers the first part of the question in a loose sense, and it does not address the second part.) So I decided to set up a small bounty to stimulate other answers.

REPLY [10 votes]: I passed on your question to John H. Conway.  Here is his response: (NB.  Everything following this line is from Conway and is written from his point of view.  Of course, in the comments and elsewhere on the site, I am not Conway.)
I think it's wrong to focus on block designs in particular.  This may not answer your question, but there are some interesting examples of theorems similar to Desargues's and Pappus's theorems.  They aren't block designs, but they do have very nice symmetries.
I call these "presque partout propositions" (p.p.p. for short) from the French "almost all".  This used to be used commonly instead of "almost everywhere" (so one would write "p.p." instead of "a.e.").  The common theme of the propositions is that there is some underlying graph, where vertices represent some objects (say, lines or points) and the edges represent some relation (say, incidence).  Then the theorems say that if you have all but one edge of a certain graph, then you have the last edge, too.  Here are five such examples:
Desargues' theorem
Graph: the Desargues graph = the bipartite double cover of the Petersen graph
Vertices: represent either points or lines
Edges: incidence
Statement: If you have ten points and ten lines that are incident in all of the ways that the Desargues graph indicates except one incidence, then you have the last incidence as well.  This can be seen to be equivalent to the usual statement of Desargues's theorem.
Pappus's theorem
Graph: the Pappus graph, a highly symmetric, bipartite, cubic graph on 18 vertices
Vertices: points or lines
Edges: incidence
Statement: Same as in Desargues's theorem.
"Right-angled hexagons theorem"
Graph: the Petersen graph itself
Vertices: lines in 3-space
Edges:  the two lines intersect at right angles
Statement:  Same as before, namely having all but one edge implies the existence of the latter.  An equivalent version is the following: suppose you have a "right-angled hexagon" in 3-space, that is, six lines that cyclically meet at right angles.  Suppose that they are otherwise in fairly generic position, e.g., opposite edges of the hexagon are skew lines.  Then take these three pairs of opposite edges and draw their common perpendiculars (this is unique for skew lines).  These three lines have a common perpendicular themselves.
Roger Penrose's "conic cube" theorem
Graph: the cube graph Q3
Vertices: conics in the plane
Edges:  two conics that are doubly tangent
Statement:  Same as before.  Note that this theorem is not published anywhere.
Standard algebraic examples
Graph:  this unfortunately isn't quite best seen as a graph
Statement: Conics that go through 8 common points go through a 9th common point.  Quadric surfaces through 7 points go through an 8th (or whatever the right number is).
Anyway, I don't know of any more examples.
Also, I don't know what more theorems one could really have about coordinatization.  I mean, after you have a field, what more could you want other than, say, its characteristic?  (Incidentally, the best reference I know for the coordinatization theorems is H. F. Baker's book "Principles of Geometry".)
In any case, enjoy!<|endoftext|>
TITLE: Grassmannian bundle theorem
QUESTION [9 upvotes]: Let's consider a vector bundle $E$ of rank $n$ over a compact manifold $X$. Consider the associated Grassmannian bundle $G$ for some $k < n$, obtained by replacing each fiber $E_x$ by $Gr(k,E_x)$.
Let's suppose that there is a full flag of subbunldes $F_1 \subset F_2 \dots \subset F_n \subset E$. I think that in this case we are able to define relative Schubert cycles on G which restrict to usual Schubert cycles on each fiber so that we can apply Leray-Hirsh theorem to deduce that $H^*(G) = H^*(X) \otimes H^*(Gr(k,n))$.

Is the reasoning above correct?
Can we still compute $H^*(G)$ in the case when the full flag of subbundles doesn't exist?

EDIT: I meant complex vector bundles and complex Grassmannians. Also the bundle can be assumed holomorphic or algebraic if it makes a difference.
EDIT: Ben in his answer mentions Serre's spectral sequence that can be used in this case. Is there a reason why it will degenerate to leave $H^*(X) \otimes H^*(Gr(k, n))$ as a result?

REPLY [3 votes]: algori's answer is completely correct. In case you want a reference, this material is in Chapter 14 of Fulton's Intersection Theory. (Except that Fulton is writing about Chow rings and you asked the question in cohomology.) In particular, the additive structure is Proposition 14.6.5 and the ring structure is 14.6.6.
I'm not sure what the best references are for the analogous statements in cohomology, but you could try tracing Fulton's references and see if they help.<|endoftext|>
TITLE: Obstructions to descend Galois invariant cycles
QUESTION [16 upvotes]: Let $X$ be a smooth projective variety over $F$, and $E/F$ - finite Galois extension.
There is an extension of scalars map $CH^\*(X) \to CH^\*(X_E)$. The image lands in the Galois invariant part of $CH^\*(X_E)$, and in the case of rational coefficients, all Galois-invariant cycles are in the image (EDIT: this follows from taking the trace argument).
With integer coefficients Galois-invariant classes don't have to descend.
For example, for $CH^1(X) = Pic(X)$ there is an exact sequence:
$$
0 \to Pic(X) \to Pic(X_E)^{Gal(E/F)} \to Br(F),
$$
so we can say that the obstruction to descend a cycle lies in a Brauer group.
Are there any known obstructions to descend elements of higher groups $CH^i$ with integer coefficients?
In my case I have a cycle in $CH^*(X_E)^{Gal(E/F)}$ and I want to find out whether or not it's coming from $CH^*(X)$. 
(The actual cycle is described in here.)

REPLY [6 votes]: Dear Evgeny Shinder,
For  CH^2(X)  one can find an "obstruction" in a non-ramified cohomology. 
Let us write H^i_nr(X, \mu_l^j) for the intersection of kernels of the residu maps d_A in Galois cohomology, where A is running over all discrete valuation rings with field of fractions F(X) containing F. For example, if X is regular, one should take in account all valuations coming from codimension one points, but also the valuations coming from exceptional divisors of various blowing-ups.
If X is a smooth geometrically rational variety defined over a finite field F (i.e. rational over an algebraic closure \bar F of F), one can show that the cokernel of the map CH^2(X)->CH^2(X_{\bar F})^G is isomorphic, up to p-torsion (p=char F), to the group of non-ramified cohomology H^3_nr(X, Q/Z(2)). One can also find examples when this last group is non zero (see http://arxiv.org/abs/1004.1897).
Is it helpful for you?
Best regards,
Alena Pirutka.<|endoftext|>
TITLE: What is the "right" universal property of the completion of a metric space?
QUESTION [22 upvotes]: I'm a little embarrassed to ask this one, but it could help for a class I'm teaching, so here goes:
Let $X$ be a metric space.  We all know that $X$ admits a completion, which is a complete metric space $\hat{X}$ together with an isometric embedding $\iota: X \hookrightarrow \hat{X}$ with dense image.  Moreover, one learns that this completion is essentially unique.
From a modern perspective, one would like to realize the completion as satisfying some universal mapping property: this makes precise the "essentially unique" above and gives some functorial properties.  But it seems to me that the completion satisfies two different such properties:
1) It is universal with respect to isometric embeddings into complete metric spaces.
2) It is universal with respect to uniformly continuous maps into complete metric spaces.  
Of course 1) is the more obvious one.  I gather from some internet research that 2) is supposed to be the "right" choice, and its usefulness is related to the fact that uniformly continuous maps have the extension property (again, I don't quite remember this from my undegraduate days; is it in Rudin's Principles, for instance?).  However, it seems strange to me that by taking 2), we also get for free that the mapping $\iota$ is an isometric embedding (in particular, from 2) it doesn't even seem completely obvious that it is injective).  Certainly one can see this by constructing the completion, but is there a more direct way?
I suspect that this is an instance when the more categorical thinkers have one up on me, and I stand ready to be enlightened.

REPLY [7 votes]: Dear Pete,
I want to give a proof that if $X$ is a metric space, and if $x_0$ and $x_1$ are two
distinct points of $X$, then there is a map
$f:X \rightarrow Y$ that is uniformly continuous, with $Y$ complete, and such that 
$f(x_0) \neq f(x_1)$.   The main point is that my proof won't refer to the completion
$\widehat{X}$ of $X$.  It will then give a proof that $X \rightarrow \widehat{X}$ is
necessarily injective, without refering to the construction of $\widehat{X}$.
The construction is simple: take $Y = {\mathbb R}$, and define
$f(x) = d(x,x_0).$
(Note: slightly edited from the first version, which had unnecessary complications
in the definition of $f$.)<|endoftext|>
TITLE: Complete graph invariants?
QUESTION [23 upvotes]: Obviously, graph invariants are wonderful things, but the usual ones (the Tutte polynomial, the spectrum, whatever) can't always distinguish between nonisomorphic graphs. Actually, I think that even a combination of the two I listed will fail to distinguish between two random trees of the same size with high probability.
Is there a known set of graph invariants that does always distinguish between non-isomorphic graphs? To rule out trivial examples, I'll require that the problem of comparing two such invariants is in P (or at the very least, not obviously equivalent to graph isomorphism) -- so, for instance, "the adjacency matrix" is not a good answer. (Computing the invariants is allowed to be hard, though.)
If this is (as I sort of suspect) in fact open, does anyone have any insight on why it should be hard? Such a set of invariants wouldn't require or violate any widely-believed complexity-theoretic conjectures, and actually there are complexity-theoretic reasons to think that something like it exists (specifically, under derandomization, graph isomorphism is in co-NP). It seems like it shouldn't be all that hard...
Edit: Thorny's comment raises a good point. Yes, there is trivially a complete graph invariant, which is defined by associating a unique integer (or polynomial, or labeled graph...) to every isomorphism class of graphs. Since there are a countable number of finite graphs, we can do this, and we have our invariant.
This is logically correct but not very satisfying; it works for distinguishing between finite groups, say, or between finite hypergraphs or whatever. So it doesn't actually tell us anything at all about graph theory. I'm not sure if I can rigorously define the notion of a "satisfying graph invariant," but here's a start: it has to be natural, in the sense that the computation/definition doesn't rely on arbitrarily choosing an element of a finite set. This disqualifies Thorny's solution, and I think it disqualifies Mariano's, although I could be wrong.

REPLY [3 votes]: The sequence of homomorphism numbers $|Hom(F_i,G)|$ for all (isomorphism types of) graphs $F_i$  is an invariant of $G$ (see Lovász, Operations with structures). 
(Does this fit your bill? Or do you want finite invariants only?)<|endoftext|>
TITLE: Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum?
QUESTION [12 upvotes]: Suppose the continuum is larger than $\aleph_2$. Does there exist a countably closed notion of forcing that collapses $\aleph_2$ to $\aleph_1$, but does not collapse the continuum to $\aleph_1$? Moreover, does there exist such a forcing notion that is separative and has size continuum? It is known (see below) that the canonical collapse Coll$(\aleph_1, \aleph_2)$ collapses the continuum. Trying something like the canonical collapse relativized to some inner model will fail to answer the question, because this forcing will not be countably closed in V. 

Background information:
This question came up as a result of my studies of the following theorem. 
Let $\kappa < \theta$ be cardinals, with $\kappa$ regular and $\theta^{<\kappa} = \theta$.  Then any forcing of size $\theta^{<\kappa}$ which is separative and $<\kappa$ closed and which collapses $\theta$ to $\kappa$ is forcing equivalent to the canonical collapse forcing Coll$(\kappa, \theta)$.
I want to know whether this theorem still holds in the case where $\theta^{<\kappa} = \theta$ fails. The question above is the simplest possible such case.
The reason why Coll$(\aleph_1, \aleph_2)$ collapses the continuum (when CH fails) is that we can think of $\aleph_1$ as $\aleph_1$ many $\aleph_0$-blocks. Consider only the elements of Coll$(\aleph_1, \aleph_2)$ such that on each $\aleph_0$ block, they are either fully defined or fully undefined. This is a dense set in Coll$(\aleph_1, \aleph_2)$, and it's isomorphic to Coll$(\aleph_1, \aleph_2^{\aleph_0})$ = Coll$(\aleph_1, \bf{c})$.

REPLY [3 votes]: It might be late, but the following paper by Todorcevic analyzes the above question for three kind of forcing notions: Proper forcings, semi-proper forcing and forcings that preserve stationary  subsets of $\omega_1$:
Collapsing $ω_2$
with semi-proper forcing<|endoftext|>
TITLE: Measure between the counting measure and the Lebegue measure
QUESTION [5 upvotes]: There are subsets of the real line that has infinite counting measure, but Lebegue measure 0, so the Lebegue measure is used for measuring larger sets than the counting measure. My question is: Is there a translation invariant measure m such that for some sets with Lebegue measure 0 the m-measure is infinite and for some sets with infinite counting measure, the m-measure is 0? 
I have found one example: m(A)=0 if A is countable, and m(A)=infinite otherwise. So I will require that the measure can take the value 1.
If such a measure exist, can we find a measure between this and the counting measure? and between this and the Lebegue measure? and so on.

REPLY [10 votes]: Hausdorff measures of dimensions between 0 and 1 are a continuous spectrum of examples.

REPLY [5 votes]: I believe what you are looking for is the $\alpha$-dimensional Hausdorff measure for some $0 < \alpha < 1$.<|endoftext|>
TITLE: Saito-Wright definition of Rickart C*-algebras
QUESTION [10 upvotes]: A C*-algebra is Rickart if for each  $x\in  A$ there is a projection $p\in A$ so that
$R(x)=pA$. 
Here the right-annihilator $R(S)$ of $S\subset A$ is defined
as $$R(S)=\{a\in A\mid xa=0\, \forall x\in S\}$$ and $R(x)\equiv R(\{x\})$.
In:

Kazuyuki Saito and J. D. Maitland Wright. $C^∗$-algebras which are Grothendieck
  spaces. Rend. Circ. Mat. Palermo (2), 52(1):141–144, 2003.

an alternative definition is studied:
define a C*-algebra to be Rickart if each maximal Abelian *-subalgebra of $A$ is Rickart (or, equivalently, monotone $\sigma$-complete).
Equivalently, one may require that every Abelian *-subalgebra is contained in an
Abelian Rickart C*-algebra.
This definition is more general and seems to be sufficient for many applications.

Is this definition in fact equivalent to the original one?

This question recently came up in our investigations in the foundations of quantum theory:
Bohrification of operator algebras and quantum logic

REPLY [10 votes]: They are equivalent. See On Defining AW*-algebras and Rickart C*-algebras by K. Saitô, and J.D.M. Wright (arXiv:1501.02434)<|endoftext|>
TITLE: current status of crystalline cohomology?
QUESTION [47 upvotes]: The great references given on Ilya's question make me wonder about the current status of the many conjectures and open questions in Illusie's survey from 1994 on crystalline cohomology. Obviously (just compare Illusie's survey from 1975 with that above or with Chambert-Loir's survey from 1998), there is very intense work on that and the connections between the various cohomology theories attacking the case "l=p". Some more recent surveys only on Fontaine's p-adic Hodge theory are already linked to in the answers to Ilya's question, Le Stum's book (Errata) covers rigid chohomology. Among the open issues mentioned in Illusie's survey are finiteness theorems, crystalline coefficients, geometric semistability, the identity of characteristic polynomials of the Frobenius of different theories,... What is the current status of these? Which new theories have been created the past decade, how fit they together and which new questions emerged?   
Edit: U. Jannsen talked recently on "a refinement of crystalline cohomology by using the theory of so-called gauges as introduced earlier by Mazur and Kato and certain syntomic sheaves." Unfortunately I found no preprint on that. Edit: Jannsen on (slides) "a cohomology theory in characteristic p which refines the crystalline cohomology – and works well for torsion" and "a sheaf theory which generalizes the Dieudonné theory – and works well for torsion."
Edit: Go Yamashita talked about "La Theorie de Hodge p-adique pour varietes ouverts" avoiding Falting's almost etale extensions. Unfortunately I found no text where one can read that.
Edit: A short note by Bhargav Bhatt and Aise Johan de Jong on a shortened proof of the comparison theorem between crystalline and de Rham cohomology.
Edit: A new proof of the semistability conjecture by Beilinson and a definition of derived crystals by Gaitsgory and Rozenblyum.
Edit: A p-adic derived de Rham cohomology by Bhargav Bhatt, giving "derived de Rham descriptions of the usual period rings and related maps in p-adic Hodge theory" and "a new proof of Fontaine's crystalline conjecture and Fontaine-Jannsen's semistable conjecture".
Edit: A  "a new cohomology theory in characteristic p>0, the so called F-gauge cohomology, a cohomology with values in the category of so-called F-gauges, which refines the cristalline cohomology" by Fontaine, Jannsen.
Edit: An other very interesting and very nice to read history of mathematics talk by Illusie  "Grothendieck at Pisa: crystals and Barsotti-Tate groups":

REPLY [9 votes]: Kedlaya gave a talk in August in which he mentioned some work of Daniel Caro on finiteness for rigid cohomology with coefficients (some of which is on the ArXiv).  On the same page, you can find notes from his talks on semistable reduction.<|endoftext|>
TITLE: complex structure on S^n
QUESTION [13 upvotes]: Using the chern character, it can be shown that there is no complex structure on $S^n$ if $n > 6$: See May's book: if $S^{2n}$ has a complex structure, let $\tau$ be the tangent bundle. $c_n(\tau) = \chi(S^{2n}) = 2$ must be divisible by $(n-1)!$ by Husemöller, Fibre bundles, chapter 20, Theorem 9.8. So the only case left is $S^4$ and $S^6$.
Is there a complex structure on $S^4$ or $S^6$?

REPLY [25 votes]: It is known that $S^4$ doesn't even have an almost complex structure, and the case for $S^6$ is open.  The lack of almost complex structure can be proved a number of ways, one way is by showing that an almost complex, compact, four manifold with $\dim_{\mathbb{Q}}H^2(X,\mathbb{Q})=0$ has $\chi(X)=0$, but the four sphere doesn't. (It follows from the index theorem, here's a quick reference, first result.)<|endoftext|>
TITLE: What is the difference between matrix theory and linear algebra?
QUESTION [13 upvotes]: Hi,
Currently, I'm taking matrix theory, and our textbook is Strang's Linear Algebra. Besides matrix theory, which all engineers must take, there exists linear algebra I and II for math majors. What is the difference,if any, between matrix theory and linear algebra?
Thanks!

REPLY [6 votes]: I'm with Jon.  Matrices don't always appear as linear transformations.  Yes, you can look at them as linear transformations, but there are times when it's better not to and study them for their own right.  Jon already gave one example.  Another example is the theory of positive (semi)definite matrices.  They appear naturally as covariance matrices of random vectors.  The notions like schur complements appear naturally in a course in matrix theory, but probably not in linear algebra.<|endoftext|>
TITLE: Can skeleta simplify category theory?
QUESTION [27 upvotes]: I am not by any means an expert in category theory. Anyway whenever I have studied a concept in category theory I have always had the feeling that most of the subtleties introduced are artificial.
For a few examples:
-one does not usually consider isomorphic, but rather equivalent categories
-universal objects are unique only up to a canonical isomorphism
-the category of categories is really a 2-category, so some natural constructions do not yield functors into categories, but only pseudofunctors
-cleavages of fibered categories do not always split
....
My question is: can skeleta be used to simplify all this stuff? It looks like building everything using skeleta from the beginning would remove a lot of indeterminacies in these constructions. On the other hand it may be the case that this subtleties are really intrinsic, and so using skeleta, which are not canonically determined, would only move the difficulties around.

REPLY [6 votes]: While I agree with the above answers, I know of some situations where it might be good to take a skeleton (or something like it). This has to do with large categories which are essentially small. You might want to take a skeleton or at least a a small category that is equivalent to the original, so as to avoid size problems.<|endoftext|>
TITLE: Borsuk pairs of Banach spaces 
QUESTION [9 upvotes]: Given $X$, $Y$ two real Banach spaces, let's say that $(X,\ Y)$
is a Borsuk pair if for any continuous mapping $T$ : {$x$ $\in$
$X$ ; $||x||\leq1$} $\rightarrow$ $Y$ s.t. $T$ is odd on {$x$
$\in$ $X$ ; $||x||=1$}, it follows that the set $T^{-1}$( { 0 } )
is nonempty.
My conjecture is : $(X,\ Y)$ Borsuk pair $\Longleftrightarrow$
dim $Y$ < $\infty$ and dim $Y$ $\leq$ dim $X$. [Here "dim"
stands for the algebraic dimension; it may be a natural number or $\infty$.]
In other words, there is no Borsuk pair with $X$, $Y$ both infinite-dimensional.
Any thoughts ?
[There exist several infinite-dimensional versions of the Classical Borsuk Theorem, e.g.,
when $X$ is reflexive, $Y$ is its dual, and $T$ is a demicontinuous mapping of monotone type.]
[I think the "big" problem (if it truly exists, of course ;-)) would be when both $X$, $Y$ are  infinite dimensional,
and the density character dens($X$) $\gg$ dens($Y$), i.e., when $X$ is "huge".]

REPLY [2 votes]: This is a  negative result, heavily relying on P. Dodos'
answer to Boundedness of nonlinear continuous functionals.
Let $\kappa$ be a measurable cardinal, let $K$ be the closed unit
ball of $\ell_{\infty}\left(\kappa\right)$, and suppose, ad absurdum,
that there exists a continuous mapping $F: K \rightarrow$ 
$\ell^{1}
$\ {${ 0} $}
that   is odd on the unit sphere of 
$\ell_{\infty}\left(\kappa\right)$.
Let $j:\{\ell}^{1}$ 
    $\rightarrow$ 
${\ell}^{2}$ 
be
the natural injection, 
and let $\rho$ be a metric generating the weak
topology on the closed unit ball of ${\ell}^{2}$. Define
next the continuous operator $T:K$ $\rightarrow$
 ${\ell}^{2}$\ {${ 0} $} by
$Tx:=$ $\dfrac{jFx}{\left\Vert jFx\right\Vert _{2}}$. 
Define also
the continuous function $f:\ K\rightarrow\mathbb{R}$,
expressed
by $f(x):=$ $\rho\left(0,Tx\right)^{-1}$. Then there is a subspace
$E$ of $\ell_{\infty}\left(\kappa\right)$ that is isomorphic
to $c_{0}\left(\kappa\right)$ s.t. $f(K$ $\cap$ $E$ $)$ is
bounded. Let $\left(P_{n}\right)$ be a sequence of ortho-projectors
on ${\ell}^{2}$, having $n$-dimensional ranges $Y_{n}$,
and s.t. $P_{n}$(x)$\ \rightarrow\ x $ 
for each $x$ in ${\ell}^{2}$.
Take also a sequence $\left(X_{n}\right)$ of $n$-dimensional subspaces
of $E$, and apply the classical Borsuk Theorem for the mappings 
$T_{n}$:=$P_{n}T:K\cap X_{n}$ $\rightarrow Y_{n}$,
in order to get a sequence $\left(x_{n}\right)$ s.t.   $T_{n}$ $x_{n}=$0$.$
Then it is easy to show that 
$Tx_{n}$ $\rightharpoonup0$ [i.e.,
weakly] in ${\ell}^{2}$, hence $f$ is not bounded on $K\cap E$,
a contradiction.
Therefore, the Conjecture is not true.<|endoftext|>
TITLE: modular eigenforms with integral coefficients [Maeda's Conjecture]
QUESTION [20 upvotes]: Are there infinitely many (linearly independent) cuspidal eigenforms for $\Gamma(1)$ with integer coefficients?
Someone told me that the Hecke algebra is conjectured to act irreducibly on the space of modular forms of level 1, so there would be no eigenforms if $\mathrm{dim} S_k > 1$, i.e. for $k \geq 12$.

REPLY [4 votes]: "He said (and I never understood this comment so feel free to fill me in) that S_k(1;Q) being irreducible as a Hecke module was related to (equivalent to?) a certain L-value not vanishing, and L-values tend to vanish occasionally when you look hard enough."
I dispute the impression of Hida with vanishing L-values. To precise this, a density statement is needed. The standard L-function technology whizzes from random matrices should expect that it doesn't vanish ever. In the same vein, Conrey conjectures that quadratic twists of weight 6+ forms never vanish aside from sign, though he kindly phrases it as "finitely many" as pointed out above.
http://www.aimath.org/~aimath/WWN/rmtapplications/rmtapplications.pdf
For weight 6 we have rank 2 vanishing for a few forms, as Dummigan lists: 95k6, 122k6, 260k6.
http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi
I expect no vanishing for weight 8+. To my knowledge, no rank 3 vanishing exists for weight 4+. My recollection (Stein 2000) is that, outside with Gamma1(N), there is one at level 122 (sic, as above) weight 2 form with quadratic sign that vanishes to order 1 with no self-dual functional equation sign (eps = -0.76822128 + 0.6401844i).
I am editing this now to explain L-function methods. The right random matrix idea is that L-values have cumulative distribution with $\sqrt t$ for small $t$. It is probably unnecessary though.
For rather look at the BSD analogue. There is $L(centre)/\Omega$ and the other side is up to few rational factors an integer. It is also a square. So it is "like" a random integral square up to size $\Omega$ as the Tamagawa and torsion and much smaller. The "probability" of an (even signed) L-function vanishing centrally can be thought as $\sqrt\Omega$ as a chance that a random integral square up to size $\Omega$ is 0 is just 1 in $\sqrt\Omega$.<|endoftext|>
TITLE: Reference for representation of Weyl group using r_ + c∂_
QUESTION [6 upvotes]: Take $W = S_n$ for simplicity, though other Weyl groups work too. Let $r_i$ denote the $i$th simple reflection acting on ${\mathbb A}^n$, and $\partial_i = 1/(x_i - x_{i+1}) ({\operatorname{Id}} - r_i)$ denote the corresponding divided difference operator.
It's easy to show that the operators $r_i + c \partial_i$ satisfy the Coxeter relations. I know I saw this in a Lascoux article, but there are so many that I'm hoping MathOverflow can tell me which one so I don't have to pore over the French, or can suggest some other canonical reference, the older the better.
Separately, I'd like to know if any author explicitly discusses these in the context of the Steinberg variety, where the $c$ should be the equivariant cohomology parameter corresponding to dilation of the cotangent bundle, I guess.

REPLY [4 votes]: The difference operators (as you presumably know) are defined in a paper of Bernstein–Gelfand–Gelfand on Schubert cells etc. which is probably roughly as old as you can get. The fact that the $r_i + c\partial_i$ satisfy the Coxeter relations is implied by (equivalent to, pretty much) the fact that the graded/degenerate affine Hecke algebra is isomorphic as a vector space to $\mathbb C[W]\otimes \mathbb C[x_1,x_2,\dotsc,x_n]$ (the operators give the action of the $\mathbb C[W]$ subalgebra on the polynomial representation of the degenerate affine Hecke algebra). The first references for the degenerate affine Hecke algebra are Drinfeld's paper Degenerate affine Hecke algebras and Yangians (for type A) and Lusztig's paper Cuspidal local systems and graded Hecke algebras, I, and the Lusztig paper that Stephen references, which is pure algebra. I think it also arises in some form in Cherednik's paper A new interpretation of Gelʹfand–Tzetlin bases.
The connection to the equivariant cohomology of the Steinberg is examined in Lusztig's cuspidal local systems papers (it is the easiest "Springer theory" case, i.e. where you don't have to worry about cuspidal local systems).<|endoftext|>
TITLE: Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?
QUESTION [25 upvotes]: How does one prove that the splitting of primes in a non-abelian extension of number fields is not determined by congruence conditions?

REPLY [11 votes]: Fix a number field $K$.  For an integer $m$, let $S_1(m,K)$ be the congruence classes $a$ mod $m$ that contain infinitely many primes $p$ such that $p \mid \mathfrak p$ for some prime $\mathfrak p$ of $K$ satisfying $f(\mathfrak p|p) = 1$. (That was a mouthful: $p$ is lying below some prime of $K$ with residue field degree 1.) 
If $K/\mathbf Q$ is Galois, then such $p$ are the primes splitting completely in $K$, up to finitely many exceptions (among the ramified primes).  That is, when $K/\mathbf Q$ is Galois, $S_1(m,K)$ is the set of congruence classes mod $m$ containing infinitely many primes which split completely in $K$. (The prime numbers that split completely in a number field are identical to the prime numbers that split completely in its Galois closure over $\mathbf Q$, so attempting to describe such "split sets" by congruence conditions could just as well assume the number field is Galois over $\mathbf Q$.  I am working over base field $\mathbf Q$ throughout.)
As Kevin has suggested, it is not obvious at first that these sets $S_1(m,K)$ have much structure, particularly that they contain $1$ mod $m$.  By the pigeonhole principle, any $S_1(m,K)$ is certainly a nonempty set, and it is a subset of the unit group $(\mathbf Z/m)^\times$ rather than just $\mathbf Z/m$, but this is kind of superficial.  
A good reason (the right reason?) that $1$ mod $m$ is in $S_1(m,K)$ is that $S_1(m,K)$ is actually a subgroup of $(\mathbf Z/m)^\times$.  In fact, under the usual identification of $\mathrm{Gal}(\mathbf Q(\zeta_m)/ \mathbf Q)$ with $(\mathbf Z/m)^\times$, $S_1(m,K)$ is the image of the restriction homomorphism $\mathrm{Gal}(K(\zeta_m)/K) \longrightarrow \mathrm{Gal}(\mathbf Q(\zeta_m)/\mathbf Q)$.  For a proof, see Theorem 3 at 
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf
and Theorem 4 there is a generalization where $(\mathbf Z/m)^\times$ is replaced with any Galois group of number fields.<|endoftext|>
TITLE: Cohomology of rigid-analytic spaces
QUESTION [12 upvotes]: Let $R$ be a complete discrete valuation ring and let $K$ be its field of fractions.  Suppose $X$ is a smooth rigid-anaytic space over $K$.  Often it is convenient to have a model of $X$ whose reduction has singularities which are as mild as possible--a semistable model.  This amounts to having an admissible covering of $X$ by open affinoids $X_i$, each of which has good reduction, such that the reductions of any pair $X_i$ and $X_j$ meet transversally, if at all.  (See the paper of Bosch/Lütkebohmert for definitions.)   Let us assume such a semi-stable model of $X$ exists.  Then the étale cohomology of $X$ can be computed from the combinatorics of the covering $X_i$, together with the étale cohomology of each $X_i$, via the weight spectral sequence of Rapaport-Zink.  
Now suppose I have an open affinoid $Z\subset X$ which happens to have good reduction.  My question is:  Does there admit a semi-stable model of $X$ for which $Z$ belongs to the covering?   Failing this, is there some sense one can make of my intuition that the cohomology of the reduction of $Z$ ought to contribute to the cohomology of $X$?
Feel free to edit/criticize my question to smithereens if you like.

REPLY [7 votes]: In dimension 1, the answer should be yes because the semi-stable reduction is well understood. The only problem would be the difference between the rigid analytic reduction and the algebraic reduction as a formal covering defines a reduced analytic reduction but non necessarily reduced algebraic reduction (= special fiber of the formal scheme associated to the formal covering). The answer should be in Bosch-Lütkebohmert's paper on the stable reduction of curves and in Fresnel-van der Put's book (the second one). 
In higher dimension, I am a litte skeptical. As Emerton pointed out, you could construct a model containning $\overline{Z}$ in its reduction at least birationally (result of Raynaud, explained in Melhmann's thesis in Münster). But it is not clear for me whether a semi-stable model dominating a given one model exist. Another difficulty in higher dimension is that the semi-stable reduction is not unique even in good reduction case (non-uniqueness of minimal birational model). However, you have two stable reductions whose irreducible components are all not uniruled, then there is a one-one birational correspondance between the components of two stable reductions (Abhyankar's lemma, I don't remember whether this requires desingularization).<|endoftext|>
TITLE: Asymptotics of iterated polynomials
QUESTION [5 upvotes]: Let the sequence $u_1, u_2, \ldots$ satisfy $u_{n+1} = u_n - u_n^2 + O(u_n^3)$.  Then it can be shown that if $u_n \to 0$ as $n \to \infty$, then $u_n = n^{-1} + O(n^{-2} \log n)$.  (See N. G. de Bruijn, Asymptotic methods in analysis, Section 8.5.)
This can be used to obtain asymptotics for $v_{n+1} = Av_n - Bv_n^2 + O(v_n^3)$, where $A$ and $B$ are constants.  Let $w_n = A^{-n} v_n$; this gives
$$ A^{n+1} w_{n+1} = A^{n+1} w_n - B A^n w_n^2 + O(A^n w_n^3)$$
and so
$$ w_{n+1} = w_n - BA^{-1} w_n^2 + O(w_n^3). $$
Then let $w_n = Ax_n/B$ to get
$$ Ax_{n+1}/B = Ax_n/B - B/A \cdot (Ax_n/B)^2 + O(x_n^3) $$
and after simplifying
$ x_{n+1} = x_n - x_n^2 + O(x_n^3)$.  This satisfies the initial requirements for $u_n$ (with some checking of the side condition); then substitute back.
But say I actually know that $u_{n+1} = P(u_n)$ for some polynomial $P$, with $P(z) = z - z^2 + a_3 z^3 + \cdots + a_d z^d.$   In this case it seems like it should be possible to get more explicit information about $u_n$.  Is there a known algorithm for computing an asymptotic series for $u_n$ as $n \to \infty$?

REPLY [2 votes]: As explained above by 002, solving Abel equation is indeed the key. Using the language of holomorphic dynamics, people would say that you are studying the dynamics of a polynomial near the parabolic fixed point $z=0$.
By a simple linear change of variables, the study of any such parabolic fixed point can be reduced to the study of $z \mapsto z+z^{2}+O(z^3)$. Then you can apply another change $w=-\frac{1}{z}$. Thus you are reduced to the study of $f(w)=w+1+O(1/w)$. If the real part $Re(w)$ is large enough you will obtain $f^{n}(w)=w+n+O(\log n)$. This will give you what you want (when going back to the z-variable).
The domain $Re(w)>R$ (for large $R$) looks like some kind of cardioid (in your particular case) when you visualize it in the z-variable (it's poetically called an attracting petal). 
All this material is explained in details in several books. One great example is the book by John Milnor on Complex dynamics. Here is a free (earlier) version:
http://www.math.sunysb.edu/cgi-bin/preprint.pl?ims90-5
So in summary the key words you should look for are: parabolic fixed point, Fatou coordinates, Leau-Fatou flower theorem (the last one describing those "petals")...<|endoftext|>
TITLE: Where are we working when we prove metamathematical theorems?
QUESTION [12 upvotes]: I am posting my comment from this question as a separate question, as was recommended to me. 
(EDIT: I'm sorry if it ended up being too similar a question, I just wanted to phrase it in the terminology I think of it in)
First, let me mention that I agree with Hilbert's formalism - when we make the claim "X is true", it is really a shorthand for: "If we label certain strings of symbols 'true', and we label certain rules for symbol manipulation 'truth-preserving', then we can manipulate our initial strings using our rules to reach X". If someone disagrees with our choice of initial strings, or our choice of manipulation rules (which, together, I'll call a "system"), then they will obviously come to some different conclusions. Now, mathematicians happen to have preferences about which systems to study, and those preferences are often motivated by apparent analogies between those systems and the real world, but no system is "right" or "wrong". 
I realize that this is not everyone's view of mathematics, but I wanted to explain my reasoning before asking my question:
In what system do logicians usually "prove" metamathematical theorems - i.e., claims about systems? As I mentioned, all theorems, metamathematical or otherwise, are really of the form "In the system Y, X is true." The potential problem I feel with some results like Godel's theorems is that X is a claim about systems which are not necessarily consistent with Y - how does this make sense?

REPLY [11 votes]: There are many flavors of "meta" in logic. Most make very minimal use of the metatheory. For example, the Montague Reflection Principle in Set Theory says the following: 

Metatheorem. For every formula $\phi(x)$ of the language of Set Theory, the following is provable: If $\phi(a)$ is true then there is an ordinal $\alpha$ such that $a \in V_\alpha$ and $V_\alpha \vDash \phi(a)$.

This is in fact an infinite collection of theorems, one for each formula $\phi(x)$. Gödel's theorem prevents ZFC from proving all of these instances at the same time. The proof of this metatheorem is by induction on the length of $\phi(x)$. The requirements on the metatheory are very minimal, all you need is enough combinatorics to talk about formulas, proofs, and enough induction to make sense of it all. This is by far the most common flavor of "meta" in logic. There really is not much to it. Most logicians don't worry about the metatheory in this context. Indeed, it is often irrelevant and, like most mathematicians, logicians usually believe in the natural numbers with full induction.
However, sometimes there is need for a much stronger metatheory. For example, the metatheory of Model Theory is usually taken to be ZFC. Gödel's Completeness Theorem is a nice contrasting example.

Completeness Theorem. Every consistent theory is satisfiable.

Recall that a theory is consistent if one cannot derive a contradiction form it, while a theory is satisfiable if it has a model. The consistency of a theory is a purely syntactic notion. When the theory is effective, consistency can be expressed in simple arithmetical terms in a manner similar to the one discussed above. However, satisfiability is not at all of this form since the models of a theory are typically infinite structures. This is why this theorem is usually stated in ZFC. For the Completeness Theorem, one can often get by with substantially less. For example, the system WKL0 of second-order arithmetic suffices to prove the Completeness Theorem for countable theories (that exist in the metaworld in question), but such a weak metatheory would be nothing more than a major inconvenience for model theorists.
Sometimes there are multiple choices of metatheory and no consensus on which is most appropriate. This happens in the case of Forcing in Set Theory, which has three common points of view:

Forcing is a way to extend a countable transitive model of ZFC' to a larger model of ZFC' with different properties. Here ZFC' is a finite approximation to ZFC, which makes this statement non-vacuous in ZFC because of the Reflection Principle above.
Forcing is a way to talk about truth in an alternate universe of sets. Here, the alternate universe is often taken to be a Boolean-valued model, formalized within the original universe.
Forcing is an effective way to transform a contradiction some type of extensions of ZFC into contradictions within ZFC itself. Here, ZFC could be replaced by an extension of ZFC.

The last point of view, which is the least common, is essentially arithmetical since it only talks about proofs. The second point of view is well suited for hard-core platonists who believe in one true universe of sets. The first point of view, which is probably most popular, essentially takes ZFC as a metatheory much like model theorists do.
Since there is no consensus, set theorists will often talk as if both $V$ and its forcing extension $V[G]$ are absolutely real universes of sets. Although this point of view is hard to justify formally, it is possible to makes sense of it using any one of the three viewpoints above and it has the advantage that it makes it easier to express ideas that go into forcing constructions, which is what set theorists really want to talk about.
Finally, the idea of analyzing which systems prove the consistency of other systems is very common in logic. In Set Theory, these systems often take the form of large cardinal axioms. In Second-Order Arithmetic these systems often take the form of comprehension principles. In First-Order Arithmetic these systems often take the form of induction principles. Together these form an incredibly long consistency strength hierarchy stretching from extremely weak basic arithmetical facts to incredibly deep large cardinal axioms. A very significant part of logic deals with studying this hierarchy and, as Andrej Bauer commented, logicians are usually very aware of where they are sitting in this hierarchy when proving metatheorems.<|endoftext|>
TITLE: Cohomology of Lie groups and Lie algebras
QUESTION [25 upvotes]: The length of this question has got a little bit out of hand. I apologize.
Basically, this is a question about the relationship between the cohomology of Lie groups and Lie algebras, and maybe periods.
Let $G$ be a complex reductive (connected) Lie group and let $T$ be a maximal torus of $G$. Set $\mathfrak{g}=Lie(G)$ and $\mathfrak{t}=Lie(T)$. Notice that $\mathfrak{t}$ has a natural integral structure: $\mathfrak{t}=\mathfrak{t}(\mathbf{Z})\otimes_\mathbf{Z}\mathbf{C}$ where $\mathfrak{t}(\mathbf{Z})$ is formed by all $x$ such that $\exp(2\pi ix)$ is the unit $e$ of $G$.
All there is to know about $G$ can be extracted from the (covariant) root diagram of $G$, which is formed by $\mathfrak{t}(\mathbf{Z})$, the sublattice $M$ corresponding to the connected component of the center of $G$ (this is a direct summand) and the coroot system $R$ of $G$, which is included in some complementary sublattice of $\mathfrak{t}(\mathbf{Z})$. For example, $\pi_1(G)$ is the quotient of $\mathfrak{t}(\mathbf{Z})$ by the subgroup spanned by $R$. See e.g. Bourbaki, Groupes et alg`ebres de Lie IX,  4.8-4.9 (Bourbaki gives a classification in terms of compact groups, but this is equivalent).
The question is how to extract information on the cohomology of $G$ (as a topological space) from the above.
For the complex cohomology there are no problems whatsoever. We only need $\mathfrak{g}$: restricting the complex formed by the left invariant forms to the unit of $G$ we get the standard cochain complex of $\mathfrak{g}$.
The next step would be the rational cohomology. One possible guess on how to get it would be to notice that $\mathfrak{g}$ is in fact defined over $\mathbf{Q}$. So one can find an algebra $\mathfrak{g}(\mathbf{Q})$ such that $\mathfrak{g}=\mathfrak{g}(\mathbf{Q})\otimes_\mathbf{Q}\mathbf{C}$. We can identify $H^{\bullet}(\mathfrak{g},\mathbf{C})=H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})\otimes\mathbf{C}$ and so we get two rational vector subspaces in the complex cohomology of $G$. One is the image of of $H^{\bullet}(G,\mathbf{Q})$ and the other is the image of $H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})$ under $$H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})\to H^{\bullet}(\mathfrak{g},\mathbf{C})\to H^\bullet(G,\mathbf{C})$$ where the last arrow is the comparison isomorphism mentioned above.
1). What, if any, is the relationship between these subspaces? More precisely, apriori the second subspace denends on the choice of $\mathfrak{g}(\mathbf{Q})$ (I don't see why it shouldn't, but if in fact it doesn't, I'd be very interested to know) and the question is if there is a $\mathfrak{g}(\mathbf{Q})$ such that the relationship between the above subspaces of $H^{\bullet}(G,\mathbf{C})$ is easy to describe.
Notice that this is somewhat similar to what happens when we compare the cohomology of the algebraic de Rham complex with the rational cohomology. Namely, suppose we have a smooth projective or affine algebraic variety defined over $\mathbf{Q}$; its algebraic de Rham cohomology (i.e. the (hyper)cohomology of the de Rham complex of sheaves) sits inside the complex cohomology, but this is not the same as the image of the topological rational cohomology. Roughly speaking, the difference between the two is measured by periods, e.g. as defined by Kontsevich and Zagier.
2). If question 1 has a reasonable answer, then what about the integral lattice in $H^{\bullet}(G,\mathbf{C})$? Again, a naive guess would be to take a $\mathfrak{g}(\mathbf{Z})$ such that $\mathfrak{g}(\mathbf{Z})\otimes\mathbf{C}=\mathfrak{g}$. At present, I'm not sure whether such an integral form exist for any reductive $G$ (and I'd be very interested to know that), but in any case it exists for $SL(n,\mathbf{C})$. By taking the standard complex of $\mathfrak{g}(\mathbf{Z})$ and extending the scalars we get a lattice in $H^{\bullet}(\mathfrak{g},\mathbf{C})\cong H^{\bullet}(G,\mathbf{C})$. Is there a choice of $\mathfrak{g}(\mathbf{Z})$ for which this lattice is related to the image of the integral cohomology in $H^\bullet(G,\mathbf{C})$ in some nice way?
3). If even question 2. has a reasonable answer, then what about the integral cohomology itself? Here, of course, the answer is interesting even up to isomorphism. A very naive guess would be to take an appropriate integral form $\mathfrak{g}(\mathbf{Z})$ as in question 2 and compute the integral cohomology of the resulting standard complex.

REPLY [6 votes]: This is rather late, and not a full answer to your question, but may be interesting nonetheless. 
Concerning point 1), I would like to propose the following natural refinement which avoids the problems mentioned in Pavel Etingof's answer. If you want to know the relation between cohomology of the group and the Lie algebra over $\mathbb{Q}$, you should work with $\mathbb{Q}$-forms of both. Take $G_{\mathbb{Q}}$ a form of $G$ defined over $\mathbb{Q}$, and take $\mathfrak{g}_{\mathbb{Q}}$ the associated Lie algebra (in the sense of algebraic groups). Then you want to compare the algebraic de Rham cohomology of $G_{\mathbb{Q}}$ with the Lie algebra cohomology of $\mathfrak{g}_{\mathbb{Q}}$. As Pavel Etingof pointed out, these $\mathbb{Q}$-vector subspaces of the cohomology of $G$ over $\mathbb{C}$ depend on the choice of form. Maybe there is an explicit formula which from a cocycle in $H^1_{\text{ét}}(\mathbb{Q},\operatorname{Aut}G)$ produces the factors relating the subspaces - the factor $i$ in Pavel Etingof's answer is most certainly related to defining $SU(3)$ in terms of the complex conjugation involution.
Concerning point 2), the same as in point 1) applies to $\mathbb{Z}$-forms: choose a $\mathbb{Z}$-form of $G$ and take its associated Lie algebra. You will most likely end up getting Kostant's $\mathbb{Z}$-forms mentioned in Scott Carnahan's comment.
In the special case $GL_n$ and cohomology in top degree, your question 2) and 3) are discussed in a preprint of Annette Huber and Wolfgang Soergel, arXiv version can be found here.
The goal of that paper is to compare different integral structures in the top cohomology of $GL_n$: one integral structure comes from the natural $\mathbb{Z}$-form of $\mathfrak{gl}_n$, another from the suspensions of Chern classes in de Rham cohomology, and the third one from the dual of the fundamental class in singular cohomology. They work out the explicit comparison factors. The comparison between de Rham and singular cohomology involves the usual period $2\pi i$, and the comparison between de Rham cohomology and the subspace coming from the Lie group cohomology is in fact a rational factor (so no further periods). 
The rational factor in the comparison between Lie algebra and de Rham cohomology that is worked out in the Huber-Soergel paper is $\prod_{j=1}^n(j-1)!$. It is interesting to note that these same factors appear when comparing homotopy and homology of $GL_n(\mathbb{C})$ - the Hurewicz map sends a generator of $\pi_{2j-1}GL_n(\mathbb{C})$ to some multiple of a generator of $H_{2j-1}(GL_n(\mathbb{C}),\mathbb{Z})$, and this multiple is $(j-1)!$. I am fairly convinced now that this is the way the subspaces are related in case $GL_n$ - the group cohomology classes in degree $2j-1$ should be the $(j-1)!$-th multiple of the generators of Lie algebra cohomology.
One would hope that such a description applies  to other Lie groups and that the rational factor can be explained by some Weyl group combinatorics in general, but there is some work to be done for that...  
Concerning periods there is a philosophical reason why  the comparison between de Rham cohomology and singular cohomology should only involve rational multiples of powers of $2\pi i$. If $G$ is a reductive group over $\mathbb{Q}$, we can view it as a variety and as such it has a mixed Tate motive; therefore, all periods should be rational multiples of powers of $2\pi i$.<|endoftext|>
TITLE: Points of a variety defined by Galois descent
QUESTION [8 upvotes]: Let k be a perfect field.  By a k-variety, I shall mean a geometrically reduced separated scheme of finite type over k.  I think that is enough conditions that the following data determine an affine k-variety:

A subset $X(\bar{k})$ of $\bar{k}^n$ which is defined by polynomials
A continuous action of $\mathop{\mathrm{Gal}}(\bar{k}/k)$ on $X(\bar{k})$, such that each $\sigma \in \mathop{\mathrm{Gal}}(\bar{k}/k)$ acts as $\sigma \circ f$ where f is a $\bar{k}$-regular map

When I say that these data determine an affine k-variety, I mean that there is a unique affine k-variety X whose $\bar{k}$-points are $X(\bar{k})$ with the correct Galois action.
Given these data, I want to work out the functor of points of X (which I consider to have domain the category of k-algebras).  You can do that by following through the proof that these data determine a k-variety: first construct the coordinate ring A of X, as the Galois-fixed points of the ring of regular functions $X(\bar{k}) \to \bar{k}$; then $X(R) = \mathop{\mathrm{Hom}}(A, R)$ for any k-algebra R.
But if L is an algebraic extension of k, then there is a much simpler way of working out the L-points of X: just take the subset of $X(\bar{k})$ fixed by $\mathop{\mathrm{Gal}}(\bar{k}/L)$.
If L is a transcendental extension of k (or even a k-algebra which is not a field), is there a direct way of writing down the L-points of X which does not require going through the coordinate ring (or essentially equivalently, going through defining equations for X)?

REPLY [12 votes]: The following seems to give a reasonable affirmative answer which avoids computing the coordinate ring directly, and replaces condition (2) with the more natural condition that the subset $\Sigma := X(\overline{k})$ in (1) is stable under the action of the Galois group on $\overline{k}^n$. 
Let's be cleaner by working more generally over an arbitrary (not necessarily perfect) field $k$ and with geometrically reduced closed subschemes $X$ in a fixed separated $k$-scheme $Y$ locally of finite type.  (Note: now affine schemes are gone; can take $Y$ to be an affine space, but this is irrelevant.)  The 
${\rm{Gal}}(k_s/k)$-stable set $\Sigma = X(k_s)$ in $Y(k_s)$ recovers $X$ as follows. For a $k$-algebra $A$, $X(A)$ is the ${\rm{Gal}}(k_s/k)$-invariants in $X(A_{k_s})$, so we just need to describe $X(A_{k_s})$ as a ${\rm{Gal}}(k_s/k)$-stable subset of $Y(A_{k_s})$. The description in this latter case will be in terms of $\Sigma$, and the ${\rm{Gal}}(k_s/k)$-stability of $\Sigma$ inside of $Y(k_s)$ will ensure that the description we give for  $X(A_{k_s})$ is ${\rm{Gal}}(k_s/k)$-stable inside of $Y(A_{k_s})$.  That being noted, we rename $k_s$ as $k$ so that $k$ is separably closed and $\Sigma$ is simply a set of $k$-rational points of $Y$ (so the notation is now marginally cleaner). 
First assume $A$ is geometrically reduced in the sense that $A_K$ is reduced for any extension field $K/k$.   Since $X(A)$ is the direct limit (inside $Y(A)$) of the $X(A_i)$ as $A_i$ varies through $k$-subalgebras of finite type in $A$ (all of which are geometrically reduced), we may assume $A$ is finitely generated over $k$. 
Then the $k$-points are Zariski-dense (as $k = k_s$) and so the condition on $y \in Y(A)$ that it lies in $X(A)$ is that $y(\xi) \in \Sigma$ for all $k$-points $\xi$ of $A$.  That describes $X(A)$ for any (possibly not finitely generated) $k$-algebra $A$ that is geometrically reduced. In general, to check if $y \in Y(A)$ lies in $X(A)$ amounts to the same for each local ring of $A$, so we can assume $A$ is local.  Then the condition for $y$ to be in $X(A)$ is exactly that there is a local map of local $k$-algebras $B \rightarrow A$ with $B$ geometrically reduced such that $y$ is in the image of $X(B)$ under the induced map $Y(B) \rightarrow Y(A)$.  I don't claim this formulation is the best way to think about it, but it "works".  
Of course, one can apply this process to any ${\rm{Gal}}(k_s/k)$-stable subset $\Sigma$ of $Y(k_s)$ provided that we first replace $\Sigma$ with with the set of $k_s$-points of its Zariski-closure in $Y_{k_s}$. Then we just obtain the Galois descent $X$ of the Zariski closure in $Y_{k_s}$ of $\Sigma$.  In general $X(k_s)$ may be larger than $\Sigma$, but nonetheless $\Sigma$ is Zariski-dense in $X_{k_s}$.  This is perfectly interesting in practice, regardless of whether or not $\Sigma$ is equal to $X_{k_s}$, since it is what underlies the construction of derived groups, commutator subgroups, images, orbits, and related things in the theory of linear algebraic groups over a general field.  For example, the $k$-group ${\rm{PGL}}_n$ is its own derived group in the sense of algebraic groups, but the commutator subgroup of ${\rm{PGL}}_n(k_s)$ is a proper subgroup whenever $k$ is imperfect and ${\rm{char}}(k)|n$. 
To give a nifty application, suppose one begins with an arbitrary closed subscheme $X'$ in $Y$ (such as $X' = Y$!), then forms the ${\rm{Gal}}(k_s/k)$-stable set $X'(k_s)$ (which could well be empty, or somehow really tiny), and then applies the above procedure to get a geometrically reduced closed subscheme $X$ in $X'$.  What is it?  It is the maximal geometrically reduced closed subscheme of $X'$, and one can check its formation is compatible with products (as well as separable extensions $K/k$, such as completions $k_v/k$ for a global field $k$).  If $k$ is perfect then $X = X'_{\rm{red}}$, so this is more interesting when $k$ is imperfect.  It is especially interesting in the special case when $X'$ is equipped with a structure of $k$-group scheme.  Then $X$ is its maximal smooth closed $k$-subgroup, since geometrically reduced $k$-groups locally of finite type are smooth.  So what?  If one is faced with the task of studying the Tate-Shararevich set for such an $X'$ (e.g., maybe $X'$ is a nasty automorphism scheme of something nice) then all that really intervenes is $X$ since it captures all of the local points, so for some purposes we can replace the possibly bad $X'$ with the smooth $X$. (This trick is used in the proof of finiteness of Tate-Shafarevich sets for arbitrary affine groups of finite type over global function fields.)  But beware: if the $k$-group $X'$ is connected (and $k$ is imperfect) then $X$ may be disconnected and have much smaller dimension; see Remark C.4.2 in the book "Pseudo-reductive groups" for an example.<|endoftext|>
TITLE: slice-ribbon for links (surely it's wrong)
QUESTION [11 upvotes]: The slice-ribbon conjecture asserts that all slice knots are ribbon.   
This assumes the context:  
1) A `knot' is a smooth embedding $S^1 \to S^3$. We're thinking of the 3-sphere as the boundary of the 4-ball $S^3 = \partial D^4$.  
2) A knot being slice means that it's the boundary of a 2-disc smoothly embedded in $D^4$. 
3) A slice disc being ribbon is a more fussy definition -- a slice disc is in ribbon position if the distance function $d(p) = |p|^2$ is Morse on the slice disc and having no local maxima.  A slice knot is a ribbon knot if one of its slice discs has a ribbon position.
My question is this.  All the above definitions have natural generalizations to links in $S^3$.  You can talk about a link being slice if it's the boundary of disjointly embedded discs in $D^4$.  Similarly, the above ribbon definition makes sense for slice links.   Are there simple examples of $n$-component links with $n \geq 2$ that are slice but not ribbon?  Presumably this question has been investigated in the literature, but I haven't come across it.  Standard references like Kawauchi don't mention this problem (as far as I can tell).

REPLY [12 votes]: Ryan, I think this is an open problem. The best related result I know is a theorem of Casson and Gordon [A loop theorem for duality spaces and fibred ribbon knots. Invent. Math. 74 (1983)] saying that for a fibred knot that bounds a homotopically ribbon disk in the 4-ball, the slice complement is also fibred. 
More precisely, they are assuming that the knot K bounds a disk R in the 4-ball such that the inclusion 
$S^3 \smallsetminus K \hookrightarrow D^4 \smallsetminus R$ 
induces an epimorphism on fundamental groups. If one glues R to a fibre of the fibration $S^3 \smallsetminus K \to S^1$ to obtain a closed surface F, then the statement is that the monodromy extends from F to a solid handlebody which is a fibre of a fibration $D^4 \smallsetminus R \to S^1$ extending the given one on the boundary.<|endoftext|>
TITLE: Mirror symmetry mod p?! ... Physics mod p?!
QUESTION [21 upvotes]: In his answer to this question, Scott Carnahan mentions "mirror symmetry mod p". What is that? 
(Some kind of) Gromov-Witten invariants can be defined for varieties over fields other than $\mathbb{C}$. Moreover other things that come up in mirror symmetry, like variation of Hodge structure, and derived categories of coherent sheaves, also make sense. (Though I can't imagine that it's possible to talk about Fukaya categories...) Can we formulate any sort of sensible mirror symmetry statement, similar to say that of Candelas-de la Ossa-Green-Parkes relating Gromov-Witten invariants of a quintic threefold to variation of Hodge structure of the mirror variety, when the varieties are over some field other than $\mathbb{C}$? In particular, can we do anything like this for fields of positive characteristic?
I googled "arithmetic mirror symmetry" and "mirror symmetry mod p", and I found some stuff about the relationship between the arithmetic of mirror varieties, but nothing about Gromov-Witten invariants. I did find notes from the Candelas lectures that Scott referred to, but I wasn't able to figure out what was going on in them.
More generally, there are many examples of mathematical statements about complex algebraic varieties which come from physics/quantum field theory/string theory. Some of these statements (maybe with some modification) can still make sense if we replace "variety over $\mathbb{C}$ with "variety over $k$", where $k$ is some arbitrary field, or a field of positive characteristic, or whatever. Are there any such statements which have been proven?
Edit: I'm getting some answers, and they are all sound very interesting, but I'm still especially curious about whether anybody has done anything regarding Gromov-Witten invariants over fields other than $\mathbb{C}$.

REPLY [9 votes]: For fixed integers $g,n$, any projective scheme $X$ over a field $k$, and a linear map $\beta:\operatorname{Pic}(X)\to\mathbb Z$, the space $\overline{M}_{g,n}(X,\beta)$ of stable maps is well defined as an Artin stack with finite stabilizer, no matter the characteristic of $k$. You can even replace $k$ by $\mathbb Z$ if you like.
Now if $X$ is a smooth projective scheme over $R=\mathbb Z[1/N]$ for some integer $N$, then $\overline{M}_{g,n}(X,\beta) \times_R \mathbb Z/p\mathbb Z$ is a Deligne-Mumford stack for almost all primes $p$. For such $p$,  $\overline{M}_{g,n}(X,\beta) \times_R \mathbb Z/p\mathbb Z$ has a virtual fundamental cycle, and so you have well-defined Gromov-Witten invariants. This holds for all but finitely many $p$. Nothing about $\mathbb C$ here, that is my point, the construction is purely algebraic and very general.
It is when you say "Hodge structures" then you better work over $\mathbb C$, unless you mean $p$-Hodge structures. 
As far as mirror symmetry in characteristic $p$, much of it is again characteristic-free. For example Batyrev's combinatorial mirror symmetry for Calabi-Yau hypersurfaces in toric varieties is simply the duality between reflexive polytopes. You can do that in any characteristic, indeed over $\mathbb Z$ if you like.<|endoftext|>
TITLE: Is every lattice the fixed-point set of an order endomorphism of ⋄^n?
QUESTION [5 upvotes]: (Asked by Nathaniel Hellerstein on the Q&A board at JMM)
Let ⋄ be the 4 element lattice

  τ
 / \
i   j
 \ /
  f

Is every lattice isomorphic to the fixed point lattice of some order-preserving function from ⋄n→⋄n?

REPLY [5 votes]: For all finite lattices, the answer is Yes. 
More generally, for all complete lattices, the answer is Yes, and for all incompleteness lattices, the answer is  No.
(Complete = every set has a LUB and GLB.)
Let me first give the positive result. Suppose that L is a complete lattice. Every lattice is naturally a sub-order of the power-set lattice P(L), by associating each point b with it's lower cone b* = { a in L |  a <= b }. This map is clearly order-preserving. (Note: it is not a lattice embedding, however, since (b* v c*) is the union of two cones, rather than the cone of (b v c). ) Thus, L is order-isomorphic to the set of lower cones. Define f:P(L) to P(L) by 

f(A) = (sup A)*. 

That is, f(A) is the lower cone of (sup A). This is the smallest lower cone containing A. Note that (sup A) is an element of L, since L is complete. The map f is order preserving, since if A is a subset of B, then sup A <= sup B.
Clearly, f(b*) = b* for any b in L, since b is the sup of b*. Conversely, if F(A) = A, then A = b* for b = sup A. That is, A is a lower cone. Thus, the fixed points of f are exactly the lower cones of L, which are order-isomorphic to L, as desired. 
Finally, to make the connection with your Diamond lattice, note that P(L) is simply 2L, a power of the 2 element Boolean algebra. Since Diamond is 22, we can view P(L) as a power of Diamond. (Add a dummy coordinate if L is odd finite size.)
Now, let's consider the negative result. Every power of Diamond is isomorphic as I mentioned earlier to a power set P(J) for some set J. Suppose that f:P(J) to P(J) is an order-preserving map from P(J) to P(J). I claim that the set of fixed points of f must have a smallest element. To see this, let B be the intersection of all A such that f(A) subset A. For any such A it follows that B subset A, so f(B) subset f(A), and so f(B) subset B. So B is the smallest with f(B) subset B. But since applying f gives f(f(B)) subset f(B), it follows that f(B)=B, as desired. By working above any given collection of fixed points, the same argument shows that the collection of fixed points is complete. This establishes:
Theorem. A lattice is complete if and only if it is isomorphic to the set of fixed points of an order-preserving endomorphism of a power set lattice P(J). 
Note that many lattices are not complete. For example, the positive integers, as Neel mentioned in his answer. So these lattices are never the set of fixed points of an order-preserving map on a power set lattice, and consequently the same for the powers of Diamond.<|endoftext|>
TITLE: Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?
QUESTION [12 upvotes]: Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?
(Asked by bcross at math.iuiui.edu on the Q&A board at JMM.)

REPLY [9 votes]: This is an elaboration of Qiaochu Yuan's prior comment: there are complex solutions (in fact, infinitely many) to $e^t-1 = t$, and then $e^{tx}$ is a solution.<|endoftext|>
TITLE: Does ⬦ generate all De Morgan algebras?
QUESTION [7 upvotes]: (Asked by Nathaniel Hellerstein on the Q&A board at JMM)
This question is about De Morgan algebras (see also Wikipedia), which are something like Boolean algebras, but with a different weaker sense of the complement ∼. Namely, a De Morgan algebra is a bounded distributive lattice with an involution ∼ satisfying De Morgan's laws.
Let ⬦ be the four element De Morgan algebra that is not a Boolean algebra, pictured below.

    1

i      j

    0

where ∼0 = 1, ∼1 = 0, but ∼i = i and ∼j = j, so i and j and self-dual with respect to ∼. This algebra seems to express one of the fundamental differences between De Morgan algebras and Boolean algebras.
Question. Does the algebra ⬦ generate all De Morgan algebras, in the sense that every De Morgan algebra is a subalgebra of a homomorphic image of a product of ⬦?
Please see the related Birkhoff's HSP Theorem in universal algebra, concerning varieties of algebras closed under H, S, and P (homomorphic image, subalgebra and product).
(Edit: I edited the question to express the question as I understood it. I'm not sure whether the OP intended SHP as stated or HSP, which would conform with Birkhoff's theorem. Probably it was intended to take the variety generated by ⬦, that is, close  {⬦} under H, S and P. The question then is whether this is equal to the class of all De Morgan algebras. Please revert if my edits are off-base.-JDH]
The ⬦ algebra can also be defined in terms of the usual 2 element Boolean algebra { f, τ } by using pairs denoted a/b, with the ∧ and ∨ operations defined coordinate-wise, but where, as mentioned by Dorais, the operation ∼ exchanges coordinates in addition to negating them, making for a "twisted square".

     1 = τ/τ

i = τ/f    j = f/τ

     0 = f/f


~(a/b)=(~b/~a)
(a/b)∧(c/d) = (a∧c)/(b∧d)
(a/b)∨(c/d) = (a∨c)/(b∨d)

REPLY [3 votes]: Yes, see http://www.math.uic.edu/~kauffman/DeMorgan.pdf<|endoftext|>
TITLE: What is the standard reference on "infinitesimal space" in algebraic geometry??
QUESTION [7 upvotes]: infinitesimal 'spaces' is a serious issue in noncommutative (and commutative) geometry: they serve as a base of a Grothendieck-Berthelot crystalline theory and are of big importance
for the D-module theory.
Can anybody point out the standard reference for this topic? I tried to look for it at nLab, but it seems it did not tell the reference in the language of algebraic geometry. 
I am not very familiar with French,so the English manuscript is better, however, French one is fine.

REPLY [5 votes]: I think that the best reference for the infinitesimal site, especially if one is motivated to learn crystalline cohomology, is found in Grothendieck's lectures "Crystals and the De Rham Cohomology of Schemes", notes by Coates and Jussila.  This appears, in English, as number 9 of 10 in the famous "Dix Exposes".  It is an absolutely beautiful, and very readable article.  The entire Dix Exposes can be obtained freely online, as a PDF file, from Leila Schnepps's Grothendieck Circle website.<|endoftext|>
TITLE: Database of polyhedra
QUESTION [11 upvotes]: As part of many hobbies (origami, sculpting, construction toys) I often find myself building polyhedra from regular polygons.  I am intimately familiar with all of the Archimedean and Platonic solids, and can construct most of the other isohedra, deltahedra, and Johnson solids from memory. The smaller prisms, antiprisms, and trapezohedra are of course trivial.  However, I often forget the precise arrangement of faces and vertices for some of the Johnson solids and most of the Catalan solids.  Thus, the question that I pose is this:
Where can I find the most complete, robustly indexed, and searchable database of polyhedra?
I would like to use such a database to answer, in short order, questions of the following nature:
Which solid is comprised of exactly eight hexagons and six squares?  Which solids are comprised of less than 10 triangles, eight squares, and six hexagons?  How many solids can be constructed with exactly 24 edges?
What solid with 12 vertices has the most edges (or faces)?  etc...
I imagine that such a database does not exist and I am going to be forced to create one, so answers suggesting features for such a database (likely to be web-based) are welcome as well.

REPLY [4 votes]: If you are interested in abstract polyhedra you can consult the atlas of small regular polytopes put together by Michael Hartley, however some of the links seem to be broken right now (I'll follow up with him, so you don't have to).
http://www.abstract-polytopes.com/<|endoftext|>
TITLE: Is there Grothendieck Riemann Roch for abelian category?
QUESTION [16 upvotes]: From the answers in noncommutative algebraic geometry, one can take abelian category as a scheme(commutative or noncommutative). So I wonder whether anyone ever developed the Grothendieck Riemann Roch theorem for abelian category. I think if such notion exists, it should be Grothendieck Riemann Roch for noncommutative scheme. 
The motivation for this question is my interest in representation theory and the question asked here:Grothendieck Riemann Roch for flag variety of Lie algebra. According to David's answer.One can interpreter this formula as Weyl-Character formula. So I wonder ask more questions.
What is the Grothendieck Riemann Roch for category of D-modules on flag variety of Lie algebra?
More generally, what is the Grothendieck Riemann Roch for category of D-module on commutative scheme?
For the second question, because quasi coherent sheaf on DeRham stack is D-module on a scheme. So one could interpreter the DeRham stack as category of D-modules on a scheme. Then we obtain the Grothendieck Riemann Roch for DeRham stack.

REPLY [23 votes]: More general setting
The question actually fits into the more general setting that is described in David Ben-Zvi's answer to this MO question (which is about categorification of the Chern character). 
HRR for dg algebras
Shklyarov is the one who developped RR theorem for noncommutatve derived schemes (by this one should understand smooth proper DG algebras): http://arxiv.org/abs/0710.1937.
Shklyarov's result has been improved recently by Petit, Lunts, and also Polishchuk-Vaintrob in the context of matrix factorizations. 
Let me explain what the statement is. Let $A$ be a proper and homologically smooth dg algebra $A$ (proper means that $\sum_n dim(H^n(A))<+\infty$, and homologicaly smooth means that $A$ is perfect in $D(A\otimes A^{op})$). Let $M$ be a perfect $A$-module. There is a trace map $ch:Hom_{D^{perf}(A)}(M,M)\to HH_0(A)$ (see e.g. this paper of Caldararu-Willerton for a very nice description of the kind of traces I am speaking about), which you can consider as being the Chern character. 
Now for an $A$-module $M$ and an $A^{op}$-module $N$ we can consider the $k$-module $N\otimes_AM$ (all my tensor products are derived). Then for $f:M\to M$ and $g:N\to N$ we can consider $ch(g\otimes f)=str(g\otimes f)\in HH_0(k)=k$. Finally the formula is 
$$
ch(g)\cup ch(f)=ch(g\otimes f)
$$
where $\cup:HH_*(A^{op})\otimes HH_*(A)\to HH_*(k)=k$ is the so-called (categorical) Mukai pairing. This is actually more a Lefshetz type formula. 
The Todd class is actually hidden in the Mukai pairing (the point is that for associative algebras there is no analogon neither for the Todd class, nor for the usual pairing given by integration). 
RR for D-modules
To my knowledge the first one who proved a RR Theorem for D-modules is Laumon (Sur la categorie derivee des D-modules filtres, Algebraic Geometry, M. Raynaud and T. Shioda eds, Lecture Notes in Math. Springer-Verlag 1016 pp. 151–237, 1983). Then Schapira and Scheinders also considered it (Index theorem for elliptic pairs II. Euler class and relative index theorem, Asterisque 224 Soc. Math. France, 1994) and made a very important conjecture which has been proved by Bressler-Nest-Tsygan using methods of deformation quantization ( http://arxiv.org/abs/math/9904121 and http://arxiv.org/abs/math/0002115) developped by Fedosov. 
There is also a paper of Engeli and Felder that gives a Lefschetz type formula. Their approach has been later clarified by Ramadoss (he has many paper on this subject that you can find on arXiv). 
The subject really moved to deformation quantization. You can learn a lot about all this (with also more details on who one should credit for what) in Section 4, 5 and 6 of this book by Kashiwara and Schapira. 
If you have a formal noncommutative deformation $A_\hbar$ of the structure sheaf $\mathcal O_X$ (maybe as a twisted resheaf, or algebroid stack - this is what Kashiwara and Schapira call a DQ algebroid), then you can play the same game as with a smooth proper dg algebras : define a trace with values in Hochschild homology, and state a HRR type Theorem (better, Lefshetz formula) about the compatibility of the cup product on Hochschild homology with the composition of kernels. The reason for that relies on some finiteness and duality properties for cohomologicall complete $A_\hbar$-modules. The main difficulty is then to prove such a result. 
To conclude this paragraph, let me observe that (the Rees algebra of) $\mathcal D_X$ can be viewed as a deformation of $\mathcal O_{T^*X}$. 
Relation between the question for nc schemes and D-modules
Last but not least, the relation between the derived non-commutative geometry and deformation quantization stuff is adressed in a very recent preprint of Petit: his strategy to prove HRR for DQ algebroids is to use the result for smooth proper DG algebras. Namely, he proves that some derived category of cohomologically complete modules over a DQ-algebroid on a projective variety has a compact generator. 
This subject is currently very active. 
I apologize for that this answer reduces to a (non-exhaustive) list of references. If you have a more specific question I can tell you where to go in these references in order to (hopefully) find an answer.<|endoftext|>
TITLE: Galoisian sets of prime numbers
QUESTION [67 upvotes]: The question is about characterising the sets $S(K)$ of primes which split completely in a given galoisian extension $K|\mathbb{Q}$.  Do recent results such as Serre's modularity conjecture (as proved by Khare-Wintenberger), or certain cases of the Fontaine-Mazur conjecture (as proved by Kisin), have anything to say about such subsets, beyond what Class Field Theory has to say ?
I'll now introduce some terminology and recall some background.
Let $\mathbb{P}$ be the set of prime numbers.  For every galoisian extension $K|\mathbb{Q}$, we have the subset $S(K)\subset\mathbb{P}$ consisting of those primes which split (completely) in $K$.  The question is about characterising such subsets; we call them galoisian subsets.
If $T\subset\mathbb{P}$ is galoisian, there is a unique galoisian extension $K|\mathbb{Q}$ such that $T=S(K)$, cf. Neukirch (13.10).  We say that $T$ is abelian if $K|\mathbb{Q}$ is abelian.
As discussed here  recently, a subset $T\subset\mathbb{P}$ is abelian if and only if it is defined by congruences.  For example, the set of primes $\equiv1\pmod{l}$  is the same as $S(\mathbb{Q}(\zeta_l))$.  "Being defined by congruences" can be made precise, and counts as a characterisation of abelian subsets of $\mathbb{P}$.
Neukirch says that Langlands' Philosophy provides a characterisation of all galoisian subsets of $\mathbb{P}$.  Can this remark now be illustrated by some striking example ?
Addendum (28/02/2010) Berger's recent Bourbaki exposé 1017 arXiv:1002.4111 says that cases of the Fontaine-Mazur conjecture have been proved by Matthew Emerton as well.  I didn't know this at the time of asking the question, and the unique answerer did not let on that he'd had something to do with Fontaine-Mazur...

REPLY [23 votes]: In order to add my bit to the already rich content on this site, here is a
nice family of galoisian extensions $K_l|{\bf Q}$ with group ${\rm GL}_2({\bf
  F}_l)$ (indexed by primes $l\neq5$) for which a ``reciprocity law'' can be
written down explicitly.  I've come across this family recently while writing
an expository article.
Let $E$ be the elliptic curve (over $\bf Q$) of conductor $11$ defined by
$y^2+y=x^3-x^2$, with associated modular form
$$
\eta_{1^2,11^2}=q\prod_{k>0}(1-q^k)^2(1-q^{11k})^2=\sum_{n>0}c_nq^n.
$$
Let $K_l={\bf Q}(E[l])$, which is thus galoisian over $\bf Q$ and unramified at
every prime $p\neq11,l$.
One can deduce from cor.1 on p.308 of Serre (Inventiones 1972) that for every
prime $l\neq5$, the representation 
$$
\rho_{E,l}:{\rm Gal}(K_l|{\bf Q})\rightarrow{\rm GL}_2({\bf F}_l)
$$
we get upon choosing an ${\bf F}_l$-basis of $E[l]$ is an isomorphism;
cf. the online notes on Serre's conjecture by Ribet and Stein.  Shimura did
this for $l\in[9,97]$ (Crelle 1966).
Suppose henceforth that $l$ is a prime $\neq5$ and that $p$ is a prime
$\neq11,l$.  The characteristic polynomial of $\rho_{E,l}({\rm Frob}_p)\in{\rm
  GL}_2({\bf F}_l)$ is
$$
T^2-\bar c_pT+\bar p\in{\bf F}_l[X].
$$
The prime $p$ splits completely in $K_l$ if and only if ${\rm Frob}_p=1$ in
${\rm Gal}(K_l|{\bf Q})$, which happens if and only if
$$
\rho_{E,l}({\rm Frob}_p)=\pmatrix{1&0\cr0&1}.
$$
If so, then $p,c_p\equiv1,2\pmod l$ but not conversely, for the matrix
$\displaystyle\pmatrix{1&1\cr0&1}$ also has the characteristic polynomial
$T^2-\bar2T+\bar1$. But these congruences on $p,c_p$ do rule out an awful lot
of primes as not splitting completely in $K_l$.
In summary, we have the following ``reciprocity law" for $K_l$ :
$$
\hbox{($p$ splits completely in $K_l$)}
\quad\Leftrightarrow\quad
E_p[l]\subset E_p({\bf F}_p),
$$ 
where $E_p$ is the reduction of $E$ modulo $p$.  Indeed, reduction modulo $p$
identifies $E[l]$ with $E_p[l]$ and the action of ${\rm Frob}_p$ on the former
space with the action of the canonical generator $\varphi_p\in{\rm
Gal}(\bar{\bf F}_p|{\bf F}_p)$ on the latter space.  To say that $\varphi_p$
acts trivially on $E_p[l]$ is the same as saying that $E_p[l]$ is contained in
the ${\bf F}_p$-rational points of $E_p$.  The analogy with the multiplicative
group $\mu$ is perfect: 
$$
\hbox{($p\neq l$ splits completely in ${\bf Q}(\mu[l])$)}
\quad\Leftrightarrow\quad
\mu_p[l]\subset \mu_p({\bf F}_p)
$$
($\Leftrightarrow l|p-1\Leftrightarrow p\equiv1\pmod l$), where $\mu_p$ is not the $p$-torsion of $\mu$ but the reduction of $\mu$ modulo $p$.
I requested Tim Dokchitser to compute the first ten $p$ which split completely
in $K_7$, and his instantaneous response was 4831, 22051, 78583, 125441,
129641, 147617, 153287, 173573, 195581, and 199501.
It is true that all this (except the list of these ten primes) was known before Serre's conjecture was proved
(2006--9) or even formulated (1973--87), but I find this example a very good
illustration of the kind of reciprocity laws it provides.
I hope you enjoyed it as much as I did.<|endoftext|>
TITLE: Cutting a rectangle into an odd number of congruent non-rectangular pieces
QUESTION [28 upvotes]: We are interested in tiling a rectangle with copies of a single tile (rotations and reflections are allowed). This is very easy to do, by cutting the rectangle into smaller rectangles.
What happens when we ask that the pieces be non-rectangular?
For an even number of pieces, this is easy again (cut it into rectangles, and then cut every rectangle in two through its diagonal. Other tilings are also easy to find).
The interesting (and difficult) case is tiling with an odd number of non-rectangular pieces.
Some questions:

Can you give examples of such tilings?
What is the smallest (odd) number of pieces for which it is possible?
Is it possible for every number of pieces? (e.g., with five)

There are two main versions of the problem: the polyomino case (when the tiles are made of unit squares), and the general case (when the tiles can have any shape). The answers to the above questions might be different in each case.
It seems that it is impossible to do with three pieces (I have some kind of proof), and the smallest number of pieces I could get is $15$, as shown above:
     (source)
This problem is very useful for spending time when attending some boring talk, etc.

REPLY [12 votes]: I am posting the 11 pieces solution shown in the article cited by Michael (it is not freely available online).
     (source)
This is the smallest known number of pieces. Some remarks:

The question is open for 5, 7, or 9 pieces. Get your pencils!
Everything so far is with polyominoes. Any suggestion with more complicated shapes?
Unlike the other solution I posted, this one cannot be resized along the $x$ or $y$ axis.<|endoftext|>
TITLE: Algorithm for decomposing permutations
QUESTION [10 upvotes]: Is there an algorithm for solving the following problem: let $g_1,\ldots,g_n$ be permutations in some (large) symmetric group, and $g$ be a permutation that is known to be in the subgroup generated by $g_1,\ldots,g_n$, can we write $g$ explicitly as a product of the $g_i$'s?
My motivation is that I'm TAing an intro abstract algebra course, and would like to use the Rubik's cube to motivate a lot of things for my students, and would, in particular, like to show them an algorithm to solve it using group theory. (That is, I can write down what permutation of the cubes I have, and want to decompose it into basic rotations, which I then invert and do in the opposite order to get back to the solved state.) Though I'm interested in the more general case, not just for the Rubik(n) groups, if a solution works out.
Note: I don't really know what keywords to use for solving this problem, if someone can point me to the right search terms to google to get the results I'm looking for, I'll gladly close this.

REPLY [11 votes]: Yes. The general rule of thumb is that groups described by permutations are computationally easy, groups described by generators and relations have computational problems that are generally undecidable, and matrix groups are somewhere in between.
There's a whole book "Permutation group algorithms" by Seress, Cambridge University Press, 2003.
The main technique for permutation groups is called the Schreier–Sims algorithm; there's a survey here, for instance. The rough idea is to stabilize the permuted elements one at a time. As Mitch already said, though, this doesn't find the shortest word in the generators that produces any particular group element, which is a more difficult problem.<|endoftext|>
TITLE: Why sin and cos in the Fourier Series?
QUESTION [13 upvotes]: Is there any special reason that we use the sines and cosines functions in the Fourier Series, while we know that if we chose any maximal orthonormal system in L2, we would get the same result? 
Is it something historical or what?
Thanks in advance.

REPLY [2 votes]: Nobody mentioned that $e^{2\pi inz}$ are the characters of $S^1$.<|endoftext|>
TITLE: Infinite tensor products
QUESTION [34 upvotes]: Let $A$ be a commutative ring and $M_i, i \in I$ be a infinite family of $A$-modules. Define their tensor product $\bigotimes_{i \in I} M_i$ to be a representing object of the functor of multilinear maps defined on $\prod_{i \in I} M_i$ (this exists by the usual construction). Thus there is a universal multilinear map $\otimes : \prod_{i \in I} M_i \to \bigotimes_{i \in I} M_i$. Some years ago, I wanted to examine this infinite tensor product, but in the literature I could not find anything going beyond some natural isomorphisms (e.g. associativity) or the submodule consisting of tensors which become eventually constant a specific element of $\prod_{i \in I} M_i$ which yields a colimit of finite tensor products (denoted $U_x$ below). In general, it seems to be quite hard to describe $\bigotimes_{i \in I} M_i$. For example, for a field $K$, $K \otimes_K \otimes_K ...$ has dimension $|K^*|^{\aleph_0}$ (see below) and you cannot write down a basis, which might be scary when you see it the first time. The point is that multilinear relations cannot be applied infinitely many times at once: For example in $K \otimes_K \otimes_K ...$, we have $x_1 \otimes x_2 \otimes ... = y_1 \otimes y_2 \otimes ...$ if and only if $x_i = y_i$ for almost all $i$ and for the rest we have $\prod_i x_i = \prod_i y_i$.
Before posing my question, I provide some results.
1.1. Assume that $M_i$ are torsionfree $A$-modules (meaning $am=0 \Rightarrow a=0 \vee m=0$). In this case, we may decompose $\bigotimes_{i \in I} M_i$ as follows: Define $X = \prod_{i \in I} M_i \setminus \{0\}$ and let $x \sim y \Leftrightarrow \{i : x_i \neq y_i\}$ finite. Then $\sim$ is an equivalence relation on $X$. Let $R$ be a set of representatives (this makes this description ugly!). Then there is a canonical map
$H : \text{Mult}(\prod_{i \in I} M_i,-) \to \prod_{x \in R} \lim_{E \subseteq I \text{~finite}} \text{Mult}^x(\prod_{i \in E} M_i,-)$,
where $\text{Mult}^x$ indicates that the transition maps of the limit are given by inserting the entries of $x$, and it is not hard to show that $H$ is bijective. Thus
$\bigotimes_{i \in I} M_i = \bigoplus_{x \in R} U_x$,
where $U_x = \cup_{E \subseteq I \text{~finite}} \bigotimes_{i \in E} M_i \otimes \otimes_{i \notin E} x_i$ is the colimit of the finite tensor products $\otimes_{i \in E} M_i$ (the transition maps given by tensoring with entries of $x$). The canonical maps $\otimes_{i \in E} M_i \to U_x$ don't have to be injective; at least when $A$ is a PID, this is the case. Remark that $U_x$ only depends on the equivalence class of $x$, so that the decomposition into the $U_x$ is canonical, whereas the representation of $U_x$ as direct limit (including the transition maps!) depends really on $x$.
1.2 If $M_i = A$ is an integral domain, we get $\bigotimes_{i \in I} A = \oplus_{x \in R} U_x$, where $U_x$ is the direct limit of  copies $A_E$ of $A$, for every finite subset $E \subseteq I$, and transition maps $\prod_{i \in E' \setminus E} x_i : A_E \to A_{E'}$ for $E \subseteq E'$. $U_x$ is just the localization of $A$ at the $x_i$.
1.3 If $A$ is a field, and $M_i$ has basis $B_i$, then $B_x = \cup_{E \subseteq I} \bigotimes_{i \in E} B_i \otimes \otimes_{i \notin E} x_i$ is a basis of $U_x$ and thus $\cup_{x \in R} B_x$ is a basis of $\bigotimes_{i \in I} M_i$. According to this question, this has cardinality $\max(|X|,|I|,\max_i(\dim(M_i)))$.
1.4 If $A_i$ are $A$-algebras, then $\bigotimes_{i \in I} A_i$ is a $A$-algebra. If the $A_i$ are integral domains, then it is a graded algebra by the monoid $X/\sim$ with components $U_x$.
If $A=A_i=K$ is a field with $U=K^x$, then there is a vector space isomorphism between $\bigotimes_{i \in I} K$ and the group algebra $K[U^I / U^{(I)}]$. A sufficient, not neccessary, condition for the existence of a $K$-algebra isomorphism is that $U^{(I)}$ is a direct summand of $U^I$, which is quite rare (see this question). Nevertheless, we can ask if these $K$-algebras isomorphic. In some sense I have proven this already locally (subalgebras given by finitely generated subgroups of the group $U^I / U^{(I)}$ are isomorphic, in a terribly uncanonical way). Many questions I'm currently posing here are addressed to this problem.
2.1 What about interchanging tensor product with duals? Let $(V_i)_{i \in I}$ be a family of vector spaces over a field $K$. For elements $\lambda_i \in K$, define their infinite product $\prod_{i \in I} \lambda_i$ to be the usual product if $\lambda_i=1$ for almost all $i$, and otherwise to be $0$. This yields a multilinear map $\prod : K^I \to K$ and thus a linear map
$\delta : \bigotimes_{i \in I} V_i^* \to (\bigotimes_{i \in I} V_i)^*, \otimes_i f_i \mapsto (\otimes_i x_i \mapsto \prod_{i \in I} f_i(x_i)).$
Then it can be shown that $\delta$ is injective, but the proof is pretty fiddly.
2.2 Let $W_i$ be another family of vector spaces over a field $K$. Then there is a canonical map
$\alpha : \bigotimes_{i \in I} \operatorname{Hom}(V_i,W_i) \to \operatorname{Hom}(\bigotimes_{i \in I} V_i, \bigotimes_{i \in I} W_i).$
Is $\alpha$ injective? This is known when $I$ is finite.
3 What about other properties of finite tensor products, do they generalize? For example let $J_i, i \in I$ be a family of index sets and $M_{i,j}$ be a $A$-module where $i \in I, j \in J_i$. Then there is a canonical homomorphism
$\delta : \bigoplus_{k \in \prod_{i \in I} J_i} \bigotimes_{i \in I} M_{i,k(i)} \to \bigotimes_{i \in I} \oplus_{j \in J_i} M_{i,j}$.
It can be shown that $\delta$ is injective, but is it also bijective (as in the finite case)?
4 The description of the tensor product given in 1.1 depends on a set of representatives and is not handy when you want to prove something. Are there better descriptions?
Remark that in this question I'm not interested in infinite tensor products defined in functional analysis or just colimits of finite ones. I'm interested in the tensor product defined above (which probably every mathematician regards as "the wrong one"). Any hints about their structure or literature about it are appreciated.

REPLY [8 votes]: Coincidentally, I started a work in 2009 concerning infinite tensor products of complex vector spaces, $^*$-algebras and Hilbert spaces. The resulting article has just been accepted for publication and I put it in the arxiv: https://arxiv.org/abs/1112.3128. In the article, I got a similar result as your 1.1 (for vector spaces) and obtained a positive answer to your question 2.2 (when $K$ is the complex filed).<|endoftext|>
TITLE: Reference for equivalent definitions of the genus
QUESTION [7 upvotes]: Let $X$ be a (edit: nonsingular) projective complex algebraic curve.  The genus of $X$ can be defined as the dimension of the space of holomorphic $1$-forms on $X$, which in turn can be defined either analytically or algebraically in terms of Kahler differentials.  It can also be defined as the topological genus of $X$ considered as a surface, which in turn can be defined either topologically as the number of tori in a connected sum decomposition of $X$ or homologically in terms of the Betti numbers of $X$.  Does anyone know of a reasonably self-contained reference where some or all of these equivalences are proven?
(There is a related question about computing the genus of a curve from its function field as well as a nice post by Danny Calegari explaining the relationship to the Newton polygon, but I am mostly interested in the algebraic-to-topological step of going from Kahler differentials to the number of tori in a connected sum decomposition.)

REPLY [7 votes]: For $\mathrm{dim} H^0(X, \Omega^1_X) = \dim H^1(X, \mathbb{Q})$ see http://en.wikipedia.org/wiki/Hodge_theory. For $\dim H^1(X, \mathbb{Q}) =$ number of tori use induction and the Mayer-Vietoris sequence.
(And for $\mathrm{dim} H^0(X, \Omega^1_X) = \mathrm{dim} H^1(X, \mathcal{O}_X)$ see http://en.wikipedia.org/wiki/Serre_duality.)<|endoftext|>
TITLE: difference between equivalence relations on algebraic cycles
QUESTION [35 upvotes]: For the definitions of the equivalence relations on algebraic cycles see http://en.wikipedia.org/wiki/Adequate_equivalence_relation.
I want to know how far away from each other the equivalence relations on algebraic cycles are and what the intuition is for them.
My impression is that rational equivalence gives much bigger Chow groups than algebraic equivalence, and that algebraic equivalence, homological equivalence and numerical equivalence are quite tight together.
Take for example an elliptic curve. We have $CH^1(E) = \mathbb{Z} \times E(K)$, algebraic equivalence (take $C = E$) $\mathbb{Z}$ = numerical equivalence.

REPLY [4 votes]: A good reference is also Fulton, Intersection Theory, Chapter 19.<|endoftext|>
TITLE: Cohomology of quaternions on an abelian variety
QUESTION [9 upvotes]: Given two non-isogenous elliptic curves $E_1$ and $E_2$ over $\mathbb{C}$.
Set $A:=E_1 \times E_2$. Given a nontrivial sheaf of quaternion algebras $D$ over $A$, what is the dimension of the vector space $H^1(A,D)$?
If one thinks of $D$ as an element in the Brauer group $Br(A)$, then it is $2$-torsion, hence belongs to $Br(A)[2]$. Since the curves are non-isogenous there is an isomorphism $Pic(E_1)[2] \otimes Pic(E_2)[2] \to Br(A)[2]$. So there should be a connection between such quaternions and $2$-torsion line bundles on the curves, but i cannot find an explicit description for this isomorphism. If there is one, i thought one could use the Künneth formula to compute $H^1(A,D)$ in terms of the cohomology of the line bundles on the curves.
For now i could only work out the bound $d=dim(H^1(A,D)) \geq 2$: using Hirzebruch-Riemann-Roch and simplifying terms one gets $d=c_2(D)+2$. After a result of M.Lieblich one has $c_2(D)\geq 0$. 
Does anyone see/have an explicit description of the isomorphism mentioned above? Is the idea using Künneth a promising approach to this problem at all? Or does anyone have another approach? Are there some calculations regarding this in the literature (i couldn't find one)?
Another question in this context is: what is the image of such an algebra under the map $Br(A) \rightarrow Br(\mathbb{C}(A))$. This should be nontrivial $\mathbb{C}(A)$-quaternions, since the map "looking at the genric point $\eta$" is injective, i.e. $D_{\eta}$ is generated by elements $i,j$ with $i^2=a, j^2=b and ij=-ji$. But what are a resp. b? I think they should have something to do with functions h such that 2*Y=div(h), where Y defines one of the line bundles. Is this true?

REPLY [2 votes]: For the description of the quaternion algebra associated to a pair of torsion line bundles, try the following.  Take line bundles ${\cal L}_i$ on $E_i$ equipped with isomorphisms ${\cal L}_i^{\otimes 2} \to {\cal O}$, and pull these back to $A$.  Define

$$D = {\cal O} \oplus {\cal L}_1 \oplus {\cal L}_2 \oplus {\cal L}_1 \otimes {\cal L}_2$$

with multiplication induced by the maps ${\cal L}_i^{\otimes 2} \to {\cal O}$, ${\cal O}$ being the unit, and the elements of ${\cal L}_1$ and ${\cal L}_2$ anticommuting.
ADDED: I wish I had a more conceptual explanation for why this represents the cup-product in the Brauer group, but here is a cocycle description along the lines of what Oren suggested.
Suppose $X$ is given with 2-torsion line bundles $\cal L$ and $\cal M$.  Choose cocycles representing these, in the form of a cover (either open in the analytic case, or etale in the algebraic case) $U_\alpha$ of $X$ together with sections $s_\alpha$ of $\cal L$ and $t_\alpha$ of $\cal M$ on $U_\alpha$ such that $s_\beta / s_\alpha = u_{\alpha \beta} \in \{\pm 1\}$ and similarly $t_\beta / t_\alpha = v_{\alpha \beta}$; these latter two are the representing cocycles.
Then D has basis $\{1,s_\alpha, t_\alpha, s_\alpha t_\alpha\}$ on $U_\alpha$, where $s_\alpha^2 = t_\alpha^2 = 1$, and you can explicitly make this isomorphic to a matrix algebra.  The change-of-basis sends $s_\alpha$ to $s_\beta = u_{\alpha \beta} s_\alpha$ and similarly for $t$.  This can be achieved by conjugation by the element $t_\alpha^{(1 - u_{\alpha \beta})/2} s_\alpha^{(1 - v_{\alpha \beta})/2} = g_{\alpha\beta} \in D \cong M_2(\mathbb{C})$.  These change-of-basis matrices reduce to a cocycle in $PGL_2(\mathbb{C})$ representing the algebra, and so the image in the Brauer group is represented by the coboundary $(\delta g)_{\alpha \beta \gamma} \in \{\pm 1\}$.
EDIT: fixed up following description of the coboundary so that it correctly described where the cup product lands.
Explicit computation finds $(\delta g)_{\alpha \beta \gamma}$ is equivalent to the cocycle $v_{\alpha \beta} \otimes u_{\beta \gamma}$ with coefficients in $\{\pm 1\} \otimes \{\pm 1\} \cong \{\pm 1\}$ (I may have mixed the indices, if I did please let me know and I'll correct it), which is precisely the formula for the cup product of the cocycles $u$ and $v$.
So based on your description of the cup product inducing an isomorphism between 2-torsion in the Brauer group and cup products of 2-torsion elements in the Picard group, this genuinely should provide you with the bundles you're looking for.<|endoftext|>
TITLE: Fun applications of representations of finite groups
QUESTION [65 upvotes]: Are there some fun applications of the theory of representations of finite groups? I would like to have some examples that could be explained to a student who knows what is a finite group but does not know much about what is a repersentation (say knows the definition). The standard application that is usually mentioned is Burnside's theorem http://en.wikipedia.org/wiki/Burnside_theorem.  The application may be of any kind, not necessarely in math. But math applications are of course very wellcome too!!! It will be very helpfull also if you desribe a bit this application.

REPLY [9 votes]: Let $G$ be a finite group with $n$ elements and $k$ conjugacy classes. Denote by $m=|G:[G,G]|$ the index of the commutator. Then $n+3m\geq 4k$. 
It is less impressive than many other answers, but I find this inequality particularly nice, especially having in mind that there are some nontrivial examples of equality, all are explicitly listed. I do not know the proof whithout using representations.<|endoftext|>
TITLE: Mathematics for machine learning
QUESTION [21 upvotes]: I would like to know what mathematics topics are the most important to learn before actually studying the theory on neural networks.
I ask that because I will start to learn about neural networks and machine learning on my own to help in the analysis I am doing on my PhD about patterns of genome evolution. 
Thank you in advance.

REPLY [3 votes]: As a deep learning practitioner with mathematical background I was yearning to have some satisfying mathematical framework of what I do in my every day job. In my opinion, very well fitted mathematical foundations of deep learning principles are captured simply by Empirical risk minimization (ERM) concept. I encourage everyone to read just Chapter 1 from The Nature of Statistical Learning Theory of V. Vapnik. In my opinion it is an eye opener.<|endoftext|>
TITLE: What is the probability distribution function for the product of two correlated Gaussian random variable?
QUESTION [11 upvotes]: Suppose we have pair $(X,Y)\sim Normal([\mu_x,\mu_y],{{\sigma_x^2\atop\rho \sigma_x\sigma_y } {\rho \sigma_x\sigma_y \atop \sigma_y^2} }] $
How is $U=X\cdot Y$ distributed?
I've tried to compute this by substituting y=u/x  in the bivariate normal pdf and taking integral(from $-\infty$ to $\infty$ with respect to x. I find the pdf of U as the sum of two  exponential distributions(one for U<0 and one for U>0) that are weighted unequally. Is my method valid, or do I have to deal with cumulative distribution functions instead?

REPLY [4 votes]: LaGatta's answer nails it, and may be useful for drawing simulations, etc.
This is just a note to remind that if one is only interested in the mean of the product of normally-distributed (possibly correlated) random variables, then the answer is straightforward, using the identity $\operatorname{E}XY= \operatorname{Cov}(X,Y) + (\operatorname{E}X)(\operatorname{E}Y)$.<|endoftext|>
TITLE: Why is GL(n,C)/U(n) a CAT(0) space?
QUESTION [6 upvotes]: The title says it all. In one of his answers to the question "Convex hull in CAT(0)" (I don't have the points to post a link, if someone doesn't mind link-ifying this that would be cool), Greg Kuperberg said that GL(n,C)/U(n) is a CAT(0) space. I was wondering why this is true, or if there's a reference for this.

REPLY [2 votes]: It also suffices to check that the sectional curvature of this space, using the Riemannian metric induced by the Killing form, is nonpositive. I recommend that you both figure out how to do the explicit calculation of the sectional curvature for this particular example and learn the general theory referred to in Andy's answer. They are, of course, essentially the same answer.<|endoftext|>
TITLE: two conjugate subgroups and one is a proper subset of the other? plus, a covering space interpretation.
QUESTION [15 upvotes]: Recently I've been reading J.P. May's A Concise Course in Algebraic Topology.  In the section on the classification of covering groupoids, he mentions that sometimes a group G may have two conjugate subgroups H and H' such that H is properly contained in H' (on pp. 26-27, according to his numbering).  This seems bizarre to me, and I'm pretty sure I've seen an example before, but I'm having trouble coming up with one now.
Anyways, he continues by saying that it is possible to have an endomorphism of a covering groupoid which is not an isomorphism.  I'd like to come up with an example of this, and I'm pretty sure that for me obstruction lies in failing to completely grasp the group-theoretic statement above.
(Of course, when I think of a covering of groupoids I'm secretly thinking about a covering space, partly because this is his motivation for introducing groupoids and partly because it's just easier for me, so ideally but not necessarily the example would really just be a map of covering spaces over the same base space.)

REPLY [11 votes]: If there is a group G and an injective endomorphism $\sigma$ of G, there is a group K containing G such that $\sigma$ extends to an inner automorphism of K.
Even more generally, if two subgroups of a group are isomorphic, the group can be embedded in a bigger group where those two subgroups are conjugate via a conjugation map that extends the isomorphism. The construction used is called a HNN-extension, and it basically adjoins elements to the group that act by conjugation as the isomorphism. (This generalizes even further: given any number of isomorphisms between pairs of subgroups of a group, there is a group containing the group such that all these isomorphisms become conjugations in that bigger group).
Thus, to find examples that answer your question, it is enough to find examples of a group that is isomorphic to a proper subgroup. For instance, if we consider the example of the group of integers isomorphic to the subgroup of even integers, the corresponding HNN-extension is the Baumslag-Solitar group mentioned above. 
Incidentally, the statement above (that any two isomorphic subgroups of a group become conjugate in some bigger group) is also true when we restrict to finite groups, though this does not give any examples for the question you are interested in because no finite subgroup can be isomorphic to a proper subgroup.
See this and http://groupprops.subwiki.org/wiki/Isomorphic_iff_potentially_conjugate_in"finite">this for more notes on these.<|endoftext|>
TITLE: Simple show cases for the Yoneda lemma
QUESTION [18 upvotes]: I've been given a very simple motivating and instructive show case for the Yoneda lemma:
Given the category of graphs and a graph object $G$, seen as a quadruple $(V_G,\ E_G,\ S_G:E\rightarrow V,\ T_G:E \rightarrow V)$. 
Consider $K_1$ and $K_2$, the one-vertex and the one-edge graph and the two morphisms $\sigma$ and $\tau$ from $K_1$ to $K_2$.
Now consider the graph $H$ with

$V_H = Hom(K_1,G)$
$E_H = Hom(K_2,G)$
$S_H(e) = e \circ \sigma: K_1 \rightarrow G$ for $e \in E_H$
$T_H(e) = e \circ \tau: K_1 \rightarrow G$ for $e \in E_H$

It can be easily seen that $H$ is isomorphic to $G$.
I have learned that a) the category of graphs is a presheaf category and that b) $K_1$, $K_2$ are precisely the representable functors.
Now I am looking for other simple motivating and instructive show cases.
By the way: Shouldn't such an show case be added to the Wikipedia entry on Yoneda's lemma?

REPLY [3 votes]: Not so much a showcase maybe but rather a connection to an algebraist's intuition. A particular case of Yoneda is that a monoid $M$ acting on itself (by (say) left multiplication) is a free $M$-set on one generator, $\hom_M(M,X)\approx X$ naturally in $X$.
This can be in fact of course extended to the full Yoneda lemma - viewing set-valued functors on a category $\mathcal C$ as algebras for a multi-sorted equational theory (objects of $\mathcal C$ for sorts, unary operations only), Yoneda lemma becomes the statement that representables are single-generator free algebras.<|endoftext|>
TITLE: Improving a sequence of 1s and -1s
QUESTION [29 upvotes]: Suppose you take a $\pm 1$ sequence and you want to "improve it" by taking pointwise limits of translates. What properties can you guarantee to get in the limit? 
Two examples illustrate what I think should be the extremes. If your sequence is random, then every finite subsequence occurs infinitely often, which means that it is easy to find translates that converge pointwise to any sequence you like. In particular, you can make the final sequence constant, though what interests me is that it is highly structured.
By contrast, let's suppose that the sequence is quasiperiodic, or more concretely that you take a real number $\alpha$ and define $x_n$ to be 1 or -1 according to whether the integer part of $\alpha n$ is even or odd. In that case, it is not hard to prove that every pointwise limit is quasiperiodic as well. (Unless I've made a mistake, the result is that there must be some $\beta$ such that $z_n$ is 1 or -1 according to whether the integer part of $\alpha n+\beta$ is even or odd.)
What can be said in general? Can we always "get rid of the randomness" and find a highly structured limit? And what does "highly structured" mean? Perhaps that an associated dynamical system is compact, though I'd ideally like a characterization in terms of the sequence itself.
This question ought to be meat and drink to ergodic theorists. Indeed, I feel slightly guilty for not knowing the answer already. It arises naturally in a Polymath project (which can be found by searching for "Erdos discrepancy problem").
Added later: I've just realized one simple lemma. Suppose you want to get rid of some finite subsequence. Then you can do it if you can find arbitrarily long subsequences that avoid that finite subsequence. So either you can get rid of the subsequence, or it occurs in the original sequence with bounded gaps. I think that means you can reach a sequence such that every finite subsequence that appears appears with bounded gaps.

REPLY [8 votes]: Here are some other comments in the spirit of Terry's and Anton's answers.
There is nothing special about the symbols $\pm 1$ in the question; the spaces of binary strings $(\mathbb{Z}/2)^\mathbb{N}$ and $(\mathbb{Z}/2)^\mathbb{Z}$ are a more standard notation.  As Terry says, you can switch between singly infinite and doubly infinite sequences.
The process of taking pointwise limits of translates over and over again until no more data can be removed is exactly equivalent to finding a minimal set with respect to the dynamical action of the shift map.  If $x$ does not lie in a minimal set, then the closure of its orbit contains a $y$ whose orbit has a smaller closure; thus $y$ is a limit of translates of $x$ in which something has been lost.  On the other hand, all points in a minimal set are limits of translates of each other, so once you are in a minimal set, there is no way to erase any more data.
Birkhoff established that a symbolic sequence $x$ is minimal, or lies in a minimal set, if and only if it is "almost periodic".  Anton used the term "quasiperiodic", but this is a confusing term that sometimes means almost periodic and sometimes means other things.  It is true that a quasiperiodic pattern such as a Penrose tiling or a quasicrystal is almost periodic.  (These are higher-dimensional examples, but the issues are the same for all locally compact abelian groups.)  Sometimes quasiperiodic examples have the additional property that the recurrence length $L(\ell)$ is linear in $\ell$ or $O(\ell)$.     Penrose tilings are already related to one interesting class of examples in one dimension:  If $\alpha$ is an irrational number between 0 and 1, then the sequence $a_n = \lfloor (n+1)\alpha \rfloor - \lfloor n \alpha \rfloor$ is an almost periodic binary sequence.
What Anton could mean by the statement "you will not get more" is two things.  First, that every almost periodic sequence is minimal, and therefore that you can't simplify it further by taking a limit of its translates.  Second, that there isn't any simpler characterization of minimal shifts than that they are almost periodic.
Section 13.7 of the book "An introduction to symbolic dynamics and coding" by Douglas Lind gives a survey of properties of minimal shifts in symbolic dynamics.  Lind says that that a widely studied example is the Morse-Thue sequence, which is an example of an almost periodic sequence obtained by substitution rules.  Lind says that any substitution-type almost periodic sequence, or substitution shift, has zero entropy, but that Furstenburg found an example of a minimal shift with positive entropy.  This suggests that it is not always possible to "get rid of the randomness", as Tim (?) asks, even though all limits are highly structured in the sense of being almost periodic.<|endoftext|>
TITLE: Does formally etale imply flat?
QUESTION [14 upvotes]: Formally étale means that the infinitesimal lifting property is uniquely satisfied. If the map is also locally of finite presentation, then it is called étale. One of many characterizations (see EGA 4.5.17) of étale is flat and unramified. So my question is whether the weaker condition of formally étale still implies flatness?

REPLY [14 votes]: It seems that Anton Geraschenkos answer to a previous question Is there an example of a formally smooth morphism that is not smooth does the trick here as well. His example of a formally smooth map that is not flat is indeed formally etale. So, formally etale does not imply flat.<|endoftext|>
TITLE: Burnside ring and zeroth G-equivariant stem for finite G
QUESTION [5 upvotes]: Let $G$ be a finite group. The theorem that the Burnside ring $A(G)$ is isomorphic to the zeroth stable stem $\pi^{G}_0(S)$ is usually said to originate from Segal. I search for a reference of a proof this theorem, which is not obscured by tom Diecks generalisation to compact Lie groups. Who knows a good reference? I can not even find the right paper of Segal. Where did the original proof appear?

REPLY [4 votes]: Here's a conceptual answer, which can be filled in to give a proof.  First, go back to the non-equivariant setting: why is $\pi_0(S) = \lim \pi_N(S^N) \cong {\mathbb Z}$?  because one can use transversality arguments (originally due to Pontryagin) to show that group is the same as the cobordism group of ``oriented'' zero-dimensional manifolds, which is then obviously the integers.  Similarly, the cobordism of oriented zero-dimensional $G$-manifolds is isomorphic to the Burnside ring, so the result you seek follows once you can establish some transversality for maps $S^V \to S^V$.  (These transversality results are notoriously complicated and not always valid in the equivariant setting, but this one doesn't present much trouble).<|endoftext|>
TITLE: What primes divide the discriminant of a polynomial?
QUESTION [11 upvotes]: Given a monic polynomial $p(t) = t^n + ... + c_1 t + c_0$ with integer (or rational) coefficients and with roots $a_1, \dots a_n$, we can compute its discriminant, which is defined to be $\prod_{i< j}(a_i - a_j)^2$.
In my case, I have a polynomial which is the characteristic polynomial of some invertible matrix $T$.  It is palindromic -- i.e., $c_{n-i} = c_i$ for all $0 \leq i \leq n$ -- so the roots come in inverse pairs $a$ and $\frac{1}{a}$.  There are no repeated roots, so the discriminant is non-zero.
My question is: is there any way of knowing which primes divide this discriminant, i.e. from the coefficients of the polynomial or from the matrix $T$?

REPLY [5 votes]: I disagree with the definition of the discriminant as the resultant of $P$ and $P'$.
When $P$ is a polynomial with integer coefficients, then a prime $q$ should divide the discriminant of $P$ if and only if the reduction of $P$ modulo $q$ has a multiple root (possibly at infinity, when the degree decreases by at least 2 under reduction). But now consider $P=2X^2+ 3X+1$. The resultant of $P$ and $P'$ is $-2$, and the reduction of $P$ modulo 2 has no multiple root. In this case, the well known discriminant $b^2-4ac$ is actually 1. The correct relation between the discriminant and the resultant for a polynomial $P(t)=a_nt^n+\cdots+a_1t+a_0$ is $\mathrm{disc}(P)= (-1)^{n(n-1)/2}\mathrm{res}(P,P')/a_n$.<|endoftext|>
TITLE: Definition of sheaves in wikipedia
QUESTION [6 upvotes]: In wikipedia, sheaves were first defined in the case of concrete categories (with usual identity and gluing axioms), then in the general case. (writing it as an "exact" sequence)
Do these two definitions agree? I find the definition for concrete categories case very strange, for if we consider a topological space of two points a,b with discrete topology, and let us consider a sheaf of topological spaces on it that assigns A to a, B to b. According to the "concrete-category-case" definition, we need the global sections to look like $A \times B$ such that the projection maps are continuous and nothing else. But if we look at the "equalizer" definition, we would require the global sections to carry the product topology as well.
So is wikipedia wrong? Or am I misunderstanding something? Thanks!
Edit: There is another not-quite-related question. In wikipedia, for the "equalizer" definition they require the category, where the sheaf's taking values in, to have products. Is this really necessary? In EGA Chapter 0 p.23 for example, the product is just "splitted", and we consider the large family of maps all together. It seems that these two approaches are just the same. Or am I wrong?

REPLY [9 votes]: I think you're quite right; Wikipedia's "concrete definition" is only correct for concrete categories whose underlying-set functor is (not just faithful but) conservative, i.e. such that any morphism which is a bijection on underlying sets is an isomorphism in the category.  The page does say that the concrete definition "applies to the most common examples such as sheaves of sets, abelian groups and rings," all of which have this property, but it ought to be fixed to make clear in exactly what situations this definition applies.
Secondly, I observe that the "normalisation" condition in the Wikipedia concrete definition is also odd.  Since the empty set is covered by the empty family, the "local identity" and "gluing" conditions already imply that the underlying set of $F(\emptyset)$ is terminal.  Saying that in addition, $F(\emptyset)$ itself is terminal is an additional condition, which is in fact a special case of the second, more generally applicable, definition.
Thirdly, I think you're also right that for the correct general definition, the category doesn't need to have any limits a priori; you can just assert that $F(U)$ is the limit of the appropriate diagram of the $F(U_i)$ and $F(U_i\cap U_j)$.
Finally, let me go out on a limb and say that it seems to me that defining "sheaves with values in an arbitrary category" is often a misguided thing to do.  More often, it seems like rather than "a sheaf with values in the category of X," the important notion is "an internal X in the category of sheaves of sets."  For familiar cases such as groups, abelian groups, rings, small categories—in fact, for any finite limit theory—the two are the same, which may be what leads to the confusion.  But the good notion of "sheaf of local rings," for instance, is not a sheaf with values in the category of local rings, but rather a sheaf of rings whose stalks are local (at least, when there are enough points), and that's the same as an internal local ring in the category of sheaves of sets.  The situation is similar, I think, for "sheaves of topological spaces" (or locales).  I'd be happy for people to point out where I'm wrong about this, though.

REPLY [6 votes]: I agree, the "definition" for a concrete category is wrong, and your example with a two-point discrete space shows that it's wrong.  
Now, if only one could edit Wikipedia...
(I make the following conjecture.  Category theory beginners are often more keen on so-called concrete categories than is entirely healthy.  Some such person may have written that passage.)

REPLY [2 votes]: Short answer: You can't take sheaves of topological spaces.  This is because a "sheaf of abelian groups" is in fact an abelian group object in the category of sheaves.  This equivalence of concepts does not hold for topological spaces because the forgetful functor adjunction of Top and Set is not monadic.  If you read Mac Lane's book Sheaves in Geometry and Logic, they explain precisely what this means and why it's important.
Long answer (I am not fully competent to answer this part): taking sheaves of topological spaces can be done, but the enrichment must be in the category of compactly generated (weak hausdorff) spaces or else we won't have some things that we want like being cartesian closed.<|endoftext|>
TITLE: A polynomial map from $\Bbb R^n$ to $\Bbb R^n$ mapping the positive orthant onto $\Bbb R^n$?
QUESTION [8 upvotes]: Question: Is there a polynomial map from $\Bbb R^n$  to $\Bbb R^n$ under which the image of the positive orthant (the set of points with all coordinates positive) is all of $\Bbb R^n$ ?

Some observations:
My intuition is that the answer must be 'no'... but I confess my intuition for this sort of geometric problem is not very well-developed.
Of course it is relatively easy to show that the answer is 'no' when $n=1$.  (In fact it seems like a nice homework problem for some calculus students.)  But I can't seem to get any traction for $n>1$.
This feels like the sort of thing that should have an easy proof, but then I remember feeling that way the first time I saw the Jacobian conjecture... now I'm wary of statements about polynomial maps of $\Bbb R^n$ !

REPLY [16 votes]: The map $z\in\mathbb C\mapsto z^4\in\mathbb C$, when written out in coordinates, is a polynomial map which sends the closed first quadrant to the whole of $\mathbb R^2$---and by considering cartesian products you get the same for $\mathbb R^{2n}=\mathbb C^n$.
Later: as observed in a comment by Charles, this can be turned into a solution for the open quadrant by composing with a translation, as in $z\in\mathbb C\mapsto (z-z_0)^4\in\mathbb C$ with $z_0$ in the open first quadrant.<|endoftext|>
TITLE: Is functional programming a branch of mathematics?
QUESTION [13 upvotes]: In Theory mainly concerned with lambda-calculus?, F. G. Dorais wrote, of the idea that the lambda-calulus defines a domain of mathematics:

That would never stick unless there's another good reason. Besides, the schism between cs and math is very recent, I would contend that "functional programming" is actually a math term, historically speaking. More importantly, it would be wrong to use a term different than those who use it most, namely theoretical computer scientists, who are very competent mathematicians by the way. 

The idea, I think, is that the overlap between the kind of constructive mathematics that follows the formulae-as-types correspondence, and pure functional programming is so substantial that the core of the two topics is essentially the same subject.
Is this true?

REPLY [3 votes]: I think most people here would agree that Category Theory is part of mathematics.
The study of strongly-typed functional programming languages is really just the study of cartesian closed categories, so I think that this particular part of functional programming is legitimate mathematics.  And Domain Theory is the study of the category of complete partial orders with bottom, so I would include that too.
I don't think I would extend that to untyped or dynamically-typed languages (LISP).  Also, I'd probably pick a term other than "functional programming" since subfields of math are rarely named with gerunds ("strongly typed functional languages" is probably the most accurate, but a bit verbose).<|endoftext|>
TITLE: Alternate expresion of L-series coefficients
QUESTION [6 upvotes]: I was hoping that someone could help clarify a source of confusion for me, I must be doing and saying something wrong but I just don't know what:
Let $E$ be an elliptic curve over $\mathbb{Q}$ and let 
$$L(s,E)=\sum_{n=1}^{\infty}a_n(E)n^{-s}$$
 be the Hasse-Weil $L$-function of $E$. Finally, let $\tilde{E}$ be the reduction of $E$ mod $p$ and assume that $p$ is a prime for which $E$ has good reduction. 
Then
$$a_p(E)=p+1-|\tilde{E}(\mathbb{F}(p))|$$
and setting $a_1(E)=1$ the $p$ power coefficients are given by
$$a_{p^e}(E)=a_p(E)a_{p^{e-1}}(E)-pa_{p^{e-2}}(E).$$
Now looking at Diamond and Shurman, for instance, I find that also we can write
$$a_{p^e}(E)=p^e+1-|\tilde{E}(\mathbb{F}(p^e))|$$
but when I use this expression as a "definition"  of $a_{p^e}(E)$ and do some explicit calculations I don't get the right recursion, for instance I seem to get in practice
$$a_{p^2}(E)=a_p(E)^2 - 2p$$
instead of 
$$a_{p^2}(E)=a_p(E)^2-p.$$
I must be misunderstanding something, but I can't figure out what. Any help?

REPLY [3 votes]: The mistake is acknowledged and corrected in the errata for the third printing:
http://people.reed.edu/~jerry/MF/mferrata3.pdf<|endoftext|>
TITLE: References for Artin motives
QUESTION [17 upvotes]: I find the following description of Artin motives in Wikipedia. Since these seem to be quite related to number theory, I am interested to learn more in that context. I request the experts available in MO to provide me a reference which can be more useful to me than this terse description. Any comments and explanations also will be helpful. I apologize for asking a seemingly basic question; but I find it impossible to wade through the numerous references available on motives.
Fix a field $k$ and consider the functor
finite separable extensions $K$ of $k$ → finite sets with a (continuous) action of the absolute Galois group of $k$
which maps $K$ to the (finite) set of embeddings of $K$ into an algebraic closure of $k$. In Galois theory this functor is shown to be an equivalence of categories. Notice that fields are $0$-dimensional. Motives of this kind are called Artin motives. By $\mathbb Q$-linearizing the above objects, another way of expressing the above is to say that Artin motives are equivalent to finite $\mathbb Q$-vector spaces together with an action of the Galois group.
I have some idea of what are pure motives and mixed motives, in the context of algebraic varieties. What I exactly have in mind is to understand the modern statement of equivariant Tamagawa number conjecture. This would appear to be the simplest instance to keep in mind, if I go ahead.

REPLY [16 votes]: A motive is a chunk of a variety cut out by correspondences.  (If you like, it is something of which we can take cohomology.)  
Artin motives are what one gets by restricting to zero-dimensional varieties.   If the ground
field is algebraically closed then zero-dimensional varieties are simply finite unions of points, so there is not much to say; the only invariant is the number of points.
But if the ground field $K$ is not algebraically closed (but is perfect, e.g. char $0$,
so that we can describe all finite extensions by Galois theory), then there are many
interesting $0$-dimensional motives, and in fact the category of Artin motives (with
coefficients in a field $F$ of characteristic $0$, say) is equal to the category of continuous
representations of $Gal(\overline{K}/K)$ on $F$-vector spaces (where the $F$-vector spaces are given their discrete topoogy; in other words, the representation must factor through $Gal(E/K)$ for some finite extension $E$ of $K$).
Perhaps from a geometric perspective, these motives seem less interesting than others.  On the other hand, number theoretically, they are very challenging to understand.  The Artin conjecture about the holomorphicity of $L$-functions of Artin motives, which is the basic reciprocity conjecture regarding such motives, remains very wide open, with very few non-abelian cases known.  (Of course, for representations with abelian image, these 
conjectures amount to class field theory, which is already quite non-trivial.)<|endoftext|>
TITLE: Magnitude of Graham's Number?
QUESTION [15 upvotes]: I recently stumbled across this number, and then (foolishly, most likely) decided to try to describe it in a blog post
http://frothygirlz.com/2010/01/14/big-numbers-part-2/
Q - Are there any comparisons of Graham's Number, hell, even G1, to other well known "big" numbers, such as googolplex?
I'd just like to have some way, however abstract, to be able to pretend that I have some sort of idea of the magnitude of this number.
Any help and/or tips would be much appreciated.

REPLY [6 votes]: I realize this thread is from a year and a half ago (a possible spam post moved it into the current thread titles), but if anyone is still interested in these issues, about 9 years ago I posted in sci.math a lengthy essay on Graham's number and then followed it up with three more essays (and a promise for more, which I never got around to).
GRAHAM'S NUMBER AND RAPIDLY GROWING FUNCTIONS [2 March 2002]
http://groups.google.com/group/sci.math/msg/0f3c8bab92145996
BIG NUMBERS #1 [8 April 2002]
http://groups.google.com/group/sci.math/msg/403051f310ff3dfc
BIG NUMBERS #2 [8 April 2002]
http://groups.google.com/group/sci.math/msg/d12962e3af2c74b7
BIG NUMBERS #3 [8 April 2002]
http://groups.google.com/group/sci.math/msg/4f2ed8e0385b72f2<|endoftext|>
TITLE: Isolated hypersurface singularities, Chow groups and D-branes
QUESTION [9 upvotes]: Say a ring $R$ is an isolated hypersurface singularity if $R = k[x_1, \ldots, x_n]_{(x_1, \ldots, x_n)}/(W)$, where $k$ is a field and $W \in k[x_1, \ldots, x_n]$ is such that the ideal $(\partial_1 W, \ldots, \partial_n W)$ is $(x_1, \ldots, x_n)$-primary. For finitely generated $R$-modules $M$ and $N$ define the function $\theta$ to be $\theta(M, N) =  \lambda( \operatorname{Tor}^R_{2i} (M,N)) - \lambda( \operatorname{Tor}^R_{2i-1}(M,N))$ for any $i \gg 0$. Here $\lambda$ denotes the length of a module. The definition makes sense because all modules over $R$ have eventually 2-periodic resolutions (since $R$ is a hypersurface) and the $\operatorname{Tor}$'s have finite length for $i \gg 0$ (since $R$ is an "isolated singularity"). Hochster made this definition in his 1981 paper "The dimension of an intersection in an ambient hypersurface."
I'm looking for examples (or preferably a family) of isolated hypersurface singularity rings with $n \geq 4$ and modules $M$ over such rings with $\theta(M, - )$ non-zero. See Hailong's answer below for an equivalent formulation of this in terms of certain Chow groups when $k = \mathbb{C}$ and $W$ is homogeneous. I would prefer if $W$ were not homogeneous but am interested in all cases.
Of course any insight as to when $\theta$ is non-zero would be great but in general this is hard. For instance in the paper above Hochster showed that the direct summand conjecture is true if $\theta$ is non-zero for an explicit family of modules and rings. It is conjectured that $\theta$ is zero when $n$ is odd (this is known when $W$ is homogeneous, see http://arxiv.org/abs/0910.1289v1), and it is known that when $n=4$ the function $\theta$ is nonzero if and only if the class group of $R$ is nonzero.
EDIT: There is a physical interpretation of the above in the spirit of this post: Matrix factorizations and physics. My knowledge of physics is limited so I apologize in advance for any mistakes.
D-branes in a B-twisted topological Landau-Ginzburg models with potential $W$ are given by matrix factorizations of $W$. We only care about values of $\theta$ on maximal Cohen-Macaulay (MCM) $R$-modules, and all such modules are given by matrix factorizations of $W$. Thus MCM modules over $R$ can be thought of as D-branes. Now physicists talk about the BRST-cohomology of two branes $M,N$ (which I don't understand) but it seems that it is given by $\operatorname{Ext}_R^2(M \oplus \Omega M, N \oplus \Omega N)$ (or equivalently the stable homomorphisms between these modules) where $\Omega( - )$ denotes the first syzygy; see for instance http://arxiv.org/abs/0802.1624. It is not hard to see, viewing the modules as matrix factorizations, that for two MCM modules $M$ and $N$ we have
$\theta(M, N) = \lambda( \operatorname{Ext}^1_R( M^*, N) ) - \lambda( \operatorname{Ext}^2_R(M^*, N) )$,
where $M^*$ is the MCM module given by $\operatorname{Hom}_R(M, R)$. Thus to find an example of modules with nonzero $\theta$ is equivalent to finding branes whose "even" and "odd" BRST cohomology have different dimensions over $k$.

REPLY [8 votes]: Assume $k= \mathbf C$ and $W$ homogeneous. Let $X=Proj  (k[x_1,\cdots,x_n]/(W))$. $X$ is then a smooth hypersuraface in $\mathbb P_{n-1}$.
Assume $n=2d$ is even. Corollary 3.10 of the paper you quoted says that $\theta=0$ for all pairs iff the homological Chow group $CH^{d-1}_{hom}$ modulo $[h]^{d-1}$ is not torsion (here $[h]$ is the class of the hyperplane section). So your question, in this case, is equivalent to
($l=d-1$):
 Examples of smooth hypersurfaces of dimension $2l$ such that  $CH^{l}_{hom}/([h]^{l})$ is not torsion ?  
(By the way, I think if you phrased your question this way, it probably would become more popular, consider how many geometry-inclined people visit this site! So if you want more and better answers, consider changing the title.) 
Now, a cheap way to get examples you want is to take $W= x_1x_{d+1} + \cdots + x_dx_{2d}$. Then the cycle defined by $(x_1,...,x_d)$ will not be a multiple of a power of the hyperplane section. Why? Because, I am waving my hand a bit here, if it is then the intersection with the cycle $(x_{d+1},\cdots, x_{2d})$ would be positive. But they are disjointed in $X$! 
The same trick works for generalized quadrics, i.e. if $W = f_1g_1 +\cdots +f_dg_d$. 
EDIT: Let me give more details here. In this situation you can easily make $W$ non-homogeneous as you desire. But the trouble is you can't use my argument above as there is no longer a projective variety $X$. But one can get around this. Let $S=k[x_1,\cdots,x_{2d}]_{m}$  
here $m$ is the irrelevant ideal. Suppose $W = f_1g_1 +\cdots +f_dg_d$ and assume that $(f_1,\cdots, f_d, g_1,\cdots, g_d)$ is a full system of parameters in $S$. Let $R=S/(W)$, $P=(f_1,\cdots,f_d)$ and $Q=(g_1,\cdots,g_d)$. I claim that $\theta^R(R/P,R/Q) \neq 0$.
The reason is that $\theta^R(R/P,R/Q) = \chi^S(S/P,S/Q)$, the Serre's intersection multiplicity (see Hochster's original paper). Because $dim S/P + dim S/Q = d+d =dim S$, we must have $\chi^S(S/P,S/Q)>0$ by Positivity, which is known in this case. 
More exotic examples should be abound, and I am sure people who know more intersection theory can provide some, once they are aware of  what this question is about. I would be interested in hearing more answers along that line.<|endoftext|>
TITLE: What is the history of the name "Chinese remainder theorem"?
QUESTION [27 upvotes]: I'd be particularly interested in who first used the name in a European language and whether it was used in a non-European language such as Arabic, Persian, or an Indian language before that.
[Edit 2010/01/22: Thanks to everyone who responded. It took me a few days to check Jonas Meyer's references. (The discussion of the CRT is on pp 175-176 of Part III of Wylie's book.) As JM said, they seem to narrow the appearance of the name in a European language to 1853-1929, which is hundreds of years later than I expected, and it now wouldn't be so surprising if it first appeared in English, maybe even in Dickson's book. So,
Question: Are there any European languages in which the CRT has a name that is not a direct translation of "Chinese remainder theorem"? 
One more point: Wylie says,
'In examining the productions of the Chinese one finds considerable difficulty in assigning the precise date for the origin of any mathematical process; for on almost every point, where we consult a native author, we find references to some still earlier work on the subject. The high veneration with which is has been customary for them to look upon the labours of the ancients, has made them more desirous of elucidating the works of their predecessors than of seeking fame in an untrodden path; so that some of their most important formulae have reached the state in which we now find them by an almost innumerable series of increments. One of the most remarkable of these is the Ta-yen, "Great Extension," a rule for the resolution of indeterminate problems. This rule is met with in embryo in Sun Tsze's Arithmetical Classic under the name of Wuh-puh-chi-soo, "Unknown Numerical Quantities," where after a general statement in four lines of rhyme the following question is proposed: ... 
In tracing the course of this process we find it gradually becoming clearer till towards the end of the Sung dynasty, when the writings of Tsin Keu-chaou put us in full possession of the principle, and enable us to unravel the meaning of the above mysterious assemblage of numerals....'
The Song dynasty apparently ended in 1279, which gives an interval of several hundred years. So, it seems that the name Chinese Remainder Theorem is not completely unreasonable, since according to Wylie, it's not clear when the general form was discovered, or at least might not have been at the time the theorem got its name. ]

REPLY [2 votes]: The Chinese Remainder Theorem first appears in "Sun Zi's Calculation Classic" between the 3rd and 5th century AD (http://en.wikipedia.org/wiki/Sun_Tzu_(mathematician)).
There is a website about the Chinese Remainder Theorem (http://www.cut-the-knot.org/blue/chinese.shtml), where the author refers to a similar puzzle described by Indian mathematician Brahmagupta in 598 AD. I think it's possible that the Chinese Remainder Theorem became well-known early on among mathematicians elsewhere in Asia after Sun Zi published his book.<|endoftext|>
TITLE: transcendental Galois theory
QUESTION [41 upvotes]: Suppose we define an arbitrary field extension $K/F$ to be Galois if, for all subextensions $L$ of $K/F$, we have $K^{\operatorname{Aut}(K/L)} = L$. In words: for any element $x$ of $K \setminus L$, there exists an automorphism $s$ of $K$ such that $s(l) = l$ for all $l$ in $L$, but $s(x) \neq x$. (Note that in case $K/F$ is algebraic, this is indeed a characteristic property of Galois extensions.) What are the transcendental Galois extensions? 
In my rough notes Transcendental Galois Theory, I show that if $F$ has characteristic $0$ and $K$ is algebraically closed, then $K/F$ is Galois in the above sense.  [Actually, these notes are somewhat incomplete.  Having been unable to complete the proof of the conjecture below, I left out some of the more straightforward details.  If anyone wants to see more detail on anything in these notes, please let me know.]
I also conjectured: if $K/F$ is Galois, then either $K/F$ is algebraic, normal and separable, or $F$ has characteristic $0$ and $K$ is algebraically closed. Is this true?
Comment: It is easy to see that if $K/F$ is not algebraic, then $K$ must have characteristic $0$.  It is possible to modify the question a bit so that the positive characteristic case is not ruled out, but I would like to understand what's going on in characteristic $0$ first!
In my notes, I show that an affirmative answer follows from a certain (arguably) less weird conjecture about Galois closures of subfields of rational function fields.  If there is any interest, I will reproduce this conjecture here explicitly.

REPLY [13 votes]: Even in our day of sophisticated search engines, it still seems that the success of a search often turns on knowing exactly the right keyword.
I just followed up on Sylvain Bonnot's comment above.  The property of a field extension $K/F$ that for all subextensions $L$ we have $K^{\operatorname{Aut}(K/L)} = L$ is apparently most commonly called Dedekind.  This terminology appears in Exercise V.9 of Bourbaki's Algebra II, where the reader is asked to show that if $L/K$ is a nonalgebraic Dedekind extension and $T$ is a transcendence basis, then $L/K(T)$ must have infinite degree.  Ironically, this is exactly what I could show in my note.  One can (in the general case, even...) immediately reduce to the case $T = \{t\}$ and then the exercise is saying that the function field $K(C)$ of an algebraic curve (again, it is no loss of generality to assume the function field is regular by enlarging $K$) is not Dedekind over $K$.  This is kind of a strange coincidence!  [However, the proof I give is openly geometric so is probably not the one that N.B. had in mind...]
It also appears in 

MR0067098 (16,669f)
  Barbilian, D.
  Solution exhaustive du problème de Steinitz. (Romanian. Russian, French summary)
  Acad. Repub. Pop. Române. Stud. Cerc. Mat. 2, (1951). 195–259 (misprinted 189–253).

In this paper, the author shows that $L/K$ is a Dedekind extension iff for all subextensions $M$, the algebraic closure $M^*$ of $M$ in $L$ is such that $M^*/M$ is Galois in the usual sense: i.e., normal and separable.  (This is a nice fact, I suppose, and I didn't know it before, but it seems that the author regarded this as a solution of the problem of which extensions are Dedekind.  I don't agree with that, since it doesn't answer my question!)
Apparently one is not supposed to read the above paper but rather this one:

MR0056588 (15,97b)
  Krull, Wolfgang
  Über eine Verallgemeinerung des Normalkörperbegriffs. (German)
  J. Reine Angew. Math. 191, (1953). 54–63.

Here is the MathSciNet review by E.R. Kolchin (who knew something about transcendental 
Galois extensions!):

The author reviews a definition and some results of D. Barbilian [Solutia exhaustiva a problemai lui Steinitz, Acad. Repub. Pop. Române. Stud. Cerc. Mat. 2, 189--253 (1950), unavailable in this country], providing proofs which are said to be simpler, and further results. Let L be an extension of a field K. Then L is called normal over K if for every intermediate field M the relative algebraic closure M∗ of M in L is normal (in the usual sense) over M. If L has the property that every M is uniquely determined by the automorphism group U(M) of L over M, then L is normal over K and, if the characteristic p=0, conversely; if p>0 the converse fails but a certain weaker conclusion is obtained. Various further results are found, and constructive aspects of normal extensions are explored. Some open questions are discussed, the most important one being: Do there exist transcendental normal extensions which are not algebraically closed? 

So it seems that my question is a nearly 60 year-old problem which was considered but left unsolved by Krull.  I am tempted to officially give up at this point, and perhaps write up an expository note informing (and warning?) contemporary readers about this circle of ideas.  Comments, suggestions and/or advice would be most welcome...
P.S.: Thanks very much to M. Bonnot.<|endoftext|>
TITLE: Strong induction without a base case
QUESTION [42 upvotes]: Strong induction proves a sequence of statements $P(0)$, $P(1)$, $\ldots$ by proving the implication 
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true." 
for every nonnegative integer $n$.  There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously.  But most strong induction proofs nevertheless seem to involve a separate argument to handle the base case (i.e., to prove the implication for $n=0$).
Can you think of a natural example of a strong induction proof that does not treat the base  case separately?  Ideally it should be a statement at the undergraduate level or below, and it should be a statement for which strong induction works better than ordinary induction or any direct proof.

REPLY [3 votes]: One may transform any strong induction proof into one which (legalistically speaking!) doesn't explicitly treat the base case.
Proving $P(n)$ by induction, one assumes $ \forall k<n \ P(k)$ which weakens to $\forall k<n\ P(0)\implies P(k)$.`  Now one may adapt whatever proof one had for the induction step to establish from this that $P(0)\implies P(n)$.  Now one proves $P(0)$ and concludes $P(n)$ by modus ponens.
Yes, I know, morally the base case still got special treatment, but formally now that happens in the induction step.  Thus finding a formal distinction between theorems that require special treatment for the base case and those that don't seems impossible.
In any case, if you can prove $\forall n\geq 0\ P(n)$ by strong induction, you can prove $\forall n>0 \ P(0)\implies P(n)$ by strong induction with no special treatment for the base case, morally or formally.  So examples abound.<|endoftext|>
TITLE: Is there a relationship between model theory and category theory?
QUESTION [23 upvotes]: According to Chang and Keisler's "Model Theory", Model Theory = Universal Algebra + Logic. Model theory generalized Universal Algebra in the sense that we allow relations while in Universal Algebra we only allow functions.
Also, we know that Category Theory generalized Universal Algebra. From wikipedia:

Blockquote Given a list of operations and axioms in universal algebra, the corresponding algebras and homomorphisms are the objects and morphisms of a category. Category theory applies to many situations where universal algebra does not, extending the reach of the theorems. Conversely, many theorems that hold in universal algebra do not generalise all the way to category theory.

So this suggest there might be some overlapping between Model Theory and Category Theory. I hope some one can elaborate about the relation (if there is)?

REPLY [2 votes]: The theory of Abstract Elementary Classes, which were introduced by Shelah as an abstract axiomatization of elementary classes was recently connected with Accessible categories.

Beke, Rosicky. Abstract elementary classes and accessible categories.
M. Lieberman's PhD Thesis.

This connection was better understood since the recent join of Sebastian Vasey.

Rosicky Lieberman Vasey. Universal abstract elementary classes and locally multipresentable categories
Rosicky Lieberman Vasey. Internal sizes in μ-abstract elementary classes
Rosicky Lieberman Vasey. Forking independence from the categorical point of view<|endoftext|>
TITLE: Heuristically false conjectures
QUESTION [43 upvotes]: I was very surprised when I first encountered the Mertens conjecture. Define 
$$ M(n) = \sum_{k=1}^n \mu(k) $$
The Mertens conjecture was that $|M(n)| < \sqrt{n}$ for $n>1$, in contrast to the Riemann Hypothesis, which is equivalent to $M(n) = O(n^{\frac12 + \epsilon})$ .  
The reason I found this conjecture surprising is that it fails heuristically if you assume the Mobius function is randomly $\pm1$ or $0$. The analogue fails with probability $1$ for a random $-1,0,1$ sequence where the nonzero terms have positive density. The law of the iterated logarithm suggests that counterexamples are large but occur with probability 1. So, it doesn't seem surprising that it's false, and that the first counterexamples are uncomfortably large. 
There are many heuristics you can use to conjecture that the digits of $\pi$, the distribution of primes, zeros of $\zeta$ etc. seem random. I believe random matrix theory in physics started when people asked whether the properties of particular high-dimensional matrices were special or just what you would expect of random matrices. Sometimes the right random model isn't obvious, and it's not clear to me when to say that an heuristic is reasonable. 
On the other hand, if you conjecture that all naturally arising transcendentals have simple continued fractions which appear random, then you would be wrong, since $e = [2;1,2,1,1,4,1,1,6,...,1,1,2n,...]$, and a few numbers algebraically related to $e$ have similar simple continued fraction expansions. 
What other plausible conjectures or proven results can be framed as heuristically false according to a reasonable probability model?

REPLY [39 votes]: This is quite elementary, but surprised me when I first saw it, and I still think it's remarkable. 
The number of pairs of integers $(x, y)$ such that $x^2 + y^2 \leq n$ is asymptotically $\pi n$, since they are the lattice points inside a circle of radius $\sqrt{n}$. Therefore the average number of ways of writing a positive integer as a sum of two squares is $\pi$. Or $\pi/8$ if we regard solutions as the same when they differ only in signs or the order of the terms.  
One would therefore expect a positive proportion of the natural numbers to have a representation as a sum of two squares. Not a $\pi/8$-fraction, since some integers have several representations, but some slightly smaller positive density, since identities like $4^2 + 7^2 = 1^2 + 8^2$ look pretty much like random coincidences.
But actually almost no numbers are sums of two squares. Whenever the prime factorization of $n$ contains some prime $p\equiv 3$ (mod 4) to an odd power, $n$ cannot be a sum of two squares, as is easily seen by considering the equation modulo powers of $p$. And by Dirichlet's theorem, almost all numbers have some such prime to power 1 in their factorization.<|endoftext|>
TITLE: Are there elements of fixed weight in a crystal not killed by too many Kashiwara operators?
QUESTION [6 upvotes]: I've come across an annoying lemma trying to finish up an argument, and I was hoping one of you guys knew about it.  
Question: Given

a weight $\lambda$ of a simple Lie algebra $\mathfrak g$, and 
integers $n_\alpha$ for each simple root $\alpha$, 

Is there a highest weight $\nu$, such that in the crystal of with highest weight $\nu$ there is an element $x$ of weight $\lambda$ such that $\tilde{F}_\alpha^{n_\alpha}x\neq 0$? 
This is true in $\mathfrak{sl}_2$, which makes me hopeful about other Lie algebras, but the argument isn't coming together for me.

REPLY [3 votes]: OK, I think I see the answer: by whatever character formula you like, you can see that for any fixed weight spaces (say, the $\lambda$ and $\lambda-n_\alpha\alpha$ for all $\alpha$), you can choose $\nu$ so that the multiplicities at all these points are very large compared to the differences between the those multiplicities.  Since the number of elements in the crystal killed by $\tilde{F}_{\alpha}^{n_\alpha}$ is just the difference in the weight multiplicities (or 0), the result follows by pigeonhole.<|endoftext|>
TITLE: Covering maps of Riemann surfaces vs covering maps of $k$-algebraic curves
QUESTION [5 upvotes]: In going from Riemann surface theory to the theory of algebraic curves over fields $k$ that are not necessarily $\mathbb{C}$, I would like to understand more about how the notion of a covering map carries over.
If I have a compact, connected Riemann surface $M$, a cover of $M$ by another such Riemann surface, say $N$, then I am taking this to mean a holomorphic map $f:N\rightarrow M$ of finite degree $m>0$ (that is, the generic fiber of $f$ consists of $m$ points).  At a $p\in M$ that is a regular value of $f$ (i.e. $p$ is not a branch point), then there is a open neighborhood $U$ of $p$ such that $f^{-1}(U)$ is the disjoint union of $m$ copies of $U$, and 'open' refers to the topology determined by the complex analytic structure on $M$.
If I have $M$ and $N$ nonsingular algebraic curves over a field $k$, then what can be said about $f^{-1}(U)$ when $f:N\rightarrow M$ is a finite regular map?  What I mean by this is when the topology is the Zariski topology, I assume that the statement "$f^{-1}(U)$ is the disjoint union of $m$ copies of $U$" translates to open sets in that topology.  When we regard a compact Riemann surface as a nonsingular curve over $\mathbb{C}$, then will these notions coincide?
Sorry if this is a trivial/ill-posed question.  (My experience so far is more with differential geometry and complex analytic geometry....)
Many thanks.

REPLY [10 votes]: No, the property of having small neighbourhoods whose preimage is a disjoint union
of $n$ homeomorphic open sets does not hold in the Zariski topology (once $n > 1$, i.e.
the cover is non-trivial).  The reason is 
that non-empty Zariski open sets are always very big; in the case of a curve,
their complement is always just a finite number of points.  In particular, two different
non-empty Zariski opens are never disjoint. 
There are two ways that one rescues the situation: the first is to use the differential
topology view-point on covers: they are proper submersions between manifolds of the same dimension.
In the context of compact Riemann surfaces, both source and target have the same dimension,
and maps are automatically proper, so it is just the submersion property that is left
to think about.  It is a property about how tangent spaces map, which can be translated into
the algebraic context (e.g. using the notion of Zariski tangent spaces).
So if $f: X \rightarrow Y$ is a regular morphism (regular morphism is the algebraic geometry terminology for an everywhere defined map given locally by rational functions) of projective curves, we can say that
$f$ is unramified at a point $p \in X$ if $f$ induces an isomorphism from the Zariski tangent
space of $X$ at $p$ to the Zariksi tangent space to $Y$ at $f(P)$.
For historical reasons, if $f$ is unramified at every point in its domain, we say that
$f$ is etale (rather than a cover), but this corresponds precisely to the notion of a covering map when we pass to Riemann surfaces.
This leads to the more sophisticated rescue: one considers all the etale maps from (not necessarily projective or connected) curves $X$ to $Y$, and considers them as forming a topology on $Y$, the so-called etale topology of $Y$.  This leads to many important notions
and results, since it allows one to transport many topological notions (in particular, fundamental groups and cohomology) to the algebraic context.<|endoftext|>
TITLE: Introduction to deformation theory (of algebras)?
QUESTION [28 upvotes]: So I know that the idea of deformation theory underlies the concept of quantum groups; I haven't found any single introduction to quantum groups that makes me fully satisfied that I have some kind of idea of what it's all about, but piecing together what I've read, I understand that the idea is to "deform" a group (Hopf) algebra to one that's not quite as nice but is still very workable.
To a certain extent, I get what's implied by "deformation"; the idea is to take some relations defining our Hopf algebra and introduce a new parameter, which specializes to the classical case at a certain point. What I don't understand is:

How and when we can do this and have it still make sense;
Why this should "obviously" be a construction worth looking at, and why it should be useful and meaningful.

The problem is when I look for stuff (in the library catalogue, on the Internet) on deformation theory, everything that turns up is really technical and assumes some familiarity with the basic definitions and intuitions about the subject. Does anyone know of a more basic introduction that can be understood by the "general mathematical audience" and answers (1) and (2)?

REPLY [8 votes]: I like "Why Deformations are Cohomological" by M Anel<|endoftext|>
TITLE: Is there a slick proof of the classification of finitely generated abelian groups?
QUESTION [67 upvotes]: One the proofs that I've never felt very happy with is the classification of finitely generated abelian groups (which says an abelian group is basically uniquely the sum of cyclic groups of orders $a_i$ where $a_i|a_{i+1}$ and a free abelian group).
The proof that I know, and am not entirely happy with goes as follows: your group is finitely presented, so take a surjective map from a free abelian group.  The kernel is itself finitely generated (this takes a little argument in and of itself; note that adding a new generator to a subgroup of free abelian group either increases dimension after tensoring with $\mathbb{Q}$ or descreases the size of the torsion of the quotient), so our group is the cokernel of a map between finite rank free groups.  Now, (and here's the part I dislike) look at the matrix for this map, and remember that it has a Smith normal form.  Thus, our group is the quotient of a free group by a diagonal matrix where the non-zero entries are $a_i$ as above.
I really do not think I should have to algorithmically reduce to Smith normal form or anything like that, but know of no proof that doesn't do that.
By the way, if you're tempted to say "classification of finitely generated modules for PIDs!" make sure you know a proof of that that doesn't use Smith normal form first.

REPLY [3 votes]: Here is a proof of the finite case that I think is pretty different from the others. It is inspired by the proof of Lemma B.13 in Pete Clark's algebra notes.
Let $G$ be a finite group, written additively. Let $n$ be the highest order of any element of $G$ and let $g$ be an element of order $n$. We will show that $\langle g \rangle$ splits off as a direct summand of $G$; then we induct.
Let $H$ be a subgroup of $G$ which is maximal with respect to the condition that $\langle g \rangle \cap H = \{ 0 \}$. We can always take $H = \{ 0 \}$, so at least one such subgroup exists. We will show that $G = \langle g \rangle \oplus H$.
If $\langle g \rangle \oplus H$ is not all of $G$, then we can find some $u$ in $G$ not lying in $\langle g \rangle \oplus H$. Let $b$ be the lowest positive integer such that $bu \in \langle g \rangle \oplus H$; say $bu = ag+ h$. Write $a = qb+r$ with $0 \leq r < b$. Putting $v:=u-qg$, we have $bv = rg + h$, and $b$ is the lowest positive integer such that $bv \in \langle g \rangle \oplus H$.
Case 1: $r=0$. In this case, put $H' = H + \langle v \rangle$; we have $H' \supsetneq H$ and $\langle g \rangle \cap H' = \{ 0 \}$, contradicting the maximality of $H$.
Case 2: $r>0$. In this case, we claim that the order of $v$ is greater than $n$, contradicting the maximality of $n$. Indeed, the order of $v$ must be divisible by $b$, and the order of $bv$ is at least $\tfrac{n}{r} > \tfrac{n}{b}$. $\square$
I feel like this should be adaptable to cover the case of finitely generated abelian groups, but it keeps not coming out slick when I try. Maybe someone can show me how!<|endoftext|>
TITLE: Why are some tilings introduced as geometrical objects, not graphs?
QUESTION [9 upvotes]: Let's say you have a planar tiling. Quite often these tilings are introduced as geometrical objects with metrics; each tile having coordinates assigned to its vertices.
The tiling has an associated graph: the nodes of the graph are the vertices of the tiles, etc.
What kind of additional value is in general provided by insisting the tile vertices to have coordinates? Does not discussing the properties of the tiling graph topology give enough information?
I understand that the answer depends on the type of the tiling in question.
If one takes graphs as the starting point, then what would be some natural ways to define infinite planar graph periodicity so that there would exist periodic planar tilings corresponding to a given graph?
Can Penrose tiling be defined by its graph topology and can some of its general properties, like aperiodicity, be described purely by its tiling graph without relying on the angles and edge length of the tiles?
(If You find this post fuzzy or ignorant, that would be because the poster is not a 
professional mathematician.)
Pontus

REPLY [3 votes]: We keep the geometric structure on tilings mainly because the tilings are generated with that structure, often from lattices or geometric group actions. 
It's quite nontrivial that if you take some types of nice tilings, and forget the geometric structure, then you can indeed recover important information about the tiling from the graph alone. You can recover whether the tiling is periodic by looking at the group of symmetries of the graph. You can also recover information about the space, not just the tiling. You can recover whether the tiling is of the Euclidean or hyperbolic plane by looking at the growth rate of the perimeter of a ball. You can recover whether the tiling was on a topological cylinder vs. the plane, as you can still define the "end" of a graph, and see that a tiling of the cylinder will have two ways to go off to infinity rather than 1 in the plane. 
This is a start of what is knwon as geometric group theory. Given a group and some finite set of generators for that group, you can construct a Cayley whose vertices are the elements of that group, whose edges connect an element $g$ with $gg_i$ and $gg_i^{-1}$ for each generator $g_i$. Then you can try to recover information about the group from the geometric properties of the graph.
There is a natural metric $d$ on the Cayley graph, so that each edge has length 1. From one perspective, it's bad that we are getting different graphs from different sets of generators. To identify these as essentially the same, we consider quasi-isometries, maps $f$ from one space to another such that there are constants $C_0$ and $C_1$ so that for every $x,y$, $\frac1{C_1} d(x,y) - C_0 \le d(f(x),f(y)) \le C_1 d(x,y) + C_0$. Changing from one set of generators to another is a quasi-isometry, since we can express each generator as a finite word in the other set of generators. Thus, many people study finitely generated groups up to quasi-isometry. 
Choices for sets of relations may correspond to tilings. You can attach a 2-cell to the graph along the word of a relation. Topological and geometric properties of this complex have meaning in group theory. 
Anyway, back to tilings of the plane. There are more reasons to keep the geometry. This picks out a few graphs among the many which embed in the plane. We also get convenient ways to compare tilings. For example, we can look at the vertices of a second tiling which are near a vertex in the first tiling. We can more easily consider entire families of tilings to try to classify all tilings of a type.
For the Penrose tilings in particular, I would hate to ignore the nice lift from number theory of a tiling to a map from the plane to $\mathbb R^4$. If you consider a tiling by rhombuses so that each edge is $\pm \zeta_5^i$, where $\zeta_5$ is a 5th root of unity, then you see that you can give each vertex a 5-dimensional set of coordinates as an integer linear combination of the 5th roots of unity. Of course, since the sum of the 5th roots of unity and 1 is 0, you can drop the dimension to 4 by considering sums of fifths of integers which add up to 0. There are nice ways to generate Penrose tilings by 2-dimensional planes in that 4-dimensional space. I don't know the classification of Penrose tilings, but I bet it has something to do with that lift, which is not obvious from the graph.<|endoftext|>
TITLE: Algebraic/Categorical motivation for the Chevalley Eilenberg Complex
QUESTION [9 upvotes]: Is there a purely algebraic or categorical way to introduce the Chevalley-Eilenberg complex in the definition of Lie algebra cohomology?
In group cohomology, for example, the bar resolution of a group is equal to the chain complex associated to the nerve of the group when considered as a category; I wonder if something like this is also possible for the Chevalley-Eilenberg complex?
I know that the Chevalley-Eilenberg complex arises as the subcomplex of left-invariant differential forms on a Lie group, but apart from this geometric origin, its definition doesn't seem very natural to me, so it would be nice to have some algebraic or categorical construction as well.

REPLY [11 votes]: Lie algebras are algebras over an operad, usually denoted $\mathscr{L}\mathit{ie}$. This is a quadratic operad, which happens to be Koszul. It therefore comes with a prefered cohomology theory (there is an analogue of Hochschild cohomology of algebras over a Koszul operad) which is defined in terms of a certain canonical complex —unsurprisingl called the Koszul complex. If you work out the details in this general construction, you obtain the Chevalley-Eilenberg resolution.
Alternatively, if $\mathfrak{g}$ is a Lie algebra, we can view $U(\mathfrak g)$, its enveloping algebra, as a PBW deformation of the symmetric algebra $S(\mathfrak g)$. The latter is just a polynomial ring, so we have a nice resolution for it, the Koszul complex, and there is a more or less canonical way of deforming that resolution so that it becomes a resolution for the PBW deformation. Again, working out the details rapidly shows that the deformed complex is the Chevalley-Eilenberg complex.
Finally (I haven't really checked this, but it should be true :) ) if you look at $U(\mathfrak g)$ as presented by picking a basis $B=\{X_i\}$ for $\mathbb g$ and dividing the free algebra it generates by the ideal generated by the relations $X_iX_j-X_jX_i-[X_i,X_j]$, as usual, you can construct the so called Annick resolution. Picking a sensible order for monomials in the free algebra (so that standard monomials are precisely the elements of the PBW basis of $U(\mathfrak g)$ constructed from some total ordering of $B$), this is the Chevalley-Eilenberg complex again.
These three procedures (which are of course closely interrelated!) construct the resolution you want as a special case of a general procedure. History, of course, goes in the other direction.

REPLY [5 votes]: Consider the universal enveloping algebra $U(\mathfrak g)$ of your Lie algebra $\mathfrak g$. Since it is a Hopf algebra, then you can construct a filtered simplicial cocommutative coalgebra $A_\bullet$: 

$A_i=U(\mathfrak g)^{\otimes i}$
face maps are given by applying the product
degeneracy maps are given by applying the unit

The $E_1$ term of the associated spectral sequence is precisely the Chevalley-Eilenberg chain complex*. 
In other words, the Chevalley-Eilenberg complex of $\mathfrak g$ is a by-product of the Bar complex of $U(\mathfrak g)$. And the Bar complex of an augmented unital algebra $A$ arises as the chain complex associated to the simplicial set $Nerve(A)$ (where I view $A$ as a linear category  with one object). 

this is another way of saying what Mariano Suárez-Alvarez says in his answer.<|endoftext|>
TITLE: What are fixed points of the Fourier Transform
QUESTION [62 upvotes]: The obvious ones are 0 and $e^{-x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?

REPLY [29 votes]: $\bf{1.}$ A more complete list of particular self-reciprocal Fourier functions of the first kind, i.e. eigenfunctions of the cosine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$:
$1.$ $\displaystyle e^{-x^2/2}$ (more generally $e^{-x^2/2}H_{2n}(x)$, $H_n$ is Hermite polynomial)
$2.$ $\displaystyle \frac{1}{\sqrt{x}}$ $\qquad$ $3.$ $\displaystyle\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $4.$ $\displaystyle \frac{\cosh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}$ $\qquad$$5.$ $\displaystyle\frac{1}{1+2\cosh \left(\sqrt{\frac{2\pi}{3}}x\right)}$
$6.$ $\displaystyle \frac{\cosh\frac{\sqrt{3\pi}x}{2}}{2\cosh \left( 2\sqrt{\frac{\pi}{3}} x\right)-1}$ $\qquad$ $7.$ $\displaystyle \frac{\cosh\left(\sqrt{\frac{3\pi}{2}}x\right)}{\cosh (\sqrt{2\pi}x)-\cos(\sqrt{3}\pi)}$ $\qquad$ $8.$ $\displaystyle \cos\left(\frac{x^2}{2}-\frac{\pi}{8}\right)  $
$9.$ $\displaystyle\frac{\cos \frac{x^2}{2}+\sin \frac{x^2}{2}}{\cosh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $10.$ $\displaystyle \sqrt{x}J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)$ $\qquad$ $11.$ $\displaystyle  \frac{\sqrt[4]{a}\ K_{\frac{1}{4}}\left(a\sqrt{x^2+a^2}\right)}{(x^2+a^2)^{\frac{1}{8}}}$
$12.$ $\displaystyle \frac{x e^{-\beta\sqrt{x^2+\beta^2}}}{\sqrt{x^2+\beta^2}\sqrt{\sqrt{x^2+\beta^2}-\beta}}$$\qquad$ $13.$ $\displaystyle \psi\left(1+\frac{x}{\sqrt{2\pi}}\right)-\ln\frac{x}{\sqrt{2\pi}}$,  $\ \psi$ is digamma function.
Examples $1-5,8-10$ are from the chapter about self-reciprocal functions in Titschmarsh's book "Introduction to the theory of Fourier transform". Examples $11$ and $12$ can be found in Gradsteyn and Ryzhik. Examples $6$ and $7$ are from this question What are all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ self-reciprocal under Fourier transform?. Some other self-reciprocal functions composed of hyperbolic functions are given in Bryden Cais's paper On the transformation of infinite series. Discussion of $13$ can be found in Berndt's article. 
$\bf{2.}$ Self-reciprocal Fourier functions of the second kind, i.e. eigenfunctions of the sine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin ax dx=f(a)$:
$1.$ $\displaystyle \frac{1}{\sqrt{x}}$ $\qquad$ $2.$ $\displaystyle xe^{-x^2/2}$ (and more generally $e^{-x^2/2}H_{2n+1}(x)$)
$3.$ $\displaystyle \frac{1}{e^{\sqrt{2\pi}x}-1}-\frac{1}{\sqrt{2\pi}x}$ $\qquad$ $4.$ $\displaystyle \frac{\sinh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}$ $\qquad$ $5.$ $\displaystyle \frac{\sinh\sqrt{\frac{\pi}{6}}x}{2\cosh \left(\sqrt{\frac{2\pi}{3}}x\right)-1}$
$6.$ $\displaystyle \frac{\sinh(\sqrt{\pi}x)}{\cosh \sqrt{2\pi} x-\cos(\sqrt{2}\pi)}$ $\qquad$ $7.$ $\displaystyle \frac{\sin \frac{x^2}{2}}{\sinh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $8.$ $\displaystyle  \frac{xK_{\frac{3}{4}}\left(a\sqrt{x^2+a^2}\right)}{(x^2+a^2)^{\frac{3}{8}}}$
$9.$ $\displaystyle \frac{x e^{-\beta\sqrt{x^2+\beta^2}}}{\sqrt{x^2+\beta^2}\sqrt{\sqrt{x^2+\beta^2}+\beta}}$$\qquad$ $10.$ $\displaystyle \sqrt{x}J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)$$\qquad$ $11.$ $\displaystyle e^{-\frac{x^2}{4}}I_{0}\left(\frac{x^2}{4}\right)$
$12.$ $\displaystyle \sin\left(\frac{3\pi}{8}+\frac{x^2}{4}\right)J_{0}\left(\frac{x^2}{4}\right)  $$\qquad$ $13.$ $\displaystyle \frac{\sinh \sqrt{\frac{2\pi}{3}}x}{\cosh \sqrt{\frac{3\pi}{2}}x}$
Examples $1-5,7$  can be found in Titschmarsh's book cited above. $8-12$ can be found in Gradsteyn and Ryzhik. $13$ is from Bryden Cais, On the transformation of infinite series, where more functions of this kind are given.<|endoftext|>
TITLE: Representing numbers in a non-integer base with few (but possibly negative) nonzero digits
QUESTION [12 upvotes]: Background
In a recent question about Fibonacci numbers, it was claimed that

every integer can be written in the form $\sum_{i=1}^6 \epsilon_i F_{n_i}$ with $\epsilon_i \in \{0,-1,1\}$. The upper limit on the summation isn't a typo: every number is the sum/difference of at most 6 fibonacci's.

I believe this is false, even for larger (but still finite) values of $6$:
First of all, without loss of generality, we may assume that the representations do not repeat any Fibonacci number (i.e., the $n_i$s are distinct) and moreover, do not contain any two consecutive Fibonacci numbers (i.e., $n_i \ne n_j+1$).  We may arrive at such a representation by using the following simplifications repeatedly:

If two consecutive Fibonacci numbers appear with opposite signs, simplify the expansion with the identity $F_n - F_{n-1} = F_{n-2}$.
If two consecutive Fibonacci numbers appear with the same sign, simplify the expansion with the identity $F_n + F_{n-1} = F_{n+1}$.
If the same Fibonacci number appears with opposite signs, simply cancel the two terms.
If the same Fibonacci number appears with the same sign, then use the identity $F_n + F_n = F_{n-2} + F_{n-1} + F_n = F_{n-2} + F_{n+1}$ to replace them with two non-identical Fibonacci numbers.

The first three operations reduce the number of terms in the expansion and thus strictly simplify the expression (in terms of how many terms there are), but the last may need to be used several times before it "simplifies" the expression (for example, in terms of how many repeated terms there are).  Nonetheless, this simplification procedure terminates, as it is impossible to get stuck in an infinite loop using the last operation alone.  (Proof: we may assume that the $n_i$ are positive.  Then all of the operations either reduce the number of terms, or leaves that unchanged and reduces the sum of the $n_i$.)
Now, assume we have such a representation (no identical terms, no consecutive terms) and suppose the largest Fibonacci number appearing is $F_n$.  Then the next largest term (in absolute value) that may appear is $F_{n-2}$, the next largest after that $F_{n-4}$, and so on.  All in all, the sum of the terms excluding $F_n$ is at most $F_{n-2} + F_{n-4} + \cdots \le F_{n-1}$ (proof by induction: add $F_n$ to both sides).  By the triangle inequality, the sum of all the terms must be at least $F_n - F_{n-1} = F_{n-2}$.  The point of this calculation is that if you want to represent a number that's less than $F_{n-2}$, you can't use terms that are $F_n$ or greater.
This leads us to our contradiction.  Consider the integers between $0$ and $F_{n-2}-1$.  How many possible representations are there of numbers in this range?  Well, we have six terms all of which are 0 or $\pm F_k$ for $k\lt n$ (from the above discussion), so we have at most $(2n+1)^6$ representations that could possibly fall into the range.  (We're over-counting here because it won't matter and this is easier.)  However, there are clearly $F_{n-2}$ different integers in the given range.  Assume for contradiction that it were always possible to represent numbers as the sum/difference of at most 6 Fibonacci's.  Then we would have
$$ (2n+1)^6 \ge F_{n-2}. $$
Finally, because the left side grows polynomially while the right side grows exponentially, a large enough value of $n$ will produce a contradiction.
My questions

Is the proof above correct?  (If not, and the original claim is correct, can you give me a representation of the number 5473?) Edit: Please see Michael Lugo's answer for a paper which finds the representation with the fewest nonzero digits in this "signed Fibonacci base".  Please consider the following the actual question here:
Assuming the proof is correct, is the original claim true for other non-integer bases?  What I mean is the following:


Does there exist a natural number $k$ and a real number $b>1$ such that every integer has a representation as $\sum_{i=1}^k \epsilon_i \lfloor b^{n_i} + \frac12 \rfloor$?  That is, does every number have a representation in "base $b$" (because $b$ is probably irrational, we round $b^n$ to the nearest integer) with at most $k$ non-zero "digits", but where the "digits" may be $\pm 1$?

Note that the original claim is an instance of this: $k=6$ and $b=\varphi = \frac{1+\sqrt 5}{2}$.
I don't think my proof works directly because I used special properties of $\varphi$/the Fibonacci numbers.  Is it possible to remove this reliance?  In particular, the second step of the proof shows that it's not "useful" to have large Fibonacci numbers in the representation of a small number.  Is the same true for every base $b$?
Note: if the digits were not allowed be negative, then my proof would go through.  The main issue is whether or not $\pm b^n$ for large $n$ can cancel and produce small numbers.
Thanks!

REPLY [8 votes]: This is an answer to your "actual question" (2), building on some of the ideas in Douglas Zare's answer.
Lemma 1: Suppose that $0 < r < 1$.  Let $S=\lbrace \epsilon r^i : \epsilon = \pm 1 \text{ and } i \in \mathbb{Z}_{\ge 0} \rbrace$.  Fix $k \ge 1$.  Let $S_k$ be the set of sums of the form $s_1+\cdots+s_k$ such that $s_i \in S$ and $|s_1|=1$ and there is no nonempty subset $I \subset \lbrace 1,\ldots,k \rbrace$ with $\sum_{i \in I} s_i = 0$.  Then $0$ is not in the closure of $S_k$.
Proof: Use induction on $k$.  The base case is trivial: $S_1=\lbrace -1,1\rbrace$.  Now suppose $k \ge 2$.  If a sequence $(x_i)$ in $S_k$ converges to $0$, then the smallest summand in the sum giving $x_i$ must tend to $0$, since a lower bound on the absolute values of the summands rules out all but finitely many elements of $S_k$, which are all nonzero.  Discarding the finitely many $x_i$ for which the smallest summand is $\pm 1$ and removing the smallest summand from each remaining $x_i$ yields a sequence $(y_i)$ in $S_{k-1}$ tending to $0$, contradicting the inductive hypothesis.
Now fix $b>1$ and $k$.  Let $T=\lbrace \epsilon \lfloor b^n + 1/2 \rfloor : \epsilon = \pm 1 \text{ and } n \in \mathbb{Z}_{\ge 0} \rbrace$.  Let $T_k$ be the set of sums of the form $t_1+\cdots+t_k$ with $t_i \in T$.
Lemma 2: Each $t=t_1+\cdots+t_k \in T_k$ equals $u_1+\cdots+u_\ell+\delta$ for some $\ell \le k$ and some $u_i \in T$ with $u_i = O(t)$ and $\delta = O(1)$.
Proof: Examine the powers of $b$ used in the $t_i$.  If any nonempty subsum (with signs) of these powers equals $0$, the corresponding $t_i$ sum to $O(1)$.  If $b^n$ is the largest power that remains after removing all such subsums, divide all the remaining $t_i$ by $b^n$, and apply Lemma 1 with $r=1/b$ to see that $|t|/b^n$ is bounded away from $0$, so all these remaining $t_i$, which are $O(b^n)$, are $O(t)$.
Corollary: The number of elements of $T_k$ of absolute value less than $B$ is $O((\log B)^k)$ as $B \to \infty$.  
Corollary: $T_k \ne \mathbb{Z}$.<|endoftext|>
TITLE: When is the canonical divisor of an algebraic surface smooth?
QUESTION [10 upvotes]: Is there some condition on a complex algebraic surface that implies it has a smooth canonical divisor?  I am searching for the sharpest possible condition, but sufficient criteria would  be nice as well.
For example, the question is quite simple for surfaces that can be embedded in $P^3$.  Roughly, just notice that in this case the linear system of the canonical line bundle is base point free and hence, by Bertini's theorem, we can find a smooth canonical divisor.     
When we cannot avoid having a singular canonical divisor, then we are left with some singular curves.  I am interested in computing the Euler characteristic of their symmetric product.  I will post a question about this in a follow up enquire.

REPLY [8 votes]: Any smooth projective surface with nonempty $K_X$ is obtained by blowing up finitely many points on its unique minimal model. From the formula $K_X=f^*K_Y+E$ for the blowup, you see that the exceptional divisors of the blowup are always in the base locus of $|K_X|$. Thus, the problem is reduced to the minimal model.
Then you have to go through the classification of the minimal models of surfaces, that's been known for a hundred years now. (Using a book such as van de Ven "Complex surfaces", or Shafarevich et al, or Beauville...) 
For a minimal surface $Y$ Kodaira dimension 0 for example, $12K_Y=0$. So either $K_Y\ne 0$ and then $|K_X|=\emptyset$, or $K_Y=0$ and then any divisor in $|K_X|$ is $\sum a_i E_i$, where $E_i$ are the exceptional divisors of the blowups. 
For a minimal surface $Y$ of Kodaira dimension 2, the question is still somewhat tricky. If looking at higher multiples $|mK_Y|$ suffices, then by a well known theorem (Bombierri? certainly I. Reider gave a very nice proof), $|5K_Y|$ is free, so a general element is smooth (in characteristic 0). For $|K_Y|$ I don't think the answer is known but why not search mathscinet.
Finally, for Kodaira dimension 1, an elliptic surface $\pi:Y\to C$, there is a well-known Kodaira's formula for the canonical class $K_Y=\pi^*K_C + R$ with explicit rational coefficients in $R$. I'd play with that. Again, for higher multiples I think $|12K_Y|$ works. 
Of course, to your example of a hypersurface in $\mathbb P^3$ you can add the case of complete intersections, and other surfaces for which $K_X$ is either zero or $\pm K_X$ is very ample.<|endoftext|>
TITLE: What is the Euler characteristic of a Hilbert scheme of points of a singular algebraic curve?
QUESTION [17 upvotes]: Let $X$ be a smooth surface of genus $g$ and $S^nX$ its n-symmetrical product (that is, the quotient of $X \times ... \times X$ by the symmetric group $S_n$).  There is a well known, cool formula computing the Euler characteristic of all these n-symmetrical products:
$$\sum_{d \geq 0} \ \chi \left(X^{[d]} \right)q^d \ \ = \ \ (1-q)^{- \chi(X)}$$ 
It is known that $S^nX \cong X^{[n]}$, the Hilbert scheme of 0-subschemes of length n over $X$.  Hence, the previous formula also computes the Euler characteristic of these spaces. 
What about for singular surfaces?  More precisely, if $X$ is a singular complex algebraic curve, do you know how to compute the Euler characteristic of its n-symmetrical powers $S^nX$? More importantly: what is the Euler characteristic of $X^{[n]}$, the Hilbert scheme of 0-schemes of length n over $X$?
I guess it is too much to hope for a formula as neat as the one given for the smooth case.  Examples, formulas for a few cases or general behaviour (e.g. if for large n, $\chi\left(X^{[n]}\right) = 0)$ are all very welcome!

REPLY [15 votes]: For singular plane curves, there is a conjectural formula (due to Alexei Oblomkov and myself) in terms of the HOMFLY polynomial of the links of the singularities.  For curves whose singularities are torus knots, i.e. like x^a = y^b for a,b relatively prime, and for a few more singularities, the conjecture has been established.  See this preprint.
Edit: More recently I have given a different characterization of these numbers in terms of multiplicities of certain strata in the versal deformation of the singular curve.<|endoftext|>
TITLE: What are the advantages and disadvantages of the Moore method?
QUESTION [26 upvotes]: Describe your experiences with the Moore method. What are its advantages and disadvantages?

REPLY [3 votes]: I am a very latecomer to the question. Thus, I doubt my contribution will do any good. I try. The Moore Method is one of my favorite "teaching" methods. I have used a variant of it several times and in particular in two courses: Multivariable Calculus and Number Theory. For the former, I designed the course myself (for details see my paper "Moore and Less" : http://www.tandfonline.com/eprint/jGE3QNxcuGzUGj273smp/full), For the latter, I used "Number Theory Through Inquiry" (http://www.maa.org/ebooks/textbooks/NTI.html). Generally speaking, the advantages have been mentioned more or less in the previous answers/comments. Thus, I focus on potential disadvantages. 
Playfulness: The Moore Method strictly used is not that much playful. The point is you have a setting in which the materials have been prearranged. This forces you and your students to stay on a predesigned track, and as a result, hinders useful and playful jumps. Let me give an example.  Suppose you start with some examples of primitive Pythagorean triples. The most famous ones are (3, 4, 5) and (5, 12, 13).  Observation: the difference between two of the numbers is one.   Having characterized Pythagorean triples, it would be natural to move to Pell equation.  However, since the materials have been prearranged, you should continue with Lemmas ., Theorems  . , all directly related to Pythagorean triples. 
Naturalness:  Again, this is somehow related to the prearrangement.  As a designer, you feel that you should provide some backgrounds to help students to prove a certain theorem. What you provide is natural for you since you already know the proof of the theorem.  However, it is not so natural for most of the learners. Let me go with the previous example.  Moving towards the theorem characterizing Pythagorean triples you write (your students read): “It turns out that there is a method for generating infinitely many Pythagorean triples in an easy way. It comes from looking at some simple algebra from high school. Remember that … “.   It gives me a very bad feeling to behave with my students in this way, to say the least.  The problem arises even for the lecturer when others have designed the course.  An arrangement that is natural for somebody else is not necessarily natural for you. 
Forcefulness:   Your students are forced into forward thinking.  There is a theorem (again take the previous one as an example). You are somehow sure that your students need some help to prove it. Where do you provide that help?  As a Lemma before the theorem! 
Connectedness: This one is very strange and paradoxical. While you are connected with individuals and/or small group of students working together, you lose your connection with the class as a whole.  
There are some other points. I stop here since my answer is already too long. Moreover, I couldn’t find suitable words ended with “ness”  to describe  the other points ☺<|endoftext|>
TITLE: What is the prime spectrum of a Cauchy series ring?
QUESTION [14 upvotes]: Let $k$ be a field, and let $| \ |$ be a norm on $k$.  The norm induces a metric.  To construct the completion $\hat{k}$ as a normed field, the standard recipe is to take the quotient of the ring $\mathcal{C}(k)$ of all Cauchy sequences in $k$ -- viewed as a subring of $k^{\infty} = \prod_{i=1}^{\infty} k$ -- by the maximal ideal $\mathfrak{m}_0$ of all sequences converging to $0$.
This brings up the following [idle] question: what are the maximal ideals of $\mathcal{C}(k)$?  The prime ideals?
My vague recollection had been that $\mathfrak{m}_0$ was the unique maximal ideal of $\mathcal{C}(k)$, but this is evidently not the case: for every $n$, there is a maximal ideal $\mathfrak{m}_n$ consisting of sequences whose $n$th coordinate is $0$, the residue field being $k$ again.  It is easy to see though that $\mathfrak{m}_0$ is the unique maximal ideal containing the prime ideal $\mathfrak{c} = \bigoplus_{i=1}^{\infty} k$.  (Edit: $\mathfrak{c}$ is prime iff the norm is trivial.)
Now this reminds me of filters.  The prime ideals of the (zero-dimensional) ring $k^{\infty}$ correspond precisely to the ultrafilters on $\mathbb{Z}^+$.  The principal ultrafilter of all sets containing $n$ pulls back to the maximal ideal $\mathfrak{m}_n$.  Since every nonprincipal ultrafilter contains the Frechet filter of cofinite sets, it follows that it pulls back to $\mathfrak{m}_0$.  But is it true that every maximal ideal of $\mathcal{C}(k)$ is pulled back from a prime (= maximal) ideal of $k^{\infty}$?  If so, is this an instance of a general theorem? 

Addendum: 
Note that in the case that the norm is trivial -- so that the induced metric is the discrete metric -- a sequence converges iff it is eventually constant, so a sequence converges to $0$ iff it has only finitely many nonzero terms: $\mathfrak{c} = \mathfrak{m}_0$.  The converse also holds: for any nontrivial norm there exist nowhere zero sequences converging to $0$, e.g. $\{x^n\}$ for any $x \in k$ with $0 < |x| < 1$.

Once the original question is worked out, I am also curious about generalizations.  What is the analogue for the ring of minimal Cauchy filters in an arbitrary topological ring?

REPLY [5 votes]: In this answer I will treat the case in which $|\text{ }|$ is not discrete.
I first claim that $\mathfrak m_0$ is not the restriction of any proper ideal in
$k^{\infty}.$  Indeed, choose $x \in k$ such that $0 < |x| < 1$.  Then $(x^i)$
is an element of $\mathfrak m_0$ which is invertible in $k^{\infty}$ (with
inverse equal to $(x^{-i})$, and so $\mathfrak m_0$ generates the unit ideal
of $k^{\infty}$. 
This doesn't contradict anything; the maximal ideals of $k^{\infty}$ pull-back
to prime ideals in $\mathcal C(k)$ which are simply not maximal (as often happens
with maps of rings).
Furthermore, this pull-back is injective.
To see this, we first introduce some notation; namely, we
let $\mathfrak m\_{\mathcal U}$ denote the prime ideal of $k^{\infty}$ corresponding to
the non-principal ultra-filter ${\mathcal U}$,and recall that $\mathfrak m\_{\mathcal U}$ is defined as follows: an element $(x_i)$ lies in $\mathfrak m\_{\mathcal U}$ if and only
if $\{i \, | \, x_i = 0\}$ lies in in $\mathcal U$.
Now suppose that $\mathcal U_1$ and $\mathcal U_2$ are two distinct non-principal ultra-filters.  Let $A$ be a set lying in $\mathcal U_1$, but not in $\mathcal U_2$.
Then $A^c$, the complement of $A$, lies in $\mathcal U_2$.
Choose $x \in k$ such $0 < | x | < 1,$ and let $x_i = x^i$ if $i \in A$ and
$x_i = 0$ if $i \not\in A$.  Then $(x_i)$ is an element of $\mathcal C(k)$,
in fact of $\mathfrak m_0$, and it lies in $\mathfrak m\_{\mathcal U_2}$
but not in $\mathfrak m\_{\mathcal U_1}$.
Thus $\mathfrak m\_{\mathcal U_1}$ and $\mathfrak m\_{\mathcal U_2}$ have distinct pull-backs.  
So the map 
Spec $k^{\infty} \rightarrow $ Spec $\mathcal C(k)$
is injective
and dominant (since it comes from an injective map of rings), but is not surjective.
Choosing the valuation $|\text{ }|$ allows us to add to Spec $k^{\infty}$ (which is the
Stone-Cech compactification of $\mathbb Z\_+$) an extra point dominating all the
other points at infinity (i.e. all the non-principal ultrafilters), because the valuation now gives
us a definitive way to compute limits (provided we begin with a Cauchy sequence).<|endoftext|>
TITLE: A question about fibrations of simplicial sets and their fibers
QUESTION [5 upvotes]: I couldn't think of a title for this, but here we go:
Fix $p:S\rightarrow T$, a left fibration of simplicial sets, and an edge $f:\Delta^1 \rightarrow T$.  Let $t$ be the first vertex of $f$, and $t'$ be the second vertex.
We name the induced map, $q: S^{\Delta^1}\rightarrow S^{\{1 \}}\times_{T^{\{1 \}}} T^{\Delta^1}$.
Now let $X$ be the simplicial set of sections of the projection $S\times_T \Delta^1\rightarrow \Delta^1$, where the pullback is taken with respect to the map $p:S\rightarrow T$ and the fixed edge $f:\Delta^1 \rightarrow T$.
More notation: We'll denote by $S_{t'}$ the fiber $S \times_T \Delta^0$ where $\Delta^0 \rightarrow T$ is given by the inclusion of the vertex $t'$ and $S\rightarrow T$ is given again by p.  We give a fiber $q':X \rightarrow S_{t'}$ of $q$ over the edge $f$  (This is about where I stop understanding what's going on). 
What we'd like to show is: $q$ and $q'$ have the same fibers over points of $S^{\{1 \}}\times_{T^{\{1 \}}} T^{\Delta^1}$ where the second projection is the edge f.  Remember that exponentiation denotes the internal Hom.
The problem is this simplicial set of sections.  What are its maps out, and why do they naturally go to $S_{t'}$ and agree with q?  I feel like the key to this is understanding how the exponential is mapping into the pullback, but it's not really clear to me how that should work.
This fact is stated in HTT by Lurie in the proof of proposition 2.1.3.1, but I don't really see how it's obvious.  
A link to the relevant proof/page: http://books.google.com/books?id=CTe68E8wK4QC&lpg=PP1&ots=o8qYDiX4mt&dq=lurie%20higher%20topos%20theory&pg=PA67#v=onepage&q=&f=false
Update:
"Work" I've done thusfar:
$$\ \ \matrix{&S^{\Delta^1}_f &\to & S^{\Delta^1}&
\cr &\downarrow &Pb &\downarrow
\cr S_{t'}\cong &L_f & \to & L & \to & S^{\{1\}} &
\cr &\downarrow &Pb&\downarrow&Pb&\downarrow p
\cr &\Delta^{0} & \to & T^{\Delta^1} & \to & T^{\{1\}}
\cr &&f&&d_1}\ \ $$
Note that $d_1$ denotes the face map at the vertex 1.  Also, $L:= S^{\{1 \}}\times_{T^{\{1 \}}} T^{\Delta^1}$, and 
The point here is that it should be "morally" the same to give a pullback of $\Delta^1\to T \leftarrow S$ as giving a pullback $\Delta^0\to T^{\Delta^1} \leftarrow S^{\Delta^1}$ with respect to the edge f.  So we'd like to show that $X$, the simplicial set of sections of the projection noted before is somehow isomorphic to $S^{\Delta^1}_f$, since this shows that $q$ and $q'$ agree where they're needed to.   So what I'm struggling with at this point is showing this last idea.

REPLY [5 votes]: I've got to agree with you; it's not the best-written proof in HTT. Let me go through it glacially slowly (for my own sake!) to see if I can write something that will help clarify the role of $X$.
Let me write $Map(U,V)$ instead of $V^U$. I find it easier to parse on the internet.
First, what is our $X$? It's the simplicial set of sections of $S\times_T\Delta^1\to\Delta^1$, that is, it is the fiber product
$Map(\Delta^1,S\times_T\Delta^1)\times_{Map(\Delta^1,\Delta^1)}\Delta^0$, where $\Delta^0$ maps in by inculsion of $id$. So (since $Map(\Delta^1,-)$ is a right adjoint) we get:
$$X=Map(\Delta^1,S\times_T\Delta^1)\times_{Map(\Delta^1,\Delta^1)}\Delta^0=Map(\Delta^1,S)\times_{Map(\Delta^1,T)}Map(\Delta^1,\Delta^1)\times_{Map(\Delta^1,\Delta^1)}\Delta^0$$
$$=Map(\Delta^1,S)\times_{Map(\Delta^1,T)}\Delta^0$$
where $\Delta^0$ is mapping in by inclusion $f$.
Now we've got our map
$$q:Map(\Delta^1,S)\to Map(\{1\},S)\times_{Map(\{1\},T)}Map(\Delta^1,T),$$
whose fibers over $f$ we seek. That is, we just include
$$S_{t'}=Map(\{1\},S)\times_{Map(\{1\},T)}\Delta^0\to Map(\{1\},S)\times_{Map(\{1\},T)}Map(\Delta^1,T)$$
(which is itself the pullback of $f:\Delta^0\to Map(\Delta^1,T)$ along the projection, of course) and we pull back to get $q'$.
So the result is a pullback of a pullback. (If I knew how, I'd draw the two pullback squares here.) The composite pullback is the pullback of $f:\Delta^0\to Map(\Delta^1,T)$ along $Map(\Delta^1,S)\to Map(\Delta^1,T)$. But this is what we called $X$. So our result is a map $q':X\to S_{t'}$, and its fiber over any vertex of $S_{t'}$ must coincide with the fiber over the corresponding vertex of $Map(\{1\},S)\times_{Map(\{1\},T)}Map(\Delta^1,T)$, since that will be a pullback of a pullback as well.
Edit (Harry): I typed up the final version of the diagram. If the letters aren't explained, you can deduce what they are just by either looking at Clark's argument or just tracing the pullbacks.  Every square is a pullback, so everything is very easy to deal with.
$$\ \ \matrix{
X&\cong&S^{\Delta^1}_f &\to &Y^{\Delta^1}&\to& S^{\Delta^1}&
\cr &\searrow&\downarrow &Pb &\downarrow&Pb&\downarrow
\cr L_f&\cong &S_{t'} & \to &L'&\to& L & \to & S^{\{1\}} &
\cr &&\downarrow &Pb&\downarrow&Pb&\downarrow&Pb&\downarrow p
\cr &&\Delta^{0} & \to &(\Delta^1)^{\Delta^1} &\to& T^{\Delta^1} & \to & T^{\{1\}}
\cr &&&id&&(f)^{\Delta^1}&&d_1}\ \ $$<|endoftext|>
TITLE: Gamma function versions of combinatorial identites?
QUESTION [6 upvotes]: We can extend the binomial coefficient $\binom{n}{k}$ to $\mathbb{R}$ or $\mathbb{C}$ by defining $\binom{x}{y}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}$. Do any the standard binomial coefficient identities have generalizations to this setting? Just as two simple examples, we have
$\sum_{k=0}^n \binom{n}{k} = 2^n$ and $\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$
What are $\int_0^x \binom{x}{y} dy$ and $\int_0^x \binom{x}{y}^2 dy$, and are the answers analogous to the discrete case? Is there any combinatorial significance we can give to these integrals? Has this already been tried?

REPLY [6 votes]: Chapter 5.5 of Concrete Mathematics discusses generalizing binomial coefficient identities to the Gamma function. It doesn't discuss the two integrals you mention, though.
Doing a bit of thinking on my own, if $n$ is a positive integer then
$$\int_{z=0}^n \binom{n}{z} dz = \int_{z=0}^n \frac{n! dz}{\Gamma(1+z) \Gamma(n+1-z)}$$
$$\int_{z=0}^{n} \frac{n! dz}{(n-z)(n-1-z) \cdots (1-z) \Gamma(1-z) \Gamma(1+z)}.$$
We have $\Gamma(1+z) \Gamma(1-z) = \pi z/\sin (\pi z)$, if I haven't made any dumb errors, so this is
$$\int_{0}^n \frac{ n! \sin (\pi z) \ dz}{\pi z (n-z)(n-1-z) \cdots (1-z)}.$$
I suspect this integrand does not have an elementary anti-derivative, because it reminds me of $\int \sin t \ dt/t$. But there might be some special trick which would let you compute the integral between these specific bounds.<|endoftext|>
TITLE: Can an infinite sequence of integers generate integer-area triangles?
QUESTION [15 upvotes]: (asked by Shanzhen Gao, shanzhengao at yahoo.com, on the Q&A board at JMM)
Does there exist an infinite, monotonically increasing sequence of integers $\{ a_n \}_{n \geq 0}$ such that for any $n$, the three integers $(a_n, a_{n+1}, a_{n+2})$ are the side lengths of a plane triangle with integer area?

REPLY [4 votes]: I'll throw out a dumb idea: can anyone find a rational point on
$$y^2 = - (x^2-x+1)(x^2+x+1)(x^2-x-1)(x^2+x-1)?$$
UPDATE: The above formula used to have a sign error, which I have just fixed, and Bjorn's reponse was to the version with the sign error. Thanks to Kevin Buzzard for pointing this out to me.
Because, if so, $a_n=x^n$ gives triangles with rational area. Of course, this still wouldn't give an integer solution, but it would rule out a number of easy arguments against one existing.
I did a brute force search of values of $x$ with numerator and denominator under 5000 and didn't find any, but I don't think that is large enough to even count as evidence against one existing.<|endoftext|>
TITLE: Experimental mathematics leading to major advances
QUESTION [149 upvotes]: I would like to ask about examples where experimentation by computers has led to major mathematical advances.
A new look
Now as the question is five years old and there are certainly more examples of mathematical advances via computer experimentation of various kinds, I propose to consider contributing new answers to the question.
Motivation
I am aware about a few such cases and I think it will be useful to gather such examples together. I am partially motivated by the recent polymath5 which, at this stage, have become an interesting experimental mathematics project. So I am especially interested in examples of successful "mathematical data mining"; and cases which are close in spirit to the experimental nature of polymath5. My experience is that it can be, at times, very difficult to draw useful insights from computer data.
Summary of answers according to categories
(Added Oct. 12, 2015)
To make the question a useful resource (and to allow additional answers), here is a quick summery of the answers according to categories. (Links are to the answers, and occasionally to an external link from the answer itself.)
1) Mathematical conjectures or large body of work arrived at by examining experimental data - Classic
The Prime Number Theorem;  Birch and Swinnerton-Dyer conjectures; Shimura-Taniyama-Weil conjecture; Zagier's conjectures on polylogarithms; Mandelbrot set; Gosper Glider Gun (answer), Lorenz attractor;  Chebyshev's bias (answer) ; the Riemann hypothesis; the discovery of the Feigenbaum constant; (related) Feigenbaum-Coullet-Tresser universality and Milnor's Hairiness conjecture; Solving numerically the so-called Fermi--Pasta--Ulam chain and then of its continuous limit, the Korteweg--de Vries equation
2) Mathematical conjectures or  large body of work arrived at by examining experimental data - Current
"Maeda conjecture"; the work of Candès and Tao on compressed sensing; Certain Hankel determinants; Weari-Phelan structure; the connection of multiple zeta values to renormalized Feynman integrals; Thistlethwaite's discovery of links with trivial Jones polynomial; The Monstrous Moonshine; McKay's account on experimentation leading to mysterious "numerology" regarding the monster.  (link to the answer); Haiman conjectures on the quotient ring by diagonal invariants
3) Computer-assisted proofs of mathematical theorems
Kepler's conjecture ; a new way to tile the plane with a pentagon: advances regarding bounded gaps between primes following Zhang's proof; Cartwright and Steger's work on fake projective planes; the Seifert-Weber dodecahedral space is not Haken; the four color theorem, the proof of the nonexistence of a projective plane of order 10; Knuth's work on a class of projective planes; The search for Mersenne primes; Rich Schwartz's work; The computations done by the 'Atlas of Lie groups' software of Adams, Vogan, du Cloux and many other; Cohn-Kumar proof for the densest lattice pacing in 24-dim; Kelvin's conjecture; (NEW) $R(5,5) \le 48$ and $R(4,5)=25$;
4) Computer programs that interactively or automatically lead to  mathematical conjectures.
Graffiti
5) Various computer programs which allow proving automatically theorems or generating automatically proofs in a specialized field.
Wilf-Zeilberger formalism and software; FLAGTOOLS
6) Computer programs (both general purpose and special purpose) for verification of mathematical proofs.
The verification of a proof of Kepler's conjecture.
7) Large databases and other tools
Sloane's  online encyclopedia for integers sequences; the inverse symbolic calculator.
8) Resources:
Journal of experimental mathematics; Herb Wilf's  article: Mathematics, an experimental science in the Princeton Companion to Mathematics, genetic programming applications a fairly comprehensive website experimentalmath.info
; discovery and experimentation in number theory; Doron Zeilberger's classes called "experimental mathematics":math.rutgers.edu/~zeilberg/teaching.html;
V.I. Arnol'd's two books on the subject of experimental mathematics in Russian, Experimental mathematics, Fazis, Moscow, 2005, and Experimental observation of mathematical facts, MCCME, Moscow, 2006
Answers with general look on experimental mathematics: 
Computer experiments allow new avenues for productive strengthening of a problem (A category of experimental mathematics).

Bounty:
There were many excellent answers so let's give the bounty to Gauss...
Related question: Where have you used computer programming in your career as an (applied/pure) mathematician?, What could be some potentially useful mathematical databases? Results that are easy to prove with a computer, but hard to prove by hand ; What advantage humans have over computers in mathematics? Conceptual insights and inspirations from experimental and computational mathematics

REPLY [4 votes]: The Robbins conjecture from 1933, asked whether two sets of Boolean algebras were equivalent. The solution has been found only in 1996 by William McCune, using the automated theorem prover EQP.<|endoftext|>
TITLE: Is there triangulated category version of Barr-Beck's theorem?
QUESTION [11 upvotes]: It is well known that Beck's theorem for Comonad is equivalent to Grothendieck flat descent theory on scheme. 
There are several version of derived noncommutative geometry. I wonder whether someone developed the triangulated version of Beck's theorem. And What does it mean,if exists?

REPLY [9 votes]: There is a Beck's theorem for Karoubian triangulated categories, proposed by Konstevich and Rosenberg in July 2004, which is proved by using the Verdier's abelianization functor and graded monads; see page 36 of A. Rosenberg, Topics in noncommutative algebraic geometry, homological algebra and K-theory, preprint MPIM Bonn 2008-57, pdf. 
In fact they got simulatenously (July 15, 2004) both versions: A-infinity and triangulated. The reference to the triangulated is above: while for A-infinity there is no write up, but Kontsevich gave a talk (I think Nov 2004, van den Bergh birthday conference) where he formulated and used the result; one of nice applications was to glue certain ordinary commutative schemes to get certain formal schemes. I remember very well the weeks preceding the result when we discussed possible shape of the result seeked at IHES. Later at the conference in Split, Kontsevich gave a talk where he gave some usages in noncommutative algebraic geometry. 
I disagree with the statement: "Beck's theorem for comonad is equivalent to Grothendieck flat descent theory on scheme". Namely Grothendieck gave both the flat descent theory for quasicoherent sheaves (SGA I.8.1) which is a special case of Beck's theorem (though it has some symmetries which general noncommutative case does not have), but also the (stronger) flat descent theorem for affine schemes (SGA I.8.2) and for morphisms (cf. SGA I.8.5), which unlike the descent for quasicoherent sheaves, does not generalize to the noncommutative algebras and consequently to categorical setup either.<|endoftext|>
TITLE: Cobordisms of bundles?
QUESTION [8 upvotes]: Is there a notion of a cobordism which is compatible with bundle structure?
That is, if I have bundles $E$ and $F$, is it the case that the manifold $W$ with $E$ and $F$ as boundary components, can be made into a bundle whose bundle structure, when restricted to $E$ or $F$, is the bundle structure of $E$ or $F$.
And, particularly, when can I connect $E$ and $F$ this way (not just when they're cobordant, but when this cobordism is compatable with this structure)?  And what can I say about the bundle structure of $W$, knowing what $E$ and $F$ look like?  (e.g., if $E$ and $F$ are G-bundles what can I say about the group action on $W$?)
Also, can anyone point me to any particular references which discuss this?  I spent a few hours in our (fairly small) math library looking for something like this, but haven't been able to find anything that seems to discuss this.  But I may just not know the right catch phrases to search for!

REPLY [5 votes]: I'll assume you're talking about principal G-bundles. These are classified by maps into $BG$, the base of the universal $G$-bundle, so if we have bundles classified by $f:E \to BG$ and $g:F \to BG$, you are looking for a bordism between $f$ and $g$ - whether there exists a $h : W \to BG$ connecting these classifying maps. So there is a bundle cobordism between the two bundles iff the bordism classes of $f$ and $g$ in  $\mathfrak{N}n(BG)$ coincide, and if they do coincide, then the choice of $W$ is parametrized by the bordism group $\mathfrak{N}_{n+1}(BG)$. I don't know an algorithmic way to obtain the class $[f]$ from $E$, but there is a splitting $\mathfrak{N}_n(BG) = \oplus H_j(BG) \otimes \mathfrak{N}_{n-j}$ which can help identify some bundles' classes.<|endoftext|>
TITLE: Why Weil group and not Absolute Galois group?
QUESTION [29 upvotes]: In many formulation of Class Field theory, the Weil group is favored as compared to the Absolute Galois group. May I asked why it is so? I know that Weil group can be generalized better to Langlands program but is there a more natural answer?
Also we know that the abelian Weil Group is the isomorphic image of the reciprocity map of the multiplicative group (in the local case) and of the idele-class group (in the global case). Is there any sense in which the "right" direction of the arrow is the inverse of the reciprocity map?
Please feel free to edit the question into a form that you think might be better.

REPLY [14 votes]: Perhaps something worth pointing out, relating to how the Weil group appears naturally: it arises from a compatible system of group extensions at finite levels. Indeed, one of the "axioms" of class field theory, is the existence of a "fundamental class" uL/*K* in $H^2(\operatorname{Gal}(L/K),C_L)$ for each finite Galois extension $L/K$ (where $C_L$ is the class module). Each of these gives a group extension
$$ 1\rightarrow C_L\rightarrow W_{L/K}\rightarrow \operatorname{Gal}(L/K)\rightarrow 1.$$
The projective limit of these gives the absolute Weil group fitting in
$$ 1\rightarrow C\rightarrow W_K\rightarrow G_K$$
with the rightmost map having dense image (and $C$ is the formation module of the class formation). Thus, you can think of the Weil group as arising canonically out of the results of class field theory, thus making it a natural replacement of $G_K$ in questions related to CFT. I like section 1 of chapter III of Neukirch–Schmidt–Wingberg's Cohomology of number fields and the last chapter of Artin–Tate for this material.<|endoftext|>
TITLE: How natural is the reciprocity map?
QUESTION [8 upvotes]: For local field, the reciprocity map establishes almost an isomorphism from the multiplicative group to the Abelian Absolute Galois group. (In global case the relationship is almost as nice). It is tempted to think that there can be no such nice accident. 
Do we know any explanation which suggest that there "should be" a relationship between the multiplicative group and the Galois group?
Actually, my current belief is that the reciprocity map is half accidental. I think that there is a natural extension where we can define a natural action of the multiplicative group. In the local case this is the Lubin-Tate extension (a generalization of cyclotomic extension). The fact that this Lubin-Tate extension is the Abelian Absolute Galois group is an accident. 
Do we know something that might support/ reject this view?
Please feel free to edit the question into a form that you think might be better.

REPLY [11 votes]: The reciprocity map is completely natural (in the technical sense of category theory).  For example, if $K$ and $L$ are two local fields,
and $\sigma:K \rightarrow L$ is an isomorphism, then $\sigma$ induces an isomorphism of 
multiplicative groups $K^{\times} \rightarrow L^{\times}$ and also of abelian absolute Galois groups $G\_K^{ab} \rightarrow G\_L^{ab}$.  The reciprocity laws for $K$ and $L$ are then compatible with these two isomorphisms induced by $\sigma$.  
On the other hand the factorization $K^{\times} = {\mathbb Z} \times {\mathcal O}\_K^{\times}$ is not canonical (it depends on a choice of uniformizer), and the identification of
${\mathcal O}\_K^{\times}$ with the Galois group of a maximal totally ramified abelian extension of $K$ also depends on a choice of uniformizer (which goes into the construction of the Lubin--Tate formal group, and hence into the construction of the totally ramified extension; different choices of uniformizer will give different formal groups, and different
extensions).   
As others pointed out, the local reciprocity map is also a logical consequence of the global Artin map and global Artin reciprocity law (which makes no reference to local multiplicative groups, but simply to the association $\mathfrak p \mapsto Frob\_{\mathfrak p}$ of Frobenius elements to unramified prime ideals; see the beginning of Tate's article in Cassels--Frohlich for a nice explanation of this).  Thus it is natural in a more colloquial sense of the word as well.
Indeed, the idelic formulation of the glocal reciprocity map and the formulation of the local reciprocity map in terms of multiplicative groups are not accidental or ad hoc inventions; they were forced on number theorists as a result of making deep investigations into the nature of global class field theory.<|endoftext|>
TITLE: Reduced scheme and closed points
QUESTION [24 upvotes]: In The Geometry of Schemes by Eisenbud and Harris, Exercise I-32 asks one to show that a scheme $X$ is reduced if and only if every local ring $\mathcal{O}_{X,p}$ is reduced for closed points $p \in X$.  However, this does not seem to work in general, since $X$ may not have enough closed points.  What additional hypotheses on $X$ do I need for such an assertion to hold?

REPLY [4 votes]: Brunoh: 
1) If $X$ is a quasi-compact scheme such that $\mathscr O_{X,x}$ is reduced for every closed point $x$, then $X$ is reduced. Indeed, let $y\in X$. The scheme $\overline{\{y\}}$ is a closed subscheme of $X$, hence is quasi-compact, and non-empty because it contains $y$. It thus has a closed point $x$, which is closed in $X$ as well. Now $\mathscr O_{X,y}$ is a localization of $\mathscr O_{X,x}$, hence is reduced because so is $\mathscr O_{X,x}$ by assumption. 
2) Let $k$ be a field and let $v$ be the valuation on $k(X_i)_{i\in \mathbb Z_{> 0}}$ defined by the composition of the successive discrete valuations provided by the $X_i$'s. Let $X$ be the spectrum of the corresponding valuation ring. Then topologically, $X=\{x_0,\ldots, x_n,\ldots\}\bigcup \{x_\infty\}$ where every $x_i$ specializes to $x_{i+1}$, and where $x_\infty$ is the unique closed point (the point $x_0$ is the generic one, and $x_i$ corresponds to the prime ideal generated by $X_i$). Now if you remove $x_\infty$ you get an open subscheme $U$ of $X$, without any closed point. 
Of course, $U$ is reduced, but $U\times_k \mathrm{Spec}\; k[\epsilon]$ (with $\epsilon\neq 0$ and $\epsilon^2=0$) is not reduced, and homeomorphic to $U$.<|endoftext|>
TITLE: Archaeogenetics
QUESTION [13 upvotes]: This question is meant to be applied to recover historic information from genetic data. 
The following model, is (probably) the simplest possible which takes recombinations into account.
First, let us introduce some terms:

human is a finite set of numbers in (0,1); further these numbers will be called "scars".
population is a finite set of humans

One can perform one operation with a population:

recombination; i.e. take two humans $A$ and $B$ from the population, choose randomly a number $x\in(0,1)$ and produce a new human $C$ which has scar $x$, all scars in $A$ which are $ < x$ and all scars in $B$  which are $ > x$.

Assume someone starts with a population formed by two empty sets  and does these operations for a while then stop.
Assume you have all information about a portion of the population, BUT you do not know how this portion had been chousen.
What one can say (even in which terms) about the history of population.
Comments:

It is not exactly mathematical problem, an answer might be something like "colored graph".
Clearly one can not say everything; yet there are many questions which can be answered. For example, assume you want to know if it is likely that at some moment your population was divided in two and there were no interbreeding between these groups for a while. You can even estimate "time" when they split. BUT I'm interested in a more abstract way to describe the history --- I want to say something without any assumptions.
Something is possible: First note that with probability 1, there is one-to-one correspondence between humans and scars. Given scar $x$, let us denote by $H_x$ the corresponding human. Assume in the portion of population you have two humans: human $A$ who has scars $x < y$ and nothing in between $x$ and $y$ and human $B$ who has scars  $x < z$ and nothing in between. Assume $z < y$ then $H_z$ is a descendant of $H_x$. That gives a partial order on all such humans. Is it all one can do?

REPLY [3 votes]: This is not an answer, just an elaboration on the comment about associating a simplicial complex to the pattern of recombinations.
Recombination means if $A$ and $B$ recombine, the possibilities might be parametrized by $A \times B \times [0,1]$, although if the parameter is close to 0 or close to 1, almost no information is kept from one parent.
What geometric operation makes a child of parents $A$ and $B$ the structure $A \times B \times [0,1]$ with $A\times B \times \{0\}$ identified with $A$ and $A\times B \times \{1\}$ identified with $B$? The convex hull does that when $A$ and $B$ are in skew affine spaces. If $A$ and $B$ are simplices of dimensions $\alpha$ and $\beta$, that makes the child a simplex of dimension $\alpha+\beta+1$. If the points of $A$ are given barycentric coordinates, $(a_0,...,a_\alpha)$ with $\sum a_i = 1$ and similarly for $B$, then we can coordinatize the child as $((1-x)a_0,...,(1-x)a_\alpha,xb_0,...,xb_\beta)$. Inductively, we can specify a simplicial complex so that each person as a simplex, the initial population consists of points, and the recombination of two people is a person whose simplex is attached to the parent simplices at opposite facets.
The (point $p$ $\in$ simplex of person $P$) pairs correspond to choices of recombination points for $P$ and each ancestor of $P$. These choices may be inconsistent if there are  multiple paths of descent from the same ancestor. Also, this doesn't fully take into account the way that recombination erases some of the information from each parent.<|endoftext|>
TITLE: An arithmetic highest weight theory?
QUESTION [11 upvotes]: I apologize if these questions seem naive or loaded.
Is there an analogous theory of highest weights for irreducible finite-dimensional representations of Lie algebras of algebraic group (or perhaps group schemes) over a non-algebraically closed field (resp. a "nice" ring, say a Dedekind domain). 
Are there analogous results to Lie's theorems in the case of algebraic groups (perhaps even arbitrary group schemes)? I am aware of Jantzen's book on representations of algebraic groups, but if I remember correctly, he does everything over an algebraically closed base field. I have not studied the book in detail to convince myself that the arguments there will carry over to the non-algebraically closed case.
I suppose the Borel-Bott-Weil-(Schmidt) construction of highest weights using sections of cohomology groups of line bundles may be generalized to a more arithmetic setting (as Jantzen has done in his book). Is there any progress in this direction beyond algebraic groups, say to include a "nice" class of group schemes? I am more curious of the case of classical groups.
Concerning more general group schemes, I have looked up parts of SGA3, but I did not find any clearly stated results connecting the Lie algebra of a group scheme (as defined there using universal properties) to the underlying group scheme.
A more general and more loaded question: to what extent is a smooth scheme determined by its tangent space at a distinguished point. I am aware of the notion of jet schemes, are there some important or at least neat results in this area anyone would like to share?
Thanks in advance.

REPLY [6 votes]: Johnson, you have one of the foremost experts in the world on such matters (over general fields) just upstairs from your office.  Make use of that.  
Many of the basic constructions work for split groups over fields, but proving good properties (such as irreducibility and classification results) requires being over a field of characteristic 0.  (Once constructions are made, to prove things one can extend scalars to an algebraic closure, or even reduce to the familiar case over $\mathbf{C}$ if so inclined, by the "Lefschetz Principle".) The Lie algebra is a good invariant (e.g., faithful!) over fields of characteristic 0, but even then it only retains at best information about groups up to isogeny.  Another case where it gives a useful invariant is over $\mathbf{Z}/p\mathbf{Z}$-algebras where, together with the $p$-Lie algebra structure, it gives an equivalence with the category of finite locally free groups $G$ with vanishing relative Frobenius morphism such that the sheaf of invariant 1-forms is locally free over the base (loosely speaking because such vanishing allows one to get by with truncated exponential in degrees $< p$); this is explained in SGA3, VII$_{\rm{A}}$, 7.2, 7.4
More generally one cannot expect to get mileage out of the Lie algebra alone (but it is still perfectly useful via its role in classification by means of root systems, among other things).<|endoftext|>
TITLE: What is an algebraic group over a noncommutative ring?
QUESTION [15 upvotes]: Let $R$ be a (noncommutative) ring.  (For me, the words "ring" and "algebra" are isomorphic, and all rings are associative with unit, and usually noncommutative.)  Then I think I know what "linear algebra in characteristic $R$" should be: it should be the study of the category $R\text{-bimod}$ of $(R,R)$-bimodules.  For example, an $R$-algebra on the one hand is a ring $S$ with a ring map $R \to S$.  But this is the same as a ring object in the $R\text{-bimod}$.  When $R$ is a field, we recover the usual linear algebra over $R$; in particular, when $R = \mathbb Z/p$, we recover linear algebra in characteristic $p$.
Suppose that $G$ is an algebraic group (or perhaps I mean "group scheme", and maybe I should say "over $\mathbb Z$"); then my understanding is that for any commutative ring $R$ we have a notion of $G(R)$, which is the group $G$ with coefficients in $R$.  (Probably there are some subtleties and modifications to what I just said.)

My question: What is the right notion of an algebraic group "in characteristic $R$"?

It's certainly a bit funny.  For example, it's reasonable to want $GL(1,R)$ to consist of all invertible elements in $R$.  On the other hand, in $R\text{-bimod}$, the group $\text{Aut}(R,R)$ consists of invertible elements in the center $Z(R)$.
Incidentally, I'm much more interested in how the definitions must be modified to accommodate noncommutativity than in how they must be modified to accommodate non-invertibility.  So I'm happy to set $R = \mathbb H$, the skew field of quaternions.  Or $R = \mathbb K[[x,y]]$, where $\mathbb K$ is a field and $x,y$ are noncommuting formal variables.

REPLY [7 votes]: Sorry for arriving here so late... I hope somebody will still notice my answer:
A possible definition would be to take a co-group in an appropriate category of noncommutative rings (or algebras), i.e. an object in such a category of noncommutative rings or algebras that represents a functor from this category into a category of groups (see also S. Carnahan's answer).
I know of several papers that follow this approach and study such co-groups, here is what I have been able to find:

Israel Berstein, On co-groups in the category of graded algebras,  Trans. Amer. Math. Soc. 115 (1965), 257–269.
Dan Voiculescu, Dual algebraic structures on operator algebras related to free products. J. Operator Theory 17 (1987), no. 1, 85–98. 
James J. Zhang, H-algebras. Adv. Math. 89 (1991), no. 2, 144–191. 
George Bergman, Adam Hausknecht,
Co-groups and co-rings in categories of associative rings.
Mathematical Surveys and Monographs, 45. American Mathematical Society, Providence, RI, 1996.
Benoit Fresse, Cogroups in algebras over an operad are free algebras, Comment. Math. Helv. 73 (1998) 637-676
Hiroshi Kihara, Cogroups in the category of connected graded algebras whose inverse and antipode coincide, arXiv:1303.7350
Loïc Foissy, Claudia Malvenuto, and Frederic Patras, B-infinity algebras, their enveloping algebras, and finite spaces. arXiv:1403.7488

Six papers and one book over a span of almost fifty years, citing each other rather sparsely (my list is certainly not complete, though). Compared to the vast literature on algebraic groups and group schemes, this does not seem to be a lot.
Question: Is there an explanation why noncommutative algebraic group theory (in the sense of the OP, maybe one should say noncocommutative?) is getting so little attention? E.g., lack of applications, technical difficulties, lack of interesting results?<|endoftext|>
TITLE: Reverse mathematics of (co)homology?
QUESTION [25 upvotes]: Background
Exercise 2.1.16b in Hartshorne (homework!) asks you to prove that if $0 \rightarrow F \rightarrow G \rightarrow H \rightarrow 0$ is an exact sequence of sheaves, and F is flasque, then $0 \rightarrow F(U) \rightarrow G(U) \rightarrow H(U) \rightarrow 0$ is exact for any open set $U$.  My solution to this involved the axiom of choice in (what seems to be) an essential way.  
Essentially, you are asking to $G(U) \rightarrow H(U)$ to be surjective when you only know that $G \rightarrow H$ is locally surjective.  Ordinarily, you might not be able to glue the local preimages of sections in $H(U)$ together into a section of $G(U)$,  but since $F$ is flasque, you can extend the difference on overlaps to a global section. This observation deals with gluing finitely many local preimages together.  Zorn's lemma enters in to show that you can actually glue things together even if the open cover of $U$ is infinite.
Now, I have not really studied sheaf cohomology, but the idea I have is that it detects the failure of the global sections functor to be right exact.  So if you can't even show sheaf cohomology vanishes for flasque sheaves without the axiom of choice, it seems like a lot of the machinery of cohomology would go out the window.
Now, just on the set theoretic level, it seems like there is something interesting going on here.  Essentially the axiom of choice is a local-global statement (although I had never thought of it this way before this problem), namely that if $f:X \rightarrow Y$ is a surjection you can find a way to glue the preimages $f^{-1}(\{y\})$ of a surjection together to form a section of the map $f$.
This brings me to my 
Questions
Can the above mentioned exercise in Hartshorne be proven without the axiom of choice?
How much homological machinery depends on choice?
Have any reverse mathematicians taken a look at sheaf cohomology as a subject to be "deconstructed"?
Have any constructive set theorists thought about using cohomological technology to talk about the extent to which choice fails in their brand of intuitionistic set theory? (it seems like topos models of such set theories might make the connection to sheaves and their cohomology very strong!)
My google-fu is quite weak, but searches for "reverse mathematics cohomology" didn't seem to bring anything up.

REPLY [10 votes]: Have any reverse mathematicians taken a look at sheaf cohomology as a subject to be "deconstructed"?

Colin McLarty has made a study of what it takes to define derived functor cohomology (with sheaf cohomology as a special case given a topos of sheaves). He finds that finite-order arithmetic (the union of $Z_n$ for $n=1,2,\ldots$) suffices

The large structures of Grothendieck founded on finite order arithmetic 
  Colin McLarty
(Submitted on 9 Feb 2011 (v1), last revised 30 Apr 2014 (this version, v4))
Abstract: Such large-structure tools of cohomology as toposes and derived categories stay close to arithmetic in practice, yet existing foundations for them go beyond the strong set theory ZFC. We formalize the practical insight by founding the theorems of EGA and SGA, plus derived categories, at the level of finite order arithmetic. This is the weakest possible foundation for these tools since one elementary topos of sets with infinity is already this strong. 
http://arxiv.org/abs/1102.1773

For Zariski cohomology of Noetherian schemes, one can use second order arithmetic

Zariski cohomology in second order arithmetic
  Colin McLarty
(Submitted on 2 Jul 2012 (v1), last revised 25 Jul 2012 (this version, v2))
Abstract: The cohomology of coherent sheaves and sheaves of Abelian groups on Noetherian schemes are interpreted in second order arithmetic by means of a finiteness theorem. This finiteness theorem provably fails for the etale topology even on Noetherian schemes. 
http://arxiv.org/abs/1207.0276<|endoftext|>
TITLE: Enumeration of graphs arising in invariant theory
QUESTION [8 upvotes]: I've been working on a talk based on some stuff in Olver's "Classical Invariant Theory" book and have been wondering about a related graph enumeration problem.
Start with a triple $(n,v,e)$ of natural numbers.  Take all $\mathbb{Q}$-linear combinations of directed graphs (allowing multiple edges, but no loops) with $v$ vertices, $e$ edges, and each vertex has at most $n$ edges going to it or coming from it.  Now, take three relations (images scanned from Olver)
Rule 1: 
Rule 2: 
Rule 3: 
Where the function $v$ with a vertex as subscript next to a graph means that graph multiplied by $n$ minus the number of edges attached to that vertex. (So, for instance, an isolated vertex gets multiplied by $n$)
Denote the space after quotienting by these relations by $V_{n,v,e}$.  And so, in final form, my question:

What is $\dim V_{n,v,e}$? Or at least, can we find relatively effective upper bounds?

EDIT: Some clarifications.  The colorings on the vertices are just to mark them in the pictures to keep track of where everything goes, the graphs are not marked themselves.  Additionally, as Rule 2 is slightly unclear from the scan, the $v$ function is always the vertex not attached to the arrow in the configuration.

REPLY [2 votes]: The number ${\rm dim}\ V_{n,v,e}$ is the number of linearly independent covariants of degree $v$ and weight $e$ of a binary form of degree $n$. This is the multiplicity of the irreducible module $Sym^k(\mathbb{C}^2)$ in the plethysm $Sym^v(Sym^n(\mathbb{C}^2))$, where
$k=nv-2e$. There is formula for that involving counting integer partitions (Cayley-Sylvester formula). The easiest way to derive it that I know of is by computing the
character which is a Gaussian polynomial. I think the book by Mukai does that. You can also
look up http://arxiv.org/abs/math/0110224<|endoftext|>
TITLE: Category of graphs.
QUESTION [5 upvotes]: Hello, I'm writting something about Malcev categories and monadicity. The fact is that I need to know if Graph is or not complete (have all finite limits). It seems easy but I would like a real answer (not my feelings saying that it is) and I don't find that information anywhere.
Thank you for your answers.

REPLY [3 votes]: Consider the category G that has exactly two distinct objects (call them a and b) and two distinct arrows a->b. A graph is precisely a functor G -> Set. This means that the category of graphs is a presheaf category. The rest follows.<|endoftext|>
TITLE: On Category O in positive characteristic
QUESTION [15 upvotes]: Let $G$ be a semisimple algebraic group over an algebraically closed field $k$. In the case that $k$ has characteristic 0, there has been intensive study of the BGG category O of representations of the enveloping algebra of the Lie algebra of $G$ (which is also, in this case, the hyperalgebra of $G$). When $k$ is a field of positive characteristic, though, the enveloping algebra and the hyperalgebra are different, and there has been a lot of study of representations of the enveloping algebra. On the other hand, it seems that there has been much less study of representations of the hyperalgebra, perhaps because it is more complicated (for example, it is not finitely generated).
So let's now assume that the characteristic of $k$ is positive. A seminal paper of Haboush, "Central Differential Operators of Split Semisimple Groups Over Fields of Positive Characteristic," set up the foundations of category O for hyperalgebras in this setting. More precisely, let $U$ denote the hyperalgebra of $G$. In that paper Haboush defined Verma modules for $U$ (which are defined analogously to the characteristic 0 case) and proved that many of the properties one would expect, including relationships with simple modules and certain "integrality" properties, hold. However, this is just the beginning; there are many questions one could ask about whether or not category O in positive characteristic behaves like category O in characteristic 0. E.g, what is the structure of the blocks in this category? What about the structure of projective generators? Are there translation functors? If so, how do they behave? I.e., how much of the (huge) characteristic 0 story carries over to positive characteristic? Haboush's paper is from 1980, and I haven't been able to find any papers that carry on the study started in the paper -- does anyone know of any?
(As a sidebar, I would note that the hyperalgebra/enveloping algebra dichotomy in positive characteristic is mirrored in the study of quantum groups at roots of 1: the De Concini-Kac algebra is the analog of the enveloping algebra, and the Lusztig algebra is the analog of the hyperalgebra).

REPLY [11 votes]: Maybe I can answer the original question more directly, leaving aside the interesting recent geometric work discussed further in later posts like the Feb 10 one by Chuck: analogues of Beilinson-Bernstein localization on flag varieties and consequences for algebraic groups (Bezrukavnikov, Mirkovic, Rumynin). 
The 1979 conference paper by Haboush may be hard to access and also hard to read in detail, but it raises some interesting questions especially about centers of certain hyperalgebras.   I tried to give an overview in Math Reviews:   MR582073 (82a:20049) 20G05 (14L40 17B40)
Haboush,W. J.,
Central differential operators on split semisimple groups over fields of positive
characteristic.
Séminaire d’Algèbre Paul Dubreil et Marie-Paule Malliavin, 32ème 
année (Paris, 1979), pp.
35–85, Lecture Notes in Math., 795, Springer, Berlin, 1980.
The hyperalgebra here is the Hopf algebra dual of the algebra of regular functions on a simply connected semisimple algebraic group $G$ over an algebraically closed field of characteristic $p$, later treated in considerable depth by Jantzen in his 1987 Academic Press book Representations of Algebraic Groups (revised edition, AMS, 2003).   After the paper by Haboush, for example, Donkin finished the determination of all blocks of the hyperalgebra.
While the irreducible (rational) representations are all finite dimensional and have dominant integral highest weights (Chevalley), the module category involves locally finite modules such as the infinite dimensional injective hulls (but no projective covers).    The role of the finite Weyl group is now played by an affine Weyl group relative to $p$ (of Langlands dual type) with translations by $p$ times the root lattice.    In fact, higher powers of $p$ make life even more complicated.   
The older work of Curtis-Steinberg reduces the study of irreducibles to the finitely many "restricted" ones for the Lie algebra $\mathfrak{g}$.  For these and other small enough weights,  Lusztig's 1979-80 conjectures provide the best
hope for an analogue of Kazhdan-Lusztig conjectures when $p>h$ (the Coxeter number).   The recent work applies for $p$ big enough": Andersen-Jantzen-Soergel, BMR, Fiebig.    
Anyway, the hyperalgebra involves rational representations of $G$ including restricted representations of $\mathfrak{g}$, while the usual enveloping algebra of the Lie algebra involves all its representations.   But the irreducible ones are finite dimensional. I surveyed what was known then in a 1998 AMS Bulletin paper.   Lusztig's 1997-1999 conjectures promised more insight into the non-restricted irreducibles and are now proved for large enough $p$ in a preprint by Bezrukavnikov-Mirkovic.   This and their earlier work with Rumynin use a version of "differential operators" on a flag variety starting with the usual rather than divided-power (hyperalgebra) version of the universal enveloping algebra of $\mathfrak{g}$.     
To make a very long story shorter, Haboush was mainly looking for the center of the hyperalgebra (still an elusive beast unlike the classical enveloping algebra center, due to the influence of all powers of $p$).   His weaker version of Verma modules may or may not lead further.   But there is no likely analogue of the BGG category for the hyperalgebra in any case.    That category depended too strongly on finiteness conditions and well-behaved central characters. 
ADDED: It is a long story, but my current viewpoint is that the characteristic $p$ theory for both $G$ and $\mathfrak{g}$ (intersecting in the crucial zone of restricted representations of $\mathfrak{g}$) is essentially finite dimensional and requires deep geometry to resolve.   True, the injective hulls of the simple $G$-modules with a highest weight are naturally defined and infinite dimensional (though locally finite), but the hope is that they will all be direct limits of finite dimensional
injective hulls for (the hyperalgebras of) Frobenius kernels relative to powers of $p$.   Shown so far for $p \geq 2h-2$ (Ballard, Jantzen, Donkin).   In particular, the universal highest weight property of Verma modules in the BGG category (and others) is mostly replaced in characteristic $p$ by Weyl modules (a simple consequence of Kempf vanishing observed by me and codified by Jantzen).    Then the problems begin, as Lusztig's conjectures have shown.   The
Lie algebra case gets into other interesting territory for non-restricted modules.<|endoftext|>
TITLE: Classifying Space of a Group Extension
QUESTION [27 upvotes]: Consider a short exact sequence of Abelian groups -- I'm happy to assume they're finite as a toy example:
$$
0 \to H \to G \to G/H \to 0\ .
$$
I want to understand the classifying space of $G$. Since $BH \cong EG/H$, $G/H$ acts on $BH$ and we can write $BG \cong E(G/H) \times_{G/H} EG/H$. Thus, we have a fiber bundle (which I'll write horizontally)
$$
BH \to BG \to B(G/H)
$$
On the other hand, the central extension is classified by an element of the group cohomology $H^2(G/H,H)$ which is the same as $H^2(B(G/H),H)$. The latter is an element in the homotopy class of maps $[B(G/H),K(H,2)]$ and $K(H,2)\cong BBH$. This map looks like it classifies a principal $BH$ bundle over $B(G/H)$. I find it hard to imagine that this 'principal' $BH$ bundle is not 'the same' as the bundle above, so the question is, how do you see that? From this construction, it's not even obvious to me that the above bundle is a principal bundle.
I would guess (and being a poor physicist, I'm not so up on my homotopy theory), there's a sense that the classifying space of an Abelian group is an 'Abelian group', and taking classifying spaces of an exact sequence gives you back an 'exact sequence'. That gets you a 'principal bundle' (aren't quotation marks fun?), but even then I'm not sure how to see that the classifying map of this bundle is the same as the class in group cohomology.
Any references to the needed background would also be greatly appreciated.

REPLY [4 votes]: @There is an equivalence between grouplike homotopy commutative spaces and double loop spaces
No, grouplike homotopy commutative is not enough. Double loop spaces have much more in thee way of higher homotopies, even if the space were to have a strictly associative homotopy commutative structure. That was one of the motivations for operads.  Also see JF Adams: 10 types of H-spaces.<|endoftext|>
TITLE: Local view of setting p*n out of n bits to 1
QUESTION [7 upvotes]: For p a constant in (0,1) and n going to infinity such that pn is an integer,
consider the distribution on n bits that selects a random subset of pn bits, sets those to 1, and sets the others to 0.
What is the largest k = k(n,p) so that the induced distribution on any k bits is 1/10 close in total variation distance (a.k.a. statistical distance) to the distribution that sets each bit to 1 independently with probability p?
For every p I would like to know k up to a sublinear (i.e. o(n)) additive term.
(For starters, p = 1/8 is good too.)
Does anybody know of a place where this is worked out?
Thanks!
Emanuele

REPLY [3 votes]: Hi Emanuele.  Short answer: take a look at Theorem 3.2 in this paper by Diaconis and Holmes:
http://www-stat.stanford.edu/~susan/papers/steinbirthdeath.pdf
as well as its reference to Diaconis and Freedman (1981).  It seems the optimal k is known to be $\Theta(n)$, independent of $p$.
I have some questions for you though:  Your choice of 1/10 seems a bit "arbitrary", which makes me curious to know whether you really want the correct value of k up to o(n)...  I think changing 1/10 to 1/20, say, would change k by a linear amount.  So if the answer is k = cn, you really want to know how c depends on 1/10?
Another question: Perhaps another way to attack this problem is to identify the event A on which the hypergeometric and binomial random variables have the most differing probabilities.  Is it possible to compute this exactly, or at least decide whether it is an event of the form $A = \{u : a \leq u \leq b\}$?<|endoftext|>
TITLE: What are some good beginner graph theory texts?
QUESTION [16 upvotes]: What are your experiences with them?

REPLY [3 votes]: I learned graph theory from John Kennedy and Christopher Hanusa, the former an extremely  well respected graph theorist and the latter a rising young combinatorialist. There's a lot of good graph theory texts now and I consulted practically all of them when learning it.
  The first edition of Adrian Bondy and U.S.R Murtry's Graph Theory is still one of the best introductory courses in graph theory available and it's still online for free, as far as I know. The second edition is more comprehensive and up-to-date, but it's more of a problem course and therefore more difficult. Jonathan Gross and Jay Yellen's Graph Theory With Applications is the best textbook there is on graph theory PERIOD. Rigorous and as comprehensive as it gets. The section on topological graph theory is particularly good. (I HATE their combinatorics text–it's a hodgepodge text that's nowhere near as well written and organized.) 
  There are several other good books. Chartrand et. al isn't as comprehensive as Gross and Yellen, but quite good and in the same spirit. Douglas West's book is considered by many to be the preeminent graph theory text. I own it–it's pretty good, but not as careful and comprehensive as Gross and Yellen. If you can get a cheap copy, by all means, get West–but if you're gonna end up spending THAT much money, might as well go a little more and get the Ferrari. 
 There's my 2 cents for what it's worth.<|endoftext|>
TITLE: Why are abelian groups amenable?
QUESTION [58 upvotes]: A (discrete) group is amenable if it admits a finitely additive probability measure (on all its subsets), invariant under left translation.  It is a basic fact that every abelian group is amenable.  But the proof I know is surprisingly convoluted.  I'd like to know if there's a more direct proof.
The proof I know runs as follows.

Every finite group is amenable (in a unique way).  This is trivial.
$\mathbb{Z}$ is amenable.  This is not trivial as far as I know; the proof I know involves choosing a non-principal ultrafilter on $\mathbb{N}$.  This means that $\mathbb{Z}$ is amenable in many different ways, i.e. there are many measures on it, but apparently you can't write down any measure 'explicitly' (without using the Axiom of Choice).   
The direct product of two amenable groups is amenable.  This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures.
Every finitely generated abelian group is amenable.  This follows from 1--3 and the classification theorem.  
The class of amenable groups is closed under direct limits (=colimits over a directed poset).  This is like step 2: it seems that there's no canonical way of constructing a measure on the direct limit, given measures on each of the groups that you start with; and the proof involves choosing a non-principal ultrafilter on the poset.
Every abelian group is amenable.  This follows from 4 and 5, since every abelian group is the direct limit of its finitely generated subgroups.

Is there a more direct proof?  Is there even a one-step proof?  

Update  Yemon Choi suggests an immediate simplification: replace 1 and 4 by
1'. Every quotient of an amenable group is amenable.  This is simple: just push the measure forward.
4'.  Every f.g. abelian group is amenable, by 1', 2 and 3.
This avoids using the classification theorem for f.g. abelian groups.
Tom Church mentions the possibility of skipping steps 1--3 and going straight to 4.  If I understand correctly, this doesn't use the classification theorem either.  The argument is similar to the one for $\mathbb{Z}$: one still has to choose an ultrafilter on $\mathbb{N}$. (One also constructs a Følner sequence on the group, a part of the argument which I didn't mention previously but was there all along).  
Yemon, Tom and Mariano Suárez-Alvarez all suggest using one or other alternative formulations of amenability.  I'm definitely interested in answers like that, but it also reminds me of the old joke:

Tourist: Excuse me, how do I get to Edinburgh Castle from here?
Local: I wouldn't start from here if I were you.

In other words, if a proof of the amenability of abelian groups uses a different definition of amenability than the one I gave, then I want to take the proof of equivalence into account when assessing the simplicity of the overall proof.
Jim Borger points out that if, as seems to be the case, even the proof that $\mathbb{Z}$ is amenable makes essential use of the Axiom of Choice, then life is bound to be hard.  I take his point.  However, one simplification to the 6-step proof that I'd like to see is a merging of steps 2 and 5.  These are the two really substantial steps, but they're intriguingly similar.  None of the answers so far seem to make this economy.  That is, every proof suggested seems to involve two separate Følner-type arguments.

REPLY [4 votes]: The way I've seen it was also with the Markov-Kakutani fixed point theorem. The steps are these:
First define
$$ K= \lbrace \Phi\in l^{\infty}(G)^\ast \mid \Phi(1) = 1 ,\Phi(F)\ge0\text{ whenever }F(g)\ge 0\text{ for all }g\in G\rbrace. $$

Equip $l^{\infty}(G)^\ast $ with the weak$^\ast$-topology. Prove that $K$ is weak$^\ast$ compact and convex.
Define for every $g\in G$ the map
$$ T_g:K\to K:(T_g(\Phi)(F)=\Phi(F\cdot g). $$
Prove that every $T_g$, $g\in G$, is weak$^\ast$ continuous and affine.
Since $G$ is a commutative group, we can apply the Markov-Kakutani fixed point theorem to the family $\lbrace T_g \mid g\in G\rbrace$ of affine maps from $K$ to $K$. 
So, we get $\Phi\in K$ such that $T_g\Phi=\Phi$ for all $g\in G$. Define $m(A)=\Phi(\chi_A)$ whenever $A\subset G$ and check that $m$ is an invariant mean on $G$.<|endoftext|>
TITLE: Balls in boxes (partition)
QUESTION [8 upvotes]: Given 100 boxes.  Each contains arbitrary number of red, blue and green balls, i.e., 100 non-negative integer triples $(r_i,b_i,g_i)$.
Prove it's always possible to find 51 boxes so that the total number of balls of each color in these boxes is no less than the ones from the rest 49 boxes.
For n boxes, replace 51 with $\lfloor(n+3)/2\rfloor$, and prove that is the best lower bound.
This is a generalization of a high-school Olympiad question, which I was told to use pigeonhole principle.  Could anyone shed light on how to apply it?

EDITED: Since this is not really related to the pigeon-hole principle, I have edited the title and the tag.
Besides the solution provided by domotorp, darij grinberg pointed to an elementary proof on mathlinks.
Also domotorp has found the origin of the problem (which was sort of buried in the comments).

REPLY [3 votes]: This proof uses a combinatorial equivalent of the Borsuk-Ulam theorem. I think that the proof is a little more complicated than the average proofs here, so please check my related paper if you have difficulties to understand.
Octahedral Tucker's lemma. If for any set-pair $A, B\subset [n], A\cap B=\emptyset, A\cup B\ne\emptyset$ we have a $\lambda(A,B)\in\pm[n-1]$ color, such that $\lambda(A,B)=-\lambda(B,A)$, then there are two set-pairs, $(A_{1},B_{1})$ and $(A_{2},B_{2})$, such that $(A_{1},B_{1})\subset (A_{2},B_{2})$ and $\lambda(A_{1},B_{1})=-\lambda(A_{2},B_{2})$.
We will use this lemma for n=100. If for the boxes in A, the sum of the red balls is more than half of the total number of red balls, then we set $\lambda(A,B)=+red$. If it is more than half in B, then we set $\lambda(A,B)=-red$. We do similarly for blue and green (if they are not set yet to red). We also set $\lambda(A,B)=\pm (|A|+|B|)$ if $|A|+|B|\le 96$   (if they are not set to anything else yet). This way the cardinality of the range of lambda is 99, just as in the lemma. It is easy to verify that it satisfies the conditions of the lemma, thus there must be a set-pair for which we did not set any value. But in that case either A or B must be bigger than 96/2=48, thus at least 49. We can put the remaining boxes to the other part and we are done.
Note that this proof easily generalizes to more colors.<|endoftext|>
TITLE: Applications of infinite Ramsey's Theorem (on N)?
QUESTION [18 upvotes]: Finite Ramsey's theorem is a very important combinatorial tool that is often used in mathematics. The infinite version of Ramsey's theorem (Ramsey's theorem for colorings of tuples of natural numbers) also seems to be a very basic and powerful tool but it is apparently not as widely used.
I searched in the literature for applications of infinite Ramsey's theorem and only found 

straight forward generalization of statements that follow from finite Ramsey's theorem (example: Erdos-Szekeres ~> every infinite sequence of reals contains a monotonic subsequence) and some other basic combinatorial applications,
Ramsey factorization for \omega-words,
the original applications of Ramsey to Logic.

Where else is infinite Ramsey's theorem used?
Especially are there applications to analysis?

REPLY [8 votes]: I think it gives the most beautiful proof of the Bolzano–Weierstrass theorem.  It's a very easy  but beautiful application of Ramsey's theorem.  
Given a sequence $x=(x_n)$ of real numbers, colour the pairs of naturals $i < j$ by whether $x_i < x_j$ or $x_i \geq x_j$.  Ramsey's theorem guarantees an infinite monochromatic set. This corresponds to a monotonic subsequence of $x$; if $x$ is bounded, then this subsequence converges.<|endoftext|>
TITLE: How seriously do professors take teaching evaluations?
QUESTION [13 upvotes]: Do they ever know who writes them? How seriously do departments take teaching evaluations? If a professor knows which student wrote a particular evaluation....would they be biased (e.g. be nicer, etc...)?
Please include in your answer what country you are in and a short description of what type of school you are at, and what you do there.  Answers presumably vary between research universities and four-year colleges, for example.  Also, professors and students could reasonably have different access to information about how teaching evaluations are used.

REPLY [3 votes]: Read Peter Sacks's book Generation X Goes to College (Open Court, Chicago, 1996), at least through chapter 9.  Case study of negative correlation between eval scores and teaching quality, and a how-to manual on manipulating your ratings.  Despite coming from a different field, Sacks's story in chapter 2 about his threatening student "Pete" almost exactly mirrored some of my experiences pre-tenure.<|endoftext|>
TITLE: Functional calculus for direct integrals
QUESTION [5 upvotes]: Suppose I have a direct integral of Hilbert spaces $H = \int^\oplus H_x dx $, and suppose I have an operator $T: H \to H$ which is decomposable, and so it can be written as 
$T = \int^\oplus T_x$ for some measurable field of operators $T_x$. Suppose furthermore that every $T_x$ is self-adjoint (and so also $T$ is self-adjoint), and let $f$ be a bounded measurable function on $\mathbb R$.
Under what conditions $f(T)$ is decomposable (I guess always) and equal to the integral of the field $f(T_x)$ ?
One paper which says something about this problem is Chow, Gilfeather, "Functions of direct integrals of operators". It actually states that the only necessary condition is that $T_x$ are contractions. But unfortunately I don't understand this paper, since it doesn't state its assumptions very precisely - for example, it doesn't seem to be assumed that the operator $T$ (or operators $T_x$) is (are) normal, and so I don't what kind of functional calculus is considered.

REPLY [3 votes]: If you want just to give a brief argument with possible references to known results, you can proceed in the following way: one picks a sequence $p_n$ of polynomials converging to $f$ in the weak-measure topology on the Borel functions; then $p_n(T)$ converges to $f(T)$ even strongly (see e.g. Helemski. Lectures and exercises on functional analysis, p. 388). As $p_n(T)$ commuted with every diagonal operator, $f(T)$ does commute as well, and therefore is decomposable (Dixmier. Les algèbres d'opérateurs dans l'espace hilbertien, Thm. II.2.5.1), say, as $\int^\oplus S_x d\nu(x)$. Now, there is a subsequence $p_{n_k}$ such that $p_{n_k}(T_x)$ converges strongly to $S_x$ $x$-almost everywhere (Dixmier, Prop. II.2.3.4), so $S_x=f(T_x)$ almost everywhere.<|endoftext|>
TITLE: Serre intersection formula and derived algebraic geometry?
QUESTION [41 upvotes]: Let $X$ be a regular scheme (all local rings are regular). Let $Y,Z$ be two closed subschemes defined by ideals sheaves $\mathcal I,\mathcal J$. Serre gave a beautiful formula to count the intersection multiplicity of $Y,Z$ at a generic point $x$ of $Y\cap Z$ as:
$$\sum_{i\geq 0} (-1)^i\text{length}_{\mathcal O_{X,x}} \text{Tor}_i^{\mathcal O_{X,x}}(\mathcal O_{X,x}/\mathcal I_x, \mathcal O_{X,x}/\mathcal J_x)$$  
It takes quite a bit of work to show that this is the right definition (even that the sum terminates is a non-trivial theorem of homological algebra): it is non-negative,  vanishes if the dimensions don't add up correctly, positivity etc. In fact, some cases are still open as far as I know. See here for some reference. 
I have heard one of the great things about Lurie's thesis is setting a framework for derived algebraic geometry. In fact, in the introduction he used Serre formula as a motivation (it is pretty clear from the formula that a "derived" setting seems natural). However, I could not find much about it aside from the intro, and Serre formula was an old flame of mine in grad school. So my (somewhat vague): 
Question: Does any of the desired properties of Serre formula follow naturally from Lurie's work? If so (since things are rarely totally free in math), where did we actually pay the price (in terms of technical work to establish the foundations)? EDIT: Clark's answer below greatly clarifies and gives more historical context to my question, highly recommend!)

REPLY [8 votes]: As Kevin points out, this is discussed in the introduction to DAG V.
A beautiful lecture of Jacob's on the subject (Bezout's theorem as an introduction to DAG) is available to view at the GRASP site. This doesn't fully answer your questions (it's basically an expository version of what Clark explained)..  though from what I understand some derived intersection theory does follow very nicely and easily from the DAG language, specifically the theory of virtual fundamental classes, and this is supposed to appear in one of the forthcoming DAG volumes.<|endoftext|>
TITLE: Gauge connections and Lie algebras?
QUESTION [6 upvotes]: I'm probably missing something obvious, but I've been wondering what the motivation is for requiring the components $A_\mu$ in a local trivialization of a gauge connection on a smooth principal $G$-bundle to lie in $\mathfrak{g}$, the Lie algebra of $G$. I can see that this gives a couple of nice properties; for example, in a local trivialization it ensures that under a gauge transformation $A'_\mu=gA_\mu g^{-1}+g\partial_\mu g$ lies in $\mathfrak{g}$, and that the  curvature form $F=dA+A\wedge A$ lies in $\mathfrak{g}$ (since $\mathfrak{g}$ is closed under the Lie bracket). But is there a more intrinsic or geometric reason that $A_\mu$ must be in $\mathfrak{g}$? Thanks.

REPLY [2 votes]: I always find it helpful to think about Cartan geometries first - they are less "abstract" than principal bundles and shed new light on things like Riemannian geometry.
For a nice introduction see
http://www.emis.de/journals/SIGMA/2009/080/sigma09-080.pdf
(look for the hamster on page 4!)
or the following nice book
http://www.amazon.com/Differential-Geometry-Generalization-Erlangen-Mathematics/dp/0387947329<|endoftext|>
TITLE: When does the group of invertible ideal quotients = the free abelian group on the prime ideals?
QUESTION [6 upvotes]: I haven't learned that much about primary decomposition, but from I understand about Dedekind domains, we have that all fractional ideals are invertible and all (plain old) ideals factor uniquely into a product of prime ideals, so that Dedekind domains should satisfy this condition. Are these the only such rings, or is there a weaker condition we can give?

REPLY [4 votes]: When one considers the ring of algebraic integers R of a number field, the Picard group Pic(R)=Inv(R)/Prin(R) of invertible fractional ideals modulo principal fractional ideals is a measure of how far the ring is from being a UFD (the finiteness of the group is a classical theorem of algebraic number theory). In particular, Pic(R)=0 precisely when R is a UFD. If the cardinality of Pic(R) is h>1, then the hth power of every invertible fractional ideal factors uniquely into the product of prime ideals. 
Perhaps unfortunately, there exist rings R which are not UFDs but for which the Picard group is trivial. An example of such a ring is $\mathbb{Q}+x\mathbb{R}[x]$. This is not to say that the Picard group does not convey important information when one leaves the world of Dedekind Domains. 
A ring is a Bezout domain if it is an integral domain in which every finitely generated ideal is principal. For example, if the ring is Noetherian, then being a Bezout domain is equivalent to being a PID.
Now define a ring R to be a Prufer domain if it is an integral domain in which every non-zero finitely generated ideal is invertible. It is a theorem that for a Prufer ring R, Pic(R) is trivial if and only if R is a Bezout domain.
Let T be a non-empty set of indeterminates and R be an integral domain. Mathematicians have long been interested in exploring the relationshipa between (1) Pic(R) and Pic(R[T]) and (2) Cl(R) and Cl(R[T]). It was proven by Gabelli that Cl(R)=C(R[T]) if and only if R is integrally closed. Call an arbitrary commutative ring seminormal if $b^2=c^3$ implies that there exists an element $a$ such that $a^3=b$ and $a^2=c$. Then  Pic(R)=Pic(R[T]) precisely when R is seminormal.
Finally, it makes sense - and can in fact be incredibly fruitful - to define the Picard group of a noncommutative ring. The most common set up in which this group is defined is as follows. Let R be an integral domain with quotient field K, let A be a finite-dimensional semisimple algebra over K and O be an order of A. Then one can define a group $Pic_R(O)$, which is a certain quotient of the group of invertible fractional ideals, and a homomorphism $\varphi: Aut_R(O)\rightarrow Pic_R(O)$. One can adapt techniques  from K-theory in order to study $Pic_R(O)$, and hence the automorphism group of the order O as well. 
An excellent reference for this is the follwoing paper of Frolich.
A. Frolich The Picard Group of Noncommutative Rings, in Particular of Orders, Transactions of the AMS, Vol 180 (Jun 1973), pp. 1-45<|endoftext|>
TITLE: Frobenius Theorem for subbundle of low regularity?
QUESTION [13 upvotes]: Frobenius Theorem says that a subbundle $E$ of the tangent bundle $TM$ of a manifold $M$ is tangent to a foliation if and only if for any two vector fields $X, Y \subset E$ the bracket $[X,Y]\subset E$. Bracket is a second order operator, hence subbundle $E$ needs to be $C^2$.
Are there any generalizations for subbundle which is $C^1$, $C^{1+smth}$?
Thank you,
Z.

REPLY [16 votes]: You can even get a Frobenius theorem for Lipshitz vector fields,
which need not  span everywhere (the rank can drop along closed subsets).
The state of the art in this domain is done by the control theorists, particularly
Sussmann and Agrachev.  Here is an abstract for a recent article which will have
other earlier references:
Journal of Differential Equations
Volume 243, Issue 2, 15 December 2007, Pages 270-300. F. Rampazzo and H.J. Sussmann
Abstract: We generalize the classical Frobenius Theorem to distributions that are spanned by locally Lipschitz vector fields. The various versions of the involutivity conditions are extended by means of set-valued Lie derivatives—in particular, set-valued Lie brackets—and set-valued exterior derivatives. A PDEs counterpart of these Frobenius-type results is investigated as well.<|endoftext|>
TITLE: Introductory text for the non-arithmetic moduli of elliptic curves
QUESTION [15 upvotes]: I'm looking for an introduction to the non-arithmetic aspects of the moduli of elliptic curves. I'd particularly like one that discusses the $H^1$ local system on the moduli space (whether it's $Y(1)$ or $Y(2)$ or whatever doesn't matter) from the Betti point of view ($SL_2(Z)$, representations of the fundamental group, etc) and the de Rham point of view (Picard-Fuchs equation, hypergeometric functions, etc). This is for a student who has taken classes in algebraic topology and complex analysis and who is just learning algebraic geometry, so I would prefer something that's as down to earth as possible, not a full-blown sheaf-theoretic treatment (i.e. with $R^1f_*$, D-modules, etc). This is a beautiful classical subject, so I can't believe there aren't really great expositions out there, but I can't think of even one!
[Edit 2010/01/21: Thanks to every who suggested references below. I'll probably suggest Clemens's book and Hain's notes. But after looking at them, I realize that I'd really like something even more basic, with no algebraic geometry (cubic curves) and no holomorphic geometry (Riemann surfaces). I just want the moduli spaces of homothety classes of lattices in C (maybe plus some level structure), viewed first as a topological space and later as a differentiable manifold, together with the H^1 local systems, viewed first as a representation of the fundamental group and later as a vector bundle with connection. Probably this is so easy that no one ever bothered to write it down, but on the chance that that's not the case, consider this a renewed request in more precise form.]

REPLY [5 votes]: A follow up to this question:
I spent the last year or so working on the project James outlined above. I ended up getting fairly sidetracked trying to figure out precisely what is meant by a moduli space, and more generally by a family of elliptic curves. The most helpful text I found for this was Kodaira's Complex Manifolds and Deformation of Complex Structures.
Although I didn't end up doing too much work with it, some useful references that discuss the main problem of the $H^1$ local system on the moduli space are the introduction to Period Mappings and Period Domains by Carlson et al, Yoshida's Hypergeometric Functions, My Love (and also other books/papers by Yoshida), and Holzapfel's Geometry and Arithmetic Around Euler Partial Differential Equations. The last book mainly deals with the genus two case, but does have a few pages on the genus one case. None of these references are perfect for the topic, at least coming from my level, but they all helped a fair bit.<|endoftext|>
TITLE: What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations?
QUESTION [20 upvotes]: In modern valuation theory, one studies not just absolute values on a field, but also Krull valuations.  The motivation is easy enough:
If $k$ is a field, a valuation ring of $k$ is a subring $R$ such that for every $x \in k^{\times}$, at least one of $x, x^{-1}$ is an element of $R$.   (It follows of course that $k$ is the fraction field of $R$.)  If $| \ |$ is a non-Archimedean norm on a field, then the set $\{x \in k \ | \ |x| \leq 1 \}$ is a valuation ring.  However, the converse does not hold, since if $R$ is a valuation ring, then $k^{\times}/R^{\times}$ need not inject into $\mathbb{R}$: rather it is (under a straightforward extension of the divisibility relation on $R$) a totally ordered abelian group.  Moreover, a certain formal power series construction shows that for any totally ordered abelian group $\Gamma$, there exists $k$ and $R$ with $k^{\times}/R^{\times} \cong \Gamma$.
My question is this: what are some instances where having the generality of Krull valuations is useful for solving some problem (which is not a priori concerned with valuation theory)?  How do Krull valuations arise in algebraic geometry?
I can almost remember one example of this.  I believe it is possible to give a quick proof of the Lang-Nishimura Theorem -- that having a smooth $k$-rational point is a birational invariant among complete [hmm, valuative criterion!] $k$-varieties.  I think I saw this in some of Bjorn Poonen's lecture notes, but I forget where.  [Last year at this time, I would have emailed Bjorn.  I am trying out this new approach on the theory that Bjorn can reply if he wishes, and if not someone else will surely be eager to tell me the answer.]  
Are there other nice examples?  Maybe something to do with resolution of singularities?

REPLY [2 votes]: For a different flavor from the answers to date: valuations beyond the classical discrete valuations arising from absolute values have proved extremely useful in analysing structures of certain division algebras finite-dimensional over their centers.  Division algebras (even when finite-dimensional over their centres) are notoriously difficult to get a handle on: for instance, it is almost always impossible to describe their subfields.  However, when the centre $F$ is Henselian with respect to a general valuation (Henselian simply means that Hensel's lemma holds: this substitutes for the notion of completeness that is used in the classical case), then it is easy to see that the valuation on $F$ extends uniquely to $D$, and if further the division algebra $D$ is "tame" (i.e., the characteristic of $\overline{F}$ does not divide $[D:F]$), then one can use the valuations on $F$ and $D$ to obtain significant information about $D$.  
As an example: if further $D$ is totally ramified over $F$, i.e., $|V(D)/V(F)| = [D:F]$, where $V()$ stands for the value group, then it turns out that there is a nondegenerate alternating pairing from $V(D)/V(F) \times V(D)/V(F)$ to the group of $m$-th roots of unity in $\overline{F}$, where $m$ is the period of $D$ in the Brauer group of $F$.  This pairing completely determines $D$: $D$ decomposes as a tensor product of "symbol algebras" (i.e., generalized quaternions), where the tensor factors correspond to a symplectic base for $V(D)/V(F)$.  The isomorphism classes of $F$-subalgebras of $D$ are determined by subgroups of $V(D)/V(F)$, and the $F$-isomorphism classes of subfields of $D$ are determined by totally isotropic subgroups of $V(D)/V(F)$.  This in turn completely determines  tame division algebras over fields of the form $k((x_1))\cdots ((x_n))$ where $k$ is separably closed.
For a survey see the paper: Wadsworth, A. R., Valuation theory on finite dimensional division algebras. Valuation theory and its applications, Vol. I (Saskatoon, SK, 1999), 385--449, Fields Inst. Commun., 32, Amer. Math. Soc., Providence, RI, 2002. (Math Reviews: MR1928379 (2003g:16023) )<|endoftext|>
TITLE: Topologically distinct Calabi-Yau threefolds
QUESTION [28 upvotes]: In dimensions 1 and 2 there is only one, respectively 2, compact Kaehler manifolds with zero first Chern class, up to diffeomorphism. However, it is an open problem whether or not the number of topological types of such manifolds of dimension 3 (Calabi-Yau threefolds) is bounded. I would like to ask what is known in this direction. In particular, is it known that the Euler characteristic or the total Betti number of Calabi-Yau threefolds can't be arbitrarily large? Are there any mathematical (or physical?) reasons to expect either answer?
As a side question: I remember having heard several times something like "Calabi-Yau 3-folds parametrize (some kind of) vacua in string theory" but was never able to make precise sense of this. So any comments on this point or references accessible to mathematicians would be very welcome.

REPLY [10 votes]: I am going to give a rough account of the physical interpretation of Calabi-Yau manifolds in string theory.  
We are studying particles moving around in 10 spacetime dimensions.  We want to pick these 10 dimensions so that spacetime looks realistic, so we choose our Lorentzian 10-manifold to be $R^{1,3}\times X$ where $X$ is a compact 6-manifold.  If we choose $X$ to be sufficiently small then at low energies $X$ itself is not directly observable.  Thus to an observer who studies the world with low energy probes spacetime will have just the familiar 4 dimensions.
Nevertheless, one cannot chose $X$ in a completely arbitrary way.  Einstein's theory of general relativity tells us that the metric on $X$ carries energy measured by its Ricci curvature.  In the absence of matter, that is $in  \ the  \ vacuum,$ the spacetime must be in a lowest energy Ricci flat configuration.  So $X$ must be such that it admits a metric of vanishing Ricci curvature.  
To get to the Calabi-Yau condition we need to add one more physical criterion and this is called supersymmetry. Supersymmetry is an extremely attractive conjectural symmetry of nature that says that bosons (e.g. photons), and fermions (e.g. electrons) are paired together and related.  Mathematically fermions are described by spinor valued fields on spacetime, while bosons are described by functions or one-forms on spacetime.  To relate these objects, and thus have a theory with supersymmetry, what is required is a covariantly constant spinor field on $X$.  This constraint on $X$ reduces its holonomy from that of a general Riemanian 6-manifold, $SO(6)$, to $SU(3)$.  Indeed $SO(6)$ is locally isomorphic to $SU(4)$ and the spinor representation is $4 \oplus \bar{4}$ where 4 denotes the fundamental representation of $SU(4)$.  Since $X$ admits a covairiantly constant spinor its holonomy must be contained in the subgroup of $SU(4)$ that stabilizes a spinor, and it is clear form the decomposition above that this subgroup is $SU(3)$.  Hence for supersymmetry, $X$ must be Calabi-Yau.
Thus we can now clearly state exacty what it is that Calabi-Yau threefolds describe physically.  They are lowest energy vacuum configurations of string theory where the physics is supersymmetric.  For more details one might want to take a look at the book "Mirror Symmetry" by Vafa and Zaslow.<|endoftext|>
TITLE: Find a "natural" group that contains the quotient of the infinite symmetric group by the alternating subgroup
QUESTION [18 upvotes]: Let $S_\infty$ the group of permutations of $\mathbb{N}$. It can be shown that there is no homomorphism $S_\infty \to \mathbf{Z}/2$ extending the sign on the finite symmetric groups. Is it possible to write down a homomorphism (an unexplicit one won't be useful in my application) of $S_\infty$ into another (infinite) group, which restricts to the sign? Perhaps we should also require that the homomorphism somehow also reminds of the sign in the infinite case. Thus perhaps we should formalize something like $(-1)^M$, where $M$ is an infinite set (as you might guess, this is related with my question about Infinite Tensor Products).
EDIT: As was pointed out by Pete, the question is equivalent to: Find a nice, "natural" group which contains $S_\infty / \cup_n A_n$.

REPLY [9 votes]: This group is quite fascinating and I've thought quite a lot about this question. Although I'm unable to answer it in a sensible way (an example of non-sensible way would be to embed this group in the group of permutations on itself through left translation), let me say a few things.
Let $S$ be the whole symmetric group (on an infinite set $X$), $F$ its finitary subgroup (finitely supported permutations), and $A$ its subgroup of index 2 of even permutations. The question is about $S/A$, which lies in a central extension
$$ 1\to F/A(\simeq \mathbf{Z}/2\mathbf{Z}) \to S/A\to S/F\to 1;$$ 
let $\omega^X\in H^2(S/F,\mathbf{Z}/2\mathbf{Z})$ be the cohomology class of this extension. It follows from Vitali's result $\mathrm{Hom}(S,\mathbf{Z}/2\mathbf{Z})=0$ that $\omega\neq 0$. Sergiescu observed, using acyclicity of $S$ (la Harpe- McDuff) that $H^2(S/F,\mathbf{Z}/2\mathbf{Z})$ is reduced to $\{0,\omega^X\}$, and indeed that $H_2(S/F)\simeq\mathbf{Z}/2\mathbf{Z}$.
Given a group $G$ and a homomorphism $f:G\to S/F$ (I call this a balanced near action), one thus gets, by pullback, a cohomology class $f^*\omega^X\in H^2(G,\mathbf{Z}/2\mathbf{Z})$, whose nonvanishing is on obstruction for $f$ to lift to a homomorphism $G\to S$, i.e., a genuine action on $X$. 
The first nontrivial explicit computations of such classes were done by Kapoudjian in the context of Higman-Thompson and Neretin's groups and their natural near actions on trees. For this reason, I call $f^*\omega^X$ Kapoudjian class of the near action, and I address it here, §8.6 (self-advertisement warning, so I cw this answer).
Concerning the initial question, it has a kind of easier analogue which deserve some comment, namely finding a "natural" embedding of $S/F$ (rather than $S/A$): the simplest answer seems to be to embed it into the group of self-homeomorphisms of the Stone-Cech boundary of $X$. For $X$ countable, there is a significant literature around how large is the image within the whole self-homeomorphism group (Rudin-Shelah problem). In any case such ideas do not seem to provide embeddings of $S/A$. I insist anyway, because any embedding of $S/A$ kind of induces an embedding of $S/F$ (not literally speaking, but after slightly modifying the target group), so first a good understanding of how to embed $S/F$ would be useful (there are not so many understood ways), and second a good understanding of the central extension would be useful too, and this is well encoded in the Kapoudjian class.

Added: I made Vitali's 1915 rare paper (in Italian) available here. Reference info: G. Vitali. Sostituzioni sopra una infinità numerabile di elementi. Bollettino Mathesis 7: 29-31, 1915. (Any suggestion for a more standard repository is welcome.)<|endoftext|>
TITLE: Why is one interested in the mod p reduction of modular curves and Shimura varieties?
QUESTION [16 upvotes]: Why is one interested in the mod p reduction of modular curves and Shimura varieties?
From an article I learned that this can be used to prove the Eichler-Shimura relation which in turn proves the Hasse-Weil conjecture for modular curves. Are there similar applications for Shimura varieties?

REPLY [27 votes]: Kevin's answer gives a very good explanation of the role of mod $p$ reduction in the theory of Galois representations and automorphic forms.  In this answer I will try to say something a little more technical about one way in which understanding the mod $p$ reduction of modular curves can be applied in arithmetic.  The precise application that I will discuss is that of constructing congruences of modular forms.
If $f$ is a Hecke eigenform (of weight 2, to fix ideas), then associated to $f$ is a Galois representation $\rho_f:G_{\mathbb Q} \to GL_2(\overline{\mathbb Q}_{\ell})$ (for any prime $\ell$). Say the level
of $f$ is equal to $N p$, where $p$ is a prime not dividing $N$.  One can ask: is there an eigenform $g$ of level $N$ such that $f \equiv g \bmod \ell$.  (Here congruence means congruence
in $q$-expansions.)  This is the question that Ribet solved in his famous Inventiones 100 paper (the paper which reduced FLT to Shimura--Taniyama).  
Note that since $p$ is not in the level of $g$, the representation $\rho_g$ will be
unramified locally at $p$.  (This comes from knowing that the modular curve of level $N$
has good reduction at $p$, since $p$ does not divide $N$ --- a first application of the theory of reduction of modular curves.)  (If $p = \ell$ one must be more careful here,
but I will suppress this point.)
Thus if $f \equiv g \bmod \ell,$ so that $\rho_f$ and $\rho_g$ coincide mod $\ell$,
we see that $\rho_f$, when reduced mod $\ell$, must be unramified at $p$.  So this
is a necessary condition for the existence of $g$.
It turns out (and Ribet proved) that (under some additional technical hypotheses) this
necessary condition is also sufficient.  The way the argument goes is the following:
the modular curve of level $N p$ has semi-stable singular reduction: it is two smooth
curves (coming from level $N$) crossing each other a bunch of times (this is the contribution from the $p$-part of the level).  Now the mod $\ell$ Galois representation
$\overline{\rho}_f$ (the reduction of $\rho_f$ mod $\ell$) is consructed out of
the $\ell$-torsion subgroup of the Picard group of this singular curve.   Since it
is unramified at $p$, it can't be entirely explained by the singularities; some part of it
must be arising from the smooth curves, which are of level $N$.  (If you like, this is an application of the a certain form of the criterion of Neron--Ogg--Shafarevic.) The Eichler--Shimura
relations then show that the system of Hecke eigenvalues attached to $f$, when reduced
mod $\ell$, must arise at level $N$: in other words, there is an eigenform $g$ of level $N$ that is congruent to $f$ mod $\ell$.
This is just one typical argument that uses a detailed knowledge of the good and bad reduction of modular curves in various situations.   Since the Galois representation attached to modular forms are constructed geometrically from the modular curves, tools 
like the Neron--Ogg--Shafarevic criterion, and variants thereof, show that there are 
very close ties between the local properties of the Galois representations at a prime $p$,
and the reduction properties of modular curves mod $p$.<|endoftext|>
TITLE: Co-Objects are better
QUESTION [8 upvotes]: This is a rather vague question, but perhaps we can talk about it.
There are two types of mathematical objects (which don't exclude each other):
A) There is a good description of morphisms defined on this object.
B) There is a good description of morphisms defined into this object.
Thus A) means that the covariant hom-functor is understood, and B) means that the contravariant hom-functor is understood. This applies most notably to universal objects. Within category theory, the concepts are just dual to each other and so the "theory" of A) is essentially the same as the theory of B). But most categories studied in mathematics don't come together with their dual, so that this categorical argument is not really good. In fact, I have the feeling that in 'daily mathematics', A) appears much more often than B). And that it is easier to work with them. Of course, we could argue about that. For example, I have a better feeling with colimits than with limits. [perhaps I will add examples here]
If you have the same feeling: Can we give reasons for this?
I think that the basic principle of gluing, which appears in many geometric categories, always belongs to A). This could be a reason. What do you think?

REPLY [12 votes]: Dear Martin,
As Harry points out in his comments, in certain settings (e.g. moduli spaces) an object is characterized by maps in.  In others (e.g. the free abelian group on one generator), the characterization is by maps out.  
Certainly in algebra, injective objects (characterized by maps in) are typically regarded
as more mysterious and black-box like than projectives (where one can typically think of
free modules, which are quite concrete).  I know several situations in which someone made real progress by judicious use of injective objects, and I'm sure part of the obstruction to previous researchers was just that injectives are not as familiar; in short, there
is probably arbitrage to be gained for some (myself, at least) by learning more about injectives in various contexts,
and trying to use them as fluently as one uses free objects.  
In topology and geometry, perhaps there is more fluidity between the two characterizations. 
E.g. maps into the circle make it the Eilenberg-Maclane space $K({\mathbb Z},1)$, while
maps out define the fundamental group.
You are correct that quotienting by an equivalence relation (gluing) is related to
maps out.  Perhaps this is one reason why the construction of moduli spaces (e.g. Picard
schemes) can be quite involved; they are characterized by maps in, but are often
constructed by a gluing procedure, which creates conflict; thus one finds oneself working locally, and is led into sheaf/stack-theoretic issues.  
Certainly, the tension between the two characterizations has been a fertile source for
good mathematics.<|endoftext|>
TITLE: Teichmuller theory and moduli of Riemann surfaces
QUESTION [12 upvotes]: This is a sequel to my earlier question asking for references for Teichmuller theory and moduli spaces of Riemann surfaces.
In this connection, I have read Chapter 11 of the book Primer of mapping class groups by Dan Margalit and Benson Farb.
So I have understood that the moduli space of a Riemann surface is the quotient of the Teichmuller space by the mapping class group, the action is properly discontinuous, the quotient is an orbifold, but it is not in general compact(Mumford's compactness criterion), it has "only one end", etc..
Other than these facts, does Teichmuller theory simplify the study of moduli spaces of Riemann surfaces in any way? Can we do something using Teichmuller theory which we can't do, say, using algebraic geometry? Are we able to prove theorems about moduli spaces, using Teichmuller theory methods? I would be grateful for any examples.

REPLY [9 votes]: I'll discuss things which are more applications of the mapping class group to moduli space rather than Teichmuller theory per se, but of course this is all tightly connected.
One of the big applications of this point of view is to the cohomology of moduli space.  The moduli space of curves is not quite a classifying space for the mapping class group because the action of the mapping class group on Teichmuller space is not free, but the problem all comes from finite order elements.  One can think of moduli space as a "rational classifying space" or an "orbifold classifying space" for the mapping class group.  The upshot is that the group cohomology of the mapping class group is identical to the cohomology of moduli space with $\mathbb{Q}$ coefficients.
I will try to give a brief survey of this field, but it is huge and I will omit a lot of important work.
There is now a lot known about the group cohomology of the mapping class group.  The most spectacular is the resolution by Madsen-Weiss of the Mumford conjecture giving the rational cohomology ring in a stable range.  This is certainly not known via algebro-geometric methods.
This was proceeded by many older results.  The most germane come from a series of papers by Harer in the '80's which (among other things) do the following:
1) Show that the cohomology stabilizes as the genus increases.
2) Calculate the Euler characteristic.  This really is not a theorem about the mapping class group, as the proof uses a certain triangulation of moduli space rather than group theory.  However, this triangulation definitely comes from Teichmuller theory rather than algebraic geometry, and it is still part of this same circle of ideas.
3) Make a number of low-dimensional calculations (up to degree 3 in published work and 4 in unpublished work).
The calculation of $H_2$ by Harer in particular is the key to calculating the Picard group of moduli space.
These low-dimensional cohomology calculations can now be (basically) done via algebraic geometry.  See the paper "Calculating cohomology groups of moduli spaces of curves via algebraic geometry" by Arbarello and Cornalba.  Thus the Picard group of moduli space can now be calculated via algebraic geometry.
A more recent application of this point of view comes from work of myself which calculates the Picard groups of the moduli spaces of curves with level structures (see my paper of the same title).  I think it would be very interesting to try to make this same calculation using algebro-geometric methods, but I have no idea how to do so.<|endoftext|>
TITLE: Jacobi fields on a "bump surface"
QUESTION [5 upvotes]: Consider a "bump surface" which looks like the following:

Such a surface is rotationally symmetric, $C^2$-smooth, has positive curvature in the middle and negative curvature along the ring (the orange region in the picture).  I don't really care what happens past that (it could flatten out, or oscillate, etc.)
Here are two examples, as surfaces of revolution in $\mathbb R^3$ in cylindrical coordinates:
$z(r) = e^{-r^2/2}$ and $z(r) = \tfrac{2}{\pi} \cos(\tfrac{\pi}{2} r)$.
I need to do some Riemannian geometry on a bump surface; in particular, analyze a Jacobi field along a radial geodesic $\gamma$.  I don't care what bump surface I use; it only has to feature both positive and negative curvature.  For any surface of revolution, it's easy to write down a formula for the scalar curvature $K$ (see p. 142 of McCleary's Geometry from a differentiable viewpoint), and the Jacobi equation takes the form $J'' + KJ|\dot\gamma|^2 = 0$.  Thus, if the scalar curvature has a simple form, then the Jacobi equation should be easy to solve.  In the case of these two examples, the scalar curvature isn't particularly pretty, hence analyzing the Jacobi equation is difficult (though not intractable).
My question to the MathOverflow community:  is there a better bump surface than the two examples I gave above, for which the scalar curvature has a particularly simple form?  
Edit:  The curvatures for the surfaces given above are
$K(r) = \frac{2 (1 - r)}{(e^{r^2/2} + r^2 e^{-r^2/2})^2}$ and $K(r) = \frac{\pi \sin(\pi r)}{2 r (1 + \sin^2(\pi r/2))^2}$,
respectively.  As you can see, they're not the worst expressions possible, but they're also not as simple as I'd like them to be.

REPLY [5 votes]: Actually, I think you'd be better off just starting with an abstract surface of revolution and doing the calculation you want.  An abstract surface of revolution in polar coordinates looks like
$$
g = \mathrm{d}r^2 + \bigl(f(r)\,\mathrm{d}\theta\bigr)^2
$$
where $f$ is an odd function of $r$ that satisfies $f'(0)=1$.  Then the Gauss curvature is 
$$
K = -\frac{f''(r)}{f(r)},
$$
so $J_1 = f(r)$ is the odd solution of the Jacobi equation $J''+K J = 0$.  (Note that $r$ is already arclength in the radial direction.)  The even solution satisfying $J_0(0)=1$ is then given by
$$
J_0(r) = -f(r)\int_0^r \frac{\mathrm{d}\rho}{f(\rho)^2}.
$$
Now you just have to choose $f$ so that $K$ starts out positive and then goes negative while the integral for $J_0$ can be performed explicitly.
It turns out that, taking
$$
f(r) = \frac{r}{1{+}r^2}
$$
gives
$$
K = \frac{2(3{-}r^2)}{(1{+}r^2)^2},\qquad\text{and}\qquad
J_0(r) = \frac{3{-}6r^2{-}r^4}{3(1{+}r^2)},\qquad
J_1(r) = \frac{r}{1{+}r^2},\quad
$$
which has all of your desired properties.
Now, the question is whether you want to see this isometrically embedded into $\mathbb{R}^3$ as a surface of revolution.  This you can do in the form
$$
(x,y,z) = \left(\frac{r\,\cos\theta}{1{+}r^2},\ \frac{r\,\sin\theta}{1{+}r^2},\ b(r^2)\right)
$$
where $b$ is the function of $t\ge0$ given by the elliptic integral
$$
b(t) = \frac12\int_0^{t}\frac{\sqrt{6{+}5\rho{+}4\rho^2{+}\rho^3}}{1+\rho}\,\mathrm{d}\rho.
$$
Thus, this surface has the shape of a 'bubble' with a long tail.
If you want a surface that is asymptotic to a plane with a bump of positive curvature in the middle, take, for example,
$$
f(r) = \frac{r(1{+}r^2)^2}{1{+}6r^2{+}r^4},
$$
which gives $J_1(r) = f(r)$ while
$$
K = \frac{8(3{-}6r^2{-}12r^4{-}18r^6{+}r^8)}{(1{+}r^2)^2(1{+}6r^2{+}r^4)^2}\quad\text{and}\quad
J_0(r) = \frac{3{-}23r^4{-}12r^6-15r\arctan(r)(1{+}r^2)^3}{3(1{+}r^2)(1{+}6r^2{+}r^4)},
$$
and has all of your desired properties.<|endoftext|>
TITLE: Reductio ad absurdum or the contrapositive?
QUESTION [60 upvotes]: From time to time, when I write proofs, I'll begin with a claim and then prove the contradiction.  However, when I look over the proof afterwards, it appears that my proof was essentially a proof of the contrapositive, and the initial claim was not actually important in the proof.  
Can all claims proven by reductio ad absurdum be reworded into proofs of the contrapositive?  If not, can you give some examples of proofs that don't reduce?  If not all reductio proofs can be reduced, is there any logical reason why not?  Is reductio stronger or weaker than the contrapositive?
Edit: Just another minor question (of course this is optional and will not affect me choosing an answer):  If they are equivalent, then why would you bother using reductio?
And another bonus question (Like the above, does not influence how I choose the answer to accept.)
Are the two techniques intuitionistically equivalent?

REPLY [3 votes]: There is yet another reason for preferring proofs of the contrapositive to proofs by contradiction that has not been mentioned so far: It is harder to come up with wrong arguments that appear to be correct proofs when one tries to prove the contrapositive directly. When one attempts to prove something by contradiction, it is easy to declare victory if one can produce nonsense- even if the nonsense is the result of a mistake.
For an extreme point of view based on this perspective, Halsey L. Royden writes the following in the "Prologue to the Student" in his analysis textbook:

All students are enjoined in the strongest possible terms to eschew
proofs by contradiction! There are two reasons for this prohibition:
First, such proofs are very often fallacious, the contradiction on the
final page arising from an erroneous deduction on an earlier page,
rather than from the incompatibility of $A$ with $\neg B$. Second,
even when correct, such a proof gives little insight into the
connection between $A$ and $B$, whereas both the direct proof and the
proof by contraposition construct a chain of argument connecting $A$
with $B$. One reason that mistakes are so much more likely in proofs
by contradiction than in direct proofs or proofs by contraposition is
that in a direct proof (assuming the hypothesis is not always false)
all deductions from the hypothesis are true in those cases where the
hypothesis holds, and similarly for proofs by contraposition (if the
conclusion is not always true) the deductions from the negation of the
conclusion are true in those cases where the conclusion is false.
Either way, one is dealing with true statements, and one's intuition
and knowledge about what is true help to keep one from making
erroneous statements. In proofs by contradiction, however, you are
(assuming the theorem true) in the unreal world where any statement
can be derived, and so the falsity of a statement is no indication of
an erroneous deduction.

This is of course mostly didactic; Royden is clearly no constructivist. A related didactic point is that students might be able to visualize and imagine the antecedent of a proof of the contrapositive. Since the antecedent of a proof by contradiction is logically impossible, this does not work with proofs by contradiction. At least not if you do not have the skills of the queen in Alice in Wonderland, who in her younger years was able to imagine six impossible things before breakfast.<|endoftext|>
TITLE: Algebraic versus Analytic Brauer Group
QUESTION [19 upvotes]: Let $X$ be a smooth projective algebraic variety over $\mathbb{C}$.  Then I think that someone (Serre?) showed that the Cohomological Etale Brauer Group agrees with the torsion part of the Analytic Brauer Group $H^{2}(X,\mathcal{O}^{\times})$.  This latter group is calculated in the classical (metric) topology on the associated complex manifold with the sheaf of nowhere vanishing holomorphic functions. 
However there can easily be non-torsion elements in $H^{2}(X,\mathcal{O}^{\times})$: for instance consider the image in $H^{3}(X,\mathbb{Z}) \cap (H^{(2,1)}(X) \oplus H^{(1,2)}(X))$.  
Could there be a topology more refined than etale but defined algebraically which can see these non-torsion classes?  Notice that one can also ask the question for any $H^{i}(X,\mathcal{O}^{\times})$.  For $i=0,1$ the Zariski and etale work fine.  
Why do things break down for $i>1$?

REPLY [5 votes]: I think the article by B. Toen "Derived Azumaya algebras and generators for twisted derived categories", arXiv:1002.2599, gives a pointer to a possible answer to your question.
EDIT I have weakened the assertion...<|endoftext|>
TITLE: Context for "Coronidis Loco" from Weil's Basic Number Theory
QUESTION [20 upvotes]: In Samuel James Patterson's article titled Gauss Sums in The Shaping of Arithmetic after C. F. Gauss’s Disquisitiones Arithmeticae, Patterson says
"Hecke [proved] a beautiful theorem on the different of k, namely that the class of the absolute different in the ideal class group is a square. This theorem - an analogue of the fact that the Euler characteristic of a Riemann surface is even - is the crowning moment (coronidis loco) in both Hecke's book and Andre Weil's Basic Number Theory."
About the same matter, J.V. Armitage says (in his review of the 1981 translation of Hecke's book):
"That beautiful theorem deservedly occupies the 'coronidis loco' in Weil's Basic number theory and was the starting point for the work on parity problems in algebraic number theory and algebraic geometry, which has borne such rich fruit in the past fifteen years."

What is a reference for learning about the parity
  problems that Armitage alludes to?

It can be impossible to verbalize the reasons for aesthetic preferences, but

Why might Weil, Patterson and Armitage
  have been so favorably impressed by
  the theorem that the ideal class of
  the different of a number field is a
  square in the ideal class group?

Weil makes no comment on why he chose to end Basic Number Theory with the above theorem. It should be borne in mind that Weil's book covers the class number formula and all of class field theory, so that the standard against which the above theorem is being measured in the above quotes is high!

REPLY [11 votes]: The parity problems Armitage alludes to includes Hecke's theorem,
as well as to other (and related) parity problems brought fourth
by Froehlich in his theory of Galois modules. For a couple of
references, see Chapter 11 of "Reciprocity Laws", e.g. at
link text
franz lemmermeyer<|endoftext|>
TITLE: Bizarre operation on polynomials
QUESTION [54 upvotes]: There I was, innocently doing some category theory, when up popped a totally outlandish operation on polynomials.  It seems outlandish to me, anyway.  I'd like to know if anyone has seen this operation before, in any context.
The categorical background isn't relevant to the question, so I'll skip it.  All I want to emphasize is that a priori, it has nothing to do with polynomials.  It's just some universal property, which produces this in a special case.  (For the curious, the categorical connection is that some functors $\mathbf{Set}^n \to \mathbf{Set}$ can be viewed as "polynomial", in the sense that they're built up from products, $\times$, and coproducts, $+$.)  
By a polynomial I mean a polynomial in commuting variables $X_1, \ldots, X_n$, with coefficients in the natural numbers $\mathbb{N}$ (which include $0$).
Here's the operation.  Given a polynomial $f = f(X_1, \ldots, X_n)$, define a new polynomial $f^*$ as follows.

Write $f$ as a sum of products of $X_i$'s.  
Change every occurrence of $+$ to $\times$, and every occurrence of $\times$ to $+$.  Call the resulting polynomial $f^*$.  

Examples:

Take $f(X, Y) = (X + Y)^2$.  Step 1 writes $f$ as 
$$
f(X, Y) = (X \times X) + (X \times Y) + (X \times Y) + (Y \times Y).
$$
Step 2 then gives
$$
f^*(X, Y) = (X + X) \times (X + Y) \times (X + Y) \times (Y + Y) = 4XY(X + Y)^2.
$$
Now let's calculate $f^{**}$.  Step 1:
$$
f^*(X, Y) = 4X^3 Y + 8X^2 Y^2 + 4X Y^3.
$$
Step 2:
$$
f^{**}(X, Y) = (3X + Y)^4 (2X + 2Y)^8 (X + 3Y)^4.
$$
Generally, if 
$$
f(X_1, \ldots, X_n) = A X_1^{a_1} \cdots X_n^{a_n} + B X_1^{b_1} \cdots X_n^{b_n} + \cdots
$$
($A, a_i, B, b_i, \ldots \in \mathbb{N}$) then 
$$
f^*(X_1, \ldots, X_n) = (a_1 X_1 + \cdots a_n X_n)^A (b_1 X_1 + \cdots + b_n X_n)^B \cdots.
$$
By the previous example, $f^{**} = f$ if $f$ is a monomial ($X_1^{a_1} \cdots X_n^{a_n}$) or linear ($a_1 X_1 + \cdots + a_n X_n$).  
Since the empty sum is 0 and the empty product is 1, it's meant to be implicit in (2) that 0s become 1s and 1s become 0s.  E.g. if $f = 0$ then $f^* = 1$, and if $f(X) = X^2 + 1$ then $f^*(X) = 2X \times 0 = 0$.  Edit: Similarly, if $f$ has nonzero constant term then $f^* = 0$. 

I'm interested to hear about anywhere that anyone has seen this operation.  
Feel free to add tags as appropriate.

REPLY [4 votes]: I don't remember any reference at the moment but I have seen this kind of "dualization" in majorization related topics. For example let
$$f(\mathbf{x,a})=f(x_1,x_2,\dots x_n\; a_1,a_2,\dots,a_n)=\sum_{\sigma \in S_n}x_{1}^{\sigma (a_1)}\cdots x_{n}^{\sigma (a_n)}$$
A classical result of Muirhead is that $f$ is Schur-convex with respect to $\mathbf{a}$ (with $\mathbf{x}$ in the positive orthant). There is a result that $f(\mathbf{x,a})^{*}$ will be Schur-concave (dual taken with respect to the x's, of course), thus $*$ sends Schur-convex polynomials to Schur-concave ones.<|endoftext|>
TITLE: Integer subset that only occupies (p-1)/2 equivalence classes mod p?
QUESTION [5 upvotes]: I'm not quite sure the best way to ask this, so bear with me: Does anyone know of a subset of integers such that, for any odd prime p, the subset only occupies (p-1)/2 equivalence classes mod p (and does so uniformly)?
For example, take the subset of squares. Elementary number theory shows that they (as quadratic residues) occupy (p+1)/2 equivalence classes mod p. But the answer to the above is not to take the non-residues since being a non-residue is a local property, not a property of an integer.
It is possible to construct such a set of integers one element at a time in an ad hoc manner using some initial members, a whole lot of CRT, and making a somewhat arbitrary choice at each step. But is there a more ``well-known'' set that has this property?

REPLY [6 votes]: See section 4.3 of Helfgott and Venkatesh, "How small must ill-distributed sets be?" 
for an example of a subset of [1..N] of size about log N with small projections onto Z/pZ,
and section 4.2 for a "guess" about what such subsets might look like in general.  They speculate that such a set might have to be either very small (say, of size N^eps) or highly correlated with a "thin set," say, the values of a polynomial (i.e. x^2, as in the first case you describe.)<|endoftext|>
TITLE: How many primes stay inert in a finite (non-cyclic) extension of number fields?
QUESTION [9 upvotes]: In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic,  there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8, 
In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves. 
So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic. 
I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?

REPLY [5 votes]: Does "inert" mean that there is only one prime over $p$, or does it mean that there is only one prime over $p$ and that prime is unramified? As the other answers have explained, unramified primes can not remain inert. However, it is possible that there is only prime over $p$. For example, let $p$ be an odd prime, take $K = \mathbb{Q}$ and $L= \mathbb{Q}(\sqrt{p}, \sqrt{a})$ where $a$ is not a quadratic residue modulo $p$.<|endoftext|>
TITLE: first-order definability transitive closure operator
QUESTION [8 upvotes]: I know this sounds dumb, but I can't for the life of me remember how to expand "TC(x)" into a first-order term in the language of set theory (ZFC, not NBG) where epsilon is the only nonlogical symbol.
The obvious definition is an $\omega$-long sentence $x\cup (\bigcup x)\cup (\bigcup\bigcup x)...$, but that isn't in $L_{\omega\omega}$.
The definition given in Jech, p64 appeals to "the intersection of any class with a set is a set" (p8), which is really expressible only in NBG, right?  I'm at a loss to figure out how to turn this into simple ZFC using separation and replacement.
I also don't have much trouble proving that for every set there exists some other set which is its transitive closure; I just can't seem to turn this proof of $(\exists y)\phi$ (for $\phi$ being "y is the transitive closure of x") into an explicit description of the $y$.
I'm starting to suspect that TC(x) isn't definable in ZFC, but that it can be defined as a class-function in NBG (which is a conservative extension of ZFC, so being able to define TC(x) doesn't actually get you any new theorems about sets).
Thanks!

REPLY [6 votes]: As an addendum to Joel Hamkins answer: the weaker assertion (`Transitive Containment')
that every set x is contained in a transitive set 
(not necessarily its transitive closure) is not provable
in ZF - Replacement (sometimes known as Z Zermelo-set theory).
As Joel says in his answer we need to collect together the results of taking successive
$\bigcup$. For this $\Sigma_1$-Replacement is more than enough (if AxFoundation is formulated
in the right way for $\Pi_1$ classes).
As a second addendum (as the discussion continues) one should remark that we have in full ZF the $\in$-recursion theorem: thus we may define the class function $x\rightarrow TC(x)$  within ZFC (no need for NBG) via the recursion scheme
                  $$TC(x)= \bigcup [TC(y) : y\in x] \cup x $$.
The function defined has just the same status as $\alpha \rightarrow V_\alpha$ (which is also defined by such a recursion scheme). The former guarantees the existence of $TC(x)$ for any $x$, the latter of $V_\alpha$ for any $\alpha$. Neither function is problematic for the full ZF-set theorist. 
The original question seems to be spurred on by the fact that we do not have an obvious {\em explicit  definition} for $TC(x)$ in the way that we do for say ordered pairs. However that is often the case: turning recursive definitions into explicit definitions
may not be possible.<|endoftext|>
TITLE: Decomposing tensor products of irreducible representations of reductive groups over a finite field
QUESTION [11 upvotes]: Let $G$ be a reductive group over a finite field (i.e. finite groups over lie type). The case I am most interested in is $G=GL_{n}(\mathbb{F}_{q})$; other classical groups are also interesting I think. 
Deligne-Lusztig theory has a lot to say about the irreducible representations and characters of these groups. For $G=GL_n(\mathbb{F}_q)$, Green's paper from the 1940's gives the characters explicitly also. The following question I guess, is in part a reference request, since the question has probably been examined in the literature somewhere, but I am unable to find a reference. 
Question: Let $V$ and $W$ be two irreducible representations of $G$. What can be said about the decomposition of $V \otimes W$ into irreducibles? Specifically:

Are there any special cases of $V$ for which the decomposition of $V \otimes W$ into irreducibles can always be explicitly determined? (for instance, with the symmetric group $S_n$, there is some theory which does this for the regular representation of dimension $n-1$, and also I believe work which does this for representations corresponding to two-row partitions). 
Is there anything that can be said for the decomposition of $V \otimes V$ in general?
What about, if $V$ and $W$ are not actually irreducibles, but instead representations obtained from $l$-adic cohomology; for instance the virtual representation $R_{T, \theta}$ is defined as alternating sums of various cohomological representations. As an example, consider the representation of $G$ acting on the $i$-th cohomology of the Deligne-Lusztig variety $X_{T}$ corresponding to a fixed torus $T$; if we tensor together two different cohomological representations corresponding to different tori, and cohomology for different values of $i$, what can we say? Since the $R_{T, \theta}$ are defined as alternating sums of these, perhaps this question will help with our original problem.

REPLY [8 votes]: Both Tamas and Victor have pointed to the best current literature for general linear groups or others of Lie type, but maybe it's useful to add a series of small comments to supplement their answers.
1) The finite general linear groups, like their classical counterparts, are by far the best-behaved groups of Lie type for representation theory (both ordinary and modular) and related combinatorics.   Even so, all aspects of the theory involve deep ideas and indirect approaches as Lusztig's work over many decades demonstrates.  Moreover, even related groups like $GL, PGL, SL, PSL$ have character theories with varying degrees of difficulty.  While Green's 1955 paper provides an algorithmic way to work out characters of finite general linear groups, the special linear groups have required much more sophisticated methods developed first by Deligne-Lusztig.    It's usually not easy for groups of Lie type so say clearly what it means to "know" the characters of the group. 
2) For any finite group, tensor products of irreducible representations (or products of characters) can be arbitrarily difficult to work out in detail even after character tables are in hand.    It's much harder for groups of Lie type, where knowledge of characters is usually algorithmic.   The best hope of getting uniform answers is to ask about tensor products of "generic" irreducibles, which predominate when both the underlying prime $p$ and its power $q$ grow large.
Deligne-Lusztig theory constructs virtual characters, but "most" of these tend to be genuine characters and already show nice patterns in their distribution into series correlated with the sizes of classes in the Weyl group.   Thus for $PGL_2(\mathbb{F}_q)$ roughly half of the characters have degree $q+1$ (principal series) and half have degree $q-1$ (discrete series).   At the other extreme are the Steinberg character of degree $q$ and the trivial character (these are the "unipotent" characters).   
3) As Victor points out, Lusztig's summary in rank 1 shows indirectly how to answer a natural qualitative question: in a "typical" tensor product of two irreducible characters, how often do principal series and discrete series characters appear?   (Answer: about equally often.)   Similar questions for other groups get much more complicated to settle. 
4) Lusztig's character results for Lie types B, C, D show how dramatically the complexity increases, even though the results can be organized combinatorially.   To study tensor products requires asking the right questions.
5) In the original question, Vinoth comments: but I'm not really interested in exceptional groups of Lie type (for those, this problem should be a standard mindless computation).    Actually, to say anything interesting about tensor products here would require a creative approach.   For example, what is shown by Lusztig about the characters of $E_8(q)$ is subtle and computationally not so easy to work with.   There are 166 unipotent characters, whose degrees are polynomials in $q$, but roughly a third of them fail to occur as constituents of the character induced from the trivial character of a Borel subgroup.   Knowing these characters is essential to producing the full character table, etc.      
6) Since the Steinberg character occurs at the other extreme from "generic" characters, it's interesting to ask how its product with other irreducible characters will decompose.    This has echoes in modular representation theory, as seen in recent work by Hiss and Zalesski, and is difficult to sort out even for type A.   Lots of questions out there about tensoring.<|endoftext|>
TITLE: moduli space and modularity
QUESTION [5 upvotes]: I recently realized some kind of analogy when considering modularity results (such as the modularity of elliptic curves over Q). The analogy comes from algebraic groups. Take one point (say, the origin) of an algebraic group, then something over the one point (for example, a tangent vector) can be extended to the whole algebraic group by group translations (so we get a translation invariant vector field). Now, modular curves are moduli spaces for elliptic curves. One elliptic curve is like one point on the moduli space, so probably things over that one point (for example, the first etale cohomology, which is the same as the Tate module Galois representation) could be extended to the whole modular curve, and indeed, that is the modular form, which also lives in some cohomology of the modular curve.
The modularity results have always been very mysterious and surprising for me, since it links two very "natural" and "intuitive" but "seems-far-away" objects together (the Tate module, and a modular form). But the above point of view might be a good reason for such things to be true.
This leads me to think more about the general theory of moduli spaces, which I don't know very much. It seems that there is a quite well developped theory of moduli spaces of curves with fixed genre, and also there is some special kind of moduli spaces such as Hilbert moduli space. There is also a very important moduli space in number theory,namely, moduli space of p-divisible groups, which recently has an important work by Mark Kisin. But it seems to be that the techniques and ideas used in Kisin's work (of course as well as in Breuil and other people's work) look quite different from that of traditional geometrical moduli spaces. 
So my question is, can someone give some motivation about studying p-divisible groups and their moduli spaces? Also, is the study of such objects analogus to that of the traditional moduli spaces?
references: 
Breuil  ''Groupes p-divisibles, groupes finis et modules filtrés'', Annals of Math. 151, 2000, 489-549.
http://www.ihes.fr/~breuil/PUBLICATIONS/p-divisibles.pdf
Kisin, Moduli of finite flat group schemes and modularity -- Annals of Math. 170(3) (2009), 1085-1180.
http://www.math.harvard.edu/~kisin/dvifiles/bt.dvi

REPLY [7 votes]: Kisin's work is fairly technical, and is devoted to studying deformations of Galois representations which arise by taking $\overline{K}$-valued points of a finite flat group 
over $\mathcal O_K$ (where $K$ is a finite extension of $\mathbb Q_p$).  
The subtlety of this concept is that when $K$ is ramified over $\mathbb Q_p$ (more precisely,
when $e \geq p-1$, where $e$ is the ramification degree of $K$ over $\mathbb Q_p$), there 
can be more than one finite flat group scheme modelling a given Galois represenation.
E.g. if $p = 2$ and $K = {\mathbb Q}\_2$ (so that $e = 1 = 2 - 1$), the trivial character
with values in the finite field $\mathbb F_2$ has two finite flat models over $\mathbb Z_2$;
the constant etale group scheme $\mathbb Z/2 \mathbb Z$, and the group scheme $\mu_2$
of 2nd roots of unity.
In general, as $e$ increases, there are more and more possible models.  Kisin's work shows that they are in fact classified by a certain moduli space (the "moduli of finite flat group schemes" of the title).  He is able to get some control over these moduli spaces, and hence prove new modularity lifting theorems; in particular, with this (and several other fantastic ideas) he is able to extend the Taylor--Wiles modularity lifting theorem to the context of arbitrary ramification at $p$, provided one restricts to a finite flat deformation problem.
This result plays a key role in the proof of Serre's conjecture by Khare, Wintenberger, and Kisin.
The detailed geometry of the moduli spaces is controlled by some Grassmanian--type structures that are very similar to ones arising in the study of local models of Shimura varieties.  However, there is not an immediately direct connection between the two situations.  
EDIT: It might be worth remarking that, in the study of modularity of elliptic curves,
the fact that the modular forms classifying elliptic curves over $\mathbb Q$ are themselves
functions on the moduli space of elliptic curves is something of a coincidence.
One can already see this from the fact that lots of the other objects over $\mathbb Q$ that are not elliptic curves are also classified by modular forms, e.g. any abelian variety
of $GL_2$-type.   
When one studies more general instances of the Langlands correspondence, it becomes increasingly clear that these two roles of elliptic curves (providing the moduli space,
and then being classified by modular forms which are functions on the moduli space) are independent of one another.
Of course, historically, it helped a lot that the same theory that was developed to study the Diophantine properties of elliptic curves was also available to study the Diophantine properties of the moduli spaces (which again turn out to be curves, though typically not elliptic curves) and their Jacobians
(which are abelian varieties, and so can be studied by suitable generalizations of many of the tools developed in the study of elliptic curves).   But this is a historical relationship between the two roles that elliptic curves play, not a mathematical one.<|endoftext|>
TITLE: Terminology: Is there a name for a category with biproducts?
QUESTION [5 upvotes]: Many people are familiar with the notion of an additive category. This is a category with the following properties:
(1)  It contains a zero object (an object which is both initial and terminal).
This implies that the category is enriched in pointed sets. Thus if a product $X \times Y$ and a coproduct $X \sqcup Y$ exist, then we have a canonical map from the coproduct to the product (given by "the identity matrix"). 
(2) Finite products and coproducts exist.
(3) The canonical map from the coproduct to the product is an equivalence.
A standard exercise shows this gives us a multiplication on each hom space making the category enriched in commutative monoids (with unit). 
(4) An additive category further requires that these commutative monoids are abelian groups.

I want to know what standard terminology is for a category which satisfies the first three axioms but not necessarily the last. 

I can't seem to find it using Google or Wikipedia. An obvious guess, "Pre-additive", seems to be standard terminology for a category enriched in abelian groups, which might not have products/coproducts.

REPLY [7 votes]: One name that I have seen used is semiadditive category.<|endoftext|>
TITLE: The difficulties in proving modularity lifting theorems over non-totally real fields
QUESTION [22 upvotes]: First of all, let me apologize in advance for the terseness of this question.
It seems that by now there are well-developed techniques (the "Taylor-Wiles-Kisin" method) for proving modularity lifting theorems over totally real fields.  For example, it is now known that many elliptic curves over totally real fields are modular.  I am curious: what exactly is the stumbling block to proving such results over non-totally real fields?  It seems to be common knowledge among experts that the usual techniques for comparing a universal deformation ring and a Hecke algebra in this situation break down quite badly, but I cannot find a reference for this in print.  It would be great if someone could illustrate the problem here.
I understand that there are some obvious difficulties.  For example, for $GL_2$ over an imaginary quadratic field, the locally symmetric spaces on which the relevant modular forms live is $SL_2(\mathbb{C})/SU(2)$, and arithmetic quotients of this are certainly not Shimura varieties.  I am asking about more fundamental obstructions to applying the Taylor-Wiles-Kisin method, i.e strange behavior in the Hecke algebra and the universal deformation ring...

REPLY [21 votes]: Note: This is a fairly precise and detailed question about an important but technical aspect of algebraic number theory.  My answer is written at a level that I think is appropriate for the question; it assumes some familiarity with the topic at hand.

The most basic difficulty is that there is not a map $R \rightarrow {\mathbb T}$ in general
(i.e. one typically doesn't know how to create Galois representations attached to automorphic forms).
The second difficulty is that in the TWK method, one must argue with auxiliary primes (the primes typically labelled $Q$), and show that as you add these primes, ${\mathbb T}$ grows in a reasonable way (basically, is free over $\mathcal O[\Delta_Q],$ where $\Delta_Q$ is something like the $p$-Sylow subgroup of $({\mathbb Z}/Q{\mathbb Z})^{\times}.)$
One shows this (or some variant of it) by considering the analogous queston about cohomology of the arithmetic quotients.  Suppose for a moment we are in the Shimura variety context, or perhaps the compact at infinity context.  Then it will be the middle dimensional cohomology that is of interest, and if we localize at a non-Eisenstein maximal ideal we might hope to kill all other cohomology.  Then we can replace a computation of middle dimensional cohomology by an Euler characteristic computation, and its easy to see that the Euler char. will multiply by $|\Delta_Q|$ when we add the auxiliary primes $Q$.
But in more general contexts, there won't be a single middle dimension in which the maximal ideal of interest is supported (even if it is non-Eisenstein), and computing Euler characteristics will just give $0$, which is not much use.  It's not clear that it's even true that adding the auxiliary primes forces the approriate growth of cohomology, and possible torsion in the cohomology just adds to the complication.
There is much current work, by various groups of researchers, with various different approaches, aimed at breaking this barrier.

I should add that one can now handle certain questions about non-totally real field,
say question related to conjugate self-dual Galois reps. over CM fields, because 
these are still related to a Shimura variety context.  This plays a role in the recent
progress on Sato--Tate for higher weight forms by Barnet-Lamb--Geraghty--Harris--Taylor
and Barnet-Lamb--Gee--Geraghty, and is also the basis for a recent striking theorem
of Calegari showing that if $\rho:G_{\mathbb Q} \to GL_2({\mathbb Q}_p)$ is ordinary
at $p$ and de Rham with distinct Hodge--Tate weights (and probably $\overline{\rho}$ should satisfy some technical conditions), then $\rho$ is necessarily odd!<|endoftext|>
TITLE: Canonical bases for modules over the ring of symmetric polynomials
QUESTION [10 upvotes]: The ring $S=\mathbb{C}[x_1,x_2,\dots,x_n]^{S_n}$ of symmetric polynomials has a number of commonly used bases, but the undisputed world champion of these is the basis consisting of Schur polynomials $s_\lambda$, where $\lambda$ ranges over non-increasing sequences $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n \geq 0$ of non-negative integers.  For a partition $\mu$ of $n$, let $V_\mu$ be the corresponding irreducible $S_n$-module, and let $M(\mu)=(\mathbb{C}[x_1,x_2,\dots,x_n] \otimes V_\mu)^{S_n}$ be the ($S$-module of) $S_n$-invariant polynomial functions on $\mathbb{C}^n$ with values in $V_\mu$.  Is there a $\mathbb{C}$-basis of $M(\mu)$ that deserves top billing?
(A bit of background: the dimensions of the homogeneous components of $M(\mu)$ can be computed from the exponents of $V_\mu$, that is, the degrees in which it appears in the coinvariant algebra.  There are combinatorial expressions known for these numbers---see e.g. Stembridge's paper "On the eigenvalues of representations of reflection groups and wreath
products", Pacific J. Math. 140 (1989), 353--396 and the references therein, but they are not obtained by writing down a particularly nice basis.)

REPLY [3 votes]: In my paper Cyclage, catabolism, and the affine Hecke algebra 
http://arxiv.org/abs/1001.1569
I exhibit a canonical basis for $\mathbb{C}[x_1,x_2,\dots,x_n]$ and more generally a canonical basis for $\mathbb{C}[x_1,x_2,\dots,x_n] \otimes V_\mu$ coming from the extended affine Hecke algebra of type A.  The subset of this canonical basis corresponding to cells of shape $(n)$ is a basis for the $S_n$-invariants in this module.  For the special case $M(n)$, these canonical basis elements do correspond to Schur functions -- they are Schur functions in the Bernstein generators times $e^+$, where $e^+$ spans the trivial representation for the finite Hecke algebra (see Theorem 6.1).
I have not thought too much about the combinatorics of the canonical basis for $M(\mu)$, but it may be possible to work this out explicitly, including an explicit description of the $S_n$ invariants (see Example 9.21 for a little bit about this).  This paper is long and may take some time to get through.  Feel free to contact me at the email address at the very bottom of the paper if you have any questions or want to discuss this in detail.<|endoftext|>
TITLE: Is there a computable model of ZFC?
QUESTION [35 upvotes]: Background
Assuming ZFC is consistent, then by downward Löwenheim–Skolem, there is a countable model (M,$\in$) of ZFC. Since the universe M is countable, we may as well think of it as actually being the set of natural numbers, so $\in$ will be some binary relation on the natural numbers.

Can such a relation ever be computable?

Partial results
One can show that the class of binary relations $R$ on the natural numbers such that $(\mathbb{N},R) \models ZFC$ forms a $\Pi_0^1$ class, and will be nonempty so long as ZFC is consistent. This already gives us some interesting results. For example, by the low basis theorem, there is a low $R$ such that $(\mathbb{N},R) \models ZFC$. But I have been unable to determine whether such a function can be made computable; the best I can do is show that if such a function is computable, then there is no effective way of finding, given a finite set D of natural numbers, the element n such that D={m : mRn}.

REPLY [15 votes]: This can also be done using Godel-Roesser instead of Tennebaum.  Suppose M is a  model of ZFC.  There is an element $a$ of that M believes is the set of Godel codes for the sentences true of the integers of M. Of course $a$ may contain nonstandard Godel codes, but if we let T be the set of standard sentences with Godel codes in $a$, then T will be a complete consistent extension of Peano Arithmetic.  If M is computable, then T would be computable, but there are no computable complete consistent extensions of Peano Arithmetic.
Of course the appeal to Godel-Roesser just hides Joel's argument about recursive inseparable r.e. sets.<|endoftext|>
TITLE: How does one get the short exact sequence in a two-column spectral sequence?
QUESTION [8 upvotes]: In a two-column double complex, one gets from the associated spectral sequence short exact sequences $0\to E_2^{1,n-1}\to H^n\to E_2^{0,n}\to 0$, where $H^n$ is the cohomology of the total complex, but I have never seen the construction of this sequence. Any text I've seen merely states it as a fact, or leaves it as an exercise which I have had no luck trying to solve. Can anyone give a construction or good reference?

REPLY [14 votes]: This follows precisely from the very definition of convergence of the spectral sequence, once one has identified the $\infty$-term. It is done with some details in McLeary's User Guide---which is, in my opinion, a very good reference for both the technicalities and the pragmatics of dealing with spectral sequences.
Now, if you are starting with a two column double complex (as opposed to starting with an arbitrary double complex whose spectral sequence has two contiguous columns), you can get the short exact sequences very much 'by hand'. 
Indeed, suppose your double complex is $T^{\bullet,\bullet}=(T^{p,q})_{p,q\geq0}$ and that $T^{p,q}\neq0$ only if $p\in\{0,1\}$. If we define complexes $X^\bullet$ and $Y^\bullet$ with $X^q=T^{0,q}$ and $Y^q=T^{1,q}$, with differentials coming from the vertical differential $d$ of $T^{\bullet,\bullet}$, then the horizontal differential of $T^{\bullet,\bullet}$ can be seen as a map of complexes $\delta:X^\bullet\to Y^{\bullet}$.
Now, in the spectral sequence induced by the filtration by columns we clearly have $E_1^{0,q}=H^q(X^\bullet)$, $E_1^{1,q}=H^q(Y^\bullet)$ and the differential on the $E_1$ page is induced by the horizontal differential in $T^{\bullet,\bullet}$. In other words, the $E_1$ page is more or less the same thing as the map $H(\delta):H(X^\bullet)\to H(Y^\bullet)$. It follows that we have short exact sequences $$0\to E_2^{0,q}\to H^q(X^\bullet)\xrightarrow{H^q(\delta)} H^q(Y^\bullet)\to E_2^{1,q}\to 0,$$ and the spectral sequence dies at the second act for degree reasons.
On the other hand, there is a short exact sequence of complexes $$0\to Y[-1]^\bullet\to\mathrm{Tot}\;T^{\bullet,\bullet}\to X^\bullet\to 0,$$
from which we get a long exact sequence $$\cdots\to H^{q-1}(Y^\bullet)\to H^q(\mathrm{Tot}\; T^{\bullet,\bullet})\to H^q(X)\to H^q(Y)\to\cdots,$$ in which you can compute directly that the map $H^q(X)\to H^q(Y)$ is precisely $H^q(\delta)$. Since the first four-term exact sequence identifies for us the kernel and the cokernel of $H^q(\delta)$, exactness of the second exact sequence provides the short exact sequence $$0\to E_2^{1,q-1}\to H^q(\mathrm{Tot}\; T^{\bullet,\bullet})\to E_2^{0,q}\to 0$$ that you wanted.
(It is an extremely instructive exercise to try to see what can one do in this spirit for a three-column double complex, and fighting with this is a great prelude to an actual exposition to spectral sequences...)<|endoftext|>
TITLE: Meaning of orientation/orientability over rings other than the integers
QUESTION [11 upvotes]: This was asked as part of an earlier question. But since this part did not attract many answers, I am asking it separately.
We consider the homology definition of an orientation for a manifold, as you define fundamental class., ie as some generator of some homology modules, satisfying some compatibility conditions. See for instance the book of Greenberg and Harper. What does it mean to say that a manifold is orientable, over rings other than $\mathbb Z$? 
It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable here, and has a unique orientation. And thus you can do Poincare duality, etc.. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring for homology to be $\mathbb{Z}/5\mathbb{Z}$?
Maybe it is just a formalism; maybe we do not really have to bother about orientations except the ones given by $+1$ and $-1$ in a ring, and the rest are just matters of additional generators giving some extra vacuous information. But I keep wondering. I hope somebody can clarify.

REPLY [9 votes]: I just wanted to mention that while orientability for cohomology with arbitrary coefficients is governed solely by cohomology with coefficients in ℤ, there are other cohomology theories for which is is not true.  For example, if you have an action of $\pi_1(X)$ on an abelian group M, then you can talk about (co)homology with twisted coefficients in M.  For any vector bundle there is a coefficient module such that the bundle is orientable with respect to these twisted coefficients (or, to paraphrase Matthew Ando, "every bundle is orientable if you're twisted enough").
Also, one can ask whether a vector bundle is orientable with respect to topological K-theory, real or complex, or many other generalized cohomology theories, which capture interesting information about the manifold.
So while ℤ/m-coefficients may not be the most interesting coefficient systems to study orientability in, they're part of a larger systematic family of questions (and they don't take much extra work if you're already doing ℤ and ℤ/2-coefficients).
Finally, for something like real coefficients you might think of orientations that differ by a real scalar geometrically, e.g. according to some volume form.<|endoftext|>
TITLE: Motivation for concepts in Algebraic Geometry
QUESTION [47 upvotes]: I know there was a question about good algebraic geometry books on here before, but it doesn't seem to address my specific concerns.
**
Question
**
Are there any well-motivated introductions to scheme theory?
My idea of what "well-motivated" means are specific enough that I think it warrants a detailed example.
**
Example of what I mean by well motivated
**
The only algebraic geometry books I have seen which cover schemes seem to leave out essential motivation for definitions.  As a test case, look at Hartshorne's definition of a separated morphism:
Let $f:X \rightarrow Y$ be a morphism of schemes. The diagonal morphism is the unique morphism $\Delta: X \rightarrow X \times_Y X$ whose composition with both projection maps $\rho_1,\rho_2: X \times_Y X \rightarrow X$ is the identity map of $X$.  We say that the morphism $f$ is separated if the diagonal morphism is a closed immersion.
Hartshorne refers vaguely to the fact that this corresponds to some sort of "Hausdorff" condition for schemes, and then gives one example where this seems to meet up with our intuition.  There is (at least for me) little motivation for why anyone would have made this definition in the first place. 
In this case, and I would suspect many other cases in algebraic geometry, I think the definition actually came about from taking a topological or geometric idea, translating the statement into one which only depends on morphisms (a more category theoretic statement), and then using this new definition for schemes. 
For example translating the definition of a separated morphism into one for topological spaces, it is easy to see why someone would have made the original definition.  Use the same definition, but say topological spaces instead of schemes, and say "image is closed" instead of closed immersion, i.e.
Let $f:X \rightarrow Y$ be a morphism of topological spaces. The diagonal morphism is the unique morphism $\Delta: X \rightarrow X \times_Y X$ whose composition with both projection maps $\rho_1,\rho_2: X \times_Y X \rightarrow X$ is the identity map of $X$.  We say that the morphism $f$ is separated if the image of the diagonal morphism is closed.
After unpacking this definition a little bit, we see that a morphism $f$ of topological spaces is separated iff any two distinct points which are identified by $f$ can be separated by disjoint open sets in $X$.  A space $X$ is Hausdorff iff the unique morphism $X \rightarrow 1$ is separated.
So here, the topological definition of separated morphism seems like the most natural way to give a morphism a "Hausdorff" kind of property, and translating it with only very minor tweaking gives us the "right notion" for schemes.
Is there any book which does this kind of thing for the rest of scheme theory?
Are people just expected to make these kinds of analogies on their own, or glean them from their professors?
I am not entirely sure what kind of posts should be community wiki - is this one of them?

REPLY [21 votes]: Dear Steven, I think Mumford's notes of the mid 60's, the first ever explaining schemes to ordinary mortals, are still the closest to what you want. They have become a book in 1988: The Red Book of Varieties and Schemes, published by Springer (LNM 1358).
After a first chapter on classical algebraic varieties, Mumford introduces schemes by quoting Felix Klein [in the 1880's!] and amazingly commenting "It is interesting to read Felix Klein describing what to all intents is nothing but the theory of schemes". 
And then Mumford brilliantly motivates the necessity of schemes and their nilpotents for a more refined study of varieties. He illustrates his text with wonderful little drawings, among which his great picture of $Spec \mathbb Z [X]$,  still admired today. For example  Lieven le Bruyn has a series of very interesting articles in his blog "Never ending books" based on that drawing (and as a  bonus you can see both the picture of  $Spec \mathbb Z [X]$ and that of Mumford...):
http://www.neverendingbooks.org/index.php/mumfords-treasure-map.html
PS Although it is the exact opposite of what you are asking for (!), let me mention that conversely the notion of proper map in Algebraic Geometry seems to have influenced Bourbaki's point of view on proper maps in General(= point-set) Topology. He defines them as universally closed maps and, almost as an afterthought, mentions that in the case of locally compact spaces they are characterized by the property that compact subsets have compact inverse images .<|endoftext|>
TITLE: Understanding iterated integrals
QUESTION [18 upvotes]: I have encountered iterated integrals on papers dealing with multizeta values, polylogarithms etc.. Since then I am trying to figure out the motivations and purpose of the theory.
It seems the defintions and methods go back to K.-T. Chen. The integrals seem to converge like an exponential series. He published many papers on this topic. Some of these(as seen in his collected works) seem to relate to path spaces, loops spaces etc., and their homology/cohomology. Many notions in algebraic topology seem to be carried out in this context using the analytic tool of iterated integral. He calls it a "de Rham theoretical approach" to the fundamental group, etc.. Is this a "de Rham homotopy theory"? Are we able to capture topological properties by repeated integration? In particular I have in mind the "iterated path integrals" paper of K.-T. Chen in mind. There are lots of others too, and some of them are in the Annals; so one cannot question the mathematical importance of the topic.
I am sorry for asking a vague question. I am a beginner on a topic struggling to understand the concepts and motivations behind them. I will be grateful for any pointers towards more understanding, so that I can get started.

REPLY [11 votes]: One topological significance of these iterated integrals is that they can be used to model the bar complex on the de Rham cochains of a manifold, and thus in some sense the "de Rham complex" of the manifold's loop space since $H_*(Bar(\Lambda^* X)) \cong H^*(\Omega X)$ [Here I am using $\Lambda$ for the de Rham complex and $\Omega X$ to denote loops].  For a simply-connected space, this bar complex gives a reasonable way to encode the rational homotopy type, though Sullivan models have been more popular.  So in particular, one can understand a complete set of homotopy functionals (that is, the linear dual of homotopy groups) using iterated integrals: Milnor and Moore showed that $\pi_*(X) \otimes {\mathbb Q}$ is isomorphic to the Lie algebra of indecomposibles of $H_*(\Omega X; {\mathbb Q})$; thus the linear dual of homotopy is the "Lie coalgebra of coindecomposibles" of the rational loopspace cohomology of $X$ (again given by Chen integrals if you wish).  Chen showed early on that these integrals give information about $\pi_1$ as well (through its nilpotent completion).  
My coauthor Ben Walter and I have developed a model for rational homotopy types which is close to both the Chen and Quillen models (and compatible with the Sullivan model as well), and clarified the story of functionals on homotopy groups as well.  Because of Chen's work, we know that we will have plenty of information to mine in the non-simply connected setting.<|endoftext|>
TITLE: definition of the end of a manifold?
QUESTION [7 upvotes]: I was hoping if somebody could help me out with the terminology. I've found that the "end of a manifold" is a function assigning to each compact set K a conected component e(K) of the complement of K. I am trying to understand Gompf's article "three exotic $R^{4}$ 's and other anomalies" and he quotes a theorem of Freedman (Corollary 1.2 in "The topology of 4-dimensional manifolds) saying "Any open 4-manifold M with $\pi_{1}(M)$, $H_{1}(M)$ and end collared (topologically) by $S^{3}\times R$ is homeomorphic to $R^{4}$". How can a function be homeomorphic to something? Im interpreting this as end meaning the "hypothetical boundary" of the manifold. Thanks in advance.

REPLY [9 votes]: As algori points out in the comments, the definition of end may be found here.  Also there you will find the definition of the neighborhood of an end.  
When he says "end collared (topologically) by $S^3 \times R$" he means that the end has a neighborhood homeomorphic to $S^3 \times R$.  Since he's assuming that $M$ has only one end, this simply means that there is a compact set whose complement is homeomorphic to $S^3 \times R$ (as Mariano said in the comments).
For future reference, it's pretty common in the literature to blur the distinction between an end and a neighborhood of an end.  Especially if when there are neighborhoods of the end that are products, as is the case here.<|endoftext|>
TITLE: Moshe Rosenfeld's Salmon Problem
QUESTION [7 upvotes]: As an amusement at the start of this talk, Moshe Rosenfeld poses the following question.

Suppose that there are n salmon which
  begin at distinct points on a unit
  circle, each facing either clockwise
  or counterclockwise. On a signal, each salmon moves around the circle in its chosen direction at a constant speed (the same for all salmon).  When two salmon meet, they both instantly reverse directions.  If any salmon ever returns to its starting point, it dies. (If two salmon meet at one of their starting points, there is a death and no change of direction; as Rosenfeld says, "Death comes first.")

Is it true that all the salmon will eventually die?
(assuming the answer to part 1 is yes) Give an algorithm to find the last survivor.


I spent a certain amount of time on buses and planes tinkering with this. It's quite easy to show that every configuration is preperiodic, as a start. I have some ideas about how one might finish. Eventually I decided just to look for more information on the problem, with no real success. 
One of the themes of his talk is how some problems become popular and some gather dust on the shelf.  Is the latter what happened to this problem?  
His second question is a bit mysterious.  The problem setup itself is algorithmic in nature, so what does it mean?  Is there anything besides "elegance" that would distinguish the kind of answer we should have in mind from a stupid answer like "just watch the salmon"?  ("Running time" could be an answer, but it seems likely that just letting the salmon swim wouldn't take all that long.)
I am really asking three subquestions on this topic.


Did this question ever get solved or taken up seriously?  If so, where?
Is there a natural, nonvacuous interpretation of the second part of
  the question?
What is the solution?  (This is actually the subquestion I am least
  interested in, but it felt wrong not
  to ask it.)


(Please feel free to re-tag, still getting used to things here.)

REPLY [3 votes]: Ah, irony.  Now that I've publicly asked the question, I practically trip over a reference.
I just found this paper of Rosenfeld from 2008, which has large overlap with the talk mentioned in my original post.  In the paper it is shown that there are initial configurations which do not lead to extinction, though little progress is made on how one might recognize these special configurations in advance.
(I'm still interested in answers to subquestion 2 above.)<|endoftext|>
TITLE: references for models of stable infinity categories
QUESTION [5 upvotes]: There's a fair amount of literature comparing different models for the homotopy theory of homotopy theories, or the homotopy theory of $(\infty,1)$-categories.  Julie Bergner has a survey of this literature at http://www.math.ucr.edu/~jbergner/OneInfty.pdf. 
Does anyone know if similar comparisons have been made for different models of stable $(\infty,1)$-categories?  In particular, have $A_{\infty}$-categories (in characteristic 0) been compared to the stable infinity categories that are conceived of as infinity categories with extra properties?  (For example, to stable quasicategories?)

REPLY [6 votes]: There seems to be a little confusion in your question about what stable ($\infty$,1)-categories are, so I want to address that first. 
The definition of a stable infinity category that I'm familiar with is Jacob Lurie's notion (definition 29 DAGI:Stable Infty-Categories). The prototypical examples of stable infinity categories which this definition is designed to capture are the category of spectra (in the topological setting) and the category of chain complexes in an abelian category (in the algebraic setting). Rather this notion is designed to also capture the derived localization of the category of chain complexes, which is a robust version of the derived category. 
The advantage of Jacob's succinct definition is that it is expressed categorically. This means that if you have equivalent notions of ($\infty$, 1)-category, then they will yield equivalent notions of stable infinity category. So in this sense Julie Bergner's is also comparing stable infinity categories. 
The only other model of stable ($\infty$,1)-categories that I know which doesn't quite fit this idea but is close is that of stable model categories. These are to stable ($\infty$,1)-categories as ordinary model categories are to ordinary ($\infty$,1)-categories. That is they are roughly equivalent to a particularly nice class of stable ($\infty$,1)-categories.
Your question also asks about $A_\infty$-categories as stable ($\infty$,1)-categories, and that is the part which is confusing me. 
On the one hand, an infinity category can heuristically be defined as a category with topological spaces of morphisms and where composition is associative only up to higher coherence. All the various notions of $\infty$-category make this precise in one way or another. Since you mention characteristic zero, I take you are thinking of the algebraic/chain complex version of $A_\infty$-category. This can be related to topological $A_\infty$-categories via the Dold-Kan correspondence (assuming your complexes are connective).
One the other hand being enriched in chain complexes is like being enriched in topological abelian groups (or in HZ-spectra in the non-connective case). So maybe the concept you are after is that of spectral category, or category enriched in spectra? I know this has been studied, but I'm lean on references.
One last comment. Later in DAGI, Lurie shows that stable infinity categories are automatically enriched in the stable infinity category of spectra. This is totally analogous to the category of chain complexes in an abelian category being enriched in the category of chain complexes of abelian groups. However, just as being enriched in abelian groups is not enough for your category to be abelian, being enriched in spectra is not enough for your infinity category to be stable.<|endoftext|>
TITLE: Maximal Ellipsoid
QUESTION [7 upvotes]: John's Theorem can be stated as "To every compact, convex body, there is a unique inscribed ellipsoid, whose volume is maximal among all inscribed ellipsoids." It goes on to classify this maximal ellipsoid. By using this theorem, one can prove that the ellipsoid of maximal volume which is contained in a square is a circle. 
This strikes me as a problem which was probably studied well before Fritz John, and yet I have been unable to prove the statement about squares and circles in an elegant, but low-brow manner. Any thoughts?

REPLY [12 votes]: Here's an attempt at a low-brow proof. Take a max area ellipse. Apply an affine transform to
make it a circle; then the problem becomes to show that a minimal area parallelogram containing a circle is a square. It is easy to see that both the height of the parallelogram
and its base are at least the diameter. Q.E.D.<|endoftext|>
TITLE: Changing coordinates so that one Riemannian metric matches another, up to second derivatives
QUESTION [7 upvotes]: Let $g$ and $g'$ be two $C^2$-smooth Riemannian metrics defined on neighborhoods $U$ and $U'$ of $0$ in $\mathbb R^2$, respectively.  Suppose furthermore that the scalar curvature at the origin is $K$ under both metrics.  
My question:  Is there a coordinate transformation taking one metric to the other, such that they agree up to second derivatives at the origin?  i.e., if $x : U \to U'$ is the transformation, we have
$g_{ij}' = g_{ab} ~x_i^a x_j^b$,
evaluating everything at $0$; there are similar equations for the first and second derivatives.  Clearly this is false if the scalar curvatures aren't equal.  I don't care what happens away from the origin.
In the excellent thread When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?, Greg Kuperberg says:If remember correctly, there is a more general result due to somebody, that any two Riemannian manifolds are locally isometric if and only if their curvature tensors are locally the "same".If "local isometry" means that the metrics are equal on a neighborhood of the origin, then the metrics I have in mind are not locally isometric, since the only information I have is that their curvatures match at one point.  
Edit:  I'm pretty sure that Deane answered my question, but let me clarify.  Let $g_{ij}$ be some "reasonable" metric, e.g. a bump surface metric, and consider a point $p$ where the scalar curvature is $K$.  Let $g_{ij}'$ be an arbitrary metric on a neighborhood $U$ of the origin in $\mathbb R^2,$ with scalar curvature $K$ at $0$.
Then the question becomes:  does there exist a coordinate change on the bump surface such that the equation $g_{ij}'(0) = g_{ab}(p) ~x_i^a x_j^b$ is satisfied, as well as the corresponding equations for the first and second derivatives?  That is, there are $18$ pieces of pertinent information 
$(\*)~~~g_{11}', g_{12}', g_{22}'; g_{11,1}', g_{12,1}', g_{22,1}', g_{11,2}', g_{12,2}', g_{22,2}'; g_{11,11}', g_{12,11}', g_{22,11}', g_{11,12}', g_{12,12}', g_{22,12}', g_{11,22}', g_{12,22}', g_{22,22}'$.
I want to change coordinates on my nice surface such that the metric and its derivatives line up with $(\*)$.

REPLY [2 votes]: As far as I understand, your question and its natural generalizations belongs to the very developed theory (now it is a part of singularity theory, I think). This theory was started by Tresse:
Tresse, A., Sur les Invariants Differentiels des Groupes Continus des Transformations, Acta Mathematica, 1894, vol.18, pp.1-88.
See also an introduction and references to  S. Dubrovskiy's paper "Moduli space of symmetric connections" http://arxiv.org/abs/math/0112291<|endoftext|>
TITLE:  limsup and liminf for a sequence of sets
QUESTION [8 upvotes]: how does limsup and liminf for a sequence of sets, apply to probability theory. any real world examples would be much appreciated

REPLY [3 votes]: Here's another simple example, in a similar vein as has2's above. Let $X_n$ be a sequence of independent, identically-distributed exponential variables, i.e., $$\mathbb P(X_n > u) = e^{-\lambda u},$$ for some real $\lambda > 0$.  Let $E_n$ be the event that $X_n > n$, and let $E = \limsup E_n$, that is, the event $E$ occurs if there's an infinite (random) subsequence $n_k$ such that $X_{n_k} > n_k$.
We compute $$\sum \mathbb P(E_n) = \sum e^{-\lambda n} < \infty,$$ thus by the Borel-Cantelli lemma, $E$ has probability zero.  With probability one, there exists a (random) number $N$ such that for all $n \ge N$, $X_n \le n$.
Let's analyze this graphically.  Make a plot with the horizontal axis representing time $n$, and the vertical axis $x = X_n$.  Draw the line $x = n$.  For small times $n$, these random points might jump above the line $x = n$.  But the argument above shows that there is some (random) time $N$ after which the points $X_n$ all lie below the line $x = n$.<|endoftext|>
TITLE: adding an n-th root to Q_p
QUESTION [9 upvotes]: What can be said about extensions à la $\mathbb{Q}_p(\sqrt[n]{a})/\mathbb{Q}_p$? Ramification behaviour, valuation ring, ...? 
I find it hard to say anything general - for example, as a function of the $p$-adic valuation of $n$ and/or $a$. Of course some special cases are rather easy to handle, and I understand what happens when $v_p(a) = 0$. This might be a hard question, or a question for which there is a standard reference - I didn't find one - or something rather easy, in which case I'm just missing something.

REPLY [5 votes]: It's possible to delve more deeply into the ramification structure in the more complicated wild case referred to in Emerton's answer. If $n=p^m$, then Coleman calculated the conductor of the Kummer extension 
${\mathbb Q}_p(\zeta_n,\sqrt[n]{a})/{\mathbb Q}_p(\zeta_n)$ 
using his reciprocity law and one should be able to use this to determine the ramification filtration of 
${\mathbb Q}_p(\sqrt[n]{a})/\mathbb{Q}_p$.
Romyar Sharifi does just that 
here
to determine the ramification groups of the maximal (nonabelian) Kummer extension ${\mathbb Q}_p\left(\sqrt[p^{\infty}]{{\mathbb Q}_p^\times}\right)/\mathbb{Q}_p$.
In particular, one finds that the (upper) ramification jumps are of the form $i$, $i+\frac{1}{p-1}$ and $i+\frac{1}{p(p-1)}$ for integers $i$.
In a similar vein and also by computing conductors, Viviani 
finds the ramification groups of the extension 
${\mathbb Q}_p(\zeta_{p^m},\sqrt[p^m]{a})/{\mathbb Q}_p$.
as long as $p^2$ doesn't divide $a$.
One trick he uses is to notice that for instance if $p$ exactly divides $a$ then
$$\frac{(1-\zeta_p)}{\sqrt[p]{a}\;\sqrt[p^2]{a}\cdots\sqrt[p^m]{a}}$$
is a uniformizer of 
${\mathbb Q}_p(\zeta_p,\sqrt[p^m]{a})$ 
and observes that the proofs could be simplified and generalized if one was able to write down uniformizers in further extensions.
So does anyone know how to do this ? Can one explicitly write down a uniformizer for the field 
${\mathbb Q}_p(\zeta_{p^m},\sqrt[p^m]{a})$ in a similar way ?
At a stretch, this might have some application in integral $p$-adic Hodge theory. The arithmetically profinite extension $K(\zeta_{p^\infty},\sqrt[p^\infty,]{\pi})/K$ where $\pi$ is a prime in $K$ makes an appearance in Tong Liu's extension of Breuil and Kisin's work and the more one knows about this extension and it's field of norms the better for calculations/proofs.<|endoftext|>
TITLE: Group Structure on CP^infinty
QUESTION [16 upvotes]: I was inspired by the following algebraic topology orals question:
"Is $S^1$ the loop space of another space?"
This is easy to see if you recognize that $S^1$ is a $K(\mathbb{Z},1)$, and the loop space of any $K(G,n)$ is a $K(G,n-1)$. 
I then also remembered that the loop space functor is a functor from pointed topological spaces and continuous maps to the category of H-spaces and continuous homomorphisms. H-spaces being topological spaces that satisfy the axioms of a group up to homotopy (see Spanier, Chapter 1, Section 5).
I have three questions:

Is there a useful criterion for when an H-space is actually a topological group?
Seeing that $S^1$,$S^3$, and $S^7$ are the only spheres that support group structures, it doesn't seem coincidental that $S^1$ is a loop space, because it is in fact an H-space. Since $CP^{\infty}$ is the loop space of $K(Z,3)$ it too is an H-space, but is it known if it is a topological group? 
Even if not, is there a way (other than concatenation of loops) to "see" this structure on $CP^{\infty}$?

Thanks!

REPLY [3 votes]: (Re: 2 & 3)
Since $\operatorname{Sym}^n\mathbb CP^1=\mathbb CP^n$ («by Viète's formulas»), $\mathbb CP^\infty$ is the free abelian monoid generated by $\mathbb CP^1=S^2$ (with some fixed point as a unit). (This is exatly the operation «representing» tensor product of $U(1)$-bundles aka addition in $H^2$ on $[-,\mathbb CP^\infty]=\operatorname{Bun}_{U(1)}(-)=H^2(-;\mathbb Z)$ — cf. other answers.)
Unfortunately, this operation lacks (stict) inverse. But by Dold-Thom theorem $\widetilde{\mathbb Z[S^2]}:=\mathbb Z[S^2]/\mathbb Z[pt]$ (the free abelian group generated by $S^2$ with some (fixed) point as a unit) has homotopy type of $K(\mathbb Z,2)$. Moreover, the natural map $\mathbb CP^\infty=\operatorname{Sym}^\infty(S^2)\to\widetilde{\mathbb Z[S^2]}$ is a homotopy equivalence.
(And any $K(G,n)$ can be made an abelian topological group in analogous way: $\widetilde{G[S^n]}$ has desired homotopy type.)
/* Essentially x-posted from math.SE */<|endoftext|>
TITLE: Is the inertia stack of a Deligne-Mumford stack always finite?
QUESTION [7 upvotes]: Let X be a DM stack over a field k. We follow the definition in Laumon and Moret-Bailly's book, so that its diagonal is quasi-compact (and hence diagonal is of finite type). Then is the diagonal necessarily finite?
Edit: I meant to ask if the inertia $I\to X$ is finite. Recall that the inertia stack $I$ is defined to be the 2-fiber product of $X$ with $X$ over $X\times X,$ where the two maps $X\to X\times X$ are both the diagonal map. This question is equivalent (I think) to the following. Let $G\to S$ be an etale $S$-group scheme of finite type, where $S$ is a $k$-scheme of finite type. Then $G$ is finite over $S.$

REPLY [4 votes]: The inertia stack of a separated Deligne Mumford stack is finite. In general algebraic stacks with finite inertia have coarse moduli space. Some references:

Gerd Faltings and Ching-Li Chai, Degeneration of abelian varieties, Ergebnisse der Mathematik und ihrer
Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], vol. 22, Springer-Verlag, Berlin, 1990.
Sean Keel and Shigefumi Mori, Quotients by groupoids, Ann. of Math. (2) 145 (1997), no. 1, 193-213.
Dan Abramovich, Martin Olsson, and Angelo Vistoli, Tame stacks in positive characteristic, Ann. Inst. Fourier (Grenoble) 58 (2008), no. 4, 1057-1091.<|endoftext|>
TITLE: Teaching and students
QUESTION [13 upvotes]: Sometimes I get stumped by students' questions in my classes I teach. I am an algebraist by training and have just started teaching. Sometimes I have to teach analysis courses. My question is: Is it normal to get stumped by questions from students in fields that are not your expertise?
Are there any ways to prevent this from happening? Or does it just come from teaching experience?

REPLY [19 votes]: Pete Clark has done a nice job answering Rachel J's question. I'll reiterate one of his points with an anecdote of my own. When I began teaching at the undergraduate level 20 years ago, I asked one of my professors for general advice. She said, "If a student asks you a question you can't answer straightaway, say, 'I don't know. But this is how I would go about answering your question.'" It is quite natural to be "stumped." It may be helpful to think of the classroom as a place of learning rather than as a place of instruction. One of the ways students learn is by observing someone who has a bit more experience answer difficult questions.<|endoftext|>
TITLE: Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $
QUESTION [32 upvotes]: EDIT, 9 March 2014: when I asked this in 2010, I did not have the courage of my convictions, and so did not ask for an if and only if proof, as Kevin Buzzard quite properly pointed out. Such problems are now somewhat known open problems, as I  told them to some experts in 2011. Probably the easiest of the bunch: It is easy to describe a set of integers that are not represented by $4 x^2 + 2 x y + 7 y^2 - z^3;$ I even sent that in as a Monthly problem (December 2010, problem 11539), only one solver, Robin Chapman(December 2012). Open problem: can we prove that the polynomial integrally represents every other number? There is a similar open problem for each discriminant of positive binary quadratic forms with class number three, including the other direction for Kevin Buzzard's answer below.
ORIGINAL: The following problem is my variant of something Irving Kaplansky noticed when we worked together. I do not think it is by nature a difficult problem, it is simply too hard for me to finish.
Suppose we have an integer  $C > 0$ that is not divisible by 2 or 3, while there is another integer  $F > 0$ such that 
$ 27 C^2 - 23 F^2 = 4.$
For any integers $x,y,z,$ 
is it true that $  2 x^2 + x y + 3 y^2 + z^3 - z \neq  C  $ and
 $  2 x^2 + x y + 3 y^2 + z^3 - z \neq  -C  ,$ or together
 $  2 x^2 + x y + 3 y^2 + z^3 - z \neq \pm C  $ ?
I have proved it for the four smallest values of $C,$ those being 
 1,  599,  14951,  9314449. The case $C=1$ comes directly from the 
Hudson and Williams paper below. Note that the even values of $C$ fail miserably, they seem to all be values of $ z^3 - z.$ The polynomial $ g(x,y,z) = 2 x^2 + x y + 3 y^2 + z^3 - z $
represents every other number $n$ with 
$ -10,000,000 \leq n \leq 10,000,000,$ according to my computer.
The Spearman and Williams article (see below) is explicit class field theory, not a topic I know.
I should point out that, in retrospect, what I proved for the four smallest (odd) $C$ seems to amount to the statement that $z^3 - z + C$ is irreducible $\pmod q$ for any prime $ q =  2 u^2 + u v + 3 v^2.$
(27 January 2010) I finally got smart and decided to do a part check of the "irreducible" version. For the next three values of $C,$ those being 232488049, 144839681351, 3615189146999, I factored $z^3 - z + C \pmod q$ and found it to be irreducible for primes $q  < 1000$ and    $ q =  2 u^2 + u v + 3 v^2.$
The two main references are:
Blair K. Spearman and Kenneth S. Williams,
    "The Cubic Conguence $x^3 + {A} x^2 + {B} x + {C} \equiv 0 \bmod p $
             and Binary Quadratic Forms",
    Journal of the London Mathematical Society,
    volume 46,
    1992,
 pages 397-410
Richard H. Hudson and Kenneth S. Williams",
 "Representation of primes by the principal form of
             discriminant $-{D}$ when the classnumber $h(-{D})$ is $3$",
 Acta Arithmetica,
    volume 57,
   1991,
    pages 131-153.
One needs this Lemma: if an integer $n$ has an integer representation as 
$ n = 2 x^2 + x y + 3 y^2,$ then $n$ is divisible by some prime 
$ q = 2 u^2 + u v + 3 v^2.$
Everything I know about this problem is in pdf's at (Feb. 2018):
http://zakuski.utsa.edu/~jagy/inhom.cgi
including a proof of the preceding Lemma in jagy_division.pdf 
and the proof for the four smallest $C$ in jagy_conjecture_23.pdf
and a list of intimately related problems in jagy_list.pdf
.
I welcome individual responses to this along with posted answers or comments. One of my email addresses can be found using the search feature at 
http://www.ams.org/cml

REPLY [34 votes]: EDIT: Hendrik Lenstra emailed me a proof of Conjecture 2. I'll append it below. So Jagy's question is now solved.

OK so I think that Jagy wants to make the following conjecture:
CONJECTURE 1: an integer $C$ is not representable by the form F(x,y,z)=2x^2+xy+3y^2+z^3-z if, and only if, $C$ is odd and $27C^2-4=23D^2$ with $D$ an integer.
[EDIT/clarification: Jagy only asks one direction of the iff in his question, and this answer below gives a complete answer to the question Jagy asks. I came back to this question recently though [I am writing this para a year after I wrote the original answer] and tried to fill in the details of the argument in the other direction (proving that if C was not an odd integer solution to $27C^2-4=23D^2$ then $C$ was represented by the form) and I failed. So the "hole" I flag in the answer below still really is a hole, and this post still remains an answer to Jagy's question, but not a complete proof of Conjecture 1, which should still be regarded as open.]
I have a proof strategy for this. I am too lazy to fill in some of the details though, so maybe a bit of it doesn't work, but it should be OK. However, I am also reliant on a much easier-looking conjecture (which I've tested numerically so should be fine, but I can't see why it's true):
CONJECTURE 2: if $C$ is odd and $27C^2-4=23D^2$, then there's no prime p
dividing D of the form $2x^2+xy+3y^2$.
So I am claiming Conj 2 implies the "only if" version of Conj 1. I don't know how to prove Conj 2
but it looks very accessible [edit: I do now; see below]. Note that the Pell equation is related to units
in $\mathbf{Q}(\sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization
in $\mathbf{Q}(\sqrt{-23})$. I've seen other results relating the arithmetic
of $\mathbf{Q}(\sqrt{D})$ and $\mathbf{Q}(\sqrt{-3D})$.

Ok, so assuming Conjecture 2, let me sketch a proof of the "only if" part of Conjecture 1.
The Pell equation is intimately related to the recurrence relation
$$t_{n+2}=25t_{n+1}-t_n$$
with various initial conditions. For example the positive $C$s which
are solutions to $27C^2-4=23D^2$ are all generated by this recurrence
starting at $C_1=C_2=1$, and the $D$s are all generated by the same
recurrence with $D_1=-1$ and $D_2=1$. Note that $C_n$ is even iff $n$
is a multiple of 3, and (by solving the recurrence explicitly) one
checks easily that $C_{3n}=(3C_{n+1})^3-(3C_{n+1})$, so we've represented
the even solutions to the Pell equation as values of $F$ (with $x=y=0$).
Let's then consider the odd solutions to the Pell equation. Say $C$
is one of these. We want to prove that there is no solution in
integers $x,y,z$ to 
$$2x^2+xy+3y^2=z^3-z+C.$$
Let's do it by contradiction. Consider the polynomial $Z^3-Z+C$. First
I claim it's irreducible. This is because it is monic, of degree 3,
and has no integer root, because $C$ is odd. Next I claim that
the splitting field contains $\mathbf{Q}(\sqrt{-23})$. This is
because of our Pell assumption and the fact that the discriminant
of $Z^3-Z+C$ is $4-27C^2$. Next I claim that the splitting
field of $Z^3-Z+C$ is in fact the Hilbert class field of
$\mathbf{Q}(\sqrt{-23})$. I only know an ugly way of seeing this:
if $\theta$ is a root of $Z^3-Z+1=0$ then I know recurrence relations
$e_n$, $f_n$ and $g_n$ (all defined using the relation above but with
different initial conditions) with $e_n\theta^2+f_n\theta+g_n$ a root of
$Z^3-Z+C_{3n+1}$, and other relations giving roots of $Z^3-Z+C_{3n+2}$
and $Z^3-Z-C_{3n+1}$ and $Z^3-Z-C_{3n+2}$. Most unenlightening but it
does the job because it embeds $\mathbf{Q}(\theta)$ into the splitting
field, and the Galois closure of $\mathbf{Q}(\theta)$ is the Hilbert
class field of $\mathbf{Q}(\sqrt{-23})$. 
Right, now for the contradiction, assuming Conjecture 2. Let's assume
that $C$ is a solution to the Pell, and $z^3-z+C$ can be written $2x^2+xy+3y^2$.
Now $C$ is odd so $z^3-z+C$ isn't zero, and hence it's positive,
so it's the norm of a non-principal ideal~$I$ in the integers $R$ of
$\mathbf{Q}(\sqrt{-23})$. This ideal $I$ is a product of prime ideals,
and $I$ isn't principal, so one of the prime ideals had better also not
be principal. Say this prime ideal has norm $p$. We conclude that $p$
divides $z^3-z+C$ and $p$ is of the form $2x^2+xy+3y^2$. Note in
particular that this implies $p\not=23$. Also $p\not=3$, because $C$
is odd and (because of general Pell stuff) hence prime to 3.
CASE 1: $p$ is coprime to $D^2$ (with $27C^2-4=23D^2$). In this
case the polynomial $Z^3-Z+C$ has non-zero discriminant mod $p$
(because $p\not=23$) and furthermore has a root $Z=z$ mod $p$.
Hence mod $p$ the polynomial either splits as the product of a linear
and a quadratic, or the product of three linears. This tells us
something about the factorization of $p$ in the splitting field
of $Z^3-Z+C$: either $p$ remains inert in $\mathbf{Q}(\sqrt{-23})$,
or it splits into 6 primes in the splitting field and hence splits
into two principal primes in $\mathbf{Q}(\sqrt{-23})$ (because the
principal primes are the ones that split completely in the Hilbert
class field). In either case $p$ can't be of the form $2x^2+xy+3y^2$,
so this case is done.
CASE 2: This is simply Conjecture 2.
In both cases we have our contradiction, and
so we have proved, so far, assuming Conjecture 2, that a solution $C$ to $27C^2-4=23D^2$
is representable as $2x^2+xy+3y^2+z^3-z$ iff it's even. 
Note that Conjecture 2 can be verified by computer for explicit values
of $C$, giving unconditional results---for example I checked in just
a few seconds that any odd $C$ with $|C|<10^{72}$ and satisfying the
Pell equation was not representable by the form, and that result
does not rely on anything. At least that's something concrete for Jagy.

OK so what about the other way: say $27C^2-4$ is not 23 times a square.
How to go about representing $C$ by our form? Well, here I am going to
be much vaguer because there are issues I am simply too tired to deal
with (and note that this is not the question that Jagy asked anyway).
Here's the idea. Look at the proof of Theorem 2 in Jagy's pdf Mordell.pdf.
Here Mordell gives a general algorithm to represent certain integers
by (quadratic in two variables) + (cubic in one variable). If you
apply it not to the form we're interested in, but to the following
equation:
$$x^2+xy+6y^2=z^3-z+C$$
then, I didn't check all the details, but I convinced myself that they
could easily be checked if I had another hour or two, but I think that
the techniques show that whatever the value of $C$ is, this equation
has a solution. The idea is to fix $C$, let $\theta$ be a root of
the cubic on the right (which we can assume is irreducible, as if it
were reducible then we get a solution with $x=y=0$), to rewrite the right
hand side as $N_{F/\mathbf{Q}}(z-\theta)$, with $F=\mathbf{Q}(\theta)$
and now to try and write $z-\theta$ as $G^2+GH+2H^2$ with
$G,H\in\mathbf{Z}[\theta]$. Mordell does this explicitly (in a slightly
different case) in the pdf. The arguments come out the same though,
and we end up having to check that a certain cubic in four variables
has a solution modulo~23 with a certain property. I'll skip the painful
details. The cubic depends on $C$ mod 23, and so a computer calculation
can deal with all 23 cases.
Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$,
so we have written $z^3-z+C$ as the norm of a principal ideal in
the integers of $\mathbf{Q}(\sqrt{-23})$. What we need to do now is
to write it as the norm of a non-principal ideal, and of course we'll
be able to do this if we can find some prime $p$ dividing $z^3-z+C$
which splits in $\mathbf{Q}(\sqrt{-23})$ into two non-principal
primes, because then we replace one of the prime divisors above $p$
in our ideal by the other one. What we need then is to show that
if the discriminant of $z^3-z+C$ is not $-23$ times a square,
then there is some prime $p$ of the form $2x^2+xy+3y^2$ dividing
some number of the form $z^3-z+C$ which is the norm of a principal
ideal. This should follow from the Cebotarev density theorem, because
Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$
which are "only constrained modulo 23", and so one should presumably
be able to find a prime which splits in $\mathbf{Q}(\sqrt{-23})$,
splits completely in the splitting field of $z^3-z+C$ and doesn't
split completely in the splitting field of $z^3-z+1$. I have run out
of energy to deal with this point however, so again there is a hole here.
This issue seems analytic to me, and I am not much of an analytic guy.
[edit: I came back to this question a year later and couldn't do it,
so this should not be regarded as a proof of the "if" part of Conj 1]

EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes
Conjecture 2. 
(EDIT: dollar signs added - GM)
Fact. Let  $\theta$  be a zero of  $X^3-X+1$,  let  $\eta$ in ${\bf Z}[\theta]$  be
a zero of  $X^3-X+C$  with  $C$ in $\bf Z$  odd,  and let  $p$  be a prime
number that is inert in  ${\bf Z}[\theta]$.  Then  $p$  does not divide
index$({\bf Z}[\theta]:{\bf Z}[\eta])$.
Proof. By hypothesis,  ${\bf Z}[\theta]/p{\bf Z}[\theta]$  is a field of size  $p^3$.
Let  $e$  be the image of  $\eta$  in that field. Since  $X^3-X+C$  is
irreducible in  ${\bf Z}[X]$  (even mod 2), it is the characteristic
polynomial of  $\eta$ over $\bf Z$.  Hence its reduction mod $p$ is the
characteristic polynomial of  $e$  over  ${\bf Z}/p{\bf Z}$.  If now  $e$  is in
${\bf Z}/p{\bf Z}$,  then that characteristic polynomial also equals  $(X-e)^3$,
so that in  ${\bf Z}/p{\bf Z}$  we have  $3e = 0$  and  $3e^2 = -1$,  a contradiction.
Hence  $e$  is not in  ${\bf Z}/p{\bf Z}$,  so  $({\bf Z}/p{\bf Z})[e] = {\bf Z}[\theta]/p{\bf Z}[\theta]$,
which is the same as saying  ${\bf Z}[\theta] = {\bf Z}[\eta] + p{\bf Z}[\theta]$.  Then
$p$  acts surjectively on the finite abelian group  ${\bf Z}[\theta]/{\bf Z}[\eta]$,
so the order of that group is not divisible by  $p$.  End of proof.<|endoftext|>
TITLE: coarse moduli space of DM stacks
QUESTION [9 upvotes]: This is related to another one of my questions on DM stacks. In Brian Conrad's article 'The Keel-Mori Theorem via Stacks', a sufficient condition on for an Artin stack to have coarse moduli space is that it has finite inertia stack. This does not include DM stacks without finite inertia. My question is that, does every DM stack of finite type over a field have a coarse moduli space? And what's the reference? Thanks.

REPLY [13 votes]: No, not every DM-stack has a coarse moduli space. The following is a counter-example (see my paper on geometric quotients):
Let X be two copies of the affine plane glued outside the y-axis (a non-separated scheme). Let G=Z2 act on X by y → –y and by switching the two copies. Then G acts non-freely on the locally closed subset {y=0, x ≠ 0}. The quotient [X/G] is a DM-stack with non-finite inertia and it can be shown that there is no coarse moduli space (neither categorical nor topological) in the category of algebraic spaces.<|endoftext|>
TITLE: (nontrivial) isotrivial family of elliptic curves
QUESTION [8 upvotes]: I think it should be a standard procedure to construct such things, can anyone give a reference or give a hint? Can this be done over any base scheme?

REPLY [4 votes]: (Edit: added motivation for my "answer" even though it isn't exactly what was asked for. It shows the moduli space of genus 1 curves cannot be fine - see last paragraph.) 
The Hopf surface is a (non-algebraic) example of a non-trivial family of isomorphic elliptic curves. Here I use "elliptic curve" to mean "smooth complex curve of genus 1", which is probably not what you mean. (In particular my curves have no distinguished base point.)
The Hopf surface $X$ is a quotient of $\mathbb{C}^2\setminus0$ by the action of $\mathbb Z$ generated by $z \mapsto 2z$. Since this action commutes with the action of $\mathbb{C}^*$ there is a map $X \to \mathbb{CP}^1$. Each fibre is a copy of $\mathbb{C}\setminus\{0\}$ divided by the action $z\sim 2z$. So all the fibres are isomorphic elliptic curves. (If you want base points on each curve, this example doesn't work because the fibration $X \to \mathbb{CP}^1$ doesn't have a section.) On the other hand, its easy to see that $X$ is diffeomorphic to $S^1\times S^3$ and so you can't trivialise the fibration even topologically. Replacing $z\mapsto 2z$ by $z\mapsto \lambda z$ for $\lambda\in \mathbb{C}$ non-zero, I guess you can get any genus-1 curve to appear as all fibres of a topologically non-trivial family. 
Even though this is not eaxctly what you're looking for, I thought it worth giving the example anyway, because it's a very simple way to see that the moduli space of genus 1 curves cannot be fine. (Fine moduli spaces carry a universal family and all other families are pulled back from the universal family via the map to moduli space - in particular a family of isomorphic objects in such a moduli space is pulled back by the constant map and so  must be trivial. One often sees that objects cannot be parametrised by a fine moduli space by giving examples of special objects with "extra" automorphisms. The Hopf surface is an alternative to that approach in this situation.)<|endoftext|>
TITLE: (Z/n)^(I) is a direct summand of (Z/n)^I
QUESTION [5 upvotes]: Dear group theorists,
Let $n \geq 1$ and $I$ be an infinite set (you may assume $I$ to be countable). Is the abelian group $(\mathbb{Z}/n)^{(I)}$ (direct sum of copies $\mathbb{Z}/n$) a direct summand of $(\mathbb{Z}/n)^I$ (direct product of copies $\mathbb{Z}/n$)? This question is motivated by that one.
If $n$ is prime, this follows from Linear Algebra. Of course, this is not constructive at all. Thus it's also true when $n$ is squarefree (use the Chinese Remainder Theorem). What happens otherwise? The smallest example is $n=4$. I don't know how to start ...

REPLY [12 votes]: Yes.  The ring ${\mathbb Z}/n {\mathbb Z}$ is injective over itself, and over a Noetherian
ring, direct limits of injectives are again injective; thus ${\mathbb Z}/n{\mathbb Z}^{(I)}$
is injective over ${\mathbb Z}/n{\mathbb Z}$.  Finally, any embedding of an injective splits,
as follows directly from the property of being injective.  We can now apply this to
the embedding $({\mathbb Z}/n{\mathbb Z})^{(I)} \hookrightarrow ({\mathbb Z}/n{\mathbb Z})^I.$<|endoftext|>
TITLE: Examples of completions and algebraic closures
QUESTION [7 upvotes]: It is widely known that the algebraic closure of the $p$-adic completion $\mathbb{Q}_p$ of $\mathbb{Q}$ isn't complete anymore. It's completion is complete and known as $\mathbb{C}_p$. 
I have read in a book about non-archimedean analysis that in this case the process ends, which means that $\mathbb{C}_p$ is also algebraically closed.
My question is: is there an example of a field K, in which the algebraic closure $K^{alg}$ isn't complete, and the completion of $K^{alg}$ isn't algebraically closed ? And how do I construct such an example.

REPLY [7 votes]: No, there is not.
If the valuation is archimedean, by Ostrowski the field is isomorphic to the real or complex numbers, so the algebraic closure will already be complete.
If the valuation is non-archimedean, the completion of the algebraic closure will always be algebraically closed. See for example here: http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf

REPLY [3 votes]: There is a theorem of Kurschák which asserts that the completion of a valued algebraically closed fied is algebraically closed. This is proved in Paulo Ribenboim's The theory of classical valuations.<|endoftext|>
TITLE: What is Yoneda's Lemma a generalization of?
QUESTION [49 upvotes]: What is Yoneda's Lemma a generalization of?
I am looking for examples that were known before category theory entered the stage resp. can be known by students before they start with category theory.
Comments are welcome why the following candidates are good or bad ones.
Further examples are welcome!
Candidate #1: Axiom of extensionality (for sets)
A set is uniquely determined/can be recovered from its elements.
Candidate #2: Dedekind completions (for posets)
A completion of a poset S is the set of its downwardly closed subsets, ordered by inclusion. S is order-embedded in this lattice by sending each element x to the ideal it generates.
Candidate #3: Stone's representation theorem (for Boolean algebras)
Every Boolean algebra B is isomorphic to the algebra of clopen subsets of its Stone space S(B).
Candidate #4: Cayley's theorem (for groups)
Every group G is isomorphic to a subgroup of the symmetric group on G.

REPLY [2 votes]: Tannaka duality is essentially applying Yoneda twice. So a special case of Tannaka duality that doesn't require the Yoneda lemma would be Pontryagin duality.<|endoftext|>
TITLE: Reconstruction puzzles
QUESTION [6 upvotes]: [Added: This is a follow-up of an earlier post.]
Consider the following "reconstruction puzzle", stated informally:
Given a concrete poset, e.g. the poset of undirected unlabeled finite graphs without isolated vertices, ordered by embeddability (arrow heads, identities and composition omitted in the diagram):
    
 (source)
Now forget about the inner structure of the objects and consider only the corresponding abstract poset:
    
 (source)
The "reconstruction puzzle" is to reconstruct the inner structure of the objects unambiguously from their "positions" in the poset.
It's obvious what a solution of this puzzle is and that it can be solved "by hand" for at least some of the smaller objects, just carefully considering the in- and out-arrows.

Question #1: How can such a puzzle be
  stated formally?



Question #2: How can it be solved
  "algorithmically"?



Question #3: What's the mathematics
  behind this kind of puzzle?

I suppose it's not category theory, since category theory is not concerned with the inner structure of objects.


Question #4: Can it be shown - at least for this concrete
  example - that every object is
  reconstructible up to isomorphism?

Since for this concrete example there are no isomorphic objects we could have omitted "up to isomorphism".


Question #5: What's an obvious way to
  generalize this kind of puzzles (from posets to what?)

REPLY [3 votes]: For your question #5, this can be generalized to: given two categories $A$ and $B$, describe the functors $A\to B$. For a closer match, you can restrict to the functors which map irreducible maps (those that cannot be written as a composition of two non-identity maps) to irreducible maps. 
You example corresponds to letting $A$ be the poset you drew considered as a category, and $B$ the category of graphs and embeddings.  This provides a formal statement of your example.
You can generalize a bit differently by asking: given a category $A$, describe the pairs $(B,F)$ with $B$ a category and $F:A\to B$ a functor (preserving irreducibility of maps, if you want...). It would not be without interest to know of another such functor from your poset to a category which is not graphs---as that would, I think, give a category equivalent to (Finite graphs and embeddings).
From this point of view, we get the following answer to your question #3: this is just representation theory.
It is clear that your puzzle can be solved algorithmically (in so far as infinite puzzles can be...): starting from the root, and proceeding level by level (where levels are defined by counting vertices) just try assigning graphs to vertices, and backtracking when you hit an inconsistency. Provided you know the puzzle is solvable (and in this case you do know!)
As Harrison notes, your question #4 is equivalent to Harary's Set Reconstruction Conjecture: the levels up to 4 vertices you work out by hand, and then use the conjecture to check that a node in a level is determined by those in the level right below it from which there is an arrow coming in.<|endoftext|>
TITLE: Question about polynomials with coefficients in Z
QUESTION [16 upvotes]: Let $f = a_0 + a_1 x + \ldots + a_n x^n$ ($f \ne 0$), where $a_i \in \{-1, 0, 1\}$. Let $p(f)$ be the largest number such that $f(x)$ is divisible by $y$ for any integer $x$ and for any $1 \leq y \leq p(f)$. Let $g(n)=max_f\; p(f)$. Is it true that $g(n) = o(n)$? What is the best upper or lower bound on $g(n)$ can be derived?
For my application it would be great to prove that $g(n) = o(n)$ in order to obtain something non-trivial, or $g(n) = o(n^{2/5})$ in order to improve the best known result. Do you think it is real?
UPD It is an obvious consequence of Bertrand's postulate and Schwartz–Zippel lemma that $g(n) \leq 2n$. Using bruteforce I've got the following values:
$g(10) = 7$, $f = x^{10} + x^8 - x^4 - x^2$.
$g(15) = 10$, $f = x^{15} + x^{13} + x^{12} + x^{11} + x^{10} - x^7 - x^6 - x^5 - x^4 - x^3$.
$g(17) = 10$, $f = x^{16} + x^{15} + x^{14} + x^{13} + x^{12} + x^{11} - x^8 - x^7 - x^6 - x^5 - x^4 - x^3$.

REPLY [11 votes]: The point of this answer is to point out that Kevin Costello's heuristic can be made rigorous. For any positive $\epsilon$, if $y=O(n^{1/2-\epsilon})$ then such a polynomial exists for large $n$.
Lemma: Let $G$ be a finite abelian group and let $g_1$, $g_2$, ..., $g_n$ be elements of $G$. If $2^n > |G|$ then there are integers $\epsilon_i \in \{ -1, 0, 1 \}$, not all zero, such that $\sum \epsilon_i g_i =0$.
Proof: Consider the $2^n$ sums $\sum a_i g_i$ with $a_i \in \{ 0, 1 \}$. By the pigeonhole principle, two of these are equal. Subtracting them, we get the claimed relation. QED
Now, consider the abelian group
$$G:=\bigoplus_{k=1}^y (\mathbb{Z}/k)^{\oplus k}.$$
Let $g_i$ be the element of $G$ whose $k$-th component is $(0^i, 1^i, 2^i, 3^i, \ldots, (k-1)^i)$, for $i=0$, $1$, ..., $n$. The order of $G$ is $\exp( \sum k \log k) = \exp( O(y^2 \log y))$. So, if $y=O(n^{1/2-\epsilon})$, then $2^{n+1} > |G|$ and the lemma tells us that there are $\epsilon_i$ such that $\sum \epsilon_i g_i=0$. Then $\sum \epsilon_i x^i$ is the required polynomial.
There is a lot of slack in this argument, but Bjorn's argument shows that we can't improve the exponent of $n$ by tightening it.<|endoftext|>
TITLE: Essential theorems in group (co)homology
QUESTION [21 upvotes]: I'm trying to fill in the gaps in my understanding of group (co)homology and I'm wondering what are considered the "must know" theorems and concepts. I'm thinking of things along the lines of 

Hopf's formula - If $G$ has presentation $F/R$, then $H_2(G)=R \cap [F,F]/[F,R]$
If $G$ has torsion then $H_n(G)$ has no top dimension
$H_n = Tor_n$ so is the left derived functor of $\otimes$
$H^n = Ext ^n$ so is the right derived functor of $Hom$
If $G$ is discrete, then $H_n(G)=H_n(K(G,1))$

REPLY [8 votes]: Here's one which is key for calculations: Let $H$ be a subgroup of $G$ and $W_G(H) = N_G(H)/H$.  Then the restriction map $H^*(BG) \to H^*(BH)$ maps to the invariants $(H^*(BH))^{W_G(H)}$.  
When $H$ is abelian, its cohomology is well-known (polynomial tensor exterior) and thus the cohomology of $G$ is mapping to something which can in principal be computed by invariant theory. 
Follow this with Quillen's theorem that the sum of these maps over all abelian subgroups has kernel which contains only nilpotent elements, and special cases such as Milgram's theorem that this is injective for symmetric groups, and you have a powerful computational tool.
Also, here is a nice survey by Alejandro Adem (whose book with Milgram is a good reference, complementary in many ways to Brown's).  It is intended for a graduate student summer school audience:
http://www.math.uic.edu/~bshipley/ConMcohomology1.pdf

REPLY [3 votes]: It will be pretty nice to learn some Galois cohomology and Class field theory, so that you can see these machineries put to good use.<|endoftext|>
TITLE: What is a good way to think about a fundamental field on a principal G-bundle?
QUESTION [10 upvotes]: Let $\pi:P \rightarrow M$ be a principal $G$-bundle, and let $A \in \mathfrak{g}$, where the Lie algebra of $G$ is indicated. The fundamental field $A$# used to define connections is given by
$A$#$(p) := \frac{d}{dt}(\exp(At)p)|_{t=0}$.
$A$# is well defined since $e^{At}p$ can be regarded as a vector in $\pi^{-1}(\pi(p))$. Intuitively, I try to think of $A$# as the (vertical) direction of the displacement on the fiber generated by $A$.  
By the defining properties of principal bundles (in particular, the free action of $G$), we have $\{A$# $: A \in \mathfrak{g}\} \simeq \mathfrak{g} \simeq \mathcal{V}(p)$, where $\mathcal{V}(p)$ is the vertical subspace at $p$. Something like 
$A$#$(p) = \lim_t t^{-1}(e^{At}p - p) = A \cdot p$ 
would be (to me) a nice way of thinking about $A$#$(p)$, except that this is (at best) a formal equality. More precisely, $L_{\exp(At)}$ is the 1-parameter group of diffeomorphisms generated by $A$#, where $L_g$ denotes left multiplication by $g$.
My question: is there a better way of thinking about a fundamental field than (either the definition itself) or the notional equation above?

REPLY [2 votes]: I find that the notion of fundamental vector field is well defined not only for $G$-principal bundle but even for any $G$-manifold, i.e. a manifold with an action by a Lie group $G$.
About your notional equation, I would say that the fundamental vector fields effectively arise from an action of $\frak{g}$ on $M$. However some clarifications are needed.
Let $\Psi:M\times G\to M$ be a right action of a Lie group $G$ on a manifold $M$.
Let $\frak{g}$ be the Lie algebra of $G$, viewed as formed by the left invariant vectorfields on $G$.
Then there exists a unique map $\zeta^{\Psi}\equiv\zeta:X\in\frak{g} \mapsto $$\zeta_X\in \mathfrak{X}$$(M)$ such that $(T\psi)\circ(0_M+X)=\zeta_X\circ\Psi$, $\zeta_X$ and $0_M+X$ are $\Psi$-related, for any $X\in\frak{g}$.(Above $0_M$ denoted the zero vectorfield on $M$.)
For any $X\in\frak{g}$, the vector field $\zeta_X$ on $M$ is called the fundamental vectorfield corresponding to $X$ w.r.t. the right action $\Psi$.
The definition of $\zeta$ is well posed just because, for any $X\in\frak{g}$, the map $T\Psi\circ(0_M+X)$ is constant on the fibers of $\Psi$; and this holds being $\Psi$ a right action and $X$ a left invariant vectorfield. 
Obviously the following properties are satisfied:

$\zeta_{aX+bY}=a\zeta_X+b\zeta_Y,\zeta_{[X,Y]}=[\zeta_X,\zeta_Y]$, for any $a,b\in\mathbb{R}$, and $X,Y\in\frak{g}$,  i.e. $\zeta:\mathfrak{g} \to \mathfrak{X} (M)$ is a Lie algebra homomorphism;
$\zeta_X$ is complete and its $t$-time flow is $\Psi^{\exp{tX}}$, for any $X\in\frak{g}$ and $t\in\mathbb{R}$.

For an abstract Lie algebra $\frak{g}$, an action of $\frak{g}$ on a manifold $M$ is defined to be a Lie algebra homomorphism from $\frak{g}$ to $\frak{X}$$ (M)$.
In such a way for any right action $\Psi$ of a Lie group $G$ on $M$, we have that $\zeta^{\Psi}$ is an action on $M$ by the Lie algebra of $G$.<|endoftext|>
TITLE: Should Krull dimension be a cardinal?
QUESTION [30 upvotes]: A totally ordered finite set  $\quad \mathcal P_0 \varsubsetneq \mathcal P_1\varsubsetneq  \dots \mathcal \varsubsetneq \mathcal P_n \quad$ of prime ideals of a ring $A$ is said to be a chain of length $n$.
As is well known, the supremum of the lengths of such chains is called the Krull dimension $\dim(A)$ of the ring $A$.
If the lengths of these chains are not bounded, the ring is said to be infinite dimensional: $\dim(A)=\infty$.This can happen, surprisingly, even for a Noetherian ring $A$.
But in the infinite dimensional case we could consider arbitrary totally ordered subsets $\Pi \subset Spec(A)$ of prime ideals, their cardinality $card(\Pi)$ and then take the sup of all those cardinals. Let us call this sup  the cardinal Krull dimension of  the ring $A$.
An equality $\dim(A)=\aleph$ would then be a more quantitative measure of the infinite dimensionality of $A$ than just $\dim(A)=\infty$
My question is whether results are known related to that cardinal Krull dimension. For example: for X a topological space, has the cardinal Krull dimension of
 $\mathcal C(X)$ (the ring of continuous functions on $X$) been calculated? I don't find this trivial, even for $X=\mathbb R$. There are obvious variants of this question concerning rings of differentiable functions on manifolds, etc.
Thanks in advance for any information on this topic.

REPLY [3 votes]: I realize this question was asked some years ago, but some coauthors and I recently published a paper on this topic in J. of Algebra: "An infinite cardinal-valued Krull dimension for rings". You may find it on my webpage:
https://www.uccs.edu/goman/sites/goman/files/inline-files/JA%20Submission%20-%20Loper%2C%20Mesyan%2C%20Oman%20%28second%20revision%29.pdf<|endoftext|>
TITLE: References for logarithmic geometry
QUESTION [15 upvotes]: Hi everyone,
I'm looking for a systematical introduction to (or treatment of) logarithmic structures on schemes. I am reading Kato's article ("Logarithmic structures of Fontaine-Illusie") at the moment, but I would like to have a more detailed source that goes through or gives an overview of the constructions of classical scheme theory that have analogs in the log-setup.
Are there any articles/books that in your opinion are required reading if I want to learn about log-geometry? What are beautiful examples of applications of this machinery?

REPLY [4 votes]: Danny Gillam at Brown also has some nice notes on his webpage.<|endoftext|>
TITLE: When does collection imply replacement?
QUESTION [22 upvotes]: In ordinary membership-based set theory, the axiom schema of replacement states that if $\phi$ is a first-order formula, and $A$ is a set such that for any $x\in A$ there exists a unique $y$ such that $\phi(x,y)$, then there exists a set $B$ such that $y\in B$ if and only if $\phi(x,y)$ for some $x\in A$.  That is, $B$ is the "image" of $A$ under the "definable class function" $\phi$.
The related axiom schema of collection modifies this by not requiring $y$ to be unique, but only requiring $B$ to contain some $y$ for each $x$ rather than all of them.  However, there are at least two different versions of this.

If for all $x\in A$ there exists a $y$ with $\phi(x,y)$, then there exists a set $B$ such that for all $x\in A$ there is a $y\in B$ with $\phi(x,y)$ (this is Wikipedia's version; I'll call it "weak collection").
If for all $x\in A$ there exists a $y$ with $\phi(x,y)$, then there exists a set $B$ such that (1) for all $x\in A$ there is a $y\in B$ with $\phi(x,y)$, and (2) for all $y\in B$ there is an $x\in A$ with $\phi(x,y)$ (I'll call this "strong collection").

The third possibly relevant axiom is the axiom schema of separation, which states that for any $\phi$ and any set $A$ there exists a set $B\subseteq A$ such that $x\in B$ if and only if $x\in A$ and $\phi(x)$.
I know the following implications between these axioms:

Strong collection implies weak collection, since it has the same hypotheses and a stronger conclusion.
Strong collection implies replacement, since it has a weaker hypothesis and the same conclusion.
Replacement implies separation (assuming excluded middle): apply replacement to the formula "($\phi(x)$ and $y=\lbrace x\rbrace$) or ($\neg\phi(x)$ and $y=\emptyset$)" and take the union of the resulting set.
Together with AC and foundation, replacement implies weak collection: let $\psi(x,V)$ assert that $V=V_\alpha$ is the smallest level of the von Neumann hierarchy such that there exists a $y\in V_\alpha$ with $\phi(x,y)$, apply replacement to $\psi$ and take the union of all the resulting $V_\alpha$.
Weak collection and separation together imply strong collection: separation cuts out the subset of $B$ consisting of those $y$ such that $\phi(x,y)$ for some $x\in A$.

My question is: does weak collection imply replacement (and hence also separation and strong collection) without assuming separation to hold a priori?  Feel free to assume all the other axioms of ZFC (including $\Delta_0$-separation).  I'm fairly sure the answer is "no," but several sources I've read seem to assume that it does.  Can someone give a definitive answer, and ideally a reference?

REPLY [33 votes]: The answer is no, if you allow me to adopt some weak-but-equivalent forms of the other axioms. And the reason is interesting: 

A shocking number of the axioms of set theory are true in the non-negative real line R+, with the usual order < being used to interpret set membership. (!) 

Let's just check. For example, Extensionality holds, because if two real numbers have the same predecessors, then they are equal. The emptyset axiom holds, since there are no non-negative reals below 0. The Union axiom holds, since for any real x, the reals less than x are precisely the reals that are less than something less than x. (Thus, every set is its own union.) A weakened version of the Power set axioms holds, the Proper Power set, which asserts that for every x, there is a set p whose elements are the strict substs of x. This is because for any real number x, the reals below x are precisely the reals y (other than x), all of whose predecessors are less than x. (Thus, every real is its own proper power set.) An alternative weakening of power set would say: for every x, there is p such that y subset x implies y in p. This is true in the reals by using any p > x. A weakened pairing axiom states similarly: for every x,y, there is z with x ε z and y ε z. This is true in the reals by using any z above both x and y. The Foundation axiom is no problem, since 0 is in every nonempty set. Also, similarly AC holds in the form about families of disjoint nonempty sets, since this never occurs in this model. The Weak Collection Axiom holds since if every y < x has phi(x,y,w), then in fact any y in the same interval will work (since this structure has many automorphisms), and so we may collect witness with any B above x and w. Note that Separation fails, since, for example, {0} does not exist in this model. Also, Replacement fails for the same reason.
Similar interesting models can be built by considering the structure (ORD,<) built using only ordinals, or the class { Vα | α in ORD }. These also satisfy all of the weakened forms of the ZFC axioms without Separation, using Collection in place of Replacement.
Thus, part of the answer to your question is that it depends on what you mean by the "other axioms of ZFC".
Apart from this, however, let me say that the term Weak Collection is usually used to refer to the axioms that restrict the complexity of the formulas in the usual Collection scheme, rather than the axiom that you state. For example, in Kripke Platek set theory KP, a weak fragment of ZFC, one has collection only for Sigma_1 formulas, and this is described as a Weak Collection axiom. (What you call Weak Collection is usually just called Collection.) And there is a correspondingly weakened version of Separation in KP.
But I am happy to adopt your terminology here. You stated that AC plus Replacement implies Weak Collection, but this is not quite right. You don't need any AC. Instead, as your argument shows, what you need is the cumulative Vα hierarchy, which is built on the Power set axiom, not AC. That is, If you have Replacement and Power set and enough else to build the Vα hierarchy, then you get Collection as you described, even if AC fails. For example, ZF can be axiomatized equivalently with either Collection or Replacement.
Your question is a bit unusual, since usually Separation is regarded as a more fundamental axiom than Replacement and Collection, and more in keeping with what we mean by set theory. After all, if one has a set A and a property phi(x), particularly when phi is very simple, it is one of the most basic set theoretic constructions to be able to form { x in A | phi(x) }, and any set theory violating this is not very set-like. We don't really want to consider models of set theory where many instances of Separation fail (for example, Separation for atomic formula is surely elemental).
Incidentally, there is another version of a weakening of Collection in the same vein as what you are considering. Namely, consider the scheme of assertions, whenever phi(x,y) is a property, that says for every set A, there is a set B such that for every a in A, if there is b with phi(a,b), then there is b in B such that phi(a,b).

OK, let me now give a positive answer, with what I think is a more sensible interpretation of your question. I take what I said above (and Dorais's comment) to show that we shouldn't consider set theories where the Separation Axiom fails utterly. Rather, what we want is some very weak set theory, such as the Kripke Platek axioms, and then ask the relationship between Weak Collection and Replacement over those axioms. And here, you get the postive result as desired.
Theorem. If KP holds, then Weak Collection implies Replacement.
Proof. Assume KP plus (Weak) Collection. First, I claim that this is enough to prove a version of the Reflection Principle, since that proof amounts to taking successive upward Skolem hulls, which is what Collection allows. That is, I claim that for every set x and any formula phi, there is a transitive set Y such that x in Y and phi(w) is absolute between Y and the universe V. This will in effect turns any formula into a Delta0 formula using parameter Y. 
Applying this, suppose we have a set A and every a in A has a unique b such that phi(a,b). By the Reflection Principle, let Y be a large set transitive containing A such that phi(a,b) and "exists b phi(a,b)" are absolute between Y and V. So Y has all the desired witnesses b for a in A. But also, now { b | exists a in A phi(a,b) } is a Delta0 definable subset of Y, since we can bound the quantifier again by Y. So the set exists. So Replacement holds. QED
I think we can get away with much less than KP. Perhaps one way to do the argument is to just prove Separation by induction on the complexity of formulas. One can collect witnesses by (weak) Collection, and this turns the formulas into lower complexity, using the new bound as a bound on the quantifiers.<|endoftext|>
TITLE: Dual of $\mathbb Z^I$ for uncountable $I$
QUESTION [7 upvotes]: Let $I$ be an infinite set. There is a homomorphism of abelian groups $\mathbb{Z}^{(I)} \to \hom(\mathbb{Z}^I,\mathbb{Z})$ which sends the basis element $e_i$ to the projection $p_i$. If $I$ is countable, it's a famous result of Specker1 that this is actually an isomorphism. But what happens when $I$ is uncountable?

Clearly it is injective. Surjectivity means that $\phi \in \hom(\mathbb{Z}^I,\mathbb{Z})$ is determined by the values $\phi(e_i)$ and that these values vanisch for almost all $i$. I can't copy the proof for the countable case.
1 Ernst Specker, Additive Gruppe von Folgen ganzer Zahlen, Portugaliae Math. 9 (1950), 131-140. MR0039719 (12,587b)

REPLY [9 votes]: With regard to Mariano's answer, I believe some clarification is in order. A closely related question was asked by Michael Barr and answered by user Ralph here. In brief, the homomorphism named in Martin's question is in fact an isomorphism, provided that $I$ has cardinality less than the first measurable cardinal. 
Shelah and Strüngmann (accessible here) refer to this result as well, using the same source given by Ralph, just before Definition 2.1: 

For generalizations to products of larger cardinalities and the resulting definition of slenderness for abelian groups we refer to [EM] or [F1] 

where [EM] is the text by Eklof and Mekler. It seems that Shelah and Strüngmann are talking about something slightly different: homomorphisms out of free complete products (but using a notation which could unfortunately suggest direct products).<|endoftext|>
TITLE: Variants of Waring's problem
QUESTION [13 upvotes]: Waring's problem (previously asked about here) asks, for each integer $k \ge 2$, what is the smallest integer $g(k)$ such that any positive integer can be written as a sum of $g(k)$ kth powers.  We have $g(2) = 4$, $g(3) = 9$, etc.  A (harder) variant asks what the smallest integer $G(k)$ is such that all sufficiently large integers can be written as a sum of $G(k)$ kth powers.  
I have two related questions:

What is known if we relax the condition ``any positive integer'' and only require a positive-density subset?  More precisely, we look for the smallest $g'(k)$ for which there is some $S \subset \mathbb{Z}_{>0}$ of positive density such that any $x \in S$ can be written as $g'(k)$ $k$th powers.  Then we have $g'(2) = 3$, while $G(2) = 4$; and $g'(3) = 4$, while it is only known that $4 \le G(3) \le 7$.  Is anything known about $g'(k)$ for k = 4,5, or larger?
For fixed k, is there an efficient algorithm that, given n, writes n as a sum of $g(k)$ kth powers?  What about decomposing n into the minimal number of kth powers for that n?   (Here `efficient' means polynomial in log(n).)  

Edit: Wikipedia says that ``In the absence of congruence restrictions, a density argument suggests that G(k) should equal k + 1.''  So perhaps this is the answer to (1)?

REPLY [15 votes]: Regarding question 1, it is easy to prove that $g'(k) \ge k$, and the expectation is that $g'(k)=k$ for all $k \ge 3$.  But the equality is an open question even for $k=3$.  See Deshouillers, Jean-Marc; Hennecart, François; Landreau, Bernard; Sums of powers: an arithmetic refinement to the probabilistic model of Erdös and Rényi.  Acta Arith. 85 (1998), no. 1, 13-33, and other papers by these authors.<|endoftext|>
TITLE: A topological consequence of Riemann-Roch in the almost complex case
QUESTION [18 upvotes]: This question originated from a conversation with Dmitry that took place here
Is there a complex structure on the 6-sphere?
The Hirzebruch-Riemann-Roch formula expresses the Euler characteristic of a holomorphic vector bundle on a compact complex manifold $M$ in terms of the Chern classes of the bundle and of the manifold. On the other hand, for complex manifolds the $\bar\partial$ operator is a differential (i.e. it squares to zero) and hence, the complex of sheaves of smooth $(p,q)$ forms for a fixed $p$ equipped with the $\bar\partial$ differential is a resolution of $\Omega^p$; this is a complex of soft sheaves and so, by taking global sections we can compute the cohomology of $\Omega^p$. Moreover, since the de Rham complex resolves the constant sheaf, the alternating sum of the Euler characteristics of $\Omega^p$'s is the Euler characteristic of $M$.
For an arbitrary pseudo-complex manifold the only part of the above that makes sense is the "right hand side" of the Riemann-Roch formula (i.e. the one involving Chern classes) and the (topological) Euler characteristic of the manifold itself. So it seems natural to ask whether the relation between the two that is true in the complex case remains true in the almost complex case. In other words, is it true that for a compact almost complex manifold $M$ of dimension $2n$ we have $$\chi(M)=\sum_{p=0}^n (-1)^p\sum_{i=0}^{n\choose{p}}\mathrm{ch}_{n-i}(\Omega^p)\frac{T_i}{i!}$$ where $\chi$ is the topological Euler characteristic, $\Omega^p$ is the $p$-th complex exterior power of the cotangent bundle (i.e., the complex dual of the tangent bundle), $\mathrm{ch}$ is the Chern character and $T_i$ is the $i$-th Todd polynomial of $M$?
In general, are there topological consequences of the existence of the Dolbeault resolution that would be difficult to prove (or, more ambitiously, would fail) for arbitrary pseudo-complex manifolds?

REPLY [3 votes]: I just wanted to point out how this question is related to (spin${}^c$) Dirac operators and their indicies since this was alluded to in the comments to the question.

Let $(M, g)$ be an $2n$-dimensional closed Riemannian manifold. Given a spin${}^c$ structure, one can form the complex spin${}^c$ bundles $\mathbb{S}^+_{\mathbb{C}}$ and $\mathbb{S}^-_{\mathbb{C}}$. For any hermitian vector bundle $E \to M$, there is a twisted spin${}^c$ Dirac operator $D^c_E : \Gamma(\mathbb{S}^+_{\mathbb{C}}\otimes E) \to \Gamma(\mathbb{S}^-_{\mathbb{C}}\otimes E)$ which has index
$$\operatorname{ind}(D^c_E) = \int_M\exp(c_1(L)/2)\operatorname{ch}(E)\hat{A}(TM)$$
where $L$ is the complex line bundle associated to the spin${}^c$ structure.
Suppose now that $M$ admits an almost complex structure $J$ and $g$ is hermitian. Then there is a canonical spin${}^c$ structure which has associated line bundle $L = \det_{\mathbb{C}}(TM)$, so $c_1(L) = c_1(M)$. Using the fact that $\exp(c_1(M)/2)\hat{A}(TM) = \operatorname{Td}(M)$, the index becomes
$$\operatorname{ind}(D^c_E) = \int_M \operatorname{ch}(E)\operatorname{Td}(M).$$
Moreover, $\mathbb{S}^+_{\mathbb{C}} \cong \bigwedge^{0,\text{even}}M$ and $\mathbb{S}^-_{\mathbb{C}} \cong \bigwedge^{0, \text{odd}}M$.
When $J$ is integrable and $E$ is holomorphic, $D^c_E = \sqrt{2}(\bar{\partial}_E + \bar{\partial}^*_E)$ and the above equality becomes the statement of the Hirzebruch-Riemann-Roch Theorem. In particular, if $E = \Omega^p$, then $D^c_p := D^c_{\Omega^p}$ has index $\operatorname{ind}(D^c_p) = \chi(M, \Omega^p)$.
In the general case, Hirzebruch defined $\chi^p(M) := \operatorname{ind}(D^c_p)$ and introduced the Hirzebruch $\chi_y$ genus $\chi_y(M) = \sum_{p=0}^n\chi^p(M)y^p$. With this notation, your question is whether or not the equality $\chi_{-1}(M) = \chi(M)$ holds. Using Chern roots $x_i$ of $TM$, one finds that
$$\chi_y(M) = \int_M\prod_{i=1}^n\frac{x_i(1+ye^{-x_i})}{1-e^{-x_i}}.$$
Setting $y = -1$, we conclude $\chi_{-1}(M) = \chi(M)$ exactly as in David E Speyer's answer.
See this note for more details on Hirzebruch's $\chi_y$ genus and its properties, as well as references for the claims made above.<|endoftext|>
TITLE: Noether's theorem in quantum mechanics
QUESTION [17 upvotes]: In classical mechanics:
If a Lagrangian $\mathcal{L}$ is preserved by an infinitesimal change in the state space variables $q_i \to q_i + \varepsilon K_i(q)$, this leads to only second order change in the Lagrangian:
$$ 0 = \frac{d\mathcal{L}}{d\varepsilon} = \sum_i \left( \frac{\partial \mathcal{L}}{\partial q_i}K_i + \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \dot{K}_i \right) = \frac{d}{dt}\left(\sum_i \frac{\partial \mathcal{L}}{\partial \dot{q}_i} K_i \right). $$
Then we get our conserved momentum because the rate of change on the right side is $0$.
In quantum mechanics, an observable $A$ commuting with the Hamiltonian, i.e. with $[\hat{H},A] = 0$, corresponds to a symmetry of the time-independent Schrödinger equation $\hat{H}\Psi = E \Psi$.  How do we compute the conserved quantity related to $A$?  In particular, what is the conserved quantity associated with the identity operator?

REPLY [4 votes]: (The laws of physics are supposedly invariant under time translations, so I hope the lateness of this answer can be forgiven.)
Both the question and the previous answers seem to have overlooked the fact that, in quantum mechanics, symmetries correspond to unitary operators on Hilbert space while observables (such as conserved charges) correspond to self-adjoint operators.
Supposing the Hilbert space to be finite-dimensional for simplicity, given any unitary matrix $A$, there exists an hermitian matrix $Q$ such that $A = e^{iQ}$, but $[H,A]=0$ (i.e. that $A$ is a symmetry of the hamiltonian) does not necessarily imply that $[H,Q]=0$ (i.e. that $Q$ is a conserved charge). Counterexamples may readily be found by considering $Q = \begin{pmatrix} 0 & -i\pi \\ i\pi & 0 \end{pmatrix}$, for which $A = -\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ commutes with any $H$.
But suppose, as for Noether's theorem in classical mechanics, that we have a smooth family $\theta \mapsto A_\theta$ of symmetry transformations such that $A_0$ is the identity transformation. Then the exponential map furnishes us with a $Q$ such that $A_\theta = e^{i\theta Q}$ and the Baker-Campbell-Hausdorff lemma in the form $[H,A_\theta] = i\theta [H,Q] + O(\theta^2)$ guarantees that $Q$ is conserved.<|endoftext|>
TITLE: Universal Covering Space of Wedge Products
QUESTION [19 upvotes]: Today I was studying for a qualifying exam, and I came up with the following question;

Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?

This question came about after calculating universal covers of the wedge of spheres ($\mathbb{S}^1 \vee\mathbb{S}^1$ and $\mathbb{S}^1 \vee\mathbb{S}^n$) and the wedge of projective space with spheres. In these cases, the universal cover looks like the cross product of the sheets of the universal covers of each space in the wedge. 
For the case of wedging two spheres, we can use the fact that $\pi_{n\geq2}\left(U\right)$ is isomorphic to $\pi_{n\geq2}\left(X\right)$ for $U$ covering $X$.
I googled around a bit to try and find something, but nothing appeared.
Thanks in advance!

REPLY [23 votes]: If $X$ and $Y$ are two reasonable spaces with universal covers $\tilde{X}$ and $\tilde{Y}$, there is a nice picture of the universal cover $\widetilde{X \vee Y}$ which has the combinatorial pattern of an infinite tree.   The tree is bipartite with vertices labeled by the symbols $X$ and $Y$.  The edges from an $X$ vertex are bijective with the fundamental group $\pi_1(X)$, and likewise for $Y$ vertices and $\pi_1(Y)$.  To make $\widetilde{X \vee Y}$, replace each $X$ vertex by $\tilde{X}$ and each $Y$ vertex by $\tilde{Y}$.  The base point of $X$ lifts to $|\pi_1(X)|$ points in $\tilde{X}$, and likewise for $Y$.  In $\widetilde{X \vee Y}$, copies of $\tilde{X}$ are attached to copies of $\tilde{Y}$ at lifts of base points.  For example, if $X = Y = \mathbb{R}P^2$, then the tree is an infinite chain and $\widetilde{X \vee Y}$ is an infinite chain of 2-spheres.
This tree picture nicely and dramatically generalizes to Bass-Serre theory.<|endoftext|>
TITLE: Applications of Euler-Cauchy ODEs
QUESTION [9 upvotes]: The Euler-Cauchy ODE (2nd order, homogeneous version) is:
$$
x^2 y'' + a x y' + b y = 0
$$
Looking in various books on ODEs and a random walk on a web search (i.e. I didn't click on every link, but tried a random sample) came up with no actual applications but just lots of vague "This is really important."s.  The closest actual application was on Wikipedia's page, which says:

The second order Euler–Cauchy equation appears in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates.

Is there a more direct application of this ODE?  Ideally, I'd like something along the lines of deriving the corresponding ODE with constant coefficients from considering springs or pendula.
My motivation is pure and simple that I'd like to be able to say something in class a little more motivating than: "We study this ODE simply because we can actually write down a solution, and it's quite amazing that we can do so."

REPLY [3 votes]: A very late answer.
I think I have a nice application which came out of a different question. I was looking for a nice 2nd order example coming from finance for my ODE class. The one I got is the equation for the Perpetual American Option (Black-Scholes time independent) which is an Euler equation.<|endoftext|>
TITLE: Taking lecture notes in lectures
QUESTION [47 upvotes]: Do you find it a good idea to take lecture notes (even detailed lecture notes) in mathematical lectures? 
Related question: Digital Pen for Math: Your Experiences?

REPLY [20 votes]: Another way of taking notes in lectures:

This is taken from The Mathematical Gazette
Vol. 25, No. 267 (Dec., 1941), p. 287 . Taken from "Memoirs of Archbishop Fredrick Temple, I".<|endoftext|>
TITLE: Loop Spaces as Generalized Smooth spaces or as Infinite dimensional Manifolds?
QUESTION [24 upvotes]: There are two ways to define smooth mapping spaces and I want to know how they compare.
Let's take the concrete special case of free loops spaces. I think this is the most studied example so will probably have the best chance of an answer. Roughly the free loop space is a space which is supposed to be the space of maps from the circle to a given space X. It is usually denoted LX. This is all fine and good for topological spaces. You get a nice mapping space LX by equipping the set of maps with the compactly generated compact open topology. It even satisfies the adjunction:
$ Map(Y, LX) = Map( Y \times S^1, X)$
However things get complicated when we want to work with manifolds. The first thing is that we want something that represents smooth maps from the circle into the manifold X. 
There are basically two approaches to making such an object precise, and I want to know how they compare. 
The first approach actually tries to construct an actual space of smooth maps. Here you start with the set of smooth maps LX, and with analytical muscle you give it the structure of an infinite dimensional Fréchet manifold. When X = G is a Lie group, this is an inifinite dimensional Lie group and is the thing whose (projective) representations make an appearence in conformal field theory.
The second approach is to study the loop space as a generalized smooth space. What is a generalized smooth space, you ask? Well there was a lot of discussion about it at the N-category Cafe, here and here. Roughly LX is thought of as a kind of sheaf via the formula:
$LX(Y) = Maps(Y, LX) = Maps(Y \times S^1, X)$
In some models it must be a concrete sheaf (i.e. it has a underlying set of points and every map from it to anything else which is concrete is a particular set map. The technical details appear in the Baez-Hoffnung paper in the second link). Clearly this model has its own desirable properties. 

How does the manifold version of loop space compare to the sheaf theoretic version?

Presumably the Fréchet manifold model gives a sheaf (since we can just map into it). Does this agree with the sheaf defined by the adjunction formula? If they are not the same sheaf, they do seem to have the same points, right? And I think there is a comparison map from the manifold LX to the sheaf LX, which should be helpful. Can anyone explain their relationship? How similar/different are they?

REPLY [12 votes]: Andrew is right! Let me amend my personal point of view, where generalized smooth manifolds are diffeological spaces.
The Fréchet manifold structure of LX agrees with the diffeology on LX in the following sense. 
There is a functor 
$$F: Frech \to Diff$$
from Fréchet manifolds to diffeological spaces defined in the same way as the well-known functor $M: Man \to Diff$ from smooth manifolds to diffeological spaces: a plot is precisely a smooth map $c: U \to LX$, where $U$ is an object in the domain category, e.g. an open subset of some $\mathbb{R}^n$. 
(By the way: a theorem of M. Losik says that the functor $F$ is full and faithful, just like the functor $M$!)
Now there are two diffeologies on LX: the first is the one described in Chris' question. A map $c:U \to LX$ is a plot if and only if the associated map $U \times S^1 \to X$ is smooth. The second diffeology is the one obtained from the functor $F$. 
These two diffeologies coincide in the sense that every plot of one is a plot of the other. In particular, they have the same sets of smooth functions.
If you want to see a pedestrian's proof I advertise Lemma A.1.7 in my paper "Transgression to Loop Spaces and its Inverse I".<|endoftext|>
TITLE: Relative version of sheaf cohomology?
QUESTION [8 upvotes]: Is there a relative version of sheaf cohomology?
EDIT: I rather mean the cohomology of pairs.

REPLY [6 votes]: It turns out that my previous answer dealt with the wrong question.  The answer to the
new question is also yes:
local cohomology $H\_Z(X,\mathcal F)$ corresponds to cohomology of the pair $(X,X\setminus Z)$.<|endoftext|>
TITLE: Switching Research Fields
QUESTION [28 upvotes]: I am a recent PhD specializing in algebraic geometry. But I also want to do research in some other areas of math (e.g. quantum computation, compressed sensing, and PDE's). What would be a good way of learning these so that they can become a research interest?
I have very little background in physics. My plan/goal is to begin research in one of these areas by next February. I do not want to be limited in the areas in which I do my research (e.g. only doing research in algebraic geometry).

REPLY [7 votes]: While it seems a bit late for the OP, I'd like to add some remarks that might be helpful to the subsequent readers.
Having completed my PhD in Combinatorics, I seem to be inevitably converting to Computational Biology.  Let me point out some of the most significant aspects that I have encountered:
Obstacles:

The letters of recommendation.  Although mentioned by Deane Yang, I feel this was not mentioned strongly enough.  I find this to be a major obstacle to my future in computational biology.  I have lots of people who would be willing to vouch for me for my expertise in combinatorics, but very few who would vouch for my expertise in computational biology.  Those who can vouch for me are only able to make limited comments due to only working in the area a short time.  [PS. One tip -- make sure the referees in your previous field have some idea of the significance of your work in the new field (thereby reducing the problem raised by Deane Yang)]
Lack of publications.  And moreover, the overall lack of relevant brownie points -- e.g. I've refereed papers in combinatorics, I'm a member of the AMS and other societies, and so on, which are not very relevant.

Advantages:

It is multidisciplinary.  So most people who enter this area have a PhD in Biology, Computer Science, Mathematics, Statistics, etc., but not in Computational Biology itself.  So most people are in the same boat.
These are neighbouring fields.  The advantages of this have already been discussed.
There is significantly more funding in computational biology.  This is sheer numbers -- there are more jobs available, so they are easier to get.
There are real world applications.  It makes it much easier to argue that this research is worth funding (thinking research grants).

Finally, one tip: try not to "switch" fields, but gradually change from one to another.<|endoftext|>
TITLE: Nonsingular/Normal Schemes
QUESTION [21 upvotes]: I always had trouble remembering this. Is it true that a curve over a non-algebraically-closed field is normal implies that it's non-singular? How about a 1 dimensional scheme? How about dimension 2? I think I heard once that surfaces over a non-algebraically closed field is normal does imply that it's non-singular. Is it true for2 dimensional schemes? What is the reason that these theorems are true for small dimensions, but fail for higher dimensions?

REPLY [33 votes]: This is not the point of the original question, but since I am an arithmetic geometer, I do find it interesting.
Over a nonperfect ground field, one needs to distinguish between regular (all the local rings are regular) and smooth (Jacobian condition; equivalently, the base change to the algebraic closure is regular).
Emerton's example of a regular but not smooth curve is "somewhat cheap" because it is not geometrically integral.  Here is a geometrically integral example which shows that the phenomenon is not so mysterious, but firmly connected to the existence of inseparable field extensions:
Let k be an imperfect field of odd characteristic p, so that there exists an inseparable field extension l/k of degree p, necessarily monogenic (since the degree is prime): say 
$l \cong k[t]/(P(t))$, where $P(t)$ is an inseparable polynomial.  Specifically, we may take $P(t) = t^p - y$ for some element $y \in k$.  Then the (unique regular projective model of) the hyperelliptic curve
$y^2 = P(x)$ 
is regular but not smooth, since upon base change to $l$ the polynomial $P(t)$ has multiple roots.

REPLY [6 votes]: There are two very nice proofs of "normal implies regular in codimension 1" in section 1.4 of "Lectures on Resolution of Singularities", by Kollar. One is said to be the most motivated proof Kollar can give, the other the slickest.<|endoftext|>
TITLE: Weil's theorem about maps from a discrete group to a Lie group.
QUESTION [8 upvotes]: Let K be a group (with discrete topology), G be a Lie group. Let $\operatorname{Hom}(K,G)$ be the space of all group homomorphisms from K to G. This is a closed subvariety of the space of all the maps from the generators of K to G, and as such has a topology.
Andre Weil's paper "On Discrete Subgroups of Lie Groups" proves that an important subset  $U \subset \operatorname{Hom}(K,G)$ is open. U is defined as the set of all homomorphism $K\to G$ such that the homomorphism is injective, the image is discrete, and the quotient $G/image(K)$ is compact.
Questions:

What happens if you remove the condition that the quotient is compact?  
How often/where is this taught? What kinds of books would it be in, what kind of courses would have it? This looks like a basic result that could be taught anywhere,
but it's completely new to me (not that I know much about representation theory). 
while Weil's paper fortunately seems very readable, I couldn't easily find any other
source that would such questions. 

Motivation:
In the case where $K=\pi_1 (S)$ is the fundamental group of a surface and $G=PSL_2(\mathbb R)$, the space $\operatorname{Hom}(K,G)$ is very closely related to the Teichmuller space of S. Every Riemann surface is a quotient of $PSL_2(\mathbb R)$ by a discrete subgroup. So, for an element of $\operatorname{Hom}(K,G)$, the quotient $G/image(K)$ corresponds to a Riemann surface and the data of the actual map $K\to G$ gives a marking on it.
Not every homomorphism $K\to G$ corresponds to a point of Teichmuller space. For example, the map that sends all of K to the identity is clearly no good, as the quotient $K/G$ is not topologically the same as the surface S. However, if the map is injective and the image of K in G is discrete, all will be well. So, Weil's theorem basically says that the Teichmuller space of S is an open subset of $\operatorname{Hom}(\pi_1(S),PSL_2(\mathbb R))$.
However, since Weil's theorem requires the quotient to be compact, this won't work if S is a non-compact Riemann  surface. I wonder how much more difficult life becomes in this case.
Disclaimer/Another question:
The above has a small lie in it. To get the Teichmuller space, you actually need to look at the quotient $\operatorname{Hom}(K,G)/G$ where G acts on $\operatorname{Hom}(K,G)$ by conjugation of the target. In the case of compact surfaces, this is not supposed to mess up the fact that the subset is open; this seems to be a result of William Goldman but I don't have the exact reference. If you can say anything about this, I'd appreciate it too.
Thank you very much!

REPLY [2 votes]: Some tangential comments:
$\textrm{PSL}_{2} \mathbb{R}$ is 3-dimensional; you get a Riemann surface by taking a lattice $\Gamma$ in $\textrm{PSL}_2\mathbb{R}$ and taking the double quotient $\Gamma \backslash \textrm{PSL}_2\mathbb{R} / \textrm{SO}(2)$. (That is, it's $\Gamma \backslash \mathbb{H}^2$ where $\mathbb{H}^2$ is the hyperbolic plane.) But this is compact iff $\Gamma \backslash$ $\textrm{PSL}_2\mathbb{R}$ is, since $\textrm{SO}(2)$ is compact.
If your Riemann surface is non-compact but of finite type, then when you uniformize, the complete hyperbolic metric will have infinite cusps at the punctures. This means that your representation $\Gamma\to$$\textrm{PSL}_2\mathbb{R}$ must send the elements $\gamma\in\Gamma$ corresponding to loops around the punctures to parabolic elements of $\textrm{PSL}_2\mathbb{R}$. Putting this restriction on a representation will force the quotient to have finite volume, and then you have the same theorem that the discrete faithful representations are open in the representation variety.
You might also look at Section 4 of Peter Shalen's paper "Representations of 3-manifold groups". For representations of hyperbolic 3-manifold groups into $\textrm{PSL}_2\mathbb{C}$ we have Mostow rigidity, which says that any two discrete faithful representations are conjugate; thus the appropriate subpace of $\textrm{Hom}(\Gamma,\textrm{PSL}_2\mathbb{C})/\textrm{PSL}_2\mathbb{C}$ is just a point (in stark contrast to the case for surface groups which you attribute to Goldman). But you still have algebraic deformations in the character variety in the case of manifolds with cusps, and these were analyzed by Thurston. In particular, Shalen says that Thurston generalized Weil's results to finite-volume cusped hyperbolic 3-manifolds, by imposing the condition mentioned above that cusp subgroups map to parabolic subgroups of $\textrm{PSL}_2\mathbb{C}$.<|endoftext|>
TITLE: Are there any books that take a 'theorems as problems' approach?
QUESTION [57 upvotes]: Are there any books that present theorems as problems? To be more specific, a book on elementary group theory might have written: "Theorem: Each group has exactly one identity" and then show a proof or leave it as an exercise. The type of book that I am imagining would have written "Problem: How many unit elements can a group have?" and similarly for all other theorems.

REPLY [3 votes]: "Selected Problems in Real Analysis" by M. G. Goluzina, A. A. Lodkin, B. M. Makarov, A. N. Podkorytov. (The authors are modest with the title, I would rather say that this book contains all problems in real analysis.) This is a perfect complement to Polya and Szego, whose book is rather on complex analysis.<|endoftext|>
TITLE: Points and DVR's
QUESTION [11 upvotes]: In the familiar case of (smooth projective) curves over an algebraically closed fields, (closed) points correspond to DVR's.
What if we have a non-singular projective curve over a non-algebraically closed field? The closed points will certainly induce DVR's, but would all DVR's come from closed points? Is there a characterization of the DVR's that aren't induced by closed points?
And how about for a general projective variety that is regular in codimension 1 (both for algebraically closed and non-algebraically closed)? Point of codimension 1 induce DVR's. Do they induce all of them? What is the characterization of the ones they do induce?
How about complete integral schemes that are regular in codimension 1?

REPLY [7 votes]: To complete partly the answer of Emerton, the picture for DVR is relatively clear. Let $X$ be an integral noetherian scheme and let $R$ be a DVR with field of fractions equal to the field of rational functions $k(X)$ on $X$. Suppose that $R$ has a center $x\in X$ (e.g. if $X$ is proper over a subring of $R$). Let $k_R$ be the residue field of $R$. Then $k_R$ has transcendental degree over $k(x)$ bounded by $\dim O_{X,x} -1$. Suppose further that $X$ is universally catenary and Nagata (e.g. $X$ is excellent), then the equality holds if and only if the center of $R$ in some $X'$ proper and birational over $X$ is a regular point of codimension 1. This is a theorem of Zariski. See M. Artin: ''Néron Models'', § 5, in Cornell & Silverman: "Arithmetic Geometry".<|endoftext|>
TITLE: Is the theory of categories decidable?
QUESTION [27 upvotes]: There are a lot of theorems in basic homological algebra, such as the five lemma or the snake lemma, that seem like they'd be more easily proven by computer than by hand.  This led me to consider the following question: is the theory of categories decidable?
More specifically, I was wondering whether or not statements about abelian categories can be determined true or false in finite time.  Also, if they can be determined to be false, is it possible to explicitly describe a counterexample?  If it is known to be decidable, is anything known about the complexity?  (Other decidable theories often have multiply-exponential time complexities.)  If it is known to be undecidable, say by embedding the halting problem, then can I change my assumptions a bit and make it decidable?  (For example, maybe I shouldn't be looking at abelian categories after all.)
Thanks in advance.
Edit: It appears a clarification is needed.  My goal was to consider the minimal theory that could state things like the five lemma, but not necessarily prove them.  For example, I want to say:

If in an abelian category, you have a bunch of maps $0\to A \to B \to C\to 0$ and $0\to A' \to B' \to C'\to 0$ which make up two short exact sequences and some more maps $a:A\to A'$, $b:B\to B'$, $c:C\to C'$ which commute with the previous maps, and $a$ and $c$ are isomorphisms, then $b$ is an isomorphism, too.

Sentences of this form would be inputs to a program, which decides if this statement is in fact true in ZFC (or your other favorite axiomatization of category theory).  The point here is that I am restricting the sentences one can input into the program, but keeping ZFC or whatnot as my framework.
I hoped (perhaps naively) that if I restricted the class of sentences, it might be decidable whether or not these statements were true.  For example, I imagined that every such theorem is either proven by diagram chasing, or it is possible to find a concrete example of maps among, say, R-modules that contradict the result.

REPLY [23 votes]: This answer builds on those of F. G. Dorais and Joel David Hamkins to answer your "specific question", the question left open by them, namely whether the theory of abelian categories is decidable.
The answer is still no.
Even the following more limited family of problems is undecidable:

Given words $r, r_1,\ldots,r_m$ in $x_1,\ldots,x_n$ (i.e., each $r_i$ is a finite product of the $x_i$ and their inverses), decide whether it is true that whenever the $x_i$ are interpreted as automorphisms of an object $M$ in an abelian category, $r_1=\cdots=r_m=1_M$ implies $r=1_M$.

If the answer to the corresponding instance of the word problem for finitely presented groups is yes, then the answer to this abelian category question is yes.
Conversely if the answer to the word problem instance is no, then we can construct the finitely presented group $G = \langle x_1,\ldots,x_n | r_1,\ldots,r_m \rangle$, form the group ring $\mathbb{Z}G$, and let $M$ be $\mathbb{Z}G$ as a module over itself, which shows that the answer to the abelian category question is no too.
So if there were an algorithm to decide this family of abelian category problems, there would also be an algorithm to decide the word problem for finitely presented groups.  But P. S. Novikov proved in 1955 that the latter algorithm does not exist.<|endoftext|>
TITLE: Variant of Fermat's last theorem
QUESTION [7 upvotes]: By Fermat's last theorem, the equation $u^3+v^3=w^3$ has no solutions
in positive integers $u,v,w$. Now consider the following variant : call $\rho(x)$
the distance between $x$ and the nearest integer, for any real number $x$
(thus $\rho(3)=0,\rho(3.2)=0.2,\rho(3.5)=0.5, \rho(3.9)=0.1$ etc). 
An "approximate" version of the Fermat equation is to ask
for $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ to be arbitrarily small. A trivial
way to achieve this is to make one of the variables very small
compared to the other, say $v$ very small compared to $u$, so that
${( u^3+v^3 )}^{\frac{1}{3}}$ is very near to $u$.
It is therefore natural to ask if there is an absolute constant $C>0$
such that $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ can be made arbitrarily small
with $u,v$ positive and $u \leq C v, v \leq C u$ (so that neither of
$u$ or $v$ dominates).
Can a (reasonably) explicit $(u_n,v_n)$ sequence be found, such that
$\rho(u_n^3+v_n^3)$ tends to $0$ as $n$ goes to infinity and
$u_n \leq C v_n, v_n \leq C u_n$ ?  "Closed-form" formula
would be the best, of course, but even a simple recursion would be nice.

REPLY [15 votes]: Elkies' paper "Rational points near curves and small nonzero |x^3-y^2| via lattice reduction" is relevant to your question:  he discusses the efficient computation of solutions to |x^3 + y^3 - z^3| < M, and remarks that there are conjectured to be infinitely many of these "near-Fermat" triples. Indeed, his argument is very general and is not restricted to algebraic curves:  he finds, for instance, that
2063^pi + 8093^pi = 8128^pi + .019....!<|endoftext|>
TITLE: What is an exponential?
QUESTION [5 upvotes]: Is there a notion of exponentiation that subsumes the well known versions, and in particular the versions on

tangent spaces (e.g., of Lie groups and Riemannian manifolds), in which the exponential map sends a vector to a point on a curve naturally defined in terms of the vector;
unital Banach algebras?

(NB. I am not conversant with category theory beyond the words "morphism" and "functor". But a categorically flavored answer that takes my limited knowledge base into account would be preferable. An internet search led me to the notion of a "Cartesian closed category", which doesn't seem to be the sort of thing I have in mind.)

REPLY [6 votes]: I think the exponential function for unital Banach algebras is a special case of the exponential functions for Lie groups modeled on topological vector spaces: the set of invertible elements in an unital Banach algebra naturally is a Banach Lie group, and the exponential function of this Lie group is the "classical" exponential function for the Banach algebra.<|endoftext|>
TITLE: transfinite composition of weak equivalences in sSet
QUESTION [5 upvotes]: Weak equivalences in the standard model structure on simplicial sets are allegedly closed under transfinite composition.
What's a reference for that?

REPLY [4 votes]: This answer serves to record two explicit proofs of this fact in the literature:
Corollary 5.1 in Raptis and Rosický, “The accessibility rank of weak equivalences”, arXiv:1403.3042v2.
Theory and Applications of Categories 30:19 (2015), 687—703.
Theorem 4.6 in Barnea and Schlank, “Model structures on Ind categories and the accessibility rank of weak equivalences”, arXiv:1407.1817v6.<|endoftext|>
TITLE: Is set-induction relatively consistent?
QUESTION [10 upvotes]: One way to state the axiom of foundation is that the $\in$ relation on any transitive set is well-founded in the following sense:

A relation $(X,\prec)$ is well-founded if for any subset $S\subseteq X$ which is inductive, in the sense that $y\in S$ for all $y\prec x$ implies $x\in S$, in fact $S=X$.

This is classically equivalent to the more usual definition, but constructively more reasonable.  Now in constructive set theories such as CZF, one often finds a different axiom of "foundation" called set-induction or epsilon-induction:

For any formula $\phi$, if $\phi(x)$ for all $x\in y$ implies $\phi(y)$ for any set $y$, then in fact $\phi(x)$ is true for all sets $x$.

This certainly implies that $\in$ is well-founded on any transitive set in the sense above, and the converse is true if every set has a transitive closure and you have the full axiom of separation, since from any transitive set $x$ you can form $\lbrace y\in x \mid \phi(y) \rbrace$ and conclude that it is all of $x$.
Now classically, one can show that the axiom of foundation is consistent, relative to the other axioms of set theory, by simply restricting any model to the submodel of well-founded sets.  I think you can do this in CZF for the version of "foundation" I stated above, that every transitive set is well-founded.  My question is, can you do anything similar with set-induction?  I care most about CZF, but I'd also be interested to know about results in any other theory lacking the full axiom of separation.
(BTW, it's possible one might have to fiddle a little bit with what I mean by "the other axioms of CZF" to make this nontrivial.  For one thing, let's drop the axiom of infinity, since I see no guarantee that $\omega$ would be "strongly well-founded" in the necessary sense.)

REPLY [2 votes]: Late answer: NO. At least with respect to CZF.
CZF with Set Induction has proof-theoretic ordinal the Bachmann-Howard ordinal.
CZF without Set Induction is weaker. Can't think of a direct reference, but:

Crosilla & Rathjen "Inaccessible Set Axioms may have little consistency strength" 

It is shown that CZF without set induction but with an inaccessible set axiom has proof-theoretic ordinal $\Gamma_0$. 
This is strictly less than the Bachmann-Howard ordinal, so no relative consistency proof is possible.<|endoftext|>
TITLE: Algebraic stacks from scratch
QUESTION [30 upvotes]: I have a pretty good understanding of stacks, sheaves, descent, Grothendieck topologies, and I have a decent understanding of commutative algebra (I know enough about smooth, unramified, étale, and flat ring maps).  However, I've never seriously studied Algebraic geomtry.  Can anyone recommend a book that builds stacks directly on top of CRing in a (pseudo)functor of points approach?  Typically, one builds up stacks segmentwise, first constructing Aff as the category of sheaves of sets on CRing with the canonical topology, which gives us CRing^op.  Then, one constructs the Zariski topology on Aff, and from that constructs Sch, then one equips Sch with the étale topology and constructs algebraic stacks above that.  (I assume that one gets Artin stacks if one replaces the étale topology there with the fppf topology?)
Does anyone know of a book/lecture notes/paper that takes this approach, where everything is just developed from scratch in the language of categories, stacks, and commutative algebra?
Edit: Some motivation: It seems like many of the techniques used to build the category of schemes in the first place are just less generalized versions of the constructions for algebraic stacks.  So the idea is to develop all of algebraic geoemtry in "one fell swoop", so to speak.
Edit 2: As far as answers go, I'm not really interested in seeing value judgements about this approach.  I know that it's at best a controversial approach, but I've seen all of the arguments against it before. 
Edit 3: Part of the motivation for this question comes from a (possibly incorrect) footnote on Wikipedia: 

One can always assume that U is an affine scheme. Doing so means that the theory of algebraic spaces is not dependent on the full theory of schemes, and can indeed be used as a (more general) replacement of that theory.

If this is true, then at least we can avoid most of the trouble Anton says we'll go through in his comment below.  However, this being true seems to indicate that we should be able to do the same thing for algebraic stacks.  
Edit 4: Since Felipe made his comment on this post, everyone has just been "voting up the comment".  Since said comment was a question, I'll just post a response.  
Mainly because I study category theory on my own time, and I've taken commutative algebra courses.
Now that that's over and done with, I've also added a bounty to this question.

REPLY [2 votes]: A old text on AG with functorial approach is Demazure-Gabriel (Introduction to algebraic geometry and algebraic groups) :
http://books.google.it/books?id=RDKRyP00aoMC&printsec=frontcover&dq=Algebraic+groups,+demazure&source=bl&ots=bINVNX5u5x&sig=PA40Oq8LOQdfBZgci6hFtXM-HGA&hl=it&ei=nQbfTJiCDs-fOqKP4esO&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBsQ6AEwAA#v=onepage&q&f=false<|endoftext|>
TITLE: When two k-varieties with the same underlying topological spaces isomorphic?
QUESTION [9 upvotes]: I have a little problem. I'm probably being just so careless..... Here k-varieties are all integral separated k-schemes of finite type over k, where k is a field.
Suppose $X, Y$ are $k$-varieties, and let $f :X \to Y$ be a morphism of $k$-varieties that is one to one and onto. Then, when can we say this $f$ is an isomorphism of $k$-varieties?
If this is too vague, let me add that the case I would like to see is when each fibre of $f$ (which is a singleton) is reduced. Under this assumption, would this give an isomorphism?

REPLY [12 votes]: The condition you are looking for is seminormality. A variety (or a reduced scheme) $Y$ is seminormal if any proper bijective morphism $f:X\to Y$, with $X$ reduced, inducing isomorphisms on residue fields $k(y)=k(x)$ for points $x\in X$, $y=f(x)\in Y$, is an isomorphism. A basic fact is that any variety has a unique seminormalization. 
A related notion which differs only in positive characteristic is weak normality for which $k(y)\to k(x)$ is required to be purely inseparable and an isomorphism for each generic point $x\in X$.
One basic reference for this notion is the appendix to Chapter 1 of Koll'ar's "Rational curves on algebraic varieties", where you will find many standard facts and examples such as: normal implies seminormal; in dim 1 seminormal means analytically isomorphic to the $n$ axes in $A^n$; irreducible components of seminormal schemes need not be normal, etc. You will also find references to many papers where this notion was comprehensively investigated.
For clarity, let me add the standard fact: $f$ is proper and bijective $\iff$ it is finite and bijective (as opposed to quasifinite = finite fibers, which of course follows from bijective).<|endoftext|>
TITLE: What, precisely, is the relationship between "fields of moduli" and "moduli spaces"?
QUESTION [12 upvotes]: Notation
The term "field of moduli" comes in up in different scenarios, but let's consider the following: Let X->ℙ1 be a G-Galois cover, where everything is over the algebraic closure of some field L. Assume that X->ℙ1 descends (without group action -- as a cover) to XL->ℙL1. Then I define the field of moduli to be the intersection of all finite extensions of L for which base change of XL->ℙL1 becomes G-Galois.
Question
There is the saying that the field of moduli is the function field of the (coarse?) moduli space of when you let the branch points vary. What is the precise statement of that? (and why is it true?)
Thoughts
It would seem that we should fix a dedekind ring whose quotient field is L (ℤ if L is ℚ), and call it D. Then descend to a D-model of ℙ1 (for a D-model of X take the integral closure of ℙ1 in the function field of X). Then do something like look at the moduli space of all covers of ℙ1 with that number of (distinct) branch points, and in it look at the subscheme of all covers that can be achieved by deforming any of the fibers of our XD->ℙD1 (pick a fiber such that there's no coalescence of branch points) by a family. But there's a lot missing here, even in terms of making this precise. For example: IS there a coarse moduli space of all covers with n branch points over ℙD1 (where by n branch points, I mean n branch point on each geometric fiber)? What does it look like? Why should the function field of said subscheme be the field of moduli?
Thanks in advance.

REPLY [7 votes]: A few responses to different parts of your question.

In my experience, the phrase "field of moduli" doesn't usually refer to the function field of a coarse moduli space.  Rather:  the base change of your cover to Lbar corresponds to a point of M(Lbar), where M/L is the coarse moduli space.  This point has a well-defined field of definition, which is by definition the field of moduli of your cover.  The phrase "field of moduli" is usually used in distinction with "field of definition" -- if your cover is actually defined over L', then the field of moduli is certainly contained in L', but it may not be equal to L'.  This phenomenon isn't restricted to Hurwitz spaces; there are abelian varieties over Qbar whose field of moduli is Q (that is, they correspond to points of A_g(Q)) but which don't descend to Q.  This can only happen when g is even.  Off the top of my head I don't remember a reference for an example, nor for the assertion of the previous sentence; maybe somebody can help me out in comments.  Certainly when g=1 you don't have this problem; given a rational number j, there is an elliptic curve E/Q with j(E) = j.  But you prove this by writing it down -- it's not completely obvious "by pure thought" that it should be so.
The most complete description of the Hurwitz stack (the moduli stack of finite covers with fixed combinatorial invariants) its associated coarse moduli space, etc., is in the Ph.D. thesis of Stefan Wewers, which is unfortunately not available online.  However, the survey paper of Romagny and Wewers should give you most of what you need.<|endoftext|>
TITLE: Large cardinal axioms and Grothendieck universes
QUESTION [23 upvotes]: A cardinal $\lambda$ is weakly inaccessible, iff a. it is regular (i.e. a set of cardinality $\lambda$ can't be represented as a union of sets of cardinality $<\lambda$ indexed by a set of cardinality $<\lambda$) and b. for all cardinals $\mu<\lambda$ we have $\mu^+<\lambda$ where $\mu^+$ is the successor of $\mu$. Strongly inaccessible cardinals are defined in the same way, with $\mu^+$ replaced by $2^\mu$. Usually one also adds the condition that $\lambda$ should be uncountable.
As far as I understand, a "large cardinal" is a weakly inaccessible cardinal with some extra properties. In set theory one considers various "large cardinal axioms", which assert the existence of large cardinals of various kinds. Notice that these axioms are quite different from, say the Continuum Hypothesis. In particular, one can't deduce the consistency of ZFC + there exists at least one (uncountable) weakly inaccessible cardinal from the consistency of ZFC, see e.g. Kanamori, The Higher Infinite, p.19. I.e., assuming ZFC is consistent, these axioms can not be shown independent of ZFC.
The "reasonable" large cardinal axioms seem to be ordered according to their consistency strength, as explained e.g. here http://en.wikipedia.org/wiki/Large_cardinal. This is not a theorem, just an observation. A list of axioms according to their consistency strength can be found e.g. on p. 472 of Kanamori's book mentioned above. (Noticeably, it starts with "0=1", which is a very strong axiom indeed.)
Large cardinals appear to occur seldom in "everyday" mathematics. One such instance when they occur is when one tries to construct the foundations of category theory. One of the ways to do that (and the one that seems (to me) to be the most attractive) is to start with the set theory and to add Grothendieck's Universe axiom, which states that every set is an element of a Grothendieck universe.
(As an aside remark, let me mention another application of large cardinal axioms: incredibly, the fastest known solution of the word problem in braid groups originated from research on large cardinal axioms; the proof is independent of the existence of large cardinals, although the first version of the proof did use them. See Dehornoy, From large cardinals to braids via distributive algebra, Journal of knot theory and ramifications, 4, 1, 33-79.)
Translated into the language of cardinals, the Universe axiom says that for any cardinal there is a strictly larger strongly inaccessible cardinal. I have heard several times that this is pretty low on the above consistency strength list, but was never able to understand exactly how low. So I would like to ask: does the existence of a (single) large cardinal of some kind imply (or is equivalent to) the Universe axiom?

REPLY [30 votes]: A Grothendieck universe  is known in set theory as the set Vκ for a (strongly) inaccessible cardinal κ. They are exactly the same thing. Thus, the existence of a Grothendieck universe is exactly equivalent to the existence of one inaccessible cardinal. These cardinals and the corresponding universes have been studied in set theory for over a century.
The Grothendieck Universe axiom (AU) is the assertion that every set is an element of a universe in this sense. Thus, it is equivalent to the assertion that the inaccessible cardinals are unbounded in the cardinals. In other words, that there is a proper class of inaccessible cardinals. This is the axiom you sought, which is exactly equivalent to AU. In this sense, the axiom AU is a statement in set theory, having nothing necessarily to do with category theory.
The large cardinal axioms are fruitfully measured in strength not just by direct implication, but also  by their consistency strength. One large cardinal property LC1 is stronger than another LC2 in consistency strength if the consistency of ZFC with an LC1 large cardinal implies the consistency of ZFC with an LC2 large cardinal.
Measured in this way, the AU axiom has a stronger consistency strength than the existence of any finite or accessible number of inaccessible cardinals, and so one might think it rather strong. But actually, it is much weaker than the existence of a single Mahlo cardinal, the traditional next-step-up in the large cardinal hierarchy. The reason is that if κ is Mahlo, then κ is a limit of inaccessible cardinals, and so Vκ will satisfy ZFC plus the AU axiom. The difference between AU and Mahloness has to do with the thickness of the class of inaccessible cardinals. For example, strictly stronger than AU and weaker than a Mahlo cardinal is the assertion that the inaccessible cardinals form a stationary proper class, an assertion known as the Levy Scheme (which is provably equivconsistent with some other interesting axioms of set theory, such as the boldface Maximality Principle, which I have studied a lot). Even Mahlo cardinals are regarded as rather low in the large cardinal hierarchy, far below the weakly compact cardinals, Ramsey cardinals, measurable cardinals, strong cardinals and supercompact cardinals. In particular, if δ is any of these large cardinals, then δ is a limit of Mahlo cardinals, and certainly a limit of strongly inaccessible cardinals. So in particular, Vδ will be a model of the AU axiom. 
Rather few of the large cardinal axioms imnply AU directly, since most of them remain true if one were to cut off the universe at a given inaccessible cardinal, a process that kills AU. Nevertheless, implicit beteween levels of the large caridnal hiearchy are the axioms of the same form as AU, which assert an unbounded class of the given cardinal. For example, one might want to have unboundedly many Mahlo cardinals, or unboundedly many measurable cardinals, and so on. And the consistency strength of these axioms is still below the consistency strength of a single supercompact cardinal. The hierarchy is extremely fine and intensely studied.  For example, the assertion that there are unboundedly many strong cardinals is equiconsistent with the impossiblity to affect projective truth by forcing. The existence of a proper class of Woodin cardinals is particularly robust set-theoretically, and all of these axioms are far stronger than AU. 
There are natural weakenings of AU that still allow for almost all if not all of what category theorists do with these universes. Namely, with the universes, it would seem to suffice for almost all category-theoretic purposes, if a given universe U were merely a model of ZFC, rather than Vκ for an inaccessible cardinal κ. The difference is that U is merely a model of the Power set axiom, rather than actually being closed under the true power sets (and similarly using Replacement in place of regularity). The weakening of AU I have in mind is the axiom that asserts that every set is an element of a transitive model of ZFC. This assertion is strictly weaker in consistency strength thatn even a single inaccessible cardinal. One can get much lower, if one weakens the concept of universe to just a fragment of ZFC. Then one could arrive at a version of AU that was actually provable in ZFC, but which could be used for most all of the applications in cateogory theory to my knowledge. In this sense, ZFC itself is a kind of large cardinal axiom relative to the weaker fragments of ZFC.

REPLY [10 votes]: It is very near the bottom of Kanamori's chart. The very bottom of the chart is the level of a (strongly) inaccessible cardinals, which is the smallest large cardinal axiom. Right above the inaccessibles in the chart are the α-inaccessible cardinals. It turns out that the Universe Axiom (UA) is strictly weaker than the existence of a 2-inaccessible cardinal. In fact, κ is 2-inaccessible if and only if κ is regular and Vκ ⊧ ZFC + UA.
Specifically, a cardinal κ is:

0-inaccessible iff κ is regular,
1-inaccessible iff κ is a regular strong limit of 0-inaccessibles,
2-inaccessible iff κ is a regular strong limit of 1-inaccessibles,
etc.

So an inaccessible cardinal is exactly the same as a 1-inaccessible cardinal, which are also precisely the regular cardinals κ such that Vκ ⊧ ZFC. If κ is 2-inaccessible, then there are unboundedly many inaccessibles λ < κ. These are inaccessible in Vκ too, which is why Vκ satisfies UA.
Note that the existence of a 2-inaccessible cardinal κ does not directly imply the Universe Axiom. Indeed, κ may well be the last inaccessible cardinal, which means that there may be no universe that contains κ itself. However, if κ is 2-inaccessible then the universe Vκ does satisfy UA, which means that the existence of a 2-inaccessible proves the consistency of ZFC + UA.
Although UA is indeed a large cardinal axiom, there is no way to formulate UA as the existence of a single large cardinal. However, morally speaking, you can think of UA as saying "the class of all ordinals (viewed as a cardinal number) is 2-inaccessible." Of course, this doesn't make sense since the class of all ordinals is not a set, but this is exactly what κ looks like when viewed from inside Vκ.<|endoftext|>
TITLE: Help with a double sum, please
QUESTION [36 upvotes]: Here is a double series I have been having trouble evaluating:
$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}}\text{.}$$
I am confident that $S=0$ for any $m>0$. In fact, I have no doubt. I have done lots of algebraic manipulation, attempted to "convert" it to a hypergeometric series, check tables (Gradshteyn and Ryzhik), etc., but I have not been able to get it into a form from which I can prove zero equivalence.
Here is another form of the sum (well, I hope at least) that might be easier to work with:
$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}\frac{m!}{k!(m-k)!}\frac{(k+m-1-j)!}{(k+m)!}\frac{(k+m-1/2)!}{(k+m-1/2-j)!}\frac{1}{2^{k-j+m+1}}\text{.}$$
I have read Concrete Mathematics and $A=B$, and looked at Gosper's and Zeilberger's work for some hints, but no cigar.
Note: $0!=1$ and $n!=n(n-1)!$ for $n\in\mathbb{N}\cup\{0\}$. For $n\in\mathbb{R}^+$, $n!=n\Gamma(n)$ where $\Gamma\colon\: \mathbb{C}\to\mathbb{C}$ and, for $\Re z>0$ and $z\notin\mathbb{Z}^{-}$, $$\Gamma\colon\: z\mapsto  \int_0^\infty t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t\text{.}$$
which can be analytically extended to $\mathbb{C}$ via the recurrence $\Gamma(z+1)=z\Gamma(z)$.

REPLY [44 votes]: Olivier Gerard just told me about this wonderful website!
Regarding the question it can be done in one nano-second using the Maple 
package 
http://www.math.rutgers.edu/~zeilberg/tokhniot/MultiZeilberger
accopmaying my article with Moa Apagodu
http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/multiZ.html
Here is the command:
F:=(-1)^k*binomial(m,k)*(k+m-1-j)!/(k+m)!*simplify((k+m-1/2)!/(k+m-1/2-j)!)/2^(k-j):  lprint(MulZeil(F,[j,k],m,M,{})[1]);
and here is the output:
-1/4*(2*m+1)/(m+1)+M
(Note that I had to divide the summand by 1/2^(m+1) if you don't you get FAIL,
the prgram does not like extraneous factors)
Translated to humaneze we have that (my S(m) is hte original S(m) times 2^(m+1))
S(m+1)=(2m+1)/(m+1)S(m)
Since S(1)=0 (check!) This is a completely rigorous proof.
P.S. The proof can be gotten by finding the so-called multi-certificate
lprint(MulZeil(F,[j,k],m,M,{})[2]);
-Doron Zeilbeger<|endoftext|>
TITLE: Why does the group law commute with morphisms of elliptic curves?
QUESTION [18 upvotes]: I know this should be pretty simple, but right now the only way I can see how to prove it is to sit down and write out explicit formulae for the group law, and see that everything works out. What's the geometric or abstract-nonsense reason why the abelian group structure of elliptic curves behaves nicely under homomorphisms?

REPLY [12 votes]: There is also a simple analytic proof which is enlightening in a different way.
Let $f : E \to E'$ be a morphism of elliptic curves. Both E and E' have $\mathbb C$ as their universal covering space. The composition $\mathbb C \to E \to E'$ lifts uniquely to a continuous map $\overline f$ from $\mathbb C$ into the the universal covering of $E'$ (since the source is simply connected), and $\overline f$ is holomorphic because the map $\mathbb C \to E'$ is locally biholomorphic. 
Let $E = \mathbb{C}/\Lambda$, $E' =  \mathbb{C}/\Lambda'$. We know that for $\lambda \in \Lambda$ we have $\overline f(z+\lambda) - \overline f(z) \in \Lambda'$, hence $\overline f(z+\lambda) - \overline f(z)$ is constant as a function of z. Differentiating, the derivative of $\overline f$ is a doubly periodic entire function, so it is a constant. Since $f$ should preserve the basepoint on the elliptic curves, which is the image of 0, we can assume that $\overline f (0) = 0$ and so $\overline f (z) = cz$ for some constant $c$. But the group laws on E and E' are induced by the group law on $\mathbb C$, so the fact that f preserves the group structure reduces to the distributive law!
The same proof holds also for any complex torus.<|endoftext|>
TITLE: Proof of Bondy and Chvátal Theorem
QUESTION [9 upvotes]: A graph is Hamiltonian if and only if
  its closure is Hamiltonian.

I am looking for a simple (i.e. short) proof of the theorem, that I can use as part of an article on topological sorting.
I've not been able to find one in the literature I have access to. Any help would be appreciated.

REPLY [12 votes]: Let $G=G_0, G_1, G_2$, etc. be a sequence of graphs where each  $G_i$ is formed by performing a single closure step to $G_{i-1}$ — that is, add an edge $uv$ to $G_i$ when $u$ and $v$ together have at least $n$ neighbors. If any graph in this sequence is Hamiltonian, let $k$ be the minimum $k$ such that $G_k$ is Hamiltonian. Then I claim that $k=0$. For, otherwise let $uv$ be the edge added to form $G_k$ from $G_{k-1}$ and let $C$ be a Hamiltonian cycle in $G_k$. There are $n-1$ other edges in $C$, and $n$ edges going out from $u$ and $v$ together, so by the pigeonhole principle there exists an edge $pq$ in $C$ (with the vertex labeling chosen so that $u$ is clockwise of $v$ and $p$ is clockwise of $q$) such that $G_{k-1}$ contains edges $up$ and $vq$. But then $C + up + vq - pq - uv$ is a Hamiltonian cycle in $G_{k-1}$ contradicting the assumption that $k>0$.

REPLY [6 votes]: Let me restate the theorem for variables.
Notation Read $p\bowtie q$ as "$p$ is adjacent to $q$", and $p\not\bowtie q$ as "$p$ is not adjacent to $q$".
Theorem Let $G$ be a graph whose order is greater than $2$. Let $v_1$ and $v_2$ be vertices such that $v_1\neq v_2 \land v_1\not\bowtie v_2$ and $\deg v_1 + \deg v_2 \ge n\in\mathbb{N}$. Then $G$ is Hamiltonian iff $G' = G + v_1 v_2$ is Hamiltonian.
Proof Assume $G'$ is Hamiltonian but not $G$. Therefore, by definition, there exists a cycle $(p_1,\ldots,p_n)$ in $G$ connecting $v_1$ (at $p_1$) to $v_2$ (at $p_n$) visiting all of $G$'s vertices. If $p_k\bowtie p_1$ then $p_{k-1} \not\bowtie p_n$, since $(p_1,p_k,p_{k+1},\ldots,p_n,p_{k-1},p_{k-2},\ldots,p_1)$ is a Hamiltonian cycle of $G$. As such, $\deg p_n \le n  - (1 + \deg p_1)$, a contradiction.<|endoftext|>
TITLE: cotangent bundle symplectic reduction and fibre bundles
QUESTION [15 upvotes]: Suppose a compact Lie group $G$ acts on a manifold $M$ with only one orbit type $G/H$ ($H$ denotes the stabiliser group). Then the manifold $M$ becomes a fibre bundle over the quotient manifold $X:=M/G$ with typical fibre $G/H$ and structure group $G$.
On the one hand one could look at the cotangent bundle $T^* X$ of the quotient (which carries a natural symplectic structure).
On the other hand consider the lifted action of $G$ on the cotangent bundle $T^* M$ with moment map $\mu: T^* M\to \mathfrak{g}^* .$ The symplectic quotient $T^* M//G:=\mu^{-1}(0)/G$ inherits the structure of a symplectic manifold.  Here comes the question: Are $T^* X$ and $T^* M//G$ (canonically) symplectomorphic?

REPLY [4 votes]: This is more a comment towards Gourishankar than an answer to the original question. 
It was part of my thesis, (UCB, about 1986), so, apologies,  I chime in. For simplicity,
I take the case $G$ Abelian, and $H = $ trivial.
 To map $J^{-1} (\mu)$ equivariantly to $J^{-1}(0)$ subtract $\mu \cdot A$ where $A$ is any $G$-connection for $\pi: X \to X/G$.  $J^{-1} (0)/G = T^* (X/G)$
canonically, independent of connection.    The map `momentum shift map
of subtracing  $\mu \cdot A$ from co-vectors  is not symplectic,
relative to the standard structure, but it    becomes symplectic if you subtract
$\mu \pi^* F_A$, where  $F_A = curv(A)$, from the standard structure. So the reduced space at  $\mu$ is $T^*(X/G)$ with the standard structure minus the ``magnetic term''
$\mu F_A$.  
For non-Abelian $G$ ($H$ still trivial),  it is easier to explain things in Poisson terms.
$T^* X/G$ is a Poisson manifold whose symplectic leaves
are the reduced spaces in question. The momentum
shift trick turns it  into  $T^* (X/G) \oplus Ad^* (X)$  where $Ad^* (X) \to X/G$
is the   co-adjoint bundle associated to $X \to X/G$ -- its fibers are the 
dual Lie algebras for $G$. This direct sum bundle admits coordinates $s_i, p_i, \xi_a$ where $s_i, p_i$ are canonical coordinates on $T^*(X/G)$ induced by  coordinates $s_i$ on $X/G$
and where $\xi_a$ are fiber-linear coordinates on the co-adjoint bundle induced
by a choice of local section of $\pi$. Then the main tricky part of the bracket is that
the bracket of $p_i$ with $p_j$ is  $\Sigma \xi_a F^a _{ij}$,  $F$ being the curvature of the connection
relative to the choice of local section.  The symplectic leaves = reduced spaces
are of the form $T^*(X/G) \oplus $(co-adjoint orbit bundle).<|endoftext|>
TITLE: Representations in characteristic p
QUESTION [13 upvotes]: Let G be a finite group and let F be an algebraically closed field.  If the characteristic of F is 0, then the number of irreducible F-representations of G is given by the number of conjugacy classes of elements of G.  A paper I'm reading says that if the characteristic of F is p>0, then the number of F-irreps of G is the same as the number of conjugacy classes of elements whose order is not divisible by p.  
If G is abelian, it seems to me that this should say that the p-sylow subgroup of G acts trivially on every characteristic p irrep.  This is because I can split G into G'x P (non-p and p-sylow subgroups), and then any irrep of G' extends to one of G by letting P act trivially.  Since the formula mentioned above would say that they have the same number of irreps these must be all of them.
My question is: Is this true?  If not, then where is my reasoning going off track?

REPLY [5 votes]: Brauer's proof that the number of similarity classes of irreducible representations of $G$ over an algebraically closed field of characteristic $p$ is equal to the number of $p$-regular conjugacy classes of $G$ is ring-theoretic in flavor, and rather tricky. There is also an easy character theoretic proof based on the following ideas. First, the set IBr$(G)$ of irreducible Brauer characters is in bijective correspondence with the irreducible representations, and this set of functions lives in the space $V$ of all complex-valued class functions defined on the set of p-regular elements. Since $\dim(V)$ equals the number of $p$-regular classes, it suffices to show that IBr($G$) is a basis for $V$. The linear independence of IBr$(G)$ is a standard result. To see that IBr$(G)$ spans, use the facts that Irr$(G)$ spans the space of all class functions and that on each $p$-regular class, the value of an ordinary character is a linear combination (and in fact, a sum) of values of Brauer characters.<|endoftext|>
TITLE: Using consistency to create new axioms in set theory
QUESTION [26 upvotes]: As everybody knows, the ZFC axioms may serve as a foundation for (almost)
all of contemporary mathematics, and it is also well-known that several results
are "indecidable" in ZFC, which means that they cannot be proved or disproved within 
ZFC.
It is therefore natural to look for "new axioms" to add to ZFC and make it a
stronger system. But by Godel's second incompleteness theorem, the consistency of
ZFC cannot be deduced from ZFC itself.
Therefore, we may add the axiom "ZFC is consistent" and obtain 
a new system $ ZFC_1 $ consisting of "ZFC+(ZFC is consistent)". We may
iterate this, and define $ ZFC_2 $ as "$ZFC_1$+($ZFC_1$ is consistent)", etc,
and we may even define $ZFC_{\omega}$, or $ZFC_{\alpha}$ for any
ordinal $\alpha$.
This seems a little too easy, so my question to logicians is : is this construction
completely irrelevant to logic ans set theory questions ? If so why? Is it true that the results which are classically independent of ZFC are also
 independent of   $ZFC_1$,  $ZFC_2$,  $ZFC_{\omega}$ etc ?

REPLY [18 votes]: What you propose is very reasonable, of course, since when we believe in a theory T, then it is natural for us also to believe that T is consistent. And the axioms that you propose to add to ZFC formalize this process. The (philosophical) question here is, does this process somehow find a completion? 
(Let me quibble with your remark that we can formalize ZFCα for any ordinal α. We need that the ordinal α is somehow representable in the theory in order for the assertion Con(ZFCα) to be expressible. Of course, in a countable language, we have only countably many statements, and so we must eventually run out of representable ordinals.)
The answer is that your axioms are the pre-beginnings of the large cardinal hierarchy, as hinted at by Kristal Cantwell and Dorais. If there is a (strongly) inaccessible cardinal κ, then Vκ is a model of ZFC, and so your theory ZFC1 holds. 
But I claim much more, and from a weaker hypothesis.  One doesn't need an inaccessible cardinal even to know that all the expressible ZFCα are consistent. 
I claim that if there is an ω model of ZFC, then all the expressible ZFCα are true and consistent. 
To see this, suppose that M is an ω model of ZFC. This means that M has the standard natural numbers. From this, it follows that the ordinals of M are well-founded for some distance above ω, but may become ill-founded much higher up. Since M has the same natural numbers as we do in the meta-theory, it follows that M has exactly the same formulas in the language of set theory and, more importantly, exactly the same proofs. Thus, for any theory T that exists in M, it will be consistent in M if and only if it is consistent. 
This is enough to perform an interesting ramping-up argument. Namely, since M is a model of ZFC, it follows that ZFC is consistent for us, and so M agrees, and so M is a model of ZFC+Con(ZFC), which is to say, of ZFC1. Thus, ZFC1 is consistent, and so M agrees that ZFC1 is consistent, and so M is a model of ZFC2. Thus, ZFC2 is consistent, and so M agrees, and so ZFC3 is consistent, and so on. Do you see how it works? If ZFCα is consistent, then M will agree (if α is in M), and so ZFCα+1 is also consistent. (And limit stages are basically free, since proofs are finite.)
So the scheme of theories ZFCα forms a hierarchy of consistency strength that sits very low below the beginning of the large cardinal hierarchy. I think much of the sense of your question is this:  

We know by the Incompleteness theorem that no theory can prove its own consistency, and so we want to consider theories that transcend this consistency in the way you describe. 

And this is exactly what the large cardinal hierarchy provides. Each level of the large cardinal hierarchy implies the consistency of the lower levels, and the consistency of the consistency and so on, iterating in the style of your questions. But the large cardinals are able to jump higher than these small steps of consistency, by finding natural axioms that imply the consistency of all iterations of the consistency process that you describe for the lower levels. 

I just noticed the bit at the end of your question, about whether independence results also hold for ZFCα. This is a very interesting question, and the answer is Yes, they all work just the same. The reason is that all the independence results, proved either by forcing or by the method of inner models, have the property that the resulting models have the same arithmetic truths as the original model. Since the consistency statements you are considering are arithemtic statements, they are not affected by forcing or inner models. In particular, Cohen's  proof that Con(ZFC) implies Con(ZFC+¬CH) turns directly into a proof that Con(ZFCα) implies Con(ZFCα+¬CH). If one formalizes a version of (ZFC+¬CH)α, it follows that it will be equivalent to ZFCα+¬CH. And the same holds for all the other indpendence results of which I am aware.<|endoftext|>
TITLE: When are all characteristic l representations liftable
QUESTION [7 upvotes]: Suppose $G$ is a finite group, and $l$ is a prime, with $l$ coprime to the order $|G|$. (Thus we have complete reducibility for $G$ representations.) Is there a straightforward condition on $l$ which ensures that every irreducible representation of $G$ is liftable to a characteristic zero representation? (For instance, does the fact that we assume $l$ coprime to $|G|$ suffice?)

REPLY [7 votes]: Another sufficient condition is that if $G$ is solvable, then for every prime $l$, every absolutely irreducible characteristic $l$ representation can be lifted to the complex numbers. In fact, solvability is not really necessary; $l$-solvability suffices. 
This is the Fong-Swan theorem.

Added later:  Since groups with order not divisible by $l$ are trivially $l$-solvable,
this sufficient condition includes, and is more general than the the condition stated by Pete L. Clark.<|endoftext|>
TITLE: Split powers of the multiplicative group of a field
QUESTION [6 upvotes]: Let $K$ be a field, $K^\times$ its multiplicative group and $I$ an infinite set. Is then $(K^\times)^{(I)} \subseteq (K^\times)^I$ a direct summand? If not, is it possible to characterize the fields for which this is true?
In any case, it's a pure subgroup. If $K$ is finite, the answer is yes. If $K$ has arbitrary roots, that is $K^\times$ is divisible, then it's also true. If $K^\times$ is the additive group of a vector space (i.e. it's elementary abelian for some prime or uniquely divisible), you can use linear algebra. If $K^\times$ is a finite direct sum of these types, then it also works; e.g. $\mathbb{R}^\times = \mathbb{Z}/2 \times \mathbb{R}^+$.
Now what about $K = \mathbb{Q}$. Here $K^\times = \mathbb{Z}/2 \oplus \mathbb{Z}^{(\mathbb{P})}$. If $I=\mathbb{N}$ and $\hom((\mathbb{Z}^{(I)})^I,\mathbb{Z})$ is countable, then it's false. But I don't know if this is true, the argument of Specker computing $\hom(\mathbb{Z}^\mathbb{N},\mathbb{Z})$ does not seem to take over. Another case would be that $K^\times$ is torsion, i.e. $K$ is an algebraic extension of $\mathbb{F}_p$ for some prime $p$, e.g. $K = \mathrm{colim}\_s \mathbb{F}\_{p^{q^s}}$ for some prime $q$ and $K^\times = \mathrm{colim}\_s \mathbb{Z}/(p^{q^s}-1)$. This is a subgroup of $\mathbb{Q}/\mathbb{Z}$, which does not have to be divisible.
I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one. But does this $G$ also arise as $K^\times$? (EDIT: I know that $G=\mathbb{Z}, I = \mathbb{N}$ does it, but $\mathbb{Z}$ is no $K^x$.) There are several characterizations1 when $G$ has the form $K^\times$ for some field $K$. Perhaps this is useful here. The whole question is motivated by the study of $K \otimes_K \otimes_K ...$ as defined here.
1R.M. Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field, Proc. London Math. Soc., 18, 1968, p.114 - 124

REPLY [5 votes]: Re: "I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one."
Let $G$ be the the integers, and $I$ a countable indexing set.  If $G^{(I)}$ were a direct summand, let $P$ be a complement summand.  
We arrive at a contradiction as follows: First, $P$ is isomorphic to $G^{I}/G^{(I)}$ which contains the element $(2,4,8,16,32,\ldots)$ modulo $G^{(I)}$, which (as we can peal off any of the initial terms) is a non-zero element which is infinitely divisible by 2.  Second, $P$ is a subgroup of $G^{I}$, which has no infinitely divisible elements (other than zero).

I think this argument might be modified to show that the algebraic closure of a finite field will give you the counter-example you need (changing "divisibililty" to some sort of degree consideration), but I don't have a lot of time to think about it right now.  I'll come back later if someone else doesn't answer your question fully.

Back now. Try the following.  Let $K$ be the field obtained by adjoining to $\mathbb{Q}$ the $2^{p}$th root of each prime prime $p$ (in $\mathbb{Z}$).  Let $G$ be the multiplicative group of $K$.  Suppose by way of contradiction that $G^{(I)}$ is a direct summand of $G^{I}$, and let $C$ be a complement.  As before, $C$ is isomorphic to $G^{I}/G^{(I)}$, and the element $(2,3,5,\ldots )$ modulo $G^{I}$ is infinitely divisible by $2$ (thinking of ``divisibility'' multiplicatively in this case--in other words, after chopping off the front, we can take square roots as many times as we want).  However, I believe it is the case that there is no element of $G^{I}$ which is infinitely divisible in this sense.  (I'll leave it to the experts to prove this, but I think some form of Kummer theory would suffice.  But it may be difficult to prove it.)  [One last edit: I think it may even be easier to look at $\mathbb{Q}(x_1,x_2,\ldots)$ and adjoin a $2^{n}$th root of $x_{n}$, and modify the example accordingly.]<|endoftext|>
TITLE: What is an immersed submanifold?
QUESTION [15 upvotes]: An  immersed submanifold is by definition the image of a smooth immersion. I know some examples but I lack general understanding of what immersed submanifolds look like. For example, can one characterize subsets of a manifold $M$ that are immersed submanifolds of a given dimension? 
For example, subsets of $M^n$ that are embedded smooth $k$-dimensional submanifolds $S$ are those for which $(M, S)$ is locally diffeomorphic to $(\mathbb R^n, \mathbb R^k)$, so by looking at $S$ one can instantly tell if it is a smooth submanifold. How does one do the same for immersed submanifolds? Is the union of countably many embedded submanifolds an immersed one? Are there any immersed submanifolds that cannot be decomposed as  the union of countably many embedded submanifolds?

REPLY [7 votes]: I think Anton said basically the same thing, but I'll expand a bit. When I think of an immersed submanifolds, two reasonable definitions come to my mind:

A map $f:N \to M$ such that N, M are both differential manifolds, $\dim M >\dim N$, and the map is locally an embedding, i.e. the derivative matrix at each point has no kernel.
The same as above, but with the additional requirement that the map be transverse to itself.

(In fact, for me an immersion is almost always number 2, but 1 might make more sense sometimes. In general, all books I've seen say that there is no universally agreed upon definition of immersed/embedded submanifolds)
I do not think it makes sense to think of the submanifold as just the image of that map. In particular, the main reason to have submanifolds is to talk about tangent vectors to the submanifolds, and this makes no sense unless you have the map. (When you imagine a tangent vector to the image, what you are actually taling about is a tangent vecotr to N).
If you accept 2 as the definition, it's an interesting question of whether you can reconstruct the map f from just the image in any reasonable unique way. I think the answer should be yes for reasonable examples, but there might be a weird counterexample. If 1 is the definition, the answer is certainly "no" (just imagine a figure eight where the self-intersection is a small interval rather than just a point). In any case, I don't think you'll be able to do anything with your immersed submanifold unless you have the map.

My answers to the specific questions of the original poster:
1) Union of countably many submanifolds is an immersed submanifolds iff you consider a disjoint union of countably many abstract manifolds a manifold. Note that for embedded submanifolds, it's always possible to construct a map from the corresponding abstract manifold to M.
2) This depends on whether you require the embedded submanifolds to be closed. A figure eight cannot be decomposed into a union of embedded closed differentiable manifolds. If they don't have to be closed, as Andrey said in the comments, you can cover N by open sets small enough that the map is an embedding on each.<|endoftext|>
TITLE: Why can't subvarieties separate?
QUESTION [5 upvotes]: I'm posting my answer to this question as its own question:
Let $V$ be an irreducible projective variety over $\mathbb{C}$. Let $U$ be a Zariski open set in $V$.  I'll use $V(\mathbb{C})$ and $U(\mathbb{C})$ to mean $V$ and $U$ equipped with their Euclidean topologies, respectively.
What is the easiest proof that $U(\mathbb{C})$ is connected?
Here's the proof I know:  Suppose that $U(\mathbb{C})$ can be written as a disjoint union of two open sets $A$ and $B$.  Since the complement of $U$ in $V$ is a variety of smaller dimension than $V$, a theorem of Remmert and Stein implies that the closures $\overline{A}$ and $\overline{B}$ of $A$ and $B$ in $V(\mathbb{C})$ are projective analytic sets. By Chow's theorem that projective analytic sets are algebraic, $\overline{A}$ and $\overline{B}$ are subvarieties of $V$.  Since they're proper, $V$ is not irreducible, and we have a contradiction.
I guess I'm really asking for the most elementary argument, as I think the above argument is nice intuitively.  A reference would be fine.
(To avoid going through the same discussion in the comments that happened at the other question, let me point out that I am aware that irreducible varieties are connected and that $U$ is itself a variety in the sense that it is locally affine.  It is just not obvious to me that it is irreducible (without appealing to the above argument).)

REPLY [10 votes]: [This has been completely rewritten at the request of Autumn Kent.]
Let $X$ be an irreducible topological space and $U$ a non-empty open subset of $X$.  Then $U$ is also irreducible -- see e.g. Proposition 141 on page 88 here.  (Surely it's also in Hartshorne and lots of other places, but one of the advantages of typing up your own notes is to be able to easily point to a reference because you know exactly where it is.)
Thus the question reduces to the fact that if $X_{/\mathbb{C}}$ is an irreducible complex variety, then $X(\mathbb{C})$ with its "Euclidean topology" is connected.  For this, see e.g. Section VII.2 of Shafarevich's Basic Algebraic Geometry II.  (Again, there are other places, but I think his discussion is especially good.) He gives two different proofs, one of which is a simple induction on the dimension.<|endoftext|>
TITLE: Submitting to arXiv when unaffiliated
QUESTION [40 upvotes]: I am writing a short paper in the area of combinatorics.
When the paper is complete, I would like to be able to submit it to arXiv.
The reasons that I would like to submit to arXiv are:

To obtain a date and time stamp from a central authority so that I can prove the work is mine.
To promote access to the paper under the Creative Commons Attribution-Noncommercial-ShareAlike license

Although I have created an account on arXiv, because I am not affiliated to any academic institution, it appears that it might be difficult to get the paper accepted by arXiv.
For example see,
arxiv.org/help/submit

The following information is also required for submission:
Institutional affiliation for the
  author(s) must be provided. Official
  report number(s) from the author(s)
  institution(s) must be provided.

Also the following blog article does not appear encouraging:
Does The Arxiv Blacklist Authors ?
Does anyone on mathoverflow have any advice on how to submit to arXiv (or a similar database) without an academic affiliation?
Thanks in advance for any advice.

REPLY [5 votes]: Official Report Number tends to stand for internal numbers assigned by the University or Research Institution for Technical Reports.  For example, ATT Bell Labs often published their internal research documents as Technical Reports, and the Computer Science department at M.I.T. also tends to publish internal research findings, often supported by government sources or sponsored-research funded by corportations, as "white papers" or technical reports.
MIT's library is pretty good at finding these things: http://libguides.mit.edu/techreports, and their webpage defines technical reports as
What is a technical report?
Technical reports:

Are written to convey new developments or final results of scientific and technical research.
Are usually funded by government departments or corporate bodies.
Deliver technical information to the funding organization.
Provide a forum for peer information exchange.
Are not easy to find


Also, a lot of research funded by the military or DARPA or the OSD in the United States has a Final Report as its ultimate end-result and the means of disseminating the findings and recommendations, rather than a peer-reviewed journal article.
It is a shame that have an educational e-mail address for affiliation gets you a bye for submitting to arxiv; perhaps even academics should need an endorsement prior to being allowed to commit an article to arxiv.  The seed recommenders would have to have been planted earlier, anyway.<|endoftext|>
TITLE: Digital pen for math: your experiences?
QUESTION [32 upvotes]: This question is similar to this question but with a slightly different purpose.
For many of the reasons presented in that question, note taking can be useful but tedious. Similarly, this question addresses the difficulty of keeping track of notes(research notes in that case). Myself, I have been amassing pages and pages of notes from courses and seminars. The really important ones I attempt to Latex but often get lazy. My question:

Have you tried using technology like the digital pen for taking notes during math lectures? If so, what do you think? If not, discuss your thoughts on this.

In general, I am just trying to get a feel if it would be worth buying one for my classes (particularly seminar).
A few notes for those of you too lazy to look at the website of the pen:

The pen records everything you write and uploads it as a image/video file into a computer.
The pen also records audio synced with your writing, so you can play the notes as a movie with the audio accompaniment.
The pen also writes in regular ink in a notebook, so you also have a hard copy.

While fairly different, those with experience using a tablet PC should also feel welcome to post their experience, since the outcome of the digital pen produces something like that of the tablet, just with some added functionality.
Thanks ahead of time, today I nearly bought one, but I thought I would ask here first.

REPLY [2 votes]: EDIT: After buying a neo smartpen and using it for 6 months or so I have a few more comments.  First, the smartpen eventually broke like every other one I've ever owned so if you aren't super careful with your pens be on guard.  Secondly, it just wasn't worth the trouble.  None of the software accurately transcribes math symbols, the tagging and searching systems suck for math so at the end of the day I just paid more than a hundred bucks for a system which was about as good as just photographing my pages with my cellphone once I write them.  I was a big fan of this idea so it's too bad but I'm going to try one of those digital pads for drawing next.
I used to use a livescribe pen years ago (perhaps an echo) and it wasn't very convenient.  The biggest problem at that time was that handwriting recognition was utterly terrible and thus the resulting notes were completely unsearchable and there was no real convenient way to synchronize in a way that didn't perfectly match up with the notebooks (if I accidentally did work on Martin-Lof randoms in the notebook for REA sets I couldn't move those two pages over to the appropriate place in a way that didn't get messed up on next sync).  Not to mention the app sucked.  However, both the actual writing capture and the audio capture worked quite well.
My conclusion was that for me it was more a way to make my research notebooks extra cool and special and it wasn't actually useful but that had I been the sort of person who took notes at talks or classes and went back to look at them it would have been mildly useful and could have been super useful with better handwriting recognition.  A pen which combined handwriting recognition with some kind of latex recognition engine would have been very useful.
I'm thinking of buying another smart pen now that the tech has had some time to mature and my advice to anyone else considering this is make your buying choice based on the app, third party tools/api, notebooks/printable paper and pen quality rather than the pen tech.  At least assuming it has basic functionality.

Note that after doing a bit more research I see that livescribe hasn't changed or even doubled down on many of the things I found frustrating about their system.  Their notebooks might be fine for taking notes in class but I want something that's both nicer and full sized for my research and the best they can do is their executive notebook which is both unimpressive and comes in only two colors.  To be fair to livescribe their pens seem better equipped (memory) to record audio but I don't use that feature.  Also they support export directly to onenote (as pdfs) but only seem to support auto-sync to evernote and even there you seem to need to tap sync.
Worse, rather than fix the issues with their API they've removed developer access entirely and (though I haven't fully checked this).  I'd like a smartpen which I can at least dream of having math recognition added.
OTOH I've been quite impressed with what I've seen of the neo smartpens.  They may lack a mac desktop app but they have a published API on github, a very nice selection of notebooks, seem to support more options for exporting content (e.g. svg) syncing (auto google drive) and advertise the ability of their software to regroup pages in dynamic notebooks.  I'm going to purchase one today and if anyone wants to know how it worked out comment on this answer to remind me to come back and update it.
Moleskin doesn't look bad either but my understanding is it's basically a neo smartpen underneath and I couldn't figure out from the website if it's fully compatible or you are stuck with a few (really nice!!) smart notebooks from moleskin and if it's not fully neo compatible I don't want to be locked in to such a boutique option.<|endoftext|>
TITLE: Stokes theorem for manifolds with corners?
QUESTION [21 upvotes]: Maybe this is an elementary question, but I'm unable to find the appropriate reference for it. The Stokes theorem tells us that, for a $n+1$-dimensional manifold $M$ with boundary $\partial M$ and any differentiable $n$-form $\omega$ on $M$, we have $\int_{\partial M} \omega = \int_M d\omega $.
But Stokes theorem is also true, say, for a cone $M = \{(x,y,z) \in \mathbb{R}^3 \ \vert\ \ x^2 + y^2 = z^2, 0\leq z \leq 1 \}$, or a square in the plane, $M =\{(x,y) \in \mathbb{R}^2 \ \vert\ 0 \leq x, y\leq 1 \}$ which are not manifolds. So my questions are:

Are these cone and square examples of what I think are called "manifold with corners"?
If this is so, where can I find a reference for a version of Stokes' theorem for manifolds with corners?
If "manifold with corners" is not, which is the appropriate setting (and a reference) for a Stokes' theorem that includes those examples?

Any hints will be appreciated.
EDIT: Since thanking individually everyone would be too long, let me edit my question to acknowledge all of your answers. Thank you very much: I've found what I was looking for and more.

REPLY [10 votes]: If you are looking for an online reference, you can check out Brian Conrads course notes on differential geometry.  Near the bottom of that page, you can find the handout with Stokes theorem for manifolds with corners.<|endoftext|>
TITLE: Are there Néron models over higher dimensional base schemes?
QUESTION [19 upvotes]: Are there Néron models for Abelian varieties over higher dimensional ($> 1$) base schemes $S$, let's say $S$ smooth, separated and of finite type over a field?
If not, under what additional conditions?

REPLY [3 votes]: See my preprint http://arxiv.org/abs/1410.5293 Theorem 3.2 on page 7ff.:
Let $S$ be a regular, Noetherian, integral, separated scheme, and $g: \{\eta\} \hookrightarrow S$ the inclusion of the generic point.  Let $\mathcal{A}/S$ be an Abelian scheme.  Then 
$$
    \mathcal{A} = g_*g^*\mathcal{A}
$$
as sheaves on $S_{\mathrm{sm}}$. (This is the "Néron mapping property" for $\mathcal{A}/S$.)<|endoftext|>
TITLE: Non Lebesgue measurable subsets with "large" outer measure
QUESTION [6 upvotes]: It is well known that for any set A in R^d there exists a measurable set E such that E contains A and m*(A)=m*(E). Is it possible to go the other direction?
In other words, is it true that  for any measurable set E (such that m(E)>0) there is a non-measurable subset A such that m*(A)=m*(E)?

REPLY [14 votes]: A set $E$ with positive Lebesgue measure can be decomposed as a union $E = A \cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A) = m^*(B) = m(E)$.  
An example for this construction is a Bernstein set.<|endoftext|>
TITLE: Can the symmetric groups on sets of different cardinalities be isomorphic?
QUESTION [55 upvotes]: For any set X, let SX be the symmetric group on
X, the group of permutations of X.
My question is: Can there be two nonempty sets X and Y with
different cardinalities, but for which SX is
isomorphic to SY?
Certainly there are no finite examples, since the symmetric
group on n elements has n! many elements, so the finite
symmetric groups are distinguished by their size.
But one cannot make such an easy argument in the infinite
case, since the size of SX is 2|X|,
and the exponential function in cardinal arithmetic is not
necessarily one-to-one.
Nevertheless, in some set-theoretic contexts, we can still
make the easy argument. For example, if the Generalized
Continuum Hypothesis holds, then the answer to the question
is No, for the same reason as in the finite case, since the
infinite symmetric groups will be characterized by their
size. More generally, if κ < λ implies
2κ < 2λ for all
cardinals, (in another words, if the exponential function
is one-to-one, a weakening of the GCH), then again
Sκ is not isomorphic to
Sλ since they have different
cardinalities. Thus, a negative answer to the question is
consistent with ZFC.
But it is known to be consistent with ZFC that
2κ = 2λ for some
cardinals κ < λ. In this case, we will
have two different cardinals κ < λ, whose
corresponding symmetric groups Sκ and
Sλ nevertheless have the same cardinality. But can we still 
distinguish these groups as groups in some other (presumably more group-theoretic) manner? 
The smallest instance of this phenomenon occurs under Martin's Axiom plus ¬CH, which implies 2ω =
2ω1. But also, if one just forces ¬CH by adding Cohen reals over a model of GCH, then again 2ω =
2ω1. 
(I am primarily interested in what happens with AC. But if
there is a curious or weird counterexample involving
¬AC, that could also be interesting.)

REPLY [4 votes]: Maybe too late as for a question asked 11 years ago, but I would like to inform that this paper contains a topological proof of the following theorem which is related to the answer of Mariano Suárez-Álvarez.

Theorem. For any cardinals $\kappa<\lambda$, the group $Alt(\lambda)$ of even finitely supported permutations of $\lambda$ is not isomorphic to a subgroup of the permutation group $Sym(\kappa)$.<|endoftext|>
TITLE: Are there as many real-closed fields of a given cardinality as I think there are?
QUESTION [17 upvotes]: Let $\kappa$ be an infinite cardinal.  Then there exists at least one real-closed field of cardinality $\kappa$ (e.g. Lowenheim-Skolem; or, start with a function field over $\mathbb{Q}$ in $\kappa$ indeterminates, choose an ordering and a real-closure).  
But I think there are many more, namely $2^{\kappa}$ pairwise nonisomorphic real-closed fields of cardinality $\kappa$.  This is equal to the number of binary operations on a set of infinite cardinality $\kappa$, so is the largest conceivable number.
As for motivation -- what can I tell you, mathematical curiosity is a powerful thing.  One application of this which I find interesting is that there would then be $2^{2^{\aleph_0}}$ conjugacy classes of order $2$ subgroups of the automorphism group of the field $\mathbb{C}$.   
Addendum: Bonus points (so to speak) if you can give a general model-theoretic criterion for a theory to have the largest possible number of models which yields this result as a special case.

REPLY [11 votes]: For real closed fields this is fairly easy.
First show that for any infinite cardinal k there are 2^k nonisomorphic linear orders
of cardinality k
For example if X is a subset of k let A_x be Q+2+Q if x is in k and Q+3+Q if x is not in X.  Let L_X be the sum of the A_x for x in k.  It is easy to see that L_X is isomorphic
to L_Y if and only if X=Y.
If F is a real closed and x and y are infinite element of R we say that x and y are 
comparable if and only if there are natural numbers m and n such that x is less than y^m and y is less than x^n.
The ordering of R induces a linear order L_R of the comparability classes, which we call the ladder of R.
Suppose L is a linear order. Let F be the real algebraic numbers. Let R_L be the real closure of the transcendental extension of the real algebraic numbers F(x_l:l\in L) ordered such that if i is less than j then x_i^n is less than x_j for all n. It's not hard to show that the ladder of R_L is isomorphic to L.
Thus if we start with nonisomorphic orders A and B then the fields R_A and R_B will be 
nonisomorphic.<|endoftext|>
TITLE: étale cohomology with G_m coefficients
QUESTION [9 upvotes]: Most calculations of étale cohomology in Milne's book deal with constructible or torsion sheaves. Are there references where the cohomology of varieties with $\mathbf{G}_m$ coefficients are calculated? I'm especially interested in the top dimension $2\mathrm{dim}(X)$ ($+ 1$). I found some calculations in Le groupe de Brauer (In: Dix Exposés sur la Cohomologie des Schémas), but they don't help me.
Edit: Assume $X$ is a variety over a finite field.

REPLY [5 votes]: I found calculations in S. Lichtenbaum, Zeta functions of varieties over finite fields at s = 1, Arithmetic and geometry, Vol. I, 173–194 Progr. Math., 35, Birkhauser Boston, Boston, MA, 1983, especially Proposition 2.1 and Theorem 2.2--2.4.<|endoftext|>
TITLE: Mode of convergence of a power series
QUESTION [5 upvotes]: I am looking for a power series $\displaystyle f(z) = \sum_{n=0}^{+\infty} a_n z^n$ that converge uniformly on $\mathcal{D} = \Big\{ z \in \mathbb{C} \ / \  \vert z \vert \leq 1 \Big\}$ but not normally. Of course, a proof that this is impossible would be even better. It seems close to this question, but it's not quite the same.
It is well-known that normal convergence implies uniform convergence, and that the converse is false but I haven't found yet a counterexample in the form of a power series on $\mathcal{D}$.
In other words, I would like a complex sequence $(a_n)_{n \in \mathbb{N}}$ satisfying the three conditions :

$\displaystyle\sum_{n=0}^{+\infty} a_n z^n$ converges for all $z$ such that $\vert z \vert \leq 1$
$\displaystyle \sup_{\vert z \vert \leq 1} \left\vert \sum_{n=N}^{+\infty} a_n z^n \right\vert \longrightarrow 0$ when $N\to+\infty$
$\displaystyle\sum_{n=0}^{+\infty} \vert a_n \vert = +\infty$

REPLY [8 votes]: An example, due to Fejér, appears in Hille's Analytic function theory on page 122 of the second edition, volume 1.
Erdős wrote a paper showing the existence of classes of examples with many zero coefficients, and in that paper he states that Hardy was the first to find an example of what you're looking for.  However, Erdős refers to Landau's book for evidence, and I don't read German, so someone more capable and willing may check that out.  
Edit: I initially misnamed the author and book containing the example, as pointed out by Harald Hanche-Olsen.
Update: Coincidentally, while skimming through Sasane's Algebras of holomorphic functions and control theory, I came upon another reference to Hardy's example.  This time the reference was to Dienes's The Taylor series.  I checked, and sure enough there is Hardy's example in the chapter "The Taylor series on its circle of convergence."   I scanned it, along with a page containing a result used in the example.  Here it is.  (The example starts two thirds of the way down page 464, and the cited "105.IV" is the "IV" that appears on page 441.)<|endoftext|>
TITLE: non-Dedekind Domain in which every ideal is generated by at most two elements
QUESTION [14 upvotes]: Does anyone know of such a domain?

REPLY [8 votes]: By way of comparison, Dedekind domains are characterized by an even stronger property, sometimes referred to colloquially as "$1+\epsilon$''-generation of ideals.  Namely:
Theorem: For an integral domain $R$, the following are equivalent:
(i) $R$ is a Dedekind domain.
(ii) For every nonzero ideal $I$ of $R$ and $0 \neq a \in I$, there exists $b \in I$ such that $I = \langle a,b \rangle$.  
The proof of (i) $\implies$ (ii) is such a standard exercise that maybe I shouldn't ruin it by giving the proof here.  That (ii) $\implies$ (i) is not nearly as well known, although sufficiently faithful readers of Jacobon's Basic Algebra will know it: he gives the result as Exercise 3 in Volume II, Section 10.2 -- "Characterizations of Dedekind domains" -- and attributes it to H. Sah.  (A MathSciNet search for such a person turned up nothing.)  The argument is as follows: certainly the condition implies that $R$ is Noetherian, and a Noetherian domain is a Dedekind domain iff its localization at every maximal ideal is a DVR.  The condition (ii) passes to ideals in the localization, and the killing blow is dealt by Nakayama's Lemma.<|endoftext|>
TITLE: Does a triangulation without fixed simplex property always exist?
QUESTION [12 upvotes]: Suppose we are given a triangulable topological space $X$. If $X$ has the fixed point property (FPP), then obviously for every triangulation $K$ of $X$ and every simplicial map $f:K\to K$ a simplex $\sigma\in K$ exists such that $f(\sigma)=\sigma$. This will be called the fixed simplex property (FSP). 
One can give examples of simplicial complexes (included below) with FSP whose geometric realizations do not have FPP. Moreover, triangulations of spaces without FPP exist whose iterated barycentric subdivisions all have FSP.
Is it true that if $X$ is a triangulable space without FPP, then a triangulation of $X$ exists that does not have FSP? A related question: given a triangulation $K$ of $X$, can we find a (non-barycentric) subdivision of $K$ that does not have FSP?

Example (taken from J.A. Barmak's thesis (p. 101), who in turn cites K.Baclawski and A.Björner "Fixed points in partially ordered sets" (Adv. Math. 1979)):
Consider the regular CW-complex (square + 4 triangles) $C$ which is the border of a pyramid with square base. $C$ is homeomorphic to $S^2$, so it doesn't have FPP. Let $K$ be the following subdivision of $C$: divide each of the 3 triangular sides into 3 triangles by adding a vertex in the middle and divide the bottom square into 4 triangles the same way.
Name the top vertex of the pyramid $x$. Let $f:K\to K$ be simplicial. If $f$ is onto (on the vertices), then, by finiteness of $K$, $f$ is an automorphism. Since $x$ is the uniqe vertex that belongs to exactly eight 1-simplices, it is a fixed point. If $f$ is not onto, then $|f|$ is nullhomotopic and thus the Lefschetz number $\lambda(f)=1$. $|f|$ has a fixed point, so $f$ fixes a simplex.
The same argument (just count the 1-simplices that $x$ and other vertices belong to) applies to arbitrary barycentric subdivisions of $K$.
However, if in $C$ we subdivide only the square base into 4 triangles by adding a vertex in the middle, then the obtained simplicial complex doesn't have FSP.

REPLY [11 votes]: EDITED. The arugment related to Mostov rigidity is completed according to a nice suggestion of Tom Church
The answer to the first question is no. There exsit manifolds of dimension 3 such that every simlicial map of the manfiold to itself (for any simplicial decomposition) has a fixed point (and hence a fixed simplex). At the same time every 3-mafiold adimits a smooth  self-map without fixed points. 
Namely, take $M^3$ with vanishing first and second homology ($H_1(M^3,R)=H_2(M^3,R)=0$) and that is a hyperbolic 3-manifold.  Moreover take such $M^3$ that does not have isometries. Existence of such manifolds is a standard result of 3-dimensional hyperbolic geometry. Let us prove that every such manifold gives us an example.
Proof. All compact 3-manifolds have zero Euler characteristics, so on $M^3$ there is a non-vanishing vector field $v$. Take the flow $F_t$ generated by $v$ in small time $t$. This will give us a family of diffeo $F_t$ of $M^3$ that don't have fixed points for small $t$. So $M^3$ in not FPP. 
Now, let us show that $M^3$ has FSP. Take any simplicial decomposition of $M^3$. First we state a simple lemma (without a a proof)
Lemma. Consider a simplicial decomposition of a compact orienable manifold. Suppose we have a simpicial map from it to itself, that send simplexes of highest dimensions to simplexes of higher dimentions (i.e. don't collapse them) and don't indentify them. Then this is an automorphism of finite order. 
Corollary. Every non-identical simplicial map $\phi$ from $M^3$ to itself either collapses a simplex of dimension 3 or identifies two such simplexes. In paricular the generator of $H^3(M^3,\mathbb Z)$ is sent to zero by this map. 
This corollary together with Lefshetz fixed point theorem implies immediately that $\phi$
has a fixed point, and so it proves FSP for $M^3$ (we use that $H_1(M^3)=H_2(M^3)=0$).
Proof of corollary. If $\phi$ does not collapse 3 simplexes of $M^3$ or identify them, then it is  a homeomorphism of $M^3$ of finite order (Lemma above). From Mostov rigidity it follows that this automorphism is homotopic to the identity. In order to show that it IS in fact the identity we need to use a more involved statement suggseted by Tom Church below. Namely, a paritial case of Proposition 1.1 in  http://www.math.uchicago.edu/~farb/papers/hidden.pdf
says that for a hyperbolic 3-manifold the group of isometries of any Riemanninan metric on it is isomorphic to a subgroup of the group of hyperbolic isometries. By our choice the group of hyperbolic isometries of $M^3$ is trivial. It is clear that $\phi$ preserves a Riemannian metric on $M^3$. So by Prop 1.1 it is the identity.
From this it immediately follows that $\phi$ sends $H^3(M^3, Z)$ to zero (since the volume is contacted). End of proof.<|endoftext|>
TITLE: Does every non-empty set admit a group structure (in ZF)?
QUESTION [107 upvotes]: It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary operation of symmetric difference forms a group, and in ZFC there is a bijection between $S$ and the set of finite subsets of $S$, so the group structure can be taken to $S$. However, the existence of this bijection needs the axiom of choice.
So my question is

Can it be shown in ZF that for any non-empty set $S$ there exists a binary operation $\ast$ on $S$ making $(S,\ast)$ into a group?

REPLY [164 votes]: In ZF, the following are equivalent:
(a) For every nonempty set there is a binary operation making it a group
(b) Axiom of choice
Non trivial direction [(a) $\to$ (b)]:
The trick is Hartogs' construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from $\aleph(X)$ into $X$. Assume for simplicity that $X$ has no ordinals. Let $\circ$ be a group operation on $X \cup \aleph(X)$. Now for any $x \in X$ there must be an $\alpha \in \aleph(X)$ such that $x \circ \alpha \in \aleph(X)$ since otherwise we get an injection of $\aleph(X)$ into $X$. Using $\circ$, therefore, one may inject $X$ into $(\aleph(X))^{2}$ by sending $x \in X$ to the $<$-least pair $(\alpha, \beta)$ in $(\aleph(X))^{2}$ such that $x \circ \alpha = \beta$. Here, $<$ is the lexicographic well-ordering on the product $(\aleph(X))^{2}$. This induces a well-ordering on $X$.<|endoftext|>
TITLE: When does a quasicoherent sheaf vanish?
QUESTION [11 upvotes]: Let $F$ be a quasi-coherent sheaf on a scheme $X$.  To check that $F$ vanishes it suffices to check that all the stalks of $F$ vanish.  I would like to know whether it suffices to check that all the fibers of $F$ vanish.
(I think I am using standard terms: the tensor product over $O_X$ with the local ring at $x$ is the stalk, and the tensor product over $O_X$ with the residue field at $x$ is the fiber.)
It suffices to answer the question on an affine scheme.  Let $R$ be a commutative ring.  For each prime ideal $p$ of $R$ let $k(p)$ be the residue field of the local ring $R_p$.  Let $M$ be an $R$-module, and suppose that $\mathrm{Tor}_i(k(p),M) = 0$ for all $i$ and all $p$.  (Tor taken in the category of $R$-modules).  Does it follow that $M = 0$?
If $M$ is finitely generated, the answer is yes.  In that case $M$ vanishes even when $\mathrm{Tor}_0(k(p),M) = 0$ for all maximal $p$, by Nakayama's lemma.  My question is whether there is a good replacement for Nakayama's lemma when $M$ is not finitely generated.

REPLY [8 votes]: If the scheme is locally noetherian, this is true and can be proved by noetherian induction. In fact, you can even replace $M$ with an object of bounded derived quasi-coherent category, if you are interested in such things.
The proof is relatively straightforward: For a complex of modules $M$ over the ring $R$, we may assume that any non-zero $f\in R$ acts by a quasi-isomorphism $f:M\to M$ (by the induction hypothesis), and then the cohomology of $M$ are defined over the field of fractions of $M$.

REPLY [5 votes]: What about $R=\mathbb Z$ and $M=\mathbb Q/\mathbb Z$ ? All fibers are zero.<|endoftext|>
TITLE: Errata to "Principles of Algebraic Geometry" by Griffiths and Harris
QUESTION [70 upvotes]: Griffiths' and Harris' book Principles of Algebraic Geometry is a great book with, IMHO, many typos and mistakes. Why don't we collaborate to write a full list of all of its typos, mistakes etc?  My suggestions:

Page 10 at the top, the definition of $\mathcal{O}_{n,z}$ is wrong (or at least written in a confusing way).
Page 15, change of coordinates given for the projective spaces only work when $i < j$.  It states that the given transitions also work in the case when $j< i$.
Page 27, there needs to be a bar on the second entry of the $h_ij(z)$ operator defined.  Also, shouldn't the title of this section be geometry of complex manifolds, instead of calculus on complex manifolds?
Page 35, the definition of a sheaf is wrong.  The gluing condition should be for any family of open sets, not just for pairs of open sets! I've seem PhD students presenting this definition of sheaf on pg seminars.
Page 74, writes $D(\psi \wedge e)$, but $\psi$ and $e$ are in two different vector spaces, and one cannot wedge vectors in different vector spaces... I guess they mean tensor product.
Page 130, the definition of divisor says it's a linear combination of codimension 1 irreducible subvarieties. By linear it means over $\mathbb{Z}$ not over the complex numbers (better should say, like Hartshorne, that $\operatorname{Div}$ is the free abelian group generated by the irreducible subvarieties).
Page 180, equation $(\ast)$ has target a direct sum of line bundles, not tensor.
Page 366, when it says "supported smooth functions over $\mathbb{R}^n$", are these complex valued or real valued functions?
Page 440 top equation, Is it really correct?
Page 445, the second phrase of the hypercohomology section, it says sheaves of abelian sheaves. Probably means set of abelian sheaves.

REPLY [4 votes]: http://www.math.stonybrook.edu/~azinger/mat545-fall19/GHnotes.pdf
Found an online document by Aleksey Zinger on Griffith and Harris errata which is by far the most comprehensive one I have seen.<|endoftext|>
TITLE: What is 'formal' ?
QUESTION [29 upvotes]: The key step in Kontsevich's proof of deformation quantization of Poisson manifolds is the so-called formality theorem where 'a formal complex' means that it admits a certain condition. I wonder why it is called 'formal'. I only found the definition of Sullivan in Wikipedia: 'formal manifold is one whose real homotopy type is a formal consequence of its real cohomology ring'. But still I am confused because most of articles I found contain the same sentence only and I cannot understand the meaning of 'formal consequence'. Does anyone know the history of this concept?

REPLY [12 votes]: Paraphrasing Groucho Marx: if you don't like my first answer..., well I have another one. :-)
Here it is: let $X$ be a simply connected differentiable manifold.
Rational homotopy theory tells us that the rational homotopy type of $X$ (that is, its homotopy type modulo torsion) is contained in its minimal model, $M_X$, which is a commutative differential graded (cdg) algebra.
By definition, this means that you have a quasi-isomorphism (quis, a morphism of cdg algebras inducing an isomorphism in cohomology)
$$
M_X \longrightarrow \Omega^*(X)  \ .
$$
Here, $\Omega^* (X)$ is the algebra of differential forms of $X$ and the minimality of $M_X$ means that, in a certain, but precise, sense, it is the smallest cdg algebra for which such a quis exists.
The fact that $M_X$ contains the rational homotopy type of $X$ implies, for instance, that you can obtain the ranks of the homotopy groups of $X$ from it:

rank $\pi_n(X) =$ number of degree n generators (as an algebra) of $M_X$, for $n \geq 2$.

Nice, isn't it?  :-)
The problem is that the algebra $\Omega^*(X)$ is, in general, not computable, so you can not obtain from it the minimal model $M_X$. And here is where formality comes to help you.
Almost by definition, $X$ is a formal space if there exists two quis
$$
\Omega^*(X) \longleftarrow M_X \longrightarrow H^*(X;\mathbb{Q})  
$$
Hence, if $X$ is formal you can compute its minimal model $M_X$, and hence its rational homotopy type, directly from the cohomology algebra $H^*(X; \mathbb{Q})$, which is nicer (smaller, more computable) than $\Omega^*(X)$.
And the final point is that there are plenty of examples of spaces which are known to be formal.
(Final remark: Actually, you'd have to put $A_{PL}^*(X;\mathbb{Q})$ instead of $\Omega^*(X)$ to work over the rationals, but this you can find it explained in the references we have provided for you.)<|endoftext|>
TITLE: Expected determinant of a random NxN matrix
QUESTION [20 upvotes]: What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?

REPLY [17 votes]: It is a little more convenient to work with random (-1,+1) matrices.  A little bit of Gaussian elimination shows that the determinant of a random n x n (-1,+1) matrix is $2^{n-1}$ times the determinant of a random n-1 x n-1 (0,1) matrix.  (Note, for instance, that Turan's calculation of the second moment ${\bf E} \det(A_n)^2$ is simpler for (-1,+1) matrices than for (0,1) matrices, it's just n!.  It is also clearer why the determinant is distributed symmetrically around the origin.)
The log $\log |\det(A_n)|$ of a (-1,+1) matrix is known to asymptotically be $\log \sqrt{n!} + O( \sqrt{n \log n} )$ with probability $1-o(1)$; see this paper of Vu and myself.  A more precise result should be that the logarithm is asymptotically normally distributed with mean $\log \sqrt{(n-1)!}$ and variance $2 \log n$.  This result was claimed by Girko; the proof is unfortunately not quite complete, but the result is still likely to be true.<|endoftext|>
TITLE: probability in number theory
QUESTION [15 upvotes]: Hi,
I am hearing that there are some great applications of probability theory (or more general measure theory) to number theory. Could anyone recommend some good book(s) on that (or other types of references)?
Please one source per answer as I would like to make this community wiki.
Thanks in advance!

REPLY [3 votes]: I recently noticed a connection, while looking at a campy sort of problem.
The problem goes like this 
A strange sort of prison has 1200 cells and 1200 guards (each numbered 1-1200). Whenever a guard turns his key in a lock it either locks the cell or unlocks the cell. Every night guard 1 goes through and turns his key in each cell, locking all of them. Then guard 2 turns his key in each cell that is divisible by 2 (which unlocks each of these) and so on until all the guards have gone through their round. So the question is at the end of the night how many cells are locked, which cells are they. 
So you can figure out pretty easily that if a cell has an even number of divisors then it will be unlocked at the end of the night. Whereas if the cell has an odd number of divisors then it will end up locked. You can use the tau function to think about when a number will have an even number of divisors and when it will have an odd number of divisors. (I won't ruin the solution for anyone) While I was working on this I noticed that the probability of an integer having an odd number of divisors decreases by a factor of 1/2 whenever a new prime factor is added to the prime factorization of the integer. In other words to compute the probability that an integer has an odd number of divisors you can raise 1/2 to the number of distinct primes in the prime factorization. 
Once you figure out which cells are locked at the end of the night this conclusion will probably seem pretty worthless but it got me interested in the connection between number theory and probability<|endoftext|>
TITLE: Which commutative groups are the group of units of some field?
QUESTION [16 upvotes]: Inspired by a recent question on the multiplicative group of fields.  Necessary conditions include that there are at most $n$ solutions to $x^n = 1$ in such a group and that any finite subgroup is cyclic.  Is this sufficient?  (Edit:  Well, no, it's not, since the only such groups which are finite are the cyclic groups of order one less than a prime power.  Hmm.)

REPLY [7 votes]: Another characterization is theorem 2.1 in this paper on the field with one element:
http://arxiv.org/pdf/0911.3537
If H is a commutative group, let H+ be H together with a new element 0.  To give a field structure on H+ is equivalent to giving a bijection s:H+ --> H+ that commutes with all of its conjugates-by-H.
Maybe this is similar to Dicker's characterization?  Dicker mentions the operation x --> 1-x, while the s in Connes-Consani is meant to be x --> x + 1.<|endoftext|>
TITLE: Is there a name for this differential operator and/or its corresponding spectrum?
QUESTION [5 upvotes]: Let $\mathcal{M}$ be a real, compact, orientable manifold and let $X$ be a vector field on $\mathcal{M}$.  Consider the functional
$$E(f) = \int_{\mathcal{M}} X_p(f)^2 dV$$
where $X_p(f)$ is the directional (Lie) derivative of $f$ along $X$ at the point $p$ and $dV$ is a volume form on $\mathcal{M}$ -- this functional essentially measures the total amount of change in $f$ along $X$ over all of $\mathcal{M}$ in the $L^2$ sense.  Then $\delta E(f)$ is a differential operator whose eigenspectrum
$$\delta E(f) = \lambda f$$
(for $\lambda \in \mathbb{R}$) yields the critical points of $E$ over the set of functions with unit norm.  Is there an established name for this operator (or functional) and/or its corresponding eigenspectrum?
The prototype for this operator is Dirichlet's energy
$$\int_{\mathcal{M}} ||\nabla f||^2 dV$$
which has as its (unit-norm) critical points the Laplacian eigenspectrum
$$\nabla^2 f = \lambda f,$$
the main difference being that Dirichlet's energy measures the total gradient, i.e., the change in all directions, rather than just the change along a particular direction at each point.

REPLY [5 votes]: There is an simple explicit formula for your operator in terms of known operators. To see this, note that $\delta E_f(g)$ (the differential of $E$ at the function $f$ in the direction $g$) is equal to
$$
2 \int X(f) X(g) dV = 2\int \left[L_X(g X(f) dV )-g\left(X(X(f)) dV+X(f)L_X(dV)\right)\right]
$$
where $L_X$ is the Lie deriviative with respect to $X$. Now, $\int L_X(\alpha) = 0$ for any top degree form $\alpha$, so we get
$$
\delta E_f(g) = - 2 \int_X g \left[X(X(f)) + X(f) div(X)\right]dV
$$
where the divergence of $X$ is defined by $div(X)= L_X(dV)/dV$. So, using the $L^2$-inner product on functions, we can interpret the 1-form $\delta E$ as the differential operator
$$
D :f \mapsto  - X(X(f)) - X(f) div(X).
$$
(This is the same way that the differential of Dirichlet energy is seen as the Laplacian.) 
Note that the leading order part of D is just $X^2$ and so, in particular, $D$ is not elliptic. You'd expect this, of course, because $E$ only sees the change of $f$ in the $X$-direction.<|endoftext|>
TITLE: Weighted Regular Graphs
QUESTION [8 upvotes]: The following graph theoretic notion appeared in an economics paper entitled: "Prize competition under limited comparability, by Michele Piccione and Ran Spiegler which studies models of economics were the firms are rational but the consumers are not. 
A graph is weighted-regular if vertices (nodes) can be assigned positive weights, such that each node has the same total weighted-expected-number of edges (links). Every regular graph is, of course, weighted regular (give all vertices weight 1). A star, tree with center vertex connected to k leaves is weighted-regular: Give the center weight 1 and every leaf weight 1/k.
(In the paper, every vertex is also contained in a loop. Therefore its weight also counts when we compute its weighted-expected-number of edges. The star is still weighted-regular: give weight 1 to the center and weight 0 to the leafs.)
Is there a characterization of weighted-regular graphs? Is this notion known? Does it have known applications or connections?

REPLY [2 votes]: Denoting by $A$ the $n\times n$ adjaceny matrix (with loops contributing $1$ 
on the diagonal) of a finite graph $\Gamma$ with $n$ vertices, a necessary condition for $A$ to be weighted-regular is the fact that the all $1$ vector 
$I_n=(1,1,\dots,1)$ of dimension $n$ is in the image of $A$. This is the case if and
only if $I_n$ is orthogonal to the kernel $\ker(A)$ of $A$.
If this necessary condition holds, the vector $I_n$ can be written uniquely
as
$$I_n=\sum_{\lambda\in{\mathrm Spec}(A)\setminus\{0\}}v_\lambda$$
where the sum is over all distinct non-zero eigenvalues of $A$ with 
$v_\lambda$ denoting the orthogonal projection of $I_n$
onto the eigenspace of eigenvalue $\lambda$.
The graph $\Gamma$ is now weighted-regular if and only if 
the affine subspace $\sum_{\lambda}\frac{1}{\lambda}v_\lambda+\ker(A)$
(corresponding to the set of preimages of $I_n$) 
intersects the open cone $({\mathbb R}_{>0})^n$
of vectors having only strictly positive coordinates.
This last condition can be checked by linear programming by 
optimizing an arbitrary linear functional (eg. the coordinate sum)
on $\ker(A)$ subject to the condition that all coefficients of 
the solution are strictly greater than the corresponding coefficients of 
$I_n-\sum_{\lambda}\frac{1}{\lambda}v_\lambda$.
In particular, a graph $\Gamma$ with invertible adjacency matrix $A$ is 
weighted-regular if and only if all coordinates of the vector 
$\sum_\lambda \frac{1}{\lambda}v_\lambda$ are strictly positive and these coordinates
give the unique set of weights (up to multiplication by a positive constant) turning 
$\Gamma$ into a weighted-regular graph.
This is of course nothing else than Kristall Cantwell's remark, fleshed out.<|endoftext|>
TITLE: Rational exponential expressions
QUESTION [7 upvotes]: Consider the following extension of polynomials.  The rational exponential expressions (REXes) are given by:

The leaves 1 and $x$ for $x$ drawn from a class of variables; and
Closed under the binary functions of addition, multiplication, exponentiation, and division.

This is a restriction to positive constants of a class of expressions studied in Brown, 1969, "Rational Exponential Expressions and a Conjecture Concerning π and e".  Buchberger & Loos, 1982, "Algebraic Simplification", sect. 6, mention the existence of algorithms for finding canonical forms for these expressions, but say their correctness depends on number-theoretic conjectures that have not been settled.
I'm interested in a decision procedure for dominance, where for two REXes in one variable, $f$ and $g$, $f \leq g$ if $\exists N \in {\mathbb N}.\forall n \in {\mathbb N}. N \leq n \implies f(n) \leq g(n)$.  Do we have either the existence of canonical forms, or decidability for this problem?

REPLY [12 votes]: Yes, there are algorithms to decide asymptotic dominance - in fact for a much wider class of elementary functions. I discovered the first such algorithm circa 1980 while an undergrad member of the MIT Mathlab group researching effective algorithms for computing limits for the Macsyma symbolic computation system. Another different algorithm was discovered independently a handful of years later by John Shackell. You should be able to find references to the literature by googling the more recent buzzword "transseries".
Many computer algebra systems have (partial) implementations of these algorithms. Here's an example of my algorithm on (my generalization of) an example Rich Schroeppel proposed to attempt to stump my algorithm (he was convinced no such algorithm existed). It shows that $\rm{\: lim_{x\to\infty}\ d40 = e^a}$

Don't dare try L'Hopital's rule on that monster!
For further discussion (and the text form of the above image) see my post on sci.math, 1996/03/20, L'Hopital's rule question:
http://groups.google.com/group/sci.math/msg/05298104ac44efd2
http://groups.google.com/groups?selm=WGD.96Mar20231913%40berne.ai.mit.edu<|endoftext|>
TITLE: How to define tuples?
QUESTION [15 upvotes]: As you probably know, you can define $2$-tuples $(x_1,x_2)$ as $\{\{x_1\},\{x_1,x_2\}\}$; then you can define $n$-tuples $(x_1,x_2\ldots,x_{n})$ as $((x_1,x_2\ldots,x_{n-1}),x_n)$.
In alternative, you can define ordered pairs $\langle x_1,x_2\rangle$ as $\{\{x_1\},\{x_1,x_2\}\}$ (please notice the use of "ordered pairs" instead of "$2$-tuples" and the use of angular brackets instead of round ones); then you can see $n$-tuples as finite sequences, that is functions whose domain is the set of natural numbers from $1$ to $n$ and whose codomain is the set $\{x_1,\ldots,x_n\}$. So $n$-tuples are sets such as $\{\langle 1,x_1\rangle,\ldots\langle n,x_n\rangle\}$; $0$-tuples are defined to be the empty set.
The first definition is not so rigorous (see the use of dots) and works only for $n\geq 2$. The second definition is rigorous and works for every $n$, but then you end up having ordered pairs and $2$-tuples being different objects; this also implies that you have two kind of cartesian products, two kind of binary relations, two kind of functions and so on.
Is there a way to avoid such problems? Is there another better definition for $n$-tuples?
Thanks.

REPLY [2 votes]: Some people do have to care about such details, at least in unusual contexts, and I do think it’s generally worth being aware of your foundations.  The details of the definition of ordered pairs is crucial in Quine’s New Foundations (e.g., http://en.wikipedia.org/wiki/New_Foundations#Ordered_pairs), and taking it as primitive can have actual set-theoretic consequences in NF.  In Church’s unpublished supplement to his “Set Theory with a Universal Set,” he uses a deliberately ugly [my interpretation] definition of m-tuple to avoid collisions.  In my follow-on work, I use the usual Kuratowski definition of ordered pairs, since their internal structure allowed me to model the singleton function as a set, since it’s a 2-equivalence class, for a generalization of Church’s definition of j-equivalence relations.<|endoftext|>
TITLE: Obstruction bundle for spaces with Kuranishi structure
QUESTION [13 upvotes]: In the symplectic topology view on Gromov-Witten-Invariants some authors use what they call a Kuranishi structure on the moduli of stable maps. These were introduced by Fukaya and Ono and are also used in their big book on Fukaya categories.They are also used a lot in recent papers by Joyce. 
The key feature of the Kuranishi structure is that there locally exists a homeomorphism or a diffeomorphism, depending on whether you follow Fukaya or Joyce, to a zero set of a section of the "obstruction bundle".
On the algebraic geometry side there also is something involving the term obstruction, namely the perfect obstruction theory on the moduli space of maps. This is a morphism [E-1 -> E0] -> LX, where L is the cotangent complex.
Here's my question: To what does the obstruction bundle on the symplectic topology side correspond on the algebraic side?
There are three candidates I can think of:

E-1 
the kernel of E-1 -> E0
the thing the kernel of E-1 -> E0 maps surjectively to, mostly called the T2

The definition of the obstruction bundle on the symplectic side is a finite dimensional supspace of the cokernel of the linearized operator of the pseudoholomorphic curve equation, p.978 of Fukaya-Ono:"Arnold Conjecture and Gromov Witten Invariant." As a die hard algebraic geometer, that's just to hard to digest...

REPLY [9 votes]: By the way, the notion of Kuranishi structure is an attempt to 
formalize ideas that had been used in the gauge theory literature somewhat eariler.
See for example Friedman and Morgan's book Smooth Four Manifolds and Complex Surfaces.
In that book the obstruction bundle idea is used in the definition of the Donaldson
invariants.
The obstruction bundle idea for studying  nonlinear PDE problems of course goes back 
at least to  Kuranshi.  Taubes used this idea to study solutions to the Yang-Mills equations
near bubbling in the 1982  paper "Self-dual Yang-Mills connections on non-self-dual 4-manifolds" in way that lead to the modern use.<|endoftext|>
TITLE: Is there a Morse theory for sections of bundles or more generally for maps?
QUESTION [9 upvotes]: This question was prompted by my interpretation of a question by cosmologist Berian James.
Background
Some cosmologists have suggested using the cosmological dark matter density, which defines a function $f:M\to \mathbb{R}$ with $M$ the spatial universe, in order to probe the topology of $M$. (edit: Berian comments below that this may not be what this is about!  Listen to him... not to me!)  The original reference seems to be this paper by Gott et al..  Although the paper does not mention it explicitly, it seems that the natural mathematical framework for this proble is Morse theory.
Berian is interested not just in functions, but also e.g., vector fields or more generally sections of bundles on $M$.  Hence the following
Question
Can one extend Morse theory beyond functions to sections of bundles?  or perhaps to differentiable maps $f: M \to X$?
Pointers to the literature would be most welcome.
Cheers.

REPLY [4 votes]: There is some really cool recent work of Gay and Kirby on "Morse 2-functions", ie generic maps $f$ from an $n$-manifold $X$ to a surface $\Sigma$ (both compact, connected, oriented, smooth) (see http://arxiv.org/abs/1102.0750).
In the $n=4$, $\Sigma = S^2$ case, they have a reconstruction theorem (http://arxiv.org/abs/1202.3487) for recovering $X$ and $f$ from a combinatorial diagram on $S^2$, which in particular records the critical values (the "Cerf graphic").  They even understand what happens in families of Morse 2-functions, ie the analogue of birth-death moves.<|endoftext|>
TITLE: Stokes' theorem etc., for non-Hausdorff manifolds
QUESTION [12 upvotes]: This question is prompted by another one. 
I want to motivate the definition of a scheme for people who know about manifolds(smooth, or complex analytic). So I define a manifold in the following way.
Defn: A smooth $n$-manifold is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ is a sheaf of rings on it, such that, every point $x \in X$ has a neighborhood $U_x$ which is homeomorphic to an open set $V_x$ in $\mathbb{R}^n$ and $\mathcal{O}_X$ restricted to  $U_x$ is isomorphic to the sheaf of ring of smooth functions on $V_x$ and its open subsets.
This agrees with the usual definition using charts and atlases, for all except the requirement that a manifold is a (separable) Hausdorff space. But indeed it seems that many things in differential topology can be proven without using the Hausdorff property. In a fleeting conversation in a brief encounter, a personal I shall refer to as O., informed me that even Stokes' theorem can be done this way. But I am unable to ask O. again about this, as he is not physically around.
If the above is true, then this is a really good point to mention when introducing schemes to a person who knows about differentiable manifolds.
So my main question: 

Is it true that the proof of Stokes' theorem for smooth manifolds can be proven without the Hausdorff condition on the manifold? If so, is it done in any well-known reference?

Aside question:
What are some crucial propositions/theorems in differential topology that use Hausdorff condition, except those involving imbedding in some $\mathbb{R}^m$ for high enough $m$, for which of course Hausdorffness is a necessary condition(together with separable property)?
Tom Church answers below that partitions of unity does not work, for instance on the example of the line with the double point. However I believe that one can still make sense of integration of differential forms even with such pathologies, because by introducing a measure for instance, we can ignore sets of measure $0$.

REPLY [2 votes]: If you are a little bit careful as to what you mean by a "non-Hausdorff manifold" then I believe that Stokes' theorem will go through.
The crucial issue is that the smooth functions cannot detect the fact that the manifold is not Hausdorff.  Thus when you try to integrate a smooth function over the manifold, the function that you are integrating does not know that the manifold was not Hausdorff.  Thus the fact that partitions of unity don't work doesn't matter because they do work "up to Hausdorffness".
Thus if "non-Hausdorff manifold" essentially means "becomes a manifold upon Hausdorffification" then you ought to be alright.  I can't think of a counterexample off the top of my head,  but that doesn't mean that one doesn't exist (i.e. an example of a "locally Euclidean space" that does not quotient down to a "Hausdorff locally Euclidean space" - ignore paracompactness for this, that's cheating).
A slightly tangential point is that if you have a "locally Euclidean space" that is not Hausdorff then you probably have the wrong topology on it.  Take, again, the double pointed line.  You probably think that the correct topology on this is the one with basis either $(a,b)$ or $(-a,a)\setminus \{0\}\cup \{\ast\}$.  Wrong!  By doing so, you are artificially imposing the condition that $0$ and $\ast$ can be distinguished without justifying that assertion.  If you try by differential topological means (i.e. not topological, but differential topological) to separate $0$ and $\ast$ you find that you cannot tell the difference between them.  So the correct topology has basis $(a,b)$ with $a$ and $b$ of the same sign, and $(-a,a) \cup \{\ast\}$.  That is, the induced topology on the subset $\{0,\ast\}$ is the indiscrete topology.
At this point I should come clean (especially in the light of Emerton's comment to the original question) and say that when thinking of things that are like-but-not-manifolds then I think of Froelicher spaces.  So a non-Hausdorff manifold really means a non-Hausdorff Froelicher space with nice local properties.  And for Froelicher spaces, the difference between Hausdorff and non-Hausdorff is extremely small.
But if what you are interested in is smooth functions, then that's the right view to take.  And if you are integrating, then you are using smooth functions.  Of course, in other contexts you may want to remember more structure than just what the smooth functions can detect, but that's a different story.<|endoftext|>
TITLE: How many labelled disconnected simple graphs have n vertices and floor((n choose 2)/2) edges?
QUESTION [15 upvotes]: I would like to know the asymptotic number of labelled disconnected (simple) graphs with n vertices and $\lfloor \frac 12{n\choose 2}\rfloor$ edges.

REPLY [11 votes]: The vast majority of disconnected graphs have a single isolated vertex.
Let $A$ be a nonempty proper subset of $\{1,...,n\}$ of size $a$. Let $s(a)$ be the number of graphs with 
$e=\lfloor \frac12 {n \choose 2}\rfloor$ edges which have no edges from $A$ to $A^c$.
We want to count the union of all of these. Inclusion-exclusion works, with the dominant terms coming from when $a=1$.
An upper bound is the sum of $s(a)$ over all $A$ of size at most $n/2$, which is at most 
$n ~s(1)$ + ${n\choose 2}s(1)$ + $2^ns(3)$.
To get a lower bound, subtract the number of graphs with no edges connecting $A$ to $A^c$ or edges connecting $B$ to $B^c$ for all disjoint $\{A,B\}$. Denote this by $s(\#A,\#B)$. So, subtract 
${n\choose2}s(1,1) + 3^ns(1,2)$ from $n~s(1)$. 
The rest should be routine estimates on $s(1)$, $s(2)$, $s(3)$, $s(1,1)$, and $s(1,2)$.
$s(a,b) \le s(a+b)$.
$s(a) = ({n\choose 2} -a(n-a))$ choose $e$. 
Let the total number of graphs with $e$ edges be $\#G = s(0)$.
$$s(a)/\#G = \prod_{i=0}^{a(n-a)-1} \frac{\lceil{n\choose2}/2\rceil-i}{{n\choose2}-i}.$$
$s(2)/s(1) \le 2^{-n+3}$.
$s(3)/s(1) \le 2^{-2n+8}$.
The dominant term in both the upper bound and the lower bound is $n~s(1)$.
If I calculated correctly, that's asymptotic to $\frac 2 e n 2^{-n} ~\#G$.<|endoftext|>
TITLE: Why do so many textbooks have so much technical detail and so little enlightenment?
QUESTION [311 upvotes]: I think/hope this is okay for MO.
I often find that textbooks provide very little in the way of motivation or context. As a simple example, consider group theory. Every textbook I have seen that talks about groups (including some very basic undergrad level books) presents them as abstract algebraic structures (while providing some examples, of course), then spends a few dozen pages proving theorems, and then maybe in some other section of the book covers some Galois Theory. This really irks me. Personally I find it very difficult to learn a topic with no motivation, partly just because it bores me to death. And of course it is historically backwards; groups arose as people tried to solve problems they were independently interested in. They didn't sit down and prove a pile of theorems about groups and then realize that groups had applications. It's also frustrating because I have to be completely passive; if I don't know what groups are for or why anyone cares about them, all I can do is sit and read as the book throws theorems at me.
This is true not just with sweeping big picture issues, but with smaller things too. I remember really struggling to figure out why it was supposed to matter so much which subgroups were closed under conjugation before finally realizing that the real issue was which subgroups can be kernels of homomorphisms, and the other thing is just a handy way to characterize them. So why not define normal subgroups that way, or at least throw in a sentence explaining that that's what we're really after? But no one does. 
I've heard everyone from freshmen to Fields Medal recipients complain about this, so I know I'm not alone. And yet these kinds of textbooks seem to be the norm. 
So what I want to know is:

Why do authors write books like this?

And:

How do others handle this situation?

Do you just struggle through? Get a different book? Talk to people? (Talking to people isn't really an option for me until Fall...) Some people seem legitimately to be able to absorb mathematics quite well with no context at all. How?

REPLY [5 votes]: I hope no one will object my raising this question from the dead...
One point which has been alluded to by Tracer Tong but which is worth emphasizing is that it is sometimes very difficult to justify the usefulness of a fundamental concept without starting a whole new book. Just saying "This gets very important later on" may satisfy the lecturer/writer who knows what he is talking about but will leave the student with an aftertaste of argument by authority.
This happens most often with exercises : it is very tempting for the author to take an example or a theorem from a more advanced corner of his subject and strip it down of its fancy apparel. 
I'll list a few examples of mathematical concepts I encountered in this way "before their times" and came out with the first impression that those were silly and unmotivated - and changed my mind when I learned about them in a more thorough manner :

Hyperbolic geometry (!!)
p-adic numbers (!!!)
Dirichlet series
Milnor K-theory

I don't know the best option here... It is nice to see glimpses of more exciting subjects, but sometimes it is more a way to satisfy the (quite natural) inclination of the teacher for what lays further down the road.<|endoftext|>
TITLE: Condition for existence of certain lattice points on polytopes
QUESTION [10 upvotes]: Let $a_1,\cdots, a_n$ be integers such that $a_i\geq 2$ for all $i$ and $k>0$ another integer. 
I am interested in whether there exist integers $x_1,\cdots, x_n$ with $0<x_i<a_i$ satisfying:
$$ \frac{x_1}{a_1}+ \cdots + \frac{x_n}{a_n}=k  \ \ (*)$$ 
For example, if $n=2,k=1$, there exist a solution in the specified range iff  $\text{gcd}(a_1,a_2) \neq 1$, in other words $\text{lcm}(a_1,a_2)-a_1a_2 \neq 0$.
Question: For what pairs $n,k$ is there a polynomial $F$ whose inputs are $\text{lcm}$ of various subsets of $\{a_i\}_{1\leq i\leq n}$ such that (*) has a solution in the specified range if and only if $F\neq 0$ (see some examples I have in mind below)?  
Some observations:  If $k\geq n$ then $F\equiv 0$ works, trivially.
Some calculations for small values of $(n,k)$ suggest the following strange function: for each subset $I \subset \{1,\cdots,n\}$ define 
$$f_I(a_1,\cdots,a_n):= \frac{\prod_{i\in I}a_i}{\text{lcm}\{a_i\}_{i\in I}}$$
For $I=\varnothing$ we set $f_I=1$. Let:  
$$F(a_1,\cdots,a_n):= \sum_{I\subset \{1,\cdots,n\}}(-1)^{|I|}f_I(a_1,\cdots,a_n) $$
(You can replace $F$ by a polynomial of the $\text{lcm}$s which vanishes at the same time). This works for $(n,k)=(2,1)$ and probably $(3,1)$ (which would imply $(3,2)$). It does not work for $(4,1)$ but seems to work for some sequence with $(n,k)= (4,2)$ (may be the function depends on  $n/k$?). Have anyone seen this kind of formula before in other contexts?  
PS: I am not sure what tags should be used. Please feel free to re-tag.
EDIT: Apparently this question is related to existence of lattice points on a polytope. I checked through some of the references pointed to in this question on MO, but could not find the exact answer to what I wanted.

REPLY [3 votes]: I don't know if you're still interested in this problem Hailong, but here is a partial result. I make two natural restrictions :


Restriction 1 : avoid small primes. Let $B_n$ denote the (infinite) set of integers all of whose prime factors are $>n$. I assume that the all the $a_i$  are in $B_n$ (this is to avoid the difficulty that for example when $n=3$ and $k=1$, (*) has a solution
when $a_1=a_2=a_3=p$ for  $p$ a prime $ >2 $ , but not  when
$a_1=a_2=a_3=2$ ).
Restriction 2 : avoid small $k$. I assume that $k \geq \frac{n}{2}$ (this is to avoid
the difficulty that for example when $n=4$ and $a_1=a_2=2,a_3=a_4=3$, (*) has a solution
for $k=2$ but not for $k=1$).
Under those restrictions, the following conditions are equivalent :

(i) (*) has a solution in the desired range.
(ii) No $a_i$ is prime to all the others $a_j$.
(iii) The polynomial $F=\prod_{i=1}^{n}G_i$ is nonzero, where $G_i$ is
the polynomial $\sum_{j\neq i}(a_ia_j-\text{lcm}(a_i,a_j))$.
Note that the polynomial is independent of $k$.

The only difficult implication is $(ii) \rightarrow (i)$. To show this, consider
the undirected graph $G$ whose vertices are the integers from $1$ to $n$ and such that 
there is an edge joining $i$ to $j$ iff $gcd(a_i,a_j)>1$. Then condition (ii) says
that $G$ is connected. By a straightforward graph-theoretic lemma, there is
a subgraph of $G$ which is a disjoint union of stars. Thus, we can write
$\lbrace 1,2, \ldots n\rbrace$ as a disjoint union $A_1 \cup A_2 \cup \ldots \cup A_t (t \geq 1)$
such that for each $l$ between $1$ and $t$ we have $|A_l| \geq 2$ and there is a distinguished vertex $u_l$ in $A_l$ that is connected to all the other vertices in $A_l$. Restriction 2 ensures that we can find a decomposition
$k=\sum_{l=1}^{t}\alpha_l$ where each $\alpha_l$ is an integer
with $0<\alpha_l < |A_l|$. Restriction 1 ensures that we may find, for each $l$
$(x_i)_{i\in A_l}$ such that $0<x_i<a_i (i\in A_l)$ 
and $\sum_{i\in A_l}\frac{x_i}{a_i}=\alpha_l$. So we are done.<|endoftext|>
TITLE: Good books on arithmetic functions?
QUESTION [5 upvotes]: As I was studying the Möbius $\mu$ function and Gram series,
I got myself some pretty nice books:
Ribenboim - The New Book of Prime Number Records
Apostol - Introduction to Analytic Number Theory
Niven, Zuckerman, Montgomery - An Introduction to the Theory of Numbers
Iwaniec and Kowalski - Analytic Number theory
All of them deal with the Möbius $\mu$ function.
But none of them dealt with the subject in details other than giving a few theorems and problems...
So I would like to know, If you guys know of some good books that deal exclusively with arithmetic functions?

REPLY [2 votes]: (As I can't comment)
If you understand German, Springer published last year a translated version of Paul McCarthy's book - "Arithmetische Funktionen" (ISBN: 978-3-662-53731-2).<|endoftext|>
TITLE: Where does the Givental reconstruction formula come from?
QUESTION [15 upvotes]: In (for example) Semisimple Frobenius structures at higher genus (section 1.2) and Gromov-Witten invariants and quantization of quadratic Hamiltonians (section 6.8), Givental gives a conjectural formula for the higher genus Gromov-Witten potentials in terms of data coming from the Frobenius manifold of quantum cohomology, assuming that the quantum cohomology is (generically) semisimple. (The formula was proven to be correct by Teleman.) Givental does not give much explanation as to how or where he obtained these mysterious formulas. Can anybody here give some explanation or background?
Another question, more general, that I have is: Where does the quantized quadratic Hamiltonians formalism come from? How does it naturally arise? Presently (and still now, several months later after first asking this question...) everything just seems to me like a bunch of magical formulas that are pulled out a hat. I'd like to have this magic explained...

REPLY [49 votes]: OK, I guess, your first question is addressed to me. The answer is: fixed point localization. In my paper "Elliptic Gromov-Witten invariants and the mirror conjecture", a formula is found for the genus-1 (no descendants) potential of a semisimple target. It is a theorem, discovered and proved by fixed point localization when a torus acts on the target with isolated fixed points, and the GW-invariants are understood as equivariant ones. Since the answer is expressed in genus-0 data making sense for any semisimple Frobenius manifold, the conjectural extension to all such manifold is immediate.  (The conjecture was proved by Dubrovin-Zhang in the sense that they showed my formula being the only candidate that would satisfy Getzler's relation.) The paper of mine you are asking about, "Semisimple Frobenius structures at higher genus", does exactly the same that the elliptic paper, but for higher genus GW-invariants, first without, and then with gravitational descendants. 
After the fact, there is a more satisfying description of how that formula could have been invented. Dubrovin's connection of a semisimple Frobenius manifod allows for an asymptotical fundamental solution (which looks like the complete stationary phase asymptotics of oscillating integrals on the mirror theory). It's construction ("the $R$-matrix") is contained in the key lemma in that elliptic paper I've mentioned. Another way to interpret this solution is to say - in terms of overruled Lagrangian cones in symplectic loop spaces as the objects that describe genus-0 theory in lieu of Frobenius structures - that the overruled Lagrangian cone of a semisimple Frobenius manifold is isomorphic to the Cartesian product of several such cones corresponding to the one-point target space, and moreover, the isomorphism is accomplished by transformations from the twisted loop group: $$ L = M (L_{pt}\times \cdots \times L_{pt}).$$ The "mysterious" conjectural higher genus formula simply says that the same relation persists for the total descendant potentials of higher genus theory:$$D \sim \hat{M} (D_{pt}\otimes\cdots\otimes D_{pt}),$$ where the elements of the loop group are quantized, and the equality is replaced by proportionality up to a non-zero "central constant".<|endoftext|>
TITLE: Are Fukaya categories Calabi-Yau categories?
QUESTION [9 upvotes]: Let X be a compact symplectic manifold. There is an idea, I think probably originally due to Kontsevich, that we should be able to get Gromov-Witten invariants of X out of the Fukaya category of X. One possible approach to doing this is via the theorem proved by Costello (I think there is also a similar(?) result of Kontsevich-Soibelman?) that a Calabi-Yau category determines a TCFT, which then should determine the Gromov-Witten invariants of X --- or at least something like the Gromov-Witten invariants of X. But in order for this to even get started, we need the Fukaya category of X to be a Calabi-Yau category (you can find the definition of CY category in Costello's paper, at the beginning of section 2).
Hence: Is the Fukaya category of a compact symplectic manifold known to be a Calabi-Yau category? What is the trace map supposed to be?

REPLY [10 votes]: At first sight, the Fukaya category has obvious cyclic symmetry, because the $A_\infty$ structure maps count points in spaces of rigid pseudo-holomorphic polygons subject to Lagrangian boundary conditions, and these spaces depend only on the cyclic order of the Lagrangians. This indeed proves that the cohomological Fukaya category, in which the hom-spaces are Floer cohomology spaces, is cyclically symmetric.
The trouble comes when these Lagrangians don't intersect one another transversely - for instance, the same Lagrangian occurs more than once - because then the morphism spaces and structure maps invoke Hamiltonian perturbations which need not be cyclically symmetric. The problem which Fukaya has solved over the reals (see Matthew Ballard's answer) is, I presume, to find a way to make these perturbations cyclically symmetric whilst also achieving the necessary coherence between them, as well as transversality for compactified moduli spaces of inhomogeneous pseudo-holomorphic polygons (or worse, their abstract perturbations). These are the things which actually define the $A_\infty$-structure.
FOOO worked extremely hard to get geometrically-meaningful units in their Fukaya endomorphism algebras, where other authors are content to define units by tweaking the $A_\infty$-structure algebraically. My hope would be that algebra will also give a cheaper approach to cyclic symmetry, particularly since I'm told that for Costello's theorem to hold, one only needs "derived" cyclic symmetry.
By the way, let's be clear that Costello's theorem, suggestive as it may be, is not about GW invariants! It's about theories over $M_{g,n}$, not over $\overline{M}_{g,n}$.<|endoftext|>
TITLE: Testing for Riemannian isometry
QUESTION [5 upvotes]: In most physics situations one gets the metric as a positive definite symmetric matrix in some chosen local coordinate system.
Now if on the same space one has two such metrics given as matrices then how does one check whether they are genuinely different metrics or just Riemannian Isometries of each other (and hence some coordinate change can take one to the other).
If in the same coordinate system the two matrices are different then is it proof enough that they are not isometries of each other?  (doing this test over say a set of local coordinate patches which cover the manifold) 
Asked otherwise, given two ``different" Riemannian Manifolds how does one prove the non-existence of a Riemannian isometry  between them?  
There have been two similar discussions on mathoverflow at this and this one. 
And this article was linked from the later. 
In one of the above discussions Kuperberg had alluded to a test for local isometry by checking if the Riemann-Christoffel curvatures are the same locally. If the base manifold of the two riemannian manifolds is the same then one can choose a common coordinate system in which to express both the given metrics and then given softwares like mathtensor by Mathew Headrick this is probably not a very hard test to do. 
So can one simply patch up such a test through out a manifold to check if two given riemannian manifolds are globally isometric? 
How does this compare to checking if the metrics are the same or not in a set of common coordinate patches covering the manifold? 
I somehow couldn't figure out whether my query above is already getting answered by the above two discussions.

REPLY [2 votes]: (I wanted to make a comment but it is bit too long)
The answer depends very much on what you are able to calculate. So try to understand first what is calculable in your case... 
For example: Assume you can calculate all eigenvalues of Laplacian for both manifolds (assume they are compact) --- if the two obtained sequences are different then manifolds are also nonisometric. If they all the same you might have different manifolds but it is very unlikely (it means that usually, one can hear shape of drum).
On the other hand --- I do not think you are able to perform these calculations.<|endoftext|>
TITLE: Historical question in analytic number theory
QUESTION [14 upvotes]: The analytic continuation and functional equation for the Riemann zeta function were proved in Riemann's 1859 memoir "On the number of primes less than a given magnitude." What is the earliest reference for the analytic continuation and functional equation of Dirichlet L-functions?  Who first proposed that they might satisfy a Riemann hypothesis?  Dirichlet did none of these things; his paper dates from 1837, and as far as I know he only considered his L-functions as functions of a real variable.

REPLY [25 votes]: To extend on Matt's comment about Euler, here is something I wrote up some years ago about Euler's discovery of the functional equation only at integral points. I hope there are no typos.
Although Euler never found a convergent analytic expression for 
$\zeta(s)$ at negative numbers, in 1749 he published a method of 
computing 
values of the zeta function at negative integers by a precursor of 
Abel's Theorem applied to a divergent series.  The 
computation led him to the asymmetric functional equation of 
$\zeta(s)$.
The technique uses the function 
$$
\zeta_{2}(s) = \sum_{n \geq 1} \frac{(-1)^{n-1}}{n^s} = 
1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \dots.
$$
This looks not too different from $\zeta(s)$, but has the advantage 
as an alternating series of 
converging for all positive $s$.  For $s > 1$, 
$\zeta_2(s) = (1 - 2^{1-s})\zeta(s)$.
Of course this is true for complex $s$, but 
Euler only worked with real $s$, so we shall as well.
Disregarding convergence issues, Euler wrote 
$$
\zeta_{2}(-m) = \sum_{n \geq 1} (-1)^{n-1}n^m = 1 - 2^m + 3^m - 4^m + 
\dots, 
$$
which he proceeded to evaluate as follows.  Differentiate 
the equation
$$
\sum_{n \geq 0} X^n = \frac{1}{1-X}
$$
to get 
$$
\sum_{n \geq 1} nX^{n-1} = \frac{1}{(1-X)^2}.
$$
Setting $X = -1$, 
$$
\zeta_{2}(-1) = \frac{1}{4}.
$$
Since $\zeta_{2}(-1) = (1-2^2)\zeta(-1)$, $\zeta(-1) = -1/12$.  Notice we can't set $X = 1$ in the second power series and compute $\sum n = \zeta(-1)$ directly.  So $\zeta_2(s)$ is nicer than $\zeta(s)$ in this Eulerian way.
Multiplying the second power series by $X$ and then differentiating, we get
$$
\sum_{n \geq 1} n^2X^{n-1} = \frac{1+X}{(1-X)^3}.
$$
Setting $X = -1$, 
$$
\zeta_{2}(-2) = 0.
$$
By more successive multiplications by $X$ and differentiations, we get
$$
\sum_{n \geq 1} n^3X^{n-1}  = \frac{X^2+4X+1}{(1-X)^4},
$$
and
$$
\sum_{n \geq 1} n^4X^{n-1} = \frac{(X+1)(X^2+10X+1)}{(1-X)^5}.
$$
Setting $X = -1$, we find $\zeta_{2}(-3) = -1/8$ and 
$\zeta_{2}(-4) = 0$.  Continuing further, with the recursion
$$
\frac{d}{dx} \frac{P(x)}{(1-x)^n} = \frac{(1-x)P'(x) + nP(x)}{(1-x)^{n+1}}, 
$$
we get
$$
\sum_{n \geq 1} n^5X^{n-1}  =  \frac{X^4+26X^3+66X^2 + 26X +1}{(1-X)^6}, 
$$
$$
\sum_{n \geq 1} n^6X^{n-1} =  \frac{(X+1)(X^4 + 56X^3 + 246X^2 + 56X+1)}
{(1-X)^7},
$$
$$
\sum_{n \geq 1} n^7X^{n-1} =  \frac{X^6 + 120X^5 + 1191X^4 + 2416X^3 + 
1191X^2 + 120X + 1}{(1-X)^8}. 
$$
Setting $X = -1$, we get $\zeta_{2}(-5) = 
1/4, \ \zeta_{2}(-6) = 0, \ \zeta_{2}(-7) = -17/16$. 
Apparently $\zeta_{2}$ vanishes at the negative even integers, while 
$$
\frac{\zeta_{2}(-1)}{\zeta_{2}(2)} = \frac{1}{4}\cdot\frac{6\cdot 2}{\pi^2} = 
\frac{3\cdot 1!}{1\cdot \pi^2}, \ \ \ \
\frac{\zeta_{2}(-3)}{\zeta_{2}(4)} = 
-\frac{1}{8}\cdot\frac{30\cdot24}{7\pi^4} = -\frac{15\cdot 3!}{7\cdot \pi^4}, 
$$
$$
\frac{\zeta_{2}(-5)}{\zeta_{2}(6)} = \frac{1}{4}\cdot 
\frac{42\cdot 6!}{31\pi^6} = 
\frac{63 \cdot 5!}{31\cdot \pi^6}, \ \ \ \ 
\frac{\zeta_{2}(-7)}{\zeta_{2}(8)} = -\frac{17}{16}\cdot 
\frac{30\cdot 8!}{127\cdot \pi^8} 
 = -\frac{255\cdot 7!}{127\pi^8}.
$$
The numbers $1, 
3, 7, 15, 31, 63, 127, 255$ are all one less than a power of 2, 
so Euler was led to the observation that for $n \geq 2$, 
$$
\frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = 
\frac{(-1)^{n/2+1}(2^n-1)(n-1)!}{(2^{n-1}-1)\pi^n}
$$
if $n$ is even and
$$
\frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = 
0
$$
if $n$ is odd. Notice how 
the vanishing of $\zeta_{2}(s)$ 
at negative even integers nicely compensates for 
the lack of knowledge of $\zeta_2(s)$ at positive odd integers $> 1$ (which 
is the same as not knowing $\zeta(s)$ at positive odd integers $> 1$).
Euler interpreted the $\pm$ sign at even $n$ and the vanishing 
at odd $n$ as the single factor $-\cos(\pi n/2)$, and with 
$(n-1)!$ written as $\Gamma(n)$ we get
$$
\frac{\zeta_{2}(1-n)}{\zeta_{2}(n)} = -\Gamma(n)\frac{2^n-1}{(2^{n-1}-1)\pi^n}
\cos\left(\frac{\pi n}{2}\right). 
$$
Writing $\zeta_{2}(n)$ as $(1 - 2^{1-n})\zeta(n)$ gives the 
asymmetric functional 
equation
$$
\frac{\zeta(1-n)}{\zeta(n)} = \frac{2}{(2\pi)^n}
\Gamma(n)\cos\left(\frac{\pi n}{2}\right).
$$
Euler applied similar ideas to $L(s,\chi_4)$ and found its 
functional equation.  You can work this out yourself in 
Exercise 2 below.
Exercises

Show that Euler's computation of zeta values at negative 
integers can be put in the form
$$
(1 - 2^{n+1})\zeta(-n) = 
\left.\left(u\frac{d}{du}\right)^{n}\right\vert_{u=1}\left(\frac{u}{1+u}
\right) = \left.\left(\frac{d}{dx}\right)^{n}\right\vert_{x=0}
\left(\frac{e^x}{1+e^x}\right).
$$
To compute the divergent series
$$
L(-n,\chi_4) = \sum_{j \geq 0} (-1)^{j}(2j+1)^n = 
1 - 3^n + 5^n - 7^n - 9^n + 11^n - \dots
$$
for nonnegative integers $n$, begin with the formal identity
$$
\sum_{j \geq 0} X^{2j} = \frac{1}{1-X^2}.
$$
Differentiate and set $X = i$ to show $L(0,\chi_4) = 1/2$.
Repeatedly multiply by $X$, differentiate, and set 
$X = i$ in order to compute $L(-n,\chi_4)$ for $0 \leq n \leq 10$. This computational technique is not rigorous, but the answers are correct. Compare with the values of $L(n,\chi_4)$ for positive $n$, if you know those, to get a formula for $L(1-n,\chi_4)/L(n,\chi_4)$. Treat alternating signs like special values of a suitable trigonometric function.<|endoftext|>
TITLE: Can we realize Weyl group as a subgroup?
QUESTION [18 upvotes]: Given a semisimple Lie group G, let T be a maximal torus, W be the Weyl group defined as the quotient N(T)/C(T), where N(T) denotes the normalizer of T and C(T) denotes the centralizer. 
Two questions are: 

How many ways are there we can realize W as a subset of G?
Can we always realize W as a subgroup of G?

REPLY [28 votes]: In general it is not possible to embed the Weyl group $W$ in the group $G$: already you can see this for $SL_2(\mathbb C)$, where the Weyl group has order $2$: if the torus fixes the lines spanned by $e_1$ and $e_2$ respectively, you want to pick the linear map taking $e_1$ to $e_2$ and $e_2$ to $e_1$, but this has determinant $-1$. A lift of $W$ to $N(T)$ must be an element of order $4$ not $2$, say $e_1 \mapsto -e_2$ and $e_2 \mapsto e_1$. 
In fact, Tits has shown (Normalisateurs de Tores I. Groupes de Coxeter Étendus (Journ. Alg. 4, 1966, pp. 96-116) that this is essentially the only obstruction: the Weyl group can always be lifted to a group $\tilde{W}$ inside $G$ which is an extension of $W$ by an elementary abelian $2$-group of order $2^l$ where $l$ is the number of simple roots. If I recall correctly, this lift is then unique up to conjugation.<|endoftext|>
TITLE: Dihedral extensions and the Ankeny - Artin - Chowla conjecture
QUESTION [11 upvotes]: Jensen and Yui (Polynomials with Dp as Galois group
J. Number Theory 15, 347-375 (1982)) proved that if p = 4n+1
is a regular prime, then there is no normal extension of the
rationals with Galois group Dp (dihedral of order 2p)
ramified only at p. When I first read it I noticed that such an
extension exists if and only if p divides u, where $t+u\sqrt{p}$
is the fundamental unit of the real quadratic number field with
discriminant p (Ankeny, Artin and Chowla conjectured that this
never happens; it is known that this property is equivalent to
the divisibility of the Bernoulli number B(p-1)/2 by p,
hence implies that p is irregular). 
I recall having seen this result in print a few years later, 
but can't find it anymore. Can anyone help me?

REPLY [2 votes]: As I am currently writing up my thesis in English I noticed that I had referred to this result in the preface. It appears as Proposition 1 in an article by Nakagoshi. See also the recent article by Cohen and Thorne.<|endoftext|>
TITLE: Frobenius Descent
QUESTION [15 upvotes]: Let $S$ be a scheme of positive characteristic $p$ and $X$ a smooth $S$-scheme. Let $F:X\rightarrow X^{(p)}$ denote the relative Frobenius. A result by Cartier (often called Cartier descent or Frobenius descent) then states that the category of quasi-coherent $\mathcal{O}_{X^{(p)}}$-modules is equivalent to the category of quasi-coherent $\mathcal{O}_X$-modules $(E,\nabla)$ with integrable connection of $p$-curvature $0$ (which means $\nabla(D)^p-\nabla(D^p)=0$ for all $S$-derivations $D:\mathcal{O}_X\rightarrow \mathcal{O}_X$).
The equivalence is given by
$$ (E,\nabla)\longmapsto E^\nabla$$
and
$$ E\mapsto (F^*E,\nabla^{can})$$
where $\nabla^{can}$ is the canonical connection locally given by $f\otimes s\mapsto (1\otimes s)\otimes df$, for 
$$f\otimes s\in \mathcal{O}_X(U)\otimes E(U).$$
(tensor over the sections of the structure sheaf of $X^{(p)}$, somehow jtex can't handle that)
The proof of this theorem can be found in 5.1. in Katz' paper "Nilpotent connections and the monodromy theorem"
My question is: As $X/S$ is smooth, the relative Frobenius is faithfully flat (at least it is if $S$ is the spectrum of a perfect field), can the above theorem be interpreted as an instance of faithfully flat descent along $F$? In other words, does the connection $\nabla$ give rise to a descent datum for $E$ with respect to $F$? 
I know that connections are "first-order descent data", i.e. modules with connection descend along first order thickenings, but I don't see how this applies here.

REPLY [13 votes]: I believe that the answer is yes, and that this may have been one of Grothendieck's motivations for developing the general theory of flat descent.  (If I remember correctly,
in the first (?) expose of FGA, in which he explains flat descent, Grothendieck has a reference to work of Cartier involving descent in the context of inseparable extensions,
and I would guess that it is a reference to this Cartier, or Frobenius, descent.  Can
anyone cofirm this?)
To put a connection on $E$ is to extend the action of $\mathcal O_X$ to an action
of $\mathcal D_X$, the ring of differential operators generated (in local coordinates) by
$\partial /\partial x_1,\ldots,\partial/\partial x_n.$  (This is not the same as the full ring
of differential operators in charateristic $p$.)  The $p$-curvatures generate an ideal
in this ring; I think it is just the ideal generated by $(\partial/\partial x_i)^p$.
So if $E$ has vanishing $p$-curvature, the action of $\mathcal D_X$ factors through
the quotient by this ideal.  One can now interpret this information in terms
of descent data.  
A precise description is given in Prop. 2.6.2 of Berthelot's book
D-modules arithmetiques II; Descente par Frobenius.<|endoftext|>
TITLE: Lawvere's "Some thoughts on the future of category theory."
QUESTION [51 upvotes]: In Lecture Notes in Mathematics 1488, Lawvere writes the introduction to the Proceedings for a 1990 conference in Como.  
In this article, Lawvere, the inventor of Toposes and Algebraic Theories, discusses two ancient philosophical "categories": that of BEING and that of BECOMING.  And he's serious.  While some of the motivation for this article is to understand these two ancient even mystical topics, the actual content is almost purely mathematical.  Lawvere makes definitions and claims in the manner of a serious mathematician engaged in deliberate but casual explanation of ideas.
I want to understand this article, but it is difficult.  The definitions seem to be written for someone with a bit more background or expertise on topos theory and its application.
Q: I am writing to ask whether anyone here has read or understood this article (or parts of it).  I'm interested in your thoughts on it.  Have the ideas been written formally?

REPLY [2 votes]: As a prelude to the answer above:  When Lawvere wanted a mathematics of Becoming and not just Being, one thing he wanted is to work in categories that have exponentials.
For example, in the category of manifolds, we can express the motion of a body as a map $T \times B \rightarrow E$, where $T$ is an object of times, $B$ is a body, and $E$ is Euclidean space.  Lawvere wants to work in bigger categories where we can also express this as $B \rightarrow E^T$ (the possible paths of an individual part of the body) or $T \rightarrow E^B$ (the collection of instantaneous descriptions of a body's location).
A category which works for these purposes is $\mathbf{Sets^{Z^{\large Op}}}$, where $Z$ is a category of loci that extends the category of manifolds.  This leads to topoi.<|endoftext|>
TITLE: How many trial picks expectedly sufficient to cover a sample space?
QUESTION [5 upvotes]: Consider a sequence of independent events where an $r$ element subset of an $n$ element set is picked uniformly randomly (ie. any of the $\begin{pmatrix}n\newline r\end{pmatrix}$ possibilities being equally likely).
What is the expected number of subsets one has pick to cover the whole set?
Here the terminology means: a sequence of picks $A_1,A_2,\ldots,A_n$ covers the whole set if
 $|A_1 \cup \cdots \cup A_n| = n$. A sequence $A_1, A_2,\ldots$ succeeds to cover the whole set in $n$ steps, if $A_1,\ldots,A_n$ covers the whole set but $A_1,\ldots, A_{n-1}$ does not.
The expected numbers seems to be much higher than one would imagine. But I could not quite come up with a closed form. But chances are, its always a rational number.

REPLY [2 votes]: The expected number of picks needed equals the sum of the probabilities that at least $t$ picks are needed, which means that $t-1$ subsets left at least one value uncovered. We can use inclusion-exclusion to get the probability that at least one value is uncovered.
The probability that a particular set of $k$ values is uncovered after $t-1$ subsets are chosen is 
$$\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1}$$
So, by inclusion-exclusion, the probability that at least one value is uncovered is 
$$ \sum_{k=1}^n  {n \choose k}(-1)^{k-1}\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg) ^{t-1} $$
And then the expected number of subsets needed to cover everything is 
$$ \sum_{t=1}^\infty \sum_{k=1}^n {n \choose k}(-1)^{k-1} \Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1} $$
Change the order of summation and use $s=t-1$:
$$ \sum_{k=1}^n {n \choose k}(-1)^{k-1} \sum_{s=0}^\infty  \Bigg( \frac{n-k \choose r}{n \choose r}\Bigg)^s$$
The inner sum is a geometric series.
$$ \sum_{k=1}^n {n \choose k} (-1)^{k-1}\frac{n \choose r}{{n \choose r}-{n-k \choose r}}$$ 
$$ {n \choose r} \sum_{k=1}^n (-1)^{k-1}\frac{n \choose k}{{n \choose r}-{n-k \choose r}}$$
I'm sure that should simplify further, but at least now it's a simple sum. I've checked that this agrees with the coupon collection problem for $r=1$.
Interestingly, Mathematica "simplifies" this sum for particular values of $r$, although what it returns even for the next case is too complicated to repeat, involving EulerGamma, the gamma function at half-integer values, and PolyGamma[0,1+n].<|endoftext|>
TITLE: Do h-coequalizers and coproducts give all h-colimits?
QUESTION [11 upvotes]: It is well known that if a category has all coequalizers and all (small) coproducts then in fact it has all (small) colimits. More important is the proof which shows that every colimit can be built by using coproducts and coequalizers. This implies that if a functor commutes with coproducts and coequalizers, then it must commute with all (small) colimits as well. 

Is there homotopical analog of this? If I have a functor which commutes with all small (homotopy) coproducts and all homotopy coequalizers, does it necessarily commute with all homotopy colimits in general?

This question makes sense for general model categories, but I am particularly interested in the usual model structure on spaces.

REPLY [13 votes]: There is an analogue, but one should replace coequalizers by geometric realizations (homotopy colimits over Δop).  If $F : I \to M$ is a diagram in a model category, one has
$$\operatorname{hocolim}_I F = \operatorname{hocolim}_{k \in \Delta^{\operatorname{op}}} \coprod_{i_0 \to \cdots \to i_k \in I} F(i_0).$$
For instance, see section 2 of http://www.math.uchicago.edu/~eriehl/hocolimits.pdf for the simplicial model category case.<|endoftext|>
TITLE: Is every flat unramified cover of quasi-projective curves profinite?
QUESTION [6 upvotes]: When I first learned about the etale fundamental group, there was a mythical theorem going around that in the algebraic case all we need to look at is the finite covers, because the infinite degree algebraic covers are inverse limits of the finite ones (obviously unlike the topological case). But I've never seen a convincing source for this theorem.
It seems reasonable that the statement would be: "every flat unramified map of a connected scheme onto a quasi-projective curve is an inverse limit of finite etale covers". Is this true? Do you have a reference for this?

REPLY [9 votes]: Any modification of the theorem where the definition of "cover" you give is local on the base and contains inverse limits of finite etale covers (e.g. flat plus unramified as in the original question) will also be false because the property of being an inverse limit of finite etale covers is not local on the base.
To see this, proceed similarly to Scott Carnahan's example, but instead of gluing a chain of $\mathbb{P}^1$'s together, glue together $\mathbb{P}^1$'s "indexed by $\operatorname{Spec} \mathbb{C}[...,x_{-1},x_0,x_1,...]/\langle x_i^2-1\rangle$." Explicitly, let the base curve $B$ be two $\mathbb{P}^1$'s glued together at two distinct points. Over each $\mathbb{P}^1$ consider the affine morphism corresponding to the sheaf of algebras $\mathcal{O}\_{\mathbb{P}^1}[...,x_{-1},x_0,x_1,...]/\langle x_i^2-1\rangle$. At one of the points, glue together the two possible $x_i$'s. At the other, glue $x_i$ to  $x_{i-1}$. Over each $\mathbb{P}^1$, the resulting morphism is an inverse limit of finite covers, but over all of $B$, it is not. This is written down fully in Warning 2.5b of http://math.harvard.edu/~kwickelg/papers/VW.pdf -- Kirsten Wickelgren<|endoftext|>
TITLE: Do hyperKahler manifolds live in quaternionic-Kahler families?
QUESTION [22 upvotes]: A geometry question that I thought about more seriously a few years ago... thought it'd be a good first question for MO.
I'm aware that there are a number of Torelli type theorems now proven for compact HyperKahler manifolds.  Also, I think that Y. Andre has considered some families of HyperKahler (or holomorphic symplectic) manifolds in some paper.
But, when I see such a moduli problem studied, the data of a HyperKahler manifold seems to include a preferred complex structure.  For example, a HyperKahler manifold is instead viewed as a holomorphic symplectic manifold.  I'm aware of various equivalences, but there are certainly different amounts of data one could choose as part of a moduli problem.
I have never seen families of HyperKahler manifolds, in which the distinction between hyperKahler rotations and other variation is suitably distinguished.  Here is what I have in mind, for a "quaternionic-Kahler family of HyperKahler manifolds:
Fix a quaternionic-Kahler base space $X$, with twistor bundle $Z \rightarrow X$. Thus the fibres $Z_x$ of $Z$ over $X$ are just Riemann spheres $P^1(C)$, and $Z$ has an integrable complex structure.
A family of hyperKahler manifolds over $X$ should be (I think) a fibration of complex manifolds $\pi:  E \rightarrow Z$, such that:

Each fibre $E_z = \pi^{-1}(z)$ is a hyperKahler manifold $(M_z, J_z)$ with distinguished integrable complex structure $J_z$. 
For each point $x \in X$, let $Z_x \cong P^1(C)$ be the twistor fibre.  Then the family $E_x$ of hyperKahler manifolds with complex structure over $P^1(C)$ should be (isomorphic to) the family $(M, J_t)$ obtained by fixing a single hyperKahler manifold, and letting the complex structure vary in the $P^1(C)$ of possible complex structures.  (I think this is called hyperKahler rotation).

In other words, the actual hyperKahler manifold should only depend on a point in the quaternionic Kahler base space $X$, but the complex structure should "rotate" in the twistor cover $Z$.  
This sort of family seems very natural to me.  Can any professional geometers make my definition precise, give a reference, or some reason why such families are a bad idea?  I'd be happy to see such families, even for hyperKahler tori (which I was originally interested in!)

REPLY [11 votes]: What you suggested makes sense. You propose to replace the $P^1$ fibre by the twistor space of an HK manifold M, so that the big total space would not only display separately the complex structures of M, but allow deformations of M to be parametrized by X. I think the real question is whether there exist sensible examples over a compact QK base like X$=S^4$ in which a consistent choice of complex structure on the varying HK manifolds is therefore not possible. I am not sure. The problem is that the construction looks a bit unwieldly, and experience dictates that it is more natural to look for bundles whose fibres are HK. In this sense, your idea is very close to a known (but in some sense simpler) construction that goes under the heading "Swann bundle" or "C map".
Let me add two comments in support of your question. First, the concept of a manifold foliated by HK manifolds (like $T^4$ or K3) is very powerful. This is most familiar in work on special holonomy, but here's a more classical construction: the curvature tensor at each point of a Riemannian 4-manifold can be used to construct a singular Kummer surface and an associated K3 (the intersection of 3 quadrics in $P^5$), but the complex structure is fixed so not twistorial. Second, escaping from quaternions, one sees twistor space fibres in the following situation: each fibre of the twistor space $SO(2n+1)/U(n)$ parametrizing a.c.s.'s on the sphere $S^{2n}$ can be identified with the twistor space of $S^{2n-2}$!<|endoftext|>
TITLE: Different definitions of Novikov ring?
QUESTION [7 upvotes]: Following, e.g., Wikipedia's definitions, the (small) quantum cohomology ring of $X$ is defined over a "Novikov ring" consisting of formal power series of the form $$ \sum_{\beta \in H_2(X;\mathbb{Z})} a_\beta e^\beta,$$
where the $a_\beta$ are in some fixed ring (which is probably usually $\mathbb{Q}$ or $\mathbb{C}$).
On the other hand, in the papers of Fukaya and company for instance, there seems to be a different "Novikov ring", consisting of power series of the form $$\sum_{i=1}^\infty a_i T^{\lambda_i},$$
where $a_i \in \mathbb{Q}$, $\lambda_i \in \mathbb{R}$, and $\lim_{i \to \infty} \lambda_i = \infty$.
What's up with this? Why are there apparently two different Novikov rings? How do I reconcile this? I'm guessing that the $\lambda_i$ in the second definition should correspond to something like $\int_\beta \omega$ where $\beta$ is as in the first definition (and where $\omega$ is the symplectic form), but beyond this I'm not sure...

REPLY [5 votes]: We must think of $T$, the generator of the Novikov ring, as a coordinate on the moduli space of symplectic structures $\mathcal M_{sym}(X)$. We can define $<\cdots>_g^\psi:=\sum_\beta<\cdots>_{g,\beta}^\psi e^{-\omega(\beta)}$ , but we don't know if the right hand side converges and so we put $T^{\omega(\beta)}=e^{-(-\log T\omega)(\beta)}$ and we can think of this as large volume limit, i.e. neighborhood of singular point in $\mathcal M_{sym}(X)$ where $\omega\to \infty$
For a $J$-holomorphic curve $u$ we have $E(u)=ω([u])=∫_Σu^∗ω≥0$, and we must take $λ_i≥0$. Moreover if $ω$ belongs to an integral cohomology class, then $ω(β)∈\mathbb Z_{≥0}$ and we can replace the Novkov ring with $\mathbb Q[[T]]$
See Denis Auroux's
paper https://arxiv.org/pdf/0902.1595.pdf .<|endoftext|>
TITLE: Can minimal surfaces be characterized by some universal property?
QUESTION [8 upvotes]: As objects which are minimal (in some respect), this seems entirely plausible, but I'm not sure what category we should be working in, and what restrictions we would need, to actually have a situation where minimal surfaces would be characterized by a universal property, if they ever can be. An uneducated guess on one possible setup where minimal surfaces would be universal: the objects are surfaces whose boundary is a given simple closed curve, and the morphisms are the area-decreasing isometries - it seems like a minimal surface should be a final object, though we would probably need to introduce an equivalence relation on the morphisms to get the maps to be unique?
I'm also curious about the same question, but for geodesics. Perhaps for them, we would use the collection of paths from point $x$ to point $y$ on a given surface, and use the length-decreasing homotopies?
Being a final object isn't the only option - maybe, for any surface, some kind of map will factor through a minimal surface associated to it?
EDIT: I'm worried this is perhaps too soft a question for MathOverflow - I'm not sure there's really a "right" answer.

REPLY [10 votes]: I'm not sure if this answer provides you with the universal property that you desire, but there is such a category that unifies these concepts that you are after.
Cohen, Jones and Segal introduced a concept known as the "Flow Category" in the paper Morse Theory and Classifying Spaces, which associates to any manifold with a Morse Function a category whose objects are the critical points of the Morse function and whose morphisms are the gradient trajectories of some gradient-like vector field. Here is the reference:
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.5003
You can get the paper on Ralph Cohen's page if you don't have university access:
http://math.stanford.edu/~ralph/papers.html
Recall that Morse Theory was invented by Marston Morse to study geodesics on manifolds. Geodesics correspond precisely to critical points of the Energy functional. I imagine that any variational problem fits into this framework. 
As a word of caution, understanding the space of gradient trajectories lies at the heart of Floer Theory, so if you want to understand Morse Theory on infinite dimensional spaces, prepared to get your hands dirty with some serious analysis. Comment if you want more references. Also, most of the above article is concerned with proving a very elegant result about the classifying space of this category for certain situations. It is very slick!<|endoftext|>
TITLE: Approximation of a Sobolev function that has vanishing trace on the reduced boundary of a Caccioppoli (i.e. finite perimeter) set
QUESTION [6 upvotes]: For $\Omega\subset\mathbb{R}^N$ open and bounded, let $W^{1,p}(\Omega)$ denote the usual Sobolev space of $L^p(\Omega)$ functions with weak partial derivatives in $L^p(\Omega)$ and $W_0^{1,p}(\Omega)$ the closure of $\mathcal C^\infty_c(\Omega)$ in this space.

Let $E\subset \Omega$ be a Caccioppoli set (i.e. a set of finite perimeter in $\mathbb{R}^N$), $U\subset\Omega$ be open and $w\in W_0^{1,p}(\Omega)$. Suppose that
  $$ \int_E \operatorname{div}(\eta w) = 0 \qquad\forall\;\eta\in\mathcal C^1_c(U;\mathbb R^N).$$
  Question: Does there exist a sequence $w_n\in\mathcal C^1_c(\Omega)$ with $\mathcal H^{N-1}(U\cap\partial^*E \cap [w_n\neq 0])=0$ that converges to $w$ in $W^{1,p}(\Omega)$? (Here $[w_n\neq 0]=\{x\in\Omega:w_n(x)\ne 0\}$ and $\mathcal H^{N-1}$ is the $(N-1)$-dimensional Hausdorff-measure).


The assertion seems natural to me because for $w\in\mathcal C_c^1(\Omega)$ the equality implies (as $E$ is Caccioppoli)
$$ \int_{\partial^*E} w\eta \nu_E = 0 \qquad\forall\;\eta\in\mathcal C^1_c(U;\mathbb R^N),$$
where $\nu_E$ is the inner normal of $E$ which exists $\mathcal{H}^{N-1}$-a.e. on the reduced boundary $\partial^*E$. This then implies that $w=0$ $\mathcal H^{N-1}$-a.e. on $\partial^*E$. 
The same holds by the divergence theorem if $E$ has Lipschitz boundary and $w$ is Sobolev because then the trace of $w$ on $\partial E$ is well defined, and hence there exists a sequence of $\mathcal C_c^1(\Omega)$ functions vanishing on $\partial E\cap U$ that approximate $w$ in $W^{1,p}(\Omega)$.
I couldn't find much on other cases, but maybe the following reference helps for the case when $w\in W_0^{1,p}(\Omega)\cap L^\infty(\Omega)$:
Chen, Gui-Qiang; Torres, Monica, Divergence-measure fields, sets of finite perimeter, and conservation laws, Arch. Ration. Mech. Anal. 175, No. 2, 245-267 (2005). ZBL1073.35156.
Are the above arguments for $w\in\mathcal C^1_c(\Omega)$ or if $E$ has Lipschitz boundary correct? If so, what about the general case where $w$ is only in $W^{1,p}_0(\Omega)$ and $E$ is merely Caccioppoli?

REPLY [2 votes]: Firstly, the condition in your question is sensible as it implies that $w$ vanishes $\mathcal{H}^{N-1}$-a.e. on $U\cap\partial^*E$. Moreover, your arguments for the two cases where $w\in C^1_\mathrm{c}(\Omega)$ or $E$ has Lipschitz boundary are correct. I expect that also the case $p>N$ where $w$ has a continuous representative works out. However, in general it is too much to ask that the approximating sequence is both smooth and vanishing on $U\cap\partial^*E$ up to a $\mathcal{H}^{N-1}$-nullset.

No, you cannot expect such a sequence to exist if $p\le N$ and $w\in W^{1,p}_0(\Omega)\cap L^\infty(\Omega)$ and $E$ has finite perimeter in $\mathbb{R}^N$ (i.e. $E$ is Caccioppoli). 


As the construction of the example is somewhat technical, I shall first sketch the argument. An open set of finite perimeter $E$ can have massive amounts (say positive Lebesgue measure) of topological boundary $\partial E$ while the essential or reduced boundary $\partial^*E$ is small and very regular. In fact, one can arrange that there exists a nontrivial Sobolev function $w\in W^{1,p}_0(\Omega)$ supported in $\partial E\setminus E$ that vanishes $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$ as $\partial^*E$ is regular.
Then $w$ satisfies the assumption in the question as it vanishes on $E$ (along with its gradient).
 Any continuous function that vanishes $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$ actually vanishes everywhere on $\partial E$. So one cannot use such functions to approximate $w$ in $W^{1,p}(\Omega)$. 
Example: It suffices to consider the case where $\Omega$ is the unit ball in $\mathbb{R}^N$ and $U=\Omega$. For simplicity let $p=2$ and $N\ge 2$, but the argument works without change for $p\in(1,N]$. We make use of the same construction as in my answer https://mathoverflow.net/a/295459.
Let $E:=\bigcup_k B_k$ be a countable union of open balls with pairwise disjoint closures contained in $\Omega$, such that $E$ is dense in $\overline{\Omega}$ and $$\operatorname{cap}(E)\le\sum_k\operatorname{cap}(B_k)<\operatorname{cap}(\Omega).$$ Here for $A\subset\mathbb{R}^N$ we denote by $\operatorname{cap}(A)$ the Sobolev capacity $$
\operatorname{cap}(A)=\inf\{ \|u\|^p_{W^{1,p}} : u\in W^{1,p}(\mathbb{R}^N)\text{ and }u\ge1\text{ a.e. on neighbourhood of }A\}.
$$
Then $\sum_k\mathcal{H}^{N-1}(\partial B_k)<\infty$ because of the above estimate on the capacity. (Alternatively, just choose the balls small enough.) It is readily checked that $E$ has finite perimeter (i.e. is Caccioppoli) and $\partial^*E = \partial^*E\cap U = \bigcup_k\partial B_k$, possibly up to a $\mathcal{H}^{N-1}$-nullset.
As in the linked answer (the function is called $u$ there), there exists a nontrivial bounded $w\in W^{1,p}(\mathbb{R}^N)$ supported in $K := \overline{\Omega}\setminus E$. As $\Omega$ has a regular boundary, it follows by standard trace theory that $w\in W^{1,p}_0(\Omega)$. Observe that
$$\int_E\operatorname{div}(\eta w) = \int_\Omega \mathbf{1}_E(w\operatorname{div}(\eta) + \nabla w\cdot \eta)=0$$
for all $\eta\in C^1_\mathrm{c}(\Omega;\mathbb{R}^N)$ as both $w$ and $\nabla w$ vanish on $E$ (recall that $w$ is supported in $K$).
Suppose now that $w_n\in C^1_\mathrm{c}(\Omega)$ with $w_n=0$ $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$. Then due to continuity and the structure of $\partial^*E$ one has $w_n=0$ everywhere on $\partial^*E\cup K$. As $w$ is nontrivial on $K$, the sequence $(w_n)$ cannot possibly converge to $w$ in $W^{1,p}(\Omega)$.

The above example shows that for general finite perimeter sets $E$ it is too much to ask that the approximating sequence is continuous and vanishes $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$. I expect, however, that everything works out if one replaces the latter condition by something like $\mathcal{H}^{N-1}(U\cap\partial^*E\cap [w_n\ne 0])<\frac{1}{n}$ or $\|w_n\|_{L^1(U\cap\partial^*E,\mathcal{H}^{N-1})}<\frac{1}{n}$.
The assumptions in the question imply that $w$ (actually its approximately continuous representative) vanishes $\mathcal{H}^{N-1}$-a.e. on $U\cap\partial^* E$. I shall not go into details, but that claim follows from 
$$
\int_U \mathbf{1}_E w\operatorname{div}(\eta) = - \int_U \mathbf{1}_E\nabla w\cdot \eta,
$$
taking the supremum over $\eta\in C^1_\mathrm{c}(U;\mathbb{R}^N)$ with $|\eta(x)|\le 1$ for all $x\in U$ on both sides, and using that $\mathbf{1}_E w\in\mathrm{SBV}(\mathbb{R}^{N})$ with the jump behaviour that one would expect.<|endoftext|>
TITLE: Quillen's Morphism Inverting Functors
QUESTION [5 upvotes]: In "Higher algebraic K-theory I" Quillen defines a morphism inverting functor to be a functor from a category C to the category Sets which maps "arrows" in C to isomorphisms in Sets.
Proposition 1:
The category of covering spaces of BC is canonically isomorphic to the category of morphism-inverting functors $F: C\rightarrow Sets$.
[For $C$ a small category, its classifying space $BC$ is the geometric realization of its nerve, $NC$]
This proposition plays an essential role in Quillen's Theorem 1 showing that his Q-construction agrees with Grothendieck's construction for $K_0$.
Theorem 1:
$\pi_1(B(QC))$ is canonically isomorphic to the Grothendieck group $K_0(M)$
Questions:  Have morphism-inverting functors played an important role in other contexts?  Is there a more modern incarnation of morphism-inverting functors related to the fundamental groupoid of an infinity-category?

REPLY [2 votes]: Proposition 1 is extremely straightforward to prove (provided you have some facts like the quillen adjunction between SSet and CGWH).  Sing(|S|) gives you a simplicial set where all of the edges "forget" their direction, and when you apply the inverse of the nerve functor, you get back a copy of C with all of its arrows as isomorphisms.  Covering spaces are equivalent to (etale) bundles (of sets) on a topological space, which by a theorem in Mac Lane (Sheaves in Geometry and Logic) is equivalent to taking sheaves on the space, so by unraveling these equivalences, you get your result.  The last equivalence is probably one you're familiar with as the espace \'etal\'e.  (While in general, the nerve functor does not have an inverse, the nerve of a category has some nice properties that make the total singular complex (the $Sing$ functor) pull back intact, modulo directedness of edges.  If you think about the actual graph of the nerve of an ordinary category, it's not hard to see why this is true. This is precisely because the geometric realization "forgets" some information.)  
The construction you're describing is generalized by a functor in HTT called the unstraightening functor, which you can read about in HTT Ch 2.2.  With a number of more sophisticated results, we can generalize the adjunction between $Sing$ and $| \cdot |$  to a Quillen equivalence between SSet-Cat and CGWH-Cat.  
HTT is Higher topos theory by J. Lurie.<|endoftext|>
TITLE: How would you compute that "average" ?
QUESTION [8 upvotes]: I created a DJ-ing application that allows you to mix your MP3s with a real turntable.
So I generated an audio timecode to burn on a CD, left channel is the absolute position, right channel is a synchronization sine which frequency is 2205 Hz. Left channel is same frequency except that it represents binary sequences.
Precision is an absolute pre-requisite, the pitch on my Pioneer CDJ-1000 has a precision of 0.02% which is good enough to properly beatmatch songs. I have no problem on getting the absolute position but with the right channel.
After some searching, I found the Zero Crossing Rate algorithm to try detect the pitch. When I process data in real-time however, the pitch moves constantly.
After some searching I found out that as samples are discrete values, so not continuous if I'm right, there's not enough precision with a sample rate of 44100 Hz. Period of 20 samples is 2205 Hz, period of 21 samples is 2100 Hz; so the problem needs something more high-level.
I found then harmonic average, and moving average, results are somewhat more stable. FFT is the best but costs a lot of CPU, using a size of 65536 and time to compute it, whereas with the zero-crossing rate, I can update the value very frequently. The latest candidate, I still need to test is the Standard Deviation.
There must be some math formula that helps for that particular problem, but as I know not much about maths, I am somewhat lost.
Do you have any ideas ?
Thanks a lot :-)))

REPLY [4 votes]: I'm chanelling a response from an electrical engineer friend who works in a related area:

The thing he's trying to do is to match the beats and you can't just use zero crossing algorithms or "averages"... The code from the the Sethares paper mentioned (above) works pretty well, but it's in matlab, which means that it doesn't work in real time (because matlab has no real-time processing capabilities, not because the algorithm is inherently noncausal). So I'm doubtful whether it would be of any immediate use to the questioner.
There are some commercial products that are starting to appear, which mostly work by assuming that there is a steady drum beat underlying the music (this makes the problem a lot easier -- for instance, if you know there is a bass drum that regularly hits on 1 and 3, you can synchronize to it). So here it would depend on what style of music he is working with.

From his response it sounds like if you're sticking to something like electronic dance music then there are readily-available commercial products (I find some by googling).  But if you're trying to automate beat-matching of Indonesian gamelan music to an acoustic guitar solo, you'd have more troubles.
edit: The Sethares paper provides routines but it also provides a reference to the beat-tracking literature.   If your main interest is to have a convienient way to beat-track, it would seem like the best thing to do would be to avoid real-time computations. Precompute the relevant data using an algorithm like the ones Sethares (and all) propose.  Store that information beside your music so that you'd have all the information available and not need any further computations or fancy hardware.<|endoftext|>
TITLE: Where can I find a catalog of known Ramsey numbers?
QUESTION [5 upvotes]: Is there an online catalog available of Ramsey numbers, preferably one that for unknown values documents the known upper/lower bounds?

REPLY [12 votes]: See "Small Ramsey Numbers" at http://www.combinatorics.org/ojs/index.php/eljc/article/view/DS1<|endoftext|>
TITLE: Erdos Conjecture on arithmetic progressions
QUESTION [26 upvotes]: Introduction:
Let A be a subset of the naturals such that $\sum_{n\in A}\frac{1}{n}=\infty$. The Erdos Conjecture states that A must have arithmetic progressions of arbitrary length. 
Question:
I was wondering how one might go about categorizing or generating the divergent series of the form in the introduction above. I'm interested in some particular techniques and I list some examples below:
If we let $S$ be the set of such divergent series: $S=\left[ A: \sum_{n\in A}\frac{1}{n}=\infty, \ A\in\mathbb{N} \right]$, what kind of operations are there that would make S a group, or at the very least a semigroup? I'm rather vague on what the operatons should be for a reason, because although I presume trivial operations exist, their usefulness in understanding the members of $S$ would be questionable. 
Alternately, can one look at these divergent sums through the technique of Ramanujan summation (think: $1+2+3+\ldots =^R -\frac{1}{12}$, $R$ emphasizing Ramanujan summation)? The generalizations of Ramanujan summation (a good reference  here ) allow one to assign values to some of these series and give some measure of what kind of divergence is occurring. Moreover, basic series manipulations that hold for convergent series tend to carry over to Ramanujan summation, so can one perhaps look at the set $S$ above as a set of equivalence classes in the sense of two elements being equivalent if they share the same Ramanujan summation constant. 
Thanks in advance for any input!

REPLY [3 votes]: An idea that I came up with is that if the conjecture is true, then there would be a finite upper bound on how big the sum could be if the set had no arithmetic progression of length k, lets call this upper bound f(k). Obviously f(1)=0, as if there are any members then there is an arithmetic progression of that length. f(2)=1 (using the set {1}), as adding more numbers would result in an arithmetic progression of length two or more. I am currently working on f(3)<|endoftext|>
TITLE: remark in milne's class field theory notes
QUESTION [7 upvotes]: In the introduction of his class field theory notes Milne mentions that some famous mathematicians failed to ask if the Artin isomorphism is canonical (between $Gal(L/K)$ and $C_m/H$ where $H$ is generated by the split primes in $L$). Does this mean:
1)in category theory terms: there is a natural transformation between the functors from abelian extensions over K to abelian groups given by $Gal(?/K)$ and $C_m/H?$ (where H? is generated by the primes split over $?/K$).
2)or some kind of vaguer statement about whether we need to make choices along the definition of the map.
or maybe 2) is precisely encoded in the definition of 1).

REPLY [14 votes]: 1.  Here is what Tate says in his account of the General Reciprocity Law in the AMS volume on Hilbert's problems :

With this work of Takagi the theory of
  abelian extensions --- "class field
  theory" --- seemed in some sense
  complete, yet there was still no
  general reciprocity law.  It remained
  for Artin to crown the edifice with
  such a theorem.  He conjectured in
  1923 and proved in 1927 that there is
  a natural isomorphism $$
> C_K/N_{L|K}C_L\buildrel\sim\over\to\operatorname{Gal}(L|K)
> $$ which is characterised by the fact
  that...

And a little later :

How did Artin guess his reciprocity
  law ?  He was not looking for it, not
  trying to solve a Hilbert problem. 
  Neither was he, as would seem so
  natural to us today, seeking a
  canonical isomorphism, to make
  Takagi's theory more functorial.  He was led to the law by trying to show...

Read him.
2. Here is a toy example --- not unrelated to class field theory --- of how a bijection can be more natural than others.  Let $p$ be a prime number and let $K$ be finite extension of $\mathbb{Q}_p$ containing a primitive $p$-th root of $1$.  There are only finitely many degree-$p$ cyclic extensions $L|K$, and there are only finitely many vectorial lines in the $\mathbb{F}_p$-space $K^\times/K^{\times p}$.  In fact the two sets have the same number of elements, but the only natural bijection is 
$$
L\mapsto\operatorname{Ker}(K^\times/K^{\times p}\to L^\times/L^{\times p}),
$$
of which the reciprocal bijections can be written $D\mapsto K(\root p\of D)$.
It follows that the number of degree-$p$ cyclic extensions $L|K$ is the same as the number of hyperplanes in $K^\times/K^{\times p}$.  But is there a natural bijection between these two sets ?  You will agree that $L\mapsto N_{L|K}(L^\times)/K^{\times p}$ is as natural a bijection as there can be.
One last point : Given a hyperplane $H\subset K^\times/K^{\times p}$, how do you recover the degree-$p$ cyclic extension $L|K$ such that $H=N_{L|K}(L^\times)/K^{\times p}$ ?  Answer : use the natural reciprocity isomorphism $K^\times/K^{\times p}\to\operatorname{Gal}(M|K)$, where $M|K$ is the maximal elementary abelian $p$-extension, to identify $H$ with a subgroup of $\operatorname{Gal}(M|K)$, and take $L=M^H$.
Addendum (2011/11/21)  In Recountings (edited by Joel Segel, A K Peters Ltd, Natick, Mass.), Arthur Mattuck recounts a conversation with Emil Artin about his reciprocity law:

I will tell you a story about the
  Reciprocity Law.  After my thesis, I
  had the idea to define $L$-series for
  non-abelian extensions.  But for them
  to agree with the $L$-series for
  abelian extensions, a certain
  isomorphism had to be true.  I could
  show it implied all the standard
  reciprocity laws.  So I called it the
  General Reciprocity Law and tried to
  prove it but couldn't, even after many
  tries.  Then I showed it to the other
  number theorists, but they all laughed
  at it, and I remember Hasse in
  particular telling me it couldn't
  possibly be true.
Still, I kept at it, but nothing I
  tried worked.  Not a week went by ---
  for three years ! --- that I did not try to prove the Reciprocity Law.  It
  was discouraging, and meanwhile I
  turned to other things.  Then one
  afternoon I had nothing special to do, so
  I said, `Well, I try to prove the
  Reciprocity Law again.'  So I went out
  and sat down in the garden.  You see,
  from the very beginning I had the idea
  to use the cyclotomic fields, but they
  never worked, and now I suddenly saw
  that all this time I had been using
  them in the wrong way --- and in half
  an hour I had it.

REPLY [9 votes]: The point is that it is one thing to show that two mathematical objects are isomorphic; it is another (stronger) thing to give a particular isomorphism between them.  A rather concrete instance of this is in combinatorics, where if $(A_n)$ and $(B_n)$ are two families of finite sets, one could show that $\# A_n = \# B_n$ by finding formulas for both sides and showing they are equal, but it is preferred to find an actual family of bijections $f_n: A_n \rightarrow B_n$.
This is not just a matter of fastidiousness or a general belief that constructive proofs are better.  When considering functorialities between various isomorphic objects, the choice of isomorphism matters.  For instance, often one wants to put various isomorphic objects into a diagram and know that the diagram commutes: this of course depends on the choice of isomorphism.
In the case of class field theory, these functorialities take the form of maps between the abelianized Galois groups / norm cokernel groups / idele class groups of different fields.  The isomorphisms of class field theory can be shown to be the unique ones which satisfy various functoriality properties (and some "normalizations" involving Frobenius elements), and this uniqueness is often just as useful in the applications of CFT as the existence statements.
All of this, by the way, is explained quite explicitly in Milne's (excellent) notes: you just have to read a bit further.  See for instance Theorem 1.1 on page 20: "There exists a unique homomorphism...with the following properties [involving Frobenius automorphisms and functoriality]..."
As a final remark: it is important to note that the word "canonical" in mathematics does not have a canonical meaning.  To say that two objects are canonically isomorphic requires further explanation (as e.g. in the Theorem I mentioned above).  Even the "unique isomorphisms" that one gets from universal mapping properties are not unique full-stop [generally!]; they are the unique isomorphisms satisfying some particular property.<|endoftext|>
TITLE: What does primary mean geometrically?
QUESTION [17 upvotes]: Given a primary ideal I in a ring A, we can consider the subscheme V(I) of Spec(A).
It is a nilpotentification (?) of the integral subscheme V(rad(I)) given by the radical rad(I) of I.
My question is what kind of nilpotentifications you get that way.
It is a vague question, but I'm asking because i am completely unable to understand the algebraic meaning of "primary ideal" (altho I learned it by heart)

REPLY [20 votes]: As Harry suggests in his answer, it is probably more intuitive to work with associated 
primes, rather than the slightly older language of primary decompositions.  
If $I$ is an ideal in $A$, an associated prime of $A/I$ is a prime ideal of $A$ which
is the full annihilator in $A$ of some element of $A/I$.  A key fact is that for any
element $x$ of $A/I$, the annihilator of $x$ in $A$ is contained in an associated prime.
The associated primes are precisely the primes that contribute to the primary decomposition
of $I$.  Geometrically, $\wp$ is an associated prime of $A/I$ if there is a section
of the structure sheaf of Spec $A/I$ that is supported on the irreducible closed set $V(\wp)$.  E.g. in the example given in Cam's answer, the function $x^2 - x$ is not identically
zero on $X:=$ Spec ${\mathbb C}[x,y]/(x y, x^3-x^2, x^2 y - xy),$ but it is annihilated
by $(x,y)$, and so is supported at the origin (if we restrict it to the complement of
$(0,0)$ in $X$ then it becomes zero).
The non-minimal primes of $I$ that play a role in the primary decomposition of $I$
(i.e. appear as associated primes of $A/I$) are the generic points of the so-called
embedded components of Spec $A/I$: they are irreducible closed subset of Spec $A/I$
that are not irreducible components, but which are the support of certain sections
of the structure sheaf.
An important point is that if $I$ is radical, so that $A/I$ is reduced, then
there are no embedded components: the only associated primes are the minimal primes
(for the primary decomposition of $I$ is then very simple, as noted in the question:
$I$ is just the intersection of its minimal primes).
There is a nice criterion for a Noetherian ring to be reduced: Noetherian $A$ is
reduced if and only if $A$ satisfies $R_0$ and $S_1$, i.e. is generically reduced,
and has no non-minimal associated primes.  Geometrically, and applied to $A/I$
rather than $A$, this says that if $A/I$ is generically reduced, then the embedded
components are precisely the irreducible closed subsets of Spec $A/I$ over which
the nilpotent sections of the structure sheaf are supported.  This may help
with your ``nilpotentification'' mental image.<|endoftext|>
TITLE: A closed subscheme of an open subscheme that is not an open subscheme of a closed subscheme?
QUESTION [41 upvotes]: A morphism $f: V \rightarrow X$ of schemes is a locally closed immersion if it can be factored into a closed immersion followed by an open immersion.   It is not hard to show that if $f$ is an open immersion followed by a closed immersion, then it is a locally closed immersion, but the converse is at the very least not clear (to me).  For a number of reasons, this choice as the definition of locally closed immersion (rather than the opposite) is the right one (e.g. it is then not hard to see that compositions of locally closed immersions are locally closed immersions). 

Is there some $f:  V \rightarrow X$ that can be factored into a closed immersion followed by an open immersion, that cannot be factored into an open immersion followed by a closed immersion?

Warning:  it isn't too hard to show that there is no example with $V$ reduced or with $f$ quasicompact, so any example has to be a little strange-looking.  Back-story:  I've been confronted with this question when learning algebraic geometry with a class (conventionally known as  "teaching"); it seems a natural question.  And any counterexample would likely be a handy example to have for other reasons as well:  a very limited stock of counterexamples tends to refine my intuition, and to warn me what can go wrong.

REPLY [23 votes]: Hi Ravi,
There is an example in Tag 01QW in Johan's stacks project. 
Jarod<|endoftext|>
TITLE: Understanding the product in topological K-theory
QUESTION [6 upvotes]: I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...
The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_0)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.
If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\left[H\right]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\left[H-1\right]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.
The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?
Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map
$K(X)\otimes K(X)\to K(X\times X)\to K(X)$.

REPLY [16 votes]: Reduced $K$-groups are ideals of the standard $K$-groups.  $\tilde K(X) \subset K(X)$ is the ideal of virtual-dimension-zero elements.
In particular, the reduced K-theory $\tilde K(S^2)$ is not $\mathbb{Z}[H]/(H-1)^2$, but rather the ideal of this generated by $(H-1)$.  In particular, any element in this group does square to zero.
Additionally, the "exterior product" isomorphism $f: \tilde K(X) \to \tilde K(X \wedge S^2)$, which is an isomorphism, is not a ring map: it takes an element $x$ to the exterior product $x \wedge (H-1)$.  Instead, it satisfies $f(x) f(y) = (H-1) f(xy) = 0$.  This is because the suspension is covered by two contractible open subsets, and so all products must vanish.

REPLY [6 votes]: The map $K(X)\otimes K(X)\to K(X)$ induced by $\otimes$, and the composition $K(X)\otimes K(X)\to K(X\times X)\to K(X)$ of the external product and the diagonal map coicinde. Just seeing what both do to a pair of vector bundles shows this.<|endoftext|>
TITLE: Special values of $p$-adic $L$-functions.
QUESTION [22 upvotes]: This is a very naive question really, and perhaps the answer is well-known. In other words, WARNING: a non-expert writes.
My understanding is that nowadays there are conjectures which essentially predict (perhaps up to a sign) the value $L(M,n)$, where $M$ is a motive, $L$ is its $L$-function, and $n$ is an integer. My understanding of the history is that (excluding classical works on rank 1 motives from before the war) Deligne realised how to unify known results about $L$-functions of number fields and the B-SD conjecture, in his Corvallis paper, where he made predictions of $L(M,n)$, but only up to a rational number and only for $n$ critical. Then Beilinson extended these conjectures to predict $L(M,n)$ (or perhaps its leading term if there is a zero or pole) up to a rational number, and then Bloch and Kato went on to nail the rational number.
Nowadays though, many motives have $p$-adic $L$-functions (the toy examples being number fields and elliptic curves over $\mathbf{Q}$, perhaps the very examples that inspired Deligne), and these $p$-adic $L$-functions typically interpolate classical $L$-functions at critical values, but the relationship between the $p$-adic and classical $L$-function is (in my mind) a lot more tenuous away from these points (although I think I have seen some formulae for $p$-adic zeta functions at $s=0$ and $s=1$ that look similar to classical formulae related arithmetic invariants of the number field).
So of course, my question is: is there a conjecture predicting the value of $L_p(M,n)$, for $n$ an integer, and $L_p$ the $p$-adic $L$-function of a motive? Of course that question barely makes sense, so here's a more concrete one: can one make a conjecture saying what $\zeta_p(n)$ should be (perhaps up to an element of $\mathbf{Q}^\times$) for an integer $n\geq2$ and $\zeta_p(s)$ the $p$-adic $\zeta$-function? My understanding of Iwasawa theory is that it would only really tell you information about the places where $\zeta_p(s)$ vanishes, and not about actual values---Iwasawa theory is typically only concerned with the ideal generated by the function (as far as I know). Also, as far as I know, $p$-adic $L$-functions are not expected to have functional equations, so the fact that we understand $\zeta_p(s)$ for $s$ a negative integer does not, as far as I know, tell us anything about its values at positive integers.

REPLY [6 votes]: Others have hinted at it, but let me emphasize the point. At least if you are happy to assume all conjectures (and perhaps that your motive has good reduction at $p$), the conjectural landscape for $p$-adic $L$-functions is as complete as that for usual $L$-functions. Namely: there is a conjectural description of the value of the cyclotomic $p$-adic $L$-function at any integer (in fact any character $\eta\chi_{cyc}^{s}$ with $\eta$ finite).
This can either be done in B.Perrin-Riou's style, see Fonctions $L$ $p$-adiques des représentations $p$-adiques, or in K.Kato's style, in which case it follows from the conjectures on special values of $L$-functions taking into account the action of a group algebra (the so-called equivariant conjectures). In fact, I exaggerate slightly here: at some special values, there could be an exceptional zero, in which case the leading term should incorporate an $\mathcal{L}$-invariant, and I don't think this has been (conjecturally) defined in all generality.
Also, as Rob H. wrote, $p$-adic $L$-function are indeed expected to satisfy a functional equation. This can be seen either from Perrin-Riou's conjectural construction from motivic elements, in which case the functional equation follows from the explicit reciprocity law of Perrin-Riou (and Colmez in the de Rham case) or via Iwasawa main conjectures, in which case it follows from duality results for cohomology complexes.
So everything you could wish for is conjectured. Not much, of course, is actually known.<|endoftext|>
TITLE: Is it always possible to compute the Betti numbers of a nice space with a well-chosen Lefschetz zeta function?
QUESTION [8 upvotes]: Let $X$ be a smooth projective variety.  If I've understood correctly, the Weil conjectures imply that it is possible to compute the Betti numbers of $X(\mathbb{C})$ by computing the local zeta function associated to $X(\overline{\mathbb{F}_p})$ for $p$ a prime of good reduction.  This is because one can identify the factors coming from different $\ell$-adic cohomology groups by the absolute value of their roots (the "Riemann hypothesis," by analogy with the case of curves).
Unfortunately, this doesn't seem to be true for the analogous situation with compact triangulable spaces $Y$ and the Lefschetz zeta function $\zeta_f$, at least not for an arbitrary choice of function $f : Y \to Y$.  For example, if $f$ is homotopic to the identity, then $\zeta_f$ can only see the Euler characteristic of $Y$.  Even if $f$ acts "generically" (i.e. none of the factors of $\zeta_f$ cancel), there doesn't seem to be a way to distinguish which factors are associated to which degree.  (Of course, I would love if I were wrong about this.)
Question 1:  When is there an analogue of the geometric Frobenius for compact triangulable spaces $Y$?  By this I mean a more-or-less canonical function $f : Y \to Y$ such that some analogue of the Riemann hypothesis holds for $\zeta_f$.  
Question 2:  Regardless of the answer to Question 1, is it always possible to choose $f$ such that $\zeta_f$ can tell you which of its factors are associated to which homology groups?
(Side question: is it true that given any $f : Y \to Y$ there is always some $f'$ homotopic to $f$ which has finitely many fixed points?)

REPLY [4 votes]: Having resolved my ignorance concerning surface groups I can now answer question 1 negatively (or at least some formulation thereof).  It is impossible if $Y$ is an oriented surface of genus at least $2$.
Suppose that $f: Y \to Y$ is a self map of the surface such that the eigenvalues of $f^*$ acting on each $H^i(Y)$ are all nonzero (otherwise we can't "detect" the betti numbers), and such that $H^i(Y)$ and $H^j(Y)$ do not have eigenvalues of common magnitude for $i \neq j$.  Then in particular $f^*$ acts on $H^2(Y)$ nontrivially, say by multiplication by some integer $d$.  This integer cannot be $\pm 1$ since then $H^0(Y)$ and $H^2(Y)$ would contain eigenvectors with eigenvalues of equal magnitude.
Consider the subgroup $H = f_*(\pi_1(Y))$ inside $G = \pi_1(Y)$.  If this had infinite index, then $f$ would lift to a map to some infinite covering of $Y$, so it would induce a trivial map of $H^2$.  So $H$ has finite index in $G$.  Let $X \to Y$ be the corresponding covering space.  Then $\pi_1(X)$ is a quotient of $\pi_1(Y)$, hence its abelianization has rank $\leq 2g$ where $g$ is the genus of $Y$.  This implies that $X$ is a closed surface of genus at most $g$.  But its Euler characteristic is precisely $[G:H]$ times the Euler characteristic of $Y$, so $X = Y$.  Thus $f$ induces a surjection on $\pi_1(Y)$.  By the post cited above, $f$ actually induces an isomorphism on $\pi_1(Y)$, so it is a homotopy equivalence.  In particular, $d = \pm 1$, contrary to assumption.  
After writing this it occurs to me that you might object to me ruling out the case $d = -1$...  At any rate, this shows that the eigenvalues can't ever look like they do in the case of the Riemann hypothesis, with magnitude $q^{i/2}$ on $H^i$ for some $q>1$.<|endoftext|>
TITLE: Conditional probabilities are measurable functions - when are they continuous?
QUESTION [15 upvotes]: Let $\Omega$ be a Banach space; for the sake of this post, we will take $\Omega = {\mathbb R}^2$, but I am more interested in the infinite dimensional setting.  Take $\mathcal F$ to be the Borel $\sigma$-algebra, and let $\mathbb P$ be a probability measure on $(\Omega, \mathcal F)$.
Denote by $\vec x = (x,y)$ a point in $\Omega$, and let $\mathcal F_1$ be the $\sigma$-algebra generated by the first coordinate.  Fix $y_0 \in \mathbb R$ and 
$\eta > 0$, and consider $$f(\vec x) = \mathbb P( ~|y - y_0| \le \eta~ |\mathcal F_1).$$  (More generally, one can consider $f(\vec x) = \mathbb E( \varphi(\vec x) | \mathcal F_1)$ for some suitable $\varphi : {\mathbb R}^2 \to \mathbb R$.)
The function $f$ is measurable; that comes from the definition of conditional expectations.  I would like to find some reasonable sufficient conditions such that $f$ is continuous and positive.  
I feel like this should be relatively elementary material, but unfortunately I'm having trouble finding any references.  How should I approach this?
I've included the [fa.functional-analysis] tag because in general I want to consider $\Omega$ to be a space of smooth functions.  I'm guessing that to give some additional structure, I'll need to assume that $\mathbb P$ is absolutely continuous with respect to a Gaussian measure, because I don't know any other reasonable measures on function space.

REPLY [3 votes]: Tue Tjur studied the existence of continuous disintegrations in a 1975 preprint "A Constructive Definition of Conditional Distributions," Issue 13, Copenhagen Universitet.  He gives necessary and sufficient conditions for their existence.  He also discusses sufficient structure, and there the question of the existence of joint densities arises.  The article is a bit hard to track down, so let me know if you need help finding it.  The existence of continuous disintegrations arises also in the study of the computability of conditional probability, which is my interest.<|endoftext|>
TITLE: Fundamental groups of the spaces of rational functions
QUESTION [18 upvotes]: Here is a question which I asked myself (and couldn't answer) while reading "The topology of spaces of rational functions" by G. Segal.
Let $X$ be a smooth complete complex curve (=a compact Riemann surface) of genus $g$ and let $Rat(X,d)$ be the space of all regular (=holomorphic) maps from $X$ to $\mathbf{P}^1(\mathbf{C})$ of degree $d$. In this question I'm interested in the fundamental group of the open subset $U(X,d)$ of $Rat(X,d)$ formed by all $f$ such that all critical points of $f$ are simple  and all critical values are distinct. (A critical point is a point at which the derivative of $f$ vanishes; a critical value is the image of a critical point.) To be more specific, let's say I'd like to

find a "nice" system of generators of $\pi_1(U(X,d))$;
to describe, for each of these generators, its image under the map induced by the map $G$ from $U(X,d)$ to the configuration space $B(\mathbf{P}^1(\mathbf{C}),k)$ of unordered subsets of $\mathbf{P}^1(\mathbf{C})$ of cardinality $k=2(d+g-1)$ that takes $f$ to its branch divisor (i.e. the divisor of the critical points).

Here are some remarks that may be useful (or may not):
First, here is how one can think of the fundamental group of $Rat(X,d)$. By associating to every function its divisor of poles we get a map $F$ from $Rat(X,d)$ to the $d$-th symmetric power $S^d(X)$ of $X$.
Assume $d> 2g-2$. By the Riemann-Roch theorem, for any degree $d$ divisor $D$ the linear space ${\cal{L}}(D)=H^0(X,{\cal{O}}(D))$ (which is formed by all rational functions $f$ such that for any $x\in X$ the order of the pole of $f$ at $x$ is at most the multiplicity of $x$ in $D$) is $d-g+1$. So $F$ is surjective and a fiber of $F$ is $\mathbf{C}^{d-g+1}$ minus some number of hyperplanes (these are given by the condition that order the pole of $f$ at a point $x$ of $D$ is less then the multiplicity of $x$ in $D$).
The map $F$ is probably not a fibration. However, the fundamental group of $Rat(X,d)$ is spanned by the loops in a general fiber of $F$ going around one of the hyperplanes, and lifts of the loops in $S^d(X)$ (these are all of the form "one of the points moves along a loop in $X$ and the other stand still").
Second, recall that the Jacobian $J(X)$ of $X$ is defined as follows. Integration along cycles gives an injective map $H_1(X,\mathbf{Z})\to\mathbf{C}^g=Hom(H^0(X,\Omega_X),\mathbf{C})$ and the Jacobian of $X$ is the quotient. Moreover, once we have chosen a base point $x$ in $X$, we get a natural map $j:X\to J(X)$ defined as follows: for any $x'\in X$ take a path $\gamma$ from $x$ to $x'$ and set $j(x')$ to be the image in $J(X)$ of the "integration along $\gamma$ function". This is well defined map that can be extended by $\mathbf{Z}$-linearity to $S^d(X)$.
Abel's theorem says that two disjoint effective divisors are the divisors of the zeros and the poles of a rational function if and only if their images under $j$ coincide. This may be useful in this problem, but I don't see how.

REPLY [11 votes]: From a topological point of view, rational functions are branched coverings of $S^2$.  The fundamental group of the space of branched coverings is the group of "liftable braids".  This group was calculated for $d=3$ by Birman and Wajnryb [1] and for $d=4$ by myself [2].  I have recently calculated the general case, the results should be published Any Time, Really Soon Now.

 Birman, Wanryb,  3--fold branched coverings and the mapping class group of a surface, LNM 1167, 24-46
    Apostolakis, On 4--fold covering moves, Algebraic and Geometric Topology 3 (2003), 117-145. 
    Mullazzani, Piergallini, Lifting Braids, arXiv:math/0107117<|endoftext|>
TITLE: Is there an elementary proof of a result about the parity of the period of the repeating block in the continued fraction expansion of square roots
QUESTION [15 upvotes]: It is a known fact that for a Prime $P$, $P\equiv 1$ mod $4$ iff the length of the period in the repeating block for the continued fraction expansion of $\sqrt{P}$ is odd.  I have an elementary proof of this using the classical result: $P\equiv 1$ mod $4$ iff $x^2-Py^2=-1$ has integer solutions and a proof that $x^2-Py^2=-1$ has integer solutions iff the length of the period in the repeating block for the continued fraction expansion of $\sqrt{P}$ is odd.  
I have tried repeatedly to give a direct elementary proof that $P\equiv 1$ mod $4$ implies that the length of the period in the repeating block for the continued fraction expansion of $\sqrt{P}$ is odd, but cant seem to figure it out.  (I have a direct elementary proof of the converse)
Does anyone know of an elementary proof of this result or where I may find one?

REPLY [3 votes]: See Nagell, Introduction to number theory, Theorem 107. Originally it was proved by Legendre, see Dickson, L. E. History of the theory of numbers. Vol. II, p. 365. There are more examples due to Dirichlet, see Dirichlet's Werke, Bd. 1 (1889), pp. 219-236.
An extended bibliography on negative Pell’s equation can be found in Gerasim, I.-Kh.I.
On the genesis of Redei’s theory of the equation x 2 -Dy 2 =-1. (Russian) Zbl 0731.01014
Istor.-Mat. Issled. 32/33, 199-211 (1990) (available in electronic form).<|endoftext|>
TITLE: Hodge Index theorem for Complex Manifolds
QUESTION [7 upvotes]: The Riemann-Roch theorem is a result about Riemann Surfaces that was extended to the Hirzebruch–Riemann–Roch theorem, a result about compact complex manifolds. The Hodge Index theorem is a result about Riemann surfaces (I'm just worried about the complex case) that is proved using Riemann-Roch. Has the Hirzebruch–Riemann–Roch theorem been used to extend the Hodge Index theorem to a result about compact complex manifolds.

REPLY [3 votes]: For a compact Kähler manifold $X$ of complex dimension $2m$, the signature is given by
$$\sigma(X) = \sum_{p=0}^{2m}\sum_{q=0}^{2m}(-1)^ph^{p,q}(X).$$
As algori points out, this formula does not hold for all even-dimensional compact complex manifolds. However, for compact Kähler manifolds, $h^{p,q}(X) = h^{q,p}(X)$ so the signature is also given by
$$\sigma(X) = \sum_{p=0}^{2m}\sum_{q=0}^{2m}(-1)^qh^{p,q}(X).$$
Surprisingly, this formula does hold for all even-dimensional compact complex manifolds (check algori's example). This can be deduced using the properties of Hirzebruch's $\chi_y$ genus, as is shown here.<|endoftext|>
TITLE: Hilbert scheme of points on a complex surface
QUESTION [8 upvotes]: I don't know about schemes and every definition of a Hilbert scheme (quite naturally!) involves schemes. But, the Hilbert scheme of points on a complex surface is known to be smooth (Fogarty). So is there a concrete description of it as a complex manifold? (For instance in the case of n=2 it is a blowup of XxX along the diagonal)

REPLY [4 votes]: Here is a geometric description in the case of $H_n(\mathbb{C}^2)$. This is meant to be a geometric rewrite of Proposition 2.6 in Mark Haiman's "(t,q)-Catalan numbers and the Hilbert scheme",
Discrete Math. 193 (1998), 201-224.
Let $S= (\mathbb{C}^2)^n/S_n$; notice that this is an orbifold. Let $S_0$ be the open dense set where the $n$ points are distinct. For $D$ an $n$-element subset of $\mathbb{Z}_{\geq 0}^2$, let $A_{D}$ be the polynomial $\det( x_i^{a} y_i^{b})$, where $(a, b)$ ranges over the elements of $D$ and $i$ runs from $1$ to $n$. For any $D$ and $D'$, the ratio $A_D/A_{D'}$ is a meromorphic function on $S$, and is well defined on $S_0$.
Map $S_0$ into $S_0 \times \mathbb{CP}^{\infty}$ where the homogenous coordinates on $ \mathbb{CP}^{\infty}$  are the $A_{D}$'s. (Only finitely many of the $A_D$'s are needed, but it would be a little time consuming to say which ones.) The Hilbert scheme is the closure of $S_0$ in $S \times \mathbb{CP}^{\infty}$. 
Algebraically, we can describe this as the blow up of $S$ along the ideal generated by all products $A_D A_{D'}$. Haiman points out that the reduction of this ideal is the locus where two of the points collide and speculates that this ideal may be reduced. If his speculation is correct, then we can describe $H_n(\mathbb{C}^2)$ geometrically as the blow up of $(\mathbb{C}^2)^n/S_n$ along the reduced locus where at least two of the points are equal.<|endoftext|>
TITLE: Reference for Tate vector spaces
QUESTION [5 upvotes]: ... aka locally linear compact vector spaces. The one reference I know is http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Nov3-10(CentExt).pdf. Does anyone know another good reference?

REPLY [6 votes]: Beilinson, Drinfeld.  Chiral Algebras section 2.7 (I think)
Beilinson, Feigin, Mazur.  Notes on Conformal Field Theory (Incomplete) available on Mazur's web page.

Also: Tate, Residues of differentials on curves<|endoftext|>
TITLE: Cool problems to impress students with group theory
QUESTION [162 upvotes]: Since this forum is densely populated with algebraists, I think I'll ask it here.
I'm teaching intermediate level algebra this semester and I'd like to entertain my students with some clever applications of group theory. So, I'm looking for problems satisfying the following 4 conditions
1) It should be stated in the language having nothing whatsoever to do with groups/rings/other algebraic notions.
2) It should have a slick easy to explain (but not necessarily easy to guess) solution using finite (preferrably non-abelian) groups.
3) It shouldn't have an obvious alternative elementary solution (non-obvious alternative elementary solutions are OK).
4) It should look "cute" to an average student (or, at least, to a person who is curious about mathematics but has no formal education).
An example I know that, in my opinion, satisfies all 4 conditions is the problem of tiling a given region with given polyomino (with the solution that the boundary word should be the identity element for the tiling to be possible and various examples when it is not but the trivial area considerations and standard colorings do not show it immediately) 
I'm making it community wiki but, of course, you are more than welcome to submit more than one problem per post.
Thanks in advance!

REPLY [3 votes]: How about Galois theorem that a prime degree equation is solvable if and only if all the roots are rational functions in two of the roots. I have read that Galois stated it this way, so that the statement does not involve the notion of a  solvable group, but the proof uses the result that a transitive subgroup of $S_p$ ($p$ prime), is contained in the normalizer of the group generated by a $p$-cycle.<|endoftext|>
TITLE: Intersection of finitely generated subalgebras also finitely generated?
QUESTION [6 upvotes]: Let $k$ be a field and $A$ be a finitely generated (commutative) algebra over $k$. If $A_1$ and $A_2$ are finitely generated $k$-subalgebras of $A$, is it true that $A_1 \cap A_2$ is also finitely generated (as an algebra) over $k$? What if $A$ is a polynomial ring?
Update (for the sake of completeness, April 1, 2017): This paper (disclaimer: it's mine) describes the smallest dimensional counterexample to the second question in zero characteristic. In positive characteristic the answer seems to be unknown.

REPLY [2 votes]: Here is a variant of Mariano's construction that answers the question. We shall construct a commutative semigroup $S$ and its subsemigroups $S_1,S_2$ such that 

(*) $S,S_1,S_2$ are finitely generated, but $S_1\cap S_2$ is not. 

Then passing to the semigroup algebras $A=k[S]$, $A_1=k[S_1]$, $A_2=k[S_2]$, which satisfy $A_1\cap A_2=k[S_1\cap S_2]$ and have the same finite generation properties as the corresponding semigroups, would yield a counterexample to the question.
Let $T$ be the commutative subsemigroup of $\Bbb{Z}^2$ generated by $x_i=(i,1)$ for $i\geq 0$. Thus $$T=\{(a,b)\in\Bbb{Z}^2: a\geq 0, b\ge 1 \text{ or } a=b=0\}$$ consists of the integer points in the semiopen real cone $\{(a,b)\in \Bbb{R}^2: a\geq 0, b>0\}$ together with the origin. It is easy to see that $T$ is not f.g. Let $S$ be formed by adjoining commuting elements $y,z$ to $T$ subject to the relations $yx_i=x_iy=x_{i+1}$ and $z x_i=x_i z=x_{i+1}$. Set $S_1=\langle T,y\rangle$ and $S_2=\langle T,z\rangle$. Clearly, $S_1$ is generated by $x_0$ and $y$, $S_2$ is generated by $x_0$ and $z$ and $S$ is generated by $x_0, y$ and $z$. Moreover, $S_1\cap S_2=T$ is not f.g., meaning (*) holds.    
Where does this come from? First, $T$ is the standard example of a non-f.g. commutative semigroup embedded into an f.g. one. In fact, both $S_1$ and $S_2$ are naturally isomorphic to $\Bbb{Z}_{+}^2$, with $y$, resp. $z$, corresponding to $(1,0)$, and $S$ is the amalgamated product of $S_1$ and $S_2$ over $T$ in the category of commutative semigroups. However, I do not see how to modify this construction to make $S$ an affine semigroup (so that $A=k[S]$ is a domain). In particular, it appears impossible to get a counterexample with $A$ a polynomial ring in this fashion.<|endoftext|>
TITLE: What characterizes rational functions with nonnegative integer Taylor coefficients?
QUESTION [8 upvotes]: I believe that there is a statement along the following lines (I would, of course, love to be corrected): a formal power series is the Taylor expansion of a rational function if and only if the coefficients eventually satisfy a linear relationship.
Let's suppose that I understand what "satisfy a linear relationship" means, because it's not the part I actually want to ask about (although clarifications are very welcome!).  What I would like to know is what conditions on a rational function are equivalent to all the Taylor coefficients being nonnegative integers.  For example, I happen to know that $1/(1-kx) = \sum (kx)^n$, and so any sum or product of such functions works.  In particular, I can try playing around with partial-fraction decompositions to see if I can write a given rational function in this way.  But I have no idea if this is all of them.
Put another way, there is a map $\mathbb R(x) \to \mathbb R[x^{-1},x]]$ (rational functions to Laurent series).  I would like to understand the inverse image of $\mathbb N[x^{-1},x]]$.
(Oh, also, I have no idea how to tag this, and I think "general mathematics" is probably an inappropriate tag for MO.  So please re-tag as you see fit.)

REPLY [3 votes]: This paper (?) of Gessel might help you out, although it is mostly about combinatorics.  There are two natural ways to write down rational functions with non-negative integer coefficients in combinatorics, one coming from transfer matrices / finite automata and one coming from regular languages.  The two give the same class of rational functions, but there exist rational functions with non-negative integer coefficients which provably don't arise in this way, so the situation seems complicated.
Your question seems to indicate you're not familiar with this class of rational functions, so here are two equivalent definitions: it is, on the one hand, the class of all non-negative linear combinations of entries of matrices of the form $(\mathbf{I} - \mathbf{A})^{-1}$ where $\mathbf{A}$ is a square matrix with entries in $x \mathbb{N}[x]$, and on the other hand the minimal class of rational functions containing $1, x$, and closed under addition, multiplication, and the operation $f \mapsto \frac{1}{1 - xf}$.
Edit:  One reason there isn't likely to be a particularly nice classification is that one can start with any rational function with integer coefficients and add a polynomial and a term of the form $\frac{1}{1 - kx}$ for $k$ such that $\frac{1}{k}$ is smaller than the smallest pole.<|endoftext|>
TITLE: Dependence of error on mesh for Riemann sums
QUESTION [6 upvotes]: Suppose $f$ is continuous on $[a,b]$ with $I = \int_a^b f(x)\: dx$,
and for every $\epsilon > 0$ let $\delta(\epsilon)$ be the largest
$\delta > 0$ such that every Riemann sum arising from a partition
of $[a,b]$ with mesh less than $\delta$ differs from $I$ by less
than $\epsilon$.
Is it true that (leaving aside the case where $f$ is constant)
$\delta(\epsilon)$ goes to zero like $\epsilon^2$, in the sense
that $\delta(\epsilon)/\epsilon^2$ is bounded above and below
by constants as $\epsilon$ goes to zero?

REPLY [4 votes]: It is possible to choose continuous functions $f$ with really bad approximations, i.e., so that $\delta(\epsilon)$ drops to zero arbitrarily quickly as a function of $\epsilon$.  For example, let $f_0$ denote the periodic function of period 1 that linearly interpolates between $(0,0)$, $(1/2,2)$, and the horizontal translates of these points by integers.  For each $n>0$, let $f_n(x) = f_0(2^n x)$, and let $f(x) = \sum_{n\geq 0} a_n f_n(x)$, for a summable sequence of positive reals $a_n$.  $f$ is integrable on $[0,1]$, with integral $\sum a_n$, and meshes with spacing $2^{-N}$ yield the first N summands.  However, if the partial sums converge very slowly, we can have, e.g., $\delta(1/N) \leq 2^{-N!}$.<|endoftext|>
TITLE: Algebraicity of the completion of a field?  Finiteness?
QUESTION [11 upvotes]: At the end of my 8410 class today (see http://www.math.uga.edu/~pete/MATH8410.html if you care), one of my students asked me the following very interesting question:
Let $(K,|\ |)$ be a normed field, with completion $(\hat{K},| \ |)$.  Suppose $\hat{K}$ is algebraic over $K$.  Must we then have $\hat{K} = K$?
As I have mentioned here before, I feel very lucky to be getting such penetrating questions.  This one I was not able to answer on the spot, although I remarked that it is true in all of the most familiar examples and that the (possible) lack of algebraicity of the completion is a key motivation for considering the Henselization instead.
Edit: the answer is no, as I have just heard from one of my students.  I have encouraged him to come to this site and register the answer.
To make the question more interesting, suppose we ask whether $\hat{K}/K$ can be finite and nontrivial?

REPLY [8 votes]: I had to think for a while to understand Scott's answer (or at least, what I suspect he meant by his answer), and in the end there were enough details to sort out that I thought they were worth posting.  It ended up being too long to post as a comment, so here it is as a separate answer.  Unless it's all nonsense, of course....
Let {$x_{\alpha}$} be a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, and let $L$ be the intermediate field that they generate, so that $\mathbb{C}$ is the algebraic closure of $L$ in $\mathbb{C}$.   Take also a collection of open disks $D_{\alpha}$ in $\mathbb{C}$ such that any collection of points $y_{\alpha} \in D_{\alpha}$ is dense in $\mathbb{C}$ in the usual topology.  Now for each $\alpha$, take $x_\alpha$ and multiply it by an appropriate root of unity and a rational number so that the result $y_\alpha$ lies in $D_\alpha$.  The collection {$y_{\alpha}$} is still algebraically independent over $\mathbb{Q}$, because a dependence gives an algebraic dependence of {$x_\alpha$} over some finite extension of $\mathbb{Q}$, which implies the existence of an algebraic dependence over $\mathbb{Q}$ as well.
So there exists $\sigma : L \to \mathbb{C}$ sending $x_{\alpha} \mapsto y_{\alpha}$.  Now by the usual fact that field embeddings into algebraically closed fields can be extended across algebraic extensions, $\sigma$ extends to a map $\mathbb{C} \to \mathbb{C}$.  But note that by construction $\sigma$ is surjective!  The image contains each $y_\alpha$, and it contains all the roots of unity, so it contains all the $x_\alpha$'s; thus the image is an algebraic closure of $L$ in $\mathbb{C}$, hence all of $\mathbb{C}$.  
In particular $\mathbb{C}$ is a quadratic extension of $\sigma(\mathbb{R})$, obtained by adjoining $\sigma(i)$.  But finally $\sigma(\mathbb{R})$ is dense in $\mathbb{C}$ since its image contains all the $y_\alpha$'s, and so giving $\sigma(\mathbb{R})$ the norm induced from the usual norm on $\mathbb{C}$, we get a normed field $\sigma(\mathbb{R})$ whose completion is exactly $\mathbb{C}$, i.e., a quadratic extension of it.  Thus the answer to your second question actually yes.<|endoftext|>
TITLE: Tate module of CM elliptic curves
QUESTION [9 upvotes]: This is an exercise in Silverman's book "the arithmetic of elliptic curves". Ex 3.24, page 109.
E/K CM elliptic curve, Prove for $\ell \neq char(K)$, the action of $Gal(\bar{K}/K)$ on $T_{\ell}(E)$ is abelian.

REPLY [3 votes]: You can find the answer to your question (and learn a whole lot more about complex multiplication) in another book by Joe Silverman, "Advanced Topics in the Arithmetic of Elliptic Curves". See Chapter II, and in particular read the proof of Theorem 2.3.<|endoftext|>
TITLE: Orthogonal matrices with small entries
QUESTION [15 upvotes]: Is it true that for any $n$, there exists a $n \times n$ real orthogonal matrix with all coefficients bounded (in absolute value) by $C/\sqrt{n}$, $C$ being an absolute constant ?
Some remarks :

If we want $C=1$, the matrix must be a Hadamard matrix.
The complex analogue has an easy answer: the Fourier matrix $(\exp(2\pi \imath jk/n)/\sqrt{n})_{(j,k)}$. Forgetting the complex structure gives a positive answer to the question in the real case when $n$ is even.
A random matrix doesn't work (the largest entry is typically of order $\sqrt{\log(n)}/\sqrt{n}$).

REPLY [6 votes]: Take $A$ to be the $n\times n$ matrix with $A_{jk}=\sqrt{\frac{2}{n+1}}\sin(\frac{jk\pi}{n+1})$. This is a variant of the answer of jj-joerg-arndt.<|endoftext|>
TITLE: Sparse approximate representation of a collection of vectors
QUESTION [6 upvotes]: Suppose I have a collection of $n$ vectors $C \subset \mathbb{F}_2^n$. They are of course spanned by the canonical set of $n$ basis vectors. 
What I would like to find is a much smaller (~ $\log n$) collection of basis vectors that span a collection of vectors which well approximate $C$. That is, I would like basis vectors $b_1,\ldots,b_k$ such that for every $v \in C$, there exists a $u \in span(b_1,\ldots,b_k)$ such that $||u-v||_1 \leq \epsilon$. 
When is this possible? Is there a property that $C$ might posses to allow such a sparse approximation?

REPLY [4 votes]: By triangle inequality, preserving the property you wish for means that you can find "representatives" for each $v$ so that the $\ell_1$ distances between any $v, v'$ are preserved to within 2$\epsilon$ additive error. 
There is a general result by Brinkman and Charikar that says that in general, for a collection of $n$ vectors in an $\ell_1$ space, there's no way to construct a set of $n$ vectors in a smaller (e.g $\log n$) dimensional space) that preserves distances approximately even multiplicatively (let alone additively). This distinction is important if you have vectors in the original space that are $O(\epsilon)$ apart, but otherwise doesn't matter greatly. 
Brinkman, B. and Charikar, M. 2005. On the impossibility of dimension reduction in l1. J. ACM 52, 5 (Sep. 2005), 766-788. DOI= http://doi.acm.org/10.1145/1089023.1089026 
So I'm guessing that the answer to your first question should be no.<|endoftext|>
TITLE: Eigenvalues of an element in a Weyl algebra
QUESTION [6 upvotes]: I have an operator acting on the polynomial algebra $\mathbb{C}[x,y,z]$ that I would like to find the eigenvalues/eigenvectors of.  More specifically, let $P(x_1, \ldots, x_6)$ be a homogeneous polynomial, my operator has the form $P(x,y,z, \frac{\partial}{\partial x},  \frac{\partial}{\partial y},  \frac{\partial}{\partial z})$.  Are there any general strategies that could help me?  For instance, say my operator were: $$z^2y\frac{\partial^3}{\partial x^2 \partial y} + y^3z\frac{\partial^4}{\partial y^2 \partial z^2} + xz^2\frac{\partial^3}{\partial x \partial z^2}.$$  My actual operator is degree 6 and more complicated, but other than that the same type of object. Any thoughts or references on how to attack this type of problem will be very welcome. Thanks a lot!
EDIT: My first example-operator was very poorly chosen, since every monomial would automatically be an eigenvector. I have now altered it a little to avoid this. Keep in mind this was only an example to show what type of object I am considering, and I am looking for general strategies. None the less, thanks for the quick response. 
EDIT 2: I had misread the degree of my actual operator - it is 6.

REPLY [4 votes]: If the polynomial $P(x_1, \ldots, x_6)$ is homogeneous of degree zero (where $x_1, x_2, x_3$ are considered of degree 1 and $x_4, x_5, x_6$ of degree -1), the computation of the eigenvalues/eigenvectors reduces to linear algebra. Indeed, since the operator $P(x,y,z, \frac{\partial}{\partial x},  \frac{\partial}{\partial y},  \frac{\partial}{\partial z})$ preserves total degree, it is enough to consider each graded piece $\mathbb{C}[x,y,z]_d$ separately, and the latter are finite-dimensional vector spaces on which the operator acts linearly. Your explicit example is of this kind.
If $P$ is of homogeneous of negative degree, then it can only have zero as an eigenvalue, since the associated operator is nilpotent. For $P$ of positive degree it also seems clear that the only possible is zero.<|endoftext|>
TITLE: Nonalgebraic complex manifolds
QUESTION [11 upvotes]: I'm turning here (a variation of) a question asked by a friend of mine. For the purposes of this question I will say that a compact complex manifold is projective if it is isomorphic to a subvariety of $\mathbb{P}^n$ and algebraic if it is isomorphic to the complex analytic space associated to a scheme.
One often finds many examples of compact complex manifolds which are not projective. For instance Hopf and Inoue surfaces, some K3, some tori... Usually the proof shows that either this manifold is not Kahler, or the Kahler cone does not intersect the lattice of integral cohomology. But of course this does not tell us anything about their being algebraic in the sense outlined above.
So there are two questions. First, what is an example of a compact complex manifold which is not algebraic? Probably this is standard, but I don't have any reference in mind. I think that the complex analytic space associated to a smooth algebraic space which is not a scheme will do, but I'm not expert of algebraic spaces, so I don't have even such an example in mind.
The second, subtler, question is: are the examples above of nonprojective complex manifold also non algebraic? How one can prove such a statement?

REPLY [2 votes]: These issues are discussed to some extent at this previous MO question.
For instance, if a compact complex manifold is algebraic then its field of meromorphic functions has transcendence degree over $\mathbb C$ equal to its dimension. So any compact complex manifold without as many meromorphic functions as one usually expects has no chance of being algebraic.<|endoftext|>
TITLE: An integral that somehow equals pi^2/6 and involves dilogarithms?
QUESTION [8 upvotes]: I am attempting to show that
$$ \sum_{k \ge 1}^\infty {k^2 x^k \over (1+x^k)^2} \sim (1-x)^{-3} {\pi^2 \over 6} $$
as $x$ approaches 1 from below.  The sum can be approximated by the integral
$$ \int_0^\infty {k^2 x^k \over (1+x^k)^2} \: dk $$
and making the change of variable $u = x^k$ this is equal to
$$ {1 \over \log^3 x} \int_1^0 {\log^2 u \over (1+u)^2} \: du. $$
Maple tells me that this integral is equal to $\pi^2/6$.  Moreover if I try to find the same integral but with arbitrary rational numbers as bounds, it is successful: for example
$$ \int_{1/2}^2 {\log^2 u \over (1+u)^2} \: du = -4 \log 2 \log 3 + {7 \over 3} \log^2 2 - 2 dilog(3) + 2 dilog(3/2). $$
Specifically, I'd like to know:

How can I find the integral in the third displayed equation (or, really what I'm interested in, the sum in the first displayed equation)? 
How can I find the indefinite integral $\int \log^2 u / (1+u)^2 \: du$?
Various integrals of this type (roughly speaking, logarithms divided by polynomials) are recurring in my work.  Presumably they can be computed with the ordinary techniques of calculus and the correct identities involving polylogarithms.  What are these identities, or where can I find them?

REPLY [3 votes]: This problem can also be approached by rewriting the sum. I'll show a lot of details, probably too many, here.
Use the binomial series to write
$ \frac{1}{(1+x^k)^2} = \sum_{m=0}^{\infty} \binom{m+1}{1} (-1)^m x^{km} $
and insert this into the original sum $S$ to obtain
$ S = \sum_{k=1} k^2 x^k \sum_{m=0} (-1)^m (m+1) x^{km}. $
Interchange the sums to obtain
$ S = \sum_{m=0} (m+1) (-1)^m \sum_{k=1} k^2 x^{k(m+1)} $
$ = \sum_{m=0} (m+1) (-1)^m \frac{x^{m+1} (1+x^{m+1})}{(1-x^{m+1})^3} $
$ = -\sum_{m=1} m (-1)^m \frac{x^{m} (1+x^{m})}{(1-x^{m})^3}. $
Now use the factorization $1-x^m = (1-x)(1+x+x^2 +x^3 + \cdots + x^{m-1}) $ to write
$ S = \frac{-1}{(1-x)^3} \sum_{m=1} (-1)^m m \frac{x^m (1+x^m)}{(1+x+x^2 + \cdots + x^{m-1})^3}. $
We have now factored out the most singular term as $x \rightarrow 1.$ We can take the most rough approximation of the sum by finding its limit as $x \rightarrow 1$, which gives
$ S \approx \frac{-1}{(1-x)^3} \sum_{m=1} (-1)^m m \frac{2}{m^3}.$
Using the known sum $ \sum_{m=1} \frac{ (-1)^m }{m^2} = -\frac{\pi^2}{12} $ now gives 
$S \approx \frac{\pi^2}{6} \frac{1}{(1-x)^3} $ as desired....
NOTE: the last sum over $m$, if you don't just take $x=1$, but keep the powers $x^m$ in the numerator, gives dilogarithms of $x$ and $x^2$. However, I've probably thrown away too much in the denominator for that to be useful!!<|endoftext|>
TITLE: Interpretation of elements of H^1 in sheaf cohomology.
QUESTION [19 upvotes]: Given a variety V and a locally free (coherent) sheaf $\mathcal{F}$ of rank 1 (equivalently a line bundle $L$), I can do a Cech cohomology on it. Then $H^0(V; \mathcal{F})$ are just global sections. Is there a similarly understandable meaning to elements of $H^1(V; \mathcal{F})$?
Thanks!

REPLY [15 votes]: $H^1(V;\mathcal{F})$ is the space of bundles of affine spaces modeled on $\mathcal{F}$.  An affine bundle $F$ modeled on $\mathcal{F}$ is a sheaf of sets that $\mathcal{F}$ acts freely on as a sheaf of abelian groups (i.e., there is a map of sheaves $F\times \mathcal{F}\to F$ which satisfies the usual associativity), and on a small enough neighborhood of any point, this action is regular (i.e., the action map on some point gives a bijection).  You should think of this as a sheaf where you can take differences of sections and get a section of $\mathcal{F}$.  
This matches up with what Anweshi said as follows: given such a thing, you can try to construct an isomorphism to $\mathcal{F}$.  This means picking an open cover, and picking a section over each open subset and declaring that to be 0.  The Cech 1-cochain you get is the difference between these two sections on any overlap, and if an isomorphism exists, the difference between the actual zero section and the candidate ones you picked is the Cech 0-chain whose boundary your 1-cochain is.
Another way of saying this is that a Cech 1-cycle is exactly the same sort of data as transition functions valued in your sheaf, so if you have anything that your sheaf acts on (again, as an abelian group), then you can use these transition functions to build a new sheaf; a homology between to 1-cycles (i.e. a 0-cycle whose boundary is their difference) is exactly the same thing as an isomorphism between two of these.  
I'll note that there's nothing special about line bundles; this works for any sheaf of groups (even nonabelian ones).  For example, if you take the sheaf of locally constant functions in a group, you will classify local systems for that group.  If you take continuous functions into a group, you will get principal bundles for that group.  If you take the sheaf $\mathrm{Aut}(\mathcal{O}_V^{\oplus n})$, you'll get rank $n$ locally free sheaves.  A particularly famous instance of this is that line bundles are classified by $H^1(V;\mathcal{O}_V^*)$.<|endoftext|>
TITLE: Kähler manifold which is not algebraic
QUESTION [6 upvotes]: Can someone provide examples of Kähler manifolds which are not algebraic?
This question came to my mind seeing the post of Andrea Ferretti.

REPLY [3 votes]: You might want to take a look at this previous MO question.
There, I mentioned Voisin's results disproving Kodaira's conjecture (every Kahler manifold 
is deformation equivalent to a projective manifold).<|endoftext|>
TITLE: Please check my 6-line proof of Fermat's Last Theorem.
QUESTION [72 upvotes]: Kidding, kidding.  But I do have a question about an $n$-line outline of a proof of the first case of FLT, with $n$ relatively small.
Here's a result of Eichler (remark after Theorem 6.23 in Washington's Cyclotomic Fields):  If $p$ is prime and the $p$-rank of the class group of $\mathbb{Q}(\zeta_p)$ satisfies $d_p<\sqrt{p}-2$, then the first case of FLT has no non-trivial solutions.  Once you know Herbrand-Ribet and related stuff, the proof of this result is even rather elementary.
The condition that $d_p<\sqrt{p}-2$ seems reminiscent of rank bounds used with Golod-Shafarevich to prove class field towers infinite.  More specifically, a possibly slightly off (and definitely improvable) napkin calculation gives me that for $d_p>2+2\sqrt{(p-1)/2}$, the $p$-th cyclotomic field $\mathbb{Q}(\zeta_p)$ has an infinite $p$-class field tower.  It's probably worth emphasizing at this point that by the recent calculation of Buhler and Harvey, the largest index of irregularity for primes less than 163 million is a paltry 7.  
So it seems natural to me to conjecture, or at least wonder about, a relationship between the unsolvability of the first case of FLT and the finiteness of the $p$-class field tower over $\mathbb{Q}(\zeta_p)$.  Particularly compelling for me is the observation that regular primes (i.e., primes for which $d_p=0$) are precisely the primes for which this tower has length 0, and have obvious historical significance in the solution of this problem.  In fact, the mechanics of the proof would probably/hopefully be to lift the arithmetic to the top of the (assumed finite) p-Hilbert class field tower, and then use that its class number is prime to $p$ to make arguments completely analogously to the regular prime case.  
I haven't seen this approach anywhere.  Does anyone know if it's been tried and what the major obstacles are, or demonstrated why it's likely to fail?  Or maybe it works, and I just don't know about it?
Edit:  Franz's answer indicates that even for a relatively simple Diophantine equation (and relatively simple class field tower), moving to the top of the tower introduces as many problems as it rectifies.  This seems pretty compelling.  But if anyone has any more information, I'd still like to know if anyone knows or can come up with an example of a Diophantine equation which does benefit from this approach.

REPLY [12 votes]: W. McCallum wrote a couple of papers in the early 90's connecting the index of irregularity and the rank of the Mordell-Weil of the Jacobian of the Fermat curve over the rationals. Then he used the (Skolem-)Chabauty-Coleman method to show that if $d_p<(p+5)/8$ then the Fermat curve had at most $2p-3$ rational points. No explicit connection with class field towers though.<|endoftext|>
TITLE: Kuranishi structures vs polyfolds
QUESTION [10 upvotes]: Moduli spaces of pseudoholomorphic curves do not carry the structure of a (compact) differentiable manifold in general (due to transversality issues). Nevertheless one would like to at least associate a fundamental class to the moduli space in question.
It looks like two approaches dominate: Kuranishi structures and polyfolds.
Both seem to be mammoth projects. Before diving seriously into one of them I may ask:
What are the advantages/drawbacks of these approaches? What is their motivation? Are they rigourosly proved? Are there reasonable alternatives?

REPLY [4 votes]: Perhaps you would like to take a look at a recent preprint by McDuff and Wehrheim 
http://arxiv.org/abs/1208.1340
This rather long paper seeks to explain the topological and analytic issues involved and provide the beginnings of a framework for resolving them.
These authors are well known for their attention to detail and expository ability, so it might be worthwhile to take a look.
I've heard rumors that an eventual Polyfold approach and the one taken by McDuff and Wehrheim share quite a bit of overlap in the end, but am not enlightened enough to say more on this matter.<|endoftext|>
TITLE: Linear algebra lemma
QUESTION [5 upvotes]: The following Lemma is in Beauville-Donagi, and I always took it for granted. Now I've tried to find a proof, but got stuck. They say it is a really simple lemma, so I may just be overlooking something easy.
Let $V$ be a vector space of dimension $6$, and let $W \subset \bigwedge^2 V^*$ be a subspace of dimension $2$. Assume that every form in $W$ is degenerate. Then there is a subspace $K \subset V$ of dimension $4$ such that each form in $W$ restricts to $0$ on $K$.

REPLY [3 votes]: First note that if a 2-form is degenerate, it is 0 on some 4-subspace (take a lagrangian subspace of the quotient by the kernel).
Now, assume not. Pick two elements that span $W$.  If either of them has 4-d kernel, it is 0 on any 4-d subspace, and we can use whichever on the other vanishes on.
Thus, every element of $W$ has 2-d kernel.  If two elements had different kernels, then one of their linear combinations would have no kernel.  Thus, they all kill the same 2-d subspace.  Thus, we've reduced to the statement that any two 2-forms on a 4-d space $Z$ have a common Lagrangian subspace.  Pick any line $L$; this is isotropic for both, since all lines are.  Consider the intersections of the symplectic orthogonals of $L$ for the two 2-forms. These are 3-d, so their intersection is a 2-space.  Now you win.<|endoftext|>
TITLE: Is a Poisson Group a group object in the category of Poisson Manifolds?
QUESTION [7 upvotes]: I realized that I am very confused about a certain sign in the definition of a Poisson group.  I will give some definitions, and then point out my confusion.
Definitions
Group objects
Let $\mathcal C$ be a category with Cartesian products.  Recall that a group object in $\mathcal C$ is an object $G \in \mathcal C$ along with chosen maps $e: 1\to G$ and $m: G\times G \to G$ (choose an initial object $1$ and a particular instance of the categorical product, and they imply all the others), such that (i) the two maps $G^{\times 3} \to G$ agree, (ii) the three natural maps $G\to G$ agree, and (iii) the map $p_1 \times m: G^{\times 2} \to G^{\times 2}$ is an isomorphism, where $p_1$ is the "project on the first factor" map $G^{\times 2}\to G$.
You may be used to seeing axiom (iii) presented slightly differently.  Namely, if $p_1 \times m: G^{\times 2} \to G^{\times 2}$ is an isomorphism, then consider the map $i = p_2 \circ (p_1\times m)^{-1} \circ (e\times \text{id}) : G = 1\times G \to G$.  It satisfies the usual axioms of the inverse map.  Conversely, if $i: G\times G$ satisfies the usual axioms, then $p_1 \times (m \circ (i\times \text{id}))$ is an inverse to $p_1 \times m$.  I learned this alternate presentation from Chris Schommer-Pries.
Poisson manifolds
A Poisson manifold is a smooth manifold $M$ along with a smooth bivector field, i.e. a section $\pi \in \Gamma(\wedge^2 TM)$, satisfying an axiom.  Recall that if $v\in \Gamma(TM)$, then $v$ defines a (linear) map $C^\infty(M) \to C^\infty(M)$ by differentiating in the direction of $v$.  Well, if $\pi \in \Gamma(\wedge^2 TM)$, then it similarly defines a map $C^\infty(M)^{\wedge 2} \to C^\infty(M)$.  The axiom states that this map is a Lie brackets, i.e. it satisfies the Jacobi identity.
A morphism of Poisson manifolds is a smooth map of manifolds such that the induced map on $C^\infty$ is a Lie algebra homomorphism.
The category of Poisson manifolds has products (Wikipedia).
Poisson groups
A Poisson Group is a manifold $G$ with a Lie group structure $m : G\times G \to G$ and a Poisson structure $\pi \in \Gamma(\wedge^2 TG)$, such that $m$ is a morphism of Poisson manifolds.  Recall that a Lie group $G$ is a group object in the category of smooth manifolds.
Recall also that a Lie group is almost entirely controlled by its Lie algebra $\mathfrak g = T_eG$.  Then it is no surprise that the Poisson structure can be described infinitesimally.  Indeed, by left-translating, identify $TG = \mathfrak g \times G$.  Consider the adjoint action of $G$ on the abelian Lie group $\mathfrak g$.  Then we can define a Lie group structure $\mathfrak g^{\wedge 2} \rtimes G$ on $\wedge^2 TG$.  Recall that a section $\pi \in \Gamma(\wedge^2 TG)$ is just a manifold map $G \to \wedge^2 TG$ that splits the projection $\wedge^2 TG \to G$.  Then a Poisson manifold $(G,\pi)$ is a Poisson group if and only if $\pi : G \to \wedge^2 TG$ is a map of Lie groups.
Thus, a Poisson group structure is precisely the same as a Lie algebra $d\pi : \mathfrak g \to \wedge^2 \mathfrak g \rtimes \mathfrak g$ splitting the obvious projection (here $\wedge^2 \mathfrak g$ is an abelian Lie algebra, and $\mathfrak g$ acts on it via the adjoint action), and such that $(d\pi)^* : \wedge^2\mathfrak g^* \to \mathfrak g^*$ satisfies the Jacobi identity.  (Any failure of $G$ to be simply-connected, which might prevent such a map from lifting, also fails in $\wedge^2 TG$, so this really is a one-to-one identification of Poisson group structures on $G$ and "Lie bialgebra" structures on $\mathfrak g$.)
From this perspective, then, it is more or less clear that the inverse map $i:G\to G$ is not a morphism of Poisson manifolds (Wikipedia).  Indeed, infinitesimally, $di = -1: \mathfrak g \to \mathfrak g$, which takes $d\pi$ to $-d\pi$ (as $d\pi$ has one $\mathfrak g$ on the left and two on the right).  Instead, $i$ is an "anti-Poisson map".  The monoid $(\mathbb R,\times)$ acts on the category of Poisson manifolds by doing nothing to the underlying smooth manifolds and rescaling the Poisson structures; a smooth map is anti-Poisson if it becomes Poisson after twisting by the action of $-1$.
The unit map $e: 1 \to G$, on the other hand, is Poisson; it follows from the axioms of a Poisson group that $\pi(e) = 0$, and the terminal object in the category of Poisson manifolds is $1 = \{\text{pt}\}$ with the trivial Poisson structure.  ($C^\infty(\{\text{pt}\}) = \mathbb R$ can only support this Poisson structure.)
My question
Suppose that $G$ is a Poisson group.  Then $p_1 \times m: G^{\times 2} \to G^{\times 2}$ is a Poisson map, and an isomorphism of smooth manifolds.  Thus, I would expect that it is an isomorphism of Poisson manifolds.  On the other hand, in the first section above I constructed the inverse map $i : G\to G$ out of this isomorphism and the other structure maps, all of which are Poisson when $G$ is a Poisson group.  And yet $i$ is not a Poisson map.  So where am I going wrong?

REPLY [6 votes]: I think the problem is that the product of Poisson manifolds is not actually a categorical product. This is due to the fact (if I remember correctly) that two Poisson maps $f: X \to Y$ and $g: X \to Z$ give a Poisson map $f \times g: X \to Y \times Z$ only when the images of $f^*$ and $g^*$ Poisson commute in $C^\infty(X)$. In particular, $p_1 \times m$ doesn't seem to be Poisson.<|endoftext|>
TITLE: citation for first statement of the Re(s) = 6 conjecture on zeros of Ramanujan L function
QUESTION [11 upvotes]: Hi, for the bibliography of a paper I'm writing I seek a citation for the first statement of the conjecture that the nontrivial zeros of $F(s) = \sum_n\tau(n)n^{-s}$ all lie on the line Re(s) = 6. (Here tau is the Ramanujan tau function.)
Thanks.
Barry Brent
Edit:
I feel (and people have said in print) that Ramanujan must have made the conjecture himself, but I haven't been able to locate the statement. As far as I can see it isn't in the paper (#18 of the Collected Papers, "On certain arithmetical functions") in which he defines the series and conjectures the Euler product.

REPLY [2 votes]: Perhaps not an answer to your question, but certainly related.
In his Twelve Lectures, Hardy discusses the "Ramanujan hypothesis" to the effect that 
$$
|\tau(p)|\le2p^{11\over 2}
$$
for every prime $p$, and says that 

this is the most fundamental of the unsolved problems
  presented by the function.

He must have been talking about Ramanujan's 1916 paper in which he also conjectured the multiplicativity and the congruences for the $\tau$-function.
The multiplicativity was established by Mordell (1918) using what we would today call Hecke operators, the congruences were studied by Swinnerton-Dyer and Serre in the early 70s, ultimately leading Serre to his modularity conjecture as proved recently by Khare--Wintenberger (2009), and the estimate $|\tau(p)|\le2p^{11\over 2}$ followed from Deligne's proof (1973) of the Weil conjectures.
Not bad as far as the mathematics inspired by a single paper goes.
Addendum.  The estimate $|\tau(p)|\le2p^{11\over 2}$ appears as formula (104) on page 153 of Ramanujan's Collected Papers as being ``highly probable''.<|endoftext|>
TITLE: Fermat over Number Fields
QUESTION [6 upvotes]: If $n>2$, does the impossibility of solving $x^n +y^n=z^n$ with $x, y, z$ rational integers imply the same with $x, y, z$ algebraic integers?  
Rather, If insolvability in algebraic integers does follow, then does it follow from simple considerations, or is it still an interesting question?

REPLY [2 votes]: A part of the results of Kolyvagin was already proven by Peter Dénes,"Über den ersten Fall des letzten Fermatschen Satzes", Monatshefte für Mathematik, Bd. 54/3, 1950.
Dénes proved : let p be an odd prime, let K denote the p-th cyclotomic field, let P denote the prime ideal $(1 - \zeta)$ of K (where $\zeta$ denotes a primitive p-th root of unity); then in every solution (if any) of the equation $x^p + y^p + z^p = 0$ where x, y and z are P-integral elements of K not divisible by P, the rational integer t congruent to x/y (resp. y/z, resp. z/x) modulo P is a root of Kummer's system of congruences $B_{2i}l^{p-2i}(t + \zeta) \equiv 0 \pmod{p}$ for i = 1 to (p-3)/2, where $l^{j}$ denotes the j-th Kummer logarithmic function (with respect to $\zeta$).
It is Satz 1 of his paper, p. 162. You can read this paper here :
http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN362162050_0054&DMDID=DMDLOG_0017&LOGID=LOG_0017&PHYSID=PHYS_0166
The paper of Kolyvagin is here :
http://www.mathnet.ru/links/025935a4111ecc8ca5d60949a1020353/into103.pdf
The beginning of the English translation is here :
http://link.springer.com/article/10.1023/A%3A1017955309615#page-1
Kolyvagin's theorem 1 is the same as Bénes's Satz 1.
As you can see from the bibliography, Kolyvagin doesn't seem to know the paper of Dénes.<|endoftext|>
TITLE: Minimal size of an open affine cover
QUESTION [7 upvotes]: This may be a silly question - but are there interesting results about the invariant: the minimal size of an open affine cover? For example, can it be expressed in a nice way? Maybe under some additional hypotheses?

REPLY [3 votes]: I think there are some relevant comments in the work of Roth and Vakil, The affine stratification number and the moduli space of curves.<|endoftext|>
TITLE: Set-theoretic forcing over sites?
QUESTION [18 upvotes]: All texts I have read on set-theoretic independence proofs consider several different sorts of constructions separately, such as Boolean-valued models (equivalently, forcing over posets), permutation models, and symmetric models.  However, the topos-theoretic analogues of these notions—namely topoi of sheaves on locales, continuous actions of groups, and combinations of the two—are all special cases of one notion, namely the topos of sheaves on a site.  Is there anywhere to be found a direct construction, in the classical world of membership-based set theory, of a "forcing model" relative to an arbitrary site?

REPLY [6 votes]: At  http://www.mathematik.tu-darmstadt.de/~streicher/forcizf.pdf  one can find a paper describing a forcing semantics for IZF in Grothendieck toposes Sh(C,J)
in terms of the site (C,J) where C is a Grothendieck topology on a small category C.
As mentioned in the paper for the case of presheaf toposes this was done already by D. Scott before. In my forcing interpretation the "names" for sets
are coming from the cumulative hierarchy in the presheaf topos. Sheafification is built into the clauses of the forcing semantics. Instantiating C by a poset gives the usual "names" and when taking for J the
double negation topology on C we get Cohen forcing.
In this context it might be worthwhile to mention that boolean Grothendieck
toposes are precisely as subtoposes of presheaf toposes via the double negation topology. This is due to Freyd and is also described in Blass and Scedrov's AMS Memoir. In this booklet they relate Freyd's models refuting AC
to the more traditional symmetric boolean valued models as employed by Cohen for refuting (even countable) AC. Freyd's first model coincides with the one devised by Cohen but Freyd's second modelwasn't considered before in the set-theoretic literature.
It might be worth mentioning in this context the famous result of Peter Freyd that every Grothendieck topos appears as an exponential variety within
a topos boolean over the Schanuel topos (the generic permutation model used
by A. Pitts in his work on "Nominal Sets"). So Grothendieck toposes are not that different from models of set theory...
Thomas
PS My above mentioned article has appeared in the Festschrift for my good old friend Mamuka Jibladze.<|endoftext|>
TITLE: Is every Adams ring morphism a lambda-ring morphism?
QUESTION [8 upvotes]: A lambda-ring $R$ is called "special" if it satisfies the $\lambda^i\left(xy\right)=...$ and $\lambda^i\left(\lambda^j\left(x\right)\right)=...$ relations, or, equivalently, if the map $\lambda_T:R\to\Lambda\left(R\right)$ given by $\lambda_T\left(x\right)=\sum\limits_{i=0}^{\infty}\lambda^i\left(x\right)T^i$ (where the $\sum$ sign means addition in $R\left[\left[T\right]\right]$, not addition in $\Lambda\left(R\right)$) is a morphism of lambda-rings. If you are wondering what the hell I am talking about, most likely you belong to the school of algebraists that denote only special lambda-rings as lambda-rings at all.
Anyway, let $A$ and $B$ be two special lambda-rings, and for every $i>0$, let $\Psi_A^i$ and $\Psi_B^i$ be the $i$-th Adams operations on $A$ and $B$, respectively. Let $f:A\to B$ be a ring homomorphism such that $f\circ\Psi_A^i=\Psi_B^i\circ f$ for every $i>0$. Does this yield that $f$ is a lambda-ring homomorphism, i. e. that $f\circ\lambda_A^i=\lambda_B^i\circ f$ for every $i>0$ ?
Note that this is clear if both $A$ and $B$ are torsion-free as additive groups (i. e., none of the elements $1$, $2$, $3$, ... is a zero-divisor in any of the rings $A$ and $B$), but Hazewinkel, in his text Witt vectors, part 1 (Lemma 16.35), claims the same result for the general case. I am writing a list of errata for his text, and I would like to know whether this should be included - well, and I'd like to know the answer anyway, as I am writing some notes on lambda-rings as well.
For the sake of completeness, here is a definition of Adams operations: These are the maps $\Psi^i:R\to R$ for every integer $i>0$ (where $R$ is a special lambda-ring) defined by the equation
$\sum\limits_{i=1}^{\infty} \Psi^i\left(x\right)T^i = -T\frac{d}{dT}\log\left(\lambda_{-T}\left(x\right)\right)$ in the ring $R\left[\left[T\right]\right]$ for every $x\in R$.
Here, even if the term $\log\left(\lambda_{-T}\left(x\right)\right)$ may not make sense (since some of the fractions $\frac{1}{1}$, $\frac{1}{2}$, $\frac{1}{3}$, ... may not exist in $R$), the logarithmic derivative $\frac{d}{dT}\log\left(\lambda_{-T}\left(x\right)\right)$ is defined formally by
$\displaystyle \frac{d}{dT}\log\left(\lambda_{-T}\left(x\right)\right)=\frac{\frac{d}{dT}\lambda_{-T}\left(x\right)}{\lambda_{-T}\left(x\right)}$.

REPLY [12 votes]: Here is a more general point of view on Charles's example, which someone might find helpful.
Let $M$ be an abelian group, and let $\mathrm{Z}[M]$ denote $\mathrm{Z}\oplus M$ with the usual split ring structure $(a,m)(a',m')=(aa',am'+a'm)$. There is a simple description of the (special) $\lambda$-ring structures on such rings. For any prime number $p$, let $\theta_p$ denote the symmetric function $(\psi_p-e^p)/p$, where $e=x_1+x_2+\cdots$ and $\psi_p=x_1^p+x_2^p+\cdots$. Observe that $\theta_p$ has integral coefficients and therefore defines a natural operation on any $\lambda$-ring, and for any element $x$ in any $\lambda$-ring, we have $\psi_p(x)=x^p+p\theta_p(x)$. In particular, for any $\lambda$-structure on $\mathrm{Z}[M]$, the Adams operation $\psi_p$ satisfies $\psi_p(m)=p\theta_p(m)$ for all $m\in M$.
Exercise: Given a commuting family of additive maps $\theta_p:M\to M$, there is a unique $\lambda$-ring structure on $\mathrm{Z}[M]$ whose $\theta_p$ operators on $M$ are the given ones. Conversely, suppose $\mathrm{Z}[M]$ has a $\lambda$-ring structure. Then each $\theta_p$ preserves the ideal $M$, and resulting map $\theta_p:M\to M$ is additive. 
Now it is a general fact that a ring map between two $\lambda$-rings is a $\lambda$-ring map if and only if it commutes with the $\theta_p$ operators. (This is because such symmetric functions and all their compositions under plethysm generate the whole ring of symmetric functions.) Thus, in our case, a $\lambda$-ring map $\mathrm{Z}[M]\to\mathrm{Z}[N]$ is the same as a linear map $M\to N$ that commutes with each $\theta_p$. But that is not the same as commuting with each $p\theta_p$, i.e. the Adams operations, and it is easy to make counterexamples. For instance Charles's counterexample is with $M=N=\mathrm{Z}/2\mathrm{Z}$, where $\theta_p$ on $M$ is the identity for all $p$, but $\theta_p$ on $N$ is the identity for all odd $p$ but is zero for $p=2$. The identity map $M\to N$ therefore commutes with each $p\theta_p$ and so is an Adams map. It commutes with each $\theta_p$ with $p$ odd, but it doesn't commute with $\theta_2$. Therefore it is not a $\lambda$-map.<|endoftext|>
TITLE: Which R-algebras are the group ring of some group over a ring R?
QUESTION [7 upvotes]: This question was inspired by Qiaochu's recent question, Which commutative groups are the group of units of some field? - my question is close to being the inverse of it. 
As mentioned here, given a ring $R$, the functor $GrpRing:Grp\rightarrow R$-$Alg$ taking a group $G$ to the group ring $R[G]$ is left adjoint to the functor $GrpUnits:R$-$Alg\rightarrow Grp$ taking an $R$-algebra to its group of units. What is the essential image of $GrpRing$, i.e., which $R$-algebras are isomorphic to the group ring of some group over $R$?
One might ask more generally, when is a ring $R$ a group ring over some ring, not fixed at the outset? (Obviously, any ring $R$ is isomorphic to $R[1]$, the group ring of the trivial group over itself, but let's exclude this trivial case.)

REPLY [6 votes]: Reid Barton's very nice answer to Computing the structure of the group completion of an abelian monoid, how hard can it be? contained a pointer to the wikipedia page for the Eilenberg-Mazur swindle.  Towards the bottom of that page, one finds the following relevant paragraph.
Example: (Lam 2003, Exercise 8.16) If $A$ and $B$ are any groups then the Eilenberg swindle can be used to construct a ring $R$ such that the group rings $R[A]$ and $R[B]$ are isomorphic rings: take $R$ to be the group ring of $A + B + A + B + \ldots$<|endoftext|>
TITLE: Explicit Spin Structures on the Torus
QUESTION [10 upvotes]: Basically, I am trying to build explicit examples of Dirac operators.  To this end, I'm looking at the surface E = C/(Z + λZ) - for some λ in H \ SL(2,Z) - with the Euclidean metric and the flat connection.  The, since the torus has genus 1, there are 22=4 spin structures on the tangent bundle of this elliptic curve.  
What are the four ways do define representations of SU(2) Spin(2)=U(1) on TpE for each $p \in$ E?  There is probably one spin structure for each element of the homology ring with coefficients in Z2.
For reference: A spin structure on E is an open covering $\{ U_\alpha : \alpha \in A\}$ and transition functions $g_{\alpha\beta}: U_\alpha \cap U_\beta \to Spin(2) = U(1)$ satisfying a cocycle condition gαβgβγ=gαγ on $U_\alpha \cap U_\beta \cap U_\gamma$.  Perhaps there are other definitions better for explicit computations.

REPLY [3 votes]: On oriented surfaces $F$, spin structures are in one-to-one correspondence with quadratic forms on the group $H_1(F;\mathbb{Z}/2\mathbb{Z})$ refining the intersection form. This correspondence is explicit and constructive -- see Atiyah, Riemannian surfaces and spin structures, Ann. Sci. Ec. Norm. Sup. (4) 4 (1971) and Johnson, Spin structures and quadratic forms on surfaces, Proc. London Math. Soc. 22 (1980). If you pick a basis of $H_1$, specifying your quadratic form on the basis elements defines it uniquely, so there are $2^{2g}$ spin structures. "One for each element of $H_1$" gives you this same number, but it's not really the natural way of counting them.<|endoftext|>
TITLE: Question concerning the arithmetic average of the Euler phi function:
QUESTION [8 upvotes]: Let $\varphi(n)$ denote Euler's phi-function. If we let 
$$ \sum_{n\leq x} \varphi(n) = \frac{3}{\pi^2}x^2+R(x),$$
then it is not hard to show that $R(x)=O(x\log x)$. What is the best known bound for $R(x)$ assuming the Riemann Hypothesis?

REPLY [8 votes]: There is information on page 68 of Montgomery and Vaughan's book, and also on page 51 of "Introduction to analytic and probabilistic number theory" by Gérald Tenenbaum. Briefly, Montgomery has established that 
$$
\limsup_{x \rightarrow +\infty}\frac{R(x)}{x\sqrt{\log\log(x)}} > 0
$$
and similarly with the limit inferior. So there is only modest room for improvement. Unfortunately I cannot find any reference to an upper bound conditional on RH. On page 40 Tenenbaum has a reference to page 144 of Walfisz' book on exponential sums. Walfisz uses Vinogradov's method to show that 
$$
R(x) = O\left(x\log^{2/3}(x)(\log\log(x))^{4/3}\right).
$$
I don't own a copy of Walfisz' book, so I have no further details.<|endoftext|>
TITLE: Malfatti Circles - Limiting point
QUESTION [5 upvotes]: "Three circles packed inside a triangle such that each is tangent to the other two and to two sides of the triangle are known as Malfatti circles" (for a brief historical account on this topic, see here and here on MathWorld).
Consider the triangle formed by the centers of these circles, one can draw a new set of smaller Malfatti circles in this triangle.  What is the limiting point of this process?
One thing sort of discouraging is that I tried on an isosceles triangle, unfortunately did not find the limiting point matching any of the known relevant points (e.g., incenter or the first Ajima-Malfatti point).

REPLY [5 votes]: I don't know the answer to your question, but it should be easy enough to compute this limit point numerically for an arbitrary triangle and use the result to search the Encyclopedia of Triangle Centers.<|endoftext|>
TITLE: Is ΩΣ in {simplicial commutative monoids} group completion?
QUESTION [10 upvotes]: Let C be the model category of simplicial commutative monoids (with underlying weak equivalences and fibrations), or equivalently the (∞,1)-category PΣ(Top), where T is the Lawvere theory for commutative monoids.  In C, as in any pointed (∞,1)-category with finite limits and colimits, we can define adjoint functors ΣC and ΩC as ΣCX = hocolim [• ← X → •] and ΩCX = holim [• → X ← •].
The category CommMon of commutative monoids sits inside C as a full subcategory (as the constant objects, or the objectwise-discrete presheaves).  Consider the two functors CommMon → C given by sending M to ΩCΣCM and to the group completion of M, respectively.  Is there a natural equivalence between these functors?
(This question is closely related to Chris's question here.  A thorough answer to that question would probably yield this immediately.)

REPLY [10 votes]: I think an answer is given by the arguments that Segal gives in Section 4 of his paper on "Categories and Cohomology Theories" (aka, the $\Gamma$-space paper), in Topology, v.13.  I'll try to sketch the main idea, translated into the context of simplicial commutative monoids.  I'll show that if $M$ is a discrete simplicial commutative monoid, then it's group completion is homotopically discrete; according to the comments, this should answer the question.
Given a commutative monoid $M$, we can define a simplicial commuative monoid $M'$ as the nerve of the category whose objects are $(m_1,m_2)\in M\times M$, and where morphisms $(m_1,m_2)\to (m_1',m_2')$ are $m\in M$ such that $m_im=m_i'$.  We can prolong this to a functor on simplicial commutative monoids.
Let $H=H_*|M|=H_*(|M|,F)$ (the homology of the geometric realization of $M$, with coefficients in some field $F$), viewed as a commutative ring under the pontryagin product.  Then Segal shows that $H_*|M'|\approx H[\pi^{-1}]$, where $\pi$ denotes the image of $\pi_0|M|$ in $H_0|M]$.  His proof amounts to computing the homology spectral sequence for a simplicial space whose realization is $M'$, and whose $E_2$-term is $\mathrm{Tor}_i^H(H\otimes H,F)$, and observing that the higher tor-groups vanish.
This means that if $M$ is discrete, then $H_*|M'|$ is concentrated in degree $0$.  Since $|M'|$ is a grouplike commutative monoid, the Hurewicz theorem should tell us that $|M'|$ is weakly equivalent to a discrete space, namely the group completion of the monoid $M$.
Segal goes on to show that $BM\to BM'$ is a weak equivalence, using the above homology calculation and another spectral sequence.  Since $M'$ is weakly equivalent to a group, $\Omega BM\approx \Omega BM'\approx M'$.<|endoftext|>
TITLE: Geometric interpretation of trace
QUESTION [329 upvotes]: This afternoon I was speaking with some graduate students in the department and we came to the following quandary;

Is there a geometric interpretation of the trace of a matrix?

This question should make fair sense because trace is coordinate independent.
A few other comments. We were hoping for something like: 
"determinant is the volume of the parallelepiped spanned by column vectors."
This is nice because it captures the geometry simply, and it holds for any old set of vectors over $\mathbb{R}^n$.
The divergence application of trace is somewhat interesting, but again, not really what we are looking for.
Also, after looking at the wiki entry, I don't get it. This then requires a matrix function, and I still don't really see the relationship.
One last thing that we came up with; the trace of a matrix is the same as the sum of the eigenvalues. Since eigenvalues can be seen as the eccentricity of ellipse, trace may correspond geometrically to this. But we could not make sense of this.

REPLY [3 votes]: For me trace of a matrix always was analogous to the real part of a complex number. As such, I consider trace divided by the matrix' rank as the "scalar part" of the matrix.
There are other analogies:

For analytic functions trace is analog of the value of the function at zero or other central point of their domain (which is the constant term in the functions' Taylor expansion).

For divergent integrals and series, trace is analog of the regularized value.

For vectors in spacetime, it is analog of the time component of the vector.<|endoftext|>
TITLE: Symmetric tensor products of irreducible representations
QUESTION [8 upvotes]: I wonder if there is a way to compute the symmetric tensor power of irreducible representations for classical Lie algebras: $\mathfrak{so}(n)$, $\mathfrak{sp}(n)$, $\mathfrak{sl}(n)$. 
The question is motivated by reading
"Introduction to Quantum Groups and Crystal Bases" by Hong, J. and Kang, S.-J.
The book provides an algorithm for computing the tensor product of any two irreducible representations for classical Lie algebras. Could it be generalized to symmetric parts of tensor products? Any references are very much appreciated!

REPLY [4 votes]: Here's a very special case for $\mathfrak{gl}_n$ in characteristic 0 (which I have found useful in my work). Let $V$ be the vector representation, and for a partition $\lambda$ with at most $n$ parts, let ${\bf S}\_{\lambda}(V)$ denote the corresponding highest weight representation. Then $Sym^n(Sym^2 V) = \bigoplus\_{\lambda} {\bf S}\_{\lambda}(V)$ where the direct sum is over all partitions $\lambda$ of size $2n$ with at most $n$ parts such that each part of $\lambda$ is even. Similarly, $Sym^n(\bigwedge^2 V) = \bigoplus\_{\mu} {\bf S}\_{\mu}(V)$ where the direct sum is over all partitions $\mu$ of $2n$ with at most $n$ parts such that each part of the conjugate partition $\mu'$ is even. If you want the corresponding result for $\mathfrak{sl}\_n$ we just introduce the equivalence relation $(\lambda\_1, \dots, \lambda\_n) \equiv (\lambda\_1 + r, \dots, \lambda\_n + r)$ where $r$ is an arbitrary integer.
One reference for this is Proposition 2.3.8 of Weyman's book Cohomology of Vector Bundles and Syzygies (note that $L\_\lambda E$ in that book means a highest weight representation with highest weight $\lambda'$ and not $\lambda$).
Another reference is Example I.8.6 of Macdonald's Symmetric Functions and Hall Polynomials, second edition, which proves the corresponding character formulas.<|endoftext|>
TITLE: Transmutation versus operads
QUESTION [8 upvotes]: A while ago, I was reading Majid's book Foundations of quantum group theory, and Section 9.4 has a rather fascinating description of a Tannaka-Krein reconstruction result for quantum groups.  In particular, there seems to be the claim that if $H$ is a quasi-triangulated quasi-Hopf algebra, then the braided endomorphisms of the identity functor on (a suitably large category of) $H$-comodules form a Hopf algebra object $H'$ in $H$-comodules, in a way that identifies $H'$-comodules with $H$-comodules.  Furthermore, $H'$ is commutative and cocommutative with respect to the braided structure on the category of $H$-comodules, and under some nondegeneracy assumptions, it is self-dual.  This is called "transmutation" because $H'$ appears to have nicer properties than $H$ (although it may live in a strange category).  Some examples are given, e.g., $U_q(g)$ and quantum doubles of finite groups.  Unfortunately, the arguments in the proof are given in a diagrammatic language that I was unable to fathom.

Why does this result seem problematic?

The first problem comes from reasoning by analogy.  If I want to describe a Hopf algebra object in a monoidal category, I need some kind of commutor transformation $V \otimes W \to W \otimes V$ to even express the compatibility between multiplication and comultiplication, e.g., that comultiplication is an algebra map.  In operad language, I need (something resembling) an E[2]-structure on the category to describe compatible E[1]-algebra and E[1]-coalgebra structures.  If you think of the spaces in the E[k] operad as configurations of points in $\mathbb{R}^k$, this is roughly saying that you need two dimensions to describe compatible one-dimensional operations.  In the above case, the category of $H$-comodules has an E[2]-structure, but I'm supposed to get compatible E[2]-algebra and E[2]-coalgebra structures.  Naively, I would expect an E[4]-category to be necessary to make sense of this, but I was unable to wrestle with this successfully.
The second problem comes from a construction I've heard people call Koszul duality, or maybe just Bar and coBar.  If we are working in an E[n]-category for n sufficiently large (like infinity, for the symmetric case), then there is a "Bar" operation that takes Hopf algebras with compatible E[m+1]-algebra and E[k]-coalgebra structures, and produces Hopf algebras with compatible E[m]-algebra and E[k+1]-coalgebra structures.  There is a "coBar" operation that does the reverse, and under some conditions that I don't understand, composing coBar with Bar (or vice versa) is weakly equivalent to the identity functor.  In the above case, I could try to apply Bar to $H'$, but the result cannot have an E[3]-coalgebra structure, since E[3] doesn't act on the category.  Applying Bar then coBar implies the coalgebra structure on $H'$ is a priori only E[1], and applying coBar then Bar implies the algebra structure on $H'$ is a priori only E[1].  It is conceivable (in my brain) that the E[2]-structures could somehow appear spontaneously, but that seems a little bizarre.

Question

Am I talking nonsense, or is there a real problem here? (or both?)

REPLY [6 votes]: Scott,  I believe the source of your confusion is that Majid doesn't claim that the braided Hopf algebra he constructs is both braided commutative and braided co-commutative in C.  Just as in the usual case, the Hopf algebra one constructs is braided co-commutative (like U(g)) or braided commutative (like O(G)) if you work dually, but not both.  I think your arithmetic about E[n]-operads is correct as to why that would be unusual.
Can you point where in Chapter 9.4 is it claimed that the resulting Hopf algebra is both commutative and co-commutative?  For instance in my copy on page 481, he explains U(C) is braided co-commutative but doesn't mention commutative anywhere.  On page 477 of my copy, he says "One can use the term 'braided group' more strictly to apply to braided-Hopf algebras which are 'braided-commutative' or 'braided-cocommutative' in some sense." (keyword "or").
By the way, in the Section 3 of http://arxiv.org/abs/0908.3013 is an exposition (not original) reconstructing the algebra A (transmutation of O_q(G)), which uses language probably more familiar to you.  Also there is a Remark 3.3 (explained to us by P. Etingof) which gives a more concise description of A (and can be used to derive its key properties like braided-commutativity) by considering module categories and internal homs.
(apologies in advance if i'm incorrect; I read that book awhile ago and opened it up briefly to address your question)<|endoftext|>
TITLE: Quantum channels as categories: question 1.
QUESTION [6 upvotes]: A quantum channel is a mapping between Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$.  In general, we are interested in CPTP maps.  The operator spaces can be interpreted as $C^{*}$-algebras and thus we can also view the channel as a mapping between $C^{*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$.  Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$ where $C(X)$ is the space of continuous functions on some set $X$ and is also a $C^{*}$-algebra.  In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{*}$-algebras.  Note, however, that these are not necessarily the same $C^{*}$-algebras.  Since the channels are represented by square matrices, the input and output $C^{*}$-algebras must have the same dimension, $d$.  Thus we can consider them both subsets of some $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.
Proof: Consider the quantum channels
$\begin{eqnarray*}
r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) &
\qquad \textrm{where} \qquad &
\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\
t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) &
\qquad \textrm{where} \qquad &
\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} 
\end{eqnarray*}$
where the usual properties of such channels are assumed (e.g. trace preserving, etc.).  We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ where
$\begin{align}
\tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\
& = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\
& = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag
\end{align}$
and the $A_{i}$, $B_{i}$, and $C_{i}$ are Kraus operators.
Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. $$\sum_{k} C_{k}^{\dagger}C_{k}=\mathbf{1}.$$  For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041).
We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state $\rho$ in the absence of any channel.  Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$.
Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$.  We have
$\begin{align}
v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\
& = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\
& = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag
\end{align}$
and thus we have associativity.  Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible).  $\square$
Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here?  Is there a clearer or more concise proof?  I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.

REPLY [8 votes]: As I see it, this posted question and some aspects of the answers turn an important but straightforward fact into something needlessly complicated and less general.
Let $\mathcal{A}$ (Alice) and $\mathcal{B}$ (Bob) be $C^*$-algebras of observables, or better yet, von Neumann algebras of observables.  Let $\mathcal{A}^\#$ denote the dual space of (finite but not necessarily positive) states on $\mathcal{A}$, and in the von Neumann algebra case let ${}^\#\mathcal{A}$ denote the predual space of normal states.  Then a quantum channel, to model a message from Alice to Bob, is a completely positive, unital map 
$$E:\mathcal{B} \longrightarrow \mathcal{A}.$$
The corresponding CPTP map on states is the transpose:
$$E^\#:\mathcal{A}^\# \longrightarrow \mathcal{B}^\#$$
in the von Neumann algebra case, $E$ should be normal and have a pre-transpose:
$${}^\#E:{}^\#\mathcal{A} \longrightarrow {}^\#\mathcal{B}$$
Yes, quantum channels should form a category, and yes they do.  Yes, you can restrict to the one-object subcategory where the object is $B(\mathcal{H})$.  You need to check that quantum channels include the identity (they do) and you need to check that they are closed under composition.  It is immediate that preserving 1 (the unital condition) is closed under composition.  As for complete positivity, the condition is that
$$E \otimes I_\mathcal{C}:\mathcal{B} \otimes \mathcal{C} \longrightarrow \mathcal{A} \otimes \mathcal{C}$$
preserves positive states for all $\mathcal{C}$.  Closure of this condition under composition isn't quite immediate, but it's still very easy.
Associativity is immediate because quantum channels are functions.<|endoftext|>
TITLE: Translation of "le nilradicalisé de g"
QUESTION [5 upvotes]: I apologize for asking something that might well be found in a mathematical dictionary, but the similarity of the French word to an English one is frustrating my attempts to Google the answer (and the library is shut at time of typing). I suspect the answer should be obvious to those who, unlike me, know some basic Lie group/Lie algebra terminology.
Some context: I am reading an old paper of Dixmier from 1969, which has the following construction/definition. Let $\mathfrak g$ be a Lie algebra (characteristic zero, finite-dimensional), let $\mathfrak n$ be its largest nilpotent ideal -- the nilradical -- and put ${\mathfrak h}=[{\mathfrak g},{\mathfrak g}]+{\mathfrak n}$. Dixmier calls ${\mathfrak h}$ "le nilradicalisé de ${\mathfrak g}$".
Literal translation would surely be "the nilradicalised", but that sounds more like a mopey university indie band than a mathematical object. So what is the usual name for this object in English?

REPLY [4 votes]: I suspect pretty strongly that this is idiosyncratic terminology; I've never seen that subalgebra used, and the term has no google hits other than this post.<|endoftext|>
TITLE: Is every monomorphism of commutative Hopf algebras (over a field) injective?
QUESTION [5 upvotes]: Is it true that any monomorphism of commutative Hopf algebras over a field is injective? Moreover, is it true that any epimorphism of commutative Hopf algebras over a field is surjective?

REPLY [2 votes]: I haven't read through yet but the following paper by Chirvasitu seems to answer your second question and discusses closely related problems. See discussion in page 7 after Proposition 2.5.
http://arxiv.org/abs/0907.2881
Namely an epimorphism of Hopf algebras over a field
$$f : H \longrightarrow K$$
is necessarily surjective if $K$ is commutative.<|endoftext|>
TITLE: Continuous automorphism groups of normed vector spaces?
QUESTION [11 upvotes]: Consider the metric space on, say, ℝ2 induced by the various $L^p$ norms, and the group of isometries from that space into itself that preserve the origin. When $p=2$ I get the continuous group of rotations, but when $p\in\{1,3,4,5,...\infty\}$ it looks like I just get $D_8$, the symmetry group of the square. Question: what's going on here? Why is 2 so special? Are there other natural norms on ℝ2 (or on ℝn) besides the euclidean one that give interesting isometry groups?

REPLY [3 votes]: Consider the following norm on $\mathbb{R}^{2}$:                                    $||(x,y)||$ := $|x|+|y|$
if $xy\leq0$; $||(x,y)||$ := $|y|$
if $xy\geq0$ and $|y|$ $\geq3|x|$; 
$||(x,y)||$ := $|x|+\frac{2}{3}|y|$
if $xy$ $>0$ and $|y|$ $\leq3|x|$. Then the group of isometries
is { $\pm I\ $}.<|endoftext|>
TITLE: a weird sequence with a non-integral term
QUESTION [15 upvotes]: Define a sequence $(a_n)_{n \geq 1}$ by $$na_n = 2 + \sum_{i = 1}^{n - 1} a_i^2.$$ 
(In particular, $a_1 = 2$.)
How can you show - preferably without using a pc! - that not all terms of the sequence are integral?
And which will be the first such term?
Motivation: nothing interesting to say, it's a random problem which I got from someone - I have no reference - and which interested me. Usually one has to prove that all terms are integral :)
Thoughts: nothing interesting. The terms are quickly getting enormous...

REPLY [13 votes]: Sequences like this are sometimes called Somos sequences (and sometimes Gobel sequences) and you can find information about them at Problem E15 in Guy, Unsolved Problems in Number Theory and in the references Guy gives; also I'm sure typing Somos or Gobel into your favorite search engine will turn up something.<|endoftext|>
TITLE: Models of ZFC Set Theory - Getting Started
QUESTION [10 upvotes]: For just any first-order theory: What are the sets I am  supposed/allowed to think of when thinking of models as sets (of something + additional structure)?
Provided:

I can think of models of any theory (other than set theory) as of sets from
the (ZFC-based) von Neumann universe.
I can think of models of any theory as of sets of terms and formulas.

But what are the sets I am supposed/allowed to think of when thinking of models of (ZFC) set theory itself?

REPLY [6 votes]: This answer is going to be a bit too informal, but I hope it helps.
Imagine we have the collection of all sets. Let us call them the real sets, and their membership relation the real set membership. The empty set is "actually" empty, and the class of all ordinals is "actually" a proper class.
Now that we have the real sets we can use them as the "ontological substratum" upon which everything else will be built from. And this, of course, includes formal theories and their models. 
A model of any first-order theory is then only a real set. This applies to your favorite set theory too. So the models of your set theory are only real sets (but the models don't know it, just as they don't know if their empty sets are actually empty or if their set membership is the real one).
This view fits well, for example, with the idea of moving from a transitive model to a generic extension of it or to one with a constructible universe: we are simply moving from a class of models to another one, each one consisting of real sets.
But this view also leaves us with too many entities, and maybe here we have an opportunity to apply Occam's razor. It looks like we have two kind of theories: one for the real sets, which is made of things that are not sets (we can formalize our informal talk about them, but that does not make essentially any difference), another one for the models of set theory, which is made of sets.
The real sets and the theory of the real sets belong to a world where there are real sets, but there are also pigs and cows, and human languages and many other things. We don't need all that to do mathematics, do we? So why not diving into the wold of the real sets and ignore everything else?
If this story sounds too platonistic, I am sure it must have a formalistic counterpart.
With my question:
How to think like a set (or a model) theorist.
I expected to obtain an official view about all this stuff. I somehow succeeded on this, but as you can see, I'm still working on it.
Here is a related answer to a related question which I also find useful:
Is it necessary that model of theory is a set?<|endoftext|>
TITLE: Is there a good (co)homology theory for manifolds with corners?
QUESTION [21 upvotes]: Recall that a (smooth) manifold with corners is a Hausdroff space that can be covered by open sets homeomorphic to $\mathbb R^{n-m} \times \mathbb R_{\geq 0}^m$ for some (fixed) $n$ (but $m$ can vary), and such that all transition maps extend to smooth maps on open neighborhoods of $\mathbb R^n$.
I feel like I know what a "differential form" on a manifold with corners should be.  Namely, near a corner $\mathbb R^{n-m} \times \mathbb R_{\geq 0}^m$, a differential form should extend to some open neighborhood $\mathbb R^{n-m} \times \mathbb R_{> -\epsilon}^m$.  So we can set up the usual words like "closed" and "exact", but then Stokes' theorem is a little weird: for example, the integral of an exact $n$-form over the whole manifold need not vanish.
In any case, I read in D. Thurston, "Integral Expressions for the Vassiliev Knot Invariants", 1999, that "there is not yet any sensible homology theory with general manifolds with corners".  So, what are all the ways naive attempts go wrong, and have they been fixed in the last decade?
As always, please retag as you see fit.
Edit: It's been pointed out in the comments that (1) I'm not really asking about general (co)homology, as much as about the theory of De Rham differential forms on manifolds with corners, and (2) there is already a question about that.  Really I was just reading the D. Thurston paper, and was surprised by his comment, and thought I'd ask about it.  But, anyway, since there is another question, I'm closing this one as exact duplicate.  I'll re-open if you feel like you have a good answer, though.  -Theo  Edit 2: Or rather, apparently OP can't just unilaterally close their own question?

REPLY [10 votes]: A manifold with corner is a diffeological space modeled on orthants, as such it has a very well defined De Rham cohomology.
Edit : With Serap Gürer, we just wrote a paper (will appear in Indag. Math.) about differential forms on manifolds with corners. Here:
http://math.huji.ac.il/~piz/documents/DFOMWBAC.pdf<|endoftext|>
TITLE: A book on locally ringed spaces?
QUESTION [27 upvotes]: Are there enough interesting results that hold for general locally ringed spaces for a book to have been written?  If there are, do you know of a book?  If you do, pelase post it, one per answer and a short description.
I think that the tags are relevant, but feel free to change them.
Also, have there been any attempts to classify locally ringed spaces?  Certainly, two large classes of locally ringed spaces are schemes and manifolds, but this still doesn't cover all locally ringed spaces.

REPLY [7 votes]: An Introduction to Families, Deformations and Moduli, by T.E. Venkata Balaji, has a beautiful little appendix that introduces smooth manifolds, complex manifolds, schemes, and complex analytic spaces in a unified way as locally ringed spaces. Although it doesn't say much in general about arbitrary locally ringed spaces, I enjoyed reading it and seeing how the stuff I knew about particular classes of locally ringed spaces fit into the general framework. One thing that particularly struck me, although it's obvious in hindsight, was the remark (A.5.5) that for all the categories of spaces mentioned above, a morphism as defined classically is the same thing as a morphism of locally ringed spaces.<|endoftext|>
TITLE: Is a free alternative to MathSciNet possible?
QUESTION [142 upvotes]: How could a free (i.e. free content) alternative for MathSciNet and Zentralblatt be created?

Comments

Some mathematicians have stopped writing reviews for MathSciNet because they feel their output should be freely available. (The Pricing for MathSciNet is not high, but it is not the point.)
Two related questions: Are there any good websites... and Errata database?;
see also r-forum, nLab and wikademic.

What can be done (based on answers below)

One thing that can be really useful and doable is to create (and maintain) a database of articles (and maybe abstracts), where you can find all the articles that were referring to a given one. 
Once it is done we can add lists of errors --- it will add something new and valuable for the project (but this will take a while).
The above two things might be already enough for practical purpose. It will be even better if it will attract enough reviewers to the project.

REPLY [21 votes]: According to this, zbmath will go open access by 2021. I haven't seen the fine print, though.<|endoftext|>
TITLE: Why does the Riemann zeta function have non-trivial zeros?
QUESTION [101 upvotes]: This is a very basic question of course, and exposes my serious ignorance of analytic number theory, but what I am looking for is a good intuitive explanation rather than a formal proof (though a sufficiently short formal proof could count as an intuitive explanation). So, for instance, a proof that estimated a contour integral and thereby showed that the number of zeros inside the contour was greater than zero would not count as a reason. A brief glance at Wikipedia suggests that the Hadamard product formula could give a proof: if there are no non-trivial zeros then you get a suspiciously nice formula for ζ(s) itself. But that would feel to me like formal magic. A better bet would probably be Riemann's explicit formula, but that seems to require one to know something about the distribution of primes. Perhaps a combination of the explicit formula and the functional equation would do the trick, but that again leaves me feeling as though something magic has happened. Perhaps magic is needed.
A very closely related question is this. Does the existence of non-trivial zeros on the critical strip imply anything about the distribution of prime numbers? I know that it implies that the partial sums of the Möbius and Liouville functions cannot grow too slowly, and it's really this that I want to understand.

REPLY [41 votes]: Rather than the problem of why the zeta function has non-trivial zeros, let me address 
 Gowers's question of why the error term in the prime number theorem needs to be large.   The short answer that I propose is: because the integers are so well distributed. To make this precise, I shall prove a general result on semigroups, showing that 
 either the "integers" in the semigroup or the "primes" must be poorly distributed -- one may think of this as an "uncertainty principle" that both primes and integers cannot be simultaneously smoothly behaved. This has the flavor of  Beurling generalized primes, but I don't recall this result in the literature; maybe it exists already (indeed it does, see the edit below). Also note that the proof will not make any use 
 of the functional equation for $\zeta(s)$ as this does not exist in a general semigroup. 
EDIT:  Indeed doing a literature search a few hours after posting this, 
I found a paper of Hilberdink http://www.sciencedirect.com/science/article/pii/S0022314X04002069 which proves the 
Theorem below, and with a similar method of proof. 
Suppose that $1<p_1< p_2 <\ldots$ is a sequence of real numbers (the "primes"), and 
 $1=a_1 \le a_2 \le \ldots$ is the semigroup generated by them (the "integers").  Initially I made the assumption that the "integers" are distinct, and satisfy a mild spacing 
condition $a_{n+1}-a_n \gg n^{-1}$ (so that in particular there is unique factorization), but this is not necessary. Let $N(x)$ denote the number of 
"integers" below $x$, and 
$$
P(x) = \sum_{p_n^{k} \le x} \log p_n,
$$ 
which is the analog of the usual $\psi(x)$ (counting prime powers with weight $\log p$).  Assume that for some $\delta>0$ 
$$
N(x) = Ax +O(x^{\frac 12-\delta}), 
$$ 
for some non-zero constant $A$, and that 
$$ 
P(x) = x + O(x^{\frac 12-\delta}). 
$$ 
Theorem.   Either the asymptotic formula for $N(x)$ or the asymptotic 
formula for $P(x)$ must fail. 
Put 
$$ 
\zeta_A(s) = \sum_{n=1}^{\infty} a_n^{-s} = \prod_{n} \Big(1-\frac{1}{p_n^s}\Big)^{-1}. 
$$
By our assumptions on $N$ and $P$, the sum and product above converge absolutely in Re$(s)>1$. By the assumption on $N(x)$, $\zeta_A(s)$ extends to an analytic function in Re$(s)>1/2-\delta$ except for a simple pole at $s=1$ with residue $A$.  By the assumption on $P$, we see that the logarithmic derivative 
$$ 
-\frac{\zeta_A^{\prime}}{\zeta_A}(s) = \sum_{n, k} \frac{\log p_n}{p_n^{ks}} 
$$ 
extends analytically to Re$(s)>1/2-\delta$ except for a simple pole at $1$.  Thus 
$\zeta_A(s)$ has no zeros in Re$(s)>1/2-\delta$.  
New edit: Sketch of a second proof.  Adapting the argument that the Riemann hypothesis implies the Lindelof hypothesis (see below), we obtain that $|\zeta_A(s)| \ll (1+|s|)^{\epsilon}$ provided $s$ is not close to the pole at $1$, and that Re$(s)>1/2-\delta/2$.  From this and a standard contour shift argument we find that for large $N$ and any $t$, 
$$ 
\sum_{a_n\le N} a_n^{it} = \frac{N^{1+it}}{1+it} +O(N^{1/2-\delta+\epsilon} (1+|t|)^{\epsilon}). 
$$
What is used here is that we have Lindelof even a little to the left of the half line.  
But the above identity can be seen to contradict the Plancherel formula.  More precisely, let $T$ be a large power of $N$, and let $\Phi$ be a non-negative function supported in $[-1,1]$ with non-negative Fourier transform.  Then we see that (discarding all but the diagonal terms)
$$ 
\int_{-\infty}^{\infty} \Big| \sum_{a_n\le N} a_n^{it}\Big|^2 \Phi(t/T) dt 
\ge T{\hat \Phi}(0) \sum_{a_n\le N} 1 \sim TN {\hat \Phi}(0).
$$ 
On the other hand, if we use our identity then the above is seen to be 
$$ 
\ll N^2 + T^{1+\epsilon} N^{1-2\delta}. 
$$ 
This is a contradiction.
Original Proof:  Below let's assume always that we are in the region Re$(s)>1/2-\delta$, and that 
the imaginary part is large so that we are not near the pole at $1$.  From 
the analytic continuation of $\zeta_A$ (using that $N(x)$ is very regular), it follows 
that there is an a priori polynomial bound $|\zeta_A(s)|\ll |s|^{B}$ in the region Re$(s)>1/2-\delta/2$.  Thus there is a bound for the real part of $\log \zeta_{A}(s)$, and by the Borel-Caratheodory lemma (standard complex analysis) one can bootstrap this to a bound for $|\log \zeta_A(s)|$.  Then applying the Hadamard three circle theorem to $\log \zeta_A(s)$ one obtains a much better bound: $|\zeta_A(s)| \ll |s|^{\epsilon}$.   This is the usual proof that Riemann implies Lindelof.  (At this stage, if we knew a "functional equation" we'd be done, as the usual $\zeta(s)$ is large when Re$(s)<1/2$.  This point appeared in Matt Young's answers earlier.)
Knowing the Lindelof hypothesis for $\zeta_A(s)$ in Re$(s)> 1/2-\delta/2$, we can 
show the following approximate formula: for $\sigma > 1/2- \delta/4$ 
$$ 
\zeta_A(\sigma +it) = \sum_{a_n \le N} a_n^{-\sigma-it} + O(|t|^{\epsilon} N^{-\delta/4}). 
$$ 
The proof is standard;  see the penultimate chapter of Titchmarsh for the real $\zeta(s)$ where this holds when $\sigma$ is strictly bigger than $1/2$, and our stronger result is true because we have Lindelof in a wider region.  
Now we are ready to get our contradiction.  Consider for large $T$ 
$$ 
\int_T^{2T} |\zeta_A(1/2+it)|^2  dt. 
$$ 
To do this carefully it may be helpful to put in a smooth weight $\Phi(t/T)$ above 
(but this is not a paper!).  Using the approximate formula derived above, and our mild spacing condition $a_{n+1}-a_n \gg n^{-1}$, 
we may see that for any $T^{\epsilon} \le N \le T^{1/10}$ we have 
$$ 
\int_T^{2T} |\zeta_A(1/2+it)|^2 dt \sim T \sum_{a_n \le N} a_n^{-1} \sim AT \log N.
$$ 
But that's absurd! This completes our sketch proof.<|endoftext|>
TITLE: Is the matrix ring $\mathrm{Mat}_n(\mathbb{C})$ "algebraically closed"?
QUESTION [8 upvotes]: In spite of the fact that the matrix ring $\mathbb{C}^{n \times n}$ is not a field, is it still possible to talk about it being 'algebraically closed' in the sense that $\forall f \in \mathbb{C}^{n \times n}[x]$ does $\exists A \in \mathbb{C}^{n \times n}$ such that $f(A) = 0$? If so, then is it 'algebraically closed'?
Are there any other non-field sets that this idea can be extended to?

REPLY [11 votes]: I'd like to add that a nice theory of roots of polynomials over noncommutative rings was developed by I. Gelfand, V. Retakh, and R. Wilson, see the paper arXiv:math/0208146 and references therein (in particular, the earlier paper by Gelfand and Retakh on the noncommutatoive Vieta theorem).<|endoftext|>
TITLE: Infinite electrical networks and possible connections with LERW
QUESTION [6 upvotes]: I've been exposed to various problems involving infinite circuits but never seen an extensive treatment on the subject. The main problem I am referring to is

Given a lattice L, we turn it into a circuit by placing a unit resistance in each edge. We would like to calculate the effective resistance between two points in the lattice (Or an asymptotic value for when the distance between the points gets large). 

I know of an approach to solve the above introduced by Venezian, it involves superposition of potentials. An other approach I've heard of, involves lattice Green functions (I would like to read more about this). My first request is for a survey/article that treats these kind of problems (for the lattices $\mathbb{Z}^n$, Honeycomb, triangular etc.) and lists the main approaches/results in the field.
My second question (that is hopefully answered by the request above) is the following:
I noticed similarities in the transition probabilities of a Loop-erased random walk and the above mentioned effective resistances in $\mathbb{Z}^2$. Is there an actual relation between the two? (I apologize if this is obvious.)

REPLY [2 votes]: If you are still interested in this, you may want to have a look in Section 6 of http://www.sciencedirect.com/science/article/pii/0095895690900658 by Thomassen. He proves for example that the effective resistance between adjacent vertices of $Z^2$ is 1/2. I don't think there is mention to LERW though.<|endoftext|>
TITLE: Proof of Borel-Weil-Bott Theorem
QUESTION [6 upvotes]: Is there any purely algebraic proof of Borel-Weil-Bott theorem. I mean only techniques from Algebraic group. In each and every proof I have seen so far they use Lie group techniques and then translate to Algebraic group version. I need a proper reference which is easily readable.
Thanks in advance.

REPLY [3 votes]: See J.C. Jantzen "Representations of Algebraic Groups" II.5 especially II.5.5. 
There you will find an algebraic proof of the result in char. 0 (probably more-or-less
Demazure's proof, mentioned in another answer). And you'll find a proof of some of
what remains true in positive char (due to Henning Anderson).<|endoftext|>
TITLE: A comprehensive functor of points approach for manifolds
QUESTION [12 upvotes]: This seems unrealistic, because the topology on a manifold doesn't have anything to do with the properties its structure sheaf, but I figured I might as well ask.  This wouldn't be the first time I was pleasantly surprised about something like this.  If not, is there any sort of way to attack differential geometry with abstract nonsense?
Even though schemes have singularities, "it's better to work with a nice category of bad objects than a bad category of nice objects".  Manifolds seem to be perfect illustration of this fact.
Edit: Apparently my question wasn't clear enough.  The actual question here is if we can define manifolds entirely as "functors of points" like we can with schemes (sheaves on the affine zariski site).  There is no fully categorical and algebraic description of the category of smooth manifolds.  When I say a "comprehensive functor of points approach", I mean a fully categorical description of the category of smooth manifolds.

REPLY [10 votes]: Here are two things that I think are relevant to the question.
First, I want to support Andrew's suggestion #5: synthetic differential geometry.  This definitely constitutes a "yes" to your question 

is there any sort of way to attack differential geometry with abstract nonsense? 

--- assuming the usual interpretation of "abstract nonsense".  It's also a "yes" to your question 

Can we describe it as some subcategory of some nice grothendieck topos?

--- assuming that "it" is the category of manifolds and smooth maps.  Indeed, you can make it a full subcategory.
Anders Kock has two nice books on synthetic differential geometry.  There's also "A Primer of Infinitesimal Analysis" by John Bell, written for a much less sophisticated audience.  And there's a brief chapter about it in Colin McLarty's book "Elementary Categories, Elementary Toposes", section 23.3 of which contains an outline of how to embed the category of manifolds into a Grothendieck topos.
Second, it's almost a categorical triviality that there is a full embedding of Mfd into the category Set${}^{U^{op}}$, where $U$ is the category of open subsets of Euclidean space and smooth embeddings between them.  
The point is this: $U$ can be regarded as a subcategory of Mfd, and then every object of Mfd is a colimit of objects of $U$.  This says, in casual language, that $U$ is a dense subcategory of Mfd.  But by a standard result about density, this is equivalent to the statement that the canonical functor Mfd$\to$Set${}^{U^{op}}$ is full and faithful.  So, Mfd is equivalent to a full subcategory of Set${}^{U^{op}}$.
There's a more relaxed explanation of that in section 10.2 of my book Higher Operads, Higher Categories, though I'm sure the observation isn't original to me.<|endoftext|>
TITLE: Free action of SL_2(F_p) on a sphere
QUESTION [10 upvotes]: Let $p>2$ be prime. Then for abstract reasons the special linear group $\text{SL}_2({\mathbb F}_p)$ possesses a free action on some sphere (one has to check that any abelian subgroup of $\text{SL}_2({\mathbb F}_p)$ is cyclic and that there's at most one element of order $2$).
Does somebody know a concrete example for such a free action for general $p$? (For $p=5$, for example, $\text{SL}_2({\mathbb F}_p)$is the binary icosahedral group which is a subgroup of ${\mathbb S}^3$ thus acting freely on it by multiplication; I'd like to know if there's one single action that can be written down for all $p$ simultaneously).

REPLY [13 votes]: Apparently, a linear free action exists only for $p=5$ (if $p\ge 5$), see paper by C. Thomas "Almost linear actions by $SL_2(p)$ on $S^{2n-1}$". There is a weaker notion of an "almost linear" action, and it seems that constructing such actions is a fairly complicated business, using state-of-the-art differential geometry and topology; see    arXiv:math/9911250. It seems that simple explicit actions for higher $p$ are not expected (also, of course, the sphere must be odd dimensional, since an orinentation preserving self-map of an even dimensional sphere has a fixed point by the Lefschetz theorem).<|endoftext|>
TITLE: Polynomial with two repeated roots
QUESTION [5 upvotes]: I have a polynomial of degree 4 $f(t) \in \mathbb{C}[t]$, and I'd like to know when it has two repeated roots, in terms of its coefficients.
Phrased otherwise I'd like to find the equations of the image of the squaring map
$sq \colon \mathbb{P}(\mathbb{C}[t]^{\leq 2}) \rightarrow \mathbb{P}(\mathbb{C}[t]_{\leq 4})$.
(for some reason the first lower index wouldn't parse, so I put it on top).
Of course I can write the map explicitly and then find enough equations by hand, but this looks cumbersome. I'm not an expert in elimination theory, so I wondered if there is some simple device to find explicit equations for this image. For instance one can detect polynomials with one repeated root using the discriminant, but I don't know how to proceed from this.

REPLY [7 votes]: It seems to me that this example is easy to do by hand. By the standard tricks, we can assume your polynomial is of the form 
$$x^4+ c x^2 + dx +e.$$
A polynomial of this form is a square if and only if $d=0$ and $4e=c^2$.<|endoftext|>
TITLE: Which mathematical ideas have done most to change history?
QUESTION [68 upvotes]: I'm planning a course for the general public with the general theme of "Mathematical ideas that have changed history" and I would welcome people's opinions on this topic.  What do you think have been the most influential mathematical ideas in terms of what has influenced science/history or changed the way humans think, and why?
I won't expect my audience to have any mathematical background other than high-school.
My thoughts so far are: non-Euclidean geometry, Cantor's ideas on uncountability, undecidability, chaos theory and fractals, the invention of new number systems (i.e. negative numbers, zero, irrational, imaginary numbers), calculus, graphs and networks, probability theory, Bayesian statistics.
My apologies if this has already been discussed in another post.

REPLY [7 votes]: Probably it can be viewed as a variant on already posted answers (cryptography etc.), but the study of permutation groups and its application in cracking the Enigma code literally changed history (namely, the outcome of World War II). Here is an article by Marian Rejewski, one of the people involved in the code-cracking, explaining what was done and how:
http://www.impan.pl/Great/Rejewski/article.html
Rejewski and his achievements were also mentioned in answers to the following MO questions:
Real-world applications of mathematics, by arxiv subject area?
Notable mathematics during World War II<|endoftext|>
TITLE: Functions separting points in Hausdorff spaces
QUESTION [13 upvotes]: A colleague in algebra asked me this, and I couldn't answer it.  On the Wikipedia page for "epimorphism" it is claimed that in the category of Hausdorff spaces and continuous maps, a function is epi if and only if it has dense range.  The "if" case is easy, but I couldn't justify the "only if" case.
This boils down to: let Y be a Hausdorff space, and let X in Y be a closed subset not equal to Y, and not empty.  Can you find a Hausdorff space Z and functions f,g:Y->Z such that f and g agree on X, but are not equal.  I think, by using a quotient argument, you can assume that X is just a point.

REPLY [9 votes]: This is really a comment on t3suji's answer, but it's too long to be a comment as such.  
t3suji's answer is the canonical one in the following precise sense.  Let $e: X \to Y$ be a morphism in any category.  It's an elementary exercise to show that the following conditions on $e$ are equivalent:

$e$ is an epimorphism
the square 
$$
\begin{array}{ccc}
X    &\stackrel{e}{\to} &Y \\
e\downarrow &                  &\downarrow 1_Y \\
Y    &\stackrel{1_Y}{\to} &Y
\end{array}
$$
is a pushout
for some morphism $f: Y \to Z$, the square
$$
\begin{array}{ccc}
X    &\stackrel{e}{\to} &Y \\
e\downarrow &                  &\downarrow f \\
Y    &\stackrel{f}{\to} &Z
\end{array}
$$
is a pushout.

I'll only use the equivalence 1 $\iff$ 3 here.  The other implications are just scene-setting.
Suppose we want to show that a particular morphism $e$ is not epi.  Assuming that there are enough pushouts around, we can argue as follows.  Form the pushout square
 $$
\begin{array}{ccc}
X    &\stackrel{e}{\to} &Y \\
e\downarrow &                  &\downarrow f \\
Y    &\stackrel{g}{\to} &Z.
\end{array}
$$
If $f \neq g$ then the implication 1 $\Rightarrow$ 3 tells us that $e$ is not epi.  Moreover, this strategy is bound to work, in the sense that if $f = g$ then the implication 3 $\Rightarrow$ 1 tells us that $e$ is epi after all.
It only remains to see that this is indeed what t3suji did.  In his/her situation, $e$ was the inclusion $X \to Y$.  He/she then took the coequalizer of the two obvious maps $X \to Y + Y$ (where $+$ means coproduct, i.e. disjoint union).  For elementary and totally general reasons, this is the same thing as taking the pushout just mentioned.  The morphisms that t3suji called $\iota_1$ and $\iota_2$, I called $f$ and $g$.  Finally, although t3suji's pushout is in the category of all topological spaces, he/she then verified that the space $Z$ is indeed Hausdorff, from which it follows that it's also a pushout in Hausdorff spaces.
So now you know, in principle, how to answer any question of the form "prove that such-and-such a morphism isn't epi".<|endoftext|>
TITLE: Finding unknown integer-valued polynomials using inequalities
QUESTION [5 upvotes]: I've come across this interesting inequalities problem recently, which seemed straight-forward at first glance but has proven interesting enough to ask about it here. 
Suppose you are given the degree of an unknown polynomial, and are told that all the coefficients are integers and are all within $min \le a\_k \le max$. Also, you have access to an oracle who will evaluate p(q), the unknown polynomial, and compare it to your guess $g\_q$, where q is the number of previous guesses you have made, and determine if your guess was less, greater than, or exactly equal to the unknown polynomial.
What is the optimal method of choosing guesses, given previous results, in order to make a correct guess as soon as possible in the worst case? 
If the degree is d and the coefficients are bounded by [min, max] then it would seem the best possible method would require slightly less than $ \frac {d \log (max - min + 1)}{\log 2}$ by binary search. Its slightly less because with an answer of > from the oracle, you exclude both the values less than and those equal to the guess, which could be more than 50% of possibilities.
If the median of the evaluated values at q of all of the remaining possible polynomials can be found, than guessing that value would be guaranteed to eliminate at least half of the possibilities. But is there any efficient way to find the median of a function of a set of polynomials that are only identified by inequalities? 
For the first guess, q is zero and therefore all but the constant term of the polynomial are irrelevant. It makes sense then to make g(0) to be $\frac {min + max}{2}$. But after that, the best way of finding the median quickly seems elusive.
As an example, make min = -100, max = 100 and d = 1. $\frac {-100 + 100}{2}$ = 0, so g(0) should be 0. If the oracle returns > then we know $0 < a\_0 \le 100$ or $1 \le a\_0 \le 100$ since we are dealing with integer coefficients.
An ideal method would include an efficient way to find the median and characterize the set of possibilities given the previous answers from the oracle. But medians can be hard to calculate so an efficient method to calculate the mean of p(q) for the set of possibilities would be close enough if an efficient method doesn't exist for medians.

REPLY [2 votes]: Just determine the coefficient at the highest power first by plugging in huge numbers and doing binary search (comparing to what you'd get for the half-integer coefficients). Once you know it, you can easily figure out the coefficient at the next power in the same way and so on.
Now, if you also have a bound on $q$ you can plug in, it becomes interesting.  
Oops, sorry for misunderstanding.
OK, you can easily do $Cd^M\log N$ then. The trick is that no matter how you split the d-dimensional simplex of volume 1 by a hyperplane through its center and no matter which piece you'll take, you'll be able to find a simplex of volume $1-d^{-m}$ conatining this piece where $m$ is some fixed number, which I will need some time to compute precisely (my educated guess would be m=4). Now just use this fact to obtain a simplex of either volume less than $N^{-2d}$ or with one vertex outside the ball of radius $N^{3d}$ containing the set of remaining polynomials after just $O(d^M\log N)$ steps. In both cases, you'll be able to find a linear dependence between the coefficients that is precise up to $N^{-2}$, which means that you can eliminate one coefficient from the polynomial entirely (the $d$-th power of the variable is still well below $N$, so the precision is enough to distinguish the integer values and each exactly attained equality which will not allow you to tell for sure which half you are in gives you food for interpolation).
Sorry if it is too sketchy. I'll try to edit it into something more reasonable later unless somebody gives a better solution.<|endoftext|>
TITLE: Number of subgroups in a Bieberbach group.
QUESTION [24 upvotes]: Assume $\Gamma$ be a Bieberbach group which acts on $\mathbb R^n$
(i.e. a discrete subgroup of isometries of $n$-dimensional Euclidean
space with a compact fundamental domain).
Denote by $M(\Gamma)$ the number of maximal finite subgroups (up to
conjugation) in $\Gamma$.
Is it true that $M(\Gamma)\le 2^n$?
Things I can do:
There is a simple geometric observation (due to Perelman) which shows
that if $N(\Gamma)$ is the number of orbits of isolated fixed point of
some subgroups of
$\Gamma$ then
$N(\Gamma)\le2^n$.
Clearly, each such point corresponds to a maximal finite subgroup.
Thus, $N(\Gamma)\le M(\Gamma)$, but in all examples I know I still
have $M(\Gamma)\le 2^n$ (and I believe it is allways true).
The formulation is completely algebraic so maybe it has a completely
algebraic solution...

REPLY [2 votes]: That is not an answer.
I want to give an example where the argument of Erdős does not work directly.
Consider an action of group $\Gamma$ on $\mathbb R^3$ generated by the reflections $r_1, r_2$ and $r_3$ correspondingly in the lines $x=z=0$ and $x+1=z=0$ and $x-y=z-1=0$.
Each of the reflections $r_i$ generate a maxiamal $\mathbb Z_2$-subgroups, all of them are  nonconjugate. 
These groups corespond to three singular circles, say $\Sigma_i$ in the factor $X=\mathbb R^3/\Gamma$. 
($X$ is homeomorphic to $S^3$ and $\Sigma_1$, $\Sigma_2$, $\Sigma_3$ form Borromean rings, but all this is not important.)
Let us try to mimic argument of Erdős.
Take subsets $X_i$ of $X$ of midpoints $m$ between $x\in X$ and a closest $x_0\in\Sigma_i$ to $x$.
As in the argument of Erdős we have $\mathrm{vol}\, X_i>\tfrac{1}{2^3}\cdot\mathrm{vol}\, X$.
BUT $X_1\cap X_3$ has interior points and here argument brakes into parts.
Comments 

Since fixed point sets are 1-dimensional, it would be enough to take $m\in [xx_0]$ such that $\tfrac{|mx_0|}{|xx_0|}=\tfrac1{2\sqrt[3]{2}}$.
But even in this case one has interior points in $X_1\cap X_2$ (the borderline in this example seems to be $\tfrac13$).
There is a natural bisecting hyperplane for any two affine subspaces. We may use it to cut a cylinder domain around each fixed point set of a maximal subgroup. The projection of these cylinders in $X$ gives Voronoi-like domains, but they do not cover whole space in general --- that is OK as far as we have lower bound on their volumes...<|endoftext|>
TITLE: How can generic closed geodesics on surfaces of negative curvature be constructed?
QUESTION [6 upvotes]: As far as I understand it the closing lemma implies that closed geodesics on surfaces of negative curvature are dense. So: how can they be constructed in general? 
A concrete answer that dovetails with the construction of such surfaces with constant negative curvature and genus $g$ from regular hyperbolic $(8g-4)$-gons along lines indicated by Adler and Flatto and gives the endpoints of the geodesics in the Poincaré disk model would be ideal. More useful still would be a way to construct all the closed geodesics that cross the boundaries of translates of the fundamental $(8g-4)$-gon some specified number of times (I am pretty sure this ought to be a finite set, but I couldn't say why off the top of my head).

REPLY [6 votes]: Are you willing to buy that the set of  non-closed geodesics are dense?  If so, here is an argument that goes back to Birkhoff and Hadamard.  Take the surface's 
universal cover -- which is to say the upper half plane.  Tile it with fundamental domains for said surface's fundamental group (relative to a fixed const neg curv metric).  These
have some number of edges, a, b, c, ... . Now count how a geodesic crosses the edges:
acbaf... thus getting an (infinite) word -- or symbol sequence--  in the edges.  Theorem: the symbol sequence  is periodic if and only if the geodesic is. Theorem:  if we are given a (variable) negatively curved metric on the surface
then the symbol sequence uniquely determines the geodesic.  Theorem: if a sequence $s_N$ of 
symbol sequences converges to a symbol sequence $s$, in the sense that for any sufficiently large `window' L of word length centered around 0, the finite word of length L of arbitrary
length eventually agree, then the corresponding geodesics also converge.   Now,
approximate your given geodesic -- ie symbol sequence -- by a periodic sequence. 
symbol sequence by longer and longer periodic sequences.<|endoftext|>
TITLE: What is the right definition of the Picard group of a commutative ring?  
QUESTION [32 upvotes]: This is a rather technical question with no particular importance in any case of actual interest to me, but I've been writing up some notes on commutative algebra and flailing on this point for some time now, so I might as well ask here and get it cleared up.
I would like to define the Picard group of an arbitrary (i.e., not necessarily Noetherian) commutative ring $R$.   Here are two possible definitions:
(1) It is the group of isomorphism classes of rank one projective $R$-modules under the 
tensor product.
(2) It is the group of isomorphism classes of invertible $R$-modules under the tensor product, where invertible means any of the following equivalent things [Eisenbud, Thm. 11.6]:
a) The canonical map $T: M \otimes_R \operatorname{Hom}_R(M,R) \rightarrow R$ is an isomorphism.
b) $M$ is locally free of rank $1$ [edit: in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}$.]
c) $M$ is isomorphic as a module to an invertible fractional ideal.
What's the difference between (1) and (2)?  In general, (1) is stronger than (2), because projective modules are locally free, whereas a finitely generated locally free module is projective iff it is finitely presented.  (When $R$ is Noetherian, finitely generated and finitely presented are equivalent, so there is no problem in this case.  This makes the entire discussion somewhat academic.)
So, a priori, if over a non-Noetherian ring one used (1), one would get a Picard group that was "too small".  Does anyone know an actual example where the groups formed in this way are not isomorphic?  (That's stronger than one being a proper subgroup of the other, I know.)  
Why is definition (2) preferred over definition (1)?

REPLY [7 votes]: 1) About the second definition:
$\alpha$) It is not true  that for an arbitrary ring a) is equivalent to c):
Indeed Bourbaki in  Algèbre commutative, Chapitre II, Exercices §5, 12) c)  exhibits a ring $B$ and a projective module of rank $1$ over $B$ which is not isomorphic to an invertible fractional ideal of $B$. This does not contradict Eisenbud's Theorem 11.6 because $R$ is explicitly supposed noetherian there.
  $\beta$) It is also not true that b) is equivalent to c):
Take $R=\mathbb Z$ and $\mathbb Z\subsetneq M=\bigcup \frac {1}{p_1\cdots p_i}\mathbb Z\subsetneq \mathbb Q$ where $p_i$ is the $i$-th prime.
 Then for all prime $\mathfrak p\subset \mathbb Z$ the $\mathbb Z_\mathfrak p$-module $M_\frak p$ is free of rank $1$, since $$M_{(0)}=\mathbb Z_{(0)}(=\mathbb Q) \quad  \operatorname {and}\quad M_{(p_i)}=\frac {1}{p_i}\mathbb Z_{(p_i)}$$ However the $\mathbb Z$-module $M$ is neither finitely generated nor projective (over $\mathbb Z$, projective=free)
2) I think the only reasonable definition of $\operatorname {Pic(R)}$ valid for any commutative ring is to define it as the Picard group of the affine scheme $X=\operatorname {Spec}(R)$.
As is the case for any locally ringed space $(X,\mathcal O_X)$ the Picard group consists of isomorphism classes of locally free $\mathcal O_X$-Modules of rank one.
This is exactly the definition used with much success for general, non-affine, schemes but also for topological spaces, differential manifolds, etc.  
3) In our special case $X=\operatorname {Spec}(R)$ the definition in 2) translates in purely  algebraic terms to:
$\operatorname {Pic(R)}$ consists of isomorphism classes of $R$-modules $M$ such that there exist finitely many elements $f_1,\cdots,f_n\in R$ with:
$\alpha$)  $\sum Rf_i=R$
$\beta$)   $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.  
These modules are called locally free of rank one.
Remark: $\beta$) implies that locally free modules of rank one  are finitely generated over $R$, since "finitely generated" is a local condition   
4) The locally free modules of rank one defined in 3) can also be characterized as the  modules $M$ over $R$ such that  equivalently:
 i) The module $M$ is finitely generated, projective and for all primes $\mathfrak p\subset R$ the (necessarily!) free $R_\mathfrak p$- module $M_\mathfrak p$   has  rank $1$
ii) The module $M$ is finitely generated and the modules $M_\frak m$ are free of rank $1$ over $R_\frak m$ for all maximal ideals $\mathfrak m\subset R$
iii) The canonical $R$-linear map $M\otimes_RM^*\to R:m\otimes \phi\mapsto\phi(m)$ is bijective
Note that these are pleasant algebraic characterizations, but the conceptual definition is that given in 2) and 3).  
Edit: WARNING
The confusion is made worse by Bourbaki's unfortunate decision to define  a projective module of rank $1$ as a finitely generated module $P$ for which $M_\mathfrak p$  is free of rank $1$ over $R_\mathfrak p$ for all primes $\mathfrak p\subset R$.
As my example 1) $\beta$) shows, omitting to require that $P$ be finitely generated [as is done in Pete's condition (2) b)] means accepting modules which aren't even projective, and which  don't satisfy (2) a) nor (2) c) of the question.<|endoftext|>
TITLE: A question on ultrapower
QUESTION [7 upvotes]: Suppose $\kappa_0$ is a measurable cardinal and $\mu_0$ is a normal measure on $\kappa_0$. $M_1$ is the transitive collapse of $Ult(V,\mu_0)$, $j_{0,1}:V\rightarrow{M_1}$ is the elementary embedding induced by the ultrapower. In $M_1$, $\kappa_1=j_{0,1}(\kappa_0)$ is a measurable cardinal and $\mu_1$ is a normal measure on $\kappa_1$ in $M_1$ such that $\mu_1$ is not in the range of $j_{0,1}$. $M_2$ is the transitive collapse of $Ult(M_1,\mu_1)$, $j_{1,2}:M_1\rightarrow{M_2}$ is the elementary embedding induced by the ultrapower. $j_{0,2}=j_{1,2}\circ{j_{0,1}}$.
Is it true that: ``Suppose $N$ is an inner model, $i:V\rightarrow{N}$ and $k:N\rightarrow{M_2}$ are elementary embeddings such that $k\circ{i}=j_{0,2}$. Then $k''N=j_{0,2}''V$ or $k''N=j_{1,2}''M_1$ or $k''N=M_2$''?

REPLY [6 votes]: This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.
The answer is Yes.
Let me denote κ0 just by κ and j02 by j. Since
μ1 is a measure in M1, it has the
form j01(m)(κ), where m =
(να | α < κ). Since you
have said that μ1 is not in
ran(j01), we may choose the
να to be all different, and different
from μ0. In this case, there is a partition
of κ as the disjoint union of Xα,
with Xα in να and none
in μ0. Let x = (Xα | α
< κ). Note that κ is not in
j01(Xα) for any α <
κ, and similarly κ1 is not in
j(Xα). But κ is in
j01(x)(β) for some β <
κ1, since this is a partition of
κ1. Apply j12 to conclude that
κ1 is in j(x)(β) for this β.
Thus, there is some β in the interval [κ,
κ1) having the form β =
j(f)(κ1) for the function f that picks the
index. From this, it follows from normality of
μ0 that we can write κ =
j(g)(κ1) for some function g, since any
β < κ1 generates κ via
j01. In my favored terminology, the seed
κ1 generates κ via j and in fact
generates all β in [κ,κ1) via
j.
Similarly, suppose that δ is in the interval
[κ1,j(κ)). We know δ =
j12(f)(κ1) for some function f
on κ1 in M1. We also know f =
j01(F)(κ) for some F in V. Thus, δ =
j(F)(κ, κ1). In Y, let
(α,β) be the smallest pair with δ =
j(F)(α,β). It cannot be that both are below
κ1, since this would be inside
ran(j12) and so the least pair must have β
= κ1. Thus, δ generates
κ1, which we already observed generates
κ.
To summarize, every ordinal in the interval
[κ1,j(κ)) generates
κ1, which generates all the ordinals
β in [κ,κ1), any of which
generate κ and all the other such β.
This is enough to answer your question. The k " N in your
question is just an arbitrary elementary substructure of
M2 containing ran(j), so suppose we have Y
elementary in M2 and ran(j) subset Y. The case Y
= ran(j) is one of your cases. Otherwise, Y has something
not in ran(j). Every object in M2 has form
j(h)(κ,κ1) for some function h, so
by looking at the smallest pair of ordinals to generate a
given object with j(h), we see that there must be ordinals
below j(κ) in Y. If Y contains any ordinal δ in
the interval [κ1,j(κ)), then it will
contain both κ and κ1, since we
observed that any such δ generates these ordinals. In
this case, Y = M2, since those two ordinals
generate everything. So we assume that Y contains no such
δ. In this last case, Y must contain some ordinal
β in the interval [κ,κ1). Since
any such β generates κ, Y contains all such
ordinals. It follows that ran(j12) subset Y and
in fact = Y, since if Y contained anything more it would
have to have an additional ordinal δ in
[κ1,j(κ)).
So we've seen that your three cases are the only
possibilities. And like your previous question, there is no
need to assume that Y or N is somehow internally definable.
By the way, this was a problem that I had solved many years
ago for my dissertation, although perhaps other people had
also thought about it. I was interested in understanding
which pairs of ordinals (α,β) generate product
measures via an embedding j, and this question is very much
related to that.
(Click the edit history to see my previous answer, which was just about the case when μ1 is in the range of j01, a case for which the answer is no.)<|endoftext|>
TITLE: Interaction of topology and the Picard group of Algebraic surfaces
QUESTION [6 upvotes]: It is well known that a smooth cubic surface $X\subset \mathbb{P}^3$ has exactly 27 lines in it. Furthermore, it is easy to check that Picard group $$Pic(X)\cong \mathbb{Z}^7$$ Here the generators are lines which are $\mathbb{P}^1$ topologically. Furthermore, it is easy to check that $$\chi_{top}(X)=2+H_2(X)=9$$Topologically speaking, notice that as smooth manifold $X$ has no 1-skeleton. This makes the 2-dimensional cells glue to points along their bounday, getting spheres $\mathbb{P}^1$ as the result of this gluing process. 
My first question is why the other 20 lines do not contribute to the Euler characteristic of $X$.
Going further, if $X\subset \mathbb{P}^3$ has degree 4, it is known that $X$ sometimes has lines, sometimes it does not. However, $$\chi_{top}(X)=2+H_2(X)=24$$ meaning that, despite the fact that $X$ can perfectly have no lines, we still have homology $H_2$ which are spheres topologically! meaning, there are in fact spheres (due to the argument above which says that $X$ has no 1-skeleton). Besides $\chi_{top}$ is constant even though $X$ may have $64$, $32$ or even $0$ number of lines in it. There are spheres whose existence is not being noticed by $\chi_{top}$ at all. Here let me be vague please. What is going on!?
Any type of editing to make this clearer will be welcome.
References highly appreciated.

REPLY [4 votes]: I worry that the other answers may be giving too much detail. What is going on is that there are linear relations between the classes of the lines so that, although they are 27 lines in the cubic surface, the lattice they span in H^2 only has dimension 7. If you want to know exactly how these 27 vectors sit in a 7 dimensional vector space, read Emerton's answer.<|endoftext|>
TITLE: Wants: Polynomial Time Algorithm for Decomposing a Multiset of Rationals into Two Additive Subsets.
QUESTION [7 upvotes]: First, allow me to say that this problem was posed to me by a professor in the department. It is related to his research in a way that I do not know. However, since I couldn't come up with anything novel, I decided to ask here.
Alright, let $S$ be a multiset of $n$ rational numbers mod 1. Assume that $0\in S$. Define a additive decomposition of the set $S$ as two sets $A$ and $B$ such that

Both have elements rational numbers mod 1 and contain 0.
For all $a\in A$ and $b\in B$ the sum, $(a+b)\mod{1}\in S$
Every $s\in S$ is the sum of an element from each of $A$ and $B$.

Just to be perfectly clear, lets consider an example. Let $S:=\lbrace 0,\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{5}{6} \rbrace$, then the only additive decompositions are 

$A=\lbrace 0\rbrace$, $B=\lbrace 0,\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{5}{6} \rbrace$
$A=\lbrace 0, \dfrac{1}{2}\rbrace$, $B=\lbrace 0,\dfrac{1}{3} \rbrace$
$A=\lbrace 0, \dfrac{1}{2}\rbrace$, $B=\lbrace 0,\dfrac{5}{6} \rbrace$

Second Example: 
If $A=\lbrace 0, \dfrac{1}{2}, \dfrac{1}{3}\rbrace$, $B=\lbrace 0,\dfrac{1}{2}, \dfrac{1}{3} \rbrace$, they would be a decomposition of the set $S=\lbrace 0, 0, \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{5}{6}, \dfrac{5}{6}\rbrace$
At this point there are a few things to mention. First, we quickly reduce the problem to looking at subsets whose orders are $\alpha$ and $\beta$ s.t. $\alpha\beta=n$. Additionally, we can see that by the additive structure splitting into these two subsets is adequate in the sense that we can get a complete decomposition recursively by breaking the set into two.
Question:

What is the fastest algorithm you can come up with to find all additive decompositions of a multiset $S$ of order $n$?

A computer has already been used to attack the problem. In small cases, the problem is not too bad. The situation arises in the fact that in the largest cases necissary $n\sim 10^5$. The professor said that an algorithm of polynomial time with respect to $n$ would be a great improvement from this current.
A word on the current algorithm. Look at the factorization of $n$. Pick $\alpha\mid n$. Select $\alpha$ elements of $S$ and let them be $A$. Then for each $s_i\in S$ remove $A+s_i\mod{1}$ from $S$. After running through $s_i$, the remaining elements for a candidate for $B$. If the cardinality of $B$ is $\beta$ for $\alpha\beta=n$ then we have a decomposition.
In addition to searching for a solution, I want to encourage discussion of other aspects of this problem as they may yield some interesting observations not noticed before.
Thanks in advance!

REPLY [11 votes]: Any general algorithm for this problem will require exponential time.  In fact, just writing down the answer can take exponential time in some cases.
For example, suppose that $n=2m$ for some odd $m$, and let $S=\lbrace 0,\frac{1}{n},\frac{2}{n},\ldots,\frac{n-1}{n}\rbrace$.  Start with $A=\lbrace 0,\frac{1}{2} \rbrace$ and $B=\lbrace 0,\frac{1}{m},\frac{2}{m},\ldots,\frac{m-1}{m} \rbrace$.  Adding $0$ or $1/2$ to each nonzero element of $B$ independently results in $2^{m-1}$ other sets $B'$ that when matched with $A$ still form an additive decomposition of $S$.  So there are at least $2^{m-1} = 2^{n/2-1}$ additive decompositions of $S$.

REPLY [4 votes]: Going off of fedja's comment I'll assume you want unique representations.  In that case, one small observation is as follows: if $d$ is the least common denominator of the elements of $S$ and $S = \{ \frac{s_1}{d}, ... \frac{s_n}{d} \}$, then the problem is equivalent to determining the possible factorizations of $x^{s_1} + ... + x^{s_n}$ into polynomials with coefficients zero or one in $\mathbb{Z}[x]/(x^d - 1)$.  These factorizations are, in turn, at least controlled by factorization in $\mathbb{Z}[\zeta_d]$, so it's possible that known algorithms in algebraic number theory might be of use.<|endoftext|>
TITLE: Construction of the Stiefel-Whitney and Chern Classes
QUESTION [36 upvotes]: I've seen two constructions of these characteristic classes.  The first comes from Milnor and Stasheff's book and involves the Thom isomorphism and (at least for me) the rather mysterious Steenrod squaring operations.  
The other construction comes from Hatchers Vector Bundles and K-Theory book. There Hatcher uses the Leray-Hirsh theorem to pick out specific classes for the tautological line bundle over $\mathbb{P}_{\mathbb{R}}^\infty$ for the Stiefel-Whitney classes and does the analogous thing over $\mathbb{C}$ for the Chern classes. 
Does anyone know of a good way of comparing these constructions i.e. verifying that they pick out the same classes?  Or is there a good source where this is discussed?
Also the cohomology rings of the infinite Grassmanians $G_n(\mathbb{R}^\infty), G_n(\mathbb{C}^\infty)$ have nice descriptions as polynomials rings in the respective characteristic classes, is there a similar description for the cohomology ring of the Thom space $T = T(\gamma_n)$ associated to the tautological vector bundle $\gamma_n$ over $G_n(\mathbb{R}^\infty), G_n(\mathbb{C}^\infty)$?

REPLY [12 votes]: Let me offer another definition not far from obstruction theory (as Ilya gave), but without referring to obstruction theory and thus more elementary.
Suppose for simplicity that $X$ is a simplicial complex, and the bundle $E$ is piecewise-linear (trivialized over each simplex, with transition maps over common faces which are linear) and of dimension $n$.  First consider a piecewise-linear section on the $n$-skeleton of $X$ with isolated zeros (such sections are dense in the space of all sections).  Then the Euler class, which is the $n$th SW class $w_n(E)$, is represented by the cochain whose value on an $n$-simplex $\sigma$ is the (mod-two) count of the zeros of that section on $\sigma$.  Fun exercises: show this is a cocycle, and that different choices of sections give rise to cohomologous cocycles.
More generally, consider $i$ different sections over the $n-i+1$ skeleton which are linearly dependent at only a finite collection of points.  The SW class $w_{n-i}$ evaluates on some $n-i$ simplex $\sigma$ as the count of the points of dependence of these sections.
I like to teach SW classes from this perspective first because it is an explicit, cochain-level definition and thus illustrates that there are good geometric reasons to consider cochains.  But then I do like to go ahead and develop the classifying space perspective as well, using Milnor's axioms and uniqueness theorem to connect them.  I conjecture (but cannot be sure) that at Milnor's time this kind of "dependence of sections" approach was widely known, so he could assume some of that familiarity as he stressed axiomatics.<|endoftext|>
TITLE: A finitely generated, locally free module over a domain which is not projective?
QUESTION [35 upvotes]: This is a followup to a previous question
What is the right definition of the Picard group of a commutative ring?  
where I was worried about the distinction between invertible modules and rank one projective modules over an arbitrary commutative ring.  I was worrying too much, because of the following theorem [Bourbaki, Commutative Algebra, Section II.5.2, Theorem 1]:
Let $R$ be a commutative ring and $M$ a finitely generated $R$-module.  The following are equivalent:
(i) $M$ is projective.
(ii) $M$ is finitely presented and locally free in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{r(\mathfrak{p})}$.
(iii) $M$ is locally free in the weaker sense and its rank function $\mathfrak{p} \mapsto r(\mathfrak{p})$ is locally constant on $\operatorname{Spec}(R)$.
(iv) $M$ is locally free in the stronger sense: there exist $f_1,\ldots,f_n \in R$, generating the unit ideal, such that for each $i$, $M_{f_i}$ is a free $R_{f_i}$-module.
(v) For every maximal ideal $\mathfrak{m}$ of $R$, there exists $f \in R \setminus \mathfrak{m}$ such that $M_f$ is a free $R_f$-module.  
This answers my previous question, because the rank function of an invertible module is identically one.
In order to really feel like I understand what's going on here, I would like to see an example of a finitely generated locally free [in the weaker sense of (ii) above] module which is not projective.  Thus $R$ must be non-Noetherian.  The wikipedia article on projective modules contains some nice information, in particular sketching an example of such a module over a Boolean ring.  For a Boolean ring though the localization at every prime ideal is simply $\mathbb{Z}/2\mathbb{Z}$, so it is not too surprising that there are more locally free modules than projectives.
I would like to see an example with $R$ an integral domain, if possible.  It would be especially nice if you can give a reference to one of the standard texts on commutative algebra which contains such an example or at least a citation of such an example.

REPLY [14 votes]: After mulling things over for a while, I realize that assuming the results from Bourbaki which I mentioned in the statement of the question, there is a very straightforward answer to my question.  (I mean this as no slight to Clark Barwick's excellent answer, which came instantaneously and contains lots of other valuable information.  Rather, I mean that had I thought more carefully I would not have needed to ask the question at all.)
The key is the following simple result:
Lemma: Let $R$ be a[n always commutative] ring, $\mathfrak{p}_1 \subset \mathfrak{p}_2$ prime ideals of $R$, and $M$ a finitely generated locally free (in the weaker sense) $R$-module.  Then $r(\mathfrak{p}_1) = r(\mathfrak{p}_2)$.
The proof is obvious, once you realize that localizing at $\mathfrak{p}_1$ is the same as localizing at $\mathfrak{p}_2$ and then localizing at (the ideal in $R_{\mathfrak{p}_2}$ naturally corresponding to) $\mathfrak{p}_1$.  (This is the same argument that allows you to see that is enough to require $M_{\mathfrak{p}}$ to be free at every maximal ideal $\mathfrak{p}$.)  Also I am using that the rank of a finitely generated free module over a [commutative!] ring is well-determined, as one sees by tensoring to the quotient field of some maximal ideal.
[I had some kind of psychological block coming from a vague memory that the rank function was merely semicontinuous.  As far as I can see now, semicontinuity does not come up anywhere in the study of the rank function.  I had to see essentially this argument in print -- in Milnor's Introduction to Algebraic K-Theory just this evening -- in order to become unblocked.]
Corollary: Let $R$ be a ring with a unique minimal prime ideal (e.g. an integral domain).  Then any finitely generated locally free $R$-module has constant rank function and therefore (by the Bourbaki result above) is projective.
This also illustrates why the most natural counterexamples come from zero-dimensional rings: the rank function is determined by its behavior on the minimal primes, so you might as well look at the zero-dimensional case.<|endoftext|>
TITLE: 1 rectangle <= 4 squares
QUESTION [41 upvotes]: Almost 25 years ago a professor at Indiana U showed me the following problem:
given a map $\mathbb{Z}^2\rightarrow\mathbb{R}$ such that the sum inside every square (parallel to the axes) is $\leq1$ in absolute value, prove that the sum inside every rectangle (parallel to axes) is $\leq4$ in absolute value.
It's fun and not too hard to prove.
I believe that at the time I was able to show that the upper limit can be improved to 3.975 - but that was a lot harder and I can't say now that this is for sure the case.  Also, with a computer search (old TRS 80) I produced an example containing a rectangle of area $3\frac{1}{3}$.
These are some of the questions that come to mind:

can the upper limit of 4 (or 3.975?) be improved?
can the lower limit of $3\frac{1}{3}$ be improved?
any proof/conjecture about the optimal limit?
do the results extend to maps $\mathbb{R}^2\rightarrow\mathbb{R}$, provided they are "nice" enough?
are any other generalizations of this problem possible (eg. different tilings of the plane or of other manifolds, or higher dimensions)?

 Update 1 (updated 7th March 2010). See answers and comments below  for examples achieving ratios as high as 181/48 = 3.7708333...!
 Update 2. Here is a sketch of the proof that 4 is an upper limit.
A limit of 254/67=3.79104477... is now known (see answers below), but the proof for that needs to be seeded with at least some known limit.
Given a rectangle R of size AxB, with A < B, call it "thin" if $B\geq2A$ or "fat" if $B\leq2A$ (the case B=2A is irrelevant as it is the union of 2 squares).
One can draw the 4 squares on the sides of R, either facing outwards (size of envelope = (2B+A)x(2A+B)), or inwards (some spilling out on the opposite sides, size of envelope = (2B-A)xB) - call these the "big-envelope" and the "small-envelope".
Assume that R has sum 4+$\epsilon$ and that every square has sum between -1 and 1. We have 3 cases, all easy exercises to work out:
(1) for any R, the fat (2A+B)x(2B+A) big-envelope will have sum $\leq-4-3\epsilon$.
(2) for a fat R, a (2A-B)x(2B-A) sub-rectangle of the small-envelope will have sum $\leq-4-3\epsilon$;
(3) for a thin R, a thin (B-2A)x(2B-A) sub-rectangle of the small-envelope, will have sum $\geq4+3\epsilon$;
Applying any of (1)+(2), (2)+(1) or (3)+(3) produces a 3Ax3B rectangle with
sum $\geq4+9\epsilon$.  Iterating n times produces a $3^{n}A \times 3^{n}B$ rectangle with sum $4+9^{n}\epsilon$.  Such rectangle is made of no more than AxB squares (each of size $3^{n} \times 3^{n}$) and therefore, for large enough n, one of the squares will have sum >1. $\square$
 Reformulation . Given an abelian group G and a map
f: GxGxGxG -> $\mathbb{R}$  such that 
1) -1<=f(a,b,c,d)<=1  if d*a=c*b
(boundedness of squares),
2) f(a,b,c,d)+f(c,b,e,d)=f(a,b,e,d)  for all a, b, c, d, e in G
(horizontal additivity of rectangles),
3) f(a,b,c,d)+f(a,d,c,e)=f(a,b,c,e)  for all a, b, c, d, e in G
(vertical additivity of rectangles),
can we find a universal best bound b(G) such that -b(G) <= f <= b(G)?
All the previous work on this question amounts to the result:
181/48 <= b($\mathbb{Z}$) <= b($\mathbb{Z}x\mathbb{Z}$) <= 254/67
For non-abelian groups one could perhaps generalize the notion of "square"
by lifting it from G/[G,G].

REPLY [3 votes]: There is a new upper bound of 254/67 (= 3.79104477...).
Define 6 sets of cardinality 4:
X1={-B+A, 0, A, B}
Y1={0, A, B-A, B}
X2={-B, -B+3A, B-2A, B+A}
Y2={-2B+A, -A, B+A, 3B-A}
X3={-6B+2A, -2B-2A, 2B+3A, 6B-A}
Y3={-2B-2A, -B+6A, 2B-6A, 3B+2A}
then we already know that in the in the grid
X1 x Y1
if the sum in the central
AxB is $4+\epsilon$
the sum in the surrounding
(2B-A)x(B-2A) is $\geq4+3\epsilon$,
similarly in the in the grid
(X1 $\cup$ X2) x (Y1 $\cup$ Y2)
if the sum in the central
AxB is $19/5+\epsilon$
then the sum in the surrounding
(2B-5A)x(5B-2A) is $\leq-19/5-21\epsilon$,
last, in the in the grid
(X1 $\cup$ X2 $\cup$ X3) x (Y1 $\cup$ Y2 $\cup$ Y3)
if the sum in the central
AxB is $254/67+\epsilon$
then the sum in the surrounding
(12B-3A)x(3B-12A) is $\geq254/67+135\epsilon$.
All of the above claims are easily verifiable with the tools already described in the previous answers and comments. I wonder if one can find sets X4 and Y4 (with 4 elements each?) to further improve the bound and maybe spot a general pattern.<|endoftext|>
TITLE: Real-analytic manifolds in real-analytic sets
QUESTION [6 upvotes]: Let $U\subset \mathbb{R}^n$ be open, and let $f:U\to\mathbb{R}$ be real-analytic. We consider the zero set $Z:=f^{-1}(\{0\})$. 
For a paper I am writing, I am looking for the best reference to the following basic fact:
If $Z$ has topological dimension equal to $d$, then $Z$ contains a real-analytic manifold of dimension $d$.

I can get this from Lojasiewicz's theorem or similar results, but that is a slightly unwieldy reference, and something probably needs to be said about how exactly one deduces it. Given that the statement is rather simple, I was wondering if someone knows of a more direct reference to this fact. 
And to add a mathematical question: This result is obviously much weaker than Lojasiewicz's theorem. Is there a proof that doesn't require developing the full structure theorem?
Many thanks for any pointers!

REPLY [3 votes]: By now, your paper is probably out, but if you need the result in the future, it can be found in:
B. Malgrange: Ideals of differential functions, Oxford University Press, 1966
In Russian translation (by A. Gabrielov, Mir 1968), this is Proposition VI.3.11:
Let $X_0$ be an analytic germ at $0 \in \mathbb{R}^n$, dim $X_0=k$. Assume that $X_0$ contains a germ $V_0$   of a variety of class $\mathcal{C}^\infty$ of dimension $k$. Then $V_0$ is a germ of an analytic variety (which is an irreducible component of the germ $X_0$).
The proof uses properties of ideals generated by germs of analytic functions. Lojasiewicz' s theorem is not used directly.<|endoftext|>
TITLE: Analogies between analogies
QUESTION [24 upvotes]: "A mathematician is a person who can find analogies between theorems; a better mathematician is one who can see analogies between proofs and the best mathematician can notice analogies between theories. One can imagine that the ultimate mathematician is one who can see analogies between analogies."
--Stefan Banach
See also here: Famous mathematical quotes
So, can someone give an example of an analogy between analogies?

REPLY [2 votes]: Quoting Final dialgebras : from categories to allegories, the authors write that, according to Freyd and Scedrov:

Allegories are to binary relations between sets as categories are to
  functions between sets.

Assuming functions between sets can be regarded as analogies between sets, and assuming that relations between sets can also be regarded as analogies between sets, then the above quote is expressing, in a pretty direct way, an analogy between analogies.<|endoftext|>
TITLE: What do intermediate Jacobians do?
QUESTION [14 upvotes]: On a smooth complex projective variety of $\dim X=n$, we have $n$ complex tori associated to it via $J^k(X)=F^kH^{2k-1}(X,\mathbb{C})/H_k(X,\mathbb{Z})$ (assuming I've got all the indices right) called the $k$th intermediate Jacobian.
If $k=1$, we have $J^1(X)=H^{1,0}/H_1$, and so $J^1(X)\cong Jac(X)$ is an abelian variety (the bilinear form is a polarization because it has to be definite on each piece of the Hodge decomposition (I think) ) and is in fact isomorphic as PPAV's to the Jacobian of the variety.
If $k=n$, we have $H^{2n-1,1}/H_{2n-1}$, which is also a PPAV, and is in fact the Albanese of $X$.
The ones in the middle, however, the "true" intermediate Jacobians, are generally only complex tori.  One example of an application is that Clemens and Griffiths proved that cubic threefolds are unirational but not rational using $J^2(X)$ for $X$ a cubic threefold.
So, what information do the intermediate Jacobians contain? I've been told that we don't really know much about that, but what is known, beyond Clemens/Griffiths?

REPLY [11 votes]: Recently I learned from a talk of Nick Addington one beautiful  classical example where intermediate Jacobians contain all information about the variety. Namely, if we consider an intersection of two quadrics in $\mathbb CP^n$ then to such a variety we can associate a hyperelliptic curve.  To do this we consider the corresponding pencil of quadrics. Singular quadrics in the pencil define a finite number of points on $CP^1$ and we take the double cover of $CP^1$ with the ramification at these points. The Jacobian of the hyperelliptic curve is isomorphic to the intermediate Jacobian of the intersection of to quadrics (this is the PhD of Miles Reid). The consruction was first considered by Weil. All this is very well explained in the introduction of the article of Nick http://arxiv.org/abs/0904.1764. 
You can reconstruct the hyperelliptic curve from its Jacobian and you can reconstructs the intersection of two quadrics from the curve.
Intermediate Jacobians of the intersection of 3 quadrics were considered by Turin. He proved the corresponding Torelli theorem: On intersections of quadrics. Russ. Math. Surv., 30, 51–105, 1975.<|endoftext|>
TITLE: How to quickly determine whether a given natural number is a power of another natural number?
QUESTION [16 upvotes]: We have a natural number $n>1$. We want to determine whether there exist
natural numbers $a, k>1$ such that $n = a^k$. 
Please suggest a polynomial-time algorithm.

REPLY [4 votes]: The computer algebra system GAP performs this test and determines a
smallest root $a$ of a given integer $n$ quite efficiently.
The following is copied directly from its source code (file gap4r6/lib/integer.gi),
and should be self-explaining:
#############################################################################
##
#F  SmallestRootInt( <n> )  . . . . . . . . . . . smallest root of an integer
##
InstallGlobalFunction(SmallestRootInt,

  function ( n )

    local   k, r, s, p, l, q;

    # check the argument
    if   n > 0  then k := 2;  s :=  1;
    elif n < 0  then k := 3;  s := -1;  n := -n;
    else return 0;
    fi;

    # exclude small divisors, and thereby large exponents
    if n mod 2 = 0  then
        p := 2;
    else
        p := 3;  while p < 100  and n mod p <> 0  do p := p+2;  od;
    fi;
    l := LogInt( n, p );

    # loop over the possible prime divisors of exponents
    # use Euler's criterion to cast out impossible ones
    while k <= l  do
        q := 2*k+1;  while not IsPrimeInt(q)  do q := q+2*k;  od;
        if PowerModInt( n, (q-1)/k, q ) <= 1  then
            r := RootInt( n, k );
            if r ^ k = n  then
                n := r;
                l := QuoInt( l, k );
            else
                k := NextPrimeInt( k );
            fi;
        else
            k := NextPrimeInt( k );
        fi;
    od;

    return s * n;
end);<|endoftext|>
TITLE: Heaviest Convex Polygon
QUESTION [5 upvotes]: Suppose we have an arbitrary function $f : \mathbb{R}^2 \to \mathbb{R}$.  For any subset $s \subseteq \mathbb{R}^2$, we can define $g_f(s)$ as the integral* of $f$ over the region $s$.  Suppose further that we have access to an oracle that will tell us the value of $g_f(s)$ for any $s$.
Now, restrict our attention to subsets of $s$ that are the convex hull of a given subset of points $\bar x_c \subseteq \{x_1, \ldots, x_N \}$ with $x_i \in \mathbb{R}^2$.  Assuming calls to the oracle are O(1), what is the complexity (in terms of $N$) of finding $\bar x_c^* = \arg \max_{\bar x_c} g_f(conv(\bar x_c))$?  Is there a known algorithm or reduction to a known problem?
EDIT: *Previous statement that Scott answered said "average value" here.

REPLY [6 votes]: I'm assuming the N points are fixed ahead of time.  In that case, it seems to me that you can just use the oracle on each triple of points, since any convex polygon with more than three sides will have average at most the maximum of the averages over triangles in any triangulation.  This gives you O(N^3) at worst.<|endoftext|>
TITLE: Research Experience for Undergraduates: Summer Programs
QUESTION [18 upvotes]: Some time ago, I found this list of REU programs held in 2009. 
The main aspects that characterize such programs are: (a) a great deal of lectures on specific topics; and, admittedly more importantly, (b) the chance to gain some hands-on experience with research projects.
I think that these programs are extremely interesting and are precious opportunities for undergraduates to gain a deeper understanding of specific mathematical topics as well as of the "work of the mathematical researcher".  
One should note, however, that most of these programs (if not all of them) are not open to European citizens (or, at least, in general non-American applicants do not receive funding).

Q: So, I would really like to hear if you know any similar programs. More
  specifically, I would like to know there are any such programs outside the U.S. (or any programs in the U.S. that accept also non-American applicants).


Remark 1: A similar question was asked on Mathematics Stack Exchange.
Remark 2: Both questions have been updated in 2015. It would be nice to receive some answers which are up-to-date.

REPLY [2 votes]: The Fields institute in Toronto offers summer research for undergraduates
http://www.fields.utoronto.ca/programs/scientific/15-16/summer-research15/
When I was there the topics were mostly linked to the thematic program at the time and there were at least a couple of students from abroad (from Hungary in that case).<|endoftext|>
TITLE: The inverse Galois problem and the Monster
QUESTION [72 upvotes]: I have a slight interest in both the inverse Galois problem and in the Monster group.  I learned some time ago that all of the sporadic simple groups, with the exception of the Mathieu group $M_{23}$, have been proven to be Galois groups over $\mathbb{Q}$.
In particular, the Monster group has been proven to be a Galois group over $\mathbb{Q}$.  What techniques are used to prove such an assertion?
Is proving that $M_{23}$ is also Galois over $\mathbb{Q}$ within reach?  I assume that the same techniques do not apply, for it is a much more manageable group than the Monster.

REPLY [21 votes]: I'll add a brief commment to Arne Semeets's thorough and useful answer.  If I fix three rational conjugacy classes c_0, c_1, c_infty in a finite group G, then there are finitely many isomorphism classes of unramified G-coverings
X -> P^1 - 0,1,infty /Qbar
with the property that the image of tame inertia at 0 (resp 1,infty) lies in the class c_0 (resp c_1,c_infty) of G.
Call the set of such covers H.  What is |H|?  One can check (by comparison with the complex case) that the number of such covers is the number of conjugacy classes of triples (g_0,g_1,g_infty) with g_i in c_i and g_0 g_1 g_infty = 1.  To say that (c_0,c_1,c_infty) is rigid is just to say that there is precisely ONE such triple.
In that case, H consists of just one cover.  But H is evidently preserved by Galois conjugacy.  So this unique cover is defined over Q.  (One has to be slightly more careful when G has nontrivial center, in which case what I've really proved is something more like "there's a cover whose isomorphism class is defined over Q," not quite the same in general as "there's a cover defined over Q."<|endoftext|>
TITLE: References on functorially-defined subgroups
QUESTION [7 upvotes]: I'm interested in results about functorially-defined subgroups (in a loose sense), especially in the non-abelian case, and would like to know about references I may have missed.
The question, it seems, comes up in its simplest form when noticing a number of common subgroups (the center, commutator subgroup, Frattini subgroup, etc) are characteristic. The characteristicity can be justified by the fact that the object mappings that define of those subgroups give rise to subfunctors of the identity functor, on the core category of Grp.
Hence I'm interested in functors F in Grp (or a carefully chosen sub-category) such that 
∀ A F(A) ⊆ A and ∀ A,B ∀ f ∈ hom(A, B), f(F(A)) ⊆ F(B)
Does that ring a bell ?
The topic was mentioned a couple of months ago on Mathoverflow, and can be traced back to (at least) 1945, where Saunders MacLane explains it in some detail in the third chapter of  A General Theory of Natural Equivalences.
In between, it seems that those functors have been baptised radicals, pre-radicals or subgroup functorials, and studied mostly in the framework of ring theory, notably by A. Kurosh. Among a number of not-so-recent (and therefore quite-hard-to-find) papers mostly dealing with rings, semigroups, or abelian groups, I came across a single reference mentioning the non-abelian case, by B.I. Plotkin : Radicals in groups, operations on classes of groups, and radical classes. Connections seem have been made with closure operators¹, but do not focus much on Grp.

Do you have ideas of connections from those functors to other parts of algebra or category theory, other than (pre-)radicals ?
Do you have some pointers to material I may have missed, specially if they mention non-abelian groups ?

¹: Categorical structure of closure operators with applications to topology 
 By N. Dikranjan, Walter Tholen, p.51

REPLY [3 votes]: After some research, I asked on the Group-pub mailing list, where very knowledgeable people roam (the University of Bath, which hosts the mailing-list, also hosted the 'Groups St Andrews' conference in 2009). Jan Krempa pointed me to a recent seminar on radicals, that included a very useful survey paper:
B. J. Gardner, Kurosh-Amitsur radicals of groups: something for overyone?
Its lists of references is a goldmine, particularly the included book by the same author (Radical Theory). It includes general radical theory that applies to the non-commutative case better than his later book (written with R. Wiegandt :Radical theory of rings).
I also got a nice answer from Mike Newman (presumptively this one), who told me:

** An early interest occurs in
MR0002876 (2,125i)
  Hall, P.
  Verbal and marginal subgroups.
  J. Reine Angew. Math. 182, (1940). 156--157.
This is a brief report of a lecture given at a meeting in Goettingen
  in 1939.
** There is recent work which hinges on these sorts of ideas in
MR2276769 (2008f:20052)
  Nikolov, Nikolay(4-OXNC); Segal, Dan(4-OXAS)
  On finitely generated profinite groups. I. Strong completeness and uniform bounds. (English summary)
  Ann. of Math. (2) 165 (2007), no. 1, 171--238.
  20E18 (20E32 20F12)

The link with verbal and marginal subgroups was already hinted at in the previous mathoverflow question I mentioned.<|endoftext|>
TITLE: Proper definition of a moduli problem
QUESTION [5 upvotes]: This question arose after I thought about Ben Webster's comments to this question.
There he asked me what was my definition of a moduli problem. When I came to think of it, I never saw a precise definition like that. My understanding is along the following lines. 
Roughly, say, we want to describe the moduli problem classifying objects of a certain type. In the functorial formulation, we would have a functor
$$Schemes \rightarrow Sets  $$
$$ X \mapsto \{ Iso.\ Classes\ of\ some\ objects\ of\ a\ certain\ type\ defined\ over\ X.\}$$
And if this functor is representable, we say that a fine moduli space exists for this moduli problem, and even if it is not, if a certain one-one correspondence between points and objects is true over algebraically closed fields, and if a certain universal property for this is satisfied, then a coarse moduli space exists.
I suppose the above is the agreed standard terminology. Please correct me if I am wrong.
Now, the problem is that in the above definition of a moduli problem, the notion of a "functor classifying isomorphism classes of a  certain type of object" is vague. We could have curves with marked points, other types of varieties with extra conditions, bundles, and so on. If we on the other hand relax the criteria and allow just any functor, then the definition becomes too broad,and any scheme will be a fine moduli space for its functor of points. 
So is there a better definition, or is this all one can say? Pardon me if this was a stupid question.

REPLY [7 votes]: Since not many people have had anything to say, I thought I might make a few remarks. But beware that this is all what I've passively picked up over the years---it's not the result of an actual study of things.
I think the right definition of a moduli problem is a fibered category $p:E\to B$. This should be thought of as a family of categories parametrized by $B$, just like you think of a map $X \to S$ of spaces as being a family of spaces parametrized by $S$. Here is an example: $E$ is the category whose objects are maps of schemes $X\to S$ making $X$ a family of elliptic curves over $S$ (i.e. an abelian scheme of relative dimension 1), and whose maps are the (hopefully) evident cartesian squares; $B$ is the category of schemes, and the functor $E\to B$ sends an object $X\to S$ to $S$. The fibered structure is given by pull back: given $X\to S$ in $E$ and a map $S'\to S$ in $B$, we get the object $X\times_S S'\to S'$ (which maps under $p$ to $S'$). If we let $E_S$ denote the fiber of this fibered category over an object $S$, then in this example, $E_S$ is the category of families of elliptic curves parametrized by $S$. Thus the fibered category encodes the data of all possible families of elliptic curves and how they behave under base change. So I hope my point that the fibered category is the moduli problem seems reasonable.
You then say the fibered category is representable if there is an object $U$ of $E$ such that for any $S$ in $B$, the pull back functor from the discrete category consisting of the set $\mathrm{Hom}(S,p(U))$ to the category $E_S$ is an equivalence. This is pretty unlikely. For instance, it implies that each category $E_S$ is discrete---all maps are isomorphisms and all automorphisms are the identity. This is certainly not the case with the elliptic curve example. Every elliptic curve has a nontrivial automorphism given by the inverse map with respect to the group structure.
Another version of representability of a fibered category is the following. Let $F:B\to\mathrm{Sets}$ denote the functor which sends an object $S$ to the set of isomorphism classes of objects of $E_S$. Then the moduli problem is (weakly?) representable if the functor $F$ is representable. This definition is surely weaker in general than the one above, but it is often equivalent in the examples that people look at in algebraic geometry. For example, the two are probably equivalent if each $E_S$ is a discrete category.
If the moduli problem is not representable, then you get into other issues, such as whether you have effective descent with respect to some topology on $B$ (i.e. $E$ is a "sheaf of categories" over $B$), and if so, stack-theoretic issues, such as whether $E$ can be represented by a category object in $E$, and if so, whether it's a groupoid object.<|endoftext|>
TITLE: Equivalence relations on permutations and pattern avoidance
QUESTION [13 upvotes]: I'm working on the interaction between equivalence relations on permutations and pattern avoidance. I've only considered Knuth equivalence and cyclic shifts until now and I'm looking for other equivalence relations to test some conjectures on. So my question is:

What interesting equivalence relations on permutations are there?

Background/Motivation
By a permutation of $n$ I mean a bijection $\lbrace 1,2,\dots,n\rbrace \to \lbrace 1,2,\dots,n \rbrace$. The set of all permutations of $n$ will be denoted $S_n$. I'll use the one-line notation for permutations, e.g., $132$ means the permutation $1\to1$, $2\to3$, $3\to2$.
A pattern will also be a permutation, but we are interested in how patterns occur in permutations. E.g., the pattern $132$ occurs in the permutation $215314$ as the letters
$2$, $5$, $3$, because these have the same relative order as the pattern. Note that the letters do not have to be adjacent in the permutation. When a pattern does not occur in a permutation we say that the permutation avoids that pattern.
Now, fix an equivalence relation on $S_n$. For any pattern $p$ we define two subsets of $S_n$:
$X_n(p) = \lbrace \sigma \in S_n \phantom{i}|\phantom{i} \sigma
\text{ and every equivalent permutation avoids } p \rbrace$
$Y_n(p) = \lbrace \sigma \in S_n \phantom{i}|\phantom{i} \sigma
\text{ avoids } p \text{ and every equivalent pattern}\rbrace$
What I am mostly interested in is how these two sets are related.
Example: Assume our equivalence relation is cyclic shifts, meaning that two permutations (or patterns) $p$, $q$ are equivalent if we can write $p = r_1 * r_2$ and $q = r_2 * r_1$ where $*$ means concatenation. E.g., $1234$ is equivalent to $3412$. Here the two sets $X_n(p)$ and $Y_n(p)$ are equal for any $p$. (This is also true if we generalize our patterns to bivincular patterns, whose definition I'll omit here)
[Note considering permutations up to cyclic shifts is equivalent to looking at circular permutations.]
Example: If our equivalence relation is Knuth equivalence, meaning that two permutations (or patterns) are equivalent if they have the same insertion tableaux under the RSK-correspondence, then the two sets are not always the same. They are the same for any pattern of length $3$, but $X_4(1324) \neq Y_4(1324)$.
When the two sets are equal then counting $Y_n(p)$ can sometimes be done more easily by looking at $X_n(p)$. When the relation is Knuth equivalence one can use the hook-length formula for instance.
So I would very happy if someone could tell me about other interesting equivalence relations on permutations so I could see how $X_n(p)$ and $Y_n(p)$ are related in other cases.

REPLY [4 votes]: Over the past year Steve Linton, Tom Roby, Julian West and I have looked at equivalence relations on permutations arising from "constrained swapping" or more generally "pattern replacement" (Knuth equivalence is a special case).  If http://julianwest.ca/swopp.htm looks interesting I can tell you more.<|endoftext|>
TITLE: Explicit description of a fibered category
QUESTION [8 upvotes]: I found the following exercise in Vistoli's notes. He proves a theorem (Theorem $3.45$, page number $64$) stating that any category $\mathcal{F}$ fibered over $\mathcal{C}$ is equivalent, as a fibered category, to a split one. Namely $\mathcal{F}$ is equivalent to the category $\mathcal{F}' = Hom_\mathcal{C}(\cdot, \mathcal{F})$, which is the fibered category associated to the following functor $F : \mathcal{C}^{op} \rightarrow Cat$.
For every $U \in \mathcal{C}$ we set $F(U) = Hom_\mathcal{C}(\mathcal{C}/U, \mathcal{F})$, where $\mathcal{C}/U$ is the comma category and $Hom_\mathcal{C}$ denotes the category of morphism of fibered categories. (In particular an arrow in this category is a morphism of functors over the identity of $\mathcal{C}$). The action of $F$ on arrows is the obvious one: an arrow $U \rightarrow V$ in $\mathcal{C}$ gives a functor $\mathcal{C}/U \rightarrow \mathcal{C}/V$, and $F$ acts by composition with this functor.
The exercise requires to carry out the construction explicitly for the following situation. A group $G$ can be seen as a category with a single object. If $G \rightarrow H$ is a surjective homomorphism, then we can see $G$ as a category fibered over $H$, and the exercise is to work out what $\mathcal{F}'$ is in this case.
I am able to do this exercise, but I think I am missing something. Vistoli says that this is a nice exercise, so I guess I should obtain as a result something which I can recognize, but I don't. If needed I can post here my answer, but it is not very enlightening.
I was tempted to write here the relevant terminology, but it is pointless, as everything is clearly defined in chapter 3 of the above mentioned notes. If you need any clarification, I'll be happy to provide more details.

REPLY [4 votes]: This construction may not be the most natural (or general) one, but I find it reasonably enlightening.
Let $\mathcal F$ and $\mathcal C$ denote the categories with one object associated to $G$ and $H$, respectively.  Notice that if $\mathcal F'$ is any category equivalent to $\mathcal F$, then in particular it admits a fully faithful functor to $\mathcal F$.  Since $\mathcal F$ has only one object, with isomorphisms in bijection with $G$, this implies that every hom-set in $\mathcal F'$ must be in bijection with $G$ as well.  It's easy to check that every morphism in $\mathcal F'$ must be an isomorphism, so this proves that $\mathcal F'$ is a groupoid, with exactly $|G|$ isomorphisms between any two objects.
I claim that we can choose $\mathcal F'$ to have objects indexed by $H$.  To be explicit, let's say that the morphisms between any two objects $h_1, h_2$ are identified with $G$, and that the composition of $g_1:  h_1 \to h_2$ and $g_2:  h_2 \to h_3$ is $g_1 g_2:  h_1 \to h_3$.  Then this admits a natural "projection" functor to $\mathcal F$, by sending every object to the unique object $*$ of $\mathcal F$ and sending each morphism to the morphism of the same name.  We get a functor in the other direction by sending $*$ to the object labeled by the identity of $H$, and preserving names of morphisms.  The composition $\mathcal F \to \mathcal F' \to \mathcal F$ is literally the identity functor, and $\mathcal F' \to \mathcal F \to \mathcal F'$ is easily seen to be naturally isomorphic to the identity functor via a base-preserving natural transformation.  So $\mathcal F'$ and $\mathcal F$ are equivalent fibered categories over $\mathcal C$.
Now let's construct a splitting of $\mathcal F' \to \mathcal C$.  Fix a representative $\widetilde h \in G$ for each element $h \in H$.  Consider the subcategory of $\mathcal F'$ that includes all objects, but only the morphisms of the form $\widetilde h_1 \widetilde h_2^{-1}:  h_1 \to h_2$.  (Note that this does contain identities and compositions.)  For any given morphism $h$ in $\mathcal C$ and any object $h_2 \in \mathcal F'$, our chosen subcategory contains a unique pullback $h_1 \to h_2$ of $h$, namely the morphism $\widetilde h_1 \widetilde h_2^{-1}:  h_1 \to h_2$ with $h_1$ chosen so that $h_1 h_2^{-1} = h$.
To make this more concrete, let's look at the simplest possible non-split group extension:  $\mathbb Z/4\mathbb Z \twoheadrightarrow \mathbb Z/2\mathbb Z$.  Here, the category $\mathcal F'$ has two objects, with four morphisms between any pair, all of them isomorphisms.  This category deserves to be equivalent to $\mathcal F$:  it has two objects, which both look exactly like the object of $\mathcal F$ and are isomorphic to each other.  Make $\mathcal F'$ into a fibered category over $\mathcal C$ by composing the "projection" functor to $\mathcal F$ with the given functor $\mathcal F \to \mathcal C$.  Recall that we can't construct a splitting of the original fibered category $\mathcal F \to \mathcal C$ precisely because we would need to choose a lift of the morphism $1 \in \mathbb Z/2\mathbb Z$ to $\mathbb Z/4\mathbb Z$, and neither of the two choices gives something that respects composition.  But in our new fibered category $\mathcal F'$, we need to choose a lift of $1 \in \mathbb Z/2\mathbb Z$ to some morphism between the two objects of $\mathcal F'$, instead of an automorphism of one of the objects.  So we don't need to worry about composing the lift with itself, and the problem is avoided.<|endoftext|>
TITLE: Closed vs Rational Points on Schemes
QUESTION [20 upvotes]: Background: When Ueno builds the fully faithful functor from Var/k to Sch/k he mentions that the variety $V$ can be identified with the rational points of $t(V)$ over $k$. I know how to prove this on affine everything and will work out the general case at some future time.
The question that this got me thinking about was if $X$ is a $k$-scheme where $k$ is algebraically closed, then are the $k$-rational points of $X$ just the closed points? This is probably extremely well known, but I can't find it explicitly stated nor can I find a counterexample.
For $k$ not algebraically closed, I can come up with examples where this is not true. So in general is there some relation between the closed points and rational points on schemes (everything over $k$)?
This would give a bit more insight into what this functor does. It takes the variety and makes all the points into closed points of a scheme, then adds the generic points necessary to actually make it a legitimate scheme. General tangential thoughts on this are welcome as well.

REPLY [15 votes]: The following result deals with the case of finite type affine schemes over an arbitrary field $k$.
Theorem: Let $A$ be a finitely generated algebra over a field $k$.  Let $\iota: A \rightarrow \overline{A} = A \otimes_k \overline{k}$.
a) For every maximal ideal $\mathfrak{m}$ of $A$, the set $\mathcal{M}(\mathfrak{m})$ of 
maximal ideals $\mathcal{M}$ of $\overline{A}$ lying over $\mathfrak{m}$ is finite and 
nonempty.
b) The natural action of $G = \operatorname{Aut}(\overline{k}/k)$ on $\mathcal{M}(\mathfrak{m})$ is transitive.  Thus $\operatorname{MaxSpec}(A) = G \backslash 
\operatorname{MaxSpec}(\overline{A})$.
c) If $k$ is perfect, the size of the $G$-orbit on $\mathfrak{m} \in \operatorname{MaxSpec}(A)$ is equal to the degree of the field extension of $k$ generated by 
the coordinates in $\overline{k}^n$ of any $\mathcal{M}$ lying over $\mathfrak{m}$.  
In brief, the closed points correspond to the Galois orbits of the geometric points.  
This is Theorem 8 in http://www.math.uga.edu/~pete/8320notes3.pdf.
The proof is left as an exercise, with some suggestions.
Exactly where this result came from, I cannot now remember.  The text for the course that these notes accompany was Qing Liu's Algebraic Geometry and Arithmetic Curves (+1!), so it's a good shot that there is at least some cognate result in there.<|endoftext|>
TITLE: What are some famous rejections of correct mathematics?
QUESTION [133 upvotes]: Dick Lipton has a blog post that motivated this question. He recalled the Stark-Heegner

Theorem: There are only a finite
number of imaginary quadratic fields
that have unique factorization. They
are $\sqrt{d}$ for $d \in \{-1,-2,-3,-7,-11,-19,-43,-67,-163 \}$.

From Wikipedia (link in the theorem statement above):

It was essentially proven by Kurt Heegner in 1952, but Heegner's proof had some minor gaps and the theorem was not accepted until Harold Stark gave a complete proof in 1967, which Stark showed was actually equivalent to Heegner's. Heegner "died before anyone really understood what he had done".

I am also reminded of Grassmann's inability to get his work recognized.
What are some other examples of important correct work being rejected by the community?
NB. There was a complementary question here before.

REPLY [20 votes]: René Schoof once told me that when he submitted his PhD Thesis, the chapter containing his algorithm to compute the number of points of elliptic curves over finite fields did not appeal at all to the referee, who wondered whether such questions had some interest at all... History decided otherwise!<|endoftext|>
TITLE: Uses of the holomorph, Hol($G$) = $G \rtimes $ Aut($G$)
QUESTION [30 upvotes]: In every group theory textbook I've read, the holomorph has been defined, and maybe a few problems done with it. I've also seen papers focusing on computing Hol($G$) for a specific class of $G$.
One thing I have never seen is any actual use for it. Are there major results using the holomorph of a group? Does it occur in the proof of any useful theorems? 
It seems intrisically interesting to me since it allows you to treat automorphisms of a group and elements of a group uniformly, and I would definitely like to learn more about it.

REPLY [6 votes]: Another extensive use of the holomorph is in the study of crossed modules /  2-groups (in the categorification sense).  Any group yields a one object groupoid. Any groupoid $G$ yields an endomorphism gadget $G^G$, that is the category of functors from $G$ to itself. This is at one and the same time a groupoid and a monoid (under composition of functors). Now look at the functors that are isomorphisms. That gives you a category which is also a group.  (Yes I do mean that! It is the subgroup of the monoid structure.) That makes it a 2-group in that categorified sense.  Any 2-group yield a crossed module and the crossed module here is just the inner automorphism homomorphism $G\to Aut(G)$.  The 2-group is the holomorph!
This basic construction is central to many treatments of non-Abelian cohomology (of groups, and of sheaves of groups.) See Larry Breen's papers, and more recently papers by Aldrovandi and Noohi. (You can find stuff on this in the n-Lab and also in my Menagerie notes, a version of which is on my n-Lab homepage.) It is also central to attempts to implement Grothendieck's Pursuing Stack programme although that needs an enormous amount more input as well!
(I can give detailed references if they are of interest.)<|endoftext|>
TITLE: Signed and unsigned Hecke algebra canonical basis
QUESTION [6 upvotes]: Consider the Hecke algebra $H_n$ of type $A_{n-1}$ with standard basis $T_w$, $w \in S_n$ with the quadratic relations $(T_s - u) (T_s + u^{-1}) = 0$ and braid relations. The unsigned canonical basis $C'_w$, $w \in S_n$ gives rise to a basis for the irreducible $H_n$-module $M_\lambda$ of shape $\lambda$: fix some SYT $T$ of shape $\lambda$; then this basis is {$C'_w : P(w) = T$} after quotienting by cells lower down in the Kazhdan-Lusztig preorder and $P(w)$ is the insertion tableau of $w$.
Similarly, $M_\lambda$ has a basis coming from the signed canonical basis $C_w$, $w \in S_n$.

What is known about the transition matrix between these two bases? Does it become the identity matrix at $u=0$?  

This seems to be trickier to understand than the transition matrix between all the $C$s and all the $C'$s. I expect I will be able to prove the second question on my own, but I'd rather cite it if it's in the literature somewhere.

REPLY [2 votes]: Not much seems to be known about this matrix in general, and it does become the identity at $u=0$.  I show this in the paper Quantum Schur-Weyl duality and projected canonical bases http://arxiv.org/abs/1102.1453 
using quantum Schur-Weyl duality and its compatibility with canonical bases.  This only proves it in type A.  I do not know what happens in other types.<|endoftext|>
TITLE: What is the Alexander polynomial of a point?
QUESTION [8 upvotes]: According to the Baez-Dolan cobordism hypothesis, an extended TQFT is determined by its value on a single point. This value a fully dualizable object of a symmetric monoidal $n$ category (a fully dualizable object is a higher categorical analogue of a finite dimensional vector space). The Alexander polynomial is a quantum invariant, and comes from a TQFT. 

How can an "Alexander polynomial" TQFT be put into an extended TQFT, and what is its value at a single point?
The question I just asked is closely related to this question. I also asked the question on the ldtopology blog here, and Theo Johnson-Freyd suggested that MO might be the place to ask it.

Briefly, I will summarize what an extended TQFT is. A TQFT as a symmetric monoidal functor Z:Cob(n)-> Vect(k) from the tensor category of $n-1$ dimensional manifolds and cobordisms between them to the tensor category of vector spaces over a field k. An extended TQFT is a symmetric monoidal functor Z:Cobk(n)->C from the n-category of cobordisms to a symmetric monoidal n-category C. I vote for the introduction to Lurie's expository account of his work proving the Baez-Dolan cobordism hypothesis as the best place to read about why extended TQFT's are natural objects, to understand their motivation, and to understand why people are so excited about them. An extended TQFT assigns a fully dualizable object to a point, and a higher “trace” on this object to a closed n-dimensional manifolds.

REPLY [5 votes]: I agree with Noah, and I disagree.  Recent investigations into the notion of extended TQFT have led to a very rigid notion of what TQFT is.  So rigid in fact, that almost nothing is a TQFT anymore, so
the answer is as above: There is no answer. 
However, maybe you could restate the question so that it does have an answer.

Turaev torsion acts somewhat like a TQFT, and somewhat like the Alexander invariant. There are papers where people work out the Turaev torsion
of manifolds with boundary and the associated gluing rules. Can you extend down to get an invariant that looks like a 2-extended TQFT?  If so what is the category assigned to a one manifold.  Can you build down to an invariant that satisfies some axioms that you could claim correspond to 3-extended
TQFT?  If so what are you assigning a to a 0-manifold?  
Heegaard Floer homology acts like the Alexander invariant, in fact there is a version that
categorifies Turaev torsion.  There are now bordered versions of Heegaard Floer.  Can you keep
extending these theories down? If so what do they assign to lower dimensional manifolds?<|endoftext|>
TITLE: Stacks and sheaves
QUESTION [23 upvotes]: I'm a bit confused by the double role which sheaves play in the theory of stacks.
On the one hand, sheaves on a site are the obvious generalization of a sheaf on a topological space. On the other hand a sheaf on a site is (or better its associated category fibered in sets is) a very particular stack itself, so a generalization of a space. This is not completely confusing: more or less it amounts (I believe) to identifying a space X with the sheaf of continuos functions with values in X.
But now my question is the following. An equivalent condition for a fibered category to be a prestack is that for any two objects (over the same base object), the associated functor of arrows should be a sheaf. In particular this is true for a stack, so for any stack and any two objects in it we have a sheaf, and so a stack (over a comma category).
What is the meaning of this geometrically?
For instance take the stack $\mathcal{M}_{g,n}$.
Giving two objects in the stack (over the same base object) means giving two families $X$ and $Y$ of stable pointed curves over the same scheme $S$, and the associated functor of arrows maps every other scheme $f \colon T \rightarrow S$ to the set of morphism between $f^* X$ and $f^* Y$. How should I think of the associated stack as a space?
To avoid misunderstandings I give the defition of the functor of arrows. Let $\mathcal{F}$ be a fibered category over $\mathcal{C}$. Take $U \in \mathcal{C}$ and $\xi, \eta \in \mathcal{F}(U)$. Then there is a functor $F \colon \mathcal{C}/U \rightarrow Set$ defined as follows. For a map $f \colon T \rightarrow U$ we put $F(f) = Hom(f^* \xi, f^* \eta)$. The action on arrows requires some diagrams to be described, but it's really the only possible one.

REPLY [2 votes]: As far as the first part of your question goes, I have exactly the same confusion (and it will probably get worse once I start thinking about sheaves on stacks...).
There are two things that give me the illusion of some comprehension.

Thinking of sheaves as (generalized) spaces is probably not terrible, in the same way you think of a bundle just as its total space. It is indeed true that any sheaf (with values in a reasonable category I guess) is the sheaf of sections of its etale space. (Although I must confess I don't particularly like etale spaces and I'm aware that this is probably not the right picture, as we shouldn't think of sheaves on a site in this fashion, but hey).
If we think of fibred categories (+cleavage) as presheaves with values in the 2-category Cat, then there is a generalization of the sheaf condition which yields the notion of a stack. Of course, arrows being a sheaf is a consequence of this, so if you find the generalization of the sheaf condition more natural, then arrows-being-a-sheaf might be viewed as a formal consequence I guess.

Anyway it's just a thought.<|endoftext|>
TITLE: Computing the structure of the group completion of an abelian monoid, how hard can it be?
QUESTION [11 upvotes]: Cherry Kearton, Bayer-Fluckiger and others have results that say the monoid of isotopy classes of smooth oriented embeddings of $S^n$ in $S^{n+2}$ is not a free commutative monoid provided $n \geq 3$.   The monoid structure I'm referring to is the connect sum of knots.
Bayer-Fluckiger has a result in particular that says you can satisfy these equations $$a+b=a+c, \ \ \ \  b \neq c$$ 
where $a,b,c$ are isotopy classes of knots and $+$ is connect sum.
When $n=1$ it's an old result of Horst Schubert's that the monoid of knots is free commutative on countably-infinite many generators. 
What I'm wondering is, does anyone have an idea of how difficult it might be to compute the structure of the group completion of the monoid of knots, say, for $n \geq 3$?  That's not really my question for the forum, though.  
It's this: Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious?  Meaning, one where rather minimal amounts of information are required to compute the group completion?  Presumably there are examples where it's painfully difficult to say anything about the group completion?  For example, can it be hard to say if there's torsion in the group completion?

REPLY [10 votes]: An ultra-classical example: the failure of unique factorization in algebraic number fields. Here one looks at the multiplicative monoid of nonzero algebraic integers in a finite extension field of $\mathbb Q$. Factoring out units gives a quotient monoid $M$, and this is free (abelian) on the irreducible elements exactly when the unique factorization property holds. The monoid $M$ embeds in the ideal group, the free abelian group generated by the prime ideals, and the group completion of $M$ is the finite-index subgroup of the ideal group generated by the principal ideals. This subgroup is also free, but without a canonical basis. One can think of this subgroup as a somewhat skewed lattice in the ideal group, and $M$ is the intersection of this lattice with the positive "orthant" of the ideal group. 
I'm not an expert on this stuff, so please correct any inaccuracies in what I said above. I gather that the structure of $M$ as a monoid can be rather complicated, even though it's a submonoid of a free monoid.  Perhaps this complexity is why the monoid structure seems rarely to be discussed explicitly. Does anyone know any references describing the monoid structure?<|endoftext|>
TITLE: Ordered geometries from convex subsets of the plane
QUESTION [10 upvotes]: Motivation
In the Klein disk model of the hyperbolic plane, the points are the interior of the disk, and the lines in $H^2$ correspond to lines intersecting the interior.

Similarly, the Euclidean plane can be modeled by the interior of a hemisphere of $S^2$ (or $\mathbb {RP}^2$ minus a line) so that lines in $\mathbb R^2$ are the intersections of geodesics of the sphere with the hemisphere. 
In both cases, the angles aren't preserved, but the orderings of points on lines are preserved.
Definitions
For any open convex set in $\mathbb R^2$, consider the nonempty intersections of lines with the set as lines in an ordered geometry with the induced ordering from $\mathbb R^2$. Two such geometries are equivalent if there is a bijection between them preserving lines and the ordering on each line.


Which convex open sets produce
  geometries equivalent to $H^2$? 
Which pairs of convex open sets
  produce equivalent geometries?


Either line-segment preserving maps are quite flexible, or else there should be ways to recover much of the information about convex sets from their incidence geometries.
Some weak results
You can distinguish the interior of a triangle from $H^2$ (or any other bounded convex open set) through the incidence relations. In the triangle, there are three lines such that every other line intersects at least one of the three. Any three lines through the vertices work. In $H^2$, you can always find a line disjoint from any finite collection of lines. 
Similarly, if a line segment makes up part of the boundary of a set in the plane, then the incidence geometry is not $H^2$.
The incidence relation plus ordering is enough to construct ideal points of the boundary of the set. These correspond to maximal sets of rays so that for any two disjoint rays $R_1$ and $R_2$ in the set, the set of points $p$ so that for some $x_1 \in R_1$ and $x_2 \in R_2$, $p$ is between $x_1$ and $x_2$, is a triangle subgeometry.

REPLY [7 votes]: Convex sets give the same geometries if and only if they are projectively equivalent. In particular, it is only conics that give $H^2$.
It is more natural to work in the projective plane $P^2(\mathbb{R})$. Then we define a set to be convex if its intersection with any line is empty or connected. We are given two open convex sets $C_1$ and $C_2$ in the plane with a bijection $\phi:C_1\to C_2$ which is order preserving in the sense that $a$ and $b$ separates $c$ and $d$ (on some projective line) if and only if $\phi(a)$ and $\phi(b)$ separates $\phi(c)$ and $\phi(d)$ (on some projective line).
We would like to prove that $\phi$ is a projective transformation.
The following attempt uses the theorem of Desargues together with the fundamental theorem of projective geometry.
Claim. $\phi$ can be extended to the whole of $P^2(\mathbb{R})$ such that lines are mapped to lines.
Consider any point $x\notin C_1$. We locate $\phi(x)$ by using the Theorem of Desargues as follows. Choose three lines through $x$ that intersect $C_1$ in three connected sets $c_1$, $c_2$, $c_3$. Choose points $a_i$ and $b_i$ on chord $c_i$. Then the triangles $\triangle a_1 a_2 a_3$ and $\triangle b_1 b_2 b_3$ are in perspective, so by the theorem of Desargues, the intersection points $p_{ij}$ of the lines $a_i a_j$ and $b_i b_j$ are collinear. This can be done in such a way that the $p_{ij}$ are all in $C_1$. For instance, the $c_i$ has to be chosen sufficiently close together, and each triangle has to be chosen so that its points are "almost collinear", with the two lines of collinearity intersecting inside $C_1$. 
Then this picture of the triangles in perspective, without the point $x$, can be transferred to $C_2$ using $\phi$. All incidences are preserved, so by the converse of the theorem of Desargues, the three connected sets $\phi(c_i)$ lie on concurrent lines. Define $\phi(x)$ to be the point of concurrency. It is easy to see that the definition is independent of which chords through $x$ are used.
We are halway there. It remains to prove that the extended $\phi$ preserves collinearity.
So let $x,y,z$ be collinear in $\mathbb{R}^2$. We would like to show that $\phi(x), \phi(y), \phi(z)$ are collinear.
If at least two of $x,y,z$ are in $C_1$, it is clear that their images will also be collinear. So assume without loss of generality that $y,z\notin C_1$.
The case $x\in C_1$ is simple: the chord of $C_1$ through $x,y,z$ maps to the chord of $C_2$ through $\phi(x)$ and $\phi(y)$, and also to the chord through $\phi(x)$ and $\phi(z)$. Thus $\phi(x),\phi(y),\phi(z)$ are collinear.
If on the other hand, $x\notin C_1$, we again use the theorem of Desargues. Find triangles inside $C_1$ such that the points in which their corresponding sides intersect, all lie on the line through $x,y,z$. (As before, it is easy to see that this is possible.) By the converse of Desargues, the triangles are in perspective. Transfer this picture with $\phi$ to the plane in which $C_2$ lives. We again get two triangles in perspective, and by Desargues, $\phi(x)$, $\phi(y)$, $\phi(z)$ are collinear.
We have shown that lines are mapped onto lines. By (a very special case of) the fundamental theorem of projective geometry, $\phi$ is a projective transformation.<|endoftext|>
TITLE: Is there something like Čech cohomology for p-adic varieties?
QUESTION [13 upvotes]: Suppose that I have a nice variety X over ℚp, with good reduction if you like, and a nice sheaf on X, say coming from a smooth group scheme G.  I can cover X by some p-adic open sets Uα, for example the mod-p neighbourhoods coming from some model $\mathcal{X}$ of X.  Clearly I can't expect to use Čech cohomology in a naïve way to compute the Hi(X,G) in terms of the cohomology of the Uα, because they don't overlap.  But the information about how they fit together to make X is instead contained in the geometry of the special fibre of $\mathcal{X}$.

Is there a spectral sequence which calculates Hi(X,G) in terms of some sort
  of cohomology of the Uα, and some information about how they 
  fit together?

Motivation
In the situation described above, the Leray spectral sequence gives
$$ 0 \to H^1(\mathcal{X},j_*G) \to H^1(X,G) \to H^0(\mathcal{X}, R^1 j_* G)$$
where $j\colon X \to \mathcal{X}$ is the inclusion of the generic fibre.
So in this situation I can compute H1(X,G) in terms of: $R^1 j_* G$, which I want to think of as somehow being the cohomology of the p-adic discs covering X; and the cohomology of $\mathcal{X}$, which I want to think of as saying how those discs fit together.
I would like to see how to generalise this to smaller p-adic neighbourhoods.
Supplementary question:

Should I just go away and read a book on rigid cohomology?

REPLY [13 votes]: The first comment to make is that Cech theory is really extremely general, and can be set up to compute the cohomology of any complex of abelian sheaves on any site (provided you have coverings that are cohomologically trivial).  This is explained at least somewhat in SGA4, Expose 5 and EGA III, Chap 0, section 12.  
I think you should be working with the rigid analytic space attached to $X$, and not with the $\mathbf{Q}_p$-points of $X$, and the latter really has no good topology on it besides the totally disconnected one induced from the topology on $\mathbf{Q}_p$.  
Let's assume that $X$ has a model 
$\mathcal{X}$ over $\mathbf{Z}_p$ that is smooth and proper and write $\widehat{\mathcal{X}}$
for the formal completion of $\mathcal{X}$ along its closed fiber.  Then the (Berthelot) generic fiber $\widehat{\mathcal{X}}^{rig}$ of $\widehat{\mathcal{X}}$ is a rigid analytic space that is canonically identified with the rigid analytification of $X$ (using properness here).  Moreover, one has a "specialization morphism" of ringed sites
$$sp:X^{an}\simeq \widehat{\mathcal{X}}^{rig}\rightarrow \widehat{\mathcal{X}}$$
with the property that for any (Zariski) locally closed subset $W$ of the target, the inverse image $sp^{-1}(W)$ is an admissible open of the rigid space $X^{an}$ (called the open tube over W).  In this way, coverings of the special fiber by locally closed subsets give coverings of the rigid generic fiber by admissible opens, and you can use Cech theory with these coverings and or your favorite spectral sequence to compute sheaf cohomology in the rigid analytic world.  Again using properness, by rigid GAGA this cohomology agrees with usual (Zariski) cohomology on the scheme $X$ (provided your sheaf is a coherent sheaf of $\mathcal{O}_X$-modules, say).  
This idea of computing cohomology using admissible coverings of the associated rigid space is a really important one as it allows you to use the geometry of the special fiber.  It occurs (allowing $\mathcal{X}$ to have semistable reduction) in the work of Gross on companion forms, of Coleman on $\mathcal{L}$-invariants and most prominently in Iovita-Coleman (see their article on "Frobenius and Monodromy operators").  This latter article might be a good place to start.
I would also highly recommend the articles of Berthelot:
http://perso.univ-rennes1.fr/pierre.berthelot/publis/Cohomologie_Rigide_I.pdf
http://perso.univ-rennes1.fr/pierre.berthelot/publis/Finitude.pdf
I'd also suggest the AWS 2007 notes by Brian Conrad for learning about rigid geometry, which seems generally quite pertinent to your situation.
For etale cohomology of rigid spaces, you might want to look at the article of Berkovich, though this would require learning about his analytic spaces. 
In any case, I hope this is a good start.<|endoftext|>
TITLE: Why are cohomologically trivial cycles abundant?
QUESTION [5 upvotes]: Suppose X is a smooth projective variety, say over $\mathbb{Q}$ for simplicity.  Let $F$ be a finite extension of $\mathbb{Q}$. Let $\mathrm {Ch}^{r}(X/F)$ denote the Chow group of codimension $r$ algebraic cycles defined over $F$.  A conjecture of Tate asserts that the cycle class map from $\mathrm{Ch}^r(X/F)$ to $H^{2r}_{et}(X)(r)$ is injective, with image in the subspace fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/F)$.  In particular, the dimension of $\mathrm{Ch}^r(X/F)$ should be uniformly bounded (by the $2r^{\mathrm{th}}$ Betti number of $X(\mathbb{C})$) as $F$ varies.
On the other hand, let $\mathrm{Ch}^{r}(X/F)_0$ denote the kernel of the cycle class map, which is to say the group of homologically trivial cycles of codimension $r$ modulo rational equivalence.  The dimension of this guy is predicted by Beilinson and Bloch to be given as the order of vanishing of $L(s,H^{2r-1}(X/F))$ at its central critical point.  Now, the order of vanishing of this L-function can by made to increase very rapidly as $F$ varies; for example, one could (in some circumstances) choose $F$ to be an abelian extension such that each of the twists $L(s,H^{2r-1}(X/\mathbb{Q})\times \chi)$ has root number $-1$ for $\chi$ varying over characters of $\mathrm{Gal}(F/\mathbb{Q})$.  When $X$ is an elliptic curve and $r$=1, this phenomenon has been confirmed in a variety of situations: for $\mathbb{Z}/p^{n}\mathbb{Z}$-towers over imaginary quadratic fields (Cornut-Vatsal), for Hilbert class fields of imaginary quadratic fields (Templier), and for towers of Kummer extensions (Darmon-Tian).  However, for higher dimensional varieties and higher codimension cycles, the relevant L-functions aren't even well understood.
My question: is there a "conceptual" reason why there should have lots of homologically trivial cycle classes over extensions of the base field?  In other words, if you believe certain conjectures about L-functions, then this is not hard to guess, but I am looking for some motivation which avoids L-functions.
(Edited in response to a comment of moonface.)

REPLY [4 votes]: I believe that the question is still slightly mis-stated; see my comment above.  (I'm sorry if I've blundered here, but will be set straight if I have, I'm sure!)
To make your question more concrete, one should take $X$ to be an elliptic curve, $r = 1$,
and then $Ch(X/F)_0  = E(F)$.  So your question becomes, why does $E(F)$ achieve large
rank over various extensions.  You discussed this in your question, but I don't your discussion fully describes the situation.
There is work of many people, especially Mazur and Rubin (as well as the others you
mention) showing that Selmer groups grow in something like the expected way as you
enlarge $F$ appropriately.  The reason I single out Mazur and Rubin is that they work in
a very general context, but they only deal with Selmer groups.  One expects that the
Mordell--Weil groups grow in the same manner (since one expects Sha to be finite), but
this is not known, and no-one knows how to exihibit any points on elliptic curves by theoretical means, other than via Heegner points constructions.  
Given this, I think the answer to the higher codimension question will be similar: no-one
knows how to make cycles (other than Heegner cycles in certain special contexts).   If you replaced the Chow group by some Galois-cohomological stand-in
(i.e. some kind of Selmer group), perhaps one could then say something.  I don't know of work of that kind, but it may well exist.<|endoftext|>
TITLE: Evidences on Hartshorne's conjecture? References?
QUESTION [18 upvotes]: Hartshorne's famous conjecture on vector bundles say that any rank $2$ vector bundle over a projective space $\mathbb{P}^n$ with $n\geq 7$ splits into the direct sum of two line bundles.
So my questions are the following:
1) what is an evidence for this conjecture?
2)why is the condition on $n\geq 7$, but not other numbers?
3)any recent survey or reference on this conjecture?

REPLY [7 votes]: A remark similar to Hailong Dao's comment under his answer:
Let $E$ be a vector bundle on $\mathbb{P}^n$. A cohomological criterion (Horrocks' criterion) states that $E$ splits if and only if $H^i(\mathbb{P}^n, E(t))=0$ for $0 < i < n$ and all $t$. 
There is a little less well known criterion, due to Evans and Griffiths, which says that we only need to check the vanishing of $H^i(\mathbb{P}^n, E(t))$ for $0 < i < \min(n, rank(E))$ and all $t$.
In particular, in the rank two case, the whole conjecture boils down to the simple claim that $H^1(\mathbb{P}^n, E) = 0$. Since $E$ is trivial on each "standard open" $U_i$, we can describe cohomology classes in this $H^1$ group using explicit Cech cocycles in this covering.
In summary, it is surprising how little we know about such a simple situation!<|endoftext|>
TITLE: Hilbert $C^*$-modules and approximate units
QUESTION [5 upvotes]: Hi,
Given a $\sigma$-unital $C^*$-algebra $A$ and a full Hilbert $A$-module $E$, is it possible to find an approximate unit $  \{\epsilon_i\}, i\in I$ in $A$ such that each $\epsilon_i$ is of the form $< e_i,e_i>_A$, where $e_i \in E, \forall i\in I$? If not, what are the conditions on $E$ and $A$ for which this might be possible?
Thanks in advance.

REPLY [2 votes]: No, there is not always such an approximate unit.  (This will be easier to formulate in terms of left modules, and with inner products linear in the first entry. The warning seems necessary due to the common convention in C*-module theory to do the opposite.)
Example
Let $A=B(\mathbb{C}^2)$ (linear operators), $E=\mathbb{C}^2$, with the module action given by operators acting on vectors (on the left) and the $A$-valued inner product of $x$ and $y$ in $E$ given by $<x,y>_A(z)=<z,y>_\mathbb{C}x$.  Then $<x,y>_A$ has rank at most one for all $x$ and $y$, so no such approximate identity exists. $\square$

Any finite dimensional Hilbert space with dimension at least 2 would give a slight modification of this example.  Or, let $E=H$ be a separable, infinite dimensional Hilbert space, and let $A=\mathcal{K}(H)$ be the algebra of compact operators on $H$.  (Added: Note that fullness follows from the fact that the span of the range of the inner product is the set of finite rank operators.)  Or, given a C*-algebra $B$, one could form $H_B=B\oplus B\oplus\ldots$ with its usual right $B$-module structure and consider the analogous construction with $A=\mathcal{K}(H_B)$ and $E=H_B$.  If $B$ is $\sigma$-unital, then so is $A$, but there will be no approximate identity of the desired form.
I do not have anything useful to say about formulating sufficient conditions for such an approximate identity to exist, but this simple example shows that lack of existence is common.<|endoftext|>
TITLE: Is this a well-known probabilistic model?
QUESTION [10 upvotes]: While I was thinking about the Erdős discrepancy problem, the following random walk model arose rather naturally. You fix a positive integer k, and you take a random step of 1 or -1 at each stage, except if that stage is a multiple of k, in which case you get to choose. Your object is to keep the walk as close to the origin as you can.
An obvious strategy is a greedy one: each time you have the choice, if the walk is currently positive you choose -1 and if it's currently negative you choose +1. (Clearly it makes no difference to the outcome what you do if it is zero.) This is like giving your random walk a drift term of size 1/k, but the drift is always directed towards the origin.
A slightly different question would be if you are presented with the entire walk up to kN-1 and you then choose the signs at k, 2k, 3k, ... , Nk, attempting to minimize the maximum distance it ever goes from the origin.
It is not hard to see that it must go at least logarithmically far, since by the time you get to M you will probably have had a run of log M 1s in the random walk, which you can do almost nothing about. I have thought about the problem in a non-rigorous way and it feels to me as though it shouldn't be too hard to prove that by the time you get to M you won't have gone further than $(\log M)^2$ or $(\log M)^3$ or something like that. (The rough argument is that if you look at a run of $k^2$ steps, it won't tend to drift more than about k, and you'll have k steps that you can choose so as to cancel out the drift. That is what usually happens, and the exceptions ought to be exponentially rare, so to speak.)
My main question is this: are these, or something like them, well-known questions? (Or do answers to them follow almost instantly from known results?)

REPLY [6 votes]: If you apply the greedy algorithm, that is very close to asking for the maximum drawdown of a random walk with positive drift. 
The distribution of maximum drawdowns of a Brownian motion with constant drift have been studied, and some answers are on page 2 here. The expected maximum drawdown with positive drift grows asymptotically like 
$c \log M$. I'm not sure how the tails look.
I asked how badly the Brownian approximation behaves for discrete walks here. It can be far off for some skewed distributions, but it should be off by a small constant factor for these nearly symmetric steps.
The globally optimal adjustments can't reduce the maximum excursion by more than a factor of 2 compared with the greedy algorithm, since on the largest (without loss of generality) positive excursion to $d$, the greedy algorithm was all -1s. Then any change made would still mean the walk increases by at least $d$, so at best a global optimization could move the start to $-d/2$ which would make the maximum at least $d/2$ in both directions.<|endoftext|>
TITLE: Degree 2 branched map from the torus to the sphere
QUESTION [8 upvotes]: Algebraic geometry predicts a degree 2 branched cover from an elliptic curve to the projective line. What does this map look like topologically?

REPLY [6 votes]: The cover of the first edition of Francis' book: "A topological picturebook".<|endoftext|>
TITLE: Where can I find a comprehensive list of equations for small genus modular curves?
QUESTION [25 upvotes]: Does there exist anywhere a comprehensive list of small genus modular curves $X_G$, for G a subgroup of GL(2,Z/(n))$? (say genus <= 2), together with equations? I'm particularly interested in genus one cases, and moreso in split/non-split cartan, with or without normalizers.
Ken Mcmurdy has a list here for $X_0(N)$, and Burcu Baran writes down equations for all  $X_{ns}^+(p)$ of genus <=2 in this preprint.

REPLY [2 votes]: For N <= 37, I computed a birational map from a simple equation to X0(N).
http://www.math.fsu.edu/~hoeij/files/X0N/Parametrization
None of the data in that file is novel, but I decided to recompute this data because I needed it in machine-readable form, and because, as pointed out in an earlier answer, these things are spread out over a number of papers.<|endoftext|>
TITLE: What's algebraic approach to QM good for?
QUESTION [11 upvotes]: The algebraic formulation of quantum mechanics (and related stuff, like quantum thermodynamics & dynamical systems etc.) via C*-algebras provides a viewpoint based mostly on abstract functional analysis. However, I haven't really seen a working application of this approach, i.e. an example of a (preferably physical) problem which is difficult to solve or even to formulate in the standard formalism, while considerably easier to tackle with all this algebraic stuff. Any ideas? Without such concrete examples the whole field seems to be interesting mathematically, perhaps, but lacking any physical substance (even if it's sometimes masqueraded as having connections with physics).
edit: by "standard formalism" I mean "observables, i.e. operators, acting on a Hilbert space of physical states", either in Schroedinger picture (time dependence of states) or Heisenberg (time dependence of operators). C*-algebraic approach starts from a C*-algebra and defines states as positive functionals on the elements of algebra, time evolution as a *-automorphism etc. (for an introduction to this formalism, see http://hal.archives-ouvertes.fr/docs/00/12/88/67/PDF/qds.pdf)
This might be of less interest to non-physicists than a typical MO post, but I hope it's still relevant.

REPLY [5 votes]: The algebraic approach subsumes the standard formulation of QM (everything that can be done in the latter can be done in the former).
One important feature of the algebraic framework is that it allows you to handle inequivalent representations of the algebra of observables (which invariably come up in field theory, or any theory with an infinite number of degrees of freedom).
An example of a problem tackled only in the algebraic framework is the perturbative construction of interacting quantum field theory on arbitrary globally hyperbolic spacetimes. The construction is perturbative because every quantity of interest is taken to be a formal power series in the interaction strength. AFAIK, this construction has only been done using algebraic methods.<|endoftext|>
TITLE: How do you switch between representations of an algebraic group and its Lie algebra?
QUESTION [11 upvotes]: I'm interested in the structures of categories like $Rep(GL_n), Rep(SL_n)$, etc. of algebraic representations of  an algebraic group.  I understand that there should be some relation between these and the categories of representations of the corresponding Lie algebras. 
However, it's not as intuitive to me what's going on here as with the case of, say, a Lie group, perhaps because the notion of a "tangent vector" is somewhat different.
So, how does one switch between the categories $Rep(G)$ and $Rep(\mathfrak{g})$ for $G$ an algebraic group and $\mathfrak{g}$ its Lie algebra---are there functors in each direction?  Can this be used to prove that $Rep(G)$ is semisimple when $G$ is reductive?
In another direction, can the structure of $Rep(\mathfrak{g})$ as known from the representation theory of, say, semisimple Lie algebras give the structure of $Rep(G)$?

REPLY [2 votes]: One point which seems not to have been mentioned (but is implicit in Pavel's answer) is that the small question posed at the end of the main question, whether $\mathrm{Rep}(\mathfrak{g})$ semisimple implies $\mathrm{Rep}(G)$ semisimple, has a simple answer regardless of most technical conditions.  In fact, in characteristic zero it is true.
The relevant theorem is:

Theorem: Suppose $k$ has characteristic zero, let $G$ be a connected affine algebraic $k$-group, and let $\mathfrak{g}$ be its Lie algebra.  If $V$ is a $G$-representation and $W \subset V$ is a subspace, then $W$ is $G$-stable if and only if it is $\mathfrak{g}$-stable.

The proof for faithful representations is in Humphreys' "Linear Algebraic Groups", Theorem 13.2, and to extend it to any representation one need only show that the Lie algebra is compatible with fibered products, which is tautological.  (Possibly this result must be stated over an algebraically closed field; this is so in the book, and I am badly acquainted with rationality properties.)
As a consequence, $V$ is irreducible or completely reducible for $G$ if and only if it is for $\mathfrak{g}$.<|endoftext|>
TITLE: Irreducibility of polynomials in two variables
QUESTION [74 upvotes]: Let $k$ be a field. I am interested in sufficient criteria for $f \in k[x,y]$ to be irreducible. An example is Theorem A of this paper (Brindza and Pintér, On the irreducibility of some polynomials in two variables, Acta Arith. 1997).
Does anyone know of similar results in the same vein? How about criteria over fields other than the complex numbers?

REPLY [21 votes]: Quite a useful result can be found in Schmidt's lecture notes on Equations over finite fields (Theorem III.1B in SLNM 536). Suppose that $K$ is any field and let $f(x,y)=c_0y^d+c_1(x)y^{d-1}+
\cdots+c_d(x) \in K[x,y]$, with $c_0 \neq 0$.  Let 
$$
\psi(f)=\sup_{1\leq
i\leq d}\frac{\deg c_i}{i}.
$$
Then $f$ is absolutely irreducible over $K$ provided that that $\psi(f)=m/d$ with $\gcd(m,d)=1$.
This shows, for example, that the polynomial $f(x,y)=g(x)-h(y)$ is irreducible when  $\deg g$ and $\deg h$ are coprime.<|endoftext|>
TITLE: Modular forms and the Riemann Hypothesis
QUESTION [35 upvotes]: Is there any statement directly about modular forms that is equivalent to the Riemann Hypothesis for L-functions? 
What I'm thinking of is this: under the Mellin transform, the Riemann zeta function $\zeta(s)$ corresponds to a modular form $f$ (of weight 1/2). The functional equation of $\zeta(s)$ follows from the transform equation of $f$. So what is the property of $f$ that would be equivalent to the (conjectural) property that all the non-trivial zeros of $\zeta(s)$ lie on the critical line? Or perhaps is there any statement about some family of modular forms that would imply RH for $\zeta(s)$?

@Hansen and @Anonymous: your answers are appreciated. I want to know why people almost never discuss this question, so even the answer that the question is not a good one is appreciated, provided it also gives a reason, like you did.
As Emerton suggested, I want to know whether RH could be stated for eigenforms directly, instead of the L-functions. I'm no expert in this field, but it seems to me that analytic properties of modular forms are easier to understand (than those of L-functions), so why not expressing RH in the space of modular forms and working with them? 
@Anonymous: do you know of any readily accessible source for statements about families of modular forms that imply RH for zeta? I don't have access to MathSciNet.

REPLY [12 votes]: I know two statements about modular forms that are Riemann Hypothesis-ish. 
First, note that the constant term of the level-one non-holomorphic Eisenstein series $E_s$ is $y^s+c(s)y^{-s}$, and that the poles of $c(s)$ are the same as the poles of $E_s$. We can directly calculate that $c(s)={\Lambda(s)\over\Lambda(1+s)}$ (this depends on your precise normalization of the Eisenstein series), where $\Lambda$ is the completed zeta function. We can actually say something about the location of the poles of $E_s$ (using the spectral theory of automorphic forms). Unfortunately, we only know how to control poles for ${\rm Re}(s)\ge 0$. This does give an alternate proof of the nonvanishing of $\zeta(s)$ at the edge of the critical strip (from the lack of poles of ${\Lambda(it)\over\Lambda(1+it)}$), but it doesn't seem possible to go further to the left (though it does generalize to other $L$-functions appearing as the constant term of cuspidal-data Eisenstein series).
Second, the values of modular forms at certain (Heegner) points in the upper-half plane can be related to zeta functions. For example, $E_s(i)={\Lambda_{{\mathbb Q}(i)}(s)\over \Lambda_{\mathbb Q}(2s)}$. The general statement is simple to express adelically. Take a quadratic extension $k_1$ of $k$, and let $H$ denote $k_1^\times$ as a $k$-group and $E_s$ the standard level-one Eisenstein series on $G=GL_2(k)$. Take a character $\chi$ on $Z_{\mathbb A}H_k\backslash H_{\mathbb A}$ then 
$$\int_{Z_{\mathbb A}H_k\backslash H_{\mathbb A}}E_s(h)\chi(h)\ dh={\Lambda_{k_1}(s,\chi)\over \Lambda_k(2s)}$$
where $Z$ denotes the center of $G$, and we have normalized the measure on the quotient space to be 1. Note that since $H$ is a non-split torus in $G$, the quotient is compact, so the integral is finite. In fact, the integrand is invariant (on the right) under a compact open subgroup $K$ of $H_{\mathbb A}$, so the integral is actually over the double coset space $Z_{\mathbb A}H_k\backslash H_{\mathbb A}/K$, which is actually a finite group.
In order to get the Riemann zeta function in the numerator on the right-hand-side, you would need to integrate over a split torus, which is precisely the Mellin transform, and you would have convergence issues. Note that if it did converge, the Mellin transform of $E_s$ would be
$$\int_{Z_{\mathbb A}M_k\backslash M_{\mathbb A}} E_s(a)|a|^v\ da={\Lambda(v+s)\Lambda(v+1-s)\over\Lambda(2s)}$$
The second idea is more commonly discussed in the context of subconvexity problems for general $L$-functions. (See Iwaniec's Spectral Methods of Automorphic Forms, especially Chp 13.) A class of subconvexity results is the Lindelof Hypothesis, which is one of the stronger implications of the Riemann Hypothesis.<|endoftext|>
TITLE: Hyperelliptic loci in Teichmueller spaces
QUESTION [13 upvotes]: Let ${\cal M}_g$ be the moduli space of smooth complex genus $g$ curves, let ${\cal H}_g\subset {\cal M}_g$ be the hyperelliptic locus and set ${{\cal H}}'_g$ to be the preimage of ${\cal H}_g$ in the Teichmueller space.
While working on a problem I arrive at two results that can't be reconciled unless ${\cal H}'_3$ is disconnected.
While it seems a bit strange to me that ${\cal H}'_3$ should be disconnnected, I don't see why it should't be. So I'd like to ask whether this is known or known to be false.
[sorry, had to cut this into small paragraphs, otherwise the tex part wouldn't show properly.]

REPLY [7 votes]: There's a slight issue I believe with the other answers. If we consider moduli space as an orbifold (of complex dimension $3g-3$), and the hyperelliptic locus an immersed suborbifold (of complex dimension $2g-1$ or so), then we may (essentially) identify the hyperelliptic locus with the orbifold of $2g+2$ points on $S^2$, obtained by quotienting each Riemann surface by the hyperelliptic involution. However, how does one know that this space doesn't "cross" itself? Imagine by analogy an immersed geodesic curve on a hyperbolic surface, such that each complementary component is a disk: the preimage in the universal cover is connected, when taken as a union of geodesics, even though each geodesic lift is embedded. 
This sort of crossing does not occur for the hyperelliptic locus. If two branches of the hyperelliptic universal cover in Teichmuller space were to intersect, then there would be a single Riemann surface fixed by two distinct hyperelliptic involutions. But a hyperelliptic involution fixes precisely the $2g+2$ Weierstrauss points of the surface, and is therefore uniquely determined, a contradiction. So in fact the hyperelliptic locus is "embedded", in the sense that each lift corresponds to a fixed set of a hyperelliptic involution, and distinct hyperelliptic involutions give distinct components in Teichmuller space.<|endoftext|>
TITLE: Cyclic extensions coming from E[p] \equiv F[p],
QUESTION [7 upvotes]: Let p be a prime and let K be a field containing the p'th roots of unity. Let E be an elliptic curve over K. We consider the the moduli problem $Y_E(p)$, which sends L to set of elliptic curves F/L, and symplectic isomorphisms $\phi:E[p] \rightarrow F[p]$. We know that this moduli problem is representable by a curve over $K$, and we let the compactification of this curve be $X_E(p)$. We know $X_E(p)$ is a twist of $X(p)$. Similarly, we can construct $X_E(p^2)$, and we see that $X_E(p^2)$ is a normal cover of $X_E(p)$. I think the Galois group of $X_E(p^2)/X_E(p)$ is $(Z/pZ)^3$. If that is the case, then given any K point of $X_E(p)$, we can look at fiber over this point. This fiber is defined over K, hence it will define a field extension of K, with Galois group a subset of $(Z/pZ)^3$.
This means, if we have E and F defined over K, with $E[p] \equiv F[p]$, then we should be able to construct a cyclic extension of K of order p. What is that extension?

REPLY [9 votes]: The answer is that in fact this construction does not produce cyclic extensions!  The problem is that $X_E(p^2) \to X_E(p)$ is not generically Galois; it is so only after extension of the ground field.
Here is a more detailed explanation.  Assume that $p \ge 3$ and that $p \nmid \operatorname{char} K$.  Then $Y_E(p^2) \to Y_E(p)$ is a torsor not under $(\mathbb{Z}/p\mathbb{Z})^3$, but under the étale group scheme $G$ corresponding to the Galois module of $\mathbb{F}_p$-linear endomorphisms $g \colon E[p] \to E[p]$ of trace zero.  Explicitly, $g \in G(\overline{K})$ maps $(F,\Phi) \in Y_E(p^2)(\overline{K})$ to $(F,\Phi')$ where $\Phi'(x):=\Phi(x+g(px))$.  (Any $\Phi'\colon E[p^2] \to F[p^2]$ with the same restriction to $E[p]$ as $\Phi$ arises from some $g \in \operatorname{End} E[p]$ in this way, and the trace-zero condition is what guarantees that $\Phi'$ is symplectic.)  Finally, this Galois module is typically irreducible, in which case it does not have $\mathbb{Z}/p\mathbb{Z}$ as a quotient.<|endoftext|>
TITLE: Approximation to divergent integral
QUESTION [10 upvotes]: Hi everyone,
I'm a physicist working on stochastic processes and I've come up against an integral that I'm not able to approximate using steepest descent (I don't have a large or small parameter), integration by parts, or any of the other common techniques. So I'd very much appreciate the input of any applied mathematicians!
The integral is 
$$f(x) = \int_{x}^{\infty} \frac{\Phi(t)}{t^{5}}dt$$
 with $\Phi(t) = e^{i \pi t^{2} / 2}[C(t) + i S(t)]$. Here, $C(t)$ and $S(t)$ are the Fresnel integrals defined by 
$$C(t) + i S(t) = \int_{0}^{t} e^{i \pi u^{2} / 2} du\ .$$
What I really want is the behaviour of $f(x)$ for small $x$. But, the integral is formally divergent if $x = 0$.
Made a little progress with integration by parts, but I wasn't able to entirely separate my integral into convergent pieces.

REPLY [15 votes]: First, cut off the tail towards infinity:
$$f(x) = \int_{x}^1 \frac{\Phi(t)}{t^5} dt + \int_1^{\infty} \frac{\Phi(t)}{t^5} dt.$$
The second term is a constant, so you can compute it numerically once and for all.
Write 
$$e^{i \pi u^2/2} = 1 + \frac{i \pi}{2} u^2 +R(u)$$
and
$$\int_{0}^t e^{i \pi u^2/2} du = t + \frac{i \pi}{6} t^3 + \int_{0}^t R(u) du.$$
So
$$\frac{\Phi(t)}{t^5} = \left( t^{-4} + \frac{i \pi}{6} t^{-2} + t^{-5} \int_{0}^t R(u) du \right) \left( 1 + \frac{i \pi}{2} t^2 + R(t) \right)=$$
$$t^{-4} + \frac{2 \pi i}{3} t^{-2} + \left( t^{-4} R(t) - \frac{\pi^2}{12} + \int_{0}^t R(u) du \right).$$
So 
$$\int_{x}^1 \frac{\Phi(t)}{t^5} dt = \frac{1}{3}\left( x^{-3} - 1 \right) + \frac{2 \pi i}{3} \left( x^{-1} -1 \right) + \int_{x}^1 \left( t^{-4} R(t) - \frac{\pi^2}{12} + \int_{0}^t R(u) du \right) du.$$
The integrands in the last term are bounded functions, and they are being integrated over bounded domains, so there is no problem approximating them numerically. 
If you want an asymptotic formula, instead of a numerical approximation, you should be able to keep taking more terms out to get a formula like 
$$f(x) = \frac{1}{3} x^{-3} + \frac{2 \pi i}{3} x^{-1} + C + a_1 x + a_2 x^2 + \cdots + a_n x^n + O(x^{n+1}) \quad \mathrm{as} \ x \to 0.$$
You probably won't be able to get the constant $C$ in closed form, because it involves all those convergent integrals. The other $a_i$ will be gettable in closed form, although they will get worse and worse as you compute more of them.<|endoftext|>
TITLE: Is there a category of non-well-founded sets?
QUESTION [12 upvotes]: Is there a category (in the category theory sense) of non-well-founded sets (something analogous to Set, the category of sets), and has it been (well-)studied?  Any references are appreciated.

REPLY [2 votes]: According to this doctoral thesis http://www.andrew.cmu.edu/~awodey/students/hughes.pdf by Jesse Hughes (supervised by Steve Awodey), cogebras = coalgebras are the appropriate category to study non-wellfounded sets.<|endoftext|>
TITLE: Is there a Whitney theorem type theorem for projective schemes?
QUESTION [29 upvotes]: We know that any smooth projective curve can be embedded (closed immersion) in $\mathbb{P}^3$. By definition a projective scheme over $k$ admits an embedding into some $\mathbb{P}^n$. Can we create an upper bound for the $n$ required (perhaps by strengthening the hypotheses) necessary to create an embedding of a smooth projective dimension $k$ scheme into $\mathbb{P}^n$ much like Whitney's theorem tells us we can embed an $n$ dimensional manifold in $\mathbb{R}^{2n}$?

REPLY [5 votes]: To contradict what I said on embedding of singular varieties, here is a theorem of Kleiman and Altman "Bertini theorems for hypersurface  sections containing a subscheme". Comm. Algebra 7 (1979), no. 8, 775--790. 
Let $X$ be an algebraic variety over a field $k$. For any $x\in X$, define the local embedding dimension $e(x)$ of $X$ at $x$ by $e(x)=\dim (\Omega_{X/k}^1\otimes k(x))$ (so if $k$ is perfect, then $e(x)$ is just the dimension of the Zariski tangent space at $x$). It is easy to see that for any integer $e$, the set $X_e$ of $x\in X$ such that $e(x)=e$ is contructible. By convention $\dim\emptyset = -\infty$. 
 Theorem (Kleiman-Altman)  Suppose $k$ is infinite and $X$ is quasi-projective (resp. projective) over $k$. Let $r$ be the maximum of $\dim (X_e) +e$ for all $e\ge 0$. Then $X$ can be embedded in a smooth quasi-projective (resp. projective) variety $Z$ of dimension $r$ over $k$. 
For example, a reduced projective curve over an infinite perfect field can be embedded in a smooth projective surface if and only if the tangent space at every point has dimension at most 2. 
In general, combining with the result on embedding of smooth projective varieties, one gets an embedding of $X$ in a projective space of dimension bounded by $\dim X$ and local embedding dimensions of $X$.<|endoftext|>
TITLE: constants in Gamma factors in functional equation for zeta functions.
QUESTION [19 upvotes]: Usually the Riemann zeta function $\zeta(s)$ gets multiplied by a "gamma factor" to give a function $\xi(s)$ satisfying a functional equation $\xi(s)=\xi(1-s)$. If I changed this gamma factor by a non-zero constant complex number, the functional equation would still hold.
More generally, for zeta functions of number fields, one gets a gamma factor for each real place and a gamma factor for each complex place. In a parallel universe, we could have defined the gamma factor for each real place to be (say) $\pi^{1/2}$ times  what we usually use, and the gamma factor at the complex places to be (say) $1/2$ of what we usually use (and I think I was taught to use $2.(2\pi)^s$ so we could just knock off that first 2) and all functional equations would still be exactly the same (because the extra factors would be the same on both sides).
More generally, for Dirichlet $L$-functions and Hecke $L$-functions: I now need a gamma factor for the sign function on the non-zero reals, and again we could use a different choice.
Now some general yoga of gamma factors tells us that really there are only 3 choices to be made (because Hodge structures basically always decompose into the types covered above)---and I've mentioned them all already.
Upshot: did we, at some point, make 3 arbitrary choices of constants, and the entire theory of $L$-functions of motives wouldn't care what choices we made, so we could have made other choices? Note for example that conjectures on special values of $L$-functions don't take the gamma factors into account (well, the ones I know don't; they predict values of the incopleted $L$-function without the gamma factors). Note also that when defining these things for Hecke characters a la Tate, again pretty arbitrary choices are made at the infinite places---there is no one canonical function that is its own Fourier transform, because we can change things by constants again.
Am I totally wrong here or are there really 3 arbitrary choices that we have made, and we could have made others?

REPLY [2 votes]: In Deninger's first paper on zeta-regularized determinants, his construction of the Archimedian L-factor at the real place for Q (i.e., the gamma factor for Riemann zeta) is off by a factor of sqrt(2) in comparison to Riemann's gamma factor.<|endoftext|>
TITLE: Tiling a rectangle with a hint of magic
QUESTION [34 upvotes]: Here's a a famous problem:
If a rectangle $R$ is tiled by rectangles $T_i$ each of which has at least one integer side length, then the tiled rectangle $R$ has at least one integer side length.
There are a number of proofs of this result (14 proofs in this particular paper by Stan Wagon). One would think this problem is a tedious exercise in combinatorics, but the broad range of solutions which do not rely on combinatorial methods makes me wonder what deeper principles are at work here. In particular, my question is about the proof using double integrals which I sketch out below:
Suppose the given rectangle $R$ has dimensions $a\times b$ and without loss of generality suppose $R$ has a corner at coordinate $(0,0)$. Notice that $\int_m^n\sin(2\pi x)dx=0$ iff $m\pm n$ is an integer. Thus, for any tile rectangle $T_i$, we have that:
$$\int\int_{T_i}\sin(2\pi x)\sin(2\pi y)dA=0$$
If we sum over all tile rectangles $T_i$, we get that the area integral over $R$ is also zero:
$$\int\int_{R}\sin(2\pi x)\sin(2\pi y)dA=\sum_i\int\int_{T_i}\sin(2\pi x)\sin(2\pi y)dA=0$$
Since the corner of the rectangle is at $(0,0)$, it follows that either $a$ or $b$ must be an integer.
My question is as follows: where exactly does such a proof come from and how does it generalize to other questions concerning tiling? There is obviously a deeper principle at work here. What exactly is that principle?
One can pick other functions to integrate over such as $x-[x]-1/2$ and the result will follow. It just seems like black magic that this works. It's as if the functions you are integrating over tease out the geometric properties of your shape in an effortless way.
EDIT: It's likely that one doesn't necessary need integrals to think in the same flavor as this solution. You're essentially looking at both side lengths in parallel with linear test functions on individual tiles. However, this doesn't really explain the deeper principles here, in particular how one could generalize this method to more difficult questions by choosing appropriate "test functions."

REPLY [3 votes]: IMHO the "deeper principle at work here" is assigning a convenient measure to the objects being combinatorially combined. The closest example that comes to mind is this problem:
Is it possible to cover a 2-D disk of diameter 10 with 9 rectangles of length 10 and width 1?
One solves the above by projecting the disk cover onto 1/2 sphere above it, noting that each rectangle's intersection with the disk projects to the area of the 1/2 sphere equal to 1/10th of the 1/2 sphere area, this proving that 9 rectangles won't be enough.
Both problems - the one in the question and the above one - chose a measure density (a signed one in the case of the rectangle R cover) that would have a convenient integration property.<|endoftext|>
TITLE: how do you evaluate the p-adic modular form E_p-1 in the region |j|<1
QUESTION [9 upvotes]: background/motivation
let Ek denote the modular form of level one and weight k with q-expansion given by $E_k(q)=1- \frac{2k}{b_k}\sum_n \sigma_{k-1}(n)q^n$ where σi   is the divisor sum and bk is the k-th bernoulli number. for k = p – 1 the the denominators of this series don't contain any powers of  p and hence by the q-expansion principle Ep–1 defines a modular form over ℤp. that is,  Ep–1 ∈ H0(X0(1),  ℤp ⊗ ω⊗k) where X0(1) is the compactified modular curve for the full modular group, and ω is the pushforward of the canonical bundle on the universal elliptic curve over  ℤ. 
now i can use p-adic uniformisation to evaluate Ep–1 via its q-expansion at elliptic curves with j-invariant satisfying | j | > 1; in concrete terms this means formally inverting the power series for 1/j(q), and then evaluating *Ep–1(q) on the result.
i don't suppose one should expect to get a well-defined value in ℂp for Ep–1 evaluated at elliptic curves for which | j | < 1 (as you would for an modular form over ℂ) since this would mean choosing a non-vanishing differential on the universal elliptic curve over ℤp. the norm |Ep–1 |, however, is well defined, since any two choices of basis for the canonial bundle on a given elliptic curve can only differ by a unit.
of course the context i am interested in is when |Ep–1| is used to define the "overconvergent" region of the modular curve. this is (roughly) defined as the region of the $j$-line satisfying |Ep–1|>r for |r|<1. this region will (always) include part of the |j|<1 region.  in order to understand what |Ep–1|>r "means" in concrete terms, i was hoping to numerically compare |E4| and |j| for some specific ℂp-values of j near 0 (for primes p = 5, 7, 11 etc. for which j = 0 is supersingular).
anyway, my question is: 


how can i explicitly evaluate |Ep–1| at elliptic curves  with | j | < 1?

REPLY [5 votes]: One has $j  = E_4^3/\Delta$.  In the region $|j|\leq 1$, one is parameterizing elliptic curves with good reduction, and so $\Delta$ is a unit.  Thus $|j| = |E_4|^3$.  This will help you
when $p = 5$.  
When $p = 7,$ one can write $j = 1278 + E_6^2/\Delta,$ hence $|E_6|^2 = | j - 1728|$ on
the region $|j| \leq 1$.  
For $p = 11$, these sort of explicit computations are harder (but maybe not much; see 
the added material below), because there are two supersingular $j$-invariants.  But the $p = 5$ and 7 cases will already be illustrative.
In the case when $p = 2$, I wrote something about this once, which appeared in an
appendix an article by Fernando Gouvea in a Park City proceedings volume.  A slightly
butchered version (missing figures, among other things) can be found on my
web-page (near
the bottom).  You might also look at the papers of Buzzard--Calegari
for related computations, as well as my thesis (available on my web-page) and later
work by Kilford and Buzzard--Kilford.  (There is, or at least once was, a cottage industry
based on combining these sorts of explicit computations with some more theoretical
estimates, to compute information about slopes of overconvergent $p$-adic modular forms
for various small primes $p$.)
Added in response to the comment below: For $p = 11$, one has 
$E_{10}  = E_4 E_6,$
so $E_{10}^6 = j^2(j-1728)^3 \Delta^5,$ and so when $|j| \leq 1,$ one has
$|E_{10}|^6  = |j|^2|j-1728|^3.$ Perhaps this will help?<|endoftext|>
TITLE: Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian?
QUESTION [41 upvotes]: I'm looking for a "conceptual" explanation to the question in the title. The standard proofs that I've seen go as follows: use the Schubert cell decomposition to get a basis for cohomology and show that the special Schubert classes satisfy Pieri's formula. Then use the fact that basic homogeneous symmetric functions $h_n$ are algebraically independent generators of the ring of symmetric functions, to get a surjective homomorphism from the ring of symmetric functions to the cohomology ring. Pieri's rule can be shown with some calculations, but is there any reason a priori to believe that tensor product multiplicities for the general linear group should have anything at all to do with the Grassmannian? 
Maybe a more specific question: is it possible to prove that these coefficients are the same without calculating them beforehand?
One motivation for asking is that the cohomology ring of the type B and type C Grassmannians ${\bf OGr}(n,2n+1)$ and ${\bf IGr}(n,2n)$ are described by (modified) Schur P- and Q-functions which seem to have nothing(?) to do with the representation theory of the orthogonal and symplectic groups ${\bf SO}(2n+1)$ and ${\bf Sp}(2n)$. So as far as I can tell the answer isn't just because general linear groups and Grassmannians are "type A" objects.

REPLY [4 votes]: There is some additional information on page 399 of Enumerative Combinatorics, vol. 2. The first conceptual explanations are due to Horrocks in 1957 and Carrell in 1978. See also pages 278-279 of Fulton's Intersection Theory.<|endoftext|>
TITLE: How bad can the recursive properties of finitely presented groups be?
QUESTION [8 upvotes]: Any finitely presented group naturally gives rise to an edge-labeled graph (the Cayley graph) and I am considering paths through this graph.  Paths correspond to infinite sequences of generators, so we can ask questions about the recursiveness (or not) of these sequences*.  For example, is there a group where any recursive sequence of generators necessarily corresponds to a self-intersecting path?
Has anyone seen any questions like this answered anywhere?  Any good resources for the recursive properties of finitely presented groups?  Maybe just an example or two of a really badly behaved (in a recursion-theoretic sense) group?
*To alleviate any potential confusion, an infinite sequence, {s_i} is recursive if there is a Turing Machine that, on input n, produces s_n.

REPLY [4 votes]: Your question is very interesting. I don't have a complete
answer.
First, let me note that in a finitely presented group, the
Cayley graph itself may not be a decidable graph, since to
know whether or not a node in the graph, which is a word in
the presentation, is trivial or not amounts exactly to the
word problem for that presentation. And since there are
finitely presented groups having an undecidable word
problem, there are finitely presented groups having an
undecidable Cayley graph.
This suggests an interesting sub-case of your problem, the
case when the Cayley graph is decidable. And in this case,
one can at least find a non-intersecting infinite path that
is low.
Let me explain. For any group presentation, one may form
the tree T of all finite non-intersecting paths. This is
the tree of attempts to build an infinite non-intersecting
path. Your question is equivalent to asking, when the group
is infinite, whether or not this tree has a computable
infinite branch. If the Cayley graph is decidable, then
this tree will be decidable. Thus, by the Low Basis
Theorem,
it follows that there is a branch b through the tree which
is low, meaning that the halting problem relative to b is
Turing equivalent to the ordinary halting problem. In
particular, this branch b is strictly below the halting
problem in the Turing degrees. This shows that any
computably-presented group with a decidable Cayley graph
admits a low non-intersecting path. Such a path is close to
being computable, but perhaps not quite computable. (Even in
this very special case when the Cayley graph is decidable,
I'm not sure whether there must be a computable
non-intersecting path.)
In the general case, the argument shows that for any group presentation of an infinite group, we can find an infinite non-intersecting path s, which has low degree relative to the Turing degree of the Cayley graph (low in the technical sense). I suspect that this is the best that one can say.
There is another classical theorem in computability that
seems relevant to your question, namely, the fact that
there are computable infinite binary branching trees T,
subtrees of 2ω, having no computable
infinite branches. These trees therefore constitute
violations of the computable analogue of Konig's lemma.
This problem is very like yours, isn't it? I am intrigued
by the possibility of using that classical construction to
build an example of the kind of group you seek.
There is a natural generalization of your question beyond
the finitely presented groups, to the class of
finitely-generated but computably presented groups. That
is, consider a group presentation with finitely many
generators and a computable list of relations. It may be
easier to find a instance of the kind of group you seek
having this more general form, since one can imagine a
priority argument, where one gradually adds relations as
the construction proceeds in order to meet various
requirements that diagonalize against the computable paths.

Here are some resources to general decidability questions in
finite group presentations: the undecidability of the word
problem,
the  conjugacy
problem,
and the  group isomorphism
problem.<|endoftext|>
TITLE: Why does the Gamma function satisfy a functional equation?
QUESTION [16 upvotes]: In question #7656, Peter Arndt asked why the Gamma function completes the Riemann zeta function in the sense that it makes the functional equation easy to write down.  Several of the answers were from the perspective of Tate's thesis, which I don't really have the background to appreciate yet, so I'm asking for another perspective.
The perspective I want goes something like this: the Riemann zeta function is a product of the local zeta functions of a point over every finite prime $p$, and the Gamma function should therefore be the "local zeta function of a point at the infinite prime."
Question 1:  Can this intuition be made precise without the machinery of Tate's thesis?  (It's okay if you think the answer is "no" as long as you convince me why I should try to understand Tate's thesis!)
Multiplying the local zeta functions for the finite and infinite primes together, we get the Xi function, which has the nice functional equation.  Now, as I learned from Andreas Holmstrom's excellent answer to my question about functional equations, for the local zeta functions at finite primes the functional equation
$$\zeta(X,n-s) = \pm q^{ \frac{nE}{2} - Es} \zeta(X, s)$$
(notation explained at the Wikipedia article), which for a point is just the statement $\frac{1}{1 - p^s} = -p^{-s} \frac{1}{1 - p^{-s}}$, reflects Poincare duality in etale cohomology, and the hope is that the functional equation for the Xi function reflects Poincare duality in some conjectural "arithmetic cohomology theory" for schemes over $\mathbb{Z}$ (or do I mean $\mathbb{F}_1$?).
Question 2:  Can the reflection formula for the Gamma function be interpreted as "Poincare duality" for some cohomology theory of a point "at the infinite prime"?  (Is this question as difficult to answer as the more general one about arithmetic cohomology?)

REPLY [12 votes]: Your questions are a part of what Deninger has been writing about for 20 years. He's proposed a point of view that sort of explains a lot of things about zeta functions. It's important to say that this explanation is more in a theoretical physics way than in a mathematical way, in that, as I understand it, he's predicted lots of new things which he and other people have then gone on to prove using actual mathematics. I guess it's kind of like the yoga surrounding the Weil conjectures before Dwork and Grothendieck made actual cohomology theories that had a chance to do the job (and eventually did). It's pretty clear to me that he's put his finger on something, but we just don't know what yet. 
Let me try to say a few things. But I should also say that I never worried too much about the details, because the details he has are about a made up picture, not the real thing. (If he had the real thing, everyone would be out of a job.) So my understanding of the actual mathematics in his papers is pretty limited.
Question 1: He gives some evidence that Euler factors at both finite and infinite places should be seen as zeta-regularized characteristic polynomials. For the usual Gamma function, see (2.1) in [1]. For the Gamma factors of general motives, see (4.1) in [1]. For the Euler factors at the finite places, see (2.3)-(2.7) in [2]. He gives a description that works simultaneously at the finite and infinite places in (0.1) of [2]. Beware that some of this is based on an artificial cohomology theory that is designed to make things uniform over the finite and infinite places. (Indeed, at the risk of speaking for him, probably the whole point was to see what such a uniform cohomology theory would look like, so maybe one day we'll be able to find the real thing.)
Question 2: He expects his cohomology theory to have a Poincare duality which is "compatible with respect to the functional equation". See the remarks and references in [3] between propositions 3.1 and 3.2.
I'd recommend having a look at [3]. It's mainly expository. Also, I remember [4] being a good exposition, but I don't have it in front of me now, so I can't say much. He also reviews things in section 2 of his recent Archive paper [5].
[1] "On the Gamma-factors attached to motives", Invent. Math. 104, pp 245-261
[2] "Local L-factors of motives and regularized determinants", Invent. Math. 107, pp 135-150
[3] "Some analogies between number theory and dynamical systems on foliated spaces", Proceedings of the ICM, Vol. I (Berlin, 1998), pp 163-186
[4] "Evidence for a cohomological approach to analytic number theory",  First ECM, Vol. I (Paris, 1992), pp 491-510
[5] "The Hilbert-Polya strategy and height pairings", arxiv.org<|endoftext|>
TITLE: Asymptotics of symmetry types of tensors
QUESTION [6 upvotes]: Introduction
Let's fix $m\in \mathbb N$. For each n, the unitary group $\mathbf U(m)$ is represented in the space of tensors of rank $n$ over $\mathbb C^m$
$$V_{n,m}=\bigotimes_{k=1}^n \mathbb C^m$$ and the symmetric group $S_n$ acts on $V_{n,m}$ by permutation of factors. Now the space $V_{n,m}$ breaks into the direct sum of subspaces $V_{n,m}(\lambda)$ which are primary with respect to each of these actions and irreducible with respect to the joint action of $S_n\times \mathbf U(m)$
$$V_{n,m}=\bigoplus_{\lambda \in \mathbb{Y}(n,m)}V_{n,m}(\lambda)$$
Where $\lambda$ ranges over all Young diagrams of size $n$ with at most $m$ rows (So $\lambda \in \mathbb{Y}(n,m)$ means $\lambda=(\lambda_1,\dots ,\lambda_k)$ is a partition of $n$ with $k\le m$). The tensors from $V_{n,m}(\lambda)$ are said to have symmetry type $\lambda$. We define the relative dimensions
$$d_{n,m}(\lambda)=\frac{\dim V_{n,m}(\lambda)}{\dim V_{n,m}}$$
which tell us how tensors are distributed into symmetry types.
Motivation and Question
I was reading Kerov's "Asymptotic representation theory of the symmetric group and it's aplications in analysis" and was trying to provide proofs for some of the results stated there. (He does give references, which I can't reach at the moment.)
The following two theorems are due to Kerov

Theorem 1 If for each $\lambda$ we associate $x=(x_1,\dots,x_m)$ with $x_k=\frac{\lambda ^{(n)}_k-n/m}{\sqrt{n}}$. The joint distribution of $x_k$'s as $n\to \infty$ with respect to the measure $d_{n,m}$ on $\mathbb{Y}(n,m)$ weakly converges to an absolutely continuous measure on the cone $C_m=\{x: x_1\geq x_2\geq\cdots \geq x_m \;;\;\sum x_k=0\}$
with density
$$\phi _m(x)=c \prod _{i < j}(x_i-x_j)^2 e^{-m/2 \sum x_k^2}$$ where
$$c=\frac{m^{(m-1)m/2}}{1!2!\cdots (m-1)!}\left(\frac{m}{2\pi}\right)^{(m-1)/2}$$

.

Theorem 2 Let $\lambda ^{(n)}\in \mathbb{Y}(n,m)$ be the Young diagram for which the tensors of type $\lambda ^{(n)}$ are most probable. Then $$\lim_{n\to \infty} \frac{\lambda ^{(n)}_k-n/m}{\sqrt{n/m}}=z_k$$
for each $k=1,2,\dots,m$ where $z_1,z_2,\dots z_m$ are the roots of the Hermite polynomial $H_m(z)$

I can prove Theorem 2 assuming Theorem 1, but I don't see a nice argument for proving the first theorem itself. Can anyone provide a sketch of the proof, or some hint how to approach Theorem 1?

REPLY [6 votes]: Kuperberg in Random words, quantum statistics, central limits, random matrices attributes Theorem 1 to Johansson and he gives two additional interesting proofs of this result.
A proof of a more general result is presented in my joint work with Benoit Collins Representations of Lie groups and random matrices.
Asymptotics when both $n$ and $m$ tend to infinity was considered by Biane in Approximate factorization and concentration for characters of symmetric groups.<|endoftext|>
TITLE: Yet more on distortion
QUESTION [7 upvotes]: I would like to elaborate a little bit on my previous question which can be found 
here.
Firstly, let me recall that a separable Banach space $(X, \| \cdot \|)$ is said to be
arbitrarily distortable if for every $r > 1$ there exists an equivalent norm $| \cdot |$
on $X$ such that for every infinite-dimensional subspace $Y$ of $X$ we can find a pair of
vectors $x, y$ in $Y$ such that $\|x\|=\|y\|=1$ and $|x| / |y| >r$. If $X$ has a Schauder
basis $(e_n)$, then this definition is equivalent to the following:

For every $r > 1$ there exists an equivalent norm $| \cdot |$ on $X$ such that for every
  normalized block sequence $(v_n)$ of $(e_n)$ there exist a non-empty finite subset $F$
  of $\mathbb{N}$ and a pair of vectors $x, y$ in $span\{v_n: n\in F\}$ such that $\|x\|=\|y\|=1$
  and $|x| / |y| >r$.

This equivalence gives us no hind of where the finite set $F$ is located. In other words:
if a Banach space $X$ with a Schauder basis is arbitrarily distortable, then where do
we have to search in order to find the vectors verifying that $X$ is arbitrarily distortable?
Now, there are various ways of quantifying Banach space properties and my question
is towards understanding who the "difficulty" for finding these vectors can be quantified.
The main tool will be certain families of finite subsets of $\mathbb{N}$. These families were discovered (independently) by two groups of researchers: Banach space theorists (Schreier families; see 1 below) and Ramsey theorists (uniform families; see 2 below). In particular, for the discussion below we need for every countable ordinal $\xi\geq 1$ a family $F_\xi$ of finite subsets of $\mathbb{N}$ such that:

$F_\xi$ is regular (i.e. compact, hereditary and spreading; I am sorry for not giving
the precise definition of these notions but this would make the post too long; but I
will be happy to answer to any comment).
The families are increasing (with respect to $\xi$) both in size and complexity. That is, the "order" of $F_\xi$ is at least $\xi$ and if $\zeta<\xi$ there there exists $k$ such that all subsets of $F_\zeta$ whose minimum is greater than $k$ belong to $F_\xi$.
For $\xi=1$, let us take the Schreier family consisting of all finite subsets of
$\mathbb{N}$ whose size (or cardinality if you prefer) is less than or equal to their minimum.

There many examples of such families, all constructed using transfinite induction. 
Some of them have extra important properties. For concreteness (and to simplify things)
let us work with the Schreier families.
Now we come to the following:

Definition: Let $(X,\| \cdot \|)$ be a Banach space with a Schauder basis $(e_n)$ and $\xi$ be a countable ordinal with $\xi\geq 1$. Let us say that $X$ is $\xi$-arbitrarily
  distortable if for every $r > 1$ there exists an equivalent norm $| \cdot |$ on $X$
  such that for every normalized block sequence $(v_n)$ of $(e_n)$ there exist a non-empty
  set $F$ belonging to the family $F_\xi$ and a pair of vectors $x, y$ in $span\{v_n: n\in F\}$ such
  that $\|x\|=\|y\|=1$ and $|x| / |y| >r$.

In other words, if $X$ is $\xi$-arbitrarily distortable, then we have narrow down the
search for the critical set $F$; it has to belong to an a priori given "nice" family
of finite subsets of $\mathbb{N}$.
For every Banach space $(X,\| \cdot \|)$ with a Schauder basis $(e_n)$ define

$$ AD(X)=\min\{ \xi: X is $\xi$-arbitrarily distortable\} $$
  if $X$ is $\zeta$-arbitrarily distortable for some $1\leq \zeta< \omega_1$. Otherwise set $AD(X)=\omega_1$.

One can prove the following equivalence:
Let $X$ be a separable Banach space with a Schauder basis.
Then $X$ is arbitrarily distortable if and only if $AD(X)<\omega_1$.
This leaves open a number of interesting questions.
Question 1: Is it true that $AD(\ell_2)>1$? This is just a restatement of my 
previous question.
Question 2: Can we compute $AD(\ell_p)$ for every $1 < p < +\infty$?
Question 3: Can we find for every countable ordinal $\xi\geq 1$ an arbitrarily distortable
Banach space $X_\xi$ such that $AD(X_\xi)>\xi$. The answer is yes for $\xi=1$; any arbitrarily
distortable asymptotic $\ell_1$ space $X$ satisfies $AD(X)>1$.
Notice that an affirmative answer to Question 3 leaves no hope for a "uniform" approach
to distortion on general separable Banach spaces. 

Some references:

D. Alspach and S. Argyros, Complexity of weakly null sequences, Dissertationes Math. 321 (1992), 1-44.
P. Pudlak and V. Rodl, Partition theorems for systems of finite subsets of integers,
Discrete Math. 39 (1982), 67-73.

REPLY [4 votes]: This is not an answer to your question but it's a very similar question and an observation concerning it. 
Suppose we ask ourselves the following question: is it true that for every equivalent norm on $\ell_2$ and every ε>0 there exists a block basis $v_1&lt;v_2&lt;\dots&lt;v_n$ such that $n$ is greater than the maximum of the support of $v_1$ and $v_1,\dots,v_n$ span a subspace $(1+&epsilon;)$-isomorphic to $\ell_2^n$?
This is a bit like asking for a block basis "of size ω". Dvoretzky's theorem says we can get $n$ for arbitrarily large $n$, and the negative solution of the distortion problem says that we can't find an infinite sequence (and therefore that there must be some countable ordinal that we can't reach, in a certain obvious sense).
Now there is reason to suppose that this question could be very very hard or perhaps even undecidable (though I am less sure of the latter). The reason is that it is very similar to the Paris-Harrington theorem. The Paris-Harrington theorem asks for a set $X$ such that its cardinality exceeds its minimal element and all its subsets of size $r$ have the same colour (and it starts after $m$, say). It is known to "require" the axiom of infinity in a certain sense: one can prove it trivially if one applies the infinite Ramsey theorem, but there is no proof within PA. But with this distortion variant, we don't have the infinite Ramsey theorem to apply as it's false! So how does one go about thinking about this? I think the example of Odell and Schlumprecht shows that you can't get to $&omega;^2$.
I wonder whether your question could also be affected by concerns of this kind.
Added slightly later: I now see that this might have been more appropriate as an answer to your previous question, which is one that I have wondered about myself (as the above makes clear).<|endoftext|>
TITLE: A Peculiar Model Structure on Simplicial Sets?
QUESTION [22 upvotes]: I'm wondering if there is a Quillen model structure on the category of simplicial sets which generalizes the usual model structure, but where every simplicial set is fibrant? I want to use this to do homological algebra for commutative monoids, but first let me explain some background, my motivation, and articulate more precisely what I am after.
Background
A Quillen Model structure on a category has three classes of morphisms: fibrations, weak equivalences, and cofibrations. This structure allows one to do many advanced homotopical constructions mimicking the homotopy theory of (nice) topological spaces. There is a notion of Quillen equivalence between model categories which consists of a particular adjunction between the two model categories in question. This gives you "equivalent homotopy theories" for the two model categories in question.
The usual Model structure on simplicial sets has fibrations the Kan fibrations, the weak-equivalences are the maps which induces isomorphisms of homotopy groups, and the cofibrations are the (levelwise) inclusions. This is equivalent to the usual model category of topological spaces, and the Quillen equivalence is realized by the adjoint pair of functors: the geometric realization functor and the singular functor.
Quillen model categories are also useful for doing homological algebra, and particularly for working with derived categories. For reasonable abelian categories there are several nice (Quillen equivalent) model category structures on the category of (possibly bounded) chain complexes which one can use which reproduce the derived category. More precisely the homotopy category of the model category is the derived category and the "homotopical constructions" I mentioned above, in this case, correspond to the notion of (total) derived functor.
This story is further enriched by the Dold-Kan correspondence which is an equivalence between the categories of positively graded chain complexes of, say, abelian groups and the category of simplicial abelian groups, a.k.a. simplicial sets which are also abelian group objects. This in turn is Quillen equivalent to a model category of topological abelian groups.
Previous MathOverflow Question and Progress
Previously I asked a question on MO about doing homological algebra for commutative monoids. I got many fascinating and exciting answers. Some were more or less "here is something you might try", some were more like "here is a bit that people have done, but the full theory hasn't been studied". After hearing those answers I'm much more excited about the field with one element. But still, in the end none of the answers really had what I was after. Then Reid Barton asked his MO question.
This got me thinking about commutative monoids and simplicial commutative monoids again. I had some nice observations which have lead me to the question at hand.

The first is that while a simplicial abelian group is automatically a Kan simplicial set  (i.e. if you forget the abelian group structure what have sitting in front of you is a Kan simplicial set) this is not the case for simplicial commutative monoids. A simplicial commutative monoid does not have to be a Kan simplicial set.

Next, I realized that the (normalized!) Dold-Kan correspondence still seems to work. You can go from a simplicial commutative monoid to a "complex" of monoids, ( and back again It is just an adjunction. Thanks, Reid, for pointing this out!).

If your simplicial commutative monoid is also a Kan complex, then under the Dold-Kan correspondence you get a complex where the bottom object is a commutative monoid, but all the rest are abelian groups (this was pointed out to me by Reid Barton in a conversation we had recently). Thus the theory of topological commutative monoids (which was one of the suggested answers to my previous question), which has a nice model category structure (see Clark Barwick's answer to an MO question), should be equivalent to a theory of chain complexes of this type. It shouldn't model arbitrary complexes of commutative monoids.

If you have a simplicial set which is not necessarily a Kan complex you can still define the naive simplicial homotopy "groups" $\pi_iX$. I put "goups" in quotes because for a general simplicial set these are just pointed sets. If your complex is Kan, these are automatically groups (for $i>0$). If your simplicial set is a simplicial abelian group, these are abelian groups (for all $i$) and they are precisely the homology groups of the chain complex you get under the Dold-Kan correspondence. If your simplicial set is a commutative monoid, but not Kan, then they are commutative monoids and are again the "homology" monoids of the chain complex you get under the Dold-Kan correspondence.


The Question
All this suggests that there should be a Quillen model structure on simplicial commutative monoids in which the weak equivalences are the $\pi_{\bullet} $-isomorphisms, where here $\pi_{\bullet} $ denotes the naive simplicial version, i.e. these are commutative monoids, not groups. I'm sure if such a thing was well known then it would have been mentioned as an answer to my previous question.  I'd really like to see something that generalizes the usual theory of abelian groups. That way if we worked with simplicial abelian groups and construct derived functors we would just reproduce the old answers. As a stepping stone, there should be a companion model structure on simplicial sets which, I think, is more likely to be well known.
One of the properties that I think this hypothetical model structure on simplicial sets should have is that every simplicial set is fibrant, not just the Kan simplicial sets.

Question: Is there a model structure on simplicial sets in which every simplicial set is fibrant, and such that the weak equivalences between the Kan complexes are exactly the usual weak equivalences? Specifically can the weak equivalences be taken to be those maps which induce $\pi_*$-isomorphism, where these are the naive simplicial homotopy sets?

If this model structure exists, I'd like to know as much as possible about it. If you have any references to the literature, I'd appreciate those too, but the main question is as it stands.

REPLY [5 votes]: I've gotten a bit lost with the revisions and subquestions, but let me try to summarize some of the discussions of the Bourbon Seminar about this question.
First, note that the usual argument that abelian group objects are automatically fibrant falls apart entirely in the context of commutative monoids. More specifically, the "least fibrant" simplicial sets out there admit a commutative monoid structure.
Example. Consider the "principal spine" $P:=\Delta^1\sqcup^{\Delta^0}\Delta^1\sqcup^{\Delta^0}\cdots\sqcup^{\Delta^0}\Delta^1\subset\Delta^n$. The set $P_0=\{0,1,\dots,n\}$ of $0$-simplices admit a commutative monoid structure given by taking the maximum. This extends in a canonical fashion to the simplicial set $P$ itself. Observe that this kind of example satisfies no inner horn-filling condition.
Second, to enlarge slightly on Reid's comment above, note that, unlike the situation with abelian group objects, the homotopy groups of a commutative monoid based at the identity cannot characterize the weak equivalences, even between fibrant objects. That is, the functor $\pi_{\ast}:s\mathbf{Comm}\to\mathbf{Ab}^{\mathrm{gr}}$ is not conservative in the sense that a map $X\to Y$ between Kan-fibrant commutative monoids is a weak equivalence if and only if $\pi_{\ast}(X)\to\pi_{\ast}(Y)$ is an isomorphism. (Of course it would be if we restricted to the class of connected objects.) This leads to the following, which should be at the heart of any Dold-Kan correspondence.
Challenge. Find an "algebraic" category $\mathbf{A}$ and a conservative functor $\Pi:s\mathbf{Comm}\to\mathbf{A}$.
Of course there are plenty of pairs $(\mathbf{A},\Pi)$ of this kind. (After all, the initial one is the homotopy category.) But it is not obvious how algebraic these can be made.<|endoftext|>
TITLE: Reference for cohomology vanishing
QUESTION [6 upvotes]: If $U \subset \mathbb C^n$  is the complement of a closed analytic subset of codimension at least three then there is a result of Cartan which says that $H^1(U,\mathcal O^{analytic}_U)=0$, see page 133 of "Theory of Stein Spaces" by Grauert and Remmert.
Does anyone know a reference for the analogous result in the algebraic category ?

REPLY [4 votes]: To expand on Emerton's answer: Using the excision sequence, Cartan's result in the algebraic case boils down to showing the following: Let $R$ be a regular local ring, and $I$ and ideal of height at least $3$, then $H^i_I(R)=0$ for $i\leq 2$.This follows because: $$H^i_I(R) = lim \ Ext^i(R/I^n,R)$$  
And $I^n$, being height $3$, always contains a regular sequence of length $2$, so the $Ext^i$ vanishes for $i\leq 2$ by standard result (see Bruns-Herzog Cohen Macaulay book, Proposition 1.2.10 for example). This argument extends to the case of codimension at least $n$ and vanishing of $H^{n-2}$. 
Incidentally, a pretty non-trivial question is to find upper bound for the vanishing of local cohomology modules, in other words, the cohomological dimension of a subvariety $Z$. Many strong results have been obtained after SGA, by Hartshorne, Ogus, Faltings, Huneke-Lyubeznik, etc. All those references can be found in
Lyubeznik's paper (they were mentioned in the very first page) which primarily treated the vanishing of etale cohomology.<|endoftext|>
TITLE: PNT for general zeta functions, Applications of.
QUESTION [16 upvotes]: When I read it for the first time, I found the whole slog towards proving the Prime Number Theorem and the final success to be magnificent. So I am curious about more general results.
We talk of complex $\zeta$-functions and $L$-functions. As a preliminary list, we fix the following list. But feel free to add to it.
$1$. Riemann $\zeta$-function.
$2$. Dedekind $\zeta$-function for a number field.
$3$. Artin $L$-functions for a character of the Galois group of some number field.
$4$. Zeta functions of algebraic varieties over number fields; for getting analytic continuation up to $s = 0$, we for instance fix the zeta function of an elliptic curve defined over $\mathbb{Q}$ which is modular in the sense of Eichler-Shiumura.
I am worried about $4$ here, since the requirement that the zeta function has a pole at $s=1$ is not fulfilled, and thus we fail to capture the main term from the residue there. Is something possible in this case? If so, we can consider that and also more general zeta functions of algebraic varieties over any finitely generated field, zeta functions of Galois representations, etc.. which have analytic continuation up to $s=0$ and the question could be extended to those cases as well.
We have the original prime number theorem, in the following form,
$\pi (x) = \frac{x}{log  x} + O(error\ term)$.
which is proved by integration of the Riemann zeta function along a rectangular contour including $s = 1$, and letting the vertical edges get longer and longer, and estimating the integrals. The reference I have in mind is Ram Murty, Problems in Analytic Number Theory, or, J. Ayoub's book.
Now the questions:

What is the known about a similar prime number theorem for more general zeta and $L$-functions?

I could imagine that it will be pretty straightforward for the Dedekind zeta function. But then I am curious about the error term.
For the rest of the cases, does a similar proof of the PNT carry over? More importantly, 

What is the "meaning" of the prime number theorem in these general cases?

Finally, 

What are some applications(to other problems) of such a prime number theorem proved with a good error term? 

The applications of the original PNT are of course well-known, as for instance given in the books of Titchmarsch and Heath-Brown, or Ivic, or in the more modern book of Iwaniec and Kowalski.

REPLY [8 votes]: Hi Anweshi,
Since Emerton answered your third grey-boxed question very nicely, let me try at the first two.  Suppose $L(s,f)$ is one of the L-functions that you listed (including the first two, which we might as well call L-functions too).  (For simplicity we always normalize so the functional equation is induced by $s\to 1-s$.)  This guy has an expansion $L(s,f)=\sum_{n}a_f(n)n^{-s}$  as a Dirichlet series, and the most general prime number theorem reads
$\sum_{p\leq X}a_f(p)=r_f \mathrm{Li}(x)+O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
Here $\mathrm{Li}(x)$ is the logarithmic integral, $r_f$ is the order of the pole of $L(s,f)$ at the point $s=1$, and the implied constant depends on $f$ and $\varepsilon$.
Let's unwind this for your examples. 
1) The Riemann zeta function has a simple pole at $s=1$ and $a_f(p)=1$ for all $p$, so this is the classical prime number theorem.
2) The Dedekind zeta function (say of a degree d extension $K/\mathbb{Q}$) is a little different.  It also has a simple pole at $s=1$, but the coefficients are determined by the rule: $a(p)=d$ if $p$ splits completely in $\mathcal{O}_K$, and $a(p)=0$ otherwise.  Hence the prime number theorem in this case reads
$|p\leq X \; \mathrm{with}\;p\;\mathrm{totally\;split\;in}\;\mathcal{O}_K|=d^{-1}\mathrm{Li}(x)+O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
This already has very interesting applications: the fact that the proportion of primes splitting totally is $1/d$ was very important in the first proofs of the main general results of class field theory. 
3) If $\rho:\mathcal{G}_{\mathbb{Q}}\to \mathrm{GL}_n(\mathbb{C})$ is an Artin representation then $a(p)=\mathrm{tr}\rho(\mathrm{Fr}_p)$.  If $\rho$ does not contain the trivial representation, then $L(s,\rho)$ has no pole in  neighborhood of the line $\mathrm{Re}(s)\geq 1$, so we get
$\sum_{p\leq X}\mathrm{tr}\rho(\mathrm{Fr}_p)=O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
The absence of a pole is not a problem: it just means there's no main term!  In this particular case, you could interpret the above equation as saying that "$\mathrm{tr}\rho(\mathrm{Fr}_p)$ has mean value zero.
4) For an elliptic curve, the same phenomenon occurs.  Here again there is no pole, and $a(p)=\frac{p+1-|E(\mathbb{F}_p)|}{\sqrt{p}}$.  By a theorem of Hasse these numbers satisfy $|a(p)|\leq 2$, so you could think of them as the (scaled) deviation of $|E(\mathbb{F}_p)|$ from its 
"expected value" of $p+1$. In this case the prime number theorem reads
$\sum_{p\leq X}a(p)=O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$
so you could say that "the average deviation of $|E(\mathbb{F}_p)|$ from $p+1$ is zero."
Now, how do you prove generalizations of the prime number theorem?  There are two main steps in this, one of which is easily lifted from the case of the Riemann zeta function.

Prove that the prime number theorem for $L(s,f)$ is a consequence of the nonvanishing of $L(s,f)$ in a region of the form $s=\sigma+it,\;\sigma \geq 1-\psi(t)$ with $\psi(t)$ positive and tending to zero as $t\to \infty$.  So this is some region which is a very slight widening of $\mathrm{Re}(s)>1$.  The proof of this step is essentially contour integration and goes exactly as in the case of the $\zeta$-function.
Actually produce a zero-free region of the type I just described.  The key to this is the existence of an auxiliary L-function (or product thereof) which has positive coefficients in its Dirichlet series.  In the case of the Riemann zeta function, Hadamard worked with the auxiliary function $ A(s)=\zeta(s)^3\zeta(s+it)^2 \zeta(s-it)^2 \zeta(s+2it) \zeta(s-2it)$.  Note the pole of order $3$ at $s=1$; on the other hand, if $\zeta(\sigma+it)$ vanished then $A(s)$ would vanish at $s=\sigma$ to order $4$.  The inequality $3<4$ of order-of-polarity/nearby-order-of-vanishing leads via some analysis to the absence of any zero in the range $s=\sigma+it,\;\sigma \geq 1-\frac{c}{\log(|t|+3)}.$  In the general case the construction of the relevant auxiliary functions is more complicated.  For the case of an Artin representation, for example, you can take $B(s)=\zeta(s)^3 L(s+it,\rho)^2 L(s-it,\widetilde{\rho})^2 L(s,\rho \otimes \widetilde {\rho})^2 L(s+2it,\rho \times \rho) L(s-2it,\widetilde{\rho} \times \widetilde{\rho})$.  The general key is the Rankin-Selberg L-functions, or more complicated L-functions whose analytic properties can be controlled by known instances of Langlands functoriality.

If you'd like to see everything I just said carried out elegantly and in crystalline detail, I can do no better than to recommend Chapter 5 of Iwaniec and Kowalski's book "Analytic Number Theory."<|endoftext|>
TITLE: Central simple algebras approach to classfield theory, Merits of.
QUESTION [14 upvotes]: As noted earlier, I found reading Weil's book "Basic Number Theory" to be a harrowing experience, and I find his writing to be intrinsically hard to understand, though it is perfectly rigorous and clean.
In any case, the second part of the book is "thoroughly unmodern", in the author's own terms. He dispenses with cohomology and builds the whole theory based on central simple algebras. And he advises the reader to make it an exercise to himself for working out the "hidden cohomology" in it.
I personally felt this as some kind of perversion(just as in the fixation for Haar measure in the first part) to show the reader that he is a powerful mathematician could do things precisely the way he wants, and my feeling was strengthened by the title "Basic Number Theory".
However since he is a great mathematician, and I am nobody, I feel unqualified dismiss him like that. There must be some merits/uses of doing it this way, rather than using cohomology. If somebody can enlighten me, I would be very grateful. 
Again, is it the case that classfield theory can be done with just $H^1$? I wanted to study the subject and attempted, but was put off for a long time by the wrong book choice, reading Weil which felt like drinking stone soup, and it takes me time to take up the subject again.

REPLY [13 votes]: If you are looking for a more conceptual way of understanding the central simple algebras approach to class field theory, I think that you would be well advised to look at Roquette's book The Brauer-Hasse-Noether Theorem in Historical Perspective. The Brauer-Hasse-Noether Theorem, oftentimes referred to as the Albert-Brauer-Hasse-Noether theorem (Adrian Albert, an american, discovered part of the theorem independently), states that a central simple algebra defined over a number field splits globally if and only if it splits everywhere locally and is intimately connected with class field theory. 
The entire book is only 80 pages long, and takes only a few hours to read. You should especially look at Chapter 6: The Brauer group and class field theory. The last few pages of this chapter make explicit the application of work done with central simple algebras to class field theory and the connection between this approach and the more modern cohomological approach.
[Edit 1] As Franz has pointed out, applying the theory of central simple algebras to class field theory is not just another approach, but rather was the original approach. The connection with cohomology arises as follows: 
For a field extension $K/k$, denote by $Br(K/k)$ the Relative Brauer group of $k$ with respect to $K$. Explicitly, it is the kernel of the homomorphism $Br(k) \rightarrow Br(K)$. Then $Br(k)=\cup Br(K/k)$. It isn't too hard to show that the elements of these relative Brauer groups (for $K/k$ Galois) are in one to one correspondence with certain factor sets relative to $K$. These factor sets naturally arise as the elements of  $H^2(K/k)$, ultimately yielding an isomorphism $Br(K/k)\cong H^2(K/k)$. This isomorphism allows one to translate results like the Albert-Brauer-Hasse-Noether theorem into the perhaps more familiar cohomological language.
[Edit 2] Apparently Roquette's book is available online (along with many other gems).<|endoftext|>
TITLE: MaxSpec, Spec, ... "RadSpec"? Or, why not look at all radical ideals?
QUESTION [27 upvotes]: I was reading this question on why algebraic geometry looks at prime ideals instead of only maximal ideals, and I understand Anton's answer, but I'm a little confused as to how this fits with Hilbert's Nullstellensatz - affine algebraic sets are in bijection with radical ideals, not prime ideals, and it seems like we'd want the extra information we'd get by looking at "RadSpec(R)" (my own imagined notation). Also, the preimage of a radical ideal is radical, so there isn't the same objection as to maximal ideals - "RadSpec" would also be a contravariant functor.
So, why not radical ideals instead of only prime ideals, and what kinds of things could we say about RadSpec(R) even if they aren't very interesting?

REPLY [3 votes]: The category of reduced $k$-algebras over a field $k$ is an example of a so-called Zariski category; see the book "Categories of commutative algebras" by Yves Diers. In every Zariski category, basic notions of commutative algebra can be developped and over every Zariski category $A$ we can consider the category of $A$-schemes. The theory shows that, in fact, most of the basic algebraic geometry takes over to this general setting. And if $A$ is a so-called rational reduced Zariski category, it turns out that the basic notions of varieties take over (such as birational equivalence is detected by the function field object, and Hilbert's Nullstellensatz). If $A$ denotes the category of reduced $k$-algebras, we get the usual category of reduced $k$-schemes. So in some sense, this is an algebraic geometry in its own right which you can study. The functor of points approach also works with functors $A \to \text{Sets}$. Note, however, that here the spectrum of a reduced $k$-algebra also consists of the usual prime ideals, not all radical ideals. Nevertheless, you may read Section 2.4 in Dier's book. There radical congruences in Zariski categories are studied in general.<|endoftext|>
TITLE: "Algebraic" topologies like the Zariski topology?
QUESTION [51 upvotes]: The fact that a commutative ring has a natural topological space associated with it is still a really interesting coincidence.  The entire subject of Algebraic geometry is based on this simple fact.
Question: Are there other categories of algebraic objects that have interesting natural topologies that carry algebraic data like the Zariski topology on a ring (spectrum)?  If they exist, what are they and how are they used?

REPLY [6 votes]: carry algebraic data like the Zariski topology on a ring (spectrum)? If they exist, what are they and how are they used? 

In model theory they define and study the 'algebraic data like the Zariski topology' irrespectively of where these data come from. These data are called Zariski geometries, and e.g. admit some intersection theory, can be used to prove Chow's lemma, admit some classification results in dim 1 etc.  You may want to have a look on 
the recent book of Zariski geometries and references therein (or the actual book Zariski Geometries : Geometry from the Logician's Point of View, by Boris Zilber). 
Also, the book has some examples which sometimes need some work.<|endoftext|>
TITLE: Is the Burnside ring a lambda-ring? + conjecture in Knutson p. 113
QUESTION [8 upvotes]: Warning: I'll be using the "pre-$\lambda$-ring" and "$\lambda$-ring" nomenclature, as opposed to the "$\lambda$-ring" and "special $\lambda$-ring" one (although I just used the latter a few days ago on MO). It's mainly because both sources use it, and I am (by reading them) slowly getting used to it.
Let $G$ be a finite group. The Burnside ring $B\left(G\right)$ is defined as the Grothendieck ring of the category of finite $G$-sets, with multiplication defined by cartesian product (with diagonal structure, or at least I have difficulties imagining any other $G$-set structure on it; please correct me if I am wrong).
For every $n\in\mathbb{N}$, we can define a map $\sigma^n:B\left(G\right)\to B\left(G\right)$ as follows: Whenever $U$ is a $G$-set, we let $\sigma^n U$ be the set of all multisets of size $n$ consisting of elements from $U$. The $G$-set structure on $\sigma^n U$ is what programmers call "map": an element $g\in G$ is applied by applying it to each element of the multiset. This way we have defined $\sigma^n U$ for every $G$-set $U$; we extend the map $\sigma^n$ to all of $B\left(G\right)$ (including "virtual" $G$-sets) by forcing the rule
$\displaystyle \sigma^i\left(u+v\right)=\sum_{k=0}^i\sigma^k\left(u\right)\sigma^{i-k}\left(v\right)$ for all $u,v\in B\left(G\right)$.
Ah, and $\sigma^0$ should be identically $1$, and $\sigma^1=\mathrm{id}$. Anyway, this works, and gives a "pre-$\sigma$-ring structure", which is basically the same as a pre-$\lambda$-ring structure, with $\lambda^i$ denoted by $\sigma^i$. Now, we turn this pre-$\sigma$-ring into a pre-$\lambda$-ring by defining maps $\lambda^i:B\left(G\right)\to B\left(G\right)$ by
$\displaystyle \sum_{n=0}^{\infty}\sigma^n\left(u\right)T^n\cdot\sum_{n=0}^{\infty}\left(-1\right)^n\lambda^n\left(u\right)T^n=1$ in $B\left(G\right)\left[\left[T\right]\right]$ for every $u\in B\left(G\right)$.
Now, let me quote two sources:
Donald Knutson, $\lambda$-Rings and the Representation Theory of the Symmetric Group, 1973, p. 107: "The fact that $B\left(G\right)$ is a $\lambda$-ring and not just a pre-$\lambda$-ring - i. e., the truth of all the identities - follows from [...]"
Michiel Hazewinkel, Witt vectors, part 1, 19.46: "It seems clear from [370] that there is no good way to define a $\lambda$-ring structure on Burnside rings, see also [158]. There are (at least) two different choices giving pre-$\lambda$-rings but neither is guaranteed to yield a $\lambda$-ring. Of the two the symmetric power construction seems to work best."
(No, I don't have access to any of these references.)
For a long time I found Knutson's assertion self-evident (even without having read that far in Knutson). Now I tend to believe Hazewinkel's position more, particularly as I am unable to verify one of the relations required for a pre-$\lambda$-ring to be a $\lambda$-ring:
$\lambda^2\left(uv\right)=\left(\lambda^1\left(u\right)\right)^2\lambda^2\left(v\right)+\left(\lambda^1\left(v\right)\right)^2\lambda^2\left(u\right)-2\lambda^2\left(u\right)\lambda^2\left(v\right)$ for $B\left(G\right)$.
What also bothers me is Knutson's "conjecture" on p. 113, which states that the canonical (Burnside) map $B\left(G\right)\to SCF\left(G\right)$ is a $\lambda$-homomorphism, where $SCF\left(G\right)$ denotes the $\lambda$-ring of super characters on $G$, with the $\lambda$-structure defined via the Adams operations $\Psi^n\left(\varphi\left(H\right)\right)=\varphi\left(H^n\right)$ (I think he wanted to say $\left(\Psi^n\left(\varphi\right)\right)\left(H\right)=\varphi\left(H^n\right)$ instead) for every subgroup $H$ of $G$, where $H^n$ means the subgroup of $G$ generated by the $n$-th powers of elements of $H$. This seems wrong to me for $n=2$ and $H=\left(\mathbb Z / 2\mathbb Z\right)^2$ already. And if the ring $B\left(G\right)$ is not a $\lambda$-ring, then this conjecture is wrong anyway (since the map $B\left(G\right)\to SCF\left(G\right)$ is injective).
Can anyone clear up this mess? I am really confused...
Thanks a lot.

REPLY [6 votes]: I've just gone and looked up [158] (Gay, C. D.; Morris, G. C.; Morris, I.
Computing Adams operations on the Burnside ring of a finite group.
J. Reine Angew. Math. 341 (1983), 87--97.
On p. 90, at the end of section 2, they say: "Knutson conjectured that the Adams operations on SCF(G) inherited from A(G) [=Burnside ring of G] are given by [the formula you mentioned, involving the subgroup generated by nth powers of a subgroup $K$].  We will show that this is correct if $K$ is cyclic, but not true in general."
I haven't looked at it carefully, but they appear to give some more complicated looking formulas for the action of Adams operations on super-characters, valid in some cases.
They don't seem to mention Knutson's claim that the Burnside ring is a lambda-ring (not merely pre-lambda).<|endoftext|>
TITLE: What is the general opinion on the Generalized Continuum Hypothesis?
QUESTION [37 upvotes]: I'm community wikiing this, since although I don't want it to be a discussion thread, I don't think that there is really a right answer to this.  
From what I've seen, model theorists and logicians are mostly opposed to GCH, while on the other end of the spectrum, some functional analysis depends on GCH, so it is much better tolerated among functional analysts.  In fact,   I considered myself very much +GCH for a while, but Joel and Francois noted some interesting stuff about forcing axioms, (the more powerful ones directly contradict CH).  
What is the general opinion on GCH in the mathematical community (replace GCH with CH where necessary)? Does it happen to be that CH/GCH doesn't often come up in algebra?  
Please don't post just post "I agree with +-CH".  I'd like your assessment of the mathematical community's opinion.  Maybe your experiences with mathematicians you know, etc. Even your own experiences or opinion can work.  I am just not interested in having 30 or 40 one line answers. Essentially, I'm not looking for a poll.  
Edit: GCH=Generalized Continuum Hypothesis
CH= Continuum Hypothesis
CH says that $\aleph_1=\mathfrak{c}$.  That is, the successor cardinal of $\aleph_0$ is the continuum.  The generalized form (GCH) says that for any infinite cardinal $\kappa$, we have $\kappa^+=2^\kappa$, that is, there are no cardinals strictly between $\kappa$ and $2^\kappa$.
Edit 2 (Harry): Changed the wording about FA.  If it still isn't true, and you can improve it, feel free to edit the post yourself and change it.

REPLY [34 votes]: Here is a historical answer of sorts. I'm looking at a copy of a spirit-duplicated questionnaire, dated August 1, 1967, which was circulated at the AMS-ASL 1967 Summer Institute in Axiomatic Set Theory. The notation "80 ballots cast" is pencilled in, rather sloppily. The tally of votes for each answer is inked in by someone with neat handwriting. From the numbers, I surmise that IC was only answered by those who answered "meaningless" to IA.
It would be interesting to know if this survey has been published somewhere.

AMS-ASL Summer Institute
  in
  Axiomatic Set Theory
OFFICIAL BALLOT
[pencilled in: "80 ballots cast"]
I. A. I believe that the proposition
$\ \ \ \ \ \ \ \ \ $'The axiom MC of measurable cardinals is true in the real universe of sets'
is
$\quad$(38) meaningful $\quad$ (38) meaningless
$\ \ \ \ $B. (To be answered only if your answer to A is "meaningful")
$\ \ \ \ \ \ \ \ \ $(8) I think that MC is almost certainly true
  $\ \ \ \ \ \ \ \ \ $(7) I think MC is more likely true than false
  $\ \ \ \ \ \ \ \ \ $(7) I think MC is more likely false than true
  $\ \ \ \ \ \ \ \ \ $(2) I think MC is almost certainly false
  $\ \ \ \ \ \ \ $(14) I have no idea whether MC is true or false
$\ \ \ \ $C. Regarding the prediction that MC will someday be refuted in ZF,
$\ \ \ \ \ \ \ \ \ $(0) I think this prediction is almost certainly true
  $\ \ \ \ \ \ \ \ \ $(2) I think this prediction is more likely true than false
  $\ \ \ \ \ \ \ $(16) I think this prediction is more likely false than true
  $\ \ \ \ \ \ \ \ \ $(8) I think this prediction is almost certainly false
  $\  \ \ \ \ \ \ \ \ $(4) I have no idea whether this prediction is true or false
II. A. I believe that the proposition
'The continuum hypothesis CH is true in the real universe of sets'
is
$\quad$(42) meaningful $\quad$ (35) meaningless
$\ \ $B. (To be answered only if your answer to IIA is 'meaningful')
$\ \ \ \ \ \ $(2) I think CH is almost certainly true
  $\ \ \ \ \ \ $(2) I think CH is more likely true than false
  $\ \ \ \ $(12) I think CH is more likely false than true
  $\ \ \ \ $(14) I think CH is almost certainly false
  $\ \ \ \ $(12) I have no idea whether CH is true or false.
$\ \ $B'. (To be answered only if your answer to IIA is 'meaningless')
$\ \ \ \ $(1) My position on IIA
$\quad\quad$(2) does$\quad$(33) does not
$\ \ \ \ \ \ $cast doubt in my own mind on the value of set theory.
$\ \ \ \ $(2) I am inclined to think that set theory based on the continuum
  $\ \ \ \ \ \ \ \ \ $hypothesis is destined to play in the long-range future develop-
  $\ \ \ \ \ \ \ \ \ $ment of mathematics a
$\ \ \ \ \ \ \ \ \ $(11) more important role than
  $\ \ \ \ \ \ \ \ \ $(13) role of equal importance with
  $\ \ \ \ \ \ \ \ \ $(11) less important role than
$\ \ \ \ \ \ \ \ \ $set theory based on the denial of the continuum hypothesis.
$\ \ $ C. Assuming that human mathematicians still exist then, I believe that
  $\ \ \ \ \ \ \ $in 2067 the prevailing opinion among them will be that the continuum
  $\ \ \ \ \ \ \ $problem:
$\ \ \ \ $(4) has been settled by the discovery of generally accepted new
  $\ \ \ \ \ \ \ \ \ $axioms or methods of proof of which the continuum hypothesis is
  $\ \ \ \ \ \ \ \ \ $a consequence
  $\ \ $(18) has been settled by the discovery of generally accepted new
  $\ \ \ \ \ \ \ \ \ $axioms or methods of proof of which the denial of the continuum
  $\ \ \ \ \ \ \ \ \ $hypothesis is a consequence
  $\ \ $(37) has been settled by the general acceptance of the belief that
  $\ \ \ \ \ \ \ \ \ $there is no one true set theory and that the continuum hypothesis
  $\ \ \ \ \ \ \ \ \ $simply holds in some theories and fails in others
  $\ \ $(11) is still unsettled
III. A. I believe that there is an absolute sense in which every sentence of
  $\ \ \ \ \ \ \ \ \ $first-order number theory based on addition, multiplication, and
  $\ \ \ \ \ \ \ \ \ $exponentiation is either true or false.
$\quad$(54) yes $\quad$ (26) no
$\ \ \ \ $B. I believe that there is an absolute sense in which every $\underline{\text{universal}}$
  $\ \ \ \ \ \ \ \ \ $sentence of first-order number theory based on addition, multiplication,
  $\ \ \ \ \ \ \ \ \ $and exponentiation is either true or false.
$\quad$(62) yes $\quad$ (18) no
Please do not sign your ballot.
August 1, 1967
  University of California, Los Angeles<|endoftext|>
TITLE: Existence of non-commutative desingularizations
QUESTION [11 upvotes]: Let $R$ be normal, local ring of dimension at least $2$. Let $M$ be a reflexive $R$-module and let $A=Hom_R(M,M)$. Suppose $A$ has finite global dimension. Then one can view $A$ as a weak non-commutative desingularization of $R$ (note that, a) there is a natural map $R\to A$ and b) in the commutative case, finite global dimension implies regularity, hence the name).
This concept imitates Van den Bergh's definition of non-commutative crepant resolution (NCCR). His definition arises from a proof of dimension $3$ case of Bondal-Orlov conjecture. This is a long story, but an excellent account of the reasons behind the definition can be found in Section 4 of this paper. For existence of NCCR in some high dimensions case, check out this.
Now, non-commutative crepant resolution does not always exist (the above papers proves the equivalence, in some cases, with existence of projective crepant resolutions, which exists rarely in high dimensions). What we do know from Hironaka, in characteristic $0$ at least, is that resolution of singularity exist. So:
Question: Does weak non-commutative desingularizations of $R$ (as specified in the first paragraph) always exist?
Some discussions (I am a beginner in this, so feel free to correct me):

Why weak? By Morita equivalence, to ensures the desingularization is an isomorphism on the regular locus one needs $M$ to be free on that locus of $R$.

If one requires extra conditions (like $M$ being an generator-cogenerator) then there are examples when such desingularization does not exist (in some paper by Iyama which I forgot the name). I have not seen an example in the generality above.

There are positive results in diemension $0,1$, but I care about normal rings, so we start in dimension $2$.

Some people like Van den Bergh or Lieven le Bruyn probably know. May be they are even on MO!


I would appreciate even heuristic reasons for one way or another.
EDIT: The question is now resolved, by Lieven's answer below (as expected (:). I will provide a little bit more details in case someone is interested: Van den Bergh and Stafford proved that, in characteristic $0$, if $A$ is a non-commutative crepant resolution, then $R$ has rational singularity. The definition of NCCR is stronger, but if, for example, $R$ is Gorenstein of dimension $2$, it coincides with my version. So a counterexample is something like $R=k[x,y,z]/(x^3+y^3+z^3)$, which is a non-rational hypersurface.

REPLY [8 votes]: There are already counter-examples in dimension 2. If you take a 2-dml non-rational singularity, then there cannot exist a non-commutative resolution in your sense. In fact, any 2-dml nc-resolution your sense is also an nc-resolution in Michel's sense and so must have rational singularities by a result of Toby Stafford and Michel.
In dimension two you can resort to ancient stuff such as the book 'Graded orders' by Fred, Michel and me, an online scanned version can be found here (Wayback Machine). Lemma IV.2.3 says that a tame order (such as your End(M)) of global dim 2 is 'moderated regular' (defn IV.1.4) and hence a tame order of finite representation type which are classified further in the book to yield rational central singularities. That lemma (and dfn IV.1.1 and thm IV.1.2) also shows that your and Michel's dfns coincide in dim 2.
Further, I didn't understand your remark about an Iyama counterexample in the case M is projective? In that case End(M) is Azumaya, so if it has finite gldim, then the center has to be regular too.<|endoftext|>
TITLE: groups as categories and their natural transformations
QUESTION [6 upvotes]: If one views a group as a one object category with the elements of the group as morphisms then a natural transformation between functors of such categories is an inner automorphism, i.e. if we have two group homomorphisms $f,g: A\to B$ then a natural transformation $\eta :f\to g$ is just an element $b\in B$ such that $f(a)\cdot b = b \cdot g(a)$ which can be rewritten as $f(a)=b \cdot g(a)\cdot b^{-1}$. This isn't the only way to turn groups into categories. Another way is to take the elements of the group as objects and to have a morphism $h_a:a\to b$ if $h\cdot a=b$. If we view groups in this way then are the natural transformations again something nice like inner automorphisms?

REPLY [4 votes]: I like the notation $\mathcal{B}G$ and $\mathcal{E}G$ for the two constructions of a category out of a group $G$ in David's question. $\mathcal{E}G$ is what Tom calls the codiscrete category $C(G)$. 
Of course there is a third construction: it has $G$ as the objects, and only identity morphisms. Let's denote this category again by $G$.
The notation is nice because you can take the nerve of any category $\mathcal{C}$, and then geometrically realize. If we denote the resulting space by $|\mathcal{C}|$, 

$|\mathcal{B}G|$ is a classifying space for $G$.
$|\mathcal{E}G|$ is a universal principal $G$-bundle  
$|G|$ is just $G$.<|endoftext|>
TITLE: What do the local systems in Lusztig's perverse sheaves on quiver varieties look like?
QUESTION [10 upvotes]: In "Quivers, perverse sheaves and quantized enveloping algebras," Lusztig defines a category of perverse sheaves on the moduli stack of representations of a quiver.  These perverse sheaves are defined as summands of the pushforwards of the constant sheaves on stacks of quiver representations along with a choice of invariant flag (and thus, by definition are supported on the nilpotent locus in the moduli stack).  They're mostly of interest since they categorify the canonical basis.
My question is: Is there a stratum in this stack where the pull-back of one of these sheaves is not the trivial local system?
Now, in finite type, this is not a concern, since each stratum is the classifying space of a connected algebraic group, and thus simply connected.  But I believe in affine or wild type this is no longer true;  this was at least my takeaway from the latter sections of "Affine quivers and canonical bases."  However, I got a little confused about the relationship between the results of the two papers mentioned above, since they use quite different formalisms, so I hold out some hope that the local systems associated to symmetric group representations aren't relevant to the perverse sheaves for the canonical basis.  Am I just hoping in vain?

REPLY [3 votes]: Another thing perhaps worth mentioning is that, outside the finite type case, you have to do some real work to find a stratification with respect to which the perverse sheaves constituting the canonical basis are constructible: for affine types, such a stratification is pretty much implicit in Lusztig's Publ IHES paper which you cite, and it uses the classification of representations of tame quivers which he recovers via the McKay correspondence.
In general it is known that the characteristic cycles of the sheaves in the canonical basis lie in a certain Lagrangian variety (this is already established in the paper on quivers and canonical bases). This doesn't give you a stratification however, because they components are conormals to locally closed subvarieties of the moduli space whose union is not the whole space. The same phenomenon happens for character sheaves, though there Lusztig did produce a stratification of the group and show character sheaves have locally constant cohomology on the strata. 
You see papers studying this sort of problem in terms of quiver representations at the level of functions on $\mathbb F_q$-points when people try and generalize the "existence of Hall polynomials" outside of finite type quivers and on the quantum group side when people look for "PBW" bases.<|endoftext|>
TITLE: Integral expression for zeta(2)
QUESTION [7 upvotes]: By computing the sum of all Bernoulli numbers via Borel summation (I learned this technique from Varadarajan's excellent book Euler through time. A new look at old themes, 2006) I found that $$\sum B_n = \int_0^\infty \frac{t}{e^{2t}-e^t} dt$$ and discovered numerically that this expression equals $\zeta(2)-1$. The web is not very good for finding out where this can be found in print. Where should I look, and how can equations such as $$\zeta(2) = 1 + \int_0^\infty \frac{t}{e^{2t}-e^t}\ dt$$ be proved?

REPLY [13 votes]: The starting point is the integral 
$$
\Gamma(s) = \int_{0}^{\infty}e^{-x}x^{s-1}dx
$$
for the gamma function. Make the change of variable $x = nu$ with $n$ an arbitrary positive integer. Then 
$$
\Gamma(s)n^{-s} = \int_{0}^{\infty}e^{-nu}u^{s-1}du
$$
and summing over $n$ from $n = 1$ yields
$$
\Gamma(s)\zeta(s) = \int_0^{\infty}\frac{1}{e^u - 1}u^{s-1}du.
$$
This formula was the starting point of one of Riemann's two proofs of the functional equation. I am not certain who discovered it first, but it may have been Abel.
Substituting $s = 2$ gives 
$$
\zeta(2) = \int_{0}^{\infty}\frac{u}{e^u - 1}du
$$
and so
$$
\zeta(2) = 
\int_{0}^{\infty}\frac{ue^u}{e^{2u} - e^u}du = 
\int_{0}^{\infty}\frac{u(e^u - 1) + u}{e^{2u} - e^u}du =
\int_{0}^{\infty}\left(ue^{-u} + \frac{u}{e^{2u} - e^u}\right)du = 
1 + \int_{0}^{\infty}\frac{u}{e^{2u} - e^u}du.
$$<|endoftext|>
TITLE: Can the failure of the multiplicativity of Euler factors at bad primes be corrected?
QUESTION [15 upvotes]: Warning: This one of those does-anyone-know-how-to-fix-this-vague-problem questions, and not an actual mathematics question at all.
If $X$ is a scheme of finite type over a finite field, then the zeta function $Z(X,t)$ lies in $1+t\mathbf{Z}[[t]]$. We can calculate the zeta function of a disjoint union by the formula $Z(X\amalg Y,t)=Z(X,t)Z(Y,t)$. There is also a formula for $Z(X\times Y,t)$ in terms of $Z(X,t)$ and $Z(Y,t)$, but this is slightly more complicated. In fact, these two formulas are precisely the standard big Witt vector addition and multiplication law on the set $1+t\mathbf{Z}[[t]]$. (Actually, there's more than one standard normalization, so you have to get the right one. I believe this ring structure was first written down by Grothendieck in his appendix to Borel-Serre, but I don't know who first made the connection with the ring of Witt vectors as defined earlier by Witt.) If we let $K_0$ be the Grothendieck group on the isomorphism classes of such schemes, where addition is disjoint union and multiplication is cartesian product, then we get a ring map $K_0\to 1+t\mathbf{Z}[[t]]$. We could also do all this with the L-factor $L(X,s)=Z(X,q^{-s})$ (where $q$ is the cardinality of the finite field) instead of the zeta function. This is because they determine each other.
This is all good. The problem I have is when there is bad reduction. So now let $X$ be a scheme of finite type over $\mathbf{Q}$ (say). Then the L-factor $L_p(X,s)$ is defined by 
$$L_p(X,s)=\mathrm{det}(1-F_p p^{-s}|H(X,\mathbf{Q}_{\ell})^{I_p}),$$
where $I_p$ is the inertia group at $p$. (Sorry, I'm not going to explain the rest of the notation.) If $I$ acts trivially (in which case one might say $X$ has good reduction), then taking invariants under $I$ does nothing, and so as above, the L-factor of a product and sum of varieties is determined by the individual L-factors. If $I$ does not act trivially, then the L-factor of a sum is again the product of the individual L-factors, but for products there is no such formula! (The following should be an example showing this. Take $X=\mathrm{Spec}\ \mathbf{Q}(i)$, $Y=\mathrm{Spec}\ \mathbf{Q}(\sqrt{2})$. The we have the following Euler factors at 2: $L_2(X,s)=L_2(Y,s)=L_2(X\times Y,s)=1-2^{-s}$ and $L_2(X\times X,s)=(1-2^{-s})^2$. So the L-factors of two schemes do not determine that of the product.) Therefore the usual Euler factor cannot possibly give a ring map defined on the Grothendieck ring of varieties over $\mathbf{Q}$.
So, is there a way of fixing this problem? I would guess the answer is No, because while some people might allow you to scale Euler factors by numbers, I don't think anyone will let you change them by anything else. But maybe there is some "refined L-factor" that determines the usual one (and maybe incorporates the higher cohomology of the inertia group?) Assuming there is no known way of repairing things, I have a follow-up question: Is there some general formalism that handles this failure? And if so, how does that work?

REPLY [4 votes]: When I asked Niranjan Ramachandran this question a few days ago, he pointed out that you can indeed fix the problem if you work with integral models instead of varieties over $\mathbf{Q}$: Let $X$ be a scheme of finite type over $\mathbf{Z}$ and define the Euler factor $L_p(X,s)$ to be $P_p(X,p^{-s})$, where $P_p(X,t)$ is the Zeta-function of the fiber of $X$ over $p$. Then the multiplicativity of the product $\prod_p L_p(X,s)$ follows immediately from that of Zeta-functions for varieties over finite fields. Lo and behold, in the example I gave above, the products of the minimal integral models are different from the minimal integral models of the products. (This is all a little embarrassing, not just because the solution is really easy, but because I take a certain amount of pleasure in telling people that it's better to work directly with integral models! Also, apologies to the people who spent time thinking about this. I probably gave the impression that working with varieties over $\mathbf{Q}$ was non-negotiable!)<|endoftext|>
TITLE: Semiadditivity and dualizability of 2
QUESTION [22 upvotes]: Short version: Let (C, ⊗, 1) be a locally presentable closed symmetric monoidal category with a zero object, and write 2 = 1 ∐ 1.  Suppose the object 2 has a dual.  Does it follow that C is a category with biproducts?
Longer version, with motivation: Let (C, ⊗, 1) be a locally presentable closed symmetric monoidal category.  If you don't know what "locally presentable" means, you can replace these conditions with "complete and cocomplete symmetric monoidal category in which ⊗ commutes with colimits in each variable".  Familiar examples include (Set, ×, •), (Set*, ∧, S0) (the category of pointed sets with the smash product), and (Ab, ⊗, ℤ).  Any such category C has a unique "unit" functor FC : Set → C preserving colimits and the unit object: the set S is sent to the coproduct in C of S copies of 1.  For a nonnegative integer n, let me also write n for the image under this functor of the n-element set.  For instance, 0 represents the initial object of C.
A dual for an object X of C is another object X* together with maps 1 → X ⊗ X* and X* ⊗ X → 1 which satisfy the triangular identities; see wikipedia for more details.  The data of X* together with these maps is unique up to unique isomorphism if it exists, so it makes sense to ask whether an object has a dual or not.
I'm interested in the relationship between which objects in the image of FC have duals and the existence of more familiar structures on C.  In our examples,

C = Set: Only 1 has a dual.
C = Set*: Only 1 and 0 = • have duals.
C = Ab: n has a dual for any nonnegative integer n.

It's easy to show that 1 is always its own dual, and slightly less trivially, that 0 has a dual iff 0 is also a final object, i.e., C has a zero object, or equivalently C is enriched in Set*.  Moreover, if C is semiadditive, i.e., enriched in commutative monoids, or equivalently has biproducts, then n has a dual (in fact, n is its own dual) for every nonnegative integer n.  Conversely, if 0 has a dual, so that C is pointed, and 2 also has a dual, then there is a canonical map 2 = 1 ∐ 1 → 1 × 1 = 2*.  My question, then, is: is this map is always an isomorphism?  Or, could it happen that 2* exists but is not isomorphic to 2 via this map?

REPLY [12 votes]: It is true that in a cosmos $\mathcal{V}$ (= a complete and cocomplete symmetric monoidal closed category) with zero object where $2$ has a dual, the canonical map $1+1 \rightarrow 1\times 1$ is invertible. In other words, you don't even need to assume that $\mathcal{V}$ is locally presentable. I think that this implies that the cosmos in question is semiadditive, which in turn implies that coproducts/products are biproducts (the only thing one needs to observe is that $C\otimes 1\times 1 \cong C\times C$, which follows from the fact that $1\times 1$ is the dual of $1+1$).
Proving this is a bit too involved for MO (because there is no good way of drawing string diagrams here). I have typed up an argument, which can be found here (Wayback Machine). I first prove that an autonomous symmetric monoidal category (autonomous means that all objects have duals) where the coproduct $1+1$ exists is semiadditive.
I can try to give some intuition for the argument in question. The key idea is that one wants to construct a diagonal map $1 \rightarrow 1+1$. The way to do this was inspired by the following paper:
André Joyal, Ross Street and Dominic Verity (1996). Traced monoidal categories. Mathematical Proceedings of the Cambridge Philosophical Society,   119 , pp 447-468
doi:10.1017/S0305004100074338
In my writeup, this "diagonal map" is the map $1\rightarrow 1+1$ in the string diagrams which is built out of a "loop" and the map $1+1 \rightarrow (1+1) \otimes (1+1)$ which I called $h$. From the formula given in the introduction of the paper by Joyal, Street and Verity it follows that my construction does indeed give the diagonal map in the case where $\mathcal{V}$ is the cosmos of vector spaces over some field.
Edit: updated expired link.<|endoftext|>
TITLE: Serre's FAC in English
QUESTION [49 upvotes]: Has somebody translated J.-P. Serre's "Faisceaux algébriques cohérents" into English? At least part of it?
In a fit of enthusiasm, I started translating it and started TeXing. But after section 8, I got tired and stopped.
However if somebody else already took the trouble, I would be most grateful. I do not know a word of French(except maybe faisceau), and forgot whatever I learned in the process of translation very quickly.
This is made community wiki, as I do not want to get into rep issues. Please feel free to close this if you think this qn is inappropriate for MO(I have added my own vote for closing, in case this helps). I would be happy to receive answers in comments.

REPLY [12 votes]: There is another translation by Andy McLennan that comes with a lot of background material, the actual translation starts at page 235. I'm not really competent to make any comparisons.<|endoftext|>
TITLE: Function field of projective space
QUESTION [6 upvotes]: I'm reading the book of Silverman about elliptic curves.  It describes the function field of a variety defined over K to be the quotient field of $K[X]/I(V)$, then says that we may look at the function field of $
\mathbf P^n$ (projective $n$-space) as the subfield of $\overline K(X_0...X_n)$ consisting of rational functions $F(X)=f(X)/g(X)$ for which $f,g$ are homogeneous with the same degree.  How do those two definitions agree, and how are such functions a subfield of $\overline K(X_0...X_n)$?

REPLY [6 votes]: Where Silverman defines the function field of $V$ as the fraction field of the intergral domain $K[X]/I(V)$ he is talking about an integral affine variety $V$. In this case, $I(V)$ is a prime ideal so that $K[X]/I(V)$ is a domain. 
If you look at Silverman's definition of function field for a projective integral variety $V$ (see page 10 in the Arithmetic of Elliptic Curves), you'll note that he defines it as the function field of $V \cap \mathbb{A}^n$ where $\mathbb{A}^n$ is one of the standard affine patches of $\mathbb{P}^n$ that intersects $V$. The (minor) subtlety is that more than one standard affine patch might intersect $V$ so you might think you would get different function fields by picking different affine patches. But this doesn't happen, and the remark you mention (remark I.2.9) describes the function field canonically as a subfield of $K(X_0,\ldots,X_n)$. The equivalence of the two definitions then follows immediately from the definition of the different standard open affine patches of $\mathbb{P}^n$ (i.e. de-homogenizing by dividing by a non-zero homogeneous coordinate).

REPLY [3 votes]: The function field of $V$ is defined as the field of fractions of $K[X]/I(V)$ for affine varieties $V$. In the case of projective varieties, Silverman chooses a Zariski-dense affine open subset $V$ of the variety and defines the function field of the variety as the function field of the subset $V$. Of course, one can prove it is independent of the choice of $V$. When the variety is $\mathbb{P}^n$, choose $V = \mathbb{A}^n$ and so $I(V)=\{ 0 \}$, so $K(\mathbb{P}^n) = K(V) = K[X]/\{ 0 \} = K[X]$, where $X$ is shorthand for $X_1,\ldots,X_n$. Finally to get an isomorphism of the subfield of $K(X_0,\ldots,X_n)$ you described with the field I just described, just set $X_0=1$.<|endoftext|>
TITLE: Understanding moment maps and Lie brackets
QUESTION [18 upvotes]: I'm trying to learn about moment maps in symplectic topology (suppose our Lie group is $G$ with Lie algebra $\mathfrak g$, acting on the symplectic manifold $(M,\omega)$ by symplectomorphisms). I'm having a hard time, and I've realized this is because I don't have a good conceptual understanding of the Lie bracket, either on the Lie algebra $\mathfrak g$, or on the group of symplectomorphisms of $(M,\omega)$, or on the space of functions $\mathcal C^\infty(M,\mathbb R)$. Therefore I can't "visualize" the Hamiltonian condition, which requires that the linear map $\mathfrak g \rightarrow \mathcal C^\infty(M,\mathbb R)$, which exists when the action by $G$ is "exact," be a Lie algebra homomorphism.
Please tell me how you personally understand/intuit/conceptualize this situation, both the Lie bracket stuff and moment maps more generally! Any help is greatly appreciated.
EDIT: I didn't realize how non-standard some of this terminology is, so my question might be confusing. I call the action $\rho: G \rightarrow {\rm Symp}(M,\omega)$ "exact" if the image of the induced map $\rho: {\rm Lie}(G) \rightarrow {\rm Lie}({\rm Symp}(M,\omega))$ is contained in the sub-lie-algebra of Hamiltonian vector fields. The condition that was confusing me, I now realize, is just a technical point: that we choose a set of representative Hamiltonian functions for the image $\rho({\rm Lie}(G))$ which is a sub-Lie-algebra of $\mathcal C^\infty(M,\mathbb R)$ with its Poisson bracket. Thanks to all the helpful answers I think I understand this much better now.
In particular, if we present ${\rm Lie}(G)$ (assumed finite dimensional, semi-simple, etc) by Lie algebra generators (with some relations), then we can probably just choose appropriate elements in $\mathcal C^\infty(M,\mathbb R)$ for these generators, and then the rest of the map from ${\rm Lie}(G)$ to $\mathcal C^\infty(M,\mathbb R)$ is just forced on us, and this gives a Hamiltonian action? Is that right?

REPLY [15 votes]: I believe the following way (Kostant's, 1970) to be the best way to think about the Hamiltonian condition.
First, "why" is there a central extension $H^0(M; {\mathbb R}) \to C^\infty (M) \to symp(M)$ of Lie algebras? Of what is $C^\infty (M)$ supposed to be the Lie group? For $symp(M)$, the Lie algebra of vector fields annihilating the symplectic form $\omega$, it's clear it should the Lie algebra of the group $Symp(M)$ of symplectomorphisms.
Assume now that $[\omega]$ is integral. Then it is $c_1$ of some "prequantization" line bundle $\mathcal L$, and $\omega$ is the curvature of some Hermitian connection $\alpha$ on that line bundle. Let $Aut(M,{\mathcal L},\alpha)$ denote the group of Hermitian bundle automorphisms (moving the base around) of $\mathcal L$ preserving $\alpha$. This group obviously maps to $Diff(M)$, forgetting the action on the fibers, but because it preserves $\alpha$ on $\mathcal L$ it preserves $\omega$ on $M$, so the image lies inside $Symp(M)$. The kernel consists of bundle automorphisms that only act fiberwise, and for them to preserve the flat connection they must, on each component, rotate all the fibers by the same element of $U(1)$.
The Hamiltonian condition, then, is about whether one can lift the action of $G$ on $M$ to an action on the line bundle over $M$. It's very easy, given such a lift, to write down a moment map. (Basically, now that you're dealing with a $1$-form $\alpha$ instead of a $2$-form $\omega$, you can pair vector fields from $\mathfrak g$ with it.)
One example I find instructive is ${\mathbb R}^{2n}$ acting on itself by translation, with the space given the usual symplectic structure. That's acting as symplectomorphisms, and the space is simply connected, so there's no $H^1$ obstruction (as when $T^1$ acts on $T^2$). But one can't lift the action to preserve the (non-flat) connection on the (trivial) line bundle; it only lifts to an action of the Heisenberg group.
Another subtle example is $SO(3)$ acting on $S^2$ with the area $1$ symplectic structure. On the Lie algebra level, yes, the action is Hamiltonian. But actually $SO(3)$ doesn't act on the line bundle; only its double cover $SU(2)$ does.
Finally, think about the case that $G$ acts algebraically on $X \subseteq {\mathbb P}V$. I like to say that $X$ is "equivariantly projective" if $G$ acts on ${\mathbb P}V$ preserving $X$, and this is pretty nearly an algebro-geometric replacement for the Hamiltonian condition. (Non-example: $X$ is a nodal cubic curve, whose smooth locus is ${\mathbb C}^\times$, acted on by ${\mathbb C}^\times$.)<|endoftext|>
TITLE: Pigeonhole Principle for infinite case 
QUESTION [5 upvotes]: Suppose $X_n$ are finite sets for any natural integer $n$. let $Y$ be an infinite subset of $\prod_n X_n$. Do there exist $y$ and $y'$ in $Y$  and an infinite subset $S$ of $\mathbb N$ such that $y_n=y'_n$ for all $n$ in $S$?

REPLY [6 votes]: First, let me improve upon the countable counterexample of
Darij Grinberg by giving an uncountable counterexample Y.
Indeed, I shall give finite sets Xn and a subset Y of the product ΠXn having size continuum (that is, as large as possible), such that any two distinct y, y' in Y
have only finitely many common values.
Let Xn have 2n elements, consisting
of the binary sequences of length n. Now, for each infinite
binary sequence s, let ys be sequence in the
product ΠXn whose nth value is
s|n, the length n initial segment of s. Let Y consist of
all these ys. Since there are continuum many s,
it follows that Y has size continuum.
Note that if s and t are distinct binary sequences, then eventually
the initial segments of s and t disagree. Thus, eventually,
the values of ys and yt are
different. Thus, ys and yt have only
finitely many common values. So Y is very large counterexample, as
desired.
A similar argument works still if the Xn grow
more slowly in size, as long as liminf|Xn| =
infinity. One simply spreads the construction out a bit
further, until the size of the Xi is large
enough to accommodate the same idea. That is, if the liminf
of the sizes of the Xn's is infinite, then one
can again make a counterexample set Y of size continuum.
In contrast, in the remaining case, there are no infinite
counterexamples. I claim that if infinitely many
Xn have size at most k and Y is a
subset of ΠXn having k+1 many elements, then
there are distinct y,y' in Y having infinitely many common
values. To see this, suppose that Y has the property that
distinct y, y' in Y have only finitely many common values.
In this case, any two y, y' must eventually have different
values. So if Y has k+1 many elements, then eventually for
sufficiently large n, these k+1 many sequences in Y must be
taking on different values in every Xn. But
since unboundedly often there are only k possible values in
Xn, this is impossible.
In summary, the situation is as follows:
Theorem. Suppose that Xn is finite and
nonempty.

If liminf |Xn| is infinite, then there is Y
subset ΠXn of size continuum, such that
distinct y, y' in Y have only finitely many values in
common.
Otherwise, infinitely many Xn have size at most k for some k, and in this case, every Y subset
ΠXn of size k+1 has distinct y,y' in Y with infinitely
many common values.

In particular, if the Xn become increasingly
large in size, then there are very bad counterexamples to the
question, and if the Xn are infinitely often
bounded in size, then there is a very strong positive answer to
the question.<|endoftext|>
TITLE: How can I conclude that I live in a solar system?
QUESTION [15 upvotes]: Well, this is an awkward question and I don't know if it is mathematical enough for MO (I'm sorry if not) but I'll try it: What observations in the coordinate system centered in my fixed position on earth are necessary to conclude that the earth (and the planets) move (approximately) in ellipses around the sun and that earth is rotating around itself?

REPLY [2 votes]: An interesting (and very old) argument in favour of heliocentrism is based on estimates of the relative sizes of the Earth and Sun.  
Actually,  Aristarchus of Samos estimated that the Sun is six to seven times wider than the Earth (and therefore over 200 times more voluminous). These calculations arguably led him to conclude that it made more sense for the Earth to be moving than for the huge Sun to be moving around it.<|endoftext|>
TITLE: Normal bundle to a curve in P^2
QUESTION [8 upvotes]: Let $C$ be a smooth curve of degree $d$ in $\mathbb{P}^2$ over $\mathbb{C}$. Say $C$ is defined by $p(x,y,z)=0$, with $p$ a homogeneous degree $d$ polynomial.
In vector calculus one learns that the gradient of $p$ is normal to $C$ at every point of the curve.
In algebraic geometry, the invertible sheaf associated to the normal bundle $N_{C|\mathbb{P}^2}$ to $C$ in $\mathbb{P}^2$, is given by $\mathcal{O}_{\mathbb{P}^2}(d) _{|C}$.
Is there any relationship between the gradient and the bundle or the sheaf?

REPLY [9 votes]: Yes, there is a strong relationship between the two.
First, let's work locally in affine space rather than in projective space (it makes more
sense to work locally just because we are dealing with a sheaf, which is defined locally).
So I will consider a non-homogen
Working without a metric (as one does in at least the algebraic aspects of algebraic geometry),
it is perhaps better to talk not about the gradient of $f$, but its exterior derivative
$df$, given by the same formula: $df = f_x dx  + f_y dy.$  Since this is differential form
valued, we will compare it with the conormal bundle to the curve $C$ cut out by $f = 0$.
Now the exterior derivative can be thought of simply as taking the leading (i.e. linear) term of $f$.
On the other hand, if $\mathcal I$ is the ideal sheaf cutting out the curve $C$, then the 
conormal bundle is $\mathcal I/\mathcal I^2$. (If $f$ is degree $d$, then $\mathcal I = \mathcal O(-d)$, and so this can be rewritten as $\mathcal O(-d)\_{| C}$, dual to the normal
bundle $\mathcal O(d)\_{| C}$.) Now $f$ is a section of $\mathcal I/\mathcal I^2$ (over the affine patch on which we are working), so we may certainly regard it as a section of $\mathcal I/\mathcal I^2$; this section
is the (image in the conormal bundle to $C$ of) the exterior derivative of $f$.
The formula $\mathcal I/\mathcal I^2$ for the conormal bundle is thus simply a structural
interpretation of the idea that we compute the normal to the curve by taking the leading term
of an equation for the curve.<|endoftext|>
TITLE: Is there a canonical Hopf structure on the center of a universal enveloping algebra?
QUESTION [11 upvotes]: Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$.  Define $\mathcal Z(\mathfrak g)$ to be the center of the universal enveloping algebra $\mathcal U\mathfrak g$, and define $(\mathcal S\mathfrak g)^{\mathfrak g}$ to be the ring of invariant elements of the symmetric algebra $\mathcal S\mathfrak g$ under the induced adjoint action of $\mathfrak g$.  (Clearly $\mathcal Z(\mathfrak g) = (\mathcal U\mathfrak g)^{\mathfrak g}$ via the adjoint action.)  The Duflo isomorphism is an isomorphism of algebras $\mathcal Z(\mathfrak g) \cong (\mathcal S\mathfrak g)^{\mathfrak g}$.  At the level of vector spaces, the trick is to realize that the PBW map $\mathcal U\mathfrak g \to \mathcal S \mathfrak g$ is an isomorphism of $\mathfrak g$-modules.  For the isomorphism of algebras in the semisimple case, see for example my unedited notes on the class by V. Serganova.
(I read here that this isomorphism can be realized as a composition of the PBW vector-space isomorphism $\mathcal U\mathfrak g \to \mathcal S\mathfrak g$ with the map $\mathcal S\mathfrak g \to \mathcal S\mathfrak g$ given by $x \mapsto \sinh(x/2)/(x/2)$. But it's not at all obvious that this composition is even well-defined or linear when restricted to $\mathcal Z(\mathfrak g)$.  I should mention that $\mathcal Z$ is not a functor, I think.  The PBW isomorphism is non-canonical, although a canonical version can be given via the symmetrization map, and I guess on the center it is canonical.)
When $\mathfrak g$ is semisimple of rank $n$, at least, one can further show that $(\mathcal S\mathfrak g)^{\mathfrak g} \cong \mathbb C[x_1,\dots,x_n]$, although you have some choice about how to make this isomorphism.  Thus, at least when $\mathfrak g$ is semisimple, $\mathcal Z(\mathfrak g)$ is a polynomial ring.
But any polynomial ring can be given a Hopf structure.  By choosing an isomorphism with $\mathbb C[x_1,\dots,x_n]$, we can take the Hopf structure generated by $\Delta: x_i \mapsto x_i \otimes 1 + 1 \otimes x_i$.  In fact, this structure doesn't depend quite on the full choice of isomorphism.  A Hopf structure on a commutative algebra $R$ is by definition the same as an algebraic group structure on $\text{Spec}(R)$.  But $\text{Spec}(\mathbb C[x_1,\dots,x_n])$ is $n$-dimensional affine space — the algebra isomorphisms of $\mathbb C[x_1,\dots,x_n])$ are precisely the affine maps — so picking a commutative group structure is the same as picking an origin.  (For certain values of $n$ there are also non-commutative group structures on affine $n$-space, and so non-cocommutative Hopf structures on the polynomial ring.  For example, the group of upper-triangular matrices with $1$s on the diagonal is affine.)
My question is whether this Hopf structure can be picked out canonically.

Question: If $\mathfrak g$ is a finite-dimensional Lie algebra over $\mathbb C$, can the center $\mathcal Z(\mathfrak g)$ of the universal enveloping algebra be given a canonical (cocommutative) Hopf algebra structure?  If no, how much extra structure on $\mathfrak g$ is needed?

Here by "canonical" I of course don't mean that there is a unique one, so you may make choices once and for all.  But there should be some definition/construction that does not require the user to make any choices to implement it.  By "extra structure" I mean either extra structure (an invariant metric, for example) or extra properties (semisimplicity, for example).
My suspicion is that the answer is "yes" for a metric Lie algebra, which is a Lie algebra $\mathfrak g$ along with a choice of an invariant nondegenerate metric, i.e. a chosen isomorphism of $\mathfrak g$-modules $\mathfrak g \cong \mathfrak g^*$.  Metric Lie algebras include the semisimples and the abelians, and certain extensions of these (in fact, I believe that there is a structure theorem that any metric Lie algebra is a metric extension of semisimples and abelians, but don't quote me), but generally there are many choices of metric (e.g. any metric on an abelian Lie algebra $\mathfrak a$ is invariant, so there are $\mathfrak{gl}(\dim \mathfrak a)$ many choices).
The motivation for my question is this: by studying Vassiliev invariant and/or perturbative Chern-Simons theory, Bar Natan and others have defined a certain commutative and cocommutative Hopf algebra $A$ of "diagrams".  Any choice of metric Lie algebra $\mathfrak g$ determines an algebra homomorphism $A \to \mathcal Z(\mathfrak g)$.  I would like to know if this can be made into a Hopf algebra homomorphism.

REPLY [4 votes]: I would be very surprised if the answer is YES, in general. This would turn the spectrum of Z(g) into an algebraic group, and, in particular, force it to be smooth (if Z(g) is finitely generated). Quick search through Dixmier's book gives a reference to his 1957 paper where he computes Z(g) for nilpotent Lie algebras of dimension up to 5. All of them are polynomial except Z(g_{5,5}). Good luck checking that SPEC(Z(g_{5,5})) is not a 3-dimensional algebraic group.<|endoftext|>
TITLE: Why is the concept of topos a "metamorphosis" of the concept of space?
QUESTION [7 upvotes]: Hi,
I recently started studying topos theory, and I am puzzled by the Grothendieck's claim that topos is a "metamorphosis" of the concept of space.  Can somebody explain what he means by this?  
Thanks,
Alexander

REPLY [2 votes]: The answers already provided are very  good and informative, so I just wish to add something concerning the "metamorphosis" of the very notion of space of which Grothendieck speaks in Semailles. 
Every space has its associated topos, but there are topoi which are NOT spatial. You can define categorically the notion of point of a topos, and this definition corresponds to the usual notion  of points when one restricts to  spatial topoi.
Now, the fact that there are plenty of topoi with no points basically means that one can do topology in a pointless world: you can still formally define notions of compactness, coverings, and well as most of the standard topological (and even homotopical) machinery,  directly in a given topos, regardless of its having  points or not. 
As it turns out, the passage from point-set to pointless topology is not just an idle game: for instance in physics at the Planck level you may still want to talk of topological and geometric properties of space-time, and yet you have no well-defined points.<|endoftext|>
TITLE: sum of squares in ring of integers
QUESTION [17 upvotes]: Lagrange proved that every positive integer is a sum of 4 squares. 
Are there general results like this for rings of integers of number fields? Is this class field theory? 
Explicitly, suppose a number field is formally real. Denote its ring of integers by $Z$. Is it true that for every algebraic integer $x$ in $Z$ either $x$ or $-x$ is a sum of squares?

REPLY [50 votes]: To address the particularities of this question for number fields, the basic theorem is attributed to Hilbert, Landau and Siegel.  First of all, any nonzero sum of squares in a number field has to be totally positive (that is, it is positive in all real embeddings).  Hilbert (1902) conjectured that in any number field, a totally positive element is a sum of 4 squares in the number field. This was proved by Landau (1919) for quadratic fields and by Siegel (1921) for all number fields. 
This sounds superficially like a direct extension of Lagrange's theorem, but there is a catch: it is about field elements, not algebraic integers as sums of squares of algebraic integers. A totally positive algebraic integer in a number field $K$ need not be a sum of 4 squares of algebraic integers in $K$. The Hilbert-Landau-Siegel theorem only says it is a sum of 4 squares of algebraic numbers in $K$. 
For instance, in $\mathbf{Q}(i)$ all elements are totally positive in a vacuous sense (no real embeddings), so every element is a sum of four squares.  As an example,
$$
i = \left(\frac{1+i}{2}\right)^2 + \left(\frac{1+i}{2}\right)^2.
$$
This shows $i$ is a sum of two squares in $\mathbf{Q}(i)$. 
It is impossible to write $i$ as a finite sum of squares in ${\mathbf Z}[i]$ since 
$$
(a+bi)^2 = a^2 - b^2 + 2abi
$$
has even imaginary part when $a$ and $b$ are in $\mathbf{Z}$.  Thus any finite sum of squares in $\mathbf{Z}[i]$ has even imaginary part, so such a sum can't equal $i$.
Therefore it is false that every totally positive algebraic integer in a number field is a sum of 4 squares (or even any number of squares) of algebraic integers.
Here are some further examples:

In $\mathbf{Q}(\sqrt{2})$, $5 + 3\sqrt{2}$ is totally positive since 
$5+3\sqrt{2}$ and $5-3\sqrt{2}$ are both positive.  So it must be a sum of at most four squares in this field by Hilbert's theorem, and with a little fiddling around you find 
$$
5 + 3\sqrt{2} = (1+\sqrt{2})^2 + \left(1 + \frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2.
$$
It is impossible to write $5 + 3\sqrt{2}$ as a sum of squares in the ring of integers $\mathbf{Z}[\sqrt{2}]$ because of the parity obstruction we saw for $i$ as a sum of squares in $\mathbf{Z}[i]$: the coefficient of $\sqrt{2}$ in $5 + 3\sqrt{2}$ is odd.
In $\mathbf{Q}(\sqrt{2})$, $\sqrt{2}$ is not totally positive (it becomes negative when we replace $\sqrt{2}$ with $-\sqrt{2}$), so it can't be a sum of squares in this field. But in the larger field $\mathbf{Q}(\sqrt{2},i)$, everything is totally positive in a vacuous sense so everything is a sum of at most four squares in this field by the Hilbert-Landau-Siegel theorem.  And looking at $\sqrt{2}$ in $\mathbf{Q}(\sqrt{2},i)$, we find 
$$
\sqrt{2} = \left(1 + \frac{1}{\sqrt{2}}\right)^2 + i^2 + \left(\frac{i}{\sqrt{2}}\right)^2.
$$

Hilbert made his conjecture on totally positive numbers being sums of four squares as a theorem, in his Foundations of Geometry. It is Theorem 42. He says the proof is quite hard, and no proof is included. A copy of the book (in English) is available at the time I write this as http://math.berkeley.edu/~wodzicki/160/Hilbert.pdf. See page 83 of the file (= page 78 of the book).
Siegel's work on this theorem/conjecture was done just before the Hasse-Minkowski theorem was established in all number fields (by Hasse), and the former can be regarded as a special instance of the latter. 
Indeed, for nonzero $\alpha$ in a number field $K$, consider the quadratic form $$Q(x_1,x_2,x_3,x_4,x_5) = x_1^2+x_2^2+x_3^2+x_4^2-\alpha{x}_5^2.$$ To say $\alpha$ is a sum of four squares in $K$ is equivalent to saying $Q$ has a nontrivial zero over $K$. (In one direction, if $\alpha$ is a sum of four squares over $K$ then $Q$ has a nontrivial zero over $K$ where $x_5 = 1$. In the other direction, if $Q$ has a nontrivial zero over $K$ where $x_5 \not= 0$ then we can scale and make $x_5 = 1$, thus exhibiting $\alpha$ as a sum of four squares in $K$. If $Q$ has a nontrivial zero over $K$ where $x_5 = 0$ then the sum of four squares quadratic form represents 0 nontrivially over $K$ and thus it is universal over $K$, so it represents $\alpha$ over $K$.) By Hasse-Minkowski, $Q$ represents 0 nontrivially over $K$ if and only if it represents 0 nontrivially over every completion of $K$. 
Since any nondegenerate quadratic form in five or more variables over a local field or the complex numbers represents 0 nontrivially, $Q$ represents 0 nontrivially over $K$ if and only it represents 0 nontrivially in every completion of $K$ that is isomorphic to ${\mathbf R}$. The real completions of $K$ arise precisely from embeddings $K \rightarrow {\mathbf R}$. For $t \in {\mathbf R}^\times$, the equation 
$x_1^2+x_2^2+x_3^2+x_4^2-t{x}_5^2 =0$ has a nontrivial real solution if and only if $t > 0$, so $Q$ has a nontrivial representation of 0 in every real completion of $K$ if and only if $\alpha$ is positive in every embedding of $K$ into ${\mathbf R}$, which is what it means for $\alpha$ to be totally positive. (Strictly speaking, to be totally positive in a field means being positive in every ordering on the field. The orderings on a number field all arise from embeddings of the number field into $\mathbf R$, so being totally positive in a number field is the same as being positive in every real completion.)
Siegel's paper is "Darstellung total positiver Zahlen durch Quadrate, Math. Zeit. 11 (1921), 246--275, and can be found online at http://gdz.sub.uni-goettingen.de/en/dms/loader/img/?PPN=PPN266833020_0011&DMDID=DMDLOG_0022.

REPLY [17 votes]: K. Conrad's answer shows that one must, in general, make a distinction between rational representations of sums of squares and integral representations of sums of squares and that the latter is significantly more subtle.
Still, a lot of work has been done.  (I myself am familiar with only a little of it.)  The following classic paper gives comprehensive results for imaginary quadratic fields:

Niven, Ivan
  Integers of quadratic fields as sums of squares.
  Trans. Amer. Math. Soc. 48, (1940). 405--417. 

http://math.uga.edu/~pete/Niven40.pdf
Niven shows that the obstruction pointed out by Conrad is essentially the only one to representing integers in an imaginary quadratic field as sums of squares.  More precisely:
Let $m$ be a squarefree positive integer, and put $K = \mathbb{Q}(\sqrt{-m})$, $\mathbb{Z}_K$ the ring of integers of $K$.
Case 1: $m \equiv 1 \pmod 4$.  In this case $\mathbb{Z}_K = \mathbb{Z}[\sqrt{-m}]$ and there is an obstruction as above.  Namely, an element $a + b \sqrt{-m}$ is a sum of squares in $\mathbb{Z}_K$ iff it is a sum of $3$ squares in $\mathbb{Z}_K$ iff $b$ is even.
Case 2: $m \equiv 3 \pmod 4$.  In this case $\mathbb{Z}_K = \mathbb{Z}[\frac{1+\sqrt{-m}}{2}]$ and the obstruction of the previous case disappears: every element of 
$\mathbb{Z}_K$ is a sum of 3 squares in $\mathbb{Z}_K$.

REPLY [8 votes]: For explicit results, see

Cohn, Harvey: 
Decomposition into four integral squares in the fields of 
$2^{1/2}$ and $3^{1/2}$, Amer. J. Math. 82, 301-322 (1960) 
(A different proof was given by J. Deutsch,
An alternate proof of Cohn's four squares theorem;
J. Number Theory 104, No. 2, 263-278 (2004)) 
Colliot-Thélène, Jean-Louis; Xu, Fei: 
 Brauer-Manin obstruction for integral points of homogeneous spaces and 
representation by integral quadratic forms
Compos. Math. 145, No. 2, 309-363 (2009)
(This one is a lot deeper and connects the representability with the Brauer-Manin
obstruction; I have only seen the review.)<|endoftext|>
TITLE: congruent to 1 mod p
QUESTION [7 upvotes]: This is a somewhat vague question: for a prime number p, we often see that various counts come out to be 1 modulo p. What are the possible reasons for this?
Here are some I've encountered:

For some reason, the counting can be reduced modulo p, or in the field of p elements, and in that setting, the answer turns out to be 1.
There is a group action of a p-group on the set with exactly one fixed point. Hence, all the other orbits have sizes equal to a power of p. This is used, for instance, in proving that the number of p-Sylow subgroups is congruent to 1 mod p.
A kind of inductive counting argument, that is based on the following observation: the sum of a numbers, each of which is 1 mod p, is congruent to a mod p. Thus, the sum is 1 mod p iff a is 1 mod p. This comes up, for instance, in inductive proofs that the number of subgroups of order $p^k$ in a group of order $p^n$ is congruent to 1 mod $p$. A more general version of this is Phillip Hall's Enumeration Theorem. (This argument can be viewed as a variant of (1), but I find it sufficiently distinctive to mention separately). 
The number comes up as the number of one-dimensional subspaces of a finite-dimensional vector space over a field with p elements, which we know is a polynomial in p with constant term 1.
More sophisticated combinations of (3) and (4) are used to prove that the number of subgroups of various kinds in p-groups is congruent to 1 mod p. See, for instance, the work of Jonah and Konvisser: Counting abelian subgroups of p-groups: a projective approach, Journal of Algebra, Page 309-330, 1975.
An application of Fermat's little theorem or a related "order in the multiplicative group" result, wherein a prime q distinct from p can divide $(a^p - 1)/(a - 1)$ only if it is congruent to 1 mod p.
Some results on Euler characteristics in combinatorics. I don't really understand how they're proved, or why the "congruent to 1 mod p" comes up. 

Looking forward to more situations where "1 mod p" comes up, and/or further insights into the ways I've already mentioned above.

REPLY [3 votes]: If $p\lt q$ are primes, then there is a non-abelian group of order $pq$ if and only if $q$ is 1 (mod $p$).<|endoftext|>
TITLE: Are there pairs of consecutive integers with the same sum of factors?
QUESTION [18 upvotes]: Background/Motivation
I was planning to explain Ruth-Aaron pairs to my son, but it took me a few moments to remember the definition.  Along the way, I thought of the mis-definition, a pair of consecutive numbers with the same sum of divisors.  Well, that's actually two definitions, depending on whether you are looking only at proper divisors.  Suppose all divisors.  I quickly found (14,15) which both have a divisor sum (sigma function) of 24.  Some more work provided (206,207) and then a search on OEIS gave sequence A002961.  
What about proper divisors only?  (2,3) comes quickly, but then nothing for a while.  Noting that the parity of this value ($\sigma(n) -n$) is the same as that of $n$ unless $n$ is a square or twice a square, any solution pair must include one number of that form.  With that much information in hand, I posted this problem at the reference desk on Wikipedia.  User PrimeHunter determined that there were no solutions up to $10^{12}$, but there were no general responses.  
Aside from the parity issue, I haven't found other individual constraints that would filter the candidates--the number of adjacent values identical modulo $p$ for other small primes is at least as great as would be expected by chance, and there are a fair number of pairs that are arithmetically close.  

Other than (2,3), are there pairs of consecutive integers such that $\sigma(n)-n = \sigma(n+1)-(n+1)$?

REPLY [2 votes]: Apologies that this isn't a complete answer. The condition $\sigma(n+1)=\sigma(n)+1$ means that $\sigma(n),\sigma(n+1)$ are relatively prime. In the question, you've taken care of the divisibility by $2$ part. 
Let $p$ be an odd prime and $q>p$ prime. Let $a_{q,p}=a_q=p$ if $q\equiv 1 \bmod p$ and $a_{q,p}=a_q$ equals the order of $q$ modulo $p$ otherwise. If $\nu_q(n)$, the highest power of $q$ dividing $n$ is, $-1$ mod $a_q$, then $p$ divides $\sigma(n)$. So it is necessary that if $q_1,q_2$ are some primes, then one musn't simultaneously have $\nu_{q_1}(n)\equiv -1 \bmod a_{q_1,p}$ and $\nu_{q_2}(n+1)\equiv -1 \bmod a_{q_2,p}$ for any $p,q_1,q_2$.
For example, if $n=q^2$, where $q$ is prime and $1$ mod $3$, then $n+1$ is $2\bmod 4$ and $3$ divides both $\sigma(n), \sigma(n+1)$, so this is an example of $n$ that must be excluded.<|endoftext|>
TITLE: How much of scheme theory can you visualize?
QUESTION [33 upvotes]: I am just starting to learn about schemes and algebraic geometry in general, but I am finding it very hard to visualize things. For example, affine schemes that look like varieties are easily visualized. But how about infinite dimensional or nonproper things? Or fibre products of schemes? So just throwing out this question to all algebraic geometers: When you are doing your research, how much of your results come from the geometric intuition? If one were to start the research in algebraic geometry, what would you say the most important things is?

REPLY [4 votes]: I've found this set of notes http://math.mit.edu/~poonen/papers/Qpoints.pdf by Bjorn Poonen very helpful in learning arithmetic geometry. I'd read everything very carefully and do all the exercises there. Bjorn is a master of exposition.<|endoftext|>
TITLE: The (n+1)-st cohomology of K(Z/p,n).
QUESTION [6 upvotes]: I was looking through my notes for a homotopy theory course and found the following mysterious statement (K is of course the Eilenberg-Maclane space): 
$$H^{n+1}(K(\mathbb Z_p,n);\mathbb Z_p) \cong \mathbb Z_p.$$
(This would be obvious if n+1 were replaced with n. This is supposed to imply that the natural transformations $H^n(X; \mathbb Z_p)\to H^{n+1}(X; \mathbb Z_p)$ are all multiples of the Bockstein homomorphism).
I'm at a loss trying to understand why. Spectral sequences haven't been covered yet, so there should be some simple reason. Also, is there a way to see the Bockstein in all this?
Thank you!

REPLY [12 votes]: Universal coefficient theorem + $H_{n+1}(K(Z_p,n);Z)=0$. An elementary way to see the latter is that the single $n+1$ cell added to $S^n$ to kill $p$ times the generator is not a cellular cycle.<|endoftext|>
TITLE: The continuous as the limit of the discrete
QUESTION [6 upvotes]: Reading this documment: www.math.ucla.edu/~tao/preprints/compactness.pdf, I got interested in the following thing: "One can also use compactifications to view the continuous as the limit of the discrete; for instance, it is possible to compactify the sequence Z/2Z, Z/3Z, Z/4Z, etc. of cyclic groups, so that their limit is the circle group T = R/Z.".
Could you give me a point of start to understand what idea of compactification is being used there? Where could I find an sketch of proof for that fact?

REPLY [2 votes]: I haven't looked at the link, but it seems likely that the author of said link is discussing
convergence in the Hausdorff topology.  In this context the idea of taking limits is due to Gromov, I believe.  (There is a wikipedia entery on Gromov--Hausdorff convergence which seems to be
the relevant one.)  It is a common technique in parts of geometric topology and geometric group theory.  
If you ask another question on Gromov--Hausdorff convergence, I'm sure it would draw the attention of the (at least) several experts on the topic who I know read MO.<|endoftext|>
TITLE: Irrational logs and the harmonic series
QUESTION [20 upvotes]: Consider the series
$$ S_f = \sum_{x=1}^\infty \frac{f}{x^2+fx}. $$
Goldbach showed that, for integers $f \ge 1$, 
$$ S_f = 1 + \frac12 + \frac13 + \ldots + \frac1f $$
(this follows easily by writing $S_f$ as a telescoping series).
Thus $S_f$ is rational for all natural numbers $f \ge 1$.
Goldbach claimed that, for all nonintegral (rational) numbers $f$,
the sum $S_f$ would be irrational.
Euler showed, by using the substitution 
$$ \frac1k = \int_0^1 x^{k-1} dx, $$
that 
$$ S_f = \int_0^1 \frac{1-x^f}{1-x} dx. $$
He evaluated this integral for $f = \frac12$ and found
that $S_{1/2} = 2(1 - \ln 2)$ (this also follows easily from 
Goldbach's series for $S_f$). Thus Goldbach's claim holds for all 
$f \equiv \frac12 \bmod 1$ since $S_{f+1} = S_f + \frac1{f+1}$.
Here are my questions: 

The irrationality of $\ln 2$ was established by Lambert, who 
proved that $e^r$ is irrational for all rational numbers 
$r \ne 0$. Are there any (simple) direct proofs?
Has Goldbach's claim about the irrationality of $S_f$ for 
nonintegral rational values of $f$ been settled in other cases?

REPLY [5 votes]: Please allow me to put my question on top of the list again by turning my comment into an answer. FC's remarks led me to the article "Transcendental values of the digamma function", J. Number Theory 125, No. 2, 298-318 (2007) by Ram Murty and N. Saradha, where Thm. 9 states that the values of S_f are transcendental for rational numbers 0 < f < 1. I apologize for not having asked this question in 2006, which is why I have only a bounty to offer (and a reference to FC from MO in Euler's OO).<|endoftext|>
TITLE: Is any true sentence in the second-order Peano Axioms provable
QUESTION [11 upvotes]: Forgive the elementary nature of the question. I understand that the second order Peano Axioms are categorical in the sense that all their models are isomorphic. This equivalence class of models is taken to be the definition of natural numbers. 
My question is that: Is the theory defined by those second order axioms complete? i.e. is every second-order statement in this theory provable or disprovable from the axioms?
For the first order Peano Arithmetic axioms any statement that is true in all models is provable or disprovable by the completeness of first-order logic, but of course this is in applicable to the second-order axioms.

REPLY [4 votes]: Between them, the previous comments and replies have already covered all of the relevant points, but here in any case is a direct, unsophisticated answer:
Any effective procedure (i.e. anything you could program up) for proving statements of second-order logic must be incomplete (unless it proves false statements). Otherwise, because second-order arithmetic is categorical, we would have an effective procedure for determining the arithmetic truth. But this is impossible by Gödel's first incompleteness theorem.<|endoftext|>
TITLE: alternative construction of the quotient group
QUESTION [9 upvotes]: The background for this question is that I know that many students starting to learn algebra focus on the standard construction of the quotient group $G/N$ instead of working with the universal property. Therefore it would be nice to give another construction, thus avoiding cosets, but so that the universal property is clear. This idea came to me years ago, but I've never found something.
For another example, there is a very nice and intuitive construction of the localization of a ring: $S^{-1} A := A[\{X_s\}_{s \in S}] / (s X_s = 1)$. The idea is: Invent new elements, and force them to be inverses to your elements of $S$. The universal property follows from the universal property of quotient and free algebra.
Perhaps quotient groups are a bit too elementary so that there could be another nice construction (please don't answer when you've just got this to say) ... anyway, any ideas are welcome :). I'm pretty sure that in this case there is no better one, but perhaps another one. I play around with the Cayley representation of $G$ ...
Sorry if this is too elementary for you ;-)

REPLY [3 votes]: In order to make the construction of the quotient, it seems that it would be helpful to have at hand some quotients of the group $G$, described in some way other than formally via cosets.
Here is one possible approach:
Generalizing the regular action of $G$ on itself, we have the action of $G$ on its
power set (given by translating a set).
If $S \subset G$ is any subset, we can look at the orbit of $S$ under $G$, which is
some subset $\mathcal O_S$ of the power set of $G$, and we are given a homomorphism
$G \to Perm(\mathcal O_S)$ (the group of permutations of $\mathcal O_S$).  Let $G'$ be
the image of $G$ under this homomorphism.
If we now take $S$ to be a normal subgroup $N$ of $G$, then $G' = G/N$.  I wonder how hard this is to show directly, in terms of the universal property?
It is easy to see at least that $G \rightarrow G'$ has the property that $N$ is in its kernel.  Can one show directly in this set-up that any map $G \rightarrow G''$ having
$N$ in its kernel factors through the explicitly constructed quotient $G'$?
I don't see a pithy argument straight away, but it doesn't seem too unfeasible to me.<|endoftext|>
TITLE: Galois representations attached to newforms
QUESTION [18 upvotes]: Suppose that $f$ is a weight $k$ newform for $\Gamma_1(N)$ with attached $p$-adic Galois representation $\rho_f$.  Denote by $\rho_{f,p}$ the restriction of $\rho_f$ to a decomposition group at $p$.  When is $\rho_{f,p}$ semistable (as a representation of 
$\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)$?
To make things really concrete, I'm happy to assume that $k=2$ and that the $q$-expansion of $f$ lies in $\mathbf{Z}[[q]]$.  
Certainly if $N$ is prime to $p$ then $\rho_{f,p}$ is in fact crystalline, while
if $p$ divides $N$ exactly once then $\rho_{f,p}$ is semistable (just thinking about the Shimura construction in weight 2 here, and the corresponding reduction properties of $X_1(N)$
over $\mathbf{Q}$ at $p$).  For $N$ divisible by higher powers of $p$, we know that these representations are de Rham, hence potentially semistable.  Can we say more?  For example,
are there conditions on "numerical data" attached to $f$ (e.g. slope, $p$-adic valuation of $N$, etc.) which guarantee semistability or crystallinity over a specific
extension?  Can we bound the degree and ramification of 
the minimal extension over which $\rho_{f,p}$ becomes semistable in terms of numerical
data attached to $f$?  Can it happen that $N$ is highly divisible by $p$ and yet $\rho_{f,p}$ is semistable over $\mathbf{Q}_p$?
I feel like there is probably a local-Langlands way of thinking about/ rephrasing this question, which may be of use... 
As a possible example of the sort of thing I have in mind: if $N$ is divisible by $p$ and $f$ is ordinary at $p$ then $\rho_{f,p}$ becomes semistable over an abelian extension of
$\mathbf{Q}p$
and even becomes crystalline over such an extension provided that the Hecke eigenvalues
of $f$ for the action of $\mu_{p-1}\subseteq (\mathbf{Z}/N\mathbf{Z})^{\times}$ via the diamond operators
are not all 1.

REPLY [15 votes]: The right way to do this sort of question is to apply Saito's local-global theorem, which says that the (semisimplification of the) Weil-Deligne representation built from $D_{pst}(\rho_{f,p})$ by forgetting the filtration is precisely the one attached to $\pi_p$, the representation of $GL_2(\mathbf{Q}_p)$ attached to the form via local Langlands. Your suggestions about the $p$-adic valuation of $N$ and so on are rather "coarse" invariants---$\pi_p$ tells you everything and is the invariant you really need to study.
So now you can just list everything that's going on. If $\pi_p$ is principal series, then $\rho$ will become crystalline after an abelian extension---the one killing the ramification of the characters involved in the principal series. If $\pi_p$ is a twist of Steinberg by a character, $\rho_{f,p}$ will become semistable non-crystalline after you've made an abelian extension making the character unramified. And if $\pi_p$ is supercuspidal, $\rho_{f,p}$ will become crystalline after a finite non-trivial extension that could be either abelian or non-abelian, and figuring out which is a question about $\pi_p$ (it will be a base change from a quadratic extension if $p>2$ and you have to bash out the possibilities).
Seems to me then that semistable $\rho$s will show up precisely when $\pi_p$ is either unramified principal series or Steinberg, so the answer to your question is (if I've got everything right) that $\rho_{f,p}$ will be semistable iff either $N$ (the level of the newform) is prime to $p$, or $p$ divides $N$ exactly once and the component at $p$ of the character of $f$ is trivial. Any other observations you need should also be readable from this sort of data in the same way.
One consequence of this I guess is that $\rho_{f,p}$ is semi-stable iff the $\ell$-adic representation attached to $f$ is semistable at $p$.

REPLY [2 votes]: Since f is potentially semi-stable, you can look at its attached filtered (φ, N, Gal(L/Qp))-representation (where ρf,p becomes semi-stable when restricted to GL). If its N is zero, then it is potentially cristalline, otherwise it is not.
As for the ordinary case, I'm not sure what definition you're using. Under Greenberg's definition, an ordinary p-adic Galois representation is semi-stable (see Perrin-Riou's article in the Bures). Also, the Tate curve is ordinary at p, but not potentially cristalline (once something is semi-stable and non-cristalline, it can't be potentially cristalline).<|endoftext|>
TITLE: Various concepts of "closure" or "completion" in mathematics
QUESTION [6 upvotes]: Out of idle curiosity, I'm wondering about all the various idempotent constructions we have in mathematics (they seem to be generally referred to as a "closure" or "completion"), and how some of them are related (e.g., the radical of an ideal and the closure of a subset of $k^n$ in the Zariski topology, via the Nullstellensatz - the radical and the topological closure both being idempotent). 
So, one answer per post, but if you have two concepts which are related, I guess it'd be okay to put them together. For the sake of the completeness (ha ha) of this list, I'll add "radical" and "topological closure".
EDIT: My bad - I should have looked around more first. There's this list at Wikipedia and this list at nLab. Well, I'm sure there's plenty more concepts out there, so if you think of any more, feel free to add them. But let's focus on how some of these concepts are related - e.g., does one kind of completion arise in terms of another? What are some general ways in which completions and closures arise?

REPLY [3 votes]: The coolest closure operation I know occurs in Razborov's lower bound for the monotone circuit complexity of the clique function. In that proof he needs a class of "simple" set systems, and to get it he defines a very ingenious k-ary operation on sets and defines a simple set system to be one that is closed under that operation. I don't know what general moral to draw from that, but it's fairly different from the examples mentioned so far.<|endoftext|>
TITLE: How big can the irreps of a finite group be (over an arbitrary field)?
QUESTION [5 upvotes]: Let G be a nontrivial finite group.  Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G?
[Edited for clarity. -- PLC]

REPLY [5 votes]: The $\sqrt{|G|}$ bound is true [for algebraically closed fields -- PLC] in any characteristic. In fact if $R$ is the
Jacobson radical of $k[G]$, then $k[G]/R\cong\prod_iM_{n_i}(k)$ where $i$ runs over
the irreducible representations and $n_i$ is the degree of the corresponding
representation. This gives $\sum_in_i^2=\dim k[G]/R\leq|G|$. In the modular case we
always have that $R\neq0$ so that in particular we have strict inequality. Also for
every irreducible representation in charateristic $p$ there is an irreducible
representation in characteristic $0$ whose degree is $\ge$ than the degree of
the characteristic $p$ representation: Choose a number field $K$ which is a
splitting field for $G$ and let $R$ be its ring of integers and let further $P$
be a maximal ideal of $R$ dividing $p$. We may filter $K[G]$ by a Jordan-Hölder
filtration $W'_i$ so that $W'_i/W' _{i-1}$ is irreducible. Put $W_i:=W'_i\cap R[G]$
so that in particular $W_i/W_{i-1}$ is $R$-torsion free. Hence reducing modulo
$P$ we get a filtration $\overline{W}_i$ of $k[G]$. This filtration can be
refined to a Jordan-Hölder filtration showing that every irreducible
$k[G]$-module which appears in any Jordan-Hölder filtration of $k[G]$ must
appear in any Jordan-Hölder filtration of some
$\overline{W}_i/\overline{W} _{i-1}$ and thus its degree is $\leq
\dim(\overline{W}_i/\overline{W} _{i-1})=\mathrm{rank}(W_i/W _{i-1})=\dim(W'_i/W' _{i-1})$.
Hence the degree of any $k[G]$-representation is $\le$ the degree of some
$K[G]$-representation. It is rare (but does happen) that
$\overline{W}_i/\overline{W} _{i-1}$ is irreducible in the modular case.<|endoftext|>
TITLE: Applications of noncommutative geometry
QUESTION [34 upvotes]: This is related to Anweshi's question about theories of noncommutative geometry.
Let's start out by saying that I live, mostly, in a commutative universe.  The only noncommutative rings I have much truck with are either supercommutative, almost commutative (filtered, with commutative associated graded), group algebras or matrix algebras, none of which really show many of the true difficulties of noncommutative things.
So, here's my (somewhat pithy) question: what's noncommutative geometry good for?
To be a bit more precise, I have a vague sense that $C^*$ stuff is supposed to work well in quantum mechanics, but I'm somewhat more interested in more algebraic noncommutative geometry.  What sorts of problems does it solve that we can't solve without leaving the commutative world? Why should, say, a complex algebraic geometer learn some noncommutative geometry?

REPLY [25 votes]: Charles,
a couple of reasons why a complex algebraic geometer (certainly someone who is 
interested in moduli spaces of vector bundles, as your profile tells me) might 
at least keep
an open verdict on the stuff NC-algebraic geometers (NCAGers from now on) are trying to do.
in recent years ,a lot of progress has been made towards understanding moduli spaces of
semi-stable representations of 'formally smooth' algebras (think 'smooth in the NC-world).
in particular when it comes to their etale local structure and their rationality.
for example, there is this book, by someone.
this may not seem terribly relevant to you until you realize that some of the more
interesting moduli spaces in algebraic geometry are among those studied. for example, the
moduli space of semi-stable rank n bundles of degree 0 over a curve of genus g is the moduli
space of representations of a certain dimension vector over a specific formally smooth algebra,
as Aidan Schofield showed. he also applied this to rationality results about these spaces.
likewise, the moduli space of semi-stable rank n vectorbundles on the projective plane with Chern 
classes c1=0 and c2=n is birational to that of semi-simple n-dimensional representations of the free algebra
in two variables. the corresponding rationality problem has been studied by NCAG-ers (aka 'ringtheorists'
at the time) since the early 70ties (work by S.A. Amitsur, Claudio Procesi and Ed Formanek). by their results, we NCAGers,
knew that the method of 'proof' by Maruyama of their stable rationality in the mid 80ties, couldn't possibly work.
it's rather ironic that the best rationality results on these moduli spaces (of bundles over the
projective plane) are not due to AGers but to NCAGers : Procesi for n=2, Formanek for n=3 and 4 and
Bessenrodt and some guy for n=5 and 7. together with a result by Aidan Schofield these results show
that this moduli space is stably rational for all divisors n of 420.
further, what a crepant resolution of a quotient singularity is to you, is to NCAGers the moduli space of certain
representations of a nice noncommutative algebra over the singularity.
likewise, when you AGers mumble 'Deligne-Mumford stack', we NCAGers say 'ah! a noncommutative algebra'.

REPLY [3 votes]: There is a lecture course by M. Kontsevich (ENS, 1998) which has a chapter entitled "algebraic geometry from a noncommutative viewpoint". The notes are available here
www.math.uchicago.edu/~mitya/langlands/kontsevich.ps
It mentions in particular Bondal-Orlov's theorem that if a variety is Fano or of general type, then one can reconstruct it from its coherent derived category. There is also a section on Mirror symmetry. Proofs are only sketched, but these sketches can be most illuminating (or sometimes not).
In general, the "noncommutative geometry" section of
http://www.math.uchicago.edu/~mitya/langlands.html
can be useful, but some of the references there (Keller, Lefevre) do not specifically deal with applications to algebraic geometry. (And maybe someone else here can elaborate on those that do.)<|endoftext|>
TITLE: What bijection on permutations corresponds under RS to transpose?
QUESTION [15 upvotes]: The Robinson-Schensted correspondence is a bijection between elements of the symmetric group and ordered pairs of standard tableaux of the same shape.
Some simple operations on tableaux correspond to simple operations on the group: switching the tableaux corresponds to inverse on the group.

What about taking the transpose of the tableaux?  Does that correspond to something easily described on permutations?

In order to rule out easy guesses, let me describe this on $S_3$: 

the identity switches with the involution 321.
the transpositions 213 and 132 switch.
the 3-cycles 312 and 231 switch.

In general, this operation preserves being order $\leq 2$ (since this is equivalent to the P- and Q-symbols being the same).

REPLY [10 votes]: When you conjugate diagrams and apply RSK or other Young tableau bijections, the answer is typically bad, with some rare exceptions.  The right way to think of RSK is to think of row length being continuous while columns still integer (see e.g. my "Geometric proof of the hook-length formula" paper and refs therein).  
An exception: there is a "hidden symmetry" for LR-coefficients when you conjugate all three diagrams - see Hanlon-Sundaram paper (1992).  Once you know this bijection, there are natural connections to RSK as described in the long Pak-Vallejo paper and a followup by Azenhas-Conflitti-Mamede (search the web for a paper and a ppt presentation).  Together, these do give a complete description of your involution, but making sense of it might require quite a bit of work.<|endoftext|>
TITLE: Visualization of Riemann–Stieltjes Integrals
QUESTION [17 upvotes]: The Riemann–Stieltjes integral $\int_a^b f(x)\,dg(x)$ is a generalization of the Riemann integral. It is e.g. heavily used as a starting point for stochastic integration. The approximating Riemann–Stieltjes sums are analogous to the Riemann sums $\sum_{i=0}^{n-1} f(c_i)(g(x_{i+1})-g(x_i))$ where $c_i$ is in the $i$-th subinterval $[x_i,x_{i+1}]$.
The Riemann-sums can be very intuitively visualized by rectangles that approximate the area under the curve.
See e.g. Wikipedia:Riemann sum.
Unfortunately, I cannot find respective intuitive visualizations of the Riemann–Stieltjes sums.
My question: Could anyone provide me with some literature, pictures, links, or especially tools (e.g. Mathematica or even Excel) with which I could play around to get a similar intuition for this more general kind of integral?

REPLY [2 votes]: This article seems useful to me, gives a good overview of existing interpretations:
https://maa.tandfonline.com/doi/abs/10.1080/07468342.2019.1580109?journalCode=ucmj20#.XPJU16WxVkA<|endoftext|>
TITLE: First-order UFD (factorial ring) condition / pre-Schreier rings
QUESTION [8 upvotes]: All rings in this post are commutative and with $1$.
Everyone knows the definition of a factorial ring, a. k. a. unique factorization domain (UFD). I have been wondering about some variations regarding this notion.
(a) A ring $R$ is called pre-pre-Schreier (this is my nomenclature) if and only if for any four elements $a$, $b$, $c$, $d$ of $R$ satisfying $ab = cd$, we can find four elements $x$, $y$, $z$, $w$ in $R$ such that $a = xy$, $b = zw$, $c = xz$, $d = yw$.
(b) A ring $R$ is called pre-Schreier if it is pre-pre-Schreier and an integral domain. (This is not my nomenclature.)
It is easy to see that a Noetherian ring is a UFD if and only if it is pre-Schreier; on the other hand, the condition on a ring to be pre-Schreier is a first-order logic formula (if I'm right; I'm not an expert in logic). This was actually my motivation to consider pre-Schreier rings: to first-orderize the UFD condition. (Is there a first-order logic formula that is equivalent to UFD always, no matter whether the ring is Noetherian or not?) As for pre-pre-Schreier rings, I was just trying to see what happens if we leave out the domain condition.
According to this paper (Remark 4.6. (1)), the polynomial ring $R\left[X\right]$ over a pre-Schreier ring $R$ doesn't have to be pre-Schreier. My questions are now:
(1) If a Noetherian ring $R$ is pre-pre-Schreier, then what can be said about $R\left[X\right]$ ?
(2) Can a pre-pre-Schreier ring contain nilpotents $\neq 0$ ? I used to think I have proven that it can't if it is Noetherian, but now I see flaws in my argument.

REPLY [16 votes]: This is an answer to your parenthetical question:

Is there a first-order logic formula that is equivalent to UFD always, no matter whether the ring is Noetherian or not?

No. In fact, if T is any extension of the first-order theory of integral domains such that every model of T is a UFD, then every model of T is a field.
I think this is well known, but I couldn't think of a reference. I'll sketch a quick proof. Let φn(x) be the standard first-order formula which says that x is a product of  n irreducible elements (and φ0(x) says that x is a unit). We argue that for some fixed n,
T ⊦ ∀x (x ≠ 0 → φ0(x) ∨ ... ∨ φn(x)).
If not, by the Compactness Theorem, T has a model R with a distinguished nonzero element x such that R ⊧ ¬φn(x) for each n; so x does not have a factorization into irreducibles. To conclude, show that every UFD which satisfies ∀x (x ≠ 0 → φ0(x) ∨ ... ∨ φn(x)) must be a field.<|endoftext|>
TITLE: Geometric Intuition for Big Monodromy
QUESTION [8 upvotes]: In various contexts, I have come across results referred to as "big monodromy."  A standard arithmetic example is the open image theorem for the image of Galois action on non-CM elliptic curves.  A general setup for such a result in algebraic geometry is:
Given a proper, generically smooth map $\pi:X \rightarrow S$  of relative dimension d, say S is connected.  This gives rise to an $l$-adic representations of the etale fundamental group $\pi_1(U)$  where $U$ is smooth locus of $\pi$ corresponding to higher pushforward $R^d \pi_* Q_l$.  One might say it has "big monodromy"  if the Zariski closure of the image is as big as it can be given that it has to respect cup-product, etc.
My specific question is what are the geometric consequences of big monodromy? If we know such a result for $\pi$, what does that say about the geometry of the fibration or at the very least is there geometric intuition for what it should mean?
I welcome intuition from number theory, algebraic geometry, or complex geometry.
I have also heard that "one should expect big monodromy unless there is a reason not to" (for example, complex multiplication).  What are other examples of things which inhibit big monodromy?

REPLY [3 votes]: I found this question a little vague, but let me at least remark on "other examples of things which inhibit big monodromy."  Mumford gives an example in section 4 of  
D. Mumford, “A note of Shimura’s paper “Discontinuous groups and abelian varieties”,” Math. Ann. 181 (1969), 345–351.
of an abelian variety A whose Galois representation has image strictly smaller than Sp_{2g}(Z_p), despite the fact that End(A) = Z.  The keyword to look up is "Mumford-Tate group", which is in some sense the answer to the question 
How big COULD the Galois representation on an abelian variety be, subject to all geometric 'things which inhibit big monodromy'?
Reference comes from a paper of Chris Hall which shows how to prove big monodromy results in many cases.<|endoftext|>
TITLE: Truth of the Poisson summation formula
QUESTION [39 upvotes]: The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). Here is the wikipedia link.
For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true.
But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.

REPLY [3 votes]: The Poisson formula is just the decomposition in Fourier series of the Dirac measure on the circle. 
It happens that the theory of distributions on the circle is far more simpler than the one on the real line since everything is compactly supported. In his book, Laurent Schwartz starts the chapter about the Fourier transform with the circle and shows that any distribution has a convergent Fourier series (VII.1.3 "theorie des distributions").
Moreover the space of distributions is the union of all the negative Sobolev spaces and these spaces can be caracterized in terms of the decay of the Fourier coefficients. I think that the Dirac is in $H^{-1}$. Anyway the usual formula holds and we have
$$
\delta_0(x) = \sum_n \langle \delta_0, e^{inx} \rangle e^{inx} = {1\over 2\pi}\sum_n e^{inx}, \quad x\in {\bf R}/{\bf Z}.
$$
Note that with this choice of normalization, the scalar product is ${1\over 2\pi}\int_0^{2\pi} f(x)\overline{g(x)} \,dx$ so that the $e^{inx}$ are unit vectors.This gives the ${1\over 2\pi}$ on the right hand side.
Now we compose this with the projection $\pi : {\bf R} \rightarrow {\bf R} /2\pi{\bf Z}$ to get the Poisson formula on ${\bf R}$. One must be a bit cautious because although we can always compose a function with another as soon as the domains match, it is not true that composition is always possible for distributions. 
In our particular case, the Dirac measure is the limit of an approximate identity, and so it is easy to guess what is happening just by composing the approximate identity and passing to the limit. Without surprise, the pullback of the Dirac measure at $\{0\}$ is given by the Dirac measure on the pullback of $\{0\}$, which is just the sum of all the Dirac measures on the points $2\pi{\bf Z}$.
$$
\sum_{n\in {\bf Z}} \delta_{2\pi n}(x) = {1\over 2\pi} \sum_{n\in {\bf Z}} e^{inx}, \quad x\in {\bf R}.
$$
We can evaluate this against a function in Schwartz space or use a convolution to obtain the original formula of Poisson. And we got the $2\pi$ at the correct locations without even thinking about it. So that's the truth.<|endoftext|>
TITLE: Understanding the definition of the Lefschetz (pure effective) motive
QUESTION [40 upvotes]: For all those who are unlikely to have answers to my questions, I provide some
Background:
In some sense, pure motives are generalisations of smooth projective varieties. Every Weil cohomology theory factors through the embedding of smooth projective varieties into the category of pure Chow motives.
Pure effective motives
In the definition of pure motives (say over a field k), the last step is to take the category of pure effective motives and formally invert the Lefschetz motive L.
The category of pure effective motives is the pseudo-abelian envelope of a category of correspondence classes, which has as objects smooth projective varieties over k and as morphisms X → Y cycle classes in X×Y of dimension dim X (think of it as a generalisation of morphisms, where morphisms are included as their graphs), where an adequate equivalence relation is imposed, to have a well-defined composition (therefore the word "classes"). When the adequate relation is rational equivalence, the resulting category is called the category of pure effective Chow motives.
In each step of the construction, the monoidal structure of one step defines a monoidal structure on the next step.
For more background see Ilya's question about the yoga of motives.
Definition of the Lefschetz motive
The Lefschetz motive L is defined as follows:
For each point p in P¹ (1-dimensional projective space over k), there is the embedding morphism Spec k → P¹, which can be composed with the structural morphism P¹ → Spec k to yield an endomorphism of P¹. This is an idempotent, since the other composition Spec k → Spec k is the identity.
The category of effective pure motives is pseudo-abelian, so every idempotent has a kernel and thus [P¹] = [Spec k] + [something] =: 1+L, where the summand [something] is now named Lefschetz motive L.
Properties
The definition of L doesn't depend on the choice of the point p.
From nLab and Kahn's leçons I learned that the inversion of the Lefschetz motive is what makes the resulting monoidal category a rigid monoidal category - while the category of pure effective motives is not necessarily rigid.
In the category of pure motives, the inverse $L^{-1}$ is called T, the Tate motive.
Questions:
These questions are somehow related to each other:


Why is this motive L called "Lefschetz"?
Why is its inverse $L^{-1}$ called "Tate"?
Why is it exactly this construction that "rigidifies" the category?
Would another construction work, too - or is this somehow universal?
How should I think of L geometrically?


I have almost no background in number theory, so even if you have good answers, it may remain totally unclear to me, why the name "Tate" intervenes. I expect however, that the name "Lefschetz" has something to do with the Lefschetz trace formula. I guess that the procedure of inverting L is the only one which makes the category rigid, in some universal way, but I have no idea, why. In addition, I guess there is no "geometric" picture of L.
If I made any mistakes in the background section, feel free to edit. As I'm currently taking a first course on motives, I may now have asked something completely stupid. If this is the case, please point me politely to some document which will then enlighten me or, at least, let me ascend to a higher level of confusion.
UPDATE: Thanks so far for the answers, questions 1-4 are now clear to me. It remains, if the "rigidification" could be accomplished by another construction - maybe some universal way to turn a monoidal category into a rigid one? Then one could later identify the Lefschetz motive as some kind of a generator of the kernel of the rigidification functor.
The geometric intuition, to think of L as a curve and of $L^{\otimes d}$ as a d-dimensional manifold, remains fuzzy, but I have the hope that this becomes clear when I have worked a little bit on the classical Lefschetz/Poincare theorems and the proof of Weil conjectures for Betti cohomology (is this hope justified?).

REPLY [10 votes]: A little more on questions 3 and 4.
The basic property of the Lefschetz motif (in this aspect) is the Cancellation Theorem: $Hom(X,Y)\cong Hom(X\otimes L,Y\otimes L)$ for any effective motives X and Y. If you don't have this property for an object, then inverting it is a 'bad' operation: the functor from the 'old' category to the 'new' one is very far from being a full embedding.
So, one should not invert any L if it does not satisfy the cancellation theorem. If it does, then its image in the ('usual') Chow motives is invertible (with respect to the tensor product). Conjecturally, any invertible Chow motif is of the form $M(n)$ where $M$ is an (effective) zero-dimensional motif which becomes isomorphic to 1 after a finite base field extension.
We obtain: there IS some flexibility in the choice of a motif which you want to invert, but they yield the same result. I do not know how to define 'universal rigidifying' for a tensor category; probably there does not exist such a construction that is nice enough.<|endoftext|>
TITLE: When to start reviewing
QUESTION [32 upvotes]: Recently I had a few papers published, and I suppose in response to this, I received a request from Zentralblatt to be a reviewer. They ask some general questions about what I would feel comfortable reviewing, and if I would be willing to receive papers electronically. My concern about just signing up stems from the fact that I am only a second year graduate student. So my question is

When is the right time to start reviewing?

and

When did you start reviewing for journals/reviewing organizations?

Thanks ahead of time!

REPLY [26 votes]: Actually, all reviews are written by research mathematicians, since MR/Zentralblatt typically only invite those who have published papers.
In fact, it is not a great burden. As everyone knows, the best way to understand some mathematics is to explain it others. If you get a paper to review that you have read, or were planning to read, then writing a review doesn't take long. If you get a paper that you weren't planning to read, then reviewing it broadens your understanding.
So my advice is to begin reviewing when invited, although no harm will be done if you prefer to wait until you are more experienced.
Since all of us benefit from reviews, all of us have an ethical obligation to contribute for at least part of our careers.<|endoftext|>
TITLE: Cartan involution for finite W-algebras
QUESTION [12 upvotes]: Does anybody know if there is an analog of the Cartan (anti)involution for W-algebra
associated to a nilpotent element e, which is principal in some Levi subalgebra
of semi-simple Lie algebra g? Actually, I am more interested whether there exists an analog of the Shapovalov form on a Verma module for such a W-algebra.

REPLY [2 votes]: In the affine case, there is a related discussion in a paper by Kac-Wakimoto; 
see pp. 23-25 in arXiv:math-ph/0304011v2.<|endoftext|>
TITLE: Invariant polynomials under a group action (hidden GIT)
QUESTION [30 upvotes]: Let's say I start with the polynomial ring in $n$ variables $R = \mathbb{Z}[x_1,...,x_n]$ (in the case at hand I had $\mathbb{C}$ in place of $\mathbb{Z}$).
Now the symmetric group $\mathfrak{S}_n$ acts by permutation on the indeterminates.
The subring of invariant polynomials $R^{\mathfrak{S}_n}$ has a nice description (by generators and relations) in terms of symmetric functions.
What if I only consider the action of the cyclic group $Z_n$? 
Does anyone know if the ring $R^{Z_n}$ admits a nice presentation?
(in the case at hand I had $\mathbb{C}[x_1,x_2,x_3]$ and the action of the cyclic group $Z_3$. Maybe in this case we can use some formula (assuming there is one) for groups splitting as semidirect products?)

REPLY [12 votes]: Even without knowing an explicit set of generators, you can compute the Hilbert series with very little work as follows. In general, suppose a finite group $G$ acts on a vector space $V$ over a field $k$ of characteristic not divisible by $|G|$ via an action map $\rho : G \to \text{GL}(V)$. Then $G$ acts on the symmetric algebra $S(V^{\ast})$ (a coordinate-independent description of the polynomial functions on $V$), and the ring of functions on the quotient $V/G$ is the invariant subalgebra
$$S(V^{\ast})^G \cong \bigoplus_{n \ge 0} S^n(V^{\ast})^G.$$
The Hilbert series of this subalgebra can be computed using Molien's formula, which gives
$$\sum_{n \ge 0} t^n \dim S^n(V^{\ast})^G = \frac{1}{|G|} \sum_{g \in G} \frac{1}{\det (1 - t \rho(g))}.$$
In particular the Hilbert series is invariant under extension of scalars (so e.g. the answer doesn't depend on whether we take $k = \mathbb{Q}$ or $k = \mathbb{C}$), which is not completely obvious. 
In this case $G = \mathbb{Z}_n$ and $V$ is the regular representation. If an element $g \in G$ has order $d \mid n$, its action in the regular representation consists of $\frac{n}{d}$ cycles of length $d$, so the determinant above is $(1 - t^d)^{n/d}$. Altogether this gives the Hilbert series
$$\frac{1}{n} \sum_{d \mid n} \frac{\varphi(d)}{(1 - t^d)^{n/d}}.$$<|endoftext|>
TITLE: Freedman's work on non-simply-connected 4-manifolds
QUESTION [8 upvotes]: In the late 1970's and in the 1980's, Michael Freedman showed a relationship between the topological surgery problem in 4-dimensions, the slice problem for links, and the classification of non-simply-connected 4-manifolds. He also showed the failure of 4-dimensional homology surgery and the homology splitting theorem via a construction I don't really follow (because I haven't read and don't know the original reference, for one thing). I know of these results vaguely but do not understand them (certainly not the proofs). In particular, I don't know the "canonical" references, and MathSciNet doesn't seem to be helping me. They look like basic results in 4-manifold topology, which should surely be standard. Is there a survey paper on this stuff, or is it in some book? What is the best reference?
I should know this (or look it up myself), but perhaps it is more useful to post it here, because perhaps I am not alone in my confusion.

REPLY [6 votes]: I would recommend you to look at the reference Freedman-Quinn book Topology of 4-manifolds , it might be helpful.<|endoftext|>
TITLE: How can an ultrapower of a model of ZFC be "ill-founded" yet still satisfy ZFC?
QUESTION [6 upvotes]: My understanding (please correct me if I'm wrong) is that if you have some transitive set M which is an $\epsilon$-model of ZFC, and you take an ultrapower of it using an approprate ultrafilter, you wind up with a new model whose membership relation is not the $\epsilon$ relation of the ambient set theory, but still satisfies ZFC.  Furthermore, if the membership relation of the ultrapower is well-founded, one can always use the Mostowski collapse theorem to produce an isomorphic $\epsilon$-model.
My question is this: how could one possibly end up with a model of ZFC which satisfies the axiom of regularity ("every set is disjoint from one of its members"), yet whose membership relation isn't well-founded?
I'm struggling to imagine this; the most I can come up with is that for no set in the infinite chain is its transitive closure also a set (of the model).  But I'm skeptical about whether or not that can be the case, because it seems like you ought to be able to construct the transitive closure using definition by transfinite recursion, letting $f(0)$ be any set in the chain and $f(n+1)=\bigcup f(n)$ (axiom of union).  Then $f(\omega)$ (axiom of infinity) ought to contain all the sets needed to build a contradiction to regularity.
Sorry if this question sounds like I'm arguing with myself.  This has been bothering me for a few days now.

REPLY [17 votes]: For an ultrapower of $V$ by an ultrafilter $\mathcal{U}$ there is an exact characterization of when the ultrapower will be well-founded: precisely when $\mathcal{U}$ is closed under countable intersections.
As for how a model $M$ could possibly satisfy regularity but not be wellfounded, the problem is that there may be infinite descending chains that $M$ cannot 'see': each individual object may belong to $M$ but the chain itself may not. I can be a little more precise. Let's say $R$ is what $M$ understands to be the $\in$-relation. There may well be $x_0,\ldots x_n,\ldots $ belonging to $M$ so that $x_{n+1}Rx_n$ for all $n\in\omega$; as long as the sequence $\langle x_n:n\in\omega\rangle$ does not belong to $M$ the axiom of regularity from $M$'s point of view need not be violated.
You don't need ultrapowers to construct such models. Assuming CON(ZFC) you can build one using the compactness theorem for first order logic.<|endoftext|>
TITLE: The work of E. Artin and F. K. Schmidt on (what are now called) the Weil conjectures.
QUESTION [34 upvotes]: I was reading Dieudonne's "On the history of the Weil conjectures" and found two things that surprised me. Dieudonne makes some assertions about the work of Artin and Schmidt which are no doubt correct, but he doesn't give references, and the thought of ploughing through Artin's collected works seems a bit daunting to me, so I thought I'd ask here first.
Background.
If $V$ is a smooth (affine or projective) curve over a finite field $k$ of size $q$, then $k$ has (up to isomorphism) a unique extension $k_n$ of degree $n$ over $k$ (so $k_n$ has size $q^n$) and one can define $N_n$ to be the size of $V(k_n)$. Completely concretely, one can perhaps imagine the case where $V$ is defined by one equation in affine or projective 2-space, for example $y^2=x^3+1$ (note that this equation will give a smooth curve in affine 2-space for $p$, the characteristic of $k$, sufficiently large), and simply count the number of solutions to this equation with $x,y\in k_n\ $to get $N_n$. Let $F_V(u)=\sum_{n\geq1}N_nu^n$ denote the formal power series associated to this counting function.
Now it turns out from the "formalism of zeta functions" that this isn't the most ideal way to package the information of the $N_n$, one really wants to be doing a product over closed points of your variety. If $C_d$ is the number of closed points of $V$ of degree $d$, that is, the number of closed points $v$ of (the topological space underlying the scheme) $V$ such that $k(v)$ is isomorphic to $k_d$, then one really wants to define
$$Z_V(u)=\prod_{d\geq1}(1-u^d)^{-C_d}.$$
If one sets $u=q^{-s}$ then this is an analogue of the Riemann zeta function, which is a product over closed points of $Spec(\mathbf{Z})$ of an analogous thing.
Now the (easy to check) relation between the $C$s and the $N$s is that $N_n=\sum_{d|n}dC_d$, and this translates into a relation between $F_V$ and $Z_V$ of the form
$$uZ_V'(u)/Z_V(u)=F_V(u).$$
This relation also means one can compute $Z$ given $F$: one divides $F$ by $u$, integrates formally, and then exponentiates formally; this works because $f'/f=(\log(f))'$.
The reason I'm saying all of this is just to stress that this part of the theory is completely elementary.
The Weil conjectures in this setting.
The Weil conjectures imply that for $V$ as above, the power series $Z_V(u)$ is actually a rational function of $u$, and make various concrete statements about its explicit form (and in particular the location of zeros and poles). Note that they are usually stated for smooth projective varieties, but in the affine curve case one can take the smooth projective model for $V$ and then just throw away the finitely many extra points showing up to see that $Z_V(u)$ is a rational function in this case too.
How to prove special cases in 1923?
OK so here's the question. It's 1923, we are considering completely explicit affine or projective curves over explicit finite fields, and we want to check that this power series $Z_V(u)$ is a rational function. Dieudonne states that Artin manages to do this for curves of the form $y^2=P(x)$ for "many polynomials $P$ of low degree". How might we do this? For $P$ of degree 1 or 2, the curve is birational to projective 1-space and the story is easy. For $V$ equals projective 1-space, we have
$$F_V(u)=(1+q)u+(1+q^2)u^2+(1+q^3)u^3+\ldots=u/(1-u)+qu/(1-qu)$$
from which it follows easily from the above discussion that
$$Z_V(u)=1/[(1-u)(1-qu)].$$
For polynomials $P$ of degree 3 or 4, the curve has genus 1 and again I can envisage how Artin could have approached the problem. The curve will be birational to an elliptic curve, and it will lift to a characteristic zero curve with complex multiplication. The traces of Frobenius will be controlled by the corresponding Hecke character, a fact which surely will not have escaped Artin, and I can believe that he was now smart enough to put everything together. 
For polynomials of degree 5 or more, given that it's 1923, the problem looks formidable. 
Q1) When Dieudonne says that Artin verified (some piece of) the Weil conjectures for "many polynomials of low degree", does he mean "of degree at most 4", or did Artin really move into genus 2?
How much further can we get in 1931?
Now this one really surprised me. Dieudonne claims that in 1931 F. K. Schmidt proved rationality of $Z_V(u)$, plus the functional equation, plus the fact that $Z_V(u)=P(u)/(1-u)(1-qu)$, for $V$ an arbitrary smooth projective curve, and that he showed $P(u)$ was a polynomial of degree $2g$, with $g$ the genus of $V$. This is already a huge chunk of the Weil conjectures. We're missing the statement that $P(u)$ has all its rots of size $q^{-1/2}$ (the "Riemann hypothesis") but this is understandable: one needs a fair amount of machinery to prove this. What startled me (in my naivity) was that I had assumed that all this was due to Weil in the 1940s and I am obviously wrong: "all Weil did" was to prove RH. So I have a very basic history question:
Q2) However did Schmidt do this?

EDIT: brief summary of answers below (and what I learned from chasing up the references):
A1) Artin didn't do anything like what I suggested. He could explicitly compute the zeta function of an arbitrary given hyperelliptic curve over a given finite field by an elegant application of quadratic reciprocity. See e.g. the first of Roquette's three papers below. The method in theory works for all genera although the computations quickly get tiresome.
A2) Riemann-Roch. Express the product defining $Z$ as an infinite sum and then use your head.

REPLY [17 votes]: Roquette's articles contain the whole story; briefly, here are the main facts:

Artin's collected works are quite small, and his thesis (two parts) is put
right at the beginning.
Artin looked at quadratic extensions of the rational funtion field and
distinguished between real and imaginary extensions; geometrically, these
are affine pieces of hyperelliptic curves. The birational point of view,
in which these distinctions disappear, was introduced later by F.K. Schmidt. 
Moreover, Artin talked about ideal classes instead of points on the curves.
The rationality of the zeta functions follows from the quadratic reciprocity law
in the rational function field. 
Given a specific extension, it is easy to compute the zeta function as well as
its zeroes, and verify the Riemann hypothesis. This is what Artin did for 
extensions of small degree. 
F.K. Schmidt introduced "birational invariance" by looking at the projective 
curves and realized that the rationality can be proved by Riemann-Roch instead
of the reciprocity law.
I think Artin started reading Hecke only after his thesis, so he did not know about
Hecke characters at the time. 

Thus Artin verified the Riemann hypothesis for specific examples of quadratic extensions, F.K. Schmidt dervied the functional equation and the rationality for general extensions,
Hasse proved the RH for curves of genus 1, and Weil gave the proof for curves (and extended the conjecture to varieties of arbitrary dimension).<|endoftext|>
TITLE: What is the relationship between t-structure and Torsion pair?
QUESTION [13 upvotes]: I am away from Torsion theory in abelian category for some while. So it might be a stupid question. 
The definition of a torsion pair in the category of modules is as follows:
Definition:
A pair $(\mathcal T,\mathcal F)$ of full subcategories of $A-\mathrm{mod}$ is called a torsion pair if following conditions hold:

$\mathrm{Hom}_{A}(M,N)=0$ for all $M \in \mathcal T,
    N \in \mathcal F$.
$\mathrm{Hom}_{A}(M,-)|_{\mathcal F}=0 \Rightarrow M \in \mathcal T$.
$\mathrm{Hom}_{A}(-,N)|_{\mathcal T}=0 \Rightarrow N \in \mathcal F$.

Condition 2) and 3) means that the pair $(\mathcal T,\mathcal F)$ is maximal for $\mathrm{Hom}_{A}(M,N)=0$.
This definition is from the book elements of representation theory of associative algebras
I found this definition is similar to the definition of t-structures in derived category. I just quote the definition from Dimca, Sheaves in topology as follows:
A t-structure on a triangulated category $\mathcal D$ consists in two strictly full subcategories: $\mathcal D^{\leq 0}, \mathcal D^{\geq 0}$ such that the following conditions hold:

$\mathrm{Hom}(X,Y)=0$ if $X \in \mathcal D^{\leq 0}, Y \in \mathcal D^{\geq 1}$.
$\mathcal D^{\leq 0} \subseteq \mathcal D^{\leq 1}, \mathcal D^{\geq 1} \subseteq \mathcal D^{\geq 0}$.
For any $X \in \mathcal D$, there is a distinguished triangle
$$A\rightarrow X\rightarrow B\rightarrow A[+1], \qquad A \in \mathcal D^{\leq 0}, B \in \mathcal D^{\geq 1}.$$

Although the axiom 3) for t-structures looks different to the axioms of torsion pairs. However, there is a proposition of torsion pairs establishing the similar formula:
Let a pair $(\mathcal T,\mathcal F)$ be a torsion pair in $A-\mathrm{mod}$, and let $M$ be an $A$-module. Then there exists a short exact sequence 
$$0 \rightarrow tM \rightarrow M \rightarrow M/tM \rightarrow 0, \qquad tM \in \mathcal T, M/tM \in \mathcal F,$$
where $t$ is the idempotent radical (it behaves like radical of module).
My questions

Is there any relationship between these two constructions?
Is there a definition of torsion theory in triangulated categories? If there exists, does it coincide with t-structures in triangulated categories?
t-structures played important roles in reconstruction schemes (or go back to abelian category) from derived category. So, is torsion theory in abelian category playing similar roles? (I suspected very much, so it might be stupid.)

Thank you in advance!

REPLY [5 votes]: As I understand it, all torsion classes correspond to t-structures in the way described by Greg, but there is almost always more t-structures than torsion classes (even taking into account the shifts).
I think t-structures are closer to some kind of filtration on the abelian category. 
There's a paper on stability conditions (Gorodentsev, A. L.; Kuleshov, S. A.; Rudakov, A. N. $t$-stabilities and $t$-structures on triangulated categories. (Russian) Izv. Ross. Akad. Nauk Ser. Mat. 68 (2004), no. 4, 117--150; translation in Izv. Math. 68 (2004), no. 4, 749--781) that talks about this a bit.<|endoftext|>
TITLE: Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?)
QUESTION [38 upvotes]: In this question, Harry Gindi states:

The fact that a commutative ring has a natural topological space associated with it is a really interesting coincidence.

Moreover, in the answers, Pete L. Clark gives a list of other "really interesting coincidences" of algebraic objects having naturally associated topological spaces.
Is there a deeper explanation of the occurrence of these "really interesting coincidences"? It seems to suggest that the standard definition of "topological space" (collection of subsets, unions, intersections, blah blah), which somehow always seemed kind of a weird and artificial definition to me, has some kind of deeper significance or explanation, since it pops up everywhere...
The (former) title of this question is meant to be provocative ;-)

See also:
What are interesting families of subsets of a given set? 
How can I really motivate the Zariski topology on a scheme? --- particularly Allen Knutson's answer

Edit 1: I should clarify a bit. Let me be more explicit: Is there a unified explanation (mathematical ... or perhaps not) for why various algebraic (where "algebraic" is loosely defined) objects should have naturally associated topological spaces? Pete in the comments notes that he does not like the use of the word "coincidence" here --- but if these things are not coincidences, then what's the explanation?
Of course I do understand the intuitive idea behind the definition of "topological space", and how it abstracts for example the notions of "neighborhood" and "near" and "far". It is not surprising that the formalism of topological spaces is useful and ubiquitous in situations involving things like R^n, subsets of R^n, manifolds, metric spaces, simplicial complexes, CW complexes, etc. 
However, when you start with algebraic objects and then get topological spaces out of them --- I find that surprising somehow because a priori there is not necessarily anything "geometric" or "topological" or "shape-y" or "neighborhood-y" going on.

Edit 2: Somebody has voted to close, saying this is "not a real question". I apologize for my imprecision and vagueness, but I still think this is a real question, for which real (mathematical) answers can conceivably exist.
For example, I'm hoping that maybe there is a theorem along the lines of something like: 

Given an algebraic object A satisfying blah, define Spec(A) to be the set of blah-blahs of A such that blah-blah-blah. There is a natural topology on Spec(A), defined by [something]. When A is a commutative ring, this agrees with the Zariski topology on the prime spectrum. When A is a commutative C^* algebra, this agrees with the [is there a name?] topology on the Gelfand spectrum. When A is a Boolean algebra... When A is a commutative Banach ring... etc.

Of course, such a theorem, if such a theorem exists at all, would also need a definition of 'algebraic object'.

REPLY [3 votes]: I know this is very late, I just happened to run into this question. 
I think Von Neumann might answer this question like this (as he once has): "One does not understand anything in mathematics one simply gets used to it." Then he may add, we have a natural functor from the category of topological spaces to the category of rings, it is an interesting natural phenomenon that this functor is strictly invertible if we restrict to suitable subcategories of the category of top spaces, and of the category of rings. This is Gelfand-Naimark.
One does not "understand" phenomena anymore then one understands why the universe exists.
By the way a topological space is also naturally an algebraic object, we can take its partially ordered set of open subsets. This often completely determines the space: http://math.nie.edu.sg/dszhao/Research%20papers/conference%20proceeding/posetmodels.pdf
(Although I didn't really read this.)<|endoftext|>
TITLE: Integrable dynamical system -  relation to elliptic curves
QUESTION [10 upvotes]: From seminar on kdV equation I know that for integrable dynamical system its trajectory in phase space lays on tori. In wikipedia article You may read (http://en.wikipedia.org/wiki/Integrable_system):

When a finite dimensional Hamiltonian
  system is completely integrable in the
  Liouville sense, and the energy level
  sets are compact, the flows are
  complete, and the leaves of the
  invariant foliation are tori. There
  then exist, as mentioned above,
  special sets of canonical coordinates
  on the phase space known as
  action-angle variables, such that the
  invariant tori are the joint level
  sets of the action variables. These
  thus provide a complete set of
  invariants of the Hamiltonian flow
  (constants of motion), and the angle
  variables are the natural periodic
  coordinates on the torus. The motion
  on the invariant tori, expressed in
  terms of these canonical coordinates,
  is linear in the angle variables.

As I also know that elliptic curve is in fact some kind of tori, then there natural question arises: Are tori for quasi-periodic motion in action-angle variables of some dynamical systems related in any way to algebraic structure like elliptic curve? Maybe some small dynamical systems and some elliptic curves are related in some way?
The most interesting in this matter is for me the size of space of elliptic functions: its quite small, every elliptic curve is rational function of  Weiestrass function, and its derivative. Has this property any analogy in integrable dynamical systems theory?
As isomorphic elliptic curves shares some invariants, it is also interesting it they have any "dynamical meaning".

REPLY [2 votes]: "The de-geometrisation of mathematical education and the divorce from physics sever these ties. For example, not only students but also modern algebro-geometers on the whole do not know about the Jacobi fact mentioned here: an elliptic integral of first kind expresses the time of motion along an elliptic phase curve in the corresponding Hamiltonian system. " 
From A.I.Arnold, here: http://pauli.uni-muenster.de/~munsteg/arnold.html
Definitely I should learn more in this area....<|endoftext|>
TITLE: Intuition for Primitive Cohomology
QUESTION [32 upvotes]: In complex projective geometry, we have a specified Kähler class $\omega$ and we have a Lefschetz operator $L:H^i(X,\mathbb{C})\to H^{i+2}(X,\mathbb{C})$ given by $L(\eta)=\omega\wedge \eta$.  We then define primitive cohomology $P^{n-k}(X,\mathbb{C})=\ker(L^{k+1}:H^{n-k}(X)\to H^{n+k+2}(X))$, and we even have a nice theorem, the Lefschetz decomposition, that says $H^m(X,\mathbb{C})=\oplus_k L^kP^{n-2k}$.  Often, in papers, people just prove their result for primitive classes, as they seem to be easier to work with.
So, what exactly ARE primitive classes? Sure, they're things that some power of $\omega$ kills, but what's the intuition? Why are they an interesting distinguished class? Is there a good reason to expect this decomposition?

REPLY [16 votes]: If you take Poincare duals, then the wedge product becomes intersection, and the Kahler form $\omega$ becomes a hyperplane $H$. Then, the primitive cohomology in dimension $n - k$ consists exactly of those classes whose duals (which are homology classes in dimension $n + k$), when we intersect with $k + 1$ generic hyperplanes gives us zero. In other words, the homology classes which don't intersect with some $(n - k - 1)$-dimensional linear subspace of $\mathbb{CP}^n$. The Lefschetz theorem in this context just says that we can get all of the homology by taking those classes which don't intersect with a linear subspace of complementary dimension and then intersecting with hyperplanes.
In the case $k = 0$, we can choose the hyperplane to be the one at infinity, so the primitive homology consists of the $n$-dimensional homology classes which sit in the "$\mathbb{C}^n$" part of $\mathbb{CP}^n$, which I think is a pretty natural set of classes to look at.<|endoftext|>
TITLE: The 'real' use of Quantum Algebra, Non-commutative Geometry, Representation Theory, and Algebraic Geometry to Physics
QUESTION [43 upvotes]: In this question, Orbicular made the following comment to Feb7 and my own answers;

Please keep in mind that - even though it is stated very often - noncommutative geometry does not give "real" insight to physics. The reason is that they only have toy models, all of which are unphysical (in the sense that they predict things which differ from real world measurements). Furthermore even the toy models are usually extremely complicated, killing most expectations to get a "real" model (which is not toyish).

First, I want to thank Orbicular for pointing this out, as it is something that I 'kinda' knew, but often forget. The purpose of this question, is to ask for a deeper explanation, either from Orbicular or someone else. In particular

to what degree does Quantum Algebra, Non-commutative Geometry, Representation Theory, and Algebraic Geometry influence/assist 'real' models and actual physics related to the physical world?

I don't wish for this question to turn into a debate about whether or not these maths will later be applied in some beautiful stringy-quantum-symmetry theory; I would much rather it be some explanation of the real use of these things. Specifically, I am interested in hearing about the use of Quantum Groups and their representations to Physicists along with some thoughts on the actual usefulness of the results in NC Algebraic Geometry of those articles I posted over here. Another particularly interesting subject I would like to hear about is the usefulness of commutative algebraic geometry in physics.
Some things I have found
Just two references that I have found that at least address these things to some degree are Peter Woit's lecture notes on Representation Theory, and in Shawn Majid's book on Quantum Groups he discusses some definite physical motivation for studying quantum groups.
Thanks!

REPLY [3 votes]: I might be wrong, but as far as I remember, in his ICM paper Drinfeld provides some motivation from physics for quantum groups.<|endoftext|>
TITLE: Can the Quantum Torus be realized as a Hall Algebra?
QUESTION [11 upvotes]: Background
The Quantum Torus
Let $q$ be an arbitrary complex number, and define (the algebra of) the quantum torus to be
$$T_q:=\mathbb{C}\langle x^{\pm 1},y^{\pm 1}\rangle/xy-qyx$$
For $q=1$, this is the commutative ring of functions on the torus $\mathbb{C}^\times\times \mathbb{C}^\times$; hence, for general $q$, this is regarded as a quantization of the torus.
Hall Algebras
Consider a small abelian category $A$, with the property that $Hom_A(M,N)$ and $Ext^i_A(M,N)$ are always finite sets for any $M,N\in A$ and $i\in \mathbb{Z}$.  Let $\overline{A}$ denote the set of isomorphism classes in $A$, and let
$$H(A)=\oplus_{[M]\in \overline{A}}\mathbb{C}[M]$$
denote the complex vector space spanned by $\overline{A}$.  Endow $H(A)$ with a multiplication by the formula
$$ [M]\cdot [N]=\sqrt{\langle [M],[N]\rangle)}\sum_{[R]\in \overline{A}}\frac{a_{MN}^R}{|Aut(M)||Aut(N)|}[R]$$
where $a_{MN}^R$ is the number of short exact sequences
$$0\rightarrow N\rightarrow R\rightarrow M\rightarrow 0$$
and
$$\langle [M],[N]\rangle = \sum (-1)^i |Ext^i_A(M,N)|$$
is the Euler form.  This multiplication makes $H(A)$ into an associate algebra called the Hall algebra of $A$; the proof can be found e.g. here.
Finite Fields and Quantization
The categories $A$ appearing in the construction of a Hall algebra are usually linear over some finite field $\mathbb{F}_q$.  Often, it is possible to simultaneously define a category $A_q$ for each finite field $\mathbb{F}_q$; usually by considering modules on the $\mathbb{F}_q$-points of some scheme over $\mathbb{Z}$.  The corresponding Hall algebras $H(A_q)$ will then usually be closely related, and can often be defined by relations that are functions in $q$.
The Question
I know that there are cases where an algebra is deformed by a parameter $q$, and then the resulting family of algebras `magically' coincides with a family of Hall algebras $H(A_q)$ in the special cases when $q$ is a prime power.  I think this happens in the case of the Hecke algebra (discussed here), and the case of quantum universal enveloping algebras (discussed here).  I somewhat understand that this is a symptom of a related convolution algebra on the scheme used to define $A_q$.
Is there a family of categories $A_q$ such that the corresponding Hall algebras $H(A_q)$ are isomorphic to the Quantum Torus $T_q$ for all $q$ a prime power?  If so, is there a convolution algebra realization of the Quantum Torus?

REPLY [10 votes]: As far as I understand, the Hall algebra of a category (say, with finite length of objects) is graded by the Grothendieck monoid of this category, spanned by simple objects over $\Bbb Z_+$, and it must have the ground field in degree $0$. The quantum torus algebra does not seem to have such a grading (it has a $\Bbb Z^2$-grading, not a $\Bbb Z_+^m$-grading). Maybe one should ask this question for the q-Weyl algebra $xy=qyx$ (not allowing negative powers of x and y)? Note that this algebra appears as a subalgebra of a Hall algebra (the Hall algebra for the quiver $A_2$ is $U_q(n_+)$, where $n_+$ is the nilpotent subalgebra of $sl(3)$; the q-Weyl algebra is generated by $e_{12}$ and $e_{13}$ inside this algebra).<|endoftext|>
TITLE: Almost but not quite a homomorphism
QUESTION [15 upvotes]: I'm interested in general heuristics where, for specific algebraic structures, we introduce new maps that are "almost" homomorphisms (or "almost" isomorphisms) but not quite so. Here are some that I have encountered in group theory (and may also be used in ring theory and commutative/noncommutative algebra):

A "pseudo-homomorphism" (sometimes also called "quasi-homomorphism") which is a set map for groups whose restriction to any abelian subgroup is a homomorphism. In other words, if two elements commute, then the image of the product is the product of the images. General idea: require the composition with certain kinds of injective maps to be homomorphisms.
A "1-homomorphism" which is a set map for groups whose restriction to any cyclic subgroup is a homomorphism. General idea: require that the restriction to any subalgebra generated by at most $k$ elements is a homomorphism. Note that for algebras defined using at most 2-ary operations, the only interesting case is $k = 1$.
An element map that sends subgroups to subgroups. The induced map on the lattice of subgroups is termed a "projectivity". General idea: Require the map to induce a map on some derivative structure (e.g., the lattice of subalgebras) that looks like it could have come from a homomorphism.

My main interest is from a group theory perspective but I'd also be interested in constructions for other algebraic structures.
ADDED LATER: There have been a lot of interesting examples here. My original focus was to look at properties of maps that could be considered, at least in principle, between two arbitrary objects. Preferably something that could be composed to give a new category-of-sorts. But there've been some interesting examples of maps that go to fixed target groups and whose definition uses additional information about the structure of those target groups. These could also be of potential interest, so please feel free to give such examples too.

REPLY [2 votes]: I know this answer is late, but just recently someone has done exactly what you asked about. The quasihomomorphisms $\mathbb{Z}$ to $\mathbb{Z}$ that are used in the construction of $\mathbb{R}$   (A'Campo has done this) are generalised to (equivalence classes of) maps between arbitrary abelian groups. This actually results in a category, as can be seen here: 
https://www.universiteitleiden.nl/binaries/content/assets/science/mi/scripties/bachelor/2017-2018/hermans-bscthesis.pdf<|endoftext|>
TITLE: Galois group of a product of polynomials
QUESTION [13 upvotes]: How can I compute the Galois group of the polynomial $fg\in K[x]$ assuming that I know the Galois groups of $f\in K[x]$ and $g\in K[x]$? Let's suppose for simplicity that the field $K$ is perfect.

REPLY [3 votes]: Here you can find the proof of $Gal(E_1 E_2) = Gal(E_1) \times_{Gal(E_1 \cap E_2)} Gal(E_2)$ and an example how this can be used to compute the Galois group of a product of two polynomials.<|endoftext|>
TITLE: Primes of the form a^2+1
QUESTION [17 upvotes]: The fact that the Riemann zeta function $\zeta(s)$ and its brethren have a pole at $s=1$ is responsible for the infinitude of large classes of primes (all primes, primes in arithmetic progression; primes represented by a quadratic form). We cannot hope proving the infinitude of primes $p = a^2+1$ in this way because the series $\sum 1/p$, summed over these primes, converges. This implies that the corresponding Euler product
$$ \zeta_G(s)= \prod_{p = a^2+1} \frac1{1 - p^{-s}} $$
converges for $s = 1$. But if we could show that $\zeta_G(s)$ has a pole at, say, 
$s = \frac12$, then the desired result would follow. Now I know that there are heuristics on the number of primes of the form $p = a^2+1$ below $x$ (by Hardy and Littlewood?)
 Can these heuristics  be explained by hypothetical properties of $\zeta_G(s)$ (or a related Dirichlet series), or can the domain of convergence of $\zeta_G(s)$ be derived from such asymptotics? 
BTW, here's a little known conjecture by Goldbach on these primes: let $A$ be the set of all numbers $a$ for which $a^2+1$ is prime ($A = ${1, 2, 4, 6, 10, $\ldots$}). Then every 
$a \in A$ ($a > 1$) can be written in the form $a = b+c$ for $b, c \in A$. I haven't seen this discussed anywhere.

REPLY [8 votes]: Franz, I wrote a paper related to an analytic heuristic on Dirichlet series associated to these prime counting problems. See http://www.math.uconn.edu/~kconrad/articles/hlconst.pdf.<|endoftext|>
TITLE: What manifold has $\mathbb{H}P^{odd}$ as a boundary?
QUESTION [18 upvotes]: This question is motivated by What manifolds are bounded by RP^odd? (as well as a question a fellow grad student asked me) but I can't seem to generalize any of the provided answers to this setting.
Allow me to give some background.  Take all (co)homology groups with $\mathbb{Z}_2$ coefficients.
Given a smooth compact manifold $M^n$, let $w_i = w_i(M)\in H^i(M)$ denote the Stiefel-Whitney classes of (the tangent bundle of) M.  Let $[M]\in H_n(M)$ denote the fundamental class (mod 2).  Consider the Stiefel-Whitney numbers of $M$, defined as the set of all outputs of $ \langle w_{i_1}...w_{i_k} , [M] \rangle$.  Of course this is only interesting when $\sum i_{j} = n$.
Pontrjagin proved that if $M$ is the boundary of some compact n+1 manifold, then all the Steifel-Whitney numbers are 0.  Thom proved the converse - that if all Stiefel-Whitney numbers are 0, then $M$ can be realized as a boundary of some compact n+1 manifold.
As a quick aside, the Euler characteristic $\chi(M)$ mod 2 is equal to $w_n$.  Hence, we see immediately that if $\chi(M)$ is odd, then $M$ is NOT the boundary of a compact manifold.
As an immediate corollary to this, none of $\mathbb{R}P^{even}$, $\mathbb{C}P^{even}$, nor $\mathbb{H}P^{even}$ are boundaries of compact manifolds.
Conversely, one can show that all Stiefel-Whitney numbers of $\mathbb{R}P^{odd}$, $\mathbb{C}P^{odd}$ and $\mathbb{H}P^{odd}$ are 0, so these manifolds can all be realized as boundaries.
What is an example of a manifold $M$ with $\partial M = \mathbb{H}P^{2n+1}$ (and please assume $n>0$ as $\mathbb{H}P^1 = S^4$ is obviously a boundary)?
The question for $\mathbb{R}P^{odd}$ is answered in the link at the top.  The question for $\mathbb{C}P^{odd}$ is similar, but slightly harder:
Consider the (standard) inclusions $Sp(n)\times S^1\rightarrow Sp(n)\times Sp(1)\rightarrow Sp(n+1)$.  The associated homogeneous fibration is given as
$$Sp(n)\times Sp(3)/ Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times S^1\rightarrow Sp(n+1)/Sp(n)\times Sp(1),$$ which is probably better recognized as
$$S^2\rightarrow \mathbb{C}P^{2n+1}\rightarrow \mathbb{H}P^{n}.$$
One can "fill in the fibers" - fill the $S^2$ to $D^3$ to get a compact manifold $M$ with boundary equal to $\mathbb{C}P^{2n+1}$.
I'd love to see $\mathbb{H}P^{odd}$ described in a similar fashion, but I don't know if this is possible.
Assuming it's impossible to describe $\mathbb{H}^{odd}$ as above, I'd still love an answer along the lines of "if you just do this simple process to this often used class of spaces, you get the manifolds you're looking for".
Thanks in advance!

REPLY [6 votes]: Jason, this not an answer, just an observation. Using your formula for $p_1$, $< p_1^{2n+1}, [\mathbb{H}P^{2n+1}]> = (2n-2)^{2n+1} < u,[\mathbb{H}P^{2n+1}]> \neq 0$ if $n>1$, so $\mathbb{H}P^{2n+1}$ cannot be the boundary of an oriented manifold, unlike the examples you give for $\mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{2n+1}$. The point is that filling spherical fibres in oriented bundles will not work.
By the way, this is my first post in Math Overflow. Yay!!!
Note: this post has been edited because the original was very false. I claimed that $\sigma(\mathbb{H}P^{2n+1})=1$ which is silly because the middle cohomology is $H^{4n+2}(\mathbb{H}P^{2n+1}) = 0$. Also the signature being odd would have contradicted the fact that $\chi (\mathbb{H}P^{2n+1})$ is even, which is stated in the question.<|endoftext|>
TITLE: PDE on manifolds
QUESTION [18 upvotes]: I am currently in a PDE course where one of the requirements is to present a paper in PDE. I am wondering if anyone can suggest an early (read foundational, first introductory) paper talking about PDE on manifolds. I am a topologist (homotopy theorist) by training so i prefer things to be coordinate free, but this may not be possible. For example something relating various notions of curvature to PDE, or something on viewing the PDE globally in terms of acting on sections would be great.
If this isn't the best forum my apologies.

REPLY [2 votes]: All previous replies have their own merit, however I am surprised not to find the first thing that popped in my head: That Frobenius' theorem concerning the integrability of a differential system on the tangent bundle. And maybe as a homotopy-oriented researcher (kudos to that) you could relate a simple result and use it as an intro to the Atiyah-Singer index theorem.
Or maybe not. Frobenius' theorem is pretty enough to hold the show on its own.<|endoftext|>
TITLE: What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem?
QUESTION [104 upvotes]: I know the following facts.  (Don't assume I know much more than the following facts.)

The Atiyah-Singer index theorem generalizes both the Riemann-Roch theorem and the Gauss-Bonnet theorem.
The Atiyah-Singer index theorem can be proven using heat kernels.

This implies that both Riemann-Roch and Gauss-Bonnet can be proven using heat kernels.  Now, I don't think I have the background necessary to understand the details of the proofs, but I would really appreciate it if someone briefly outlined for me an extremely high-level summary of how the above two proofs might go.  Mostly what I'm looking for is physical intuition: when does one know that heat kernel methods are relevant to a mathematical problem?  Is the mathematical problem recast as a physical problem to do so, and how?  
(Also, does one get Riemann-Roch for Riemann surfaces only or can we also prove the version for more general algebraic curves?)
Edit:  Sorry, the original question was a little unclear.  While I appreciate the answers so far concerning how one gets from heat kernels to the index theorem to the two theorems I mentioned, I'm wondering what one can say about going from heat kernels directly to the two theorems I mentioned.  As Deane mentions in this comments, my hope is that this reduces the amount of formalism necessary to the point where the physical ideas are clear to someone without a lot of background.

REPLY [5 votes]: This is a wonderful question, but I think people are having a hard time answering it because even in the simplest settings there is a lot going on and it is not clear what not a lot of background means.  The simplest case would have to be proving GBC for a surface (nothing I say will be any easier for a surface, unfortunately!).  First you have to know that heat kernel here means heat kernel for the Hodge Laplace operator, which is the Laplace operator on forms $(d+d^*)^2.$  Second it involves the supertrace of the heat kernel, which means the trace, except you add a minus sign for the parts taking odd forms to odd forms.  A previous post mentions McKean and Singer, whose argument I will not repeat because it cannot be improved upon, but it argues that the eigenforms of this Laplacian with nonnegative eigenvalue cancel out, so only the $0$ eigenfunctions contribute, so the quantity is $t$-independent and in fact the Euler number.  But if it is $t$-independent, we may as well look at this quantity as $t \to 0.$  Since the heat kernel is approaching a delta function on the diagonal, the small $t$ behavior is a local question, and can be addressed by comparing it to the flat space heat kernel.  This reasonably straightforward calculation gets you the curvature, or in higher dimensions the Pfaffian of the of the curvature (the supertrace itself contained an integral which I have not mentioned explicitly). If you are looking for a principle or technique to bring with you to other problems, I would say it is this:  Show some quantity you can compute from the time evolution kernel is somehow topological and therefore time independent, then equate the small time calculation of it to the large time calculation and see a connection between two things that look unrelated.<|endoftext|>
TITLE: Mittag-Leffler condition: what's the origin of its name?
QUESTION [17 upvotes]: Why the Mittag-Leffler condition on a short exact sequence of, say,  abelian groups, that ensures that the first derived functor of the inverse limit vanishes, is so named?

REPLY [13 votes]: The wording of your question suggests that you're familiar with the "classical" Mittag-Leffler theorem from complex analysis, which assures us that meromorphic functions can be constructed with prescribed poles (as long as the specified points don't accumulate in the region).
It turns out - or so I'm told, I must admit to never working through the details - that parts of the proof can be abstracted, and from this point of view a key ingredient (implicit or explicit in the proof, according to taste) is the vanishing of a certain $\lim_1$ group -- as assured by the "abstract" ML-theorem that you mention.
I'm not sure where this was first recorded - I hesitate to say "folklore" since that's just another way of saying "don't really know am and not a historian". One place this is discussed is in Runde's book A taste of topology: see Google Books for the relevant part.
IIRC, Runde says that the use of the "abstract" Mittag-Leffler theorem to prove the "classical" one, and to prove things like the Baire category theorem, can be found in Bourbaki. Perhaps someone better versed in the mathematical literature (or at least better versed in the works of Bourbaki) can confirm or refute this?<|endoftext|>
TITLE: An algebraic proof of Mumford's smoothness criterion for surfaces?
QUESTION [7 upvotes]: (Disclaimer: I'm a beginner in this area, so welcome corrections.)
Let $(X,x)$ be a germ of a complex surface (i.e. locally the zero set of some holomorphic functions) and assume that $x$ an isolated singular point.  Mumford proved that if the local fundamental group of $X$ at $x$ is trivial, then in fact $x$ is smooth.  
All the critters in the above paragraph have algebraic analogues, and the conversion was carried out (I believe) by Flenner: Let $A$ be a two-dimensional complete local normal domain containing an algebraically closed field of characteristic zero; if the \'etale fundamental group of [EDIT: the punctured spectrum of] $A$ is trivial, then $A$ is regular.
However, Flenner's proof is essentially by reduction to Mumford's theorem [as far as I, a non-German-speaker, can tell], rather than a new algebraic (or algebro-geometric) proof.  So:

Does there exist a purely algebraic or algebro-geometric proof of Mumford's theorem?


Motivations include: (1) Mumford's proof is completely opaque to me; (2) No, I mean really really opaque; (3) I'm curious about extensions of the theorem to non-isolated singularities [which should probably be another question].

REPLY [3 votes]: I found what I think is the answer, in a paper by Cutkosky and Srinivasan called "Local fundamental groups of surface singularities in characteristic $p$".  They prove, as Corollary 5: Suppose that $(A, m)$ is a complete normal local domain of dimension two, with algebraically closed resicue field $k$ of characteristic zero.  (Slightly surprising, given the title of the paper.)  Then $\pi_1(\operatorname{Spec} A -m)=0$ if and only if $A$ is smooth over $k$.    They say that this gives "an arithmetic proof of the theorem of Mumford and Flenner."
The proof apparently uses Flenner's paper, but I don't think it uses Mumford's result.  They get an expression for the local fundamental group in terms of a tree, and appeal to Flenner's Theorem 2.7 to know that the group is trivial iff $A$ is smooth.  I haven't tried to read that section of Flenner's paper yet, but it seems to be independent of Mumford.<|endoftext|>
TITLE: Outer automorphisms of simple Lie Algebras
QUESTION [27 upvotes]: There is, of course, a complete classification for simple complex Lie algebras. Is there a good reference which lists the group of outer automorphisms for each?

REPLY [8 votes]: Aut(\g) is the semidirect product of Inn(\g) and Out(\g) in the complex AND in the real case. However, this is a rather recent result:
tinyurl.com/68748hn
Edit by jc: The link goes to a PDF abstract of: 
Hasan Gündogan, The Component Group of the Automorphism Group of a Simple Lie Algebra and the Splitting of the Corresponding Short Exact Sequence, Journal of Lie Theory 20 (2010), No. 4, 709--737.
Abstract:
Let $\frak g$ be a simple Lie algebra of finite dimension over $\mathbb K \in \{\mathbb R,\mathbb C\}$ and $\mathop{\rm Aut}(\frak g)$ the finite-dimensional Lie group of its automorphisms. We will calculate the component group $\pi_0(\mathop{\rm Aut}(\frak g)) = \mathop{\rm Aut}(\frak g)/\mathop{\rm Aut}(\frak g)_0$ and the number of its conjugacy classes, and we will show that the corresponding short exact sequence $$ {\bf1}\to\mathop{\rm Aut}(\frak g)_0\to\mathop{\rm Aut}(\frak g)\to\pi_0(\mathop{\rm Aut}(\frak g))\to{\bf1} $$ is split or, equivalently, there is an isomorphism $\mathop{\rm Aut}(\frak g)\cong \mathop{\rm Aut}(\frak g)_0 \rtimes\pi_0(\mathop{\rm Aut}(\frak g))$. Indeed, since $\mathop{\rm Aut}(\frak g)_0$ is open in $\mathop{\rm Aut}(\frak g)$, the quotient group $\pi_0(\mathop{\rm Aut}(\frak g))$ is discrete. Hence a section $\pi_0(\mathop{\rm Aut}(\frak g))\to\mathop{\rm Aut}(\frak g)$ is automatically continuous, giving rise to an isomorphism of Lie groups $\mathop{\rm Aut}(\frak g)\cong\mathop{\rm Aut}(\frak g)_0 \rtimes\pi_0(\mathop{\rm Aut}(\frak g))$.<|endoftext|>
TITLE: Can injective modules over R give non-injective sheaves over Spec R?
QUESTION [9 upvotes]: In [Hartshorne, III.3] he proves that injective modules over $R$ give flasque sheaves over $Spec\ R$. I presume that's because they don't give injective sheaves, and flasque is the consolation prize. Is there an easy counterexample?
EDIT: in III.3 he's assuming Noetherian. And he's already proved in II.5.5 the equivalence
of categories of $R$-modules and quasicoherent ${\mathcal O}_{Spec\ R}$-modules. (And that injective sheaves are flasque, in III.2.)
EDIT: his proof that injectives are flasque uses some non-quasicoherent sheaves. So the ingredients "injective R-modules give injective objects in the category of quasicoherent sheaves [II.5.5]" plus "injective objects in the category of sheaves are flasque [III.2]" isn't enough for the result he gets in III.3, that injective R-modules give flasque sheaves.

REPLY [3 votes]: I wonder why no one has mentioned this yet: A counterexample by Verdier can be found in SGA 6, Exp. 2, App. I. You can read it here. It is also shown that the forgetful functor $\mathrm{Qcoh}(X) \to \mathrm{Mod}(X)$ is ill-behaved on derived categories. Another example can be found in the Stacks Project's Examples (#26).<|endoftext|>
TITLE: How can I define the product of two ideals categorically?
QUESTION [35 upvotes]: Given a commutative ring $R$, there is a category whose objects are epimorphisms surjective ring homomorphisms $R \to S$ and whose morphisms are commutative triangles making two such epimorphisms surjections compatible, and the skeleton of this category is a partial order that can be identified with the lattice of ideals of $R$.  Now, I have always been under the impression that anything one can say about ideals one can phrase in this purely arrow-theoretic language: most importantly, the intersection of ideals is the product in this category and the sum of ideals is the coproduct.  (Since we're working in a partial order, product and coproduct are fancy ways to say supremum and infimum.  The direction of the implied ordering on ideals may differ here from the one you're used to, but that's not important.)
However, Harry's made some comments recently that made me realize I don't know how to define the product of two ideals purely in terms of this category, that is, via a universal construction like the above.  It would be really surprising to me if this were not possible, so maybe I'm missing something obvious.  Does anyone know how to do this?

REPLY [24 votes]: Bjorn's answer shows that the product of two ideals cannot be defined by means of the partially ordered set of ideals. However, there is a category-theoretic definition of the product of two ideals if we allow us to use the category of modules. This works because of the Rosenberg reconstruction Theorem.
Explicitely, there is a bijection between ideals of $R$ and reflective, topologizing subcategories of $\text{Mod}(R)$, which is given by $I \to \{M \in \text{Mod}(R) : I M = 0\}$. If $T$ is such a subcategory with reflector $F : \text{Mod}(R) \to T$, then the corresponding ideal is $I=\ker(R \to F(R))$. Now, the multiplication of ideals corresponds to the so called Gabriel product: If $S,T$ are subcategories of an abelian category, then $S \bullet T$ is the subcategory which consists of those objects $M$ for which there is an exact sequence $0 \to M' \to M \to M'' \to 0$ with $M' \in T, M'' \in S$. The idea for the notation is $(S \bullet T)/T = S$.<|endoftext|>
TITLE: Connection between bi-Hamiltonian systems and complete integrability
QUESTION [6 upvotes]: As I understand, the lack of indication on how to obtain first integrals in Arnol'd-Liouville theory is a reason why we are interested in bi-Hamiltonian systems.
Two Poisson brackets
$\{ \cdot,\cdot \} _{1} , \{ \cdot , \cdot \} _{2}$ on a manifold $M$ are compatible if their arbitrary linear combination
$\lambda \{ \cdot , \cdot \} _1+\mu\{\cdot,\cdot\} _2$ is also a Poisson bracket. A bi-Hamiltonian system is one which allows Hamiltonian formulations with respect to two compatible Poisson brackets. It automatically posseses a number of integrals in involution.
The definition of a complete integrability (à la Liouville-Arnol'd) is:
Hamiltonian flows and Poisson maps on a $2n$-dimensional symplectic manifold $\left(M,\{ \cdot, \cdot \}_M\right)$ with $n$ (smooth real valued) functions $F _1,F _2,\dots,F _n$ such that: (i) they are functionally independent (i.e. the gradients $\nabla F _k$ are linearly independent everywhere on $M$) and (ii) these functions are in involution (i.e. $\{F _k,F _j\}=0$) are called completely integrable.
Now, I would like to understand the connections between these two notions, and because I haven't studied the theory, any answer would be helpful. I find reading papers on these subjects too technical at the moment. Specific questions I have in mind are:
Does completely integrable system always allow for a bi-Hamiltonian structure? Is every bi-Hamiltonian system completely integrable? If not, what are examples (or places where to find examples) of systems that posses one property but not the other?
I apologize for any stupid mistakes I might have made above. Feel free to edit (tagging included).

REPLY [11 votes]: Your understanding is essentially correct. There are three basic (and closely related) approaches to
constructing the integrals of motion required for complete integrability: through separation of variables, through the Lax representation, and through the bi-Hamiltonian representation. The relationship among them is not yet fully understood. See, however, this paper by M. Błaszak, which, in essence, states that any Hamiltonian system that admits separation of variables is (or, rather, can be extended to) bi-Hamiltonian, and this survey paper by G. Falqui and M. Pedroni on separation of variables for bi-Hamiltonian systems. As for the relationship among the Lax representation and bi-Hamiltonian property, see this paper by F. Magri and Y. Kosmann-Schwarzbach and references therein. Now to your questions.
First of all, the bi-Hamiltonian property as you state it, without further restrictions, does not necessarily lead to integrability, and the claim that a bi-Hamiltonian system automatically possesses some integrals of motion does not hold in full generality, as far as I know. I can't think of a specific example right now, but, roughly speaking, if both your Poisson structures are too degenerate (their rank is too low), the recursion can break down and you will not get enough integrals of motion. An example of this for the infinite-dimensional case can be found in the paper Is a bi-Hamiltonian system necessarily integrable? by B.A. Kupershmidt. However, if you put in some additional nondegeneracy assumptions, the answer is yes, and dates back to Magri, Morosi, Gelfand and Dorfman. It is nicely summarized e.g. in Theorem 1.1 of this paper by R.G. Smirnov. The idea behind this is that the integrals of motion are provided by the traces of powers of the ratio of your Poisson structures.
As for the second question, not any Liouville integrable system is bi-Hamiltonian, at least if you impose some fairly reasonable technical assumptions, see the paper Completely integrable bi-Hamiltonian systems by R.L. Fernandes; cf. also the above Smirnov's paper.<|endoftext|>
TITLE: Checking if two graphs have the same universal cover
QUESTION [16 upvotes]: It's possible I just haven't thought hard enough about this, but I've been working at it off and on for a day or two and getting nowhere.
You can define a notion of "covering graph" in graph theory, analogous to covering spaces. (Actually I think there's some sense -- maybe in differential topology -- in which the notions agree exactly, but that's not the question.) Anyway, it behaves like a covering space -- it lifts paths and so on.
There's also a "universal cover," which I think satisfies the same universal property as topological universal covers but I'm not sure. Universal covers are acyclic (simply connected) in graph theory, so they're trees, usually infinite. The universal cover doesn't determine the graph; for instance, any two k-regular graphs (k > 1) have the same universal cover. You can construct plenty of other pairs, too. 
I'm interested in necessary and sufficient conditions for two graphs $G, H$ to have the same universal cover. One such condition (I'm pretty sure!) is whether you can give a 1-1 correspondence between trails in $G$ and trails in $H$ that preserves degree sequences. Unfortunately this doesn't help me much, since this is still basically an infinite condition. Is there some less-obvious but more easily checkable condition? In particular is it possible to determine if two (finite) graphs have the same universal cover in polynomial time?

REPLY [4 votes]: Maybe this is of interest to you. In section 3.3 of the below paper (see also 2.2), the universal covers are extended from graphs to degree matrices, and some conditions are given under which two degree matrices (graphs) have the same universal cover: 
Locally constrained graph homomorphisms and equitable partitions 
J.Fiala, D.Paulusma and J.A.Telle
European Journal of Combinatorics. Volume 29.(4) p. 850-880<|endoftext|>
TITLE: What are the local Langlands conjectures nowadays, for connected reductive groups over a $p$-adic field?
QUESTION [80 upvotes]: Let me stress that I am only interested in $p$-adic fields in this question, for reasons that will become clear later. Let me also stress that in some sense I am basically assuming that the reader knows what the "1970s version of the local Langlands conjectures" are when writing this question---there are plenty of references that will get us this far (I give one below that works in the generality I'm interested in).
So let $F$ be a finite extension of $\mathbf{Q}_p$, let $G$ be a connected reductive group over $F$, let $\widehat{G}$ denote the complex dual group of $G$ (a connected complex Lie group) and let ${}^LG$ denote the $L$-group of $G$, the semi-direct product of the dual group and the Weil group of $F$ (formed using a fixed algebraic closure $\overline{F}$ of $F$).
Here is the "standard", or possibly "standard in the 1970s", way of formulating what local Langlands should say (for more details see Borel's paper "Automorphic $L$-functions", available online (thanks AMS) here at the AMS website. One defines sets $\Phi(G)$ ($\widehat{G}$-conjugacy classes of admissible Weil-Deligne representations from the Weil-Deligne group to the $L$-group [noting that "admissible" includes assertions about images only landing in so-called "relevant parabolics" in the general case and is quite a subtle notion]) and $\Pi(G)$ (isomorphism classes of smooth irreducible admissible representations of $G(F)$), and one conjectures:
LOCAL LANGLANDS CONJECTURE (naive form): There is a canonical surjection $\Pi(G)\to\Phi(G)$ with finite fibres, satisfying (insert list of properties here).
See section 10 of Borel's article for the properties required of the map.
Now in recent weeks I have had two conversations with geometric Langlands type people both of whom have mocked me when I have suggested that this is what the local Langlands conjecture should look like. They point out that studying some set of representations up to isomorphism is a very "coarse" idea nowadays, and one should reformulate things category-theoretically, considering Tannakian categories of representations, and relating them to...aah, well there's the catch. Looking back at what both of them said, they both at a crucial point slipped in the line "well, now for simplicity let's assume we're in the function field/geometric setting. Now..." and off they went with their perverse sheaves. The happy upshot of all of this is that now one has a much better formulation of local Langlands, because one can demand much more than a canonical surjection with finite fibres, one can ask whether two categories are equivalent.
But I have been hoodwinked here, because I am interested in $p$-adic fields. So yes yes yes I'm sure it's all wonderful in the function field/geometric setting, and things have been generalised beyond all recognition. My question is simply:
Q) Can we do better than the naive form of Local Langlands (i.e. is there a stronger statement about two categories being equivalent) when $F$ is a p-adic field?
The answer appears to be "yes" in other cases but I am unclear about whether the answer is yes in the $p$-adic case. Even if someone were to be able to explain some generalisation in the case where $G$ is split, I am sure I would learn a lot. To be honest, I think I'd learn a lot if someone could explain how to turn the surjection into a more bijective kind of object even in the case of $SL(2)$. Even in the unramified case! That's how far behind I am! As far as I can see, the Satake isomorphism gives only a surjection in general, because there is more than one equivalence class of hyperspecial maximal compact in general.

REPLY [27 votes]: Now that our paper Geometrization of the local Langlands correspondence with Fargues is finally out (ooufff!!), it may be worth giving an update to Ben-Zvi's answer above. In brief: we give a formulation of Local Langlands over a $p$-adic field $F$ so that it is finally

an actual conjecture, in the sense that it asks for properties of a given construction, not for a construction;
of a form as in geometric Langlands, in particular about an equivalence of categories, not merely a bijection of irreducibles.

First, I should say that in the notation of the OP, we construct a canonical map $\Pi(G)\to \Phi(G)$, and prove some properties about it. However, we are not able to say anything yet about its fibres (not even finiteness).
Moreover, we give a formulation of local Langlands as an equivalence of categories, and (essentially) construct a functor in one direction that one expects to realize the equivalence. In particular, this nails down what the local Langlands correspondence should be, it "merely" remains to establish all the desired properties of it.
Let me briefly state the main result here. Let $\mathrm{Bun}_G$ be the stack of $G$-bundles on the Fargues--Fontaine curve. We define an ($\infty$-)category $\mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$ of $\ell$-adic sheaves on $\mathrm{Bun}_G$. The stack $\mathrm{Bun}_G$ is stratified into countably many strata enumerated by $b\in B(G)$, and on each stratum, the category $\mathcal D(\mathrm{Bun}_G^b,\overline{\mathbb Q}_\ell)$ is the derived ($\infty$-)category of smooth representations of the group $G_b(F)$. In particular, for $b=1$, one gets smooth representations of $G(F)$.
Moreover, there is an Artin stack $Z^1(W_F,\hat{G})/\hat{G}$ of $L$-parameters over $\overline{\mathbb Q}_\ell$.
Our main result is the construction of the "spectral action":

There is a canonical action of the $\infty$-category of perfect complexes on $Z^1(W_F,\hat{G})/\hat{G}$ on $\mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$.

The main conjecture is basically that this makes $\mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)^\omega$ a "free module of rank $1$ over $\mathrm{Perf}(Z^1(W_F,\hat{G})/\hat{G})$", at least if $G$ is quasisplit (or more generally, has connected center).
More precisely, assume that $G$ is quasisplit and fix a Borel $B\subset G$ and a generic character $\psi$ of $U(F)$, where $U\subset B$ is the unipotent radical, giving the Whittaker representation $c\text-\mathrm{Ind}_{U(F)}^{G(F)}\psi$, thus a sheaf on $[\ast/G(F)]$, which is the open substack of $\mathrm{Bun}_G$ of geometrically fibrewise trivial $G$-bundles; extending by $0$ thus gives a sheaf $\mathcal W_\psi\in \mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$, called the Whittaker sheaf.

Conjecture. The functor
$$ \mathrm{Perf}(Z^1(W_F,\hat{G})/\hat{G})\to \mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$$
given by acting on $\mathcal W_\psi$ is fully faithful, and extends to an equivalence
$$\mathcal D^{b,\mathrm{qc}}_{\mathrm{coh}}(Z^1(W_F,\hat{G})/\hat{G})\cong \mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)^{\omega}.$$

Here the superscript $\mathrm{qc}$ means quasicompact support, and $\omega$ means compact objects. As $Z^1(W_F,\hat{G})$ is not smooth (merely a local complete intersection), there is a difference between perfect complexes and $\mathcal D^b_{\mathrm{coh}}$, and there is still a minor ambiguity about how to extend from perfect complexes to all complexes of coherent sheaves. Generically over the stack of $L$-parameters, there is however no difference.
It takes a little bit of unraveling to see how this implies more classical forms of the correspondence, like the expected internal parametrization of $L$-packets; in the case of elliptic $L$-parameters, everything is very clean, see Section X.2 of our paper.
(There are related conjectures and results by Ben-Zvi--Chen--Helm--Nadler, Hellmann and Zhu; see also the work of Genestier--Lafforgue in the function field case. And this work is heavily inspired by previous work in geometric Langlands, notably the conjectures of Arinkin--Gaitsgory, and the work of Nadler--Yun and Gaitsgory--Kazhdan--Rozenblyum--Varshavsky on spectral actions.)
PS: It may be worth pointing out that this conjecture is, at least a priori, of a quite different nature than Vogan's conjecture, mentioned in the other answers, which is based on perverse sheaves on the stack of $L$-parameters; here, we use coherent sheaves.<|endoftext|>
TITLE: What is the recent development of D-module and representation theory of Kac-Moody algebra?
QUESTION [15 upvotes]: I just started to collect the papers of this field and know little things. So if I make stupid mistake, please correct me.
It seems that there are several approaches to localize Kac-Moody algebra(in particular, affine Lie algebra). I just took look at several papers by
Kashiwara-Tanisaki:(1989)
They constructed the flag variety of symmetrizable Kac-Moody algebra as ind-scheme.
Edward Frenkel-B.Feign:
They constructed the semi-infinite flag manifold and introduce the semi-infinite cohomology.
Edward Frenkel-Dennis Gaitsgory:
It seems that they dealt with two kinds of things:
1 Affine Grassmannian
2 Affine flag variety
Olivier Mathieu
It seems that he gave the general definition of flag variety of arbitrary Kac-Moody algebra as a stack.(I can read French but not very quickly, so there might be possibilities that I made a mistake to describe his work)
My Question

Is there any other definition of flag variety of Kac-Moody algebra(at least for affine case)? What are the relationship between these definitions I mentioned above?

What is the relationship between the D-module theory on affine flag variety and D-module theory on affine grassmannian? (Frenkel-Gaitsgory) Why did they consider these two ways to localize affine Lie algebra?

It seems all of the construction above are not very easy to deal with(ind-scheme,group ind-scheme which are not locally affine). Is there any existent work to define it as a locally affine space(classical scheme,algebraic space or at least locally affine stack with smooth topology)?

From the work of Frenkel-Gaitsgory, they built the derived equivalence between the category of D-modules and full subcategory of modules over enveloping algebra of affine Lie algebra. They claimed in their paper that one can not obtain the equivalence in abelian level. Is there any intuitive explanation for this?

I am looking forward to getting some guy who can explain the work of Oliver to me.


Thank you in advance!

REPLY [7 votes]: I am not an expert in this but I would of course expect something like ind-scheme approach to be natural. Gerd Faltings used I think ind-schemes to treat Sugawara construction, algebraic loop groups and Verlinde's conjecture in 
Gerd Faltings, Algebraic loop groups and moduli spaces of bundles. 
J. Eur. Math. Soc. (JEMS) 5 (2003), no. 1, 41--68. 
You might also like to work with versions of Kac-Moody GROUPS in analytic approaches. 
You could also consult comprehensive and not that old Kumar's book (Kac-Moody groups, their flag varieties and representation theory, Birkhauser) which is written in geometric language. 
As far as Frenkel is concerned, not only his work with Feigin but even more I think his paper with Gaitsgory must be relevant (see arxiv:0712.0788). 
Semi-infinite cohomologies are important but still misterious thing. Some related homological algebra has been recently studied by Positelskii in great generality. Another important thing is relation between the geometry of representations of quantum groups at root of unity and of affine Lie algebras, like in the book of Varchenko and many papers later.
Edit: Frenkel himself I think does not claim (I talked to him at the time) to have intuitive explanation why only derived equivalence. But you should not expect for more: by the correspondence with quantum groups the situation should be like in affine case where one has problems with non-closedness of diagonal in noncommutative geometry what has repercussions on the theory of D-modules. How this reflects in the case of relevant ind-schemes for affine side I do not know but somehow it does.<|endoftext|>
TITLE: Do you know any good introductory resource on sequent calculus?
QUESTION [14 upvotes]: I'm looking for a good introductory resource on sequent calculus suitable for someone who has studied natural deduction before. Books and online resources are both OK, as long as each rule of inference and any notational convention is explained. Thanks.

REPLY [2 votes]: I second here the recommendations of Troelstra and Schwichtenberg's Basic Proof Theory for learning sequent calculus because of its exercises; focus on chapter 3 and do the exercises to get a good grip on standard sequent calculi for classical and intuitionistic logic, as well as get some exposure to a few variations (e.g., single-sided calculi).  von Plato and Negri's Structural Proof Theory is also very good, but it is more of a monograph than a textbook (there are no exercises).  In von Plato and Negri you will find detailed and systematic discussion of the relationship between natural deduction and sequent calculi throughout the book, a discussion that is is largely missing from Troelstra and Schwichtenberg (though it must be said that T and S do prove that various sequent calculi are equivalent with natural deduction and with certain Hilbert/Frege-style proof systems).<|endoftext|>
TITLE: Why do flag manifolds, in the P(V_rho) embedding, look like products of P^1s?
QUESTION [20 upvotes]: Bert Kostant mentioned an odd fact to me some time ago. As usual (with such statements), fix a 
complex, connected, reductive) Lie group $G$, with maximal torus $T$, and Weyl vector $\rho$ equal to half the sum of the positive roots. Let $L_\beta := T \cdot $ the root $SL_2$ subgroup corresponding to the positive root $\beta$.
Quoth Bert: the character of the irrep $V_{n\rho}$ is $T$-isomorphic to the tensor product over all positive roots of the $L_\beta$-irrep with highest weight $n\beta$. (The latter is a $T$-representation by sticking $T$ diagonally into $\prod_{\Delta_+} L_\beta$.)
Once someone tells you, it's very easy to prove from the Weyl character formula.
Geometrically, this says the following. Inside ${\mathbb P}^* (V^G_\rho)$, we have a copy of the flag manifold $G/B$ as the orbit of the highest weight vector. (Indeed, this is the smallest embedding by a complete linear series.) Identifying 
$V^G_\rho \cong \bigotimes_{\Delta_+} V^{L_\beta}_\beta$ 
as $T$-representations, we also have in this projective space a Segre-embedded $\prod_{\Delta_+} {\mathbb P}^* (V^{L_\beta}_\beta),$
 a product of ${\mathbb P}^1$s. Kostant's observation is that these two subvarieties have the same $T$-equivariant Hilbert series.

"Why" is the flag manifold masquerading as a product of ${\mathbb P}^1$s?
More concretely, we know that the two varieties lie in the same connected component of the Hilbert scheme of this projective space, by Hartshorne's thesis. Can one connect them without breaking the $T$-action? (As far as I know, Hartshorne connectivity doesn't hold in general if you keep track of the multigrading, not just the single grading.)
Since the two varieties are both smooth, (EDIT:) and have different topology, there won't be a flat family over an irreducible base in which one is a general fiber, one the special. Do they have a common degeneration? 
Hartshorne's thesis just guarantees that we can degenerate, deform, degenerate, deform, ... to get from one to the other, not that they will be on adjacent components of the Hilbert scheme.

REPLY [4 votes]: Your first question is about two objects becoming isomorphic after quantization, and you're asking "Why?"
Here, the relevant quantum object is the spin representation of $\mathfrak g$, which is a representation of $\mathfrak g\ltimes \mathit{Cliff}(\mathfrak g)$,
where $\mathit{Cliff}(\mathfrak g)$ is the Clifford algebra of (the underlying vector space of) $\mathfrak g$, with respect to some invariant inner product; a $\mathbb Z/2$-graded algebra.
Let $S$ be the unique (up to grading reversal) irreducible $\mathbb Z/2$-graded representation of $\mathit{Cliff}(\mathfrak g)$. It has a graded-commuting action of
$$C:=\begin{cases}\mathbb C&\text{ if }\quad \dim(\mathfrak g) \text{ is even}
\\
\mathit{Cliff}(1)&\text{ if }\quad \dim(\mathfrak g) \text{ is odd.}
\end{cases}
$$
Let $V$ be the category of modules of the above algebra, so that $S\in V$.
To make things a bit more canonical, one can use the graded Morita equivalence between $C$ and $\mathit{Cliff}(\mathfrak h)$ to identify $V$ with the category of $\mathit{Cliff}(\mathfrak h)$-modules.
Let $\alpha$ denote the adjoint action of $G$ on $\mathit{Cliff}(\mathfrak g)$. For any element $g\in G$, we can pre-compose the action of $\mathit{Cliff}(\mathfrak g)$ on $S$ by $\alpha_g$ to get a new, isomorphic $\mathit{Cliff}(\mathfrak g)$-module in $V$ [here, I'm using that $G$ is connected].
The map that sends $g$ to such an isomorphism is unique up to scalar, and so we get a projective representation of $G$ on $S$.
If $G$ is simply connected, this lifts to an honest action of $G$ on $S$, and so we get an action of $G\ltimes\mathit{Cliff}(\mathfrak g)$ on the object $S\in V$.
All in all, (the underlying vector space of) $S$ has actions of $G$, of $\mathit{Cliff}(\mathfrak g)$, and of $\mathit{Cliff}(\mathfrak h)$.
Now, we can "cancel" two $\mathit{Cliff}(\mathfrak h)$ actions to get a vector space with actions of $G$ and of $\mathit{Cliff}(\mathfrak g\ominus h)$.
That's the irreducible $G$-rep with highest weight $\rho$.
This vector space has two descriptions:
(1) The irreducible $G$-rep with highest weight $\rho$.
(2) The irreducible $\mathit{Cliff}(\mathfrak g\ominus h)$-module.
Note however that this operation of "canceling" the two $\mathit{Cliff}(\mathfrak h)$ actions is a bit unnatural.
For example, the vector space (1) is purely even, while (2) is genuinely $\mathbb Z/2$-graded. You probably see that same weirdness on the classical side of the problem when you try to identify the flag variety with a product of $\mathbb P^1$s.<|endoftext|>
TITLE: How does $\pi_1(SO(3))$ relate exactly to the waiters trick?
QUESTION [19 upvotes]: I hope this is serious enough. It is a well-known fact that $\pi_1(SO(3)) = \mathbb{Z}/(2)$, so $SO(3)$ admits precisely one non trivial covering, which is 2-sheeted.
Another well known fact is that you can hold a dish on your hand and perform two turns (one over the elbow, one below) in the same direction and come back in the original position.
These facts are known to be related, and I more or less can guess why. Some configuration of the system (hand + dish) must draw a path in $Spin(3)$ whose projection in $SO(3)$ is the closed non trivial loop pointed at the identity.
The problem is that I cannot make this precise, since it is not clear to me which is the variety which parametrizes the position of the elbow and the hand. Is there a clean way to see how $Spin(3)$ comes into play?

REPLY [11 votes]: Spin(3) comes into the play only as the covering space of SO(3), I think. You do everything in SO(3). Draw a curve through your body from a stationary point, like your foot, up the leg and torso and out the arm, ending at the dish. Each point along the curve traces out a curve in SO(3), thus defining a homotopy. After you have completed the trick and ended back in the original position, you now have a homotopy from the double rotation of the dish with a constant curve at the identity of SO(3). You can't stop at the halfway point, lock the dish and hand in place, now at the original position, and untwist your arm: This reflects the fact that the single loop in SO(3) is not null homotopic.

REPLY [2 votes]: See p. 164-167 of Bredon's Topology and Geometry. Page 166 (not viewable online) is just a sequence of nine pictures demonstrating the waiter's trick.

REPLY [2 votes]: (This is a bit long for a comment but isn't intended as an answer.)
I've broken glasses in the past demonstrating this trick so I strongly advise doing it with empty plastic mugs until you have it down pat.  For those who can't quite work it out, here's a simpler way to do it: take an elastic band and two rods that have distinguishable "up" and "down".  Loop the band around the two rods and keep it fairly taut (just so it doesn't fall off).  Ascii picture:
   |     |
 /-|-----|-\   
/  |     |  \   
\  |     |  /   
 \---------/   
   |     |   

Now turn one of the rods upside-down.  By moving the rod through space but without turning it (or the system falling apart), try to untangle the elastic band.
Can't do it?  Good.  Can?  Whoops!  Either you've done something wrong or Whitehead did.
Start again from the beginning.  Now turn one of the rods through a full twist (so it's right-side-up again).  Now try to untangle the elastic band as before.
Can't do it?  Try again!  It's possible.
So in mathematical terms, what you are looking for is the position of the second rod plus it's "up-down"ness.  In terms of the waiter's arm, you want the position of the hand and the expression of agony on his face: if his face is in agony, his arm is twisted; if his face is calm, his arm isn't twisted.

Mnemonic: twisting someone's arm twice gets you nowhere.<|endoftext|>
TITLE: Primes $p$ for which $p-1$ has a large prime factor
QUESTION [20 upvotes]: What are the best known density results and conjectures for primes $p$ where $p-1$ has a large prime factor $q$, where by "large" I mean something greater than $\sqrt{p}$.
The most extreme case is that of a safe prime (Wikipedia entry), which is a prime $p$ such that $(p - 1)/2$ is also a prime (the smaller prime is called a Sophie Germain prime). I believe it is conjectured (and not yet proved) that infinitely many safe primes exist, and that the density is roughly $c/\log^2 n$ for some constant $c$ (as it should be from a probabilistic model).
For the more general setting, where we are interested in the density of primes $p$ for which $p-1$ has a large prime factor, the only general approach I am aware of is the prime number theorem for arithmetic progressions, and some of its strengthenings such as the Bombieri-Vinogradov theorem (conditional to the GRH), the (still open) Elliott-Halberstam conjecture, Chowla's conjecture on the first Dirichlet prime, and some partial results related to this conjecture. All of these deal with the existence of primes $p \equiv a \pmod q$ for arbitrary $q$ and arbitrary $a$ that is coprime to $q$.
My question: can we expect qualitatively better results for the situation where $q$ is prime and $a = 1$? Also, I am not interested in specifying $q$ beforehand, so the existence of a $p$ such that there exists any large prime $q$ dividing $p-1$ would be great. References to existing conjectures, conditional results, and unconditional results would be greatly appreciated.

REPLY [14 votes]: See "On the number of primes $p$ for which $p+a$ has a large prime factor." (Goldfeld, Mathematika 16 1969 23--27.) Using Bombieri-Vinogradov he proves, for a fixed integer $a$, that
$$\sum_{p \leq x} \sum_{\substack{ x^{1/2}< q \leq x  \\ q | p+a}} \ln q = \frac{x}{2} + O\left(\frac{x \ln \ln x}{ \ln x} \right)$$
where the summation is over $p$ and $q$ prime. Note that this implies that the number of primes $p$ less than $x$ such that $p-1$ has a prime factor greater than $p^{1/2}$ is asymptotically at least $\frac{x}{2\ln x}$.<|endoftext|>
TITLE: Riemann hypothesis generalization names: extended versus generalized?
QUESTION [8 upvotes]: This is a "names" question. There are two standard directions of generalization of the Riemann hypothesis: one to L-functions (which is used quite a bit in analytic number theory, and for extending density results for primes to Dirichlet primes) and another to Dedekind zeta-functions. Wikipedia says that the version for L-functions is called the GRH and the version of zeta-functions is called the ERH, and other generalizations are called the GRH. But I have seen conflicting terminology in some books and papers, which refer to the L-function version as the ERH and the other version as the GRH.
Which is the more standard convention?

REPLY [3 votes]: I also agree that the literature is not quite consistent on this topic. I tried to find a published reference on this question and found the following article:
Link
which is Chapter 5 of a book "The Riemann hypothesis : a resource for the afficionado and virtuoso alike", by Peter Borwein, Stephen Choi, Brendan Rooney and Andrea Weirathmueller, published by the Canadian Mathematical Society.
In the reference above, they use GRH for Dirichlet L-series (section 6.2), and ERH for Dedekind zeta functions (section 6.5). However, to make things more complicated:

They mention (in section 6.3) that ERH may be referring to the conjecture for L series of the form
$$\sum_{n=1}^\infty \frac{\left(\frac{n}{p}\right)}{n^s}$$
where $p$ is a prime and $\left(\frac{n}{p}\right)$ is the Legendre symbol. In fact they call this version ERH, and the one for Dedekind zeta functions is called another extended Riemann hypothesis.

They mention that the "Grand Riemann hypothesis" (which I had never heard of) refers to L functions of automorphic cuspidal representations.


Alvaro<|endoftext|>
TITLE: products and smooth/étale/unramified morphisms
QUESTION [6 upvotes]: Let $X$, $Y$ and $Z$ be Noetherian schemes. 
If $f: Y \to X$ and $g: Z \to X$ are morphisms of finite type, such that at each point of $X$, at least one of the two morphisms is smooth/étale/unramified (at all points of its inverse image), can we conclude that the induced morphism $Y \times_X Z \to X$ is smooth/étale/unramified everywhere? 
If not, which results can we obtain?
(In his textbook on Algebraic Geometry, Liu asks to prove that the answer is always "yes"...)
EDIT. So, indeed, the problem statement in the book is wrong...

REPLY [5 votes]: Smooth, unramified, and etale morphisms are stable under base change, so $Y\times_X Z \to Z$ is {smooth, unramified, etale} if $Y\to X$ is {smooth, unramified, etale}.<|endoftext|>
TITLE: Taking roots in simple linear algebraic groups 
QUESTION [7 upvotes]: Suppose $G$ is a simple (linear) algebraic group over an algebraically closed field of characteristic zero, that $n$ is a positive natural number, and that $g\in G$. Can we always find an $h\in G$ such that $h^n=g$?
(It appears to be possible to check this for the classical algebraic groups by direct computations in each case, but covering the exceptional Lie Algebras this way seems like it might be tricky, and anyhow, I'm inclined to think that a case analysis is probably not the optimal way of approaching this problem!)
Note added: Kovalev made a comment showing that the answer is `no' in general. The counterexamples appear to revolve around non-semisimple elements. I wonder whether the answer becomes positive if one restricts oneself to $g$ of finite order?

REPLY [7 votes]: A semisimple element lies in a maximal torus, so you can extract any root from it inside this torus.

REPLY [5 votes]: As Pavel says, any semisimple element lies in a maximal torus, where you can take any root. On the other hand, if $g$ is unipotent, then it is in the image of the exponential map, so you can make sense of $g^\lambda$ for any $\lambda \in \mathbb C$, so counterexamples must have nontrivial semisimple and unipotent parts. 
Suppose $g=s.u=u.s$ is the Jordan decomposition of an element. Then $s^n$ and $u^n$ are semisimple and unipotent respectively, so they are the Jordan decomposition of $g^n$. Thus the existence of roots is compatible with Jordan decomposition. 
Now take a semisimple element $s$ such that $Z_G(s)^0$ does not contain a central torus (so its center is finite) -- if $G$ is simply connected then in fact $Z_G(s)$ is connected, so I'll assume that. Now pick a regular unipotent element $u$ in $Z_G(s)$ and consider 
$g =s.u$. I want to claim $g$ is a counterexample. Indeed suppose for each $n$ we have $h_n$ an $n$-th root, and $h=s_nu_n$ is its Jordan decomposition. Then $s_n^n =s$ and $u_n^n=u$, and both $s_n$ and $u_n$ lie in $Z_G(s)$. Then we see that $s_n$ centralizes $u$ for all $n$, but since $u$ is regular in $Z_G(s)$ and $s_n$ semisimple it follows that $s_n$ must lie in the centre of $Z_G(s)$. But then taking, say, $n$ equal to the order of that centre (which is finite) we get a contradiction, as $s_n^n$ must then be $1$.
Semisimple elements $s$ such that $Z_G(s)$ does not contain a central torus exist, but there are only finitely many conjugacy classes of them, as was essentially shown in the paper of Borel and de Siebenthal. In fact I think that paper establishes that there are $r+1$ such classes where $r$ is the rank of $G$, so this would give a negative answer for the exceptional groups also. I suspect that these might somehow be the only counterexamples?<|endoftext|>
TITLE: Is there a finitely complete category with terminal object but NO subobject classifier?
QUESTION [5 upvotes]: This came up today while thinking about topoi in seminar, as the title suggests my question is;

Is there a finitely complete category with terminal object but NO subobject classifier?

Hopefully if the answer is yes, you can give an interesting example that is useful in some way. One of my professors has a saying that 

for every definition you should have an interesting example and an interesting non-example

So I am trying to stick to this. 
Thanks!

REPLY [3 votes]: I'm terribly late to this party, but the topic came up at the nForum, and so I might add some more observations. 
In any category with a subobject classifier, all monomorphisms are regular monomorphisms, which has some further consequences: the category is balanced (monic epis are isos), any two epi-mono factorizations of a morphism are canonically isomorphic, etc. Right away this kills off many examples such as the category of posets, the category of topological spaces, the category of commutative monoids, the (1-) category of small categories, etc. 
But what I think is even more of a killer is that in such a category, for any object $X$ the poset of subobjects $Sub(X)$ is a cartesian closed poset, so in particular has the property that meets distribute over any joins that exist (finitary or infinitary). So $Sub(X)$ is a distributive lattice (more specifically a Heyting algebra), provided that joins exist (which is usually true for categories of structured sets "in the wild"). "Most" categories do not have distributive subobject lattices (and it's usually not a hard condition to check, using the familiar "forbidden sublattice" criterion). For example, this kills off the category of groups, since subgroup lattices usually are not distributive; the same is true for abelian groups, rings, and so on. 
Details have been written up in the nLab article subobject classifier.<|endoftext|>
TITLE: Are the strata of Nakajima quiver varieties simply-connected?  Do they have odd cohomology?
QUESTION [10 upvotes]: Nakajima defined a while back a nice family of varieties, called "quiver varieties" (sometimes with "Nakajima" appended to the front to avoid confusion with other varieties defined in terms of quivers).  These are most concisely defined as the moduli of certain representations of certain preprojective algebras. 
I'll be interested in the affine (in the sense of "affine variety" not "affine Lie algebra") version of these, which is the moduli space of semi-simple representations of a certain preprojective algebra.  This variety is singular, but can be divided into smooth strata which correspond to fixing the size of the automorphism group of the representation (i.e. one stratum for simple representations, one for sums of pairs of non-isomorphic simples, etc.).  I would like to know a bit about the geometry of these strata.  One basic (and important for me) question is

Are these strata simply connected or equivariantly simply connected for the action of a group?

Actually, I know that they are not simply connected from some very simple examples (such as the nilcone of $\mathfrak{sl}_2$), but in those examples there is an action of a group such that there are no equivariant local systems (if you're willing to think of the quotient as a stack, the quotient is simply connected), so they are "equivariantly simply connected."
Similarly, I'm interested in the cohomology of these strata; I'd like its odd part to vanish.  Again, there's no hope of this in the most obvious way.  The only way it could happen is equivariantly.

Does the odd (equivariant) cohomology of these strata vanish for any group action?

let me just note in closing: I would be entirely satisfied if these results were only true in the finite type case; I know a lot of geometric statements for quiver varieties go a little sour once you're outside the finite type case.
If anyone knows these or any other results in the literature about the geometry of these strata, I would be very happy to hear them.

REPLY [19 votes]: As Ben wrote, there are examples of non-simply connected strata.
For a quiver variety corresponding to a fundamental representation, it, a priori, only has a small group action. So it is usually far away from homogeneous.
So I suspect the answer is NO, though I do not know a concrete example.
But, I think the question itself is not right:
It is true that perverse sheaves appearing in the push-forward from `natural resolutions' are IC sheaves associated with constant sheaves. But it is not because stratum are simply-connected, or equivariantly simply connected. It is from a very different reason:
This reason cannot be seen if one treat an individual quiver variety separately. One can see it only when one consider various quiver varieties simultaneously, and relate them to a representation. It is basically because there is a component consisting of a single point, and other components are `connected' to it in some sense.
(Each component corresponds only to a weight space, and considering several components simultaneously, one get the structure of a representation. The single point corresponds to the highest weight vector.)
In summary, quiver varieties are not homogeneous in a conventional sense, but have substitutive property (I do not know how to call it) when one treat several components simultaneously.<|endoftext|>
TITLE: When a formal power series is a rational function in disguise
QUESTION [18 upvotes]: Given a formal power series $f \in k[[X]]$, where $k$ is a commutative field, is there any good way to tell whether or not $f\in k(X)$?
Edit: To clarify, "good way to tell" means "computable algorithm to tell".
Edit 2: I really screwed up this question, so I am recusing myself from accepting an answer.  I will accept an answer after a sufficient number of votes have been cast for the best answer.

REPLY [34 votes]: Continued fractions!  
To motivate this answer, first recall the continued fraction algorithm for testing whether a real number is rational.  Namely, given a real number $r$, subtract its floor $\lfloor r \rfloor$, take the reciprocal, and repeat.  The number $r$ is rational if and only if at some point subtracting the floor gives $0$.  
Of course, an infinite precision real number is not something that a Turing machine can examine fully in finite time.  In practice, the input would be only an approximation to a real number, say specified by giving the first 100 digits after the decimal point.  There is no longer enough information given to determine whether the number is rational, but it still makes sense to ask whether up to the given precision it is a rational number of small height, i.e., with numerator and denominator small relative to the amount of precision given.  If the number is rational of small height, one will notice this when computing its continued fraction numerically, because subtracting the floor during one of the first few steps (before errors compound to the point that they dominate the results) will give a number that is extremely small relative to the precision; replacing this remainder by $0$ in the continued fraction built up so far gives the small height rational number.
What is the power series analogue?  Instead of the field of real numbers, work with the field of formal Laurent series $k((x))$, whose elements are series with at most finitely many terms with negative powers of $x$: think of $x$ as being small.  For $f = \sum a_n x^n \in k((x))$, define $\lfloor f \rfloor = \sum_{n \le 0} a_n x^n$; this is a sum with only finitely many nonzero terms.  Starting with $f$, compute $f - \lfloor f \rfloor$, take the reciprocal, and repeat.  The series $f$ is a rational function (in $k(x)$) if and only if at some point subtracting the floor gives $0$.
The same caveats as before apply.  In practice, the model is that one has exact arithmetic for elements of $k$ (the coefficients), but a series will be specified only partially: maybe one is given only the first 100 terms of $f$, say.  The only question you can hope to answer is whether $f$ is, up to the given precision, equal to a rational function of low height (i.e., with numerator and denominator of low degree).  The answer will become apparent when the continued fraction algorithm is applied: check whether subtracting the floor during one of the first few steps gives a series that starts with a high positive power of $x$.
Bonus: Just as periodic continued fractions in the classical case correspond to quadratic irrational real numbers, periodic continued fractions in the Laurent series case correspond to series belonging to a quadratic extension of $k(x)$, i.e., to the function field of a hyperelliptic curve over $k$.  Abel in 1826 exploited this idea as an ingredient in a method for determining which hyperelliptic integrals could be computed in elementary terms!<|endoftext|>
TITLE: What would a "moral" proof of the Weil Conjectures require?
QUESTION [33 upvotes]: At the very end of this 2006 interview (rm), Kontsevich says

"...many great theorems are originally proven but I think the proofs are not, kind of, "morally right." There should be better proofs...I think the Index Theorem by Atiyah and Singer...its original proof, I think it's ugly in a sense and up to now, we don't have "the right proof." Or Deligne's proof of the Weil conjectures, it's a morally wrong proof. There are three proofs now, but still not the right one."

I'm trying to understand what Kontsevich means by a proof not being "morally right." I've read this article by Eugenia Cheng on morality in the context of mathematics, but I'm not completely clear on what it means with respect to an explicit example. The general idea seems to be that a "moral proof" would be one that is well-motivated by the theory and in which each step is justified by a guiding principle, as opposed to an "immoral" one that is mathematically correct but relatively ad hoc.
To narrow the scope of this question and (hopefully) make it easier to understand for myself, I would like to focus on the second part of the comment. Why would Kontsevich says that Deligne's proof is not "morally right"? More importantly, what would a "moral proof" of the Weil Conjectures entail? 
Would a morally proof have to use motivic ideas, like Grothendieck hoped for in his attempts at proving the Weil Conjectures? Have there been any attempts at "moralizing" Deligne's proof? How do do the other proofs of the Weil Conjectures measure up with respect to mathematical morality?

REPLY [7 votes]: This is both an answer and a question:
As part of a response to a previous question of mine, David Speyer wrote that:

... it is known how to adapt Weil's proof of the Riemann hypothesis to higher dimensional S, if one had an analogue of the Hodge index theorem for $S \times S$ in characteristic p. I've been told that a good reference for this is  Kleiman's Algebraic Cycles and the Weil Conjectures...

So perhaps a "moral" proof would require a Hodge index theorem in characteristic p.
However, David later writes that Grothendieck's standard conjectures assert that the Hodge  theorem holds. So is this possible proof the same as "Grothendieck's envisaged" one?<|endoftext|>
TITLE: Normal bundle of $CP^n$ in $CP^{n+1}$
QUESTION [6 upvotes]: Let $S^1$ act on $S^{2n+1}$ via Hopf action and $S^1$ also acts on $\mathbb{R}^2$ via rotation about the origin.
Then $S^1$ acts on $S^{2n+1}\times \mathbb{R}^2$ diagonally. 
Let $M$ be the quotient of this diagonal action.
My question is why $ M$ can be viewed as the normal bundle of $\mathbb {CP}^n$ in $\mathbb {CP}^{n+1}$.
I have a feeling that it must be related to the fact that: after removing a $(2n+2)$ disk in $\mathbb {CP}^{n+1}$, the boundary $S^{2n+1}$ is fibered over the $\mathbb {CP}^n$. 
But where can I find the proof of the statement.

REPLY [8 votes]: $S^{2n+1}$ sits in $S^{2n+3}$ with a trivial normal bundle.  So the quotient map $S^{2n+3} \to \mathbb CP^{n+1}$ carries the normal bundle of $S^{2n+1}$ in $S^{2n+3}$ to the normal bundle of $\mathbb CP^n$ in $\mathbb CP^{n+1}$.<|endoftext|>
TITLE: Is there a refinement of the Hochschild-Kostant-Rosenberg theorem for cohomology?
QUESTION [16 upvotes]: The HKR theorem for cohomology in characteristic zero says that if $R$ is a regular, commutative $k$ algebra ($char(k) = 0$) then a certain map $\bigwedge^* Der(R) \to CH^*(R,R)$  (where $\wedge^* Der(R)$ has zero differential) is a quasi-isomorphism of dg vector spaces, that is, it induces an isomorphism of graded vector spaces on cohomology.
Can the HKR morphism be extended to an $A_\infty$ morphism? Is there a refinement in this spirit to make up for the fact that it is not, on the nose, a morphism of dg-algebras?

REPLY [20 votes]: Yes there is. It was noted by Kontsevich long time ago that the HKR quasi-isomorphism on cochains can be corrected to give a quasi-isomorphism of dg-algebras  and thus induce an $A_\infty$ quasi-isomorphism of minimal models. The correction is very natural - one needs to compose the HKR map it the contraction by the square root of the Todd class, where the latter is understood as a polynomial of the Atiyah class. This story has been studied in great detail in the past few years and has been generalized further to give Tsygan formality which is a quasi-isomorphism of $\infty$-calculi. This was proven by Dolgushev-Tamarkin-Tsygan and also by Calaque-Rossi-van den Bergh.
The literature on the subject is huge but you should get a good sense of the results if you look at
this survey  by  Dolgushev-Tamarkin-Tsygan and at this paper of Calaque-Rossi-van den Bergh. There are also many interesting references listed in these papers, for instance the works of Caldararu on the Mukai pairing.<|endoftext|>
TITLE: Random Alternating Permutations
QUESTION [27 upvotes]: An alternating permutation of {1, ..., n} is one were π(1) > π(2) < π(3) > π(4) < ... For example: (24153) is an alternating permutation of length 5.
If $E_n$ is the number of alternating permutations of length n, the $\sec x + \tan x = \sum_{n \geq 0} E_n \frac{x^n}{n!}$ is the exponential generating function.
How does one sample the alternating permutations uniformly at random?  Rejection sampling would quickly become inefficient since $\frac{E_n}{n!} \approx \frac{4}{\pi} \left(\frac{2}{\pi} \right)^n$ decays exponentially in n.

REPLY [11 votes]: An $O(n \log n)$ algorithm to uniformly sample alternating permutation is described by P. Marchal in *Generating alternating permutations in time $O(n \log n)$." (by contrast, Pak's answer is $O(n^3),$ Boltzmann sampling is $O(n^2).$<|endoftext|>
TITLE: slice=ribbon generalization to higher genus + potential counterexamples to slice=ribbon.
QUESTION [10 upvotes]: I have two questions about the slice=ribbon conjecture.
(1) If a knot $K \hookrightarrow S^3$ has smooth slice genus $g$, you can ask if it bounds a smooth genus $g$ surface in $S^3 \times [0, -\infty)$, with the function defined by restriction to $[0, -\infty)$ being Morse on the surface without index=0 critical points (maximal points).  When $g=0$ this is just asking if the slice knot $K$ has a ribbon disc.  I was wondering if there are any knots known with $g \geq 1$ for which such a surface cannot exist.  If there are none such known, is there a topological reason why the truth of the slice=ribbon conjecture would also imply the existence of such surfaces?
(2) Are there any potential counterexamples to slice=ribbon (in the same way that there are potential counterexamples to smooth 4-d Poincare [until Akbulut kills them])?
Thanks,
Andrew.

REPLY [3 votes]: There is a paper by Gompf and Scharlemann: Fibered knots and Property 2R, II, which gives an infinite family of two component links which are smoothly slice but not obviously ribbon.<|endoftext|>
TITLE: How much of differential geometry can be developed entirely without atlases?
QUESTION [62 upvotes]: We may define a topological manifold to be a second-countable Hausdorff space such that every point has an open neighborhood homeomorphic to an open subset of $\mathbb{R}^n$.  We can further define a smooth manifold to be a topological manifold equipped with a structure sheaf of rings of smooth functions by transport of structure from $\mathbb{R}^n$, since $\mathbb{R}^n$ has a canonical sheaf of differentiable functions $\mathbb{R}^n\to \mathbb{R}$, with a canonical restriction sheaf to any open subset.    This gives a manifold as a locally ringed space. (Of course this definition generalizes to all sorts of other kinds of manifolds with minor adjustments).   
Then the questions:
If we totally ignore the definition using atlases, will we at some point hit a wall? Can we fully develop differential geometry without ever resorting to atlases?  
Regardless of the above answer, are there any books that develop differential geometry primarily from a "locally ringed space" viewpoint, dropping into the language of atlases only when necessary?  I looked at Kashiwara & Schapira's "Sheaves on Manifolds", but that's much more focused on sheaves of abelian groups and (co)homology.  
Edit:
To clarify (Since Pete and Kevin misunderstood): It's easy to show that the approaches are equivalent, but proofs using charts don't always translate easily to proofs using sheaves.

REPLY [10 votes]: There is another way to develop differential geometry without atlases, and even without charts, that is Diffeology. I'm not sure this is the right answer to your question but it worths looking at.
Comment: There are many ways to develop differential geometry without atlases. You may change the category of differentiable manifolds for a larger one. This is the case for Diffeology, or Differential Spaces à la Sikorski. These two approaches correspond to the two ways you may interpret the smooth structure: in the first case the smooth structure is characterized by the smooth parametrizations in the space (called plots), this is the "in-way", in the second case the smooth structure is characterized by the smooth functions from the space into the field of real numbers $\bf R$, the "out-way". In some sense they are "dual" but not equivalent approaches. The "intersection" of these two approches gives the so-called Frölicher spaces (reflexive diffeological spaces). The diffeological way gives a richer category than the Sikorski's one. For example it gives a non trivial structure for quotients (like spaces of leaves of a dense foliation, for example) where Sikorski structure is trivial. The two of them gives access to infinite dimensional spaces, without necessarily modeling these spaces on topological vector spaces. You can develop a whole theory of homotopy, cohomology, differential calculus and De-Rham cohomology, groups, fiber bundles etc. in diffeology without loosing much of what you have learned in manifold differential geometry.  
Well, there is a lot say, may be too much for this discussion :-)<|endoftext|>
TITLE: a question about diagonal prikry forcing
QUESTION [5 upvotes]: Suppose <\kappa_n|n<\omega> is a strictly increasing sequence of measurable cardinals, 
\kappa is the limit of this sequence. For each n<\omega, U_n is a normal measure on 
\kappa_n. P is the diagonal Prikry forcing corresponding to \kappa_n's and U_n's. 
Suppose g is P-generic sequence over V. We have known that for each strictly increasing 
sequence x of length \omega such that each x(i)<\kappa_i and x\in{V}, x is eventually 
dominated by g. In V[g], suppose A is a subset of \kappa, A is not in V. Is there a strictly 
increasing sequence y of length \omega such that each y(i)<\kappa_i and y\in{V[A]}, y is not 
eventually dominated by g?
(g can eventually dominate all such sequences in V, V[A] is greater than V, I feel g can not 
eventually dominate all such sequences in V[A].)

REPLY [8 votes]: Yes.    The answer is obviously "yes" if $A$ is a subset of $g$, so it is sufficient to show for any subset $A\subset\kappa$ there is $A'\subset g$ such that $V[A']=V[A]$.
I assume that this is known, and it struck me at the start as obviously true, but I can't recall having seen it.  Here is an outline of a proof.
Write the conditions (following Gitik) as $x=\langle x_i\mid i \in\omega\rangle$, with
$x_i\in\kappa_i\cup U_i$.    Write $A_n=A\cap \kappa_n$.    There is
$x\leq^* 1^P$ which forces that $A_n$ is decided by conditions $y$
with $y_i\in U_i$ for $i>n$.   It follows in particular that $A_n\in V$.
Find $x'\leq^* x$ which decides, for each $n$, the sentence "there is
$z\in \dot G$  such that $z_i$ decides the value of $\dot A_n$ and
$z_i\in U_i$."   Let $b_n$  be the (finite) set of $i$ for which this
sentence is forced to be false.  Then there is $x''\leq^* x'$ and 
1--1 functions $h_n\colon \Pi_{i\in b_n} x_i''\to \kappa_n$ such that
$x''$ forces that $A_n=h_n(g\upharpoonright b_n)$.   
Set $A'=\bigcup_n b_n$.  Then $V[A] = V[A']$.<|endoftext|>
TITLE: Looking for reference talking about relationship between descent theory and cohomological descent
QUESTION [14 upvotes]: I am now taking a course focusing on triangulated geometry. The professor has formulated the Beck's theorem for Karoubian triangulated category. The proof is very simple. Just using the universal homological functor(equivalent to Verdier abelianization)back to abelian settings(in particular, Frobenius abelian category), then using the Beck's theorem for abelian category to get the proof. 
When he finished the proof, he made a remark that the cohomological descent theory can be taken as a consequence of triangulated version of Beck's theorem. 
As we know, the Beck's theorem for abelian category is equivalent to Grothendieck flat descent theory(Beck's theorem may be more general). I have two questions:

Is there any reference(other than SGA 4)in English explaining the relationship of usual descent theory and cohomolgical descent theory? What I am looking for is not a very thick book but a lecture notes with some examples. 
I know Jacob Lurie developed the derived version of  Beck's theorem for his infinity category(correct me if I make mistake). But I have never read his paper very carefully. I wonder whether he explained the relationship of Beck's theorem and "cohomogical descent in his settings"(if there exists such terminology).  

All the other comments are welcome.
Thank you

REPLY [5 votes]: For 1 -- Brian Conrad's notes here are great and in english. Lemma 6.8 and the discussion preceding it explain the relation to descent theory.<|endoftext|>
TITLE: at which rational points does the Hypergeometric function take rational values
QUESTION [7 upvotes]: A generic example is ${}_2 F_1(\frac{1}{3},\frac{2}{3},\frac{5}{6};\frac{27}{32})=\frac{8}{5}$. So my question: Is there any description of the set of rational points at which the hypergeometric function ${}_m F_n$ takes rational values?
The references to the literature where a lot of such examples listed are apreciated very much too.

REPLY [7 votes]: There are theorems that give conditions for the set to be finite even when "rational" is replaced by "algebraic". See the work of Paula Cohen Tretkoff. E.g the following paper is a survey and Theorem 2 is about this:
http://www.math.tamu.edu/~ptretkoff/martinpub_final.pdf<|endoftext|>
TITLE: Every Manifold Cobordant to a Simply Connected Manifold
QUESTION [20 upvotes]: I am wondering if it is true that every compact, connected, oriented manifold is cobordant to a simply connected manifold.
I believe that some sort of surgery will do the trick. Roughly speaking, I want to add handles so that I can kill representative loops. However, I don't know if my surgery process builds a cobordism and it is hard to for me to see what the new boundary is. Another possibility is to build a Morse function that constructs the cobordism for free.
It is hard for me to get an intuition for what is going on, because all the compact, oriented 2-manifolds are boundaries and consequently cobordant to the empty set. CP^2 seems like the "easiest" test case, but it is already simply-connected for cellular reasons. Lens Spaces might be a nice candidate, but the 3 dimensional ones can be realized as boundaries of some disc bundle on S^2.
If possible, I'd prefer a constructive procedure, but any answer that helps elucidate the material is welcome.

REPLY [20 votes]: Assume that $M^n$ has $\pi_1$ finitely generated (Edit: and n>3). Choose a generator. We will construct (using surgery) a cobordism to $M'$ which kills that generator, and by induction we can kill all of $\pi_1$. Choose an embedded loop which represents the generator, and choose a tubular neighborhood of the loop. We can view this as a (n-1)-dimensional vector bundle over $S^1$, the normal bundle. Since $M$ is oriented, this is a trivial vector bundle so we can identify this tubular neighborhood with $S^1 \times D^{n-1}$. 
Now we build the cobordism. We take $M \times I$, which is a cobordism from $M$ to itself. To one end we glue $D^2 \times D^{n-1}$ along the boundary piece $S^1 \times D^{n-1}$ via its embedding into $M$. This is just attaching a handle to $M \times I$. This new manifold is a cobordism from $M$ to $M'$, where $M'$ is just $M$ where we've done surgery along the given loop.
A van Kampen theorem argument shows that we have exactly killed the given generator of $\pi_1$. Repeating this gives us a cobordism to a simply connected manifold. 
Note that it is essential that our manifold was oriented. $\mathbb{RP}^2$ is a counter example in the non-oriented setting, as all simply connected 2-manifolds are null-cobordant, but $\mathbb{RP}^2$ is not.

[I was implicitly thinking high dimensions. Thanks to Tim Perutz for suggesting something was amiss when n=3]
If n=3 then this is "surgery in the middle dimension" and it is more subtle. First of all the normal bundle is an oriented 2-plane bundle over the sphere, so there are in fact $\mathbb{Z} = \pi_1(SO(2))$ many ways to trivialize the bundle (these are normal framings). Ignoring this, if you carry out the above construction, you will see that (up to homotopy) M' is the union of $M - (S^1 \times D^2)$ and $D^2 \times S^1$ along $S^1 \times S^1$. This can (and does) enlarge the fundamental group. 
However a different argument works in dimensions n=1,2,3. The oriented bordism groups in those dimensions are all zero (see the Wikipedia entry on cobordism), so in fact every oriented 3-manifold is cobordant to the empty set (a simply connected manifold). The fastest way to see this is probably a direct calculation of the first few homotopy groups of the Thom spectrum MSO.<|endoftext|>
TITLE: Is this naive test to tell whether a complex elliptic curve has complex multiplication effective?
QUESTION [9 upvotes]: I have a question about a naive test to tell whether a complex elliptic curve $E$ has complex multiplication.
Recall that the endomorphism ring $End(E)$ of $E$ is isomorphic to either $\mathbb{Z}$ or an order in an imaginary quadratic field $K$. In the latter case we say that $E$ has CM by $K$.
Suppose that we are given an elliptic curve over $\mathbb{C}$, say $$E: y^2=x^3+17x^2-19.$$ One wants to know if there is an imaginary quadratic field $K$ for which $E$ has CM by $K$.
Let $j(E)$ be the j-invariant of our elliptic curve (for example, the above curve has j-invariant $\frac{6179217664}{363641}$) and recall that the j-invariant, viewed as a modular function, gives a surjective map from the upper half plane to $\mathbb{C}$. Let $\omega$ be any element in the preimage of $j(E)$. We now define a second elliptic curve: $$E_\omega: y^2=4x^3-g_2(\omega)-g_3(\omega),$$ where $g_2=60G_4$ and $g_3=140G_6$ are multiples of the appropriate Eisenstein series. Both of these elliptic curves are defined over $\mathbb{C}$ and have the same j-invariant. They are therefore isogenous. It is known that the elliptic curve $E_\omega$ is isomorphic to the complex torus $\mathbb{C}/\Lambda_\omega$ where $\Lambda_\omega=\mathbb{Z}+\mathbb{Z}\omega$. 
It is easy to show that $\mathbb{C}/\Lambda_\omega$ has CM by some imaginary quadratic field if and only if $\omega$ is an imaginary, quadratic number. In this case the endomorphism ring of $\mathbb{C}/\Lambda_\omega$ will be an order in the field $\mathbb{Q}(\omega)$.
This suggests a test for CM: given an elliptic curve $E$ defined over $\mathbb{C}$ with j-invariant $j(E)$, find a preimage of $j(E)$ under the modular function $j:\mathbb{H}\rightarrow\mathbb{C}$ and determine whether or nor the preimage generates an imaginary quadratic extension of $\mathbb{Q}$. 
Now for my question: can this test actually be performed? Wikipedia tells me that the inverse of the j-invariant can be computed in terms of hypergeometric functions, but I don't know if one could use this inverse to determine whether a given j-invariant was associated to a curve with CM.

REPLY [2 votes]: Unlike Pete Clark, I have not spent any real time working on this sort of problem, so you should take all of this with a heavy grain of salt. But I think this algorithm is quite reasonable as a way to rule out CM, although it would be diffciult to use it to prove CM exists. EDIT: As FC points out, this algorithm only finds the lattices of the form $\mathcal{O}_K$, not the lattices of the form $I$ where $I$ is an ideal in $\mathcal{O}_K$. The latter may not even have real $j$-invariant. I don't see a way to get around this.
Here is how I would do it: Start with your $j$ invariant, $j_0$. If it is not real, the curve does not have CM . If $j_0$ is real, compute $\tau$ such that $j(\tau) = j_0$ and such that $\tau$ is in
$$\{ i t: t \geq 1 \} \cup \{ z : |z|=1, \ 0 \leq \Re(z) \leq 1/2 \} \cup \{ 1/2 + i t: t \geq \sqrt{3}/2 \}.$$
I'll describe the case where $\tau$ is on the imaginary axis; the other cases are similar. Here is the point which no one has made yet: The pure imaginary $it$ generates an imaginary order  if and only if $t^2$ is an integer. So we only need to perform the computation with enough precision to determine whether $t$ is an integer. Of course, a floating point computation can never be known to converge to an integer, but it will very soon either do so or clearly fail to do so.<|endoftext|>
TITLE: Non-constructive proofs of decidability?
QUESTION [17 upvotes]: Are there examples of sets of natural numbers that are proven to be decidable but by non-constructive proofs only?

REPLY [3 votes]: Indeed, there exists some mathematical theorems with non-constructive proofs which cannot have any constructive proof (now and ever). Here is an example:
Definition. Let $\boldsymbol\varphi_1,\boldsymbol\varphi_2,\boldsymbol\varphi_3,\cdots$ be a list of all unary partial computable functions (from $\mathbb{N}$ to $\mathbb{N}$). Denote the Chaitin-Kolmogorov Complexity by $\mathcal{C}$ which is defined by $\mathcal{C}(n)=\min \{e\mid\boldsymbol\varphi_e(0)\!\downarrow = n\}$.
Let us note that $\mathcal{C}$ is a total function ($\mathcal{C}:\mathbb{N}\rightarrow\mathbb{N}$).
The following theorem has a non-constructive proof:
Theorem 1. For every natural $n$ there exists some $m$ such that $\mathcal{C}(m)\!>\!n$. That is $\forall y \exists x \mathcal{C}(x)\!>\!y$.
Proof: For each $i\leqslant n$ put $\alpha_i=\boldsymbol\varphi_i(0)$ if the value exists, i.e., $\boldsymbol\varphi_i(0)\!\downarrow$; otherwise let $\alpha_i=0$. Take $m$ to be any natural number greater than all $\alpha_i$'s.
 $\texttt{QED}$
As a matter of fact, the above theorem cannot have any constructive proof:
Theorem 2. There exists no computable function $f$ such that for every natural $n$ one has $\mathcal{C}\big(f(n)\big)\!>\!n$.
Proof: If there were such a computable (and total) $f$ then by Kleene's Recursion (aka Fixedpoint) Theorem there would exist some $\mathbf{e}$ such that $\boldsymbol\varphi_{\mathbf{e}}(0)=f(\mathbf{e})$. For this $\mathbf{e}$ then we would have $\mathcal{C}\big(f(\mathbf{e})\big)\leqslant\mathbf{e}$ contradicting the assumption of $\forall n: \mathcal{C}\big(f(n)\big)\!>\!n$.  $\texttt{QED}$
Now, we have an example of a decidable set whose decidability cannot be proven constructively:
Let $\mathfrak{B}=\{n\in\mathbb{N}\mid \exists m: \mathcal{C}(m)>n\}$. We know that $\mathfrak{B}$ is decidable just because $\mathfrak{B}=\mathbb{N}$. But if anyone (now or in future) can have any constructive proof of its decidability, s/he should be able to compute the characteristic function $\chi_{\mathfrak{B}}$ of $\mathfrak{B}$ (defined by $\chi_{\mathfrak{B}}(x)=0$ if $x\not\in\mathfrak{B}$ and $\chi_{\mathfrak{B}}(x)=1$ if $x\in\mathfrak{B}$). That would amount to showing that $\chi_{\mathfrak{B}}=\mathbf{1}$ the constant function $1$. Thus any constructive proof of the decidability of $\mathfrak{B}$ would show (constructively) that $\mathfrak{B}=\mathbb{N}$ and thus would provide a computable function $f$ such that $\forall n\!\in\!\mathbb{N}: \mathcal{C}\big(f(n)\big)\!>\!n$; contradicting Theorem 2. 
So, I believe, $\mathfrak{B}$ is a decidable set such that no proof of its decidability can be constructive.<|endoftext|>
TITLE: How many ways are there to globalize Harish Chandra modules?
QUESTION [20 upvotes]: Suppose $G$ a reductive Lie group with finitely many connected components, and suppose in addition that the connected component $G^0$ of the identity can be expressed as a finite cover of a linear Lie group. Denote by $\mathfrak{g}$ the complexified Lie algebra, and denote by $K$ a maximal compact in the complexification of $G$.
Denote by $\mathbf{HC}(\mathfrak{g},K)$ the category of admissible $(\mathfrak{g},K)$-modules or (Harish Chandra modules), and $(\mathfrak{g},K)$-module homomorphisms. Denote by $\mathbf{Rep}(G)$ the category of admissible representations of finite length (on complete locally convex Hausdorff topological vector spaces), with continuous linear $G$-maps.
The Harish Chandra functor $\mathcal{M}\colon\mathbf{Rep}(G)\to\mathbf{HC}(\mathfrak{g},K)$ assigns to any admissible representation $V$ the Harish Chandra module of $K$-finite vectors of $V$. This is a faithful, exact functor. Let us call an exact functor $\mathcal{G}\colon\mathbf{HC}(\mathfrak{g},K)\to\mathbf{Rep}(G)$ along with a comparison isomorphism $\eta_{\mathcal{G}}\colon\mathcal{M}\circ\mathcal{G}\simeq\mathrm{id}$ a globalization functor.
Our first observation is that globalization functors exist.
Theorem. [Casselman-Wallach] The restriction of $\mathcal{M}$ to the full subcategory of smooth admissible Fréchet spaces is an equivalence. Moreover, for any Harish Chandra module $M$, the essentially unique smooth admissible representation $(\pi,V)$ such that $M\cong\mathcal{M}(\pi,V)$ has the property $\pi(\mathcal{S}(G))M=V$, where $\mathcal{S}(G)$ is the Schwartz algebra of $G$.
If we do not restrict $\mathcal{M}$ to smooth admissible Fréchet spaces, then we have a minimal globalization and a maximal one.
Theorem. [Kashiwara-Schmid] $\mathcal{M}$ admits both a left adjoint $\mathcal{G}_0$ and right adjoint $\mathcal{G}_{\infty}$, and the counit and unit give these functors the structure of globalization functors.
Construction. Here, briefly, are descriptions of the minimal and maximal globalizations. The minimal globalization is
$$\mathcal{G}_0=\textit{Dist}_c(G)\otimes_{U(\mathfrak{g})}-$$
where $\textit{Dist}_c(G)$ denotes the space of compactly supported distributions on $G$, and the maximal one is
$$\mathcal{G}_{\infty}=\mathrm{Hom}_{U(\mathfrak{g})}((-)^{\vee},C^{\infty}(G))$$
where $M^{\vee}$ is the dual Harish Chandra module of $M$ (i.e., the $K$-finite vectors of the algebraic dual of $M$).
For any Harish Chandra module $M$, the minimal globalization $\mathcal{G}_0(M)$ is a dual Fréchet nuclear space, and the maximal globalization $\mathcal{G}_{\infty}(M)$ is a Fréchet nuclear space.
Example. If $P\subset G$ is a parabolic subgroup, then the space $L^2(G/P)$ of $L^2$-functions on the homogeneous space $G/P$ is an admissible representation, and $M=\mathcal{M}(L^2(G/P))$ is a particularly interesting Harish Chandra module. In this case, one may identify $\mathcal{G}_0(M)$ with the real analytic functions on $G/P$, and one may identify $\mathcal{G}_{\infty}(M)$ with the hyperfunctions on $G/P$.
[I think other globalizations with different properties are known or expected; I don't yet know much about these, however.]
Consider the category $\mathbf{Glob}(G)$ of globalization functors for $G$; morphisms $\mathcal{G}'\to\mathcal{G}$ are natural transformations that are required to be compatible with the comparison isomorphisms $\eta_{\mathcal{G}'}$ and $\eta_{\mathcal{G}}$. Since $\mathcal{M}$ is faithful, this category is actually a poset, and it has both an inf and a sup, namely $\mathcal{G}_0$ and $\mathcal{G}_{\infty}$. This is the poset of globalizations for $G$.
I'd like to know more about the structure of the poset $\mathbf{Glob}(G)$ — really, anything at all, but let me ask the following concrete question.

Question. Does every finite collection of elements of $\mathbf{Glob}(G)$ admit both an inf and a sup?


[Added later]
Emerton (below) mentions a geometric picture that appears to be very well adapted to the study of our poset $\mathbf{Glob}(G)$. Let me at introduce the main ideas of the objects of interest, and what I learned about our poset. [What I'm going to say was essentially outlined by Kashiwara in 1987.] For this, we probably need to assume that $G$ is connected.
Notation. Let $X$ be the flag manifold of the complexification of $G$. Let $\lambda\in\mathfrak{h}^{\vee}$ be a dominant element of the dual space of the universal Cartan; for simplicity, let's assume that it is regular. Now one can define the twisted equivariant bounded derived categories $D^b_G(X)_{-\lambda}$ and $D^b_K(X)_{-\lambda}$ of constructible sheaves on $X$. Now let $\mathbf{Glob}(G,\lambda)$ denote the poset of globalizations for admissible representations with infinitesimal character $\chi_{\lambda}$, so the objects are exact functors $\mathcal{G}\colon\mathbf{HC}(\mathfrak{g},K)_{\chi_{\lambda}}\to\mathbf{Rep}(G)_{\chi_{\lambda}}$ equipped with natural isomorphisms $\eta_{\mathcal{G}}:\mathcal{M}\circ\mathcal{G}\simeq\mathrm{id}$.
Matsuki correspondence. [Mirkovic-Uzawa-Vilonen] There is a canonical equivalence $\Phi\colon D^b_G(X)_{-\lambda}\simeq D^b_K(X)_{-\lambda}$. The perverse t-structure on the latter can be lifted along this correspondence to obtain a t-structure on $D^b_G(X)_{-\lambda}$ as well. The Matsuki correspondence then restricts to an equivalence $\Phi\colon P_G(X)_{-\lambda}\simeq P_K(X)_{-\lambda}$ between the corresponding hearts.
Beilinson-Bernstein construction. There is a canonical equivalence $\alpha\colon P_K(X)_{-\lambda}\simeq\mathbf{HC}(\mathfrak{g},K)_{\chi_{\lambda}}$, given by Riemann-Hilbert, followed by taking cohomology. [If $\lambda$ is not regular, then this isn't quite an equivalence.]
Now we deduce a geometric description of an object of $\mathbf{Glob}(G,\lambda)$ as an exact functor $\mathcal{H}\colon P_G(X)_{-\lambda}\to\mathbf{Rep}(G)_{\chi_{\lambda}}$ equipped with a natural isomorphism $\mathcal{M}\circ\mathcal{H}\simeq\alpha\circ\Phi$, or equivalently, as a suitably t-exact functor $\mathcal{H}\colon D^b_G(X)_{-\lambda}\to D^b\mathbf{Rep}(G)_{\chi_{\lambda}}$ equipped with a functorial identification between the (complex of) Harish Chandra module(s) of $K$-finite vectors of $\mathcal{H}(F)$ and $\mathrm{RHom}(\mathbf{D}\Phi F,\mathcal{O}_X(\lambda))$ for any $F\in D^b_G(X)_{-\lambda}$. In particular, as Emerton observes, the maximal and minimal globalizations can be expressed as
$$\mathcal{H}_{\infty}(F)=\mathrm{RHom}(\mathbf{D}F,\mathcal{O}_X(\lambda))$$
and
$$\mathcal{H}_0(F)=F\otimes^L\mathcal{O}_X(\lambda)$$
Note that Verdier duality gives rise to an anti-involution $\mathcal{H}\mapsto(\mathcal{H}\circ\mathbf{D})^{\vee}$ of the poset $\mathbf{Glob}(G,\lambda)$; in particular, it exchanges $\mathcal{H}_{\infty}$ and $\mathcal{H}_0$.
I now expect that one can show the following (though I don't claim to have thought about this point carefully enough to call it a proposition).
Conjecture. All globalization functors are representable. That is, every element of $\mathbf{Glob}(G,\lambda)$ is of the form $\mathrm{RHom}(\mathbf{D}(-),E)$ for some object $E\in D^b_G(X)_{-\lambda}$.

Question. Can one characterize those objects $E\in D^b_G(X)_{-\lambda}$ such that the functor $\mathrm{RHom}(\mathbf{D}(-),E)$ is a globalization functor? Given a map between any two of these, under what circumstances do they induce a morphism of globalization functors (as defined above)?

In particular, note that if my expectation holds, then one should be able to find a copy of the poset $\mathbf{Glob}(G,\lambda)$ embedded in $D^b_G(X)_{-\lambda}$.

REPLY [6 votes]: Dear Clark,
This answer is addressed primarily at the final paranthetical comment.  There is a "dual" geometric description of $(\mathfrak g, K)$-modules to the Beilinson--Bernstein picture,
using orbits of the real group $G(\mathbb R)$ on the flag variety.  It is employed by
Schmid and Vilonen in many of their papers, and there is an expository article about it
by Vilonen in one of the Park City proceedings.
In this dual picture, when one takes sections of the sheaves, one really gets $G(\mathbb R)$-representations; I'm not sure off the top of my head which models you get though (the smooth ones or some other ones, or whether you get an option depending on the particular
brand of sheaves you use).  
Regarding your broader question, I've always imagined that one can find a model for a $(\mathfrak g, K)$-module using any brand of regularity you choose (analytic, smooth, distributional, hyperfunctions, and other brands in between, whatever they might be (and perhaps answering that is part of the point of the question!)), and that these will be ordered in the obvious way.
But this is certainly not the precise answer you want, and is only a vague intuition.<|endoftext|>
TITLE: Have people successfully worked with the full ring of differential operators in characteristic p?
QUESTION [26 upvotes]: This question is inspired by an earlier one about the possibility of using the full ring of differential operators on a flag variety to develop a theory of localization in characteristic $p$.  (Here by the full ring of differential operators I mean the same thing as the ring of divided-power differential operators, which is the terminology used in the cited question.)
My question is:

Do people have experience using the full ring of differential operators successfully in characteristic $p$ (for localization, or other purposes)?

I always found this ring somewhat unpleasant (its sections over affines are not Noetherian,
and, if I recall correctly a computation I made a long time ago, the structure sheaf
${\mathcal O}_X$ is not perfect over ${\mathcal D}_X$).  Are there ways to get around
these technical defects? (Or am I wrong in thinking of them as technical defects, or am
I even just wrong about them full stop?)
EDIT: Let me add a little more motivation for my question, inspired in part by Hailong's answer and associated comments. A general feature of local cohomology in char. p is that you don't have the subtle theory of Bernstein polynomials that you have in char. 0. See e.g. the paper of Alvarez-Montaner, Blickle, and Lyubeznik cited by Hailong in his answer. What I don't understand is whether this means that (for example) localization with the full ring of differential ops is hopeless (because the answers would be too simple), or a wonderful prospect (because the answers would be so simple).

REPLY [4 votes]: Certainly the fact that the ring of differential operators is non-Noetherian is an inconvenience but it is not clear if it is more than that. For instance one can define the notion of holonomic module. It is not a direct translation of the characteristic zero definition (and this is certainly related to this inconvenience) but once given it seems to work as well as in characteristic zero:
MR1918185 (2003h:14030)
Bögvad, Rikard(S-STOC)
An analogue of holonomic D-modules on smooth varieties in positive characteristics. (English summary)
The Roos Festschrift volume, 1.
Homology Homotopy Appl. 4 (2002), no. 2, part 1, 83–116.
14F10 (16S32 32C38)<|endoftext|>
TITLE: extending cusp forms
QUESTION [5 upvotes]: Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.
Let $\tilde{G} := GU(V)$ and $G := U(V)$.  Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$
Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$?  Note that here, the restriction is that of cusp forms, not of the representation itself.

REPLY [3 votes]: I believe the answer should be yes, by some version of the following sketch of an argument:
(Note: by restriction of scalars, I regard all groups as being defined over $\mathbb Q$,
and I write ${\mathbb A}$ for the adeles of $\mathbb Q$.)
We are given $V_{\pi} \subset Cusp(G(F)\backslash G({\mathbb A}_F)).$
Let $\tilde{C}$ denote the maximal $\mathbb Q$-split torus in the centre of $\tilde{G}$
(this is just a copy of $\mathbb G_m$), and write $C = \tilde{C}\cap G$.  (I guess
this is just $\pm 1$?)
Now $C(\mathbb A)$ acts on $V_{\pi}$  through some character $\chi$ of $(\mathbb A)/C(\mathbb Q)$.  Choose an extension
$\tilde{\chi}$ of $\chi$ to a character of $\tilde{C}(\mathbb A)/\tilde{C}(\mathbb Q)$, 
and regard $V_{\pi}$ as a representation of $\tilde{C} G$ by have $\tilde{C}$ act
through $\tilde{\chi}$.  Since $\tilde{C} G$ is normal and Zariksi open in $\tilde{G}$,
we should be able to further extend the $\tilde{C} G(\mathbb A)$-action on $V_{\pi}$
to an action of $\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A).$
Now if we consider $Ind_{\tilde{G}(\mathbb Q)\tilde{C} G(\mathbb A)}^{\tilde{G}(\mathbb A)} V_{\pi},$ we should be able to
find a cupsidal representation $V_{\tilde{\pi}}$ of the form you want (with $\tilde{C}(A)$ acting via $\tilde{\chi}$).
The intuition is that automorphic forms on $G(\mathbb A)$ are $Ind_{G(\mathbb Q)}^{G(\mathbb A)} 1,$
and similarly for $\tilde{G}$.  We will consider variants of this formula that takes into account central characters, and think about how to compare them for $G$ and $\tilde{G}$.
Inside the automorphic forms, we have those where $C(\mathbb A)$ acts by $\chi$; this we can
write as $Ind_{G(\mathbb Q)C(\mathbb A)}^{G(\mathbb A)} \chi$, and then rewrite as
$Ind_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)} \tilde{\chi}.$  This is where $V_{\pi}$ lives, once we extend it to a repreresentation
of $\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)$ as above.
Now the automorphic forms on $\tilde{G}(\mathbb A)$ with central character $\tilde{\chi}$
are $Ind_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb A)} \tilde{\chi},$
which we can rewrite as
$Ind_{\tilde{G}(\mathbb Q) \tilde{C}G(\mathbb A)}^{\tilde{G}(\mathbb A)}
Ind_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)}
\tilde{\chi}.$  This thus contains $Ind_{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)}^{\tilde{G}(\mathbb A)}V_{\pi}$ inside it, and so an irreducible constituent of the latter
should be a $V_{\tilde{\pi}}$ whose restriction (as a space of functions) to $G(\mathbb A)$ contains $V_{\pi}$. 
What I have just discussed is the analogue for $G$ and $\tilde{G}$ of the relation between automorphic forms on $SL_2$ and $GL_2$ discussed e.g. in Langlands--Labesse.  Hopefully I haven't blundered!<|endoftext|>
TITLE: Dense inclusions of Banach spaces and their duals
QUESTION [6 upvotes]: This seems like a really simple question, but I'm struggling with it.  Let $X$ be a separable Banach space, $H$ be a separable Hilbert space, and suppose $i : H \hookrightarrow X$ is a dense, continuous embedding of $H$ into $X$.  (This is the abstract Wiener space construction due to Gross, hence the [pr.probability] tag)  If we associate $H$ with its dual $H^{\star}$, we have the inclusions $$X^{\star} \hookrightarrow H^{\star} \cong H \hookrightarrow X.$$
My question:  Is $i^{\star} : X^{\star} \hookrightarrow H^{\star}$ a dense injection?

REPLY [2 votes]: Just to flesh out Bill's answer and comments thereon, we have the following facts.  Let $X,Y$ be Banach spaces and $T : X \to Y$ a bounded linear operator.

If $T$ has dense range then $T^*$ is injective.

Since this is a standard homework problem I'll just give a hint. Suppose $f \in Y^*$ with $T^*f=0$.  This means that $f(Tx)=0$ for every $x$, i.e. $f$ vanishes on the range of $T$...

Suppose further that $X$ is reflexive.  If $T$ is injective then $T^*$ has dense range.

Hint: By Hahn-Banach, to show that $T^*$ has dense range, it suffices to show that if $u \in X^{**}$ vanishes on the range of $T^*$, then $u=0$.  And by reflexivity, $u \in X^{**}$ is represented by some $x \in X$...
Note that the proofs I have in mind don't need to discuss the weak-* topology (at least not explicitly).
We cannot drop reflexivity in 2.  Consider the inclusion of $\ell^1$ into $\ell^2$.  It is injective, but its adjoint is the inclusion of $\ell^2$ into $\ell^\infty$, whose range is not dense.
(Fun fact: you can't prove 2 without Hahn-Banach or some other consequence of the axiom of choice.  If you work in ZF + dependent choice (DC), it's consistent that $\ell^1$ is reflexive, but we still have the injective map from $\ell^1$ to $\ell^2$ whose adjoint doesn't have dense range.  So under those axioms it's consistent that 2 is false.)<|endoftext|>
TITLE: How to find  examples of non-trival kernel of maps between Brauer groups Br(R) -> Br(K)
QUESTION [10 upvotes]: Background/Motivation: The facts about the Brauer groups I will be using are mainly in Chapter IV of Milne's book on Etale cohomology (unfortunately it was not in his online note). 
Let $R$ be a Noetherian normal domain and $K$ its quotient field. Then there is a natural map $f: Br(R) \to Br(K)$. In case $R$ is regular, a well known result by Auslander-Goldman says that  $f$ is injective. The natural question is when can we drop the regularity condition? But anything in the  $Cl(R)/Pic(R)$ will be in the kernel of $f$, so we need that quotient to be $0$ to make it interesting. 
Also, it is known that even assuming said quotient to be $0$, one has example like $R=\mathbb R[x,y,z]_{(x,y,z)}/(x^2+y^2+z^2)$ which is UFD, but the kernel contains (I think) the quaternion algebra over $R$. The trouble in this case is that $\mathbb R$ is not algebraically closed. In fact, if $R$ is local, then $ker(f)$ injects into $Cl(R^{sh})/Cl(R)$, here $R^{sh}$ is the strict henselization. Most of the examples with non-trivial kernel I know seem a little ad-hoc, so:
Question: Are there more general methods to generate examples of $R$ such that $f$ is not injective? 
I would love to see answers with more geometric/arithmetic flavors. I am also very interested in the positive and mixed charateristics case. Thanks in advance.

REPLY [6 votes]: As R is normal all localizations at height one primes are DVR whence regular and so the question is really one of describing the kernel of the morphism
Br(R) ---> beta(R) = \cap_P Br(R_P)   (intersection over all ht 1 primes)
beta(R) is called the 'reflexive Brauer group' (see for example ancient papers by Orzech and Hoobler, in a Brauer-group proceedings of a conference in Antwerp, in the Springer LNM 917 begin 80ties)
anyway, I've once given a cohomological description of beta(R) in this paper and also one for the exact sequence related to your question
0-->Pic(RT)--->Cl(R)-->BCl(R)--->Br(R)--->beta(R)
here BCl(R) is the 'Brauer-class group'. It consists of reflexive R-modules M such that End(M) is a projectie R-module, made into a group under the modified tensor product (see paper mentioned before).
as you will see in that paper, these cohomological descriptions have everything to do with the smooth locus of R. For example, beta(R) is the Brauer group of the regular open piece and BCl(R) can 'in principle' be calculated from cohomology with support on the singular locus.<|endoftext|>
TITLE: Estimate population size based on repeated observation
QUESTION [6 upvotes]: I take the bus to work every day.  Every bus has a serial number, but unlike in the German Tank Problem, I don't know if they are numbered uniformly $1...n$.
Suppose the first $k$ buses are all different, but on day $k+1$ I take one I've been on before.  What is the best estimate for the total number of buses?

REPLY [5 votes]: Maximum likelihood estimate is the smallest $n$ for which
$$\left( 1+\frac{1}{n} \right)^k \leq \frac{n}{n-k+1},$$
that gives a value of $n$ asymptotically equal to $\frac{k^2}{2}$, consistently with the Birthday Paradox. Not sure whether an unbiased estimate would be better for any practical purpose; maybe you do have an a priori distribution for which a Bayesian estimate makes sense?<|endoftext|>
TITLE: Effects of "weak" vs. "strict" categories in Eckmann-Hilton arguments
QUESTION [9 upvotes]: A standard example for demonstrating the need for genuinely weak n-categories is that a weak 3-category with unique 0- and 1-cells amounts to the same thing as a braided monoidal category (by an Eckmann-Hilton argument), but were one to use a strict 3-category instead, this would automatically become a fortiori symmetric. In trying to get a better intuition for "the right notion" of weak categories (this is still unsettled, right?), I was wondering if anyone could give me a good intuition for what step in the argument for symmetry in the strict case we want to fail in the weak case and why. [I suppose this amounts to more generally giving a good intuition for what higher-dimensional coherence isomorphisms we should refrain from demanding to exist in the definition of weak categories, and why, but this example seems in particular like a usefully illustrative introductory context]

REPLY [6 votes]: Here is a nice pictorial proof of the Eckmann-Hilton argument in a higher category (made by Eugenia Cheng).  One way to read this is as a proof that in a weak 2-category with one 0-cell and one 1-cell, the composition of 2-cells is commutative: in this case each diagram is equal to the two diagrams next to it, either by a unit law or by an interchange law.
Alternately, we can read it as a proof that in a weak 3-category with unique 0- and 1-cells, the composition on 2-cells is braided.  In this case each diagram is isomorphic to those next to it, via either a unit isomorphism or an interchange isomorphism.  The braiding is the composite isomorphism along the top half (say), and the fact that it isn't a symmetry arises because the composite all the way around need not be the identity.
Of course, if both interchange and unit isomorphisms were to be identities, the braiding would be a symmetry.  It is known in dimension 3 that it's actually good enough (i.e. things don't collapse and become too strict) if either the unit or interchange law is weak, and the other is strict.  If interchange is weak but the unit is strict, we get Gray-categories as in the original Gordon-Power-Street paper on tricategories, while if the interchange is weak but the unit is strict, this is the subject of Simpson's conjecture as mentioned by David (for the n=3 case, see here).  Note that associativity never enters the picture at all, and at least in dimension 3 it can also be made strict.
There isn't really a "one step" that we want to fail.  In a "fully weak" higher category, all of the associativity, unit, and interchange laws would hold only weakly.  It so happens that such fully weak categories can be "semi-strictified" to make some, but not (in general) all, of these laws hold strictly, while still being equivalent to the original category in a suitable sense.  Such semistrictification can often be technically useful, but I don't regard it as really of foundational importance.
Regarding "intuition for what higher-dimensional coherence isomorphisms we should refrain from demanding to exist," I think it's misleading to view this example in that way.  There isn't really a coherence isomorphism that we're refraining from demanding to exist---in a weak 3-category, it's still the case that all coherence isomorphisms exist and all "formally describable" diagrams commute.  Rather, what's happening is that accidentally, if we happen to be considering cells whose source and target are both identities, then there are some structural isomorphisms that we happen to be able to compose in a way "unforeseen" by the general theory of weak n-categories.  Therefore, in this particular case, the assertion of that general theory that "all diagrams commute" doesn't apply, since the diagram we're looking at is only well-defined by accident.  This is kind of vague, but it can be made precise by the theory of contractible globular operads, where it is true that "all diagrams commute" in the formal, operadic, sense, but in some particular algebra over such an operad, there may be accidental "composites" which do not commute.  See also this question.<|endoftext|>
TITLE: Can the algebraic closure of a complete field be complete and of infinite degree?
QUESTION [7 upvotes]: Yes, this is yet another "foundational" question in valuation theory.
Here's the background: it is a well known classical fact that the dimension (in the purely algebraic sense) of a real Banach space cannot be countably infinite.  The proof is a simple application of the Baire Category Theorem: see e.g. PlanetMath.
Suppose now that $(K,| \ |)$ is a complete non-Archimedean (edit: nontrivial) normed field.  One has the notion of a $K$-Banach space, and the Baire Category Theorem argument works verbatim to show that such a thing cannot have countably infinite $K$-dimension.
Now let $\overline{K}$ be an algebraic closure of $K$.  Then $\overline{K}$, by virtue of being a direct limit of finite-dimensional normed spaces over the complete field $K$, has a canonical topology, and indeed a unique multiplicative norm which extends $|\ |$ on $K$.
My question is: does there exist a complete normed field $(K, | \ |)$ such that:

$[\overline{K}:K] = \infty$ and
$\overline{K}$ is complete with respect to its norm?

As with a previous question, it is not too hard to see that this does not happen in the most familiar cases.  Indeed, by the above considerations this can only happen if $[\overline{K}:K]$ is uncountable.  But $[\overline{K}:K]$ will be countable if $K$ has a countable dense subfield $F$ [to be absolutely safe, let me also require that $F$ is perfect].  Indeed, the algebraic closure of any infinite field has the same cardinality of the field, so $\overline{F}$ can be obtained by adjoining roots of a countable collection of separable polynomials $P_i(t) \in F[t]$.  It follows from Krasner's Lemma that by adjoining to $K$ the roots of these polynomials one gets $\overline{K}$.
What about the general case?

REPLY [5 votes]: No,  there exists no such field (with a non-trivial norm).  A proof can be found in Bosch, Güntzer, Remmert: Non-Archimedean Analysis, Lemma 1, Section 3.4.3.<|endoftext|>
TITLE: Commutator of Lie derivative and codifferential?
QUESTION [15 upvotes]: Let $(M,g)$ be some smooth, Riemannian manifold. Let $d$ be the exterior derivative and $\delta$ the codifferential on forms. For a smooth vector field $X$, let $L_X$ be the Lie derivative associated to $X$. We know from Cartan formula that $L_X = d \iota_X + \iota_X d$ where $\iota_X$ is the interior derivative associated to the vector field $X$. So it is well-known that $L_X$ and $d$ commute: for any arbitrary form $\omega$, we have that $L_Xd\omega = dL_X\omega$. 
This is, of course, not true for codifferentials. In general $[L_X,\delta]\neq 0$. For certain cases the answer is well known: if $X$ is a Killing vector ($L_Xg = 0$) then since it leaves the metric structure in variant, it commutes with the Hodge star operator, and so $L_X$ commutes with $\delta$. Another useful case is when $X$ is conformally Killing with constant conformal factor ($L_X g = k g$ with $dk = 0$). In this case conformality implies that the commutator $[L_X,*] = k^\alpha *$ where $\alpha$ is some numerical power depending on the rank of the form it is acting on (I think... correct me if I am wrong), so we have that $[L_X,\delta] \propto \delta$. 
So my question is: "Is there a general nice formula for the commutator $[L_X,\delta]$?" If it is written down somewhere, a reference will be helpful. (In the Riemannian case, by working with suitable symmetrisations of metric connection one can get a fairly ugly answer by doing something like $\delta \omega \propto \mathop{tr}_{g^{-1}} \nabla\omega$ and use that the commutators $[L_X, g^{-1}]$ and $[L_X,\nabla]$ are fairly well known [the latter giving a second-order deformation tensor measuring affine-Killingness]. But this formula is the same for the divergence of arbitrary covariant tensors. I am wondering if there is a better formula for forms in particular.) 
A simple explanation of why what I am asking is idiotic would also be welcome.

REPLY [2 votes]: Closely related to your question is what is the commutator of Lie derivative and Hodge dual *. I recently cam across a nice answer to that question, in a 1984 article by Trautman, which is referenced here: http://inspirehep.net/record/206126?ln=en
The answer is $[L_X,*]\alpha = [i(h) - \frac12 Tr(h)]*\alpha$, where $\alpha$ is a form, $h$ is the 1-1 tensor defined by the Lie derivative of the metric, contracted on one index with the inverse metric (i.e., $\nabla_a X^b$, where $\nabla_a$ is the derivative operator determined by the metric) and $i(h)$ is the derivation generated by $h$ sending 1-forms to 1-forms.<|endoftext|>
TITLE: Why do people "forget" Verdier abelianization functor?(Looking for application)
QUESTION [16 upvotes]: I am now learning localization theory for triangulated catgeory(actually, more general (co)suspend category) in a lecture course. I found Verdier abelianization which is equivalent to universal cohomological functor) is really powerful and useful formalism. The professor assigned many problems concerning the property of localization functor in triangulated category.He strongly suggested us using abelianization functor to do these problems
If we do these problems in triangulated category, we have to work with various axioms TRI to TRIV which are not very easy to deal with. But if we use Verdier abelianization functor, we can turn the whole story to the abelian settings. Triangulated category can be embedded to Frobenius abelian category(projectives and injectives coincide). Triangulated functors become exact functor between abelian categories. Then we can work in abelian category. Then we can easily go back(because objects in triangulated category are just projectives in Frobenius abelian category, we can use restriction functor). In this way, it is much easier to prove something than Verider did in his book. 
My question is:

What makes me surprised is that Verdier himself even did not use Abelianization in his book to prove something. I do not know why?(Maybe I miss something) 
I wonder whether there are any non-trivial application of Verdier abelianization functor in algebraic geometry or other fields? 

Thank you

REPLY [16 votes]: There is a proof of Brown representability using the abelianisation. One can identify the abelianisation with the category of coherent functors (in the sense of M. Auslander). It is an amusing fact that Auslander and Brown were both colleagues at Brandeis in the early 1960s, and most likely not aware about the close relationship of their work.<|endoftext|>
TITLE: Is the group von Neumann algebra construction functorial?
QUESTION [16 upvotes]: Let $G$ be a group and $CG$ the complex group algebra over the field $C$ of complex number. The group von Neumann algebra $NG$ is the completion of $CG$ wrt weak operator norm in $B(l^2(G))$, the set of all bounded linear operators on Hilbert space $l^2(G)$. Let $f:G \to H$ be any homomorphism of groups. My question is: is there a homomorphism of the group von Neumann algebra $NG \to NH$ induced from $f$?.
If $NG$ is replaced with $CG$, it's obvious true. If $NG$ is replaced with $C^\ast_r(G)$, the reduced group $C^\ast$ algebra, it's not necessary true.

REPLY [13 votes]: Let $f : G \to H$ be a homomorphism of discrete groups.

The homomorphism $f$ extends to a homomorphism of reduced group $C^{\ast}$-algebras if and only of $\ker(f)$ is amenable, and extends to a homomorphism of group von Neumann algebras if and only if $\ker(f)$ is finite.<|endoftext|>
TITLE: In what degrees does Ext(S/(f),S) vanish?
QUESTION [5 upvotes]: Let $S=k[x_0,...,x_n]$ be the polynomial ring over a field $k$ and $f\in S$ non-zero and homogeneous. Is it true that $Ext^m(S/(f),S)$ is zero?
This would help me to show that $Ext^m(S/fI,S)\cong Ext^m(S/I,S)(\deg f)$ for $m\geq 2$ and a homogeneous ideal $I$ of codim $\geq 2$. I tried the following approach:
Applying the long exact sequence of $Ext$ to the exact sequence of graded $S$-modules 
$$0\to S/I\xrightarrow{\cdot f} S/fI(\deg f)\xrightarrow{\tau}S/(f)(\deg f)\to 0,$$
where $\tau$ is the canonical morphism $s+fI\mapsto s+(f)$,
brings
$$\ldots\to Ext^m(S/(f)(\deg f),S)\to Ext^m(S/fI(\deg f),S)\to Ext^m(S/I,S)\to$$
$$Ext^{m+1}(S/(f)(\deg f),S)\to Ext^{m+1}(S/fI(\deg f),S)\to Ext^{m+1}(S/I,S)\to\ldots$$
and two zeros on the left would suffice.

REPLY [6 votes]: It is not true, $Ext^1(S/(f), S)\neq 0$ as Ben pointed out. However, to prove what you want $Ext^m(S/I,S)\cong Ext^m(S/fI,S)(deg(f))$ for $m\geq 2$, just note that 
$$Ext^m(S/I,S) = Ext^{m-1}(I,S) $$ for any $I$, any $m\geq 2$ (using $0 \to I \to S\to S/I \to 0$) and $I(-deg(f)) \cong fI$ as $S$-modules.<|endoftext|>
TITLE: Algebraic equivalence VS Numerical Equivalence - An Example.
QUESTION [5 upvotes]: This question is arose from the question
 Difference between equivalence relations on algebraic cycles 
and the example 3 in lecture 1 in Mumford's book Lectures on curves on an algebraic surface.
Here is the example.
Let $E$ be an elliptic curve. A curve $C$ on $\mathbb{P}^1\times E$ is said has projection degree $(d, e)$, if the projections $C\to\mathbb{P}^1$ and $C\to E$ are of degree $d$ and $e$ respectively. In that example Mumford shows that the curves with projection degree (d,e)  and $d>0$ forms a $d(e+1)$ irreducible family of curves. Clearly, these curves are algebraic equivalence. But this family is not a linear system.
I can understand why the dimension is $d(e+1)$ with the help that numerical and algebraic equivalence coincide for divisors. But why the equivalences coincide for divisors? Where is the place discussing these equivalence of algebraic cycles? Also what is the irreducible family in this example?

REPLY [7 votes]: Numerical and algebraic equivalence coincide up to torsion for codimension $1$ cycles in non-singular projective varieties over $\mathbb C$.  Also, the group $Alg_{\tau}^1(X)/Alg^1(X)$ can be identified with $H^2(X,\mathbb Z)_{tor}$ (here $Alg_{\tau}^1(X)$ is the group of cycles who multiple is in $Alg^1(X)$). This is discussed in 19.3.1 of Fulton's book "Intersection Theory". So you just need to show that $H^2(X,\mathbb Z)_{tor}=0$, i.e it is free abelian. But presumably this follows from the Kunneth formula.<|endoftext|>
TITLE: Two definitions of Calabi-Yau manifolds
QUESTION [33 upvotes]: Why is it that the vanishing of the integral first Chern class of a compact Kahler manifold is equivalent to the canonical bundle being trivial? I can see that it implies that the canonical bundle must be topologically trivial, but not necessarily holomorphically trivial. Does proving the equivalence require Yau's theorem, in order to produce a flat connection on the canonical bundle, or is there a more elementary proof?

REPLY [12 votes]: It seems worthwile to point out that it is not true that the vanishing of the integral first Chern class of a compact Kahler manifold implies that the canonical bundle is holomorphically trivial. It only implies that it is torsion, i.e. some positive multiple is holomorphically trivial, as indicated by Misha and Dmitri (an English version of Bogomolov's paper is here). 
An example where $c_1(K_M)$ vanishes in integral cohomology while $K_M$ is not holomorphically trivial is any bielliptic surface (a finite unramified quotient of a complex $2$-torus), in which case $12K_M$ is trivial. This fact is remarked for example in this paper
of Tian-Jun Li, Remark 6.4 (see also the paper of McDuff-Salamon cited there as [30]).<|endoftext|>
TITLE: Size of the smallest group not satisfying an identity.
QUESTION [38 upvotes]: Given $F = F(x_0,\ldots,x_n)$ the free group on $n+1$ generators.  Define a function $M: F\rightarrow \mathbb{N}$ such that $F(w) = l$, if the smallest group in which $w$ is not an identity is of size $l$.
My question is what the function $M$ looks like.  Are there nice bounds?
Here are some observations that have come from earlier discussions I have had about this question.
0)if there is a subset of the generators appearing in $w$ where the sum of the exponents is nonzero, then you can use a cyclic group where the order of the group does not divide this sum of exponents as an example.
1) an upper bound of $F(w)$ is $|w|!$: you can by hand construct a permutation group in which the identity is not satisfied.  (the fact that $M$ is welldefined is equivalent to the residual finiteness of the free group)
2) the function $M$ is unbounded: every finite group $G$ on $n+1$ generators corresponds to a finite index subgroup of $F$ (a group $W\subseteq F$ for which $G = F/W$; for each $G$ there are finitely many such $W$), and the intersection of finitely many finite index subgroups is still of finite index.  So take all groups of size less then $k$, every word in the intersection of their corresponding finite index subgroups requires a group of size greater than $k$ to not be satisfied.

REPLY [21 votes]: The asymptotic version of this question raised by Bjorn Poonen has been studied by Khalid Bou-Rabee for general groups, not just free groups. That is, given G a residually finite group, for each g we can ask: how large is the smallest finite group F which detects g, meaning there exists f: G -> F so that f(g) is nontrivial? Now fix a word metric on G, and ask how the maximum of this "detection number" grows as you consider words of length at most n.
See "Quantifying residual finiteness" and "Asymptotic growth and least common multiples in groups" (with Ben McReynolds) for his results. For example, as long as G is linear, the growth function is polylog, meaning asymptotically less than (log n)k for some k, if and only if G is virtually nilpotent.
To answer your question, by considering congruence quotients of SL2Z, Bou-Rabee concludes that for every word of length n in the free group F2, there is a finite group of order O(n3) where the word is not an identity.
The same bound can be obtained uniformly as follows. Ury Hadad gives a lower bound in "On the shortest identity in finite simple groups of Lie type" which implies that the shortest identity in PSL2(Fq) has length at least (q-1)/3. Since the size of PSL2(Fq) is order q3, this implies that every word of length at most n fails to be an identity in one single group PSL2(Fq) of order O(n3)!
I learned this argument from Martin Kassabov and Francesco Matucci's paper "Bounding the residual finiteness of free groups". In it, they use a nice analysis of finite groups with elements of "large order" to construct a word of length n in the free group F2 which is trivial in every finite group of order at most O(n2/3). This improved on the lower bound of n1/3 due to Bou-Rabee and McReynolds; I believe this is now the best lower bound known.<|endoftext|>
TITLE: Do you find your students are less competent in basic algebra and arithmetic, and, if so, do you believe that this is due to overuse of calculators at an early level?
QUESTION [18 upvotes]: So first I gave my class the quiz problem: Compute $$\lim_{h\rightarrow 0} \frac{\frac{1}{3+h} - \frac{1}{3}}{h}.$$ Upon finding that they could not do that (no real surprize) I asked them to compute $\frac{1}{3.01}-\frac{1}{3}$ in hopes that they would recognize the kernel of the former problem in the latter, and in hopes of indicating that it is perfectly reasonable for an entering college student to be able to add fractions. 
A disturbingly large number of students could not perform the latter arithmetical calculation even though i had made comments about how to add fractions within class. 
I imagine my experience is not unique. And I think that the current forum may have a sufficiently large readership to deliver an informed opinion about whether or not calculator use is inhibiting algebraic ability. If it is not the calculator, then what is the cause?

REPLY [4 votes]: The OP asks for comments from university-level professors on whether (a) they have seen a decreasing trend in arithmetic skills among their students over time, and whether (b) such a trend might be attributable to the use of calculators. 1973 was roughly the last year in which one could teach freshman calculus to a group of students who had not been exposed to electronic calculators. Anyone who was teaching freshman calc in 1973 is at least ~64 years old, so at most we will have a very thin cohort of teachers who can comment on how their own students in 2012 compare to their past students who used slide rules.
It's also very risky to use anecdotal or subjective evidence to measure such trends. The best objective evidence that I'm aware of is in a book called Academically Adrift, Arum and Roksa, 2011. The authors document that certain downward trends have indeed occurred over the last 50 years. Two such trends are a marked decrease in the time students spend studying and a decrease in the amount of improvement in critical thinking and writing skills that occurs while students are in college. These trends persist even when one controls for such factors as the greater percentage of the population that now attends college.
I have been teaching physics at a community college in California since 1996. In my experience the main difference between students who have taken a calculus course and those who haven't is an increased probability that they will be fluent in basic arithmetic and algebra (e.g., being able to solve a=b/c for the variable c). This may indicate that they can't pass calculus with a C without these skills, or it may be an example of self-selection.
I find that very few students who have passed calculus can do any of the following without extensive coaching and remediation: Differentiate or integrate any function that is expressed in terms of variables other than x and y. Differentiate or integrate an expression containing symbolic constants. State the geometrical interpretations of the integral and derivative. Find the value of $x$ that maximizes $-x^2+x-2$. State under what circumstances $\Delta y/\Delta x$ is a valid measure of a rate of change, and under what circumstances $dy/dx$ is needed instead. Determine whether a car's odometer performs differentiation, or integration.
In other words, if we label the courses in a college-level math sequence with successive integers, what I find is that students who have passed course $n$ can only be counted on to display some level of competence in the material covered in course $n-3$ or $n-4$.<|endoftext|>
TITLE: Kullback-Leibler divergence of scaled non-central Student's T distribution
QUESTION [8 upvotes]: What is the Kullback-Leibler divergence of two Student's T distributions that have been shifted and scaled?  That is, $\textrm{D}_{\textrm{KL}}(k_aA + t_a; k_bB + t_b)$ where $A$ and $B$ are Student's T distributions.
If it makes things easier, $A$ could be a Gaussian.  (That is, it could have infinite degrees of freedom.)
The motivation behind this question is that the scaled non-central Student's T distribution is the posterior predictive distribution of normally distributed data with unknown mean and variance.  Thus, I would like to compare the true distribution $k_aA + t_a$ with the estimate $k_bB + t_b$.

REPLY [2 votes]: I guess I can't leave comments just yet so this will have to be in the form of an answer:
Assuming that the distributions have the same number of degrees of freedom ($n$), I think the answer will look something like (with some abuse of notation)
$\mathcal{H}(k_aA + t_a \mid k_bB + t_b) = \log(\frac{k_b}{k_a}) + \frac{n+1}{2} \left( E[ \log(1 + (\frac{Yk_a + t_a - t_b}{k_b})^2 \frac{1}{n}) ] - E[ \log (1 + \frac{Y^2}{n}) ]\right) $,
where the r.v. $Y$ has the Student's t distribution with $n$ degrees of freedom. If the number of degrees of freedom are assumed to be different between the two distributions you will get additional terms that correspond to the logarithm of the "$\Gamma$-parts" of the density functions and in the above formula the $n$'s will be changed accordingly. Expectation will still be w.r.t. the Student's t with $n$ deg. of freedom (or whatever degree that is associated with the r.v. $A$).<|endoftext|>
TITLE: Is there a matrix C so that the trace of C^n is dense in R?
QUESTION [15 upvotes]: I am looking for a matrix $C$ so that the sequence $tr(C^n)$ is dense in the set of real numbers. Equivalently (in the $2 \times 2$ case), find a complex number $z$ so that the sequence $z^n+w^n$ is dense in $\mathbb{R}$ where $w$ is the conjugate of $z$.

REPLY [5 votes]: Bjorn has already answered this question in the affirmative, and shown that such matrices do exist. I'd like to add a further comment here though - 'almost no' matrices satisfy the required property. That is, the collection of 2x2 matrices such that Tr(C^n) is dense in R has zero Lebesgue measure.
We know that Tr(C^n) = a^n + b^n where a,b are the roots of the characteristic polynomial of C. If a and b are both real then it is not possible for C to have the required property. The only possibility is where they are complex conjugates, a = r exp(iθ), b = r exp(-iθ) for r >1. Then, Tr(Cn)=2rcos(nθ). Suppose that θ is uniformly distributed over [-π,π], so that exp(inθ) is uniformly distributed on the unit circle for each n. For any positive K, |Tr(C^n)|<K is equivalent to |cos(nθ)|<r-nK/2. The set of values of exp(inθ) for which this holds forms a pair of arcs of length r -nK (to leading order). So,
$$\mathbb{P}(\vert{\rm Tr}(C^n)\vert\lt K)\approx r^{-n}K/\pi$$
to leading order. Summing over n, this is finite. Then, the Borel-Cantelli lemma says that, with probability one, |Tr(Cn)|<K only finitely often. So, with probability 1, |Tr(Cn)| diverges to infinity.<|endoftext|>
TITLE: Linear Algebra Problems?
QUESTION [23 upvotes]: Is there any good reference for difficult problems in linear algebra? Because I keep running into easily stated linear algebra problems that I feel I should be able to solve, but don't see any obvious approach to get started.
Here's an example of the type of problem I am thinking of: Let $A, B$ be $n\times n$ matrices, set $C = AB-BA$, prove that if $AC=CA$ then $C$ is nilpotent. (I saw this one posed on the KGS Go Server)
Ideally, such a reference would also contain challenging problems (and techniques to solve them) about orthogonal matrices, unitary matrices, positive definiteness... hopefully, all harder than the one I wrote above.

REPLY [3 votes]: If you happen to know a little bit of Italian, another good resource is Problemi risolti di algebra lineare, by Broglia, Fortuna, Luminati.
(By the way, if you have never done it, reading a math book in another language is often easier than it seems at first sight)<|endoftext|>
TITLE: cocompact discrete subgroups of SL_2
QUESTION [8 upvotes]: How can one construct families of cocompact discrete subgroups of the topological group $\text{SL}_2(\mathbb{C})$? 
Here quaternion algebra's might help, I believe, but I have some difficulties with the construction.

REPLY [10 votes]: Let $k$ be a number field with one complex place and let $B$ be a quaternion algebra defined over $k$ which ramifies at every real place of $k$ (and perhaps some finite places as well - Hilbert reciprocity implies that the total number of ramified places must be even). Let $\mathcal{O}$ be an order of $B$ and $\mathcal{O}^1$ its elements of reduced norm $1$. Choosing an embedding $k\hookrightarrow \mathbb{C}$ induces an embedding $B\hookrightarrow M_2(\mathbb{C})$, which in turn restricts to an embedding $\Psi: \mathcal{O}^1\hookrightarrow SL_2(\mathbb{C})$. A group which is commensurable with $\Psi(\mathcal{O}^1)$ is called arithmetic. An arithmetic group is cocompact if and only if the quaternion algebra $B$ is a division algebra (by Wedderburn's Theorem this is equivalent to saying that the set of places of $k$ which ramify in $B$ is nonempty).<|endoftext|>
TITLE: Weakened conditions for étale + X implies faithfully flat.
QUESTION [5 upvotes]: Let $F:R \to S$ be an étale morphism of rings.  It follows with some work that $f$ is flat.  
However, faithful flatness is another story.  It's not hard to show that faithful + flat is weaker than being faithfully flat.  An equivalent condition to being faithfully flat is being surjective on spectra.  
The question:
Is there any further condition we can require on an étale morphism that implies faithful flatness?
"Faithfully flat implies faithfully flat" or "surjective on spectra is equivalent to faithfully flat" do not count.  The answer should in some way use the fact that the morphism is étale (or at least flat).  
As you can see by the tag, all rings commutative, unital, etc.
Edit: Why faithfully flat is weaker than faithful + flat.
Edit 2: I resent the voting down of this question without accompanying comments as well as the voting up of the glib and unhelpful answer below.  It's clear that some of you are in the habit of voting on posts based on the poster rather than the content, and I think that is shameful.  There is nothing I can do because none of you has the basic decency to at least leave a comment.  I am completely at your mercy.  You've won.  I hope it's made you very happy.
Edit 3: To answer Emerton's comment, I asked here after:
a.) Reading this post by Jim Borger 
b.) Asking my commutative algebra professor in an e-mail
Which led me to believe (perhaps due to a flawed reading of said sources) that this was a harder question than it turned out to be.

REPLY [6 votes]: so yeah, look, i was trying to be funny & also trying to highlight the absurdly haughty nature of the caveats in the question.  to be serious i would say that if F is etale then it is faithfully flat iff it is surjective on separably-closed field valued points, but also remark that the same is true with "etale" replaced by "smooth", so i guess i'm not using the full strength of the etale condition.<|endoftext|>
TITLE: Number theory textbook based on the absolute Galois group?
QUESTION [29 upvotes]: I've just finished reading Ash and Gross's Fearless Symmetry, a wonderful little pop mathematics book on, among other things, Galois representations.  The book made clear a very interesting perspective that I wasn't aware of before: that a large chunk of number theory can be thought of as a quest to understand $G = \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.  For example, part of the reason to study elliptic curves is to describe two-dimensional representations of $G$, and reciprocity laws are secretly about ways to describe the traces of Frobenius elements in various representations.  (That's awesome!  Why didn't anybody tell me that before?)
Are there number theory textbooks (presumably not introductory, but hopefully not too sophisticated either) which explicitly take this as a guiding principle?  I think this is a great idea to organize things like quadratic reciprocity around and I'm wondering if anybody has decided to actually do that at the undergraduate (or introductory graduate, maybe) level.
Edit:  In response to some comments and at least one downvote, most of the other questions on MO about the absolute Galois group that I can find are about the state of the art, and the references in them seem fairly sophisticated.  But it seems to me there are still interesting things to say along the lines of Fearless Symmetry, but directed to an undergraduate or introductory graduate-level audience as a kind of "second course in number theory."  I'm imagining a textbook like Serre's Course in Arithmetic.

REPLY [8 votes]: In the 1-dimensional case we are talking about class field theory, and I can't recall having seen anything better than Cox's book (already mentioned by Emerton) in this direction (undergrad etc.). With Artin's reciprocity law (the central role of the Frobenius elements in connection with reciprocity laws wasn't exactly kept secret 8-) under your belt you might want to check out

G. Shimura, A reciprocity in law in non-solvable extensions,
J. Reine Angew. Math. 221, 209-220 (1966)

and

J. Velu, Lois de reciprocite liees aux courbes elliptiques,
Semin. Delange-Pisot-Poitou, 14e annee 1972/73, Fasc. 1, 2, Expose 9, 5 p. (1973)

both of which are rather nice to read, and then continue with the articles by Serre and Swinnerton-Dyer that Emerton suggested (actually they were motivated by concrete problems concerning Ramanujan's $\tau$-function, and they've written quite a bit about it at the time).<|endoftext|>
TITLE: how good an approximation to the equivariant derived category is given by the Grassmannian filtration of the classifying space?
QUESTION [5 upvotes]: So, let's say one has an action of $GL_n$ on an algebraic variety $X$ over a field $k$, and two objects $F,G$ in the equivariant derived category (i.e., the derived category of constructible sheaves on the stack $X/GL_n$).
For each integer $m$, let $Y_m$ be the space of injective maps of $k^n\to k^m$ and let $X_m=(Y_m\times X)/GL_n$ (with the diagonal action, as usual).  Note that we have a map $p_m:X_m\to X/GL_n$.  
Now, it's a fact that $Hom_{X/GL_n}(F,G)$ injects into the inverse limit $\varprojlim Hom_{X_m}(p_m^*F,p_m^*G)$, but it usually isn't injective for any given $m$.

Can anything precise be said about how fast this kernel shrinks?  

The most boring case is when $F$ and $G$ are both the constant sheaf on a point.  Then $Hom_{X/GL_n}(F,G)=H^*(BGL_n)$, the cohomology of the classifying space and $Hom_{X_m}(p_m^*F,p_m^*G)=H^*(Gr(m,n))$, the cohomology of the Grassmannian of $n$-planes in $m$-space.  In this case the kernel is pretty well understood.
Ideally, the kernel in general would simply come from this case:  i.e. these cohomology rings act on the right and left no matter what $X$ is, and the kernel might be generated by multiplying maps by classes in the kernel of the map from $H^*(BGL_n) \to H^*(Gr(m,n))$, the map from the cohomology of the classifying space to the cohomology of the Grassmannian.  This seems like a reasonable statement, but I'm not sure where to look for it.

REPLY [2 votes]: Something related (but not exactly what you are asking) is covered in some detail in Bernstein-Lunts (around 2.2.3).
Call a map $\pi : P \to X$ $n$-acyclic if for any sheaf $F$ the truncated adjunction morphism $F \to \tau_{\le n} \pi_* \pi^* F$ is an isomorphism.
Given an $n$-acyclic $G$-equivariant map $P \to X$ where $P$ has free $G$-action one has a map $p: P/G \to X/G$ (keeping your notation).
Then, if $F$ and $G$ have cohomology sheaves concentrated in an interval $I$ with $|I| < n$ then the natural map from $Hom (F, G) \to Hom(p^* F, p^* G)$ is an isomorphism. (Here, in contrast to your usage in the question, $Hom$ means only degree zero homomorphisms).
Two comments:

(depending on your definition) $Hom(F,G)$ is defined to be $\varprojlim  Hom(p_m^* F, p_m^* G)$ and so the statement "is injective" is a bit misleading!
in geometric representation theory the objects on $pt/G$ that are being considered are often direct sums of equivariant constant sheaves (eg if one takes the equivariant intersection cohomology of a projective variety) in which case your description of the kernel works just fine!<|endoftext|>
TITLE: Topological description of Manifold with boundary
QUESTION [5 upvotes]: Is there any 'general' topological invariant to tell the difference between $M$ and $N$, where $M$ has homotopy type of a closed manifold and $N$ has homotopy type of a manifold with boundary. 
I meant something like homotopy group/ homology group can 'detect' the obstruction of being homotopic to a closed manifold.

REPLY [4 votes]: Perhaps the fundamental algebraic difference between a compact manifold with boundary and one without boundary is whether it satisfies Poincare duality or Lefschetz duality. Indeed, in algebraic surgery theory one defines the boundary of a chain complex as a "correction term" to a "Poincare duality" which does not hold, to turn it into a Lefschetz duality which does hold. This definition is fundamental to the discipline.
In more detail, let R be a ring with involution and consider a bounded chain complex over R
$C_\bullet:=C_0\longrightarrow C_{1}\longrightarrow\cdots\longrightarrow C_n.$
The dual cochain complex is 

$C^\bullet:=C^n\longrightarrow C^{n-1}\longrightarrow\cdots\longrightarrow C^0,$
where by definition $C^i:= Hom_R(C_i,R)$.
Poincare duality PD:Hk(X)$\longrightarrow$ Hn-k(X) is induced by taking the cap product with a fundamental class $[X]\in C_n$. On a chain level, the cap product is given by the composition
$C^\bullet\overset{\Delta}{\longrightarrow }(C^\bullet)^{op}\otimes C^\bullet(X)\overset{\setminus}{\longrightarrow} Hom_R(C^\bullet,C_\bullet),$
where the first map is induced from the diagonal map via the Eilenberg-Zilber theorem, and the second is the slant map. The traditional name for $\setminus\Delta[X]$ is $\varphi_0$, which is a chain map from $C^\bullet$ to $C_\bullet$. Algebraic surgery theory also defines chain maps $\varphi_1,\ldots,\varphi_n$, which fit together to form a collection {$\varphi_0,\ldots,\varphi_n$} called a symmetric structure on the chain complex $C_\bullet$. The boundary of a chain complex with symmetric structure is defined as the mapping cone of $\varphi_0$. That's your invariant- no boundary if and only if that mapping cone is chain homotopic to zero.
This leads to definitions of L-groups, algebraic cobordism, algebraic surgery, and all kind of good and deep mathematics which goes along with all of that.
If you want to know more, the canonical reference for this sort of stuff perhaps, which has sort of a cult status, is Andrew Ranicki's ATS1.<|endoftext|>
TITLE: fpqc covers of stacks
QUESTION [22 upvotes]: Artin has a theorem (10.1 in Laumon, Moret-Bailly) that if $X$ is a stack which has separated, quasi-compact, representable diagonal and an fppf cover by a scheme, then $X$ is algebraic.  Is there a version of this theorem that holds for fpqc covers?

REPLY [38 votes]: It is false.  I'm not sure what the comment about algebraic spaces has to do with the question, since algebraic spaces do admit an fpqc (even \'etale) cover by a scheme.  This is analogous to the fact that the failure of smoothness for automorphism schemes of geometric points is not an obstruction to being an Artin stack.  For example, $B\mu_n$ is an Artin stack over $\mathbf{Z}$ even though $\mu_n$ is not smooth over $\mathbf{Z}$ when $n > 1$.  Undeterred by this, we'll make a counterexample using $BG$ for an affine group scheme that is fpqc but not fppf.  
First, we set up the framework for the counterexample in some generality before we make a specific counterexample.  Let $S$ be a scheme and $G \rightarrow S$ a $S$-group whose structural morphism is affine.  (For example, if $S = {\rm{Spec}}(k)$ for a field $k$ then $G$ is just an affine $k$-group scheme.)  If $X$ is any $G$-torsor for the fpqc topology over an $S$-scheme $T$ then the structural map $X \rightarrow T$ is affine (since it becomes so over a cover of $T$ that splits the torsor).  Hence, the fibered category $BG$ of $G$-torsors for the fpqc topology (on the category of schemes over $S$) satisfies effective descent for the fpqc topology, due to the affineness requirement.  
The diagonal $BG \rightarrow BG \times_S BG$ is represented by affine morphisms since for any pair of $G$-torsors $X$ and $Y$ (for the fpqc topology) over an $S$-scheme $T$, the functor ${\rm{Isom}}(X,Y)$ on $T$-schemes is represented by a scheme affine over $T$. Indeed,  this functor is visibly an fpqc sheaf, so to check the claim we can work locally and thereby reduced to the case $X = Y = G_T$ which is clear.  
Now impose the assumption (not yet used above) that $G \rightarrow S$ is fpqc.  In this case I claim that the map $S \rightarrow BG$ corresponding to the trivial torsor is an fpqc cover.  For any $S$-scheme $T$ and $G$-torsor $X$ over $T$ for the fpqc topology, the functor
$$S \times_{BG} T = {\rm{Isom}}(G_T,X)$$
on $T$-schemes is not only represented by a scheme affine over $T$ (namely, $X$) but actually one that is an fpqc cover of $T$.  Indeed, to check this we can work locally over $T$, so passing to a cover that splits the torsor reduces us to the case of the trivial $G$-torsor over the base (still denoted $T$), for which the representing object is $G_T$.  
So far so good: such examples satisfy all of the hypotheses, and we just have to prove in some example that it violates the conclusion, which is to say that it does not admit a smooth cover by a scheme. Take $S = {\rm{Spec}}(k)$ for a field $k$, and let $k_s/k$ be a separable closure and $\Gamma = {\rm{Gal}}(k_s/k)^{\rm{opp}}$. (The "opposite" is due to my implicit convention to use left torsors on the geometric side.) Let $G$ be the affine $k$-group that "corresponds" to the profinite group $\Gamma$ (i.e., it is the inverse limit of the finite constant $k$-groups $\Gamma/N$ for open normal $N$ in $\Gamma$).  To get a handle on $G$-torsors, the key point is to give a more concrete description of the ``points'' of $BG$.  
Claim: If $A$ is a $k$-algebra and $B$ is an $A$-algebra, then to give a $G$-torsor structure to ${\rm{Spec}}(B)$ over ${\rm{Spec}}(A)$ is the same as to give a right $\Gamma$-action on the $A$-algebra $B$ that is continuous for the discrete topology such that for each open normal subgroup $N \subseteq \Gamma$ the $A$-subalgebra $B^N$ is a right $\Gamma/N$-torsor  (for the fpqc topology, and then equivalently the \'etale topology).
Proof: Descent theory and a calculation for the trivial torsor.  QED Claim
Example:  $A = k$, $B = k_s$, and the usual (right) action by $\Gamma$.  
Corollary: If $A$ is a strictly henselian local ring then every $G$-torsor over $A$ for the fpqc topology is trivial.
Proof:  Let ${\rm{Spec}}(B)$ be such a torsor.  By the Claim, for each open normal subgroup $N$ in $\Gamma$, $B^N$ is a $\Gamma/N$-torsor over $A$.  Since $A$ is strictly henselian, this latter torsor is trivial for each $N$.  That is, there is a $\Gamma/N$-invariant section $B^N \rightarrow A$.  The non-empty set of these is finite for each $N$, so by set theory nonsense with inverse limits of finite sets (ultimately not so fancy if we take $k$ for which there are only countably many open subgroups of $\Gamma$) we get a $\Gamma$-invariant section $B \rightarrow A$.  QED Corollary
Now suppose there is a smooth cover $Y \rightarrow BG$ by a scheme.  In particular, $Y$ is non-empty, so we may choose an open affine $U$ in $Y$.  I claim that $U \rightarrow BG$ is also surjective.  To see this, pick any $y \in U$ and consider the resulting composite map
$${\rm{Spec}} \mathcal{O}_{Y,y}^{\rm{sh}} \rightarrow BG$$
over $k$. By the Corollary, this corresponds to a trivial $G$-torsor, so it factors through  the canonical map ${\rm{Spec}}(k) \rightarrow BG$ corresponding to the trivial $G$-torsor. This latter map is surjective, so the assertion follows.  Hence, we may replace $Y$ with $U$ to arrange that $Y$ is affine.  (All we just showed is that $BG$ is a quasi-compact Artin stack, if it is an Artin stack at all, hardly a surprise in view of the (fpqc!) cover by ${\rm{Spec}}(k)$.) 
OK, so with a smooth cover $Y \rightarrow BG$ by an affine scheme, the fiber product
$$Y' = {\rm{Spec}}(k) \times_{BG} Y$$
(using the canonical covering map for the first factor) is an affine scheme since we saw that $BG$ has affine diagonal. Let $A$ and $B$ be the respective coordinate rings of $Y$ and $Y'$, so by the Claim there is a natural $\Gamma$-action on $B$ over $A$ such that the $A$-subalgebras $B^N$ for open normal subgroups $N \subseteq \Gamma$ exhaust $B$ and each $B^N$ is a $\Gamma/N$-torsor over $A$.  But $Y' \rightarrow {\rm{Spec}}(k)$ is smooth, and in particular locally of finite type, so $B$ is finitely generated as a $k$-algebra.  Since the $B^N$'s are $k$-subalgebras of $B$ which exhaust it, we conclude that $B = B^N$ for sufficiently small $N$.  This forces such $N$ to equal $\Gamma$, which is to say that $\Gamma$ is finite. 
Thus, any $k$ with infinite Galois group does the job.  (In other words, if $k$ is neither separably closed nor real closed, then $BG$ is a counterexample.)<|endoftext|>
TITLE: Computing fundamental groups and singular cohomology of projective varieties
QUESTION [35 upvotes]: Are there any general methods for computing fundamental group or singular cohomology (including the ring structure, hopefully) of a projective variety (over C of course), if given the equations defining the variety?
I seem to recall that, if the variety is smooth, we can compute the H^{p,q}'s by computer -- and thus the H^n's by Hodge decomposition -- is this correct? However this won't work if the variety is not smooth -- are there any techniques that work even for non-smooth things?
Also I seem to recall some argument that, at least if we restrict our attention to smooth things only, all varieties defined by polynomials of the same degrees will be homotopy equivalent. The homotopy should be gotten by slowly changing the coefficients of the polynomials. Is something like this true? Does some kind of argument like this work?

REPLY [25 votes]: This is an interesting question. To repeat some of the earlier answers, one should be able to get one's hands on a triangulation algorithmically using real algebro-geometric methods, and thereby compute singular cohomology and (a presentation for) the fundamental group. But this should probably be a last resort in practice. For smooth projective varieties, as people have noted, one can compute the Hodge numbers by writing down a presentation for the sheaf p-forms and then apply standard Groebner basis techniques to compute sheaf cohomology. This does work pretty well on a computer. For specific classes, there are better methods. For smooth complete intersections, there is a generating function for Hodge numbers due to Hirzebruch (SGA 7, exp XI), which is extremely efficient to use.
As for the fundamental group, if I had to compute it for a general smooth projective variety, I would probably use a Lefschetz pencil to write down a presentation.
For singular varieties, one can still define Hodge numbers using the mixed Hodge structure on cohomology. The sum of these numbers are still the Betti numbers. I expect these Hodge numbers are still computable, but it would somewhat unpleasant to write down a general algorithm. The first step is to build a simplicial resolution using resolution of singularities. My colleagues who know about resolutions assure me that this can be done algorithmically now days.
(This is my first reply in this forum. Hopefully it'll go through.)<|endoftext|>
TITLE: Database of finite presentations of used groups
QUESTION [11 upvotes]: Do You know any kind of database of presentations of groups? 
It may be on-line or off-line in form of tables, ideally case would be integrated in some Computer Algebra System. I am interested the most in infinite group presentations, but feel free to put here also information about tables of presentations of finite groups. 
Maybe this thread should be wiki-type because probably there is many good answers to this question, and it is hard to compare for example software system with some kind of book or publication about this matter? I add biglist tag in a hope that it would appear;-)

REPLY [2 votes]: In my dissertation I developed an algorithm for finding bounds on $H_2$ of a finitely presented group with finite field coefficients. I was motivated by a conjecture Quillen on the (co)homology of linear groups. As such, I included an appendix with presentations of several linear groups and the homology calculations using my algorithms. I didn't include the list of presentations for publication, but if these types of groups are of interest I could get it to you.<|endoftext|>
TITLE: Can projective hypersurfaces contain linear spaces? How big?
QUESTION [10 upvotes]: I am in this, rather friendly, situation:
I have a complex projective space $\mathbb{P}^n$, and there i have a (possibly non-smooth) hypersurface $S$ defined by one irreducible polynomial $P$ of degree $d$.
What i want is to get information about the existence or not of linear subvarieties of $S$, and their maximal dimension $m$.
I seem to remember the existence of some ways to get bounds on $m$, given $n$ and $d$, but i don't remember anymore and i don't know where to look..

REPLY [13 votes]: I think the key word you are looking for is Fano schemes of $S$. See this 
note by Jason Starr for a reference. For example, if $S$ is smooth and non-degenerate, it can not contain a linear subspace of dimension bigger than half $dim \ S$.
A very interesting related question is when can you find subvarieties of $S$ which was spanned in the Chow group by images of linear subspaces in $\mathbb P^n$. It was discussed in this paper by Levine, Esnault and Viehweg.

REPLY [9 votes]: I agree with Hailong, you want to look at Fano schemes.  Here's an article by Debarre and Manivel.  It's very good, and I've gotten a lot out of it despite note knowing much French.<|endoftext|>
TITLE: Maximum Order of elements in $GL(n,Z)$
QUESTION [8 upvotes]: Hi,
I know that $\mathrm{GL}(n,\mathbb{Z})$ has an element of order $m$ iff $\Phi(m)\leq n$, where $\Phi(m) = \varphi(m)$ if $p_1^{\alpha_1}\neq 2$ or $m=2$, $\Phi(m) = \varphi(m)-1$ if $p_1^{\alpha_1}= 2$ or $m\not=2$, and $\varphi$ is Euler totient.
From there I can show that the maximum order of a element of finite order in $\mathrm{GL}(n,\mathbb{Z})$, $f(n)$ statisfies $\displaystyle \lim_{n\to \infty} \frac{\ln f(n)}{\sqrt{n\ln n}} =1$.
Here is my question:
Can we find the asymptotic behavior of $f(n)$ (and not $\ln \bigl(f(n)\bigr)$)?

REPLY [5 votes]: See MR1655470 (99m:20111) on MathSciNet.<|endoftext|>
TITLE: Hopfian and Co-Hopfian groups (examples)
QUESTION [9 upvotes]: Hi,
I'm looking for examples of groups that are both Hopfian and Co-Hopfian. I have a non trivial (and beautiful, at least to me) example: $\mathrm{SL}(n,\mathbb{Z})$ (with $n>2$).
Do you know others (non trivial)?
Thank you.

REPLY [2 votes]: Torsion-free finitely generated nilpotent groups are Hopfian. Although the easiest (nontrivial) ones (such as abelian ones, Heisenberg, those occurring in unipotent radicals of parabolics of reductive groups...) are not cohopfian, many are cohopfian too.
An iff condition is that the Malcev completion (say, the complex one, but it works equally with the rational or real one) has a nontrivial non-negative grading. See my paper. The initial observation showing their existence, with a sufficient (but not necessary) condition, namely that the Malcev completion is characteristically nilpotent, appears in I. Belegradek arXiv link and earlier in D. Segal's MR review of Smith, Geoff C. Compressibility in nilpotent groups.
Bull. London Math. Soc. 17 (1985), no. 5, 453–457. Quoth:

It is perhaps not so well known that finitely generated torsion-free nilpotent groups need not be compressible: this follows from J. L. Dyer's construction of a nilpotent Lie algebra all of whose automorphisms are unipotent [Bull. Amer. Math. Soc. 76 (1970), 52–56.

(In turn, Dyer 1970 is not the original reference for the latter fact, but Dixmier, J.; Lister, W., Derivations of nilpotent Lie algebras.
Proc. Amer. Math. Soc. 8 (1957), 155–158.)<|endoftext|>
TITLE: Does formally etale imply flat for noetherian schemes?
QUESTION [21 upvotes]: This is a followup to an earlier question I asked: Does formally etale imply flat? After some remarks I received on MO I noticed that this was answered to the negative by an answer to an earlier question Is there an example of a formally smooth morphism which is not smooth. However, the simple example involves a non-noetherian ring (in fact, a perfect ring; these are seldom noetherian unless they are a field). 
So my challenge is to provide an example of a formally etale map of noetherian schemes which is not flat, or otherwise proof that for maps of noetherian schemes formally etale implies flat.

REPLY [41 votes]: Every formally smooth morphism between locally noetherian schemes is flat; this is a deep result of Grothendieck.  Indeed, the formal smoothness is preserved by localization on target and then likewise on the source, so we can assume we're deal with a local map between local noetherian rings. By EGA 0$_{\rm{IV}}$, 19.7.1 (also proved near the end of Matsumura's book on commutative ring theory) a formally smooth local map between local noetherian rings is flat.<|endoftext|>
TITLE: Some questions on the intersection theory on a Hilbert scheme of points of a surface.
QUESTION [6 upvotes]: If $\Sigma$ is a smooth complex curve in a smooth projective surface $X$, then we can consider the homology class represented by $\Sigma^{[n]} \subset X^{[n]}$. $\ \ $ Where, $X^{[n]}, \Sigma^{[n]}$ stand for the Hilbert scheme of $n$-points on $X$ and $\Sigma$, respectively.  Is it possible to construct a  homomorphism    function $\Phi_n: \rm{H}_2(X) \rightarrow H_w(X^{[n]})$, such that $[\Sigma] \mapsto [ \Sigma^{[n]} ]$?   
$\ \ \ $   One has the following at ones disposal:  we have the obvious quotient map $X^n \rightarrow S^nX$ (where $S^nX$ is the symmetric product of $X$).  Now, if $\beta \in H_2(X)$, then we can consider the image of $B := \beta \times \cdots \times \beta$ in $H_{2n}(S^nX)$.  If $\beta $ can be represented by an algebraic curve, we can take the proper transform of $B$ under the Chow map $X^{[n]} \rightarrow S^nX$.  If $\beta$ is not represented by such a curve, is there anything akin to proper transform that one can apply to $B$ to construct the desired homomorphism  function $\Phi_n$?   
I am interested in studying the intersection theory between the classes $\Phi_n(\beta)$.  Nakajima in his book "Lectures  on  Hilbert  schemes  of points on surfaces" states the following nice result.  If $\Sigma$ and $\Sigma'$ are two smooth curves in $X$, then (page 102 of Nakajima's book):
$$\sum_n z^n \ [\Sigma^{[n]}] \cdot [\Sigma'^{[n]}] = (1+z)^{[\Sigma] \cdot [\Sigma']}$$
Does anyone know if there are related results for singular curves?  
As a side remark. the above formula is obvious if $\Sigma$ and $\Sigma'$ are two curves intersecting transversely.  All it says is that of the set of $m = [\Sigma]\cdot [\Sigma']$ points were it intersects, we choose $n$ of them (there are $\binom{m}{n}$ of these guys, which is what the formula is giving).  But the general proof of the formula is more intricate - one uses a representation  of the Heisenberg group on the space $\oplus_n H_*(X^{[n]})$ to derive it. This fancy shmancy  approach is more helpful when computing things like the self intersection of $\Sigma^{[n]}$ when $\Sigma$ is a $(-1)$-curve in $X$.  From it we get that $[\Sigma^{[n]}] \cdot[\Sigma^{[n]}] =  \binom{-1}{n} = (-1)^n$     
EDITED:  In view of Nakajima's comment below, please replace function for homomorphism when reading the above question.  Notice that, as stated in my comment below, the extension of the map $[\Sigma] \rightarrow [\Sigma^{[n]}]$ should be a "nice" one. 
EDITED (I am copying my hidden comments here since their maths don't display well)
I can explain my motivation. I am working with some moduli spaces of objects on a surface $X$ and out of them I get a homology class $V_n$ in $X^{[n]}$. In nice cases, one can show that these homology classes are $[\Sigma^{[n]}]$, for some curve $\Sigma \subset X$. Or a sum of such classes. Using this classes $V_n$ I am trying to obtain a map $N : H_2(X) \rightarrow \mathbb{Z}$, defined by $N(\beta) := V_n \cdot \Phi_n(\beta)$. Such that, in the nice case when $V_n = [\Sigma^{[n]}]$ and $\beta = [\Sigma']$, then $$N(\beta) = [\Sigma^{[n]}] \cdot [\Sigma'^{[n]}]$$ 
Then, my problem became what should be the definition of $\Phi_n(\beta)$, when $\beta$ not represented by a curve. Presumably, we should be able to extend $\Phi_n$ to some 2-classes that are not represented by curves since, by perturbing the complex structure, we could start seeing more curves than before. I don't know what should be $\Phi_n(-2H)$. The best I could imagine is that it should satisfy the equation $$[\Sigma^{[n]}] \cdot \Phi_n(-2H) = \binom{\Sigma \cdot (-2H)}{n} $$ but I really don't know what it should be. Thanks a lot again!
EDIT  I am now assume that the formula
$$\alpha \mapsto  exp\left( \sum \frac{z_i P_\alpha[-i]}{(-1)^{i-1}i} \right) \cdot 1 $$ 
(the definition of the term $P_\alpha[-i]$ can be found in Prof. Nakajima's book "Lectures on Hilbert schemes of points on surfaces" page 84), is well defined.  By one of his results, $[\Sigma] \mapsto \sum z^i [\Sigma^{[n]}]$ (op. cit. page 99). If so, I presume this satisfy the posed question.

REPLY [2 votes]: I would like to make one naive suggestion, related to the work of
Voisin, who constructed Hilbert scheme of every almost complex manifolds $X$ of real dimension $4$. And to a more recent work of Julien Grivaux on this topic
http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.0119v1.pdf
As far as I understand Julien can express the cohomology ring of this scheme in the case when $X$ is a symplectic manifold (Theorem 1.1). Symplectic structure is used in order to construct symplectic surfaces on $X$. These surfaces are quite plentiful on $X$ by a theorem of Donaldson. For any such symplectic surface $C$ one can find an almost complex sturcture on $X$, integrable in a neighborhood of $C$. This is discussed Section 3.2 (in particular Theorem 3.5 and Corollary 3.3).
Now if we have a smooth almost complex curve $C$ inside of $X$, such that the complex structure is integrable in its neighborhood then $C^{[n]}$ seem to be well defined as a submanifold of $X^{[n]}$ (the Voisin Hilbert scheme). Moreover, Lemma 3.8 seem to suggest that the homology class of $[C^{[n]}]$ is given by a formula analgous to the one in the case when $X$ is an algebraic surface (the formula of Nakajima).
I have an extremely superficial understanding of Julien's artice and of the whole topic, so I may make some here some silly mistake. Moreover it is not clear for me if after all the class $[C^{[n]}]$ in homologies of $X^{[n]}$ depends only on the homolgy class  $[C]$ in $H_2(X)$, but will not depend on the choice of a sympectic curve $C$ that realises it.
And, of course, there are some obvious (and not obvious) restictions on $C$, such as $\int_C \omega>0$.<|endoftext|>
TITLE: subgroups of graph groups
QUESTION [6 upvotes]: Let $G$ be a graph group and let $S$ be a finitely generated subgroup of $S$.
What torsion can $H_1(S)$ have?
Let me put this in context: Let $Y$ be a graph, then the corresponding graph group has a generator for each vertex and for each edge we add the relation that the corresponding generators commute. Such groups are also known as right angled Artin groups.
If $Y$ has no edges the graph group is the free group, if $Y$ is a complete graph, then the graph group is a free abelian group.
It is known by work of Crisp and Wiest that non-orientable surface groups embed in graph groups. Such surface groups can have 2-torsion in their abelianization. Is that the only type of torsion which can appear in the abelianization of the subgroup of a graph group?
My interest in this question comes from the recent announcement by Dani Wise that `most' hyperbolic 3-manifold groups embed in a graph group.

REPLY [7 votes]: This isn't a proper answer, but here are some remarks.

Wise's work actually purports to tell you that most (all?) hyperbolic 3-manifold groups virtually embed in graph groups - ie, a finite-index subgroup embeds.  So you might lose a lot of torsion when you pass to this finite-index subgroup.
Haglund and Wise have a very nice criterion for injecting fundamental groups of cube complexes into graph groups.  It would be completely reasonable to try to use this to construct interesting examples of torsion in the homology of subgroups of graph groups.
Because the embeddings in Haglund--Wise are quasiconvex, it follows from work of Haglund that every such subgroup H is a virtual retract - that is, there is a finite-index subgroup K that contains H and the inclusion map has a left inverse K->H.  In particular, any torsion you see in the homology of H will also show up in the homology of K.  It would be easy to use a computer algebra package like GAP to look for torsion in finite-index subgroups of graph groups.

UPDATE:
In a comment below, Stefan points out a very nice way of constructing torsion in finite-index subgroups of graph groups, due to Bridson and Miller.  As I'm familiar with the argument, I'll explain it.  The graph group in question will always be a direct product of two free groups.
Let Q be any group you like and let q:F->Q be a surjection from a free group.  Now take the fibre product K of q with itself; in other words, consider the subgroup
$K= \{(g,h)\in F\times F\mid q(g)=q(h)\}~.$
There is a standard five-term exact sequence in homology that derives from the map $q:F\to Q$, which reduces to
$0\to H_2(Q)\to H_0(Q,H_1(\ker f))\to H_1(F)\to H_1(Q)$
where, by definition, $H_0(Q,H_1(\ker f))$ is the quotient of the abelianisation of $\ker f$ by the natural action of $Q$ by conjugation.  On the other hand, projecting onto a factor decomposes $K$ as $\ker f\rtimes F$ (where the action is, again, by conjugation), from which it follows that
$H_1(K) = H_0(Q,H_1(\ker f))\oplus H_1(F)~.$
Putting these together, we see that $H_2(Q)$ embeds into $H_1(K)$.
The fibre product $K$ can also be characterised as the preimage of the diagonal subgroup of $Q\times Q$ in $F\times F$, so if $Q$ is finite then $K$ is of finite index in $F\times F$.  This proves the following.
Proposition.  For any finite group $Q$ there is a finite-index subgroup $K$ of $F\times F$ with $H_2(Q)\subseteq H_1(K)$.
This shows that all sorts of torsion can arise in $H_1$ of subgroups of graph groups.
Remark.  We took $Q$ to be finite because otherwise $K$ isn't quasiconvex.  But the rest of the argument goes through.<|endoftext|>
TITLE: The maximum order of finite subgroups in $GL(n,Q)$
QUESTION [13 upvotes]: For several years people have tried to characterize finite groups of maximal order in $\mathrm{GL}(n,\mathbb{Q})$ (or $\mathrm{GL}(n,\mathbb{Z})$) and their orders.
It appear in many articles a reference to an "preprint" article from Walter Feit in 1995 that gave full characterization. And I read a quote that Feit's paper also relies on a unpublished paper from Weisfeiler.
Does anyone know of a (pucblished) paper on this?

REPLY [13 votes]: Feit published his paper in the proceedings of the first Jamaican conference, MR1484185.  He defines M(n,K) to be the group of monomial matrices whose entries are roots of unity.  M(n,Q) is the group of signed permutation matrices.
Theorem A: A finite subgroup of GL(n,Q) of maximum order is conjugate to M(n,Q) and so has order n!2^n except in the following cases... [n=2,4,6,7,8,9,10].  In all cases the finite subgroup of maximum order in GL(n,Q) is unique up to conjugacy.
He notes that the maximum order subgroups of GL(n,Z) need not be unique up to GL(n,Z) conjugacy, since Weyl(Bn) and Weyl(Cn) are GL(n,Q) conjugate but not GL(n,Z) conjugate for n>2.
Theorem B gives a similar result for the cyclotomic fields, Q(l).
Feit published other papers which were very similar, and all of them rely heavily on Weisfeiler's work.  However, I believe this is the only published account of his "here is the list" preprint.<|endoftext|>
TITLE: How many vertices of a polytope can be chopped off to produce a k-vertex facet?
QUESTION [6 upvotes]: Let P be a simple n-facet d-polytope with facet F, and let F have k vertices. Let H be a halfspace and Q be a simple (n-1)-facet polytope such that H ∩ Q = P.
In terms of k, what is an upper bound on the number of vertices of Q contained in (ℝd \ H)?  Informally: what is the largest number of vertices of Q that can be chopped off to produce a k-vertex facet F?
I've derived an algorithm to compute this value exactly in terms of the f-vector of F, but I'd like to determine a tight upper bound when the f-vector is unavailable.  
Some observations: The removed set of vertices cannot contain a facet of Q, thus there would seem to be an upper limit on its size.  k ≥ d, naturally, since cutting off a single vertex produces d vertices.
If this question is particularly difficult or contains any open problems of which I'm not aware, I'd be interested in knowing that, too!

REPLY [6 votes]: I doubt this is anywhere close to tight, but here's a proof that it's bounded. The new facet has k vertices, so by the upper bound theorem it has $O(k^{\lfloor (d-1)/2\rfloor})$ ridges. Every facet of the polytope you cut off is either $F$ itself or a facet through one of these ridges, so there are $O(k^{\lfloor (d-1)/2\rfloor})$ facets of the cut-off polytope. Applying the upper bound theorem again, it has $O((k^{\lfloor (d-1)/2\rfloor})^{\lfloor d/2\rfloor})$ vertices.
This works without the assumption of simplicity. Using the fact that all your polytopes are simple you get at most k ridges directly, and $O(k^{\lfloor d/2\rfloor})$ vertices, I think.

REPLY [3 votes]: In three dimensions look at the prism whose ends are two regular polygons with $k$ sides. Look at the plane close to one end parallel to the plane containing one regular polygon that cuts of the regular polygon. Then that would satisfy all the conditions of the problem except one it cuts off one face. The plane can be rotated through a line close to one vertex so that it cuts off all vertices except that vertex. Then it will cut off $k-1$ points and leave all the original faces intact. It will have $k+1$ vertices So in three dimensions at least $k-1$ vertices can be cut off by a facet with $k+1$ vertices without losing any of the original faces. I think this technique can be used in higher dimensions.
If $v$ vertices are cut off by a plane which intersects a polyhedron in a polygon 
with $k$ sides with no faces cut off then the cut off polyhedron will have $k+1$ faces. As David Eppstein has observed in his answer to this question every facet 
of the polytope you cut off is either $F$ the face formed by the cutting 
hyperplane here a plane which cuts the polyhedron in a polygon with $k$ edges 
itself or a facet through one of these ridges,so in the two dimensional case this 
gives $k+1$ faces. 
The edges connecting the vertices cut off must form a tree. They contain no cycle 
or else a face would be cut off. They are connected since removing the edges of a 
face of polyhedron leaves the graph of the remaining edges connected. So these edges 
form a tree.
We have edges of the graph of a polyhedron equal the sum of the faces and the 
vertices-2. We know the number of vertices and faces in the polyhedron cut of by 
the plane. The vertices are $v + k$, and there are $k+1$ faces. So we must have 
$v + 2k-1$ edges we have $k$ edges formed by the vertices of the polygon F and 
$v-1$ from the tree formed by the vertices cut off. The only possiblity left is 
vertices between the cut of vertices and the vertices of $F$. Since each point 
must be adjacent to three edges each point of the polygon must have one edge 
which touches the cut of vertices. There must be $v+2$ edges in addtion to the 
edges of the tree to make every cut off vertex adjacent to exactly three edges. 
thus $k$ edges must equal $v+2$ edges and $k=v+2$.
So in three dimensions every cut that doesn't remove a face and intersects a polyhedron in $k$ vertices cuts of $k-2$ vertices. In higher dimensions I am not sure what happens. I know that there examples when $k$ vertices cut off $k-d$ vertices but beyond that I am not sure.<|endoftext|>
TITLE: product of all F_p, p prime
QUESTION [25 upvotes]: Let $R$ be the ring $$R = \prod_{p\ \text{prime}} \mathbb{F}_p$$ where $\mathbb{F}_p$ is the field having $p$ elements. 
Is it true that $R$ has a quotient by a maximal ideal which is a field of characteristic zero and contains $\overline{\mathbb{Q}}$?
Motivation: I like the problem and I can't solve it... 
It should have something to do with the Chebotarev density theorem.

REPLY [16 votes]: The answer is Yes, and this is the ultraproduct construction. Let U be any nonprincipal ultrafilter on the set of primes. This is simply the dual filter to a maximal ideal on the set of primes, containing all finite sets of primes. (In other words, U contains the Frechet filter.) 
The quotient R/U is a field of characteristic 0. 
The ultraproduct construction is completely general, and has nothing to do with rings or fields. If Mi for i in I is any collection of first order structures, and U is an ultrafilter on the subsets of I, then we may form the ultraproduct Π Mi/U, which is the set of equivalence classes by the relation f equiv g iff { i in I | f(i) = g(i) } in U. Similarly, the structure is imposed on the ultraproduct coordinate-wise, and this is well-defined. The most important theorem is Los's theorem, which says that Π Mi/U satisfies a first order statement phi([f]) if and only if { i in I | Mi satisfies phi(f(i)) } in U. 
In your case, since every Fp is a field, the ultraproduct is also a field. And since the set of p bigger than any fixed n is in U, then the ultraproduct will have 1+...+1 (n times) not equal to 0, for any fixed n > 0. That is, the ultraproduct will have characteristic 0.

Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F_p has the property is in the ultrafilter U. 
What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in Fp. This property is equivalent to asking whether every finite list of equations over Z is true in at least one Fp, since one can always add one more equation so as to exclude any particular Fp. Is this true? (I wasn't sure.)
But according to what Kevin says in the comments below, it is true, and this is precisely what you need for the construction to go through. You can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter. In this case, any particular equation would have a solution in Fp for a set of p in U, and so the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers.

REPLY [9 votes]: How far does the logic route get you? Here's a plan of attack. Say we have such a max ideal $m$.
1) If $S$ is a subset of the primes then there's an idempotent $e_S$ in $R$, which is 1 at $p$ if $p\in S$ and $0$ at $p$ otherwise. Now $e_S$ is an idempotent so either $e_S\in m$ or $e_S-1\in m$. Say $S$ is "big" if $e_S-1\in m$ and "small" if $e_S\in m$. I suspect that 
(@) a set is big iff its complement is small (easy).
(a) $m$ is determined by the collection of $S$ such that $S$ is big
(b) the big $S$ form an ultrafilter
(c) This gives you a bijection between ultrafilters on the set of primes and maximal ideals.
The principal ultrafilters are the ones corresponding to the obvious quotients $R\to\mathbf{F}_p$. Any non-principal ultrafilter will give a quotient of $R$ of characteristic zero.
I don't know if (1) is right but I've always suspected it is. If it's true the proof will be straightforward algebra.
2) Now use the logicians point of view. A 1st order statement of field theory is true in $R/m$ iff the set of primes for which it's true in $\mathbf{F}_p$ is big. For example the statement "I am a field with a square root of 2" is true for a set of primes of density 1/2 (those which are 1 or 7 mod 8), so this set had better be big if you want the quotient to have a square root of 2.
3) Now enumerate all polynomials with integer coefficients and say "I have all my roots in $R/m$". All these sets had better be big. So now we have some sort of combinatorial question, which I think is the following. Let $K$ run through all number fields, Galois over $\mathbf{Q}$ for simplicity, and for each one let $S(K)$ denote the set of primes that split completely in $K$. Is there an ultrafilter which contains all of these $S(K)$?
4) Now a bunch of sets is contained in an ultrafilter iff it generates a filter that isn't the whole set of primes. But now this is some combinatorial statement about primes splitting completely in number fields. For example, statements of the form "if $S$ is the primes that split completely in $K$ and $T$ is the set of primes that split completely in $L$, then $S\cap T$ is the set of primes that split completely in $KL$" will be of use to you here. In fact is that all you need?? Figure this out and you're home.
It's time for bed here in the UK (well, at least if you have 3 kids it is) but I'd be interested to know what happens if you try and push this through.

REPLY [9 votes]: Combining the answers of Joel and Kevin does work.
Write $\overline{\mathbb{Q}} = \bigcup_{n=0}^\infty K_n$ where the $K_n$ are an increasing sequence of finite Galois extension of $\mathbb{Q}$. Let $S_n$ be the primes that split completely in $K_n$. By Chebotarev's Theorem, this is a descending sequence of infinite sets of primes. Therefore, there is a nonprincipal ultrafilter $\mathcal{U}$ over the set of primes which contains all $S_n$. If $f(x) \in \mathbb{Z}[x]$ has a root in $K_n$, then $f(x)$ has roots in $\mathbb{F}_p$ for all $p \in S_n$ and hence $f(x)$ has a root in the ultraproduct $\prod_p \mathbb{F}_p/\mathcal{U}$.<|endoftext|>
TITLE: Periods and L-values
QUESTION [11 upvotes]: A famous theorem of Euler is that Zeta(2n) is a rational number times pi^(2n). Work of Kummer, Herbrand, Ribet and others shows that the rational multiplier has number theoretic significance.
For more general L-functions attached to motives, the philosophy has emerged (Deligne, Beilinson, Bloch, Kato, etc.) that (in vague terms) their values at certain integers are algebraic multiples of transcendental numbers and that the particular algebraic number that's a multiple of the transcendental number contains information about the motive that the L-function is attached to.
But a (nonzero) algebraic multiple of a transcendental number is again transcendental, so an arbitrary real number does not have a well defined decomposition as a product of an algebraic number and a transcendental number.
Still, because of the theorems of Kummer, et. al. one suspects that powers of pi are (at least close to) the "right" "transcendental parts" of the L-function values to be looking at. Maybe one should really be looking a powers of 2pi? But it seems clear that one should not be looking at powers of 691*pi because otherwise the statement of Kummer's criterion for the regularity of a prime would have an exceptional clause involving the prime 691.

Is there a conceptually motivated means of picking out the
  "right" "transcendental part" of a special value of an L-function?

Presumably the reason that Euler expressed his theorem in terms of pi is because pi was a commonly used symbol. (I've heard people argue that 2 pi is more conceptually primitive and that a label should have been made for the quantity 2 pi rather than for pi and am not sure what I think about this). In any case, there should be an a priori means to pin down the relevant transcendental number down "on the nose" (not up to a rational/algebraic multiple).
I have heard that Beilinson's conjecture give the transcendental number only up to a rational multiple and that the Bloch-Kato conjecture pins down the number. But I don't know enough to understand the statement of either conjecture and so am at present ill equipped to derive insight from reading the paper of Bloch and Kato. Are there more elementary considerations that give insight into how to pick out a particular transcendental number out of the set of all of its algebraic multiples?

REPLY [14 votes]: The ingredient in the Beilinson and Bloch--Kato conjectures is a motive (over ${\mathbb Q}$,
say).  If we take the integral cohomology of this motive (mod torsion, say) we get an integral lattice.  If we take some kind of Neron model, and take the algebraic de Rham cohomology of this, we get a second integral lattice.  Now computing the determinant of the pairing of one of these on the other, we get a transcendental number, well defined up to a unit in ${\mathbb Z}^{\times}$, i.e. a sign.
This should give you an idea of how one can attach a canonical period to a motive, and is the
basic idea underlying the construction of periods for motives.  (Since one doesn't have
Neron models in general, this idea is just heuristic as it stands, but I think it gives the right idea.  If you apply it to ${\mathbb G}_m$, you should recover the period $2\pi i$.)
EDIT: I should point out that the above really is just a heuristic, explaining how
there are two ways of getting integral structures in cohomology: in singular cohomology,
one just takes integral cycles (i.e. "true" cycles on the motive, with no funny
coefficients), and in de Rham cohomology, one takes algebraic differential forms that
are defined over the integers, like the Neron differential $dx/2y$ on an elliptic curve
with minimal Weierstrass equation $y^2 = f(x)$.
To actually get the correct periods for a given $L$-function, one has to do a little more
manipulation than I indicated; e.g. for an elliptic curve over ${\mathbb Q}$,
one will integrate the Neron differential over the a basis for the real integral cycles
(i.e. the cycles that are fixed by the action of complex conjugation on $E({\mathbb C})$;
these are rank one subgroup of the cohomoloy of $E({\mathbb C})$).  But hopefully what
I wrote above gives some intuition for what is going on.<|endoftext|>
TITLE: Converse to Hilbert basis theorem?
QUESTION [6 upvotes]: Specifically, is it possible for a non-Noetherian ring $R$ to have $R[x]$ Noetherian? Every reference I've seen for the Hilbert basis theorem only states the direction "$R$ Noetherian $\Rightarrow$ $R[x]$ Noetherian", which would certainly seem to imply that the converse is false. Unfortunately, it's tough to think about non-Noetherian rings, and what I'm sure is most people's favorite example of one, $K[x_1,x_2,\ldots]$ for a field $K$, is obviously not going to help us here.

REPLY [4 votes]: Dear Zev,
There are some sources which give the converse.  See e.g. pp. 64-65 of
http://math.uga.edu/~pete/integral.pdf
For that matter, see also pp. 32-33 of loc. cit. for the Chinese Remainder Theorem and its converse.  (And I am not the only one to do this...)
Note that in both cases the converse is left as an exercise.  I think (evidently) that this is the right way to go: it may not be so easy for the journeyman mathematician to come up with the statement of the converse, but having seen the statement it is a very valuable exercise to come up with the proof.  (In particular, I believe that in a  math text or course at the advanced undergraduate level and beyond, most exercises should indeed be things that one could find useful later on, and not just things which are challenging to prove but deservedly forgettable.)  
Finally, by coincidence, just yesterday in my graduate course on local fields I got to the proof of the "tensor product theorem" on the classification of norms in a finite-dimensional field extension (which came up in a previous MO answer).  The key idea of the proof -- which I found somewhat challenging to write; I certainly admit to the possibility of improvements in the exposition -- seems to be suspiciously close to the valuation-theoretic analogue of the converse of the Chinese Remainder Theorem!  See pp. 18-19 of 
http://math.uga.edu/~pete/8410Chapter2.pdf
if you're interested.<|endoftext|>
TITLE: Why do we need admissible isomorphisms for differential Galois theory?
QUESTION [15 upvotes]: Background: In Kaplansky's Introduction to Differential Algebra, an isomorphism between differential fields $K, L$ is defined to be admissible if $K,L$ are contained in a larger differential field $M$.  This appears to be an integral part of the development in his book.   For instance, Kaplansky proves the theorem that if $K, L$ are contained in a differential field $M$ and $f: K \to L$ is a differential isomorphism, then there is an admissible isomorphism $g: M \to M'$ extending $f$, using various differential analogs of results from commutative algebra (e.g. that a radical ideal is an intersection of prime ideals), which aren't at all necessary in ordinary field theory (as far as I know, anyway).
I find the definition hard to understand partially because it's not arrow-theoretic; it would help to have a more categorical notion than thinking in terms of subsets.  Moreover, in the case of Picard-Vessiot extensions (which if I understand correctly is what differential Galois theory focuses on), any admissible isomorphism is an automorphism anyway.  
Some googling suggests that one wants to take the (differential?) compositum between a differential field and its image under an admissible isomorphism.  Also, I have a suspicion that the absence of an analog of the concept of algebraic closure may be relevant here, but I'm not at all sure.
Question: To what extent is the restriction to admissibility necessary or useful, and if there's no way to avoid it for differential fields, why can one develop ordinary Galois theory without mentioning it?
And if it's indeed unavoidable, is there any way to think of it categorically?

REPLY [6 votes]: Expanding on Gjergji Zaimi's answer, (1) an hypothesis of admissibility is unnecessary and (2) the right context for differential Galois theories based on properties of extensions of differential fields as opposed to differential equations is somewhat more general than the strongly normal differential Galois theory of Kolchin.
There is an analogue of the notion of an algebraic closure for differential fields, called a constrained closure by Kolchin and his school and a differential closure in the model theoretic literature.  
Specializing to characteristic zero, a differential closure $(L,\partial_1,\ldots,\partial_n)$ of a partial differential field $(K,\partial_1,\ldots,\partial_n)$ is a differential field extension having the property that $L$ is differentially closed which means that every finite system of differential equations over $L$ which has a solution in some differential field extension of $L$ already has a solution in $L$ and is universal for differentially closed field extensions of $K$ in the sense that if $K \hookrightarrow M$ is an embedding of $K$ into a differentially closed field, then there is an embedding of $L$ into $M$ over $K$.   The differential closure of a differential field $K$ is unique up to isomorphism over $K$, but unlike the algebraic closure it is not minimal.  That is, if $L$ is the differential closure of $K$, then there may be a differentially closed field $M$ containing $K$ and properly contained in $L$.
The more general differential Galois theory I mentioned above is explained in the paper Pillay, Anand Differential Galois theory. I.  Illinois J. Math.  42  (1998),  no. 4, 678--699 at least in the case of ordinary differential fields.  Unlike Kolchin's theory, it is possible to have differential algebraic groups (instead of merely algebraic groups) as Galois groups.<|endoftext|>
TITLE: Deriving a relation in a group based on a presentation
QUESTION [12 upvotes]: Suppose I have the group presentation $G=\langle x,y\ |\ x^3=y^5=(yx)^2\rangle$. Now, $G$ is isomorphic to $SL(2,5)$ (see my proof here).  This means the relation $x^6=1$ should hold in $G$.  I was wondering if anyone knows how to derive that simply from the group presentation (not using central extensions, etc.).  Even nicer would be an example of how software (GAP, Magma, Magnus, etc.) could automate that.

REPLY [3 votes]: It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $Ext(C_{30},G)$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$
$$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = u^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ 
Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too. 
The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, 
Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

REPLY [3 votes]: This is a very basic answer to the last part of the question. One can derive the relation $x^6=1$ in gap and magma and even identify the group in this case. As a word of caution, these methods may break down depending on the automation one has in mind.
In gap:
 gap> F:= FreeGroup(2); x:=F.1; y:=F.2;
 <free group on the generators [ f1, f2 ]>
 f1
 f2
 gap> G:= F/[x^3*y^-5,x^-3*(y*x)^2];    
 <fp group on the generators [ f1, f2 ]>
 gap> Order(Subgroup(G,[G.1]));        
 6

To identify the whole group:
gap> Order(G); # note this line is not strictly necessary
120
gap> StructureDescription(G);
"SL(2,5)"

In magma, we can do the same operations:
 > G<x,y>:=Group<x,y|x^3=y^5=(y*x)^2>; 
 > Order(sub<G|x>);
 6
 > Order(G); # again this line is not strictly necessary
 120
 > IdentifyGroup(G);
 <120, 5> 

From here, you look up the group as the 5th group of order 120 in the small group data base (see http://magma.maths.usyd.edu.au/magma/handbook/text/703). In the alternative, you could put in your favorite presentation of $SL(2,5)$ and check that is it also group "<120, 5>."<|endoftext|>
TITLE: Dividing a square into 5 equal squares
QUESTION [11 upvotes]: Can you divide one square paper into five equal squares?
You have a scissor and glue. You can measure and cut and then attach as well. Only condition is You can't waste any paper.

REPLY [15 votes]: Since $1+2i$ has length $\sqrt5$, you can lift a square fundamental domain of $\mathbb C/\mathbb Z[i]$ to $\mathbb C/(1+2i)\mathbb Z[i]$. Overlay a square fundamental domain for the larger torus to get a way to divide a square into 5 smaller squares. 
It's pretty easy to decompose any rectangle into a square geometrically, but the general decomposition is not as nice.<|endoftext|>
TITLE: Is the ratio Perimeter/Area for a finite union of unit squares at most 4?
QUESTION [27 upvotes]: Update: As I have just learned, this is called Keleti's perimeter area conjecture.
Prove that if H is the union of a finite number of unit squares in the plane, then the ratio of the perimeter and the area of their union is at most four.
Remarks. If the squares must be axis-aligned, then this is easy to prove. If we replace unit squares with unit circles, then the statement is sharp and true for two (instead of four). The best known bound (to me) is 5.551... by Zoltán Gyenes. There is much more about the problem in his master thesis which you can find here.

REPLY [16 votes]: Disproved by Viktor Kiss and Zoltán Vidnyánszky, see http://arxiv.org/abs/1402.5452<|endoftext|>
TITLE: Ghost components of a Witt vector - Motivation
QUESTION [14 upvotes]: I'd like to know if anyone has a good explanation for where the ghost components that are used to define Witt vectors come from.  A lot of sources I've read take the ghost components for their starting point.
The closest I've seen to an explanation was in Joe Rabinoff's notes on Witt vectors.  This is my paraphrasing what I understood from them.  Consider $p$-typical Witt vectors and let $k$ be a perfect field (or ring) of characteristic $p$.  We have a multiplicative Teichmüller map $[\ ]: k \rightarrow W(k)$ which is a section to the projection map.  Every element of $W(k)$ can be uniquely written as a series $$\sum_{i=0}^{\infty} p^i[a_i],$$ with $a_i \in k$.  If we know how to add witt vectors, then we will know how to multiply them as well, because the Teichmüller map is multiplicative.  
The series expression for $[a_0] + [b_0]$ is already not obvious.  Write $\sum p^i [c_i]$ for this expression.  We have $c_0 = a_0 + b_0$, but to find $c_1$ we need the following trick.  Because our field is perfect, there is no harm in replacing $a_0$ and $b_0$ by $\alpha^p$ and $\beta^p$.  Then we want to compute $[\alpha^p] + [\beta^p] = [\alpha]^p + [\beta]^p.$  We know $[\alpha] + [\beta] \equiv [\alpha + \beta] \mod p$.  Raising both sides to the $p$th power, we can find a new expression for $[\alpha]^p + [\beta]^p \mod p^2$, and this gives us $c_1$.  
The formulas that are used to find $c_1$, $c_2$, etc. look similar to the ghost polynomials.  [They're a little different because the Witt vector corresponding to a series as above is $(c_0, c_1^p, c_2^{p^2}, \ldots)$, and so we haven't found the usual Witt vector components.  On the other hand, the usual ghost components don't involve any $p$th roots, which is what $\alpha$ and $\beta$ are.  Finally, the ghost map tells us not just how to add $[a_0]$ and $[b_0]$, but how to add any two series $\sum p^i[a_i]$ and $\sum p^i[b_i]$.]
I'm willing to believe that we can find precisely the usual ghost components this way (although I haven't actually succeeded in doing this).  So it seems natural that the ghost map $w: W(k) \rightarrow k^{\mathbb{N}}$ (where the ring operations on $k^{\mathbb{N}}$ are componentwise) is an additive map.  What I find surprising is that it's also multiplicative; it seems like we didn't do anything to guarantee this. 
Am I correct that this is one way to find the ghost components?  Is this the most natural path to them?  Should it be obvious that the ghost map is multiplicative?
I'd be happy to hear how you think of the ghost components.

REPLY [3 votes]: I might probably add some hint to the above motivations. I'm trying to provide some relations, using the exponential map on power series, that allow to see the definition of the ghost map as natural.
It is actually easier to begin by motivating the "big" Witt vectors $\mathbb{W}$.
Let $A$ be a commutative ring with unit element. Everything starts with   $\Lambda(A) = 1+ t A[[t]]$. It is a group under multiplication of power series.
Now, let $f(t)$ be a power series in $\Lambda(A)$. For every $f\in\Lambda(A)$ there are (at least) two ways to write it

$f(t) = 1 + \sum_{n\geq 1} b_n t^n$ with $b_n\in A$

$f(t) = \prod_{n=1}^{\infty}(1-a_nt^n)^{-1}$, with $a_n\in A$


The first sequence $(b_n)_n$ is often what we want to really understand.
The sequence $(a_n)_n$ is the big Witt vector associated to $f$. Denote by $\mathbb{W}(A)$ the family of sequence $(a_n)_n$ as above.
Now, if $A$ is a $\mathbb{Q}$-algebra there is a third way to write such a power series

$f(t) = \exp(\sum_{n\geq 1}\phi_n\frac{t^n}{n})$

The sequence $(\phi_n)_n$ is the ghost vector associated to the big Witt vector $(a_n)_n$. Calculating $\frac{d}{dt}(\log(f(t))$ provides the relation
$$ \phi_n\;=\;\sum_{d|n}d\cdot a_d^{n/d}\;. $$
When $A$ is a $\mathbb{Q}$-algebra, the multiplication rule of $\Lambda(A)$ induces a multiplication rule on the sequences $(a_n)_n$ which is the sum in $\mathbb{W}(A)$. One proves that this addition law is defined by polynomials with coefficients in $\mathbb{Z}$ and therefore it defines a scheme $\mathbb{W}$, and hence a group law on the Witt vectors $\mathbb{W}(A)$ for every ring $A$.
The fact that every power series in $\Lambda(A)$ can be written as in 1. and 2. translates the idea that for all ring $A$ one has $\mathbb{W}(A)\cong\Lambda(A)$.
These relations can be a starting point to justify the definition of the ghost map. When $n=p^m$, you can recognize the expression
$$\phi_{p^m}\;=\;\sum_{d|p^m}d\cdot a_d^{p^m/d}\;=\;\sum_{k=0}^m p^k\cdot a_{p^k}^{p^{m-k}}$$
which define the $p$-typical ghost map associated with the $p$-typical Witt vector $(a_1,a_{p},a_{p^2},\ldots)$. This expression shows that the $p$-typical Witt vectors $W(A)$ is a sub-group of $\mathbb{W}(A)$ but unfortunately it is not a sub-ring. The $p$-typical Witt vectors $W(A)$ is actually a quotient of the ring $\mathbb{W}(A)$, and the ring $\mathbb{W}(A)$ can be expressed as product of copies of the ring $W(A)$. It is hence relatively complicated to describe it easily as above. Allow me to quote the ADDENDUM 15 of the nice survey https://www.math.nagoya-u.ac.jp/~larsh/papers/s03/wittsurvey.pdf by Lars Hesselholt, which provide an exact relation between $\mathbb{W}(A)$ and $W(A)$ (seen as $\mathbb{W}_{S}(A)$, with $S=\{1,p,p^2,p^3,\ldots\}$ using the notations of Hesselholt's paper). Unfortunately, it does not mention explicitly the relations with the exponentials. More informations using the exponentials can be found in the Exercices of Bourbaki Alg. Comm. Chapter 9.
However, let me try to sketch out a similar approach using Artin-Hasse exponential at the place of exponential (these are my personal computations since I'm unable to find references). Again, from now on $A$ is a $\mathbb{Q}$-algebra (as for the global Witt vectors, the rationality of the coefficients has to be treated in a second moment). Let us rewrite every series $f(t)\in\Lambda(A)$ as follows. First we observe that $\exp(-\log(1-a_nt^n)) = (1-a_nt^n)^{-1}$ and then 2. can be rewritten as

$f(t) \;=\; \prod_{n\geq 1} \exp(-\log(1-a_nt^n))\;=\;\prod_{n\geq 1}\exp(\sum_{k\geq 1}\frac{(a_nt^{n})^k}{k})\;=\;\prod_{n\geq 1}\varepsilon(a_nt^n)$

where
$$\varepsilon(t) \;=\; \exp(-\log(1-t)) \;=\;\exp(\sum_{n\geq 1}\frac{t^n}{n})\;=\; \sum_{m\geq 0}t^m\;.$$
If we consider only the terms that are power of $p$, we obtain the Artin-Hasse exponential
$$E(t)\;=\;\exp(\sum_{k\geq 0}\frac{t^{p^k}}{p^k})\;=\;\exp(t+\frac{t^p}{p}+\frac{t^{p^2}}{p^2}+\cdots)$$
We may then start from the expression 3. and consider the following re-summation

$f(t)=\exp(\sum_{(n,p)=1}\sum_{m\geq 0}\phi_{np^m}\frac{t^{np^m}}{np^m})\; =\; \prod_{(n,p)=1}\exp(\sum_{m\geq 0}\phi_{np^m}\frac{t^{np^m}}{p^m})^{\frac{1}{n}}\;.$

For $n=1$, the coefficients $(\phi_{p^m})_{m\geq 0}$ are the phantom components of the $p$-typical Witt vector $(a_{p^m})_{m\geq 0}$ and it is easy to prove the relation
$$\exp(\sum_{m\geq 0}\phi_{p^m}\frac{t^{p^m}}{p^m})\;=\; \prod_{m\geq 0} E(a_{p^m}t^{p^m})$$
which is reminiscent of the expression $f(t)=\prod_{n\geq 1}\varepsilon(a_nt^n)$ in item 4. On the other hand, we have a similar relation for evert $(n,p)=1$.<|endoftext|>
TITLE: Different interpretations of moduli stacks
QUESTION [24 upvotes]: I'm taking my first steps in the language of stacks, and would like something cleared up. The intuitive idea of moduli spaces is that each point corresponds to an object of what we're trying to classify (smooth curves of genus g over ℂ, for example). Fine moduli spaces are defined to be the objects that represent the functor that takes an object and gives you the [set, for schemes; groupoid, as I understand it, for stacks] of ways that that object parametrizes families of the object we want to classify. Now, for schemes - this makes sense in the following way:
Let that functor be F, and let it be represented by M. Then F(Spec ℂ) are the families of (desired) objects parametrized by Spec ℂ (a point), so it corresponds to all desired objects (the ones we want to classify). But F(Spec ℂ) is also Hom(Spec ℂ, M), and so corresponds to the closed points of M. Thus M really does, intuitively, have as points the objects it wishes to classify.
Does this idea go through to moduli stacks? Of course, it probably does, and this is all probably trivial - but I feel like I need someone to assure me that I'm not crazy. So let me put the question like this: Can you formulate how to think of a fine moduli stack (as an object that represents an F as above; also: how would you define this F in the category of stacks?) in a way that makes it clear that it parametrizes the desired objects?

REPLY [28 votes]: I'll assure you that you're not crazy. Not only does the idea go through for stacks, but it's impossible (or at least very hard) to make sense of stacks without that idea.
If you're trying to parameterize wigits, you can build a functor F(T)={flat families of wigits over T}. If there is a space M that deserves to be called the moduli space of wigits, it should represent F. It's not just that the points of M must correspond to isomorphism classes of wigits, so we must have F(Spec ℂ)=Hom(Spec ℂ,M). The points are also connected up in the right way. For example, a family of widits over a curve should correspond to a choice of wigit for every point in the curve in a continuous way, so it should correspond to a morphism from the curve to M.
It happens that if wigits have automorphisms, there's no hope of finding a geometric object M so that maps to M are the same thing as flat (read "continuous") families of wigits. The reason is that any geometric object should have the property that maps to it can be determined locally. That is, if U and V cover T, specifying a map from T to M is the same as specifying maps from U and V to M which agree on U∩V. The jargon for this is "representable functors are sheaves." If a wigit X has an automorphism, then you can imagine a family of wigits over a circle so that all the fibers are X, but as you move around the circle, it gets "twisted" by the automorphism (if you want to think purely algebro-geometrically, use a circular chain of ℙ1s instead of a circle). Locally, you have a trivial family of wigits, so the map should correspond to a constant map to the moduli space M, but that would correspond to the trivial family globally, which this isn't. Oh dear!
Instead of giving up hope entirely, the trick is to replace the functor F by a "groupoid-valued functor" (fibered category), so the automorphisms of objects are recorded. Now of course there won't be a space representing F, since any space represents a set-valued functor, but it turns out that this sometimes revives the hope that F is represented by some mythical "geometric object" M in the sense that objects in F(T) (which should correspond to maps to M) can be determined locally. If this is true, we say that "F is a stack" or that "M is a stack." Part of what makes your question tricky is that as things get stacky, the line between M and F becomes more blurred. M isn't really anything other than F. We just call it M and treat it as a geometric object because it satisfies this gluing condition. We usually want M to be more geometric than that; we want it to have a cover (in some precise sense) by an honest space. If it does, then we say "M (or F) is an algebraic stack" and it turns out you can do real geometry on it.

REPLY [4 votes]: Yeah, you've got the right idea!  Say you've got some moduli problem -- for instance you have some class of objects X over C you're interested in -- and you want to say what the moduli space of X's is, as a variety (or stack or whatever) over C.  The idea is that you can describe a variety by how other varieties map to it -- Yoneda's Lemma.  So the moduli space M will be determined if you know what to call Hom(T,M) for every variety T.  Certainly if T=Spec(C) is the point, this should be the original set (groupoid, whatever) of objects X over C; but in general Hom(T,M) should be the set of families of X's continuously parametrized by T, which usually is nicely encoded by some flat family over T whose fibers over C-points are X's.
So, to sum up, the notion of a moduli space as a functor exactly encodes the idea that the points of a moduli space should be the objects you're trying to parametrize.  One just needs to use "points" in the more general context of maps from an arbitrary variety T, and be able to say what it means to give an object of the type you're interested in over an arbitrary such T.  This is not formally determined by the case T=Spec(C) a point, but there is usually a natural extension to make.<|endoftext|>
TITLE: Space Bounded Communication Complexity of Identity
QUESTION [13 upvotes]: $\bf Definition.$ We define the space bounded communication in the following way. A and B are
supernatural beings capable of computing anything but
they only have a limited amount of memory and that is shared. The
minimum size of this common memory that they can use to evaluate
a given function $f$ for which both of them possesses one half of the input, resp. $x$ and $y$,
shall be denoted by $S(f)$. At the
beginning it is filled with zeros. Then in each step one of the
players can put there an arbitrary message depending only on the
previous message and his input. They are finished when both of
them knows the value of $f(x,y)$. We can also imagine this as two
people communicating who have no memory at all (however, they can
remember their own input) and are allowed to send each other a
rewritable disk. The question is how big the disk has to be if
both of them wants to know the value of $f(x,y)$.
Define the identity function as $I(x,y): \{0,1\}^n\times \{0,1\}^n\rightarrow \{0,1\}$ with $I=1$ if and only if $x=y$.
$\bf Question.$  How much is $S(I)$?
$\bf Remarks.$ I know it is between $\log n$ and $\log n - \log \log n$, but which? Is it possible to solve it in $\log n-\omega(1)$ space? Anyone heard of any related things?
$\bf Example/Easy Claim.$ $S(I) \le \log(n) + O(1)$.
$\bf Proof.$ We present a construction. A sends her bits one after the other along with their ordinal number and a leading 1, meaning that it is up to B to speak. B replies to each message with his bit with the same ordinal number and a leading 0. This requires $2 + \log n$ space. If in a step his bit differs from her, they know that the answer is 0, the algorithm is over. If they finish sending all their bits, the answer is 1. Therefore, $S(I) \leq \log n + O(1)$.

REPLY [8 votes]: Here's an argument that I believe shows that $\log n - \omega(1)$ is impossible.  (This argument came out of a discussion I had with Steve Fenner.)
Let Alice's input be $x\in\{0,1\}^n$ and let Bob's input be $y\in\{0,1\}^n$.  Assume the shared memory stores states in $\{0,1\}^m$, and its initial state is $0^m$.  We are interested in lower-bounding $m$ for any protocol that computes EQ$(x,y)$.
A given protocol is defined by two collections of functions $\{f_x\}$ and $\{g_y\}$, representing the functions applied to the shared memory by Alice on each input $x$ and Bob on each input $y$, respectively, along with some answering criterion.  To be more specific, let us assume that if the shared memory ever contains the string $1^{m-1}b$ then the output of the protocol is $b$ (for each $b\in\{0,1\}$).  For convenience, let us also assume that $f_x(1^{m-1}b) = 1^{m-1}b\:$ and $g_y(1^{m-1}b) = 1^{m-1}b\:$ for every $x,y\in\{0,1\}^n$ and $b\in\{0,1\}$.  In other words, Alice and Bob don't change the shared memory once they know the answer.
Now, for each $x,y\in\{0,1\}^n$, consider what happens when Alice and Bob run the protocol on the input $(x,y)$.  Define $A_{x,y}\subseteq\{0,1\}^m$ to be the set of all states of the shared memory that Alice receives at any point in the protocol, and likewise define $B_{x,y}\subseteq\{0,1\}^m$ to be the states of the shared memory that Bob receives.  Also define
$$
S_{x,y} = \{0w\,:\,w\in A_{x,y}\} \cup \{1w\,:\,w\in B_{x,y}\}.$$
We will assume Alice goes first, so $0^m \in A_{x,y}$ for all $x,y$.  Let us also make the following observations:

By the definition of $A_{x,y}$ and $B_{x,y}$, it holds that $f_x(A_{x,y}) \subseteq B_{x,y}$ and $g_y(B_{x,y}) \subseteq A_{x,y}$ for all $x,y$.
For every $x,y$ with $x\not=y$ it holds that $1^{m-1}0\in A_{x,y} \cup B_{x,y}$, because Alice and Bob output 0 when their strings disagree.
For every $x$ it holds that $1^{m-1}0\not\in A_{x,x} \cup B_{x,x}$, because Alice and Bob do not output 0 when their strings agree.

Now let us prove that $S_{x,x}\not=S_{y,y}$ whenever $x\not=y$.  To do this, let us assume toward contradiction that $x\not=y$ but $S_{x,x} = S_{y,y}$ (i.e., $A_{x,x} = A_{y,y}$ and $B_{x,x} = B_{y,y}$), and consider the behavior of the protocol on the input $(x,y)$.
Let $w_t$ be the contents of the shared memory after $t$ turns have passed in the protocol, so we have $w_0 = 0^m$, $w_1 = f_x(w_0)$, $w_2 = g_y(w_1)$, and so on.  It holds that
$$
\begin{array}{c}
w_0 = 0^m \in A_{x,x}\\
w_1 = f_x(w_0) \in f_x(A_{x,x}) \subseteq B_{x,x} = B_{y,y},\\
w_2 = g_y(w_1) \in g_y(B_{y,y}) \subseteq A_{y,y} = A_{x,x},
\end{array}
$$
and in general
$$
\begin{array}{c}
w_{2t + 1} = f_x(w_{2t}) \in f_x(A_{x,x}) \subseteq B_{x,x} = B_{y,y}\\
w_{2t+2} = g_y(w_{2t+1}) \in g_y(B_{y,y}) \subseteq A_{y,y} = A_{x,x}
\end{array}
$$
for each $t\in\mathbb{N}$.  It follows that $A_{x,y} \subseteq A_{x,x}$ and $B_{x,y} \subseteq B_{y,y}$.
But now we have our contradiction, assuming the protocol is correct:  given that $A_{x,y}\subseteq A_{x,x}$ and $B_{x,y} \subseteq B_{y,y}$, it follows that $1^{m-1}0 \in A_{x,x}\cup B_{y,y}$, so Alice and Bob output the incorrect answer 0 on either $(x,x)$ or $(y,y)$.
Each $S_{x,x}$ is a subset of $\{0,1\}^{m+1}$, so there are at most $2^{2^{m+1}}$ choices for $S_{x,x}$. Given that the sets $S_{x,x}$ must be distinct for distinct choices of $x$, it follows that $2^{2^{m+1}} \geq 2^n$, so $m \geq \log n - 1$.<|endoftext|>
TITLE: Proof of the Reidemeister theorem
QUESTION [27 upvotes]: While preparing for my introduction to topology course, I've realized that I don't know where to find a detailed proof of the Reidemeister theorem (two link diagrams give isotopic links, iff they can be connected by a sequence of Reidemeister moves). The students in my class are not really very advanced and I'm not sure all of them would be able to reconstruct all the details of the proof given in Burde and Zieschang's Knot theory or in other sources I know of. And, being a lazy man, I would like to avoid having to type a detailed proof with pictures when I can refer the students to a book. So does anybody know of a source which would have a proof with all details spelled out?

REPLY [2 votes]: Messer and Straffin's book "Topology Now!" provides most of the steps starting from their definition of a knot (able to decomposed into a finite number of linear segments), building up the ideas of general position and triangular moves, and then proving the Reidemeister moves. I say "most of" because some of the steps are left as exercises. The material is very accessible to undergraduates.<|endoftext|>
TITLE: Markov chain on groups
QUESTION [17 upvotes]: Let $G$ be a permutation group on the finite set $\Omega$. Consider the Markov chain where you start with an element $\alpha \in \Omega$ chosen from some arbitrary starting probability distribution. One step in the Markov chain involves the following:

Move from $\alpha$ to a random element $g\in G$ that fixes $\alpha$.
Move from $g$ to a random element $\beta\in \Omega$ that is fixed by $g$.

Since it is possible to move from any $\alpha$ to any $\beta$ in $\Omega$ in a single step (through the identity of $G$) the Markov chain is irreducible and aperiodic. This implies that there is a unique distribution that is approached by iterating the procedure above from any starting distribution. It's not hard to show that the limiting distribution is the one where all orbits are equally likely (i.e. the probability of reaching $\alpha$ is inversely proportional to the size of the orbit containing it).

I read about this nice construction in P.J. Cameron's "Permutation Groups", where he brings up what he calls a slogan of modern enumeration theory: "...the ability to count a set is closely related to the ability to pick a random element from that set (with all elements equally likely)." 
One special case of this Markov chain is when we let $\Omega=G$ and the action be conjugation, then we get a limiting distribution where all conjugacy classes of $G$ are equally likely. Now, except for this nice result, it would also be interesting to know something about the rate of convergence of this chain. Cameron mentions that it is rapidly mixing (converges exponentially fast to the limiting distribution) in some important cases, but examples where it's not rapidly mixing can also be constructed. My question is:

Question: Can we describe the rate of convergence of the Markov chain described above in terms of group-theoretic concepts (properties of $G$)?

While giving the rate of convergence in terms of the properties of $G$ might be a hard question, answers with sufficient conditions for the chain to be rapidly mixing are also welcome.

REPLY [3 votes]: I recently published a paper on this exact problem for some classes of groups (link below). In short for CA groups I show that the rate of convergence is controlled the size of the group's center and the size of the largest centralizer of a non-central group element.
Link to paper or arXiv (also in Journal of theoretical probability): https://arxiv.org/abs/2003.04432<|endoftext|>
TITLE: Is there an "elementary" proof of the infinitude of completely split primes?
QUESTION [47 upvotes]: Let $K$ be a Galois extension of the rationals with degree $n$. The Chebotarev Density Theorem guarantees that the rational primes that split completely in $K$ have density $1/n$ and thus there are infinitely many such primes. As Kevin Buzzard pointed out to me in a comment, there is a simpler way to see that there are infinitely many rational primes that split completely in $K$, namely that the Dedekind zeta-function $\zeta_K(s)$ has a simple pole at $s = 1$. While this result is certainly much easier to prove than Chebotarev's Theorem, it is still not an elementary proof.

Is there a known elementary proof of the fact that there are infinitely many rational primes that split completely in $K$?

Selberg's elementary proof of Dirichlet's Theorem for primes in arithmetic progressions handles the case where $\text{Gal}(K/\mathbb{Q})$ is Abelian. I don't know anything about the general case. Since Dirichlet's Theorem is stronger than required, it is possible that an simpler proof exists even in the Abelian case.
Remarks on the meaning of elementary. I am aware that there is no uniformly recognized definition of "elementary proof" in number theory. While I am not opposed to alternate definitions, my personal definition is a proof which can be carried out in first-order arithmetic, i.e. without quantification over real numbers or higher-type objects. Obviously, I don't require it to be explicitly formulated in that way — even logicians don't do that! Odds are that whatever you believe is elementary is also elementary in my sense.
Kurt Gödel observed that proofs of (first-order) arithmetical facts can be much, much shorter in second-order arithmetic than in first-order arithmetic. This observation explains some of the effectiveness of analytic number theory, which is implicitly second-order. In view of Gödel's observation, it is possible that we have encountered arithmetical facts with a reasonably short second-order proof (i.e. could be found in an analytic number theory textbook) but no reasonable first-order proof (i.e. the production of any such proof would necessarily exhaust all of our natural resources). The above is unlikely to be such, but it is interesting to know that beasts of this type could exist...

REPLY [10 votes]: If $p \mid f(n)$ precisely, then one of the ideals $(n - \sigma(\alpha))\mathcal{O}_K$ must have a factor of $p\mathcal{O}_K$ in its factorization and hence all of them do. Therefore $p\mathcal{O}_K$ splits into a product of $\deg(f)$ ideals and therefore they must all have degree one. 
One could see quite elementarily that there exist infinitely many primes that divide some value of $f$ precisely: if $p$ is not such a prime, then if $p^2 \mid f(n)$, then $p^2 \mid f(n + p) = f(n) + pf'(n) + p^2A$ for some integer $A$ and therefore $p \mid f'(n)$. Thus, $n$ is a common zero of both $f, f'$ in $\mathbb{F}_p$, which cannot happen for $p$ large enough, as $\mathrm{Res}(f, f')$ does not vanish.<|endoftext|>
TITLE: triviality of fibre bundles
QUESTION [11 upvotes]: are there some general method in judging if a fible bundle is trivial?
At least,for vector bundles,there is a well-developed theory,that is Charicteristic Classes.the triviality of vector bundles is equivalent to the vanishment of its characteristic classes.
for principal bundles,the triviality is equivalent to the exsience of a cross section.
(any good perspective on this assertion?besides,how to tell if there is a cross section)
for general fiber bundle (E,F,B,G),(here,E is total space,F the fiber,B the base space,G the structure group),we can construct its associate principal bundle (E',G,B,G),i.e.to replace the fiber F with the topological group G.there is a theorem that a fiber bundle is trivial iff its associate principal bundle is trivial.
hence the problem is reduced to find a cross section of principal bundle.
I want to know if there is some other methods that are more usable?
Thank you!

REPLY [3 votes]: Here are some thoughts on the question of the posting:

Suppose a principal $G$-bundle $E\to B$ for $G$ a topological group has a section $s:B\to E$. Then $(g,b)\mapsto g s(b)$ is a trivialization (recall the action of $G$ is global). The space of trivializations of a trivial bundle is not necessarily trivial, it is the gauge group $Map(B,G)$.
Fiber bundles usually come equipped with a structure group $G$. (If none is specified, then it is usually assumed to be the group of all homeomorphisms or diffeomorphisms of the fiber.) Whether of not the bundle is trivial, depends on $G$. For example: the isomorphism classes of principal $G$-bundles on the sphere $S^n$ are one-to-one with $\pi_{n-1}(G)$ for connected $G$ (roughly because sphere is made of two disks, the bundle is trivial on each and we have to specify the transition maps on the intersection of the disks, which is the sphere of one dimension less). The identification is natural in $G$. For example, $\pi_1(U(n))=\mathbf{Z}$ and $\pi_1(SO(n))=\mathbf{Z}/2$ for $n\geq 3$ so there are complex rank 2 vector bundles on $S^2$ that are trivial as real vector bundles.
For any topological group $G$ there is a classifying space $BG$ and the isomorphism classes of principal $G$-bundles on $X$ are one to one with the homotopy classes of maps $X\to BG$. If $G$ acts on $F$ by homeomorphisms, then the above set is one to one with the set of isomorphism classes of bundles with structure group $G$ and fiber $F$.
Here is another description of the set of the isomorphism classes of principal bundles: this set is bijective with $H^1(X,\mathcal{F})$ where $\mathcal{F}$ is the ``gauge'' sheaf of groups on $X$ obtained from the presheaf $U\mapsto Map(U,G)$. This is particularly useful when $G$ is abelian, in which case we have not only $H^1$ but the higher groups as well; a particular case is the description of the isomorphism classes of line bundles on an algebraic variety $Y$ as elements of $H^1(Y,\mathcal{O}^{\*})$.<|endoftext|>
TITLE: How to estimate the growth of a recurrence sequence
QUESTION [5 upvotes]: If we have a linear recurrence sequence where each term depends on all previous terms, say
$a_n = \sum_{k=0}^{n-1} \binom{n}{k} a_k, \quad a_0 = 1$
is there any way to estimate the growth of a_n in terms of a Big-O notation?
I suppose the growth must be super-exponential, because if $a_1, \ldots, a_{n-1}$ grows exponentially, say, $q^i$, then we have $a_n = (q+1)^n - q^n$. Hence The exponent grows from $q$ to $q+1$. But I am not sure if this serves as an argument.
Thanks!

REPLY [3 votes]: This is sequence A000670 in the On-Line Encyclopedia of Integer Sequences. There are many comments, formulas, links, and references there.<|endoftext|>
TITLE: Two [n] to [n] function families
QUESTION [8 upvotes]: Note. This question had a bounty, so at the end I accepted the best (and only) answer. However, a solution is implied by the answer to this question.
Question.
Fix n. We are interested in the biggest t for which there exist two families of functions, $P_i,Q_i$, of size t from [n] to [n] such that for any $i,j$ whenever we consider the infinite sequence $P_i(Q_j(P_i(Q_j\ldots P_i(3))\ldots)$ (where the number of iterations tends to infinity), it contains no 2's and infinitely many 1's if $i=j$ and it contains no 1's and infinitely many 2's if $i\ne j$.
A lower bound.
I know a construction that shows that $t\ge 2^{\frac n2-O(1)}.$ For every subset $S$ of [n] that contains exactly one of $2k$ and $2k+1$ for $2\le k\le \frac n2-2$ we construct a pair of functions, $P_S,Q_S$ as follows. For any number m denote by $m^+$ the smallest element of $S$ that is bigger than m or if all elements of $S$ are at most m then define it to be 1.
$P_S(1)=1, P_S(2)=2$ and for bigger $m$'s $P_S(m)=m^+$, while $Q_S(1)=1, Q_S(2)=2$ and for bigger $m$'s $Q_S(m)=m$ if $m\in S$ and $Q_S(m)=2$ if $m\notin S$. This way we go through all the elements of S and end in 1 if the functions have the same index, but we are pushed to 2 if they differ.
Upper bound.
It is of course true that $t\le n^n$. So can you do better than $2^n$?

REPLY [4 votes]: Okay, so I tried to see how this could possibly work.  After some thinking I decided that one may as well take $P_i=Q_i$, so that the orbit of 3 (under the action of $P_i$) is a cycle containing 1.  If you take the length of this cycle to be roughly $n/2$, send $2\to 3$ and everything else to 2, that's not a bad idea except that it doesn't work for all possible $n/2$-subsets; otherwise we would have roughly $\binom{n}{n/2}$ possible $i$, as you wanted to begin with.  If you now look at the orbit of 3 under $(P_iP_j)$ in this setting you pretty quickly conclude that there is an inherent "even-town theorem" (see Babai-Frankl's book) and thus $2^{n/2}$ is really the best possible.  Of course, in the full generality weird things might be possible - I have no intuition for this, but this doesn't look good and unless the difference is really really important for some applications I wouldn't recommend working on this problem.<|endoftext|>
TITLE: homogeneous forms as norms
QUESTION [5 upvotes]: Motivation/example. Consider $K = \mathbb{Q}(\sqrt[3]{2})$. This is a number field with ring of integers $O_K = \mathbb{Z}[\sqrt[3]{2}]$. We have a norm map $N_{K/\mathbb{Q}}$ which maps $x + y\sqrt[3]{2} + z\sqrt[3]{4}$ to $x^3 + 2y^3 + 4z^6 - 6xyz$; restricting to $\mathcal{O}_K$ gives of course the same form. Using standard results about factorization of prime ideals, it is not too hard to see which integers are norms of elements in $\mathcal{O}_K$. Therefore we can get the values of $n$ (with some work...) for which $x^3 + 2y^3 + 4z^3 - 6xyz = n$ has integral solutions, and I guess it is also possible to say something about the solutions for a fixed $n$ - although it is not obvious to me how to do this in general.
The same is of course true for many interesting quadratic forms - to cite a famous example: we can get all positive integers $n$ which can be written as the sum of two perfect squares, or more generally as $x^2 + \alpha y^2$ (for some interesting values of $\alpha$).
Questions. Is this a fruitful method to study diophantine equations? Are there interesting "large" classes of higher degree polynomials/diophantine equations which can be treated by this sort of argument? What is known in general about such "norm forms"? How to decide whether a polynomial is a norm form? Et cetera :) 
(I know that it is a bit vague... I didn't find any useful references.)

REPLY [5 votes]: A necessary condition: Let $f: \mathbb{Q}^d \to \mathbb{Q}$ be a homogenous map of degree $d$. If $f$ is a norm, then the corresponding map $\mathbb{C}^d \to \mathbb{C}$ will be a product of linear forms. That's always true for $d=2$ (fundamental theorem of algebra), but rare for $d=3$.

REPLY [4 votes]: Just a remark : It is certainly interesting to study the equation $N_{L|K}(x)=y$, where $L|K$ is an extension of number fields, and there must be an extensive literature on the subject.  Hasse proved that if such an equation is solvable everywhere locally (at every place of $K$), and if $L|K$ is cyclic, then there is a global solution.  This is sometimes called the Hasse Norm Principle.
Hasse also found examples where $L|K$ is galoisian and the equation is solvable everywhere locally, but not globally.  Even for $K=\mathbb{Q}$, there are infinitely many biquadratic extensions $L$ which provide counter-examples.<|endoftext|>
TITLE: Two questions about units in Number Fields
QUESTION [14 upvotes]: If $K/\mathbb{Q}$ is a number field which is not $\mathbb{Q}$ or a quadratic imaginary field then, by the Dirichlet unit theorem it has a unit of infinite order.  Is there a simple proof of this fact which doesn't haul in all the machinery of the unit theorem?  In a related question is it true that such a $K$ is always generated over $\mathbb{Q}$ by one unit (which might not be of infinite order -- cf. the case of a full cyclotomic field), and if it is, is there a simple proof of this fact?

REPLY [11 votes]: The answer to question 2 as stated is no. Let $K = \mathbb{Q}(\sqrt{p},\sqrt{-b})$, with $p$ 1 mod 4. By Frohlich and Taylor, p. 196, the fundamental unit group is generated by a unit from the real quadratic subfield. It then suffices to show that we can pick b so that K contains no roots of unity other than $\pm 1$. But the only possibility is that K contains a 3rd, 4th, or 5th root of unity, and so we just pick $b$ to avoid ramification at 2, 3, or 5.<|endoftext|>
TITLE: BU with tensor product H-space structure
QUESTION [10 upvotes]: Hi, 
I came across the space $BU_\otimes$ when struggling with twisted K-theory. Segal proved that this is an H-space, right? I have read a dozen times by now that the group $[X, BU_\otimes]$ consists of the vector bundles of virtual dimension 1. I can clearly see the semi-group structure there, but what does the inverse of such a bundle look like?
Best regards,
Ulrich Pennig

REPLY [14 votes]: I'll write $U(X)=[X,BU_\otimes]$.  So $U(X)\subset K(X)=[X,Z\times BU]$ is the multiplicatively closed subset corresponding to "virtual bundles of rank 1".
Likewise, I'll write $I(X)=[X,BU]\subset K(X)$ for the ideal of "virtual bundles of rank 0".
There's a bijection $a\mapsto 1+a$ from $I(X)$ to $U(X)$.
Here's the claim: if $X$ is compact (let's say it's a finite CW-complex of dimension $n$), then $I(X)$ is a nilpotent ideal (in fact, $I(X)^{n+1}=0$).  This is a generic fact about multiplicative cohomology theories, and it can be proved in a number of ways; you can think about it in terms of the multiplicative properties of the Atiyah-Hirzebruch spectral sequence, for instance.
For such $X$, it is clear that elements of $1+a\in U(X)$ are invertible, given by the series $(1+a)^{-1}=1+a+a^2+\cdots$ which terminates since $a\in I(X)$ and $I(X)$ is nilpotent.
For infinite dimensional $X$, you need to argue a little harder.  If $X=\lim_{\to} X_i$ where the $X_i$ are finite CW-complexes, then there is a surjection $K(X)\to \lim_{\leftarrow} K(X_i)$, which restricts to  surjections for $I(X)$ and $U(X)$. 
The kernel is a $\lim^1$-term.  If the $\lim^1$-term vanishes, then it's clear that elements of $U(X)$ are invertible, since their images in the $U(X_i)$ are invertible.  It's enough to check that $\lim^1$-vanishes in the case that $X=BU_\otimes$, which is a standard calculation.<|endoftext|>
TITLE: How cavalier can I be when demanding a category have direct sums?
QUESTION [13 upvotes]: In my meaning, a direct sum in a category should really be called a "biproduct".  If $X,Y$ are objects, then a direct sum $X \oplus Y$ is an object $Z$ along with isomorphisms $\hom(Z,A) = \hom(X,A) \times \hom(Y,A)$ and $\hom(A,Z) = \hom(A,X) \times \hom(A,Y)$ for all objects $A$.  A direct sum is unique up to canonical isomorphism if it exists, of course.  A category has (finite) direct sums if it has a zero object (an object that is both initial and terminal; i.e. "the direct sum of zero things") and if $X\oplus Y$ exists for any objects $X,Y$.  If a category has direct sums, then it is naturally enriched in abelian monoids.  I believe that an additive category is a category with direct sums in which all the hom-sets (which are already abelian monoids) are actually abelian groups.
There are many times when people say "include all direct sums".  For example:
Example:
Let $\mathcal C$ be any category (enriched over $\rm SET$).  Then I can make it enriched over $\rm ABGP$ by applying the $\rm Free: SET \to ABGP$ functor to each hom-set.  So now I have a new catefory ${\rm Free}(\mathcal C)$ in which I can add morphisms.  But often I want to add objects, too, so I do something like "take the matrix category" ${\rm Mat}(\mathcal C)$, whose objects are finite sequences of objects in $\mathcal C$ and whose morphism are matrices of morphism in ${\rm Free}(\mathcal C)$.  Then it's more or less obvious that ${\rm Mat}(\mathcal C)$ is an additive category.  If $\mathcal C$ is freely generated by some (objects and) morphisms, then ${\rm Mat}(\mathcal C)$ is presumably "the free additive category generated by those morphisms".
But often I'm not content with free additive categories.  For example, I might want to present a category by generators and relations.

Question: Is it clear that when I take the quotient of an additive category by some ideal (as an $\rm ABGP$-enriched category), that it still has direct products?

Or perhaps I really want the abelian category presented by generators and relations.  Or maybe I just want every idempotent to split, in which case I might take the Karoubi envelope.

Question: If I extend my category to split all idempotents, or to include kernels and cokernels, or ..., is it clear that it still has direct products?

A very explicit application contained in these constructions is the formation of the exterior tensor product of categories: if $\mathcal B,\mathcal C$ are additive categories, then $\mathcal B \boxtimes \mathcal C$ is the free additive category generated by $\mathcal B \times \mathcal C$ with a bunch of relations.

REPLY [14 votes]: The answer to the first question is yes.  If A and B have a direct sum A ⊕ B in C, then there are inclusions iA : A → A ⊕ B, iB : B → A ⊕ B and projections pA : A ⊕ B → A, pB : A ⊕ B → B such that pAiA = 1, pBiB = 1, and iApA + iBpB = 1.  Conversely, the existence of such maps in an Ab-enriched category make A ⊕ B a direct sum of A and B, even if we do not asssume a priori that A and B have a direct sum.  Now if we form the quotient C/I by an ideal I, and two objects A and B with a direct sum A ⊕ B in C, the image of this system of maps presents the image of A ⊕ B as the direct sum of the images of A and B.  In short, direct sums are absolute colimits, and as the quotient functor C → C/I is essentially surjective (indeed, bijective on objects), every pair of objects of C/I inherits a direct sum from C.<|endoftext|>
TITLE: Affine morphisms in different settings coincide?
QUESTION [5 upvotes]: 1.If we identify two schemes $X$ and $Y$ as two presheaves of set on category of affine schemes.($Aff:=\text{CRing}^{op}$) If there is a morphism as natural transformations $f:X\to Y$, then, how to give the definition of the affine morphism? 
2.If we identify two schemes $X$ and $Y$ as two category of quasi coherent sheaves $QCoh_{X}$ and $QCoh_{Y}$. Then morphism between this two schemes is a functor $f: Qcoh_{X}\to Qcoh_{Y}$, how to give the definition of affine morpshim ? 
3.If we just consider the classical case. X and Y are two schemes. suppose $f:X\to Y$ is an affine morphism. Does this definition of affineness coincide with the other two definitions(if there exists)?
If such definitions of affine morphisms exists, are they equivalent or not?

REPLY [5 votes]: Emerton and Zoran's answers completely answer the question as stated, but there's another way to think about affine morphisms that is worth mentioning.
Given any quasi-compact and quasi-separated morphism of schemes $f:X\to Y$, $\newcommand{\O}{\mathcal O}f_*\O_X$ is a quasi-coherent sheaf of $\O_Y$-algebras. This functor has an adjoint, called relative $Spec_Y$, or relative Spec. Given a quasi-coherent sheaf of $\O_Y$-algebras $\newcommand{\A}{\mathcal A}\A$, we get a scheme over $Y$, $\phi^\A:Spec_Y \A\to Y$, with the property that $\phi^\A_*(\O_{Spec_Y \A})=\A$ and $Hom_Y(X,Spec_Y \A)\cong Hom_{\O_Y\text{-alg}}(\A,f_*\O_X)$ for any $f:X\to Y$. A morphism $f:X\to Y$ is affine if and only if $X\cong Spec_Y(\A)$ (as a $Y$-scheme) for some $\A$ (which must be $f_*\O_X$). See EGA II §1 for this development of affine morphisms.
I find this way of thinking about affine morphisms is useful for two reasons. First, if I'm working with a bunch of schemes affine over $Y$, it's often easier for me to think about a bunch of $\O_Y$-algebras with algebra morphisms between them. Second, the adjunction in the previous paragraph tells you that any (quasi-compact quasi-separated) morphism $f:X\to Y$ has a canonical factorization through an affine morphism $X\to Spec_Y(f_*\O_X)\to Y$, called the Stein factorization (the first morphism is Stein, meaning that the structure sheaf pushes forward to the structure sheaf). This factorization is often extremely handy; for example, if a morphism is quasi-affine, the Stein factorization is a witness of its quasi-affineness.<|endoftext|>
TITLE: Why can't there be a general theory of nonlinear PDE?
QUESTION [74 upvotes]: Lawrence Evans wrote in discussing the work of Lions fils that

there is in truth no central core
  theory of nonlinear partial
  differential equations, nor can there
  be. The sources of partial
  differential equations are so many -
  physical, probabilistic, geometric etc.
  - that the subject is a confederation of diverse subareas, each studying
  different phenomena for different
  nonlinear partial differential
  equation by utterly different methods.

To me the second part of Evans' quote does not necessarily imply the first. So my question is: why can't there be a core theory of nonlinear PDE?
More specifically it is not clear to me is why there cannot be a mechanical procedure (I am reminded here by [very] loose analogy of the Risch algorithm) for producing estimates or good numerical schemes or algorithmically determining existence and uniqueness results for "most" PDE. (Perhaps the h-principle has something to say about a general theory of nonlinear PDE, but I don't understand it.)
I realize this question is more vague than typically considered appropriate for MO, so I have made it CW in the hope that it will be speedily improved. Given the paucity of PDE questions on MO I would like to think that this can be forgiven in the meantime.

REPLY [4 votes]: In my limited experience, the furthest you can carry a general theory of PDEs, assuming only smoothness of the PDEs for example, is to describe the characteristic variety and its integrability, or to determine whether the equations are formally integrable (in the sense of the Cartan-Kaehler theorem). Already you find that every real algebraic variety is the characteristic variety of some system of PDE. Even when the characteristic variety is very elementary (a sphere, for example), we know very little about the PDE (it is hyperbolic, but we don't have a complete theory of boundary value problems, initial value problems, long term existence, uniqueness). So I think that a general theory of PDE would have to be much more difficult than real algebraic geometry, which already has elementary problems that seem to be very difficult.<|endoftext|>
TITLE: Simulating Turing machines with {O,P}DEs.
QUESTION [28 upvotes]: Qiaochu Yuan in his answer to this question recalls a blog post (specifically, comment 16 therein) by Terry Tao:

For instance, one cannot hope to find an algorithm to determine the existence of smooth solutions to arbitrary nonlinear partial differential equations, because it is possible to simulate a Turing machine using the laws of classical physics, which in turn can be modeled using (a moderately complicated system of) nonlinear PDE


Is this "it is possible" an application of a Newton thesis, much as the usual Church-Turing thesis, going along the lines of «if something is imaginably doable in real life, one can simulate it with the usual equations of physics», or has the simulation actually been actually carried out? 
Does one need PDEs to simulate Turing machines, or are ODEs good enough?

REPLY [8 votes]: I believe that the reference is to seminal work of Pour-El and Richards where they measure the computational contents of various types of classical systems of partial differential equations. They have a series of papers where they carry this out; I believe much of it can be found in their book Computability in analysis and physics (Perspectives in Mathematical Logic, Springer-Verlag, 1989). There are other approaches to this, but they all reach essentially the same conclusions. Pour-El and Richards have the advantage that they concentrate on specific systems that actually arise in physics, such as the wave equation.<|endoftext|>
TITLE: Collapsible group words
QUESTION [45 upvotes]: What is the length $f(n)$ of the shortest nontrivial group word $w_n$ in $x_1,\ldots,x_n$ that collapses to $1$ when we substitute $x_i=1$ for any $i$?

For example, $f(2)=4$, with the commutator $[x_1,x_2]=x_1 x_2 x_1^{-1} x_2^{-1}$ attaining the bound.  
For any $m,n \ge 1$, the construction $w_{m+n}(\vec{x},\vec{y}):=[w_m(\vec{x}),w_n(\vec{y})]$ shows that $f(m+n) \le 2 f(m) + 2 f(n)$.
Is $f(1),f(2),\ldots$ the same as sequence A073121:
$$ 1,4,10,16,28,40,52,64,88,112,136,\ldots ?$$
Motivation: Beating the iterated commutator construction would improve the best known bounds in size of the smallest group not satisfying an identity.

REPLY [19 votes]: See the paper "Brunnian links" by Gartside and Greenwood, published in Fundamenta Mathematicae.  Theorems 8 and 7 imply that iterated commutators are optimal and the sequence you suggest gives the minimal length.<|endoftext|>
TITLE: visualizing what's going on in based homotopy theory, et al.
QUESTION [8 upvotes]: I'm reading J.P. May's Concise Course in Algebraic Topology, and I'm having a lot of trouble visualizing how things work in Chapter 8, "Based cofiber and fiber sequences".  Of course this is pretty basic stuff, but it's really cool to me that there are such clear topological analogues to the usual exact sequences in homological algebra.  Still, I can't even get a clear picture of what a smash product looks like for any but the most basic of spaces, and based cones/suspensions/loopspaces make my head hurt.
a) Will I be alright if in my head I just sort think of a smash product as a usual product (for example), with the understanding that I need to tack on an extra condition that I really shouldn't think too hard about?
b) Why all the fuss about based homotopy theory, anyways?
c) While I'm at it, can anyone suggest a book that is less terse?  I feel like this one rarely gives the motivation and visual intuition that I'd like...

REPLY [6 votes]: Another book with pictures of reduced suspension etc. is Ronnie Brown's Topology and Groupoids.
see http://www.bangor.ac.uk/~mas010/topgpds.html. Which is also excellent for non-based stuff.
Don't believe all you hear about the unbased case being grotty! It is beautiful, but it is possibly  easier to learn Alg. Top. in the based situation first, especially if it is that the someone else has decided you should do. :-)
I have put some material that might help (Abstract Homotopy...) on my n-Lab home page (follow the links from Tim Porter (found by a search)).
For a) I had something like your problem when I started, but then thought of the based cylinder as a cylinder with a long base point! Of course, you really need to squidge that line to a point. It is safe when mapping out of a smash to retain the subspace that is to be squidged just always mapping all of it to the base.  (In other words, don't agonise about the smash at this stage.  Use it as a device for the moment and after you learn to use it and see how it behaves its strangeness will probably have dissapated.)

REPLY [4 votes]: I've got answers for b.) and c.).  
b.) The reason based homotopy is so important is that in unbased homotopy theory, there's no good notion of a homotopy group.  Now you may say, "what about the fundamental groupoid?", but that is somewhat misleading.  To actually deal with the higher homotopy groupoids, you in fact need higher category theory, and all of the complicated notions of equivalence that come with it.  So if you're trying to do classical homotopy theory, you need basing to deal with the higher homotopy data.  
c.) I really like May's book, but if you want a good geometric picture as well, you might want to look through Spanier or Hatcher as a "companion" book. You may also want to look through Switzer or Whitehead, but they are much less focused on the geometric aspects of Algebraic Topology.<|endoftext|>
TITLE: Peakedness of multimodal distributions
QUESTION [6 upvotes]: In Probability theory, does there exist some measures of peaked-ness for multi-modal distributions. I guess kurtosis as such would not be a good measure of peaked-ness for multimodal distributions. Please correct me if I am wrong. Can you point me to some of them which are simpler to compute.

REPLY [3 votes]: What about information entropy?  The smaller it is, the more mass is in peaks.<|endoftext|>
TITLE: What properties "should" spectrum of noncommutative ring have?
QUESTION [20 upvotes]: There are already a lot of discussion about the motivation for prime spectrum of commutative ring. In my perspective(highly non original), there are following reasons for the importance of prime spectrum. 

$\text{Spec}(R)$ is the minimal spectrum containing $\text{Spec}_{\rm max}(R)$ which has good functoriality which means the preimage of a prime ideal is still a prime ideal.
if $p\in \text{Spec}(R)$, then $S_{p} = R-p$ is multiplicative set. Then one can localize.
$S_{p}^{-1}R$ for $p\in \text{Spec}(R)$ is a local ring (has unique maximal ideal which is equivalent to have unique isomorphism class of simple modules). Local ring is easy to deal with and the maximal ideal can be described in explicitly, i.e $m=S_p^{-1}p$

(My advisor told me P.Cartier pushed Grothendieck to built commutative algebraic geometry machinery based on prime spectrum and the reasons mentioned above are the reasons they used prime spectrum)
Addtional reason: one can have good definitions of topological space and a structure sheaf on it so that one can recover this commutative ring back as its global section
Now, my question is for the people coming from commutative world, what other properties do you expect the spectrum of a noncommutative ring should have? 
I am aware that people are coming from different branches, there might be various kinds of noncommutative ring arising in your study. Therefore, the question for people coming from different branches of mathematics is that which kind of noncommutative ring do you meet and what properties do you feel that the spectrum of noncommutative ring should have to satisfy your need?
The main motivation for me to ask this question is I am learning noncommutative algebraic geometry. In the existence work by Rosenberg, there are several kinds of spectrum(at least six different spectrum) for different purposes and they satisfy the analogue properties(noncommutative version)I mentioned above and coincide with prime spectrum when one impose the condition of commutativity. I wonder check whether these spectrum satisfied the other reasonable demand

REPLY [9 votes]: I know this question is old, and has an accepted answer, but this excellent paper by Manny Reyes gives some further thoughts about possible Spec's for noncommutative rings.<|endoftext|>
TITLE: General form of amount of triangles that can be formed in an MxN point lattice
QUESTION [5 upvotes]: I've been searching for the answer for many years, both by researching by myself and reading about the subject. Now I'm wondering if this has a solution.
The problem can be stated as follows.
Given a M x N grid of points, how many triangles with vertices in the grid can be formed?
Note that you can't select two points that coincide or three collinear points because that wouldn't conform a triangle (area would be 0)
OK, I know a bit of programming and could manage to code a program that solves this, but would REALLY like to know if there is a general form depending on M and N.
I suspect it has to do with prime numbers... (Perhaps I totally missed heheh)
Thanks for your time!
Manuel

REPLY [8 votes]: For the NxN case, at least, there is a literature. An entry point is sequence A000938 in the On-Line Encyclopedia of Integer Sequences.<|endoftext|>
TITLE: Exercises in Hodge Theory
QUESTION [20 upvotes]: I was wondering: is there a good place to find exercises in Hodge theory? Mostly computations and proving small (preferably nifty) theorems, is what I have in mind.  Something roughly like the Problems in Group Theory book by Dixon, or many other such problem books.
The biggest difficulty I'm having is that I don't have an adequate number of problems to do, as Griffiths & Harris's Principles has none, and Voisin's otherwise excellent books have only a couple of exercises each chapter, and I've not been able to find a book or paper on the arxiv that does.

REPLY [5 votes]: There are at least 20 exercises on Hodge theory of Kahler manifolds in Huybrechts book "Complex geometry", section 3, maybe this is what you are looking for. In general this book has a lot of exercises and seem to be well written.<|endoftext|>
TITLE: Is $Sym^n (V^*) \cong Sym^n (V)^\ast$ naturally in positive characteristic?
QUESTION [9 upvotes]: Background/motivation
It is a classical fact that we have a natural isomorphism $Sym^n (V^*) \cong Sym^n (V) ^\ast$ for vector spaces $V$ over a field $k$ of characteristic 0. One way to see this is the following.
On the one hand elements of $Sym^n (V^*)$ are symmetric powers of degree n of linear forms on $V$, so they can be identified with homogeneous polynomials of degree n on $V$. On the other hand elements of $Sym^n (V) ^\ast$ are linear functionals on $Sym^n V$; by the universal property of $Sym^n V$ these correspond to n-multilinear symmetric forms on $V$. The isomorphism is then as follows.
An n-multilinear symmetric form $\phi$ corresponds to the homogeneous polynomial $p(v) = \phi(v, \dots, v)$. In the other direction to a polynomial $p(v)$ we attach the multinear form obtained by polarization $\phi(v_1, \dots, v_n) = \frac{1}{n!}\sum_{I \subset [n]} (-1)^{n - \sharp I} p(\sum_{i \in I} v_i)$. Here $[n]$ is the set $\lbrace 1, \dots, n \rbrace$.
Problem
Of course this will not work for $n$ greater than the characteristic of $k$ if the latter is positive.
One can expect that an isomorphism $Sym^n (V^*) \cong Sym^n (V) ^\ast$ holds also in positive characteristic, and that this should be trivially true by using the universal properties of the symmetric powers. The problem is that if I try to define a natural map between the two spaces using the universal properties I have at some point to divide by $n!$ anyway.
Still there may be some natural isomorphism that I cannot see. Or maybe there is not a natural isomorphism, but I don't know how to prove this.

Is there a natural isomoprhism $Sym^n (V^*) \cong Sym^n (V)^\ast$ in positive characteristic?

REPLY [18 votes]: The answer is no (and well-known to people working in the representation theory of algebraic groups in positive characteristic). In fact for $V$ finite dimensional and of dimension $>1$ the two vector spaces are not 
isomorphic as $GL(V)$-modules ($GL(V)$ is either considered naively as an abstract group when the field $k$
is infinite or as an algebraic group in the general case) for $n=p$ equal to the characteristic.
Under the assumption that $V$ is finite dimensional we may instead formulate the problem as the impossibility
of having a $GL(V)$-isomorphism $Sym^n(V) \cong Sym^n(V^\ast)^\ast$. Now, we have an injective $GL(V)$-map $V^{(p)} \to Sym^p(V)$ given by $v \mapsto v^p$, where $V^{(p)}=k\bigotimes_kV$, where $k$ acts on the left hand side through the $p$'th power (concretely if we choose a basis for $V$ then the action on $V=k^m$ is given by the group homomorphism $GL_m(k) \to GL_m(k)$ which takes $(a_{ij})$ to $(a^p_{ij})$). As $\dim V > 1$ we have $\dim V^{(p)}=\dim V < \dim Sym^p(V)$ so that the inclusion is proper.
It is easily verified that $V^{(p)}$ is irreducible and it is in fact the unique irreducible submodule of $S^p(V)$.  This can be seen by starting with an arbitrary non-zero element $f$ of $S^p(V)$ and then acting on it by suitable linear combinations of the action of elementary matrices of $GL(V)$ until one arrives at a non-zero element of (the image of) $V^{(p)}$. This is more easily understood if one uses the fact that we have an action
of an algebraic and consider the induced action by its Lie algebra. Choosing again a basis of $V$ we have elements $x_i\partial/\partial x_j$ whose action on a monomial are very visible. In this way it is clear that 
starting with any monomial of degree $p$ one may apply a sequence of such operators to obtain a non-zero
multiple of a monomial of the form $x_k^p$. This plus some thought shows the statement.
Assume now that we have a $GL(V)$-isomorphism $Sym^p(V) \cong Sym^p(V^\ast)^\ast$. Dualising the inclusion $V^{\ast(p)} \hookrightarrow Sym^p(V^\ast)$ and composing with the isomorphism we got a quotient map $Sym^p(V) \to V^{(p)}$. It is easy to see that in Jordan-Hölder sequence of $Sym^p(V)$ $V^{(p)}$ so that the composite $V^{(p)} \to Sym^p(V) \to V^{(p)}$ must be an isomorphism and hence the inclusion $V^{(p)} \hookrightarrow Sym^p(V)$ is split, contradicting that $V^{(p)}$ is the unique simple submodule.

REPLY [14 votes]: The dual to symmetric powers of a projective module of finite rank over any commutative ring is most elegantly expressed in terms of divided powers (thereby also "explaining" why over a field of nonzero characteristic there are subtleties once the dimension reaches the characteristic, and why the version in characteristic zero has factorials all over the place in the denominators).  It is very nicely explained in an appendix to the book by Berthelot-Ogus on crystalline cohomology.<|endoftext|>
TITLE: primitive of an exact differential form with special properties
QUESTION [6 upvotes]: We were working on a smoothing problem and ran across the apparently simple following question:
X is a triangulated smooth manifold of dimension $n$, and $\alpha$ is an exact differential form of degree $n$. Assume moreover that its integral vanishes on each $n$-simplex. 
Is it possible to find a primitive $\beta$ of $\alpha$ (that is, $d\beta=\alpha$) such that $\beta$ vanishes tangentially to the $(n-1)$-skeleton?

The last sentence means that, for any vectors $v_1,...,v_{n-1}$ based on a same point and tangent to an $(n-1)$-simplex, we have $\beta(v_1,...,v_{n-1})=0$.

What we already know : take any primitive $\beta$ of $\alpha$, then, by stokes its integral on the boundary of each $n$-simplex is zero. However, its integral on each $(n-1)$-simplex is a priori not zero. 
Lemma 2 p.165 of Singer-Thorpe (Springer edition) implies that $\beta$ can be found to have integral 0 on each $(n-1)$-simplex. But can we then find $\beta$ to be uniformly zero tangentially to those simplices?
We know how to prove it if we replace the triangulation by a "sweet" cubic complexe.

REPLY [4 votes]: I think, the answer can depend on how you interpret the question. Let me show that the answer is negative for one of the interpretations already in the case of $2$-dimensional manifolds. We study the question locally in a neighbourhood of a vertex of a triangulation, so the condition on the integral over $n$-simplexes does not play any role. The obstruction for the existence of $\beta$ comes form the local behaviour of curves at a vertex.
Lemma. Let $\alpha=dx\wedge dy$ on $\mathbb R^2$. For $n\ge 8$ there exist $\gamma_1,...,\gamma_n$, smooth rays on $\mathbb R^2$ that meet at $0$ with different tangent vectors and such that there is no $\beta$ defined in any neighbourhood of 0 with $d\beta=\alpha$ and vanishing been restricted to $\gamma_i$.
It is clear that this lemma implies the negative answer to a version of the question, when we are not allowed to deform the triangulation. 
Proof of Lemma. Suppose by contradiction that $\beta$ exists. Then $\beta_1=\beta-\frac{1}{2}(xdy-ydx)$ is a closed 1-form. So we can write $\beta_1=dF$, where $F$ is a function defined in a neighbourhood of $0$, $F(0)=0$. Since the number $n$ of the rays is more than $2$, $dF$ should vanish at zero. Moreover, it is not hard to see, that since the number of rays is more than $4$, the quadratic term of $F$ vanishes at zero too.
Now, since $\beta$ vanishes on $\gamma_i$, the restiction of $\beta_1$ to $\gamma_i$ equals $\frac{1}{2}(ydx-xdy)$. So we get the formula for $F$, resticted to $\gamma_i$
$$F=\frac{1}{2}\int_{\gamma_i}ydx-xdy.$$
Now,  we will chose the rays $\gamma_1,...,\gamma_8$. Namely $\gamma_1(t)=(t,t^2)$, $\gamma_2(t)=(t,t-t^2)$, and take $\gamma_3,...,\gamma_8$ by consecutively rotating $\gamma_1,\gamma_2$ by $\pi/2$, $\pi$, $3\pi/2$.
It is not hard to see, that $F$ is cubic modulo higher terms in $t$ when it is restricted to $\gamma_i$. At the same time $F$ is positive on $\gamma_{1},\gamma_3,\gamma_5, \gamma_7$ and negative restricted to other rays. So it changes its sign at lest $8$ times on a little circle surrounding $0$. This is impossible for a cubic Function (in a little neighbourhood the cubic term of $F$ should be dominating). Contradiction.<|endoftext|>
TITLE: Characterizations of UFD and Euclidean domain by ideal-theoretic conditions
QUESTION [18 upvotes]: This questions is inspired by an exercise in Hungerford that I have only partially solved. The exercise reads: "A domain is a UFD if and only if every nonzero prime ideal contains  a nonzero principal ideal that is prime." (For Hungerford, 'domain' means commutative ring with $1\neq 0$ and no zero divisors). 
One direction is easy: if $R$ is a UFD, and $P$ is a nonzero prime ideal, let $a\in P$, $a\neq 0$. Then factor $a$ into irreducibles, $a = c_1\cdots c_m$. Since $P$ is a prime ideal in a commutative ring, it is completely prime so there is an $i$ such that $c_i\in P$, and therefore, $(c_i)\subseteq P$. Since $c_i$ is a prime element (because $R$ is a UFD), the ideal $(c_i)$ is prime.
I confess I am having trouble with the converse, and will appreciate any hints.
But on that same vein, I started wondering if there was a similar "ideal theoretic" condition that describes Euclidean domains. Other classes of domains have ideal theoretic definitions: PID is obvious, of course, but less obvious perhaps are that GCD domains can be defined by ideal-theoretic conditions (given any two principal ideals $(a)$ and $(b)$, there is a least principal ideal $(d)$ that contains $(a)$ and $(b)$, least among all principal ideals containing $(a)$ and $(b)$), as can Bezout domains (every finitely generated ideal is principal). Does anyone know if there is an ideal theoretic definition for Eucldean domains?

REPLY [6 votes]: Although Pete Clark's answer is great, I thought I'd post a partial answer that addresses the UFD question in a different direction. My favorite ideal-theoretic characterization of UFDs is that a domain $R$ is a UFD if and only if every $t$-closed ideal of $R$ is principal.  The $t$-closure operation on the fractional ideals of a domain $R$ is given by $$t: I \longmapsto I^t = \bigcup\{(J^{-1})^{-1}: J \subseteq I \mbox{ is a finitely generated ideal of } R\},$$
where $J^{-1} = (R :_{Q(R)} J)$ and $Q(R)$ is the quotient field of $R$.  The $t$-closure operation is a useful closure operation on the fractional ideals of a domain $R$.  Krull introduced such $'$-Operations in the 1930s in his book Idealtheorie and some subsequent papers.  Today we call them star operations.
A UFD is equivalently a Krull domain with trivial divisor class group. A Krull domain is equivalently an integral domain $R$ such that $(II^{-1})^t = R$ for every nonzero ideal $I$ of $R$.  It follows from these two well known results of multiplicative ideal theory that a UFD is equivalently a domain in which every ideal $I$ such that $I^t = I$ is principal.  I like this characterization because it doesn't mention anything about principal prime ideals or even prime ideals.  After all, it is easy to see that a domain $R$ is a UFD if and only if every nonzero principal ideal is a product of principal prime ideals.  If you're allowed to mention principal prime ideals in the characterization, then such characterizations are easy to come by.
I would note that the notion of a principal ideal doesn't have a "purely" ideal-theoretic description, since they can't be recovered from the ideal lattice alone.  In particular, in my view even the notion of a PID does not have a purely ideal-theoretic description.  However, the notion of a Dedekind domain does: a domain $R$ is a Dedekind domain if and only if $II^{-1} = R$ for every nonzero ideal $I$ of $R$.  This all hinges on what you mean by ``ideal-theoretic.'' To me it means, can the notion be defined in terms of the ordered monoid of all ideals, or fractional ideals, of the ring. For example, the finitely generated ideals of a commutative ring $R$ have a purely ideal-theoretic description, namely, as the compact elements of the poset of all ideals of $R$.
By the way, a PID is a equivalently a Dedekind domain with trivial ideal class group, and this is generalized by the fact mentioned earlier that a UFD is equivalently a Krull domain with trivial divisor class group.<|endoftext|>
TITLE: From symmetric groups to symplectic groups?
QUESTION [8 upvotes]: The question is self-contained finite group theory but the motivation requires more background.
The finite groups I am interested in are the groups $Sp(n,F_3)$. For $n$ even these are the usual symplectic groups over the field with three elements. For $n$ odd these are the odd symplectic groups. These are a semi-direct product of a symplectic group with a Heisenberg group and we have a sequence of subgroups $Sp(n,F_3)\rightarrow Sp(n+1,F_3)$.
Consider one of these inclusions and look at induction/restriction of irreducible complex representations. My question is: take an irreducible representation of the subgroup and induce. Is this representation a direct sum of pair-wise non-isomorphic irreducible representations (sometimes this is called multiplicity free)?
I can prove this for $n$ even because the irreducible representations of $Sp(2m+1,F_3)$ can be constructed using Clifford theory (known to physicists as Mackey theory). I can use the computer for small $n$. So my question is really for $n$ odd.

I can give more information which hints at my interest. Define a sequence of groups $G(n)$ by a sequence of finite presentations so that we have surjective homomorphisms $B(n)\rightarrow G(n)$ where $B(n)$ is the usual braid group. Take generators $\sigma_1,\ldots ,\sigma_{n-1}$ and the Artin relations. In addition take $\sigma_i^3$ and for $n\ge 5$ take $(\sigma_1\sigma_2\sigma_3\sigma_4)^{10}=1$. Then we have $G(n)=Sp(n-1,F_3)$.
Then we also have $G(n)\times G(m)\rightarrow G(n+m)$ compatible with $B(n)\times B(m)\rightarrow B(n+m)$. This is similar to the symmetric groups.

I am aware that this question can be generalised. I have deliberately restricted to a simple example as I would like to have one case fully worked out before generalising. If you have a proof of the above and your proof generalises, that's different!

REPLY [4 votes]: Without working it out completely, I can't give a yes or no answer. 
But here's a start:  Let $G = Sp(2n)$ (a group of $2n$ by $2n$ matrices, in the notation I use) and let $K = Sp(2n-2) \ltimes H$, where $H$ is the appropriate Heisenberg group.  Let's work over any finite field of odd characteristic -- I'd guess that if what you want is true, then it's true in this generality.  
You wish to show (by Frobenius reciprocity) that restriction from $G$ to $K$ is multiplicity-free.  A standard way to accomplish this would be to prove the following:
Claim:  The convolution ring $A = C[K \backslash G / K]$ of $K$-bi-invariant functions on $G$ is commutative.
In other words, try to prove that $(G,K)$ is a Gelfand pair.  The standard method involves accomplishing the following:
Task:  Find an anti-involution $\sigma$ of $G$ (meaning $\sigma^2 = Id$ and $\sigma(gh) = \sigma(h) \sigma(g)$, such that every $K$-double-coset $K g K$ in $G$ is stable under $\sigma$, i.e., if $\sigma(g) \in K g K$ for all $g \in G$.
Such an involution yields an anti-automorphism of $A$, making it commutative -- this is the "Gelfand-Kazhdan method".
My advice:  try something like conjugation by a matrix (the matrix $J$ defining the symplectic form, perhaps) followed by the transpose, for the anti-involution.  It's up to you to analyze the double cosets, but I bet it's been done, at least in low rank (2n = 4, perhaps).  The double cosets can get unwieldy, but in your case it almost suffices to analyze the double cosets for the "Heisenerg parabolic" $P$ containing $K$.  The $P$ double cosets in $G$ can be analyzed via the Weyl group.  
Final advice:  the "odd symplectic groups" that you refer to are often called Jacobi groups in the literature due to their relevance to Jacobi forms.  Someone may have worked out some of this already!<|endoftext|>
TITLE: How are fiber bundles, transition functions and principal bundles related?
QUESTION [11 upvotes]: Please read the edit below. 
Is my understanding of this correct? Fix a sufficiently nice and connected topological space $B$ and a topological group $G$. A principal bundle $E\to B$ with structure group $G$ is (modulo equivalence) the same as a collection of transition functions $g_{\alpha,\beta}:V_{\alpha}\cap V_{\beta}\to G$. A fiber bundle $b:Y\to B$ with fiber $F$ defines a collection of transition functions $g_{\alpha,\beta}:V_{\alpha}\cap V_{\beta}\to \mathrm{Aut}(F)$ but you cannot get the fiber bundle back from the transition functions in general. You get only a principal bundle $p:E\to B$ with structure group $G=\mathrm{Aut}(F)$ by the above mentioned equivalence. How does this principal bundle look like (in relation to the fiber bundle $b$) intuitively?
The additional information you need to get a more general fiber bundle from a principal bundle $p:E\to B$ with structure group $G$ is a space $F$ with an action of $G$. Then you define a fiber bundle
$$
q:E\times F/\sim\to B
$$
where the relation is generated by $(x,y)\sim(xg,gy)$ and the equivalence class of $(x,y)$ is maped to $p(x)$. This bundle has fiber $F$. Applying this construction to the bundle $b$ from above, I suppose that with the obvious action of $G=\mathrm{Aut}(F)$ on $F$ you get the bundle $q\cong b$ back, right? 
If the fiber bundle $b$ is a vector bundle (which is generally not a principal bundle) one does not need the second step: They are equivalent to their associated transition functions. Why is this (intuitively) true?
Can someone help me to clarify my picture? Is it really important to take care of the direction of the actions (left action on F, right action on E)? Intuitively, I think that a principal bundle does somehow encode global information where the second construction brings a local action on the fiber in.
Edit: Thank you all for the clarification. Are the following two statements correct?

Fix a topological group $G$ and a topological space $B$. Does this mean that there is an isomorphism of categories
$$
X\times Y\to Z
$$
where $X$ is the category of topological spaces with a left $G$-action (and $G$-equivariant morphisms), $Y$ is the category of principal bundles over $B$ with structure group $G$ modulo isomorphism (and only identity morphisms) and $Z$ is the category of fiber bundles over $B$ with structure group $G$ (and morphisms over $B$)?
Now fix a topological space $F$ and a topological space $B$. Does this mean that there is an isomorphism of categories
$$
X'\times Y'\to Z'
$$
where $X'$ is the category of left $\mathrm{Aut}(F)$-actions on $F$ (and $\mathrm{Aut}(F)$-equivariant isomorphisms (?)), $Y'$ is the category of principal bundles over $B$ with structure group $\mathrm{Aut(F)}$ modulo isomorphism (and only identity morphisms) and $Z'$ is the category of fiber bundles over $B$ with fiber $F$ (and isomorphisms over $B$)?

REPLY [7 votes]: For comparing vector bundles to principal GL$(n)$ bundles, there is a second, more canonical viewpoint that you may find useful.  If $E \to B$ is a vector bundle, then the bundle of $n$-frames in E is a principal GL$(n)$ bundle over $B$, denoted by Fr$(E) \to B$.  This bundle consists of all $n$-tuples $(v_1, \ldots, v_n)$ such that the $v_i$ form a basis for some fiber of $B$.  The general linear group acts on Fr$(E)$ by the following formula, which just mimics matrix multiplication:
$$((v_1, \ldots, v_n) \cdot (a_{ij}) = (\sum a_{i1} v_i, \ldots, \sum a_{in} v_i),$$
where $A = (a_{ij}) \in \textrm{GL}(n)$.
You can check that Fr$(E)$ is a principal GL$(n)$ bundle, and that for any choice of transition functions for the bundle $E$, the principal GL$(n)$ bundle built from that data is isomorphic to Fr$(E)$.<|endoftext|>
TITLE: Examples for which it is not known if Grothendieck's Standard Conjectures hold.
QUESTION [17 upvotes]: What is the simplest example of a projective variety for which it is not known if Grothendieck's Standard Conjectures hold?
What is the simplest example of a flag manifold for which it is not known if Grothendieck's Standard Conjectures hold?

REPLY [7 votes]: It seems that for flag manifolds everything is very easy.:) For other varieties much less is known. For example, recently there was an attempt to construct a counterexample to the Hodge conjecture in the Cartesian square of a K3 surface:
http://adsabs.harvard.edu/abs/2006math......8265K
It seems that a mistake was found later (I'm not quite sure); yet probably nobody at the moment knows how to prove the conjecture in this case.<|endoftext|>
TITLE: Which journals publish expository work?
QUESTION [282 upvotes]: I wonder if anyone else has noticed that the market for expository papers in mathematics is very narrow (more so than it used to be, perhaps).
Are there any journals which publish expository work, especially at the "intermediate" level?  By intermediate, I mean neither (i) aimed at an audience of students, especially undergraduate students (e.g. Mathematics Magazine) nor (ii) surveys of entire fields of mathematics and/or descriptions of spectacular new results written by veteran experts in the field (e.g. the Bulletin, the Notices).
Let me give some examples from my own writing, mostly just to fix ideas.  (I do not mean to complain.)

About six years ago I submitted an expository paper "On the discrete geometry of Chicken McNuggets" to the American Mathematical Monthly.  The point of the paper was to illustrate the utility of simple reasoning about lattices in Euclidean space to give a proof of Schur's Theorem on the number of representations of an integer by a linear form in positive integers.  The paper was rejected; one reviewer said something like (I paraphrase) "I have the feeling that this would be a rather routine result for someone versed in the geometry of numbers."  This shows that the paper was not being viewed as expository -- i.e., a work whose goal is the presentation of a known result in a way which will make it accessible and appealing to a broader audience.  I shared the news with my officemate at the time, Dr. Gil Alon, and he found the topic interesting.  Together we "researchized" the paper by working a little harder and proving some (apparently) new exact formulas for representation numbers.  This new version was accepted by the Journal of Integer Sequences:

https://cs.uwaterloo.ca/journals/JIS/VOL8/Clark/clark80.html
This is not a sad story for me overall because I learned more about the problem ("The Diophantine Problem of Frobenius") in writing the second version with Gil.  But still, something is lost: the first version was a writeup of a talk that I have given to advanced undergraduate / basic graduate audiences at several places.  For a long time, this was my "general audience" talk, and it worked at getting people involved and interested: people always came up to me afterward with further questions and suggested improvements, much more so than any arithmetic geometry talk I have ever given.  The main result in our JIS paper is unfortunately a little technical [not deep, not sophisticated; just technical: lots of gcd's and inverses modulo various things] to state, and although I have recommended to several students to read this paper, so far nothing has come of it.

A few years ago I managed to reprove a theorem of Luther Claborn (every abelian group is isomorphic to the class group of some Dedekind domain) by using elliptic curves along the lines of a suggestion by Michael Rosen (who reproved the result in the countable case).  I asked around and was advised to submit the paper to L'Enseignement Mathematique.  In my writeup, I made the conscious decision to write the paper in an expository way: that is, I included a lot of background material and explained connections between the various results, even things which were not directly related to the theorem in question.  The paper was accepted; but the referee made it clear that s/he would have preferred a more streamlined, research oriented approach.  Thus EM, despite its name ("Mathematical Education"), seems to be primarily a research journal (which likes papers taking new looks at old or easily stated problems: it's certainly a good journal and I'm proud to be published in it), and I was able to smuggle in some exposition under the cover of a new research result.

I have an expository paper on factorization in integral domains:


http://alpha.math.uga.edu/~pete/factorization.pdf
[Added: And a newer version: http://alpha.math.uga.edu/~pete/factorization2010.pdf. ]
It is not finished and not completely polished, but it has been circulating around the internet for about a year now.  Again, this completely expository paper has attracted more attention than most of my research papers.  Sometimes people talk about it as though it were a preprint or an actual paper, but it isn't: I do not know of any journal that would publish a 30 page paper giving an intermediate-level exposition of the theory of factorization in integral domains.  Is there such a journal?
Added: In my factorization paper, I build on similar expositions by the leading algebraists P. Samuel and P.M. Cohn.  I think these two papers, published in 1968 and 1973, are both excellent examples of the sort of "intermediate exposition" I have in mind (closer to the high end of the range, but still intermediate: one of the main results Samuel discusses, Nagata's Theorem, was published in 1957 so was not exactly hot off the presses when Samuel wrote his article).  Both articles were published by the American Mathematical Monthly!  I don't think the Monthly would publish either of them nowadays.
Added: I have recently submitted a paper to the Monthly:
http://alpha.math.uga.edu/~pete/coveringnumbersv2.pdf
(By another coincidence, this paper is a mildly souped up answer to MO question #26.  But I did the "research" on this paper in the lonely pre-MO days of 2008.)
Looking at this paper helps me to see that the line between research and exposition can be blurry.  I think it is primarily an expository paper -- in that the emphasis is on the presentation of the results rather than the results themselves -- but I didn't have the guts to submit it anywhere without claiming some small research novelty: "The computation of the irredundant linear covering number appears to be new."  I'll let you know what happens to it.
(Added: it was accepted by the Monthly.)

REPLY [2 votes]: SIAM Review (applied math, not pure math)
SIAM Review (SIREV) consists of the following five sections, all containing articles of broad interest:

Survey and Review features papers with a deliberately integrative and up-to-date perspective on a major topic in applied or computational mathematics or scientific computing.

...<|endoftext|>
TITLE: Tools for the Langlands Program?
QUESTION [35 upvotes]: Hi,
I know this might be a bit vague, but I was wondering what are the hypothetical tools necessary to solve the Langlands conjectures (the original statments or the "geometic" analogue). What I mean by this is the following: for the Weil Conjectures it became clear that, in order to prove them, one needed to develop a marvelous cohomology theory that would explain Weil's observations.  Of course, we all know that etale cohomology is that marvelous tool.  By analogy, what "black box" tools are necessary for the Langlands program? Broadly speaking what tools do we need for the Langlands program?
Curious grad student,
Ben

REPLY [7 votes]: I don't know a whole lot about the Langlands program, but if there is one tool that seems to come up a lot in geometric Langlands, it's perverse sheaves.  You see a lot of singular algebraic varieties in geometric Langlands, and perverse sheaves are meant as a singular generalization of a vector bundle with a flat connection.  Ordinary sheaves are already a singular generalization of vector bundles, but not the relevant one.  Perverse sheaves (which are made from sheaves but not sheaves themselves) are a more apropos generalization that incorporates and sort-of just is intersection (co)homology.
I can also say that I wasn't going to learn about perverse sheaves until I had to.  However, I have now seen several important papers, in the related categorification program, that read this way:  "Perverse sheaves + necessary restrictions = a good solution".  So now I might be slowly getting used to them.  I can also see that even the formalism perverse sheaves or intersection homology is sort-of inevitable.  In some of the simpler constructions, the varieties (over $\mathbb{C}$, say) are non-singular and certain answers arise as ordinary cohomology products or intersection products.  For instance, the Schubert calculus in a Grassmannian manifold.  What choice do you have if the Grassmannian is replaced by a singular variety $X$?  For some of these categorification/Langlands questions, you can either propose wrong answers, or ad hoc answers, or you can automatically get the right answer by using intersection homology on $X$.  (With middle perversity, as they say.)<|endoftext|>
TITLE: How do you compute the space of lifts of an E-infinity map?
QUESTION [13 upvotes]: Let X, Y and B be $E_\infty$ spaces, and let $p: X \rightarrow Y$ and $f: B \rightarrow Y$ be $E_\infty$ maps.  We can ask for the space of lifts of f across p, that is the space of $E_\infty$ maps $g: B \rightarrow X$ such that $pg = f$.

Q1: What spectral sequences or other technology exists for computing the (homotopy groups of the) space of such E-infinity lifts?

Any $E_\infty$ lift $g: B \rightarrow X$ provides a lift of the map of commutative monoids $\pi_0 B \rightarrow \pi_0 Y$ across the map $\pi_0 X \rightarrow \pi_0 Y$.  

Q2: Do all the techniques for computing E-infinity lifts in effect require that you first solve this $\pi_0$ lifting problem in commutative monoids, and begin an obstruction calculation from there, or are there techniques that solve both the $\pi_0$ problem and the E-infinity problem 'simultaneously' and perhaps in a way that eases both computations?

I am particularly interested in merely knowing if there exists an $E_\infty$ lift, thus might have asked the seemingly more basic question:

Q1': When is there an E-infinity lift of an E-infinity map?

I think the obstruction groups answering Q1' are liable to come packaged in the answers to Q1/Q2, but if there are separate techniques for the existence question, that would also be helpful.
Remark: I could imagine that one answer to Q1 involving relative Andre-Quillen cohomology might be extracted from Goerss-Hopkins, Moduli Problems for Structured Ring Spectra, but perhaps there are more elementary means.  I'd be very interested for answers to Q1 along those or especially other lines of thought, and any ideas about Q2.  Thanks!

REPLY [5 votes]: The issue with the underlying monoid seems to complicate everything, in a similar way to how Postnikov decompositions are complicated by the $\pi_1$ issue.  In that case, the common technique is to fix $G = \pi_1$ and consider the Postnikov tower as operating in the category of spaces over $K(G,1)$ rather than in the ordinary category of spaces.
So I don't see a lot of hope immediately for getting the $\pi_0$ problem out of the way at the same time; it seems like it colors the whole problem.
Once you've decided on a lifting $\pi_0 B \to \pi_0 X$, though, you can fix the underlying monoid because then you're reduced to studying lifts $B \times_{\pi_0 Y} \pi_0 X \to  X$ over $\pi_0 X$.
If you then fix $M = \pi_0 X$ then there's certainly some kind of obstruction theory, but the problem is identifying the obstruction classes as coming from something cohomological that you can actually calculate.  It seems to me that one should study the symmetric monoidal category of "spaces over $M$", with product having fibers
$$
(X \star Y)_m = \coprod_{m' m'' = m} (X_{m'} \times Y_{m''})
$$
(which is some kind of left Kan extension), and try to get some handle on it.
Even when $M = \mathbb{N}$ the bookkeeping gets complicated.  Then you're studying "graded $E_\infty$ spaces" and your obstruction theory will land in something like cohomology with coefficients in the relative homotopy groups of $Y$ over $B$, but you're taking cohomology of the "derived indecomposables" in your $E_\infty$ space.  The zero'th space of derived indecomposables of an $E_\infty$ space $B$ over $\mathbb{N}$ is the topological Andr\'e-Quillen homology object of $B_0$.  Even if $B_0$ is trivial, then the zero'th derived indecomposable space is trivial, the first is $B_1$, and the next is the homotopy cofiber of the squaring map $(B_1 \times B_1)_{h\mathbb{Z}/2} \to B_2$.
Based on this kind of futzing around I am led to believe that your obstructions may possibly occur in the relative topological Andre-Quillen cohomology of $\Sigma^\infty_+ B$ over $\Sigma^\infty_+ M$ with coefficients in the relative homotopy of $X$ over $Y$.  But the problem seems very difficult for a general monoid $M$.
(Especially evidenced by the fact that Charles Rezk hasn't popped in here with an answer yet.)<|endoftext|>
TITLE: Coboundary Representations for Trivial Cup Products
QUESTION [9 upvotes]: Suppose $G$ is a pro-$p$-group, $p$ odd, and $\mathbb{F}_p$ is given the trivial $G$-action.  By skew-symmetry of the cup-product in degree 1, given $\chi\in H^1(G,\mathbb{F}_p)$, we have $\chi\cup\chi=0\in H^2(G,\mathbb{F}_p)$.  In fact, in this case, it's even possible to explicitly write $\chi\cup\chi$ as a coboundary -- $\chi\cup\chi=d\left(\binom{\chi}{2}\right)$, the coboundary of "$\chi$ choose 2".  
In any case, my question is whether or not there anyone has seen any other tricks of this sort, i.e., for the explicit realization of a trivial cup product as a coboundary.  In my specific case, I know a particular cup product is zero since I can force it, via the $G$-equivariance of the cup-product, to land in a known-to-be-trivial eigenspace of $H^2$.  I was hoping there was some "eigenspace-averaging" trick similar to the construction of orthogonal idempotents to get my hands on an explicit pre-image, but really, I'd just like to be aware of any tricks for doing this.

REPLY [2 votes]: You can make explicit an homotopy showing that the map $(\chi,\xi)\mapsto \chi\smile\xi\pm\xi\smile\chi$ is homotopic to zero. Using it, you can generalize your formula for $\chi\smile\chi$.<|endoftext|>
TITLE: Reference for quantum Schur-Weyl duality
QUESTION [5 upvotes]: I am trying to prove a version of quantum Schur-Weyl duality. I hope to be able to generalize the proof of the Schur-Weyl duality between $U_q(\mathfrak{gl}_n)$ and the Hecke algebra $H_r$. So I am looking for a good reference for this with a careful proof. It would also be nice to see a proof that uses the quantum coordinate ring of $GL_n$ instead of the enveloping algebra (and therefore is phrased in terms of decomposing $V^{\otimes r}$ as a left-comodule for this coordinate ring and a right $H_r$-module).

REPLY [2 votes]: Too late maybe but still - "Noncommutative symmetric functions V: a degenerate version of $U_q(\mathfrak{gl}_N)$" by Krob and Thibon has the quantum coordinate ring version very carefully (I think) written. It is aimed at the $q=0$ case but the generic $q$ is treated in detail too.
Later - found the book "Algebras of Functions on Quantum Groups" by Korogodski and Soibelman (1998), it contains a thorough description of the quantization of algebras of functions on groups.<|endoftext|>
TITLE: Whitehead products on manifolds
QUESTION [7 upvotes]: What are some good examples of simply connected manifolds with interesting Whitehead Lie algebras over R? Most of the manifolds that one thinks about if one is pretty naive are not so interesting--- Lie groups have abelian Whitehead algebras and homogeneous spaces have no higher product structure.

REPLY [8 votes]: I guess that by the Whitehead Lie algebra, you mean the homotopy group Lie algebra $\pi_*(\Omega X)\simeq \pi_{*-1}(X)$ maybe tensored by the reals $R$. In that case there is a theorem of Felix-Halperin-THomas, called the dichotomy theorem which tells you that either this Lie algebra is finite-dimensional (and the space is said to be "elliptic"), or it is very big in the sense that the ranks of $\pi_k(X)$ grows exponentially with k (and the space is then called "hyperbolic"). If the Euler characteristic of the manifold is negative then the space is always hyperbolic/ Moreover when the space is hyperbolic the Whitehead Lie algebra is very far from being abelian: actually its radical is finite dimensional. Therefore any manifold with negative euler characteristic has an non abelian infinite dimensional homotopy Lie algebra.
To generalize what Ryan says, actually any  connected sum of two simply connected manifolds $M$ and $N$ is hyperbolic unless  the cohomology of both $M$ and $N$ are truncatated polynomial algebras on a single genrator (like the sphere or $CP(n)$). In particular the connected sum of 3 or more closed manifolds not having the rational homotopy type of a sphere is hyperbolic.
Another example of a non abelian Whitehead Lie algebra but finite dimensional, is the one associated to a manifold $M$ obtained as an $S^5$-bundle with base $S^3\times S^3$ and where the euler class of the bundle is the fundamental class of the base (or any non zero multiple of it). In that case the Whitehead rational Lie algebra $\pi_*(M)\otimes Q$ is of dimension $3$ with basis  $x,y,[x,y]$ where $x$ and $y$ are in degree $3$ and $[x,y]$ is in degree $5$. Thus this manifold M is elliptic. Interestingly enough, the cohomology algebra of M is isomorphic to that of the connected sum $W$ of two copies of $S^3\times S^8$, but $W$ is hyperbolic.<|endoftext|>
TITLE: An ubiquitous pattern of questions
QUESTION [9 upvotes]: There is an ubiquitous pattern of questions concerning assumedly any kind of mathematical object or structure: groups, graphs, numbers, categories, and so on. It goes like this (informally):

Can a class of objects or structures of a given kind X that is characterized by some "external condition" Y be defined by a condition Z in their respective "internal"
  language, and if so: how?

Well-known examples ("external condition" = "internal condition"):

groups $G$ isomorphic to a subgroup of the symmetric group on $G$ = all groups (Cayley's theorem)
graphs embeddable in the plane = graphs not containing a subgraph that is a subdivision of $K_5$  or $K_{3,3}$ (Kuratowski's theorem)
numbers n of trees on k labeled vertices = numbers n = kk-2 for some k > 1 (Cayley's theorem on trees)
numbers n with only one group of order n = numbers n = p1 · p2 ·  ... · pk for some k > 0, where the pi are distinct primes and no pj-1 is divisible by any pi (cyclic numbers, see Sloane's A003277)

Further examples from MO:

Which graphs are Cayley graphs?
Can we recognize when a category is
equivalent to the category of models
of a first order theory?
Can you determine whether a graph is
the 1-skeleton of a polytope?


Question #1: What's the proper way to
  characterize this pattern of
  questions? What's the common
  context / rationale?



Question #2: How is the introductory question to be
  posed properly?

REPLY [16 votes]: There are three sibling theorems of logic which guarantee that such characterizations are bound to happen.

The Craig Interpolation Theorem
The Robinson Joint Consistency Theorem
The Beth Definability Theorem

[The wikipedia entries need some work. I suggest you look up these theorems in a good logic book, for example Hodges's Model Theory or his more accessible Shorter Model Theory.]
I will focus on the Beth Definability Theorem, though the other two siblings lead to similar conclusions in slightly different contexts.
Suppose you have a first-order language L0 and a larger language L. Let T be a theory in L and let φ(x) be a formula of the larger language L with the following property. Whenever A1 and A2 are two models of T which have the same universe A and the same interpretation for all parts of the small language L0, then A1 ⊧ φ(a) iff A2 ⊧ φ(a) for all a ∈ A. Beth's Definability Theorem says that there must be a formula φ0(x) of the smaller language L0 such that T ⊦ ∀x(φ(x) ↔ φ0(x)).
The connection with your question is as follows. The base language L0 is the 'internal' language of the structures you really care about, while the larger language L has some additional 'external' data. The theory T characterizes the structures with external data that you care about, and φ(x) is a property of such structures that you are interested in. If φ(x) is sufficiently independent of the external data, then φ(x) must be equivalent to an internal formula φ0(x).
Not all of the examples you give are easily cast into this formalism, but the basic flavor is the same. Unfortunately, the Beth Definability Theorem (and its proof) does not say much on how to find the internal formula φ0(x) but, at least, it says that the search will not be in vain.<|endoftext|>
TITLE: Relationship between Line Bundles with isomorphic ring of sections
QUESTION [5 upvotes]: Suppose two positive holomorphic line bundles $L_1 \to X_1, L_2\to X_2$ over two projective complex manifold $X_1, X_2$ have isomorphic ring of sections $R=R_1=R_2$ where $R_i=\oplus_{m=0}^\infty\Gamma(X_i,mL_i)$. Isomorphism as graded ${\mathbb C}$- algebras.
Is there any relationship betweeen $X_1$ and $X_2$? Eg, some morphism between them? How about relationship to $Proj R$?
Thanks.

REPLY [4 votes]: To expand on the answer above: as B. Cais says, if the line bundles are ample (which I think follows from positivity by Kodaira), we have a canonical isomorphism $\mathrm{Proj} R_i\cong X_i$.  Thus, if the graded rings $R_i$ are isomorphic, then the induced map of Proj's gives an isomorphism $R_1\cong R_2$ carrying one line bundle to the other.<|endoftext|>
TITLE: finite generated group realized as fundamental group of manifolds
QUESTION [28 upvotes]: This is discussed in the standard textbooks on algebraic topology.
Pick a presentation of the group $G = \langle g_1,g_2,...,g_n|r_1,r_2,...r_m \rangle$
where $g_i$ are generators and $r_j$ are relations. Then we have a wedge of $n$ circles and attach two-cells to the wedge sum according to the relations $r_j$. Denote the final space $X$. Then van Kampen says $\pi_1(X)=G$.
While usually $X$ is not a manifold, it is well-known that every finitely generated group $G$ can be realized as the fundamental group of some 4-manifold $X$. Can someone sketch the proof? Also, if $X$ could not 
be some manifold of dimension $<4$, what is the obstruction?

REPLY [3 votes]: Yet another explanation of the same constructions given above is to add 1 and 2 handles to the 4 ball according t the given presentation, obtaining a 4 manifold $X$ with boundary. Now the boundary of $X\times I$ (i.e. the double of $X$) has the same fundamental group by Van Kampen and the fact that $\partial X\subset X$ induces a surjection on fundamental groups (turning $X$ upside down shows that $X$ is obtained from $\partial X$ by adding 2 and 3 handles).
Since the first homology=abelianization of $\pi_1$  of closed 1 and 2-manifolds are known, it is easy to see most groups dont occur for $n=1$ or $2$. For $n=3$, another algebraic obstruction is to observe that if $\pi=\pi_1(M^3)$, then $H_2(M)\to H_2(\pi)$ is onto, and if $M$ is orientable, then $H_2(M)=H^1(M)=H^1(\pi)$. So if $H^1(\pi)$ is smaller than $H_2(\pi)$, it cannot occur (for an oriented 3-manifold, in any case).<|endoftext|>
TITLE: Infinitely many prime numbers of the form $n^{2^k}+1$?
QUESTION [7 upvotes]: I am not a specialist of number theory, so please excuse my ignorance: is the following question still an open problem?
Let $k \in \mathbb{N}^*$, are there infinitely many prime numbers of the form $n^{2^k}+1$?

REPLY [4 votes]: It took me a while to find this: http://www.pnas.org/content/94/4/1054.full 
Anyway by Friedlander and Iwaniec (1997). They proved that there are infinitely many primes of the form $x^2 + y^4 .$ They mention near the end that they do not have a proof for primes of the form $x^2 + y^6 $ but would like one. So there is a way to go to settle $x^2 + 1.$
FYI, what I did (not remembering title, authors, anything but the result) was write a program to give the primes $x^2 + y^4 $ and put the first dozen in Sloane's sequence site search feature.<|endoftext|>
TITLE: A slick proof of the Bruhat Decomposition for GL_n(k)?
QUESTION [39 upvotes]: On one of my exams last year, we were given a problem (we chose five or six out of eight problems) on an exam, the goal of which was to prove the Bruhat decomposition for $GL_n(k)$.  I was one of the two people to choose said problem.  I gave a very long convoluted argument which although correct was really inelegant.  I proved it more than once because I wasn't satisfied with my proof, and I figured out a somewhat slick contradiction argument based on maximizing leading zeroes of rows (number of zeroes before the pivot), but the proof was still a real mess.
Statement of the problem:
Let $G:=GL(V)$ for $V$ a finite dimensional $k$ vector space.  Let $B$ be the stabilizer of the standard flag (these will be invertible upper triangular matrices), and let $W$ be the subgroup of permutation matrices.
Show that $G=\coprod_{w\in W} BwB$, where the $BwB$ are double cosets.  That is, show that $G=BWB$.  
Question:
Is there a slick proof of this fact maybe using "more machinery"? In particular, is there any sort of "coordinate-free" proof (Can we even define the Borel and Weyl subgroups without coordinates?)?

REPLY [17 votes]: One can prove the Bruhat decomposition by applying the theorem of Jordan-Hölder.
This theorem shows that two chains of submodules of a module of finite length
 whose successive quotients
are simple have the same lengths, and that the same  quotients appear.
But it is slightly more precise, because it gives a precise recipe
for a bijection between the two lists. Here we apply it for modules of finite length over a field,
aka finitely dimensional vector spaces.
Let $E=(e_1,\dots,e_n)$ and $F=(f_1,\dots,f_n)$ be two bases of  a vector space $M$
over a field $K$.
For $0\leq i\leq n$, define
 $M_i=\langle e_1,\dots,e_i\rangle $ and $N_i=\langle f_1,\dots,f_i\rangle$.
The proof of the theorem of Jordan-Hölder furnishes
a (unique) permutation $\sigma$ of $\{1,\dots,n\}$
such that $M_{i-1}+M_i\cap N_{\sigma(i)-1}=M_{i-1}$
and  $M_{i-1}+M_i\cap N_{\sigma(i)}=M_i$,
for every $i\in\{1,\dots,n\}$. 
For any $i$, let $x_i$
be a vector belonging to $M_{i}\cap N_{\sigma(i)}$
but not to $M_{i-1}$.
For every $i$, one has $\langle x_1,\dots,x_i\rangle =M_i$;
it follows that $X=(x_1,\dots,x_n)$ is a basis of $M$;
moreover, there exists a matrix $B_1$, in upper triangular form,
such that $X=E B_1$.
Set $\tau=\sigma^{-1}$.
Similarly, one has $\langle x_{\tau(1)},\dots,x_{\tau(i)}\rangle=N_i$
for every $i$. Consequently, there exists a matrix $B_2$,
still in upper-triangular form, such that
$(x_{\tau(1)},\dots,x_{\tau(n)})=F B_2$.
Let $P_\tau$ be the permutation matrix associated to $\tau$,
we have $(x_{\tau(1)},\dots,x_{\tau(n)})=(x_1,\dots,x_n)P_\tau$.
This implies that $FB_2=EB_1P_\tau$, hence
$F=E B_1 P_{\tau} B_2^{-1} $.
Therefore, the matrix $A=B_1P_{\tau}P_2^{-1}$
that expresses the coordinates of the vectors of $F$
in the basis $E$
is the product of an upper-triangular matrix, a permutation matrix
and another upper-triangular matrix.
In the group $\mathop{\rm GL}(n,K)$, let  $B$ be the subgroup consisting
of upper-triangular matrices, and let $W$ be the subgroup
consisting of permutation matrices.
We have proved that $\mathop{\rm GL}(n,K)=BWB$: this is precisely the Bruhat decomposition.<|endoftext|>
TITLE: Which spaces are inverse limits of discrete spaces ?
QUESTION [29 upvotes]: There is the following theorem:
"A space $X$ is the inverse limit of a system of discrete finite spaces, if and only if $X$ is totally disconnected, compact and Hausdorff."
A finite discrete space is totally disconnected, compact and Hausdorff and all those properties pass to inverse limits. I guess the other direction might be proved by taking the system of all decompositions of $X$ into disjoint clopen sets. The inverse limit should give $X$ back.
So what happens, if I dismiss the finiteness condition. As mentioned above every inverse limit of discrete spaces is totally disconnected, Hausdorff. So the question is:
"Which totally disconnected Hausdorff spaces are inverse limits of discrete spaces?"
For example I think it is impossible to write $\mathbb{Q}$ as an inverse limit of discrete spaces, but I don't have a proof.

REPLY [10 votes]: The rational numbers are not the inverse limit of a countable sequence of discrete spaces, but the rational numbers are in fact the inverse limit of an uncountable collection of discrete spaces. 
Another way to see that the rational numbers are not an inverse limit of a countable sequence of discrete spaces is to first take note that an inverse limit of a countable sequence of discrete spaces is metrizable by a complete metric. On the other hand, every completely metrizable subset of $\mathbb{R}$ is a $G_{\delta}$-set [DUG p. 307]. If $\mathbb{Q}$ were the intersection of countably many open sets $O_{n}$, then each $O_{n}$ would be dense making $\mathbb{Q}$ of second category. This is a contradiction. Therefore $\mathbb{Q}$ is not completely metrizable and not the inverse limit of a sequence of discrete spaces.
The rational numbers are in fact an inverse limit of discrete spaces. First take note that $\mathbb{Q}$ is
Lindelof and regular, so $\mathbb{Q}$ is realcompact [WAL p. 41]. Another way to see that $\mathbb{Q}$ is realcompact is to take note that $\mathbb{Q}$ is paracompact and of cardinality below the first measurable cardinal. Furthermore, since $\mathbb{Q}$ is Lindelof and zero-dimensional. $\mathbb{Q}$ is strongly zero-dimensional[WAL p. 85]. Therefore since $\mathbb{Q}$ is realcompact and strongly zero-dimensional, $\mathbb{Q}$ is $\mathbb{N}$-compact[WAL p. 264]. Therefore since $\mathbb{Q}$ is $\mathbb{N}$-compact, $\mathbb{Q}$ is the inverse limit of discrete spaces[CHE].
The spaces which are inverse limits of discrete spaces are precisely the spaces with a compatible complete ultrauniformity as it was pointed out earlier. We shall call a topological space ultracomplete if it can be given a compatible complete ultrauniformity. Let $X$ be a zero-dimensional space. Then let $\mathcal{U}$ be the uniformity generated by equivalence relations $E$ such that each equivalence class in $E$ is a clopen set.  Then we shall call $\mathcal{U}$ the fine ultrauniformity on the topological space $X$.  One can show that a zero-dimensional space $X$ is ultracomplete if and only if $X$ is complete in the fine ultrauniformity.
On a different note, every inverse limit of discrete spaces is a closed subspace of a product of discrete spaces [DUG p. 429]. Furthermore, one can easily show that every closed subspace of a product of discrete spaces can be given a compatible complete ultrauniformity, and hence the closed subspaces of products of discrete spaces can be written as inverse limits of discrete spaces. Therefore the spaces representable as inverse limits of discrete spaces are the spaces representable as closed subspaces of products of discrete spaces. Furthermore, for spaces of cardinality below the first measurable cardinal, the N-compact spaces correspond with the spaces representable as inverse limits of discrete spaces.
[CHE] Chew, Kim-Peu. "N-compact Spaces as Limits of Inverse Systems of Discrete Spaces." Journal of the Australian Mathematical Society 14.04 (1972): 467.
[DUG] Dugundji, James. Topology. Boston: Allyn and Bacon, 1966.
[WAL] Walker, Russell C. The Stone-Cech Compactification. Berlin: Springer-Verlag, 1974.<|endoftext|>
TITLE: Examples of eventual counterexamples
QUESTION [177 upvotes]: Define an "eventual counterexample" to be

$P(a) = T $ for $a < n$

$P(n) = F$

$n$ is sufficiently large for $P(a) = T\ \ \forall a \in \mathbb{N}$ to be a 'reasonable' conjecture to make.


where 'reasonable' is open to interpretation, and similar statements for rational, real, or more abstractly ordered sets for $n$ to belong to are acceptable answers.
What are some examples of eventual counterexamples, famous or otherwise, and do different eventual counterexamples share any common features? Could we build an 'early warning system' set of heuristics for seemingly plausible theorems?
edit: The Polya conjecture is a good example of what I was trying to get at, but answers are not restricted to number theory or any one area.

REPLY [5 votes]: I'm late to the party, but here's one from algebraic number theory.
The ring of integers of $\mathbb Q(\sqrt[n]2)$ is exactly $\mathbb Z[\sqrt[n]2]$ for $2\leq n \leq 1092$. At $n =1093$, the ring of integers is bigger. One can show that $\displaystyle\frac{(\sqrt[1093]{2}-2)^{1092}}{1093}$ is an algebraic integer, but is not in $\mathbb Z[\sqrt[1093]2]$.
Keith Conrad has a nice paper on this: https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf<|endoftext|>
TITLE: Variants of Eisenstein irreducibility
QUESTION [10 upvotes]: In his article where he stated what we know as Eisenstein's irreducibility criterion (which actually was first proved by Schönemann, as was Scholz's reciprocity law and Hensel's Lemma), he claimed that the result also holds in the following case: assume that
$$ f(x) = x^m + a_{m-1}x^{m-1} + \ldots + a_0, $$
where the $a_i$ are integers divisible by $p$, and where $p^2 \nmid a_j$ for some 
$0 \le j \le m-1$. There is an obvious counterexample provided by the quadratic polynomial $f(x) = (x-p)^2 = x^2 -2px + p^2$, so that could be the end of that story. But it isn't: I've read somewhere that some form of this criterion holds for polynomials of degree $\ge 3$, and that the degrees of the possible factors of counterexamples can be predicted in terms of this index $j$. Unfortunately, I don't remember the exact statement (those who are familiar with Newton polygons will probably be able to figure out a correct version) or where I've seen this. Can anyone help?

REPLY [2 votes]: Franz, if memory serves correct there is extensive discussion of variants of Eisenstein's criterion and relations with Newton polygon's etc in some of Filaseta's work, e.g. in his interesting book (draft) "The theory of irreducible polynomials". It was previously available at [1] but you may now need to write him for a password to access it. He also has some software available, e.g. Java applets for computing Newton polygons, etc.
[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/<|endoftext|>
TITLE: Functorial characterization of morphisms of schemes
QUESTION [7 upvotes]: This question is akin in spirit to this one:
Functorial characterization of open subschemes?
In the above MO question, a "functorial" characterization is given for closed immersions and open immersions. I am wondering if there are similar characterizations for concepts such as universally closed, separated, proper, projective etc.? Or maybe there are characterizations in a different flavor? In particular, I am wondering if we work over functors (i.e., generalization of schemes), are there things like separated/proper/etc morphism between functors?
Any reference or comments would be appreciated!

REPLY [2 votes]: Another point of view
if you follow the page you quote: Functorial characterization of open subschemes. There are also correspondence notions for separatedness and properness and so on. Let me elaborate a bit.
What you do is to identify a commutative scheme $X$ with $Qcoh_{X}$(Gabriel-Rosenberg reconstruction theorem). Let$f_{*}$=$F$
:$Qcoh_{X}\rightarrow Qcoh_{Y}$ (Assume $X,Y$ are quasi compact and quasi separated).
Affineness
$F$ is affine if $f_{*}$ is conservative(faithful in abelian case),having left adjoint functor
$f^{*}$ and having right adjoint functor $f^{!}$
closed immersion
Let $C_{X}=Qcoh_{X}$ and $C_{U}=Qcoh_{U}$.(Suppose they are abelian categories). Then $C_{U}\rightarrow C_{X}$ ($u_{*}$) is closed immersion if ($u_{*}$) is an categorical equivalence of:
$C_{U}$ and full topologizing subcategory $C_{V}$ of $C_{X}$(topologizing subcategory is full subcategory which is closed under finite direct sum and subquotient taken in $C_{X}$)
thickennings
We call a closed immersion $U\rightarrow T$ a thickenning, if the smallest saturated multiplicative system in $HomC_{T}$ containing $(u*)(HomC_{U})$ coincides with $Hom(C_{T})$
Formally smooth,formally unramified,formally etale
I will talk about these notions later. They are defined via thickennings.
separatedness and properness
Once you have the definition of closed immersion given above, then the definition of separatedness is free(follows the same pattern as EGA) Properness is similar. I will formulated later.
Notice
The reason to identify space with category of quasi coherent sheaves on it is mainly for noncommutative algebraic geometry. What I wrote here is trivial case of this consideration because we can drop the categorical language in commutative case.
Functor point of view and categorical point of view are not equivalent in general<|endoftext|>
TITLE: Algebraic characterization of transitive spaces of matrices
QUESTION [6 upvotes]: Fix an integer $d \ge 2$ and let $M_d$ be the space of real $d \times  d$ matrices. Let $E$ be a vector subspace of $M_d$. We say that $E$ is transitive if $E \cdot \mathbb{R}^d_* = \mathbb{R}^d$, that is, for all vectors $v \in \mathbb{R}^d_* = \mathbb{R}^d-\{0\}$ and $w \in \mathbb{R}^d$ there exists a matrix $A \in E$ such that $A \cdot v = w$.
The question is how to determine algebraically if a space of matrices is transitive or not.
More precisely, which algebraic (ie, polynomial) conditions on the entries of matrices $A_1,...,A_k$ express the fact that the space $E$ spanned by them is non-transitive?

Remarks:
1) Fix the number $k$ of generators of $E$. Let $Z$ be the subset of $\mathbb{R}^{kd^2}$ corresponding to the $k$-tuples of matrices that generate a non-transitive set. That $Z$ is the projection of an algebraic set, and therefore by Tarski-Seidenberg theorem, is a semi-algebraic set.
2) Consider the analogous problem with complex matrices and vectors in $\mathbb{C}^d$, and let $Z_C$ be the set corresponding to $Z$ above. Then $Z$ is algebraic (projectivize everything and apply the theorem that says that pprojection of algebraic is algebraic). Anyway what I'd like to see are the explicit equations for this algebraic set.

REPLY [4 votes]: Here is the real answer:
A space of matrices is transitive iff its "orthogonal complement" contains no matrix of rank one.
The idea was not mine; it is I found in Sec. 4 from the paper below (See also some more modern and more readable papers that cite it):
Azoff, E.A. On finite rank operators and preannihilators. Mem. Amer. Math. Soc. 64, no. 357 (1986).<|endoftext|>
TITLE: Examples of one-dimensional non-Cohen Macaulay rings
QUESTION [5 upvotes]: Can you offer some examples of such rings, other than $\frac{k[x,y]}{(x^{2}, xy)}$.  Thanks.

REPLY [9 votes]: In dimension $1$, Cohen-Macaulay just mean unmixed, so all the associated primes have the same dimension. Thus the easiest way to cook up a non-CM ring of dimension $1$ is: Pick your favorite regular ring (say $A=k[x,y,z]$). Take an ideal of dimension $1$, say $I=(x,y)$. Take another ideal of dimension $0$, say $J=(x^3,y^4,z^5)$ such that $J$ is not contained in $I$. Now take $R=A/(I\cap J)$. Geometrically we just throw 2 things of pure dimensions $1$ and $0$ together. In some sense, all non-CM rings of dimension $1$ arise this way.<|endoftext|>
TITLE: What classes am I missing in the Picard lattice of a Kummer K3 surface?
QUESTION [12 upvotes]: Constructing the Kummer K3 of an Abelian surface $A$, we have an obvious 22-dimensional collection of classes in $H^2(K3, \mathbb{Z})$ given by the 16 (-2)-curves (which by construction do not intersect each other), and the pushforward-and-pullback of the six classes generating $H^2(A, \mathbb{Z})$. However, this is clearly not all of the classes that I need to find; first of all, the intersection form is wrong---it is certainly not unimodular.
Secondly, there are a few other classes that can be constructed geometrically which are missing---for example, since $\sum_{i=1}^{16} E_i$, the sum of the exceptional divisors, is the branch locus of a 2-1 cover of K3, it must be divisible by two, which my naive description misses.
Ultimately what I am hoping to do is to produce a generating function summing over all effective curve classes in the K3 (with coefficients determined by some GW-invariants), but as stated above, I am missing some classes whose description I do not know. I've been looking through Barth, Peters, Van de Ven, and the best statement I can find is Proposition VIII 3.7:

The set of effective classes on a Kahler K3-surface is the semigroup generated by the nodal classes and the integral points in the closure of the positive cone.

That being said, is there a nice concrete description of these somewhere?

REPLY [13 votes]: The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is $2E_8+3U$, with $E_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular and has signature $(3,19)$.
The 16 (-2)-curves $E_i$ form a sublattice $16A_1$ of determinant $2^{16}$. It is not primitive in $L_{K3}$. The primitive lattice $K$ containing it is computed as follows. Consider a linear combination $F=\frac12\sum a_i E_i$ with $a_i=0,1$. Recall that $E_i$ are labeled by the 2-torsion points of the torus $A$, i.e. the elements of the group $A[2]$.
Then $F$ is in $K$ $\iff$ the function $a:A[2]\to \mathbb F_2$, 
$i\mapsto a_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5.
(The element $\frac12\sum E_i$ in your example corresponds to the constant function 1, which is affine linear).
Thus, $K$ has index $2^5$ in $16A_1$ and its determinant is $2^{16}/(2^5)^2=2^6$.
$K$ is called the Kummer lattice. By the above, it is a concrete negative-definite lattice of rank 16 with determinant $2^6$. Nikulin proved that a K3 surface is a Kummer surface iff $Pic(X)$ contains $K$.
The orthogonal complement $K^{\perp}$ of $K$ in $L_{K3}$ is $H^2(A,\mathbb Z)$ but with the intersection form multiplied by 2. As a lattice, it is isomorphic to $3U(2)$. It has determinant $2^6$, the same as $K$. The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is recovered from the primitive orthogonal summands $K$ and $K^{\perp}$.
However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,\mathbb Z)$. To begin with, it has signature $(1,r-1)$, not $(3,19)$. For a Kummer surface, it contains Kummer lattice $K$ described above, and its intersection with $K^{\perp}$ is the image of the Picard group of $A$. For a Kummer surface one has $r=17,18,19$ or 20.
For the Mori-Kleiman cone of effective curves, which you would need for Gromov-Witten theory, the description you put in a box is already the best possible.<|endoftext|>
TITLE: How strict can I be in the definition of "2-group"?
QUESTION [6 upvotes]: Recall that a group is an associative, unital monoid $G$ such that the map $(p_1,m) : G \times G \to G\times G$ is an isomorphism of sets.  Here $p_1$ is the first projection and $m$ is the multiplication, so the map is $(g_1,g_2) \mapsto (g_1,g_1g_2)$.
My question is a basic one concerning the definition of "2-group".
Recall that a monoidal category is a category $\mathcal G$ along with a functors $m : \mathcal G \times \mathcal G \to \mathcal G$ and $e: 1 \to \mathcal G$, where $1$ is the category with one object and only identity morphisms, such that certain diagrams commute up to natural isomorphism and those natural isomorphisms satisfy some axioms of their own (the natural isomorphisms are part of the data of the monoidal category).
Then a 2-group is a monoidal category $\mathcal G$ such that the functor $(p_1,m): \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ is an equivalence of categories.  I.e. there exists a functor $b: \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ such that $b\circ (p_1,m)$ and $(p_1,m) \circ b$ are naturally isomorphic to the identity.  Note that $b$ is determined only up to natural isomorphism of functors.

Question: Can I necessarily find such a functor $b$ of the form $b = (p_1,d)$, where $d : \mathcal G \times \mathcal G \to \mathcal G $ is some functor (called $d$ for "division")?  If so, can I necessarily find $d = m\circ(i \times \text{id})$, where $i: \mathcal G \to \mathcal G$ is some functor (called $i$ for "inverse")?

In any case, the natural follow-up question is to ask all these at the level of 3-groups, etc.

REPLY [4 votes]: You can always do this. Take any $b$ and define $d = p_2 b$. Then $b' = (p_1, d)$ is equivalent to the original $b$. To see this note that 
$$(p_1, m) \circ b = (p_1b, m \circ (p_1 b, d)) \simeq id = (p_1, p_2)  $$
The first component shows $p_1 b \simeq p_1$. We use this transformation $\times id$ to show that 
$b \simeq b' = (p_1, d)$. 
Now we consider the equivalence $(p_1, m)  \circ b' \simeq id$. Here we have,
$$(p_1, m) \circ (p_1, d) = (p_1, m \circ (p_1, d)) \simeq id = (p_1, p_2) $$
restricting to $G = G \times \{ 1\} \subseteq G \times G$, this gives a natural isomorphism $m(x, d(x, 1)) \simeq 1$. You can take $i(x) = d(x,1)$, and we have $x i(x) \cong 1$.
We also have 
$$(p_1, d) \circ (p_1, m) = (p_1, d \circ (p_1, m)) \simeq (p_1, p_2) $$
which gives a natural isomorphism, $ d(x, xy) \simeq y$ (writing $m(x,y) = xy$).
Thus we have,
$$d(x,y) \simeq d(x,1 y) \simeq d(x, x i(x) y) \simeq i(x) y, $$
which is the formula you were after. So we can replace $d(x,y)$ with $m(i(x), y)$ to get a third inverse functor b''. Note that this doesn't mean that we have a strict 2-group, just that we can define the inverse functors and difference functors  you asked about.
Notice also that we didn't really use anything about G being a 1-category as opposed to an n-category (except the associator and unitors) so this argument generalizes to the n-group setting basically verbatim.<|endoftext|>
TITLE: Mathematical solution for a two-player single-suit trick taking game?
QUESTION [16 upvotes]: The question on games and mathematics that appeared recently on mathoverflow
(Which popular games are the most mathematical?)
reminded me of a problem I encountered some time ago : starting with the insane
dream of completely solving the game of bridge with a nice mathematical theory,
I ended up considering extremely simplified versions of bridge. One of them was 
as follows : there are only 2 players instead of 4, and instead of the usual
deck there are only 2n cards numbered from 1 to 2n. Each player holds half 
of the deck, so this is a "complete information" game : each player knows exactly
what is in his opponent's hand. There are no bids, just a sequence of n moves
where each player drops a card ; as in bridge the strongest card wins the trick
and the winner of the game is the player with the largest number of tricks
in the end (take n odd to avoid draws). Also, the winner of the preceding
trick is the first to play (for the very first move the first player is determined
by some rule, random or other ; this is immaterial to the subsequent discussion).
This looks like a very basic kind of game, especially amenable to 
mathematization : for example the set of all initial positions is nicely indexed
by the subsets $I$ of $\lbrace 1,2, \ldots , 2n\rbrace$ whose cardinality is $n$
(say $I$ is the set of cards held by the first player). I was
however unable to answer the following questions :

Is there an algorithm which, given the initial position, finds out which player
will win if each one plays optimally ? What is the best strategy ?

Has this game already been studied by combinatorialists ?

REPLY [21 votes]: Yes, it has been studied by Johan Wästlund in A solution of two-person single-suit whist, which gives an efficient algorithm to compute the value of a position in this game (Theorem 10.1).  He has also studied the more general situation of multiple suits, but with the restriction that each suit is split evenly between the two players, in Two-person symmetric whist.  Here some familiar values from combinatorial game theory appear, reflecting the fact that breaking a new suit is generally disadvantageous to the player doing so.<|endoftext|>
TITLE: Spheres over rational numbers and other fields
QUESTION [9 upvotes]: Let K be an ordered field. Define the n-sphere:
$$S^n(K) := \{ (x_1,x_2,\dots,x_n+1) \in K^{n+1} \mid \sum_{i=1}^{n+1} x_i^2 = 1 \}$$
A set of vectors $v_1, v_2, \dots, v_r \in S^n(K)$ is orthonormal if the dot product of any two of them is zero. An orthonormal basis is an orthonormal set of cardinality $n + 1$.

Is every vector in $S^n(K)$ a member of an orthonormal basis? If not, what is the largest r such that very vector is a member of an orthonormal set of size r?
More generally, given n and s, what is the largest r such that every orthonormal set in $S^n(K)$ of size s is contained in an orthonormal set of size r?

What's known:

For $n = 1$, every vector in $S^n(K)$ is a member of an orthonormal basis, regardless of K.
If K is Pythagorean (i.e., a sum of squares is a square) every orthonormal set completes to an orthonormal basis (use Gram-Schmidt).

Can more be said? I'm most interested in the case of K the field of rational numbers or a real number field, and the case $n = 2$.
ADDED LATER: I am assuming that K is an ordered field here. Otherwise, we need to modify our definition of orthonormal set to also include the condition that the vectors are linearly independent, which is automatically true for ordered fields. Observations for fields that are not ordered (such as non-real number fields or fields of positive characteristic) would also be much appreciated. MODIFIED: As Bjorn Poonen points out below, linear independence turns out to follow automatically in this case. (Though in general, over non-ordered fields, there can exist orthogonal vectors that are linearly dependent, our condition that the vectors be "normal" rules this out).

REPLY [16 votes]: Any orthonormal set extends to an orthonormal basis, over any field
of characteristic not $2$.  This is a special case of Witt's theorem.
EDIT: In response to Vipul's comment: The proof of Witt's theorem is constructive, and leads to the following recursive algorithm for extending an orthonormal set $\lbrace v_1,\ldots,v_r \rbrace$ to an orthonormal basis.
Let $e_1,\ldots,e_n$ be the standard basis of $K^n$, where $e_i$ has $1$ in the $i^{\operatorname{th}}$ coordinate and $0$ elsewhere.  It suffices to find a sequence of reflections defined over $K$ whose composition maps $v_i$ to $e_i$ for $i=1,\ldots,r$, since then the inverse sequence maps $e_1,\ldots,e_n$ to an orthonormal basis extending $v_1,\ldots,v_r$.  In fact, it suffices to find such a sequence mapping just $v_1$ to $e_1$, since after that we are reduced to an $(n-1)$-dimensional problem in $e_1^\perp$, and can use recursion.
Case 1: $q(v_1-e_1) \ne 0$, where $q$ is the quadratic form.  Then reflection in the hyperplane $(v_1-e_1)^\perp$ maps $v_1$ to $e_1$. 
Case 2: $q(v_1+e_1) \ne 0$.  Then reflection in $(v_1+e_1)^\perp$ maps $v_1$ to $-e_1$, so follow this with reflection in the coordinate hyperplane $e_1^\perp$.
Case 3: $q(v_1-e_1)=q(v_1+e_1)=0$.  Summing yields $0=2q(v_1)+2q(e_1)=2+2=4$, a contradiction, so this case does not actually arise.<|endoftext|>
TITLE: Problem equivalent to "largest square in a cube"
QUESTION [5 upvotes]: The "largest square in a cube" problem, which asks for the largest square inside a cube, has a solution as can be seen on this page, which also says that the general problem in higher dimensions is unsolved.

 
 
 
 
 


 
 
 
 
 
MathWorld image.

The problem is equivalent to the following optimization problem: find two orthogonal unit vectors in $\mathbb{R}^3$ such that the maximum of the absolute values of all their coordinates is minimized. For the "largest square in a cube" problem to have the answer it does, this minimum should be $2/3$, i.e., it occurs when the coordinates are $(2/3,2/3,1/3),(1/3,-2/3,2/3)$ (or some coordinate permutation-cum-axis reflection of those). How would we show that this is indeed where the optimum occurs? This seems like it should be some really simple algebra, but it is eluding me.
More abstractly, this is asking for a pair of orthogonal unit vectors such that the maximum of their $\ell^\infty$-norms is minimized. What happens if we are trying to minimize the larger of the $\ell^p$-norms, $p > 2$? Does the minimizing value tend to $2/3$ as $p \to \infty$?

REPLY [3 votes]: When studying this problem in the general case, i.e. $m$-dimensional cube inside $n$-dimensional cube for $m<n$, it might make sense to look at small examples to gain some intuition. From the small examples (largest square in (hyper)cubes) one could get the impression that the coordinates of the optimal solution are always rationals and hence the optimal edge length are always square root of rationals. In this post I argue that this is not the case in general.
Here is how your formulation as an optimization problem generalizes: Let $S$ be a set of half of the $2^m$ vertices of the m-cube of diagonal lenght 2 centered at the origin such that $S \cup \{-s  : s\in S\}$ is the entire set of vertices. Now find $2^{m-1}$ unit-vectors in $\mathbb{R}^n$ such that each pair of them has the same vector product as a corresponding pair in $S$, and such that the absolute value of their coordinates in minimized.
For example for m=3 one can take as $S$ every other vertex of the cube, such that they form a regular tetrahedron (see picture by Kepler), and then for each pair there vector product would have to be $-\frac{1}{3}$

One potential problem with this formulation as optimization problem in practice is that the number of variables is $n2^{m-1}$ and there are $\frac{2^{m-1}(2^{m-1}+1)}{2}$ quadratic equations.
Together with this optimization formulation and methods described in http://arxiv.org/abs/1407.0683 I calculate some cases $f(m,n)$ for small $m$ and $n$. There is some numerical optimization involved, so it is not a formal proof that the following result is optimal, but I obtain exact symbolic values for the coordinates and it is certainly a lower bound.
On mathworld the cases $f(1,n)$ and $f(2,n)$ are explained, so the smallest cases that seem to be unknown are $f(3,n)$. The cases $f(3,4)$ and $f(3,5) $ are also mentioned in Sloane's oeis: A243309, A243313 . 


$f(3,4)$:

The 8 vertices  of a largest 3-cube inside $[0,1]^4$ have the following coordinates:
$$(1, 0, 1-a, b),(1, 0, b, 1-a), (0,c,1-d, 0), (0,c, 0, 1-d),(1, 1-c, 1,d), (1, 1-c,d, 1), (0,1, 1-b, a),(0, 1, a,1-b)$$
where $a,b,c,d$ are algebraic numbers of degree $4$, with the these minimal polynomials and decimal approximations:
$$\begin{align*}a\quad&16x^4 + 8x^3 - 23x^2 + 14x - 2& 0.204901553506651293143\\
b\quad&16x^4 - 24x^3 + 25x^2 - 14x + 1& 0.082734498297453867827\\
c\quad&8x^4 - 32x^3 + 45x^2 - 30x + 1& 0.035139649420907685891\\
d\quad&4x^4 - 4x^3 - x^2 + 4x - 1& 0.287636051804105160970
\end{align*}$$
This gives a edge length s, which hast the following minimal polynomial:
$$4x^8 - 28x^6 - 7x^4 + 16x^2 + 16$$
and $1.007434756884279376098253595231$ as decimal approximation.


$f(3,5)$:

The 8 vertices  of a largest 3-cube inside $[0,1]^5$ have the following coordinates:
$$(1, e, 0, e, 0), (1, e, 0, 1, 1-e), (0, 0, e, 0, e), (0, 0, e, 1-e, 1), (1, 1, 1-e, e, 0), (1, 1, 1-e, 1, 1-e), (0, 1-e, 1, 0, e), (0, 1-e, 1, 1-e, 1)$$
where $e=\sqrt{\frac{3}{2}}-1\approx 0.224744871391589049098$
This gives as edge length $\sqrt{11-8\sqrt{\frac{3}{2}}}\approx 1.09637631717731280407593110$.


$f(3,6)$:

The 8 vertices of a largest 3-cube inside $[0,1]^6$ all lie on the vertices of (a diagonal section of) the 6-cube:
$$(0, 1, 1, 0, 1, 0), (0, 0, 0, 1, 1, 1), (1, 0, 1, 0, 0, 1), (1, 1, 0,
1, 0, 0), (1, 0, 0, 1, 0, 1), (1, 1, 1, 0, 0, 0), (0, 1, 0, 1, 1, 0),
(0, 0, 1, 0, 1, 1)$$
and this gives an edge length of $\sqrt{2}\approx 1.414213562373095048$.<|endoftext|>
TITLE: Notions of "independent" and "uncorrelated" for subsets of the natural numbers
QUESTION [9 upvotes]: In probability/statistics, there is a notion of two things being "independent", which would basically mean that any information we can get about one thing has no effect on our (probabilistic) knowledge of the other.
What are the possible notions of "independent" for the natural numbers? Under such a notion, for instance, the properties of being "multiple of 2" and "multiple of 3" are independent, while "multiple of 4" and "multiple of 6" are not, because something being a multiple of 4 means that it is even and that makes it more likely that it is a multiple of 6.
There's no probability measure on the natural numbers where every natural number carries an equal positive weight (no uniform distribution on the natural numbers, because they're countable) so the substitute seems to be to look at all the natural numbers up to some natural number n, consider the extent of dependence there, and then take the limit as $n \to \infty$. Or, perhaps instead of looking at initial segments, we can look at segments of consecutive integers starting and ending at finite points, and measure the degree of dependence there. Are there other notions that are qualitatively different, or stronger, or weaker? What notions of independence are most useful for specific applications (such as the distribution of prime numbers, additive combinatorics)?
On a related note, is there some way of making sense of the "correlation" between two (infinite) subsets of the natural numbers, that would play some role analogous to what correlation plays in probability/statistics? Even if there isn't a numerically rigorous way, is there some way we can define a notion of "uncorrelated" for infinite subsets of the natural numbers. (Hopefully, in a way that independent subsets are uncorrelated)? My guess would be to measure correlations in some suitable way for all numbers up to $n$ and then take the limit as $n \to \infty$.

REPLY [7 votes]: There are some ways to assign probability measures to the set of natural numbers.  Consider the probability measure $P_s$ on the positive integers which assigns "probability" $n^{-s}/\zeta(s)$ to the integer $n$.  ($s$ is a constant real number greater than $1$.)$\newcommand{\lcm}{\operatorname{lcm}}$
Then under this measure being a multiple of $r$ and a multiple of $s$ are independent events, in the probabilistic sense, if $r$ and $s$ don't have a common multiple.  You can show this starting form the fact that the measure assigned to the set of multiples of $k$, for some positive integer $k$, is
$$ {1 \over \zeta(s)} \sum_{n=1}^\infty {1 \over (kn)^s} = {1 \over \zeta(s)} {1 \over k^s} \zeta(s) = {1 \over k^s}. $$
That is, the probability that a random positive integers is divisible by $k$ is $k^{-s}$.  Of course you really want all integers to be equally likely, which should correspond to $s = 1$.
(I learned this from Gian-Carlo Rota, Combinatorial Snapshots.  Link goes to SpringerLink; sorry if you don't have access.)
Under "suitable conditions", which I don't know what they are because Rota doesn't say, the density of any set of natural numbers $A$ is the limit $\lim_{s \to 1^+} P_s(A)$.
In particular it might be reasonable to define correlation between sets of natural numbers in the same way.  Let $A$ and $B$ be two sets of natural numbers. Let $X$ and $Y$ be the indicator random variables of the sets $A$ and $B$ in the measure $P_s$.  The [Pearson correlation coefficient}(https://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient) between $X$ and $Y$ is
$$ {(E(XY) - E(X) E(Y)) \over \sigma_X \sigma_Y }$$
where $E$ is expectation and $\sigma$ is standard deviation.  Of course this can be simplified in the case where $X$ and $Y$ are indicators (and thus only take the values $0$ or $1$) -- in particular it simplifies to
$$ {P_s(A \cap B) - P_s(A) P_s(B) \over \sqrt{P_s(A) P_s(B) (1-P_s(A)) (1-P_s(B))}} $$
We could then deifne the correlation between $A$ and $B$ to be the limit of this as $s \to 1+$.
In the case where $A$ is the event divisible by 2'', for example, and $B$ is the event divisible by 3'', then $A \cap B$ is the event ``divisible by 6''.  So $P_s(A \cap B) = 6^{-s}$, $P_s(A) = 2^{-s}$, and $P_s(A) = 3^{-s}$, so the numerator here is 0 and so the correlation is zero.
But in the case where $A$ is the event divisible by 4'' and $B$ is the event divisible by 6'', then $A \cap B$ is the event ``divisible by 12''.  So the correlation with respect to $P_s$ is
$$ {12^{-s} - 24^{-s} \over \sqrt{4^{-s} 6^{-s} (1-4^{-s}) (1-6^{-s})}} $$
which has the limit $1/\sqrt{15}$ as $s \to 1^+$; more generally the correlation between being divisible by $a$ and being divisible by $b$ is
$$ {ab - \lcm(a,b) \over \lcm(a,b) \sqrt{(a-1)(b-1)}} $$
and this may or may not be what you want.<|endoftext|>
TITLE: Solutions to a Monge-Ampère equation on the simplex 
QUESTION [10 upvotes]: Let $\Delta_k$ be the k-simplex and $\mu$ a non-negative measure over $\Delta_k$. I want to know if there exists a function $u : \Delta_k \to \mathbb{R}$ such that $u$ is convex, $u(e_i) = 0$ for all vertices $e_i$ of $\Delta_k$, and $M[u] = \mu$ where 
$M[u] = \det\left(\frac{\partial^2 u}{\partial x_j \partial x_k}\right)$ is the Monge-Ampère operator. Furthermore, I'd like to know if the solution is unique. Any techniques for how one might solve a specific instance of this problem would be a bonus.
My background is not in PDEs but the closest I've found to an answer seem to be in [1] and [2] where the boundary conditions are more restrictive and the domain is required to be strictly convex for uniqueness.
[1] "On the fundamental solution for the real Monge-Ampère operator", Blocki and Thorbiörnson, Math. Scand. 83, 1998
[2] "The Dirichlet problem for the multidimensional Monge-Ampère equation", Rauch and Taylor, Rocky Mountain Journal of Mathematics, 7(2), 1977.
Any other pointers to solving this type of problem would be greatly appreciated.

REPLY [2 votes]: In the spirit of 002's answer, the existence is shown by the following argument. Choose a uniformly convex domain $\Omega$ such that the vertices of $\Delta_k$ belong to the boundary $\partial\Omega$. Extend $\mu$ as a non-negative measure $\bar\mu$ over $\Omega$. Choose a continuous function $\phi:\partial\Omega\rightarrow\mathbb R$ which vanishes at the vertices. Then solve the Dirichlet boundary value problem for the Monge-Ampère equation in $\Omega$, with data $(\bar\mu,\phi)$. The restriction to $\Delta_k$ of the unique convex solution solves your problem.
The non-uniqueness is almost obvious, because of the freedom in the extension of $\mu$ and in the choice of $\phi$.<|endoftext|>
TITLE: Does a group have a unique pro-p topology?
QUESTION [8 upvotes]: If $G$ is a residually $p$ group and $G_i$ ANY filtration (i.e. $G_i\subset G_{i-1}$ and $\cap G_i=e$) of normal 
$p$-power index subgroups, is the corresponding filtration the usual pro-$p$ filtration?
Put differently, if $H$ is any normal $p$-power index subgroup, does it contain one of the $G_i$'s?
The answer to the corresponding question for finite filtrations is of course negative, e.g. the filtration $p^i\Bbb{Z}$ of $\Bbb{Z}$ does not give the profinite topology of $\Bbb{Z}$. On the other hand for abelian groups the answer to the above question is positive.

REPLY [2 votes]: Contrary to OP's claim, this is false even for finitely generated abelian groups, including $\mathbf{Z}^2$.
Indeed, fix a dense group embedding $i:\mathbf{Z}^2\to\mathbf{Z}_p$ (namely $i(n,m)=n+mt$ for some fixed irrational $t\in\mathbf{Z}_p$). Define $G_n=i^{-1}(p^n\mathbf{Z}_p)$. Then $(G_n)$ is a decreasing sequence of subgroups with intersection reduced to $\{0\}$. Moreover $G_0/G_n$ is isomorphic to $\mathbf{Z}/p^n\mathbf{Z}$. Hence no $G_n$ is contained in $p\mathbf{Z}^2$.<|endoftext|>
TITLE: topological "milnor's conjecture" on torus knots.
QUESTION [19 upvotes]: Here's a question that has come up in a couple of talks that I have given recently.
The 'classical' way to show that there is a knot $K$ that is locally-flat slice in the 4-ball but not smoothly slice in the 4-ball is to do two things

Compute that the Alexander polynomial of $K$ is 1, and so by results of Freedman's you know that $K$ is locally-flat slice.
(due to Rudolph) Somehow obtain a special diagram of $K$ (or utilize a more subtle argument) to show that you can present $K$ as a separating curve on a minimal Seifert surface for a torus knot.  Since we know (by various proofs, the first due to Kronheimer-Mrowka) that the genus of torus knot is equal to its smooth 4-ball genus (part of Milnor's conjecture), the smooth 4-ball genus of $K$ must be equal to the genus of the piece of the torus knot Seifert surface that it bounds, and this is $\geq 1$.

Boiling the approach of 2. down to braid diagrams, you come up with the slice-Bennequin inequality.
Well, here's the thing.  I have this smooth cobordism from the torus knot to $K$, and then I know that $K$ bounds a locally-flat disc.  This means that the locally-flat 4-ball genus of the torus knot must be less than its smooth 4-ball genus.  So if you were to conjecture that the locally flat 4-ball genus of a torus knot agrees with its smooth 4-ball genus, you would be wrong.
My question is - are there any conjectures out there on the torus knot locally-flat genus?  Even asymptotically?  Any results?  Any way known to try and study this?
Thanks, Andrew.

REPLY [7 votes]: I realize the following answer comes somewhat belated.
Rudolph worked on this (Some topologically locally-flat surfaces in the complex projective plane), and more recently Baader, Feller, Liechti and myself (https://arxiv.org/abs/1509.07634).
Here is a brief summary of what is known:
Levine-Tristram signatures give a lower bound for the topological 4-genus of torus knots, which is asymptotically equal to half of their 3-genus. This is the best we have; other methods, like Casson-Gordon invariants or $L^2$-signatures, might give better bounds, but they have not been computed for torus knots.
Upper bounds may be obtained by the "algebraic genus": constructing locally flat slice surface by ambient surgery, using Freedman's result that Alexander-polynomial-1 knots are topologically slice. In this way, one can prove that for small (3-genus $\leq$ 14) torus knots, the topological 4-genus equals the maximum of the Levine-Tristram signatures. In particular, for the examples Ian mentions, one has $g_4^t(T(3,7))=g_4^t(T(4,5))=5$. More generally, for $p<q$ one has $g_4^t(T(p,q)) < g_3(T(p,q))$ unless $p=2$ or $(p,q)\in\{(3,4),(3,5)\}$.
With the same method, one finds that asymptotically, the topological 4-genus of large torus knots is at most three quarters of their 3-genus.
Feeling adventurous, one could conjecture that for all torus knots, the topological 4-genus equals the maximum of the Levine-Tristram signatures.<|endoftext|>
TITLE: Asymptotic series for roots of polynomials
QUESTION [5 upvotes]: Let $f(z) = z + z^2 + z^3$.   Then for large $n$, $f(z) = n$ has a real solution near $n^{1/3}$, which we call $r(n)$.  This appears to have an asymptotic series in descending powers of $n^{1/3}$, which begins
$$ r(n) = n^{1/3} - {1 \over 3} - {2 \over 9} n^{-1/3} + {7 \over 81} n^{-2/3} + O(n^{-1}). $$
In order to derive this series I use a method of undetermined coefficients.  If we assume such a series exists, of the form
$$ r(n) = An^{1/3} + B + Cn^{-1/3} + Dn^{-2/3} + O(n^{-1}), $$
then applying $f$ to both sides gives
$$ n = A^3 n + (3 A^2 B + A^2) n^{2/3} + (3 A^2 C + 3 A B^2 + A + 2AB)n^{1/3} + (B^2 + 2AC + 3A^2 D + B + B^3 + 6 ABC + O(n^{-1/3}). $$
Thus we have $A^3 = 1$ and so $A = 1$; $3A^2 B + A^2 = 0$ and so $B = -1/3$; similarly $C = -2/9, D = 7/81$.
More generally, let $f$ be a polynomial with leading term $Ax^k$.  Then the largest real root of $f$ appears to have an asymptotic series of the form
$$ A^{-1/k} n^{1/k} + c_0 + c_1 n^{-1/k} + c_2 n^{-2/k} + \cdots $$
which can be computed by a similar method to the one I gave above.  
My question, which I would like an answer to in order to make a minor point in my dissertation: how can I prove that the solutions of equations of the form $f(z) = n$ have such series?

REPLY [4 votes]: If you have a good initial guess (such as your case), then you don't need either undetermined coefficients or Newton's method (although they both work), you can in fact get a procedure which will give you as many coefficients as you want.  Harald's first step is excellent, so we'll start from that.  Given $F(\epsilon,w) = \epsilon^2w+\epsilon w^2+w^3-1 = 0$, we can derive a differential equation for $w(\epsilon)$.  With the help of Maple's gfun[algeqtodiffeq], giving
$$ 9+ \left( 6{\epsilon}^{5}+7{\epsilon}^{2} \right) w ( \epsilon ) + \left( 27-6{\epsilon}^{6}-7{\epsilon}^{
3} \right) {\frac {d}{d\epsilon}}w(\epsilon) + \left( -3{\epsilon}^{7}-14{\epsilon}^{4}-27\epsilon \right) {\frac {d^{2}}{d{\epsilon}^{2}}}w ( \epsilon ) $$
with initial conditions $w (0) =1,w^{'''}(0) =-4/9 $ [the DE is singular at $\epsilon=0$ so the initial conditions are non-standard).  Nevertheless, from there one can continue and use gfun[diffeqtorec] to get a recurrence for the coefficients of the series.  The result is
$$ \left( 6-3n-3{n}^{2} \right) u(n) + \left( -98-77n-14{n}^{2} \right) u (n+3) + \left( -
648-27{n}^{2}-270n \right) u (n+6 ) $$
with starting conditions $u(0) =1, u(1) =-1/3,u (2) =-2/9,u (3) ={\frac {7}{81}},u(4) =0,u(5) ={\frac {14}{729}}$.  These are sufficient to allow you to 'unwind' the recurrence as much as you want.  
Bjorn's answer gives the underlying 'analytic' answer that justifies that the above gives a convergent result.  Actually, the procedure above works in other cases as well, but then the result is only valid in a sector, and computing the size of that sector can be fiendishly difficult.<|endoftext|>
TITLE: Nonlinear Nuclear Operators ?
QUESTION [5 upvotes]: Is there a "right" definition of the nuclear
operator in the nonlinear framework ? Of course, such an operator
must be compact, while a linear operator should be "nonlinearly"
nuclear iff it is nuclear as usual. [These are naive conditions, of course.]
Since here the Frechet derivative is useless. For example, the "cubing"
self-map of $\ell^{2}$, i.e., $(x_{n})$ $\rightarrow\(x_{n}^{3})$
, has nuclear derivative at each point, yet it is clearly noncompact.
OTOH, it seems that a [nonlinear] Lipschitz $p$-summing operator
need not to be power compact. Some thoughts ?

REPLY [5 votes]: Funny you should ask.  My former student, Bentuo Zheng, and my visitor, Dongyang Chen, are in the process of developing this theory.  The "right" definition involves the Pietsch factorization diagram, and is an off-shoot off what Farmer and I did for p-summing and p-integral operators (get the paper from my home page).  Send me an email and I'll put you in contact with them.  
On a related topic, my current student Alejandro ChavezDominguez is developing the operator ideal theory connected to non-linear p-summing operators and related mappings. This is something I have been interested in for a long time but did not see what to do (even the duality theory). In Javier's hands, the theory is developing very well.<|endoftext|>
TITLE: Representations of the three string braid group
QUESTION [8 upvotes]: Do you know any explicit constructions of families of irreducible finite dimensional representations of the three string braid group?
I will describe what I already know and then invite you to suggest alternatives. This is somewhat open-ended so I will also pose some more precise questions.

This problem is solved for dimension at most five in
http://arxiv.org/abs/math/9912013  by Imre Tuba, Hans Wenzl
and published in Pacific Journal of Mathematics. So I am particularly interested in six dimensional representations.

The usual presentation is generators $\sigma_1$, $\sigma_2$ and the relation $\sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2$. An alternative is generators $s$, $t$ and the relation $s^2=t^3$ where $s=\sigma_1\sigma_2\sigma_1$ and $t=\sigma_1\sigma_2$. This shows we have a central extension of the free product of $C_2$ and $C_3$.
Given a representation of dimension $N$ take the dimensions of the eigenspaces of $s$ and $t$. Then we have $n_1+n_2=N$ and $m_1+m_2+m_3=N$. If the representation is irreducible then $n_i\ge m_j$. This data labels the irreducible components of the variety of irreducible representations of dimension $N$. The dimension of the component is $N^2-n_1^2-n_2^2-m_1^2-m_2^2-m_3^2+2$. For $N=6$ we have 

(4,2) and (2,2,2) of dimension 6
(3,3) and (2,2,2) of dimension 8
(3,3) and (3,2,1) of dimension 6
(3,3) and (3,3,0) of dimension 2


The case (3,3) and (3,2,1) is interesting because these representations do not deform to representations of the symmetric group.

Can you give a representation and a braid $\alpha$ such that the trace of $\alpha$ and the trace of the braid given by reading $\alpha$ backwards are different? I have an example of dimension 12.

The description as a central extension of the free product means we have a construction which associates a representation to each invertible matrix (for each component).
I would like to improve on this in two ways. First this is unwieldy. Second I would like to be able to specify the eigenvalues of $\sigma_1$. This means passing to a covering of the moduli space.

REPLY [15 votes]: Bruce,
I can give you representants for an open piece of the moduli space (3,3) resp. (3,2,1). I'll add them in Mathematica-form so that you can plug them into any braid you like to work on. The variables a,b,d,x,y,z are the coordinates of the moduli space and l stands for a third root of unity (i dont know how to tell Mathematica to work with roots of unity, so if you know replace l by that thing). 
I obtained these representants from the hexagon of 1-dml simples of the modular group and taking the dimension vector (1,1,1,2,1,0) as required. The corresponding moduli problem is then rational (in x,y,z) over that of pairs of 2x2 matrices of rank one. these are rational (in a,b,d). Hope this helps (and sorry for the lengthy formulas below).
sigma1={{-(1 - a - b - d + a d + x - a x - b x - d x + a d x + l x - a l x - b l x - d l x + a d l x - y + a y) z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, -(-1 + l) (1 - a - b - d + a d + l - a l - b l - d l + a d l - y + a y) z, (-1 + l) z, (b + d - a d) (-1 + l) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-b - d + a d + l - a l + b x + d x - a d x - l x + a l x + y - a y + l y - a l y) z, (-1 + l) (-1 + x) z, (-1 + a) (-1 + l) (l + x) y z, -(-1 + l) (-1 + x) z, -(b + d - a d) (-1 + l) (-1 + x) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, (a + b - a d - l + d l - a x - b x + a d x + l x - d l x - a y + l y) z, -b (-1 + l) (l + x) y z, (-1 + l) (a + b - a d - a x - b x + a d x - a y) z, -b (-1 + l) (-1 + x + y) z}, {-(1 - a - b - d + a d) (-1 + l) (1 + l) x z, (-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, (-1 + l) z, -(1 - a - b - d + a d + l - a l - b l - d l + a d l + x - a x - b x - d x + a d x + l y - a l y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {-b (-1 + l) (1 + l) x y z, - b (-1 + l) (-1 + x + y + l y) z, (-1 + l) (1 - d - x + d x - y) z, b (-1 + l) (l + x) y z, -(-1 + d + a l + b l - a d l + x - d x - a l x - b l x + a d l x + y - a l y) z, b (-1 + l) (-1 + x + y) z}, {(-1 + a) (-1 + l) (1 + l) x y z, (-1 + a) (-1 + l) (-1 + x + y + l y) z, -(-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) (l + x) y z, (-1 + l) (-1 + x) z, -(-1 + a + b l + d l - a d l + x - a x - b l x - d l x + a d l x + y - a y) z}}
sigma2={{-(1 - a - b - d + a d + x - a x - b x - d x + a d x + l x - a l x - b l x - d l x + a d l x - y + a y) z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, (-1 + l) (1 - a - b - d + a d + l - a l - b l - d l + a d l - y + a y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-b - d + a d + l - a l + b x + d x - a d x - l x + a l x + y - a y + l y - a l y) z, (-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) (l + x) y z, (-1 + l) (-1 + x) z, (b + d - a d) (-1 + l) (-1 + x) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, ( a + b - a d - l + d l - a x - b x + a d x + l x - d l x - a y + l y) z, b (-1 + l) (l + x) y z, -(-1 + l) (a + b - a d - a x - b x + a d x - a y) z, b (-1 + l) (-1 + x + y) z}, {(1 - a - b - d + a d) (-1 + l) (1 + l) x z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, -(1 - a - b - d + a d + l - a l - b l - d l + a d l + x - a x - b x - d x + a d x + l y - a l y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, -(-1 + l) (1 - d - x + d x - y) z, b (-1 + l) (l + x) y z, -(-1 + d + a l + b l - a d l + x - d x - a l x - b l x + a d l x + y - a l y) z, b (-1 + l) (-1 + x + y) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-1 + a) (-1 + l) (-1 + x + y + l y) z, (-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) ( l + x) y z, (-1 + l) (-1 + x) z, -(-1 + a + b l + d l - a d l + x - a x - b l x - d l x + a d l x + y - a y) z}}
EDIT March 9th : The component (3,3;3,2,1) is not able to detect braid-reversion. On the other hand, the component of 6-dimensional representations (3,3;2,2,2) can. A minimal braid that can be separated by this family from its reversed braid is s1^-1s2^2s1^-1s2. One can show that every irreducible component of simple B(3)-representations has a Zariski dense family parametrized by a minimal rational variety. For representations of dimension <= 11 explicit such families are given in this note.
Each of these families can then be turned into a family of 3-string braid invariants over Q(rho) for rho a primitive 3-rd root of unity by specializing the parameters to random integers. For the family (3,3;2,2,2) mentioned above this can be done in SAGE as follows :

K.=NumberField(x^2+x+1)
a=randint(1,1000)
b=randint(1,1000)
c=randint(1,1000)
d=randint(1,1000)
e=randint(1,1000)
f=randint(1,1000)
g=randint(1,1000)
h=randint(1,1000)

B=matrix(K,[[1,0,0,a,0,f],[0,1,1,0,1,0],[1,1,0,1,0,0],[0,0,1,0,d,e],[0,1,0,b,c,0],[g,0,1,0,0,1]])
Binv=B.inverse()

mat2=matrix(K,[[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,1,0,0,0],[0,0,0,-1,0,0],[0,0,0,0,-1,0],[0,0,0,0,0,-1]])
mat3=matrix(K,[[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,l^2,0,0,0],[0,0,0,l^2,0,0],[0,0,0,0,l,0],[0,0,0,0,0,l]])
s1=h*Binv*mat3*B*mat2
s2=h*mat2*Binv*mat3*B

s1inv=s1.inverse()
s2inv=s2.inverse()

One can then check braid-reversion for all knots with at most 8 crossings that are closures of 3-string braids. Here are the tests to perform

test41=(s1inv*s2*s1inv*s2-s2*s1inv*s2*s1inv).trace()
test52=(s1inv**3*s2inv*s1*s2inv-s2inv*s1*s2inv*s1inv**3).trace()
test62=(s1inv**3*s2*s1inv*s2-s2*s1inv*s2*s1inv**3).trace()
test63=(s1inv**2*s2*s1inv*s2**2-s2**2*s1inv*s2*s1inv**2).trace()
test73=(s1**5*s2*s1inv*s2-s2*s1inv*s2*s1**5).trace()
test75=(s1inv**4*s2inv*s1*s2inv**2-s2inv**2*s1*s2inv*s1inv**4).trace()
test82=(s1inv**5*s2*s1inv*s2-s2*s1inv*s2*s1inv**5).trace()
test85=(s1**3*s2inv*s1**3*s2inv-s2inv*s1**3*s2inv*s1**3).trace()
test87=(s1**4*s2inv*s1*s2inv**2-s2inv**2*s1*s2inv*s1**4).trace()
test89=(s1inv**3*s2*s1inv*s2**3-s2**3*s1inv*s2*s1inv**3).trace()
test810=(s1**3*s2inv*s1**2*s2inv**2-s2inv**2*s1**2*s2inv*s1**3).trace()
test816=(s1inv**2*s2*s1inv**2*s2*s1inv*s2-s2*s1inv*s2*s1inv**2*s2*s1inv**2).trace()
test817=(s1inv**2*s2*s1inv*s2*s1inv*s2**2-s2**2*s1inv*s2*s1inv*s2*s1inv**2).trace()
test818=(s1inv*s2*s1inv*s2*s1inv*s2*s1inv*s2-s2*s1inv*s2*s1inv*s2*s1inv*s2*s1inv).trace()
test819=(s1**3*s2*s1**3*s2-s2*s1**3*s2*s1**3).trace()
test820=(s1**3*s2inv*s1inv**3*s2inv-s2inv*s1inv**3*s2inv*s1**3).trace()
test821=(s1inv**3*s2inv*s1**2*s2inv**2-s2inv**2*s1**2*s2inv*s1inv**3).trace()

The following tests should give non-zero invariants : 6.3,7.5,8.7,8.9,8.10 (which are known as 'flypes' in the theory) and 8.17 (the non-invertible knot with minimal number of crossings). I tried to include all details in the note Rationality and dense families of B(3) representations. All comments to this text are welcome. I like to thank Bruce Westbury for providing me with feedback.<|endoftext|>
TITLE: Verdier duality via Brown representability?
QUESTION [9 upvotes]: Hello,
I wonder if the techniques introduced in Neemans paper:
"The Grothendieck duality theorem via Bousfield's techniques and Brown representability "
can be used to establish Verdier duality. More precisely:
Consider the unbounded, derived category $D(M)$ of $\mathbb{Q}$ vector spaces on a compact complex manifold $M$ . I would like to show that $Rf_!$ has a right adjoint.
In order to use  Brown representability one has to show that $D(M)$ is compactly generated.
i.e. there exists a set of objects $c_i$ that commutes with direct sums:
$$Hom(c_i,\bigoplus x_j)=\bigoplus Hom(c_i,x_j)$$
and generates $D(M)$:
$$\forall c_i Hom(c_i,x)=0 \Rightarrow x=0$$
My problem is that i can't find such a set of generators. I first tried shifts of  
$$i_*\mathbb{Q}$$ where $i$ is the inclusion of an open subset. However these do neither commute with coproducts nor are they generators (they can not see sheaves without global sections).
My second try was shifts of
$$i_!\mathbb{Q}$$
these are generators, but again they do not seem to respect coproducts.
Can someone give a set of compact generators? Or is this approach to Verdier duality doomed anyway?

REPLY [9 votes]: The category of sheaves of $\mathbb{Q}$ vector spaces on $M$ is a Grothendieck abelian category. It follows that the derived category of such, $D(M)$ in your notation, is a well generated triangulated category. A proof of this can be found in Neeman's paper "On The Derived Category of Sheaves on a Manifold". In particular $D(M)$ satisfies Brown representability.
I thought I'd also comment on the proof. The point is that by a result of Alonso, Jeremías and
Souto one can extend the Gabriel-Popescu theorem to the level of derived categories. This realizes the derived category of any Grothendieck abelian category as a localization of the derived category of R-modules for some ring R. The kernel of this localization can be generated by a set of objects and so general nonsense gives the desired generating set of $\alpha$-small objects for some regular cardinal $\alpha$ for the localization.<|endoftext|>
TITLE: Zariski open sets are dense in analytic topology
QUESTION [13 upvotes]: How does one show that if $U \subseteq \mathbb{C}^n$ is nonempty and Zariski open then $U$ is also dense in the analytic topology on $\mathbb{C}^n$?

REPLY [23 votes]: It is enough to show that the complement of $U$ has empty interior. Also, that complement is contained in the zero set $Z$ of a non-constant polynomial $f$, so it is enough to show that $Z$ does not contain open sets. 
If $z\in Z$ is a point in the interior of $Z$, then the Taylor series of $f$ at $z$ is of course zero. Since $f$ is an entire function, this is absurd.<|endoftext|>
TITLE: Minimal number of generators of a homogeneous ideal (exercise in Hartshorne)
QUESTION [21 upvotes]: In the very first chapter Hartshorne proposes the following seemingly trivial exercise (ex. I.2.17(ii)):

Show that a strict complete intersection is a set theoretic complete intersection.

Here are Hartshorne's definitions:

A variety $Y$ of dimension $r$ in $\mathbb{P}^n$ is a (strict) complete intersection if $I(Y)$ can be generated by $n-r$ elements. $Y$ is a set-theoretic complete intersection if $Y$ can be written as the intersection of $n-r$ hypersurfaces.

Here $I(Y)$ is the homogeneous ideal of $Y$. The point is that the first definition seems wrong, since one would naturally require that $I(Y)$ can be generated by $n-r$ homogeneous elements (with this definition the exercise becomes trivial).
I have never made my mind if this is a misprint by Hartshorne. So the question is

Is it true that in a polynomial ring any homogeneous ideal generated by $k$ elements is also generated by $k$ homogeneous elements?

If I recall well it is not difficult to find counterexamples in graded rings which are not polynomial rings, so the point of the exercise may be to show that polynomial rings have this special property.

REPLY [21 votes]: Dear Andrea: Hartshorne was right, but we need to do some work. Let $\mu(I)$ be the minimal number of generators of $I$, and $\mu_h(I)$ be the minimal number of a homogenous system of generators of $I$. Let $R=k[x_1,\dots,x_n]$ and $\mathfrak m=(x_1,\dots,x_n)$. Suppose $\mu_h(I)=m$ and $f_1,\dots, f_m$ is a minimal homogenous set of generators. At this point we switch to the local ring $A=R_{\mathfrak m}$ (the reason: it is easier to do linear algebra over local rings, as anything not in $\mathfrak m$ is now invertible). It will not affect anything since $I\subset\mathfrak m$.
Construct a surjective map $F_0 = \bigoplus_{i=1}^m A(-\deg \ f_i) \to I \to 0$ and let $K$ be the kernel. We claim that  $K \subset\mathfrak mF_0$. If not, then one can find an element $(a_1,...,a_m) \in K$ such that $\sum a_if_i=0$ and $a_1$, say, has a degree $0$ term $u_1\neq 0$. By considering terms of same degree in the sum  one sees that there are $b_i$s such that:$$u_1f_1 = \sum_{i=2}^m b_if_i$$
so the system is not minimal, as $u_1 \in k$, contradiction.
Now tensoring the sequence $$ 0 \to K \to F_0 \to I \to 0$$ with $k=A/\mathfrak m$. By the claim $K\subset\mathfrak mF_0$, so $F\otimes k \cong I\otimes k$. It follows that  $m= \operatorname{rank} F_0 = \dim_k(I\otimes k)$. But over a local ring, the last term is exactly $\mu(I)$, and we are done.<|endoftext|>
TITLE: Self-adjoint extension of locally defined differential operators
QUESTION [5 upvotes]: The following is well known. Given a symmetric differential operator, like $\partial_x^2$, defined on smooth functions of compact support on $\mathbb{R}$, $C_0^\infty(\mathbb{R})$, one can count the number of independent $L^2$-normalizable solutions of $\partial_x\pm i$ and use the von Neumann index theorem to classify possible self-adjoint extensions of this operator on $L^2(\mathbb{R})$. This can be generalized to more complicated differential operators, to $\mathbb{R}^n$ as well as bounded open subsets thereof.
On the other hand, suppose that I have a manifold $M$ that is covered by a set of open charts $U_i$ with differential operators $D_i$ defined in corresponding local coordinates. It is easy to check if the $D_i$ are restrictions of a globally defined differential operator $D$ on $M$: the transition functions on intersections of charts $U_i\cap U_j$ must transform $D_i$ into $D_j$ and vice versa. Suppose that is the case and that I am interested in self-adjoint extensions of $D$ to $L^2(M)$ (supposing that an integration measure is given and that $D$ is symmetric with respect to it). Now, the question:

Is there way of classifying the self-adjoint extensions of $D$ on $L^2(M)$ in terms of its definition in local coordinates, the actions of $D_i$ on $C_0^\infty(U_i)$.

A simple example would be the cover of $S^1$ by two overlapping charts. I know that a self-adjoint extension of $\partial_x^2$ on $[0,1]$ with periodic boundary conditions gives the naturally defined self-adjoint Laplacian on $S^1$. Then $(0,1)$ is interpreted as a chart on $S^1$ that excludes one point. However, I don't know how to define the self-adjoint Laplacian on $S^1$ if it's given on two overlapping charts.

REPLY [4 votes]: Yes, if the manifold is compact without boundary, and the operator $D$ is elliptic (equivalently, if each of the $D_i$ are elliptic), and if $D$ is symmetric with
respect to some smooth positive density, then $D$ is essentially self-adjoint, i.e. extends uniquely from the core domain $\mathcal C^\infty$ to an unbounded self-adjoint operator on $L^2$ of the manifold $M$. The way to prove this is as follows. If $D$ has order $m > 0$, then the maximal domain of $D$ is equal to $H^m(M)$; this is proved by (local!) elliptic regularity. On the other hand, every element in this maximal domain is approximable in the graph norm by smooth functions, again this can be reduced to a local calculation in each chart. Hence the minimal and maximal domains agree (for more on this, see Reed & Simon, Volume 2, I think).
If $M$ has boundary or is otherwise noncompact or singular, symmetric elliptic operators are usually not essentially self-adjoint.<|endoftext|>
TITLE: Topologies on an infinite symmetric group
QUESTION [12 upvotes]: Let $X$ be an infinite set, and let $G$ be the symmetric group on $X$.  I want to understand $G$ by putting a topology on it, without imposing any more structure on $X$.  What 'interesting' possibilities are there and what is known about them?
In particular, I have heard of the pointwise convergence topology (an open neighbourhood of g is a set of permutations that agree on some specified finite set of points), and found some papers on this, but are there any other topologies that have been studied?
What if I take the coarsest topology compatible with the group operations such that either a) the stabiliser of any subset is closed, b) the stabiliser of any partition is closed, or c) both are closed?  I think these will be coarser than the pointwise convergence topology, because in the pointwise convergence topology the stabiliser of any first-order structure is closed.  Is there a useful characterisation of the open subgroups and/or closed subgroups?

REPLY [9 votes]: It is known that there are exactly two separable group topologies on $S_\infty$ (i.e., on the group of permutations of a countable set): one is antidiscrete and the other one is topology of pointwise convergence. This is a statement of Theorem 6.26 in
here. Hence Polish topology on $S_\infty$ is unique.
There are some abstract characterizations of closed subgroups, for example they are exactly the Polish groups with a countable basis of identity consisting of open subgroups. Another characterization is: a Polish group G is homeomorphic to a closed subgroup of $S_\infty$ if and only if it admits a compatible left invariant ultrametric. You may want to look at Section 1.5 of Becker and Kechris "The Descriptive Set Theory of Polish Group Action".<|endoftext|>
TITLE: Intrinsic metric with no geodesics
QUESTION [23 upvotes]: It seems that I have the needed example, but I want it to be simple and self-explaining...

Construct a nontrivial complete metric space $X$ with intrinsic metric which has no nontrivial minimizing geodesics.

Definitions:

A metric $d$ is called intrinsic if for any two points $x$, $y$ and any $\epsilon>0$ there is an $\epsilon$-midpoint $z$; i.e. $d(x,z),d(z,y)<\tfrac12 d(x,y)+\epsilon$. 
A minimizing geodesic is nontrivial if it connects two distinct points.
A meric space is nontrivial if it contains two distinct points.

Comments:

Clearly, $X$ can not be locally compact.

REPLY [3 votes]: This does not answer the question. The following paper contains a very natural example, namely the regular Frechet Lie group $Diff_{\mathcal S}(\mathbb R)$ of all diffeomorphisms of the real line which fall rapidly towards the identity, with a weak right invariant Riemannian metric induced by inner product
$$
G_{Id}(X,Y) = \int_{\mathbb R} X'Y'\,dx
$$
on the Lie algebra, which has the following property: 

Geodesic distance is an intrinsic metric, but there does not exist a single nontrivial geodesic. But it is not complete as a metric space.

This group is a normal subgroup (isometrically contained for the geodesic distance, by thm 4.5) in the slightly larger regular Lie group 
$Diff_{\mathcal S_1}(\mathbb R)$ where one allows shifts at $+\infty$. 
The inner product extends to the larger space of vector fields in the Lie algebra. The resulting weak right invariant Riemannian metric  (see 4.3) is flat,
has minimal geodesics between any two points (it is geodesically convex), it allows for a formula for geodesic distance, and it has a nice geodesic completion which is a monoid.
The geodesic equation is the Hunter-Saxton PDE in the real line. 
Any geodesic in  $Diff_{\mathcal S_1}(\mathbb R)$ hits the subgroup $Diff_{\mathcal S}(\mathbb R)$ at most twice.
Geodesic distance on $Diff_{\mathcal S}(\mathbb R)$ is an intrinsic metric; this follows from the proof of theorem 4.5.

Martin Bauer, Martins Bruveris, Peter W. Michor: Homogeneous Sobolev metric of order one on diffeomorphism groups on the real line. Journal of Nonlinear Science 24, 5 (2014), 769-808 (pdf)<|endoftext|>
TITLE: Is there a version of the Archimedean property which does not presuppose the Naturals?
QUESTION [9 upvotes]: All the statements of the Archimedean property with which I am acquainted fundamentally uses ℕ -- more than as a totally ordered semi-group, really being the 'standard model' of the naturals.  It is a fundamental ingredient in showing that the reals are (up to isomorphism) the only complete totally ordered field.
In this particular result, there are two non ingredients which seem to be of a fundamentally different nature: Dedekind completeness (which is about subsets, while all the other axioms are about elements), and the Archimedean property, which pre-supposes the existence of the Naturals.  But because being 'Archimedean' already makes sense in (ordered) monoids, that is the one I am most interested in.
Crazy scenario: replace the 'naturals' in the Archimedean property with a non-standard model of the 'naturals' (call this N-Archimedean).  Now the reasoning for the argument that all Archimedean totally ordered fields are sub-fields of ℝ readily lifts, but no longer proves quite the same thing.
In other words, it seems that this unique position of ℝ is in part due to ℕ already being baked in to the question.  And thus my question: is there a proper generalization of the Archimedean property which is ``properly abstract''?

REPLY [2 votes]: Tarski showed the first-order theory of the reals is decidable.  But of course the first-order theory of the naturals isn't.  So (assuming consistency, I guess) we know you cannot construct the naturals in the reals using only first-order notions.  But we can construct the natural numbers if we are allowed to quantify on sets of reals.<|endoftext|>
TITLE: What interesting class of Matroids are there that contains the class of Gammoids?
QUESTION [5 upvotes]: I am looking for a (already-studied or interesting) class of Matroids such that
- Class of Gammoids are contained in it
One example would be Strongly-base-orderable Matroids. I would also be grateful if someone knows a class of Matroids such that

Class of Gammoids are contained in it AND
It is contained in the class of "Strongly-base-orderable" Matroids.

By the way, strongly-base-orderable is a property such that :
GIven any two bases I,J of the Matroid, there exists a bijection \pi between I-J and J-I such that given any subset H of I-J, I- H +\pi(H) and J - \pi(H) + H is a base in the Matroid. (In Oxley's "Matroid theory" pp410 )
Motivation :
Well, I have something I can show for Gammoids but cannot for Strongly-base-orderable Matroids, although computational evidence suggests that it holds for general Matroids.

REPLY [4 votes]: The way I understand your question is as follows: you have a property which you believe holds for all matroids (these are rare); you can prove it for gammoids, but your proof does not extend to a certain bigger class of "strongly-base-orderable" (SBO) matroids.  I take it you have an example of a SBO matroid where your proof breaks down because some intermediate lemma explicitly fails (even if the main property still holds).  Without knowing if this is the right picture and what are all these properties and examples, it's hard to give you a good answer.  However, if that is indeed your logic you might as well consider classes of matroids which contain gammoids but are not completely included in the class of SBO matroids.  Say, Piff and Welsh proved in 1970 that every gammoid is representable over large enough field.  So, how about the class of real-representable matroids?  Not all of them are SBO, but so what - perhaps your "bad example" is not there...<|endoftext|>
TITLE: Is there a combinatorial reason that the (-1)st Catalan number is -1/2?
QUESTION [36 upvotes]: The $n$th Catalan number can be written in terms of factorials as 
$$ C_n = {(2n)! \over (n+1)! n!}. $$
We can rewrite this in terms of gamma functions to define the Catalan numbers for complex $z$:
$$ C(z) = {\Gamma(2z+1) \over \Gamma(z+2) \Gamma(z+1)}. $$
This function is analytic except where $2n+1, n+2$, or $n+1$ is a nonpositive integer -- that is, at $n = -1/2, -1, -3/2, -2, \ldots$.
At $z = -2, -3, -4, \ldots$, the numerator of the expression for $C(z)$ has a pole of order 1, but the denominator has a pole of order $2$, so $\lim_{z \to n} C(z) = 0$.  
At $z = -1/2, -3/2, -5/2, \ldots$, the denominator is just some real number and the numerator has a pole of order 1, so $C(z)$ has a pole of order $1$.
But at $z = -1$:
- $\Gamma(2z+1)$ has a pole of order $1$ with residue $1/2$; 
- $\Gamma(z+2) = 1$;
- $\Gamma(z+1)$ has a pole of order $1$ with residue $1$.
Therefore $\lim_{z \to -1} C(z) = 1/2$, so we might say that the $-1$st Catalan number is $-1/2$.
Is there an interpretation of this fact in terms of any of the countless combinatorial objects counted by the Catalan numbers?

REPLY [6 votes]: Here's an explanation for why, as far as I can tell, the answer to your question should be "no", and -1/2 isn't really a Catalan number.
The generating function $C(x)=\sum_{n=0}^\infty C_n x^n$ 
for the Catalan numbers (with the usual interpretation starting from $C_0=1$) satisfies the functional relation
$$C(x) = 1 + xC(x)^2,$$
and from this we can derive the closed-form expression $C(x)=\frac{1-\sqrt{1-4x}}{2x}$.
The formula you wrote above for the Catalan number $C_n$ in terms of factorials, or the gamma function, is really expressing the coefficients of the Taylor expansion of $F(x)= \frac{-\sqrt{1-4x}}{2x}$.
This just so happens to match up with the coefficients of $C(x)$ for non-negative powers of $x$, but not for the $x^{-1}$ term, since
$ F(x) = -\frac12 x^{-1} + C(x).$
So now the question is, are we really sure that $C(x)$ is the "right" Catalan generating function rather than $F(x)$? Well yes, because $C(x)$ is the one that satisfies the functional relation above, which leads to all the nice combinatorial interpretations. The function $F(x)$, on the other hand, doesn't satisfy any particularly nice equation---there is $F(x)^2 = \frac14 x^{-2} - x^{-1}$, which I don't see a natural interpretation for combinatorially.<|endoftext|>
TITLE: To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?
QUESTION [23 upvotes]: In his answer to a question about simple proofs of the
Nullstellensatz
(Elementary / Interesting proofs of the Nullstellensatz),
Qiaochu Yuan referred to a really simple proof for the case of an
uncountable algebraically closed field. 
Googling, I found this construction also in Exercise 10 of a 2008 homework
assignment from a course of J. Bernstein (see the last page of
http://www.math.tau.ac.il/~bernstei/courses/2008%20spring/D-Modules_and_applications/pr/pr2.pdf).
Interestingly, this exercise ends with the 
following (asterisked, hard) question: 
(*) Reduce the case of arbitrary field $k$ to the case of
an uncountable field.
After some tries to prove it myself, I gave up and returned to googling.
I found several references to the proof provided by Qiaochu Yuan, but no
answer to exercise (*) above. 
So, my question is: To prove the Nullstellensatz, how can the general case
of an arbitrary algebraically closed field be reduced to the
easily-proved case of an uncountable algebraically closed field? 
The exercise is from a course of Bernstein called 'D-modules and their
applications.'  One possibility is that the answer arises somehow when
learning D-modules, but unfortunately I know nothing of
D-modules. Hence, proofs avoiding D-modules would be particularly
helpful.

REPLY [6 votes]: The easiest way to reduce to the uncountable case may be as follows. 
Let $I$ be an ideal of $k[X_1,...,X_d]$ which does not contain $1$.
Let $P_1,\dots,P_r$ be a generating family of $I$.
Let $A=k^{\mathbf N}$ and let $m$ be a maximal ideal of $A$ which contains
the ideal $N=k^{(\mathbf N)}$ of $A$. Then $K=A/m$ is an algebraically closed field
which is has at least the power of the continuum. 
(Alternative description: let $K$ be an ultrapower of $k$, with respect to a non-principal ultrafilter.)
Lemma. For $i\in\{1,\dots,r\}$, let $a_i=(a_{i,n})\in A$. Assume that $(\bar a_1,\dots,\bar a_r)=0$ in $K^r$. Then the set of $n\in\mathbf N$ such that $(a_{1,n},\dots,a_{r,n})=0$
is infinite.
Proof. Assume otherwise. For every $n$ such that $(a_{1,n},...,a_{r,n}) \neq 0$,
choose $(b_{1,n},\dots,b_{r,n})$ such that $\sum a_{i,n}b_{i,n}=1 $,
and let $b_i=(b_{i,n})_n\in A$. Then $\sum a_i b_i - 1 $ belongs to $N^r$,
hence $\sum \bar a_i \bar b_i=1$. Contradiction.
Thanks to the lemma, one proves easily that 
the ideal $I_K$ of $K[X_1,...,X_d]$ generated by $I$
does not contain $1$. 
By the uncountable case, there exists $x=(x_1,...,x_d)\in K^d$ such 
that $P_j(x_1,...,x_d)=0$ for every $j$. 
For every $i$, let $a=(a_{n})\in A^d$ be such that $\bar a=x$.
By the lemma again the set of integers $n$ such that
$P_j(a_n) \neq 0$ for some $j$ is finite.
In particular, there exists a point $y\in k^d$ such that $P_j(y)=0$ for every $j$.<|endoftext|>
TITLE: Do plane projections determine a convex polytope?
QUESTION [33 upvotes]: Suppose a compact convex body $P \subset \Bbb R^3$ has only polygonal orthogonal projections onto a plane.  Does this imply that $P$ is a convex polytope?
This question occurred to me when I was making exercises for my book.  I figured this is probably easy and well known, but the literature hasn't been any help.  One remark:  if the number of sides of all polygons is bounded by $n$, the problem might be easier.  Furthermore, if $P$ is assumed to be a convex polytope, this elegant paper by Chazelle-Edelsbrunner-Guibas (1989) gives a (perhaps, unexpectedly large) sharp $\exp O(n \log n)$ upper bound on the number of vertices of $P$ (ht Csaba Toth who generalized this to higher dimensions).

REPLY [13 votes]: Here is a more direct proof of this statement.
A Short Proof of Klee's Theorem by John J. Zanazzi<|endoftext|>
TITLE: What is the classifying space of "G-bundles with connections"
QUESTION [38 upvotes]: Let $G$ be a (maybe Lie) group, and $M$ a space (perhaps a manifold).  Then a principal $G$-bundle over $M$ is a bundle $P \to M$ on which $G$ acts (by fiber-preserving maps), so that each fiber is a $G$-torsor (a $G$-action isomorphic, although not canonically so, to the action of $G$ on itself by multiplication).  A map of $G$-bundles is a bundle map that plays well with the actions.
Then I more-or-less know what the classifying space of $G$ is: it's some bundle $EG \to BG$ that's universal in the homotopy category of (principal) $G$-bundles.  I.e. any $G$-bundle $P \to M$ has a (unique up to homotopy) map $P\to EG$ and $M \to BG$, and conversely any map $M\to BG$ (up to homotopy) determines a (unique up to isomorphism) bundle $P \to M$ and by pulling back the obvious square.
At least this is how I think it works.  Wikipedia's description of $BG$ is here.
So, let $G$ be a Lie group and $M$ a smooth manifold.  On a $G$-bundle $P \to M$ I can think about connections.  As always, a connection should determine for each smooth path in $M$ a $G$-torsor isomorphism between the fibers over the ends of the path.  So in particular, a bundle-with-connection is a (smooth) functor from the path space of $M$ to the category of $G$-torsors.  But not all of these are connections: the value of holonomy along a path is an invariant up to "thin homotopy", which is essentailly homotopy that does not push away from the image of the curve.  So one could say that a bundle-with-connection is a smooth functor from the thin-homotopy-path-space.
More hands-on, a connection on $P \to G$ is a ${\rm Lie}(G)$-valued one-form on $P$ that is (1) invariant under the $G$ action, and (2) restricts on each fiber to the canonical ${\rm Lie}(G)$-valued one-form on $G$ that takes a tangent vector to its left-invariant field (thought of as an element of ${\rm Lie}(G)$).
Anyway, my question is: is there a "space" (of some sort) that classifies $G$-bundles over $M$ with connections?  By which I mean, the data of such a bundle should be the same (up to ...) as a map $M \to $ this space.  The category of $G$-torsors is almost right, but then the map comes not from $M$ but from its thin-homotopy path space.
Please re-tag as desired.

REPLY [7 votes]: There should be an answer to Theo's question in terms of universal connections, but I don't know it.
This universal connection is a connection on the universal principal $G$-bundle over $BG$, such that every $G$-bundle with connection over $M$ is isomorphic (as a bundle with connection) to the pullback of $EG$ along some map $M \to BG$.
I've never found an answer to the following immediate question: what is the correct equivalence relation on the space of maps from $M$ to $BG$, such that equivalence classes of maps are in one-to-one correspondence with isomorphism classes of $G$-bundles with connection over $M$. Does anybody know this?
Also, I wanted to remark that $G$-bundles with connection are not the same as smooth functors on the thin path groupoid of $M$. Assuming global smoothness, you'll only get connections on trivializable bundles. The full story is here: http://arxiv.org/abs/0705.0452.<|endoftext|>
TITLE: Surgery and homology: a reference request
QUESTION [7 upvotes]: I need a reference (or a short proof) for the following statement:
Suppose a closed manifold $N$ is the result of a surgery (along an embedded sphere) on a closed manifold $M$. Then the difference $\sum dim H_i(N) - \sum dim H_i(M)$ (the homology is taken with coefficients in a field) is at most 2.

REPLY [12 votes]: To say that a smooth, closed manifold $N$ is obtained by surgery along a (framed) sphere in $M$ is to say that there is a cobordism $P$ from $M$ to $N$ and a Morse function $f\colon P\to [0,1]$, with $f^{-1}(0)=M$, $f^{-1}(1)=N$, and exactly one critical point $c$. By Morse theory, $H_*(P,M)$ is then 1-dimensional, generated by the descending disc of $c$. Likewise, $H_*(P,N)$ is 1-dimensional, generated by the ascending disc of $c$. 
By the homology exact sequence of the pair $(P,M)$, $\dim H_*(M)$ differs from $\dim H_*(P)$ by $1$. By the homology exact sequence of the pair $(P,N)$, $\dim H_*(N)$ also differs from $\dim H_*(P)$ by $1$. Hence $|\dim H_*(M) - \dim H_*(N)|$ is $0$ or $2$.<|endoftext|>
TITLE: What is the best graph editor to use in your articles?
QUESTION [52 upvotes]: Here is the criteria for a "perfect" graph editor:

it should be able to perform an automated, but controllable layout
one is able to make "manual" enforcements to nodes and edges locations when you need it (or at least such fine automated layout so you don't need "manual" enforcements)
one could add some math symbols and formulae on a graph

Common vector graphics editors could do the trick, but there is a lot of overhead efforts to draw every node, every edge, every label.
Graphviz is good enough, but sometimes you cannot get needed layout (even if you use several tricks like additional invisible nodes etc) and you should use ladot or dot2tex for math formulae
yEd has nice layouts, but there is a problem with a math text.
This is probably not a math question, but it is common to draw graphs in articles i think.
Result graph
(Update: 27.12.2010)
Here is another candidate for the best editor in TeX - Asymptote (asymptote.sf.net).
The very powerful tool at first glance.
(Update: 28.04.2014)
The very tasty semiautomated tool to use with PGF/TiKZ is TiKZEdt. You can extend its palette with your own tools and make the process of diagram creation very simple!

REPLY [5 votes]: I personnally mostly use xfig, and start moving to inkscape. However I guess that they are not what you want. A friend of mine, working on graphs, uses metapost with great results. Last, for 3D images, I use povray which is very efficient with geometric forms. It renders natively in .png, so the best is to compile with pdflatex. You can find mathematical picture rendered with povray on my web page: http://www-fourier.ujf-grenoble.fr/~bkloeckn/images.html<|endoftext|>
TITLE: When is the composition of two totally ramified extension totally ramified?
QUESTION [6 upvotes]: If K is a finite extension of $\mathbb Q_p$ for some prime number $p$, (possibly need $p \neq 2$), $L_1$ and $L_2$ totally ramified abelian extension of $K$, $ \pi_1, \pi_2$ are respectively the uniformizer that generates each field. Is it true that $ L_1 L_2$ is totally ramified iff $Nm_{L_1 / K}(\pi_1)$, $Nm_{L_2 / K}(\pi_2)$ differs by an element in the intersection of the two norm groups.
Is there a proof of this result (or the correct version of the result) without employing big tools?
Add at 6:49 pm 18th Feb: From Class Field Theory we know that there are one maximal totally ramified abelian extension of a local number field $K$ corresponding to each uniformizer, so I would expect that some version of the above statement is true. At least when $Nm_{L_1 / K}(\pi_1)=Nm_{L_2 / K}(\pi_2)=x$, $ L_1 L_2$ is totally ramified, because both are contained in $K^{ram}_x$.

REPLY [2 votes]: This does not answer your question, but rather goes in the opposite direction. Abhyankar's lemma states that if, say, $L_1$ is tamely ramified, then $L_1L_2/L_2$ will be unramified. I doubt that you can exclude this possibility simply by looking at the norms of the uniformizers, but I might be wrong.<|endoftext|>
TITLE: An unfamiliar (to me) form of Hensel's Lemma
QUESTION [24 upvotes]: In his very nice article

Peter Roquette,
  History of valuation theory. I. (English summary) Valuation theory and its applications, Vol. I (Saskatoon, SK, 1999), 291--355,
  Fields Inst. Commun., 32, Amer. Math. Soc., Providence, RI, 2002

Roquette states the following result, which he attributes to Kurschak:
Hensel-Kurschak Lemma: Let $(K,|\ |)$ be a complete, non-Archimedean normed field.  Let $f(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \in K[x]$ be a polynomial.  Assume (i) $f(x)$ is irreducible and (ii) $|a_0| \leq 1$.  Then $|a_i| \leq 1$ for all $0 < i < n$.  
He says that this result is today called Hensel's Lemma and that Hensel's standard proof applies.
This is an interesting result: Roquette explains how it can be used to give a very simple proof of the fact that, with $K$ as above, if $L/K$ is an algebraic field extension, there exists a unique norm on $L$ extending $| \ |$ on $K$.  This is in fact the argument I gave in a course on local fields that I am currently teaching.  
It was my initial thought that the Hensel-Kurschak Lemma would follow easily from one of the more standard forms of Hensel's Lemma.  Indeed, in class last week I claimed that it would follow from
Hensel's Lemma, version 1: Let $(K,| \ |)$ be a complete non-Archimedean normed field with valuation ring $R$, and let $f(x) \in R[x]$ be a polynomial.  If there exists $\alpha \in R$ such that $|f(\alpha)| < 1$ and $|f'(\alpha)| = 0$, then there exists $\beta \in R$ with $f(\beta) = 0$ and $|\alpha - \beta| < 1$.  
Then in yesterday's class I went back and tried to prove this...without success.  (I was not at my sharpest that day, and I don't at all mean to claim that it is not possible to deduce Hensel-Kurschak from HLv1; only that I tried the obvious thing -- rescale $f$ to make it a primitive polynomial -- and that after 5-10 minutes, neither I nor any of the students saw how to proceed.)  I am now wondering if maybe I should be trying to deduce it from a different version of Hensel's Lemma (e.g. one of the versions which speaks explicitly about factorizations modulo the maximal ideal).  
This brings me to a second question.  There are of course many results which go by the name Hensel's Lemma.  Nowadays we have the notion of a Henselian normed field, i.e., a non-Archimedean normed field in which the exended norm in any finite dimensional extension is unique.  (There are many other equivalent conditions; that's rather the point.)  Therefore, whenever I state a result -- let us restrict attention to results about univariate polynomials, to fix ideas -- as "Hensel's Lemma", I feel honorbound to inquire as to whether this result holds in a non-Archimedean normed field if and only if the field is Henselian, i.e., that it is equivalent to all the standard Hensel's Lemmata.
Is it true that the conclusion of the Hensel-Kurschak Lemma holds in a non-Archimedean valued field iff the field is Henselian?
More generally, what is a good, reasonably comprehensive reference for the various Hensel's Lemmata and their equivalence in the above sense?

REPLY [17 votes]: A far more general result is the "non-archimedean inverse function theorem". I haven't looked at Roquette's reference, so maybe he is mentioning it. But it is something which I didn't really find in the standard number theory textbooks - probably you can find it in texts on $p$-adic analysis - and I learned it from my number theory professor last semester (Jean-Benoît Bost). This theorem is powerful - and I find it fascinating and surprising - and all versions of Hensel's lemma which one usually encounters while learning number theory are immediate consequences.
Let $K$ be a field, $\left| \cdot \right|$ a non-archimedean absolute value on $K$ for which $K$ is complete, $\mathcal{O}$ the associated valuation ring, $\mathcal{M}$ the maximal ideal, $\pi$ a uniformizer. Let $\Phi_i \in \mathcal{O}[X_1,\,\cdots,X_n]$ for $1 \leq i \leq n$ and consider the map $\Phi = (\Phi_1,\,\cdots,\Phi_n) : \mathcal{O}^n \to \mathcal{O}^n$. Let $J$ be the Jacobian $\det(\partial \Phi_i / \partial X_j) \in \mathcal{O}[X_1,\,\cdots,X_n]$.
Theorem. If $x_0 \in \mathcal{O}^n$, $y_0 = \Phi(x_0)$ and $J(x_0) \neq 0$, then for any $R \in (0, \left|J(x_0)\right|)$, $\Phi$ induces a bijection $$\overline{B}(x_0,R) \to y_0 + (D\Phi)(x_0) \overline{B}(0,R)$$ (where $D\Phi$ is the derivative we all know!) and furthermore we have a bijection $$B^\circ(x_0,\left|J(x_0)\right| \to y_0 + (D\Phi)(x_0) B^\circ(0,\left|J(x_0)\right|).$$
(I use the standard notations $\overline{B}$ and $B^\circ$ for closed and open balls respectively.)
The proof uses in an essential way the Picard fixed point theorem.
Corollary 1. Take $n = 1$, $\Phi_1 = P$, $x_0 = \alpha$, $\varepsilon \in (0,1)$. Suppose that $\left|P(\alpha)\right| \leq \varepsilon \left|P'(\alpha)\right|^2$. Then there exists a unique $\beta \in \mathcal{O}$ such that $P(\beta) = 0$ and $\left|\beta - \alpha\right| \leq \varepsilon \left|P'(\alpha)\right|$. (We take $R = \varepsilon \left|P'(\alpha)\right|$ in the first bijection.) 
Hence, as a special case, if $\left|P(\alpha)\right| < \left|P'(\alpha)\right|^2$, we find $\left|\beta - \alpha\right| < \left|P'(\alpha)\right|$.
As an even more special case, if $P'(\alpha) \in \mathcal{O}^\times$ and $\left|P'(\alpha)\right| <1$, there exists $\beta \in \mathcal{O}$ such that $P(\beta) = 0$ and $\left|\beta - \alpha\right| < 1$. Restating this in terms of the residue field: a simple zero in the residue field can be lifted to a real zero in $\mathcal{O}$. This is the really known version of Hensel's lemma, I guess.
[Definition: the Gauss norm of a polynomial with coefficients in $K$ is defined as the maximum of the absolute values of its coefficients. It is very easy to check that the Gauss norm is multiplicative.]
Corollary 2. Take $f,g,h \in \mathcal{O}[X]$ such that $\deg g = n$, $\deg h = m$ and $\deg f = \deg g + \deg h = n + m$. Assume that there exists $\varepsilon \in (0,1)$ such that $\left\|f - gh\right\| \leq \varepsilon\left|\text{Res}(g,h)\right|^2$ and $\deg(f - gh) \leq m + n - 1$. Then there exist $G, H \in \mathcal{O}[X]$ such that $f = GH$, $\deg(G - g) \leq n - 1$, $\deg(H - h) \leq m - 1$, and also $\left\|G - g\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$ and $\left\|H - h\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$. (Obviously $\text{Res}$ denotes the resultant here, and $\left\|\cdot\right\|$ the Gauss norm.)
To prove this: write $G = g + \xi$ and $H = h + \eta$ where $\xi$ and $\eta$ are polynomials with coefficients in $\mathcal{O}$ and have degrees $\leq n - 1$ and $\leq m - 1$ respectively. Then $f = GH$ if and only if $f = (g + \xi)(h + \eta)$. It can be seen as a map from $\mathcal{O}^n \times \mathcal{O}^m \to \mathcal{O}^{n + m}$ given by polynomials. So consider the map $\Phi: (\xi, \eta) \mapsto (g + \xi)(h + \eta) - f$. We have also  $\text{Res}(g,h) = \det((\xi, \eta) \mapsto g \xi + h \eta))$. It is easy to see that the theorem above then gives the result.
As a corollary: if $f$, $g$ and $h$ satisfy $\overline{f} = \overline{g} \overline{h}$ - where $\overline{f}$ is $f$ reduced modulo $\mathcal{M}$ et cetera - and if $\overline{g}$ and $\overline{h}$ are coprime (this is a condition on the resultant!) then there exist $G,H \in O[X]$ satisfying the following conditions: $f = GH$, $\deg(G - g) \leq n - 1$, $\deg(H - h)\leq m - 1$, $\overline{G} = g$ and $\overline{H} = h$. Hence "a factorization over the residue field lifts to a factorization over $\mathcal{O}$" (under the right conditions).
Corollary 3. Finally, let us come to the motivation for the question: the more general result is that if $P \in K[X]$ is irreducible, then $\left\|P\right\|$ (Gauss) is the maximum of the absolute values of the leading coefficient and the constant coefficient. (As a special case, we find the result which Pete L. Clark cites as the Hensel-Kurschak lemma.) 
Indeed, let $P(X) = \sum_{i = 0}^n a_i X^{n - i} \in K[X]$. Suppose WLOG that $\left\|P\right\| = 1$. Let $\mathbb{F}$ be the residue field and let $\overline{P}$ be the image of $P$ modulo $\mathcal{M}$. Set $r = \min \{n : \overline{a_{n - r}} \neq 0\}$. Then we have in the residue field the factorization $\overline{P}(X) = X^r \left(\overline{a_{n - r}} + \overline{a_{n - r - 1}}X + \cdots + \overline{a_0} X^{n - r}\right)$ and we can lift the factorization by Corollary 2, contradicting irreducibility.
I know this is quite some digression; but I find the whole discussion about the various forms of Hensel's lemma very interesting, and I thought this could add something to the discussion.<|endoftext|>
TITLE: Is it necessary that model of theory is a set?
QUESTION [13 upvotes]: From Model Theory article from wikipedia : "A theory is satisfiable if it has a model $ M\models T$ i.e. a structure (of the appropriate signature) which satisfies all the sentences in the set $T$". In structure definition there is also requirement for "container of a structure" to be set. 
As we assume then, every model have to lay inside of some set-container. This obviously give us in serious trouble, as for set theory, there is no model of this type, and even maybe cannot be. One of possible explanations why set theory cannot be closed inside set ( which will lead us to some well known paradoxes) is opinion that "there can be no end to the process of set formation" so we have "structure" which cannot be closed inside itself which is obviously rather well state.
As we know that not every theory may have a model (see set-theory) then some question arises:

What are the coses (other than pure practical -  if they are set we know how to work with them) of putting so strong
requirements for model to be set?
Is there any way to weaken this
requirement?
Are there any "explorations" of
possible extension of model theory
with fundamental objects other than
sets?

I presume that some categorical point of view may be useful here, but is there any? I am aware about questions asked before, specially here 8731, but it was asked in context of category theory which is of course valid point of view but somehow too fine. I would like to ask in general. 
And last one, philosophical question: is then justified, that condition for a theory to have model in set universe is some kind of anthropomorphic point of view -  just because we cannot build any other structures in effective way we build what is accessible for us way but it has no objective nor universal meaning? Is true that model theory is only a "universal algebra+logic" in universe of the set, or it justifications may be extended to some broader point of view? If yes: which one?
I have hope that this question is good enough for mathoverflow: at least please try to deal wit as kindly request for references.

Remark: Well formulated point from n-CathegoryCafe discussion: "In the centre of Model Theory there is " fundamental existence theorem says that the syntactic analysis of a theory [the existence or non-existence of a contradiction] is equivalent to the semantic analysis of a theory [the existence or non-existence of a model]." 
In fact the most important point is: may it be extended on non "set container" universes?

I would like to thank everyone who put here some comments or answers. In is the most interesting that in a light of answer of Joel David Hamkins it seems that for first order theories (FOT) consistency is equivalent to having set model. It is nontrivial, because it is no matter of somehow arbitrary definition of "having model" but it is related to constructive proof of Completeness Theorem of Gödel. From ontological point of view it then states that for FOT there is no weaker type of consistency than arising from model theory based on sets, and in some way it is maximal form of consistency simultaneously. Then there is no way to extend for FOT equivalence to non-set containers, which is nontrivial part -  the only theories which are consistent in FOT are those which has a set models and this statement is proved not using set theoretical constructions in nonconstructive ways. So it was important to me, and I learn a lot from this even if for specialist it is somehow maybe obvious. I have hope that I understand it right;-)
@Tran Chieu Minh: thank You for pointing to interesting discussion, I will try to understand the meaning of Your remarks here and there.

REPLY [9 votes]: The fundamental reason why models in model theory are required to be sets is that for such models $M$ there exists the satisfaction relation $M\models\phi[e]$ between formulas $\phi$ and evaluations $e$, obeying Tarski’s definition. This is not possible in general for class models in ZFC (or NBG)—for example, the existence of a satisfaction relation for $(V,\in)$ would imply the consistency of ZFC, which is not provable in ZFC.
There are other reasons for sticking to set models (for example, various constructions of models, in particular those using the compactness theorem, tend not to work for proper classes), but this is the most important one. On the other hand, in some parts of model theory it is convenient to work inside a “monster model” which may be a proper class (at least metaphorically, i.e., large enough so that it behaves as if it were a proper class for all arguments we want to employ it in), but one has to be careful when working in such a setup.<|endoftext|>
TITLE: Symmetric groups which are not quotients of Z/2Z*Z/3Z
QUESTION [15 upvotes]: Somehow this question made me think of instances of small exceptions in general, and I remembered the statement I heard once that $S_5,A_6,S_6,A_7,A_8,S_8$ are the only instances of symmetric/alternating groups that are not quotients of $Z/2Z*Z/3Z=PSL_2(Z)$ (see this MathSciNet entry, which I just found). Does anyone have an idea of a conceptual explanation for this fact? 
Edit: I also find this article which mentions the same result. It's quite interesting that the positive part (for all $n>8$ these groups are quotients) is proved using Bertrand's postulate. I think it's cool that Bertrand's postulate can be used for group theory.

REPLY [10 votes]: Yes, this is well known.  In fact, there is a name for such groups - this is a (2,3)-generation property.  And yes, by now there is a conceptual understanding why all sufficiently large finite simple groups have this property - the basic ideas are outlined in this helpful MathSciNet review gently explaining the major breakthrough by Liebeck and Shalev (1996).  There are more recent developments in the field, both in the asymptotic direction and in the explicit construction, such as figuring out which $PSL(n,q)$ are (2,3)-generated - see papers by Tamburini, etc. - the literature is too big to be reviewed here.<|endoftext|>
TITLE: Newton and Newton polygon
QUESTION [29 upvotes]: What did Newton himself do, so that the "Newton polygon" method is named after him?

REPLY [6 votes]: This was intended to be a comment on Bill Dubuque's answer, but I apparently don't yet have enough reputation points to comment, and in any event this is probably too long to appear as a comment.
Given Chrystal's intended audience, I'm surprised that he didn't mention Talbot's 1860 English translation and extensive commentary of Newton's Enumeration Linearum Tertii Ordinis. In Talbot's work, which is freely available on the internet, see the sections On the Analytical Parallelogram (pp. 88-104) and Examples (pp. 104-112). By the way, whoever scanned the book for google wasn't paying attention when the lengthy list of figures at the end of the book were being scanned, so I'm also giving the University of Michigan Historical Math Collection version, which has those figures correctly scanned.
Sir Isaac Newton's Enumeration of Lines of the Third Order, Generation of Curves by Shadows, Organic Description of Curves, and Construction of Equations by Curves, Translated from the Latin, with notes and examples, by C.R.M. Talbot, 1860.
http://books.google.com/books?id=6I97byFB3v0C
http://name.umdl.umich.edu/ABQ9451.0001.001<|endoftext|>
TITLE: Largest possible order of a nilpotent permutation group?
QUESTION [8 upvotes]: I'm trying to obtain a bound for the order of some finite groups, and part of it comes down to the order of a permutation group of degree $n$ that is nilpotent.  I imagine these have to be much smaller than the full symmetric group, and a bound that is sub-exponential in $n$ would seem reasonable (given that permutation $p$-groups fall a long way short of having exponential order), but I haven't seen this written down anywhere.
I found one reference that looks promising:
P. Palfy, Estimations for the order of various permutation groups,  Contributions to general algebra, 12 (Vienna, 1999),  37-49, Heyn, Klagenfurt, 2000.
However, I can't actually find the article anywhere online.  Any suggestions?

REPLY [7 votes]: The paper of Vdovin mentioned by Steve shows that the nilpotent subgroups of the symmetric groups of maximal order are either the Sylow 2-subgroups P(n) of Sym(n), or P(n-3) x Alt(3) when n =  2(2k+1)+1.
Vdovin, E. P. "Large nilpotent subgroups of finite simple groups."
Algebra Log. 39 (2000), no. 5, 526-546, 630; translation in Algebra and Logic 39 (2000), no. 5, 301-312.
http://www.ams.org/mathscinet-getitem?mr=1805754
http://dx.doi.org/10.1007/BF02681614
It looks at the orbits of the center of the nilpotent group, and the action of the quotient on those orbits to give a reasonable bound.  Then it shows that all nilpotent subgroups of maximal order are conjugate to the types mentioned.
The paper also handles alternating groups, groups of Lie type, and sporadic groups.  For groups of Lie type, the maximal order nilpotent is usually a Sylow p-subgroup for p the characteristic.  Sporadics are only handled briefly: the nilpotent subgroups of maximal order are always Sylows and they satisfy the same general bound as the other simple groups.<|endoftext|>
TITLE: Group cohomology vs. topological cohomology in the case of non-trivial action
QUESTION [7 upvotes]: When A is an abelian group with trivial G-action (G being a discrete group) we get that Hn(G,A)≅Hn(BG,A). Is there a similar connection between group cohomology and topological cohomology if A is a non-trivial ℤG-module? What can we say in that case?

REPLY [4 votes]: For any space $X$ with $\pi_1X=G$  and any $ZG$ module $A$, define $C_*(X;A)$ to be $C_*(\tilde{X})\otimes_{ZG}A$ and define $C^* (X;A)$ to be 
$Hom_{ZG}(C_*(\tilde{X}),A)$, where  $C_*(\tilde{X})$ is the
(cellular, or singular) chain complex of the universal cover of $X$ viewed as a free $ZG$ complex. (supressing notation involving left v.right action). Then take co/homology. It makes no difference whether the action is trivial or not. If $A$ has a trivial action, then $C_*(\tilde{X})\otimes_{ZG}A=C_*(X)\otimes A$ and $Hom_{ZG}(C_*(\tilde{X}),A)=Hom(C_*(X),A)$.  Taking $X=BG$ answers your question, assuming you like topology and  define group homology in terms of $BG$
If you prefer to start with an algebraic definition of group cohomology (eg bar construction) then you need some argument that relates the algebra to a cell structure for $BG$: but you need this argument whether or not your coefficients are twisted.<|endoftext|>
TITLE: Cryptomorphisms
QUESTION [28 upvotes]: I am curious to collect examples of equivalent axiomatizations of mathematical structures.  The two examples that I have in mind are 

Topological Spaces. These can be defined in terms of open sets, closed sets, neighbourhoods, the Kuratowski closure axioms, etc.
Matroids.  These can be defined via independent sets, bases, circuits, rank functions, etc.

Are there are other good examples? 
Secondly, what are some advantages of multiple axiomatizations?
Obviously, one advantage is that one can work with the most convenient definition depending on the task at hand.  Another is that they allow different generalizations of the object in question.  For example, infinite matroids can be axiomatized by adapting the independent set axioms, but it is unknown how to axiomatize them via the circuit axioms.  An acceptable answer to the second question would be an example of a proof in one axiom system that doesn't translate easily (not sure how to make this precise) into another axiom system.

REPLY [3 votes]: A very familiar example is given by the different ways to express the completeness property of the real line --- Cauchy sequences converge, bounded nonempty sets have suprema, etc.<|endoftext|>
TITLE: Common Computations in Group Cohomology
QUESTION [7 upvotes]: Let G=A⋊B, where A and B are abelian, and of coprime order. It seems, from my computations (and correct me if I'm wrong), that Z1(Cp,Cq) is trivial, for p and q different primes. Meaning that the automorphisms of G, if A=Cq, and B=Cp, that preserve A, and preserve the cosets G/A, are all trivial. How far can we extend this? Would it be true in general that Z1(A,B) is trivial, with the above assumptions (that A and B are abelian and of coprime order)? If not, under what assumptions is it trivial? And when can we say about it if it's not trivial?

REPLY [3 votes]: You might be considering a special case of the Schur-Zassenhaus theorem.  If A is a normal Hall-subgroup of G, then A has a complement B, and all complements B are conjugate under the action of G.  This is more properly H^1(B,A) rather than Z^1(B,A).
For Z^1(B,A) to be trivial, B^1(B,A) must be trivial, but B^1(B,A) is basically [B,A], which could be nonzero.  For instance if G is non-abelian of order 6, then A is cyclic and normal of order 3, B is cyclic of order 2, and B acts as inversion on A.  Then for B^1(B,A) should be isomorphic to A, that is have order 3.  Basically, A has 3 complements in G, so B^1(B,A) should have three elements, but all are conjugate, so H^1(B,A)=0.
I suppose there are lots of definitions of C^1(B,A), so maybe you've chosen one where B^1(B,A)=0, but I think the standard choice when looking at semi-direct products ("Crossed homomorphisms"), will not have B^1(B,A) trivial.<|endoftext|>
TITLE: A two-variable Fourier series and a strange integral
QUESTION [11 upvotes]: I have recently had occasion to investigate the Fourier series of the function $f(x,y)=\log({2+\cos 2\pi x} +\cos{2\pi y})$.  Accordingly, define
$I(m,n)=\int_{0,0}^{1,1}f(x,y)\cos{2\pi mx}\cos{2\pi ny}dxdy$ 
which is the $(m,n)$th Fourier coefficient.  If there is some easy change of variables or flick of the wrist to compute $I$ in closed terms, I haven't been able to find it. One may compute $I(0,0)=\frac{4G}{\pi}-\log 2$, where $G$ is Catalan's constant.  As for other values, Mathematica's integration routine returns the values $I(1,1)=\frac{1}{2}+\frac{2}{\pi}$ and $I(2,1)=1-\frac{10}{3\pi}$ (and takes about 10 minutes per computation).  I have tried and failed to affirm the following conjecture, which seems reasonable to present here.
Conjecture. If $(m,n)\neq(0,0)$, then $I(m,n)=a+\pi^{-1}b$ with $a,b\in \mathbb{Q}$.
Of course, if you could prove this, it would be great to give a closed formula for $a,b$.
Full disclosure: My interest in this stems from a formula of Kasteleyn for the partition function of dimer coverings of squares; in fact the evaluation of $I(0,0)$ given above can be extracted from his original paper.

REPLY [5 votes]: Mathematica gets the (1,1) case right. 
Integrate the x-variable to obtain an answer 
that involves the absolute value of Cos[Pi y].
Then split the y-integral at y=1/2 to get the 
correct value.
The answer is 1/2 - 2/Pi<|endoftext|>
TITLE: Concerning the dimension of a complex variety modulo a prime
QUESTION [6 upvotes]: Let V be a complex affine variety given as the vanishing set of a set of polynomials with integral coefficients. I have 3 questions.
1)
Under what assumption  will the dimension of V over C remain the same as the dimension of V computed over some finite field like Z_P,  assuming that the prime P does not divide any of the coefficients of the polynomials defining V?
2)
How can such a prime P be found if we have  a set of polynomials whose vanishing set is  V? 
3)
Suppose that V has a singularity over Z_P. Can we conclude that V is singular variety? In other words. Does having a singularity over a finite field immediately imply the variety is not smooth? If not, what guarantees that having  a singularity modulo p implies a singularity over C?

REPLY [6 votes]: To make more precise the answer of Felip. You have a scheme $X=Spec(A)$ over $\mathbb Z$, where $A$ is a finitely generate $\mathbb Z$-algebra such that its generic fiber $X_{\mathbb Q}$ (just consider your polynomials as polynomials with rational coefficients) gives $V$ by field extension $\mathbb{C}/\mathbb{Q}$. Of course, $X_{\mathbb Q}$ has the same dimension as $V$, and $X_{\mathbb Q}$ is smooth if and only $V$ is smooth. 
Questions 1-2. You want to compare $\dim X_p$ with $\dim X_{\mathbb Q}$. In general $\dim X_p\ge \dim X_{\mathbb Q}$. The equality holds under some flatness condition at $p$. Namely, there is rather general result: if $f: Y\to Z$ is a flat morphism of finite type between noetherian schemes and suppose that $Z$ is integral and universally catenary (e.g.any scheme of finite type over a noetherian regular scheme, so any open subset of $Spec(\mathbb Z)$ is OK), then all [EDIT: non-empty] fibers $Y_z$ of $f$ have the same dimension [EDIT: if the generic fiber of $f$ is equidimensional (i.e. all irreducibles components have the same dimension)].  
Problem: your $X$ is not necessarily flat over $\mathbb Z$. But the flatness over $\mathbb Z$ (or any Dedekind domain) is easy to detect: it is equivalent to be torsion free. So consider the ideal $I$ equal to
$$ { a\in A \mid ka=0 \text{ for some non zero } k \in \mathbb Z \}$$
(don't know how to type "{" and "}"). Then $A/I$ is flat over $\mathbb Z$ and defines a closed scheme of $X$ which coincides with $X$ over an open subset of $\mathbb Z$. Actually, as $I$ is finitely generated, there exists a positive integer $N$ such that $NI=0$. Then $I=0$ over $Spec(\mathbb Z[1/N])$. So for any prime number $p$ prime to $N$, $X_p$ will have dimension $\dim V$. Now you have to compute such a $N$... For hypersurface, $N$ is just the gcd of the coefficients (see comments in Felip's answer). I don't know whether efficient methodes exist in general. Of course if a prime $p$ does not divid any polynomial in the definning ideal of $X$, then this $p$ is OK. But you have to test this property for all polynomials and not just a set of  generators (Example: the ideal generated by $T_1+pT_2, T_1$, then $p$ is bad). 
Question 3. Suppose $V$ is smooth (and connected for simplicity), and you are looking for the $p$ such that $X_p$ is also smooth. You first proceed as in Question 1 to find an open subset $Spec(\mathbb Z[1/N]$ over which $X$ is flat. Let $d=\dim X_{\mathbb Q}=\dim X_p$ for all $p$ prime to $N$. Write 
$$ X=Spec\big(\mathbb Z[T_1, \ldots T_n]/(F_1,\ldots, F_m)\big)$$ 
Then $X_p$ is smooth is and only if the Jacobian matrix of the $F_i$'s mod $p$ has rank $n-d$ at all points of $X_p$. Equivalently, the ideal $J\subseteq \mathbb Z[T_1,\ldots, T_n]$ generated by $F_1,...,F_m$ and the rank $n-d$ minors of the Jacobian matrix is the unit ideal mod $p$. Therefore, the computation consists in determining the ideal $J$. As $X_{\mathbb Q}$ is supposed to be smooth, $J$ is generated by a positive integer $M$. Now for all $p$ prime to $MN$, $X_p$ is smooth of dimension $\dim V$. 
[EDIT]: The assertion on the dimensions of the fibers needs some hypothesis on the generic fiber of $Y\to Z$. We must assume it is equidimensional. Otherwise $Y_z$ has dimension equal to that of the generic fiber for $z\in Z$ belonging to the image of all irreducible components of $X$. For the initial question, it is better to assum the original variety $V$ is irreducible.<|endoftext|>
TITLE: The Infinitesimal topos in positive characteristic
QUESTION [17 upvotes]: This question was inspired by and is somewhat related to this question.
In his article "Crystals and the de Rham cohomology of schemes" in the collection "Dix exposes sur la cohomologie des schemas" Grothendieck defines the (small) infinitesimal site of an $S$-scheme $X$ using thickenings of usual opens. He then proceeds to prove that in characteristic $0$ the cohomology with coefficients in $\mathcal{O}_{X}$ computes the algebraic de Rham cohomology of the underlying scheme. This is remarkable, because the definition of the site does not use differential forms and it is not necessary for $X/S$ to be smooth.
This fails in positive characteristics, and as a remedy, Grothendieck suggests adding the additional data of divided power structures to the site, which he then calls the "crystalline site of $X/S$". This site then has good cohomological behaviour (e.g. if $X$ is liftable to char. $0$, then cohomology computed with the crystalline topos is what it "should be"). The theory of the crystalline topos was of course worked out very successfully by Pierre Berthelot.
My question is: Even though the infinitesimal site is in some sense not nicely behaved in positive characteristics, have people continued to study it in this context? What kind of results have been obtained, and has it still been useful? I'm particularly interested in results about $D$-modules in positive characteristic (i.e. crystals in the infinitesimal site if $X/S$ is smooth), but I am also curious to see in which other directions progress has been made.

REPLY [23 votes]: There is a paper of Ogus for 1975, "The cohomology of the infinitesimal site",
in which he shows that if $X$ is proper over an algebraically closed field $k$ of char. $p$, and embeds into a smooth
scheme over $k$, then the infinitesimal cohomology of $X$ coincides with etale cohomology with coefficients in $k$ (or more generally $W_n(k)$ if we work with the infinitesimal site of $X$ over $W_n(k)$). 
Note that, since $k$ has char. $p$,  we are talking about etale cohomology with
mod $p^n$ coefficients, so this is "smaller" than the usual etale cohomology;
it just picks up the "unit roots" of Frobenius.  So the infinitesimal cohomology
gives the unit root part of the crystalline cohomology.<|endoftext|>
TITLE: Calculating 6j-symbols (aka Racah-Wigner coefficients) for quantum groups
QUESTION [20 upvotes]: Which $6j$-symbols for quantised enveloping algebras are known explicitly?
The quantum $6j$-symbols for $sl(2)$ are well-known. The references are
Masbaum and Vogel  and 
Frenkel and Khovanov. 
What is known for other simple Lie algebras?
In case this seems somewhat vague here is a precise question. The data for a $6j$-symbol starts with a tetrahedral graph with edges labelled by highest weights. Then usually there is additional information needed at the vertices which I want to avoid. Take the example of $sl(n)$ and use partitions instead of highest weights. Label two opposite edges by a partition of the form $1^k$ (corresponding to an exterior power of the vector representation) and label the other four edges by partitions. Then associated to this data is a scalar.
Then I would expect this function to be determined by linear recurrence relations (i.e. D-finite or holonomic). Is this correct? and if so can you give recurrence relations?
Ideally we would also regard $n$ as an indeterminate.
Background In general $6j$-symbols arise for any semisimple abelian category which is also monoidal. They are the components of the natural transformation $(\otimes)(\otimes \times 1)\cong (\otimes)(1\times \otimes)$.
In more down to earth terms. If you know the Grothendieck ring of a semisimple abelian monoidal category and you attempt to construct this then the information you are missing is the $6j$-symbols. You can construct the abelian category and you can construct the tensor product functor but you don't have the associator.
For example it is well-known that the character table of a finite group does not determine the group. It does determine the category of representations as a semisimple abelian category. The $6j$-symbols are needed to make this a monoidal category.
For further discussion see:
Character table does not determine group Vs Tannaka duality

REPLY [14 votes]: We can also use partitions ($k$) (symmetric powers) instead of ($1^k$) on one or two of the edges. This still gives just scalars, but includes the full story for sl(2). 
This problem seems to be equivalent to the problem of computing the exchange operator 
in the tensor product of two (quantum) symmetric or exterior powers  of the vector representation of the quantum sl(n), e.g. $S^kV\otimes S^mV$, in the standard basis of the symmetric (resp., exterior) powers (about exchange operators see my paper with Varchenko arXiv:math/9801135 and my ICM talk arXiv:math/0207008). This can be computed if you know the fusion operator for these representations, which can be computed efficiently using the ABRR (Arnaudon-Buffenoir-Ragoucy-Roche) equation, see e.g. the appendix to arXiv:math/9801135. I am not sure if the answer is completely worked out anywhere, but there are at least some answers. For instance, see the paper arXiv:q-alg/9704005 where something is done even in the elliptic setting (which relates to elliptic 6j symbols of Frenkel-Turaev). What they do is compute the matrix elements for $m=1$, but the general $m$ can be obtained using the fusion procedure. This should be a really nice computation with a nice answer of the type you are expecting. In particular, in a special case you'll get coefficients of Macdonald's difference operators attached to symmetric powers. In the exterior powers case (or a product of a symmetric and an exterior power) the answer will be simpler, since $k$ cannot get larger than $n$ (in this case you should perhaps get a pure product, and it should be completely covered in the above paper). 
EDIT: Remark. Let $V,W$ be representations of the quantum group with 1-dimensional zero weight spaces. Then a natural basis in ${\rm Hom}(L_\lambda, (V\otimes L_\mu)\otimes W)$ is the compositions of intertwiners in one order, while a natural basis of ${\rm Hom}(L_\lambda, V\otimes (L_\mu\otimes W))$ is the composition of intertwiners in the other order. Thus, the 6j-symbol matrix (which is by definition the transition matrix between these two bases) is the exchange operator for intertwiners.<|endoftext|>
TITLE: Restriction theorems over finite fields
QUESTION [9 upvotes]: A short while ago, Dvir proved the Kakeya conjecture over finite fields. Does this have any implications for restriction theorems over finite fields? I am aware only of implications going in the opposite direction (Mockenhaupt-Tao). Is there anything new known about restriction theorems over finite fields?

REPLY [13 votes]: Not directly.  The key connection between restriction and Kakeya in Euclidean settings is that  thanks to Taylor expansion, a surface in Euclidean space looks locally flat, and so the Fourier transform of measures on that surface are a superposition of Fourier transforms of very flat measures, which by the uncertainty principle tend to propagate along tubes.  The arrangement of these tubes is then governed by the Kakeya conjecture.
In finite fields, there is no notion of Taylor expansion, unless one forces it into existence by working on the tangent bundle of a surface, rather than a surface itself.  So there is a connection between restriction and Kakeya on the former object, but not the latter. (This is discussed at the end of my paper with Mockenhaupt.)<|endoftext|>
TITLE: When does a probability measure take all values in the unit interval?
QUESTION [9 upvotes]: Let $\mathbb{P}$ be a probability measure on some probability space $(\Omega,\mathcal{A})$. Are there conditions on the $\sigma$-algebra $\mathcal{A}$ such that for every real number $c\in [0,1]$ we find a set $A\in\mathcal{A}$ with $\mathbb{P}(A)=c$. 
It is like the intermediate value theorem for continuous functions.

REPLY [4 votes]: A necessary and sufficient condition is that every atom is no larger than the sum of all smaller atoms, plus the non-atomic part.<|endoftext|>
TITLE: Confusion over a point in basic category theory
QUESTION [11 upvotes]: "Let Top be the category of topological spaces." If I see a definition like this, in which homeomorphic (isomorphic in the category) spaces are not identified together, then for each given topological space, how many times does it show up as an object in Top? Once? Countably many times? Uncountably many times? Is there a semantic crisis if we don't identify all homeomorphic topological spaces to the same object?

REPLY [4 votes]: This is not essentially different from what everyone else has said, but for some reason I feel like saying it anyway.
The number of times that a given topological space appears in the category of topological spaces is exactly once.  That's the definition of the class of objects of Top.
As for isomorphism classes: there is exactly one object in Top which is isomorphic to the empty set $\varnothing$ with its unique topology: namely $\varnothing$.  For any nonempty topological space $X$, the subclass of Top consisting of spaces $X'$ which are homeomorphic to $X$ forms a proper class: i.e., they are more numerous than any set.  This is true because the class of all sets is a proper class, and if $(X,\tau)$ is any nonempty topological space and $S$ is any set, then $X \times \{S\}$ is a set which is different from $X$ but in bijection with it: $x \mapsto (x,S)$, and the image of $\tau$ under this bijection gives a topological space which is homeomorphic to $(X,\tau)$ but with a different underlying set.
So, except for one, the homeomorphism classes in Top are "unsetically" huge.  Nevertheless, the modern perspective is that this is fine (or, perhaps, harmlessly irrelevant) whereas it is a bad idea to work with the homeomorphism class of the space instead of the space itself.  One way to think about this is that a given topological space $X$ is a relatively simple object, but the class of all topological spaces homeomorphic to $X$ is a ridiculously complicated object.  (This has not always been the received wisdom: notably Bertrand Russell's definition of the number $2$ was the class of all sets which can be put in bijection with, say, $\{\emptyset, \{ \emptyset\} \}$.)  From a less philosophical perspective, one wants to speak about sets of maps between two topological spaces $X$ and $Y$, and if $X$ and $Y$ are only well-defined up to a homeomorphism, these sets are themselves not so well defined.  As soon as one studies commutative diagrams of morphisms of objects in a category (or more generally, functorial constructions), one recognizes that the concept of "topological space up to a homeomorphism" is a painfully awkward one.  
It is also simply against the spirit of category theory to pass to isomorphism classes: many would say that this overemphasizes the somewhat quaintly philosophical notion of equality of objects.  Instead of saying "the topological space $X$ is equal to the topological space $Y$", many mathematicians now think it is both simpler and more useful to say "$\Phi: X \rightarrow Y$ is a homeomorphism".  For more on this point, I highly recommend Barry Mazur's article When is one thing equal to some other thing?<|endoftext|>
TITLE: Degrees of etale covers of stacks
QUESTION [7 upvotes]: This is probably pretty basic, but as I said before I'm just beginning my way in the language of stacks.Say you have an etale cover X->Y of stacks (in the etale site). Is there a standard way to define the degree of this cover? Here's my intuition: if X and Y are schemes, we can look etale locally and then this cover is Yoneda-trivial in the sense of http://front.math.ucdavis.edu/0902.3464 , meaning that etale locally it is just a disjoint unions of (d many) pancakes. Can we do this generally? Is there some "connectivity" conditions on Y for this to work? Is there a different valid definition for degree of an etale cover of stacks?
Yoneda-Triviality
I figured since nobody answered so far, maybe I should write down what a possible Yoneda-triviality condition could mean for stacks:
Def: Call f:X->Y (stacks) Yoneda-trivial if there exists a set of sections of f, S, such that the natural map Y(Z)xS->X(Z) is an isomorphism (or maybe a bijection?) for any connected scheme Z.
 But I'm still clinging to the hope that there's a completely different definition out there that I'm just not aware of.

REPLY [2 votes]: The question depends on which maps you want to call "etale". If one thinks that the property of $f:X \to Y$ being etale should be etale local on $X,$ then etale morphisms doesn't need to be representable; for instance $BG \to pt$ is etale and epic, if $G$ is a finite group, and its degree is....$1/|G|?$ Anyway, you can rule them out by always considering representable etale surjections. 
Also, I guess degree should better be defined for representable finite etale coverings, rather than just representable etale surjections. Imagine the disjoint union of two open sets covering a space: the degree over their overlap is 2, and is 1 elsewhere. This is not a locally constant function on the space.<|endoftext|>
TITLE: How do the compact Hausdorff topologies sit in the lattice of all topologies on a set?
QUESTION [74 upvotes]: This question is about the space of all topologies on a
fixed set X. We may order the topologies by refinement, so
that τ ≤ σ just in case every τ open set is open in σ.
Equivalently, we say in this case that τ is coarser
than σ, that σ is finer than τ or that
σ refines τ. (See wikipedia on comparison of
topologies.)
The least element in this order is the indiscrete topology and the largest topology is the discrete topology. 
One can show that the collection of all topologies on a fixed set is a complete lattice. In the downward direction, for example, the intersection of any collection of
topologies on X remains a topology on X, and this intersection
is the largest topology contained in them all. Similarly,
the union of any number of topologies generates a smallest
topology containing all of them (by closing under finite
intersections and arbitrary unions). Thus, the collection of all topologies on X is a complete
lattice. 
Note that the compact topologies are closed downward in
this lattice, since if a topology τ has fewer open sets than
σ and σ is compact, then τ is compact.
Similarly, the Hausdorff topologies are closed upward,
since if τ is Hausdorff and contained in σ, then
σ is Hausdorff. Thus, the compact topologies inhabit
the bottom of the lattice and the Hausdorff topologies the
top.
These two collections kiss each other in the compact
Hausdorff topologies. Furthermore, these kissing points,
the compact Hausdorff topologies, form an antichain in the
lattice: no two of them are comparable. To see this,
suppose that τ subset σ are both compact
Hausdorff. If U is open with respect to σ, then the
complement C = X - U is closed with respect to σ and
hence compact with respect to σ in the subspace
topology. Thus C is also compact with respect to τ in
the subspace topology. Since τ is Hausdorff, this
implies (an elementary exercise) that C is closed with respect to τ, and so U is
in τ. So τ = σ. Thus, no two distinct compact Hausdorff topologies are comparable, and so these topologies are spread out sideways, forming an antichain of the lattice.
My first question is, do the compact Hausdorff topologies
form a maximal antichain? Equivalently, is every topology
comparable with a compact Hausdorff topology? [Edit:  François points out an easy counterexample in the comments below.] 
A weaker version of the question asks merely whether every
compact topology is refined by a compact Hausdorff
topology, and similarly, whether every Hausdorff topology
refines a compact Hausdorff topology. Under what
circumstances is a compact topology refined by a unique
compact Hausdorff topology? Under what circumstances does a
Hausdorff topology refine a unique compact Hausdorff
topology?
What other topological features besides compactness and
Hausdorffness have illuminating interaction with this
lattice?
Finally, what kind of lattice properties does the lattice
of topologies exhibit? For example, the lattice has atoms,
since we can form the almost-indiscrete topology having
just one nontrivial open set (and any nontrivial subset
will do). It follows that every topology is the least upper
bound of the atoms below it. The lattice of topologies is
complemented.
But the lattice is not distributive (when X has at least
two points), since it embeds N5 by the
topologies involving {x}, {y} and the topology generated by
{{x},{x,y}}.

REPLY [6 votes]: "Under what circumstances does a Hausdorff topology [properly] refine a unique compact Hausdorff topology?"
If $X$ is locally compact (non-compact, Hausdorff), then $X$ refines infinitely many distinct compact Hausdorff topologies. However, there are also examples of (non-compact, Hausdorff) spaces $X$ that refine a unique compact Hausdorff topology.
Theorem: If $X$ is a non-compact, locally compact Hausdorff topology that refines at least one compact Hausdorff topology, then it refines at least $|X|$ compact Hausdorff topologies.
Proof: See the proof of Proposition 4.3 in this paper. The basic idea is to take a one-point compactification of $X$, then take the point at infinity and glue it back down onto any point of $X$. Now you have a topology that's compact, Hausdorff, and refined by $X$. Choosing different targets for the gluing will result in different topologies. QED. [Note: These topologies, while different, may be homeomorphic, for example if $X$ is the real line.]
In contrast, we have the following example. Let $I = [0,1]$ (the set, not the topological space). Let $\sigma$ denote the usual topology on $I$, and let $\langle \sigma,A \rangle$ denote the topology on $I$ with subbasis $\sigma \cup \{A\}$. Let $A = I \setminus \{\frac{1}{n}:n \geq 1\}$. Then I claim that $\sigma$ is the only compact Hausdorff topology refined by $\langle \sigma,A \rangle$.
Let $\tau$ be any compact Hausdorff topology that is refined by $\langle \sigma,A \rangle$.
Notice that if $a > 0$ then $\sigma$ and $\langle \sigma,A \rangle$ agree on $[a,1]$. Furthermore, since $[a,1]$ is compact Hausdorff, passing to $\tau$ will not change its topology, since any strictly coarser topology on $[a,1]$ fails to be Hausdorff. Thus $\sigma$ and $\tau$ agree on every $[a,1]$, and hence on $(0,1]$.
Thus $\tau$ is a (Hausdorff) one-point compactification of $(0,1]$ (with the usual topology). But there's only one of those! So $\langle \sigma,A \rangle$ is refined by only one compact Hausdorff topology, namely $\sigma$.<|endoftext|>
TITLE: hard diophantine equation: $x^3 + y^5 = z^7$
QUESTION [11 upvotes]: Does the equation $x^3+y^5=z^7$ have a solution $(x,y,z)$ with $x,y,z$ positive integers and $(x,y)=1$? In his book H. Cohen (Number theory,2007) said "[...] seems presently out of reach".
I couldn't find any suggestion beyond Cohen's book.
Thanks in advance,
Montanari Fabio
department of math
university of bologna
italy
e-mail montana@dm.unibo.it

REPLY [2 votes]: I may be misunderstanding the question, but I do not believe that it has any integer solutions.  At the very least, none are known to exist at the moment.  Any solutions would be counterexamples to the Fermat-Catalan conjecture with {m,n,k} = {3,5,7} (since 1/3 + 1/5 + 1/7 = 71/105 < 1).  The most I can tell you is that, for coprime {x,y,z}, there are finitely many solutions to your equation.  I think your (x,y) = 1 means that they're coprime, anyway, so it follows that z must be coprime.  Therefore, any solution at all would disprove the related Beal's conjecture.<|endoftext|>
TITLE: Classification of simply connected smooth projective varieties?
QUESTION [5 upvotes]: This question is related to this one.
I am wondering whether there is any sort of classification of simply connected smooth projective varieties, or any work in related directions.
The reason I am interested in this is because, by Deligne-Griffiths-Morgan-Sullivan, smooth projective varieties are formal. Thus, if we have a smooth projective variety that is simply connected, and if we know its cohomology ring, then we can compute its rational homotopy theory.

REPLY [7 votes]: That is way too ambitious, I think, considering what is known about classification of algebraic varieties. Ignoring the trivial, for these purposes, case of curves, the next and best studied case is algebraic surfaces. For them, the classification is classical and has been known for almost a century. So that is a good first case to consider.
So you are asking: what are the smooth projective surfaces $X$ with $\pi_1(X)=1$, and in particular with regularity $q=h^1(\mathcal O_X)=0$. Well, pick up your favorite text, Shafarevich etc. or Barth-(Hulek-)-Peters-van de Ven, or whatever, and go through the list.
Kodaira dimension $-\infty$: here you get all rational surfaces, and only these.
Kodaira dimension $0$: K3 surfaces. 
Kodaira dimension $1$, elliptic surfaces $X\to C$ (a general fiber is elliptic). Clearly, $C$ must be $\mathbb P^1$. But actually getting $\pi_1(X)=1$ seems like not a completely trivial condition, something to think about.
Kodaira dimension $2$, i.e. surfaces of general type. Well, the examples of simply connected surfaces of general type are highly prized, especially if they have $p_g=h^0(\omega_X)=0$. Many such surfaces are known (e.g. http://en.wikipedia.org/wiki/Barlow_surface, some Godeaux surfaces, some Campedelli surfaces) but a complete classification? Not even close. Like I said, this is too ambitious for the present state of knowledge.<|endoftext|>
TITLE: What to look for in applicants to graduate programs (in mathematics)?
QUESTION [15 upvotes]: Hello,
I was thinking about what should be looked at when deciding on the admission of applicants to graduate programs in mathematics and I thought MO would be a good place to get opinions. What do you think?

REPLY [13 votes]: Let me take a crack at the question, since I am currently on the graduate committee at UGA.  [The University of Georgia is about the 50th best department in the country, so just a little shy of being a research 1 university.  We are strongest in algebra, number theory, and algebraic geometry and are able to attract some excellent students in these areas.]
The two most important things for us are:
1) Very good to excellent grades, in courses which go beyond the minimum necessary for a math major and include, if possible, at least one graduate course.  We are looking for all grades B or higher and at least as many A's as B's (which implies a GPA of at least 3.5).  One or two poor grades will not concern us too much if they are in lower level courses, are followed by several years of better grades, or some explanation is given in the personal statement and/or the letters of recommmendation.  A successful applicant will probably have taken real analysis, abstract algebra and topology.  We certainly do take into consideration the student's school: e.g. a student from a liberal arts college may not have any graduate courses available.
2) High GRE scores.  We like to see at least 700 on the GRE quantitative.  It is not as criticial, but I would like to (and often do not!) see at least 600 on the GRE verbal; both scores are kept in mind by the university when it makes decisions about who will get prestigious fellowships.  As for the GRE math subject exam: I am sorry to say that as of this year we do not require it.  After looking at other universities of equal and greater status, we have decided to start requiring this exam next year (i.e., for students who are applying to start in Fall 2011), although we recognize that this may shrink our applicant pool.  A score in the top 50% on the math subject exam looks good to us.  
Next come the recommendation letters, which we use to gauge the student's enthusiasm, ability and preparedness for graduate school as compared to other aspiring graduate students.  It is much better for us if the letters come from someone that we have heard of, or whom we can verify is a successful research mathematician (I have looked some recommenders up on MathSciNet).  Good things to see in such letters are comparisons to other students who have gone on to be successful at research 1 graduate programs.  
REU experience looks good, especially if accompanied by a recommendation letter from the REU supervisor who can be specific about what is accomplished.  Sometimes students do enclose papers or preprints that are the result of REU work.  Again we like this in general (more if the paper looks interesting, less if it looks rather trivial) and may forward this along to other faculty members to see if they are especially interested in the student.
I would say that the personal statement is in fact not very important, except perhaps to address/explain weaknesses in other parts of the application.  [I accept that it might be more useful at a different institution.  I have also advised graduate students and postdocs applying for academic jobs that their cover letter is not very important, and I know that some people -- especially at liberal arts colleges -- have said exactly the opposite.]  It is useful as a writing sample, and a lack of spelling, grammatical and punctutation errors is evidence that the student is serious about their application.  
In fact it is probably more likely that you will lose points in your personal statement than gain them.  When I was a college senior, we had a Q&A session about applying to grad school.  The head of the computer science department (Lance Fortnow, I think) told us the following story: he once had an application from a candidate who had very strong grades, GRE scores and recommendation letters.  But in his personal statement he was asked "Why do you want to go to graduate school in computer science?" The candidate's response "Because I am trying to avoid working very hard" was exactly the opposite of what the department head wanted to hear.  The rest of the application was so strong that, albeit with some misgivings, this candidate was admitted.  The result was disastrous: the candidate really didn't want to do any work so was (of course!) a most unsuccessful graduate student, eventually getting kicked out of the program.  The CS department head concluded that after this experience, he would never admit a candidate who said something like that on their personal statement.  (And neither would I.)<|endoftext|>
TITLE: Disintegrations are measurable measures - when are they continuous?
QUESTION [13 upvotes]: This is a sequel to another question I have asked.
The notion of disintegration is a refinement of conditional probability to spaces which have more structure than abstract probability spaces; sometimes this is called regular conditional probability.  Let $Y$ and $X$ be two nice metric spaces, let $\mathbb P$ be a probability measure on $Y$, and let $\pi : Y \to X$ be a measurable function.  Let $\mathbb P_X(B) = \mathbb P(\pi^{-1} B)$ denote the push-forward measure of $\mathbb P$ on $X$.  The disintegration theorem says that for $\mathbb P_X$-almost every $x \in X$, there exists a nice measure $\mathbb P^x$ on $Y$ such that $\mathbb P$ "disintegrates":$$\int_Y f(y) ~d\mathbb P(y) = \int_X \int_{\pi^{-1}(x)} f(y) ~d\mathbb P^x(y) d\mathbb P_X(x)$$
for every measurable $f$ on $Y$.
This is a beautiful theorem, but it's not strong enough for my needs.  Fix a Borel set $B \subseteq X$, and let $p(x) = \mathbb P^x(B)$.  Part of the theorem is that $p$ is a measurable function of $x$.  Suppose that the map $\pi : Y \to X$ is continuous instead of simply measurable.  My question:  What is a general sufficient condition for $p(x)$ to be continuous?
To me, this is an obvious question to ask, since if $x$ and $x'$ are two close realizations of a random $x \in X$, then the measures $\mathbb P^x$ and $\mathbb P^{x'}$ should be close too, at least in many natural situations.  However, in my combing through the literature, I haven't been able to find an answer to this question.  My guess is that most people are content to integrate over $x$ when they use the theorem.  For my purposes, I need some estimates which I get by continuity.
At this point, I've managed to prove and write down a pretty good sufficient condition for the case I care about (Banach spaces), using an abstract Wiener space-type construction.  However, I am hoping that an expert can point me toward a good reference that does this in wider generality.

REPLY [6 votes]: Tue Tjur studied the existence of continuous disintegrations in a 1975 preprint "A Constructive Definition of Conditional Distributions," Issue 13, Copenhagen Universitet.  He gives necessary and sufficient conditions for their existence, at least in the setting of Radon measures.
He also discusses sufficient structure, and there considers a basic probability space that is an open subset of a finite-dimensional Euclidean space and the problem of conditioning on a  random variable taking values in an open subsets of a Euclidean space $\mathbb R^k$ such that, when the random variable is considered as a (measurable) function, it is surjective and continuously differentiable with differential of maximal rank.  This particular special case may be too narrow for you, but perhaps the general case can give you some guidance.
The article is a bit hard to track down, so let me know if you need help finding it.  The existence of continuous disintegrations arises also in the study of the computability of conditional probability, which is my interest.<|endoftext|>
TITLE: Convergence of a sequence of continuable Dirichlet series
QUESTION [5 upvotes]: Let's say $f$ is a Dirichlet series which converges on the half-plane $\text{Re }s>\sigma$ to a function $f(s)$. Suppose further that $f(s)$ admits an analytic continuation to an entire function, together with the standard sort of functional equation.  Let $g_n$ be a sequence of Dirichlet series, also convergent on $\text{Re }s>\sigma$, which each admit an analytic continuation and functional equation, though their precise FEs may vary.  We assume that $g_n$ converges to $f$ in the following sense:  for every $m>0$ there exists an $N$ for which the series $g_n$ and $f$ match on every term up to the $m$th, for all $n>N$.  Note this implies that $g_n(s)$ converges to $f(s)$ for every $\text{Re }s>\sigma$.
Can it be said that $g_n(s)$ converges to $f(s)$ for any $s$ outside the domain of convergence?
Perhaps that's too much to hope for, and you can't even expect that $g_n(s)$ converges to $f(s)$ even for the point $s=\sigma$.  I'd certainly be interested in a counterexample which does this!

REPLY [2 votes]: (This answer is a community wiki version of a comment above by FC which answered the question.)
For any integer M, there exists a prime p such that chi_p(n) = (n/p) = 1 for all n = 1...M. This means that the Dirichlet series L(s,V chi_p) (for any representation V) "converges" in your sense to L(s,V). but they do not converge at s = 0. If V is trivial, then we are comparing zeta(0) = -1/2 with L(0,chi_p) which grows without bound by Brauer-Siegel. I think in this class of examples one does get convergence at the critical point.<|endoftext|>
TITLE: Non-principal ultrafilters on ω
QUESTION [13 upvotes]: I thought I had heard or read somewhere that the existence of a non-principal ultrafilter on $\omega$ was equivalent to some common weakening of AC. As I searched around, I read that this is not the case: neither countable choice nor dependent choice are strong enough.
This leads me to two questions:
Where would I find a proof that DC is not strong enough to prove the existence of a non-principal ultrafilter on $\omega$?
Is the assumption that there exists a non-principal ultrafilter on $\omega$ strong enough to show DC or countable choice? I.e. is "there exists a non-principal ultrafilter on $\omega$" stronger than either of countable or dependent choice?

REPLY [5 votes]: There is a nice class of problems that are equivalent to the existence of a non-principal ultrafilter. One such, if I remember correctly, is the existence of a colouring of the infinite subsets of the natural numbers in such a way that no infinite set has all its infinite subsets of the same colour. The obvious proof is to colour the sets in such a way that if you add or take away a single element then you change its colour. To make this proof work, you define two sets to be equivalent if their symmetric difference is finite, and do the colouring in each equivalence class separately. But to get it started you have to pick a set in each equivalence class, and for that the obvious thing to do is use AC.
But you can in fact do it with a non-principal ultrafilter as follows. Given an infinite subset A, define its counting function f(n) to be the cardinality of the intersection of A with {1,2,...,n}. Then take a non-principal ultrafilter α and define F(A) to be the limit along α of $(-1)^{f(n)}$. If you add an element m to A, then f(n) is unchanged up to m and then adds 1 thereafter, so its parity is changed everywhere except on a finite set, which implies that F(A) changes. So F gives you your colouring.
I've never actually thought about the other direction (getting from such a colouring to a non-principal ultrafilter) so I don't know how hard it is. I'm not even 100% sure that it's true, but I'm pretty sure I remember hearing that it was.<|endoftext|>
TITLE: Editors in peer-review systems
QUESTION [12 upvotes]: This is a strictly technical question on peer-review systems currently employed in the mathematical literature, not a subjective discussion of merits/drawbacks of such systems, so I think/hope it's suitable for MO.
I have noticed that some journals (e.g. PNAS, CRAS, Nonlinearity...) always publish papers with the name of the editor who supervised the refereeing process ("Presented by X", "Recommended by X", "Communicated by X").  Most other journals, while having editors listed explicitly for each area (and hence in theory one could also know in most cases who supervised what), do not make this explicit.
I was wondering was difference it makes, as a junior author, to have an editor's name on a paper: 

is that a strong endorsement of the paper?
a way to say that the journal is ultra-strict about the refereeing process?
simply a full-disclosure practise of the journal?
an incentive to publish there (when some editor is a "big name", or to make sure a specific person read your paper) ? 

It's really not clear to me.

REPLY [17 votes]: I actually think these "communicated by" lines are very useful: they let young authors know which editors publish which topics. It's not always clear to me which editor is the right one to submit to, but I can look up similar papers and find editors that way.<|endoftext|>
TITLE: In what topology DM stacks are stacks
QUESTION [17 upvotes]: Background/motivation
One of the main reason to introduce (algebraic) stacks is build "fine moduli spaces" for functors which, strictly speaking, are not representable. The yoga is more or less as follows.
One notices that a representable functor on the category of schemes is a sheaf in the fpqc topology. In particular it is a sheaf in coarser topologies, like the fppf or étale topologies. Now some naturally defined functors (for instance the functor $\mathcal{M}_{1,1}$ of elliptic curves) are not sheaves in the fpqc topology 
(actually $\mathcal{M}_{1,1}$ is not even an étale sheaf) so there is no hope to represent them.
Enters the $2$-categorical world and we introduce fibered categories and stacks. Many functors which are not sheaves arise by collapsing fibered categories which ARE stacks, so not all hope is lost. But, as not every fpqc sheaf is representable, we should not expect that every fpqc stack is in some sense "represented by a generalized space", so we make a definition of what we mean by an algebraic stack.
Let me stick with the Deligne-Mumford case. Then a DM stack is a fibered category (in groupoids) over the category of schemes, which
1) is a stack in the étale topology
2) has a "nice" diagonal
3) is in some sense étale locally similar to a scheme.
I don't need to make precise what 2) and 3) mean.
By the preceding philosophy we should expect that DM stacks generalize schemes in the same way that stacks generalize sheaves. In particular I would expect that DM stacks turn out to be stacks in finer topologies, just as schemes are sheaves not only in the Zariski topology (which is trivial) but also in the fpqc topology (which is a theorem of Grothendieck).
Question

Is it true that DM stacks are actually stacks in the fpqc topology? And if not, did someone propose a notion of "generalized space" in the context of stacks, so that this result holds?

REPLY [22 votes]: The rule of thumb is this:  Your DM (or Artin) stack will be a sheaf in the fppf/fpqc topology if the condition imposed on its diagonal is fppf/fpqc local on the target ("satisfies descent").  
In other words, in condition 2 you asked that the diagonal be a relative scheme/relative algebraic space perhaps with some extra properties.  If there if fppf descent for morphisms of this type (e.g., "relative algebraic space", "relative monomorphism of schemes"), you'll have something satisfying fppf descent.  If there is fpqc descent for morphisms of this type (e.g., "relative quasi-affine scheme"), then you'll have something satisfying fpqc descent.
See for instance LMB (=Laumon, Moret-Bailly. Champs algebriques), Corollary 10.7.  Alternatively: earlier this year I wrote up some notes (PDF link) that included an Appendix collecting in one place the equivalences of some standard definitions of stacks, including statements of the type above.<|endoftext|>
TITLE: What does it take to run a good learning seminar?
QUESTION [50 upvotes]: I'm thinking about running a graduate student seminar in the summer. Having both organized and participated in such seminars in the past, I have witnessed first-hand that, contrary to what one might expect, they can be rather successful. However I haven't been able to quite put my finger on what makes a good seminar good.
Certainly there are obvious necessary conditions for success, such as having sufficiently many (dedicated) participants and at least some semblance of a goal. But in my experience these conditions aren’t at all sufficient.
And on the other hand there are clear pitfalls that should be avoided, such as going too fast or not going fast enough, or scheduling the seminar at 8 in the morning. But there are also more subtle pitfalls that aren't as easily avoided: for example, having consecutive speakers of a certain style that might put off or discourage other participants. (Of course a plausible solution to this specific problem is to not have such people speak one after another, but often this is infeasible.)
So I turn to the collective wisdom of MO: In your experience, what has made a specific learning seminar feel successful to you? Feel free to interpret the word "successful" any way you want. Anecdotes and horror stories welcome.
(Aside: The seminar I'm planning is a "classics in geometry and topology" type of deal. By this I mean, each participant will select a classic paper at the beginning of the summer and then briefly discuss its contents sometime during the course of the seminar. I would consider this seminar successful if, at the end, each participant walks away with a set of their own notes on each paper, explaining why it's important, and containing a sketch of its main ideas and how it fits in the grand scheme of things; the hope is that such a set of notes might prove useful if one were to take a closer look at the paper down the road. If anyone has any experience about running a seminar of this type, then I'd be especially interested in hearing their comments!)

REPLY [3 votes]: I'm currently participating in a learning seminar (a "Classics in Geometry/Topology kind of deal", on Teichmuller theory) which seems to work quite well. It is quite tightly managed, with the professor running it both picking the papers we read and giving an introductory lecture before each paper is presented. I appreciate that, since otherwise I'd have trouble knowing how to pick something interesting and understandable from the huge amount of stuff out there.
On the other hand, I can't see any overarching "goal" in the seminar we are striving towards; it's all just various interesting classical topics in the subject. IMHO, this doesn't hurt anything.<|endoftext|>
TITLE: One dimensional (phi,Gamma)-modules in char p
QUESTION [22 upvotes]: I would like to better understand the simplest case of the correspondence between Galois representations and (phi,Gamma)-modules.  Namely, consider 1-dimensional Galois representations of $G_{Q_p}$ over $F_p$ which are in correspondence with 1-dimensional etale (phi,Gamma)-modules over $F_p((T))$.  
There are finitely many such Galois representations.  Moreover, their associated (phi,Gamma)-modules are very simple -- the action of phi and Gamma can be described (on some basis element) as scaling by an element of $F_p$ (as opposed to $F_p((T))$).
My question: can one see this directly on the (phi,Gamma)-module side?  That is, given a 1-dimensional etale (phi,Gamma)-module $D$ over $F_p((T))$, find a $D'$ isomorphic $D$ such that $D'$ has a basis in which the matrices for phi and elements of Gamma are in $F_p^\times$.

REPLY [13 votes]: OK...I think I see how to do this now.  In the end, I am seeing $(p-1)^2$ distinct $(\phi,\Gamma)$-modules which matches well with the Galois side.
To do this, let $D$ be any 1-dimensional etale $(\phi,\Gamma)$-module.  Let $e$ be a basis, and set $\phi(e)=h(T)e$ with $h(T) \in F_p((T))^\times$.  Write $h(T) = h_0 T^a f(T)$ with $h_0 \in F_p^\times$ and $f(T) \in F_p[[T]]$ with $f(0)=1$.  
Changing basis from $e$ to $u(T)e$ with $u(T) \in F_p((T))^\times$ gives
$$
\phi(u(T)e) = u(T^p)h(T)e = \frac{u(T^p)}{u(T)} h(T) (u(T)e).
$$
I claim one can find $u(T)$ such that $u(T)/u(T^p)$ equals any element of $1+TF_p[[T]]$.  Indeed, for such an element $g(T)$, the infinite product $\prod_{j=1}^\infty \phi^j(g(T))$ (which hopefully converges since $g(0)=1$) works.  
Thus, we can change basis so that $\phi$ has the form $\phi(e) = h_0 T^a e$ -- i.e. we can kill off the $f(T)$ term.  Further, by making a change of basis of the form $e$ goes to $T^b e$, we may assume that $0 \leq a < p-1$.  
Now, we use the fact that the $\phi$ and $\Gamma$ actions commute (which is a strong condition even in dimension 1).  Namely, let $\gamma$ be a generator of $\Gamma$, and set $\gamma e = g(T) e$.  Then $\gamma \phi e = \phi \gamma e$ implies
$$
((1+T)^{\chi(\gamma)}-1)^a g(T) = g(T^p) T^a.
$$
Comparing leading coefficients, we see this is only possible if $a=0$ and $g(T)$ is a constant.
Thus, $D$ has a basis $e$ so that $\phi(e) = h_0 e$ and $\gamma(e) = g_0 e$ with $h_0,g_0 \in F_p^\times$ as desired.
Does this look okay?  Any takers for the 2-dimensional case?<|endoftext|>
TITLE: Whitehead Products without Base Points?
QUESTION [7 upvotes]: Let $(X, x_0)$ be a pointed space. Then we can define the homotopy groups $\pi_i(X, x_0)$ for $i \geq 1$. They are abelian groups for $i \geq 2$. It is well-known that the fundamental group $\pi_1(X, x_0)$ acts on each of the higher groups $\pi_i(X, x_0)$, and that this action generalizes to the Whitehead Products which are maps  
$$ \pi_p(X, x_0) \times \pi_q(X, x_0) \to \pi_{p+q -1}(X, x_0).$$
The details are given in the wikipedia article I linked to above. Together the Whitehead products turn the graded group $\pi_*(X, x_0)$ (for $* > 0$) into a graded (quasi-) Lie algebra over $\mathbb{Z}$, where the grading is shifted so that $\pi_i(X, x_0)$ is in degree $(i-1)$. Well, it is a little funny since the bottom group is not necessarily abelian. 

This is all well and good, but what if we don't want to pick base points? Is there a similar algebraic gadget in that situation?

If we don't pick base points, then it seems natural to consider the fundamental groupoid $\Pi_1(X)$. Then the different homotopy groups of $X$ at different base points can be assembled into local systems on $X$. That is for each $i \geq 2$ we have a functor,
$$\pi_i: \Pi_1 X \to AB$$
where $AB$ is the category of abelian groups. This already incorporates the action of $\pi_1$ on the higher homotopy groups but does it in a way which doesn't depend on the choice of base point.

Question: Can we enhance these local systems with a structure which generalizes the Whitehead product, and if so what precisely is this extra structure?

REPLY [6 votes]: As I posted in my comment, I think Paul's suggestion does work.  Here's a (sloppy) description of how I think things will work:
The local systems you describe can be obtained, by passing to homotopy groups, from a "local system of loop spaces" $$ \Omega: \Pi_{\leq \infty} X \to \Omega\mathbf{Spaces}$$ 
One can imagine that this corresponds under the Grothendieck construction to the free loop-space fibration $\Omega X \to LX \to X$.  Alternatively, if we fix a basepoint and identify $X = BG$ for a simplicial group $G$, then this is just encoding the simplicial conjugation action of $G$ on itself.
Rather than think about (strangely-graded) Whitehead products, I prefer to think about (reasonably graded) Samelson products: We think of the structure (Whitehead product) on $\pi_{*+1} X $ as really being a structure (Samelson product) on $\pi_{*} \Omega X$.  I claim that Samelson products give a functor
$$ \pi_*: \Omega\mathbf{Spaces} \to \mathbf{grqLie} $$ so that composing with the above gives our desired "local system of graded (quasi-)Lie algebras".
For convenience, I'll replace loop spaces with (strict) simplicial groups.  Then, the Samelson product comes from noticing that the commutator map $[,]: G^2 \to G$ is trivial if one of the factors is the identity, and so factors through a pointed map $[,]: G \wedge G \to G$.  This pointed map goes on to induce the (quasi-)Lie structure on homotopy.  A group homomorphism $H \to G$ preserves commutators and identities, and so induces a map $H \wedge H \to G \wedge G$ compatible with the brackets, so that this construction is indeed functorial.<|endoftext|>
TITLE: Moduli space of K3 surfaces
QUESTION [17 upvotes]: It is known that there exists a fine moduli space for marked (nonalgebraic) K3 surfaces over $\mathbb{C}$. See for example the book by Barth, Hulek, Peters and Van de Ven, section VIII.12. Of course the marking here is necessary, otherwise the presence of automorphisms can be used to construct isotrivial non trivial families of K3.
Assume that we want to construct an algebraic analogue of this. Of course some structure has to be added; I am thinking of something like torsion points for elliptic curves.

Is there a way to add some structure and actually build a fine moduli space of (K3  + structure) which is defined over $\mathbb{Z}$?

Of course this will only parametrize algebraic K3 surfaces, so it will need to have infinitely many components. A second question, if we do not want to add structure, is

Is there a fine moduli stack of algebraic K3 surfaces which is algebraic (either in the DM or Artin sense)?

It is not difficult to produce K3 surfaces with a denumerable infinity of automorphisms, so I'm not really expecting the answer to be yes, but who knows.
Edit: since there seems to be some confusion in the answers, the point is that I'm asking for a FINE moduli space. I'm aware that one can consider moduli spaces of polarized K3, and this is why I wrote "Of course this will only parametrize algebraic K3 surfaces, so it will need to have infinitely many components."

REPLY [15 votes]: This is in response to the second question: the stack of all K3 surfaces is not algebraic. If you're working with Artin's axioms, then the problem is that there exist formal deformations that are not effective, i.e., one can write down a compatible system of K3 surfaces $X_n \to \mathrm{Spec}(k[t]/(t^{n+1}))$ which do not come from an algebraic K3 surface over the power series ring, at least when the field $k$ is big enough. I learnt this fact from Jason Starr's paper www.math.sunysb.edu/~jstarr/papers/moduli4.pdf (see page 14 and 15). 
Added later: As Scott mentions in the comments, this problem is specific to the non-polarised case. Any formal deformation of a pair (X,L) (where X is a projective variety, and L is line bundle giving a projective embedding) is automatically algebraic by formal GAGA.<|endoftext|>
TITLE: Which functions are Wiener-integrable?
QUESTION [14 upvotes]: I'm looking for either a few precise mathematical statements about Wiener integrals, or a reference where I can find them.
Background
The Wiener integral is an analytic tool to define certain "integrals" that one would like to evaluate in quantum and statistical mechanics.  (Hrm, that's two different mechanics-es....)  More precisely, one often wants to define/compute integrals over all paths in your configuration or phase space satisfying certain boundary conditions.  For example, you might want to integrate over all paths in a manifold with fixed endpoints.  It's conventional to write the integrand as a pure exponential $\exp(f)$.  In statistical mechanics, the function $f$ in the exponent is generally real and decays to $-\infty$ (fast enough) in all directions.  If the path space were finite-dimensional, you would expect such integrals to converge absolutely in the Riemann sense.  In quantum mechanics, $f$ is usually pure-imaginary, so that $\exp(f)$ is a phase, and the integral should not be absolutely convergent, but in finite-dimensional integrals may be conditionally convergent in the Riemann sense.  Typically, $f$ is a local function, so that $f(\gamma) = \int L(\gamma(t))dt$, where $\gamma$ is a path and $L(\gamma(t))$ depends only on the $\infty$-jet of $\gamma$ at $t$.  In fact, typically $L(\gamma(t))$ depends only on the $1$-jet of $\gamma$, so that $f(\gamma)$ is defined on, for example, continuous piece-wise smooth paths $\gamma$.
For example, one might have a Riemannian manifold $\mathcal N$, and want to define:
$$U(q_0,q_1) = \int\limits_{\substack{\gamma: [0,1] \to \mathcal N \\ {\rm s.t.}\, \gamma(0) = q_0,\, \gamma(1)=q_1}}  \exp\left( - \hbar^{-1} \int_0^1 \frac12 \left| \frac{\partial \gamma}{\partial t}\right|^2dt \right)$$
where $\hbar$ is a positive real number (statistical mechanics) or non-zero pure imaginary (quantum mechanics).  The "measure" of integration should depend on the canonical measure on $N$ coming from the Riemannian metric, and the Wiener measure makes it precise.
On $\mathcal N = \mathbb R^n$, I believe I know how to define the Wiener integral.  The intuition is to approximate paths by piecewise linears.  Thus, for each $m$, consider an integral of the form:
$$I_m(f) =  \prod_{j=1}^{m-1} \left( \int_{\gamma_j \in \mathbb R^n} d\gamma_j \right) \exp(f(\bar\gamma)) $$
where $\bar\gamma$ is the piecewise-linear path that has corners only at $t = j/m$ for $j=0,\dots,m$, where the values are $\bar\gamma(j/m) = \gamma_j$ (and $\gamma_0 = q_0$, $\gamma_m = q_1$).  Then the limit as $m\to \infty$ of this piecewise integral probably does not exist for a fixed integrand $f$, but there might be some number $a_m$ depending weakly on $f$ so that $\lim_{m\to \infty} I_m(f)/a_m$ exists and is finite.  I think this is how the Wiener integral is defined on $\mathbb R^n$.
On a Riemannian manifold, the definition above does not make sense: there are generally many geodesics connecting a given pair of points.  But a theorem of Whitehead says that any Riemannian manifold can be covered by "convex neighborhoods": a neighborhood is convex if any two points in it are connected by a unique geodesic that stays in the neighborhood.  Then we could make the following definition.  Pick a covering of $\mathcal N$ by convex neighborhoods, and try to implement the definition above, but declare that the integral is zero on tuples $\gamma_{\vec\jmath}$ for which $\gamma_j$ and $\gamma_{j+1}$ are not contained within the same convex neighborhood.  This would be justified if "long, fast" paths are exponentially suppressed by $\exp(f)$.  So hope that this truncated integral makes sense, and then hope that it does not depend on the choice of convex-neighborhood cover.
Of course, path integrals should also exists on manifolds with, say, indefinite "semi-"Riemannian metric.  But then it's not totally clear to me that the justification in the previous paragraph is founded.  Moreover, really the path integral should depend only on a choice of volume form on a manifold $\mathcal N$, not on a choice of metric.  Then one would want to choose a metric compatible with the volume form (this can always be done, as I learned in this question), play the above game, and hope that the final answer is independent of the choice.  A typical example: any symplectic manifold, e.g. a cotangent bundle, has a canonical volume form.
One other modification is also worth mentioning: above I was imagining imposing Dirichlet boundary conditions on the paths I wanted to integrate over, but of course you might want to impose other conditions, e.g. Neumann or some mix.
Questions
Question 0: Is my rough definition of the Wiener integral essentially correct?
Question 1: On $\mathcal N = \mathbb R^n$, for functions $f$ of the form $f(\gamma) = -\hbar^{-1}\int_0^1 L(\gamma(t))dt$, for "Lagrangians" $L$ that depend only on the $1$-jet $(\dot\gamma(t),\gamma(t))$ of $\gamma$ at $t$, when does the Wiener integral make sense?  I.e.: for which Lagrangians $L$ on $\mathbb R^n$, and for which non-zero complex numbers $\hbar$, is the Wiener integral defined?
Question 2: In general, what are some large classes of functions $f$ on the path space for which the Wiener integral is defined?
By googling, the best I've found are physics papers from the 70s and 80s that try to answer Question 1 in the affirmative for, e.g., $L$ a polynomial in $\dot\gamma,\gamma$, or $L$ quadratic in $\dot\gamma,\gamma$ plus bounded, or...  Of course, most physics papers only treat $L$ of the form $\frac12 |\dot\gamma|^2 + V(\gamma)$.

REPLY [6 votes]: Hi Theo,
0) Your definition is roughly correct, yes.  For Wiener measure on paths in vector spaces, see Chapter 3 + Appendix A of the 2nd edition of Glimm & Jaffe.  On curved targets, I think Bruce Driver has some good lecture notes.  One warning:  the rough definition of Wiener measure is misleading in one way:  Wiener measure is naturally constructed as a measure on a space of distributions which contains the continuous functions.
1) I don't think there's a general theory.  Trying a Lagrangian of the form $F = (\dot{\phi})^{1000}$ will not result in happiness.  But:  You can define Wiener measure using the standard kinetic term $\int \frac{1}{2}|\dot{\phi}|^2dt$ for any $\hbar$, and you can safely add a potential $V$ which is bounded below and integrable to the kinetic term.
2) Any observable you can write down should give you a Wiener integrable function, in quantum mechanics.  This is not true in QFT, however.  Most of the work in constructive QFT is proving that the measure on the space of histories actually has moments!<|endoftext|>
TITLE: How do you become a good listener? 
QUESTION [13 upvotes]: Often I find myself just taking notes during lectures and not really following what is said. This has always been an issue for me, I do not seem to learn anything in the classroom. Instead the learning process starts right after when going through the notes. 
On the other hand, I do somewhat feel more involved when attending to seminars (I have not attended to many), even though the subject is more advanced and sometimes way above my level.
I guess this is kind of a vague question, but are there any good techniques on how to become a good listener (the question can be split in two: one where you need to take complete notes, such as during a class, and the other case, e.g. seminars)?

REPLY [2 votes]: I have the same problem when I am listening to something challenging(i.e. not the undergraduate math club).  For me, one of the most helpful things I can do is that once a definition is given, I construct both examples and constructions that almost fit the definition but do not fit one of the criteria.  I then check how these work(or fail) with the theorems presented and the claims made.  If you find a simple example and are having trouble following, the person lecturing will probably be happy explaining how the concepts presented relate to your example.  
Another idea that helps is for me to read all the material beforehand, no matter how confused I am.  Then I can pay attention to how it is structured in class.  Even if I do not understand the lecture, contrasting the structures of the reading and of the lecture can give me deeper understanding of the material.  In addition, since I (like to think I ) understand how to structure a lecture, it gives me something I can understand to pay attention to.
Lastly, I think it is important to be forgiving of oneself.  Think of paying attention to a lecture like being focused when meditating.  If you find your attention wandering, just refocus on the matter at hand.  Try not to get caught up in a cycle of being frustrated that your attention wandered.<|endoftext|>
TITLE: How difficult is Morse theory on stacks?
QUESTION [20 upvotes]: The title is a little tongue-in-cheek, since I have a very particular question, but I don't know how to condense it into a pithy title.  If you have suggestions, let me know.
Suppose I have a Lie groupoid $G \rightrightarrows G_0$, by which I mean the following data:

two finite-dimensional (everything is smooth) manifolds $G,G_0$, 
two surjective submersions $l,r: G \to G_0$,
an embedding $e: G_0 \hookrightarrow G$ that is a section of both the maps $l,r$,
a composition law $m: G \times_{G_0} G \to G$, where the fiber product is the pull back of $G \overset{r}\to G_0 \overset{l}\leftarrow G$, intertwining the projections $l,r$ to $G_0$.
Such that $m$ is associative, by which I mean the two obvious maps $G \times_{G_0} G \times_{G_0} G \to G$ agree,
$m(e(l(g)),g) = g = m(g,e(r(g)))$ for all $g\in G$, 
and there is a map $i: G \to G$, with $i\circ l = r$ and $i\circ i = \text{id}$ and $m(i(g),g) = e(r(g))$ and $m(g,i(g)) = e(l(g))$.

Then it makes sense to talk about smooth functors of Lie groupoids, smooth natural transformations of functors, etc.  In particular, we can talk about whether two Lie groupoids are "equivalent", and I believe that a warm-up notion for "smooth stack" is "Lie groupoid up to equivalence".  Actually, I believe that the experts prefer some generalizations of this — (certain) bibundles rather than functors, for example.  But I digress.
Other than that we know what equivalences of Lie groupoids are, I'd like to point out that we can work also in small neighborhoods.  Indeed, if $U_0$ is an open neighborhood in $G_0$, then I think I can let $U = l^{-1}(U_0) \cap r^{-1}(U_0)$, and then $U \rightrightarrows U_0$ is another Lie groupoid.
Oh, let me also recall the notion of tangent Lie algebroid $A \to G\_0$ to a Lie groupoid.  The definition I'll write down doesn't look very symmetric in $l\leftrightarrow r$, but the final object is.  The fibers of the vector bundle $A \to G\_0$ are $A\_y = {\rm T}\_{e(y)}(r^{-1}(y))$, the tangent space along $e(G\_0)$ to the $r$-fibers, and $l: r^{-1}(y) \to G\_0$ determines a God-given anchor map $\alpha = dl: A \to {\rm T}G\_0$, and because $e$ is a section of both $l,r$, this map intertwines the projections, and so is a vector bundle map.  In fact, the composition $m$ determines a Lie bracket on sections of $A$, and $\alpha$ is a Lie algebra homomorphism to vector fields on $G_0$.
Suppose that I have a smooth function $f: G_0 \to \mathbb R$ that is constant on $G$-orbits of $G_0$, i.e. $f(l(g)) = f(r(g))$ for all $g\in G$.  I'd like to think of $f$ as a Morse function on "the stack $G_0 // G$".  So, suppose $[y] \subseteq G_0$ is a  critical orbit, by which I mean: it is an orbit of the $G$ action on $G_0$, and each $y \in [y]$ is a critical point of $f$.  (Since $f$ is $G$-invariant, critical points necessarily come in orbits.)  If $y$ is a critical point of $f$, then it makes sense to talk about the Hessian, which is a symmetric pairing $({\rm T}\_yG\_0)^{\otimes 2} \to \mathbb R$, but I'll think of it as a map $f^{(2)}\_y : {\rm T}\_yG\_0 \to ({\rm T}\_yG\_0)^*$.  In general, this map will not be injective, but rather the kernel will include $\alpha\_y(A\_y) \subseteq {\rm T}\_yG\_0$.  Let's say that the critical orbit $[y]$ is nondegenerate if $\ker f^{(2)}_y = \alpha\_y(A\_y)$, i.e. if the Hessian is nondegenerate as a pairing on ${\rm T}\_yG\_0 / \alpha\_y(A\_y)$.  I'm pretty sure that this is a condition of the orbit, not of the individual point.
Nondegeneracy rules out some singular behavior of $[y]$, like the irrational line in the torus.
Anyway, my question is as follows:

Suppose I have a Lie groupoid $G \rightrightarrows G_0$ and a $G$-invariant smooth function $f: G_0 \to \mathbb R$ and a nondegenerate critical orbit $[y]$ of $f$.  Can I find a $G$-invariant neighborhood $U_0 \supseteq [y]$ so that the corresponding Lie groupoid $U \rightrightarrows U_0$ is equivalent to a groupoid $V \rightrightarrows V_0$ in which $[y]$ corresponds to a single point $\bar y \in V_0$?  I.e. push/pull the function $f$ over to $V_0$ along the equivalence; then can I make $[y]$ into an honestly-nondegenerate critical point $\bar y \in V_0$?

I'm assuming, in the second phrasing of the question, that $f$ push/pulls along the equivalence to a $V$-invariant function $\bar f$ on $V_0$.  I'm also assuming, so if I'm wrong I hope I'm set right, that ${\rm T}\_{\bar y}V\_0 \cong {\rm T}\_yG\_0 / \alpha\_y(A\_y)$ canonically, so that e.g. $\bar f^{(2)}_{\bar y} = f^{(2)}_y$.

REPLY [10 votes]: Yes, it's possible to find such a neighbourhood $U_0$ of $[y]$.
Here's how you do it.
Pick a submanifold $M\subset G_0$, $y\in M$, transverse to $[y]$.
By your Morse-ness assumption, the restriction $f|_M$ is Morse,
with critical point $y$. Pick a neighborhood $V_0\subset M$ of $y$, such that
$y$ is the only critical point of $f$ in $V_0$. 
Let $U_0$ be the orbit of $V_0$ under $G$. Clearly, $U_0$ is a neighborhood of $[y]$.
Now consider the restriction of $G$ to $U_0$. This is defined to be the groupoid with object space $U_0$, and morphism space $U_0\times_{G_0}G\times_{G_0}U_0$. That's your groupoid $U\rightrightarrows U_0$. Similarly, you can consider the restriction $V\rightrightarrows V_0$
of $G$ to $V_0$.
The inclusion $(V\rightrightarrows V_0) \hookrightarrow (U\rightrightarrows U_0)$ is a Morita  equivalence because it's essentially surjective and fully faithfull.

Note: The notions "essentially surjective" and "fully faithfull" for Lie groupoids are somewhat stronger than what you might initially guess. The first one also requires the existence of locally defined smooth maps $U_0\to V_0$, while "fully faithfull" means that $V$ is the pullback of $V_0\times V_0 \to G_0\times G_0\leftarrow G$.<|endoftext|>
TITLE: Propositional Logic, First-Order Logic, and Higher-Order Logics
QUESTION [20 upvotes]: I've been reading up a bit on the fundamentals of formal logic, and have accumulated a few questions along the way. I am pretty much a complete beginner to the field, so I would very much appreciate if anyone could clarify some of these points.

A complete (and consitent) propositional logic can be defined in a number of ways, as I understand, which are all equivalent. I have heard it can be defined with one axiom and multiple rules of inferences or multiple axioms and a single rule of inference (e.g. Modus Ponens) - or somewhere inbetween. Are there any advantage/disvantages to either? Which is more conventional?
Propositional (zeroth-order) logic is simply capable of making and verifying logical statements. First-order (and higher order) logics can represent proofs (or increasing hierarchial complexity) - true/false, and why?
What exactly is the relationship between an nth-order logic and an (n+1)th-order logic, in general. An explanation mathematical notation would be desirable here, as long as it's not too advanced.
Any formal logic above (or perhaps including?) first-order is sufficiently powerful to be rendered inconsistent or incomplete by Godel's Incompleteness Theorem - true/false? What are the advantages/disadvantages of using lower/higher-order formal logics? Is there a lower bound on the order of logic required to prove all known mathematics today, or would you in theory have to use an arbitrarily high-order logic?
What is the role type theory plays in formal logic? Is it simply a way of describing nth-order logic in a consolidated theory (but orthogonal to formal logic itself), or is it some generalisation of formal logic that explains everything by itself?

Hopefully I've phrased these questions in some vaguely meaningful/understandable way, but apologies if not! If anyone could provide me with some details on the various points without assuming too much prior knowledge of the fields, that would be great. (I am an undergraduate Physics student, with a background largely in mathematical methods and the fundamentals of mathematical analysis, if that helps.)

REPLY [6 votes]: Yes there are advantages/disadvantages in where the balance lies between the number of inference rules and the number of axioms when defining a logic.  As Francois Dorais says in his answer, it depends on what you want to do with the logic.
All logics are for representing proofs, including propositional logic.  The higher the order of the logic, the more powerful it is in the sense of its language being more expressive and its deduction being more general.
The criterion that determines the order of a logic relates to the kinds of value that can be quantified over.  In a zeroth-order logic, there are just values and quantification is not supported (e.g. propositional logic, where the values are boolean values).  In a first-order logic, there are functions which are distinct from values; only values can be quantified over (e.g. first-order predicate logic, natural number arithmetic).  In a second-order logic, functions may take first-order functions as arguments, and first-order functions may be quantified over.  In a third-order logic, second-order functions may themselves be arguments to functions and be quantified over, etc, etc.  In a higher-order logic (this is a distinct concept from the concept of an nth-order logic), there is no fundamental distinction between functions and values, and all functions can be quantified over.
Note that usage of predicates can be considered as equivalent to usage of sets (some predicate returning "true" for a given value can be considered as equivalent to the value being an element of some set).  Note also that a given logic may or may not fundamentally distinguish between general values and boolean values, and between functions and predicates (functions that return a boolean value).
Godel's (First) Incompleteness Theorem only relates to logics capable of at least expressing natural number arithmetic - any such logics are incomplete (unless they are inconsistent, in which case they are trivially complete).  His Second Incompleteness Theorem relates to whether such logics are capable of proving their own consistency.
The advantage of a less powerful logic is that it is easier to reason about, and that it is tends to be easier to write algorithms for, in the sense that (depending on what the algorithm is intended to do) these algorithms will tend to be more complete and/or efficient and/or to terminate (e.g. algorithms for proving statements in the logic).  The advantage of a more powerful logic is that it is more expressive and thus capable of representing and/or proving more of mathematics.
There are certainly higher-order logics that are not capable of expressing the whole of mathematics today (e.g. not capable of fully expressing category theory).  I'm afraid I don't know enough to say whether there are/aren't any formal logics that are capable of expressing all of contemporary mathematics.
Type theory is the study of type systems.  Presumably your question relates to the purpose of using a type system in a formal logic?  (Apologies if I got the wrong end of the stick.)
Whether or not a logic uses a type system is another fundamental distinguishing attribute between logics.  The alternative is to base the logic on set theory.  Either way, the logic must somehow avoid consistency problems, like Russell's Paradox that exposed the inconsistency of Frege's formal logic, or the Kleene-Rosser Paradox that exposed the inconsistency of Church's original lambda calculus.
The purpose of a type system is to impose extra well-formedness restrictions on a formal language in addition to the restrictions imposed by the language's syntax.  This is a way of ensuring that paradoxes can be avoided, as well as a practical way of helping the user of a logic avoid writing meaningless statements (such as "the number 5 is a vector space").  Russell actually invented type theory to solve the problem raised by his own paradox.  Church used type theory to come up with an alternative, consistent lambda calculus.<|endoftext|>
TITLE: Defining "average rank" when not every ranking covers the whole set
QUESTION [8 upvotes]: Here's a mathematical modeling problem I came across while working on a hobby project.
I have a website that presents each visitor with a list of movie titles. The user has to rank them from most to least favorite. After each visit, I want to create a cumulative ranking that takes into account each visitor's individual ranking. Normally I would just take the mean ordinal rank: e.g., if Person A rated "Avatar" 10th and Person B rated it 20th, its cumulative rank would be 15th. However, new movies will be added to the list as the website grows, so each person will have ranked only a subset of the full movie list.
Any thoughts on how I can define "average rank" when some rankings do not cover the whole set? My best idea so far is to model this as a directed graph, where nodes are movies and weighted edges are preferences (e.g. "10 people ranked 'Avatar' right above 'District 9'"), and then finding sinks and sources. How else could one go about this?
(Sorry if this question is too applied.)

REPLY [5 votes]: There are a few different ways of approaching the problem. A good reference for this precise problem is 'Rank Aggregation methods For The Web', by Dwork, Kumar, Naor and Sivakumar from the WWW conference in 2001. It's not the most recent work, but it lays out the mathematics nicely.
In general, the way to define an average is to define a metric, and then look at the point that minimizes the sum of distances from the individual lists. If the lists were full (i.e all defined over the same set), you could use Spearman's footrule distance, or the Kendall distance. Since they are not, the general idea is to find some global ranking that's locally optimal (i.e there's no other full ranking that yields a smaller distance to the partial lists and can be obtained by flipping rankings from the current candidate).
As an aside, you can compare a full ranking with a partial ranking by merely projecting the full ranking onto the partial ranking and then computing one of the above mentioned distances. 
Most of this is worked out in detail in the paper referenced, so your best bet is to start there.<|endoftext|>
TITLE: Is any interesting question about a group G decidable from a presentation of G?
QUESTION [22 upvotes]: We say that a group G is in the class Fq if there is a CW-complex which is a BG (that is, which has fundamental group G and contractible universal cover) and which has finite q-skeleton.  Thus F0 contains all groups, F1 contains exactly the finitely generated groups, F2 the finitely presented groups, and so forth.
My question:  For a fixed q ≥ 3, is it possible to decide, from a finite presentation of a group G, whether G is in Fq or not?  I would assume not, but am not having much luck proving it.
One approach would be to prove that, if G is a group in Fq and H is a finitely presented subgroup, then H ∈ Fq as well.  This would make being in Fq a Markov property, or at least close enough to make it undecidable.
Henry Wilton's comment below makes it clear that being Fq is not even quasi-Markov, so the above idea won't work.  I still suspect that "G ∈ Fq" is not decidable, but now my intuition is from Rice's theorem:

If $\mathcal{B}$ is a nonempty set of computable functions with nonempty complement, then no algorithm accepts an input n and decides whether φn is an element of $\mathcal{B}$.

It seems likely to me that something similar is true of finite presentations and the groups they define.
John Stillwell notes below that this can't be true for a number of questions involving the abelianization of G.  This wouldn't affect the Rips construction/1-2-3 theorem discussion below if the homology-sphere idea works, since those groups are all perfect.
Any thoughts?

REPLY [11 votes]: Sorry I came to this question late. $F_q$ is undecidable from a presentation for each fixed $3\leq q<\infty$.  First note that for finitely presented groups $F_q$ is equivalent to the homological finiteness condition $FP_q$ so it is enough to show this condition undecidable. This was done for $q =3$ in Section 5 of  Cremanns, Robert; Otto, Friedrich, For groups the property of having finite derivation type is equivalent to the homological finiteness condition FP3. J. Symbolic Comput. 22 (1996), no. 2, 155–177 https://www.sciencedirect.com/science/article/pii/S0747717196900462.  The same construction works for any fixed finite $q\geq 3$.   They use essentially the same construction as the proof that Markov properties are undecidable but different details.<|endoftext|>
TITLE: Quantum analogue of Wiener process
QUESTION [11 upvotes]: The Wiener process (say, on $\mathbb{R}$) can be thought of as a scaling limit of a classical, discrete random walk. On the other hand, one can define and study quantum random walks, when the underlying stochastic process is governed by a unitary transform + measurement (for an excellent introduction, see http://arxiv.org/abs/quant-ph/0303081).
My question is - do quantum random walks have a reasonable continuous limit, something which would give a quantum analogue of the Wiener process?

REPLY [2 votes]: In section III.B of the survey paper you cite, it describes continuous quantum walks, which are I think are a natural analogue of the Wiener process.  These are basically Hamiltonian evolution when the Hamiltonian is something like the adjacency matrix (or Laplacian) of a graph.
On the relationship between continuous- and discrete-time quantum walk has some recent developments with fascinating applications to simulating Hamiltonians on quantum computers.<|endoftext|>
TITLE: Motivation for the étale topology over other possibilities
QUESTION [14 upvotes]: In the search for a Weil cohomology theory $H$ over a field $K$ (with $\text{char}(K)=0$) for varieties in characteristic $p$, a classical argument by Serre shows that the coefficient field cannot be a subfield of $\mathbb{R}$ or of $\mathbb{Q}_p$; an obvious choice is to take $\mathbb{Q}_\ell$ for a prime $\ell \neq p$.
Now, we can try to make a Weil cohomology theory by taking the sheaf cohomology with constant sheaves with the Zariski topology, but this does not work as all cohomology vanishes.
Grothendieck's insight was that we can find a different topology, for example the étale topology. Then we can build a Weil cohomology theory with coefficients in $\mathbb{Q}_\ell$ by taking cohomology with coefficients in the constant sheaves $\mathbb{Z}/ \ell^n\mathbb{Z}$ and then taking the inverse limit with respect to $n$ and tensor with $\mathbb{Q}_\ell$: this gives $\ell$-adic cohomology.
But it is not so clear to me why the étale topology is best suited at this task. What happens if we repeat the above procedure on other sites? Does the cohomology theory we get fail to be a Weil cohomology theory?
P.S.: Information for fields other than $\mathbb{Q}_\ell$ would also be nice!

REPLY [13 votes]: As observed, the Zariski topology has too few open subsets to compute cohomology with constant coefficients. It had already been observed by Serre that etale covers were enough to trivialize principal bundles for many algebraic groups. That suggested using etale covers. The etale "topology" is the coarsest for which the inverse function theorem holds. The flat topology also gives Weil cohomologies (the same ones as the etale topology), but why use flat covers when etale covers are enough.<|endoftext|>
TITLE: If K/k is a finite normal extension of fields, is there always an intermediate field F such that F/k is purely inseparable and K/F is separable?
QUESTION [10 upvotes]: I was feeling a bit rusty on my field theory, and I was reviewing out of McCarthy's excellent book, Algebraic Extensions of Fields. Out of Chapter 1, I was able to work out everything "left to the reader" or omitted except for one corollary, stated without proof (see here for the page in the book):

Let $K/k$ be a finite normal
  extension. Then $K$ can be obtained by
  a purely inseparable extension,
  followed by a separable extension.

The text immediately preceding this implies that the intermediate field that's going to make this happen is $F=\{a\in K:\sigma(a)=a$ for all $\sigma\in Gal(K/k)\}$, and I understand his argument as to why $F/k$ is purely inseparable (in fact, that's the theorem, Theorem 21, which this is a corollary to). What I don't understand is why $K/F$ is separable; I don't see how we've ruled out it being non-purely inseparable.
Note that I will be making a distinction between non-purely inseparable (inseparable, but not purely inseparable) and not purely inseparable (either separable or non-purely inseparable).
Here are some observations / my general approach:

One big thing that seemed promising was Theorem 11 (at the bottom of this page), which is basically the reverse of the corollary I'm having trouble with:


Let $K$ be an arbitrary algebraic extension of $k$. Then $K$ can be obtained by separable extension followed by a purely inseparable extension.

(the separable extension referred to is of course the separable closure of $k$ in $K$). It seems like we want to use Theorem 11 on $K/F$, and argue that there can't be "any more" pure inseparability, but I couldn't figure out a way of doing this.

Theorem 21 is actually an "if and only if" (that is, $a\in K$ is purely inseparable over $k$ iff $\sigma(a)=a$ for all $\sigma\in Gal(K/k)$). Because this implies that any $a\in K$ with $a\notin F$ is not purely inseparable over $k$, we have that $F$ is the maximum (not just maximal) purely inseparable extension of $k$ in $K$.
If any $a\in K$ were purely inseparable over $F$, by Theorem 8 (see here), there is some $e$ for which $a^{p^e}\in F$. But by the same theorem, since $F/k$ is purely inseparable, there is some $b$ for which $(a^{p^e})^{p^b}=a^{p^{e+b}}\in k$. Thus $a$ would be purely inseparable over $k$ by the converse (Corollary 1 to Theorem 9, see here), and hence be in $F$. Thus, $K$ (and any field between $K$ and $F$, besides $F$ itself) is not purely inseparable over $F$. 

So, that's why I don't see how we've ruled out $K/F$ being non-purely inseparable. Sorry about making lots of references to the book - I'm just not sure what previously established results McCarthy intended to be used, and I wanted to point out what I saw as the important ones for people not familiar with the book. I'm sure I'm missing something obvious here. Does anyone see the last bit of the argument?

REPLY [4 votes]: Dear Zev, as LASER pointed out, Lang's Proposition 6.11 indeed solves your problem. Still, I'd like to add two remarks that might put your problem in perspective.
A) Given a field $K$ and a group of automorphisms $G$ of $K$ we get a fixed field $K^G$ and an extension $K^G \subset  K$ . This extension is algebraic if and only if all orbits of $G$ are finite and if this is the case the extension is always separable: each element of $x \in K$ has as minimal polynomial over $K^G$ the product $\prod \limits_{y \in  Orb(x)}(T-y)$.
[Let me emphasize that  i) there is no base field $k$ in this general statement and ii) the group $G$ may very well be infinite with all its orbits finite: every infinite dimensional galois extension gives an example.]
B) The statement is false if you don't assume normality of $k \subset K$ (as I'm sure you guessed!). Here is an example taken from Bourbaki's Algebra V, exercise  3) for §5.
Take a field $F$ of characteristic $p>2$ and define $k=F(x,y)$ where $x,y$ are indeterminates.
Let $\theta$ be a zero of the polynomial $P(T)=T^{2p}+xT^p+y$ (in an algebraic closure of $k$, say).Then if we put $K=k(\theta)$, we have  our counter-example: no element in $K$ is purely inseparable over $k$, except if it already is in $k$. And yet $K$ is not separable over k since the minimal polynomial of $\theta$ over $k$ is $P(T)$ , which is obviously not separable.<|endoftext|>
TITLE: What's the Hilbert class field of an elliptic curve?
QUESTION [17 upvotes]: My question points in a direction similar to Qiaochu's, but it's not the same (or so I think). Let me provide you with a little bit of background first.
Let E be an elliptic curve defined over some number field K. The Tate-Shafarevich group of E/K consists of certain curves of genus 1, isomorphic to E over some extension, with points everywhere locally. In the simplest case of an element of order 2, such a curve has the form C: y2=f4(x) for some quartic polynomial f4(x) in K[X]; here, C does not have a K-rational point, but has points in every completion of K. 
If we look at E in some extension L/K, this curve C still has points everywhere locally, but if it has a global point (in L) we say that the corresponding element in the Tate-Shafarevich group of E/K capitulates. Heegner's Lemma says that elements of order 2 cannot capitulate in extensions of odd degree, which is the analogue of the similarly trivial observation that ideals generating a class of order 2 cannot capitulate (become principal) in an extension of odd degree. 
I gave a few talks on the capitulation of Tate-Shafarevich groups more than 10 years ago. A little later the topic became almost fashionable under the name of "visualizing" elements of Sha. I discussed the following question with Farshid Hajir back then, but eventually nothing came out of it. Here it is. For capitulation of ideal classes, there is a "canonical" extension in which this happens: the Hilbert class field. So my question is: 

may we still dream about the existence of a curve with all the right properties, or are there reasons why such a thing should not exist?

We also know that capitulation is not the correct notion for defining the Hilbert class field, which is the largest unramified abelian extension of a number field. These notions do not seem to make any sense for elliptic curves, but we can characterize the Hilbert class field also in the following way: among all finite extensions L/K for which the norm of the class group of L down to K is trivial, the Hilbert class field is the smallest.    
Taking the norm of Sha of an elliptic curve defined over L down to K does make sense (just add the equivalence classes of the conjugate homogeneous spaces using the Baer-sum construction or in the appropriate cohomology group). So here's my second question:

Has this "norm map" been studied in the literature?

(I know that the norm map from E(L) to E(K) was investigated a lot, in particular in connection with Heegner points). 
Let me add that I do not assume that such a "Hilbert class curve" can be found among the elliptic curves defined over some extension field; if there is a suitable object, it might be the Jacobian of a curve of higher genus or an abelian variety coming from I don't know where.

REPLY [15 votes]: EDIT: This is a completely new answer.
I will prove that your specific suggestion of defining a Hilbert class field of an elliptic curve $E$ over $K$ does not work.  I am referring to your proposal to take the smallest field $L$ such that the corestriction (norm) map $\operatorname{Sha}(L) \to \operatorname{Sha}(K)$ is the zero map.  (I have to assume the Birch and Swinnerton-Dyer conjecture (BSD), though, for a few particular elliptic curves over $\mathbf{Q}$.)

Theorem: Assume BSD.  There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is no smallest field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.

Proof: We will use BSD data (rank and order of Sha) from Cremona's tables.  Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation
$$y^2 + y = x^3 - x^2 - 929 x - 10595.$$
Then $\operatorname{rk} E(\mathbf{Q})=0$ and $\#\operatorname{Sha}(\mathbf{Q},E)=4$.  Let $L_1 = \mathbf{Q}(\sqrt{-1})$ and $L_2 =\mathbf{Q}(\sqrt{-11})$.  It will suffice to show that the Tate-Shafarevich groups $\operatorname{Sha}(L_i,E)$ are trivial.
Let $E_i$ be the $L_i/\mathbf{Q}$-twist of $E$.  MAGMA confirms that $E_1$ is curve 9136C1 and $E_2$ is curve 69091A1.  According to Cremona's tables, $\operatorname{rk} E_i(\mathbf{Q})=2$ and $\operatorname{Sha}(\mathbf{Q},E_i)=0$, assuming BSD.  Thus $\operatorname{rk} E(L_i) = 0+2=2$ and $\operatorname{Sha}(L_i,E)$ is a $2$-group.  On the other hand, MAGMA shows that the $2$-Selmer group of $E_{L_i}$ is $(\mathbf{Z}/2\mathbf{Z})^2$.  Thus $\operatorname{Sha}(L_i,E)[2]=0$, so $\operatorname{Sha}(L_i,E)=0$.<|endoftext|>
TITLE: Lie Semigroups?
QUESTION [6 upvotes]: Why is a Lie group wanted instead of a semigroup, what does the group structure give? References on this would be much appreciated.
I'm currently pondering manifolds and lie groups and their associations with certain computer vision problems. Semigroups or other algebraic objects may be an interesting idea to study in this venture and I wanted to know whether or not a Lie Semigroup is a structure that makes sense to study.

REPLY [3 votes]: Much of the literature on the representation theory of the group $GL_n$ is really about the representation theory of the monoid $M_n$. (One might argue that it's really about the representation theory of the category Vec.)
There's plenty of literature on Lie monoids; e.g. by Lex Renner.<|endoftext|>
TITLE: What is the  relation between quantum symmetry  and  quantum groups?
QUESTION [15 upvotes]: What kind of role do quantum groups play in modern physics ? 
Do quantum groups naturally arise in quantum mechanics or quantum field theories?
What should quantum symmetry refer to ?
Can we say that the "symmetry" of a noncommutative space (quantum phase space) should be a quantum group?
Do quantum groups describe "extended symmetry" ?

REPLY [11 votes]: It is not quite true that the usual quantum groups are symmetries of WZNW model. The standard root of unity q. groups are hidden symmetries at the level of pre-Hilbert space where some spurious ghost-like norm zero states appear. The true symmetry (in the sense of axiomatic quantum field theory) includes rather certain quotient which is just a weak quasi-Hopf algebra, and whose representation automatically exclude the nonphysical representations with quantum dimension zero. See
Gerhard Mack, Volker Schomerus: QuasiHopf quantum symmetry in quantum theory. DESY-91-037, May 1991. 37pp. Nucl.Phys.B370:185-230,1992, doi, scan<|endoftext|>
TITLE: The finite subgroups of SL(2,C)
QUESTION [41 upvotes]: Books can be written about the finite subgroups of $\mathrm{SL}(2,\mathbb C)$ (and their immediate family, like the polyhedral groups...) I am about to start writing notes for a short course about them and I would like to include references to as much useful and interesting information about them as possible. Since they show up in quite different contexts, and can be looked upon from many different points of view, I am sure the very varied MO audience knows lots of things about them I don't.
So, despite this being more or less canonically too broad/vague a question for MO according to the FAQ:

Can you tell me (or at least point me to) all about the finite subgroups of $\mathrm{SL}(2,\mathbb C)$?

LATER: Thanks to everyone who answered. So far, the information is essentially of algebraic and geometric nature. I wonder now about combinatorics and such beasts. 
For example, it is a theorem of Whitney (or maybe it just follows easily from a theorem of Whitney) that a 3-connected simple planar graph with $e$ edges has an automorphism group of order at most $4e$, and that the order is $4e$ precisely when the graph comes from a polyhedron, so that the group is a polyhedral group.
Do you know of similar results?

REPLY [2 votes]: Check out Curtis' "Construction of a family of Moufang Loops" in Math Proc Camb Phil Soc for a rather interesting extension of the finite subgroups in realtion to the octonions.<|endoftext|>
TITLE: Suppose C and D are Morita equivalent fusion categories, can you say anything about R I: C->Z(C)=Z(D)->D?
QUESTION [8 upvotes]: If C and D are (higher) Morita equivalent fusion categories, then the Drinfel'd centers Z(C) and Z(D) are braided equivalent.  Given any fusion category C we have a restriction functor Z(C)->C (by forgetting the "half-braiding"), and adjoint to that an induction functor C->Z(C).
If C and D are Morita equivalent then you can compose the induction and restriction to get a functor C->Z(C)=Z(D)->D.  (Actually now that I think about you may need to fix the Morita equivalence in order to actually identify Z(C) and Z(D)?)  Is there anything nice one can say about this composition?  If C=D then Etingof-Nikshych-Ostrik says that $R \circ I(V) = \sum_X X \otimes V \otimes X^*$.
The reason that I ask is that Izumi calculated the induction and restriction graphs for the Drinfel'd center of one of the even parts of the Haagerup subfactor, and I would like to understand the same picture for the other even part.

REPLY [6 votes]: I think the answer should depend on the particular choice of Morita equivalence between C and D.
So let M be (bi)module category connecting C and D. My first guess would be that
$R\circ I(V)=\sum_X{\underline Hom}(X,V\otimes X)$ (sum over simple objects of M;
${\underline Hom}$ is the internal $Hom$).<|endoftext|>
TITLE: A reference for smooth structures on R^n
QUESTION [17 upvotes]: There is a theorem stating that there is essentially one smooth structure on $R^n$ for every n other than 4. Does anybody know where i could find the proof of this? Not so much of what happens in dimension four, where there are infinitely many, but of the uniqueness in other dimensions? Thanks!

REPLY [13 votes]: For $n \geq 5$ , this was first proven in Stallings' The piecewise-linear structure of Euclidean space. It actually proves the PL case and applies smoothing theory. Anyway, Theorem 5.1 of it says

Let $M^n$ be a contractible differentiable manifold which is 1-connected at infinity. If $n \geq 5$, then $M$ is diffeomorphic to Euclidean space $\mathbb{R}^n$.

A related result appears in Lashof's ICM address on smoothing theory. His Corollary says

Every contractible open topological manifold is smoothable.

and he follows by remarking that this smooth structure is unique if $n \geq 5$.
For $n = 2,3$, there is Moise, but his articles may be hard to read. There are now easier proofs using Kirby-Siebenmann techniques. For $n = 2$, one can use Hatcher's The Kirby torus trick for surfaces. For $n = 3$, one can use Hamilton's The triangulation of 3-manifolds in the PL case and apply smoothing theory.<|endoftext|>
TITLE: Simply connected quasi-projective varieties in positive characteristic
QUESTION [7 upvotes]: I am looking for examples of non-projective (quasi-projective) varieties $X$ defined over a field of positive characteristic, which have trivial étale fundamental group.
It is well known that the étale fundamental group in positive characteristics is a very difficult object, especially so in the non-projective case due to possibly wild ramification at infinity. I'm not even sure if there are examples of the kind above. Is this known?

REPLY [10 votes]: This is an answer to Pete's question on simply connected affine varieties (I can not put it in a comment because of space limitation). 
I think that in positive characteristic $p$, no affine irreducible variety $X$ of positive dimension is simply connected. We can assume $X=\operatorname{Spec}(A)$ integral because $\pi_1$ is insensible to nipotent elements (SGA IX.4.10). Let $k[t_1,\ldots, t_d] \subseteq A$ be a finite extension with minimal degree $[k(A):k(t_1,\ldots, t_d)]$. Consider the étale cover $Y\to \mathbb A^d_k= \operatorname{Spec}(k[t_1,\ldots, t_d])$ defined by $s^p-s=t_1$. Then $X\times_{\mathbb A^d_k} Y\to X$ is an étale cover of degree $p$. As $k(Y)$ and $k(X)$ are linearly disjoint over $k(t_1,\ldots, t_d)$ ($k(Y)$ is Galois over $k({\bf t}):=k(t_1,\ldots, t_d)$ and $k(Y)\cap k(X)=k({\bf t})$), the tensor product $k(Y)\otimes_{k({\bf t})} k(X)$ is a field. This implies that $X\times_{\mathbb A^d_k} Y$ is connected.<|endoftext|>
TITLE: Is there a name for this algebraic structure?
QUESTION [5 upvotes]: I found myself "naturally" dealing with an object of this form:
X is a complex vector space, with a "product" (a,b) → {aba} which is quadratic in the first variable, linear in the second, and satisfies some associativity conditions. These conditions are actually complicated, but more or less say that {aba} looks like the product (aba) in an alternative algebra Y containing X as a subspace.
For example, the main "associativity condition" I am interested in is:
{a{b{aca}b}a}={{aba}c{aba}}
Examples

Symmetric matrices
Octonions, or indeed any alternative algebra
Let J belong to GL(n,ℂ), with tJ=-J and J²=-Id, and W={w∈M(n×n,ℂ)|JtwJ=-w}

all with the standard product {aba}=aba.
All of these examples are Jordan algebras, with respect to the symmetrized product a∘b=½(ab+ba), but I cannot see any direct link between the Jordan product and my product.

REPLY [26 votes]: In a Jordan algebra with product $\cdot$, a triple product is defined by $$\{abc\}=(a\cdot b)\cdot c+(b\cdot c)\cdot a-(a\cdot c)\cdot b.$$
In a special Jordan algebra (constructed by symmetrising an associative product) one has $\{aba\}=aba$, and it is easy to show that in such algebras one always has the identity $$\{\{aba\}c\{aba\}\}=\{a\{b\{aca\}b\}a\}.$$ Now, there is an amazing general theorem of Macdonald's that states that any identity in three variables which is of degree at most one in one of them and which is valid in special Jordan algebras actually holds in all Jordan algebras. This is proved in Jacobson's breath-taking Structure and representations of Jordan algebras.
So your identity holds in all Jordan algebras. As a consequence, from the information you give it is more or less impossible to distinguish your structure from Jordan algebras, as far as I can see.
By the way, in his book, Jacobson notes that McCrimmon has developed the theory of Jordan algebras based exclusively on the composition $(a,b)\mapsto aba$, and gives [McCrimmon, Kevin. A general theory of Jordan rings. Proc. Nat. Acad. Sci. U.S.A.  56  1966 1072--1079. MR0202783 (34 #2643)] as reference. I do not have access to the paper, though. The paper can be gotten from this link Andrea provided in a comment below.<|endoftext|>
TITLE: How is the physical meaning of an irreducible representation justified?
QUESTION [48 upvotes]: This is maybe not an entirely mathematical question, but consider it a pedagogical question about representation theory if you want to avoid physics-y questions on MO.  
I've been reading Singer's Linearity, Symmetry, and Prediction in the Hydrogen Atom and am trying to come to terms with the main physical (as opposed to mathematical) argument of the text.  The argument posits, if I understand it correctly, that a quantum system described by a Hilbert space $H$ on which a group $G$ of symmetries acts by unitary transformations should have the property that its "elementary states" "are" irreducible subrepresentations of the representation of $G$ on $H$.  She begins this argument in section 5.1:
Invariant subspaces are the only physically natural subspaces.  Recall from Section 4.5 that in a quantum system with symmetry, there is a natural representation $(G, V, \rho)$.  Any physically natural object must appear the same to all observers.  In particular, if a subspace has physical significance, all equivalent observers must agree on the question of a particular state's membership in that subspace.
and continues it in section 6.3:
We know from numerous experiments that every quantum system has *elementary states*.  An elementary state of a quantum system should be **observer-independent**.  In other words, any observer should be able (in theory) to recognize that state experimentally, and the observations should all agree.  Secondly, an elementary state should be indivisible.  That is, one should not be able to think of the elementary state as a superposition of two or more "more elementary" states.  If we accept the model that every recognizable state corresponds to a vector subspace of the state space of the system, then we can conclude that elementary states correspond to irreducible representations.  The independence of the choice of observer compels the subspace to be invariant under the representation.  The indivisible nature of the subspace requires the subspace to be irreducible.  So elementary states correspond to irreducible representations.  More specifically, if a vector $w$ represents an elementary state, then $w$ should lie in an *irreducible* invariant subspace $W$, that is, a subspace whose only invariant subspaces are itself and $0$.  In fact, every vector in $W$ represents a state "indistinguishable" from $w$, as a consequence of Exercise 6.6.
(For people who actually know their quantum, Singer is ignoring the distinction between representations and projective representations until later in the book.)
My first problem with this argument is that Singer never gives a precise definition of "elementary state."  My second problem is that I'm not sure what physical principle is at work when she posits that physically natural subspaces and elementary states should be observer-independent (i.e. invariant under the action of $G$).  What underlying assumption of quantum mechanics, or whatever, is at work here?  Why should a mathematician without significant training in physics find this reasonable?  (I have the same question about the identification of elementary particles with irreducible representations of the "symmetry group of the universe," so any comments about this physical argument are also welcome.)
Singer goes on to use this assumption to deduce the number of electrons that fill various electron orbitals, and I won't be able to convince myself that this makes sense until I understand the physical assumption that allows us to use irreducible representations to do this.

REPLY [2 votes]: I have been struggling with very similar issues and was almost ready to ask here, but then found this question, so I'll instead post my ramblings as an "answer" to avoid duplication of questions (of course, it doesn't answer anything).
The official story seems to go something like this (for example in Weyl's classic): Look at something like $H=-\Delta + V(|x|)$ on $L^2(\mathbb R^3)$. Then the natural action of rotations $R\in SO(3)$ on $L^2$ by the unitary maps $(R\cdot f)(x) = f(R^{-1}x)$ is a symmetry of the system; this feels obvious, especially if I don't specify what I mean by symmetry. "Therefore" the irreducible representations of $SO(3)$ must be important.
I would really love to have this "therefore" explained in mathematical style, but after some reading around, my current impression is that there is nothing of the sort available. Rather, it seems to be a vaticinium ex eventu: For example, the spherical harmonics play a key role when analyzing $H$ above, and these do give us finite-dimensional representations. The representation on $\mathbb C^2$ (and for some reason we have now also found it convenient to slightly change $SO(3)$ to $SU(2)$) shows up in the description of spin. "Therefore," we conclude that representations must be relevant in general.
Since this answer is written from a position of ignorance, comments are very welcome.<|endoftext|>
TITLE: Is the cohomology of a topological operad a cooperad?
QUESTION [8 upvotes]: For cohomology with coefficients in a field $F$ the map $H^\cdot(X;F) \otimes H^\cdot(Y;F) \to H^\cdot(X \times Y;F)$ of the Kunneth theorem is an isomorphism of algebras over $F$. I am correct in thinking that this together with the fact that cohomology with a contravariant functor, implies that the cohomology of a topological operad is a cooperad, the dual of a notion of a cooperad?
If this is not true, where does my reasoning fail? Does it at least hold on the level of vectorspaces over $F$? If this is true, why aren't there many references about this construction? It seems that the additional algebra structure would give you some additional information.
Furthermore, using the Thom isomorphism for the cohomology, would this also imply that there is a notion of string cohomology dual to string topology?

REPLY [5 votes]: There is one little caveat, though: you need either to assume your spaces in the operad are of finite type, or work in the category of topological vector spaces and completed tensor products. The reason is that the Künneth theorem in cohomology only holds for spaces of finite type if you use the regular tensor product. Consider, for a counterexample, an infinite wedge of spheres.<|endoftext|>
TITLE: Localization of vanishing cycles
QUESTION [7 upvotes]: Consider a regular holonomic D-module (or a perverse sheaf) $M$ on a smooth variety $X$. Let $f:X\to A^1$ be a polynomial (or holomorphic) function. 
Question: Is it true that the $\lambda \in A^1$ such that the vanishing cycles $\phi_{f-\lambda}(M) = 0$ is a dense open set? 
Here are my thoughts:
If $M = O_X$ (or the constant perverse sheaf $A[dim X]$), this is just the fact that the critical values of $f$ are isolated.    
In the general case, we can  factor $f$ through its graph $X\to X\times A^1$,  $x\mapsto (x,f(x))$, reducing to the case where $f$ is the (smooth) projection $t:X\times A^1 \to A^1$. Our sheaf $M$ on $X\times A^1$ has a characteristic variety $\bigcup_\alpha T^*_{S_\alpha}(X\times A^1)$ for a stratification $X\times A^1 = \bigcup_\alpha S_\alpha$. My guess is that $\phi_{t-\lambda}(M) = 0$ when $\{t-\lambda  = 0 \}$ is transverse to all the $S_\alpha$ and that this is a generic condition but I'm having trouble making this intuition precise.

REPLY [6 votes]: An answer to this question was given to me by Pierre Schapira. This is known as the microlocal Bertini-Sard theorem (cf. Sheaves on manifolds cor. 8.3.12). 
Consider a map $f:X\to A^1$. It induces $f_\pi : X\times_{A^1} T^*A^1 \to T^*A^1$ and $f_d : X\times_{A^1} T^*A^1 \to T^*X$. Set $\Lambda = SS(M)$ the characteristic variety of $M$. This is a closed conic isotropic subset of $T^*X$. Now 
$$
  supp(\phi_{f-t}(M)) \subset [ x ~|~ f(x) = t,~(x,df(x)) \in \Lambda ]
$$ 
so 
$$
  [ t\in A^1 ~|~ \phi_{f-\lambda}(M) \neq 0 ] \subset 
  [ t\in A^1 ~|~ (t,dt)\in  f_\pi f_d^{-1}(\Lambda) ]
$$  
Now assume that $f$ is compactifiable as $X\overset{j}{\to} \bar{X} \overset{\bar{f}}{\to} A^1$, $j$ an open immersion and $\bar{f}$ proper. The closure $\bar{\Lambda}$ of $\Lambda$ is $T^*\bar{X}$ is a closed conic isotropic subset and since $\bar{f}$ is proper, $\bar{f}_\pi \bar{f}_d^{-1}(\bar{\Lambda})$ is a closed conic isotropic subset of $T^*A^1$. So its intersection with the nowhere vanishing section
$$
  [t \in A^1 ~|~ (t,dt) \in \bar{f}_\pi \bar{f}_d^{-1}(\bar{\Lambda})]
$$
has dimension 0. Since $f_\pi f_d^{-1}(\Lambda) \subset \bar{f}_\pi \bar{f}_d^{-1}(\bar{\Lambda})$ the same is true for 
$$
  [ t\in A^1 ~|~ \phi_{f-\lambda}(M) \neq 0 ] \subset 
  [ t\in A^1 ~|~ (t,dt)\in  f_\pi f_d^{-1}(\Lambda) ]
$$ 
and the theorem is proved.  
If $f$ is algebraic it is always compactifiable. If $f$ is analytic, I don't know if the theorem still holds in general.
PS: If $f(x) = \lambda $, the condition $(x,df(x)) \in T^*_Z X$ just says that the fiber $\{f = \lambda\}$ is tranverse to $Z$ at $x$. So when $SS(M) \subset \bigcup T^*_{S_\alpha} X$ this gives the geometric interpretation that the vanishing cycles are 0 whenever the fibers are transverse to the strata.<|endoftext|>
TITLE: Rationality of GIT quotients
QUESTION [12 upvotes]: I recently worked through most of the proof of the rationality of the moduli of genus 3 curves, which seemed to have the following structure:

Every nonhyperelliptic genus 3 curve is a smooth plane quartic.
The plane quartics form a projective space.
Apply GIT to this projective space and the $PGL(3)$ action.
Prove that this quotient is rational.

I've seen somewhat similarly structured arguments before.  So my question:

When is a GIT quotient rational?

In particular, are quotients of $\mathbb{P}^n$ by $PGL_k$ rational, under some reasonable hypotheses?
Are there any natural invariants that are preserved by quotients (again, with reasonable conditions, or of the above form)?

REPLY [2 votes]: There is a very nice (if somewhat dated - it predates Katsylo's work of M3) survey  of the problem by Dolgachev in the AG Bowdoin volume. Here is the google books link.<|endoftext|>
TITLE: Complexity of testing integer square-freeness
QUESTION [25 upvotes]: How fast can an algorithm tell if an integer is square-free?
I am interested in both deterministic and randomized algorithms.  I also care about both unconditional results and ones conditional on GRH (or other reasonable number-theoretic conjectures).
One reference I could find was on the Polymath4 wiki, where it states

No unconditional polynomial-time
  deterministic algorithm for
  square-freeness seems to be known. (It
  is listed as an open problem in this
  paper
  from 1994.)

I can't tell if that quote implies that both conditional and randomized polynomial-time algorithms exist, but it might (the exception that proves the rule?).
Thanks in advance.

REPLY [2 votes]: I was involved in the conversations about this topic on the Polymath4 blog (actually, looking back, it looks like I was the one who dug up that old paper...) and I came to believe that there was no such algorithm (randomized, conditional, whatever). Certainly I searched the literature as best I could and didn't find one. But I'm pessimistic about finding a reduction from factoring, for reasons I touched on in the linked post.
I was going to mention this beautiful argument, but actually I don't think it applies here -- you can only use squarefreeness to tell if a prime factor $p | N$ ramifies over some extension (Edit: I think this is true -- but something weird might happen if the extension isn't Galois? Maybe? I know so little algebraic number theory it's not even funny), but that's only possible if p divides the discriminant -- but you can do that already by the Euclidean algorithm. So squarefreeness would only let you maybe factor if for some reason you could do the algorithm quickly in number fields with huge discriminant, which admittedly might be possible. Edit: Although of course if the discriminant is big enough to make a difference, it's unclear how you'd extract information about p anyway. Which, modulo a whole bunch of holes and handwaving, would seem to rule out any naive attempt to adapt that "reduction."<|endoftext|>
TITLE: Which Fréchet manifolds have a smooth partition of unity?
QUESTION [16 upvotes]: A classical theorem is saying that every smooth, finite-dimensional manifold has a smooth partition of unity. My question is:

Which Fréchet manifolds have a smooth partition of unity?
How is the existence of smooth partitions of unity on Fréchet manifolds related to paracompactness of the underlying topology?

From some remarks in some literature, I got the impression that not all Fréchet manifolds have smooth partitions of unity, but some have, e.g. the loop space $LM$ of a finite-dimensional smooth manifold $M$. 
For $LM$, the proof seems to be that $LM$ is Lindelöf, hence paracompact. 
Is this true for all mapping spaces of the form $C^\infty (K,M)$ for $K$ compact?

REPLY [14 votes]: Use the source, Luke.
Specifically, chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality).  You may be particularly interested in:

Theorem 16.10 If $X$ is Lindelof and $\mathcal{S}$-regular, then $X$ is $\mathcal{S}$-paracompact.  In particular, nuclear Frechet spaces are $C^\infty$-paracompact.

For loop spaces (and other mapping spaces with compact source), the simplest argument for Lindelof/paracompactness that I know of goes as follows:

Embed $M$ as a submanifold of $\mathbb{R}^n$.
So the loop space $LM$ embeds as a submanifold of $L\mathbb{R}^n$.
$L\mathbb{R}^n$ is metrisable.
So $L M$ is metrisable.
Hence $L M$ is paracompact.

(Paracompactness isn't inheritable by all subsets.  Of course, if you can embed your manifold as a closed subspace then you can inherit the paracompactness directly.)
I use this argument in my paper on Constructing smooth manifolds of loop spaces, Proc. London Math. Soc. 99 (2009) 195–216 (doi:10.1112/plms/pdn058, arXiv:math/0612096) to show that most "nice" properties devolve from the model space to the loop space for "nice" model spaces (smooth, continuous, and others).  See corollary C in the introduction of the published version.
(I should note that the full statement of Theorem 16.10 (which I did not quote above) is not quite correct (at least in the book version, it may have been corrected online) in that the proof of the claim for strict inductive sequences is not complete.  I needed a specific instance of this in my paper The Smooth Structure of the Space of Piecewise-Smooth Loops, Glasgow Mathematical Journal, 59(1) (2017) pp27-59. (arXiv:0803.0611, doi:10.1017/S0017089516000033) (see section 5.4.2) which wasn't covered by 16.10 but fortunately I could hack together bits of 16.6 with 16.10 to get it to work.  This, however, is outside the remit of this question as it deals with spaces more general than Frechet spaces.)
On the opposite side of the equation, we have the following after 16.10:

open problem ... Is every paracompact $\mathcal{S}$-regular space $\mathcal{S}$-paracompact?

So the general case is not (at time of publishing) known.  But for manifolds, the case is somewhat better:

Ch 27 If a smooth manifold (which is smoothly Hausdorff) is Lindelof, and if all modelling vector spaces are smoothly regular, then it is smoothly paracompact.  If a smooth manifold is metrisable and smoothly normal then it is smoothly paracompact.

Since Banach spaces are Frechet spaces, any Banach space that is not $C^\infty$-paracompact provides a counterexample for Frechet spaces as well.  The comment after 14.9 provides the examples of $\ell^1$ and $C([0,1])$.
So, putting it all together: nuclear Frechet spaces are good, so Lindelof manifolds modelled on them are smoothly paracompact.  Smooth mapping spaces (with compact source) are Lindelof manifolds with nuclear model spaces, hence smoothly paracompact.
(Recall that smooth mapping spaces without compact source aren't even close to being manifolds.  I know that Konrad knows this, I merely put this here so that others will know it too.)<|endoftext|>
TITLE: Intersection form in twisted homology (homology with local coefficients)
QUESTION [7 upvotes]: The answer to this question should be obvious, but I can't seem to figure it out. Suppose we have a surface $F$, and a representation $\rho : \pi_1(F)\to SU(n)$. We can define the homology with local coefficients $H_*(F,\rho)$ straightforwardly as the homology of the twisted complex $$C_*(F,\rho):=C_*(\widetilde{F};\mathbf{Z})\otimes_{\mathbf{Z}[\pi_1(F)]} \mathbf{C}^n$$ where $\widetilde{F}$ is the universal cover, and $\mathbf{Z}[\pi_1(F)]$ acts on each side in the obvious way. 
Now, this complex is actually very easy to compute explicitly: just lift a nice basis of cells in $F$ to $\widetilde{F}$, and write down the boundary maps explicitly. For example, if $F$ is a torus and we take $n=2$, say, we can choose a natural meridian-longitude basis $(x,y)$ for $H_1(F)$, and the twisted boundary map $\partial_1:C_1(F,\rho)=\mathbf{C}^4\to C_2(F,\rho)=\mathbf{C}^2$ is $$ \left( \begin{array}{ccc}
\rho(x)-Id  \newline\rho(y)-Id\end{array} \right)$$
So, here's my question. Since $\rho$ is a unitary representation, we should get a twisted intersection form on $H_1(F)$, simply by combining the untwisted intersection form with the standard hermitian product on $\mathbf{C}^2$, right? And I would imagine this is also really easy to compute, in a similar basis, say? I can't seem to figure out how it would go. Could anyone help me, even show me how it works for the same torus example?
Or, if I've said anything wrong, tell me where?

REPLY [6 votes]: What you say is right, and makes sense on any even dimensional manifold. Computing it can be tricky: a useful approach is to use a regular cell complex and the dual complex, then on the chain level the intersection form is given by the identity matrix (see the first couple pages of Milnor's "a duality theorem for Reidemeister torsion").
One suggestion for calculation is to assume $\rho$ is irreducible,  since if $C^n$ splits invariantly under the $\pi_1F$ action so does the cohomology. In your torus example, since $\pi_1=Z\oplus Z$ is abelian, the only irreducible reps are 1-dimensional.  For Euler characteristic reasons (and Poincare duality) in this case it turns out either the rep is trivial in which case you know the answer, or else the rep is non-trivial in which case the homology vanishes and the intersection form is trivial. For higher genus surfaces you will get something non-zero, but in this dimension you get a skew-hermitian form, which is determined up to iso by its rank, if I'm thinking clearly. For dimensions divisible by 4, you can get interesting (i.e. non-zero) signature, but for a closed manifold it will just equal n times the ordinary signature by the twisted form of the Hirzebruch signature theorem. 
But the twisted intersection form is interesting when your manifold has non-empty boundary, since it gives invariants of the boundary. Hundreds of papers  are based on this observation. Even when your surface has non-empty boundary you get something interesting.

REPLY [5 votes]: For me it is easier to work with cohomology (just for psychological reasons).  Also, I will distinguish the representation $\rho$ from the local system $V$ with fibres ${\mathbb C}^2$ that it gives rise to.    So where you would write $H^1(F,\rho)$ I will write $H^1(F,V)$.
I will let $\overline{V}$ denote the complex conjugate local system to $V$.
(So it is the same underlying local system of abelian groups, but we give it the
conjugate action of $\mathbb C$.)
The Hermitian pairing on the fibres of $V$ and $\overline{V}$ gives a pairing of local
systems $V \times \overline{V} \to \mathbb R$, where $\mathbb R$ is the constant local system
with fibre the real numbers.  If you like we can think of this as an $\mathbb R$-linear map
$V\otimes_{\mathbb C}\overline{V} \to \mathbb R.$  This pairing will induce a map on
cohomology $H^2(F,V\otimes_{\mathbb C}\overline{V}) \to H^2(F,\mathbb R)$.
There will also 
be a cup product $H^1(F,V) \times H^1(F,\overline{V}) \to H^2(F, V\otimes_{\mathbb C} \overline{V})$.
Composing this with the previous map on $H^2$ gives your twisted cup product
$H^1(F,V)\times H^1(F,\overline{V}) \to H^2(F,\mathbb R)$.
This gives one perspective on your construction.  To compute it, write down the twisted cochains $C^{\bullet}(\tilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2$,
then
write down the cup-product
$$(C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2 ) \times
(C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2) 
\to C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]} \mathbb R^2
= C^{\bullet}(F,\mathbb R).$$
The cup product will just be given by the usual formula, and then you will also
pair the $\mathbb C^2$ parts of the cochains using the hermitian pairing.
Hopefully you can follow your nose and do this explicitly for the torus.  Then you can
just dualize everything to get to the homology version.<|endoftext|>
TITLE: Hamiltonian paths where the vertices are integer partitions
QUESTION [16 upvotes]: I have been working on this problem for several months now but have not made much progress. It concerns the set of all integer partitions of n. 
Let the vertices of the graph G=G(n) denote all the p(n) integer partitions of n. There is an edge between two partitions if and only if one can be transformed into another by only moving one dot between rows in their Ferrers diagram representations. 
So, for example, the partitions (3,2,1) and (3,3) of 6 are linked because we can move the dot in the last row to the second row.
OOO               OOO
OO      --------  OOO
O   

My question: for what values of n does G(n) have a Hamiltonian path from (n) to (1,1,...,1)? 
That is, is it possible to go through, without repetition, all the partitions of n by simply moving around the dots in the Ferrers diagrams? 
Is there a determinate way to construct such paths?
I have only been able to construct paths for n = 1 to 6.
n=1 (trivial)
n=2: 
(2) => (1,1)
n=3:
(3) => (2,1) => (1,1,1)
n=4:
(4) => (3,1) => (2,2) => (2,1,1) => (1,1,1,1)
n=5:
(5) => (4,1) => (3,2) => (2,2,1) => (3,1,1) => (2,1,1,1) => (1,1,1,1,1)
n=6:
(6) => (5,1) => (4,1,1) => (4,2) => (3,3) => (3,2,1) => (2,2,2) => (2,2,1,1) => (3,1,1,1) => (2,1,1,1,1) => (1,1,1,1,1,1)
None of the basic theorems about Hamiltonian paths have not helped me here.

REPLY [3 votes]: If you mean to allow any part to go down by 1 and any part (or 0) to go up by 1, while maintaining non-increasing order, then Carla Savage's 1989 paper does in fact solve your problem (without requiring "double moves" as in the 10th listed partition of 7 above).
There are refinements of this operation in the literature.  If you only allow a dot to "fall" into an adjacent part, i.e., from $(\ldots, p_i, p_{i+1}, \ldots)$ with $p_i \geq p_{i+1}+2$ to $(\ldots, p_i - 1, p_{i+1}+1, \ldots)$, that is called the "sandpile operation" or "Brylawski's vertical rule".  There is a transpose horizontal rule, and restictions of these called $\theta$ operations and "ice pile" operations, respectively.  For some of these operations, the set of partitions of $n$ is not a connected graph, much less one with a Hamiltonian path.  A survey article on these operations is available at https://www-complexnetworks.lip6.fr/~latapy/Publis/tcs04a.pdf.<|endoftext|>
TITLE: Example of connected-etale sequence for group schemes over a Henselian field?
QUESTION [13 upvotes]: Can someone give a really concrete example of such a sequence? I am looking at several notes related with such things, but haven't seen any well-calculated example. And I'm really confused at this point.
Besides asking for a good example, I am also wondering about the following two things:

There is an exact sequence for elliptic curves defined over a local field $K$, $0 \rightarrow \hat E(m) \rightarrow E(K) \rightarrow \tilde E(k) \rightarrow 0$, where $\hat E(m)$ is the formal group associated to $E$ and $\tilde E(k)$ is the reduction. (See Silverman AEC I, page 118), is this sequence related with connected-etale sequence?

2.Take the p-torsion kernel $E[p]$ of $[p]: E \rightarrow E$ for $E$ defined over $K$ a local field.Is $E[p]$ a finite flat group scheme over $R$ the valuation ring? And if so, what is its connected-etale sequence? (maybe I should change $p$ to an $n$, but I'm also curious what will happen if $p$ is the characteristic of $k$?)
Thank you.

REPLY [48 votes]: For concepts related to algebraic geometry when the base is not a field, it can be difficult for a beginner to reconcile the approach in Silverman with the approach via schemes. I wasted a lot of time as a student trying to relate the "3 points through a line" definition of the group law over fields with the concept of "reduction mod $p$" on points. Likewise, the approach with formal groups tends to make things confusing, despite their apparent "concreteness". 
This sort of stuff drove me crazy when I was a student, until I realized that the best way to understand such topics is to give up working over fields and with equations, and to work over the valuation ring and with functorial viewpoints (only translating into field language at the very end).  The relevant schemes in the question are really torsion schemes over the valuation ring, not torsion in the separate fibers (so you mean to assume $E$ has good reduction in both questions).  I address this below. The following answer is way too long, since I do not know of a suitable reference not involving EGA/SGA. Tate's article on finite flat group schemes probably explains some aspects, but I doubt it addresses the link with the concrete stuff for elliptic curves.  
If $R$ is a local ring, then an "elliptic curve over $R$" can be defined in two ways:  the concrete way is as a Weierstrass plane cubic with unit discriminant, and the right way is as a smooth proper $R$-scheme with geometrically connected fibers of dimension 1 and genus 1 and a distinguished section.  As usual, the concrete way is hard to use to actually prove anything interesting (and it is the "wrong" notion when the base is non-local and especially has non-trivial line bundles; e.g., over suitable number fields with class number $> 1$ one can make CM elliptic curves with "everywhere good reduction" which do not admit a global planar model with unit discriminant).   How to prove there is a unique $R$-group structure with the indicated section as the identity?  A real nightmare with the concrete definition, and elegantly explained in Chapter 2 of Katz-Mazur with the right definition. Likewise, that $E$ is functorial in its generic fiber when $R$ is a discrete valuation ring is a mess to prove by hand (which affine opens to use?), but has an elegant proof when approached through the "smooth and proper" viewpoint.  Of course it is important and interesting that these concrete and abstract notions agree, and that is explained in Katz-Mazur Chapter 2.   
That being said, if $E$ is an elliptic curve over any noetherian (say) scheme $S$ and $[n]_E:E \rightarrow E$ is multiplication by a positive integer $n$, then on geometric fibers this is a finite flat map, so $[n]$ is quasi-finite.  Now proper and quasi-finite maps are finite (by Zariski's Main Theorem), so $[n]_E$ is a finite map, and the fibral flatness criterion implies that it is also flat.  Being a finite flat map between noetherian schemes, it has a "degree" which is locally constant on the target and yet is $n^2$ on fibers over $S$.  Hence, we conclude that $E[n] := {\rm{ker}}([n]_E)$ is a finite flat commutative $S$-group with constant fiber rank $n^2$. Honestly, I do not know any way to prove this which avoids the serious results that I just cited.  But that's why the theorems are useful:  because we can use them to make our intuition over fields carry over to cases when the base is not a field. (The noetherian condition can be dropped if we are more careful with the phrase "finite flat". I won't dwell on it here.) This answers the first part of the 2nd question (taking the base to be spectrum of the valuation ring there).  It the notation there, the $p$-torsion of the elliptic curve over $K$ is not a finite $R$-scheme, and in general it may extend to a finite flat $R$-group in many ways.  But the elliptic curve over $K$ uniquely extends to one over $R$ by the theory of Neron models, and its torsion levels provide the "right" finite flat groups you want to use over the valuation ring. 
OK, now assume $R$ is a complete local noetherian ring (e.g., a complete discrete valuation ring).  Could even assume it is a henselian local ring, but the complete case is easier to deal with and covers the case in the question.  Let $G$ be a finite flat $R$-group, a case of interest being $E[n]$ for an elliptic curve $E$ over $R$.  Let $k$ be the residue field, and consider $G_k$.  Being a finite $k$-scheme, it has an open and closed identity component $G_k^0$ which is cut out by an idempotent.  By 8.15 (or thereabouts) in Matsumura's Commutative Ring Theory, every idempotent in the special fiber of a finite $R$-algebra uniquely lifts.  In particular, if $X$ is a finite $R$-scheme then its connected component decomposition uniquely lifts that of $X_k$.  If $X$ is $R$-flat then so is each of its connected components. This is all compatible with products, so if $X$ has a structure of $R$-group then the open and closed connected component $X^0$ containing the identity section is an $R$-subgroup.  Returning to our friend $G$, we get the so-called "relative identity component" $G^0$, an open and closed (hence finite flat) $R$-subgroup. 
Remark:  The formation of $G^0$ commutes with any flat local extension on $R$, as follows from the uniqueness!  It doesn't usually commute with non-local extension, such as inclusion of a complete dvr into its fraction field. 
Example: $G = E[n]$.  Suppose $R$ is a complete discrete valuation ring with fraction field $K$, and $n \in K^{\times}$.  What is $(G^0)_K$?  Well, each "point" occurs over a finite extension $K'/K$, say with valuation ring $R'$, and $G(K') = G(R')$ by elementary integrality considerations (or in fancy terms, valuative criterion, which is killing a fly with a sledgehammer). Since the spectrum of $R'$ is connected, a point in $G(R')$ lies in $G^0(R')$ if and only if its specialization into $G_k(k')$ vanishes ($k'$ the residue field of $R'$).  In other words, $(G^0)(\overline{K})$ consists of the $n$-torsion geometric points of $E_K$ whose specialization into geometric points of $E_k$ by valuative criterion for the $R$-proper $E$ ($E_K(K') = E(R') \rightarrow E(k') = E_k(k')$!) is 0. 
Now we need to explain the "etale quotient" in concrete terms.  This is best understood as a generalization of the following procedure over a field. 
Example:  Let $k$ be a field and $A_0$ a finite $k$-algebra.  There is a unique maximal \'etale $k$-subalgebra $A_0'$ in $A_0$: concretely, in each local factor ring of $A_0$ uniquely lift the separable closure of $k$ in the residue field up into the local factor ring via Hensel's Lemma and the primitive element theorem. Since it is uniquely characterized by lifting separable closures of $k$ in the residue fields of the factor rings, it is a good exercise to check the following crucial thing:  if $B_0$ is another finite $k$-algebra then $(A_0 \otimes_k B_0)' = A_0' \otimes_k B_0'$, and $A_0'$ is functorial in $A_0$.  Observe that $A_0' \rightarrow A_0$ is faithfully flat since at the level of factor rings of $A_0$ it is an inclusion of a field into a nonzero ring. Also observe that any \'etale $k$-algebra equipped with a map to $A_0$ uniquely factors through $A_0'$. 
Exercise: The formation of $A_0'$ commutes with any field extension on $k$. (Hint: use Galois descent to reduce to the separate cases of separable algebraic extensions and the easy case $k = k_s$.) 
In geometric terms, for a finite $k$-scheme $X_0$, the preceding Example constructs a finite \'etale $k$-scheme $X_0'$ and a faithfully flat $k$-map $f_0:X_0 \rightarrow X_0'$ which is initial among all $k$-maps from $X_0$ to finite \'etale $k$-schemes, and its formation is functorial in $X_0$ and commutes with products in $X_0$ and with any extension on $k$.  In particular, if $X_0$ is a $k$-group then $X_0'$ has a unique $k$-group structure making $f_0:X_0 \rightarrow X_0'$ a $k$-homomorphism. 
Example: Now let $R$ be a complete discrete valuation ring with residue field $k$, and let $X$ be a finite flat $R$-scheme.  (Can relax the hypothesis on $R$ if familiar with finite \'etale maps in general.) In this setting, "finite 'etale" over $R$ just means "product of finitely many unramified finite extensions".  By using Hensel's Lemma in finite local $R$-algebras, to give a map from a finite \'etale $R$-algebra $A$ to a finite $R$-algebra $B$ is the same as to give a map $A_0 \rightarrow B_0$ between their special fibers.  In particular, finite \'etale $k$-algebras uniquely and functorially lift to finite \'etale $R$-algebras, and so $X_0'$ uniquely lifts to a finite \'etale $R$-scheme $X'$ and there is a unique $R$-map $f:X \rightarrow X'$ lifting $f_0:X_0 \rightarrow X_0'$.  By fibral flatness (using $X$ is $R$-flat!), $f$ is faithfully flat since $f_0$ is.  By uniqueness of everything in sight, the formation of $f$ commutes with products and local extension on $R$ and is also functorial in $X$.  In particular, if $G$ is a finite flat $R$-group then $G'$ admits a unique $R$-group structure making $f$ an $R$-homomorphism. We call $G'$ the maximal \'etale quotient of $G$.  
Now we can put it all together and obtain the connected-etale sequence:
Proposition:  Let $G$ be a finite flat group scheme over a complete discrete valuation ring $R$. (Even ok for complete local noetherian $R$, or even henselian local $R$.) The faithfully flat $R$-homomorphism $f:G \rightarrow G'$ to the maximal \'etale quotient has scheme-theoretic kernel $G^0$.
Proof: The kernel $H = \ker f$ is a finite flat $R$-group.  To show it contains $G^0$ we have to check that the composite map $G^0 \rightarrow G \rightarrow G'$ vanishes.  Being a map from a finite $R$-scheme to a finite \'etale $R$-scheme, the map is determined by what it does on the special fiber, so it suffices to show that $G_k^0 \rightarrow G_0'$ vanishes.  This is a map from a finite infinitsimal $k$-scheme to a finite \'etale $k$-scheme which carries the unique $k$-point to the identity point.  Thus, it factors through the identity section of $G_0'$, which is open and closed since $G_0'$ is finite etale over $k$.   
Now that $H$ contains $G^0$, to prove the resulting closed immersion $G^0 \hookrightarrow H$ between finite flat $R$-schemes is an isomorphism it suffices to do so on special fibers.  But that reduces us to the variant of our problem over the residue field.  We can increase it to be algebraically closed, and so the problem is to show that if $G$ is a finite flat group scheme over an algebraically closed field $k$ then $G \rightarrow G'$ has kernel exactly $G^0$.  But $G'$ is a constant $k$-scheme since it is etale and $k$ is algebraically closed, so by construction $G'$ is just the disjoint union of the $k$-points of the connected components of $G$.  It is then physically obvious that the kernel is $G^0$. QED
Remark: If $X$ is any finite flat $R$-scheme, with $X \rightarrow X'$ the initial map to
a finite \'etale $R$-scheme, then the induced map on $\overline{k}$-points is bijective.  Indeed, we can pass to geometric special fibers and connected components to reduce to the case when $X$ is local finite over an algebraically closed field (in place of $R$), in which case the assertion is clear. 
By this Remark, the geometric points of the $n$-torsion in $E_k$ are identified with the geometric points of the special fiber of the maximal etale quotient $E[n]'$.  In particular, if $n$ is not divisible by the characteristic of $K$ and if $K'/K$ is a sufficiently big finite separable extension which splits $E_K[n]$ then the finite etale $R'$-scheme $E[n]'_{R'}$ is constant (as it may be checked on $K'$-fiber), so the map 
$$E _K[n] (\overline{K}) = E _K[n]  (K') = E[n] (R') \rightarrow E[n]' (k') \hookrightarrow
E  _k[n]' (\overline{k}) = E[n]'(R') = E[n]' (\overline{K})$$
is identified with the naive map in question 1. In other words, that step computes the "quotient" part of the connected-etale sequence of $E[n]$ after passing to $\overline{K}$-points!
Example: If $E$ has supersingular reduction then $E[p] = E[p]^0$ and the etale part of the sequence for $E[p]$ vanishes.
Example: If $E$ has ordinary reduction then working over an algebraic closure of the residue field shows that $E[p]^0$ and $E[p]'$ each have rank $p$ as finite flat $R$-groups. 
Finally, it remains to relate $E[n]^0$ to $n$-torsion in the so-called "formal group" of $E$ (not the formal group of $E_K$, which loses contact with the integral structure and for ${\rm{char}} (K) = 0$ is actually the formal additive group which has no nontrivial $n$-torsion!).  A moment's reflection on the definition of the formal group in Silverman shows that its $R'$-points for any finite local valuation ring extension of $R$ are precisely the local $R'$-points of the complete local ring  $\widehat{\mathcal{O}}_{E,0_k}$ at the origin of the special fiber (or the completion along the identity section, comes to the same since $R$ is complete). By the universal properties of local rings on schemes and completions of local noetherian rings, such $R'$-points of the latter type are simply points in $E(R')$ specializing to $0_k$ in $E_k (k')$.   But we saw earlier that $E[n]^0 (R')$ is exactly the set of points in $E[n] (R')$ specializing to $0_k$ on $E_k$.  So indeed $E[n]^0 (R')$ inside of $E[n] (R') = E_K[n] (K')$ is exactly the $n$-torsion in the $K'$-points of the "formal group" in the sense of Silverman's book. 
Voila, so that answers the questions.  The arguments used are designed to apply equally well to abelian varieties.<|endoftext|>
TITLE: Is there a combinatorial proof of Cauchy-Schwarz?
QUESTION [21 upvotes]: I've only played with this a little for the past day or so, and haven't thought about it too hard, so it might be obvious. Obviously it's not fair to ask for a "combinatorial proof" of an inequality involving real numbers, so we'll ask that the vectors be in $\mathbb{N}^n$. More concretely:
Given n boxes subdivided into a "right half" and a "left half" with $a_i$ objects in the right half of box $i$, and $b_i$ in the left half of box i, is there a natural injective function from

Two pairs (ordered, with replacement) of objects, with each pair containing one object from the left half and one object from the right half of a fixed box

to

A pair (ordered, with replacement) from the right half of some box, and a pair (O,WR) from the left half of some (possibly different) box?

(Sorry if this is a double; my wireless is being strange.)

REPLY [13 votes]: Honestly, not really, there is no interesting combinatorial proof.  Think of the most trivial case $a^2 + b^2 \ge 2 a b$.  How do you prove that combinatorially?  Well, arrange the terms into $(a+b)^2$ as in proving Pythagoras theorem and cut both squares by a diagonal.  Now observe that reflection of white triangles will cover the blue rectangles.  This proof can be stated in a purely combinatorial language, but why bother - it's a standard "book proof" you can find everywhere.  Similarly, you can take:
$$\sum_{i=1}^n \sum_{j=1}^n \left(a_i b_j + a_j b_i \right)^2$$
expand all the terms and get a similar flavor "combinatorial proof", more or less the same what you can find in standard books.  But again why bother?
P.S.  If you want to see the real power of bijective proofs, you can check out my survey of partition bijections.  Sorry for the self-promotion, but there is no other such large compendium (although Stanley's "Enumerative Combinatorics" does mention many beautiful bijections, many of which hidden in the solutions to exercises).<|endoftext|>
TITLE: Comparing two similar procedures for quantizing a Casimir Lie algebra
QUESTION [11 upvotes]: My primary reference for this question is the very good book Quantum Groups and Knot Invariants by C. Kassel, M. Rosso, and V. Turaev.  I'm also drawing from P. Etingof and O. Schiffmann, Lectures on quantum groups, another very good book.  If you want to see pictures of Lie bialgebras and quasitriangular structures, I will shamelessly self-promote my notes on Lie bialgebras, which I put together while studying for my quals last year.
Pick your favorite field of characteristic $0$, and let $\mathfrak g$ be a finite-dimensional Lie algebra.  Abusing the language a bit, let me define a (quadratic) Casimir to be any symmetric $\mathfrak g$-invariant element $t\in \mathfrak g^{\otimes 2}$.  (Equivalently, $t$ is symmetric and its image in the universal enveloping algebra $\mathcal U\mathfrak g$ is central.)  I'd like to compare two constructions of "quantum" categories that start with this Lie theoretic data.
Quantization of infinitesimally-braided categories
So pick a quadratic Casimir $t$.  Let $\mathcal S$ be the category of finite-dimensional $\mathfrak g$-modules.  For any two modules $(\pi_U,U)$ and $(\pi_V,V) \in \mathcal S$, we have an element $t\_{U,V} \in {\rm End}_{\mathfrak g}(U\otimes V)$ given by $t_{U,V} = (\pi_U \otimes \pi_V)(t)$; it is a $\mathfrak g$-morphism because $t$ is $\mathfrak g$-invariant.  Moreover, under the canonical "flip" map $\sigma_{U,V}: U\otimes V \to V\otimes U$, $t_{U,V}$ maps to $\sigma_{U,V}t_{U,V}\sigma_{V,U} = t_{V,U}$, because $t$ is a symmetric element of $\mathfrak g^{\otimes 2}$.  Along with the rule for how $\mathfrak g$ acts on tensor products (determining the action of $t_{U,V\otimes W}$ on $U\otimes V\otimes W$), it follows that $t$ defines on $\mathcal S$ the structure of a infinitesimally braided category.
Then recall the following very general construction.  Let $\Phi$ be a Drinfeld associator: i.e., $\Phi(A,B)$ is a formal power series in non-commuting variables of the form $\exp(\text{a Lie series})$, satisfying certain nonlinear conditions (a "pentagon" and two "hexagon"s) that make the following construction work.  (Only one Drinfeld associator is explicitly known, given by the solution to the Knizhnik–Zamolodchikov differential equation, and its coefficients are real but transcendental.  But the equations defining $\Phi$ at each order are an over-determined linear system in rational coefficients, so if any solution exists, a rational one does.  And by a theorem of Le and Murakami, up to equivalence of braided monoidal categories, the output of this construction does not depend on the choice of associator.)
Then define a new category $\mathcal S[[\hbar]]$.  The objects are the same as those of $\mathcal S$, and $\hom_{\mathcal S[[\hbar]]}(V,W) = \hom_{\mathcal S}(V,W)[[\hbar]]$ (formal power series of morphisms) with composition given simply by the composition in $\mathcal S$ extended $\hbar$-linearly (and adicly), i.e., following the rule for multiplication of formal power series.
Moreover, give $\mathcal S[[\hbar]]$ the tensor product inherited from $\mathcal S$.  However, give it nontrivial associativity and braiding constraints.  Namely, define the associator by $a_{123} = \Phi(\hbar t_{12}, \hbar t_{23})$ and $c = \sigma \exp(\hbar t / 2) = \exp(\hbar t / 2)\sigma$.  The axioms for the Drinfeld associator $\Phi$ imply that $\mathcal S[[\hbar]]$ is a (weak) braided monoidal category.
Quantization of quasitriangular Lie bialgebras
Again let's start with $t$ a quadratic Casimir.  But let's suppose a bit more: let's suppose that $t$ is the symmetrization of some element $r\in \mathfrak g^{\otimes 2}$ satisfying the (very over-determined) classical Yang-Baxter equation (CYBE).  Namely, let $\beta: \mathfrak g^{\otimes 2} \to \mathfrak g$ be the Lie bracket, and use the obvious index notation.  Then the CYBE says:
$$ \bigl[(\beta_{13} \otimes \mathrm{id}_2 \otimes \mathrm{id}_4) + (\mathrm{id}_1 \otimes \beta_{23} \otimes \mathrm{id}_4) + (\mathrm{id}_1 \otimes \mathrm{id}_3 \otimes \beta_{24}) \bigr] (r_{12}\otimes r_{34}) = 0.$$
A choice of such an $r$ is a quasitriangular structure on $\mathfrak g$.  Consider the map $\delta: \mathfrak g \to \mathfrak g^{\wedge 2}$ given by antisymmetrizing the output of $x \mapsto \mathrm{ad}_x(r)$, where $\mathrm{ad}_x$ is the action of $x\in \mathfrak g$ on $\mathfrak g^{\otimes 2}$.  It follows from the CYBE and the fact that $t$ is $\mathfrak g$-invariant that $\delta$ satisfies the co-Jacobi identity.
Then by a general theorem of Etingof and Kazhdan, the data $(\mathfrak g,r)$ as above along with a choice of Drinfeld associator $\Phi$ determines a (noncommutative, noncocommutative) Hopf algebra $\mathcal U_\hbar \mathfrak g$ (over power series in $\hbar$) and a nontrivial braiding on the (strongly-associative monoidal) category of (finitely-generated topologically free) $\mathcal U_\hbar\mathfrak g$-modules.  As an algebra, $\mathcal U_\hbar\mathfrak g \cong \mathcal U\mathfrak g[[\hbar]]$, I think, so, as a category (although maybe not as a braided monoidal category?), $\mathcal S[[\hbar]] \cong \text{$(\mathcal U_\hbar\mathfrak g)$-mod}$.
Questions
The two constructions above are similar: both start with a Lie algebra and a Casimir, and both end up with braided monoidal categories.  But they're not the same.  The first construction had to deform the associator to make everything work out, whereas in the second the associator is what I called strongly associative: the category embeds naturally in the category of vector spaces, and the associator is the same as the essentially-trivial associator there that we all know and love.  (So it's almost "strictly associative"; "strong" seems like a good antonym for "weak".)  On the other hand, the second construction required more data: it required choosing a classical $r$-matrix.
So: how related are these two constructions?  For example, suppose I run the first construction, but happen to know that $t$ comes from an $r$-matrix.  Does this tell me anything about more about the structure of $\mathcal S[[\hbar]]$?  Conversely, if I take the first construction, and then do some Mac Lane-style strictifying, how close can I get to the second construction?

REPLY [7 votes]: The second construction (Lie bialgebra quantization) in fact also uses a Drinfeld associator. The braided tensor categories obtained in these two ways are equivalent, since the quasitriangular QUE algebra produced by the second construction is obtained by twisting the quasitriangular quasiHopf QUE algebra produced by the first construction. The construction of this twist (which we call $J$) from the Drinfeld associator is in fact the main construction of my paper with Kazhdan "Quantization of Lie bialgebras, I". Categorically, this twist $J$ provides a tensor structure on the forgetful functor on the category produced by the first construction, and the endomorphism algebra of this tensor functor is the Hopf algebra produced by the second construction. Such a tensor functor exists once you find a classical r-matrix r such that $r+r_{21}=t$ (in fact, such functors bijectively correspond to such classical r-matrices over $k[[h]]$, up to isomorphism). 
I'd like to add two comments. 

If you want the braiding to be $e^{ht/2}$ then the KZ associator $\Phi_{KZ}$ will not be real, since the KZ equation will have an overall coefficient $1/2\pi i$. So another 
associator is the complex conjugate KZ associator $\overline{\Phi_{KZ}}$. 
There is now also a third "explicitly" known associator - the Alexeev-Torossian associator,
see e.g. arXiv:0905.1789, arXiv:0906.0187. This associator is indeed real and depends on $h^2$ (i.e., is even). I wonder if it is the "midpoint" between the KZ associator and its conjugate (the notion of a midpoint on the space of associators makes sense, since the space of associators, according to Drinfeld, has a free transitive action of the Grothendieck-Teichmuller group $GT_1$, which is prounipotent.) This midpoint is also real and depends on $h^2$. 
I think the independence of the category in the first construction on the choice of associator is due to Drinfeld, before Le and Murakami; he proves in his paper "Quasi-Hopf algebras" that the associator for any Casimir is unique up to twisting.<|endoftext|>
TITLE: Can models of set theory contain extra ordinals?
QUESTION [7 upvotes]: In the paper "Complete topoi representing models of set theory" by Blass and Scedrov, they consider a general notion of Boolean-valued model of set theory, and one of the conditions they impose is that the model contain "no extra ordinals after those of V", i.e. that for all z in the model we have
$$\Vert z \text{ is an ordinal} \Vert = \bigvee_{\alpha \text{ is an ordinal of } V} \Vert z=\check{\alpha}\Vert $$
where $\Vert-\Vert$ denotes the truth function of the model valued in some complete Boolean algebra.
My question is: do there exist models which do contain "extra ordinals" in this sense?  I presume so, or they wouldn't have needed to impose this condition.  What do such models look like?
(By way of clarification, certainly if the starting model V is a set model in some larger universe, then one can find other set models in that larger universe which contain more ordinals.  But I'm interested in just starting with a single model V and building models from it, which can be sets or proper classes.)

REPLY [6 votes]: As Joel pointed out, that requirement is true for internal Boolean valued models. It is also true for symmetric models, permutation models (Blass & Scedrov allow atoms), and a variety of mixed constructions. Blass and Scedrov were simply isolating the key common features of these constructions.
That said, there are ways to violate this condition. By an earlier requirement by Blass and Scedrov on canonical names, this would require the Boolean valued model to be an end-extension of V. For example, an old result of Keisler and Morley says that every countable models of ZF has an elementary end-extension. There is no internal way to construct such things in plain ZFC since that would easily violate Gödel's Theorem, but it is possible to have decent approximations assuming large cardinals.
There is an interesting construction by Sy Friedman where he essentially forces with a poset of size Ord+. (You have to jump through several hoops to do this, see Chapter 5 of Sy Friedman's book Fine Structure and Class Forcing.) Let loose, this forcing would naturally violate the no new ordinals condition. However, Friedman is careful to cut down the model so that no new ordinals appear.

Here are a few additional remarks on a potential construction that comes very close to what you want.
Other than the fact that we just ran out of ordinals, there is no real reason to stop the usual construction of L at Ord. We can construct LOrd+1 in the same way except that we can't replace definitions for classes by actual sets. Nevertheless, we can define elements of LOrd+1 to be (Gödel codes for) one-variable formulas with ordinal parameters, identifying formulas that define the same class. This is tricky to do internally since we have no truth definition, but we can work around that using 0#, assuming it exists in V. Once we have LOrd+1, we can similarly construct LOrd+2. In fact, with large enough cardinals, we can keep going like this for quite a while. There is no reason we will ever hit a model of ZF in this way, but I don't think it's impossible. To accomplish this, you need to have L live inside a model V where all the relevant information is packed into a set. This is where Friedman (building on seminal work of Jensen and others) picks up. His methods in Chapter 5 suggest that you could pack all of the relevant information into a single real number. Unfortunately, this is also where I stop, opening Friedman's book at a random page should explain why...<|endoftext|>
TITLE: Primes P such that ((P-1)/2)!=1 mod P
QUESTION [39 upvotes]: I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this
implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$. 
Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?
After convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS. I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$).
Thanks,
Jacob

REPLY [5 votes]: Apologies for repeating some information in my reply to question 121678, which I came across before seeing this one.
Several previous answers already explain the connection to the class number. It can be added that the value of $h(-p)$ was investigated by Louis C. Karpinski in his doctoral dissertation (Mathematischen und Naturwissenschaftlichen Facultät der Kaiser Wilhelms-Universität zu Strassburg, 1903), published as “Über die Verteilung der quadratischen Reste,” Journal für die Reine und Angewandte Mathematik 127 (1904): 1–19. Karpinski proved a collection of formulae (all of which assume $p > 3$) involving sums over Legendre symbols, and showed that the most concise sums possible contain only $\lfloor p/6 \rfloor$ terms:
\begin{equation}
\left\{ 2 - \left( \frac{2}{p} \right) \right\} h(-p) = \sum_{k=1}^{(p-1)/2} \left( \frac{k}{p} \right) \quad (p \equiv 3 \bmod{4});
\end{equation}
\begin{equation}
\left\{ 3 - \left( \frac{3}{p} \right) \right\} h(-p) = 2 \sum_{k=1}^{\lfloor p/3 \rfloor} \left( \frac{k}{p} \right) \quad (p \equiv 3 \bmod{4});
\end{equation}
\begin{equation}
\left\{ 2 - \left( \frac{2}{p} \right) \right\} h(-p) = \sum_{k=\lfloor p/4 \rfloor +1}^{(p-1)/2} \left( \frac{k}{p} \right) (p \equiv 3 \bmod{8});
\end{equation}
\begin{equation}
\left\{ 2 - \left( \frac{2}{p} \right) \right\} h(-p) = \quad \sum_{k=1}^{\lfloor p/4 \rfloor} \quad \left( \frac{k}{p} \right) (p \equiv 7 \bmod{8});
\end{equation}
\begin{equation}
\left\{ 1 + \left( \frac{2}{p} \right) + \left( \frac{3}{p} \right) - \left( \frac{6}{p} \right) \right\} h(-p) =  2 \sum_{k=1}^{\lfloor p/6 \rfloor} \left( \frac{k}{p} \right) \quad (p \equiv 7, 11, 23 \bmod{24});
\end{equation}
\begin{equation}
\left\{ 1 + \left( \frac{2}{p} \right) + \left( \frac{3}{p} \right) - \left( \frac{6}{p} \right) \right\} h(-p) = -2p  + 2 \sum_{k=1}^{\lfloor p/6 \rfloor} \left( \frac{k}{p} \right) \quad (p \equiv 19 \bmod{24}).
\end{equation}<|endoftext|>
TITLE: What is the difference between PSL_2 and PGL_2?
QUESTION [34 upvotes]: Let $K$ be a field and $G:=SL_2(K)$, then $G$ is a $K-$split reductive group (to use some big words). These groups are classified by a based root datum $(X,D,X',D')$. Let $G'$ be group associated to $(X',D',X,D)$, the so called dual group.
Is it correct that $G'=PGL_2(K)$? 
I"m wondering how $PSL_2(K)$ fits into this picture.
I'm aware of the fact that if C is algebraically closed, then $PSL_2(C) \cong PGL_2(C)$ as abstract groups; can this be made into an isomorphism of algebraic groups, i.e. is $PSL$ a $K-$form of $PGL$?

REPLY [30 votes]: As Kevin says, the "right" definition of ${\rm{PSL}}_n$ is as representing the quotient sheaf ${\rm{SL}}_n/\mu_n$, just as one defines ${\rm{PSO}}(q) = {\rm{SO}}(q)/Z_{{\rm{SO}}(q)}$ (with $Z_G$ denoting the scheme-theoretic center of a smooth group $G$).  So really, there is no difference between ${\rm{PSL}}_n$ and ${\rm{PGL}}_n$ when defined correctly, and likewise ${\rm{PSO}}(q) = {\rm{PGO}}(q)$.   Personally, I avoid the notation ${\rm{PSL}}_n$ like the plague, since it creates too much confusion.  
Lest this seem like a flippant answer, let me point out that for a general ring $R$ with nontrivial Picard group, it is likewise not true that ${\rm{PGL}}_n(R)$ is the "naive" thing either! For example, if $R$ is a Dedekind domain whose Picard group has nontrivial 2-torsion then ${\rm{PGL}} _2(R)$ is generally bigger than 
${\rm{GL}} _2(R)/R^{\times}$. And this is not a quirk with algebraic geometry.  The same thing happens with Lie groups:  if a smooth manifold $M$ has nontrivial 2-torsion line bundles it can and does happen that there are $C^{\infty}$ maps $f:M \rightarrow {\rm{PGL}} _2(\mathbf{R})$ which do not arise from a map to ${\rm{GL}} _2(\mathbf{R})$ (concretely, pulling back the quotient map ${\rm{GL}} _2(\mathbf{R}) \rightarrow {\rm{PGL}} _2(
\mathbf{R})$ along $f$ yields a line bundle on $M$ that may be non-trivial). 
And the "weirdness" of it all (based on experience over an algebraically closed field) is also seen by the fact that the concrete definition of the group scheme ${\rm{PGL}}_n$ is as a basic affine open in the projective space of $n \times n$ matrices, and we know that "points" of projective spaces are a subtle thing (compared with the case of geometric points) when the source has nontrivial line bundles (whether a scheme or manifold). 
Since one cannot get by with field-valued points when doing representability arguments, the same "problem" which one sees for the naive viewpoint on ${\rm{PSL}} _n$ is also relevant when doing proofs for ${\rm{PGL}} _n$. The difference is that for the latter one has to work more "globally" to see the surprise because the degree-1 Zariski and fppf cohomologies for $\mathbf{G} _m$ coincide (so for local rings nothing funny happens, as they have vanishing higher Zariski sheaf cohomology) whereas for the former there is already a funny thing happening for local rings and even fields (which can have nontrivial degree-1 fppf cohomology for $\mu_n$). In more concrete terms, it is equivalent to define the functor ${\rm{PGL}} _n$ as a quotient sheaf for either the Zariski or fppf topologies, whereas for ${\rm{PSL}} _n$ one has to sheafify for the fppf topology (etale ok when $n$ is a unit on the base), and in either case the naive functor on rings (inspired by the case of algebraically closed fields) is not even a Zariski sheaf and hence beyond local rings something has to be done to get the right functor (e.g., one that is representable). 
Many books on linear algebraic groups use a version of algebraic geometry that is not well-suited to the subtleties of quotient considerations (e.g., Borel uses Serre's clever method "quotient by $p$-Lie algebra" to handle quotients by infinitesimal groups without saying "infinitesimal group", and some of his quotient arguments would be much shorter if he could have used flatness systematically). In particular, Springer's book has some serious errors when the ground field is not algebraically closed (in the later parts, where he discusses $F$-reductive groups and related things).  For example, he uses the incorrect argument that surjectivity of an $F$-map between smooth $F$-varieties can be checked on $F_s$-points, which is not true when $F$ is not perfect (remove an inseparable point from the affine line and consider the inclusion into the line).  So be careful in that part of his book.  (Some statements are false, not just proofs.)<|endoftext|>
TITLE: Elementary proof that projective space is a quotient
QUESTION [7 upvotes]: Fix an algebraically closed † ground field $k$ of any characteristic.  I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.
I'm not satisfied with my understanding of why this is the same as the quotient topology from the map $\mathbb{A}^{n+1}\setminus 0\to \mathbb{P}^{n}$.  This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear.  The last part is the least trivial, meaning that:

If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f_1,\ldots,f_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some homogeneous polynomials $g_1,\ldots,g_s$.

I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.
Thanks for the help, anyone!
† Follow up: the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.

REPLY [7 votes]: Let $f$ be a polynomial which vanishes on $\hat{S}$. Write $f=\sum f_i$, where $f_i$ is homogenous of degree $i$. The set $\hat{S}$ is homogenous so, for any $\lambda \in k^*$, the polynomial $f(\lambda \cdot x) = \sum \lambda^i f_i$ also vanishes on $f$.
Since $k$ is infinite, we can find more equations of the form $\sum \lambda^i f_i=0$ than there are terms in $f$. Taking linear combinations, we deduce that each $f_i$ individually vanishes on $\hat{S}$.

REPLY [4 votes]: Look at the subspace of $\mathbf{A}^{n+1}$ cut out by your polynomials. This set is invariant under the diagonal action of $k^\times$. So the functions that vanish on it will be an ideal $I$ (the radical of the ideal generated by your $f_i$) which is invariant under this action. But because $k$ is infinite, it's now relatively easy to check that if $f\in I$ and if you write $f$ as a sum of homogeneous pieces, then each homogeneous piece will be in $I$ too (the point is that the homogeneous pieces scale differently under $k^\times$ so if the degree of f is $d$ then you just use $d+1$ elements of $k^\times$ and then put a linear combination of them together and use the fact that a certain Van der Monde determinant isn't zero). Hence $I$ is generated by homogeneous elements and now you're home, unless I missed something.<|endoftext|>
TITLE: Is there a tournament schedule for 18 players, 17 rounds in groups of 6, which is balanced in pairs?
QUESTION [13 upvotes]: We are interested in a solution to the following scheduling problem, or any information about how to find it or its existence. This one comes from real life, so you will not only be helping a mathematician quench his thirst of knowledge!

We have 18 players playing a certain sport (let's say curling) on 3 different alleys (6 players per alley) at the same time. They play 17 games and we want that every combination of 2 players play exactly 5 times together.

(As Douglas Zare points out in a comment below, this is known as a resolvable block design with t=2, v=18, k=6, lambda=5 (and b=51, and r=17)).
We asked around and someone came up with a near solution: almost every pair playing 5 times except for a few 6's and 4's. Brute force seemed too slow so we tried with a genetic algorithm, to no avail (being complete beginners in this, we could not even get close to the near-solution that we had, so we do not draw conclusions from our experiments).
I found the near-solution in my old files, in case anyone wants to tinker a bit.

{{1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18}},
{{1, 6, 10, 12, 14, 16}, {2, 3, 8, 11, 15, 17}, {4, 5, 7, 9, 13, 18}},
{{1, 5, 7, 8, 15, 16}, {2, 4, 10, 11, 13, 14}, {3, 6, 9, 12, 17, 18}},
{{1, 4, 8, 9, 14, 17}, {2, 6, 7, 10, 15, 18}, {3, 5, 11, 12, 13, 16}},
{{1, 6, 8, 11, 13, 18}, {2, 4, 9, 12, 15, 16}, {3, 5, 7, 10, 14, 17}},
{{1, 2, 7, 12, 13, 17}, {3, 4, 8, 10, 16, 18}, {5, 6, 9, 11, 14, 15}},
{{1, 3, 9, 10, 13, 15}, {2, 5, 8, 12, 14, 18}, {4, 6, 7, 11, 16, 17}},
{{1, 5, 10, 11, 17, 18}, {2, 6, 8, 9, 13, 16}, {3, 4, 7, 12, 14, 15}},
{{1, 2, 9, 11, 16, 18}, {3, 6, 7, 8, 13, 14}, {4, 5, 10, 12, 15, 17}},
{{1, 4, 8, 12, 15, 18}, {2, 3, 7, 9, 11, 14}, {5, 6, 10, 13, 16, 17}},
{{1, 3, 7, 14, 16, 18}, {2, 5, 8, 9, 10, 17}, {4, 6, 11, 12, 13, 15}},
{{1, 5, 6, 9, 12, 14}, {2, 3, 10, 13, 15, 18}, {4, 7, 8, 11, 16, 17}},
{{1, 3, 10, 11, 12, 16}, {2, 4, 5, 8, 13, 14}, {6, 7, 9, 15, 17, 18}},
{{1, 2, 3, 4, 6, 17}, {5, 7, 11, 12, 13, 18}, {8, 9, 10, 14, 15, 16}},
{{1, 4, 7, 9, 10, 13}, {2, 12, 14, 16, 17, 18}, {3, 5, 6, 8, 11, 15}},
{{1, 2, 5, 7, 15, 16}, {3, 8, 9, 12, 13, 17}, {4, 6, 10, 11, 14, 18}},
{{1, 11, 13, 14, 15, 17}, {2, 6, 7, 8, 10, 12}, {3, 4, 5, 9, 16, 18}}

REPLY [3 votes]: Others have posted that a resolvable block design will help answer the question.  If you want to chew up some computer cycles, consider the following approach.
There are 122 ways to divide a set of 6 elements into at most 3 sets.  Consider how a
solution to your problem looks on the first 6 elements:  each of the 17 sessions
produces a division of the 6 set in one of the 122 ways.  If the first session has
6 individuals together, then each of the remaining sessions produces one of the
122 partitions mentioned above.  Because of the restriction that each pair occurs
in exactly five of the sessions, in the remaining 16 sessions, you will have no more than
four instances of the same partition.  (If you do an analysis similar to one below
based on pairs, you will find that at most one more instance of the trivial partition will be allowed.)
Now generate a list of combinations of partitions that can occur while keeping the
restriction on pairs.  This involves choosing 16 items from a multiset of (at most) 488 
partitions, many of which will be imadmissible because of the pair restriction.
If we look at how many pairs are made by a partition, we have the trivial partition
making 15 pairs, 6 others making 10 pairs, 15 others making 7 pairs, 25 others making 6 pairs, on down to 15 making 3 pairs.  Since you want 60 pairs from the remaining 16 partitions, it will happen that you will need at least 4 partitions which make 3 pairs, at most 3 partitions which make 7 or more pairs, and at most 4 partitions which will make 6 or more pairs.  So of the 16 items chosen from the above multiset, at most 4 will come from a multiset of at most 204 elements, at least 4 will come from a multiset of 60 elements, and the remainder will still have some restrictions on it.  The idea is that a simple algorithm can check a partial combination and quickly weed it out, so that your search space is much smaller than (488 choose 16) items.
An exhaustive list of all such combinations of 16 partitions may be small, or it may be large; for the next step, I recommend starting with a not too large sublist:  Pick 3 candidates from the sublist and see if they can be "stitched" together.  As an example, suppose I choose a combination of 16 partitions, one of those which is the trivial partition.  Then in order to stitch a solution together, I need another combination which has a partition with at most two parts, because I can't use a combination in which all 16 items have 3 or more parts.  
As you attempt a stitching, you can see which attempts violate the condition of producing more than 5 pairs among the twelve or 18 elements used for the stitching.  As a subexercise, consider the combination consisting of the trivial partition followed by each of the 15 partitions of 6 elements into 2 element sets.  See if you can stitch 3 copies of that combination together.
Gerhard "Ask Me About System Design" Paseman, 2010.02.23<|endoftext|>
TITLE: Formalizing "no junk, no confusion"
QUESTION [7 upvotes]: Goguen has popularized the initial algebra view of semantics via his "no junk, no confusion" slogan.  By "no junk", he means that models of a theory presentation should not have unnecessary elements, and "no confusion" that terms should not be mapped to equal values unless they are provably equal.  Sometimes, "no junk" is also interpreted as every element in the model is a denotation of a term, while "no confusion" as two different terms denote different elements in the model.  [These are classically equivalent statements, but they are not intuinistically equivalent, so I mention both].
My questions are:

What is a 'good' formalization of this slogan?  By this I mean an explicit statement of "no junk, no confusion" in the meta-logic (since we're talking about models), where the logical strength of the corresponding statement is well understood.
Are there logics in which these requirements can be internalized?
What would be the corresponding slogan to "no junk, no confusion" for final coalgebras?

REPLY [3 votes]: Belatedly, an answer in set-based situations to

What would be the corresponding slogan to "no junk, no confusion" for final coalgebras?

Given an initial algebra, any algebra will have a special subobject which is the image of the initial structure. The subobject may have confused elements (terms) of the initial algebra, and the object may have extra junk. A map between objects will map the first special subobject to the second, possibly confusing more. The initial algebra has no confusion and no junk.
Given a final/terminal coalgebra, the elements of any coalgebra will have a special colouring in terms of images in the elements of the terminal structure. The object may have more than one element with the same colour, and the object may not use all the colours. A map between objects will preserve the colouring, the domain possibly using fewer colours. The terminal coalgebra colours without ambiguity and without redundancy. If two things behave the same way, they are the same; all behaviours are covered.
No junk, no confusion; No redundancy, no ambiguity.<|endoftext|>
TITLE: $(\infty,1)$-categories and model categories
QUESTION [14 upvotes]: I read several times that $(\infty,1)$-categories (weak Kan complexes, special simplicial sets) are a generalization of the concept of model categories. What does this mean? Can one associate an $(\infty,1)$-category to a model category without losing the information on the co/fibrations? How? Why is the $(\infty,1)$-category viewpoint the better one?
$(\infty,1)$-categories are equivalent to simplicial categories (categories enriched over simplicial sets). This is outlined in Lurie's higher topoi. A simplicial model category is in particular a simplicial category. Is this the way the association works? Every model category is Quillen equivalent to a simplicial model category and can thus be enriched over simplicial sets.
It would be nice if somebody could help me to clarify this.
Edit: Thank you all for the answers. It seems to me that $(\infty,1)$-categories are not a generalization of the concept of model categories. A model category is more than a category $C$ together with a class of maps $W$ such that $C[W^{-1}]$ is a category. A model structure data on a category $C$ contains the information on what a cofibration and what a fibration is. This is important for the structure. There exist different model structures with the same homotopy category as for example model structures on functor categories.
This means that if there is a kind of functor
$$
F: \{\mbox{model categories}\} \to \{\mbox{($\infty,1)$-categories}\}
$$
it is at least not an embedding. In spite of the answers, I still don't see how this functor (if it is really a functor) works. Where is a model category mapped to?

REPLY [4 votes]: The functor $F$ you are looking for can be described as follows. Given a model category $M$, with class of weak equivalences $W$, one may associate to it an $\infty$-category $M_\infty$, equipped with a map $M \to  M_\infty$, which is characterized by the following universal property: For every $\infty$-category $D$, the natural map
$$Fun(M_\infty,D) \to Fun(M,D)$$
is fully faithful, and its essential image is spanned by those functors $M \to D$ which send $W$ to equivalences. 
The $\infty$-category $M_\infty$ may be constructed follows: Construct the Hammock localization $L^H(M,W)$ of $M$ with respect to $W$, which is a simplicial category. The $\infty$-category $M_\infty$ can then be obtained by taking the coherent nerve of any fibrant model of $L^H(M,W)$ (with respect to the Bergner model structure).
We refer to
http://arxiv.org/abs/1311.4128
for the above universal property. 
Note that the above construction works for any relative category, that is, a category equipped with a subcategory of weak equivalences containing all the objects. Let us call this procedure $\infty$-localization.  
Morphisms of model categories are given by Quillen pairs:
$$F:M\rightleftarrows N:R.$$
According to  Proposition 1.5.1 in the paper above, every Quillen pair as above induces an adjoint pair of $\infty$-categories:
$$F_\infty:M_\infty\rightleftarrows N_\infty:R_\infty.$$
It can be shown that a Quillen equivalence goes to an equivalence of $\infty$-categories by this construction.
So consider the category of model categories and left Quillen functors between them. This is actually a relative category, where the weak equivalences are the Quillen equivalences. The above construction defines a relative functor from this relative category to the relative category of $\infty$-categories (weak Kan-complexes and simplicial maps between them) with the weak equivalences being the equivalences of $\infty$-categories. This relative functor should be a derived  fully faithful embedding of the $\infty$-category of model categories (in the sense of the $\infty$-localization above) into the $\infty$-category of $\infty$-categories and left adjoints between them. (Perhaps there are some delicate technical issues here that I am suppressing.)<|endoftext|>
TITLE: Binomial distribution parity
QUESTION [5 upvotes]: Let $X \text{~} \text{Binomial}(n, p)$.
What is $\text{P}[X \mod 2 = 0]$? Is it of the form $1/2 + O(1/2^n)$?

REPLY [5 votes]: The effect of adding a Bernoulli$(p)$ on the parity applies this matrix 
$$M=\left(\begin{array}{cc} 1-p & p \\\ 
p & 1-p \end{array}\right)$$
to $(prob(even), prob(odd))$.
We can decompose the initial vector of $(1,0)$ for Binomial$(0,p)$  as the sum of the eigenvectors $(1/2,1/2)$ and $(1/2,-1/2)$ with eigenvalues $1$ and $1-2p$, respectively. So, the parity vector of Binomial$(n,p)$ is 
$(1,0)M^n = 1^n (1/2,1/2) + (1-2p)^n (1/2,-1/2)$.
$1/2 + (1-2p)^n /2$ even, $1/2 - (1-2p)^n / 2$ odd.<|endoftext|>
TITLE: Value of "of course" in the mathematical literature
QUESTION [27 upvotes]: I've been thinking about the value of writing "of course" in mathematical papers (or its variants such as "obviously" etc.).  In particular, my current train of thought is, if something is obvious, then it is obvious that it is obvious (so why include it at all?).
The example that inspired this post is:  If d divides a and d divides b, then of course d also divides a+b.

Are there examples in the mathematical literature where the term "of course" is of value?

More precisely, I'm after an example (or a few), ideally by a well-known author, where "of course" or "obviously" or similar actually adds tangible value to a sentence (rather than just implying: (a) it's obvious to me, I'm so smart or (b) I can't actually be bothered working out the details)

REPLY [8 votes]: "of course", "clearly", "obviously", and other like interjections are a form of higher-order punctuation. While strictly speaking they're eliminable, good authors use them as rhythmic elements, cognitive breaths to delimit chunks of an argument. Yes they are "noise words" and phrases, but they are the ones we all know and love (in English) so probably it's best not to press this point too hard. :)<|endoftext|>
TITLE: Why can the Dolbeault Operators be Realised as Lie Algebra Actions
QUESTION [6 upvotes]: I've been looking at an example in the non-commutative geometry literature and I'm having trouble figuring out what the classical motivation is. I'll just describe the classical case here: Recall that $\mathbb{CP}^2 = SU(\mathbb{C},3)/U(\mathbb{C},2)$. Recall also that since $T_\mathbb{C}^*(\mathbb{CP}^2)$ can be viewed as a vector bundle associated to the $SU(2)$-bundle $SU(\mathbb{C},3) \to \mathbb{CP}^2$, we can view $\Omega_\mathbb{C}^1(\mathbb{CP}^2)$ as a subset  of  $\mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2)$. Now in the example, the action of each of the Dolbeault operators $\partial,\overline{\partial}$ is given in terms of two mappings, $\partial_1,\partial_2$ and $\overline{\partial_1},\overline{\partial_2}$, such that 
$$
\partial(f) = (\partial_1(f),\partial_2(f)) \in \mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2),
$$
and similarly for $\overline{\partial}$. The mappings ${\partial}_i, \overline{\partial}_i$ are constructed using an action of the Lie algebra $\mathfrak{su}(3)$ on $\mathcal{O}(SU(3))$ constructed using the canonical pairing between Lie algebras and coordinate algebras.
I have a feeling this is a complicated incarnation of a simple classical object. Does any of this ring any bells with anyone?
Please ask if you would like more details.
Edit: This question has been superseded by this question and I am voting to close. I would ask others to do likewise.

REPLY [3 votes]: The Dolbeault operators are usually defined in terms of the de Rham operator and the complex structure (see e.g. Wells' book or Griffith and Harris). The example you outline generalizes to the situation $G_{\mathbb{C}} / P = G / G_0$, where $G$ is compact, $G_{\mathbb{C}}$ is the complexification, $P$ is a parabolic subgroup, and $G_0 = G \cap P$. The holomorphic tangent bundle is the homogeneous vector bundle $G_{\mathbb{C}} \times_P \mathfrak{g}\_{\mathbb{C}} / \mathfrak{p}$, and the cotangent bundle is a homogeneous vector bundle in similar fashion.
In this case, the Dolbeault complex with coefficients in a homogeneous vector bundle $G_{\mathbb{C}} \times_P V$ translates to the Koszul complex for the relative
Lie algebra cohomology $H^*(\mathfrak{g},\mathfrak{g}\_0,V \otimes C^{\infty}(G))$. The Dolbeault operator $\overline{\partial}$ translates to the boundary operator for Lie algebra cohomology, which of course involves the action of $\mathfrak{g}$. I haven't worked out what
happens to $\partial$ in this situation, but most likely a similar expression in terms of the Lie algebra can be derived. The translation works by thinking about the smooth sections as elements of $C^{\infty}(G) \otimes V)^{\mathfrak{g}\_0}$ (for holomorphic sections use $C^{\infty}(G) \otimes V)^{\mathfrak{p}}$).<|endoftext|>
TITLE: Is there an associative metric on the non-negative reals?
QUESTION [34 upvotes]: Recall that a function $f\colon X\times X \to \mathbb{R}_{\ge 0}$ is a metric if it satisfies:

definiteness: $f(x,y) = 0$ iff $x=y$,
symmetry: $f(x,y)=f(y,x)$, and
the triangle inequality: $f(x,y) \le f(x,z) + f(z,y)$.

A function $f\colon X\times X \to X$ is associative if it satisfies:

associativity: $f(x,f(y,z)) = f(f(x,y),z)$.

If $X=\mathbb{R}_{\ge 0}$, then it might be possible for the same function to be a metric and associative.
Is there an associative  metric on the non-negative reals?
Note that these demands actually make $X$ into a group.  The element $0$ is the identity because $f(f(0,x),x) = f(0,f(x,x)) = f(0,0) = 0$ by associativity and definiteness, so again by definiteness $f(0,x) = x$.  Every element is its own inverse because $f(x,x)=0$.  
In fact, the following question is equivalent.  Is there an abelian group on the non-negative reals such that the group operation satisfies the triangle inequality?
Note also that the answer is yes if $X=\mathbb{N}$, the non-negative numbers!  Click here for a spoiler.
"Also known as nim-sum."
The question is originally due to John H. Conway.  To my knowledge, the question is unsolved even for $X = \mathbb{Q}_{\ge 0}$, but he does not seem to care about that case.  The spoiler above does extend to the non-negative dyadic rationals $\mathbb{N}[\frac 12]$, but apparently not to $\mathbb{N}[\frac 13]$.

REPLY [3 votes]: I'd like to present a proof without transfinite induction. As has been mentioned, such a commutative group is a vector space over $\mathbb{F}_2$. It seems we cannot write down a Hamel basis explicitly, which means this proof is not constructive either. But by the Axiom of Choice, a Hamel basis exists. Denote this basis by $\mathcal{B}$. By definition, each $x \in \mathbb{R_{\geq 0}}$ can be uniquely identified with a finite subset $X \subset \mathcal{B}$ such that $x = \sum X$. (Here and throughout we write $\sum X := \sum_{x \in X}x$ as a shorthand; read $\sum X$ as 'sum the elements of $X$').
Claim: Let $x = \sum X$ and $y = \sum Y$ such that $X,Y$ are finite subsets of the aforementioned Hamel basis $\mathcal{B}$. Then $f(x,y) = \sum X \triangle Y$ is an associative metric on $\mathbb{R}_{\geq 0}$ where $\triangle$ is the symmetric difference $X \triangle Y := (X \cup Y)\setminus (X \cap Y)$.
Proof: First we check associativity, which follows from the associativity of the symmetric difference: \begin{align*} f(f(x,y),z) &= f\left(\sum X \triangle Y, z\right) \\ &= \sum X \triangle Y \triangle Z \\ &= f\left(x, \sum Y \triangle Z\right) \\ &= f(x,f(y,z)).\end{align*}
To check the metric axioms is much quicker:

Definiteness: $f(x,x) = \sum X \triangle X = \sum \varnothing = 0$, and for the other direction $f(x,y) = 0 \implies X \triangle Y = \varnothing \implies X = Y \implies x=y$.
Symmetry: $f(x,y) = \sum X \triangle Y = \sum Y \triangle X = f(y,x)$.
Subadditivity: $f(x,y) = \sum X \triangle Y \leq \left(\sum X\right) + \left(\sum Y\right) = x+y$ which suffices.

And just like that, we are done. Note that although this answer is different to the accepted one, it also uses the Axiom of Choice. It would be interesting to know whether AC is not only sufficient but also necessary for such a metric to exist.


The question is unsolved even for $X = \mathbb{Q}_{\geq 0}$

Fortunately, not only is there such a metric, but it's also explicit! The Minkowski question-mark function is a continuous, strictly increasing bijection $?:\mathbb{Q}_{\geq 0} \to \mathbb{N}[\frac{1}{2}]$
given by
$$?(q) = a_0 + 2\sum_{k=1}^n \frac{(-1)^{k+1}}{2^{a_1 + \cdots + a_k}}$$ where $q = [a_0 ; a_1,...,a_n]$ is the finite continued fraction representation. Since the bitwise exclusive-or operation $\oplus$ is a well-defined metric on the dyadic rationals, the composition $$f(p,q) = ?^{-1}(?(p) \oplus ?(q))$$ is an associative metric for all $p,q \in \mathbb{Q}_{\geq 0}$.<|endoftext|>
TITLE: Where can I find online copies of class field theory publications by Kronecker, Weber, Chevalley, Hasse, Hilbert, Takagi, etc?
QUESTION [9 upvotes]: I am writing an undergraduate thesis on local and global class field theory from a classical (i.e., non-cohomological) approach and am hoping to obtain copies of the early groundbreaking publications in the field. I am primarily interested in finding English translations of  articles from Weber, Hasse, Hilbert, Kronecker, and Takagi from 1850 to around 1935 as possible. A fairly comprehensive list of them is found in Hasse's "History of Class Field Theory" in Cassels & Frohlich's ANT, and I can provide a list if needed. Any recommendations for sites that provide translated back issues of Mathematische Annalen and/or Gottinger Nachrichten from these time periods would be of great utility.
Edit:
Thank you all for responding! You're answers will definitely help me develop the historical portion of my thesis. In response to several comments, I realized when posting that it was probably naive to assume that English translations of all these works were available but presented the question in the above manner for the sake of brevity.
@KConrad:
I actually found your cfthistory.pdf file while googling the subject early in my research and have benefited from it greatly! Small world! :)
I would be fine with library copies of the works, of course, but my university would have to order copies from other institutions. As such, I thought I'd first make sure there wasn't some large repository of translated articles from Hilbert et al. online somewhere.
I can't thank you enough for this list of sources, too! I will definitely attempt to get my hands on as many as possible.
Thanks again for helping and posting your history of CFT online!
@Ben Lenowitz:
Thank you for pointing me to the Gottingen database. I haven't been in awhile and not while searching for these articles. I will give it a shot.

REPLY [4 votes]: Recently I found that some of Takagi articles were published in a Botany journal (?!) called "The journal of the College of Science, Imperial University of Tokyo, Japan".
Iyanaga, in his "The Theory of Numbers", cites three articles by Takagi published in this journal: 
[1] Über die im Bereiche der rationalen komplexen Zahlen Abelscher Zahlkörper, J. Coll. Sci. Tokyo 19 (1903), Article 5, 1-42.
[2] Über eine Theorie des relativ-Abelschen Zahlkörpers, J. Coll. Sci. Tokyo 41 (1920), Article 9, 1-133.
[3] Über das Reziprozitätsgesetz in einem beliebigen algebraischen Zahlkörper, J. Coll. Sci. Tokyo 44 (1922), Article 5, 1-50.
All three of them are available in the links above.
Actually, there are some nice illustrations in the Botany articles :)<|endoftext|>
TITLE: Spectra and localizations of the category of topological spaces
QUESTION [6 upvotes]: Can we construct the category of spectra (or maybe just its homotopy category) from the category of pointed topological spaces
using some kind of localization combined with other categorical constructions?
[The first part of the original question was wrong for a trivial reason pointed out by Reid Barton.]

REPLY [8 votes]: [Removed a paragraph relating to an earlier version of the question]
You can construct Spectra categorically by adjoining an inverse to the endofunctor Σ of Top as a presentable (∞,1)-category.  Inverting an endofunctor is a very different operation than inverting maps!  It's like the difference between forming ℤ[1/p] and ℤ/(p).
Here is one way to verify the claim.  To invert the endomorphism Σ of Top we should form the colimit, in the (∞,1)-category Pres of presentable categories and colimit-preserving functors, of the sequence Top → Top → ... where all the functors in the diagram are Σ.  A basic fact about Pres is that we can compute such a colimit by forming the diagram (on the opposite index category) formed by the right adjoints of these functors, and taking its limit as a diagram of underlying (∞,1)-categories [HTT 5.5.3.18].  The functors in the limit cone will have left adjoints which are the functors to the colimit in Pres.  In our case we obtain the sequence Top ← Top ← ... where the functors are Ω, and the limit of this sequence is precisely the classical definition of (Ω-)spectrum: a sequence of spaces Xn with equivalences Xn → ΩXn+1.<|endoftext|>
TITLE: What kinds of sets can replace lattices in the Shannon sampling theorem?
QUESTION [18 upvotes]: Background: The Shannon sampling theorem states that a bandlimited function (an $L^2$ function whose Fourier transform has compact support) can be determined uniquely from sampling it an integer lattice. More precisely, if the Fourier transform* $\hat{f}$ is supported in $[-\Omega/2, \Omega/2]$ (which we express by writing $f \in B_{\Omega}$) and $\tau>0$ satisfies $\Omega \tau \leq 1$, then $f$ can be reconstructed from the family $\{f(k\tau)\}_{k\in \mathbb Z}$.   This is proved more or less by periodizing and reducing to the theory of Fourier series.   A higher-dimensional analog using lattices in higher dimensions can be proved similarly.
Question:  What sets can these uniformly spaced lattices be replaced by?  Is there a general criterion, or some characterization of the collection of such sets (e.g. measure-theoretic)?

If $\Omega$ is fixed, then for what sets $E$ (say, countable), is the map $B_{\Omega} \to \mathbb{R}^E$ injective?

Motivation:  I overheard a conversation between a graduate student and a faculty member; the student wished to determine if one had a ``deformed lattice'' $\{ k\tau  + X_k \}$ where $\tau$ was a very small mesh relative to $\Omega$  and the $X_k$ was a sequence of reals, for what sequences $\{X_k\}$ would the samples $\{f(k\tau + X_k)\}$ determine $f \in B_{\Omega}$?  We could think of the $X_k$ as independent bounded random variables with a small variance, for instance.  He suspected (and wanted to know) whether the set of singular parameters (where this information was insufficient) would be a set of measure zero.  Intuitively, the small mesh size suggests something like this should be the case.
It was suggested that one might use some sort of finite-dimensional approximations to the full space $B_{\Omega}$ and taking finite subsets of the whole deformed lattice, and that the set of admissible $X_k$ would be some algebraic subvariety of proper codimension, therefore of measure zero, but it wasn't clear how to do this.  This led to my more general question boxed above.
Another reason this question may be of interest is that the sampling used in practical applications (e.g. CAT-scans) does not sample based upon a lattice pattern as in the Shannon theorem.
*I am using the normalization with the factor $2\pi$ for the Fourier transform.

REPLY [7 votes]: One sufficient condition is that, if $|X_k| <= L$ for some $L < 1/4$ then the sampling map would be an injection,  as well as a bounded map from $B$ to $l^2$.  This is known as Kadec's theorem.  
See 
Nonuniform Sampling and Reconstruction in Shift-Invariant Spaces, 
Akram Aldroubi and Karlheinz Gröchenig, 
SIAM Review, Vol. 43, No. 4 (Dec., 2001), pp. 585-620
for a survey and further references.<|endoftext|>
TITLE: Semi-linear operators
QUESTION [13 upvotes]: If $V_1$ and $V_2$ are finite-dimensional vector spaces over a field $E$, each equipped with an $E$-linear operator $\phi$, we can tell if $V_1$ and $V_2$ are isomorphic as $\phi$-modules by comparing the Jordan canonical form of the $\phi$-operator of each space.
If $E$ is itself equipped with a $\phi$-operator, and now each $V_i$ has a semi-linear operator $\phi$, is there some way (algorithm?) to determine whether or not $V_1$ is isomorphic to $V_2$ as a $\phi$-module?
(I would be content for an answer in the case when $E=F_p((T))$ and $\phi(f(T)) = f(T^p)$.)

REPLY [3 votes]: This question is really interesting. As Brian said, there is no doubt that things are very sensitive to the specific $E$ and $\phi$, and I'd be interested in knowing how things work for just about any example. Here are a few things that you might find helpful, if you don't know them already.
Suppose $\phi$ is an automorphism of $E$ of finite order $n$, and suppose that we look at $E$-vector spaces with semi-linear operators of order $n$. If we let $F$ denote the fixed field of $\phi$, then $E/F$ is a cyclic Galois extension of order $n$, and an $E$-vector space with a semi-linear operator $\phi$ of order $n$ is, by Galois descent, the same as an $F$-vector space. More precisely, given an $F$-vector space $N$, then $E\otimes_F N$ is an $E$-vector space with a semi-linear operator $\phi\otimes 1$ of order $n$; and given an $E$-vector space $M$ with semi-linear operator $\phi$ of order $n$, the subset of $\phi$-invariants is an $F$-vector space. These two functors induce an equivalence of categories.
But things seem to be much richer if we don't require the semi-linear endos of $M$ to be of finite order, even if the automorphism $\phi$ of $E$ does have finite order. For instance, Dieudonne-Manin theory mentioned by Brian.
A few months ago, I wondered a bit about the case where $E=\mathbf{C}$ and $\phi$ is complex conjugation. Then if we write $\phi$ as a matrix $A$ with respect to a basis, a change of basis matrix $B$ changes $A$ to $\bar{B}A B^{-1}$. (Even the case of 1-dimensional vector spaces is not completely trivial, though it's not hard.) I asked quite a few people, and eventually someone was able to tell me, after doing a bit of poking around himself, that this equivalence relation is called consimilarity by some people, and there's a paper (from 1988!) by Hong and Horn giving a Jordan theorem for it. It's called "A canonical form for matrices under consimilarity". (NB I haven't read it. The style didn't really appeal to me, and at the time, I didn't care enough to get beyond that.)<|endoftext|>
TITLE: What properties define open loci in excellent schemes?
QUESTION [17 upvotes]: Let $R$ be an excellent Noetherian ring. A property $P$ is said to be open if the set $\{q \in \operatorname{Spec}(R) \ | \ R_q \ \text{satisfies} \  (P)\}$ is Zariski open. Examples of open properties include:

Regular (well-known), complete intersections, Gorenstein, Cohen-Macaulay, Serre's condition $(S_n)$ (Matsumura's book). These imply openness of other properties, for example normality, which means $(R_1)$ and $(S_2)$.

Factorial (for $R$ of characteristic $0$, since the proof uses resolution of singularities). UPDATE: in a recent very interesting preprint, the factorial and $\mathbb Q$-factorial property are proved to be open for varieties over any algebraically closed field.

$\mathbb Q$-Gorenstein, i.e. the canonical module is torsion in the class group (I don't know a convenient reference, please provide if you happen to know one).

Being a rational singularity.


My questions are:
Question 1: Do you know other interesting class of open properties?
Question 2: Are there good heuristic reasons for why a certain property should be open? Phrased a bit more ambitiously, are there common techniques for proving openness for certain class of properties?
Some comments: The excellent condition is quite mild but crucial. Question 2 was motivated  by  another question of mine.

REPLY [9 votes]: to give a shamelessly trivial answer to your question 2, I would say that the point is usually that the failure of these properties is closed, often "obviously". I would also add one more auxiliary property, that acts as a meta property for some of these:
A coherent sheaf being locally free is an open property. This follows from Nakayama's lemma. 
So, here is your list. The properties on the left fail along the loci on the right.
[Caveat: I did not include conditions that are sometimes necessary, but I figured that this is a philosophical question and so the answer does not have to be stated in the most precise way.]
regular/smooth -------------- zero set of the Jacobian ideal, also the locus where $\Omega_X$ is not locally free
Cohen-Macaulay, $S_n$ ----- support of appropriate Ext sheaves
Gorenstein ------------------ {not CM} $\bigcup$ {CM but $\omega_X$ is not a line bundle}
$\mathbb Q$-Gorenstein --------------- $\bigcap_m$ {$\omega_X^{[m]}$ is not a line bundle}
rational singularity --------- (in char $0$) $\bigcup_{i=1}^{\dim X}{\rm supp\,}R^i\phi_*\mathcal O_{\widetilde X}$ where $\phi:\widetilde X\to X$ is a resolution.
klt singularity ---------------- zero set of the multiplier ideal.
Du Bois singularity --------- $\bigcup_{i\neq 0} {\rm supp\,} h^i(\underline\Omega_X^0)\bigcup \,{\rm supp\,}{\rm coker\,}[\mathcal O_X\to h^0(\underline\Omega_X^0)]$
(semi-)normality ------------ ${\rm supp\,}{\rm coker\,}[\mathcal O_X\to \pi_*\mathcal O_{Y}]$, where $\pi:Y\to X$ is the (semi-)normalization.
etcetera...<|endoftext|>
TITLE: Why is it useful to classify the vector bundles of a space?
QUESTION [32 upvotes]: It seems to me that vector bundles are useful because they allow us to bring to bear all of the linear algebra we know to aid in the study of topological spaces.  Now, I've read somewhere that it is an important and difficult problem to classify all of the vector bundles of a space.  I'm willing to accept that the problem is difficult, but why is it important? What are some applications of such a classification?

REPLY [4 votes]: I suppose a simpler answer to your question is that the classification of vector bundles is to vector bundles as Whitney's embedding theorem is to manifolds.  
Specifically, Whitney's weak embedding theorem says that all n-manifolds embed in $\mathbb R^{2n+1}$ and that any two embeddings of an $n$-manifold in $\mathbb R^{2n+2}$ are isotopic. So studying the "abstract" problem of $n$-manifolds up to diffeomorphism is equivalent to the "less abstract" problem of studying $n$-dimensional submanifolds of $\mathbb R^{2n+2}$ up to isotopy. 
Whitney's proof of the above fact is almost exactly how the proof of the classification of vector bundles works.  Moreover, they're philosophically almost identical proofs, as the Whitney theorem says abstract manifolds are submanifolds. The classification theorem says abstract vector bundles can be thought of as having their fibres in Euclidean space.<|endoftext|>
TITLE: Tetrahedra with prescribed face angles 
QUESTION [23 upvotes]: I am looking for an analogue for the following 2 dimensional fact:
Given 3 angles $\alpha,\beta,\gamma\in (0;\pi)$ there is always a triangle with these prescribed angles. It is spherical/euclidean/hyperbolic, iff the angle sum is smaller than /equal to/bigger than $\pi$. And the length of the sides (resp. their ratio in the Euclidean case) can be computed with the sine and cosine law.
The analogous problem in 3 dimensions would be:
Assign to each edge of a tetrahedron a number in $(0;\pi)$. Does there exists a tetrahedron with these numbers as face angles at those edges. And when is it spherical/euclidean/hyperbolic. Is there a similar Invariant to the angle sum?
And are there formulas to compute the length of the edges?

REPLY [10 votes]: The question seems a little confused, in particular since the OP is asking about dihedral angles but is calling them face angles. In any case, despite the gloom in the accepted answer, a lot is known.  In particular, the dihedral angle gram matrix comes from a Euclidean, hyperbolic, spherical tetrahedra if and only if the signature is $(0, 3)$ $(1, 3)$, $(4, 0)$ respectively. Further, in the Euclidean case, the Gauss map maps the tetrahedron onto the surface of the unit sphere. The triangles of the induced cell decomposition have sides equal to the exterior dihedral angles, and their areas can be computed using the spherical theorem of cosines. The exact analogue of the "sum of angles is $\pi$" relation is that the sum of the areas of these triangles is $4\pi.$ This is not a sufficient condition in this case. There are also side conditions to the effect that the sum of the face angles (which can be computed from the dihedral angles) of each face is $\pi.$
In the hyperbolic and spherical cases, the Gaussian image (see my thesis, or its write-up as Rivin-Hodgson, Inventiones) should be a spherical cone manifold, whose angles are either smaller (spherical) or bigger (hyperbolic) than $2\pi,$ plus, in the hyperbolci case, another condition on the length of the shortest closed geodesic (should be longer than $2\pi.$
For the generalization of the Gram matrix condition to arbitrary convex polyhedra, seem Raquel Diaz-Sanchez' thesis:
 A characterization of Gram matrices of polytopes
 R Díaz - Discrete & Computational Geometry, 1999 - Springer<|endoftext|>
TITLE: How to unify various reconstruction theorems (Gabriel-Rosenberg, Tannaka,Balmers)
QUESTION [46 upvotes]: What I am talking about are reconstruction theorems for commutative scheme and group from category. Let me elaborate a bit. (I am not an expert, if I made mistake, feel free to correct me)
Reconstruction of commutative schemes
Given a quasi compact and quasi separated commutative scheme $(X,O_{X})$ (actually, quasi compact is not necessary), we can reconstruct the scheme from $Qcoh(X)$ as category of quasi coherent sheaves on $(X,O_{X})$. This is Gabriel-Rosenberg theorem. Let me sketch the statement of this theorem which led to the question I want to ask:
The reconstruction can be taken as geometric realization of $Qcoh(X)$ (abelian category). Let $C_X$=$Qcoh(X)$. We define spectrum of $C_X$ and denote it by $Spec(X)$
(For example, if $C_X=R-mod$.where $R$ is commutative ring, then $Spec(X)$ coincides with prime spectrum $Spec(R)$). We can define Zariski topology on $Spec(X)$, we have open sets respect to Zariski topology. Then we have contravariant pseudo functor from category of Zariski open sets of the spectrum $Spec(X)$ to $Cat$,$U\rightarrow C_{X}/S_{U}$,where $U$ is Zariski open set of $Spec(X)$ and $S_U=\bigcap_{Q\in U}^{ }\hat{Q}$
(Note:$Spec(X)$ is a set of subcategories of $C_{X}$ satisfying some conditions, so here,$Q$ is subcategory belongs to open set $U$ and $\hat{Q}$=union of all topologizing subcategories of $C_{X}$ which do not contain $Q$.)
For each embedding:$V\rightarrow U$, we have correspondence localization functor: $C_{X}/S_{U}\rightarrow C_{X}/S_{V}$. Then we have fibered category over the Zariski topology of $Spec(X)$, so we have given a geometric realization of $Qcoh(X)$ as a stack of local category which means the fiber(stalk)at each point $Q$,
$colim_{Q\in U} C_{X}/S_{U}$=$C_{X}/\hat{Q}$ is a local category.
Zariski geometric center
Define a functor $O_{X}$:$Open(Spec(X))\rightarrow CRings$
$U|\rightarrow End(Id_{C_{X}/S_{U}})$
It is easy to show that $O_{X}$ is a presheaf of commutative rings on $Spec(X)$, then Zariski geometric center is defined as $(Spec(X),\hat{O_{X}})$,where $\hat{O_{X}}$ is associated sheaf of $O_{X}$.
Theorem:
Given X=$(\mathfrak{X},O_{\mathfrak{X}})$ a quasi compact and quasi separated commutative scheme. Then the scheme X is isomorphic to Zariski geometric center of $C_X=Qcoh(X)$. If we denote the fibered category mentioned above by $\mathfrak{F}_{X}$.
then center of each fiber (stalk) of $\mathfrak{F}_{X}$ recover the presheaf of commutative ring defined above and hence Zariski geometric center. Moreover, Catersion section of this fibered category is equivalent to $Qcoh(X)$ when $X$ is commutative scheme
Question
It is well known that for a compact group $G$, we can use Tanaka formalism to reconstruct this group from category of its representation $(Rep(G),\bigotimes _{k},Id)$. Is this reconstruction Morally the same as Gabriel-Rosenberg pattern reconstruction theorem?
From my perspective, I think $Qcoh(X)$ and category of group representations are very similar because $Qcoh(X)$ can be taken as "category of representation of scheme". On the otherhand, group scheme is a scheme compatible with group operations. Therefore, I think it should have united formalism to reconstruct both of them. Then I have following questions:
1 Is there a Tanaka formalism for reconstruction of general scheme$(X,O_{X})$ from $Qcoh(X)$?
2.Is there a Gabriel-Rosenberg pattern reconstruction (geometric realization of category of category of representations of group) to recover a group scheme? I think this formalism is natural (because the geometric realization of a category is just a stack of local categories and the center(defined as endomorphism of identical functor)of the fiber of this stack can recover the original scheme) and has good generality (can be extend to more general settings)
Maybe one can argue that these two reconstruction theorems live in different nature because reconstruction of group schemes require one to reconstruct the group operations which force one to go to monoidal categories (reconstruct co-algebra structure) while reconstruction of
non-group scheme doesn't (just need to reconstruct algebra structure).
However, If we stick to the commutative case. $Qcoh(X)$ has natural symmetric monoidal structure. Then, in this case, I think this argue goes away.
More Concern
I just look at P.Balmer's paper on reconstruction from derived category, it seems that he also used Gabriel-Rosenberg pattern in triangulated categories: He defined spectrum of triangulated category as direct imitation of prime ideals of commutative ring. Then, he used triangulated version of geometric realization (geometric realization of triangulated category as a stack of local triangulated category) hence, a fibered category arose, then take the center of the fiber at open sets of spectrum of triangulated categories to recove the presheaf (hence sheaf) of commutative rings. This triangulated version of geometric realization is also mentioned in Rosenberg's lecture notes Topics in Noncommutative algebraic geometry,homologial algebra and K-theory page 43-44, it also discuss the relation with Balmer's construction. But in Balmer's consideration, there is tensor structure in his derived category
Therefore, the further question is:
Is there a triangulated version of Tannaka Formalism which can recover P.Balmer's reconstruction theorem. ? I heard from some experts in derived algebraic geometry that Jacob Lurie developed derived version of Tannaka formalism. I know there are many experts in DAG on this site. I wonder whether one can answer this question.
In fact, I still have some concern about Bondal-Orlov reconstruction theorem but I think it seems that I should not ask too many questions at one time.So, I stop here.All the related and unrelated comments are welcome
Thanks in advance!
Reconstruction theorem in nLab reconstruction theorem
EDIT: A nice answer provided by Ben-Zvi for related questions

REPLY [7 votes]: I never worked hard on reconstruction and spectra (though I should and plan to). It is an interesting area of deep importance for geometry in general and it is undoubtfully related to Galois theory as well. Tomasz Maszczyk gave a seminar in Warszawa describing how the Galois theory of Grothendieck (and generalization of Joyal-Tierney) and Tannakian formalism, are both special cases of a theorem on base change in certain geometrically motivated setup of bicategories. One should also look for basic mechanisms at the categorical Galois theory of Janelidze and Galois descent:
G. Janelidze, F. Borceux, Galois Theories, Cambridge Studies in Advanced Mathematics 72, Cambridge University Press, 2001
G. Janelidze, Magid’s theorem in categories, Bull. Georgian Acad. Sci. 114, 3, 1984, 497-500 (in Russian)
G. Janelidze, The fundamental theorem of Galois theory, Math. USSR Sbornik 64 (2), 1989, 359-374
G. Janelidze, Precategories and Galois theory, Proc. Como, Springer Lect. Notes in Math. 1488, 1991, 157-173
G. Janelidze, D. Schumacher, R. Street, Galois theory in variable categories, Applied Categorical Structures 1, 1993, 103-110
My advice to myself (as I said I hopew to work more on this) and therefore to you is to look at simpler (like Barr embedding theorem, Mitchell embedding theorem, Giraud's reconstruction theorem) rather than more complicated examples of reconstruction theorems. Yoneda arguments (and sometimes also monadicity) are in basis of all such theorems at some place. A sample usage of Yoneda from hands of Urs is at this nlab page.
To David: I think Rosenberg has looked at the spectra in A-infinity version as well, if my memory does not fail (though I have not seen it written). He sometimes tells his experience about various examples outside of the realm of abelian/triangulated categories (like e.g. spectral reconstructions for smooth manifolds...), but is interested mostly in algebraic examples for his purposes in representation theory which I do not understand enough.<|endoftext|>
TITLE: Quotient of a  reductive group by a non-smooth central finite subgroup
QUESTION [6 upvotes]: I need a construction in linear algebraic groups which uses taking quotient by a central finite group subscheme.
My question is, whether it goes through in ``bad'' characteristics, when this group subscheme is not smooth.
First I write this construction in a special case, and then in the general case.
Let $G$ be a connected semisimple $k$-group over a field $k$ of characteristic $p>0$.
We may assume that $k$ is algebraically closed.
Assume that the corresponding adjoint group $G^{ad}$ is $PGL_n$.
In general, my group $G$ is not special (recall that a $k$-group $G$ is called special if $H^1(K,G)=1$ for any field extension $K/k$).
I want to construct a special $k$-group $H$ related to $G$.
For this end I consider the universal covering $G^{sc}$ of $G$, then $G^{sc}=SL_n$. Let $Z$ denote the center of $G^{sc}$, then $Z=\mu_n$.
We have a canonical epimorphism $\varphi\colon SL_n \to G$. We denote by $C$ the kernel of $\varphi$.
Then $C$ is a group subscheme of $Z$, defined over $k$.
Since $Z=\mu_n$, there is a canonical embedding $Z\hookrightarrow \mathbb{G}_m$ into the multiplicative group $\mathbb{G}_m$.
Thus we obtain an embedding $C\hookrightarrow \mathbb{G}_m$. Consider the diagonal embedding
$$C\hookrightarrow SL_n\times \mathbb{G}_m.$$
I would like to define $H:=(SL_n\times \mathbb{G}_m)/C$. Is such a quotient defined, when  char($k$) divides $n$ and  $C$ is not smooth?
Note that $SL_n$ embeds into $H$, and we have a short exact sequence
$$1\to SL_n \to H \to \mathbb{G}_m \to 1$$
In this exact sequence both $SL_n$ and $\mathbb{G}_m$ are special, and from the Galois cohomology exact sequence we see that $H$ is special as well.
In the general case I assume that $G^{ad}$ is a product of groups $PGL_{n_i}$, $i=1,\dots s$. Then $G^{sc}$ is the product of $SL_{n_i}$.
Let $C$ denote the kernel of the canonical epimorphism $\varphi\colon G^{sc}\to G$, then $C$ is contained in the center $Z$ of $G^{sc}$.
We have $Z=\prod_{i=1}^s \mu_{n_i}$.  Again we embed diagonally
$$C\hookrightarrow (\prod_{i=1}^s SL_{n_i}) \times (\mathbb{G}_m)^s $$
and denote by $H$ the quotient. Again $H$ is special (if it is defined), and again my question is, whether this construction makes sense
when char($k$) divides $n_i$ for some $i$.
Any help is welcome!

REPLY [9 votes]: The standard isomorphism theorems in abstract group theory all hold for group schemes of finite type over a field. This is implicit in SGA (a key point is the statement mentioned by Ekedahl) and is explicit in the notes on algebraic groups,... appearing on my website (Section 7 of Chapter I). This makes a lot of things obvious (including your questions).
[The isomorphism theorems fail when you don't allow nilpotents, which is why the standard expositions on algebraic groups are so complicated.]<|endoftext|>
TITLE: Algebraic number theory and applications to properties of the natural numbers.
QUESTION [15 upvotes]: Please allow me, for the purposes of this question(but only here), to exaggerate matters and state two polemic definitions. Please forget these definitions after answering this question, and pardon my silly nitpicking. 
Definition $1$: "Algebraic number theory" is the theory of algebraic numbers. We exclude arithmetic geometry and such.
Definition $2$: "Number theory" is the study of properties of natural numbers.
In the above sense, I seek examples of applications of algebraic number theory to number theory. I mean, those applications which throw light on "numbers" as we know them in primary school. There is of course enlightenment by looking at a bigger picture of so many number rings, but that is not what I mean. I have specifically the application to the down-to-earth integers in mind. What I know are the following:
$1$. The theorem that an odd prime is of the form $a^2 + b^2$ if and only if it is of the form $4n +1$, proved by looking at factorization in the Gaussian ring.
$2$. Pell's equation is solved with Dirichlet's unit theorem.
$3$. von Staudt–Clausen theorem on Bernoulli numbers, proved using cyclotomic theory.
$4$. Certain equations, like the Fermat equation $X^n + Y^n = Z^n$, may "split" in some extension field and thus it makes sense to go to bigger rings, to study diophantine equations. Here I mean the work of Kummer which started ideal theory, algebraic number theory, etc..
I exclude the following:
$5$. Arithmetic geometry can be used together with algebraic geometry, to study diophantine equations. Elliptic curves fall in here, when their geometry is used significantly(such as in the work of Katz-Mazur). That is "arithmetic geometry", for the purposes of this question. I am more interested in hearing about applications of "algebraic number theory", as defined above.
$6$. Again using algebraic geometry and also modular forms, conjectures such as the Ramanujan bound on the tau function can be proved. Here "modular forms" are "analytic", or "transcendental", and also "geometry is involved. So it goes beyond the "algebraic number theory"
$7$. Dirichlet's theorem on arithmetic progressions is "analytic number theory".
So I thus exclude any touch of "analytic number theory" and "arithmetic geometry", from "algebraic number theory" as defined above. But it can include Kummer theory, classfield theory, etc.. I do not know where to put in Dorian Goldfeld's results on the Gauss class number problem. It uses Gross-Zagier, which is significantly geometric, but gives a result expressible in terms of rational integers. Also I do not know whether Iwasawa theory is arithmetic geometry or not. Langlands theory etc., must be excluded, because it is even more abstract. I want only the "first course in algebraic number theory", "basic cyclotomic theory", "classfield theory" etc., in short only those things which are obviously the study of algebraic numbers.
So, question:

Are there other applications of "algebraic number theory" to "study of natural numbers", than the examples 1-4 above?

I tag this question "big-list" because I hope there are indeed quite a few.

REPLY [4 votes]: The Ramanujan constant $e^{\pi\sqrt{163}}$ is almost a natural number, or more generally, the so-called Heegner numbers are very close to being natural numbers. The explanation can be found here, just to cite a web-ready reference, and is based on the theory of modular functions and complex multiplication (which I consider under the umbrella of algebraic number theory).
This may be not the type of examples you are looking for (they really explain why a transcendental number is almost a natural number, not why a natural number is almost a transcendental number) but it perhaps brings home a better point: algebraic number theory can explain not only properties of natural numbers, but also properties of transcendental numbers (think of values of zeta and L-functions, etc).<|endoftext|>
TITLE: Fourier vs Laplace transforms
QUESTION [39 upvotes]: In solving a linear system, when would I use a Fourier transform versus a Laplace transform?  I am not a mathematician, so the little intuition I have tells me that it could be related to the boundary conditions imposed on the solution I am trying to find, but I am unable to state this rigorously or find a reference that discusses this.  Any help would be appreciated.  Thanks.

REPLY [6 votes]: Fourier transform (FT) - (roughly) a tool to visualize ANY signal as a sum of sinusoids.
Laplace transform (LT) - a tool to analyze the stability of systems.
Why can't we use FT to analyze systems? because it cannot handle exponentially growing signals. That in turn is because a signal needs to be absolutely integrable(necessary, but not sufficient condition) for it to have a FT.
On the other hand, LT is specially designed (by Laplace himself after he rejected the thesis of Fourier and extended FT) to handle exponentially growing signals so that now you could if a system has an exponentially growing output ( $a > 0$  where $s = a + jw$) or exponentially decaying output ($a < 0$).
Therefore if you have a square wave type signal and you wanted to break it up into sinusoids of various frequencies, you'd use FT and on the other hand, if you have a system and you want to understand how stable it is, then you'd try to get LT of it's impulse response to further analyze the systems relative stability.<|endoftext|>
TITLE: The central role of varieties (a comment from Mumford's Red Book)
QUESTION [11 upvotes]: I was reading Mumford again and I noticed a comment in the beginning of the book:
"Finally, in any study of general preschemes, the varieties are bound, for many reasons which I will not discuss here, to play a unique and central role."
My question is: is there a particular theorem (or series of theorems) Mumford might have been hinting at?

REPLY [19 votes]: Here is a really cool illustration of the principle which Emerton was outlining.  We know that the Picard group of projective $(n-1)$-space over a field $k$ is $\mathbf{Z}$ ($n \ge 2$), generated by $\mathcal{O}(1)$.  This underlies the proof that the automorphism group of such a projective space is ${\rm{PGL}}_n(k)$. But what is the automorphism group of $\mathbf{P}^{n-1}_A$ for a general ring $A$?  Is it ${\rm{PGL}}_n(A)$?  That is, there is a natural map
$${\rm{PGL}} _n (A) \rightarrow {\rm{Aut}} _A (\mathbf{P}^{n-1} _A)$$ 
(see my answer in the posting called something like "Is ${\rm{PSL}}_2 = {\rm{PGL}} _2$?") and we want to know if it is an isomorphism.  It's a really important fact that the answer is yes.  But how to prove it?  It's a shame that this isn't done in Hartshorne. 
By an elementary localization (as in the last step of Emerton's answer), we may assume $A$ is local.  In this case we claim that ${\rm{Pic}}(\mathbf{P}^{n-1}_A)$ is infinite cyclic generated by $\mathcal{O}(1)$.  Since this line bundle has the known $A$-module of global sections, it would give the desired result if true (since ${\rm{PGL}}_n(A) = {\rm{GL}}_n(A)/A^{\times}$ for local $A$) by the same argument as in the field case.  And since we know the Picard group over the residue field, we can twist to get to the case when the line bundle $\mathcal{L}$ is trivial on the special fiber and then we can formulate the problem in two equivalent ways: (I)"lift" the generating section there to a generating section over $A$, or (II) prove that $f _{\ast}(\mathcal{L})$ is invertible in $A$ with the natural map $f^{\ast}(f _{\ast} \mathcal{L}) \rightarrow \mathcal{L}$ an isomorphism.  How to do it? 
Step 0: The case when $A$ is a field.  Done. 
Step 1: The case when $A$ is artin local, via (I): this goes via induction on the length, the case of length 0 being Step 0 and the induction resting on cohomological results for projective space over the residue field. 
Step 2: The case when $A$ is complete local noetherian ring.  This goes via (I) using Step 1 and the theorem on formal functions (formal schemes in disguise).
Step 3: The case when $A$ is local noetherian.  This is faithfully flat descent for (II) from Step 2 applied over $\widehat{A}$.
Step 4: The case when $A$ is local:  descent from the noetherian local case in Step 3 via direct limit arguments.
"QED"<|endoftext|>
TITLE: Why does one invert $G_m$ in the construction of the motivic stable homotopy category?
QUESTION [16 upvotes]: Morel and Voevodsky construct the motivic stable homotopy category, a category through which all cohomology theories factor and where they are representable, by starting with a category of schemes, Yoneda-embedding it into simplicial presheaves, endowing those with the $\mathbb{A}^1$-local model structure, and then passing to $S^1 \wedge \mathbb{G}_m$-spectra. The last step ensures that smashing with $S^1$ or with $\mathbb{G}_m$ induce functors with a quasi-inverse on the homotopy category. 
Inverting $S^1$ leads to a triangulated structure on the homotopy category, which is very welcome, but I would like a motivation for inverting $\mathbb{G}_m$. Since $\mathbb{P}^1$ is $\mathbb{A}^1$-equivalent to $S^1 \wedge \mathbb{G}_m$ I would also be content with a motivation to invert $\mathbb{P}^1$.

I must admit I already know some answers which certainly are reason enough to invert $\mathbb{G}_m$, e.g. (from slides by Marc Levine, start at page 64):

Inverting $\mathbb{G}_m$ is necessary to produce a Gysin sequence
The algebraic K-theory spectrum appears naturally as a $\mathbb{P}^1$-spectrum

However, I am greedy and would like to hear a motivation like the one for inverting the Lefschetz motive in the construction of pure motives: There one could say that for all envisaged realization functors which should factor through the category of pure motives, the effect of tensoring with the Lefschetz motive can be undone (e.g. is just a change of Galois representation leaving the cohomology groups unchanged).
Or, related to this, as Emerton explained in his nice answer here one has to invert the Lefschetz motive in order to make the Pure Motives a rigid tensor category. Ideally one would like the triangulated category of motives to arise as derived category of some rigid tensor category - if this was true, would it be reflected in the fact that $\mathbb{P}^1$ or $\mathbb{G}_m$ are invertible? (in this case of course one should ensure iinvertibility when constructing a candidate for this derived category)

REPLY [12 votes]: $H^1({\mathbb G}_m)$ is the same motive as $H^2(\mathbb P^1)$, so I believe that inverting $\mathbb G_m$ is the same as inverting the Lefschetz motive.  (Topologically, fundamental
classes all ultimately arise from the $H^1$ of the circle, so we must  invert this
if fundamental classes are to be invertible.)  In the pure context, one is not allowed to talk about ${\mathbb G}_m$, and so talks of $H^2(\mathbb P^1)$ instead; but I do think it is the same process.
Note that this is compatible with Dustin Clausen's and Marty's answers: in all realizations (Galois representations, computing 
periods, ... ) of motives, we can and do invert the Lefschetz motive (since it just becomes
the inverse cyclotomic character, or $2 \pi i$, or ... ).    Incidentally, related to David Roberts's answer, smashing with
$\mathbb G_m$ is what number theorists call a Tate twist, I believe, and on the level of Galois representations it it is just tensoring with the cyclotomic character (here I am thinking of $H_1(\mathbb G_m)$).   So asking for $\mathbb G_m$ to be invertible is the same as asking that one can perform Tate twists (of arbitrary integer power) on the level of motives.
Now if $M$ is a motive over a number field, with $L$-function $L(M,s)$,
and $M(n)$ is its $n$th Tate twist, then the $L$-function of $M(n)$ is simply 
$L(M,s + n)$.  So the Tate twisting parameter is the same as the parameter $s$ in
the $L$-function (restricted to integer values, of course).  Thus the desire to have
this parameter really be an integer (and not just a natural number) also has deep roots
in the conjectured relations between special values of $L$-functions and the arithmetic geometry of motives.  (If one looks at the simplest $L$-function, namely the Riemann zeta function, it has special values at positive and negative integers, and one certainly wants to consider all of these and relate all of them to motivic considerations.)<|endoftext|>
TITLE: How helpful is non-standard analysis?
QUESTION [60 upvotes]: So, I can understand how non-standard analysis is better than standard analysis in that some proofs become simplified, and infinitesimals are somehow more intuitive to grasp than epsilon-delta arguments (both these points are debatable).  
However, although many theorems have been proven by non-standard analysis and transferred via the transfer principle, as far as I know all of these results were already known to be true.  So, my question is: 
Is there an example of a result that was first proved using non-standard analysis?  To wit, is non-standard analysis actually useful for proving new theorems?
Edit: Due to overwhelming support of Francois' comment, I've changed the title of the question accordingly.

REPLY [7 votes]: I just came across a 2013 book by F. Herzberg entitled "Stochastic Calculus with Infinitesimals", see http://link.springer.com/book/10.1007/978-3-642-33149-7/page/1 where probability and stochastic analysis are done without having to develop the complexities of measure and integration theory first.  Ever since E.Nelson, such an approach is called "radically elementary" and it really is.  What this proves is the new result that stochastic calculus can be done without measure theory.  
To give a historical parallel, recall that Leibniz's mentor in mathematics was Huygens.  When Huygens first learned of Leibniz's invention of infinitesimal calculus, Huygens was sceptical, and wrote to Leibniz that he is merely doing what Fermat and others have done before him in a different language.  What Huygens failed to recognize immediately (but did recognize later) was the generality of the methods and the lucidity of the presentation of Leibniz's new approach.  The Nelson-Herzberg approach to stochastic calculus is in a way more significant than merely a new "result", since it provides a new methodology.<|endoftext|>
TITLE: Conceptual understanding of the Gross-Zagier theorem.
QUESTION [22 upvotes]: The Gross-Zagier paper "Heegner points and derivatives of $L$-series", is really computational and hard to plow through. It seems it is futile to read it as such and one must look for a more conceptual understanding.
The more conceptual attempts I know are the following:
$1$. The work of Kolyvagin on Birch-Swinnerton-Dyer conjecture, in which he re-proves part of Gross-Zagier using Euler systems. The problem with this is that some of the original Gross-Zagier is still needed for getting the results on BSD conjecture(if I understand things correctly. Please point out if I am wrong).
$2$. The volume of Darmon and Zhang published by MSRI, in which they attempt a $p$-adic theory. Again this is going away from the original complex analytic case. Again please correct me if I am wrong.
So I am wondering whether anybody published a more conceptual approach to the complex analytic Gross-Zagier theorem. I would be grateful for any references.

REPLY [4 votes]: Indeed, there is a conceptual understanding of this via "incoherent Siegel-Weil Formula",cft S.Kudla `s papers.See also the last section of recent preprint of Gan-Gross-Prasad.

REPLY [4 votes]: In my current (no very deep) understanding, there are two possible ways to make the proof of the Gross-Zagier more conceptual.
The first is to recognize in each terms  of the equation products of local terms which are local linear functionals. Now, a famous theorem of Saito and Tunnell states that such linear functionals live in a dimension 1 vector space. So there are proportional and the Gross-Zagier amounts to specifying the factor of proportionality. This requires a large amount of representation theory, but I think now this program has been completed. Using Gross-Prasad conjecture in place of Saito-Tunnell, GZ can apparently be extended widely.
The second is to observe that the $p$-adic variant of GZ is in fact easier to prove (this is because $p$-adic heights naturally factor through the first Bloch-Kato cohomology group). Conceptually, this is maybe not so surprising because Heegner points verify the distribution relations of an Euler system, so they are naturally linked to the $p$-adic $L$-function. Hence, to prove the Gross-Zagier theorem, all there is to do is to relate the special value of the derivative of the $p$-adic $L$-function to the value of the derivative of the complex $L$-function. But here is the rub: as far as I know, proving that the derivative of the $p$-adic $L$-function interpolates $p$-adically the derivative of the $L$-function is more or less equivalent to showing that the $p$-adic height pairing is not degenerate. So this looks hopeless.<|endoftext|>
TITLE: For a given finite group G, is there a cover of P^1 over Q s.t. over C it's G-Galois?
QUESTION [7 upvotes]: For any finite group, G, we can find a cover of ℙ1ℂ which is G-Galois. The regular inverse Galois problem is equivalent to there existing such a cover that descends with action to ℚ. My question is about the easier problem: given a finite group G, can we find a cover of ℙ1ℂ such that it descends to ℚ as a mere cover (meaning not necessarily with group action)?
From the results that I know, I would be really surprised if this is solved. But what is known? And where is it written?

REPLY [4 votes]: I don't know about every finite group $G$ (I'll guess no), but there are definitely infinitely many finite groups $G$ for which the situation you describe obtains: the extension $K/\mathbb{C}(t)$ has a model over $\mathbb{Q}$ but is not Galois over $\mathbb{Q}$.  (And for most of these groups, we do not know how to realize them as Galois groups over $\mathbb{Q}$, regularly or otherwise.)
For instance, this is the situation in a work in progress of John Voight and me:
http://math.uga.edu/~pete/triangle-091309.pdf
In our slightly different language, there are plenty of situations where the covering itself is defined over $\mathbb{Q}$ but the field of definition of the automorphism group $G$ is strictly larger.  (This is equivalent to what you're asking, right?  Please let me know.)
[Warning: recently, with the help of Noam Elkies, John and I realized that our arguments as given only work when (in our notation) $a = 2$.  This is still a generalization of the setting in which I began this work some years ago: I had $a = 2$, $b = 3$, so I know for sure that there are infinitely many examples of this form.]<|endoftext|>
TITLE: A generalization of Cauchy's mean value theorem.
QUESTION [11 upvotes]: The following simple theorem is known as Cauchy's mean value theorem. Let $\gamma$ be an immersion of the segment $[0,1]$ into the plane such that $\gamma(0) \ne \gamma(1)$. Then there exists a point such that the tangent line at that point is parallel to the line passing through $\gamma(0)$ and $\gamma(1)$. So the boundary values of an immersion determine a direction such that for any immersion of a segment with given boundary values there exists a tangent line parallel to the direction.
I would like to propose the following conjecture, generalizing this statement. Instead of immersions of a segment we consider immersions of a compact manifold $M^n$ with non-empty boundary $\partial M$ into ${\mathbf R}^{n+1}$. For a map $f\colon \partial M \to {\mathbf R}^{n+1}$ we consider the space $L(f)$ of all immersions $g\colon M \to {\mathbf R}^{n+1}$ such that $g|_{\partial M}=f$.
The conjectural claim is the following: If $f$ is sufficiently generic, then for every connected component $L_0$ of the space $L(f)$ there exists a hyperplane direction $l$ such that for any immersion $g$ from $L_0$ $l$ is parallel to the tangent plane to $g(M)$ at some point.
If the conjecture is true then it is very interesting how $l$ depends on $g$.

REPLY [4 votes]: I will consider another counterexample, based on Anton's ideas. I think it looks simpler.
We will construct a closed domain $D$ in a plane and three functions on it with same restrictions to the boundary of the domain such that there is no choice of points on graphs of functions with pairwise parallel tangent planes.
First observation is the following: Let function $f$ on the coordinate plane depends on the coordinate $x$ only and function $g$ depends on the coordinate $y$ only. Then $df(a)=dg(b)$ if and only if $df(a)=dg(b)=0$. (It seems to me that the crucial Anton's idea is to consider functions with very degenerate gradient map, having value on a curve).
Now we take the functions $f(x,y)=-x^2$ and $g(x,y)=P_4(y)$ where $P_4$ is a polynomial of degree $4$ with two zeroes, three pairwise different critical values and leading term $y^4$.
We set the domain $D$ to be a set of solutions to an inequality $f \ge h$. It is a closed subset in a plane diffeomorphic to a unit disk. We define functions now: the first function is $f$, the second function is $g$. Third function is a function without critical points on $D$ coinciding with a restriction of $f$ (or $g$) to the boundary. Such a function exists! (it is an exercise from Morse theory. I mention here that we need a polynomial of degree 4 (degree 2 is unsufficient) to satisfy that extension without critical points property).<|endoftext|>
TITLE: Motivation for the covariant model structure on SSet/S
QUESTION [5 upvotes]: I was reading HTT 2.1.4, and I just totally lost what was going on.  Could someone provide some motivation for this section?  Why do we want another model structure?
I'm sorry for not providing motivation, but as you can see, I'm unable to do so.  
Here's a link to the ArXiv version: http://arxiv.org/pdf/math/0608040v4

REPLY [4 votes]: The quick answer is the covariant model structure on sSet/S is one way to build an infinity-category of infinity-copresheaves on S, when S is an infinity-category. In fact, I don't understand why the covariant model structure is introduced first rather than the contravariant one- it is the later which is used to construct infinity-presheaves, which are of course important examples of infinity-topoi.
(I am using the terminology "infinity-presheaf" to mean a contravariant infinity-functor from S to the infinity-category of infinity-groupoids) 
It would perhaps be best to recall what happens in the case of 2-categories:
Let C be a category. We can, on one hand, consider the bicategory of weak functors C^op->Gpd, where the target is the bicategory of groupoids. On the other hand, we can consider categories fibred in groupoids over C, that is a functor D->C which is a Grothendieck fibration in groupoids. Both of these objects, weak presheaves and fibred categories respectively, naturally form bicategories.
We have a 2-functor G:Gpd^{C^op}->Fib_Gpd(C) between these bicategories given by the "Grothendieck construction". It has a left 2-adjoint and together this adjoint pair forms an equivalence of bicategories.
Lurie proves the infinity-analogue of this statement. To do so, he needs to form an infinity category of "Grothendieck fibrations in groupoids", and an infinity category of "infinity presheaves" and show they are equivalent.
The infintiy-version of Grothendieck fibration in groupoids is what Lurie calls a "right fibration". In particular, C->D is a Grothendieck fibration in groupoids if and only if N(C)->N(D) is a right fibration. Also, the fibers of any right-fibration are Kan-complexes, hence, infinity groupoids.
Given a simplicial set S, the contravariant model structure on sSet/S is enriched in sSet_Quillen so we can form the associated full simplicial category on fibrant and cofibrant objects. An an object X->S is fibrant in this model structure if and only if it is a right-fibration. Hence, the homotopy-coherent nerve of this simplicial category is the infinity-category of "Grothendieck fibrations in infinity groupoids over S".
In 2.2.1, Lurie introduces a functor St:sSet/S->sSet^{C(S)^op} where C is the left-adjoint to the homotopy-coherent nerve- here we mean we are considering functors of simplicial categories (treating sSet as a simplicial category since it is enriched over itself). Since we can identify sSet/S with Set^{(\Detla/S)^op} and the functor is colimit preserving, by formal nonsense it has a right adjoint which Lurie denotes by "Un". "Un" is the "infinity-Grothendieck construction".
We can now equip sSet^{C(S)^op} with the projective model structure, and then the adjoint pair (St,Un) forms a Quillen-equivalence. 
Now, sSet^{C(S)^op} can be turned into an infinity-category by applying the same construction I said before- treat it as a simplicial category and restrict to fibrant and cofibrant objects, and take the homotopy-coherent nerve. This infinity-category is the infinity-category of infinity-presheaves on S. 
The Quillen-equivalence (St,Un) turns into an adjunction between the infinity-category of right-fibrations over S, and the infinity-category of infinity-presheaves over S, and moreover is an equivalence of infinity-categories.
The upshot is, the contravariant model structure gives us another way of describing infinity-presheaves.
As a side note, to understand why the contravariant model structure is defined the way it is, you should look at how the functor St is defined- the model structure is essentially "designed" to so that (St,Un) becomes a Quillen equivalence.<|endoftext|>
TITLE: Origins of functional field arithmetic
QUESTION [9 upvotes]: Background: By function field, we mean a finite extension of the field of rational functions of one variable over a finite field with $p$ elements. Classfield theory for function fields was established by Chevalley in an Annals paper. An axiomatic characterization for number fields and function fields was established by Artin and Whaples, thus finally putting on firm ground the analogy between function fields and number fields. I have seen allusions that the germ of the idea was coming from Gauss. However since fields were not defined then, this was not a definitive statement.
Question: When was a definitive conjecture first made in mathematical history that there is a major analogy between algebraic number fields and function fields over finite fields?

REPLY [7 votes]: One of the most early observations on the analogy between function fields and number fields is the little known work by 

E. Heine, 
 Fernere Untersuchungen über ganze Functionen,
J. Reine Angew. Math. 48 (1854), 243--266 

who started investigating quadratic forms with coefficients in polynomial rings. 
Dedekind's and Weber's contributions have already been mentioned, but not those of Robert König, who investigated the connection between quadratic function fields and quadratic forms with polynomial coefficients in various publications prior to Artin
(and perhaps just as unknown as Heine's contribution):

 Über die quadratischen Formen mit rationalen Funktionen als Koeffizienten, 
Monatsh. f. Math. Phys. 23 (1912), 321-346
Beiträge zur Arithmetik der hyperelliptischen Funktionenkörper,
J. Reine Angew. Math. 142 (1913), 191-210

He also explicitly stated the analogy between function fields and number fields in the titles (and the body) of the following two papers

Arithmetisch-funktionentheoretische Parallelen,
Jahresber. DMV 23 (1914), 181--192
Funktionen- und zahlentheoretische Analogien,
Jahresber. DMV 28 (1920), 208-213.

The constant fields investigated by König have characteristic 0, however.<|endoftext|>
TITLE: Is there a volume conjecture for closed 3-manifolds?
QUESTION [19 upvotes]: A typical statement of the volume conjecture, for instance in Murakami's survey 1002.0126, is

Conjecture: For $K$ a knot in $S^3$, the N-th colored Jones polynomials are related to the volume of the knot complement by $$ 2 \pi \lim_{N \rightarrow \infty} \frac{1}{N} \log | J_N(K; \exp(2\pi i / N)) | = Vol( S^3 \backslash K).$$

A refinement of the conjecture is $ 2 \pi \lim_{N \rightarrow \infty} \frac{1}{N} \log J_N(K; \exp(2\pi i / N)) = Vol( S^3 \backslash K) + i \; CS( S^3 \backslash K) \; (\mod \pi^2 i)$
where CS is the Chern-Simons invariant.  Both sides of the conjecture can be formulated for 3-manifolds more general than knots in S^3 and their complements.  In particular, we might ask about closed 3-manifolds without a knot at all.

Question: Is there an analogous volume conjecture for (some) closed 3-manifolds, or for closed 3-manifolds with embedded knots, and if so where in the literature are these formulations discussed?

REPLY [6 votes]: There's another volume conjecture formulated by Chen and Yang for Turaev-Viro invariants of closed manifolds. They present some evidence for the conjecture in the paper. In a second paper, Yang and collaborators formulate another volume conjecture based on the colored Jones polynomial, but evaluated at different roots of unity: see Question 1.7. They prove this version for the figure eight knot.<|endoftext|>
TITLE: Does anyone have an electronic copy of Waldspurger's "Sur les coefficients de Fourier des formes modulaires de poids demi-entier"?
QUESTION [5 upvotes]: Is there an electronic copy of Waldspurger's paper "Sur les coefficients de Fourier des formes modulaires de poids demi-entier" floating around the internet somewhere?  This appeared in J. Pures Math. Appliquees, Vol. 60 No. 4, in 1981.

REPLY [4 votes]: Here's another version, single-paged and OCRed:
http://mate.dm.uba.ar/~nsirolli/files/waldspurger.pdf<|endoftext|>
TITLE: Minimum Hamming Distance Distribution in a Random Subset of Binary Vectors+
QUESTION [6 upvotes]: Select $K$ random binary vectors $Y_i$ of length $m$ uniformly at random.
Let the following collection of random variables be defined: $X_{i,j}=w(Y_i \oplus Y_j)$ where $w(\cdot)$ denotes the Hamming weight of a binary vector, i.e., the number of the nonzero coordinates in its argument. Define $D_{min}(Y_1,\ldots,Y_K)$ as the smallest of the $X_{i,j}$ for $i \neq j.$
Thus we have $n=C(K,2)=K(K-1)/2$ non-independent random variables $X_{i,j}$ with support {$0,1,\ldots,m$} and individual distribution $Bin(m,1/2)$. It seems to me that the random variables $X_{i,j}$ will be $s$-wise negatively correlated (for $s$ large enough) if distances between pairs chosen from a subcollection of $Y_{i_1},Y_{i_2},\ldots,Y_{i_v}$  where ($v < K$) tincreases then the distances between $Y_{i_j}$ and the remaining vectors will tend to decrease. Take $s=v+1.$
It is possible to get a bound on the following quantity. Fix $w$ an integer less than $m/2.$  The Hamming sphere of radius w has "volume", i.e., contains $V_w(m)=\sum_{s=0}^w C(m,s)$ vectors and we approximately have to first order in the exponent
$$
V_w(m) =2^{m H((w+1)/2)}
$$
where $H(\cdot)$ is the binary entropy function. Then, for a random uniform choice of the $Y_i$ for $i=1,2,\ldots,K$ it is clear that if the Hamming spheres centred at these vectors are disjoint then the minimum distance is at least $2w+1$  thus 
$Pr[D_{min} \geq 2 w+1] \leq \frac{(2^m-V)}{2^m}\frac{ (2^m-2 V) }{2^m} \cdots\frac{ (2^m - (K-1)V)}{ 2^{m}}$
where $V=V_w(m).$ This means that, by replacing each fraction of the form $(1-x)$ by $exp(-x)$ where $x >0$ but small, we get the approximate upper bound
$Pr[ D_{min} \geq 2w+1] \leq exp\left[-K(K-1)V^2/(2^{m+1} \right]$ which then expresses this upper bound in terms of the entropy function, which is nice. Unfortunately this upper bound is quite loose.
I will be happy with any pointers to literature or any other suggestions.

REPLY [5 votes]: Here is a direct application of Theorem 21 from Gabor Lugosi's concentration of measure notes. Your $Y_i$ corresponds to his $X_{i,1}^m$ and your $X_{i,j}$ to his $d(X_{i,1}^m, X_{j,1}^m)$. Take his $A$ to be your $\{X_{i,j}\}_{i \neq j}$. The birthday problem gives the probability that any two of the $Y_i$ are exactly the same. That is:
$$\mathbb{P}(0^m \in A) = \mathbb{P}\left(\left\{X_{i,j} = 0^m : i \neq j\right\}\right) = \mathrm{(omitted\ for\ simplicity)} $$
Now your $D_{min}$ corresponds to his $d(0^m,A)$. By the Theorem, for any $t > 0$,
$$\mathbb{P}\left(D_{min} \geq t + \sqrt{\frac{m}{2} \mathrm{log}\frac{1}{\mathbb{P}(0^m \in A)}}\right) \leq e^{-2t^2/m}.$$
This bound may be OK for your needs. If it isn't, see Lugosi's discussion of Talagrand's convex distance inequality, which is a big improvement.<|endoftext|>
TITLE: Looking for good conference this summer for homotopy theory
QUESTION [5 upvotes]: I'm a 2nd year grad student and I'm looking for conferences/summer schools to attend this summer.  I checked out the AMS calendar but couldn't find anything I found relevant there. Anyone have any suggestions?

REPLY [4 votes]: Sarah Whitehouse maintains a list of topology conferences at
http://www.sarah-whitehouse.staff.shef.ac.uk/btconfs.html
The Talbot workshops are a great opportunity for PhD students and young researchers (this year's topic: twisted K-theory). I don't know if they still have places available.
I'd also recommend the Young Topologists' Meeting in Copenhagen (June 16-20) and, with a somewhat more focussed and advanced scope, the meeting at the Fields Institute (August 30-Sept 3)<|endoftext|>
TITLE: Kuratowski closure-complement problem for other mathematical objects?
QUESTION [16 upvotes]: The original Kuratowski closure-complement problem asks:

How many distinct sets can be obtained by repeatedly applying the set operations of closure and complement to a given starting subset of a topological space?

My question is: what is known about analogous questions in other settings? 
Here's an example of what I'm thinking of, for rings: 

How many distinct ideals can be obtained by repeatedly applying the operations of radical and annihilator to a given starting ideal $I$ of a commutative ring $R$? 

Note that $r(r(I))=r(I)$ and $I\subseteq Ann(Ann(I))=\{x\in R: x\cdot Ann(I)=(0)\}$, which are the best analogs I could think of to $\overline{\overline{S}}=\overline{S}$ and $(S^C)^C=S$.
Also: what is the structure necessary to formulate this kind of question called, and where does it occur naturally? 
It seems like we need at least a poset, but with distinguished idempotent and involution operations to generalize the closure and complement, respectively.

REPLY [7 votes]: Here's a paper that might be of interest:
D. Peleg, A generalized closure and complement phenomenon, Discrete Math., v.50 (1984) pp.285-293.
Other than what's found in the above paper I do not know of any general theory or framework specifically aimed at organizing results similar to the Kuratowski closure-complement problem, i.e., those which involve starting with a seed object (or objects) and repeatedly applying operations to generate further objects of the same type in a given space.
Here's a general sub-question I thought of recently, that might be interesting to study:
"What's the minimum possible cardinality of a seed set that generates the maximum number of sets via the given operations?"
A few years ago I proposed a challenging Monthly problem (11059) that essentially asks this question for the operations of closure, complement, and union in a topological space.  It does turn out there's a space containing a singleton that generates infinitely many sets under the three operations, but it's a bit tricky to find.  I haven't looked into the question yet for other operations.  As far as I know it hasn't been discussed yet in the literature (apart from the specific case addressed by my problem proposal).<|endoftext|>
TITLE: "Oldest" bug in computer algebra system?
QUESTION [17 upvotes]: The goal of this question is to find an error in a computation by a computer algebra system where the 'correct' answer (complete with correct reasoning to justify the answer) can be found in the literature.  Note that the system must claim to be able to perform that computation, not implementing a piece of (really old) mathematics is sad, but is a different topic.
From my knowledge of the field, there are plenty of examples of 19th century mathematics where today's computer algebra system get the wrong answer.  But how far back can we go?

Let me illustrate what I mean.  James Bernoulli in letters to Leibniz (circa 1697-1704) wrote that [in today's notation, where I will assume that $y$ is a function of $x$ throughout] he could not find a closed-form to $y' = y^2 + x^2$.  In a letter of Nov. 15th, 1702, he wrote to Leibniz that he was however able to reduce this to a 2nd order LODE, namely $y''/y = -x^2$.  Maple can find (correct) closed-forms for both of these differential equations, in terms of Bessel functions.
An example that is 'sad' but less interesting is
$$r^{n+1}\int_0^{\pi}\cos(r\rho \cos (\omega))\sin(\omega)^{2n+1}d\omega$$
with $n$ assumed to be a positive integer, $r>0$ and $\rho$ real; this can be evaluated as a Bessel functions but, for example, Maple can't.  Poisson published this result in a long memoir of 1823.
One could complain that (following Schloemilch, 1857) that he well knew that
$$J_n(z) = \sum_{0}^{\infty} \frac{(-1)^m(z/2)^{n+2*m}}{m!(n+m)!}$$
Maple seems to think that this sum is instead $J_n(z)\frac{\Gamma(n+1)}{n!}$, which no mathematician would ever write down in this manner.
Another example which gets closer to a real bug is that  Lommel in 1871 showed that the Wronskian of $J_{\nu}$ and $J_{-\nu}$ was $-2\frac{sin(\nu\pi)}{\nu z}$.  Maple can compute the Wronskian, but it cannot simplify the result to $0$.  This can be transformed into a bug by using the resulting expression in a context where we force the CAS to divide by it.
For a real bug, consider
$$\int_{0}^{\infty} t^{-\lambda} J_{\mu}(at) J_{\nu}(bt)$$
as investigated by Weber in 1873.  Maple returns an unconditional answer, which a priori looks fine.  If, however, the same question is asked but with $a=b$, no answer is returned!  What is going on?  Well, in reality that answer is only valid for one of $0\lt a\lt b$ or $0\lt b \lt a$.  But it turns out (as Watson explains lucidly on pages 398-404 of his master treatise on Bessel functions, this integral is discontinuous for $a=b$.  Actually, the answer given is also problematic for $\lambda=\mu=0, \nu=1$.  And for the curious, the answer given is
$$\frac{2^{-\lambda}{a}^{\lambda-1-\nu}{b}^{\nu}
\Gamma  \left( 1/2\nu+1/2\mu-1/2\lambda+1/2 \right)} { 
\Gamma\left( 1/2\mu+1/2\lambda+1/2-1/2\nu\right) \Gamma  \left( \nu+1 \right)}
{F(1/2-1/2\mu-1/2\lambda+1/2\nu,1/2\nu+1/2\mu-1/2\lambda+1/2;\nu+1;{\frac {{b}^{2}}{{a}^{2}}})}
$$

EDIT: I first asked this question when the MO community was much smaller.  Now that it has grown a lot, I think it needs a second go-around.  A lot of mathematicians use CASes routinely in their work, so wouldn't they be interested to know the 'age' gap between human mathematics and (trustable) CAS mathematics?

REPLY [3 votes]: In Minkowski space-time one expects $$\epsilon_{ijkl}\epsilon_{i^\prime j^\prime k^\prime l^\prime}g^{ii^\prime}g^{jj^\prime}g^{kk^\prime}g^{ll^\prime}=-24, \tag{1}$$
where $\epsilon_{ijkl}$ is the Levi-Civita tensor and $g^{ij}$ represents the metric. However, if you (uncritically) calculate the l.h.s of (1) in the symbolic manipulation system FORM using FixIndex statement to assign specific values to selected diagonal elements of the Kronecker delta, which by default represents the metric, you get +24, not -24. 
In fact, this is not a bug but a subtlety of inner workings of FORM and the users were warned that one can try to change the behaviour of the Kronecker delta a bit but "this is dangerous and needs, in addition to a good
understanding of what is happening, good testing to make sure that what the
user wants is indeed what does happen":
http://www.nikhef.nl/~form/maindir/documentation/reference/online/
Interestingly, this subtlety of manipulations with the Levi-Civita tensor in FORM led to a long lasted sign error in the calculations of the pion-pole dominant term in the hadronic light-by-light scattering contribution to the muon anomalous magnetic moment, and the source of this error was discovered only when some later time calculations of the same quantity, based on the REDUCE Computer Algebra System, gave an oposite sign: http://arxiv.org/abs/hep-ph/0112102 (Comment on the sign of the pseudoscalar pole contribution to the
muon g-2, by Masashi Hayakawa and Toichiro Kinoshita).
I think the following paragraph in the above cited FORM manual gives a very good advice how modern computer algebra systems should be used:
"As in the Zen saying: 
To the beginning student mountains are mountains and water is water. To the advanced student mountains stop being mountains and water stops being water. 
To the master mountains are mountains again and water is water again. 
Of course the modern master also checks that what he expects the system to do, is indeed what the system does."<|endoftext|>
TITLE: Least ordinal not in a countable transitive model of ZFC
QUESTION [7 upvotes]: Frequently it is useful do deal with countable transitive models M of ZFC, for example in forcing constructions.  
The notion of being an ordinal is absolute for any transitive model, so certainly if ($\alpha$ is an ordinal)M then also $\alpha$ is an ordinal.  For the same reason, M will contain successors of every ordinal in it. 
On the other hand, if M is countable then M cannot contain every countable ordinal; there must be a least (countable) ordinal not in M. 
Can anything be said about this ordinal? Does it have any special significance?

REPLY [11 votes]: The least ordinal not in any transitive model of ZFC can also be described as the supremum of the heights of transitive models of ZFC. It is natural here to consider the class S consisting of all ordinals λ for which there is a transitive model of ZFC of height λ. Thus, the ordinal of your title, when it  exists, is simply the supremum of the countable members of S. There are a number of relatively easy observations:

If M is a transitive model of ZFC, then so is LM, the constructible universe as constructed inside M, and these two models have the same height. Thus, one could equivalently consider only models of ZFC + V = L. 
It is relatively consistent with ZFC that S is empty, that is, that there are no transitive models of ZFC. For example, the least element of S is the least α such that Lα is a model of ZFC. This is sometimes called the minimal model of ZFC, though of course it refers to the minimal transitive model. It is contained as a subclass of all other transitive models of ZFC. The minimal model has no transitive models inside it, and so it believes S to be empty. 
The least element of S (and many subsequent elements) is Δ12 definable in V. This is because one can say: a real codes that ordinal iff it codes a well-ordered relation and there is a model of that order type satisfying that no smaller ordinal is in S (a Σ12 property), also iff every well-founded model of ZFC has ordinal height at least the order type of the ordinal coded by z (a Π12 property). 
If S has any uncountable elements, then it is unbounded in ω1. The reason is that if Lβ satisfies ZFC and β is uncountable, then we may form increasingly large countable elementary substructures of Lβ, whose Mostowski collapses will give rise to increasingly large countable ordinals in S. 
In particular, if there are any large cardinals, such as an inaccessible cardinal, then S will have many countable members. 
If 0# exists, then every cardinal is a member of S. This is because when 0# exists, then every cardinal κ is an L-indiscernible, and so Lκ is a model of ZFC. Thus, under 0#, the class S contains a proper class club, and contains a club in every cardinal.
S is not closed. For example, the supremum of the first ω many elements of S cannot be a member of S. The reason is that if αn is the nth element of S, and λ = supn αn, then there would be a definable cofinal ω sequence in Lλ, contrary to the Replacement axiom.
S contains members of every infinite cardinality less than its supremum. If β is in S, then we may form elementary substructures of Lβ of any smaller cardinality, and the Mostowski collapses of these structures will give rise to smaller ordinals in S. 
If β is any particular element of S, the we may chop off the universe at β and consider the model Lβ. Below β, the model Lβ calculates S that same as we do. Thus, if β is a limit point of S, then Lβ will believe that S is a proper class. If β is a successor element of S, then Lβ will believe that S is bounded. Indeed, if β is the αth element of S, then in any case, Lβ believes that there are α many elements of S. 
If S is bounded, then we may go to a forcing extension V[G] which collapses cardinals, so that the supremum of S is now a countable ordinal. The forcing does not affect whether any Lα satisfies ZFC, and thus does not affect S. 

Reading your question again, I see that perhaps you meant to consider a fixed M, rather than letting M vary over all transitive models. In this case, you will want to look at fine-structural properties of this particular ordinal. Of course, it exhibits many closure properties, since any construction from below that can be carried out in ZFC can be carried out inside M, and therefore will not reach up to ht(M).<|endoftext|>
TITLE: Proof formalization
QUESTION [5 upvotes]: I read some time ago some papers about proof formalization. Typically, I began whith this one, from Lamport.
Are there more recent works in this field ?

REPLY [5 votes]: There was a special issue of the Notices of the AMS on Formal Proof in 2008.  Freek Wiedijk, who wrote one of the Notices articles, has some good resources on his home page.<|endoftext|>
TITLE: How to interpret the Sugawara construction from a physical or mathematical viewpoint?
QUESTION [16 upvotes]: In theoretical physics, the Sugawara theory is a set of formulae and theorems that allow one to construct a stress-energy tensor of a specific type of conformal field theory from a bilinear expression involving currents.
How to interpret the Sugawara construction from a physical or mathematical viewpoint?
Sugawara construction is a kind of method to embed Virasoro algebra into completions of universal enveloping algebras of affine algebras ?
what special properties do this kind of embedding have?
In soliton theory, I know there is a boson-fermion correspondence which realize free boson algebra in the completion of free fermion algebra.
I wonder if there are some common principles under them?

REPLY [4 votes]: Just a historical comment (from my memory; I am at the moment rather far from the subject)
Ivan Todorov gave and wrote few times lectures about the history of Sugawara construction to which he himself also contributed. He even mentions some works in late 1940-s in field theory of photons among curious early occurrences and then the intricate history in 1960's and 1970's, involving Kac, Feĭgin-Fuks, Sugawara, Segal, Todorov etc. Sugawara is in any case not the first in that sequence and he had a wrong proportionality constant calculated, what is corrected in 1970s.
It is also curious that the (affine) Sugawara construction is not confined to working over complex numbers, there is a more general version over more general fields explained by Faltings. I think I have seen it in his paper (I do not have it here so I can not check)
Gerd Faltings, A proof for the Verlinde formula.  J. Algebraic Geom.  3  (1994),  no. 2, 347--374. MR1257326 (95j:14013)<|endoftext|>
TITLE: What is the relation between characters of  a group and its lie algebra?
QUESTION [5 upvotes]: What is the relation between characters of  a group and its lie algebra?
Roughly,I know that there is a one to one correspondence between representations of a lie algebra and its simple connected lie group by the exp map,and two irreducible representations of a lie group are unequivalent if and only if their characters are different. 
Now,then,I think the above statement  will still hold for representations of lie algebras, but I am not sure about it .
If someone can give some advises?
Thanks  Kevin Buzzard. For represatatons of general  lie algebras the character (trace) don't work .
But for compact lie group,characters still work,so I restrict my question on  compact lie groups and their lie algebras.
In representation theory of semisimple lie algebras ,we have formal characters/algebra characters which are sums of some formal elements over the weights.
I wonder whether we can realize these algebra characters to be real characters ?
Or can we make them to be functions on cartan subalgebras or maximal torus?

REPLY [5 votes]: Let $G$ be a compact Lie group. Then the character of a finite-dimenstional representation $R_Λ$ of $G$, is the map $\mathrm{ch}_Λ\colon G \to \mathbb{C}$ defined by
$$
\mathrm{ch}_Λ(g):=\mathrm{Tr}(R_Λ(g))\equiv \sum_{n=1}^{\dimΛ}R^{n,n}_Λ(g), \qquad\forall g\in G,
$$
and it depends  only on the isomorphism class or the representation $R_Λ$. Furthermore, because of the cyclic invariance of the trace the character is a class function. Being class functions, the characters are completely determined by their values on the maximal torus $\mathbb{T}\subset G$, for $G$ semisimple. Now, the characer of the Lie algebra $\mathfrak{g}$ of $G$ is defined as
$$
χ_Λ(h)=\mathrm{Tr}(\exp [R_Λ(h)]),
$$
for $h$ in a Cartan subalgebra $\mathfrak{g_0}$ of $\mathfrak{g}$. If $t\in\mathbb{T}$, then $t=\exp h$ with $h\in\mathfrak{g_0}$, the Cartan subalgebra associated to $\mathbb{T}$. Therefore, one has the formula
$$
χ_Λ(h)=\mathrm{ch}_Λ(\exp h).
$$
However, this does not in general work globally on $\mathbb{T}$ because a single exponential may not be sufficient to obtain all of $\mathbb{T}$ from $\mathfrak{g_0}$ [1, Sec. 21.10] . 
References 
[1] J. Fuchs & C. Schweigert, Symmetries, Lie Algebras and Representations: A graduate course for physicists, Cambridge University Press (2003).<|endoftext|>
TITLE: Count of full, binary trees with fixed number of leaves
QUESTION [5 upvotes]: How many ways is there to build an arithmetic expression with fixed number of terms and fixed order? Let’s assume we have only one distinct operation that is neither commutative nor associative. The problem can be reduced to the question of how many different, full, binary trees could be constructed with a fixed number of leaves.
Suppose we have a list of n elements  which have to become leaves in a full, binary tree. The root could be chosen between each two sequential numbers. Thus, there is (n-1)  different ways to choose the root. If the root is located after the i-th number,  we can still construct the left child as a binary tree with i leaves and the right one with (n – i) leaves.
The formula for the number of different ways to construct a full, binary tree is
$ \Phi(n) = \sum^{n-1}_{i=1}\Phi(i).\Phi(n-i)$
The first value for n=1 ist set to:
$ \Phi(1) = 1 $
Is there a closed formula for this function? Is it - maybe - a popular problem in the Graph Theory with known solutions?

REPLY [12 votes]: This is a very well-known enumeration problem.  The number of full binary rooted trees with $n+1$ leaves is just the $n$-th Catalan number $$C_n = \frac{(2n)!}{(n+1)!n!}.$$  
You can read more by looking at Wikipedia's entry on Catalan numbers, for example.<|endoftext|>
TITLE: $\operatorname{SL}_2(\mathbb R)$ Casson invariant?
QUESTION [15 upvotes]: $\DeclareMathOperator\SL{SL}$Casson's invariant is an invariant of a homology 3-sphere, obtained by
“counting” representations of the fundamental group into $\operatorname{SU}(2)$. 
I was wondering if there is an analogous invariant counting representations
into $\SL(2,\mathbb R)$? Curtis has an invariant counting representations into
$\SL(2,\mathbb C)$. These invariants are obtained by taking a Heegaard splitting
of the manifold, and considering the intersection of the representation
varieties of the two handlebodies in the representation variety of the Heegaard
surface. Casson has to perturb the resulting varieties to make them transverse,
then counts the intersections. Curtis counts only the finite points of intersection
using algebraic geometry to resolve any singularities, and ignoring any
higher dimensional components of the intersection. Then they both have
to show that this count is invariant under stabilization of Heegaard splittings,
and therefore an invariant of the manifold. I was wondering whether one
could combine the two approaches to get an analogous invariant in the 
case of $\SL(2,\mathbb R)$ representations? One would throw away higher dimensional
components of intersection of the $\SL(2,\mathbb R)$ varieties of the two handlebodies,
and perturb near the isolated intersections to get a count of intersection points.
If this works, what about making an analogous Floer theory, by counting 
holomorphic disks between finite intersection points? 
I have't done a literature search, but I suspect this is an open question.

REPLY [5 votes]: A 2020 arxiv posting of Nosaka (An $SL_2(\mathbb{R})$-Casson invariant and Reidemeister torsions) defines an $SL(2,\mathbb{R})$ Casson invariant. As Charlie's answer suggests, the approach is inspired by Johnson's unpublished work.<|endoftext|>
TITLE: Quasi-coherent sheaves in the Functor-of-points approach
QUESTION [14 upvotes]: How do we define quasi-coherent sheaves on schemes?
Say we start by defining the category of affine schemes Aff as CRing$^{op}$ (the opposite category of unitary commutative rings).
In this context we have an obvious way to define quasi-coherent sheaves:
A quasi-coherent sheaf on an affine scheme X=Spec A is just an A-module.
If we now define schemes as presheaves on Aff (satisfying some condition), how do we define what a quasi-coherent sheaf is?
The same question applies also to the operations of pushforward and pullback, which in Aff have obvious definitions.

REPLY [2 votes]: For the non-experts out there, I've found this great expository article by Gomez
http://arxiv.org/PS_cache/math/pdf/9911/9911199v1.pdf
Quasi-coherent sheaves are defined in Definition 2.45 (the same way Dinakar and Yuhao defined them)<|endoftext|>
TITLE: Using linear algebra to classify vector bundles over P^1
QUESTION [43 upvotes]: There is a theorem of Grothendieck stating that a vector bundle of rank $r$ over the projective line $\mathbb{P}^1$ can be decomposed into $r$ line bundles uniquely up to isomorphism. If we let $\mathcal{E}$ be a vector bundle of rank $r$, with $\mathcal{O}_X$ the usual sheaf of functions on $X = \mathbb{P}^1$, then we can write our line bundles as the invertible sheaves $\mathcal{O}_{X}(n)$ with $n \in \mathbb{Z}$. Thus, the decomposition can be stated as $$\mathcal{E} \cong \oplus_{i=1}^n \mathcal{O}(n_i) \quad n_1 \leq ... \leq n_r.$$
If we use the usual open cover of $\mathbb{P}^1$ with two affine lines $U_0 = \mathbb{P}^1 - \{\infty\}$ and $U_1 = \mathbb{P}^1 - \{0\}$, note that $\mathcal{O}_{U_0 \cap U_1} = k[x,x^{-1}]$ (with $\mathcal{O}_{U_0} = k[x]$ and $\mathcal{O}_{U_1} = k[x^{-1}]$). A vector bundle (up to isomorphism) $\mathcal{E}$ of rank $n$ is then a linear automorphism on $\mathcal{O}_{U_0 \cap U_1}^r$ modulo automorphisms of each $\mathcal{O}_{U_i}^r$ for $i = 0,1$. (I am looking at the definition given in Hartshorne II.5.18 where $A = k[x,x^{-1}]$, the linear automorphisms are $\psi_1^{-1} \circ \psi_0$ where $\psi_i: \mathcal{O}_{U_i}^r \rightarrow \left.\mathcal{E}\right|_{U_i}$ are isomorphisms, and the definition of isomorphism of vector bundles allows us to change bases of $\mathcal{O}_{U_i}^r$. 
Thinking of this in linear algebra terms, these linear automorphisms on $\mathcal{O}_{U_0 \cap U_1}^r$ are elements of $GL_r(k[x,x^{-1}])$, and changing coordinates in $\mathcal{O}_{U_i}^r$ are elements of $GL_r(k[x])$ for $i = 0$ and $GL_r(k[x^{-1}])$ for $i = 1$. Thus up to isomorphism, the vector bundles of rank $r$ on $\mathbb{P}^1$ are elements of the double quotient $$ GL_r(k[x^{-1}]) \left\backslash \large{GL_r(k[x,x^{-1}])} \right/ GL_r(k[x]).$$ The decomposition of vector bundles into line bundles SHOULD mean that these double cosets can be represented by matrices of the form $$\left(\begin{array}{cccc} x^{n_1} &  &  & 0 \\ & x^{n_2}&  &  \\ &  & \ddots &  \\ 0 &  &  & x^{n_r}\end{array}\right) \quad n_1 \leq n_2 \leq ... \leq n_r.$$ I want to know whether there is a way to prove this fact purely via linear algebra (equivalently, if the geometric proof [cf. Lemma 4.4.1 in Le Potier's "Lectures on Vector Bundles"] has a linear algebraic interpretation). 
[Note: For the affine case, taking the double quotient $$GL_n(k[x]) \left \backslash M_{n,m}(k[x]) \right/ GL_m(k[x])$$ gives the classification of vector bundles over $\mathbb{A}^1_k$ (and of course, when replacing $k[x]$ with an arbitrary PID, gives the usual structure theorem of finitely generated modules over PID).]

REPLY [6 votes]: I know this result as Birkhoff factorization. I was assigned it as an exercise in an algebraic geometry class and came up with the following rather tedious proof. Below I work over $\mathbb{C}$ but the proof makes no use of any facts about the base field.

Any element of $\mathbb{C}[t, t^{-1}]$ can be written uniquely in the form $t^k f(t)$ where $f$ is a polynomial with nonzero constant term. We call $k$ the $t$-adic valuation of $t^k f(t)$.
We want to classify the double cosets 
\begin{equation} \text{GL}_n(\mathbb{C}[t]) \backslash \text{GL}_n(\mathbb{C}[t, t^{-1}]) / \text{GL}_n(\mathbb{C}[t^{-1}]). \end{equation}
Observe that starting from a matrix in $\text{GL}_n(\mathbb{C}[t, t^{-1}])$ and performing $\mathbb{C}[t]$-linear row operations resp. $\mathbb{C}[t^{-1}]$-linear column operations does not change the coset to which the matrix belongs. 
Beginning from such a matrix, perform $\mathbb{C}[t]$-linear row operations as follows. First, consider the $\mathbb{C}[t]$-submodule of $\mathbb{C}[t, t^{-1}]$ generated by the elements of the first column. This submodule is $t^k$ times an ideal of $\mathbb{C}[t]$ for some $k$, which is principal, hence it is generated by one element. Using row operations, replace one of the entries of the column by a generator, place it in the top row, and eliminate the rest of the entries in the column. Repeat for each column. 
After performing the above row operations, we obtain an upper triangular matrix in the same coset as the original. Each of its diagonal entries is a unit in $\mathbb{C}[t, t^{-1}]$, hence must be an invertible constant times $t^k$ for some $k \in \mathbb{Z}$ (and by multiplying the rows or columns by suitable constants we can assume WLOG that they have the form $t^k$). 
Now perform the following sequence of row and column operations on each diagonal above the main diagonal (do not move on to a new diagonal until each entry in the current diagonal is zero). Using row and column operations, ensure that each term of the diagonal entry $p(t)$ has the property that if $t^{\ell}$ is the main diagonal entry to the left of it and $t^d$ is the main diagonal entry below it, then $p(t)$ contains no terms of degree greater than or equal to $d$ or less than or equal to $\ell$. We will abbreviate this situation using the $2 \times 2$ matrix
\begin{equation} \left[ \begin{array}{cc} t^{\ell} & p(t) \\ 0 & t^d \end{array} \right] \end{equation}
where $p(t) = \sum_{m=\ell+1}^{d-1} p_m t^m$. Let $k < 0$ be the negative integer such that the greatest power of $t$ dividing $t^k p(t)$ is $t^{\ell}$, and perform the column operation
\begin{equation} \left[ \begin{array}{cc} t^{\ell} - t^k p(t) & p(t) \\ -t^{d+k} & t^d \end{array} \right]. \end{equation}
Now perform row operations to eliminate the bottom left entry. The result will have the form
\begin{equation} \left[ \begin{array}{cc} t^{\ell'} & p'(t) \\ 0 & t^{d'} \end{array} \right] \end{equation}
where $\ell'$ is the minimum of the $t$-adic valuation of $t^{\ell} - t^k p(t)$ (which is greater than $\ell$) and $d+k$. But since $p(t)$ by assumption contains no terms of degree greater than or equal to $m$, $d+k$ is necessarily also greater than $\ell$. Hence our row and column operations have ensured that $\ell' > \ell$ and $d' < d$. If $p'(t) = 0$, we are done and can move on to a new entry in the same diagonal or to a new diagonal. Otherwise, we repeat. After a finite number of steps, we will have $\ell^{(N)} \ge d^{(N)}$ (where $N$ is the number of steps), at which point $p^{(N)}(t)$ necessarily vanishes and we can move on to a new entry or a new diagonal. 
The above algorithm eventually produces a diagonal matrix with entries of the form $t^k$ for some $k$, which shows that every double coset in
\begin{equation} \text{GL}_n(\mathbb{C}[t]) \backslash \text{GL}_n(\mathbb{C}[t, t^{-1}]) / \text{GL}_n(\mathbb{C}[t^{-1}]). \end{equation}
contains a matrix of the desired form. It remains to be shown that this matrix is unique up to permutation of its diagonal entries. So suppose two diagonal matrices $D, D'$ with diagonal entries $t^{k_i}, t^{k_i'}$ lie in the same double coset, hence there exist $A \in \text{GL}_n(\mathbb{C}[t])$ and $B \in \text{GL}_n(\mathbb{C}[t^{-1}])$ such that
\begin{equation} D' = A^{-1} DB. \end{equation}
By taking determinants we see that $\sum k_i = \sum k_i'$. Rewrite the above identity as
\begin{equation} AD' = DB. \end{equation}
Let $a_{ij}, b_{ij}$ be the entries of $A, B$. Then the above gives
\begin{equation} t^{k_j} a_{ij} = t^{k_i'} b_{ij} \end{equation}
or
\begin{equation} b_{ij} = t^{k_j - k_i'} a_{ij}. \end{equation}
Since $\sum k_i = \sum k_i'$, it follows that $\sum (k_i - k_i') = 0$. If $\sigma \in S_n$ is a permutation, it follows that 
\begin{equation} \sum_i (k_{\sigma(i)} - k_i') = 0 \end{equation}
hence that either $k_{\sigma(i)} = k_i'$ for all $i$ (in which case we are done) or that there exists an $i$ such that $k_{\sigma(i)} - k_i' > 0$. But $a_{ij} \in \mathbb{C}[t]$ and $b_{ij} \in \mathbb{C}[t^{-1}]$, so this is possible if and only if $a_{ij} = b_{ij} = 0$. Since $A, B$ are not identically zero, there must exist a permutation $\sigma$ such that $k_{\sigma(i)} = k_i'$ for all $i$, and the conclusion follows.<|endoftext|>
TITLE: Info about Elton–Odell theorem
QUESTION [5 upvotes]: Hello everyone, could anyone please tell me where can I find information about the Elton–Odell theorem?
It states:
For any infinite dimensional Banach space $X$ there is a $q > 1$ so that $X$ contains a sequence $(x_n)$ with $\|x_n\|=1$ and $\|x_n-x_m\|\ge q$, whenever $m\ne n$.
Thanks

REPLY [4 votes]: J. Diestel, Sequences and series in Banach spaces, Springer-Verlag, New York, 1984.
(Chapter XIV, page 241)<|endoftext|>
TITLE: How to see the Phase Space of a Physical System as the Cotangent Bundle
QUESTION [30 upvotes]: Two things today motivated this question.
First, the professor said that in a lecture Thurston mentioned

Any manifold can be seen as the configuration space of some physical system.

Clearly we need to be careful here, so the first question is 

1) What is a precise formulation and an argument to see why the previous statement is true.

Second, the professor went on to say that because of the Poisson Bracket, we see the phase space of a physical system as the Cotangent Bundle of a manifold. I understand that we associate a symplectic form to the cotangent bundle, and that we want to think of Phase Space with a symplectic structure, but my second question is

2) Could you provide an example of a physical system, give the associated "configuration manifold" show the cotangent space, and explain why this is the phase space of the system.

I pushed the lecturer quite a bit to get this level of detail, so pushing much further would of probably been considered rude. I should also mention he is speaking about these things because we want to quantize the geometry associated to this manifold. So he looks at the cotangent space which apparently has a symplectic structure, and sends the symplectic form to the Lie Bracket. If any of the things I have said are incorrect, please comment with corrections. I am just learning this material and trying to understand how it fits with my current understanding. 
One last bonus question(tee hee),

If your manifold is a Lie Group, we get a Lie algebra structure on the Tangent Space. Is there a relationship between this Lie Algebra structure and the one you would get by considering the cotangent space and then quantizing in the fashion of above?

Thanks in advance!

REPLY [4 votes]: I don’t have the rep to put this in a comment where it belongs, but wanted to note that TM has a canonical structure analogous to the canonical 1-form of T*M: the “Lagrangian” or vertical vector field. Rather than write a formula, let me explain it intuitively. Any vector space V when viewed as a manifold is naturally isomorphic to the tangent space TV(v) at any point v. Therefore, we can take the (vertical!) tangent space at any point in the fiber of TM over x, and identify it with the fiber itself. If at each point in the fiber TM(x) we choose the tangent vector in T(TM(x)) naturally identified with that point, we end up with a vertical vector field defined on all of TM. With this natural structure in hand, you can now proceed to geometrize Lagrangian mechanics.<|endoftext|>
TITLE: Flatness over non-reduced schemes
QUESTION [6 upvotes]: Is any family of hypersurfaces of fixed degree d (in a projective space of dimension n) over a non-reduced base flat? This is true over reduced bases (assume everything Noetherian), since the Hilbert polynomial is constant. Is there any criterion of flatness of the form: If the Hibert polynomial is constant and (extra conditions that do not assume the base is reduced) then the family is flat? 
Is there any result that guarantees that push forwards and higher push-forwards (via a proper mophism of Noetherian schemes) of a flat over the base sheaf are locally free if the base is non-reduced? This is about the Cohomology and Base Change statements that appear in Mumford (Abelian Varieties) and Hartshorne. In both places the base is assumed to be reduced. However, it seems that this is used often even when the base is non-reduced (such as when constructing the Hilbert scheme - although I must be misunderstanding this proof).  

Thanks a lot!

REPLY [3 votes]: In response to #2:  you mention that cohomology and base change in Mumford and Hartshorne requires reducedness of the target.  I don't have either with me right now, but I agree with VA:  it's possible that you saw versions in these books that had this hypothesis (the proof is easier in this case --- I've seen it called Grauert's Theorem), but that there is a later version (in the same sources) that works for a nonreduced base.  Here is one possible statement (whose hypotheses can be relaxed as usual, e.g. Noetherian hypotheses often can be traded in for finite presentation ones).  Suppose $\pi: X \rightarrow Y$ is a proper morphism to a locally Noetherian scheme, $\mathcal{F}$ is coherent and flat over $Y$, and for each point $y \in Y$, the natural map $\phi^p_y$ from the fiber of the $p$th pushforward to the $p$th cohomology of the fiber is surjective.  Then (i) for any $\rho: Z \rightarrow Y$, restriction to $Z$ commutes with the $p$th pushforward.  In particular, $\phi^p_y$ is an isomorphism.  (ii) Furthermore, $\phi^{p-1}_y$ is surjective for all $y$ if and only if the $R^p \pi \mathcal{F}$ is locally free.  This in turn implies that $h^p$ is constant.  
Better discussion and proof is given here (see for example 25.8-9 in the October 21, 2011 version).<|endoftext|>
TITLE: Topologists loops versus algebraists loops
QUESTION [29 upvotes]: Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?
Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The affine Grassmannian case also provides positive evidence.
Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. Regarding putting the classical topology of X((t))(ℂ), one should not be scared of the ind-scheminess. ℂ((t)) has a natural structure of a topological ring, and hence we topologise X(ℂ((t))) in the usual manner, taking the subspace topology using a closed embedding into affine n-space for some n.
[paragraph redacted]

REPLY [2 votes]: While the answer to your question is negative in general, as pointed out before, the answer is positive for certain type of loop groups. This can be proved using topological twin buildings. See Linus Kramer, Loop Groups and Twin Building. (It is not stated very explicitly, but the topology used on the twin building and hence the algebraic loop group is meant to be the ind-topology coming from the Bruhat cell decomposition.)<|endoftext|>
TITLE: A geometric interpretation of independence?
QUESTION [18 upvotes]: Consider the set of random variables with zero mean and finite second moment. This is a vector space, and $\langle X, Y \rangle = E[XY]$ is a valid inner product on it. Uncorrelated random variables correspond to orthogonal vectors in this space. 
Questions:
(i) Does there exist a similar geometric interpretation for independent random variables
in terms of this vector space?
(ii) A collection of jointly Gaussian random variables are uncorrelated if and only if they 
are independent. Is it possible to give a geometric interpretation for this?

REPLY [5 votes]: If you leave the realm of abstract probability spaces and focus on probability in Banach spaces, there's a lot of geometry to take advantage of.  Here's an example.
Let $X$ be a Banach space, and let $\mathbb P$ be a Radon probability measure on $X$ such that continuous linear functionals are square-integrable (i.e. $\int_X |f(x)|^2 ~d\mathbb P(x) < \infty$ for all $f \in X^*$).  For example, $X = C([0,1])$ with Wiener measure $\mathbb P$.
These are sufficient conditions for there to exist a mean $m \in X$ and covariance operator $K : X^* \to X$ such that $$\mathbb Ef = f(m) \qquad \mathrm{and} \qquad \mathbb E (fg) - f(m)g(m) = f(Kg)$$
for all $f, g \in X^*$.  One can show that
$$\mathbb P \left( m + \overline{KX^*} \right) = 1.$$
Under these very general assumptions, the probability concentrates on the affine subspace generated by the mean and covariance.
Reference:  Vakhania, Tarieladze and Chobanyan, Probability Distributions in Banach Spaces<|endoftext|>
TITLE: Context free grammar + functions = ?
QUESTION [5 upvotes]: If you start with the rules for building a context-free grammar and extend them by allowing left-hand nonterminals to be functions of one or more arguments, does that go beyond the definition of a context-free grammar?  If so (I imagine it does), what class of grammars is it?  Context-sensitive?
To clarify, suppose we have the following grammar:
$S \to F(a,b,c)$
$F(x,y,z) \to xyz$
The grammar above is equivalent to the following because x, y, and z are substituted:
$S \to abc$
Now suppose we leverage the function capability with the following grammar:
$S \to F(\epsilon)$
$F(x) \to x\ a\ |\ x\ a\ F(x\ b)$
($\epsilon$ means empty string)
That grammar produces:
a
aba
ababba
ababbabbba
...

A practical application of functions in a grammar is to express an indentation-formatted computer language.  Trivial example:
statement(indent)
    : indent basic_statement
    | indent header ':' newline statement_list(indent whitespace)
statement_list(indent)
    : statement(indent)*

This means that a statement, given some indentation string, may either be a one-liner (e.g. print 'hello') or it may be a header (e.g. if (x == 5)) followed by ':', a newline, and one or more child statements of a greater indentation level.
Note that statement_list has to be out by itself here, as statement(indent whitespace)* could yield child lines of inconsistent indentation if the whitespace expands to something not constant.  By calling statement_list once with the possibly variable whitespace, we coerce all the child statements to have the exact same indentation.  Once again, functions come to the rescue.
To be more specific on the rules of the grammar system I'm talking about:

The left hand of each rule is either a terminal, a nonterminal, or a function declaration for zero or more arguments.
A function of zero arguments is equivalent to a nonterminal.
The arguments of a function declaration must be single variables (I used $x$, $y$, and $z$ in the examples above).
The arguments of a "function call" (used on the right-hand side of a rule) may contain terminals, nonterminals, and nested function calls.
The arguments of a function call are expanded before being passed.  Thus, functions in this grammar system are no higher than first order (if I'm using the term correctly).

So, by adding "functions" to CFG, what do I get?

REPLY [2 votes]: Have a look at the following two papers of Engelfriet and Schmidt:
@article{engelfried:1977:IOandOI,
        author =        {J. Engelfriet and E. M. Schmidt},
        title =         {{IO} and {OI}. {P}art {I} and {II}},
        journal =       JCSS,
        volume =        {15 and 16},
        pages =         {328--353, 67--99},
        year =          1977
}<|endoftext|>
TITLE: Representation of Groupoids
QUESTION [15 upvotes]: The title is vague, my actuall question is the following:
Has the representations of groupoids been systematically studied? Is there any new phenomenon, compare with the representation of groups? (Either brand new things or pitfalls for those who too familiar with representations of groups). Does this point of view simplified any proof of theorems in representation of groups?
I've been only think of this for 15 mins. But I feel like it might be helpful to think of representation of groupoids for the following reasons:

When one talk about local systems, thinking of it as "representation of the fundamental groupoid" seems more natural than talking about "representation of the fundamental group".
When we talk about modules on stacks, if we choose a presentation of the stack (which is a groupoid), can we treat a given module as a representation of the groupoid? (Just as modules on BG gives representations of G.) Studying when two representations give the "same" module will be interesting.

REPLY [3 votes]: There is a theory of representation of groupoids, that runs parallel to the classical theory of locally compact groups. A locally compact Hausdorff group $G$ always has a left (as well as a right) invariant measure. Using this measure one can define the convolution *-algebra of the group. A classical result is, that there is a natural bijection between the irreducible unitary representations of $G$ and non-degenerate *-representations of the convolution algebra of $G$.
On the similar lines, J. Renault proves, that if $G$ is a locally compact, Hausdorff groupoid with a left invariant continuous family of measures, then one can form a convolution *-algebra of $G$. An irreducible unitary representation of $G$ are defined using a measurable bundle of Hilbert spaces over the space of units of $G$ and a quasi-invariant measure on the space of units of $G$. Then he proves that there is a natural bijection between the irreducible unitary representations of $G$ and the convolution *-algebra of $G$.
When $G$ is a group, this theory is same as the classical theory representation theory.
Indeed, proving this theory for groupoids is very hard and there are many issues to take care of. Groupoids behave differently than groups. For example, a locally compact Hausdorff groupoid need not carry an invariant continuous family of measures. One has to demand the family of measures, as data. As mentioned earlier, a good reference for this purpose is Renault's book,``A groupoid approach to $C^*$-algebras. This theory is very useful for studying dynamical systems and associated $C^*$-algebras.<|endoftext|>
TITLE: Using Weierstrass’s Factorization Theorem
QUESTION [6 upvotes]: I am trying to factorize $\sin(x)\over x$ which by Taylor series expansion and using the roots is $$a \cdot \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 -  \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$
Now I was told that this nasty factor $a$ conveniently becomes $1$ due to Weierstrass’s Factorization Theorem which is a transcendental generalization of the Fundamental Theorem of Algebra.
My question
Could you please show me how $a$ is being neutralized using this theorem? Or don't you even need this theorem to do so?

REPLY [17 votes]: The Weierstrass factorization theorem as usually stated tells you only that $a=e^{g(x)}$ for some entire function $g(x)$.  Hadamard's refinement says a little more, based on the growth rate of the function.  In your case, since $\left| \frac{\sin x}{x} \right| < \exp\left(|x|^{1+o(1)} \right)$ as the complex number $x$ grows, Hadamard tells you that $g(x)$ is a polynomial of degree at most $1$.  Since $\frac{\sin x}{x}$ and $\prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2} \right)$ are both even functions, so is $e^{g(x)}$.  Thus $g(-x)-g(x)$ is a (constant) integer multiple of $2\pi i$.  Hence $g(x)$ is constant, and so is $a$.  Finally, as everyone else has pointed out, taking the limit as $x$ goes to $0$ shows that $a=1$.
See Ahlfors, Complex analysis for more about Hadamard's refinement, which relates the "order" and "genus" of an entire function.<|endoftext|>
TITLE: Applications of homotopy groups of spheres
QUESTION [14 upvotes]: The study of the homotopy groups of spheres $\pi_i(S^n)$ is a major subject in algebraic topology. One knows for example that nearly all of them are finite groups. Some are explicitly known. There is a 'stable range' of indices which one understands better than the unstable part. 
I think that there is an analogy (have to be careful with that word) to the distribution of primes: It seems that there exists a general pattern but no one has found it yet. It is a construction producing an infinite list of numbers (or groups) but no numbers were put into it. Such a thing always fascinates me.
The largely unknown prime pattern leads to applications in cryptography for example. Are there similar applications of the knowledge (or not-knowledge) of the homotopy groups of spheres? Are there applications to real natural sciences or does one study the homotopy groups of spheres only for their inherent beauty?

REPLY [6 votes]: A few comments on applications that aren't covered by the above Wikipedia article.
I don't know any applications to cryptography.  Most cryptosystems require some kind of one-way lossless function and it's not clear how to do that with the complexity of the homotopy groups of spheres.  Moreover, the homotopy-groups of spheres have a lot of redundancy, there are many patterns. 
There's work by Fred Cohen, Jie Wu and John Berrick's where they relate Brunnian braid groups to the homotopy-groups of the 2-sphere.  It's not clear if that has any cryptosystem potential but it's an interesting aspect of how the homotopy-groups of a sphere appear in a natural way in what might otherwise appear to be a completely disjoint subject. 
Homotopy groups of spheres and orthogonal groups appear in a natural way in Haefliger's work on the group structure (group operation given by connect sum) on the isotopy-classes of smooth embeddings $S^j \to S^n$.  I suppose that shouldn't be seen as a surprise though.  Moreover, it's not clear to me that this is always the most efficient way of computing these groups.   But I think all techniques that I know of ultimately would require some input in the form of computations of some relatively simple homotopy groups of spheres.
I think one of the most natural applications of homotopy groups of spheres, Stiefel manifolds and orthogonal groups would be obstruction-theoretic constructions.  Things like Whitney classes, Stiefel-Whitney classes and general obstructions to sections of bundles. Not so much the construction of the individual classes, more just the understanding of the general method.<|endoftext|>
TITLE: Heuristic argument for the prime number theorem?
QUESTION [24 upvotes]: Here is a bad heuristic argument for the prime number theorem. Let n be a positive integer and assume that PNT holds up to n. Then n itself is prime if and only if for each prime p<n the event p|n fails to hold. Assuming that all these events are independent, the probability that n is prime should be around $\prod_{p<n}(1-1/p)$. Minus the logarithm of this is approximately $\sum_{p<n}1/p$. Since the rth prime is about $r\log r$ (by PNT), this is about $\sum_{r<n/\log n}1/r\log r$, which is about $\log\log n$. Therefore, the probability that $n$ is prime is about $\exp(-\log\log n)$, which is $1/\log n$. 
Here are a few criticisms of the above argument. The most glaring problem is that for the approximations to be valid, we need our estimates to be correct up to o(1), and they are not. For instance, it is known that $\sum_{p<n}1/p$ is not $\log\log n+o(1)$, but rather $\log\log n+M$, where M is the Meissel-Mertens constant. We can break this failure down into two subfailures. The first is that minus the logarithm of $\prod_{p<n}(1-1/p)$ differs from $\sum_{p<n}1/p$ by a non-zero constant (plus o(1)). The second is that minus the logarithm of $\prod_{p<n}(1-1/p)$ differs from $\log\log n$ by $\gamma+o(1)$, where $\gamma$ is the Euler-Mascheroni constant.
The second problem is more devastating, since it shows that the independence assumption is seriously flawed. (Everything I have said, by the way, is a well known and often pointed out observation.) My question is whether, despite all these problems, some kind of heuristic argument like this can be salvaged. For instance, it's clear that if p and q are two primes that are fairly large and fairly close, then there will be a repulsion between the events p|n and q|n. Can we say heuristically what the effects of these repulsions should be, and thereby understand where the $\gamma$ comes in?
Just to be clear, I am looking for a simple and non-rigorous argument that does not use the zeta function (except perhaps making use of the product formula in a very elementary way, but I'd rather avoid it altogether) that predicts that if PNT holds up to n then the probability that n is prime should be around $1/\log n$. I'm asking the question because I'm pretty sure the answer will be known, and pretty standard, to many people. It just isn't to me.

REPLY [3 votes]: You might look at Courant and Robbins in the section at the end, "The Prime Number Theorem Obtained by Statistical Methods".
Also, there is an article by Montgomery and Wagon, in which they mention the "Mertens Paradox" and extend the argument of Courant and Robbins.<|endoftext|>
TITLE: Big Picture: What is the connection of Malliavin calculus with differential geometry?
QUESTION [18 upvotes]: I know that Paul Malliavin was heavily influenced by ideas from differential geometry while developing his calculus on Wiener space. But what are the concrete analogies between both areas of mathematics? What has this to do with Hörmander's theorem (if so)?

REPLY [7 votes]: Some of Malliavin's ideas are well-explained in his own book: Stochastic Analysis. 
I think the main geometric idea behind his proof of Hormander's theorem is the idea of submersion. More precisely, if we consider a stochastic differential equation in Stratonovitch form
$dX_t=V_0(X_t) dt +\sum_{i=1}^n V_i(X_t) \circ dW^i_t
$
where $W$ is a Wiener process and the $V_i$'s vector fields on $R^n$ (or a manifold), we can see the solution $X_t$ as function of the path $\{ W_s , s \le t\}$. We thus have a map $\Pi$ from the Wiener space $\mathbb{W}$, which is the path space of $W$ and $\mathbb{R}^n$. 
A key insight of Malliavin is now to see the Wiener space $\mathbb{W}$ as an infinite-dimensional manifold with tangent space the Cameron-Martin space (space of functions with derivatives in $L^2$) and to realize that Hormander's conditions make the  map $\Pi$ a submersion. As a consequence the Wiener measure is mapped into a smooth measure, which means that if Hormander's conditions are satisfied then the distribution of $X_t$ has a smooth density. It implies the hypoellipticity of the generator of the Markov process $(X_t)_{t \ge 0}$ which is $L=V_0+\frac{1}{2} \sum_{i=1}^n V_i^2$.
Of course, this has to be made rigorous and some care has to be given because we have to use Sobolev type derivatives instead of Frechet derivatives.
This idea that there is a "natural" manifold structure on path spaces is quite fruitful and besides Malliavin's proof of Hormander's theorem explains and leads to several interesting properties of the operator $V_0+\frac{1}{2} \sum_{i=1}^n V_i^2$.<|endoftext|>
TITLE: Spins as tensor fields 
QUESTION [5 upvotes]: I have often come across this implicit translation of the classical field of a particle of a given spin into a specific tensor field. But I could not locate any literature from which I could learn this.
In a paper of Avrimidi I found this statement,
"The tensor fields describe the particles with integer spin while spin-tensor fields describe particles with half-odd spins" 
I would like to know what is the precise mathematical map that he is referring to. 
Like if I have a particle of spin s on a space-time manifold M then the classical field of this particle is a section of which bundle on the space-time? 
In the same vein I would like to know some expository mathematical reference on what is a "spin connection". I find the usual physics book definition very unconvincing where spin-connection coefficients are just 'defined" analogous to the Christoffel symbols (but with the signs flipped) when the connection is made to act on vierbeins.

REPLY [4 votes]: In a nutshell, particles "are" unitary irreducible representations of the Poincaré group, which is the isometry group of Minkowski spacetime on which it acts transitively.  Such representations can be constructed using the method of induced representations (cf. Wigner, Bargmann, Mackey,...) as classical fields on Minkowski spacetime subject to certain field equations: wave, Klein-Gordon, Weyl, Dirac, Maxwell, Rarira-Schwinger,...   Mathematically these are sections of homogeneous vector bundles associated to certain finite-dimensional representations of the "little group", which is (the maximal compact subgroup of) the stabilizer (in the spin cover of the Lorentz group)  of a point on the "mass shell" (=the momenta $p$ with $p^2 = - m^2$, where $m$ is the mass of the particle).  The little group for massive representations is isomorphic to $SU(2)$, whereas that of massless particles to a nontrivial double cover of $SO(2)$.  Hence massless particles are defined by their helicity (the label of the $SO(2)$ representation from which one induces) and massive particles by their mass and their spin (the label of the irrep of $SU(2)$ from which one induces).  The covariant field equations are (the Fourier transform of) the projectors onto irreducible components.
Let's discuss the massive case, since your question mentions spin explicitly.
There are two kinds of irreps of $SU(2)$, those where $-1$ acts trivially and those where it does not.  The former case are the integer spin representations whilst the latter are the half-integer representations.
The integer spin representations of $SU(2)$ are contained in the tensorial representations of the Lorentz group, whence this is why the fields for integer-spin massive particles are tensorial.
The irreps of $SU(2)$ with half-integer spin, those where $-1$ does not act trivially, are not contained in tensorial representations of the Lorentz group, since on these representations $SU(2)$ acts by conjugation and $-1$ acts trivially.  In order to describe such reps in terms of fields you need to consider spinor fields, which are sections of spinor bundles (possibly twisted by tensor bundles for higher spin fields).
You can read about spinor bundles on any book in Spin Geometry.  For example, there's a book by Lawson and Michelsohn of that name.  Green-Schwarz-Witten's string theory book (second volume) has a physicsy discussion of this.  They will also define the spin connection, which is a connection induced by the Levi-Civita connection on any "spin bundle", which is a principal fibre bundle lifting the bundle of oriented orthonormal frames in such a way that the bundle map between them restricts to the spin covering $\mathrm{Spin} \to \mathrm{SO}$ on the fibres.
It is not hard to show that the connection one-form for the spin connection, when pulled back to the manifold by the lift of a local frame (hence a local 1-form with values in the orthogonal Lie algebra) agrees with the similar expression for the Levi-Civita connection, which is why many books perhaps do not go through the trouble of defining it properly.  It also requires introducing quite a bit of formalism, to which many physicists are allergic; although increasingly less so.<|endoftext|>
TITLE: How to show the galois group of a polynomial is not an alternating group?
QUESTION [11 upvotes]: I am trying to prove that a certain class of polynomials have symmetric galois group.
Using the Newton polygon, I have shown that the galois groups of these polynomials are transitive on $k$-sets for all $k$ less than the degree of the polynomial.  Beaumont and Peterson proved that this condition holds only for symmetric and alternating groups, along with 4 sporadic exceptions.  I have discounted the 4 exceptions, so I am left with the task of showing that the alternating group does not occur.
I am looking for a way to do this.  I have tried without success to show that there must be a transposition in the group; likewise showing that the determinant is not a square does not seem to be possible.
Does anybody know of any tricks for showing that one of the above sufficient conditions holds?  Alternatively, do you know of any other way to prove that a group is the symmetric group, or not the alternating group?

REPLY [12 votes]: There is a version of the Chebotarev density theorem for finitely generated fields, or more precisely, after spreading out, for an étale Galois cover of schemes of finite type over a ring of $S$-integers.  This is a consequence of work surrounding the Weil conjectures.  See Lemma 1.2 in Torsten Ekedahl, An effective version of Hilbert's irreducibility theorem, Séminaire de Théorie des Nombres, Paris 1988-89, Birkhäuser 1990.
But what I would recommend trying first is to use the monodromy groups.  If you spread out your Galois extension to a finite branched cover of a dense open subscheme $U$ of $\mathbf{P}^n_{\mathbf{Q}}$, then the monodromy group (inertia group) associated to any irreducible divisor on $U$ is contained in your Galois group.  In particular, if there is a divisor that ramifies in the simplest possible way (ramification indices $2,1,1,\ldots,1$), then you know that your group contains a transposition.
The other approach to try, already mentioned by others in the comments, is specialization, which amounts to spreading out as above and then restricting to the cover above an irreducible closed subscheme.  More concretely, you could plug in rational numbers for some or all of $y_1,\ldots,y_n$ such that you get a separable polynomial (of the same degree as the original polynomial).  Or you could spread out to a scheme of finite type over $\mathbf{Z}$ and then restrict to an irreducible closed subscheme; this includes reduction modulo primes.  If specialization results in a separable polynomial of the same degree over the new function field, then the Galois group of the specialization is a subgroup of the original Galois group, so it suffices to show that the specialized Galois group is not contained in $A_n$.
For more techniques for computing Galois groups, I recommend the book Topics in Galois theory by Jean-Pierre Serre.<|endoftext|>
TITLE: Local complementation in undirected graphs
QUESTION [8 upvotes]: Problem statement
Let $G=(V,E)$ be an undirected graph whose vertices are either black or white. A local complementation of $G$ with respect to a black vertex $v$ consists in:

complementing the subgraph induced by $v$ and its neighbours,
flipping the colour of each neighbour of $v$ (i.e. black vertices become white and conversely), and finally
removing $v$ from $V$.

The goal is to delete the whole graph using only local complementations.
Questions
Given an ordering $\mathcal O$ of the vertices of $V$, can we characterise cases in which  $\mathcal O$ allows us (or not) to delete $G$?
Comments
A lot of work on local complementations (or "vertex eliminations" in some papers) concerns itself with algorithmic issues, especially with finding orderings that will work. Note that this differs from my question, since here you don't get to choose an ordering. 
Of course, verifying whether an ordering works is easy: keep complementing until you're done or stuck. Finding necessary or sufficient nontrivial structural conditions on $G$ or $\mathcal O$ seems harder. Does this problem ring any bell?
Example
Two different orderings for the same graph; the first one does not work:
http://homepages.ulb.ac.be/~alabarre/local-complementation-1.png
The second one does:
http://homepages.ulb.ac.be/~alabarre/local-complementation-2.png
References
Hannenhalli and Pevzner, starting from page 14, and Hartman and Verbin. All other authors (e.g. Sabidussi) consider variants like using directed graphs, or non-coloured vertices, or complementations which do not modify edges adjacent to $v$. Other authors whose papers I'm currently looking into are Donald J. Rose, Robert Endre Tarjan and François Genest.

REPLY [4 votes]: Warning. I just realized that my reduction is not good as if a node has two outputs, their will be new edges created between them, so we would need a more complicated gadget. I suspect this to be doable, but as meanwhile the question turned out to have a different motivation (see Parity below), I did not give it much thought.
No, it is not possible to characterize them. Of course it is possible to find some conditions for special cases but do not hope to find any simple/local dependent rule as this problem is P-complete. Below I give a sketch of how to reduce CVP (Circuit Value Problem) to it.
Before I start one simple observation that you did not mention. For every graph with ordered vertices, there is exactly one coloring that deletes the whole graph. This suggests that in the reduction true variables should correspond to certain edges in our graph rather than to colors.
There will be one main gadget that we use, which I tried to depict here with my humbling artistic laziness.
alt text http://dcg.epfl.ch/webdav/site/dcg/users/184485/public/gadget.JPG
Now I'll try to sketch the reduction. To every node of the circuit we will have a corresponding pair of vertices, $v_1$ and $v_2$, which are next to each other in the order. They will be both black and unconnected before we start deleting them if the node is evaluated as TRUE while $v_1$ will be black, $v_2$ white and they will be connected before we start deleting them if the node is evaluated as FALSE. Thus the above gadget is a negation gate.
Note that the color of 3 changes iff our first variable is true. Thus basically we can modify a later part of the graph arbitrarily by using several copies of this gadget. If we combine two gadgets, we can also simulate an AND gate. For this it is enough to show how to have an edge iff two variables are true. Let A, B, C have this order, at the beginning A being black, A being the "4" in two gadgets, the role of "3" played by B and C in the two gadgets. Then after deleting both variables and A, the color of B and C will be unchanged and there will be an edge between iff both variables were true.
I know that this is a lengthy answer and uses complexity instead of some nice combinatorial observations, but I think it helps to show why there can be no simple characterization. Let me know if you have any questions/would like a more detailed exposition.
 Parity. If a graph satisfies that every black vertex has an odd degree and every white vertex has an even degree, then it is not possible to delete it. The proof is by induction on the number of vertices. If there is only one vertex, it must be white, thus we are struck. Otherwise, whenever you delete a black vertex, the parity of the degree of all of its neighbors will change as well as their color, thus we are done by applying the induction hypothesis to this graph.<|endoftext|>
TITLE: Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
QUESTION [5 upvotes]: Max Koecher (for example, in The Minnesota Notes on Jordan Algebras and Their Applications; new edition: Springer Lecture Notes in Mathematics, number 1710, 1999), defined a domain of positivity for a symmetric nondegenerate bilinear form $B: X \times X \rightarrow \mathbb{R}$ on a finite dimensional real vector space $X$, to be an open set $Y \subseteq X$ such that $B(x,y) > 0$ for all $x,y \in Y$, and such that if $B(x,y) > 0$ for all $y \in Y$, then
$x \in Y$.  (More succinctly, perhaps, we could say it's a maximal set $Y \subseteq X$ such that $B(Y,Y) > 0$.)  Aloys Krieger and Sebastian Walcher, in their notes to chapter 1 of this book, state that "In the language used today, a domain of positivity is a self-dual open proper convex cone."  [I now believe this is wrong; see my answer below for what I think is true instead.]  It's quite easy to prove that it's an open proper convex cone.  (Proper means it contains no nonzero linear subspace of $X$, i.e. that its closure is pointed.)  But, although I have a vague recollection of having encountered a proof once in a paper on homogeneous self-dual cones, I haven't succeeded in finding it again, or in supplying it myself.  I'm pretty sure Krieger and Walcher's claim is correct—for example, the 1958 paper by Koecher that is generally cited (along with a 1960 paper by Vin'berg) for the proof of the celebrated result that the (closed) finite-dimensional homogeneous self-dual cones are precisely the cones of squares in finite dimensional formally real Jordan algebras, is titled "The Geodesics of Domains of Positivity" (but in German).
The most natural way to prove this would be to find a positive semidefinite nondegenerate $B'$, such that the cone is a domain of positivity for $B'$ as well.  In principle, $B'$ might depend on the domain $Y$.  (While maximal in the subset ordering, domains of positivity for a given form $B$ are not unique.)  But a tempting possibility, independent of $Y$, is to transform to a basis for $X$ in which $B$ is diagonal, with diagonal elements $\pm 1$, change the minus signs to plus signs, and transform back to obtain $B'$.  
To clarify the question: we will define a cone $K$ in a real vector space $X$ to be self-dual iff there exists an inner product—that is, a positive definite bilinear form $\langle . , . \rangle: X \times X \rightarrow \mathbb{R}$—such that $K = K^*_{\langle . , . \rangle}$.  Here $K^*_{\langle . , . \rangle}$ is the dual with respect to the inner product $\langle . , . \rangle$, that is $K^*_{\langle . , . \rangle} := \{ y \in X: \forall x \in X ~\langle y, x \rangle > 0 \}$.  So in asking for a proof that a domain of positivity is a self-dual cone, we are asking whether some inner product $\langle . , . \rangle$ with respect to which $K$ is self-dual exists.  Above, I considered the case $K=Y$, and called the inner product I was looking for, $B'$.
Does anyone know, or can anyone come up with, a proof?

REPLY [3 votes]: In the book by Faraut and Koranyi "Analysis on symmetric cones" (Exercise 10 on p. 21 and note on p. 23) it is stated that it was Vinberg in

Vinberg, Homogeneous cones, Soviet Math. Dokl. 1, (1960) 787-790

who discovered that there are homogeneous cones which are not self dual with respect to any scalar product.<|endoftext|>
TITLE: Does every decidable question about finitely presented groups amount to a question  about abelian groups?
QUESTION [19 upvotes]: This question is about an issue left unresolved by Chad
Groft's excellent
question and
John Stillwell's excellent
answer of
it. Since I find the possibility of an affirmative answer
so tantalizing, I would like to pursue it further here.
For background, Rice's Theorem
asserts essentially that no nontrivial question about computably enumerable sets is decidable. If We is the set enumerated by program e, then the theorem states:
Rice's Theorem.  If A is a collection of computably
enumerable sets and { e |
We ∈ A } is decidable, then either A is 
empty or A contains all computably enumerable sets.
In short, one can decide 
essentially nothing about a program e, if the answer is to
depend only on what the program computes rather than how it
computes it. 
The question here is about the extent to which a similar
phenomenon holds for finitely presented groups, using the
analogy between programs and finite group presentations:

a program e  is like a finite group presentation p
the set We enumerated by e is like the group
⟨p⟩ presented by p.

According to this analogy, the analogue of Rice's theorem
would state that any decidable collection of finitely
presented groups (closed under isomorphism) should be either trivial or everything.
John Stillwell pointed out in answer to Chad Groft's
question that this is not true, because from a presentation
p we can easily find a presentation of the abelianization of
⟨p⟩, by insisting that all generators commute, 
and many nontrivial questions are decidable about finitely
presented abelian groups. Indeed, since the theory of
abelian groups is a decidable theory, there will be many
interesting questions about finitely presented abelian
groups that are decidable from their presentations. 
My question is whether this is the only obstacle.
Question. Does Rice's theorem hold for finitely
presented groups modulo abelianization? 
In other words, if
A is a set of finitely presented groups (closed under
isomorphism) and the corresponding set of presentations { p | ⟨p⟩ ∈ A } is
decidable, then does A completely reduce to a question
about the abelianizations of the groups, in the sense that there is a set B of abelian groups such that G ∈ A iff Ab(G) ∈ B? 
Of course, in this case B consists exactly of the abelian groups in A. The question is equivalently asking whether A respects the equivalence of groups having isomorphic abelianizations. In other words, must it be that G ∈ A iff Ab(G) ∈ A? 
The question is asking whether every decidable set of finitely presented groups amounts actually
to a decidable set of abelian groups, extended to all
finitely presented groups just by saturating with respect to abelianization. 
In particular, the set A should contain either none or all perfect groups. 
An affirmative answer would seem to provide a thorough
explanation of the pervasive undecidability phenomenon in
group presentations. But perhaps this may simply be too much to hope for...
In any event, I suppose that there is an equivalence relation on finite group presentations, saying that p ≡ q just in case ⟨p⟩ and ⟨q⟩ have the same answer with repsect to any decidable question about finitely presented groups. The question above asks whether this equivalence relation is just Ab(⟨p⟩) = Ab(⟨q⟩). If this turns out not to be true, then what can be said about ≡?

REPLY [5 votes]: One more counterexample.  Makanin's algorithm solves systems of equations and inequations over any non-abelian free group F.  The statement
'$G=\langle  x_1,\ldots, x_m\mid r_1,\ldots,r_n\rangle$ surjects a non-abelian free group'
is equivalent to the negation of the system of equations and inequations 
$ (r_i=1 \mid \forall i) \Rightarrow ([x_i,x_j]=1\mid\forall i,j) $
over F, where $x_i$ are now interpreted as variables and $r_j$ are now interpreted as equations.
Therefore, the statement is decidable.  On the other hand, $F$ surjects a non-abelian free group, but its abelianisation does not.<|endoftext|>
TITLE: The urge to combine 1- and 2-morphisms in slicing a 2-category.
QUESTION [16 upvotes]: Suppose that $C$ is a 2-category, perhaps $C=\rm{Cat}$, the 2-category of small categories, functors, and natural transformations.  Let $T$ be an object in $C$.  
I form the new 1-category whose objects are morphisms $f\colon A\rightarrow T$ in $C$, and in which a morphism from $f$ to some $f'\colon A'\rightarrow T$ consists of a pair $(\phi,\phi^\sharp)$ where $\phi\colon A\rightarrow A'$ is a 1-morphism in $C$ and $\phi^\sharp\colon\phi\circ f'\rightarrow f$ is a 2-morphism between arrows $A\rightarrow T$. Call this new category the $(C\Uparrow T)$.  An obvious variation comes about by reversing the direction of the 2-morphism, i.e. we could take $\phi^\sharp\colon f\rightarrow\phi\circ f'$; perhaps I might call this variation $(C\Downarrow T)$.
What is the high-brow way to refer to these strange slice-categories?  How do you locate them within a good understanding of 2-categories?  Where are the properties of such things discussed?  What is the relation between these strange slices and the usual 2-categorical slices?
Thanks!

REPLY [4 votes]: For what it's worth, the construction you describe features prominently in http://front.math.ucdavis.edu/0807.4146, though that paper does not use 2-categorical language.
More generally, given a 2-category $C$ (which I usually assume is pivotal, though maybe that's not necessary here), one can construct a 1-category $D$ whose objects are 1-morphisms $f:a\to b$, and whose morphisms are "rectangles": the domain $f:a\to b$ along the bottom, the range $f':a'\to b'$ along the top, additional 1-morphisms (of $C$) $g:a\to a'$ and $h:b\to b'$ along the right and left sides, and a 2-morphism of $C$ filling in the rectangle.  Composition in $D$ is given by stacking the rectangles vertically.  I like to think of the pair $(g, h)$ as the (bi)grading of the morphisms of $D$.  What you describe is the case where we restrict $h$ to be an identity 1-morphism (of $C$).  In the paper linked to above we put an inner product on $D$ and complete it to a von Neumann algebra (in fact, a factor).
In response to David's comment below:
Modulo some details, a planar algebra is equivalent to a pivotal 2-category whose 2-morphisms are vector spaces and whose 1-morphisms are finitely generated.  The standard example is constructed from a pair of factors (irreducible von Neumann algebras) $N\subset M$.  From this data we construct a 2-category whose objects are $N$ and $M$, whose 1-morphisms are generated by the two bimodules $_N M_M$ and $_M M_N$, and whose 2-morphisms are intertwinors.
(So for example the 1-morphisms are 
$M\otimes_N M\otimes_N\cdots\otimes_N M$, thought of as either an $N$-$N$ or $N$-$M$ or $M$-$N$ or $M$-$M$ bimodule.)  You can think of the usual planar algebra definition as axiomatizing the "string diagrams" you would draw for this 2-category.
The diagrams in the paper I referred to are rotated 90 degrees from my explanation above.  The left and right sides of the rectangles in the paper correspond to the $f$ and $f'$ of your (David's) original question.  The tops of the rectangles corresponds to your $\phi$, and the interiors of the rectangles correspond to your $\phi^\sharp$.<|endoftext|>
TITLE: subgroup of SU(N) with maximal manifold dimension 
QUESTION [9 upvotes]: Given the group SU(N) of NxN unitary matrices, does there exist a subgroup S
with a manifold dimension larger than the SU(N-1) manifold dimension and
smaller than the SU(N) one? S should not necessarily have SU(N-1) as subgroup.

REPLY [3 votes]: Your question is equivalent to the question about maximal dimension of a proper Lie subalgebra
of $su(N)$. Clearly such subalgebra is reductive since it has a positive definite invariant form.
Thus you are looking for a reductive Lie algebra of maximal possible dimension strictly smaller
than $N^2-1$ with
faithful representation $V$ of dimension $N$. We claim that this dimension is $(N-1)^2$ for $N>4$.
Here is a sketch of argument:
1) representation $V$ is irreducible: if it splits into direct sum of subrepresentations of dimensions $a$ and $b$, then dimension of our Lie algebra is less than $a^2+b^2\le (N-1)^2$ (one needs to
consider the special case when $a=1$ separately).
2) representation $V$ is not a tensor product of irreducible representations: if it is a tensor
product of representations of dimensions $a$ and $b$, then our Lie algebra has dimension less than $a^2+b^2\le (N/2)^2+(N/2)^2=N^2/2\le (N-1)^2$.
3) it follows from 2) above that our Lie algebra is in fact simple Lie algebra. The smallest possible dimension of an irreducible representation of any simple Lie algebra is well known 
(this is always one of the fundamental representations). Now quick search gives you the result
(and the counterexample pointed out by Somnath).<|endoftext|>
TITLE: Do decidable properties of finitely presented groups depend only on the profinitization?
QUESTION [15 upvotes]: This is a just-for-fun question inspired by this one. Let $P$ be a property of finitely presentable groups. Suppose that

The truth of $P(G)$ only depends on the isomorphism class of $G$.
Given a finite presentation of $G$, the truth of $P(G)$ is computable.

Let $\hat{G}$ denote the profinite completion of $G$. Is it possible to have groups $G$ and $H$, and such a property $P$, so that $\hat{G} = \hat{H}$ but $P(G) \neq P(H)$?
For example, is there a computable property which separates Higman's group from the trivial group?

REPLY [5 votes]: OK, I think I have an example of two groups with the same profinitization and a computable property which distinguishes them. The point is that very fine detail about the commutator subgroups can't be seen in the profinitization.
Let $q$ be prime and let $K$ be the $q$-th cyclotomic field.
Choose $q$ such that the class group of $K$ is not trivial. Let $I$ be a trivial ideal of $\mathcal{O}_K$ and $J$ a nontrivial ideal. Our groups $G$ and $H$ will be $(\mathbb{Z}/q) \ltimes I$ and $(\mathbb{Z}/q) \ltimes J$.
For any group $B$, let $B' = [B,B]$ and $B'' = [B', B']$. Note that $B/B'$ acts on $B'/B''$ by conjugation. Our computable criterion is the following:

$B/B' \cong \mathbb{Z}/q \times \mathbb{Z}/q =: A$, the action of the group ring $\mathbb{Z}[A]$ on $B'/B''$ factors through a map $\mathbb{Z}[A] \to \mathcal{O}_K$ and, as such, $B'/B''$ is a free $\mathcal{O}_K$ module.

We leave it as an exercise that $G$ satisfies this condition and $H$ does not. 
I believe this condition should be computable. We can go from a finite presentation of $B$ to one of $B'$. (UPDATE I have revised this argument.) Abelianizations are computable, so we can check whether $B/B'$ has the right format. If it does, then $B'$ has finite index in $B$. I think we can use this to get a finite presentation of $B'$: Let $\Delta$ be a two-dimensional $CW$-complex with one vertex, an edge for each generator of $B$ and a two cell for each relation. Let $\Delta'$ be the cover of $B$ corresponding to $B'$. Since $B$ has finite index in $B'$, $\Delta'$ will have finitely many cells, and we get a finite presentation of $B'$.
We can the compute the abelianization of $B'$ and, I think, the action of the abelianization of $B$ on that of $B'$ should be computable. Note that there are only $q^2$ maps from $\mathbb{Z}[A]$ to $\mathcal{O}_K$, so we can just check them each in turn. The class of a finite generated module for a Dedekind domain should be computable by standard number theory methods, although I admit I couldn't describe them.
The fact that these two groups have the same profinitization is relatively well known. Let $\hat{I}$ and $\hat{J}$ denote the profinite completions of $I$ and $J$. The profinite completions of $G$ and $H$ are $\mathbb{Z}/n \ltimes \hat{I}$ and $\mathbb{Z}/n \ltimes \hat{J}$.
We can identify $\hat{I}$ and $\hat{J}$ with submodules of $\mathbb{A}^0_K$, the integral adeles of $K$. Since $I$ and $J$ are locally principal, these are principal ideals in the ring $\mathbb{A}^0_K$. They are thus equivalent as $\mathbb{A}^0_K$ modules, and thus as $\mathcal{O}_K$ modules.<|endoftext|>
TITLE: Are there positive formulae for the inner product between elements of a Lie algebra representation in the Shapovalov form?
QUESTION [6 upvotes]: On a simple representation of a simple Lie algebra, there is a unique bilinear form called the Shapovalov form for which the actions of $E_i$ and $F_i$ are biadjoint, and some fixed highest weight vector has $\langle v_h,v_h\rangle=1$.  
The representation has a distinguished collection of vectors $F_{i_1}\cdots F_{i_n}v_h$ for all sequences $\mathbf{i}$.  One can calculate any inner product $\langle F_{i_1}\cdots F_{i_n}v_h, F_{j_1}\cdots F_{j_n}v_h\rangle$, by simply moving the $F_j$'s to become $E_j$'s on the other side, and commuting them past the $F_i$'s.  This is not hard to do computationally, but the formulas one gets are not positive, which is annoying for my purposes.  

Does anyone know of positive formulae for these inner products?  What about their $q$-analogues for quantum groups?

EDIT:  I should note, following Allen's comment: I'm pretty sure that I know a vector space that has the dimension which is this inner product.  There's also a positivity proof using the canonical basis (all the elements I'm interested in are positive linear combinations of canonical basis elements).
I'm trying to show that a surjective map to this vector space is an isomorphism, and do so by finding a spanning set of the domain that has the right cardinality.

REPLY [3 votes]: Ben, my paper on the Shapovalov form does give a generating series for the entries of a Gram matrix in Corollary 3.4, and those entries are evidently positive. It is not very hard to  deduce a q-version of this generating series from the paper by Chari and Jing appearing in the references. This is not really what you want, though, since the calculations only apply to the basic representation.
In general, you are asking a hard question. The only place I can think to point you is Jantzen's thesis where he calculates the determinant of the Shapovalov form for irreducible representations of simple finite-dimensional Lie algebras. Of course, this isn't the easiest document to get and it is in German.
Otherwise, you might try to prove the Khovanov-Lauda conjecture on cyclotomic quotients of quiver Hecke algebras and see what that gets you. At any rate, there is a representation theoretic interpretation of the Shapovalov form on irreducible modules in type A in the "Graded Decomposition Numbers" paper of Brundan and Kleshchev.<|endoftext|>
TITLE: Topological HNN extensions
QUESTION [11 upvotes]: First, let me recall what an abstract HNN extension is. Let $G$ be an abstract group, $A, B < G$ be subgroups of $G$ and $\phi : A \to B$ be an isomorphisms. Then there is a group $H$ and an element $h \in H$ such that $G < H$ and for any $a \in A$ one has $\phi(a) = hah^{-1}$. Such an extension is, certainly, not unique, but there is a canonical way to construct one. The group $H$ obtained in this canonical way is called HNN extension of $G$ by $\phi$. More information can be found in a nice wikipedia article.
My question is: for what classes of topological groups HNN extension is possible? 
More formally, let $G$ be a topological group, $A, B < G$ and $\phi : A \to B$ is an isomorphism in the category of topological groups. Is there a topological group $H$ that contains $G$ and such that $A$ and $B$ are conjugated inside $H$? If yes, what properties $H$ inherits from $G$, e.g., if $G$ is metrizable can $H$ be chosen metrizable?
Let me also mention that classical HNN construction uses free products and amalgamation of abstract groups. It is known that amalgamation of two Hausdorff topological groups over a closed subgroup may not be Hausdorff (though amalgamation, that appears in HNN extension is quite special, I don't know how bad it is).
The simplest case when $A$ is generated by one element already seems interesting.

REPLY [4 votes]: Topological HNN-extensions have been considered, e.g. in that paper by S. Gal and T. Januszkiewicz: http://www.heldermann-verlag.de/jlt/jlt13/galpl.pdf<|endoftext|>
TITLE: Can a connected planar compactum minus a point be totally disconnected?
QUESTION [33 upvotes]: What the title said.  In a slightly more leisurely fashion:-

Let $X$ be a compact, connected subset of $\mathbb{R}^2$ with more than one point, and let $x\in X$.  Can $X\smallsetminus\{x\}$ be totally disconnected?

Note that the Knaster-Kuratowski fan shows that, in the absence of the compactness hypothesis, the answer can be 'yes'.
To give credit where it's due, this question was inspired by one that I was asked by Barry Simon.

REPLY [4 votes]: I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.
Proposition. Let $X$ be a connected metric space that contains more than one point. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.
Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.
Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step. 
Corollary. If $X$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.
In particular, we cannot remove a closed totally disconnected set (e.g. a single point) from a continuum to make it totally disconnected.<|endoftext|>
TITLE: An identity for the cosine function
QUESTION [5 upvotes]: Let $x = \pi/(2k+1)$, for $k>0$.
Prove that
$$
\cos(x)\cos(2x)\cos(3x)\dots\cos(kx) = \frac{1}{2^k}
$$
I've confirmed this numerically for $n$ from $1$ to $30$.
I'm finding it surprisingly difficult using standard trigonometric formula manipulation.
Even for the case $k = 2$, I needed to actually work out $\cos x$ by other methods to get the result.
Please let me know if you have a neat proof.

REPLY [2 votes]: Another proof arises from the identity
$$ \prod_{m=-k}^k\cos\left(t+\frac{m\pi}{2k+1}\right) = 2^{-2k}\cos((2k+1)t),\tag{1}$$
by putting $t=0$ and taking the square root of both sides. We can deduce (1) from 
$$ \prod_{m=-k}^k \left(z\exp\left(\frac{m\pi i}{2k+1}\right)+z^{-1}\exp\left(\frac{-m\pi i}{2k+1}\right)\right)=z^{2k+1}+z^{-2k-1}\tag{2},$$
by putting $z=\exp(it)$. Finally, we can prove (2) by multiplying both sides by $z^{2k+1}$ and examining the roots of the resulting two polynomials in $z$.<|endoftext|>
TITLE: About state-field correspondence
QUESTION [5 upvotes]: In the definition of vertex algebra, we call the vertex operator state-field correspondence, does that mean that it is an injective map??
Are there some physical interpretations about  state-field correspondence ? Or why we need state-field correspondence in physical viewpoint?? 
Does it have some relations to highest weight representations?

REPLY [12 votes]: I want to elaborate a little on Pavel's excellent answer.
We can think (very schematically) of local operators in an n-dimensional 
field theory the following way. We have an n-1 manifold M with some additional 
structures (topological, conformal, metric etc), to which our field theory assigns a vector space Z(M) of states. Given x in M and a time t in the interval,
we can ask for local operators on Z(M) at the point x and time t. This
can be visualized (following field theory axiomatics) as follows: we cross 
M with an interval, and cut out a tiny ball around the point (x,t) in
this cylinder. We obtain a cobordism (with additional structure) between
M times $S^{n-1}$ and M. We can then use the field theory axioms to turn states $Z(S^{n-1})$ into operators on $Z(M)$. Physically we think of inserting measurements on fields
on spacetime M times interval that only ask about the value of fields in a small (punctured) neighborhood of (x,t).
In general this is a very complicated structure. But if we're in a topological field theory, then this picture is independent of lots of things - such as most importantly the size and shape of the ball we removed (as well as its position). In a 2d CFT at least we know this structure is independent of size and shape of the disc we've cut out, and depends holomorphically on z=(x,t) a point in a Riemann surface. 
For simplicity though let's stick to TFT, since this picture works equally well in any dimension. If we apply this idea to the case $M=S^{n-1}$ itself, we find that $Z(S^{n-1})$ has an algebra structure --- in fact an algebra structure parametrized by cutting a ball out of a cylinder (if you look at this carefully you find the topologist's notion of $E_n$ algebra -- for $n=1$ it's simply associative, for $n=2$ it's "braided" (commutative in a coarse sense) and it gets more and more commutative as n increases).
Moreover Z(M) for ANY M is now a module over this algebra.
This is how I think of state-field correspondence: states on the n-1 sphere are equivalent to local operators in the field theory acting on any space (these are the fields). In chiral CFT we find the notion of vertex algebra immediately from this -- it's a conformal refinement of the abstract notion of $E_2$ (or "braided") algebra we derived above from TFT, where now things depend holomorphically rather than locally constantly on parameters..
EDIT: One more piece of data here is the unit - there's a canonical state on the (n-1) sphere, given by considering it as the boundary of the ball we cut out (ie doing the path integral on the ball with boundary conditions on the sphere..) This is the vacuum state. It's easy to see it corresponds to the identity operator on $Z(M)$ for any $M$, and is the unit for the algebra structure on $Z(S^{n-1})$. We now recover the injectivity of the state-field correspondence: we consider the pair of pants (punctured cylinder) picture above for $M=S^{n-1}$ itself, and apply a given operator $v\in Z(S^{n-1})$ to the vacuum incoming state, obtaining an outgoing state which is again v. Saying this more carefully in the 2d CFT case recovers the vacuum axiom of a vertex algebra, which Pavel explains gives injectivity of the state-field correspondence.<|endoftext|>
TITLE: Specializing early
QUESTION [8 upvotes]: Topic: this is a mathematics education question (but applies to other sciences too). 
Assumptions: my first assumption is that most mathematical concepts used in research are not intrinsically more complicated to grasp than high-school and undergraduate maths, the main difference is the amount of prerequisites (and hence time and experience) involved. My second assumption is that some undergraduate topics currently taught compulsarily are a bit of a burden for someone focussed on a particular topic.
Now of course cognitive development is a constraint, but upon reaching the age of high-school, I would think that a fairly large proportion of the scientifically-enclined students could really understand things usually taught much later and indeed become active at research level within a few years, provided some shortcuts are introduced. 
Early specialization: I'm wondering if a balanced curriculum already exists (or is planned) to provide such early specialization. What I'm looking for is this:  a one-week panorama of maths (or physics, or biology) would be organized at the beginning, and then the students would decide which subtopic to study. For example someone interested by group theory (or quantum optics, or genetics) would thus start with basics at the age 15 or 16, and gradually learn more stuff and skills, but for a few years with a strong emphasis on things directly relevant for the chosen subtopic. 
So for example the student specializing in group theory would only learn differential calculus and manifolds in passing in the context of Lie groups, and would skip most undergraduate real and functional analysis until it becomes relevant for his/her research topic, if at all. Of course other general courses would still be taught (history, sciences, programming, foreign languages...), but at least 50% of the student's week would be devoted to the research topic, ensuring satisfying progress.  
Question: do you know of any active or planned educative curriculum (at a high-school or university, or maybe a specific home-schooling program) as outlined above?  As an example of successful early specialization see e.g. the winners of the Siemens Foundation Prizes, but I haven't been able to learn much about their specific curriculum if any. 
Note: Skipping grades in school to enter university earlier is not the point, I'm really interested in a subtopic-oriented curriculum.

REPLY [12 votes]: I agree with José's comment above: I do not think early specialisation is a good idea.  Did I understand correctly that you want to give a one week to a 15-year old to decide on which area of mathematics to specialize?  
I want to add something different, however.  I fail to see how "some undergraduate topics currently taught compulsorily are a bit of a burden".  Mathematics is not a set of disconnected areas.  They are all highly related.  Most research problems, while staying in one area, may be related to another, motivated by another, applicable in another, or steal ideas or techniques from another.  One general course in, say, real analysis, complex analysis, abstract algebra, differential geometry, discrete mathematics, or topology is not a burden, but I dare say an actual necessity for anybody wanting to do research on any topic in pure math.  To use your own example, someone doing research in Lie theory will benefit from, rather than be burdened by, a solid understanding of basic differential geometry.  Or to use my own case, I am a Poisson geometer, but I have used ideas or results from all the above topics in my research.<|endoftext|>
TITLE: Line bundles: from transition functions to divisors
QUESTION [6 upvotes]: Recently I was thinking about how local systems are the same thing as vector bundles with flat connection, and how representations of the fundamental group gave rise to vector bundles. This got me thinking about how we get a handle on line bundles in general, and made me suspect that I don't understand them as well as I should, so I thought I'd ask a novice question about them. In brief: one has two different ways of regarding line bundles on a smooth complex algebraic variety, as a set of transition functions and as an equivalence class of Weil divisors, and I want to see how the two relate.
Let's break it down and be very explicit: $E$ is an elliptic curve over $\mathbb{C}$ with (topological) fundamental group generated by $a, b$. I've chosen $E$ because I know all about its Picard group. A one-dimensional local system on $E$ corresponds to an assignation of nonzero complex numbers $\lambda, \mu$ to $a, b$, and by tensoring this setup with $\mathcal{O}$ I get the sheaf of sections of a certain line bundle, whose transition functions in an appropriate set of trivialisations should be $\lambda$ and $\mu$. I've convinced myself that different choices of $\lambda$ and $\mu$ give me non-isomorphic bundles. Now, if I fix a distinguished point $P_0$ of $E$, choosing $d \in \mathbb{Z}$ and $P \in E$ is the same thing as choosing an isomorphism class of line bundles on $E$, whose associated divisor will be $P + (d-1)P_0$.
So how are $\lambda$, $\mu$ and $P, d$ related? Is there a nice formula? What about for higher genus curves? Am I just confused?

REPLY [4 votes]: As was noted in the comment, $d$ must be $0$ (bundles with a flat connection can have only degree $0$), and different connections can lead to the same bundle. However, every line bundle of degree zero on an elliptic (or higher genus) curve has a unique flat unitary connection, i.e. in your example with $|\lambda|=|\mu|=1$. This identifies the Jacobian of any smooth projective complex curve with a product of $2g$ circles (once you fix a basis of $H_1(C,\Bbb Z)$). For an elliptic curve, this is the usual identification with the 2-torus.<|endoftext|>
TITLE: bad reduction for elliptic curves
QUESTION [13 upvotes]: Why do elliptic curves have bad reduction at some point if they are defined over Q, but not necessarily over arbitrary number fields?

REPLY [8 votes]: It's also worth noting that "good reduction at $\mathfrak{p}$" is a local condition, so an elliptic curve may have everywhere good reduction, despite not having a Weierstrass equation that has good reduction at all primes. This is because over fields of class number greater than 1, there always exist elliptic curves that do not have global minimal Weierstrass equations. The existence, or lack of, a global minimal Weierstrass equation is governed by a certain ideal class (see Proposition VIII.8.2 in my Arithmetic of Elliptic Curves). The fact that if the class number is greater than 1, then there always exist curves with no global minimal Weierstrass equation is in the paper: "Weierstrass equations and the minimal discriminant of an elliptic curve", Mathematika 31 (1984), no. 2, 245–251. There is also a paper by Bekyel that describes the density of curves having (or not having) global minimal Weierstrass equations: "The density of elliptic curves having a global minimal Weierstrass equation", J. Number Theory 109 (2004), no. 1, 41–58. The moral is that you can produce curves with everywhere good reduction by writing down a specific Weierstrass equation, but to determine whether a given curve has everywhere good reduction is done via local calculations, and the associated elliptic scheme having everywhere good reduction may need to be patched together using more than one Weierstrass equation.<|endoftext|>
TITLE: Cohomology of a sheaf of functions locally constant along a foliation
QUESTION [17 upvotes]: Take a smooth manifold $M^n$ with a smooth foliation $F$. Consider the sheaf $\cal F$ of $C^{\infty}$ functions on $M^n$, locally constant along the foliation $F$. What is known about Chech cohomology of such a sheaf?
I am pretty sure that such  a question was studied (and maybe even has a complete answer), but I don't know a reference. 
A more specific question is: what happen when $F$ is 1-dimensional, given by integral trajectories of a  non-vanishing vector field? Or even more specifically, suppose $H^1(M^n)=0$ and we consider a Killing vector field $v$ on $M^n$ (i.e. $v$ is preserving a metric). Is it true the the sheaf of functions  $\cal F$ locally constant along trajectories of $v$ is acyclic? (we need $H^1(M^n)=0$, otherwise $S^1$ will be an obvious counterexample).
An example of a foliation. Consider the unit sphere $S^3$ in $\mathbb C^2$ and conisder the action of $\mathbb R$ via diagonal matrixes : $(z,w)\to (e^{ita}z, e^{itb}w)$ with $\frac{a}{b}$ irrational.

REPLY [2 votes]: I am joining this discussion a bit late, but let me add an example. If you consider a smooth minimal action of Z on the circle S^1 the suspension gives a flow on the torus. If the action is C^2 conjugate to an irrational rotation, then the transverse basic cohomology is finite dimensional. But if the action is only topologically conjugate to a rotation, then the basic cohomology may be infinite. The literature on this is quite a long time ago, in the 1970's perhaps. here is one reference
Haefliger, A.and Banghe, Li
Currents on a circle invariant by a Fuchsian group. Geometric dynamics (Rio de Janeiro, 1981), 369–378, Lecture Notes in Math., 1007, Springer, Berlin, 1983. 
Here is a more recent article
Avila, Artur and Kocsard, Alejandro
Cohomological equations and invariant distributions for minimal circle diffeomorphisms. Duke Math. J. 158 (2011), no. 3, 501–536. 
and there is one more artcile that is likely relevant to the question
Lott, John
Invariant currents on limit sets. Comment. Math. Helv. 75 (2000), no. 2, 319–350.<|endoftext|>
TITLE: calculating the genus of a curve using the Newton polygon
QUESTION [8 upvotes]: Given a plane affine curve $\sum_{i,j}a_{i,j}X^iY^j = 0$, its genus can be calculated as the number of integral points of the interior of the convex hull of $\{(i,j) \mid a_{i,j} \neq 0\}$. (claimed here: http://lamington.wordpress.com/2009/09/23/how-to-see-the-genus/)
How can this be proved?

REPLY [13 votes]: Here are the references I know concerning this:
H. F. Baker, Examples of applications of Newton's polygon to the theory of singular points of algebraic functions, Trans. Cambridge Phil. Soc. 15 (1893), 403-450.
A. G. Khovanskii, Newton polyhedra and the genus of complete intersections, Funct. Anal. i ego pril. English translation: Functional Anal. Appl., 12 (1978), 38-46.
V. I. Danilov and A. G. Khovanskii, Newton polyhedra and an algorithm for
computing Hodge-Deligne numbers, Izv. Akad. Nauk SSSR Ser. Mat. 50 (1986), 925-945; English translation: Math. USSR-Izv. 29 (1987), 279-298.
P. Beelen and R. Pellikaan, The Newton polygon of plane curves with many
rational points, Designs, Codes and Cryptography 21 (2000), 41-67.
(See Theorem 4.2.)
I think the statement should really be, given an irreducible curve in $\mathbf{G}_m^2$, a formula for the arithmetic genus of its closure in the 2-dimensional projective toric variety corresponding to the polygon.  This way one should not need to impose genericity hypotheses or restrictions on the characteristic.  The references above don't quite do all of this, however, so there is still room for a better reference or proof, I think.<|endoftext|>
TITLE: What do you lose when passing to the motive?
QUESTION [27 upvotes]: I contemplated about what information about a scheme we lose when passing to its motive. I came up with the following examples:

The projective bundle of a vector bundle does only depend on the rank of the vector bundle whatever it is twisted (this follows from the Mayer-Vietoris exact triangle).
The motive of the blow-up of a $\mathbf{P}^2$ in a point equals the motive of the quadric surface.

Are there further big classes of such phenomena? (Is there a class which 2. fits into?)
And conversely, what can we recover about the variety from its motive?

REPLY [10 votes]: Here the word "motive" will stand for Grothendieck pure motives modulo rational equivalence. Your point 1. is also true for Grassmann bundles. More precisely the following result holds :
Let $E\longrightarrow X$ be a vector bundle of rank $n$, $k\leq n$ and  $Gr_k(E)\longrightarrow X$ the associated Grassmann bundle. Then $M(Gr_k(E))\simeq \coprod_{\lambda}M(X)[k(n-k)-\lambda]$, where $\lambda$ runs through all partitions $\lambda=(\lambda_1,...,\lambda_k)$ satisfying $n-k\geq \lambda_1\geq...\geq \lambda_k\geq 0$.
You can prove it in the same fashion as for the projective bundle theorem, as an application Yoneda type lemma for Chow groups.
We now know many things on the motives of quadrics. For example if a quadratic form $q$ is isotropic, the motive of the associated quadric $Q$ has a decomposition $\mathbb{Z} \oplus M(Q_1) \oplus \mathbb{Z}[\dim(Q)]$, where $Q_1$ is a quadric of dimension $\dim(Q)-2$ associated to a quadratic form $q_1$ Witt equivalent to $q$. Using it inductively you get the motivic decomposition of split quadrics and for example if $\dim(q)$ is odd and $q$ is split the motive of $Q$ is $\mathbb{Z}\oplus \mathbb{Z}[1]\oplus ... \oplus \mathbb{Z}[\dim(Q)]$. Another very important result is the Rost nilpotence theorem, which asserts that the kernel of the change of field functor on Chow groups of quadrics consists of nilpotents. This result is very fruitfull because it implies that the study of the motive of quadrics can be done over a field which splits the quadric, working with rational cycles in stead of cycles over the base field. Even though these motivic results give severe restrictions on the higher Witt indices of quadrics and have very important applications, the motive does not contain "everything" about the associated quadratic forms (even in terms of higher Witt indices).
Another interesting class of varieties to motivic computations are the cellular spaces, i.e. schemes $X$ endowed with a filtration by closed subschemes $\emptyset \subset X_0\subset ... \subset X_n= X$ and affine bundles $X_i\setminus X_{i-1}\rightarrow Y_i$. In this situation the motive of $X$ is isomorphic to the direct sum of (shifts) of the motives of the $Y_i$. For example the filtration of $\mathbb{P}^n$ given by $X_i=\mathbb{P}^i$ and those affine bundles are given by the structural morphism of $\mathbb{A}^{i}$ imply the motivic decomposition $M(\mathbb{P}^n)=\mathbb{Z}\oplus ... \oplus \mathbb{Z}[n]$, and as you can see this is the same motive as odd dimensional split quadrics, so you certainely loose information.
The situation is much more complicated replacing quadratic forms by projective homogeneous varieties, but still under some assumption you can recover some results such as Rost nilpotence theorem, and we now begin to have a good description of their motive. Under these assumption the motive of projective homogeneous varieties encodes informations about the underlying variety, such as the canonical dimension, with the example of the computation of those of generalized Severi-Brauer varieties. Some works have also been done to link motives in this case with the higher Tits indices of the underlying algebraic groups.
Just to cite a few mathematicians from who we owe these great results : V. Chernousov, N. Karpenko, A. Merkurjev, V.Petrov, M. Rost, N.Semenov, A. Vishik, K. Zainoulline and probably many others that i forgot to mention.
edit : to add more precision to the nice answers of Mr. Chandan Singh Dalawat and Mr. Evgeny Shinder, motives of (usual) Severi-Brauer varieties of split algebras are indeed the same as projective space and split quadrics (in odd dimension) but it is obvious that on the base field they're are not necessarily isomorphic since the Severi-Brauer variety is totally split as long as there is a rational point, whereas an isotropic quadratic form is not completely split.<|endoftext|>
TITLE: Stiefel-Whitney Classes over Integers?
QUESTION [21 upvotes]: An interesting thing happened the other day. I was computing the Stiefel-Whitney numbers for $\mathbb{C}P^2$ connect sum $\mathbb{C}P^2$ to show that it was a boundary of another manifold. Of course, one can calculate the signature, check that it is non-zero and conclude that it can't be the boundary of an oriented manifold. I decided it might be interesting to calculate the first and only Pontrjagin number to check that it doesn't vanish. I believe Hirzebruch's Signature Theorem can be used to show that it is 6, but I was interested in relating the Stiefel-Whitney classes to the Pontrjagin classes.
I believe one relation is 

$p_i (\mathrm{mod} 2) \equiv w_{2i}^2$  (pg. 181 Milnor-Stasheff)

So I went ahead and did a silly thing. I took my first Chern classes of the original connect sum pieces say 3a and 3b, used the fact that the inclusion should restrict my 2nd second "Stiefel-Whitney Class" (scare quotes because we haven't reduced mod 2) on each piece to these two to get $w_2(connect sum)=(3\bar{a},3\bar{b})$. I can use the intersection form to square this and get $3\bar{a}^2+3\bar{b}^2=6c$ since the top dimensional elements in a connect sum are identified. Evaluating this against the fundamental class gives us exactly the first Pontrjagin number! This is false. Of course this is wrong because it should be 9+9=18 as pointed out below. This does away with my supposed miracle example. My Apologies!
This brings me to a broader question, namely of defining Stiefel-Whitney Classes over the integers. This was hinted at in Ilya Grigoriev's response to Solbap's question when he says 

On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes?

Of course the natural reason to restrict to $\mathbb{Z}/2$ coefficients is to get around orientability concerns. But it seems like if we restrict our orientation to orientable bundles we could use a construction analogous to those of the Chern classes where Milnor-Stasheff inductively declare the top class to be the Euler class, then look at the orthogonal complement bundle to the total space minus its zero section and continue. I suppose the induction might break down because the complex structure is being used, but I don't see where explicitly. If someone could tell me where the complex structure is being used directly, I'd appreciate it. Note the Euler class on odd dimensional fibers will be 2-torsion so this might produce interesting behavior in this proposed S-W class extension.
Another way of extending Stiefel-Whitney classes would be to use Steenrod squares. Bredon does use Steenrod powers with coefficient groups other than $\mathbb{Z}/2$ (generally $\mathbb{Z}/p$ $p\neq 2$), but this creates awkward constraints on the cohomology groups. Is this an obstruction to extending it to $\mathbb{Z}$ coefficients? It would be interesting to see what these two proposed extensions of S-W classes do and how they are related.

REPLY [11 votes]: The integral cohomology rings of both $BO(n)$ and $BSO(n)$ were computed by E. H. Brown, Proceedings AMS, 85, 2, 1982, p. 283-288. These rings are generated by the Pontrjagin classes, Bocksteins of monomials in even Stiefel-Whitney classes and, in the case of $BSO(2k)$, the Euler class. The description is as follows. All torsion is 2-torsion. The subalgebra generated by the Pontrjagin classes (and the Euler class in the case of $BSO(2k)$) has no torsion and is subject to just one relation: the square of the Euler class is the corresponding Pontrjagin class in the $BSO(2k)$ case. The torsion ideal can be identified with the $A$-submodule of the mod 2 cohomology generated by the image of $Sq^1$ where $A$ is the subalgebra generated by the reductions of the Pontrjagin classes, and the reduction of the Euler class in the case of $BS(2k)$. The key observation is Lemma 2.2.
The cohomology of $BO(n)\times BO(m)$ and $BSO(n)\times BSO(m)$ can be described in a similar way. E.H. Brown also computes the images of the Pontrjagin and Euler classes under the Whitney sum maps $BO(n)\times BO(m)\to BO(n+m),BSO(n)\times BSO(m)\to BSO(n+m)$. The Euler classes behave as expected; the torsion component of the images of the Pontrjagin classes is a bit more complicated. Finally, the image of the Bockstein of a monomial in the Siefel-Whitney classes can be computed using Lemma 2.2 and the action of the Steenrod algebra on the mod 2 cohomology.
So ``integral characteristic classes'' do not give any new tools for distinguishing real vector bundles up to isomorphism. However, in principle these classes may give new obstructions to representing bundles as Whitney sums and, by the splitting principle, as tensor products, symmetric or exterior powers etc.<|endoftext|>
TITLE: Products and the skeletal filtration in K-theory
QUESTION [6 upvotes]: Given a finite CW complex X, there is a filtration of the topological K-theory of X given by setting $K_n(X) = \ker \left(K(X) \to K(X^{(n-1)})\right)$, where $X^{(n-1)}$ is the (n-1)-skeleton of X.  (The choice of indexing here is from Atiyah-Hirzebruch.)
My question is:
How does this filtration interact with the external product $K(X)\times K(Y)\to K(X \times Y)$?  I believe the answer should be that $K_n (X) \cdot K_m (Y) \subset K_{n+m} (X\times Y)$.
Just to be clear, and to set notation, this external product is the one induced by sending a pair of vector bundles $V\to X$ and $W\to Y$ to the external tensor product, which I'll write $V\widetilde{\otimes} W = \pi_1^* V \otimes \pi_2^* W \to X\times Y$.
Of course, if $V\in K_n (X)$ and $W \in K_m (Y)$, then $V\widetilde{\otimes} W$ restricts to zero in both $K(X^{(n-1)} \times Y)$ and $K(X \times Y^{(m-1)})$, and $(X\times Y)^{(n+m-1)}$ is contained in the union of these two subsets.  Is there some way to deduce from this information that the class $V\widetilde{\otimes} W$ is actually trivial in $K((X\times Y)^{(n+m-1)})$?
Here's the reason I'm asking (which is really a second question, I guess).
In Characters and Cohomology Theories, Atiyah states (without comment) that for the internal product $K(X)\times K(X)\to K(X)$, one has $K_n (X) \cdot K_m (X) \subset K_{n+m} (X)$.  In Atiyah-Hirzebruch, they state this formula and say that it "admits a straighforward proof."
I thought I remembered that the straighforward proof was the following: 

Show that the external product satisfies $K_n (X) \cdot K_m (Y) \subset K_{n+m} (X\times Y)$
Observe that if $f:X\to X\times X$ is a cellular approximation to the diagonal $X\to X\times X$, then $f(X^{(n+m-1)}) \subset (X\times X)^{(n+m-1)}$.  So for any $V, W\in K(X)$, we have $V\otimes W = f^*(V\widetilde{\otimes} W)$, and if $V\in K_n (X)$ and $W\in K_m (X)$, it then follows from 1. that $V\otimes W\in K_{n+m} (X)$. 

Am I barking up the wrong tree here?
Presumably these questions will turn out to have an easy answer, but I've been thinking about them for a while now and haven't gotten any further.  Any suggestions or references would be great!  I haven't found any sources other than the two mentioned above that talk about the relation between skeleta and products, and neither of these sources mentions case of external products.

REPLY [5 votes]: Hi Dan, welcome to Math Overflow.
The group you denote $K_m(X)$ is the image of the relative K-group $K(X,X^{(m-1)})$, which for nice spaces (e.g. finite CW-complexes) consists of equivalence classes of formal differences $V - W$ of vector bundles equipped with an isomorphism $V|_{X^{(m-1)}} \cong W|_{X^{(m-1)}}$.  The product on K-groups lifts to an exterior pairing
$$
K(X,A) \otimes K(Y,B) \to K(X \times Y,A \times Y \cup X \times B).
$$
In particular, if $X$ and $Y$ are CW then using the standard CW structure on the product we have $$(X \times Y)^{(n+m-1)} \subset (X^{(n-1)} \times Y) \cup (X \times Y^{(m-1)}).$$
This gives us an exterior pairing
$$
K(X,X^{(n-1)}) \otimes K(Y,Y^{(m-1)}) \to K(X \times Y,(X \times Y)^{(n+m-1)})
$$
that lifts the ordinary K-theory product, and implies the result you want about the image of the group $K_n(X) \times K_m(Y)$.  This answers your part (1), and (2) follows just as you said.<|endoftext|>
TITLE: Smoothness of hyperplane sections
QUESTION [5 upvotes]: Suppose $X\subset \mathbb{P}^n$ is a smooth hypersurface defined over $\mathbb{Q}$.  For a "generic" prime $p$, what can be said about the set of hyperplanes $H$ in $\mathbb{P}^n(\mathbb{F}_p)$ for which $H \cap X$ is smooth over $\mathbb{F}_p$?  For $p$ fixed and $X$ varying, by contrast, the situation can be arbitrarily bad: in fact, every hyperplane section of 
${\sum_{i=1}^{n+1}X_i X_{i+n+1}^p=0} \subset \mathbb{P}^{2n+1}$ over $\mathbb{F}_p$ is singular.

REPLY [2 votes]: You mey be interested in the paper:
"Bertini Theorems over Finite Fields"(2002) Bjorn Poonen.<|endoftext|>
TITLE: Weil group, Weil-Deligne group scheme and conjectural Langlands group
QUESTION [19 upvotes]: I was reading a series of article from the Corvallis volume. There are couple of questions which came to my mind:

Why do we need to consider representation of Weil-Deligne group? That is what is an example of irreducible admissible representation of $ Gl(n,F)$ which does not correspond to a representation of $W_F$ of dimension $n$ ? An example for $ n=2 $ will be of great help.
In the setting of global Langlands conjecture, why extension of $W_F$ by $G_a$ or products of $W'_{F_v}$ does not work? 

Thank you.

REPLY [3 votes]: (I'm putting an "answer" to clarify Rob's question, and answer Dipramit's question in the comments, because I don't yet have the reputation to comment).
Let's first recall that the L-function of the Steinberg representation $\sigma = \sigma(\chi|\cdot|^{-1/2},\, \chi|\cdot|^{1/2})$ (for $\chi$ an unramified character) is $(1 - \chi(\varpi)q^{-s-1/2})^{-1}$, where $\varpi$ is a uniformizer (Bump shows this in detail in his book).  In particular, its reciprocal is a degree-one polynomial in $q^{-s}$.
By the ideas of Bernstein-Zelevinski (described in Kudla' article in "Motives"), $\sigma$ corresponds to the Weil-Deligne representation $\rho' = (\rho,\, V,\, N)$, where $\rho = \chi |\cdot|^{-1/2} \oplus \chi |\cdot|^{1/2}$, and the operator $N$ takes the first summand to the second, and the second summand to $0$.
If we only look at $\rho$, then we see that $V^I = V$ and therefore the $L$-function of $\rho$ would be the reciprocal of a degree-two polynomial.  Thus the monodromy operator becomes necessary for match of $L$-function: we see that $V^I_N \cong \chi|\cdot|^{1/2}$, which has the desired $L$-function.
This is a good specific example if you like thinking about match of $L$-functions.  More generally, you need to consider Weil-Deligne representations because if $\pi_1$ and $\pi_2$ have the same cuspidal support and correspond to $\rho_1' = (\rho_1,\, N_1,\, V_1)$ and $\rho_2' = (\rho_2,\, N_2,\, V_2)$, then $\rho_1\cong \rho_2$ as Weil representations; this follows from the ideas of Bernstein-Zelevinski.<|endoftext|>
TITLE: De Rham homology
QUESTION [10 upvotes]: Suppose M is an arbitrary smooth manifold and D is its bundle of 1-densities.
On the category of finite-dimensional vector bundles over M and linear differential operators between them
there is a contravariant endofunctor that sends a vector bundle E to E*⊗D
and a differential operator f: E→F to the adjoint differential operator f*: F*⊗D→E*⊗D.
Applying this endofunctor to the standard de Rham (cochain) complex 0→Ω^0(M)→Ω^1(M)→⋯→Ω^n(M)→0
with morphisms being de Rham differentials we obtain another (chain) complex 0←Λ^0(M)⊗D←Λ^1(M)⊗D←⋯←Λ^n(M)⊗D←0
with morphisms being codifferentials.
Here Λ^k(M) denotes the bundle of k-polyvectors (kth exterior power of the tangent bundle).
What is the exact relationship between the homology of this complex and the usual singular (co)homology of M?
Using Hodge duality we can rewrite this complex as
0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0,
where W is the orientation bundle.
It looks like the answer should be some standard fact from the 1950s,
therefore any references will be appreciated.

REPLY [12 votes]: When you dualize the bundle of differential forms and multiply it with the line bundle of top forms, you get differential forms again, and not polyvector fields.<|endoftext|>
TITLE: Convergence and non-convergence of left-point and mid-point Riemann sums
QUESTION [10 upvotes]: In standard calculus it is a well known fact that left-point and mid-point Riemann sums do become equal in the limit. When it comes to stochastic integration this is no longer the case. Taking the left-hand sums renders the Ito integral with an extra term, taking the midpoints renders the Stratonovich integral (see for example: Higham, p 531). While the Ito integral is the usual choice in applied math, the Stratonovich integral is frequently used in physics. Unlike the Itō calculus, Stratonovich integrals are defined such that e.g. the chain rule of ordinary calculus holds.
My question
1.) What is/are the deeper reason(s) that we have a convergence in ordinary calculus but have a non-convergence here?
2.) Are there examples in non-stochastic calculus where we also have a non-convergence of these limiting cases? If yes, how do they look like (and why)? Could you give a toy example of such a function?

REPLY [12 votes]: The reason that in stochastic calculus the left-hand and right-hand sums give different integrals really all boils down to quadratic variations. Processes such as Brownian motion have non-zero quadratic variation.
Suppose that you are integrating a process X with respect to some other process Y, then choosing a partition 0=t0≤...≤tn=t the approximations using left and right hand sums respectively are,
$$
\int_0^t X\ dY\approx \sum_{k=1}^nX_{t_{k-1}}(Y_{t_k}-Y_{t_{k-1}})
$$
$$
\int_0^t X\ \overleftarrow{d}Y\approx \sum_{k=1}^nX_{t_{k}}(Y_{t_k}-Y_{t_{k-1}})
$$
The difference between these can be bounded as follows
$$
\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\max_k\vert X_{t_k}-X_{t_{k-1}}\vert\sum_k\vert Y_{t_k}-Y_{t_{k-1}}\vert
$$
The final term on the right hand side converges to the variation of Y as the mesh of the partition goes to zero and, if X is continuous, the first term goes to zero. So, in standard calculus where integration is always with respect to finite variation functions, it makes no difference whether the left hand or right hand sums are used.
Alternatively, the Cauchy-Schwarz inequality can be applied to get the following bound.
$$
\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\sqrt{\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})^2\sum_{k=1}^n(Y_{t_k}-Y_{t_{k-1}})^2}.
$$
As the mesh of the partition goes to zero, the terms inside the square root converge to the quadratic variations of X and Y respectively, denoted by [X] and [Y]. Again, in standard calculus, we use (continuous) finite variation functions, which have zero quadratic variation. However, in stochastic calculus, processes such as Brownian motion have non-zero quadratic variation. Convergence to the quadratic variation along partitions occurs in the sense of convergence in probability - it does not have to converge in the usual sense with any positive probability. A Brownian motion B has [B]t=t, so the left and right hand sums can converge to different numbers.
With the Stratonovich integration the correct thing to use is the average of the left and right hand sums, not the mid-point. For Ito processes, which are integrals with respect to Brownian motion and time, it makes no difference. This is because their quadratic variations are absolutely continuous. However, for general continuous semimartingales, the mid-point sums don't actually have to converge to anything. (See Sur quelques approximations d'intégrales stochastiques by Marc Yor).
So, Stratonovich integration uses the average of the left and right hand sums, and the difference between this and the Ito integral is precisely half of what you get using the right-hand sums.
As to the question of whether this difference shows up in standard (non-stochastic) calculus, the answer is, as far as I know, hardly ever. In fact, given any continuous functions then you can choose a sequence of partitions along which the quadratic variation vanishes as the mesh goes to zero (I'll leave this as an exercise!). So, given any continuous function with non-zero quadratic variation with respect to some sequence of partitions, so that the left and right hand sums converge to different numbers then, there will be other partitions along which the quadratic variation vanishes. So the integral isn't really defined at all in this case.
However, there is one case I have seen where left and right hand sums for deterministic functions converge to different numbers. For this to happen, you have to fix some sequence of partitions with mesh going to zero, and stick to using these to define the integral. Different partitions could lead to different results. Hans Follmer published a paper using this idea in 1981 (Calcul d'Ito Sans Probabilities). There aren't any natural (and useful) cases that I know of where this occurs, but you can construct some examples.
Given a Brownian motion, the quadratic variation along a sequence of partitions where each is a refinement of the previous one will converge with probability one. So, selecting a Brownian motion sample path at random, you can then define integrals using these partitions in a pathwise sense, rather than using the machinery of stochastic integration. They will converge to the same thing though, so this is a bit of a cheat.Alternatively, you could construct a continuous function along the lines of the Weierstrass function while forcing the quadratic variation to converge to a nonzero number along a given sequence of partitions. Then, left and right hand Riemann sums along these partitions will converge to a different answer.
For example, let s(t) be the 'sawtooth' function s(t) = 1-|1-2{t/2}| ({t}=fractional part of t). Then, define the following function on the unit interval,
$$
f(t) = s(t)+\sum_{n=1}^\infty 2^{-(n+1)/2}s(2^nt).
$$
This has quadratic variation 1, calculated along dyadic partitions. The following left-hand, right-hand and `Stratonovich' integrals are easily verified,
$$
\int_0^1 f df = (f(1)^2-f(0)^2-[f]_1)/2.
$$
$$
\int_0^1 f \overleftarrow{d}f = (f(1)^2-f(0)^2+[f]_1)/2.
$$
$$
\int_0^1 f \partial f = (f(1)^2-f(0)^2)/2.
$$<|endoftext|>
TITLE: Does Con(ZF) imply Con(ZF + Aut C = Z/2Z)?
QUESTION [23 upvotes]: How many field automorphisms does $\mathbf{C}$ have?  If you assume the axiom of choice, there are tons of them -- $2^{2^{\aleph_0}}$ I believe.  And what if you don't -- how essential is the axiom of choice to constructing "wild" automorphisms of $\mathbf{C}$?  Specifically, if you assume that ZF admits a model, does that imply that ZF admits a model where $\mathbf{C}$ has no wild automorphisms: $\mathop{Aut}\mathbf{C}=\mathbf{Z}/2\mathbf{Z}$?
I suppose if that's true, then the next logical question is to construct models of ZF where $\mathop{Aut}\mathbf{C}$ has cardinality strictly between 2 and $2^{2^{\aleph_0}}$--pretty disturbing if you ask me.  Which finite groups can you hit?

REPLY [29 votes]: The use of inaccessible cardinals is not necessary here, the Baire property works just as well as Lebesgue measure. Shelah (Can you take Solovay's inaccessible away, Isr. J. Math. 48, 1984, 1-47) shows that ZF + DC + "every subset of R has the Baire property" is relatively consistent with ZF. (This is also the paper where Shelah also shows that the inaccessible cardinal is necessary for Solovay's result.)
The connection is an old theorem of Banach and Pettis which says that any Baire measurable homomorphism between Polish groups is automatically continuous. This result is provable in ZF + DC. Since C is a Polish group under addition, it follows that every additive endomorphism of C is continuous in Shelah's model. Since the continuous additive endomorphisms of C are precisely the R-vector space endomorphisms, it follows that the only field automorphisms of C in Shelah's model are the identity and conjugation.

As pointed out by Pete Clark in the comments, the Artin-Schreier Theorem goes through using only the Boolean Prime Ideal Theorem (PIT), which is significantly weaker than full AC. This shows that AC is not completely necessary to show that there is a unique conjugacy class of elements of order 2 in Aut(C) and that these correspond precisely to the finite subgroups of Aut(C).
Looking at Pete Clark's Field Theory Notes, specifically at Steps 4 and 5 of his proof of the Grand Artin-Schreier Theorem on pages 62-63, I think that it is a theorem of ZF that the only possible order for a nontrivial finite subgroup of Aut(C) is 2.<|endoftext|>
TITLE: Does any tensor category correspond to a bialgebra?
QUESTION [9 upvotes]: I wonder how strong the power of Tannaka philosophy is, and if we accept that a tensor category is a generalized bialgebra, what difficulties we will come up against ?
Edit: Whether most tensor categories are representable, or whether for every "good enough" tensor category there exist a bialgebra with its module category isomorphic to the given category?

REPLY [23 votes]: I'd like to explain Bruce's answer a bit more. The fusion categories Bruce mentioned have non-integer Frobenius-Perron dimensions, so it is very easy to see that they are not categories of finite dimensional modules over a bialgebra. E.g. one of the simplest of them, the so called Yang-Lie category, has simple objects $1,X$ with $X^2=X+1$. So if $X$ were a finite dimensional representation of a bialgebra, then the dimension of $X$ would be the golden ratio, which is absurd. 
This, however, can be fixed if we allow weak bialgebras and weak Hopf algebras. In fact, any fusion category is the category of modules over a finite dimensional weak Hopf algebra, see arXiv.math/0203060.  
As to Akhil's example (Deligne's categories), it is also true that they cannot be realized as categories of finite dimensional representations of a bialgebra (or even a weak bialgebra), but for a different reason. Namely, if X is a finite dimensional representation of a bialgebra, then the length of the object $X^{\otimes n}$ is at most ${\rm dim}(X)^n$, where ${\rm dim}$ means the vector space dimension. But in Deligne's categories, the length of $X^{\otimes n}$ grows faster as $n\to \infty$. Actually, in another paper, Deligne shows that if in a symmetric rigid tensor category over an algebraically closed field of characteristic zero, the length of $X^{\otimes n}$ grows at most exponentially, then this is the category of representations of a proalgebraic supergroup, where some fixed central order 2 element acts by parity
(so essentially this IS the category of (co)modules over a bialgebra). This is, however, violently false in characteristic $p$, since if the root of unity $q$ is of order $p$, where $p$ is a prime, the the fusion categories for $U_q({\mathfrak g})$ mentioned by Bruce admit good reduction to characteristic $p$, which are semisimple symmetric rigid tensor categories with finitely many simple objects and non-integer Frobenius-Perron dimensions.    
A third very simple example of a tensor category not coming from a bialgebra is the category of vector spaces graded by a finite group $G$ with associator defined by a nontrivial $3$-cocycle. This category, however, is the category of representatins of a quasibialgebra (and also of a weak bialgebra, as mentioned above). 
So the conclusion is as in the previous two answers: tensor categories are more general than bialgebras. More precisely, the existence of a bialgebra for a tensor category is equivalent to the existence of a fiber functor to vector spaces, which is an additional structure that does not always exist. And if it exists, it is often not unique, so you may have many different bialgebras giving rise to the same tensor category.<|endoftext|>
TITLE: Finite interpolation by a nondecreasing polynomial
QUESTION [17 upvotes]: Let $x_1 < x_2 <  \ldots  < x_n$ and  $y_1 < y_2 <  \ldots  < y_n$ be two sequences
of $n$ real numbers. It is well known that there are polynomials that "interpolate"
in that $f(x_i)=y_i$ for all $i$, and the Lagrange interpolating polynomial
even warrants a solution of degree $ < n$. Now, what happens if we want the
polynomial $f$ to be nondecreasing on the interval $[x_0,x_n]$ ? Is there always
a solution, and is there a bound on the degree also ?

REPLY [18 votes]: This problem has appeared before in literature and is now well understood, I guess. The general version is when you have no restriction on the $y_i$'s and you ask for an interpolating polynomial that is monotone on each sub-interval $[x_ix_{i+1}]$. The first paper proving the existence of such a polynomial is: 
W.Wolibner, "Sur un polynom d'interpolation", Colloq. Math (2) 1951, 136-137
but it is a non-constructive proof, as it uses the Weierstrass approximation theorem much like the answer given by Harald Hanche-Olsen above. Another proof for the case $0=y_0\le \cdots \le y_n=1$ is given in "Polynomial Approximations to Finitely Oscillating Functions" by W.J. Kammerer (Theorem 4.1) and the non-constructive aspect of his proof is the use of uniform convergence of appropriate Bernstein polynomials. In "Piecewise monotone polynomial interpolation", S.W. Young proves the same theorem and makes the final remark that the existence of such monotone interpolating polynomial is in fact equivalent to the Weierstrass theorem. On the other hand Rubinstein has some papers devoted to proving the existence of interpolating polynomials which are increasing in all of $\mathbb R$.
The first paper which gives bounds on the degrees is, I think, 
E. Passow, L. Raymon, "The degree of piecewise monotone interpolation", which is here
and an improvement is made in "Exact estimates for monotone interpolation" by G.L. Iliev.
Note that the bounds are in terms of 
$$A=\max \Delta y_i=\max (y_{i}-y_{i-1}) \qquad B=\min \Delta y_i \qquad C=\min \Delta x _i$$
And no uniform bound exists.

REPLY [13 votes]: To add to Gjergji Zaimi's informative answer: It is easy to see that the degree cannot be bounded in terms of $n$ alone, even when $n=3$.  
Suppose that we want $f$ of degree $m$ such that $f(0)=0$, $f(1)=\epsilon$, and $f(2)=1$, and $f$ is increasing on $[0,1]$, where $\epsilon>0$ is small.  Then $|f(k/m)| \le \epsilon$ for $k=0,\ldots,m$, so the Lagrange interpolation formula shows that for fixed $m$, the coefficients of $f$ are $O(\epsilon)$, so $f(2)$ is $O(\epsilon)$ and cannot be $1$ if $\epsilon$ is small enough.  In other words, the degree of any solution $f$ must grow as $\epsilon$ shrinks.

REPLY [2 votes]: I just came across the paper by Powers and Reznick "Polynomials that are Positive on an interval" http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.120.9773&rep=rep1&type=pdf
In particular they point to a theorem of Schmudgen who gives a characterization of polynomials positive on a specific compact set.<|endoftext|>
TITLE: When is the product of a set of numbers greater than the sum of them?
QUESTION [13 upvotes]: This could well be too general a question, but I'd be interested in solutions to special cases too. Say you have some finite set of positive real numbers $x_i$, when is it the case that $\sum_i x_i > \prod_i x_i$? And when are they equal?
The special case that prompted this was an argument about whether any number is equal to the sum of its prime factors.
Any references or quick proofs welcome.

REPLY [16 votes]: If you have a set of positive integers (that is, no duplicates are allowed) then the sum is greater than the product if and only if the set is of the form {1,x}. The sum is equal to the product only for singleton sets {x} and the set {1,2,3}.
For, examining the remaining cases:

If the set is empty the sum is 0 and the product is 1, so sum < product
If the set has two elements {x,y}, neither of which is 1, then $xy\ge 2\max(x,y)>x+y$.
If the set has three elements {1,2,x}, with $x>3$, the sum is $x+3$ and the product is the larger number $2x$.
If the set has any other three elements then its product is at least three times its max and its sum is less than that.
If the set has {1,2,3,x} then the product is 6x and the sum is x+6, smaller for all $x\ge 4$.
If the set has any other form with $k>3$ elements then by induction the sum of the smallest $k-1$ items is less than their product. Multiplying or adding the largest item doesn't change the inequality.

REPLY [7 votes]: The "special case" is not a special case, since only squarefree numbers equal to the product of their prime factors (I guess you forgot that primes can occur with multiplicities), and the product of a finite multiset of integers > 1 is always greater or equal to their sum, with equality only if the multiset is [2, 2] (proof by induction). So it is not really clear to me what you actually want.<|endoftext|>
TITLE: The Angel Problem - was the bet paid?
QUESTION [9 upvotes]: Did Conway pay the wager for either of the proofs to the The Angel Problem? 
I'd check in on this every now and again when it was an unsolved problem and would like to know how the story ends. Anyone know more details?

REPLY [15 votes]: Actually, until very recently, Conway didn't even believe his problem had been solved.  (This despite the fact that multiple solutions have been published, some years ago by now, and the solutions had even been exposited at seminars at Princeton.)
Only a few months ago did a few graduate students at Princeton convince him that the problem was solved.  He was particularly excited when he heard about the "nice devil" (who never kills a square that could have been visited before).
I have checked with Conway: the bet has not yet been paid.  However, it will be soon.  I will update this answer if and when it has been paid.

REPLY [3 votes]: If I remember well Andras Mathe told me that the prize was supposed to be split to four parts which would have made it 25 each, so he decided to renounce his part.<|endoftext|>
TITLE: Geometrically interpreting the answer to a vector calculus question involving tangent line segments to ellipses.
QUESTION [12 upvotes]: Let E be an ellipse centered at the origin on the x, y plane with major radius b and minor radius a. The length of the shortest line segment tangent to E that begins on the x-axis and ends on the y-axis is a+b. This can be shown using Lagrange multipliers. This answer is very simple and leads us to ask the following question:

Can you give a geometric reason for why the length is a+b?

This was originally asked to me by Frank Jones a few years ago.

REPLY [13 votes]: There is a geometric way to show that $n$-gon circumscribed around an ellipse has minimal perimeter if it is inscribed in a confocal ellipse. From Poncelet porism (and generalization of optical property) it follows that we have continuous family of "minimal" polygons.
If we know it, then it is easy to understand that the circumscribed rhomb (from your question) and the circumscribed rectangular (with perimeter $4(a+b)$) are minimal polygons. So, side of the rhomb equals $a+b$.<|endoftext|>
TITLE: Alive dynamical system
QUESTION [12 upvotes]: Intuitively, one can say that a dynamical system is alive if one can build a universal Turing machine inside.
So, Conway's Game of Life is alive and shift space should be dead.
I fail to make this definition precise. 
Please let me know if someone made it (or at least was trying to).
(It is  nice if you can make a def. 
BUT I'm mostly interested if you saw such definition in the literature.)
Comments:

There is a relateted question, and it links to this paper by Graça--Campagnolo--Buescu (thanks to Joel).

REPLY [4 votes]: I don't pretend to have an answer to this question, but let me at least try (informally and tentatively) to clarify the distinction. Suppose I have a computational problem and I want to solve it. I also happen to have the Game of Life handy, and plenty of information about it. What I presume the usual statement about the Game of Life encoding a universal Turing machine is that there is some algorithmic way of translating my algorithm (for finding a matching in a bipartite graph, say) into an initial configuration for the Game of Life such that after a certain time I can look at the output and see whether a certain position has a + or a -, and that will agree with what the algorithm gives.
Now suppose that I am presented with the shift space. This time, I run the entire computation, describe this run somehow as a sequence of 0s and 1s, and then apply the shift map repeatedly to the resulting sequence. But what have I gained from applying the shift map? Precisely nothing, because in order to work out the sequence I had to run the entire computation. 
Here perhaps is a genuine difference between the two. Suppose I have a program and I don't know whether it halts, and would like to find out. With the Game of Life, after a finite time I can translate the program into an initial configuration, and there can be some stopping rule, and then I can go off and have lunch and leave the Game of Life chugging away. With the shift space, if it happens that the program doesn't halt, I will never finish the process of creating a sequence on which the shift should act.<|endoftext|>
TITLE: homotopy associative $H$-space and $coH$-space
QUESTION [8 upvotes]: Let $[X, Y]_0$ denote base point preserving homotopy classes of maps $X\rightarrow Y$. A multiplication on a pointed space $Y$ is a map $\phi: Y\times Y\rightarrow Y.$ From this map, we can define a continuous map for each pointed space $X$, $\phi_X: [X, Y]_0\times [X, Y]_0\rightarrow [X, Y]_0,$ 
by the composition $$\phi_X (\alpha, \beta)(x)=\phi(\alpha(x), \beta(x)).$$
If $([X, Y]_0, \phi_X)$ is a group for each $X$, then $(Y, \phi)$ is called a homotopy associative $H$-space.
A $coH$-space is defined from a comultiplication, namely, a map $\psi: X\rightarrow X\vee X.$ Then, for each pointed space $Y$, we can define a function $\psi^Y: [X, Y]_0\times [X, Y]_0\rightarrow [X, Y]_0$ in this way: 
$$\psi^Y(\alpha, \beta)=(\alpha\vee\beta)\circ\psi.$$ If $([X, Y]_0, \psi^Y)$ is a group for each $Y$, then $(X, \psi)$ is called a homotopy associative $coH$-space.
So, as we can see, if we have a homotopy associative $coH$-space $(X, \psi)$ and a homotopy associative $H$-space $(Y, \phi)$, then we can define two group structures on the space $[X, Y]_0$. My question is: are they "equivalent" in some sense? Obviously, whatever $\phi$ or $\psi$ is, the zero element of the group is the constant map in $[X, Y]_0.$ However, the two group structures do depend on the choice of $\phi$ and $\psi$, which seems have little relationship with each other.

REPLY [8 votes]: I looked at my homotopy theory lecture notes and we had the following similar result:
$X$ H-CoGroup, $Y$ $H$-Group, then both group structures defined on [X,Y] agree.
The proof goes roughly as follows:
Call the upper products $\cdot$, resp. $*$. Inserting the definitions of those products, one can show the following "distributivity":
$(a\cdot b)*(c\cdot d)=(a * c)\cdot(b * d)$
Then one shows that both products have the same neutral element and finally 
$f*g=(f\cdot 1) * (1\cdot g)=(f * 1)\cdot(1 * g)=f\cdot g$,
gives the result. That's the strategy of the proof in the case of $H$-(co-)groups.<|endoftext|>
TITLE: Favourite scholarly books?
QUESTION [19 upvotes]: What are your favourite scholarly books?  My favourite is definitely G.N. Watson's "A treatise on the theory of Bessel functions" (full text). Every single page is full of extremely precise references, ranging over 300 years of mathematics and hundreds of papers.  One gets the definite impression that Watson carefully studied each and every paper he refers to.
In an entirely different field, I would say that Umberto Eco's "The search for the perfect language" (about linguistics and the evolution of language) is similarly scholarly.
Note that this question is focused on books displaying great depth and breadth of knowledge of the relevant literature, not on great mathematical writing or ``good'' undergraduate level math books (though there is not necessarily an empty intersection).

REPLY [5 votes]: Éléments de mathématique by Nicolas Bourbaki.

REPLY [3 votes]: Abramowitz and Stegun's Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, and its successor, the NIST Handbook of Mathematical Functions.<|endoftext|>
TITLE: EGZ theorem (Erdős-Ginzburg-Ziv)
QUESTION [23 upvotes]: Erdős, Ginzburg and Ziv prove the following:
Let $n \geq 1$ and $a_1,\ldots, a_{2n-1}\in \mathbb{Z}$. There exist distinct $i_1,\ldots , i_n$
such that
$$
a_{i_1} + \cdots  + a_{i_n} \equiv 0 \pmod{n}.
$$
Is there a proof that doesn't use the Chevalley–Warning theorem (or a variant of its proof)?

REPLY [10 votes]: Here is what I remember from a proof I came up with long time ago (it appeared in some competitions). I am sure it is known, but since the proof is short, I will put it here: 
The statement can be reduced to the case when $n=p$ is prime. Now it will follow from the following:
Lemma: Let $2\leq i\leq p$ and consider any set $A$ of elements $ a_1,\cdots, a_{2i-1} \in \mathbb F_p$ such that no $i$ of them are the same. Let $S$ be the set of all sums of $i$ elements of $A$. Then $\left|S\right|\geq i$.  
Proof: Induction on $i$, the case $i=2$ is easy. Suppose the result is true for $p>i=k\geq 2$. Consider the set $A'$ of $2k+1$ elements $\{a_1,\cdots,a_{2k-1},b,c\}$, WLOG we may assume that $b,c$ represent the two most frequently appearing elements in $A'$. By assumption $b\neq c$. 
The set $\{a_1,\cdots,a_{2k-1}\}$ satisfies the condition that any frequency is less than $k$, otherwise there will be $3$ elements appearing at least $k$ times in $A'$, impossible. By induction we can choose $k$ subsets of $k$ elements whose sums $s_1,\cdots, s_{k}\in \mathbb F_p$ are distinct. Let $B=\{b+s_1,\cdots, b+s_{k}\}$ and $C=\{c+s_1,\cdots, c+s_{k}\}$, obviously $\left|B\right|=\left|C\right|=k$. We will be done by showing $B \neq C$. If not, then by summing all the elements in each set one gets $k(b-c)=0$, impossible. QED. 
Applying the lemma for $i=p$, note that if there are $p$ identical elements then their sum is $0$. 
By the way, I asked some question on generalizations of this theorem, and get really good answers here.<|endoftext|>
TITLE: Is a "non-analytic" proof of Dirichlet's theorem on primes known or possible?
QUESTION [67 upvotes]: It is well-known that one can prove certain special cases of Dirichlet's theorem by exhibiting an integer polynomial $p(x)$ with the properties that the prime divisors of $\{ p(n) | n \in \mathbb{Z} \}$ must lie in certain arithmetic progressions, with a finite number of exceptions.  This is because any nonconstant polynomial must have infinitely many distinct prime divisors, which one can prove in a manner imitating Euclid's proof of the infinitude of the primes.  For example, taking $p(x) = \Phi_n(x)$, we can prove Dirichlet's theorem for primes congruent to $1 \bmod n$.  It is known (see, for example, this paper of K. Conrad) that this is possible precisely for the primes congruent to $a \bmod n$ where $a^2 \equiv 1 \bmod n$.  
However, the result about polynomials having infinitely many prime divisors has the following generalization: any sequence $a_n$ of integers which is eventually monotonically increasing and which grows slower than $O(2^{\sqrt[k]{n}})$ for every positive integer $k$ has infinitely many distinct prime divisors.  In particular, any sequence of polynomial growth (not necessarily a polynomial itself) has this property.  
Question 1:  Given an arithmetic progression $a \bmod n, (a, n) = 1$ such that $a^2 \not \equiv 1 \bmod n$, is it ever still possible to efficiently construct a monotonically increasing sequence of positive integers satisfying the above growth condition such that, with finitely many exceptions, the prime divisors of any element of the sequence are congruent to $a \bmod n$?  ("Efficiently" rules out answers like "the positive integers divisible by primes congruent to $a \bmod n$," since I do not think it is possible to write down this sequence efficiently.  On the other hand, evaluating a polynomial is very efficient.)  The idea is that such a sequence immediately gives a proof of Dirichlet's theorem for primes congruent to $a \bmod n$ generalizing the Euclid-style proofs.  
Question 2:  If the above is not possible, are there any known techniques for proving Dirichlet's theorem or at least some of the special cases not covered above without resorting to the usual analytic machinery?  For example, Selberg published an "elementary" proof in 1949, but it relies on the "elementary" proof of the prime number theorem, which to me is "finitary analytic machinery."  What is the absolute minimum amount of analysis necessary to produce a proof?  (Edit:  In response to a suggestion in the comments, one way to describe the kind of answer I'm looking for is that it would generalize to a proof of Chebotarev's density theorem that shows very clearly where the distinction between the number field and function field cases is; aside from some "essential" analytic argument there should be no difference between the two.)
This question is inspired at least in part by the following observation: Dirichlet's theorem is equivalent to the seemingly weaker statement that for every progression $a \bmod n, (a, n) = 1$ there exists at least one prime congruent to $a \bmod n$.  The reason is that if there exists some such prime $a_1$, then letting $n_1$ be the smallest multiple of $n$ greater than $a_1$, there exists a prime congruent to $a_1 + n \bmod n_1$, and so forth.

REPLY [6 votes]: I was struggling to find an elementary proof of Dirichlet's theorem using another interesting technique.
I came finally to a proof from an entirely different direction but as I found out Erdos came first before many years.
The proof is not well known (I never understood why anyone mentions it!) and it uses Chebyshev type estimates. Here is the proof: http://kam.mff.cuni.cz/~klazar/ln_antcII.pdf (Wayback Machine)
I hope this will be helpfull!
We are looking for prime numbers of the form $a+nm$.$(a,m)=1,n=1,2,...$
The general plan is:
$n!$ divides the product of $n$ consecutive terms of the arithmetic proggression $a+m$,$a+2m$,$...,$$a+nm$ with $(a,m)=1$ (if we disregard the factor of $n!$ which includes divisors of $m$) 
For example, consider the proggression $1+3m$:
$4\cdot7\cdot10\cdot13\cdot16\cdot19\cdot22$ is divided by $7!/3^2$
(The factor that we "disregard" is $3^2$) 
It is easy to prove in analog with Legendre's Lemma at binomial coefficients, that the highest power of a prime $p$ which divides $\frac{(a+m)\cdot(a+2m)...\cdot(a+nm)}{n!}$ does not exceed $a+nm$.
The problem is that such a prime $p$ could not be of the form that is wanted.For example turning back to $\frac{4\cdot7\cdot10\cdot13\cdot16\cdot19\cdot22}{7!/3^2}$ we find 11 as a divisor which is a prime of the form $2+3m$.
We wish to  simplify the unwanted primes from the "big" fraction which Erdos calls $Pn(a,m)$.
In order to do this we divide $Pn(a,m)$ with a fraction of the same kind but from another proggression $a'+nm$ with  $(a',m)=(a,m)=1$.
(The "other" proggression for $1+3m$ is $2+3m$ since  only 1 and 2 are the only numbers coprime to 3 and less than 3).
But we find that in $Pn(a,m)$  every prime of the form $a'+km$ which is greater than $n$ exists exactly once, so  (and here is the big idea) dividing $Pn(a,m)$ with $P(n/h)(a',m)$ ($h$ is the number with the property  $a'h=a \pmod m$ ) every prime of the form $a'+km$ which is greater than $n$ cancels.
Continuing like this you can cancel every unusefull prime that exceeds n and have only "small" unusefull primes whose product is significally smaller than $Pn(a,m)$.
With this,you prove that primes of the form $a+nm$ have a product which tends to infinity and so they are infinite. 
I hope this is helpfull.
(note)We use $h$ because the smallest term of the proggression $a+nm$ that is divided by a prime $p$ of the form $a'+km$ is the term $a+km=h\cdot p$ .<|endoftext|>
TITLE: Fast Fourier Transform for Graph Laplacian?
QUESTION [20 upvotes]: In the case of a regularly-sampled scalar-valued signal $f$ on the real line, we can construct a discrete linear operator $A$ such that $A(f)$ approximates $\partial^2 f / \partial x^2$.  One way to interpret this operator is via spectral decomposition of the corresponding matrix:
$$A = UVU^T.$$
If our operator $A$ has spectral accuracy, then $U^T$ is precisely the discrete Fourier transform matrix.  Hence, we could compute the Fourier transform of $x$ by computing all the eigenvectors of $A$ and sticking them in the columns of $U$.  Of course, we all know there's a quicker way to do it: use the fast Fourier transform (FFT).
What about in a more general setting?  In particular, consider the graph Laplacian $L=UVU^T$ which for a weighted, undirected graph on $n$ vertices is an $n \times n$ matrix with the weights of incident vertices on the off-diagonal and (the additive inverse of) total incident weight on the diagonal.
Question:
Can we transform a signal on vertices into frequency space without computing the entire spectrum of $L$ (using something like the FFT)?
In particular I'm interested in the case where $L$ approximates the Laplace-Beltrami operator on some manifold $M$ -- here eigenvectors of $L$ approximate an orthogonal eigenbasis for square-integrable functions on $M$.  However, pointers to nearby results (e.g., FFT for the combinatorial graph Laplacian) are appreciated.

REPLY [4 votes]: The trick to making the FFT work is factoring out a complex exponential from the sum over odd terms. For this to happen your function needs to be sampled across a uniform grid. Greengard refers to this property as "brittle" (cf math.nyu.edu/faculty/greengar/shortcourse_fmm.pdf ).
When your function is sampled over a nonuniform grid fast multipole methods or Barnes-Hut style algorithms can help.<|endoftext|>
TITLE: When is Br(X) = H^2(X,G_m)?
QUESTION [9 upvotes]: In Milne, Étale cohomology, it is proved that $\mathrm{Br}(X) = H^2(X,\mathbf{G}_m)$ for $X$ regular of dimension $\leq 2$. Are there in the meantime further results for $X$ regular?

REPLY [3 votes]: Dear norondion: It appears to me from your comments to Qing Liu's answer that you are interested in when this cohomological Brauer group is actually $0$. If that is true, then these two related MO questions (and Emerton's answer to one of them) may be of interest:
Two conjectures by Gabber.
Flat cohomology and Picard groups.
(Of course, the punctured spectrum of a regular local ring is a regular scheme. Also, you can probably get some statements for projective $X$ by looking at the local ring of the cone over $X$). My apology if this is not relevant.<|endoftext|>
TITLE: What is the motivation for maps of adjunctions?
QUESTION [20 upvotes]: In Mac Lane, there is a definition of an arrow between adjunctions
called a map of adjunctions. In detail, if a functor $F:X\to A$ is left
adjoint to $G:A\to X$ and similarly $F':X'\to A'$ is left adjoint to
$G':A'\to X'$, then a map from the first adjunction to the second is a
pair of functors $K:A\to A'$ and $L:X\to X'$ such that $KF=F'L$,
$LG=G'K$, and $L\eta=\eta'L$, where $\eta$ 
and $\eta'$ are the units of the first and second adjunction. (The
last condition makes sense because of the first two conditions; also,
there are equivalent conditions in terms of the co-units, or in terms
of the natural bijections of hom-sets).
As far as I can see, after the definition, maps of adjunctions do not
appear anywhere in Mac Lane.  Googling, I found this definition also
in the unapologetic mathematician,
again with the motivation of being an arrow between adjunctions.  
But what is the motivation for defining arrows between adjunctions 
in the first place?  I find it hard to believe that the only
motivation to define such arrows is, well, to define such arrows... 
So my question is:  What is the motivation for defining a map of
adjunctions?  Where are such maps used?
Besides the unapologetic mathematician, the only places on the web
where I found the term ''map of adjunctions'' were sporadic papers,
from which I was not able to get an answer to my question (perhaps
''map of adjunctions'' is non-standard terminology and I should have
searched with a different name?). 
I came to think about this when reading Emerton's first answer
to a question about completions of metric spaces.
In that question, $X$ is metric spaces with isometric embeddings, $A$
is complete metric spaces with isometric embeddings, $X'$ is metric 
spaces with uniformly continuous maps, $A'$ is complete metric
spaces with uniformly continuous maps, and $G$ and $G'$ are the
inclusions.  Now, if I understand the implications of Emerton's answer
correctly, then it
is possible to choose left adjoints $F$ and $F'$ to $G$ and $G'$ such
that the (non-full) inclusions $A\to A'$ and $X\to X'$ form a map of
adjunctions.  This made me think whether the fact that we have a map
of adjunctions has any added value.  Then I realized that I do not
even know what was the motivation for those maps in the first place.
[EDIT:  Corrected a typo pointed out by Theo Johnson-Freyd (thanks!)]

REPLY [2 votes]: One place a category of adjunctions is used is in proposition 3.1.5 of Hovey's book on model categories.  It says that if $\mathscr{C}$ is a category with all small colimits, then the category $\mathscr{C}^\Delta$ of cosimplicial objects is equivalent to the category of adjunctions $\mathbf{SSet}\rightarrow\mathscr{C}$.  I guess if you had a morphism between two different definitions of simplices in $\mathbf{Top}$, then you might want a morphism of adjunctions to say what this does to geometric realization.<|endoftext|>
TITLE: Scott on the consistency of the lambda calculus
QUESTION [8 upvotes]: I have twice heard it attributed to Dana Scott that he said something to the effect that the consistency of the lambda-calculus was an accident.
Does anyone have a reasonable-sounding source for this?  I find it hard to believe that Scott would talk about the Church-Rosser theorem in this way; I guess that this a mangling of something else he said, or some context is hidden.

REPLY [6 votes]: Not exactly λ-calculus, but untyped formalisms in general.
Here is an excerpt from the introduction of A type-theoretical alternative to ISWIM, CUCH, OWHY (bold emphasis is mine):

No matter how much wishful thinking we do, the theory of types is here
  to stay. There is no other way to make sense of the foundations of
  mathematics. Russell (with the help of Ramsey) had the right idea, and
  Curry and Quine are very lucky that their unmotivated formalistic
  systems are not inconsistent. <...> My point is that formalism
  without eventual interpretation is in the end useless. Now, it may turn out that a system such as the λ-calculus will have an
  interpretation along standard lines (and I have spent more days than I
  care to remember trying to find one), but until it is produced I would
  like to argue that its purposes can just as well be fulfilled by a
  system involving types. <...>

Both the introduction and the preface to this paper are very interesting to read as they somewhat explain Scott’s feelings towards typed vs. untyped systems and contain some nice bits of history.<|endoftext|>
TITLE: Equality of the sum of powers
QUESTION [9 upvotes]: Hi everyone, I got a problem when proving lemmas for some combinatorial problems,
and it is a question about integers.
Let 
$\sum_{k=1}^m a_k^t = \sum_{k=1}^n b_k^t$ 
be an equation,
where $m, n, t, a_i, b_i$ are positive integers, and
$a_i \neq a_j$ for all $i, j$,
$b_i \neq b_j$ for all $i, j$,
$a_i \neq b_j$ for all $i, j$.

Does the equality have no solutions?

For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem,
and even for $n = m$, we have solutions like 
$1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$.
For $t > 2$, similar equalities hold:
$1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$
and 
$1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$,
and we can extend this trick to all $t > 2$.

The question is, if we introduce one more restriction, that is,
  $|a_i - a_j| \geq 2$ and $|b_i - b_j| \geq 2$ for all $i, j$,
  is it still possible to find solutions for the equation?

For $t = 2$ we can combine two Pythagorean triples, say,
$5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$,
but how about the cases for $t > 2$ and $n = m$?

REPLY [3 votes]: For any $t$, if $m$ is sufficiently large relative to $t$, and $n$ is any positive integer (possibly equal to $m$), then the circle method proves that there exists an infinite sequence of increasingly large solutions such that the ratios between the $a_1,\ldots,b_n$ approach any real positive ratios you want (assuming that a real solution with those ratios exists for that $m,n,t$).  This answers your question with much stronger inequalities than the ones you imposed.  If you want, you can simultaneously specify the residues of the $a_i$ and $b_j$ modulo some fixed number $N$, provided that those residues are compatible with the equation. In fact, again when $m \gg t$, you can even fix the $b_j$ in advance: the solution to Waring's problem guarantees the existence of $a_1,\ldots,a_m$, and a slight strengthening lets you impose your inequalities too, if the $b_j$ are large enough.  (Proof: the circle method again.)<|endoftext|>
TITLE: Lower bounds on (truncated) Fourier transform of functions of constant modulus and bounded derivative
QUESTION [11 upvotes]: Let $f(x)=e^{i\phi(x)}$ define a function from $[0,1]$ to the complex unit circle through the real smooth function $\phi(x)$. Also, this function is periodic: $\phi(0)=\phi(1)=0\text{ mod }2\pi$ and has bounded derivatives for $x\in[0,1]$: $$\vert d\phi(x)/dx\vert\le \omega.$$ Consider the (truncated) Fourier transform of $f(x)$: $$\hat{f}(k)=\int_0^1 f(x) e^{ikx}dx.$$
What is a lower bound on $E(K) := \Vert f\Vert_{L^2,[0,K]}$? More explicitly what is a lower bound for $$E(K)= \int_0^K \vert \hat{f}(k)\vert^2 dk$$
A similar question was asked and answered here where there are no  restrictions on $f$ resulting on a lower bound of zero for the value of $\vert \hat{f}(k)\vert$. The case where $\phi(x)$ is a piecewise constant function is answered here where the problem is equivalent to bounding polynomials (even with non-integer exponents) and is related to the following physics paper. From physical considerations, I expect the lower bound to depend on $\omega$.
I suspect that: 

the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off. 
a lower bound might best be described by a different quantity other than the derivative that I have provided: another sort of complexity will have to go into $\phi(x)$.
This could be related to some uncertainty principle as we want to minimize the support of $\hat{f}$. Lemma 3, in Terry Tao's blog entry on Hardy's Uncertainty Principle seems to be related except that it works only for even order derivatives and the connection is vague anyways.
I am missing a trivial point. Maybe constant modulus, periodic, etc is unnecessary and something can be said directly using $\sup\vert f(x)\vert$ and  $\sup\vert f'(x)\vert$? That would somewhat annul my next question though.
My physics background automated me to discretize the problem (i.e. piece-wise linear $\phi(x)$) but I was not successful although the approach seemed promising. See 1 also.

How could I then find the functions $\phi(x)$ that saturate the lower bound?

REPLY [9 votes]: It is an interesting problem which is related to some recent work of mine. The reason for why I started to work on similar problems  is because  connections to a problem of Ramachandra on Dirichlet polynomials, connections to the nordic school of Hardy classes of Dirichlet series (Hedenmalm, Saksman, Seip, Olsen, Olofsson, Lindqvist and others), as well as universality questions for zeta-functions and their properties on the line Re(s)=1.
While my papers are not quite finished, I have put two early preprints on my homepage,  On a problem of Ramachandra and approximation of functions by Dirichlet polynomials with bounded coefficients and  On generalized Hardy classes of Dirichlet series. I have talked about some of these problems at analytic number theory conferences in India. Like in your paper I have considered Dirichlet series (it should be possible to obtain something like Theorem 2.1 in your paper by my method also, although I have not stated a direct analogue in my paper).
Now your problem in the question is rather easy for small $\omega$ so we will from now on assume that $\omega>1/2$.  In fact if $\omega<1/2$, then $|f(0)|>1/2$ and $\int_0^K |\hat f(t)^2|dt \geq \min(1/10,K/10)$ (constants not chosen in an optimal way)
In my papers on Dirichlet series I have used a somewhat different method than you use in your paper, namely the Jensen inequality on the logarithmic integral in a half-plane. This method is applicable for the problem at hand.  Lemma 7 in my paper ``On generalized Hardy classes of Dirichlet series'' can be used with  $\sigma=0$ and $L(it)=\hat f(-t)$   and we obtain
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq     \frac D \pi \int_{-\infty}^\infty \frac {\log^+  |\hat f (t)|} {D^2+t^2} dt -         \log  |\hat f(iD)|.
$$ 
For similar results see also Koosis - The logarithmic integral. (Remark Feb 16: The above inequality is an equality if the function is non-zero on a half plane. The inequality follows from Jensen's formula on a disc by mapping the half plane on the disc by the standard holomorphic bijection where $iD$ goes to $0$)
The reason why we can do this is that with the definition of the fourier-transform in your question it means that $ \hat f(z)$ will be a bounded analytic function in the half plane Im$(z) \geq 0$. 
Now in this case we also have that $\log^+  |\hat f (t)|=0$ since $ |\hat f (t)| \leq 1$. Thus the inequality simplifies to
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq   -         \log  |\hat f(iD)|.$$ 
It is not too difficult to see that for $\omega>1/2$
$$
 |\hat f(i\omega)|= \left|\int_0^1 e^{i \phi(x)-\omega x} dx \right|>\frac {1} {10 \omega}.
$$ 
(The constant $10$ not chosen optimally). Thus we can choose $D=\omega$ and it is clear that
$$
 \int_0^K \log^- |\hat f(t)| dt < \frac \pi {\omega} \left({\omega^2+K^2} \right) \frac {\omega} \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {\omega^2+t^2} dt
$$
From these estimates we see that
$$
  \frac 1 K \int_0^K  \log^- |\hat f(t)| dt< \frac {\pi(\omega^2+K^2)}{\omega K} \log (10 \omega).
$$
Now we can use the Jensen inequality
$$
  \exp\left(\frac 1 K \int_0^K  \log |\hat f(t)| dt\right)< \sqrt{\frac 1 K \int_0^K  |\hat f(t)|^2 dt}
 $$
We get the lower bound
$$
 K \left(\frac 1 {10 \omega} \right)^{2\pi (\omega^2+K^2)/(K \omega)} \leq  \int_0^K  |\hat f(t)|^2 dt
$$
for $\omega>1/2$.  If $c>2 \pi$ and $\omega/K$  is sufficiently  large this gives a lower bound
$$\omega^{-c \omega/K} \leq \int_0^K |\hat f(t)|^2 dt$$
which is  weaker than your expected $e^{-c \omega/K}$. At least we have an explicit lower bound. 
Updated Feb 16: In the case where both $\omega$ and $K$ are large but still $\omega>K$ this can be improved by the following trick. Let $g$ be the convolution of $\hat f$
 with a non negative test-function $\Phi(t/K)$, such that $\hat \Phi(0)>0$ where  $\Phi$ has support on $[0,1/2]$ . Then use Jensen's inequalities on the function $g$ instead of $\hat f$ as above. The advantage with this is that it then follows that $|\hat g(iw)| \gg K/\omega$ and thus we can get the lower bound (by using Jensen's inequality w.r.t the L^1-norm instead of the L^2-norm.)
$$(\omega/K)^{-c \omega/K} \leq \frac 1 K \int_0^{K/2} |g(t)| dt$$
for some constant $c>0$. Since
$$ g(t)=\int_0^t \Phi((t-x)/K) \hat f(x) dx$$ it is clear by the triangle inequality that
$$\frac 1 K \int_0^{K/2} |g(t)| dt = \frac 1 K \int_0^{K/2} \left|\int_0^t \Phi((t-x)/K)\hat f(x) \right| dx \leq $$ 
 $$\leq \frac 1 K \int_0^{K/2} |f(x)| dx  \int_0^{K/2}  |\Phi(x/K)| dx  \leq c \int_0^{K/2} |\hat f(x)| dx $$
The inequality
$$K^{-1} (\omega/K)^{-c \omega/K} \leq \int_0^{K/2} |\hat f(t)|^2 dt$$
follows by the Cauchy-Schwarz inequality for some constant $c>0$.
This formula just use involves dimensionless quantity $\omega/K$ as expected. Since the function $E(K)$ is increasing in $K$ it gives the lower bound $E(K) > C_0 K^{-1}>0$ for $1 \leq \omega \leq K$ for some absolute constant $C_0$.<|endoftext|>
TITLE: Generation in a group versus generation in its abelianization.
QUESTION [22 upvotes]: Background
I have been spending a lot of time in my research with subsets of groups that are very close to being generating sets. To make this precise:
Let $G$ be a group. If a subset $S$ of $G$ projects onto a generating set of $G/[G,G]$, we say that $S$ weakly generates $G$. The following fact (see page 350 in this book for a proof) shows that weak generation in nilpotent groups is a strong condition.

Fact. Let $G$ be a nilpotent group. If $S$ weakly generates $G$, then $S$ generates $G$.

In light of this result, we ask the following question:

Does there exist a finitely presented but non-nilpotent group $G$ such that every weakly generating subset of $G$ generates $G$?

If we drop the condition "finitely presented" then the first Grigorchuk group suffices. I'd be pretty surprised if no finitely presented examples exist.
Surface groups and free groups
In response to Matt's question below: For the free group $F(a,b)$, the set $\{a[[a,b],a],b \}$ weakly generates but doesn't generate (you can show this directly using uniqueness of freely reduced word form in a free group). You can use this to show that any non-abelian closed surface group has subsets that weakly generate but don't generate. For instance, in the genus two case, suppose $G$ has the standard presentation with generators $a,b,c,d$ and relation $[a,b][c,d]$. Consider the set $S = ${$a,b[[b,c],b],c,d$}. If this set generates G, then it generates the image $G/N$, where $N$ is the normal subgroup generated by $a$ and $d$. This image is a free group generated by the images of $b$ and $c$. The set S  projects to a set which does not generate.

REPLY [4 votes]: The goal of my answer is to provide references to articles that studied the class of groups under consideration. 
Let $\mathcal{C}$ be the class of the finitely generated groups in which every weakly generating subset is a generating subset. Then this class has been studied, to some extent, in [1] and [2] in the context of the Andrews-Curtis conjecture.
Indeed, the authors considered in [1] and [2] the class $\mathcal{MN}$ of finitely generated groups in which every normally generating subset is a generating subset (the notation $\mathcal{MN}$ is used here because the latter class coincides with the class of groups whose maximal subgroups are normal; this is also the class of groups $G$ for which $[G, G] \subset \text{Frattini}(G)$).
Clearly, $\mathcal{MN}$ contains $\mathcal{C}$.
If every lift in $G$ of a generating subset of $G/[G, G]$ normally generates $G$, then $G \in \mathcal{C}$ if and only if $G \in \mathcal{MN}$. Finitely generated soluble groups have this property.
Unfortunately, none of these references answer OP's question: the non-nilpotent examples given in [2] are the Grigorchuck groups and the Gupta-Sidki groups and they don't admit any finite presentation.
Still, I find these articles helpful for they certainly improve our understanding of $\mathcal{C}$. Some stability conditions for $\mathcal{MN}$ are established in [2] (the class $\mathcal{MN}$ is stable under direct product [Corollary 2.4, 2], it enjoys some stability under taking finite index subgroups [Proposition 2.3, 2]) and many examples of groups that don't belong to $\mathcal{MN}$ are given (groups with finitely many ends [Proposition 2.5, 2], groups containing a non-abelian free subgroup of finite index [Corollary 1, 2]). 
There is also this remark, which wasn't mentioned yet, see [Corollary 2, 2]: 

A finitely generated linear group lies in $\mathcal{MN}$ if and only if it is nilpotent.

In [1], the class $\mathcal{MN}$ is identified with the class of groups with zero recalcitrance (the recalcitrance of a finitely generated group, if finite, is an integer that measures how well the group complies to the generalized Andrews-Curtis conjecture). So, the class of $r$-recalcitrance groups with $r > 0$ naturally generalizes $\mathcal{MN}$.

[1] "Recalcitrance in groups", R. G. Burns et al., 1999.
[2] "The class $\mathcal{MN}$ of groups in which all maximal subgroups are normal", A. Myropolska, arXive:1509.08090 [math.GR], 2015.<|endoftext|>
TITLE: What are the possible sets of degrees of irreducible polynomials over a field?
QUESTION [38 upvotes]: Hopefully this is not too easy an exercise.
Let $F$ be a field.  Let $I \subset \mathbb{N}$ be the set of all positive integers $d$ such that there exists an irreducible polynomial of degree $d$ over $F$.  What kind of $I$ can occur?
Of course $1 \in I$, and of course we can have $I = \mathbb{N}$ or $I = \{ 1, 2 \}$ or $I = \{ 1 \}$.  The Artin-Schreier theorem implies (I think) that if $I$ is finite, then only the latter two cases occur.  So what kind of infinite $I$ can occur?  
Edit:  For example, correct me if I'm wrong, but we can get $I = \{ 1, p, p^2, ... \}$ for any prime $p$.  Start with $\mathbb{F}_l, l \neq p$, which has absolute Galois group $\hat{\mathbb{Z}} = \prod_q \mathbb{Z}\_q$ and take the fixed field $K$ of $\prod_{q \neq p} \mathbb{Z}\_q$.  Then $G_K = \text{Gal}(\overline{\mathbb{F}_l}/K) = \mathbb{Z}_p$, which (again, correct me if I'm wrong) has the property that its only finite quotients are the groups $\mathbb{Z}/p^n\mathbb{Z}$.  Does this work?

REPLY [14 votes]: This is a very interesting question, however it seems like a characterization of all subsets of $\mathbb N$ which can be such a degree sequence is very hard. I will point you to the paper "On the degrees of finite extensions of a field" by B. Gordon and E.G. Straus where they study this very problem. Unfortunately, it doesn't seem like the paper is available online so I will include here some of their results. For a field $k$, denote by $S(k)$ the sequence of degrees of irreducible polynomials in $k[x]$, since the only finite $S(k)$ are $\{1\}$ and $\{1,2\}$ by Artin-Schreier we will consider infinite sequences.

 Let $P$ be any set of prime numbers, denote by $N_P$ the set of integers whose prime divisors are in $P$. There is a field $k$ of any prescribed characteristic with $S(k)=N_P$.
For every field $k$, if $a,b\in S(k)$ and $gcd(a,b)=1$ then $ab\in S(k)$. If $K$ is algebraic over $k$ then every element of $S(K)$ divides some element of $S(k)$.
For every finite set $A\subset \mathbb{N}$ which doesn't contain $1$, there is a finite $A'\subset \mathbb N$ and a field $k$ so that $S(k)=\mathbb N - A'$ and $A\subset A'$.
If $S(k)$ consists only of powers of a prime $p$ then $S(k)=N_p$.
If $k$ is a CE field (all finite extensions are cyclic) then $S(k)$ is either equal to some $N_P$ or to all elements in some $N_P$ which are not divisible by $4$. 

There is also something you can say from the perspective of computability. If $P$ is $\Sigma_1$ then there is a field $k$ so that $S(k)=N_P$ and you can take $k$ to be computable. See "Sets of degrees of computable fields".<|endoftext|>
TITLE: Chas-Sullivan string topology
QUESTION [8 upvotes]: I recently read the original paper by Chas-Sullivan on string topology, in which they introduce some operations on homology of free loopspace LM, where M is a compact oriented manifold, giving it the structure of a (Gerstenhaber-)Batalin-Vilkovisky algebra. However, the arguments in this paper rely on some transversality assumptions, and I'm not sure whether these assumptions are justified. I know that the Chas-Sullivan operations have been constructed via homotopy theoretic methods by Cohen, Jones, Voronov (hopefully I'm not missing any names here), but I am wondering whether anybody has managed to construct the Chas-Sullivan operations in a way that more or less follows the original ideas (e.g. without using any homotopy theory).

REPLY [11 votes]: I would like to point at the Diploma thesis of my student Lennart Meier, who has given various elementary descriptions of the Chaas Sullivan product (for example using my description of singular homology in terms of bordism groups of stratifolds, see: http://www.hausdorff-research-institute.uni-bonn.de/files/kreck-DA.pdf). I'm sure he will send you an electronic version of his thesis: lennart@meier-bielefeld.de.
Matthias Kreck<|endoftext|>
TITLE: Are compact submanifolds of "S X (0,1)" with one boundary component handlebodies, where S is a closed surface?
QUESTION [5 upvotes]: Suppose I have a compact 3-dimensional submanifold N of S X (0,1) which has one boundary component, where S is a closed surface.  Must N be a handlebody?

REPLY [8 votes]: There is a class of three-manifolds which are called  Handlebodies with Wormholes.  M is a handlebody with wormholes if it is homeomorphic to the result of removing the regular neighborhood of a system of properly embedded arcs from a handlebody, that includes the example that Richard gave above.  There is a Topology paper by Menasco and Thompson
"Compressing Handlebodies with holes"  that studies this class of manifolds.
If you also allowed the removal of the regular neighborhoods of graphs that did not touch the boundary of $S$, that class of manifolds would include every compact three-manifold with boundary that can be embedded in $S\times [0,1]$.  
More simply, suppose you decided to remove the neighborhood of one properly embedded arc
in $S\times [0,1]$ , whose boundary consisted of a point in $S\times \{0\}$ and one in $S\times \{1\}$, I proved that the complement is a handlebody if and only if the arc
is isotopic to a vertical arc $\{pt\}\times [0,1]$.  That appeared in,
Minimal surfaces and Heegaard splittings of the three-torus.  Pacific J. Math.  124  (1986),  no. 1, 119--130
I then developed an unknotting lemma for systems of arcs
An unknotting lemma for systems of arcs in $F\times I$.  Pacific J. Math.  139  (1989),  no. 1, 59--66. 
There is related work of Gordon, that was done about the same time, but the referee lost my paper so that there was a two year lag publication.
On primitive sets of loops in the boundary of a handlebody.
Topology Appl. 27 (1987), no. 3, 285--299. 
appeared.<|endoftext|>
TITLE: Is an algebraic bijection from a projective variety to itself necessarily an isomorphism?
QUESTION [8 upvotes]: Let $X$ be a projective variety. Assume there is an algebraic map $f: X \rightarrow X$ that is a bijection. I am thinking of $X$ as a variety, not a scheme, so by a bijection I mean a bijection on closed points. Most likely I am working over the complex numbers, so if you like I mean a bijection on complex points. Can you conclude that $f$ has an algebraic inverse? 
I think this is not immediately obvious, since it is not true that any algebraic bijection between two projective varieties is an isomorphism. For instance, there is an algebraic bijection from ${\Bbb P}^1$ to a cuspidal cubic in ${\Bbb P}^2$ given by $[x,y] \rightarrow [x^3, x^2y, y^3]$. So if this is true one must use the fact that the map is from $X$ to itself.
I am interested in cases where $X$ is both singular and reducible (although is of pure dimension, if that helps), so a complete answer would cover any such case. Alternatively, if it is not true that such a map has an algebraic inverse, I would like an explicit counter example.

REPLY [8 votes]: I am indeed claiming that it works over any field, but with the additional assumption that the morphism f be separable (this is required to make the induced morphism between the normalizations an isomorphism, rather than simply a finite morphism).  Also, the Frobenius morphism is actually proper.
The main emphasis of the argument above though was that it did not require any form of irreducibility on the schemes, which I thought was important.  d<|endoftext|>
TITLE: blowing up, -1 curves, effective and ample divisors
QUESTION [26 upvotes]: Lets say we're on a smooth surface, and we blow up at a point.
Is there a simple explicit computation that shows to me the fact that the exceptional divisor E has self intersection -1 ?  I don't consider the canonical divisor explicit (but am open to it). I do consider power series hacking to be explicit.
I'm quite unnerved by this -1. Is E effective (seems to be, by definition?). Is E ample (seems to not be, by Nakai-Mozeishon type things)?  More generally, I used to think of effective and ampleness as both being measures of "positivity"; but perhaps this is wrong - what do effectiveness and ampleness have to do with each other.
What happens locally at a point of -1 intersection? I thought two irreducible curves on a surface should intersect either in 0 points, or in a positive number of points. To find E.E
I would have tried to move E to some other divisor, and then I would get E.E = 0 or nonnegative.
Sorry for the multiple questions, but I'm really distressed :(

REPLY [6 votes]: Write $\mathbb{CP}^1$ as two copies of $\mathbb{C}$, with coordinates $z_1$ and $z_2$ respectively, glued along the map 
$z_1  \mapsto z_2=\frac{1}{z_1}$ on $z_1\neq 0$.
Write the line bundle $\mathcal{O}(-1) \rightarrow \mathbb{CP}^1$ as two copies of $\mathbb{C} \times \mathbb{C}$, with coordinates $(z_1, v_1)$ and $(z_2,v_2)$ respectively, glued along the map $(z_1,v_1)  \mapsto (z_2, v_2) = (\frac{1}{z_1}, z_1 v_1)$ on $z_1\neq 0$. Denote the zero-section by $Z$.
By definition of a blow-up, there is a holomorphic isomorphism from a small neighborhood of E to a neighborhood of $Z$, and this isomorphism sends $E$ to $Z$. 
Therefore, it is enough to compute the self-intersection of $Z$. This is a topological notion, so a natural thing to do is to find a cycle $\gamma$ homologous to $Z$ and intersecting $Z$ transversely. (You can't ask $\gamma$ to be a divisor: $Z$ is the only compact divisor in $\mathcal{O}(-1)$.)
Construct such a $\gamma$ as continuous section of $\mathcal{O}(-1)$: 
on $\vert z_1 \vert \leq 1$, take $z_1 \mapsto (z_1,v_1=1)$; on $\vert z_2 \vert \leq 1$, take $z_2 \mapsto (z_2,v_2= \overline{z_2})$. On the overlap $\vert z_1 \vert=1= \vert z_2 \vert$, we have 
$v_2= \overline{z_2} = \frac{1}{z_2} = z_1 = z_1 v_1$, as needed. 
A homotopy from $Z$ to $\gamma$ is given by 
$z_1 \mapsto (z_1,t)$ and $z_2 \mapsto (z_2, t\ \overline{z_2})$, for $t\in [0,1]$. 
In particular (the image of) $\gamma$ is homologous to $Z$. 
The only intersection point of $\gamma$ and $Z$ is at $(z_2,v_2)=(0,0)$. There, the orientation of $Z$ given by its complex structure is represented by the vectors $(1,0)$ and $(i,0)$. Pushing this orientation with the above homotopy gives the orientation for $\gamma$ represented by $(1,1)$ and $(i,-i)$. 
The $\mathbb{R}$-basis of $\mathbb{C} \times \mathbb{C}$ given by $(1,0)$, $(i,0)$, $(1,1)$ and $(i,-i)$ has same orientation as 
$(1,0)$, $(i,0)$, $(0,1)$ and $(0,-i)$, which is negative. Conclusion: $Z.\gamma = -1$, so 
$Z.Z=-1$, so $E.E=-1$.<|endoftext|>
TITLE: What's the probability that k + n^2 is squarefree, for fixed k?
QUESTION [15 upvotes]: While playing around with this question (when is the sum of two squares squarefree?), from some experimental computations (and bolstered by the fact that the density of squarefree positive integers is known to exist), I came up with the following conjecture:  the asymptotic density of squarefree numbers in the sequence $(k+1^2, k+2^2, k+3^2, \ldots)$, for fixed k, exists and depends on k.
To give an example of what I mean, consider numbers of the form $1 + n^2$.   895 of the numbers $1+1^2, 1+2^2, \ldots, 1+1000^2$ are squarefree; 897 of the next thousand are; 895 of the third thousand; 896 of the fourth thousand.  891 of the numbers $1+1000001^2, \ldots, 1+1001000^2$ are squarefree, as are 895 of the numbers $1+2000001^2, \ldots, 1+2001000^2$.   So there seems to be some constant $C_1$, probably between 0.89 and 0.90, such that $1+k^2$ has ``probability'' $C_1$ of being squarefree.  The same thing seems to happen if we replace 1 with other integers, although with other constants.
Are such constants known to exist?  If they are, how can they be computed?

REPLY [25 votes]: More generally, suppose that $f(x) \in \mathbf{Z}[x]$ has no repeated factors.  For each prime $p$, let $c_p$ be the number of integers $x \in \{0,1,\ldots,p^2-1\}$ satisfying $f(x) \equiv 0 \pmod{p^2}$.  Heuristically, the probability that a random integer $x$ is such that $f(x)$ is not divisible by $p^2$ equals $1-c_p p^{-2}$, and these conditions should be independent by the Chinese remainder theorem, so one would conjecture that the fraction of integers $x$ in $\{1,2,\ldots,N\}$ such that $f(x)$ is squarefree should tend to $\prod_p (1-c_p p^{-2})$, where the product is taken over all primes $p$.  For large $p$, we have $c_p \le \deg f$, so this product converges.
This guess has been proved for $\deg f \le 3$ (the case $\deg f=3$ is a nontrivial result of C. Hooley).  In particular, this answers your question for $f(x)=x^2+k$.  There is no $f$ of degree $4$ or greater for which the density is known to exist (except in cases when the density is $0$ because some $c_p$ equals $p^2$).  On the other hand, A. Granville proved that the $abc$ conjecture implies that the density exists and equals the predicted value for $f$ of any degree.  For further references and a generalization to multivariable polynomials, see the papers cited in the references below.
C. Hooley, On the power free values of polynomials, Mathematika 14 (1967), 21-26.
A. Granville, $ABC$ allows us to count squarefrees, Internat. Math. Res. Notices 1998, no. 19, 991-1009.
B. Poonen, Squarefree values of multivariable polynomials, Duke Math. J. 118 (2003), no. 2, 353-373.<|endoftext|>
TITLE: longest consecutive subsequence of a random permutation
QUESTION [5 upvotes]: What is the expected length of the longest consecutive subsequence of a random permutation of the integers 1 to N? To be more precise we are looking for the longest string of consecutive integers in the permutation (disregarding where this string occurs). I believe the answer should be ~ c ln(n), but I have been unable to prove this.
Update: We are looking for the longest increasing subsequence. This is to say 9,1,5,2,7,3 has an increasing subsequence 1,2,3.

REPLY [11 votes]: The purpose of this answer is to use the second moment method to make rigorous the heuristic argument of Michael Lugo.  (Here is why his argument is only heuristic: If $N$ is a nonnegative integer random variable, such as the number of length-$r$ increasing consecutive sequences in a random permutation of $\{1,2,\ldots,n\}$, knowing that $E[N] \gg 1$ does not imply that $N$ is positive with high probability, because the large expectation could be caused by a rare event in which $N$ is very large.)
Theorem: The expected length of the longest increasing block in a random permutation of $\{1,2,\ldots,n\}$ is $r_0(n) + O(1)$ as $n \to \infty$, where $r_0(n)$ is the smallest positive integer such that $r_0(n)!>n$.  ("Block" means consecutive subsequence $a_{i+1},a_{i+2},\ldots,a_{i+r}$ for some $i$ and $r$, with no conditions on the relative sizes of the $a_i$.)
Note: As Michael pointed out, $r_0(n)$ is of order $(\log n)/(\log \log n)$.
Proof of theorem: Let $P_r$ be the probability that there exists an increasing block of length at least $r$.  The expected length of the longest increasing block is then $\sum_{r \ge 0} r(P_r-P_{r+1}) = \sum_{r \ge 1} P_r$.  We will bound the latter sum from above and below.
Upper bound: The probability $P_r$ is at most the expected number of increasing blocks of length $r$, which is exactly $(n-r+1)/r!$, since for each of the $n-r+1$ values of $i$ in $\{0,\ldots,n-r\}$ the probability that $a_{i+1},\ldots,a_{i+r}$ are in increasing order is $1/r!$.  Thus $P_r \le n/r!$.  By comparison with a geometric series with ratio $2$, we have $\sum_{r > r_0(n)} P_r \le P_{r_0(n)} \le 1$.  On the other hand $\sum_{1 \le r \le r_0(n)} P_r \le \sum_{1 \le r \le r_0(n)} 1 \le r_0(n)$, so $\sum_{r \ge 1} P_r \le r_0(n) + 1$.
Lower bound: Here we need the second moment method.  For $i \in \{1,\ldots,n-r\}$, let $Z_i$ be $1$ if $a_{i+1}<a_{i+2}<\ldots<a_{i+r}$ and $a_i>a_{i+1}$, and $0$ otherwise.  (The added condition $a_i>a_{i+1}$ is a trick to reduce the positive correlation between nearby $Z_i$.)  The probability that $a_{i+1}<a_{i+2}<\ldots<a_{i+r}$ is $1/r!$, and the probability that this holds while $a_i>a_{i+1}$ fails is $1/(r+1)!$, so $E[Z_i]=1/r! - 1/(r+1)!$.  Let $Z=\sum_{i=1}^{n-r} Z_i$, so 
$$E[Z]=(n-r)(1/r! - 1/(r+1)!).$$
Next we compute the second moment $E[Z^2]$ by expanding $Z^2$.  If $i=j$, then $E[Z_i Z_j] = E[Z_i] = 1/r! - 1/(r+1)!$; summing this over $i$ gives less than or equal to $n/r!$.  If $0<|i-j|<r$, then $E[Z_i Z_j]=0$ since the inequalities are incompatible.  If $|i-j|=r$, then $E[Z_i Z_j] \le 1/r!^2$ (the latter is the probability if we drop the added condition in the definition of $Z_i$ and $Z_j$).  If $|i-j|>r$, then $Z_i$ and $Z_j$ are independent, so $E[Z_i Z_j] = (1/r! - 1/(r+1)!)^2$.  Summing over all $i$ and $j$ shows that
$$E[Z^2] \le \frac{n}{r!} + E[Z]^2 \le \left(1 + O(r!/n) \right) E[Z]^2.$$
The second moment method gives the second inequality in
$$P_r \ge \operatorname{Prob}(Z \ge 1) \ge \frac{E[Z]^2}{E[Z^2]} \ge 1 - O(r!/n).$$
If $r \le r_0(n)-2$, then $r! \le (r_0(n)-1)!/(r_0(n)-1)$, so $r!/n \le 1/(r_0(n)-1)$, so $P_r \ge 1 - O(1/r_0(n))$.  Thus
$$\sum_{r \ge 1} P_r \ge \sum_{r=1}^{r_0(n)-2} \left( 1 - O(1/r_0(n)) \right) = r_0(n) - O(1).$$<|endoftext|>
TITLE: Are there Ricci-flat riemannian manifolds with generic holonomy?
QUESTION [24 upvotes]: This may well be an open problem, I'm not sure.
In Berger's classification (refined by Simons, Alekseevsky, Bryant,...) of the holonomy representations of irreducible non-symmetric complete simply-connected riemannian manifolds, there are some cases which imply Ricci-flatness: namely, $\mathrm{SU}(n)$ (Calabi-Yau) in dimension $2n$, $\mathrm{Sp}(n)$ (hyperkähler) in dimension $4n$, $G_2$ in dimension $7$ and $\mathrm{Spin}(7)$ in dimension $8$.
A natural question is the converse: whether Ricci-flatness implies a reduction of the holonomy.   The other holonomy representations are known not to be Ricci-flat: $\mathrm{Sp}(n)\cdot \mathrm{Sp}(1)$ (quaternionic kähler) is known to be Einstein with nonzero scalar curvature, and in the case of $\mathrm{U}(n)$ in dimension $2n$ (Kähler) it is known that if a Kähler manifold is Ricci-flat then the holonomy is contained in $\mathrm{SU}(n)$, so that it is Calabi-Yau.  So the remaining question is whether there exists any Ricci-flat riemannian manifolds with generic holonomy $\mathrm{SO}(n)$ in dimension $n$.
I would like to know the present status of this question and if it's still open what the experts think: do people expect examples of Ricci-flat riemannian manifolds with generic holonomy?
Bonus question: How about if the manifold is pseudoriemannian?

Added
Thanks to Igor's answer below, here are some further remarks.
The question needs to be refined.  The riemannian analogue of the Schwarzschild metric
on $\mathbb{R}^2 \times S^2$ is an example of a complete, simply-connected noncompact Ricci-flat metric with generic holonomy.  So the question is about compact examples.
In fact, in Berger's 2003 book A panoramic view of Riemannian geometry (page 645) one reads at the bottom of the page:

It remains a great mystery that no Ricci flat compact manifolds are known which do not have one of these special holonomy groups.

REPLY [13 votes]: I am not an expert but the question:   "Does there exist a simply-connected closed Riemannian Ricci flat $n$-manifold with $SO(n)$-holonomy?"  is a well-known open problem. Note that Schwarzschild metric is a complete Ricci flat metric on $S^2\times\mathbb R^2$ with holonomy $SO(4)$, so the issue is to produce compact examples; I personally think there should be many. The difficulty is that it is hard to solve Einstein equation on compact manifolds. If memory serves me, Berger's book "Panorama of Riemannian geometry" discusses this matter extensively.<|endoftext|>
TITLE: When are the units of R[x] exactly the units of R?
QUESTION [6 upvotes]: I (Anton) have edited this question to be the question Pete and Zeb discuss in the first few comments.

What conditions on a ring $R$ imply that the units of $R[x]$ are exactly the units of $R$?

REPLY [23 votes]: If $R$ is a commutative ring, then by the following result, the answer is "if and only if $R$ is reduced."

If $R$ is a commutative ring, then $a_0+a_1x+\cdots + a_nx^n\in R[x]$ is a unit if and only if $a_0$ a unit in $R$ and $a_i$ is nilpotent for $i>0$.

Proof. One direction is easy. Any polynomial of the given form is a unit because the sum of a unit and a nilpotent element is always a unit. 
The other direction isn't too hard if $R$ is a domain (the product of non-zero elements is always non-zero). If $g=b_0+\cdots b_mx^m$ (with $b_m\neq 0$) is the inverse of $f=a_0+\cdots+a_nx^n$ (with $a_n\neq 0$), then the highest order term of $1=f\cdot g$ is $a_nb_mx^{n+m}$, so we must have $n=m=0$ and $a_0$ invertible (with inverse $b_0$)
For the general case, suppose $a_0+\cdots +a_nx^n$ is a unit. Reducing modulo $x$, we must get a unit in $R[x]/(x)\cong R$, so $a_0$ must be a unit. Reducing modulo any prime $\mathfrak p\subseteq R$, we get a unit in $(R/\mathfrak p)[x]$. Since $R/\mathfrak p$ is a domain, the previous paragraph shows that $a_i\in \mathfrak p$ for all $i>0$ and all primes $\mathfrak p$. Since the intersection of all primes is the nilradical, each $a_i$ must be nilpotent.

A more "bare hands" elementary proof is given in Ex. 1.32 of Lam's Exercises in Classical Ring Theory. He also gives counterexamples to both implications if $R$ is not assumed commutative and mentions a really interesting related question. If $I\subseteq R$ is an ideal all of whose elements are nilpotent and $a_i\in I$, then does it follow that $1+a_1x+\cdots +a_nx^n$ is a unit in $R[x]$? If you can prove that it does, it would imply the Köthe conjecture, a famous problem in ring theory.<|endoftext|>
TITLE: How can I produce 'canonical' forms for rooted bipartite graphs?
QUESTION [7 upvotes]: The graphs I'm interested in are bipartite graphs with a specified root vertex. Because there's a root, all the vertices are 'graded' by their distance from the root. Because the graph is bipartite, vertices at depth $d$ are only ever connected to other vertices at depths $d \pm 1$ (and in particular not depth $d$).
When I represent these graphs, I order the vertices at each depth, and record the edges by a series of matrices, essentially the list of adjacency matrices from each depth to the next. (That is, the full adjacency matrix is symmetric and block tridiagonal, with zero diagonal blocks. I just write down the superdiagonal blocks.)
Now, if I reorder the vertices at some depth (this just permutes the rows of one matrix and the columns of the next), obviously I have the same underlying graph. I'd like an algorithm that picks a particular ordering at each depth, for each such graph, producing a 'canonical form', with the following properties:

the algorithm is idempotent; applying it a second time does nothing,
the algorithm is stable, in the sense that if you just look at the first $d$ depths of a graph, and see that that graph is already 'in canonical form', then when you produce a canonical form for the whole graph those first $d$ depths aren't changed, and
as many isomorphic graphs as possible are identified!

It may not be possible to satisfy 3. completely; for example the identity operation satisfies 1. and 2., but does a very bad job at 3. It's not essential for my application that every isomorphic pair of graphs are identified. (I'd be using this algorithm to speed up a combinatorial search of certain types of graphs, where I know that I'm unnecessarily producing many isomorphic copies of the same graph, but the details of the search require that I use this representation.)
Does anyone know of such an algorithm? Can anyone suggest something good?

REPLY [10 votes]: This problem is algorithmically equivalent to the general problem of finding a canonical labelling for a graph. To see that, take an arbitrary graph, add a new vertex adjacent to everything and call it the root, then subdivide every edge with a new vertex.  The result is a rooted bipartite graph. Going back to the original is easy.  Canonical labelling has unknown worst case complexity, though there are effective programs like nauty, bliss and Traces that can handle graphs with thousands of vertices.<|endoftext|>
TITLE: What are your favorite instructional counterexamples?
QUESTION [192 upvotes]: Related: question #879, Most interesting mathematics mistake.  But the intent of this question is more pedagogical.
In many branches of mathematics, it seems to me that a good counterexample can be worth just as much as a good theorem or lemma.  The only branch where I think this is explicitly recognized in the literature is topology, where for example Munkres is careful to point out and discuss his favorite counterexamples in his book, and Counterexamples in Topology is quite famous.  The art of coming up with counterexamples, especially minimal counterexamples, is in my mind an important one to cultivate, and perhaps it is not emphasized enough these days.
So: what are your favorite examples of counterexamples that really illuminate something about some aspect of a subject?  
Bonus points if the counterexample is minimal in some sense, bonus points if you can make this sense rigorous, and extra bonus points if the counterexample was important enough to impact yours or someone else's research, especially if it was simple enough to present in an undergraduate textbook. 
As usual, please limit yourself to one counterexample per answer.

REPLY [2 votes]: The group algebras of non-isomorphic groups could be isomorphic.
The simplest such example is as follows. If $G_1$ and $G_2$ are the two non-isomorphic non-abelian groups of order $p^3$ for an odd prime $p$, then $\mathbb{Q}[G_i]$ for $i=1,2$ are isomorphic. This is indirectly mentioned in an earlier answer to a question regarding counterexamples in algebra.
If one wants examples with $\mathbb{Q}$ replaced by finite fields or even integers, then such groups have been found but are significantly more complicated.<|endoftext|>
TITLE: Slightly weakened / altered concepts of a field
QUESTION [5 upvotes]: I've heard of at least three slight modifications of the standard concept of field: 
meadow, which (according to this paper) is a commutative ring with unit equipped with a total unary operation $x^{−1}$,
named inverse, that satisfies these additional equations:
$(x^{−1})^{−1} = x$ and $x·(x·x^{−1}) = x$.
wheel - I heard of these in conversation, so I'm unsure of their exact definition. I believe they have a unary "inverse" operation like meadows, but I assume something is different about them.
neofield, which (according to this paper) appear to be fields, without the associativity of addition.
(I'm not going to count $\mathbb{F}_1$, I don't think it's relevant here - though I may be totally off about this). But only having these definitions, I still feel unsatisfied with the concepts. I don't feel like I understand what's going on with them, I don't know why any of these are natural things to look at, or what important theorems there are to be had about them (I mean, other than the ones in the papers I referenced, which I would hopefully understand after getting a better grounding).
So, can anyone...

provide better/more explanatory definitions for these structures, and also - as a proper category theory student - their morphisms (the papers I linked to don't state that, I think)
provide instructive examples of each structure (i.e. examples which are not also fields, demonstrating the differences)
provide whatever are considered to be "standard" references for any of these structures (a book studying them, or paper where they were first defined, etc.)
explain why we should look at these structures (I mean, beyond just curiosity about them) - where do they naturally arise, if anywhere?
explain which classic concepts/theorems about fields carry over to each structure (do they have a notion of algebraic elements? is there a Galois-like theory for them? etc.) and which don't

and if you have even more things to say about them - even better!

REPLY [2 votes]: The meadow (as defined in the question, and in the paper linked) is an "equational theory".  
A meadow is a set $A$ together with operations $0,1,+,-,\cdot,{}^{-1}$ such that $(A,0,1,+,-,\cdot)$ is a commutative ring with unit, and identities
$$
(x^{-1})^{-1} = x
\\
x\cdot(x\cdot x^{-1}) = x
$$
hold for all $x \in A$.
As with all equational theories, this tells us what are the morphisms, subalgebras, ideals, products, quotients, and so on.  So: if $A, B$ are meadows, then a map $f : A \to B$ is a homomorphism iff
$$
f(0)=0\\
f(1)=1\\
f(x+y)=f(x)+f(y)
\\
f(-x)=-f(x)\\
f(xy) = f(x)f(y)
\\
f(x^{-1})=f(x)^{-1}
$$
(Some of these follow from the others, but abstractly you just say it preserves all the operations.)
A submeadow of a meadow $A$ is a subset $B \subseteq A$ such that
if $x,y \in B$, then
$$
0, 1, x+y, -x, x\cdot y, x^{-1} \in B
$$  
An important example of a meadow is a field, with the usual partial operation $x^{-1}$ enhanced to a total operation by defining $0^{-1}=0$.  This is called a zero totalized field.
More interesting examples are products of fields.  For example $\mathbb{Md}_6 = \mathbb F_2 \times \mathbb F_3$.
Theorem: Any meadow is (up to isomorphism) a submeadow of a product of zero totalized fields.
As Robin Chapman noted (quoted in the paper mentioned): Take a meadow and forget the inverse operation, and you have a von Neumann regular ring; start with a von Neumann regular ring with unit, there is a unique way to define the inverse making it a meadow.<|endoftext|>
TITLE: Noncommutative smooth manifolds
QUESTION [19 upvotes]: Connes defined a noncommutative analog of a closed oriented Riemannian spin^c manifold using spectral triples.
Using his definition it is unclear how to separate the smooth structure from the metric.
How can we define a noncommutative smooth manifold without the additional Riemannian and spin^c structures?
Any references on this subject will be appreciated.

REPLY [16 votes]: I'm a bit wary of resurrecting such an old question, but given that the precise content of the reconstruction theorem doesn't seem to be terribly well disseminated, please permit me to cross-post from math.SE and then make some extra comments:
"To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:

A unital Frechet pre-$C^\ast$-algebra $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact orientable $p$-manifold if and only if there exists a $\ast$-representation of $A$ on a Hilbert space $H$ and a self-adjoint unbounded operator $D$ on $H$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$.
In particular, $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact spin$^{\mathbb{C}}$ $p$-manifold if and only if there exist $H$ and $D$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$ and $A^{\prime\prime}$ acts on $H$ with multiplicity $2^{\lfloor p/2\rfloor}$.

"Once you know that $A \cong C^\infty(X)$, you can then apply the much earlier "baby reconstruction theorem" (for lack of a better phrase) announced by Connes and proved in detail by Gracia-Bondia--Varilly--Figueroa to conclude that:

In the general case, $(A,H,D) \cong (C^\infty(X),L^2(X,E),D)$ where $E \to X$ is a Hermitian vector bundle and $D$ can be interpreted as an essentially self-adjoint elliptic first-order differential operator on $E$.
In the case where $A^{\prime\prime}$ acts with multiplicity $2^{\lfloor p/2 \rfloor}$, $E \to X$ is in fact a spinor bundle (i.e., irreducible Clifford module bundle) and $D$ is Dirac-type (viz, a perturbation of a spin$^{\mathbb{C}}$ Dirac operator by a symmetric bundle endomorphism of $E$).

"So, whilst you can refine the reconstruction theorem to a characterisation of compact spin$^{\mathbb{C}}$ manifolds with spinor bundle and essentially self-adjoint Dirac-type operator, the general result is really just a statement about compact orientable manifolds. Indeed, one can even refine the reconstruction theorem to a characterisation of compact oriented Riemannian manifolds with self-adjoint Clifford module and essentially self-adjoint Dirac-type operator."
As for why you need more than just an algebra $A$, here's what I understand of the situation:

Gel'fand--Naimark says that a commutative unital $C^\ast$-algebra gives a compact Hausdorff space, no more and no less.
Going by the example of $C^\infty(X) \subset C(X)$, one might try considering commutative unital Frechet pre-$C^\ast$-algebras, but one can readily cook up examples of such algebras that aren't isomorphic to $C^\infty(X)$ for some $X$. So, if one still wishes to follow this line of inquiry, one would need still more structure.
In terms of the reconstruction theorem itself (which does tend to be treated as a black box), Connes's proof (insofar as I can understand) really makes absolutely essential use of all three parts of the spectral triple $(A,H,D)$:

the norm closure of $A$ in $B(H)$, by Gel'fand--Naimark, gives you a (canonical) compact Hausdorff space $X$;
the noncommutative integral defined by $D$ gives a Radon measure on $X$;
the Hochschild cycle of the orientability condition gives you candidates for charts;
the operator $D$ then gets used to show that you actually have a smooth atlas. 

This doesn't suggest, of course, that a spectral triple is the optimal notion of algebra + extra data qua noncommutative manifold, but it does suggest that it might well be a reasonably economical definition. In particular, it suggests that the real puzzle with the spectral triple formalism isn't the extra Riemannian data, but rather the seemingly absolute necessity of orientability.

My apologies for the long-windedness!<|endoftext|>
TITLE: Cone of curves and Mori theorem for algebraic surfaces
QUESTION [8 upvotes]: In describing part of the geometry of the cone of curves for an algebraic surface $S$, we need to find $(-1)$ curves within $S$. Once we've done that, then we can say that the "negative" part of the cone of curves has as many extremal rays as $(-1)$ curves. Here I am using the cone theorem: $$\overline{\mathrm{NE}}(S)=\overline{\mathrm{NE}}(S)_{K_S\geq 0}+\sum_i \mathbb{R}^+[C_i]$$ where $C$ are such negative self-intersection rational curves.
However, is there an intuitive argument showing that if we have a curve of negative self-intersection, then such a curve is going to generate a extremal ray in the cone of curves?
What is known about this "positive" part $\overline{\mathrm{NE}}(S)_{K_S\geq 0}$ of the cone describe in the theorem?

REPLY [6 votes]: EDIT: We may assume that the Picard number is at least two, as otherwise the cone is simply a ray generated by any effective curve. In particular, every effective curve is extremal.
I will also assume that "curve" means "effective curve".
(This edit was prompted by Damiano's comment that is now (sadly) deleted. It was a useful contribution.)
A curve on a surface is simultaneously a curve and a divisor and assuming the surface is smooth or at least $\mathbb Q$-factorial, then the curve, as a divisor, induces a linear functional on $1$-cycles. This works better if the surface is proper, so let's assume that.
So, if $C$ is such a curve, then the corresponding linear function on the space where $NE(S)$ lives is best represented by the hyperplane on which it vanishes and remembering which side is positive and which one is negative.
If $C$ is reducible, then it may have negative self-intersection, but it is not extremal. For an example, blow up two separate points on a smooth surface and take the sum of the exceptional divisors. My guess is that you meant irreducible, so let's assume that.
Now we have $3$ cases:

$C^2>0$. In this case $C$ is in the interior of the cone and it cannot be extremal, can't even be on the boundary (Use Riemann-Roch to prove this).

$C^2=0$. Since $C$ is irreducible, it follows that it is nef and hence a limit of ample classes, so it is effective, but as Damiano pointed out I have already assumed that. (It is left to the reader to rephrase this if $C$ is assumed to be nef instead of effective). In this case the hyperplane corresponding to $C$ as a linear functional is a supporting hyperplane of the cone, intersecting it at least in the ray generated by $C$. So $C$ is definitely on the boundary, but it may or may not be extremal depending on the surface. For example any curve of self-intersection $0$ on an abelian surface is extremal, but for instance a member of a fibration that also has reducible fibers is not extremal despite being irreducible. For the latter think of a K3 surface with an elliptic fibration that has some $(-2)$-curves contained in some fibers.

$C^2<0$. If $C$ is effective, then $C\cdot D>0$ for any irreducible curve $D\neq C$. This means that $C$ and all other irreducible curves lie on different sides of the hyperplane corresponding to $C$ as a linear functional, so the convex cone they generate must have $C$ generating an extremal ray.


Observe that we did not use the Cone Theorem. In fact one gets a different "cone theorem" this way:

Theorem
Let $S$ be a smooth projective surface $H$ an arbitrary ample divisor on $S$ and let
$$
Q^+=\{\sigma\in N_1(S) \vert \sigma^2 >0, H\cdot\sigma \geq 0 \}
$$
be the "positive component" of the interior of the quadric cone defined by the intersection pairing. Then
$$
\overline{NE}(S) = \overline{Q^+} + \sum_{C^2<0} \mathbb R_+[C] 
$$

There is also one for $K3$'S, using the above notation:

Theorem
Let $S$ be a smooth algebraic K3 surface and assume that its Picard number is at least $3$. (If the Picard number is at most $2$, then there are not too many choices for a cone).
Then one of the following holds:


(i)
$$
\overline{NE}(S) = \overline{Q^+}, or
$$


(ii) $$
\overline{NE}(S) = \overline{\sum_{C\simeq \mathbb P^1, C^2<0} \mathbb R_+[C]}.
$$
The two cases are distinguished by the fact whether there exists a curve in $S$ with negative self-intersection. If the Picard number is at least $12$, then only (ii) is possible.

For proofs and more details, see The cone of curves of a K3 surface.
(There is also a newer version which is less detailed, but works in arbitrary characteristic. See here or here.)<|endoftext|>
TITLE: Introduction to "commutative semialgebra"?
QUESTION [15 upvotes]: Of course, commutative algebra is a fundamental topic in algebraic geometry, number theory, representation theory, and so on.
However, there are some instances (most obviously tropical geometry) where one wishes to consider commutative rigs (or semirings) instead of commutative rings. Although most of the basic concepts of commutative algebra generalize to this setting, it's not as obvious which of commutative algebra's most famous theorems also generalize. And unfortunately there aren't canonical texts along the lines of Atiyah-Macdonald or Eisenbud or Matsumura.
Does anyone know of a good reference for learning commutative algebra over semirings?

REPLY [5 votes]: One starting point would be the paper On Ideals in semirings of K. Iseki and follow on the articles that cite it.
On the other hand "A guide to the literature on semirings and their applications in mathematics and information sciences" by K. Glazek is a guide to the extensive literature on semirings.
Among books cited there:
"Semirings and their applications", "Semirings and affine equations over them: theory and applications", "Power algebras over semirings" by J.S. Golan and "Semirings: algebraic theory and applications in computer science" by U. Hebisch, H.J. Weinert<|endoftext|>
TITLE: Family of Enriques surfaces and Grothendieck-Riemann-Roch
QUESTION [7 upvotes]: Currently I'm studying the article Moduli of Enriques surfaces and Grothendieck-Riemann-Roch by Pappas.
I am particularly interested in how he applies the GRR.
Q1. What is meant by a "family of Enriques surfaces"? I was guessing a flat morphism $f:Y\longrightarrow T$ of smooth projective varieties such that each fibre is an Enriques surface, but maybe this is too general?
Q2. In the article it says that the higher direct images $R^i f_\ast O_Y$ are zero ($i>0$). Is this an application of Grauert's theorem (III.12, Cor. 12.9, Hartshorne)? 
Q3. How to see easily that $R^0f_\ast O_Y = O_T$ ? I am guessing Grauert's theorem again.

REPLY [5 votes]: I couldn't figure out how to comment on Emerton's post, so I apologize for this non sequitur nitpick of an answer. 
"But, if $f$ is flat with geometrially connected and reduced fibres (e.g $f$ is smooth with geometrically connected fibres), then base-change for flat maps (Hartshorne III.9.3) shows that the fibre mod $P$ of $R^0fY$ is equal to $H^0(Y_P,\mathcal{O}_P)$ (the actual fibre over $P$, now, not the formal fibre), which equals $k(P)$ (the residue field at $P$), since $Y_P$ is projective, geometrically reduced, and geometrically connected over $k(P)$."
I am not sure that result from Hartshorne quite does that. It needs the map along which the base change is being made to be flat (since we need the Cech complex to remain exact under base change). In this case, we are pulling back along the decidedly unflat inclusion of the point $P$ into $Y$, and so would need the further technology of semi-continuity and base change from Hartshorne III.12 (which is unfortunately a piece of bad exposition, IMO).<|endoftext|>
TITLE: Proofs of Kirby's theorem
QUESTION [23 upvotes]: Each orientable 3-manifold can be obtained by doing surgery along a framed link in the 3-sphere. Kirby's theorem says that two framed links give homeomorphic manifolds if and only if they are obtained from one another by a sequence of isotopies and Kirby moves.
The original proof by R. Kirby (Inv Math 45, 35-56) uses Morse theory on 5-manifolds and is quite involved. There are two simpler proofs that use Wajnryb's presentations of mapping class groups. One is due to N. Lu (Transactions AMS 331, 143-156) and the other to S. Matveev and M. Polyak (Comm Math Phys 160, 537-556). I would like to ask what other proofs of Kirby's theorem are known. In particular, is there a proof that uses only Morse theory/handle decompositions of 3-manifolds?

REPLY [23 votes]: There is Bob Craggs' 1974 proof, which was never published. It relies on Wall's result, that any two 2-handle cobordisms between S^3 and itself  are stably homeomorphic if the associated bilinear forms have the same signature and type. It's very much what you are after. I have a hard-copy in my office. I do not fully follow all aspects of the argument and therefore cannot yet vouch for it. I uploaded a scan of his preprint HERE (thanks Ryan for help combining PDF files!).
I'm not positive which proof (Cerf theory or MCG) is really "simpler", because both proofs rely on a lot of background "black boxes". For Kirby's proof, later simplified by Fenn and Rourke, and then later by Justin Roberts in by Kirby calculus in manifolds with boundary, the only black box is Cerf's theorem. Surely the proof of Cerf's theorem could be tremendously simplified, and after this is eventually done by somebody, my money would be on the Cerf theory proof to be slicker. The MCG proof uses presentations of the mapping class group, which use simple connectedness of the Hatcher-Thurston complex (itself not so easy to prove), and results on buildings (a result of Brown) in order to construct presentations of the group from its action on a simply connected simplicial complex. This is actually a lot of machinery, if you think about it. Again, you can simplify the proof by using Gervais's presentation from A finite presentation of the mapping class group of an oriented surface, proven directly by Silvia Benvenuti (Finite presentations for the mapping class group via the ordered complex of curves) using an ordered complex of curves, or by Susumu Hirose Action of the mapping class group on a complex of curves and a presentation for the mapping class group of a surface using a complex of non-separating curves.
When all is said and done, I am personally not satisfied with any of the proofs out there. The Cerf theory proof, while being conceptual, takes you into deep and hard analytic terrain- while there is nothing at all conceptual about the MCG proof, despite it being "easy" (it's definitely much easier than Cerf theory, at least for me). The heart of the proof is to check that it happens to be possible to realize "relations" in your favourite finite presentation of the MCG by Kirby moves on links which generate the Dehn twists. The presentation itself is almost incidental, and the relations don't represent Kirby moves in any obvious way (the proof only goes one way- you could not prove a finite presentation of the MCG from Kirby's theorem).<|endoftext|>
TITLE: Curvature and Parallel Transport
QUESTION [24 upvotes]: Here is an updated formulation of the question, which is more precise and I think completely correct:
Suppose $M$ is a Riemannian manifold. Pick a point $p$ in $M$ and let $U$ be a neighborhood of the origin in $T_p M$ on which $exp_p$ restricts to a diffeomorphism. Let $X$ and $Y$ be tangent vectors in $T_p M$, and let $V$ be the intersection of $U$ and the plane spanned by $X$ and $Y$. Let $c(t)$ be a piecewise smooth simple closed curve in $V$. I claim that for any vector $Z$ in $T_p M$,
$R(X,Y)Z=(P_c(Z)−Z)Area(c)+o(Area(c))$
where $R$ is the Riemannian curvature tensor, $P_c$ is parallel transport around the image of $c$ under $exp_p$, and $Area(c)$ is the area enclosed by the image of $c$ under $exp_p$.
Can anyone refer me to a proof of this statement or something similar? I am fairly sure the argument has something to do with integrating the curvature 2-form over the embedded surface obtained by restricting $exp_p$ to the region enclosed by $c$, but I am having trouble with the estimates. Unfortunately I can't find anything in Kobayashi and Nimazu.
Thanks in advance!
Paul

REPLY [2 votes]: Your formula for $R(X,Y)Z$ seems to be identical to the formula on page 256 of Peter Petersen, "Riemannian Geometry", second edition, Springer 2006. Petersen gives a sketch of a proof, and calls the formula "Cartan's characterization of the curvature".<|endoftext|>
TITLE: Existence of a zero-sum subset
QUESTION [79 upvotes]: Some time ago I heard this question and tried playing around with it. I've never succeeded to making actual progress. Here it goes:
Given a finite (nonempty) set of real numbers, $S=\{a_1,a_2,\dots, a_n\}$, with the property that for each $i$ there exist $j,k$ (not necessarily distinct) so that $a_i=a_j+a_k$ (i.e. every element in $S$ can be written as a sum of two elements in $S$, note that this condition is trivially satisfied if $0\in S$ as then every $x\in S$ can be written as $x+0$).
Must there exist $\{i_1,i_2,\dots, i_m\}$ (distinct) so that $a_{i_1}+a_{i_2}+\cdots +a_{i_m}=0$?
ETA: A possible reformulation can be made in terms of graphs. We can take the vertex set $\{1,\dots ,n\}$ and for each equation $a_i=a_j+a_k$ in S add an edge $[ij]$ and its "dual" $[ik]$. The idea is to find a cycle in this graph, whose dual is a matching.

REPLY [17 votes]: The brilliant proof  of Seva, János and Péter seems pretty short for me. Here it is.
We prove the following statement which obviously implies the claim.

Let $S$ be a non-empty finite set of variables. Assume that for each $a\in S$ we have a linear form $\ell_a:=b(a)+c(a)-a$ for certain $b(a),c(a)\in S\setminus\{a\}$ (possibly $b(a)=c(a)$). Then there exists a non-empty set $A\subset S$ such that all coefficients of the linear form $\sum_{a\in A}\ell_a$ belong to $\{0,1\}$.

By induction, we may assume that this holds for smaller values of $|S|$.
Consider the multiset $T=\{b(a),c(a):a\in S\}$. We have $|T|=2|S|$. If $T$-multiplicity of every element $x\in S$ equals 2, we have $\sum_{a\in S} \ell_a=\sum_{x\in S} x$, so we may take $A=S$. If $T$-multiplicity of some $x\in S$ equals 0, just remove $x$ from $S$ ($S$ remains non-empty) and apply induction.
It remains to consider the case when $T$-multiplicity of some $z\in S$ equals 1. Say, $z=b(x)$, $\ell_x=z+y-x$ for certain $y\notin \{x,z\}$. Denote $T=S\setminus \{x,z\}$.  For $a\in T$ consider the linear form $\ell_a'$, obtained from $\ell_a$ by replacing $x$ by $y$ if necessary (note that if this happens for $\ell_y$, i.e., $\ell_y=x+t-y$, then $\ell_y+\ell_x=z+t$ and we may take $A=\{x,y\}$.) By induction, the sum of several such forms $\ell_a'$ have coefficients 0's and 1's. For the corresponding forms $\ell_a$ this means either what we need, or that the coefficient of $y$ in their sum equals -1 and the coefficient of $x$ equals 1 or 2. In the latter case add $\ell_x$ to their sum.<|endoftext|>
TITLE: Smallest permutation representation of a finite group?
QUESTION [33 upvotes]: Given a finite group G, I'm interested to know the smallest size of a set X such that G acts faithfully on X. It's easy for abelian groups - decompose into cyclic groups of prime power order and add their sizes. And the non-abelian group of order pq (p, q primes, q = 1 mod p) embeds in the symmetric group of degree q as shown here: www.jstor.org/stable/2306479.
How much is known about this problem in general?

REPLY [4 votes]: Although tighter results may be available for specific classes of groups (as in the comments and other answer), Babai, Goodman, and Pyber [1] showed that in some sense - made precise in the result below - the only obstacle to having a small faithful permutation representation is having a prime-power cyclic group of small index.
Quoting the main result from the abstract:

Let $\mu(G)$ be the smallest degree of a faithful permutation representation of $G$, and let $i(G)$ be the index of its largest cyclic subgroup of prime-power order. Then $i(G) \geq |G| / \mu(G) \geq \exp(c \sqrt{\log i(G)})$ for some constant $c > 0$.

Rephrasing in terms of bounds on $\mu(G)$, this gives
$$ |G|/\exp(c \sqrt{\log i(G)}) \geq \mu(G) \geq |G| / i(G)$$
[1] László Babai, Albert J. Goodman & László Pyber. On faithful permutation representations of small degree. Comm. Alg., 21(5):1587-1602, 1993. doi:10.1080/00927879308824639<|endoftext|>
TITLE: Different notions of associated prime (in the non Noetherian case)
QUESTION [9 upvotes]: Most books I have treat primary decomposition only in the Noetherian case. Atyiah-MacDonald goes a step further and prover the uniqueness theorems of primary decomposition without the Noetherian hypothesis. But it seems to me they get a slight different result from the usual one.
Definitions
Recall that a prime $P$ is said to be associated to the $A$-module $M$ if there exists $m \in M$ such that $P = Ann(m)$; equivalently $A/P$ injects into $M$. I denote by $Ass(M)$ the set of associated primes. If $A$ and $M$ are Noetherian, this is always not empty.
For an ideal $I$ we let $Ass(I) = Ass(A/I)$. So a prime $P$ is associated to $I$ if and only if $P$ is of the form $(I : x)$ for some $x \in A$.
For the purposes of this question let me say that $P$ belongs to $I$ if and only if $P$ is of the form $\sqrt{(I : x)}$ for some $x \in A$. We call $Bel(I)$ the set of primes belonging to $I$.
Then the result of Atyiah-MacDonald shows that if $I$ has a (minimal) primary decomposition $I = \bigcap Q_i$, and if we let $P_i = \sqrt{Q_i}$, the set of $P_i$ which appear is exactly $Bel(I)$. The usual formulation gives instead that for $A$ Noetherian this set is $Ass(I)$.
The problem
I want to understand the relationship between $Ass(I)$ and $Bel(I)$. Clearly, since prime ideals are radical, $Ass(I) \subset Bel(I)$. In general I see no reason why the opposite inclusion should be true.
Let us see how to go proving the opposite inclusion in a special case. Assume $I$ is decomposable. Then by the result of Atyiah-MacDonald it is enough to show that if we have a minimal primary decomposition $I = \bigcap Q_i$, and if we let $P_i = \sqrt{Q_i}$, then $P_i \in Ass(I)$.
Let us do this for $P_1$ and call $R = Q_2 \cap \cdots \cap Q_n$. I also call $P = P_1$, $Q = Q_1$, so $I = Q \cap R$.
Then observe that $R/I = R/(R \cap Q) \cong (R + Q) /Q \subset A/Q$. Since $Q$ is $P$-primary, $Ass(A/Q) = P$. So $Ass(R/I) \subset \{ P \}$.
If moreover $A$ is Noetherian this set has to be non empty, so $Ass(R/I) = \{ P \}$ and a fortiori it follows that $P \in Ass(A/I)$.
I don't see how to do this without the Noetherian hypothesis, though.
Questions

Is $Ass(I) = Bel(I)$ always, even if $A$ is not Noetherian?
Is $Ass(I) = Bel(I)$ if we assume that $A$ is not Noetherian, but at least $I$ is decomposable?

REPLY [4 votes]: Dear Andrea, let $A=K[X_1,X_2,\ldots ,X_n,\ldots]$, the polynomial ring in countably many variables and $I$ be the ideal $I=(X_1^2, X_2^2,\ldots )\subset A $. Then
$$\mathcal M=(X_1,X_2,\ldots)\in Bel(I) \setminus Ass(I)$$
Indeed, $(I:1)=I$ has as radical $\sqrt I=\mathcal M$, hence $\mathcal M \in Bel(I)$.
But there is no polynomial $x=P(X_1,X_2,\ldots ,X_N)\notin I$ such that $(I:x)=\mathcal M$ because $X_M$ will not satisfy $X_M.x\in I$ for $M>N$ [Of course if $x\in I$, we have $(I:x)=A\neq\mathcal M$]<|endoftext|>
TITLE: Set of vectors separated by at least a specified angle
QUESTION [8 upvotes]: Suppose $\theta$ and $d$ are given.
How big can a set of $d$-dimensional vectors be such that no pair of them are at angle less than theta?
I particularly want an upper bound; that is, an $n=n(\theta,d)$ such that given $n$ $d$-dimensional vectors, there must be at least $2$ with angle less than theta between them.
Of course, the question can be rewritten in all sorts of ways, for example, coverings of the surface of the d-dimensional sphere by $(d-1)$-dimensional caps of given radius etc. 
The bound doesn't need to be tight. Something out by a factor of $(constant)^d$ might be fine
(although something more exact would be interesting too).

REPLY [10 votes]: The subject name you are looking for is spherical codes. A good reference for this subject is Conway and Sloane's "Sphere Packings, Lattices, and Groups." In chapter 9 they give the details of the proof for the best bounds (I believe it is due to Levenstein, but don't have the book with me).
This ends up being related to density of sphere packings. There's a very elegant proof in the book which relates the answer to your question in dimension $n+1$ to the maximal density of sphere packing in dimension $n.$
Sorry I don't have my references with me, but this is all in chapter 9 of the book.<|endoftext|>
TITLE: When and why did the postdoctoral position originate?
QUESTION [22 upvotes]: Does anyone know when and how the system of post-doctoral studies after a Ph. D. originated? I've heard in a few places that there was a time when there was no such thing as a post-doc, and people used to go for faculty positions right after their Ph. D.'s. Is this true? When and why did it change? What does the post-doctoral position achieve?

REPLY [11 votes]: Another factor is that as the decades have gone by, in many branches of math the amount one has had to learn to get going researchwise has increased markedly. This of course varies a lot from subject to subject, but in many areas it can take quite a while after the PhD to develop the kind of research record that a research university wants to see before they hire you on the tenure-track and take the risks associated with that. This encourages the proliferation of postdocs as well.<|endoftext|>
TITLE: When do divisors pull back?
QUESTION [31 upvotes]: If we have a nonconstant map of nonsingular curves $\varphi:X\rightarrow Y$, then Hartshorne  defines a map $\varphi^* Div(Y)\rightarrow Div(X)$ using the fact that codimension one irreducibles are just points, and looking at $\mathcal{O}_{Y,f(p)} \rightarrow \mathcal{O}_{X,p}$. My question is if we don't have a nice map of curves, what conditions can we put on the morphism so that we may pull divisors back? Clearly it's not true in general, since we can take a constant map and then topologically the inverse image doesn't even have the right codim.
Thinking about this in terms of Cartier divisors (and assuming the schemes are integral), it seems like we just need a way to transport functions in $K(Y)$ to functions in $K(X)$. If $\varphi$ is dominant, then we'll get such a map. Is this sufficient? Also is there something we can say when $\varphi$ is not dominant? Something like we have a way to map divisors with support on $\overline{\varphi(X)}$ to divisors on $X$?

REPLY [20 votes]: It seems to me that you are approaching this in the wrong way.
First of all you have to decide whether you want to pull-back      

divisors

or

divisor classes

Pulling back divisors means pulling back cycles and you run into all sorts of problems. If your map is flat, then this is OK. See Fulton's Intersection theory book for more on this. (There is a section called "flat pull-back of cycles" somewhere early). This is actually why Hartshorne did not have to worry about pulling back divisors or divisor classes. Any non-constant morphism between non-singular curves is flat. Of course, I am not saying this is the most general condition under which you can pull-back divisors, just that this is the "natural" one. Then again, this is a little too much, because it works for any dimensional cycles, not just divisors.
Pulling back divisor classes is somewhat easier (and harder at the same time). It is easier, because then you don't have to worry about the support. As already pointed out by Frank, you just "convert" your divisor class to a line bundle, pull that back and "convert" it back to a Cartier divisor. Actually for this you need $X$ to be integral, but you seem to be OK with assuming that.
Anyway, another, perhaps more interesting question is how to pull-back Weil divisors. The main problem there is that they are not defined locally by a single equation, so their associated sheaf
 is not an line bundle, only a reflexive sheaf of rank $1$. Unfortunately pulling back those does not necessarily give you a reflexive sheaf. One solution is to still do that and then take the reflexive hull, but this will likely not give a group homomorphism. Another partial way to handle Weil divisors is to restrict to $\mathbb Q$-Cartier divisors, i.e., a divisor that itself may not be Cartier, but a multiple of which is Cartier. Then you just take that multiple, pull that back and then divide by the same number you multiplied with. Interestingly this can result in fractional coefficients, so it will only preserve linear equivalence mod $\mathbb Q$. The simplest example of a $\mathbb Q$-Cartier, but not Cartier divisor is a ruling of a quadratic cone (look at intersection numbers to prove this). A little more interesting example is included in this MO answer. For an example of a non-$\mathbb Q$-Cartier divisor consider the cone over a quadric surface in $\mathbb P^3$ and let the divisor be the cone over one line on the surface.<|endoftext|>
TITLE: The probability for a sequence to have small partial sums
QUESTION [20 upvotes]: The question
Let $a_1,a_2,\dots,a_n$ be a sequence whose entries are +1 or -1. Let t be a parameter. My question is to give an estimate for the number of such sequences so that 

$|a_1+a_2+\dots a_k| \le t$, for every $k$, $1 \le k\le n$. 

(In other words, the probability that a random sequence will satisfy the above relation.) 
I am especially interested in this probability when t is small. Either a constant, or slowly growing (say, it behaves like (log n)^s for some real number s, or slower).
variations
1) I would also like to know what is the situation if you demand that the average value of |a_1+a_2+\dots a_k| is smaller than t, rather than the maximum value. 
2) If there are more delicate estimates for the case that t itself is a function of k e.g. t itself grows as (log n)^s I would be very interested as well. 
Motivation
This question is relevant to the recent collective effort (polymath5) regarding the Erdos Discrepancy Problem (EDP). It particular it is relevant to a probabilistic heuristic regarding what the answer to EDP, and to several related questions, should be. 
It is also relevant to certain probabilistic approaches towards construction of sequences with low discrepancy.
Expectation
I would expect that the answers to the questions above are known. But they are not known to me. It is easy to be convinced, for example, that when t is bounded the number of such sequences is $c_t^{-n}$, for $c_t<2$ but I would like to know the dependence of c_t on t.

REPLY [7 votes]: Here is a useful supplement and references to the existing answers. I asked Yuval Peres a few days ago the question formulated as follows:

What is the probability that the
  simple random walk of n steps will be
  confined to the interval 
  $[-K,K]$?

Yuval's answer a few hours later was: 

The confinement probability in [-K,K]
  decays up to a constant like
  $\exp(-cn/K^2)$ where $c$ is known: it
  is $\pi/2$. This is classical and you
  can find it e.g. in Feller volume 2 or
  in Spitzer’s book. This holds for all
  $K=o(\sqrt{n})$.

I was especially interested (for polymath5 purposes) in the value of $K=K(n)$ for which this probability is $2^{-n/\log n}$. The answer to this specific query is thus $K=C\sqrt{\log n}$ for a suitable $C$.<|endoftext|>
TITLE: Prove a function is primitive recursive
QUESTION [6 upvotes]: Hey,
I'm taking a course in computability theory, but I'm struggling with primitive recursion. More specifically we are often asked to prove that some arbitrary function is primitive recursive, but I really don't know how to go about doing so.
For example:
Let $\pi (x)$ be the number of primes that are $\le x$. Show that $\pi (x)$ is primitive recursive.
How do you go about proving this formally? And, in general, what are the necessary conditions to prove if something is or is not primitive recursive.
Note: this isn't homework, I've pulled the question out from a list of examples we have been given :)

REPLY [3 votes]: There is a theorem that says that any function computable on a Turing machine in time that is a primitive recursive function of the length of the input is primitively recursive.
This just mentioned result is a really powerful theorem with numerous applications.
It easily yields results like this one.
Assume g,h,p are prim rec. Then f is prim rec where
f(x,y)=g(x,y) if x times y is 0 and
f(x+1,y+1)=h(x,y,f(x,f(x,p(x,y))),f(x+1,y)).
Trying to figure out "direct solutions" is more or less impossible.
So my recommendation: If course of values recursion and simultaneous
recursion don't work try the theorem above.<|endoftext|>
TITLE: What is a reference for profinite sets?
QUESTION [8 upvotes]: The question is in the title. The motivation behind the question is as follows: there are plenty of references about profinite groups and profinite completions of groups. It seems that their not exactly a wealth of references about profinite sets and profinite completions of sets.

REPLY [6 votes]: Johnstone's book Stone Spaces is a suitable reference for this and much more. The book by Ribes and Zalesskii Profinite Groups also has good information.  Algebre et Théories Galoisienne by Douady and Douady also gives a nice accounting.<|endoftext|>
TITLE: Terminology: lax vs. oplax colimits
QUESTION [8 upvotes]: I would like to know the standard usage of "lax colimit" and "oplax colimit" in the 2-categorical literature.  The nLab does not give an explicit definition of "lax colimit", as far as I can see, and I don't know what the most reliable source is.  I think I have seen at least one paper using each convention, but I have not encountered the notion often enough to have a good sense of whether this one of those places where the terminology is not really standardized, or there is general agreement with a few exceptions.
Concretely, given a diagram X : I → C in a 2-category C (for my purposes indexed by a 1-category I), suppose I have a cone (Y, {gi}i∈Ob I, {αf}f∈Mor I), with gi : Xi → Y, and for f : i → j in I, αf : gjf → gi (such that various diagrams commute).  Is this a lax colimit cone or a oplax colimit cone?

REPLY [8 votes]: I agree with Finn that the way to derive the correct choice of lax vs oplax is to connect it back to natural transformations.  Of course, as Finn pointed out, there is controversy over the choice for natural transformations, but my views on that are clear at the nlab page so I'll just write using that terminology.
However, in contrast to Finn, I think what you've got there is actually a lax cocone, since it is given by a lax natural transformation $* \to C(X-,Y)$, where $*:I^{op}\to Cat$ is the functor constant at the point.  It's true that, as Finn says, it is also an oplax natural transformation from $X$ to $\Delta_Y$, where $\Delta_Y$ is the functor $I\to C$ constant at $Y$.  But I think it's better to think of a cone as a transformation $* \to C(X-,Y)$, since this is the version that generalizes to weighted limits: for any weight $J:I^{op}\to Cat$, a $J$-weighted cocone is a transformation of the appropriate sort $J\to C(X-,Y)$.
The weighted-limit perspective on lax (co)limits is especially valuable because of the existence of lax morphism classifiers.  Namely, for any weight $J$ there is another weight $J^\dagger$ such that lax transformations out of $J$ are the same as strict (or pseudo) transformations out of $J^\dagger$.  Thus, lax $J$-weighted limits are the same as ordinary $J^\dagger$-weighted limits, so that lax $J$-weighted cones and limits are the same as ordinary $J^\dagger$-weighted cones and limits.  Thus a "lax limit" is really just a particular type of weighted limit, whose weight happens to be of the form $J^\dagger$.  Similarly, there is an oplax morphism classifier $J^\diamond$.
I think the choice I'm proposing is fairly widespread in Australia.  For instance, it's the one used here and here and here.  Actually, I'm not sure offhand whether I've even ever seen the other choice in print.<|endoftext|>
TITLE: Are submodules of free modules free?
QUESTION [9 upvotes]: Are all submodules of free modules free? I would like a reference to a proof or counterexample please.

REPLY [6 votes]: To answer ding8191's question: let $A$ be any valuation ring of height $1$ which is not a discrete valuation ring. That is to say, let $F$ be a field and suppose that there is a function $v : F \to \mathbb{R} \cup \{ \infty \}$ such that $v(xy) = v(x) + v(y), v(1) = 0, v(0) = \infty$ and $v(x + y) \geq \min $ $\{ v(x), v(y) \}$ (thus $v$ is a valuation), such that $v(F)$ is not a discrete subgroup of $\mathbb{R}$, and let $A = \{ x \in F :v(x)\geq 0\}$ (the valuation ring of $F$).
Then the maximal ideal $\mathfrak{m}$ in this ring is a direct limit of principal ideals and is therefore flat. If $\mathfrak{m}$ was projective, then it would have to be free by Kaplansky's Theorem (all projectives over any local ring are free), but since $\mathfrak{m}$ is contained in $F$ it has to have rank at most $1$, hence $\mathfrak{m}$ is principal. But this would force the valuation on $A$ to be discrete.
You can find this material in Chapter VI of Bourbaki's "Commutative Algebra" (Hermann, 1972). In particular, Lemma $1$ of $\S 3.6$ says that if $A$ is a valuation ring (in a more general sense than I defined above), then every torsion-free $A$-module is flat, and Proposition $9$ of the same subsection implies that $A$ is a discrete valuation ring whenever $A$ is a valuation ring of height $1$ with principal maximal ideal.
To give a concrete example, consider the field $F$ of Puiseux series and let $A$ be its valuation ring.<|endoftext|>
TITLE: Hausdorff Dimension and Hölder Continuity
QUESTION [13 upvotes]: Suppose we have a curve γ : [0,1] -> ℝn.  It is well known that if this curve is Hölder continuous for some exponent α then the Hausdorff dimension of γ[0,1] is bounded above by 1/α.
My question is: Is there a partial converse of the following form.  Suppose that the curve γ is such that γ[0,1] has Hausdorff dimension d, then is it true that for any α < 1/d we have that there exists some reparameterization of γ so that γ is Hölder continuous with exponent α?
I would like it to be true, although I doubt it is, however I have not been able to find a counterexample.  Proofs, references, or counterexamples would be appreciated.

REPLY [5 votes]: Edit:
The short answer is that there are planar curves that cannot be parametrized in a Holder continuous manner. Thus any such curve provides a counterexample for some $d \le 2$.
My original answer, showing that the curve can be chosen even of Hausdorff dimension $d=1$, follows.
I believe the answer to your question is negative. Indeed, I would think you can construct a curve in $\mathbb{R}^2$ that has Hausdorff dimension 1 but cannot be parametrized by a Holder-continuous curve.
The basic idea is that at smaller and smaller scales, the curve oscillates rather wildly in the horizontal direction, which can be done to ensure that the curve cannot be parametrized in a Holder-continuous way. However, if we make sure that the vertical extent of these oscillations is much, much smaller at each step, then the Hausdorff dimension should still be one. (Of course the linear measure of such a curve will be infinite.)
Let me try to be a bit more precise. I hope the idea is suitably clear.
Let us suppose that we construct the curve $\gamma_k$ at successive scales $\delta_k = 2^{-k}$. The curve $\gamma_k$ will consist of a number $M_k$ of horizontal segments, each of length $\delta_k$, together with some vertical segments. In the construction, we may ensure that the vertical segments have bounded total length, so we can just ignore them. 
To obtain $\gamma_{k+1}$, we split each horizontal segment in two segments of length $\delta_{k+1}$, and then replace each of these segments by a curve that oscillates, consisting of some number $m_{k+1}$ of horizontal segments of length $\delta_{k+1}$. However, the vertical extent of this oscillation should be some very small number $\varepsilon_{k+1}$. (Which in particular we may make so small that the total length of vertical segments added in this step is smaller than, say $2^{-k}$, so the total length of vertical segments stays bounded at all times.) 
Now the total number of segments at stage $k+1$ is $M_{k+1} = 2m_{k+1} M_k$. Any parametrization of this curve by the unit interval must contain two points that are at distance at most $1/M_{k+1}$ but have points whose images are at distance at least $\delta_{k+1}$. So if we choose $m_{k+1}$ sufficiently large, we can ensure that the limit curve cannot be Holder-parametrized.
On the other hand, the curve can be covered by about
$M_{k+1}\cdot\frac{\delta_{k+1}}{\varepsilon_{k+1}}$
sets of diameter $\varepsilon_{k+1}$. If we make $\varepsilon_{k+1}$ small enough, this number will be smaller than, say $\varepsilon_{k+1}^{-(1+1/k)}$, so the Hausdorff dimension of the limiting curve will be equal to one.<|endoftext|>
TITLE: Hochschild Kostant Rosenberg theorem for varieties in positive characteristic?
QUESTION [7 upvotes]: Is there is a known version of the HKR theorem as proved in say Swan's paper "Hochschild Cohomology of Quasiprojective Varieties" in positive characteristic? I assume something is known about this as in the affine case the theorem is still true so this seems like a reasonably naive question. Although there is no discussion of this in the paper, I guess the methods in the Swan paper roughly give the result as long the dimension of the variety is smaller than the characteristic (correct?) but don't generalize completely. 
If it isn't known, any ideas about the possibility of looking for a counterexample could also be helpful. If the above statement is true it seems reasonable to look at surfaces in characteristic 2 such as Enriques surfaces or K3 surfaces. I have no idea if it could be possible to do calculations in these cases.

REPLY [7 votes]: See this paper of mine and Gabriele Vezzosi. We prove that HKR holds in particular for smooth proper schemes $X$ of dimension at most $p$, the characteristic prime. In particular, it holds for smooth proper surfaces in characteristic $2$.<|endoftext|>
TITLE: Is there an additive model of the stable homotopy category?
QUESTION [17 upvotes]: Is there a model category $C$ on an additive category such that its homotopy category $Ho(C)$ is the stable homotopy category of spectra and the additive structure on $Ho(C)$ is induced from that on $C$.
Basically I want to add and subtract maps in $C$ without going to its homotopy category.  
I'm not asking for $C$ to be a derived category or anything like that. Just that it should have an additive structure. 
As John Palmieri pointed out I should really say what structure I want the equivalence (between $Ho(C)$ and the stable homotopy category) to preserve. Since I do want it to be a triangulated equivalence, Cisinski indicates why this is not possible.

REPLY [30 votes]: The answer is: no there isn't such a thing. Here is a rough argument (a full proof would deserve a little more care).
Using the main result of
S. Schwede, The stable homotopy category is rigid, Annals of Mathematics 166 (2007), 837-863
your question is equivalent to the following: does there exist a model category $C$, which is additive, and such that $C$ is Quillen equivalent to the usual model category of spectra?
In particular, we might ask: does there exist an additive category $C$, endowed with a Quillen stable model category structure, such that the corresponding stable $(\infty,1)$-category is equivalent to the stable $(\infty,1)$-category of spectra?
Replacing $C$ by its full subcategory of cofibrant objects, your question might be reformulated as: does there exist a category of cofibrant objects $C$ (in the sense of
Ken Brown), with small sums (and such that weak equivalences are closed under small sums), and such that the corresponding $(\infty,1)$-category (obtained by inverting weak equivalence of $C$ in the sense of $(\infty,1)$-categories) is equivalent to the stable $(\infty,1)$-category of spectra? If the answer is no, then there will be no additive model category $C$ such that $Ho(C)$ is (equivalent to) the category of spectra (as a triangulated category).
So, assume there is an additive category of cofibrant objects $C$, with small sums, such that $Ho(C)$ is (equivalent to) the category $S$ of spectra (as a triangulated category). Let $C_f$ be the full subcategory of $C$ spanned by the objects which correspond to finite spectra in $S$. Then $Ho(C_f)\simeq S_f$, where, by abuse of notations, $Ho(C_f)$ is the $(\infty,1)$-category obtained from $C_f$ by inverting weak equivalences, while $S_f$ stands for the stable $(\infty,1)$-category of finite spectra (essentially the Spanier-Whitehead category of finite CW-complexes). Given any (essentially) small additive category $A$ denote by $K(A)$ the "derived $(\infty,1)$-category of $A$" (that is the $(\infty,1)$-category obtained from the category of bounded complexes of $A$, by inverting the chain homotopy equivalences). Then, the canonical functor $A\to K(A)$ (which sends an object $X$ to itself, seen as a complex concentrated in degree $0$), has the following universal property: given a stable $(\infty,1)$-category $T$, any functor $A\to T$ which sends split short exact sequences of $A$ to distinguished triangles (aka homotopy cofiber sequences) in $T$ extends uniquely into a finite colimit preserving functor $K(A)\to T$. In particular, the functor $C_f\to Ho(C_f)\simeq S_f$ extends uniquely to a finite colimit preserving functor $F:K(C_f)\to S_f$. Let $Ker(F)$ be the full $(\infty,1)$-subcategory of $K(C_f)$ spanned by objects which are sent to zero in $S_f$. Then the induced functor
$$K(C_f)/Ker(F)\to S_f$$
is an equivalence of (stable) $(\infty,1)$-categories (to see this, you may use the universal property of $S_f$: given a stable $(\infty,1)$-category $T$, a finite colimit preserving functor $S_f\to T$ is the same as an object of $T$; see Corollary 10.16 in DAG I). This implies that, for any object $X$ of $S_f$, if $X/n$ denotes the cone of the map $n:X\to X$ (multiplication by an integer $n$), then $n.X/n\simeq 0$ (see Proposition 1 in Schwede's paper Algebraic versus topological triangulated categories). But such a property is known to fail whenever $X$ is a finite spectrum for $n=2$ (see Proposition 2 in loc. cit.). Hence there isn't such a $C$...<|endoftext|>
TITLE: Compact and quasi-compact
QUESTION [12 upvotes]: Why do algebraic geometers still use the term "quasi-compact" when they almost never deal with Hausdorff spaces?  They certainly use "local" rather than "quasi-local" (local = quasi-local + noetherian), so is there any reason other than historical contingency? 
Do algebraic geometers who do work in other fields still follow this convention when they write other papers?  If they do, do they write at the beginning of the paper something along the lines of "by compact, we mean quasi-compact and Hausdorff"?

REPLY [2 votes]: This is far from standard, but in my mind, when I'm doing algebraic geometry, "quasicompact" means "compact in the Zariski topology," and "compact" means "compact in the analytic topology" (or is not used at all if I'm not working over a topological field).  Thus, in principle, having explained that the terms were being used this way, one could write statements like  
"$\mathbb{A}^n_{\mathbb{C}}$ is quasicompact but not compact"
and
"Projective varieties are both compact and quasicompact."
But I would be very hesitant to do so, since I've never seen the terminology used this way so explicitly.<|endoftext|>
TITLE: Why do my quantum group books avoid homotopical language?
QUESTION [16 upvotes]: I am sitting on my carpet surrounded by books about quantum groups, and the only categorical concept they discuss are the representation categories of quantum groups.
Many notes closer to "Kontsevich stuff" discuss matters in a far more categorical/homotopical way, but they seem not to wish to touch the topic of quantum groups very much....
I know next to nothing about this matter, but it seems tempting to believe one could maybe also phrase the first chapters of the quantum group books in a language, where for example quasitriangulated qHopf algebras would just be particular cases of a general "coweak ${Hopf_\infty}$-algebra" (does such thing exist?). Probably then there should be some (${\infty}$,1) category around the corner and maybe some other person's inofficial online notes trying to rework such a picture in a dg Hopf algebra model category picture.
My quantum group books don't mention anything in such a direction (in fact ahead of a chapter on braided tensor categories the word 'category' does kind of not really appear at all). So either

[ ] I am missing a key point and just proved my stupidity to the public
[ ] There is a quantum group book that I have missed, namely ......
[ ] Such a picture is boring and/or wrong for the following reason .....

Which is the appropriate box to tick?

REPLY [13 votes]: There is certainly a natural homotopical analog of braided tensor category, namely a stable $E_2$ category (ie an $E_2$ object of the $\infty$-category of dg categories, or if you prefer or stable $(\infty,1)$-categories). Such things can be defined using Lurie's DAG I (for stable) and III (for $E_2$). Rather than trying to define versions of Hopf algebras, you can talk about fiber functors on such. In fact the general Tannakian pattern discussed in other MO posts generalizes from the symmetric monoidal setting to the braided setting -- i.e. given a braided category (say in this homotopical sense) you can define a "Spectrum", consisting of $E_2$ functors to modules over various $E_3$-algebras (which are $E_2$ categories). This defines a kind of object that you can call an $E_3$ stack (stack on $E_3$ algebras). [If you work in a nonderived setting there's no difference between $E_3$ and commutative.] This has an underlying usual stack. Anyway I learned all this from John Francis, who's been working on developing $E_n$-algebraic geometry...
anyway that was a digression, the point is you can talk about $E_2$ categories with a good fiber functor and use that as the definition of an $\infty$-quantum group..
As for other interactions, there is a very significant interplay between homotopy theory and quantum groups in the work in progress of Gaitsgory and Lurie on "quantum geometric Langlands". This was the topic of the 2008 Talbot workshop (see here). There are some related notes also by Gaitsgory and Lurie here. One awesome idea is the use of the $E_2$ perspective to explain WHY quantum groups relate to local systems on configuration spaces of points (Drinfeld-Kohno theorem and its generalizations) and in fact use it to prove the Kazhdan-Lusztig equivalence between quantum groups and affine Lie algebras. This would be a perfect topic for your dreamed-of book --- for now I'd make do with a paper or even course notes!!<|endoftext|>
TITLE: On a positivity of a matrix with trace entries.
QUESTION [6 upvotes]: Some basic observations lead me to ask the following quesiton
Let $A_1, \cdots, A_m$ be $n\times n$ complex matrices. For positive integer $k\ge 1$, show
$$\left(\begin{array}{cccc}Tr\{(A_1^*A_1)^k\}&Tr\{(A_1^*A_2)^k\}&\cdots &Tr\{(A_1^*A_m)^k\}\\Tr\{(A_2^*A_1)^k\}&Tr\{(A_2^*A_2)^k\}&\cdots &Tr\{(A_2^*A_m)^k\}\\\cdots&\cdots&\cdots&\cdots\\Tr\{(A_m^*A_1)^k\}&Tr\{(A_m^*A_2)^k\}&\cdots &Tr\{(A_m^*A_m)^k\}
\end{array}\right)$$
is positive semidefinite.
Remark
1). When $m=2$, it suffices to show $|Tr\{(A_1^*A_2)^k\}|^2\le Tr\{(A_1^*A_1)^k\}\cdot Tr\{(A_2^*A_2)^k\}$, which is a consequence of a unitarily invariant norm inequality appeared in p.81 of X.Zhan, Matrix inequalities, Springer, 2002.
2).  It is easy to show $$\left(\begin{array}{cccc}(Tr\{A_1^*A_1\})^k&(Tr\{A_1^*A_2\})^k&\cdots &(Tr\{A_1^*A_m\})^k\\(Tr\{A_2^*A_1\})^k&(Tr\{A_2^*A_2\})^k&\cdots &(Tr\{A_2^*A_m\})^k\\\cdots&\cdots&\cdots&\cdots\\(Tr\{A_m^*A_1\})^k&(Tr\{A_m^*A_2\})^k&\cdots &(Tr\{A_m^*A_m\})^k
\end{array}\right)$$
is positive semidefinite, since it is $k$ Hadamard product of a Gram matrix.

REPLY [9 votes]: It seems that this is not true. Here is a counterexample. Consider a regular $d$-gon, where $d\ge 3$ is an odd number. Let $S,T$ be the permutation matrices on the vertices of this $d$-gon, induced by reflections in two adjacent symmetry axes. Let $A_1=1, A_2=S, A_3=T$, which are $d$ by $d$ matrices. We have $S^2=T^2=1$, and $ST$ is a rotation of order $d$
(so $(ST)^2$ has no fixed points). 
Let $k=2$. Then the matrix in the question seems to be the following:
$$
\left(\begin{matrix} d & d & d\\ d & d & 0\\ d & 0 & d\\ \end{matrix}\right)
$$
The determinant of this matrix is $-d^3$.<|endoftext|>
TITLE: What are the connections between pi and prime numbers?
QUESTION [34 upvotes]: I watched a video that said the probability for Gaussian integers to be relatively prime is an expression in $\pi$, and I also know about $\zeta(2) = \pi^2/6$ but I am wondering what are more connections between $\pi$ and prime numbers?

REPLY [26 votes]: Leonard Euler discovered many years ago that
$\frac π 4 = \frac 3 4 \cdot \frac 5 4 \cdot \frac 7 8 \cdot \frac {11} {12} \cdot \frac {13} {12} \cdots$ 
where the numerators on the right-hand side are the odd prime numbers and each denominator (on both sides) is the multiple of 4 nearest to the corresponding numerator. Pretty fascinating if you ask me.<|endoftext|>
TITLE: non principally polarized complex abelian varieties 
QUESTION [19 upvotes]: I've read in (abstracts of) papers that there are abelian varieties over fields of positive characteristic that admit no prinicipal polarization.  Apparently its not the easiest thing to find an example of, but I was thinking it should be much easier over the complex numbers. 
All abelian varieties of dimension 1 are elliptic curves which always have a principal polarization.  So any example would have to be at least two dimensional.  So my question is given an abelian variety with a polarization, is there a good way of telling if there is or isn't a principal polarization?
Or if that's in general a difficult question, are there some relatively simple examples where you can really see that there are or aren't any principal polarizations?

REPLY [29 votes]: Here is another construction, followed by some comments on how to solve the existence problem in general.

If $A$ is a $g$-dimensional principally polarized abelian variety over $\mathbf{C}$ with $\operatorname{End} A = \mathbf{Z}$, and $G$ is a finite subgroup whose order $n$ is not a $g$-th power, then $B:=A/G$ is an abelian variety that admits no principal polarization.

Proof: If $B$ had a principal polarization, its pullback to $A$, given by the composition $A \to B \to B' \to A' \simeq A$ (where $A'$ is the dual of $A$ and so on) would be an endomorphism of degree $n^2$.  But this endomorphism is multiplication-by-$m$ for some integer $m$, which has degree $m^{2g}$.  So $n$ would have to be a $g$-th power. $\square$
To complete this answer, observe that most abelian varieties $A$ over $\mathbf{C}$ satisfy $\operatorname{End} A=\mathbf{Z}$.  An explicit example is the Jacobian of the hyperelliptic curve that is the smooth projective model of the affine curve 
$$y^2= a_{2g+1} x^{2g+1} + \cdots + a_1 x + a_0$$
where $a_{2g+1},\ldots,a_0 \in \mathbf{C}$ are algebraically independent over $\mathbf{Q}$.
Remarks:
1) Of course there is no reason to restrict to $\mathbf{C}$.  For instance, one can find examples over $\mathbf{Q}$ by using the fact that the endomorphism ring injects into the endomorphism ring of the reduction modulo any prime of good reduction, and combining this information for several primes.
2) For an arbitrary abelian variety $A$, if you are given a polarization $A \to A'$, then if there is a principal polarization, following the first by the inverse of the second would give you an endomorphism of $A$.  So one way of answering the existence question is to determine the endomorphism ring of $A$ and to study those endomorphisms that factor through your given polarization.  (That's not quite sufficient, but it gives you an idea of the complexity of the problem since determining the endomorphism ring can be rather difficult.)<|endoftext|>
TITLE: Reference for the Frolicher-Nijenhuis Bracket
QUESTION [9 upvotes]: I'm looking for a (comprehensible) reference for the Frolicher-Nijenhuis bracket, hopefully more down to earth than Michor's books and different from Saunders's book on Jets.
I'm interested in it as this bracket seems the appropriate means to define the curvature of a general Ehresmann connection on a bundle.
Alternatively it would be really great to have a reference (again different from Michor) that defines connections, curvature and holonomy on general bundles, later specializing to principal bundles, vector bundles and maybe even to the riemannian setting.

REPLY [6 votes]: I also only learned about Frolicher-Nijenhuis brackets from Saunders' book on jets but I doubt that there is any other authorative reference besides Michor's book and the original papers.  I don't know if this is what the original poster intended, but here's how I understand the link between FN and curvature.  Generally, I tend to stay away from the full generality of the FN bracket, and only use those (axiomatic) properties that I need...
Let $\pi: E \rightarrow M$ be a fiber bundle.  An Ehresmann connection is a subbundle $H$ of $TE$ such that $H \oplus VE = TE$, where $VE$ is the vertical bundle.  Denote the projector from $TE$ onto $H$ by $h$.  Saunders (and many other authors) define Ehresmann connections directly in terms of bundle maps $h$, since they are easier to work with, but this doesn't matter.
Now, to introduce curvature, we would like to formalize the idea that curvature is the failure of "parallel transport" to commute, where of course we haven't define parallel transport properly.  However, it is only a small step of the imagination to guess that this must be related to the integrability of $H$, i.e.whether $[h(X), h(Y)] \in H$ for arbitrary vector fields $X, Y$.  The failure of two horizontal vector fields to be horizontal again is measured by the expression
$$
  R(X,Y) = [h(X), h(Y)] - h([h(X), h(Y)]),
$$
and in fact this is nothing but Saunders' definition 3.5.13 of curvature.  Note that $R(X, Y) = 0$ if either $X$ or $Y$ is in $VE$.
The Frohlicher-Nijenhuis bracket of two linear bundle maps is in general quite complicated, but if you look at proposition 3.4.15 in Saunders, you get that for a linear bundle map $h: TE \rightarrow TE$, 
$$
  [h, h]  (X,Y) = 2(h([X,Y]) + [h(X), h(Y)] - h([h(X),Y]) - h([X,h(Y)])).
$$
Now note that the right-hand side is zero as soon as either $X$ or $Y$ is in $VE$, just as for $R$.  If both $X$ and $Y$ are horizontal, we can apply the above formula to $X = h(X)$ and $Y = h(Y)$, and conclude that $[h, h] (X, Y) = 2( [h(X), h(Y)] - h([h(X), h(Y)]))$ (where we use the identity $h \circ h = h$ liberally), and so 
$$
   R = \frac{1}{2} [h,h].
$$
If you have a principal fiber bundle, $h$ is related to the connection one-form $\mathcal{A}$, and the above formula gives the curvature of $\mathcal{A}$ in terms of the covariant differential of $\mathcal{A}$.  For a symplectic connection, something similar happens.
Edit: here's how I think it works for principal fiber bundles.  Take a connection one-form $A:  TE \to \mathfrak{g}$, and let $\sigma: \mathfrak{g} \rightarrow VE$ be the infinitesimal generator of the $G$-action.  The composition $v := \sigma \circ \mathcal{A}$ is then the vertical projector of the connection and $h := 1 - v$ is the horizontal one.  Now plug this expression for $h$ into the formula for the curvature:
$$
   [h, h] = [1, 1] - [\sigma \circ \mathcal{A}, 1]  - [1, \sigma \circ \mathcal{A}]
    + [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}].
$$
The term $[1,1]$ is zero, and you can use Saunders' proposition 3.4.15 to show that term 2 and 3 vanish as.  The last term can be written as (again using S3.4.15) as 
$$
   [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) = 
   2( (\sigma \circ \mathcal{A})^2([X, Y]) + [ (\sigma \circ \mathcal{A})(X), (\sigma \circ \mathcal{A})(Y)] - (\sigma \circ \mathcal{A})([(\sigma \circ \mathcal{A})(X), Y)]) - (\sigma \circ \mathcal{A})([X, (\sigma \circ \mathcal{A})(Y)])).
$$
Now, show that this vanishes whenever $X$ or $Y$ is vertical, so that we can take $X$ and $Y$ to be horizontal.  In that case, the above simplifies to 
$$
  [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) = 2(\sigma \circ \mathcal{A})([X, Y])
$$
or 
$$
   R(X, Y) = (\sigma \circ \mathcal{A}) ([X^h, X^h])
$$
where $X^h$ represents the horizontal part of $X$.  But $\mathcal{A} ([X^h, X^h])$ is just the negative of the curvature (as a two-form with values in $\mathfrak{g}$) of $\mathcal{A}$, so that
$$
   R(X, Y) = -\sigma ( \mathcal{B}(X, Y) ).
$$<|endoftext|>
TITLE: Complex orientations on homotopy
QUESTION [15 upvotes]: I am wondering if there is a more "geometric" formulation of complex orientations for cohomology theories than just a computation of $E^*\mathbb{C}$P$^{\infty}$ or a statement about Thom classes. It seems that later in Hopkins notes he says that the complex orientations of E are in one to one correspondence with multiplicative maps $MU \rightarrow E$, is there a treatment that starts with this perspective? How do the complex orientations of a spectrum E help one compute the homotopy of $E$, or the $E$-(co)homology of MU? Further, what other kinds of orientations could we think about, are there interesting $ko$ or $KO$ orienations? how much of these $E$-orientations of X is detected by E-cohomology of X?
I do have some of the key references already in my library, for example the notes of Hopkins from '99, Rezk's 512 notes, Ravenel, and Lurie's recent course notes. If there are other references that would be great. I am secretly hoping to get some insight from some of the experts. (I guess I should really also go through Tyler's abelian varieties paper)
(sorry for the on and off texing but the preview is giving me weird feedback.)
EDIT: I eventually found the type of answer i was looking for in some notes of Mark Behrens on a course he taught. This answer is that a ring spectrum $R$ is complex orientable is there is a map of ring spectra $MU \to R$. This also appears in COCTALOS by Hopkins but neither source takes this as the more fundamental concept. Anyway, the below answer is more interesting geometrically.

REPLY [21 votes]: The natural starting point of this story are E-orientations on, say closed, manifolds M. That's just a fundamental class $[M^n] \in E_n(M)$ such that cap product induces a (Poincare duality) isomorphism. 
Given E, the question becomes which M are E-orientable. In many cases it happens that this follows if the stable normal bundle of M admits a lift through a fibration $X\to BO$. For example, if E=HZ is ordinary Z-cohomology then X=BSO works, if E=KO then X=BSpin works, if E=KU then X=BU or X=BSpin$^c$ works etc. 
To formalize the idea that every X-manifold has an E-orientation, form the bordism groups $\Omega^X_n$ of X-manifolds and observe that the fundamental classes lead to natural maps $\Omega^X_n(Y) \to E_n(Y)$ for any space Y. In other words, there are natural transformations of cohomology theories $\Omega^X \to E$, or even better, maps of spectra $MX \to E$, where $MX$ is the Thom spectrum associated to the fibration $X\to BO$. 
In the case X=BU this is called a complex orientation of E and has been studied extensively because it simplifies computations of E-cohomology tremendously. The original and still relevant reference is Adams' little blue book.<|endoftext|>
TITLE: Why do people think that abelian varieties are the hardest case for the Hodge conjecture?
QUESTION [25 upvotes]: Today, I heard that people think that if you can prove the Hodge conjecture for abelian varieties, then it should be true in general.  Apparently this case is important enough (and hard enough) that Weil wrote up some families of abelian 4-folds that were potential counterexamples to the Hodge conjecture, but I've never heard of another potential counterexample.
Anyway, in short:
1) Does the Hodge Conjecture for abelian varieties imply the full Hodge conjecture?
2) If not, is there an intuitive reason why abelian varieties should be the hardest case?

REPLY [13 votes]: Related to Jim Milne's answer, one might mention that Deligne proved that for abelian varieties, all Hodge cycles are "absolutely Hodge" (i.e. when you think of them embedded diagonally inside the product of the algebraic de Rham cohomology and  $\ell$-adic cohomology
(for every $\ell$) and apply an automorphism of $\mathbb C$, the resulting cycles are again diagonally embedded rational cycles, and are in fact again Hodge).  Note that 
if the Hodge conjecture holds, then this is certainly true (since the conjugate under any automorphism of $\mathbb C$ of an algebraic cycle is again an algebraic cycle).
On the one hand, this is much more than is known about the Hodge conjecture for more general classes of varieties.  
On the other hand, one can't immediately extend this to other classes of varieties
 because the motives of abelian varieties don't generate all motives over a field of char. 0
(in fact, far from it, as far as I know), a fact already brought up in Donu Arapura's answer.<|endoftext|>
TITLE: Is Lang's definition of a tensor bundle nonstandard?
QUESTION [6 upvotes]: Background
Let $\mathfrak{A}$, $\mathfrak{B}$, and $\mathfrak{C}$ be subcategories of the category of Banach spaces (over $\mathbb{R}$).  Suppose we have a functor $\lambda:\mathfrak{A}^{op}\times\mathfrak{B}\to \mathfrak{C}$.  
Let $f:E'\to E$ be a morphism belonging to $\mathfrak{A}$, and let $g:F\to F'$ be a morphism belonging to $\mathfrak{B}$.  (Note: These are morphisms of topological vector spaces).  
Then we have a map $$\matrix{Hom(E',E) \times Hom(F,F')\to Hom(\lambda(E,F),\lambda(E',F'))\\
(f,g)\mapsto\lambda(f,g)}$$
We say $\lambda$ is of class $C^p$ if for all manifolds $U$, and any two $C^p$ morphisms $U\to Hom(E',E)$ and $U\to Hom(F,F')$, the composition $$U\to Hom(E',E) \times Hom(F,F')\to Hom(\lambda(E,F),\lambda(E',F'))$$
is also of class $C^p$.  (Note: We can replace $\mathfrak{A}$ and $\mathfrak{B}$ with categories of tuples to generalize this to several variables.  In fact, this is what we do below.)
It is not hard to show that this induces a unique functor $$\lambda_X:VB(X, \mathfrak{A})^{op}\times VB(X,\mathfrak{B})\to VB(X,\mathfrak{C}).$$  on vector bundles taking values in the appropriate vector bundle categories over $X$.  
We define a tensor bundle of type $\mathbf{\lambda}$ on $X$ to be $\lambda_X(TX)=\lambda_X((TX,\dots,TX),(TX,\dots,TX))$, where $TX$ is the tangent bundle.  
However, this doesn't agree with the definition given on Wikipedia or anywhere else I've looked.  
Questions

Is this terminology nonstandard?
Is the notion itself nonstandard?
If the terminology is nonstandard, but the notion is standard, does it have a different name?  
Is this definition useful?
Does this include more vector bundles as tensor bundles than the standard definition?

REPLY [5 votes]: Is the notion non-standard? As Emerton says, the answer is no, except perhaps in minor details.
Is the terminology non-standard (and more permissive than the usual notion of tensor)?  Yes, I'd say it is. 
Because Lang allows arbitrary categories of Banach spaces, his notion is very general; by taking the categories small, one needn't have much functoriality at all. For instance, in the case where $\lambda$ has just one, covariant input, we can take $\mathfrak{B}$ to be the category with one object,  $\mathbb{R}^n$, and morphisms $GL(n)$. Then $\lambda(\mathbb{R}^n)$ is a representation $V$ of $GL(n)$. On an $n$-manifold $X$, the resulting bundle $\lambda_X(TX)$ is usually known as an associated bundle. Form the principal $GL(n)$-bundle $Fr_X$ of frames (consisting of pairs of $x\in X$ with an isomorphism $T_x X \to \mathbb{R}^n$). Then $\lambda_X(TX)= Fr_X \times_{GL(n)}V$. 
The notion of associated bundle is one that differential geometers often find more convenient than Lang's, because it allows one to say exactly which structure groups of interest (here, $GL(n,\mathbb{R})$) and exactly which representations. 
A restrictive definition of tensor bundles allows those associated bundles where $V$ is a tensor product of copies of $\mathbb{R}^n$ and its dual. A broader (and I suspect fairly standard) definition allows subquotient representations of such - in particular, symmetric tensors and differential forms. But Lang apparently allows other things, e.g. $r$-densities (take the representation of $GL(n)\to \mathbb{R}^\times$ given by $g\mapsto |\det g|^r$).
Is the greater generality (compared to the standard notion of tensor) useful? In examples (e.g. densities) yes; in the abstract, not really - the interesting geometry is attached to specific classes of structure groups and their representations.<|endoftext|>
TITLE: Modular curves of genus zero and normal forms for elliptic curves
QUESTION [31 upvotes]: This is maybe the first question I actually need to know the answer to!
Let $N$ be a positive integer such that $\mathbb{H}/\Gamma(N)$ has genus zero.  Then the function field of $\mathbb{H}/\Gamma(N)$ is generated by a single function.  When $N = 2$, the cross-ratio $\lambda$ is such a function.  A point of $\mathbb{H}/\Gamma(2 )$ at which $\lambda = \lambda_0$ is precisely an elliptic curve in Legendre normal form
$$y^2 = x(x - 1)(x - \lambda_0)$$
where the points $(0, 0), (1, 0)$ constitute a choice of basis for the $2$-torsion.  When $N = 3$, there is a modular function $\gamma$ such that a point of $\mathbb{H}/\Gamma(3)$ at which $\gamma = \gamma_0$ is precisely an elliptic curve in Hesse normal form
$$x^3 + y^3 + 1 + \gamma_0 xy = 0$$
where (I think) the points $(\omega, 0), (\omega^3, 0), (\omega^5, 0)$ (where $\omega$ is a primitive sixth root of unity) constitute a choice of basis for the $3$-torsion. 
Question:  Does this picture generalize?  That is, for every $N$ above does there exist a normal form for elliptic curves which can be written in terms of a generator of the function field of $\mathbb{H}/\Gamma(N)$ and which "automatically" equips the $N$-torsion points with a basis?  (I don't even know if this is possible when $N = 1$, where the Hauptmodul is the $j$-invariant.)  If not, what's special about the cases where it is possible?

REPLY [11 votes]: I think the answer to your question is the content of Velu's thesis: Courbes elliptiques munies d'un sous-groupe $Z/NZ\times \mu_N$. In there, he explicitly writes down the universal elliptic curve over $X(p)$ for $p>3$.<|endoftext|>
TITLE: Which p-adic numbers are also algebraic?
QUESTION [25 upvotes]: What is $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ ?  
For instance, we know that $\mathbb{Q}_p$ contains the $p-1$st roots of unity, so we might say that $\mathbb{Q}(\zeta) \subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$, where $\zeta$ is a primitive $p-1$st root.  
As a more specific example, $x^2 - 6$ has 2 solutions in $\mathbb{Q}_5$, so we could also say that $\mathbb{Q}(\sqrt{6},\sqrt{-1})\subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$.  
Edit: I removed the motivation for this question (which I think stands by itself), as it will be better as a separate question once I think it through a bit better.

REPLY [6 votes]: There's a slightly subtle point near here of which some people are not aware: that it is dangerous (perhaps even nonsensical) to compare algebraic numbers under various different completions. So, to talk about $Q_p\cap \bar Q$, you should be talking about a completion of $Q$ containing $Q_p$, not, e.g., a completion of $Q$ lying inside $C$. I don't think this is what is happening here, but some people may find this interesting.
Now, there are lots of isomorphisms floating around, so usually everything turns out just fine, but sometimes not. Here are two examples.
(1) The following fallacious argument that $e$ is transcendental is from a talk by Gouvêa, "Hensel's p-adic Numbers: early history" (originally due to Hensel himself).
The series expansion of $e^p$ converges in $Q_p$, thus $e$ is a solution to the equation $X^p=1+p\epsilon$, where $\epsilon$ is a $p$-adic unit. So $[Q_p(e):Q_p]=p$ (of course you need to argue that the polynomial is irreducible), and so $[Q(e):Q]\ge p$. Since $p$ was arbitrary, $e$ must be transcendental over $Q$. 
The fallacy is that even though the series for $e$ (and $e^p$) converges in $R$ and $Q_p$, the numbers they converge to are not the same. 
(2) The following is from Koblitz's $p$-adic book, page 83 (with an example and some other fallacious arguments).
It is not true that if an infinite sum of rational numbers (a) converges $p$-adically to a rational number for some $p$ and (b) converges in the real topology to a rational number, then the rational numbers the two series converge to are the same!<|endoftext|>
TITLE: Factorials in Pascals Triangle
QUESTION [36 upvotes]: Hi,
I asked this question of Keith Conrad, and he suggested that I try posting here.  One of my students observed that the only instances of factorials in the interior of Pascal's triangle are $\binom{4}{2}=3!$ and $\binom{10}{3}=\binom{16}{2}=5!$.  I checked the first 500 rows, and he's right up to that point.
This is a special case of the apparently unsolved problem of finding non-trivial solutions to n!=a!b!c!...  The special feature here is that I need (a+b)!=a!b!c!, and that seems like a special enough case to have been treated by someone.  Unfortunately, literature searches have been fruitless, because every paper about Pascal's triangle contains the word "factorial" somewhere.
My best idea (which I can't make work) is to show that the powers of 7 in the equation (a+b)!=a!b!c! can't be made to match up unless neither side is a multiple of 7.  Then exhaustive searching can show that the above are the only non-trivial solutions.  
Thanks greatly for any ideas that anyone might have.

REPLY [8 votes]: The latest issue of the Journal of Number Theory (February 2016 issue) contains a paper by Nair and Shorey which gives a conditional resolution of this problem.  They consider the problem of finding solutions to 
$$ 
n! = a_1! a_2! \cdots a_t!,
$$ 
with $n>a_1 \ge a_2 \ge \ldots a_t>1$ and $t>1$.   Obviously we can assume that $a_1 \le (n-2)$, for if $a_1 =n-1$ we are just looking at $n=a_2!\cdots a_t!$.  
Then Theorem 4 of the cited paper shows that assuming Baker's explicit $abc$-conjecture, the only solutions to this equation are
$$ 
7!3!3!2!=9!; 7!6!=10!; 14!5!2!=16!.
$$
Here the explicit $abc$ conjecture of Baker states that if $a$, $b$ and $c$ are coprime positive integers with $a+b=c$ and $\omega$ denotes the number of distinct prime factors of $abc$ then 
$$
c< \frac{6}{5} N \frac{(\log N)^{\omega}}{\omega!},
$$ 
with 
$$ 
N = \prod_{p|abc} p 
$$ 
denoting the radical of $abc$.<|endoftext|>
TITLE: Why is "h" the notation for class numbers?
QUESTION [26 upvotes]: A student asked me why $\mathcal{O}_K$ is the notation used for the ring of integers in a number field $K$ and why $h$ is the notation for class numbers.  I was able to tell him the origin of $\mathcal{O}$ (from Dedekind's use of Ordnung, the German word for order, which was taken from taxonomy in the same way the words class and genus had been stolen for math usage before him), but I was stumped by $h$.  Does anyone out there know how $h$ got adopted?  
I have a copy of Dirichlet's lecture notes on number theory (the ones Dedekind edited with his famous supplements laying out the theory of ideals), and in there he is using $h$.  So this convention goes back at least to Dirichlet -- or maybe Dedekind? -- but is that where the notation starts? And even if so, why the letter $h$? 
I had jokingly suggested to the student that $h$ was for Hilbert, but I then told him right away it made no historical sense (Hilbert being too late chronologically).

REPLY [22 votes]: Gauss, in his Disquisitiones, used ad hoc notation for the class number when he needed it. He did not use h. Dirichlet used h for the class number in 1838 when he proved the class number formula for binary quadratic forms. I somewhat doubt that he was thinking of "Hauptform" in this connection - back then, the group structure was not as omnipresent as it is today, and the result that $Q^h$ is the principal form was known (and written additively), but did not play any role. Kummer, 10 years later, used H for the class number of the field  of p-th roots of unity, and h for the class number of a subfield generated by Gaussiam periods (and "proved" that $h \mid H$); in the introduction he quotes Dirichlet's work on forms at length.<|endoftext|>
TITLE: When do two holonomy maps determine flat bundles that are isomorphic as just bundles (w/o regard to the flat connections)?
QUESTION [13 upvotes]: Suppose we have a surface S (although the question might make as much sense in higher dimensions) and a topological group G. The data of a flat vector bundle on S (up to isomorphism) is the same as a holonomy representation $\pi_1(S) \to G$; note that this includes both the bundle and the flat structure on it. For the purpose of defining isomorphisms between flat bundles, we can also think of them in Steenrod's terminology: a flat G-bundle is the same thing as a G'-bundle, where G'=G as a group but has the discrete topology.
Is there a way to tell when two G'-bundles over S are isomorphic as G-bundles (in other words, when are two flat bundles isomorphic without regard to the flat structure)? In other words, if two flat G'-bundles are isomorphic as G-bundles, what can we say about the holonomy maps (and is there an if and only if statement)? Does it matter whether the bundles are principal or not?
UPDATE: MOTIVATION.: I'll add a couple of words about my motivation (I was hesitant, since this makes the question a bit less specific). A paper of William Goldman proves that in the cases of $G=PSL_2(\mathbb R)$ or $G=PSL_2(\mathbb C)$, the isomorphism classes of G-bundles correspond exactly to the path-components of the representation variety (see Igor's answer below for more info, Dan's comment for more examples when that happens, and Joel's comment for examples where that does not happen). Even more interestingly, the same paper proves that in the case of $G=PSL_2(\mathbb R)$ all the representations in the path-component that corresponds to the tangent bundle to the surface turn out to be faithful, and to have a discrete image, so the quotient of the corresponding $\mathbb H^2$-bundle by the action of $\pi_1(S)$ gives a nice hyperbolic surface diffeomorphic to S. This does not happen in the case of $G=PSL_2(\mathbb C)$. I am trying to understand which parts of this rather mysterious situation can be understood in general terms, and which cannot.
Thank you very much - the answers people already gave are very helpful! It seems that there might not be a good general answer, but I'm still very interested in what can be said.

REPLY [6 votes]: The usual algebraic topology way to address this is to consider the identity as a continuous map from $G$ with the discrete topology (call it $G^d$) to $G$ as a lie group. Then a homomorphism $\pi_1(X)\to G$ is the same  (modulo conjugation, which doesnt affect your question) as a homotopy class of maps $X\to BG^d$. The principal $G$ bundle that corresponds to it is given by the composite $X\to BG^d\to BG$. So the (not always useful) consequence is that your question is about understanding the kernel/image of $[X,BG^d]\to[X,BG]$. One thing you see from this is that characteristic classes you get on flat bundles factor through $H^*(BG)\to H^*(BG^d)$. As Igor B. notes, when Chern-Weil theory applies you conclude that the characteristic classes are torsion.
For example, for $G=U(1)$   (and similarly for any abelian $G$)  $BG^d=K(R/Z,1)$, $BG=BU(1)=K(Z,2)$ and $[X,BG^d]\to [X, BG]$ is just the Bockstein $H^1(X;R/Z)\to H^2(X;Z)$, which takes  $\pi_1(X)\to U(1)$   to the first Chern class of the underlying bundle. So computing with the coefficient exact sequence tells you about the kernel (what flat $U(1)$ bundles are trivial as bundles) and the image (what bundles admit flat $U(1)$ connections.)
This perspective is also useful when combined with obstruction theory, e.g you can use it to determine if a representation $\pi_1X\to SO(n)$ lifts to $Spin(n)$ by asking the same thing about the map $X\to BSO(n)^d$ and the fibration $BSpin(n)^d\to BSO(n)^d$.<|endoftext|>
TITLE: The finite subgroups of SU(n)
QUESTION [42 upvotes]: This question is inspired by the recent question "The finite subgroups of SL(2,C)". While reading the answers there I remembered reading once that identifying the finite subgroups of SU(3) is still an open problem. I have tried to check this and it seems it was at least still open in the Eighties.
Can anyone confirm or deny that the finite subgroups of SU(3) are not all known? And if this is true, then what is the source of the difficulty? 
Secondly, what is known of the finite subgroups of SU(n)  for n > 3?
UPDATE: Thanks to those below who have corrected my ignorance! It seems that I may have been tricked by some particularly sensationalised abstracts (or perhaps just misunderstood them.)

REPLY [2 votes]: Just want to point out that an algorithm to answer,
given $n$ and a presentation of a finite group $G$, whether $G$ is isomorphic to a finite subgroup of $SU(n)$,
exists as a consequence of Tarski's Theorem on the decidability of the real field $(\mathbb R,+,\times)$!
Namely, the statement that certain matrices $A_1,\dots,A_k$ exist that satisfy the corresponding equations under matrix multiplication can be written as a statement in the first-order theory of algebraically closed fields of characteristic zero, i.e., the theory of $\mathbb C$. This reduces to $\mathbb R$ as pointed out by Joel elsewhere on MO.
How much faster the Zassenhaus algorithm mentioned by José is, I don't know...<|endoftext|>
TITLE: When is a homogeneous space a variety?
QUESTION [13 upvotes]: Let $G$ be a Lie group and let $H$ be a closed subgroup of $G$.  Then $G/H$ may not be a group, but it will be a homogeneous space for $G$ with stabilizers conjugate to $H$.  Sometimes, this is a variety, for instance, when $G$ is a complex reductive group and $H$ is a complex subgroup, and it will even be projective when $H$ is parabolic (by definition).
However, when we take real Lie groups, the situation seems more subtle.  For instance, if $G=Sp(g,\mathbb{R})$ and $H=U(g)$, then $G/H$ is the Siegel upper half space, which is an (analytic) open set in a variety, namely, the variety is the space of symmetric $g\times g$ matrices and the open set is given by the ones with positive imaginary part.  Similarly, many constructions in Hodge theory, particularly that of period domains, end up coming from real Lie groups, and so may be varieties, open subsets of varieties, or not varieties at all a priori.  Clearly, the quotient being even dimensional is a necessary condition, but I'd be surprised if it were sufficient.
So the first part of the question is

When is a homogeneous space (an open subset of) a variety?

Now, additionally, these period domains often can be quotiented by a discrete (I believe Griffiths says arithmetic) subgroup of the original group to actually get a variety.  For instance (if I'm understanding right) if we take $\mathfrak{h}_g=Sp(g)/U(g)$ above, we can quotient further by $Sp(g,\mathbb{Z})$ and this gives us $\mathcal{A}_g$, the moduli space of abelian varieties.

When is there a discrete subgroup that we can take a further quotient by to get a variety?

REPLY [7 votes]: I'll try to answer both questions, though I will change the first question somewhat.  Let's work in the setting of a real reductive algebraic group $G$ and a closed subgroup $H \subset G$.  
Your first question asks when $G/H$ is an open subset of some (presumably complex) variety.  I think that this question should be modified in a few ways.
You can't really say that $G/H$ "is a subset" of a variety, since $G/H$ is not a priori endowed with a complex structure.  So you need a bit more data to go with the question -- a complex structure on the homogeneous space $G/H$.  Such a complex structure can be given by an embedding of the circle group $U(1)$ as a subgroup of the center of $H$.  Let $\phi: U(1) \rightarrow G$ be such an embedding, and let $\iota = \phi(i)$ be the image of $e^{pi i} \in U(1)$ under this map.  Such an embedding yields an integrable complex structure on the real manifold $G/H$, I believe (though I haven't seen this stated in this degree of generality).
So now one can ask if $G/H$, endowed with such a complex structure, is an open subset of a complex algebraic variety.  But again, I have some objection to this question -- it's not really the right one to ask.  Indeed, it's very interesting when one finds that some quotients $\Gamma \backslash G /H$ are (quasiprojective) varieties -- but such quotients are not obtained as quotients in a category of varieties, from $G/H$ to $\Gamma \backslash G / H$.  They are complex analytic quotients, but not quotient varieties in any sense that I know.  
So what's the point of knowing whether $G/H$ is an open subset of a variety?  Really, one needs to know properties of $G/H$ as a Riemannian manifold and complex analytic space (e.g. curvature, whether it's a Stein space).  That's the most important thing!
As Kevin Buzzard and his commentators note, under the assumption that $G$ comes from a reductive group over $Q$, and under the assumption that $H$ is a maximal compact subgroup of $G$, and under the assumption that there is a "Shimura datum" giving the quotient $G/H$ a complex structure, the quotient $G/H$ is a period domain for Hodge structures, and the quotients $\Gamma \backslash G / H$ are quasiprojective varieties when $\Gamma$ is an arithmetic subgroup of $G$.  
But these are quite strong conditions, on $G$ and on $H$!  I have also wondered about other situations when $X = \Gamma \backslash G / H$ might have a natural structure of a quasiprojective variety.  A general technique to prove such a thing is to use a differential-geometric argument.  A great theorem along this line is due to Mok-Zhong (Compactifying complete Kähler-Einstein manifolds of finite topological type and bounded curvature, Ann. of Math 1989).  The theorem, as quoted from MathSciNet, reads:
"Let $X$ be a complex manifold of finite topological type. Let $g$ be a complete Kähler metric on $X$ of finite volume and negative Ricci curvature. Suppose furthermore that the sectional curvatures are bounded. Then $X$ is biholomorphic to a Zariski-open subset $X'$ of a projective algebraic variety $M$."
Such results can be applied to prove quasiprojectivity of Shimura varieties of Hodge type.  I believe I first learned this by reading J. Milne's notes on Shimura varieties.
I tried once to apply this to an arithmetic quotient of $G/H$, where $H$ was a bit smaller than a maximal compact (when $G/H$ was the twistor covering of a quaternionic symmetric space) -- I couldn't prove Mok-Zhong's conditions for quasiprojectivity, and I still don't know whether such quotients are quasiprojective.<|endoftext|>
TITLE: Quadratic forms over finite fields
QUESTION [11 upvotes]: I'm reading some very old papers (by Birch et al) on quadratic forms and I don't get the following point: 

If $f$ is a quadratic form in $X_1,X_2,\cdots,X_n$ over a
  finite field, then one can change variables such that $f$ can be written as $\sum_{i = 1}^s Y_{2i - 1}Y_{2i} + g$, where $g$ is a quadratic form which involves
  variables other than
  $Y_1,Y_2,\cdots,Y_{2s}$ and has order at
  most 2 (i.e. can be written using at
  most two linear forms).

So either this is a well-known result - but I don't find a reference - or either this is easy to see, but in that case I'm just missing the point. By the way, is this really true in characteristic 2?  
(And in fact I'm not sure what role is played by the fact that the field is finite...)

REPLY [10 votes]: I will sketch below a standard argument to show what you need, because I find it very neat!
Let $V$ be a finite dimensional vector space over a field $k$ and let $q \colon V \to k$ be a quadratic form on $V$.  Denote by $b$ the symmetric bilinear form associated to $q$: thus for vectors $v,w \in V$ define $b(v,w) := q(v+w)-q(v)-q(w)$.  Suppose that $v$ is a non-singular zero of $q$.  Since $v$ is non-singular, it follows that there is a $w' \in W$ such that $\alpha := b (v , w') \neq 0$.  Let $w := \frac{1}{\alpha^2} (\alpha w' - q(w') v)$; it is immediate to check that $q(w)=0$ and $b (v , w) = 1$.  Observe that the ``orthogonal complement'' of $v,w$ with respect to the form $q$ has codimension two and does not contain the span of $v,w$.  Thus, we conclude that we can find a basis of $V$ such that $q(x_1,\ldots,x_n) = x_1 x_2 + q'$, where $q'$ is a quadratic form over a space of dimension two less than the dimension of $V$.
The statement about finite fields follows at once, since over a finite field, any quadratic form in three or more variables admits a non-trivial zero.  This is a consequence of the Chevalley-Warning Theorem.  More generally, any field such that quadratic forms in three variables always admit a zero has the property you need, e.g. any $C_1$-field would work.

REPLY [8 votes]: Let me try a similar explanation with different words.  (Note that my explanation does not cover characteristic $2$.)  
A quadratic form is nondegenerate if any of its associated symmetric matrices has nonzero determinant.  (Alternately, if the associated bilinear form $B(x,y) = q(x+y) - q(x) - q(y)$ is nondegenerate in the usual sense: $B(x,y) = 0 \ \forall y \in K \implies x = 0$.)
Let $K$ be a field of characteristic different from $2$.  The hyperbolic plane is the special quadratic form
H(x,y) = xy.
(As with any quadratic form over $K$, it can be diagonalized: $\frac{1}{2} x^2 - \frac{1}{2} y^2$.)  
A nondegenerate quadratic form $q(x_1,\ldots,x_n)$ is isotropic if there exist $a_1,\ldots,a_n \in K$, not all $0$, such that $q(a_1,\ldots,a_n) = 0$ and otherwise anisotropic.
Witt Decomposition Theorem: Any quadratic form $q$ can be written as an orthogonal direct sum of an identically zero quadratic form, an anistropic quadratic form, and some number of hyperbolic planes.  In particular, any isotropic quadratic form $q(x_1,...,x_n)$ can be written, after a linear change of variables, as $x_1 x_2 + q(x_3,...,x_n)$.  
For your purposes, you might as well assume your quadratic form is nondegenerate -- otherwise, it simply involves more variables than actually appear!  
Now over a finite field, the Chevalley-Warning theorem implies that any nondegenerate quadratic form in at least three variables is isotropic, so that by Witt Decomposition, you can split off a hyperbolic plane.  If you still have at least three variables, you can do this again.  Repeated application gives your result.  
References:
For Chevalley-Warning:
http://math.uga.edu/~pete/4400ChevalleyWarning.pdf
For Witt Decomposition:
http://math.uga.edu/~pete/quadraticforms.pdf<|endoftext|>
TITLE: Reference for the expected number of prime factors of n larger than n^alpha is -log alpha
QUESTION [6 upvotes]: Let $0 < \alpha < 1$ be a constant.  The expected number of prime factors of a "random" integer near $n$ which are greater than $n^\alpha$ is $-\log \alpha$.
It's my understanding that (properly formulated) this is a well-known fact in analytic number theory but I cannot find a reference for it.  Can anybody provide a reference?
Edited to add (March 28):: The asymptotic density of positive integers $n$ with $k$th largest factor smaller than $n^{1/\alpha}$ is $\rho_k(\alpha)$, where we have $L_0(\alpha) = [\alpha > 0]$ and 
$$ L_k(\alpha) = [\alpha \ge k] \int_k^\alpha L_{k-1}(t-1) \: {dt \over t}, $$
and $1-\rho_k(\alpha) = \sum_{n=0}^\infty {-k \choose n} L_{n+k}(\alpha)$.  (See Riesel, p. 162.)  The density of positive integers with $k$th largest factor larger than $n^{1/\alpha}$ is therefore $1-\rho_k(\alpha)$, and so the expected number of factors larger than $n^{1/\alpha}$ is $\sum_{k \ge 1} (1-\rho_k(\alpha))$.  Therefore the expected number of such factors is
$$ \sum_{k \ge 1} \sum_{n \ge 0} {-k \choose n} L_{n+k}(\alpha). $$
Letting $n+k = j$ we can rewrite this sum as
$$ \sum_{j \ge 1} \sum_{n=0}^{j-1} {n-j \choose n} L_j = \sum_{j \ge 1} L_j \left( \sum_{n=-0}^{j-1} (-1)^n {j-1 \choose n} \right) $$
and the inner sum is $0$ except when $j=1$, when it is $1$.  So the expected number of factors larger than $n^{1/\alpha}$ is $L_1(\alpha)$; this is $\log \alpha$.

REPLY [6 votes]: You seem to be over complicating things.  For $\alpha$ fixed, this is relatively easy to prove.  I leave it to you deduce the desired result as a consequence of the following theorem:  

Theorem:  Let $x>0$ and fix $0<\alpha<1$.  Define $\omega_{x,\alpha}(n)=\sum_{p|n,\ p>x^\alpha} 1$.  Then this function has average value $$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=-\log \alpha +O\left(\frac{1}{\log x}\right).$$

Proof:  Notice that 
$$\frac{1}{x}\sum_{n\leq x} \omega_{x,\alpha}(n)=\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left[\frac{x}{p}\right] =\sum_{x^{\alpha}<p<\leq x}\frac{1}{p}-\frac{1}{x}\sum_{x^{\alpha}<p<\leq x}\left\{\frac{x}{p}\right\}$$  
$$= \sum_{x^\alpha<p<x} \frac{1}{p} +O\left(\frac{1}{\log x}\right)=\log \log (x)-\log \log x^\alpha +O\left(\frac{1}{\log x}\right)$$ 
$$=-\log \alpha +O\left(\frac{1}{\log x}\right).$$<|endoftext|>
TITLE: Complete intersections and flat families
QUESTION [10 upvotes]: If I have a flat family $f \colon X \to T$ such that some fiber is (locally) a complete intersection, does that imply that there is an open set $U$ in $T$ such that the fibers above $U$ are (locally) complete intersections?
In general, what kind of intuition should one have about which properties are "open" with respect to flat families?

REPLY [17 votes]: EGA IV$_4$, 19.3.8 (and 19.3.6); this addresses openness upstairs without properness, and (as an immediate consequence) the openness downstairs if $f$ is proper (which I assume you meant to require).
The general intuition is that openness holds upstairs for many properties, and so then holds downstairs when map is proper.  As for proving openness upstairs, the rough idea is to first prove constructibility results, and then refine to openness by using behavior under generization.  But it's a long story, since there are many kinds of properties one can imagine wanting to deal with.  These sorts of things are developed in an extraordinarily systematic and comprehensive manner in EGA IV$_3$, sections 9, 11, 12 (especially section 12 for the niftiest stuff).

REPLY [8 votes]: I just wanted to point out that the situation for globally complete intersections is very different; the property of being a globally complete intersection is not open. For example, consider $(\mathbb{P}^2)^9$. This is an $18$-dimensional irreducible space; the locus of $9$-tuples of points which can be written as a complete intersection of two cubics only has dimension $16$, and thus is not open. 
Here is a nice trick question: Take a family of $9$-tuples of points in $\mathbb{P}^2$, indexed by $t \in \mathbb{A}^1$, which are the intersection of two cubics for $t=0$ but not for generic $t$. Cone this family to get a family of $1$-dimensional subschemes of $\mathbb{P}^3$. Why is this not an example of a singularity which is a locally complete intersection deforming to a singularity which is not?<|endoftext|>
TITLE: Restriction of a complex polynomial to the unit circle
QUESTION [28 upvotes]: I am pretty sure that the following statement is true. I would appreciate any references (or a proof if you know one).
Let $f(z)$ be a polynomial in one variable with complex coefficients. Then there is the following dichotomy. Either we can write $f(z)=g(z^k)$ for some other polynomial $g$ and some integer $k>1$, or the restriction of $f(z)$ to the unit circle is a loop with only finitely many self-intersections. (Which means, more concretely, that there are only finitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$.)
EDIT. Here are a couple reasons why I believe the statement is correct.
1) The statement is equivalent to the following assertion. Consider the set of all ratios $z/w$, where $|z|=1=|w|$ and $f(z)=f(w)$ (here we allow $z=w$). If $f$ is a nonconstant polynomial, then this set is finite.
[[ Here is a proof that the latter assertion implies the original statement. Suppose that there are infinitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$. Then some number $c\neq 1$ must occur infinitely often as the corresponding ratio $z/w$. However, this would imply that $f(cz)=f(z)$ (as polynomials). It is easy to check that this forces $c$ to be a root of unity, and if $k$ is the order of $c$, then $f(z)=g(z^k)$ for some polynomial $g(z)$. ]]
Going back to the latter assertion, note that the set of all such ratios is a compact subset of the unit circle, and it is not hard to see that 1 must be an isolated point of this set. So it is plausible that the whole set is discrete (which would mean that it is finite).
2) If I am not mistaken, experiments with polynomials that involve a small number of nonzero monomials (such as 2 or 3) also confirm the original conjecture.

REPLY [20 votes]: You're right.  Quine proved in "On the self-intersections of the image of the unit circle under a polynomial mapping" that if the degree is $n$ and $f(z)\neq g(z^k)$ with $k>1$, then the number of points with at least 2 distinct preimage points is at most $(n-1)^2$.  An example shows that this is sharp.  Here's the review in MR.
After the proof, there is a remark: 

As a simple consequence of this theorem we note that a polynomial $p$
  cannot map $|z| < 1$ conformally onto a domain with a slit, for in this case
  $p(e^{i\phi})$ would have an infinite number of vertices.

REPLY [4 votes]: The image of the unit circle is a real-algebraic curve, so the number of self-intersections should be finite. 
Addendum:
I'm not sure how to complete the argument, but here's a heuristic (following Speyer's suggestion). The curve $x^2+y^2=1$ is a rational curve (i.e., birational to $\mathbb{CP}^1$, the Riemann sphere), so it's image is also a rational curve (here, I'm thinking of the map $p:\mathbb{C}\to \mathbb{C}$ as a polynomial map $\mathbb{R}^2 \to \mathbb{R}^2$, and its extension to $\mathbb{C}^2\to \mathbb{C}^2$ and $\mathbb{CP}^2\to \mathbb{CP}^2$). Complex conjugation gives an antiholomorphic involution of $x^2+y^2=1$, fixing the circle (on the Riemann sphere, this must be conjugate to complex conjugation). The image (in $\mathbb{CP}^2$) is a singular sphere, and again complex conjugation should be an anti-holomorphic involution fixing the image of the unit circle. So the map should be a composition of a polynomial map with a map of the Riemann sphere sending the circle to the circle.  All such maps of the Riemann sphere are products of Mobius transformations of the form $\frac{z-\varphi}{1-\overline{\varphi}z}$ (this is an exercise in Ahlfors, making use of the Schwarz lemma). If the composition is to be a polynomial map, then $\varphi$ must $=0$ in each factor, and the map is of the form $z^k$. My algebraic geometry is quite weak, so I'm not sure if this argument can be made rigorous (and I probably shouldn't have posted an answer in the first place!).<|endoftext|>
TITLE: Combinatorial proof that large-girth graphs are sparse?
QUESTION [12 upvotes]: Theorem. Fix $\epsilon > 0$; for sufficiently large n, any graph with n vertices and $\epsilon \binom{n}{2}$ edges contains many (nondegenerate) cycles of length 4.

The proof is simple; put an indicator variable $\delta_{x, y}$ for each pair of vertices corresponding to whether or not there is an edge there; then start with
$n^8 \epsilon^4 = (\sum \delta_{x, y})^4$
and apply Cauchy-Schwarz twice; finally, note that there are $O(n^3)$ "degenerate 4-cycles".
A basic corollary of this is the following fact:

Corollary. Any graph with girth at least 5 and n vertices has $o(n^2)$ edges.

This seems like it should be possible to prove without resorting to "analytic" machinery like Cauchy-Schwarz; indeed, it seems like it should be weak enough to prove almost by arguing "locally." But none of the obvious lines of reasoning seem to provide a proof.
Is it possible to get a good bound on the density of large-girth graphs without using Cauchy-Schwarz or equivalent?

REPLY [20 votes]: It's known more specifically that any graph with girth ≥ 5 has $O(n^{3/2})$ edges — see e.g. Wikipedia on the Zarankiewicz problem.
Here's a combinatorial proof. Suppose that graph $G$ has $\ge kn^{3/2}$ edges for a sufficiently large constant $k$. As long as there are vertices with degree smaller than some appropriate constant times $\sqrt n$ one can remove them and get a smaller graph with the same property of having at least $kn^{3/2}$ edges, so eventually one can reach a state where every vertex has degree at least $\Omega(\sqrt n)$. Once this happens, there are $O(n^2)$ possible pairs of neighbors that a vertex might have, and each vertex has $\Omega(n)$ pairs of neighbors, so some pair of neighbors appears twice causing the graph to have a 4-cycle.<|endoftext|>
TITLE: Group completion theorem
QUESTION [21 upvotes]: Let $M$ be a topological monoid. How does the homology-formulation of the group completion theorem, namely (see McDuff, Segal: Homology Fibrations and the "Group-Completion" Theorem)

If $\pi_0$ is in the centre of $H_*(M)$ then $H_*(M)[\pi_0^{-1}]\cong H_*(\Omega BM)$

imply that $M\to \Omega BM$ is a weak homotopy equivalence if $\pi_0(M)$ is already a group? I don't see the connection to homology. Can one prove the latter (perhaps weaker) statement more easily than the whole group completion theorem? 
A topological group completion $G(M)$ of $M$ should transform the monoid $\pi_0(M)$ into its (standard algebraic) group completion. But a space with this property is not unique. Why is $\Omega BM$ the "right" choice? Perhaps this is clear when I see the connection to the homology-formulation above.

REPLY [21 votes]: The statement that $M \to \Omega BM$ is a weak equivalence when $M$ is a group-like topological monoid is indeed easier: the map $EM = B(M \wr M) \to BM$ is then a quasi-fibration, has geometric fibre $M$ over the basepoint and homotopy fibre $\Omega BM$.
However the homological group-completion theorem also implies this: if $M$ is group-like then $\pi_0(M)$ already consists of units in $H_*(M)$, so it just says that $M \to \Omega BM$ is a homology equivalence. Each of these spaces has homotopy equivalent path components, so it is then enough to observe that the map of 0 components is a homology equivalence between simple spaces, so a weak homotopy equivalence.
However it is perverse to prove the "$M \simeq \Omega BM$" result this way.

REPLY [14 votes]: Well, if $\pi_0=\pi_0(M)$ is already a group, then $H_*(M)\approx H_*(M)[\pi_0^{-1}]$.  So $M$ and $\Omega B M$ have the same homology in this case.  This isn't quite enough on its own, but if you can produce a map $M\to \Omega BM$ which induces this homology isomorphism, then the result follows using the Hurewicz theorem.
What McDuff-Segal actually do is show that if $M$ is a topological monoid which acts on a space $X$, in such a way that every $m\in M$ induces a homology equivalence $x\mapsto mx\colon X\to X$, then you can produce a "homology fibration" $f:X_M\to BM$ with fiber $X$. "Homology fibration" means that the fibers of $f$ are homology-equivalent to the homotopy fibers of $f$.  
If $\pi_0M$ is an abelian group, you can find an $X$ such that $X_M$ is contractible, and the fiber of $f:X_M\to BM$  is $X$.  This gives the homology equivalence you want, since the homotopy fibers of $f$ look like $\Omega BM$.  
Take a look at McDuff and Segal's paper, it's nice.  There is a also a treatment in terms of simplicial sets in Goerss-Jardine, *Simplicial Homotopy Theory".  
Added: The functor $M\mapsto \Omega BM$ is the "total derived functor of group completion".  The only convincing explanation of why this is so (that I'm aware of) is in Dwyer-Kan, Simplicial Localizations of Categories, JPAA (17) 267-283.  Though they work simplicially, and work more generally (with categories in place of monoids), they show that $M$ is a cofibrant simplicial monoid, then the simplicial monoid $M[M^{-1}]$ is weakly equivalent to the space $\Omega |BM|$.<|endoftext|>
TITLE: Dirichlet and the prime number theorem
QUESTION [21 upvotes]: I browsed Dirichlets Werke today and was kind of surprised by two remarks that he made on p. 354 (Über die Bestimmung ...) and p. 372 (Sur l'usage ...). In the second paper, he claims (my translation)
 I have applied these principles to a demonstration of the remarkable formula given by Legendre for expressing in an approximate manner how many prime numbers there are below an arbitrary, but very large, limit.

In a handwritten note on the reprint he sent to Gauss he remarked that $\sum 1/\log n$ (this is Gauss's version of the PNT, at least if you replace the sum by an integral) is a better estimate than Legendre's.
I am a little bit puzzled as to why Dirichlet's claim to have proved the prime number theorem is not discussed anywhere in the literature. Or is it?

REPLY [7 votes]: Dirichlet's remark from the first paper is extracted and translated on page 98 of The Development of Prime Number Theory by Narkiewicz. So this has not passed completely unnoticed. Narkiewicz remarks that Dirichlet believed that his analytic methods would enable him to prove Legendre's conjecture, and that Dirichlet never returned to the problem.
Dirichlet remained interested in the asymptotic growth laws ("Asymptotische Gesetze") of arithmetic functions for the rest of his life, as seen from his 1849 paper with the estimate 
$$
\sum_{n \leq x}d(n) = x\log(x) + (2\gamma - 1)x + O(x^{1/2}),
$$
and a couple of other estimates, and a letter of 1858 to Kronecker reprinted in Dirichlet's Werke, where he mentions having obtained a substantial improvement of the error term $O(x^{1/2})$ by a new method.
Since Dirichlet demonstrably did not lose interest in such questions, and never returned to the PNT in print, it seems reasonable to believe that he discovered that his real-variable method would not yield the PNT.<|endoftext|>
TITLE: Why is the harmonic oscillator so important? (pure viewpoint sought). How to motivate its role in Getzler's work on Atiyah-Singer? 
QUESTION [17 upvotes]: I'm in the process of understanding the heat equation proof of the Atiyah-Singer Index Theorem for Dirac Operators on a spin manifold using Getzler scaling. I'm attending a masters-level course on it and using Berline, Getzler Vergne.
While I think I could bash my way through the details of the scaling trick known as `Getzler scaling', I have little to no intuition for it.
As I understand it, one is computing the trace of the heat kernel of the ("generalized") Laplacian associated to a Dirac operator. The scaling trick reduces the problem to one about the ("supersymmetric" or "generalized") harmonic oscillator, whose heat kernel is given by Mehler's formula. I am repeatedly assured that the harmonic oscillator is a very natural and fundamental object in physics, but, being a `pure' analyst, I still can't sleep at night.

What reasons are there for describing the harmonic oscillator as being so important in physics?

Why/how might Getzler have thought of his trick? (Perhaps the answer to this lies in the older proofs?)

Is there a good way I could motivate an attempt to reduce to the harmonic oscillator from a pure perspective?

(i.e. "It's a common method from physics" is no good). I'm looking for: "Oh it's simplest operator one could hope to reduce down to such that crucial property X still holds since Y,Z"...or..."It's just like the method of continuity in PDE but a bit different because..." 
Thanks.

REPLY [4 votes]: Here is a fairly physicsy point of view.   You are calculating the supertrace of the heat operator, and because this is magically time independent, you can calculate it in the small $t$ limit.  The heat operator is the time-evolution operator (propagator) for a certain theory (supersymmetric quantum mechanics) and the supertrace amounts to considering it only on loops (I am skipping some details here!).  The small $t$ approximation means that you can consider loops that are nearly constant.  
The small $t$ limit is similar to the semiclassical (small $\hbar$) limit, so we will borrow our intuition from there.  In the semiclassical approximation you consider small perturbations from a classical solution and expand the action in powers of the perturbation.  Because classical solutions are critical points of the action, you get no linear term, so the first interesting term in quadratic in the perturbation.  So in the small $t$ limit you can ignore everything but the quadratic part of the action.  A quadratic action is basically just a harmonic oscillator.
I believe (and here I am expressing an intuition I have no concrete argument for!) that Getzler's rescaling amounts to converting between the small $t$ limit and the small $\hbar$ limit, since these are not quite the same thing.  I also think that you are actually getting a magnetic term, not a quadratic potential (which is the harmonic oscillator), but there is a standard physics trick for going from one to the other.
More generally, if you are in any physics situation where you have a minimum/critical point of a potential/action, it is probably going to make sense to expand it in powers of distance from that point, and the first term of the expansion will be quadratic and therefore a harmonic oscillator.  So you can always model systems sufficiently close to their equilibrium by a system of harmonic oscillators.  In quantum theory it is only a little bit of a lie to say that the only things we can solve are quadratic theories (harmonic oscillator) so nearly all understanding starts from there and builds out.  The magic of the Index Theorem, and most places where QFT give you wonderful results in mathematics, is that this approximation is somehow exact.<|endoftext|>
TITLE: Collecting various theories on toy examples: Projective space
QUESTION [15 upvotes]: I am looking for text books/notes/papers/documents playing with toy examples: projective space, in particular, $P^{1}$. Because I think this is really a cute example. Although algebraic geometry on $P^{1}$ is comparatively simple, it gave inspirations to treat more general situtation. Precisely, I am looking for something including following topics:(but you can add whatever you want)

algebraic geometry of $P^{n}$, say, $Coh(P^{n})$, $D^{b}(Coh(P^{n}))$, say, exceptional collection,semi-orthogonal decomposition,stability conditions
Relation to representation theory, say, Hall algebra of $Coh(P^{n})$ and $D^{b}(Coh(P^{n}))$ and its relation to affine quantum group: $U_{q}(\hat{sl_{n+1}})$, representation theory of Kronecker quiver. Tilting theory. Weighted projective line and so on.
$D-module$ on flag variety of $U(sl_{2})$(or $P^{1}$) and so on..................
Add whatever you like.
Add whatever you like.

..................................................

REPLY [8 votes]: Maybe I can answer this question by myself now. I did some literature research and find some papers and notes illustrating $P^{1}$ to establish various theory
Lectures on Hall algebras
The author talks about Hall algebra of coherent sheaves on $P^{1}$, relation with representation theory of $U_{q}(\hat{sl_{2}})$ and also The Hall algebra of the category of coherent sheaves on the projective line talks about similar facts.
 Twisted rings of differential operators on the projective line and the Beilinson-Bernstein theorem. It is a master thesis by Koushik Panda. He established $P^{1}$(flag variety of $sl_{2}$) version of Beilinson-Bernstein localization. His treatment is very detailed.
t-stabilities and t-structures on triangulated categoriesillustrates classifications of t-structures on $D^{b}(Coh{P^{1}})$
Introduction to coherent sheaves on weighted projective lines by Chen-Xiaowu and Henning Krause. Very expository notes for coherent sheaves, Tilting theory, derived category of $P^{1}$<|endoftext|>
TITLE: When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)
QUESTION [26 upvotes]: Sorry if this is a silly question.  I was wondering, under what axioms of set theory is it true that if $\alpha$,$\beta$ are cardinals, and $2^\alpha=2^\beta$, then $\alpha=\beta$? Do people use these conditions to prove interesting results?
This question is prompted from a recent perusing of Johnson's "Topics in the Theory of Group Presentations", where in the first few pages he "proves" free groups of different rank are non-isomorphic:
the number of mappings from a free group of rank $\omega$ to the group $\mathbb{Z}/2\mathbb{Z}$ is $2^\omega$, which would be invariant under isomorphism; and then he assumes the topic of my question: $2^\alpha=2^\beta$ implies $\alpha=\beta$.
But I remember reading something a few years ago about a student of R.L. Moore (I only remember his last names was Jones) "proving" the Moore Space conjecture, and using that $\alpha > \beta$ implied $2^\alpha > 2^\beta$, but that this was incorrect.
Anyway, I realized I don't know anything about when this is true or false, so I thought I'd ask.

REPLY [21 votes]: François gives the correct affirmative answer. For the negative side, the usual method of proving that the negation of the Continuum Hypothesis is consistent with ZFC is to use the method of forcing to add Aleph2 many Cohen reals, so that 2ω = ω2 in the forcing extension V[G]. In this model V[G], it is also true that 2ω1 = ω2. Thus, this model shows that it is not necessarily true that different-sized sets have different sized power sets.  The case of symmetric groups is likely more interesting than free groups, because in this model, the symmetric groups Sω and Sω1 have the same cardinality ω2. Nevertheless, these two groups are not isomorphic, as explained in this MO question. 
The general answer about what can be true for the continuum function κ --> 2κ is exactly provided by Easton's Theorem. This remarkable theorem states that if you have any class function E, defined on the regular cardinals κ, with the properties that 

κ ≤ λ implies E(κ) ≤ E(λ)
κ < E(κ) 
κ < Cof(E(κ)) 

then there is a forcing extension V[G] in which 2κ = E(κ) for all regular cardinal κ. In particular, this shows that the sizes of the power sets (on regular cardinals) are restricted to obey only and exactly the hypotheses listed explicitly above. Each of these properties corresponds to a well-known fact about cardinal exponententiation.
Using Easton's theorem, we can build models of set theory where 2κ = κ++ for all regular κ. The added power of the Woodin/Foreman result mentioned by François is that they also get this for singular cardinals. 
The point now is that there are innumerable examples provided by Easton's theorem that satisfy your hypothesis that the continuum function is one-to-one. If one begins with a model of GCH and selects any injective Easton function E, then the resulting model of set theory V[G] will have E as it's continuum function κ --> 2κ = E(κ) for regular κ, and the GCH will continue to hold at singular κ, preserving injectivity. So one is quite free to satisfy your hypothesis while having any kind of crazy failures of GCH.

REPLY [15 votes]: The Generalized Continuum Hypothesis, i.e. $2^\kappa = \kappa^+$ for every infinite cardinal $\kappa$, implies the desired result, but it can hold in other circumstances. For example, Woodin has shown that it is relatively consistent with the existence of a supercompact cardinal that $2^\kappa = \kappa^{++}$ for every infinite cardinal $\kappa$. (See Foreman and Woodin, The generalized continuum hypothesis can fail everywhere, Annals of Mathematics 133, 1991, 1-35.)
The statement $2^{\aleph_0} < 2^{\aleph_1}$ is known as the Weak Diamond Principle. This principle was introduced by Devlin and Shelah (A weak version of $\diamondsuit$ which follows from $2^{\aleph_{0}}<2^{\aleph_{1}}$, Israel J. Math. 29, 1978, 239-247). This principle has been very useful as a substitute for stronger enumeration principles such as the Continuum Hypothesis and Jensen's Diamond Principle $\diamondsuit$. I think the Weak Diamond Principle was inspired by Shelah's work on the Whitehead Problem (which may be related to your group theoretic application).

REPLY [6 votes]: One thing that is made very easy by assuming that $2^\alpha < 2^\beta$ whenever $\alpha <\beta$ is showing that for sets $S_1$ and $S_2$ of different cardinalities, the Banach spaces $\ell_\infty(S_1)$ and $\ell_\infty(S_2)$ are not isomorphic. This follows immediately from the fact that for an infinite set $S$, the smallest possible cardinality of a dense subset of $\ell_\infty(S)$ is $2^{|S|}$. 
If one does not assume that $2^\alpha = 2^\beta$ implies $\alpha =\beta$, then one must revert to some other argument. I have not thought about this problem at length, so there may be much simpler reasons why $\ell_\infty(S_1)$ and $\ell_\infty(S_2)$ cannot be isomorphic when $S_1$ and $S_2$ have different cardinalities (without assuming $2^\alpha = 2^\beta\Rightarrow\alpha =\beta$), but one way is to instead consider the least possible cardinality of a norm dense subset of an arbitrary weakly compact subset of $\ell_\infty(S)$. In this case, a result of Haskell Rosenthal asserts that for $S$ infinite and a weakly compact set $W \subseteq \ell_\infty (S)$, $W$ admits a norm dense subset of cardinality no greater than $|S|$. On the other hand, a standard argument shows that $\ell_2 (S)$ (in fact, any Banach space of density character not exceeding $|S|$) is isomorphic to a subspace of $\ell_\infty (S)$ (generalise the standard argument that every separable Banach space embeds into $\ell_\infty$). Thus, if $\alpha$ and $\beta$ are cardinals with $\alpha <\beta$, then $\ell_2 (\beta )$ is isomorphic to a subspace of $\ell_\infty (\beta )$, but not to a subspace of $\ell_\infty (\alpha)$. Hence $\ell_\infty (\alpha )$ and $ \ell_\infty (\beta)$ are not isomorphic.
The result of Rosenthal cited above is from his paper "On injective Banach spaces and the spaces $L_\infty (\mu )$ for finite measures $\mu$".

REPLY [2 votes]: François's and Joel's answers to the general question are correct.
As for the "dimension of a free group is uniquely determined, even if it's infinite", there's a better argument.  Let G be a free group on two sets of generators x and y; we wish to show |x| = |y|.  First let A = Gab be the free abelian group on generators x and y; then let V = A ⊗Z F2 be the vector space over F2 with bases x and y, so that |V| = 2|x| = 2|y| (strike that, it's false for infinite cardinalities).  (We could use any field here, but the argument is more intricate.)
If |V| is finite, then |x| = |y| follows immediately. So suppose |V| is infinite, so that |x| and |y| are infinite.  Each element of x is the sum of finitely many elements of y, and every element of y must show up in at least one such sum (why?).  This means |y| ≤ |x|•ℵ0 = |x| (cardinal arithmetic). Similarly |x| ≤ |y|, so |x| = |y|.<|endoftext|>
TITLE: Which finitely presented groups can be distinguished by decidable properties?
QUESTION [17 upvotes]: This question continues the line of inquiry
of these
three
questions.
Question. Which finitely presented groups can be
distinguished by decidable properties?
To be precise, let us say that φ is a decidable property of finitely
presented groups, if there is a class A of finitely
presented groups, closed under group isomorphisms, such that
{ p | ⟨p⟩ ∈ A } is decidable, where ⟨p⟩ denotes the group presented by p. That is, we insist that the decision procedure give the same answer for presentations leading to the same group up to isomorphism.
One extreme case, perhaps unlikely, would be that any two non-isomorphic
finitely presented groups can be distinguished by decidable
properties, so that for any two finitely presented
non-isomorphic groups ⟨p⟩ and ⟨q⟩,
there is a decidable property φ where φ(p) holds
and φ(q) fails. That would be quite remarkable. 
If this is not the case, then there would be two finite group
presentations p and q, such that the groups presented
⟨p⟩ and ⟨q⟩ are not isomorphic, but
they have all the same decidable properties. This also
would be remarkable. 
Which is the case?
Another way to describe the question is in terms of the
equivalence relation ≡, which I introduced in my
previous question, where p
≡ q if φ(p) and φ(q) have the same answer for
any decidable property φ of finitely presented groups. This
is precisely the equivalence relation of "having all the
same decidable properties". Of course, this includes the
group-isomorphism relation, and the current question is
asking: What is this relation ≡? In particular, is
≡ the same as the group isomorphism relation?
If it is, then any two non-isomorphic finitely presented
groups can be distinguished by decidable properties; if
not, then there are two finitely presented non-isomorphic
groups ⟨p⟩, ⟨q⟩ having all the same
decidable properties.
Henry Wilton has emphasized several times that there are
relatively few truly interesting decidable properties of
finitely presented groups. This may very well be true.
Nevertheless, the answers to the previous MO questions on
this topic have provided at least some decidable
properties, and my question here is asking the extent to
which these properties are able to distinguish any two
finitely presented groups.
In particular, in these previous MO questions, Chad Groft
inquired whether there were
any nontrivial decidable properties of finitely presented
groups. John Stillwell's 
answer was
that one could decide many questions about the
abelianization of the group. In a subsequent question, I
inquired whether all decidable properties were really about
the abelianization, and David Speyer's
answer was
that no, there were questions about other quotients, such
as whether the group had a nontrivial homomorphism into a
particular finite group, such as A5. In a third question, David generalized further and inquired
whether all decidable properties depended on the
profinitization, and the answer again was no (provided by David and Henry). So at least
in these cases we have been increasingly able to separate
groups by decidable properties.
A generalization of the question would move beyond the
decidable properties. For example, if we consider the
computably enumerable (c.e.) properties, then we have quite
a lot more ability to distinguish groups. A property is
c.e. if there is a computable algorithm to determine the
positive instances of φ(p), but without requiring the
negative instances to ever converge on an answer. For
example, the word problem for any finitely presented group,
or indeed, for any computably presented group, is
computably enumerable, since if a word is indeed trivial, we will eventually discover this. Using the same idea as David's answer to my question, it follows that the question of
whether a finitely presented group ⟨p⟩ admits a
nontrivial homomorphism into the integers Z, say,  or many other
groups, is computably enumerable. One may simply try out all possible maps of the generators. A generalization of
this establishes:
Theorem. The question of whether one finitely
presented group ⟨p⟩ maps homomorphically onto (or into) another ⟨q⟩ is computably enumerable.
The proof is that given p and q, one can look for a map of
the generators of p to the words of q, such that all
relations of p are obeyed by the image in q, and such that
all the generators of q are in the range of the resulting
map. This is a c.e.
property, since one can look for all possible candidates
for the map of the generators of p into words of q, and
check whether the relations are obeyed and the generators
of q are in the range of the map and so on. If they are,
eventually this will be observed, and at the point one can
be confident that ⟨p⟩ maps onto ⟨q⟩. More generally, is the isomorphism relation itself c.e.? It is surely computable from the halting problem 0', since we could ask 0' whether the kernel of the proposed map was trivial or not, and it would know the answer. 

Where does the isomorphism relation on finitely
presented groups fit into the hierarchy of Turing
degrees? Is it c.e.? Is it Turing equivalent to the Halting problem?

Once one moves to the c.e. properties, it is similarly
natural to move beyond the finitely presented groups to the
computably presented groups (those having a computable
presentation, not necessarily finitely generated). In this
context, the proof above no longer works, and the natural
generalization of the question asks:

Which computably presented groups are distinguished by
c.e. properties?

The isomorphism relation on finitely generated computably
presented groups (given the presentations) seems to be
computable from the halting problem for the same reason as
in the proof above, but now one doesn't know at a finite
stage that the proposed map of the generators will
definitely work, since one must still check all the
relations-yet-to-be-enumerated. But 0' knows the answer, so
we get it computably in 0'. In the infinitely generated case,
however, things are more complicated.

REPLY [13 votes]: The isomorphism relation for finitely presented groups is c.e., and in fact is Turing equivalent to the halting problem.  

Proof: 
To check whether two finitely presented groups $G$ and $H$ are isomorphic, search for data that might describe maps $G \to H$ and $H \to G$ and for words that show that it is a consequence of the relations that the composition in either order maps each generator to itself, and verify that the relations of $G$ map to $1$ in $H$ and vice versa so that the maps are well-defined and that the remaining data shows that they are inverse isomorphisms.  Thus the isomorphism relation is c.e.  It is also no easier than the halting problem, because an algorithm even for deciding whether a finitely presented group is trivial could be used to solve the halting problem.  (That is how it is known that triviality is an undecidable property, by reductions passing through the word problem along the way.) $\square$
EDIT: As for your "main question" asking whether decidable properties can distinguish every pair of non-isomorphic finitely presented groups, I'll prove the following negative result:

There is no c.e. set $\mathcal{F}$ of decidable properties such that any two non-isomorphic finitely presented groups can be distinguished by some $\phi \in \mathcal{F}$.

(Call a set $\mathcal{F}$ of decidable properties c.e. if there is a Turing machine that produces a sequence of algorithms, each of which is guaranteed to compute a decidable property, such that these decidable properties are exactly the ones in $\mathcal{F}$.)
Proof: Suppose that $\mathcal{F}$ exists.  Then we could decide whether an arbitrary finitely presented group $G$ is trivial as follows: By day, search for an isomorphism between $G$ and $\{1\}$ (this search is possible since the isomorphism relation is c.e.)  By night, search for a decidable property $\phi \in \mathcal{F}$ such that $\phi$ distinguishes $G$ and $\{1\}$.  If $\mathcal{F}$ does what it claims to, then one of these processes will terminate and tell you whether $G \simeq \{1\}$.  But triviality is known to be an undecidable property. $\square$
This leaves open the question of whether there is a non-c.e. $\mathcal{F}$ that does the job, but even if there were one, it wouldn't be of much use from the practical point of view!<|endoftext|>
TITLE: Is an irreducible holomorphic symplectic manifold a simple Lie algebra?
QUESTION [7 upvotes]: The tangent bundle of a hyper-Kahler manifold gives a quadratic Lie algebra in the derived category. Can this be regarded as a simple Lie algebra according to Vogel's definition?
A point of view that came from studying Rozansky-Witten invariants is that the tangent bundle of a holomorphic symplectic manifold or hyper-Kahler manifold is a Lie algebra with a non-degenerate invariant symmetric bilinear form. Here the tangent bundle is taken as an object in the derived category and then shifted. The Atiyah class is interpreted as a Lie bracket and the Bianchi identity as the Jacobi identity. The symplectic form is interpreted as a symmetric form since we shifted. Some references are (and please add or request any reference I have omitted)
MR2024627 (2004m:57026)  Roberts, Justin . Rozansky-Witten theory.
 Topology and geometry of manifolds (Athens, GA, 2001), 
 1--17, Proc. Sympos. Pure Math., 71, Amer. Math. Soc., Providence, RI,  2003. 
MR2110899 (2005h:53070)  Nieper-Wißkirchen, Marc . Chern numbers and Rozansky-Witten invariants of compact hyper-Kähler
 manifolds.
World Scientific Publishing Co., Inc., River Edge, NJ,  2004. xxii+150 pp. ISBN: 981-238-851-6 
MR2472137 (2010d:14020)  Markarian, Nikita . The Atiyah class, Hochschild cohomology and the Riemann-Roch
 theorem.
 J. Lond. Math. Soc. (2)  79  (2009),  no. 1, 129--143.
Now Vogel has constructed a universal simple Lie algebra. The question is whether the tangent bundle of an  irreducible holomorphic symplectic manifold meets Vogel's criteria for a simple Lie algebra. This question is for algebraic geometers so I will expand on this. The first condition is that End(L)=End(I) where I is the trivial representation so End(I) is the commutative ring of scalars. In this example Ext(O). This obviously fails for the product of two manifolds so I have naively excluded this by imposing the irreducible condition. The second condition is that $\mathrm{Hom}(\bigwedge^2L,L)$ is a free End(I)-module with basis the Lie bracket.
One reason I find this confusing is that End(I) has nilpotent elements whereas I am used to a field.
If the answer to both questions is Yes then we get a character of Vogel's universal ring.
I would expect this to be of interest to both subjects.
Edit The paper http://arxiv.org/abs/1205.3705 has now been posted on the arxiv and this proves that $K3$-surfaces do give a character of Vogel's ring.

REPLY [2 votes]: I believe this is a very interesting question, that I have been asking myself for quite a long time. 
Nevertheless, I have been told by Prof. Beauville that even in the irreducible case one does not have that 
$$
Ext_X(\mathcal O_X,\mathcal O_X)=Ext_X(T_X,T_X)
$$
Namely, consider $X$ being the Hilbert scheme of two points on a $K3$ surface. 
Then $Ext_X(\mathcal O_X,\mathcal O_X)=\mathbb{C}\oplus\mathbb{C}[-2]\oplus\mathbb{C}[-4]$. 
But $Ext_X(T_X,T_X)=Ext_X(\mathcal O_X,(T^*_X)^{\otimes 2})$ contains $Ext_X(\mathcal O_X,\Omega^2_X)$, which is huge ($h^{2,2}=232$). 
Anyway, I must say that this does not kill the question (this just tells we have to reformulate it). I hope to be able to write more about it soon. 
EDIT: it seems that the answer to the question is NO. The point is that 232 is also the dimension of $H^1(X,S^3(T_X))$ ($X$ is again a $K3$), therefore $Ext_X^1(S^2(T_X),T_X)=RHom_X(\wedge^2(T_X[-1]),T_X[-1]))$ has dimension $\geq232$.<|endoftext|>
TITLE: How does this relationship between the Catalan numbers and SU(2) generalize?
QUESTION [35 upvotes]: This is a question, or really more like a cloud of questions, I wanted to ask awhile ago based on this SBS post and this post I wrote inspired by it, except that Math Overflow didn't exist then.  
As the SBS post describes, the Catalan numbers can be obtained as the moments of the trace of a random element of $SU(2)$ with respect to the Haar measure.  This is equivalent to the integral identity
$$\int_{0}^{1} (2 \cos \pi x)^{2k} (2 \sin^2 \pi x) \, dx = C_k.$$
I can prove this identity "combinatorially" as follows: let $A_n$ denote the Dynkin diagram with $n$ vertices and $n-1$ undirected edges connecting those vertices in sequence.  The adjacency matrix of $A_n$ has eigenvectors $\mathbf{v}\_i$ with entries $\mathbf{v}\_{i,j} = \sin \frac{\pi ij}{n+1}$ with corresponding eigenvalues $2 \cos \frac{\pi i}{n+1}$.  If $k \le n-1$, then a straightforward computation shows that the number of closed walks from one end of $A_n$ to itself of length $2k$ is
$$\frac{1}{n+1} \sum_{i=1}^{n} \left( 2 \cos \frac{\pi i}{n+1} \right)^{2k} 2 \sin^2 \frac{\pi}{n+1} = C_k$$
by the combinatorial definition of the Catalan numbers.  Taking the limit as $n \to \infty$ gives the integral identity; in other words, the integral identity is in some sense equivalent to the combinatorial definition of the Catalan numbers in terms of closed walks on the "infinite path graph" $A_{\infty}$.  (Is this the correct notation?  I mean the infinite path graph with one end.)
Now, closed walks of length $2k$ from one end of $A_n$ to itself can be put in bijection with ordered rooted trees of depth at most $n$ and $k$ non-root vertices.  (Recall that the Catalan numbers also count ordered rooted trees of arbitrary depth.)  The generating function $P_n$ of ordered rooted trees of depth at most $n$ satisfies the recursion
$$P_1(x) = 1, P_n(x) = \frac{1}{1 - x P_{n-1}(x)}.$$
This is because an ordered rooted tree of depth $n$ is the same thing as a sequence of ordered rooted trees of depth $n-1$ together with a new root.  (Recall that the generating function of the Catalan numbers satisfies $C(x) = \frac{1}{1 - x C(x)}$.  In other words, $C(x)$ has a continued fraction representation, and $P_n$ is its sequence of convergents.)  On the other hand, since $P_n(x)$ counts walks on the graph $A_n$, it is possible to write down the generating function $P_n$ explicitly in terms of the characteristic polynomials of the corresponding adjacency matrices, and these polynomials have roots the eigenvalues $2 \cos \frac{\pi i}{n+1}$.  This implies that they must be the Chebyshev polynomials of the second kind, i.e. the ones satisfying
$$q_n(2 \cos x) = \frac{\sin (n+1) x}{\sin x}.$$
But the Chebyshev polynomials of the second kind are none other than the characters of the irreducible finite-dimensional representations of $SU(2)$!  In particular, they're orthogonal with respect to the Haar measure.  
The overarching question I have is: how does this sequence of computations generalize, and what conceptual framework ties it together?  But I should probably ask more specific sub-questions.
Question 1:  I remember hearing that the relationship between the Catalan numbers and the Chebyshev polynomials generalizes to some relation between moments, continued fractions, and orthogonal polynomials with respect to some measure.  Where can I find a good reference for this?  
Question 2:  I believe that adding another edge and considering the family of cycle graphs gives the sequence ${2k \choose k}$ and the Chebyshev polynomials of the first kind, both of which are related to $SO(2)$.  According to the SBS post, this is a "type B" phenomenon, whereas the Catalan numbers are "type A."  What exactly does this mean?  What would happen if I repeated the above computations for other Dynkin diagrams?  Do I get continued fractions for the other infinite families?  
Question 3:  Related to the above, in what sense is it natural to relate walks on the Dynkin diagram $A_n$ to representations of $SU(2)$?  This seems to have something to do with question #16026.  How do the eigenvectors of the adjacency matrices fit into the picture?  I want to think of the eigenvectors as "discrete harmonics"; does this point of view make sense?  Does it generalize?
As you can see, I'm very confused, so I would greatly appreciate any clarification.

REPLY [8 votes]: Re Question 1: For the connection between walks and continued fractions (which is closely related to orthogonal polynomials, but not, as far as I know, to root systems) see Philippe Flajolet, Combinatorial Aspects of Continued Fractions, Discrete Mathematics 32 (1980), pp. 125–161. Flajolet's paper has been very influential and many people have done further work in this direction.<|endoftext|>
TITLE: Computing (on a computer) higher ramification groups and/or conductors of representations.
QUESTION [13 upvotes]: I am supervising an undergraduate for a project in which he's going to talk about the relationship between Galois representations and modular forms. We decided we'd figure out a few examples of weight 1 modular forms and Galois representations and see them matching up. But I realised when working through some examples that computing the conductor of the Galois representation was giving me problems sometimes at small primes.
Here's an explicit question. Set $f=x^4 + 2x^2 - 1$ and let $K$ be the splitting field of $f$ over $\mathbf{Q}$. It's Galois over $\mathbf{Q}$ with group $D_8$. Let $\rho$ be the irreducible 2-dimensional representation of $D_8$. What is the conductor of $\rho$? Note that I don't particularly want to know the answer to this particular question, I want to know how to work these things out in general. In fact I think I could perhaps figure out the conductor of $\rho$ by doing calculations on the modular forms side, but I don't want to do that (somehow the point of the project is seeing that calculations done in 2 different ways match up, rather than using known modularity results to do the calculations).
Using pari or magma I see that $K$ is unramified outside 2, and the ideal (2) is an 8th power in the integers of $K$. To compute the conductor of $\rho$ the naive approach is to figure out the higher ramification groups at 2 and then just use the usual formula. But the only computer algebra package I know which will compute higher ramification groups is magma, and if I create the splitting field of $f$ over $\mathbf{Q}_2$ (computed using pari's "polcompositum" command)
Qx<x>:=PolynomialRing(Rationals());
g:=x^8 + 20*x^6 + 146*x^4 + 460*x^2 + 1681;
L := LocalField(pAdicField(2, 50),g);
DecompositionGroup(L);

then I get an instant memory overflow (magma wants 2.5 gigs to do this, apparently), and furthermore the other calculations I would have to do if I were to be following up this idea would be things like
RamificationGroup(L, 3);

which apparently need 11 gigs of ram to run. Ouch. Note also that if I pull the precision of the $p$-adic field down from 50 then magma complains that the precision isn't large enough to do some arithmetic in $L$ that it wants to do.
I think then my question must be: are there any computer algebra resources that will compute higher ramification groups for local fields without needing exorbitant amounts of memory? Or is it a genuinely an "11-gigs" calculation that I want to do?? And perhaps another question is: is there another way of computing the conductor of a (non-abelian finite image) Galois representation without having to compute these higher ramification groups (and without computing any modular forms either)?

REPLY [37 votes]: It's rather late in the day, but there's an easy way of getting the whole Hasse-Herbrand function $\varphi^K_{\mathbb{Q}_p}(x)$ if you know the minimal polynomial $F$ of a prime element $\pi$. First, you write down the copolygon (valuation function) of $F(X+\pi)$ using the valuation normalized to have $v(p)=1$, then you stretch it horizontally by a factor of $[K\colon\mathbb{Q}_p]$, then you move it down and to the left by one unit, to get the numberings consistent with Serre's convention. The vertices in KB's case are $(1,1)$, $(3,2)$ and $(5,5/2)$. I couldn't figure out the prime of the Galois closure till I saw the extension as quadratic over $\mathbb{Q}_2(\zeta_8)$. At any rate, the chain of fields corresponding to the ramification filtration is $\mathbb{Q}_2\subset\mathbb{Q}_2(i)\subset\mathbb{Q}_2(\zeta_8) \subset K$. Needless to say, you don't need any kind of powerful package to do this kind of computation.<|endoftext|>
TITLE: Teaching Methods and Evaluating them
QUESTION [12 upvotes]: Hey,
As a lowly graduate student, I'm on a committee (I'm not sure how important my role really is) trying to evaluate how effective different approaches teaching undergraduates. We are looking at all sorts of methods like the standard lecturing style seen in most schools, the Moore method, a Dewey/Montessori approach, Polya, etc. Now, my question is this: how do you go about developing some sort of effective criteria evaluating these different approaches?  
One of the reasons I'm asking is that I'm sitting on the committee with another graduate student, and we were discussing what we thought about our measure theory class. It was taught in a slightly different manner; basically the students gave the lectures to one another.  Our prof sat in the back and would occasionally ask a question directing us to a subtle point or an important topic to be covered. Now, I loved the class. It was challenging and forced me to work very hard to understand the material.  Further, I enjoyed reconstructing the entire theory with my classmates. And being asked pointed questions by my prof was traumatizing but also very good preparation for my quals.  Now, in contrast, my classmate didn't like the method at all, felt it went too slow, and he didn't think that many of the standard tricks, techniques or approaches that one ought to get out the class is what he actually got out of it.  I'm not interested in debating the merits of our respective conclusions, but what I am interested in is developing some sort of analysis that would help us objectively evaluate various approaches/methods to teaching students. Our dept is looking at implementing some of these methods next academic year.   We'd like to come up with a way of objectively evaluating the a particular method based off of firmer ground than anecdotal evidence and end of semester surveys.  In sum, I was wondering if anybody here had experience in this area of curriculum analysis/development? If so I'd love to hear about it.  
I hope this question isn't too far off from the mainstream questions of MO.
Regards,
Ben

REPLY [4 votes]: You may find it interesting to observe how Eric Mazur, professor at Physics at Harvard, determined that his own teaching methods were inadequate. The whole talk is fascinating in its description of an alternative teaching approach he has developed; However to your question I think the important point is the method used by Mazur to initially observe that something is not going well at all. 
Starting at 00:04:00, Prof. Mazur describes the initial criteria which led him to believe he was doing a good job (success at the end of term questionnaire). Then - more importantly - starting at 00:06:30, he describes how a survey of conceptual understanding helped him realize that something was very fundamentally wrong. The specific proposal is running an identical pre-course and post-course test which attempts to examine the quality of basic understanding of the concepts.<|endoftext|>
TITLE: Consequences of the Riemann hypothesis
QUESTION [146 upvotes]: I assume a number of results have been proven conditionally on the Riemann hypothesis, of course in number theory and maybe in other fields. What are the most relevant you know?
It would also be nice to include consequences of the generalized Riemann hypothesis (but specify which one is assumed).

REPLY [3 votes]: I haven't seen this application mentioned:
The closed horocycle $\gamma_l$ of length $l$ of the modular surface equidistributes to the hyperbolic volume form $\omega$ when $l\to +\infty$, and RH is equivalent to the error term:
$$\frac{1}{l} \int_{\gamma_l} f = \int f \omega  + o(l^{-3/4+\epsilon})$$
for every smooth function with compact support.
This is due to Don Zagier. https://people.mpim-bonn.mpg.de/zagier/files/scanned/EisensteinRiemannZeta/eisenstein-zeta-978-3-662-00734-1_10.pdf
Verjovsky has another reformulation of the Riemann hypothesis in terms of convergence of measures. https://projecteuclid.org/journals/kodai-mathematical-journal/volume-17/issue-3/Discrete-measures-and-the-Riemann-hypothesis/10.2996/kmj/1138040054.full<|endoftext|>
TITLE: To what extent does Spec R determine Spec of the Witt vector ring over R?
QUESTION [51 upvotes]: Let $R$ be a perfect $\mathbb{F}_p$-algebra and write $W(R)$ for the Witt ring [i.e., ring of Witt vectors -- PLC] on $R$. I want to know how much we can deduce about $\text{Spec } W(R)$ from knowledge of $\text{Spec } R$. I am especially interested in the case that $R$ is a domain.
Suppose that $R$ is a field. Then it's not hard to see that $W(R)$ is a DVR, so in this case the answer is completely known. For a general $R$, surjections $R \rightarrow R/\mathfrak{p}$ lift to surjections $W(R) \rightarrow W(R/\mathfrak{p})$, and as the latter is a domain, we get a copy of the spectrum of $R$ inside the spectrum of $W(R)$.
Sometimes, though, there are extra primes in the spectrum of $W(R)$! Here's the case that motivated my question. (These rings come up in the study of $(\phi, \Gamma)$-modules and in the construction of Fontaine's rings of periods.) Let $R$ be the "perfection" of $\mathcal{O}_{\mathbb{C}_p}/p$. That is, $R$ is the set of sequences $(x_i)$, $i \geq 0$, where each $x_i \in  \mathcal{O}_{\mathbb{C}_p}/p$ and $x_i^{p} = x_{i-1}$. Note that $R$ is a (non-discrete!) valuation ring; reducing mod the maximal ideal of $\mathcal{O}_{\mathbb{C}_p}$ and projecting to the first component gives a surjection to $\overline{\mathbb{F}_p}$. It's not hard to show $R$ is a domain.
There's three obvious prime ideals. First, as before, there's the prime ideal $(p)$, the kernel of the surjection $W(R) \rightarrow R$. Further, the universal property of Witt vectors gives a map $W(R) \rightarrow W(\overline{\mathbb{F}_p}) = \widehat{\mathcal{O}_{\mathbb{Q}_p^{ur}}}$ whose kernel is a prime. Finally there's the maximal ideal, which is the kernel of $W(R) \rightarrow R \rightarrow \overline{\mathbb{F}_p}$.  We've used only abstract reasoning about Witt vectors to find these.
But there's a fourth prime ideal! It turns out that there's a surjection $\theta: W(R) \rightarrow \mathcal{O}_{\mathbb{C}_p}$, which is critical in building Fontaine's rings. The existence of $\theta$ can't be deduced from the universal property of Witt vectors, since $\mathcal{O}_{\mathbb{C}_p}$ is not a strict $p$-ring. In the literature, all proofs that $\theta$ is a homomorphism "look under the hood" and actually think about the addition and multiplication of Witt vectors. Can this be abstracted? That is, is there a way to know which "extra" primes we'll get in $W(R)$, just from knowing $R$?

REPLY [2 votes]: This might not be the answer you expect, but there is a result saying that when $R$ is a non-discrete valuation ring, like the $\mathcal{O}_\mathbb{C_p}^\flat$ you mentioned, then there are infinitely many prime ideals in $W(R)$. The following two papers show that the Krull dimension of $W(R)$ is infinite, moreover, at least the cardinality of the continuum.

Lang and Ludwig - $\mathbb A_{\text{inf}}$  is infinite dimensional
Du - $\mathbb A_{\text{inf}}$ has uncountable Krull dimension<|endoftext|>
TITLE: Finite groups with centerless quotients
QUESTION [8 upvotes]: Is there a description of finite groups whose all quotients have trivial center? Is it true that only direct products of non-abelian simple groups have this property?

REPLY [7 votes]: The quotients of a finite group, G/N have minimal normal subgroups K/N.  These are called chief factors.  Chief factors are divided into two kinds, central chief factors and eccentric chief factors.  The central ones are precisely those such that K/N ≤ Z(G/N).  You are therefore asking for a classification of groups all of whose chief factors are eccentric.
Another way to say this that sounds clever is that G has no chief factor of prime order.
Such a group cannot have any central "top" factors, and in particular G must be a perfect group.  Perfect groups are not entirely easy to classify, but one standard view of them is to build them layer by layer, and under this view you simply never adjoin a central factor.
However, such a classification may not be very useful in concrete circumstances.  The things you can adjoin are any (fixed point free, that is, noncentral) G-module repeatedly to construct the solvable radical, and any wreath product of G with a non-abelian simple group.
For every permutation representation of A5, you can take A5 wr A5 to get such examples.  For any A5-module V (with no central A5-composition factors), you can take the semi-direct product of A5 with V.
There are 26 such groups (not counting G=1) of order at most 10,000; most are just the simple groups of those orders.  There are also 2^4:A5, 4^2:A5, 2^3:L3(2), 2^3.L3(2), A5 x A5, 3^4:A5, 3^4.A5, 2^4:A6, 5^3:A5, 5^3.A5.  The first wreath product example is A5 wr A5 of order 60^6.<|endoftext|>
TITLE: completeness axiom for the real numbers
QUESTION [7 upvotes]: Do any treatises on real analysis take the following as the basic completeness axiom for the reals?
"Let $A$ and $B$ be set of real numbers such that
(a) every real number is either in $A$ or in $B$;
(b) no real number is in $A$ and in $B$;
(c) neither $A$ nor $B$ is empty;
(d) if $\alpha \in A$, and $\beta \in B$, then $\alpha < \beta$.
Then there is one (and only one) real number $\gamma$ such that
$\alpha \leq \gamma$ for all $\alpha \in A$, and $\gamma \leq \beta$
for all $\beta \in B$."
This appears as Theorem 1.32 in Walter Rudin's "Principles of Mathematical Analysis", and can be traced back to Dedekind's "Continuity and Irrational Numbers" (section V, subsection IV).  Both Rudin and Dedekind derive this result from the construction of the reals via cuts of the rationals.
Authors who prefer to axiomatize the reals directly (instead of constructing them from the rationals) might be expected to take the above property as an axiom, but I haven't found anyone who does this.  Instead, they all assume the least upper bound property as an axiom, or the nested interval property, or the convergence of Cauchy sequences.
I personally think the way to go is to take Rudin's Theorem 1.32 as an axiom (because it is simple and compelling) and then derive the least upper bound property (since it is more useful in practice than 1.32) and then get to work building up the apparatus of real analysis.  But leaving aside the issue of whether this is the right way to go: have any authors taken this approach?
I should remark that the geometrical analogue of Theorem 1.32, characterizing the completeness of the line, appears to be well known to geometers (especially those interested in the foundations of geometry; see for instance Marvin Jay Greenberg's very nice article in the March 2010 issue of the Monthly).

REPLY [3 votes]: In his "A Course of Pure Mathematics" G H Hardy makes use of the Dedekind's theorem to prove almost all the usual theorems of real analysis and he prefers to avoid least upper bound property altogether. In my opinion this is more easily understandable and presentable. The least upper bound property looks high brow and has been championed by later authors unnecessarily.<|endoftext|>
TITLE: Is discrete mathematics mainstream?
QUESTION [18 upvotes]: Recently, the Department of Mathematics at
our University issued a recommendation encouraging its members
to publish their research in non-specialized, mainstream mathematical journals. For
numerical analysts this will make an additional obstacle for their promotions. But even
for discrete mathematicians this recommendation is causing concerns. 
For several top mainstream journals I checked with tools offered by MathSciNet what percentage
of discrete mathematical papers they published in recent years. Some statistics indicate that in some journals the number of papers with primary MSC classification, say 05 or 06 decreased significantly in the past 30 years. There are several possible explanations to this fact. 

The quality of research in DM is dropping.
The majority of research in discrete mathematics is so specialized that it is of no interest for the rest of mathematics
Some discrete mathematics journals attract even the best work of discrete mathematicians.  
Some top journals may be biased against discrete math.  
Maybe discrete math is no longer part of mainstream mathematics and will, like theoretical computer science, eventually develop into an independent body of research.

But the key issue is whether discrete math is nowadays perceived as mainstream mathematics.

REPLY [10 votes]: You mention numerical analysis explicitly. From my point of view, numerical analysis has a completely different set of top-flight generalist journals than traditional mathematics. People here care to publish in SIAM Review, in Inverse Problems, in the other slightly less generalist SIAM journals (just to list a few).
If I had written an excellent paper in my field, it wouldn't even cross my mind to submit it to Acta or Inventiones; I'd find it no more appropriate than, say, Nature or  The new England J of Medicine. Surely it would strike me as peculiar if someone asked me to publish in those.
Actually I think many people I meet in conferences don't even know which are the best pure-math journals. Just for fun, I tried looking on MathSciNet for papers by past or present big names in NA published in top-flight pure math journals; it took me a few minutes to find even one of them.
Does this make us a completely different discipline that lives under the same roof (and competes for the same funding and positions) as pure mathematicians? Maybe, I don't know. It's a matter of definitions. 
OK, now that I've written this, I'd better go and check scicomp.stackexchange.com --- there is not enough applied maths here. :)<|endoftext|>
TITLE: Permutation representation inner product
QUESTION [10 upvotes]: Let $\rho : S_n \rightarrow \text{GL}(n, \mathbb{C})$ be the homomorphism mapping a permutation $g$ to its permutation matrix. Let $\chi(g) = \text{Trace}(\rho(g))$.
What is the value of $\langle \chi, \chi \rangle = \displaystyle \frac{1}{n!} \sum_{g \in S_n} \chi(g)^2$ ? Computing this expression for small $n$ yields $2$. Is this always true?

REPLY [7 votes]: Much as I like Burnside's Lemma, induced (permutation) representations, and other parts of group theory, I can't resist pointing out an alternative argument that uses essentially no group theory but relies on the fact that expectation (of random variables) is linear.  Since $\chi(g)$ is the number of fixed-points of $g$, its square is the number of fixed ordered pairs $(x,y)$, where of course fixing a pair means fixing both its components.  So the $\langle\chi,\chi\rangle$ in the question is the average number of fixed pairs of a permutation $g$, in other words the expectation (with respect to the uniform probability measure on $S_n$) of the random variable "number of fixed pairs."  That random variable is the sum, over all pairs $(x,y)$, of the indicator variable $F_{x,y}$ whose value at any permutation $g$ is 1 or 0 according to whether $g$ fixes $x$ and $y$ or not.  So $\langle\chi,\chi\rangle$ is the sum, over all $x,y$,  of the expectations of these $F_{x,y}$, and these expectations are just the probabilities that a random permutation fixes $x$ and $y$.  For each of the $n$ pairs where $x=y$, that probability is $1/n$, so all these together contribute 1 to the sum.  For each of the remaining $n^2-n$ pairs, the probability is $(1/n)(1/(n-1))$ (namely, probability $1/n$ to fix $x$ and conditional probability $1/(n-1)$ to fix $y$ given that $x$ is fixed).  So these pairs also contribute 1 to the sum, for a total of 2.<|endoftext|>
TITLE: Which part of physical B model is not rigorous?
QUESTION [5 upvotes]: Which part of physical B model is not rigorous? the physical theory of B model,if it is not mathematical rigorous just because the Feynman integral,but it looks like for me the space is finite dimensional, what the problem is causing that it still not rigorous?

REPLY [3 votes]: To define (as Kevin Lin does above) the B-model purely as the derived category of coherent sheaves is fine and rigorous, but it ignores the higher-genus aspects of mirror symmetry -- which was the original question.  As I wrote above, Kevin Costello gives a rigorous description of the higher-genus amplitudes, but it is still conjectural whether this agrees with the physics.  The issue is that higher-genus string amplitudes depend on an integration over the moduli space of Riemann surfaces (or a space of maps from them, depending on the model), and this demands compactification.  The full, non-topological theory is of course an ordinary two-dimensional quantum field theory, with all the usual difficulties in making the path integral rigorous.<|endoftext|>
TITLE: Fundamental domains of measure preserving actions
QUESTION [5 upvotes]: Suppose a finite group $G$ acts on a standard probability space $(X, \mu)$ by measure-preserving actions (i.e. $\mu(g(A)) = \mu(A)$ for all $g \in G$ and $A \subset X$ measurable).  In addition, suppose that for $g \in G$ and $g$ not the identity then $\mu(\{x:g(x) = x\})=0$.  I am wondering if it is always possible to find a fundamental domain $D$ of the action of $G$, i.e. is there a measurable $D \subset X$ such that $G$ is the disjoint union (up to measure 0) of the sets $g(D)$?

REPLY [2 votes]: Here is another way to find a fundamental domain.
First identify $X$ with $[0,1]$. You want to pick a single point in each orbit of the action. Just take the smallest one.
Let be more specific. 
Consider the set $A$ of points $x$ for which the number of points in the orbit of $x$ is equal to the cardinal of $G$. Your assumption insures that this set is of full measure. Take some Borel subset $B$ in $A$ of full measure in $A$, such that $G$ acts on $B$ through Borel transformations.
The fundamental domain $D$ is then defined as the image of $B$ by the map $x \rightarrow  \min\lbrace\ gx\ |\ g\in G\ \rbrace$.
Now the image of a Borel set by a Borel map is always measurable. Restricting again to a Borel subset of full measure, we get  a Borel fundamental domain for the action.<|endoftext|>
TITLE: The middle eigenvalues of an undirected graph
QUESTION [17 upvotes]: Let $ \lambda_1 \ge \lambda_2 \ge  \dots \lambda_{2n} $ 
be the collection of eigenvalues of an adjacency matrix of an undirected graph $G$ on $2n$ vertices. I am looking for any work or references that would consider the middle eigenvalues $\lambda_n$ and $\lambda_{n+1}$. In particular, the bounds on  $R(G) = max(|\lambda_n|,|\lambda_{n+1}|)$ are welcome. 
For instance, a computer search showed that most connected graphs with maximum valence 3 have $R(G) \le 1$. The only known exception is the Heawood graph.
Motivation for this question comes from theoretical chemistry, where the difference $\lambda_n-\lambda_{n+1}$ in Hueckel theory is called the HOMO-LUMO gap.
Edit: Note that in the original formula for $R(G)$ the absolute value signs were missing.

REPLY [3 votes]: First, regarding the mathematics of the middle-eigenvalue problem there are results for the case of a graph G with just a single perfect matching, especially if also the graph is bipartite.  See D. J. Klein & A. Misra, Croat. Chim. Acta 77 (2002) 179-191, where a transform (called "Kekulean") from an original G with a single perfect matching is made to another graph G' with signed edge weights such that the adjacency matrix A(G') is the inverse of the original A(G). Circumstances are identified where the signs on G' may be eliminated to leave an ordinary unsigned graph G" still with eigenvalues inverse to those of G. Further circumstances are found where G & G" are isomorphic.  Techniques for dealing with maximum eigenvalues (of A(G') or A(G")) are used to give information on the "middle" eigenvalues (of A(G)).
Second, some comments might be made on the chemical context.  The adjacency-matrix eigenvalues nearest 0 are much considered in chemistry, as they locate the electrons most easily excited and the eigenvalue difference for middle eigenvalues gives an appropriate measure of the energy needed for excitation. In more detail, the eigenvalues of the adjacency matrix provide crude (Huckel-theoretic) estimates of 1-electron energies for the pi-electrons of conjugated carbon networks -- the other electrons being for the most part more tightly bound.  A full N-electron energy is then just a sum over 1-electron energies each multiplied by an occupation number n(e) for that eigenvalue e -- the occupation numbers taking values 0, 1, or 2 and summing to N.  For an electrically neutral conjugated-carbon network (as is a common circumstance), the total number N of such electrons matches the number of sites.  Then the most favored N-electron state for N=even has the N/2 largest eigenvalues e each with n(e)=2 and all other eigenvalues e' with n(e')=0.  For odd N=2k+1, the k largest e have n(e)=2, the next lowest eigenvalue e' has n(e')=1, and the k remaining lower eigenvalues e" have n(e")=0.  The gap of interest is the least energy difference between a level not doubly occupied and another not empty.  [A point of potential confusion is that the 1-electron eigenvalues are proportional to the eigenvalues with a proportionality which is negative; then the favorable ("ground-state") circumstance of occupying the largest eigenvalues corresponds to occupying the lowest energy 1-electron levels.]  From all this commentary it can be seen that for bipartite graphs the "middle" eigenvalues of interest are those nearest 0, while for nonbipartite graphs this is not necessarily so.  The chemical graphs of interest for conjugated-carbon networks have vertices just of degrees 1, 2, or 3.<|endoftext|>
TITLE: Bipartite Nim-Geography
QUESTION [13 upvotes]: Two players are playing a game on a bipartite graph where all of the edges are nim-heaps of various sizes. A token starts on one of the vertices, and on your turn you must move the token over an edge and pick up some of the matchsticks in the nim heap corresponding to that edge. If all of the edges meeting the vertex containing the vertex are empty when you start your turn, you lose.
Is there a polynomial time algorithm to decide this game?
Motivation:
I am interested in a multiplayer combinatorial game played as follows: between every pair of people sits a combinatorial game. One player starts with a "hot potato" token. Whoever has the token must make a move on one of the games incident to him to pass the token to the other player playing that game. If he can't make a move in any of the games adjacent to him, he loses and everybody else wins.
Since multiplayer games are hard to analyze, we can simplify the problem by splitting the players into two teams, where each team wants the potato to end up in the hands of the other team. To simplify the game further, we might also assume that no two players on the same team have a game between them. Even then, if all of the games are numbers then this problem is directed bipartite geography which I'm pretty sure is PSPACE-complete, so I'm assuming all of the games are impartial as well.
If all of the nim-heaps have size one or zero, though, then this problem becomes undirected bipartite geography, which can be solved in polynomial time, so I strongly suspect that bipartite nim-geography should have a polynomial time solution as well.

REPLY [4 votes]: I guess you already know this, but I just stumbled on this paper that studies the exact same game you described:
M. Fukuyama, A Nim game played on graphs, Theoret. Comput. Sci. 304 (2003), 387–399.
$\text{Section 3}$ of this paper studies the Nim game on a simple bipartite graph in which all vertices on one side are of degree (at most) $2$. At the start of the game, the token is assumed to be on a vertex on the side without the degree condition. I put the term "at most" in parentheses because this is proved to be irrelevant in the paper so that you can safely assume that all vertices on the restricted side are of degree $2$. The author gives necessary and sufficient conditions for the first player to have a winning strategy.
To decide if the first player can win by using the necessary and sufficient conditions given in this paper, if my understanding through skimming the paper is correct, you check if the following three conditions hold for each vertex $u$ on the side with the degree restriction:

The number $h$ of matchsticks in the nim heap on one edge with $u$ as its endpoint is different from that of the other edge with $u$ as its endpoint, (In the following, $h$ is assumed to be smaller.)
Split the vertex $u$ into $u_1$ and $u_2$ and w.l.o.g. assume that the unique edge of which $u_1$ is an endpoint after splitting is of weight $h$. The minimum capacity of $u_1$-$u_2$ cuts is equal to $h$.
For any minimum $u_1$-$u_2$ cut, the vertex the token is currently at is connected to $u_2$ (i.e., the one with more matchsticks on its edge).

Checking the first condition is trivial. The second one only needs to compute the minimum capacity, which can be done in polynomial time. For the third condition, while checking if two vertices are connected in a graph can be done quickly, we may have to list all minimum $u_1$-$u_2$ cuts. Google got me this paper that mentions an algorithm for listing all minimum $s$-$t$ cuts for fixed $s$ and $t$, where the run time depends on the total number of minimum $s$-$t$ cuts for all pair $s, t$ in the vertex set, which can be exponential... The details are relegated to another technical report by the same authors (Ref. [GN1] in the linked PDF), and there seems to be an improvement on this in said technical report. But I couldn't find it online.
I looked for a similar paper that doesn't impose the degree condition, but my google-fu failed me.<|endoftext|>
TITLE: Counting colored rook configurations in the cube - when is it even?
QUESTION [10 upvotes]: Informal Statement
In the $n\times n \times n$ grid, we can places rooks (those from chess) such that no two rooks can attack each other.  One way to achieve this is to place a rook in position $(i,j,k)$ if and only if $i+j+k=0\mod n$.  In general, there are "many" ways to do this.
Each such "attack-free" rook position can be colored with $c$ colors.  When we fix an $i$, we can then count the colors in the matrix $(i,.,.)$, and can do similarly for each $j$ and $k$.  Call this set of tuples of colors-counts the "color profile".  For each color profile, there is either an even or odd number of colored rook positions that achieve it.  I want to know the largest $c$ such that all color profiles have an even number of colored rook positions achieving it.  In particular, I want to say that $c=\omega(n)$. This question came up in some complexity theory research, but the question seems interesting in its own right.

Formal Statement
Define $[n]$ to be the set ${1,\ldots, n}$, and define $[n]^3=[n]\times[n]\times[n]$.  Define a $c$-coloring of a set $S\subseteq[n]^3$ to be a function $C:S\to[c]$.  We can say that this is a $c$-coloring of $[n]^3$ with the convention that $C(i,j,k)=0$ for $(i,j,k)\notin S$.  A $c$-coloring $C$ induces a color profile $P$, which is a function from $P:[n]\times[3]\times[c]\to[n]$, via the rules

$P(i,1,c)$ is the number of $(j,k)\in[n]^2$ such that $C(i,j,k)=c$.

$P(j,2,c)$ is the number of $(i,k)\in[n]^2$ such that $C(i,j,k)=c$.

$P(k,3,c)$ is the number of $(i,j)\in[n]^2$ such that $C(i,j,k)=c$.


where we keep in mind the convention above on $(i,j,k)\notin S$.
Call a set $S\subseteq [n]^3$ to be a rook set, if

for all $i,j\in[n]$, there is exactly one $k\in[n]$ such that $(i,j,k)\in S$

for all $j,k\in[n]$, there is exactly one $i\in[n]$ such that $(i,j,k)\in S$

for all $i,k\in[n]$, there is exactly one $j\in[n]$ such that $(i,j,k)\in S$


Let a colored rook set $C_S$ correspond the coloring $C$ of a rook set $S$.
Define $N(P)$ to be the number of colored rook sets $C_S$ that induce the color profile $P$.
The question is:

For each fixed $n$, what is the largest $c$ such that for all color profiles $P$, $N(P)\equiv 0\mod 2$? In particular, is the largest $c$ asymptotically $\omega(n)$?


What I know
It should be clear that this problem can be defined analogously in any dimension, and I'm interested in this more general question.  I state it with $d\ge3$ because I can solve the $d=2$ case exactly.  In particular

For each $n$, for any $c\le n-1$, and any color profile $P$ on grid $[n]^2$, $N(P)\equiv 0 \mod 2$.  For $c\ge n$, there are profiles $P$ where $N(P)\equiv 1\mod 2$.

This can be proven by exhibiting a bijection between colored rook sets (which in d=2 are just permutation matrices).  Specifically, using the pigeonhole principle $c\le n-1$ implies that there are two rooks with the same color.  If they were at positions $(i,j)$ and $(i',j')$, then we replace them with the rooks of the same color at positions $(i,j')$ and $(i',j)$. (Of course, one needs to make this well-defined to ensure a bijection.)

Possible Methods
I see two possible methods of proof

generalize the above bijection proof to the 3-dimensional case

I don't know how to use the pigeonhole principle to get such an extension, but it seems possible that there is a method to show that some motif exists in any colored rook set, and then argue that we can alter this motif to get the bijection


define a system of polynomials (over $\mathbb{F}_2$) such that the solution set corresponds to exactly the colored rook sets inducing a color profile $P$.  Then try to apply the Chevalley-Warning theorem.

I've tried this, but can't seem to get systems of polynomials where the sum of the total degrees is strictly less than the number of variables, so the C-W theorem does not apply.
One can observe the the C-W theorem is an "iff" here: if $N(P)$ is even then there is a multilinear polynomial with degree strictly less than the number of variables, such that the solution set encodes those $C_S$'s that induce $P$.


Instead of using "rook sets", one can ask the question for other classes of subsets of $[n]^3$.  I'd be happy with establishing $c=\omega(n)$ for any class of subsets (although I'd like to be able to compute at least one example of such a subset efficiently).



Are there other methods for counting modulo two that I missed?

REPLY [12 votes]: This can be phrased as a problem concerning Latin squares.  Eg. a "rook set" is equivalent to a Latin square.  For example:
123       100 010 001
231  <->  001 100 010
312       010 001 100

A colouring of the Latin square is a partition of its entries (corresponding to a coloured rook set).  
We can therefore readily construct colour profiles P with c=n2 such that N(P)=1 (that is, by partitioning some Latin square into n2 parts).  But we can do much better...
A defining set is a partial Latin square with a unique completion.  A critical set is a minimal defining set.  Let scs(n) be the size of a smallest critical set for Latin squares of order n.  From a Latin square L containing a critical set of size scs(n), we can choose a partition (i.e. a c-colouring) such that the entries in the critical set are in parts of size 1 and the remaining entries of L are in a single part.  This also will give rise to a colour profile P in which N(P)=1.  It has been shown that scs(n)≤n2/4 for all n.  (see J. Cooper, D. Donovan and J. Seberry, Latin squares and critical sets of minimal size, Australasian J. Combin 4 (1991), 113–120.)  Hence we can deduce that for some c<=n2/4+1 we have N(P)≡1 (mod 2).  But with a more intelligent choice of the partition, we can do better...
In fact, we can use the constructions in Cooper et al. to construct a colour profile P with c=n for which N(P)=1.  I will only be able to prove this by example here (but it should be clear it can be readily generalised):  Partition the "back circulant" Latin square of order 6 as follows.
123456     100000   020000   003000   000000   000000   000456
234561     000000   200000   030000   000000   000000   004561
345612 <-> 000000 + 000000 + 300000 + 000000 + 000000 + 045612
456123     000000   000000   000000   000000   000000   456123
561234     000000   000000   000000   000004   000000   561230
612345     000000   000000   000000   000040   000005   612300

Now observe that the three colour profiles are:
100005  111003  111003
020004  011004  011004
003003  001005  001005
000204  000006  000006
000015  000105  000105
000006  000114  000114

Together these form P.  Given P, we can observe that any Latin square of order 6 with colour profile P must contain the following partial Latin square:
123...
23....
3.....
......
.....4
....45

Which is a critical set -- and therefore admits a unique completion (that is, to the "back circulant" Latin square of order 6).  Therefore, if c=n there exists a color profile P for which N(P)=1, not just N(P)≡1 (mod 2).
EDIT: I presented this problem to our research group at Monash and we improved the upper bound to c=1 (which was a bit surprising!).  The idea came from the following critical set (call it C) of the back-circulant Latin square, which is related to the one above, but contains more entries than the one above (but this doesn't matter for this problem).
12345.
2345..
345...
45....
5.....
......

We observed that if every entry in the critical set above is assigned one colour, and the remaining entries another colour, then there is a unique Latin square with that colour profile.  That is, we derive the colour profile from in the following way.
123456     12345.   .....6
234561     2345..   ....61
345612 <-> 345... + ...612
456123     45....   ..6123
561234     5.....   .61234
612345     ......   612345

which gives rise to the following colour profile P:
15   51   51
24   42   42
33   33   33
42   24   24
51   15   15
06   06   06

We can deduce that any Latin square with the colour profile P will, in fact, contain the critical set C.  Hence, N(P)=1.
To see how we deduce the critical set from the colour profile, we note that for the first colour, it contains 5 entries in the first row and column, 4 entries in the second row and column, and so on (and the same for columns).  This selection of cells can take only one shape -- that is, it is unique.  Then placing the symbols 1..5 in it can only be achieved in one way (to preserve the Latin property).
Since this can be generalised for all n≥2, the answer to your question "what is the largest c such that for all colour profiles P, N(P)≡0 (mod 2)" is c=1.
EDIT 2:  We also discussed at our research meeting the complexity side of the problem.
Instance: colour profile P
Question: is N(P)≥1?
In fact, there is an easy way to see that this problem is NP-complete, since (a) we can embed instances of the problem of partial Latin square completion in the above problem and (b) a Latin square together with its colouring can be used as a certificate.
The problem of partial Latin square completion was shown to be NP-complete in:  C. J., Colbourn, The complexity of completing partial Latin squares, Discrete Appl. Math. 8 (1984), no. 1, 25--30.<|endoftext|>
TITLE: What's the current state of Yang–Mills mass gap question?
QUESTION [11 upvotes]: What's the current state of Yang–Mills mass gap question, is there any place that does this problem? Especially I want to know if there is any progress (out of that mentioned in the introduction article by Witten and Jaffe). Is it too hard for a mathematician? Thanks!

REPLY [2 votes]: There is a rather strange situation here because a russian mathematician, Alexander Dynin, is claiming to have solved it. Please, refer to this question of mine or this paper published in a reputable journal. Till now I have not seen any reaction from the community but I think that an analysis of the work of this author is overdue. Indeed, he has worked on this since 2009 as I can see from arxiv.<|endoftext|>
TITLE: $K_0$ of a non-separated scheme
QUESTION [9 upvotes]: This question is on "computing" the Grothendieck group of the projective $n$-space with $m$ origins ($m\geq 1$). For any (noetherian) scheme $X$, let $K_0(X)$ be the Grothendieck group of coherent sheaves on $X$.
Firstly, let me sketch that $K_0(\mathbf{P}^n) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^n)$.
Let $X=\mathbf{P}^n$ be the projective $n$-space. 
(I omit writing the base scheme in the subscript. In fact, you can take any noetherian scheme as a base scheme in the following, I think.)
Let $H\cong \mathbf{P}^{n-1}$ be a hyperplane with complement $U\cong \mathbf{A}^n$. By a well-known theorem on Grothendieck groups, we have a short exact sequence of abelian groups $$K_0(H) \rightarrow K_0(X) \rightarrow K_0(U) \rightarrow 0.$$ Now, let $i:H\longrightarrow X$ be the closed immersion. Then the first map in the above sequence is given by the "extension by zero", which in this case is just the K-theoretic push-forward $i_!$, or even better, just the direct image functor $i_\ast$. Now, there is a projection map $\pi:X\longrightarrow H$ such that $\pi\circ i = \textrm{id}_{H}$. 
By functoriality of the push-forward, we conclude that $\pi_! \circ i_\ast = \pi_! \circ i_! = \textrm{id}_{K_0(H)}$.
Therefore, we may conclude that $i_\ast$ is injective and that we have a split exact sequence $$0 \rightarrow K_0(H) \rightarrow K_0(X) \rightarrow K_0(U) \rightarrow 0.$$ Thus, we have that $K_0(\mathbf{P}^n) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^n)$.
Q1: Let $\mathbf{P}^{n,m}$ be the projective $n$-space with $m$ origins ($m\geq 1$). For example, $\mathbf{P}^{n,1} = \mathbf{P}^n$. (Again the base scheme can be anything, I think.) Now, is it true that  $$K_0(\mathbf{P}^{n,m}) \cong K_0(\mathbf{P}^{n-1,m}) \oplus K_0(\mathbf{A}^n)?$$ 
Idea1: Take a hyperplane $H$ in $\mathbf{P}^{n,m}$. Is it true that $H\cong \mathbf{P}^{n-1,m}$ and that its complement is $\mathbf{A}^n$? Also, even though the schemes are not separated, the closed immersion $i:H\longrightarrow \mathbf{P}^{n,m}$ is proper, right? Also, is the projection $\pi:\mathbf{P}^{n,m}\rightarrow H$ proper? If yes, the above reasoning applies. If no, how can one "fix" the above reasoning? I think that in this case one could still make sense out of $i_!$ and $\pi_!$ (even if they are not proper maps.)
Idea2: Maybe it is easier to show that $K_0(\mathbf{P}^{n,m}) \cong K_0(\mathbf{P}^{n-1})\oplus K_0(\mathbf{A}^{n,m})$, where $\mathbf{A}^{n,m}$ is the affine $n$-space with $m$ origins. Then one reduces to computing $K_0(\mathbf{A}^{n,m})$...
Idea3: One could also take $m=2$ as a starting case and look at the complement of one of the origins. Then we get a similar exact sequence as above and one could reason from there.
Which of these ideas do not apply and which do? 
Note: Suppose that the base scheme is a field. Since $K_0(\mathbf{A}^n) \cong \mathbf{Z}$ and $K_0(\mathbf{P}^n) \cong \mathbf{Z}^{n+1}$, this would show that $$K_0(\mathbf{P}^{n,m}) \cong \mathbf{Z}^{n+m}.$$ More generally, if $S$ is the base scheme, $K_0(\mathbf{P}^{n,m}) \cong K_0(S)^{n+m}$.

REPLY [6 votes]: This is exactly the sort of example where it is relevant which version of K-theory you are employing. (In particular, which theorems you can employ here depends on this!) The issue here is a tad subtle.
To illustrate, let's restrict attention to the case of affine $n$-space $X$ with a doubled origin ($n\geq 2$). (An iteration of the discussion below, combined with your localization argument in Idea2, will let you address your general multiply-origined projective spaces.) For convenience, let me assume that the base is a regular noetherian scheme $S$.
Let me first answer your question as it was asked. Quillen's localization sequence gives fiber sequences of spectra or spaces
$$G(S)\to G(\mathbf{A}_S^n)\to G(\mathbf{A}_S^n-\{0\})$$
and
$$G(S)\to G(X)\to G(\mathbf{A}_S^n)$$
Here I'm using $G$-theory is the $K$-theory spectrum (or space) of coherent sheaves. This has the property that $G(S)\simeq G(\mathbf{A}_S^n)$. Putting this together, we see that
$$G(X)\simeq G(\mathbf{A}_S^n)\times G(S)\simeq G(S)\times G(S)$$
This answers your question for $G$-theory; in particular $\pi_0$ of $G$ is the Grothendieck group you seek, and so we conclude that $G_0(X)=G_0(S)\oplus G_0(S)$. Now use your localization arguments to get that $G_0(\mathbf{P}^{n,2})$ is $n+2$ copies of $G_0(S)$.
However: for $K$-theory, the way this computation gets done depends critically on which model of $K$ you use.
(A) If we define $K$-theory as in Thomason-Trobaugh (as the Waldhausen $K$-theory spectrum of perfect complexes), then in this case $K(X)\simeq G(X)$ and $K(S)\simeq G(S)$. (This follows from "Poincaré Duality;" see the end of section 3 of Thomason-Trobaugh.)
(B) If, however, we define $K^{\mathrm{naive}}(X)$ as the Quillen $K$-theory of the category of algebraic vector bundles on $X$, then things look different. [N.B. that this is the name given by Grothendieck, Illusie, et al. in SGA 6. It is not meant to be insulting!] Indeed, the inclusion $\mathbf{A}_S^n\to X$ induces an equivalence between the catgories of algebraic vector bundles on $X$ and those on $\mathbf{A}_S^n$, since the origin has codimension at least $2$. So then
$$K^{\mathrm{naive}}(X)\simeq K^{\mathrm{naive}}(\mathbf{A}_S^n)\simeq K(\mathbf{A}_S^n)\simeq K(S).$$
The second equivalence here follows from the fact that "naive" $K$-theory agrees with $K$-theory for schemes that admit an ample family of line bundles (see section 3 of Thomason-Trobaugh). The difference between (A) and (B) here reflects the failure of our $X$ to admit such a family.
An example like this is discussed is Thomason-Trobaugh I think at the end of section 8 (wherever they talk about Mayer-Vietoris). [Sorry, this is from memory; I don't have a copy of the Festschrift handy.]
Hope this helps!<|endoftext|>
TITLE: Quick proof of the fact that the ring of integers of $\mathbb Q(\zeta_n)$ is $\mathbb Z[\zeta_n]$?
QUESTION [29 upvotes]: I cannot find a good reference for the proof that the ring of integers in a cyclotomic field $\mathbb{Q}(\zeta_n)$ is $\mathbb{Z}[\zeta_n]$. The proof I usually find does an induction on the number of prime factors of $n$, coupled with a lengthy and somewhat computational proof in the case where $n$ is the power of a prime.
Do you know a quicker and possibly more conceptual approach?

REPLY [14 votes]: The ring of integers $R_n$ of ${\mathbf Q}(\zeta_n)$ contains ${\mathbf Z}[\zeta_n]$ as a subring with finite index.  To show the containment of rings is an equality, it suffices to show the inclusion ${\mathbf Z}[\zeta_n] \rightarrow R_n$ becomes an isomorphism after tensoring with ${\mathbf Z}_p$ for all $p$, and this basically boils down to showing the ring of integers of ${\mathbf Q}_p(\zeta_n)$ is ${\mathbf Z}_p[\zeta_n]$ for all $p$.  Now you have to know something about how to compute rings of integers in unramified and totally ramified extensions of local fields.  I'm leaving off some details here, admittedly, and since I used the terrible word "compute" maybe this isn't an answer you are looking for.
You didn't tell us whether you were okay with the inductive argument but disliked (apparently) the prime-power case or you were unhappy with both aspects.<|endoftext|>
TITLE: Results about the order of a group forcing a particular property.
QUESTION [14 upvotes]: Given a group of order $n$ where $n$ is either a specific number, or a number of a particular form, e.g. square-free, when does $n$ completely determine a particular group property among all groups of that order? Vipul's group theory wiki has several stubs on this topic, and in the language of his wiki, I will call this a $P$-forcing number, where $P$ is a particular group theoretic property.
We already have quite a few easy examples, for instance, orders $pq$, $pqr$, and $p^2q$ force solvability, and $p^2$ forces abelian. Then there are more specific results like 99 is an abelian-forcing number.
I am interested in general, in any results of this flavor beyond what would be considered a common result in a standard graduate-level group theory book.

REPLY [10 votes]: My very first original group-theory theorem is of this type, although I cannot guarantee that this fact was not observed previously. The (easy) result is this:
Let $|G| = p^3q$, where $p$ and $q$ are primes. Then $G$ has a normal Sylow subgroup unless $|G| = 24$. (In fact, the symmetric group $S_4$ is the unique exceptional group of order $24$.)<|endoftext|>
TITLE: Does the beth function have fixed points of arbitrarily large cofinality?
QUESTION [6 upvotes]: Background
The beth function is defined recursively by: $\beth_0 = \aleph_0$, $\beth_{\alpha + 1} = 2^{\beth_\alpha}$, and $\beth_\lambda = \bigcup_{\alpha < \lambda} \beth_\alpha$. Since the beth function is strictly increasing and continuous, it is guaranteed to have arbitrarily large fixed points by the fixed-point theorem on normal functions. 
The cofinality of an ordinal $\alpha$ is the smallest ordinal $\beta$ such that there are unbounded increasing functions $f : \beta \to \alpha$.

REPLY [5 votes]: Yes, the κth fixed point will have the same cofinality as κ, since the κ many earlier fixed points are cofinal in κ. 
More generally, for any function f from the ordinals to the ordinals, it is not difficult to see that the collection of ordinals α for which f " α subset α, that is, the closure points of f, will form a closed unbounded class C.  And for any limit ordinal β, the β-th element of C has the same cofinality as β, since it is the limit of the β many smaller elements of C. 
In particular, if we consider the beth function, where α maps to Bethα, then β is a closure point of this function if and only if  Bethα < β for all α < β, and this implies Bethβ = β. So the club C in this case consists of the Beth fixed points.

REPLY [4 votes]: Yes. The definable class $C$ of fixed points $\beth_\kappa = \kappa$ is closed and unbounded and therefore contains ordinals of all possible cofinalities. Specifically, let $\kappa_\alpha$ denote the $\alpha$-th element of $C$. Note that $\kappa_\delta = \sup_{\alpha<\delta} \kappa_\alpha$ for every limit ordinal $\delta$. If $\lambda$ is a regular cardinal then $\kappa_\lambda$ has cofinality $\lambda$.<|endoftext|>
TITLE: Is it possible to dissect a disk into congruent pieces, so that a neighborhood of the origin is contained within a single piece?
QUESTION [30 upvotes]: Problem: is it possible to dissect the interior of a circle into a finite number of congruent pieces (mirror images are fine) such that some neighbourhood of the origin is contained in just one of the pieces?

It may be conceivable that there is some dissection into immeasurable sets that does this. So a possible additional constraint would be that the pieces are connected, or at least the union of connected spaces.
A weaker statement, also unresolved : is it possible to dissect a circle into congruent pieces such that a union of some of the pieces is a connected neighbourhood of the origin that contains no points of the boundary of the circle?
This is doing the rounds amongst the grads in my department.  So far no one has had anything particularly enlightening to say - a proof/counterexample of any of these statements, or any other partial result in the right direction would be much obliged!
Edit: Kevin Buzzard points out in the comments that this is listed as an open problem in Croft, Falconer, and Guy's Unsolved Problems in Geometry (see the bottom of page 87).

REPLY [18 votes]: A new paper was posted to the arXiv on related questions: 
"Infinite families of monohedral disk tilings," by Joel Haddley and Stephen Worsley
(arXiv abs.).
Here are tilings of the disk into congruent pieces where at least one piece
does not touch the center (a result mentioned by Anton Geraschenko):



They conjecture (with extensive support) that,

"for any monohedral tiling of the disk, the centre may only intersect a tile at a vertex."

This would answer the original posed question in the negative.<|endoftext|>
TITLE: Where to publish computer computations
QUESTION [20 upvotes]: In a paper I developed some theory; some of the applications require extensive computations that are not part of the paper. I wrote a Mathematica notebook. I want to publish a PDF and .nb version somewhere to refer to from the paper. arXiv.org seems a good choice, but they won't accept the .nb file. I do not want to put this through peer review. Where to publish it?

REPLY [16 votes]: The problem with storing the files on your personal website is that it is not clear how long that website will remain valid.  What happens if you move to a new institution and your old account gets deactivated?  The arXiv solution pointed out by Scott Morrison is better.
Here is another possibility: Some journals that publish an electronic version in addition to (or instead of) a paper version will allow you to include ancillary files with the electronic version of an article.  If you'll excuse the self-promotion, see this journal's website, for example.<|endoftext|>
TITLE: Italian school of algebraic geometry and rigorous proofs
QUESTION [56 upvotes]: Many of the amazing results by Italian geometers of the second half of the 19th and the first half of the 20th century were initially given heuristic explanations rather than rigorous proofs by their discoverers. Proofs appeared only later. In some cases, an intuitive explanation could be more or less directly translated into modern language. In some other cases, essentially new ideas were required (e.g., among others, the classification of algebraic surfaces by Shafarevich's seminar; construction of the moduli spaces of curves and their projective compactifications by Deligne, Mumford and Knudsen; solution of the Luroth problem by Iskovskikh and Manin).
I would like to ask: what are, in your opinion, the most interesting results obtained by pre-1950 Italian geometers which still do not have a rigorous proof?
[This is a community wiki, since there may be several answers, none of which is the "correct" one; however, please include as many things as possible per posting -- this is not intended as a popularity contest.]
[upd: since I'me getting much less answers that I had expected (in fact, only one so far), I would like to clarify a couple of things: as mentioned in the comments, I would be equally interested in results which are "slightly false" but are believed to be essentially correct, e.g. a classification with a particular case missing etc. I'm also interested in natural generalizations that still haven't been proven such as extending a result to finite characteristic etc.]

REPLY [2 votes]: I think that the Italians geometers conjectured a generalization of Terracini's lemma which, if I am not mistaken, might be true but is still unknown. 
Let $X \subset \mathbb{P}^N$ be a projetive variety over an algebraically closed field of characteristic $0$. Denote by $S(X) \subset \mathbb{P}^N$ the Zariski closure of the set of point lying on bisecant lines to $X$. The variety $S(X)$ is caled the secant variety to $X$. Let $x,y \in X$ be general points and let $z$ be a general point in $\langle x,y \rangle$ (the line joining $x$ and $y$), then we have:
$$ T_{S(X),z} = \langle T_{X,x}, T_{X,y} \rangle,$$
where $T_{S(X),z}$ is the embedded tangent space to $S(X)$ at $z$ and $\langle T_{X,x}, T_{X,y} \rangle$ is the linear span in $\mathbb{P}^N$ of the tangent spaces to $X$ at $x$ and $y$.
This result, a consequence of the generic smoothness Theorem, is called Terracini's lemma and is due (probably with a non-rigorous proof) to Terracini (as the name suggests!).
There is a generalization of this result which was expected to be true be the Italian school. Denote by $\delta = 2 \dim X + 1 - \dim S(X)$. The number $\delta$ is called the secant-defect and is the difference between the expected and the actual dimensions of $S(X)$. If $x,y \in X$ are general points, Terracini's lemma immediately implies that:
$$ \dim (T_{X,x} \cap T_{X,y} ) = \delta - 1.$$
We denote by $S(X)_{stat}$ the Zariski closure of the set of points in $S(X)$ lying on secants $\langle x', y' \rangle$ such that $\dim (T_{X,x'} \cap T_{X,y'}) > \delta -1$. This variety is called the variety of stationnary bisecants to $X$.
I am pretty sure the Italian school considered the following fact to be true:
Fact :
Let $x',y' \in X$ be general points such that $\langle x',y' \rangle \subset S(X)_{stat}$. Let $z$ be general in $\langle x',y' \rangle$, then:
$$ T_{S(X)_{stat},z} = \langle T_{X,x'}, T_{X,y'} \rangle.$$ 
If $X \subset \mathbb{P}^N$ is a curve, then the above result is easily shown to be true. Indeed, if $X$ is not a line, then $S(X)_{stat}$ is necessarily a surface and since the generic tangent to $X$ is not a bitangent, we deduce that the result is true. I think it is also possible to prove this generalization of Terracini's lemma if $X$ is the closed orbit of a linear algebraic group acting linearly on $\mathbb{P}^N$. There are certainly many other cases where this generalization of Terracini's lemma is known to be true. 
On the other hand, I am pretty sure this result is unknown in general. It might even drop into the box of the slightly false results, but which are still very interesting.<|endoftext|>
TITLE: What is so "spectral" about spectral sequences?
QUESTION [37 upvotes]: From recent mathematical conversations, I have heard that when Leray first defined spectral sequences, he never published an official explanation of his terminology, namely what is "spectral" about a spectral sequence.
In Timothy Chow's relevant article, he writes

"John McCleary (personal communication) and others have speculated that since Leray was an analyst, he may have viewed the data in each term of a spectral sequence as playing a role that the eigenvalues, revealed one at a time, have for an operator."

This certainly seems like a reasonable answer, but are there any other plausible explanations? Did Leray ever communicate, perhaps in personal correspondences or unpublished manuscripts, why he chose that particular term? Does anybody have a better explanation for the origin of the adjective "spectral" in spectral sequences?

REPLY [37 votes]: After my article was published, John Harper sent me email and said that when he was a graduate student back in the 1960s, he personally asked Leray about the term "spectral" and in particular asked whether it had something to do with the spectrum of an operator.  Leray began his reply by saying, "Non"; unfortunately, before he could continue, some professors approached and interrupted the conversation.
This is perhaps some weak evidence that the spectrum of an operator is not what Leray had in mind, but unfortunately gives us no more positive information about what he did have in mind.<|endoftext|>
TITLE: Finding a cycle of fixed length in a bipartite graph 
QUESTION [6 upvotes]: Is finding a cycle of fixed even length in a bipartite graph any easier than finding a cycle of fixed even length in a general graph? This question is related to the question on Finding a cycle of fixed length
Edit: There is natural refinement of this question: what happens if the graph has boundend valence, e.g. if the bipartite graph is cubic?

REPLY [11 votes]: Finding a cycle of length 2k in an arbitrary graph is the same thing as finding a cycle of length 4k in the bipartite graph formed by subdividing every edge. So in general even cycles of fixed length are no easier to find in bipartite graphs than in arbitrary graphs. But it's possible that the lengths that are 2 mod 4 are easier to find than the lengths that are 0 mod 4, in bipartite graphs.<|endoftext|>
TITLE: Serre spectral sequence with spectra
QUESTION [6 upvotes]: A friend recently asked me if i had heard anything about a stable Serre Spectral Sequence or one constructed with spectra, has any one else ever heard of this? is there any reason other than historical that we don't think of the Serre SS this way initially?
Any thoughts or references would be great!

REPLY [10 votes]: Since the ordinary Serre spectral sequence is about a fibration of spaces, I don't think you can just talk about a "fibration of spectra" and expect that to be a generalization, since the suspension spectrum functor doesn't preserve fibration sequences.  However, there is a version of the Serre spectral sequence involving parametrized spectra, which one can think of as a fibration whose base is an ordinary space and whose fibers are spectra.  It can be found in section 20.4 of May-Sigurdsson, Parametrized Homotopy Theory.<|endoftext|>
TITLE: Tensor products and two-sided faithful flatness
QUESTION [12 upvotes]: Let $f: R \to S$ be a morphism of Noetherian rings (or more generally $S$ can just be an $R-R$ bimodule with a bimodule morphism $R \to S$). Suppose $f$ is faithfully flat on both sides, so $M \to M \otimes_R S$ is injective for any right $R$-module $M$, and $N \to S \otimes_R N$ is injective for any left $R$-module $N$.
Is it then true that $M \otimes_R N \to M\otimes_R S \otimes_R N$ is injective for any pair  $(M, N)$ of a right and a left $R$-module? This seems way too optimistic, but I can't seem to find a counterexample.

REPLY [8 votes]: Here's an example.  Let $R = {\mathbb C}[x]$ and let $S = {\mathbb C}\langle x,y\rangle/(xy-yx-1)$, i.e. the first Weyl algebra $A_1$.  Then $S$ is free as both a left and right 
$R$-module, and comes equipped with the natural ($R$-bimodule) inclusion of $R$.  On the 
other hand, if you take $M = {\mathbb C}[x]/(x) = N$, you'll get for $M\otimes_R S\otimes_R M$ the zero module.  Indeed, any element of $S$, i.e. any differential operator with polynomial coefficients (writing $\partial = \partial/\partial x$ in place of $y$) can be written in the form $\sum_i p_i(x) \partial^i$, so any element in $M\otimes_R S = {\mathbb C}[x]/(x) \otimes S$ is represented by an expression $\sum_i c_i \partial^i$ where the $c_i$ are constants, and now an induction on $k$ shows (I believe, my brain is a little fuzzy at this hour) that, for the right $R$-module structure on $S/xS$, one has $\partial^k \cdot x  = k \partial^{k-1}$.  One can conclude that $M\otimes_R S\otimes_R M = 0$, whereas of course $M\otimes_R M\cong M$ in this example...<|endoftext|>
TITLE: Why do wedges of spheres often appear in combinatorics?
QUESTION [43 upvotes]: Robin Forman writes in "A User's Guide to Discrete Morse Theory":

The reader should not get the
  impression that the homotopy type of a
  CW complex is determined by the number
  of cells of each dimension. This is
  true only for very few spaces (and the
  reader might enjoy coming up with some
  other examples). The fact that wedges
  of spheres can, in fact, be identified
  by this numerical data partly explains
  why the main theorem of many papers in
  combinatorial topology is that a
  certain simplicial complex is homotopy
  equivalent to a wedge of spheres.
  Namely such complexes are the easiest
  to recognize. However, that does not
  explain why so many simplicial
  complexes that arise in combinatorics
  are homotopy equivalent to a wedge of
  spheres. I have often wondered if
  perhaps there is some deeper
  explanation for this.

The question is: "Why so many simplicial complexes that arise in combinatorics are homotopy equivalent to a wedge of spheres?"

REPLY [15 votes]: Various simplicial complexes arising in combinatorics have the Cohen Macaulay property. Such complexes are always (homologically) a wedge of spheres of the same dimension (the dimension of the complex) and also locally have this property. Shellability is an important combinatorial property that again implies the complex and all links to be wedge of spheres. There are interesting extensions of these concepts to the case of spheres of different dimension. Shifted complexes are simplicial complexes whose vertices have a total order and with the property that if S is a face and R is a face of the same dimension with smaller vertices (namely there is a bijection a:R->S so that v<=a(v) for every v) then R is a face. Shifted complexes are always wedge of spheres (of different dimensions) and there are interesting operations which associate to arbitrary simplicial complexes, such shifted complexes. 
There are also quite a few interesting classes of simplicial complexes arising in combinatorics which are far from being wege of spheres. Take for example the chessbod complex whose faces are the locations on n nonattacking rooks in an n by n+1 chessboard board. for n=2 it is a hexagon for n=3 a torus for n=4 a certain pseudomanifold all links of vertices are tori, etc.<|endoftext|>
TITLE: Regular spaces that are not completely regular
QUESTION [14 upvotes]: In the undergraduate toplogy course we were given examples of spaces that are $T_i$ but not $T_{i+1}$ for $i=0,\ldots,4$. However, no example of a space which is $T_3$ but not $T_{3.5}$ was given. Later I was told by a colleague that such examples are rare and difficult to construct.
I know there is an example of such a space, called the Tychonoff corkscrew (or the spiral staircase), in the "Counterexamples in topology" book by Steen and Seebach. I've also found the following paper, though at the moment I'm not able to view it: A.B. Raha "An example of a regular space that is not completely regular", Proceedings Mathematical Sciences 102 (1992), 49-51.
Are there any other, folklore examples of regular spaces that are not completely regular? Are there any relatively easy ones?

REPLY [2 votes]: In the "Handbook of the History of General Topology" vol. 1, M. E. Rudin in the article "The early Work of F. B. Jones", commenting the results of F. B. Jones introduced the name  "The Jones machine". The Jones machine is a simple design which, using countably many copies of a completely regular space, but not normal, creates a regular space, but not a completely regular one.
   In the article "On regular but not completely regular spaces", Topology and its Applications
Volume 252, 1 February 2019, Pages 191-197, we give a simple, i.e., available for ordinary students of mathematics, examples of applying this method.<|endoftext|>
TITLE: What happened to the Vacuum Hypothesis in TQFT?
QUESTION [19 upvotes]: I remember that in the beginning, there was an axiom for $(n+1)$-dimensional
TQFT that said that the state space $V(\Sigma)$ assigned to an $n$-dimensional
oriented manifold is spanned by the invariants of all $n+1$-dimensional oriented manifolds
$M$ with $\partial M=\Sigma$.  If we call the invariant of $M$, $Z(M)\in V(\Sigma)$, this
just says that $V(\Sigma)$ is spanned by all $Z(M)$.  Maybe it was in Segal's Swansea notes, maybe it was in an early version of Atiyah's axioms. It doesn't seem to have made it
into "The Geometry and Physics of Knots"
For instance if you read "Topological Quantum Field Theories Derived from the Kauffman Bracket" by Blanchet, Habegger, Masbaum and Vogel, the vacuum hypothesis is implicit in their constructions.
There is a theorem that the category of Frobenius algebras is equivalent to the category of 
$1+1$-dimensional TQFT's.  For instance, let $A=\mathbb{C}[x]/(x^3)$, with Frobenius map
$\epsilon(1)=\epsilon(x)=0$ and $\epsilon(x^2)=-1$.  This is the choice that Khovanov made to construct his $sl_3$-invariant of links.  At this point, no one would deny that this gives rise to a TQFT.
The state space associated to a circle is just $A$.  The $2$-manifolds with boundary
the circle are classified by genus.  Using TQFT to compute them, I get that a disk has invariant $1$, a surface of genus one with one boundary component has invariant $3x^2$, and any other surface has invariant $0$.  The invariants don't span $A$.
The problem as I learned from Chris French is that $A$ is not semisimple. In fact, $A$ being spanned by the invariants of surfaces with one boundary component is equivalent to the semisimplicity of $A$.
Here is my question.  At what point, and why was the vacuum hypothesis abandoned?

REPLY [4 votes]: At least in your example about Khovanov's TQFT for $\mathfrak{sl}_3$ link homology, part of the answer is that that theory also allows 'singular surfaces', locally modelled on a book with 3 pages. Now you can take a "punctured torus with a disk stuck in its throat" (see my paper, for example), as a singular 2-manifold with boundary that circle, and you'll see that this gives the missing element of $A$.
I'm not sure, though, what the general story is here.<|endoftext|>
TITLE: Is there a bipartite analog of graph theory?
QUESTION [5 upvotes]: I would like to compile a list of questions about graphs that have a non-trivial analogs for bipartite graphs.
Let me give the following examples:

Cycle vs Even cycle. Most questions about cycles in graphs have analogs in even cycles for bipartite graphs. For instance, it is trivial to show that a bipartite graph on an odd number of vertices cannot have a Hamilton cycle.  In such a case the bipartite analog of a Hamilton cycle is a cycle missing exactly one vertex.
Minimal Girth. For graphs, 3 is the minimal possible length of a cycle. For bipartite graphs, the analogous number is 4.
Triangular vs quadrangular embeddings. In topological graph theory, a triangular embedding of a simple graph determine its genus. For bipartite graphs, the analog is an embedding with quadrangles as faces.

REPLY [3 votes]: Some classical theorems involving complete graphs have analogues involving complete bipartite graphs.  For example, the complete graph $K_n$ has $n^{n-2}$ spanning trees, while the complete bipartite graph $K_{m,n}$ has $n^{m-1} m^{n-1}$ spanning trees.  Finding the largest complete subgraph of a graph is a standard NP-hard problem, and finding the largest (in terms of number of edges) complete bipartite subgraph of a bipartite graph is also an NP-hard problem.
But possibly the area of graph theory that has most benefited from the analogy between general graphs and bipartite graphs is matching theory.  Matching theory tends to be easier in the bipartite case, but the bipartite case often gives us clues for the non-bipartite case.  For example, finding a maximum matching in a bipartite graph is solvable in polynomial time, but nontrivially so, and this leads us to look for a polytime algorithm for maximum matching in a non-bipartite graph (which does exist, but is more complicated than in the bipartite case).  Lovász and Plummer's Matching Theory, now back in print thanks to the American Mathematical Society, gives an excellent account of the interplay between bipartite and non-bipartite matching.<|endoftext|>
TITLE: existence of a connected set with given connected projections.
QUESTION [10 upvotes]: Suppose A and B are compact connected sets in the XY plane and XZ plane respectively in R^3. Suppose further that the the range of x-values taken by A and B are the same (i.e, projections of A and B onto the x-axis are the same closed interval). Is there always a connected set in R^3 whose projections onto XY and XZ planes are A and B respectively?

REPLY [6 votes]: The answer is yes.
Let $\alpha,\beta:[0,1]\to[0,1]\times \mathbb R$ be two paths;
$\alpha(t)=\left(\alpha_1(t),\alpha_2(t)\right)$ and $\beta(t)=\left(\beta_1(t),\beta_2(t)\right)$.
Assume that $\alpha_1(0)=\beta_1(0)=0$, $\alpha_1(1)=\beta_1(1)=1$.

Claim. The points $a=\left(0,\alpha_2(0),\beta_2(0)\right)$ and  $b=\left(1,\alpha_2(1),\beta_2(1)\right)$ lie in the same connected coponent of the set $\Sigma\subset \mathbb R^3$ formed by points of the following type $ \left(\alpha_1(t),\alpha_2(t),\beta_2(\tau)\right)$ such that $\alpha_1(t)=\beta_1(\tau)$.

Proof. Note that for generic smooth choice of $\alpha$ and $\beta$ the set $\Sigma$ is a smooth 1-dimensional manifold which might be not connected, but it has only two boundary points in $a$ and $b$.
Thus, in this case one can connect these points by a curve.
The general case can be done by approximation. $\square$
The rest is easy: one can approximate $A$ and $B$ by path-connected sets $A'$ and $B'$ with the same property. 
Thus one can present $A'$ and $B'$ as a union of curves $\alpha$ and $\beta$ with the above property. 
Moreover we can assume that the ends of $\alpha$ and $\beta$ are fixed (i.e. $a$ and $b$ are fixed).
For each pair $(\alpha,\beta)$, choose the connected component of $a$ in $\Sigma$ and take the union of all of them.<|endoftext|>
TITLE: Can a group be a finite union of (left) cosets of infinite-index subgroups?
QUESTION [24 upvotes]: To be more precise (but less snappy): is there an example of a group $G$ with finitely many infinite-index subgroups $H_1,\dots, H_n$ and elements $k_1,\dots, k_n$ such that $G$ is the union of the left cosets $k_1 H_1 , ..., k_n H_n\ ?$  And what if we relax the requirement that these all be left cosets, and ask: can $G$ be the union of finitely many such cosets, some being left cosets, others being right cosets?
If $G$ is amenable then this can't happen, since any coset of an infinite-index subgroup must have measure $0$.  So this immediately rules out any abelian group $G$.
I've tried playing around with the only non-amenable groups that I'm comfortable with, the free groups on two or more generators.  A few months ago I thought I found a simple counterexample in the free group on $\aleph_0$ generators, but now I've lost my notes and am beginning to doubt I ever had such an example.
(This question was asked to me by a friend who's interested in some kind of application to model theory, but I think it's interesting as a stand-alone puzzle.)

REPLY [38 votes]: No.  This follows from a beautiful theorem of B.H. Neumann:
Let $G$ be a group.  If $\{x_iH_i\}_{i=1}^n$ is a covering of $G$ by cosets of proper subgroups, then $n \geq \min_{i} [G:H_i]$. 
Explicitly, this is Lemma 4.1 in
http://math.uga.edu/~pete/Neumann54.pdf
As Neumann remarks, the identity $gH = (g H g^{-1}) g$ shows that it is no loss of generality to restrict to coverings by left cosets.<|endoftext|>
TITLE:  What are the 'standard' applications of the duals of the adjoint functor theorems?
QUESTION [9 upvotes]: There are some 'standard' applications of the adjoint functor
theorem (AFT) and the special adjoint functor theorem (SAFT), for
example,  the existence of a free $\tau$-algebra (where
$\tau=$(operations,identities)) on a small set by the AFT,
Stone-Cech compactification by the SAFT, and, if I am not mistaken, the proof that the category of $\tau$-algebras is cocomplete (by
using the AFT to establish a left adjoint to the appropriate diagonal
functor). 
However, I was not able to find any applications of the duals of
the AFT and the SAFT, neither in MacLane, nor in the Joy of
Cats.   
The Joy of Cats contains the following intriguing
remark on p. 311: 


Since many familiar categories have separators but
    fail to have coseparators, the dual of the Special Adjoint Functor
    Theorem is applicable even more often than the theorem itself.


But what are the mentioned application of the dual of the SAFT?
So my questions is:  What are the 'standard' applications of the duals
of the AFT and the SAFT? 
Googling for combinations of phrases like ''adjoint functor theorem''
and ''dual'' is not very useful, so I have tried ''dual of the adjoint
functor theorem'' and ''dual of the special adjoint functor theorem.''
This resulted in a total of 7 papers/books, from which I was not able
to get a clear answer to the current question. I have also tried in
The Wikipedia article on adjoint
functors, in nLab's
article on the adjoint functor
theorems, and
in some
MO 
questions, but without success.

REPLY [6 votes]: One example is the construction of geometric morphisms.  Any colimit-preserving functor between Grothendieck toposes has a right adjoint, so if it also preserves finite limits, then it is part of a geometric morphism.  Of course, in many cases in practice, the right adjoint is also easy to write down explicitly.<|endoftext|>
TITLE: Criteria for accepting an invitation to become an editor of a scientific journal
QUESTION [15 upvotes]: Yesterday I receivied the following letter:

Dear Tomaz Pisanski,
I am writing to introduce the XXXX to you. XXXX is a new journal launched by the YYYY from USA. With an open access publication model of this journal, all interested readers around the world can freely access articles online at http:ZZZ without subscription.
  We are soliciting scholars to form the editorial board. If you are interested, please send your resume along with your areas of interest to WWWW. Manuscripts should be submitted to the journal online at: http:UUU. Once accepted for publication, your manuscripts will undergo language copyediting, typesetting, and reference validation in order to provide the highest publication quality possible.
Please do not hesitate to contact me if you have any questions about the journal.
Editorial Office
YYYY

I removed the actual journal title and other details since they are not important for this
question. I would like to have some general criteria or guidelines that would help me decide 
whether to accept such an offer or not.

REPLY [2 votes]: Sorry for answering my own question, but the e-mail I received today may explain that
one really has to think twice before making the decision.

Dear Tomaz Pisanski,
I am writing to introduce the Applied Mathematics (AM) to you. AM is a new journal launched by the Scientific Research Publishing from USA. With an open access publication model of this journal, all interested readers around the world can freely access articles online at http://www.scirp.org/journal/am without subscription.
We are soliciting scholars to form the editorial board. If you are interested, please send your resume along with your areas of interest to am@scirp.org. Manuscripts should be submitted to the journal online at: http://www.papersubmission.scirp.org/login.jsp. Once accepted for publication, your manuscripts will undergo language copyediting, typesetting, and reference validation in order to provide the highest publication quality possible.
  Please do not hesitate to contact me if you have any questions about the journal.
Editorial Office
  Applied Mathematics
  Scientific Research Publishing, USA
  Email: am@scirp.org
This message was sent to [tomaz.pisanski@fmf.uni-lj.si].
  Unsubscribe at any time by clicking here.<|endoftext|>
TITLE: Cohomology classes annihilated by pullbacks
QUESTION [12 upvotes]: A friend of mine is interested in examples of the following situation:
an oriented smooth fiber bundle $\pi \colon M \to B$ with $M$ and $B$ compact 
and a non-zero class $a \in H^3(B; \mathbb{Q})$
such that $\pi^* a=0$  in $H^3(M; \mathbb{Q})$. 
It is easy to construct such an example if the class $a$ is a product of a degree $1$ class and degree $2$ class; are there examples not of this kind?

REPLY [3 votes]: The Becker--Gottlieb transfer implies that $\pi^*$ is rationally a (split) monomorphism unless the Euler characteristic of the fibre is zero. Thus any proposed example must have this property.<|endoftext|>
TITLE: Tautological bundle on G$(n,k)$ and Chern classes
QUESTION [5 upvotes]: Given the complex grassmannian variety G$(n, k)$, I consider the tautological bundle $S$, i.e. the $n$-plane bundle whose fiber at each point of G$(n, k)$ is given by the corresponding $n$-plane in $\mathbf{C}^k$. I consider now the Chern polynomial of $S$, $c(S)$. How can I explicitly compute the Chern roots of $c(S)$ (i.e. cohomology classes $f_i \in H^2(G(n, k), \mathbf{Z}) , i = 1, \dots, n$ such that $c(S) = \Pi_{i = 1}^n (1+ f_i t)$)?

REPLY [4 votes]: Depending on what you need to do with the Chern roots, it may be cleaner to just ask for the Chern classes of $S$. 
In this case, let $Q$ be the quotient bundle, i.e., there is a trivial bundle ${\bf C}^k$ which contains $S$ as a subbundle, and $Q = {\bf C}^k / S$. The $i$th Chern class of $Q$ is the cohomology class of $\sigma_i$, the special Schubert class of codimension $i$ (see Proposition 3.5.5 of Manivel's book Symmetric Functions, Schubert Polynomials, and Degeneracy Loci). Using the relation $c(S)c(Q) = 1$, and knowledge of the cohomology ring of $G(n,k)$ should be enough to perform any usual calculations. 
For computing with these Schubert classes, one only needs to learn the Pieri rule (and perhaps the Littlewood-Richardson rule depending on the circumstance), both of which can be found in Manivel's book (chapter 1) or see Wikipedia.<|endoftext|>
TITLE: Quadratic forms that evaluate to zero mod p only when their input is zero.
QUESTION [7 upvotes]: Let $Q$ be a quadratic form in $n$ variables with integer coefficients.  Let us say that $Q$ has the "special property" mod $p$, if the relation $Q(x_1,...,x_n)=0$ (mod $p$) implies that $(x_1,...,x_n)=(0,...,0)$ (mod $p$).  (There must be a name for this property, but I don't know it, which is why I'm calling it "the special property".)  Let us say that $Q$ is "special infinitely often" if there are infinitely many primes $p$ such that $Q$ has the special property mod $p$.  For example, the one-variable quadratic form $Q(x)=x^2$ is special infinitely often.  Another simple example is that $Q(x,y)=x^2+y^2$ is special infinitely often because it is special mod $p$ whenever -1 is a quadratic non-residue mod $p$.  In fact, any two-variable quadratic form that is non-degenerate over the rationals is special infinitely often because $Q(x,y)=ax^2+bxy+cy^2$ is special whenever $p$ is odd and the discriminant $b^2-4ac$ is a quadratic non-residue mod $p$.
I've been trying to find a quadratic form in more than two variables that is special infinitely often, but I'm doubtful that such a thing exists.  As far as I know, each of the following statements could either be valid or invalid.  (Although obviously [c] implies [b] implies [a].)
[a]  For every integer quadratic form $Q$ in three or more variables, the set of $p$ such that $Q$ has the special property mod $p$ is finite.
[b]  The set of primes $p$ such that there exists an integer quadratic form $Q$ in three or more variables having the special property mod $p$ is finite.
[c]  The set of primes $p$ such that there exists an integer quadratic form $Q$ in three or more variables having the special property mod $p$ is empty.
Can anyone help resolve any of these questions?

REPLY [11 votes]: The Chevalley–Warning theorem (see wikipedia) implies that any quadratic form in at least three variables has a non-trivial solution modulo p.<|endoftext|>
TITLE: Homotopy Limits over Fibered Categories
QUESTION [17 upvotes]: Suppose I have a small category $ \mathcal{C} $ which is fibered over some category $\mathcal{I}$ in the categorical sense.  That is, there is a functor $\pi : \mathcal{C} \rightarrow \mathcal{I}$ which is a fibration of categories.  (One way to say this, I guess, is that $\mathcal{C}$ has a factorization system consisting of vertical arrows, i.e. the ones that $\pi$ sends to an identity arrow in $\mathcal{I}$, and horizontal arrows, which are the ones it does not.  But there are many other characterizations.)
Now let $F : \mathcal{C} \rightarrow s\mathcal{S}$ be a diagram of simplicial sets indexed by $\mathcal{C}$.  My question concerns the homotopy limit of $F$.  Intuition tells me that there should be an equivalence
$$ \varprojlim_{\mathcal{C}} F \simeq \varprojlim_{\mathcal{I}} \left (\varprojlim_{\mathcal{C}_i} F_i \right ) $$
where I write $\mathcal{C}_i = \pi^{-1}(i)$ for any $i \in \mathcal{I}$, $F_i$ for the restriction of $F$ to $\mathcal{C}_i$ and $\varprojlim$ for the homotopy limit.
Intuitively this says that when $\mathcal{C}$ is fibered over $\mathcal{I}$, I can find the homotopy limit of a $\mathcal{C}$ diagram of spaces by first forming the homotopy limit of all the fibers, realizing that this collection has a natural $\mathcal{I}$ indexing, and then taking the homotopy limit of the resulting diagram.
Does anyone know of a result like this in the model category literature?
Update: After reading the responses, I was able to find a nice set of exercises here which go through this result in its homotopy colimit version.

REPLY [7 votes]: The axiomatic way to present this is Heller's theory of homotopy theories, which is the same as Grothendieck's theory of derivators (see the references in the links below). As any model category defines a derivator (see here), what you want is precisely Lemma 2.3 in this paper.<|endoftext|>
TITLE: Does this problem have a name?  [Ducci Sequences]
QUESTION [5 upvotes]: Let $a_1, ... a_n$ be real numbers.  Consider the operation which replaces these numbers with $|a_1 - a_2|, |a_2 - a_3|, ... |a_n - a_1|$, and iterate.  Under the assumption that $a_i \in \mathbb{Z}$, the iteration is guaranteed to terminate with all of the numbers set to zero if and only if $n$ is a power of two.  A friend of mine knows how to prove this, but wants to be able to reference a source where this problem (and/or its generalization to real numbers) is mentioned, and we can't figure out what search terms to use.  Can anyone help us out?
(If someone can figure out a better title for this question, that would also be appreciated.)

REPLY [11 votes]: This is known as Ducci's problem. "Ducci map" or "Ducci sequence" as key words should let you search on most of the articles studying properties of the above map.<|endoftext|>
TITLE: Expository treatment of Schubert Cells Paper
QUESTION [9 upvotes]: I was wondering about the paper by Bernstein, Gel'fand, and Gel'fand on Schubert Cells. This paper is fairly old(and often cited) so I figured someone must have represented this material. In particular, I was wondering if this was treated in an expository paper. More generally, I was wondering if there was a paper that explained the usefulness of the Schubert Calculus for representation theory, and even better one that talked about how Schubert Calculus came into the picture for BBD, again hopefully in an expository way.
Thanks in advance!

REPLY [5 votes]: The Lecture Notes in Mathematics number 1689, "Schubert Varieties and Degeneracy Loci" by Fulton and Pragacz seems to be exactly what you're looking for. I think chapter 6 is particularly relevant.

REPLY [2 votes]: You might enjoy this article by Harry Tamvakis.<|endoftext|>
TITLE: Conductor of monomial forms with trivial nebentypus
QUESTION [5 upvotes]: Is it true that the conductor of a holomoprhic or a Maass cusp form with trivial nebentypus corresponding to a two-dimensional dihedral representation (over $\mathbb{Q}$ )is non-square-free?

REPLY [6 votes]: Thinking about Idoneal's question to Emerton about what is being used about the Steinberg, it's not just standard facts about the Steinberg one needs via this approach, but also local-global compatibility. The standard facts about the Steinberg are also easily proved if one uses local Langlands (i.e. works on the Galois side). This made me realise that in fact one can pull off the entire argument on the Galois side! Let's assume you're talking about complex 2-dimensional representations of the Galois group. Emerton has already observed that they must be ramified at some prime $p$, so it suffices to prove that $p^2$ divides the conductor. But now say
$\rho:Gal(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)\to GL_2(\mathbf{C})$
has trivial determinant and conductor $p$. One instantly gets a contradiction: the inertial invariants can't be 0-dimensional because already this would mean the conductor is at least $p^2$, and they can't be 1-dimensional because if $i\in I_p$ has one eigenvalue 1 then the other eigenvalue must be 1 too, and $\rho(i)$ is diagonalisable and hence trivial, so the inertial invariants must be 2-dimensional but now the conductor must be 1. This answers the question completely without recourse to the smooth representation theory side of things and is surely the easiest approach to the question.<|endoftext|>
TITLE: Making a non-monotone function monotone
QUESTION [15 upvotes]: Consider a function $f: \{0,1\}^n \rightarrow \{1..R\}$. This function can be interpreted as a coloring $Color(v)$ of vertices in a unit n-dimensional hypercube in $R$ colors. 
We say there is a directed edge $(v1, v2)$ in the hypercube if $v1$ and $v2$ differ in only one coordinate in n-dimensional space and this coordinate is equal to $0$ for $v1$ and to $1$ for $v2$.
Let's define $E(f)$ to be the number of non-monotone directed edges in this hypercube, i.e. edges $(v1, v2)$, such that $Color(v1) > Color(v2)$. Having a non-monotone function $f$ we want to make it monotone, changing its values in as few points of domain as possible. Let's denote $M(f)$ to be the minimal number of points where we need to change the values to make the function monotone.
There is a hypothesis that $M(f) \le E(f)$  that I'm trying to prove.
Known results:
0) There exist such $f$ that $M(f) \ge E(f)$ (if $E(f)$ is not too large, say $E(f) \le 2^{n-1}$). This one is an easy exercise.
1) For $R=2$ the hypothesis is true. The method I know is rather difficult to describe briefly. If necessary, I can give a link (see EDIT).
2) For general $R$ it can be proved that $M(f) \le E(f) \log{R}$. The proof involves construction for $R=2$ and some range reduction technique.
I am interested in what techniques can be applied to prove or disprove this hypothesis. Any result better than $M(f) \le E(f) \log{R}$ (like $M(f) = O(E(f))$) will be interesting.
EDIT: Here is a link to M.Sc. thesis by Sofya Raskhodnikova, where results 1) and 2) can be found in Chapters 3 and 5 respectively.
EDIT: Here is a link to some informal description of motivation for this problem.

REPLY [3 votes]: A combinatorial proof was found last year by Chakrabarty and Seshadhri: http://eccc.hpi-web.de/report/2012/030/download. In Theorem 3 they show that indeed $E(f) = \Omega(M(f))$.<|endoftext|>
TITLE: Derivators (in English)
QUESTION [21 upvotes]: Grothendieck, before he disappeared, was working on a manuscript called "Les Derivateurs", which detailed the theory of derivators.  Prof. Cisinski has done work with them as he mentioned in this post.  However, most of his work is in French, and I was wondering if there are any typed up references in english (the nLab has written notes from a seminar, but that's it).  I intend to read the references in French later, but could someone explain or give a reference that explains in English the definition of a derivator and the motivation for them?

REPLY [4 votes]: M Groth's book project: The theory of Derivators is a highly recommendable reading!
They are two volumes, the volume I is available as a draft version.
Enjoy!<|endoftext|>
TITLE: measure theory for regular cardinals
QUESTION [10 upvotes]: Measure theory is somewhat focused on the cardinal $\aleph_0$: First of all we have the usual $\sigma$-additivity, Polish (separable!) spaces such as $\mathbb{R}^n$, countable sequences and limits, countable recursive constructions. I know that this is a vague question, but is there a possibility to extend measure theory to an arbitrary regular cardinal $\kappa$? Thus in the definitons we replace $\aleph_0$ by $\kappa$ and try to imitate the theory? Perhaps we should also replace $\mathbb{R}$ by some other model because in $\mathbb{R}$ every convergent sum is supported on a countable index subset. Or is there already some research on it?
The reason for my question is not only just of curiosity. I want to understand in detail this "cardinality boundary" of measure theory.

REPLY [2 votes]: A related topic:  A $\sigma$-smooth linear funtional $L$ on $C(X)$, the set of continuous real-valued functions on a topological space $X$, is a linear functional such that: if $f_n$ is a sequence that decreases pointwise to $0$, then $L(f_n) \to 0$.  A generalization is $\tau$-smooth linear functional: if $\mathcal A$ is a family of nonnegative functions in $C(X)$ that is directed in the sense: for any $f_1, f_2 \in \mathcal A$ there exists $f_3 \in \mathcal A$ with $f_3 \le f_1$ and $f_3 \le f_2$, then $\inf_{f \in \mathcal A} L(f) = 0$.  For some topological spaces, every $\sigma$-smooth functional is $\tau$-smooth. For other topological spaces, the two concepts are different.<|endoftext|>
TITLE: Defining 'free monoid' without Nat?
QUESTION [5 upvotes]: Is there a definition of what is a 'free monoid' which does not pre-suppose that the natural numbers has already been defined?  The definitions that I have been able to track down all use the natural numbers (since sequences/words need them).
In other words, I am trying to define the theory of free monoids (as a signature with sorts, operations and axioms), and I would like to know if I can do this without having the theory of the naturals already defined.  For exhibiting/constructing a free monoid, I do expect to need Nat.
Note that my ambient logic is higher-order, so I am fine with a second-order axiomatization.  If dependent-types are needed, that would also be acceptable.

REPLY [3 votes]: I am not this is what you want, but there is an old theorem of Levi that a monoid is free iff 

It is cancellative.
It has only trivial group of units.
It is equidivisible (you can google the definition; the key thing is that it is first order).
There are only finitely many principal right ideals containing any given principal right ideal.

1-3 are first order.  Probably 4 is a problem for what you want.<|endoftext|>
TITLE: Examples of inequality implied by equality. 
QUESTION [8 upvotes]: It is well known Cauchy's inequality is implied by Lagrange's identity. Bohr's inequality  $|a -b|^2 \le p|a|^2 +q|b|^2$, where $\frac{1}{p}+\frac{1}{q}=1$, is implied by $|a -b|^2 +|\sqrt{p/q}a+\sqrt{q/p}b|^2= p|a|^2 +q|b|^2$. L.K Hua's determinant inequality $\det(I-A^*A)\cdot \det(I-B^*B)\le |\det(I-A^*B)|^2$ is implied by Hua's matrix equality $(I-B^*B)-(I-B^*A)(I-A^*A)^{-1}(I-A^*B)=-(A-B)^*(I-AA^*)(A-B)$. What other examples can be found?

REPLY [3 votes]: Two examples due to Hurwitz. 

The AM-GM inequality.
For the function $f=f(x_1,x_2,\dots,x_n)$ let $Pf(x_1,x_2,\dots,x_n)$ denote the sum of $f$ over the $n!$ quantities that result from all possible $n!$ permutations of the $x_i$. Then
$$\frac{x_1^n+x_2^n+\dots+x_n^n}{n}-x_1x_2\dots x_n=\frac{1}{2\ n!}(\phi_1+\phi_2+
\dots \phi_n),$$
where
$$\phi_k=P[(x_1^{n-k}-x_2^{n-k})(x_1-x_2)x_3x_4\dots x_{k+1}]=P[(x_1-x_2)^2(x_1^{n-k-1}+\dots x_2^{n-k-1})x_3x_4\dots x_{k+1}]\geq0.$$
The proof can be found in Inequalities by Beckenbach and Bellman. 
The isoperimetric inequality. Let the boundary of $\Omega\subset \mathbb R^2$ be a rectifiable Jordan curve $\partial \Omega=\{((x(s),y(s))|\ s\in[0,2\pi))\}$. Then 
$$L^2-4\pi A=2\pi^2\sum\limits_{n=1}^{\infty}\left[(na_n-d_n)^2+(nb_n+c_n)^2+
(n^2-1)(c_n^2+d_n^2)\right],$$
where
$$x(s)=\sum\limits_{n=0}^{\infty}(a_n\cos ns+b_n\sin ns),\quad  y(s)=\sum\limits_{n=0}^{\infty}(c_n\cos ns+d_n\sin ns).$$<|endoftext|>
TITLE: Two discs with no parallel tangent planes
QUESTION [14 upvotes]: Is it possible to construct smooth embedded of 2-discs $\Sigma_1$ and $\Sigma_2$ in $\mathbb R^3$ with the same boundary curve such that there is no pair of points $p_1\in \Sigma_1$ and $p_2\in \Sigma_2$ with parallel tangent planes?

Comments:

The question is inspired by this; you will find there a construction of three such discs with no triples of points.
More formally, "the same boundary curve" means that $\Sigma_1=f_1(D^2)$ and $\Sigma_2=f_2(D^2)$ for some smooth embedding $f_1$ and $f_2$ of the unit disc $D^2$ 
such that $f_1|_ {\partial D^2}\equiv f_2|_ {\partial D^2}$.

REPLY [19 votes]: Suppose two such disks Σ1 and Σ2 exist, and pull back TΣ2 to Σ1 by some homeomorphism.  Viewed as a subbundle of TR3|Σ1, this plane bundle intersects TΣ1 in a line bundle L over Σ1 since no two tangent planes are parallel.  Furthermore, L|∂Σ1 is exactly the bundle of lines tangent to ∂Σ1.
Since a line bundle over a disk is trivial, we can take a nonzero section of L, and thus we get a nonvanishing vector field on the disk Σ1 which is tangent to the boundary at every point of ∂Σ1.  But then by doubling we can construct a nonvanishing vector field on S2, and this is impossible.

REPLY [8 votes]: If I understood your question correctly, I think that the answer is no.  In fact, even more is true: if you choose any identification $\varphi \colon \Sigma_1 \to \Sigma_2$ in such a way that the boundary are compatibly identified, then for any embedding of $\Sigma_1$ and $\Sigma_2$ there is a point $p$ of $\Sigma_1$ such that the tangent plane to $\Sigma_1$ is the same as the tangent plane to $\Sigma_2$ at $\varphi(p)$.  To see this, let $\mathbb{R}P^2$ denote the real projective plane, the space parameterizing linear subspaces of dimension one in $\mathbb{R}^3$.  Suppose by contradiction that you found embeddings with the mentioned property.  Let $\gamma \colon \Sigma_1 \cup \Sigma_2 \to \mathbb{R}P^2$ be the function defined by sending $p$ to the linear subspace spanned by $N_p \wedge N_{\varphi(p)}$, where $N_p$ is the normal direction to the image of $p$ and similarly for $N_{\varphi(p)}$, and the wedge product is the usual wedge product in $\mathbb{R}^3$.  The function $\varphi$ is indeed a function, since never the normal directions at $p$ and $\varphi(p)$ are parallel.  But observe that $\gamma(p)$ is always a vector contained in the tangent plane at $p$.  Thus, $\gamma$ determines a global vector field on the image of $\Sigma_1 \cup \Sigma_2$ that is never zero.  This is obviously impossible!
Note that the above argument is essentially the one used in the Borsuk–Ulam Theorem.<|endoftext|>
TITLE: What rings/groups have interesting quaternionic representations?
QUESTION [17 upvotes]: Let $\mathbb{H}$ denote the quaternions.  Let $G$ be a group, and define a representation of  $G$ on $\mathbb{H}^n$ in the natural way; that is, its a map $\rho:G\rightarrow Hom_{\mathbb{H}-}(\mathbb{H}^n,\mathbb{H}^n)$ such that $\rho(gg')=\rho(g)\rho(g')$ (where $Hom_{\mathbb{H}-}$ denotes maps as left $\mathbb{H}$-modules).  Representations of algebras and Lie algebras can be defined in a similar way.
Any quaternionic representation of $G$/$R$/$\mathfrak{g}$ can restrict to a complex representation by choosing a $\gamma\in \mathbb{P}^1$ such that $\gamma^2=-1$, and using $\gamma$ to define a map $\mathbb{C}\hookrightarrow \mathbb{H}$.  In this way, any quaternionic representation gives a $\mathbb{C}\mathbb{P}^1$-family of complex representations which parametrize the choice of $\gamma$.  Furthermore, since every element in $\mathbb{H}$ is in the image of some inclusion $\mathbb{C}\hookrightarrow \mathbb{H}$, this family of complex representations determines the quaternionic representation.
This observation almost seems to imply that there is nothing interesting to say about quaternionic representations that wouldn't come up while studying complex representations.  However, this is neglecting the fact that there might be interesting information in how the $\mathbb{C}\mathbb{P}^1$-family of complex representations is put together.  
For instance, any finite group will have a discrete set of isomorphism classes of complex representations, and so any quaternionic representation will have all complex restrictions isomorphic.  However, the quaternion group $\mathbf{Q}:= ( \pm1,\pm i, \pm  j,\pm k)$ has an 'interesting' quaternionic representation on $\mathbb{H}$ (more naturally, it is a representation of the opposite group $\mathbf{Q}^{op}$ by right multiplication, but $\mathbf{Q}^{op}\simeq \mathbf{Q}$).
My question broadly is: What other groups, rings and Lie algebras have quaternionic representations that are interesting (in some non-specific sense)?
This question came up when I was reading a paper of Kronheimer's, where he describes a non-canonical hyper-Kahler structure on a coadjoint orbit of a complex semisimple Lie algebra.  At any point in such a coadjoint orbit determines a representation of $\mathfrak{g}$ on the tangent space to the coadjoint orbit, which by the hyper-Kahler structure is naturally a quaternionic vector space.  I wondered if this representation could be 'interesting', and then realized I had no real sense of what an interesting quaternionic representation would be.

REPLY [12 votes]: Recall that if G is a group, k a field, and V_k an irreducible representation of G over k then End_G(V_k) is a division algebra D over k.  For example, if $V_{\mathbb{R}}$ is a real representation then the endomorphism ring is $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$.  Such representations are typically called real, complex, or quaternionic.
This usage is complimentary to your definition.  Suppose that $V_{\mathbb{R}}$ is a quaternionic representation, then $\mathbb{H}$ acts on $V_{\mathbb{R}}$ on the right.  Since $\mathbb{H}$ is a division ring, it follows that $V_{\mathbb{R}} \cong \mathbb{H}^k$.  Hence a real representation which is quaternionic is canonically a quaternionic representation in your sense.
Now Frobenius-Schur indicator theory explains how to determine which representations over $\mathbb{C}$ "come from" which kind of real representations.  Explicitly, if $V_{\mathbb{C}}$ is not selfdual then it plus its dual is the complexification of a representation over $\mathbb{R}$ with $D=\mathbb{C}$; if $V_{\mathbb{C}}$ orthogonal (i.e. has an invariant bilinear form) then it is the complexification of a representation over $\mathbb{R}$ with $D=\mathbb{R}$; finally, if $V_{\mathbb{C}}$ is symplectic then it plus itself is the complexification of a representation over $\mathbb{R}$ with $D=\mathbb{H}$.
In summary, given any symplectic representation $V_{\mathbb{C}}$ the representation $V_{\mathbb{C}} \oplus V_{\mathbb{C}}$ is the complexification of a representation $W_\mathbb{R}$ which is itself a module over $\mathbb{H}$.  In particular, there is an "interesting" representation over $\mathbb{H}$ whose quaternionic dimension is half the complex dimension of $V_{\mathbb{C}}$.  Your example fits into this scheme, where you start with the 2-dimensional irrep of $\mathbf{Q}$. Another great example involves the symplectic Lie group which can also be realized as a unitary group over the quaternions (see wikipedia).<|endoftext|>
TITLE: The resultant and the ideal generated by two polynomials in $\mathbb{Z}[x]$
QUESTION [45 upvotes]: I was asked the following question by a colleague and was embarrassed not to know the answer.
Let $f(x), g(x) \in \mathbb{Z}[x]$ with no root in common. Let $I = (f(x),g(x))\cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which are linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. Then $I$ is clearly an ideal in $\mathbb{Z}$. Let $D>0$ be a generator of this ideal. The question is: what is $D$?
Now, we do have the standard resultant $R$ of $f,g$, which under our hypotheses, is a non-zero integer. We know that $R \in I$ and it's not hard to show that a prime divides $R$ if and only if it divides $D$. I thought $R = \pm D$ but examples show that this is not the case.

REPLY [16 votes]: This difference was well-known in the 19th century  when people
a) Knew about invariants, and
b) Calculated by hand.
I believe a lot of the confusion today stems from Lang's Algebra book which is at best misleading about how to interpret what the resultant and discriminant are (and the ideas of famous books, right or wrong, tend to be perpetuated in other people's books!).
As an example, the resultant of the two polynomials $3x+1$ and $3x+2$ is, according to Sylvester's matrix definition, equal to $3$. Here Voloch's $D=1$. Surely this makes no sense according to the well-known theory, that  a prime $p$ dividing the resultant of two polynomials should be interpreted to mean that these polynomials share a root when reduced mod $p$? This is evidently nonsense in this example ... unless  one re-interprets these polynomials projectively (which is what one should do).
But now if we look at the pair of polynomials $y+2$ and $3y+2$ then the resultant is $4$, and here $D=4$, but how do you interpret here $2^2$ dividing the resultant? It is not immediate from the interpretation of modern algebra books!
There are all sorts of reasons that prime powers can divide a resultant (and discriminant) and it is complicated to understand all the cases when you wish to interpret higher power divisibility. 
In Bhargava's work, he needs to understand squarefree values of a multivariable polynomial which is the  the value of a discriminant of a class of parametrized polynomials. In other words he needs to parametrize when $p^2$ divides terms in this particular class of discriminants. Even this relatively simple request breaks down into several non-trivial cases, which he handles so beautifully as if to make it look trivial, but it's not.<|endoftext|>
TITLE: Versality in deformation theory vs. versality in moduli spaces
QUESTION [5 upvotes]: As I mentioned before, I'm a novice at deformation theory. I was wondering if the definition of versality in deformation theory is related to the versality in moduli spaces:
Deformation theory
"Moduli of Curves" defines a versal deformation space as a deformation $\phi: X \rightarrow Y$ such that for any other deformation $\xi: X \rightarrow Z$ and for every point in $Z$ there exists an open set (in the complex topology) $U$ such that the pullback of $\phi$ via $f: U \rightarrow Y$ is $\xi$ restricted to $U$. (I imagine that in general instead of an open set one takes open etale covers - is this true?)
Moduli Spaces
In moduli spaces, versality has always meant a space such that instead of the geometric points being in 1-1 correspondence with the objects we're interested in (over the field of the geo. point), each object is going have several geometric points in the versal space corresponding to it.
Question
Are these two notions related? If so - how?

REPLY [2 votes]: I remember that I was going to give essentially the same answer to this question as Hartmann but apparently I forgot! :) However, it can be given in a more conceptual way (using stacks or groupoids) generalizing the above beyond curves and germs (see the reference below). 
Have a look in one of the appendices to the book "Introduction to Singularitites and Deformations" by Greul, Lossen and Shustin (Springer-Verlag). The book is however mainly concerned with "complex space germs" and their deformations but the appendices treats some more general stuff. In particular, there are references to more general treatises.
If you want to go really hi-tech, try the paper "Versal deformations and algebraic stacks" by Michael Artin (hard to read though).<|endoftext|>
TITLE: Best way to teach concept of real numbers using a hands-on activity?
QUESTION [6 upvotes]: I know a middle school math teacher looking for some suggestions for hands-on activities to teach the concept of real numbers.  I'm new to this site, so this may be a little off topic.

REPLY [3 votes]: My mind immediately jumps to grabbing a ruler and some string and trying to determine the ratio of the circumference of a circle to its diameter, or the hypotenuse of the right triangle with unit legs.  In this way, the most accessible irrational numbers to most non-math majors (i.e. up to college even for most freshman) can be explored.
It should be easy to see that numbers can be built up from the humble unit 1, and quickly overtake the natural numbers with addition, the integers with subtraction, and the fractions with division.  Informing someone that there are numbers outside these operations should be the first step.
I would then construct some irrational numbers and argue (likely without proof) that they are not rational.  0.1234567891011121314... seems like a good candidate for this.  This also gives a powerful technique to the students to construct other irrational numbers and notice how many there actually are.
We can also explore the number Pi.  Cut out in paper a square, a pentagon, a hexagon, etc, to have several shapes of about the same area.  In some sense you can consider the ratio of a length to the perimeter in relation to the number of sides each shape has (and assign it as homework).  As the number of sides goes up, you can argue that this number is approaching Pi.
I am of the impression that this question counts as community wiki since the question is subjective.<|endoftext|>
TITLE: The ring of algebraic integers of the number field generated by torsion points on an elliptic curve
QUESTION [15 upvotes]: (Warning: a student asking)
Let $E$ be an elliptic curve over $\mathbf Q$. Let $P(a,b)$ be a (nontrivial) torsion point on $E$. Is there an easy description of the ring of algebraic integers of $\mathbf Q(a,b)$? I'm curious about the answer for general elliptic curves, but I'm not sure whether such an answer is possible.
(This question is motivated by the nice description of the rings of integers of cyclotomic fields $\mathbf Q(\zeta_n)$)

REPLY [4 votes]: For further work on monogeneity questions, you might want to have a look at
some of the papers of Reinhard Schertz (I'm afraid that I don't have precise
references right now). He apparently also has a new book out entitled "Complex
Multiplication" or something like this, which I've not seen, but which probably
also discusses some of his work on this topic.<|endoftext|>
TITLE: Any reference on multilinear algebra
QUESTION [8 upvotes]: Do you know any good reference on multilinear algebra?

REPLY [2 votes]: For the tensor, exterior and symmetric algebras of a module over a commutative ring I suggest the notes by Murfet
http://therisingsea.org/notes/TensorExteriorSymmetric.pdf<|endoftext|>
TITLE: Does linearization of categories reflect isomorphism?
QUESTION [65 upvotes]: Given a category $C$ and a commutative ring $R$, denote by $RC$ the $R$-linearization: this is the category enriched over $R$-modules which has the same objects as $C$, but the morphism module between two objects $x$ and $y$ is the free $R$-module on $\operatorname{Hom}_C(x,y)$. Thus in $RC$ we allow arbitrary $R$-linear combinations of morphisms from the original category $C$.
Question: if two objects in $x$, $y \in C$ are isomorphic in $RC$, are they already isomorphic in $C$?
I do not know the answer to this question for any nontrivial ring $R$, but I'm particularly interested in $R=\mathbb{Z}$ and $R=\mathbb{Z}/2\mathbb{Z}$.
What's obviously not true is that every isomorphism in $RC$ comes from an isomorphism in $C$ (take $-id_x$). (Thus the word "isomorphism" in the title refers to a relation on objects rather than to a property of morphisms.)
Of course, it is enough to consider categories $C$ with two objects $x$, $y$, but we cannot assume that $C$ is finite.
It's fairly elementary to see that if $x$ and $y$ are isomorphic in $RC$ then in $C$, $x$ is a retract of $y$ and vice versa, but the latter does in general not imply that $x \cong y$.
A more catchy way of phrasing this problem is: can we always classify objects in a category up to isomorphism by means of functors taking values in $R$-linear categories? (The inclusion $C \to RC$ is the universal such functor.)
Edit: A lot of people have posted an "answer" that wasn't, and deleted it, so here's something that will not work, to save others going down the same road. I said that we cannot assume that the category is finite; in fact, it must be infinite. Here is an elementary argument:
Since $x$ and $y$ are mutual retracts, there are maps $f,\;f'\colon x \to y$ and $g,\;g'\colon y \to x$ with $fg=\operatorname{id}$ and $g'f'=id$. Consider the powers of $fg' \in \operatorname{End}(y)$. If $\operatorname{End}(y)$ is finite then $(fg')^n = (fg')^m$ for some $m \neq n$; since $fg'$ has a right inverse (viz, $f'g$), we must have that $(fg')^n=\operatorname{id}$ for some $n>0$. So we see that $g'$ has not only a right inverse ($f'$) but also a left inverse: $(fg')^{n-1}f$. So they are the same and $g'$ is already an isomorphism.

REPLY [28 votes]: Hi Tilman. I believe I proved that (in your language) linearization reflects isomorphism. The following is a sketch. I will send you a more detailed version. The general case may be reduced to the case of prime fields $F_p$ and certain categories $C$ with fixed objects $x$ and $y$ and morphisms $f_1,\dots,f_m\colon x\to y$ and $g_1,\dots,g_n\colon y\to x$ subject to relations which correspond to the fact that $u=f_1+\dots+f_m$ and $u^{-1}=g_1+\dots+g_n$ are mutually inverse in the $F_p$-linearization. Apart from trivial cases, we may reindex these generators such that $f_1g_1 = 1_y$ and $g_nf_m=1_x$, while the other summands in the expansion of $uu^{-1}$ and $u^{-1}u$, respectively, fall into equivalence classes whose size is a multiple of $p$. It is then possible to derive a sequence of pairs
$(i_1,j_1),(i_2,j_2),\dots,(i_k,j_k)$
such that $f_{i_r}g_{j_r} = f_{i_{r+1}}g_{j_{r+1}}$ for $r=2,3,\dots,k-1$ and $g_{j_r}f_{i_r}=g_{j_{r+1}}f_{i_{r+1}}$ for $r=1,3,\dots,k-2$. Then
$f_{i_1}g_{j_2}f_{i_3}g_{j_4}\dots f_{i_k}$ and
$g_{j_k}f_{i_{k-1}}g_{j_{k-2}}f_{i_{k-3}}\dots g_{j_1}$
are mutual inverses of $C$.<|endoftext|>
TITLE: Classifying spaces for enriched categories
QUESTION [13 upvotes]: Is there a standard construction of a classifying space $BC$ for a category $C$ which is enriched which takes into account the enrichment? 
This is of course vague... The simplest example I can think of arises when $C$ is enriched over abelian groups or $k$-modules for some commutative ring $k$.

REPLY [3 votes]: Edit: Modified in accordance with Tom Leinster's entirely reasonable objections.
Sorry to exhume this question from 5+ years ago. In case someone is still looking for an answer, note that a very specific version (restricting to $V = \text{Cat}$) is addressed up to homotopy in the paper

M Bullejos and A Cegarra, On the geometry of 2-categories and their classifying spaces.  K-theory, 29:211 – 229, (2003). 

using the Duskin/Street nerve construction. Given a $V$-enriched category $C$, one constructs the simplicial set $\Delta C$ as follows: vertices are the objects of $C$, and higher simplices spanning objects $x_0,\cdots, x_d$ consist of 

1-morphisms $f_{ij}:x_i \to x_j$ for $0 \leq i \leq j \leq d$, and
2-morphisms $\alpha_{ijk}:f_{ik} \Rightarrow f_{jk} \circ f_{ij}$ for $0 \leq i \leq j \leq k \leq d$
subject to certain associativity and identity constructions (see the Introduction of the linked paper for details). 

One can also construct the Segal nerve as outlined in Chris Schommer-Pries's answer, but it is not as directly related to the objects and morphisms of the underlying enriched category. Bullejos and Cegarra show in the linked paper that the two constructions are naturally homotopy-equivalent so it doesn't matter too much either way!<|endoftext|>
TITLE: Is there a description of sheaf cohomology in algebraic-topological terms?
QUESTION [9 upvotes]: Is there a description of of sheaf cohomology for the sheaf of sections of a continuous function in terms of common constructions in Algebraic Topology?
In more detail: Any sheaf on a space X can be described as the sheaf of sections of some continuous map from the étale space Y to X. In fact, the category of sheaves (of sets) on X is equivalent to the category of maps to X which are local homeomorphisms. A sheaf of Abelian groups is the same as an Abelian group object in the category of sheaves of sets, so instead of talking about cohomology of sheaves, we could talk about cohomology of an Abelian group object in the category of local homeomorphisms to X, that is, a local homeomorphism from some space Y to X such that, roughly, every fibre has an Abelian group structure where all the multiplications (of the fibres) put together form a continuous map from Y × X Y to Y.
It seems like there should be a simple description of cohomology of X with coefficient in a sheaf of Abelian groups in terms of the corresponding map Y → X that uses only usual constructions in Algebraic Topology and the (fibrewise) group structure of Y. Is there one?

REPLY [3 votes]: I am pretty sure that you know what I'm going to say below, if it's correct, but maybe you or someone else can set me straight if I'm wrong. 
Let $X$ be a space, and let $\mathcal{F}$ be a sheaf of abelian groups on $X$. Then $\mathcal{F}$ defines a functor 
$\mathcal{O}(X) \to \mathbf{Ab}$ from the category $\mathcal{O}(X)$ of local homeomorphisms to $X$ to the category  $\mathbf{Ab}$ of abelian groups, which satisfies descent: that is, if $U \to X$ is any local homeomorphism, then $\mathcal{F}$ can be recovered in the usual way from its pull-backs to $U, U \times_X U, \dots$ (actually, you only need the first two, and no "dots"). 
Now, $\mathcal{F}$ also defines a functor $$\mathcal{F}: \mathcal{O}(X) \to \mathbf{Sp}$$ where $\mathbf{Sp}$ is the $\infty$-category of spectra (namely, taking values in Eilenberg-MacLane spectra in degree zero). There is a well-defined notion of a sheaf of spectra: it's one which satisfies an analogous homotopy descent condition where you take the whole cosimplicial thing for the homotopy limit rather than an equalizer (and for hypercovers rather than Cech covers). 
So $\mathcal{F}$ is a sheaf of abelian groups, but it's not a sheaf of spectra. In fact,  if you take the sheafification of $\mathcal{F}$ (as a sheaf of spectra), and take its homotopy groups, you get the sheaf cohomology groups of $\mathcal{F}$. If I am not mistaken, this follows from the (degenerate) descent spectral sequence: that is, to sheafify $\mathcal{F}$, you take the inverse limit of the totalizations over each hypercover, and this gives you the sheaf cohomology groups of $\mathcal{F}$ (that is, $\pi_i$ of the sheafification is $H^{-i}$ of the sheaf over that open set). 
Another way to check this is to treat $\pi_i$ of the sheafification as a $\delta$-functor on sheaves of abelian groups. The main thing to check is that if you have a sheaf of injective abelian groups, then this sheafification business doesn't give you anything new. Again, this follows from the descent spectral sequence, but there's probably another way  to do it. 
This doesn't quite answer your question: you want to describe the higher cohomology groups of $\mathcal{F}$ in terms of its "espace etale." 
I can't see how to do this in terms of the present discussion: we really needed
to be in a stable context, as the sheaf cohomology groups occur in negative
degrees. But again, I haven't thought too much about this.<|endoftext|>
TITLE: Sheaves and bundles in differential geometry
QUESTION [44 upvotes]: Because the theory of sheaves is a functorial theory, it has been adopted in algebraic geometry (both using the functor of points approach and the locally ringed space approach) as the "main theory" used to describe geometric data.  All sheaf data in the LRS approach can be described by bundles using the éspace étalé construction.  It's interesting to notice that the sheafification of a presheaf is the sheaf of sections of the associated éspace étalé.  
However, in differential geometry, bundles are for some reason preferred.  Is there any reason why this is true?  Are there some bundle constructions which don't have a realization as a sheaf?  Are there advantages to the bundle approach?

REPLY [3 votes]: In differential geometry one often also has connections on the bundles, e.g. the Levi-Cevita connection on the tangent bundle. Many concepts of differential geometry use connections, such as holonomy or geodesics. 
I'd love to learn the opposite to the following statement, but I have the feeling that there is no "nice" definition of a connection in the sheaf-theoretical approach. So I think this is another aspect why bundles are often preferred. 
As a little side remark, I'd like to point out that in higher differential geometry the story continues. Curiously, the word "gerbe" - which comes originally from the sheaf-theoretical side - is often at the same time used for the higher analog of a bundle. This is one of the reasons why different people associate so diverse meanings with "gerbe".<|endoftext|>
TITLE: Orthogonal complements in Hilbert bundles
QUESTION [5 upvotes]: It's a standard fact that for a finite-dimensional vector bundle with an inner product, the othogonal complement of any subbundle is itself a locally trivial vector bundle.
What is known about the corresponding question for infinite-dimensional bundles?  
Specifically, say $X$ is a smooth Riemannian manifold (modeled on a separable Hilbert space) and we are given a smoothly varying inner product (i.e. a metric) on the tangent bundle $TX$, which induces the correct topology on each tangent space (i.e. the metric is strong).  If $Y\subset X$ is a locally closed, smooth submanifold, then $TY$ is a subbundle of $TX|_Y$.  Is it known whether this guarantees that $TY^\perp$ is locally trivial?  What if $Y$ has finite codimension?  
Note that if the orthogonal projection operators associated to $T_y Y \subset T_y X$ vary smoothly, in an appropriate sense, we obtain a smooth surjection of vector bundles $TX|_Y \to TY$ and hence its kernel, $TY^\perp$, is locally trivial.
Of course one could ask similar questions over any base space.
Here is some background:
David Ebin has a paper from 1970 (The manifold of Riemannian metrics, MR0267604) in which he considers an orbit $O$ of the diffeomorphism group on the manifold $M$ of Riemannian metrics (on some underlying finite-dimensional smooth manifold).  (To be precise, Ebin works with $L^2_s$ versions of these spaces, rather than the usual $C^\infty$ versions.)  He considers a weak metric on $TM$ (one that does not induce the correct topology on $T_m M$), and goes to some effort to show that the (weak) orthogonal complement to $TO$ is a smooth subbundle of $TM$.  It seems to me that he's doing this because when working with the weak metric, it's easier to see that the orthogonal projections vary smoothly.  There is a strong metric in the picture here, and if the orthogonal complement with respect to the strong metric were automatically a smooth subbundle, then I don't know why Ebin would be using the weak metric (but I may be missing something).  In this situation, the orbits have infinite codimension, so I'm hoping the assumption of finite codimension is helpful.
Kondracki and Rogulski go through a similar process to Ebin's in their article MR0866577.

REPLY [3 votes]: I shall prefix this with my standard "rubber stamps":

This is not really an answer, but is a bit longer than a comment allows.
You should read "A Convenient Setting of Global Analysis" by Kriegl and Michor.  In particular, section 45 (Manifolds of Riemannian Metrics) and section 27.11ff (Submanifolds, in particular there's a good discussion of the necessity of the splitting condition).

I should also say that, as I mentioned in the comments above, I don't have access to the two articles cited so I can only speculate on what they are trying to achieve.
Firstly, many times in infinite dimensional analysis one wants to work with a space $X$ but it's tricky, so we work instead with a space $Y$.  Only there's not one particular choice for $Y$, there's lots.  And sometimes if we can say something for every such $Y$ in a compatible way then we can deduce that it also holds for our original $X$.  The key here is the "in a compatible way".  A common example is studying some infinite dimensional Frechet manifold (such as here) by expressing it as an inverse limit of Hilbert manifolds.  Now each of those Hilbert manifolds has its own Hilbertian structure, and so (for example) is diffeomorphic to an open subset of some Hilbert space.  But just because these exist, doesn't mean that they exist nicely with respect to each other.  So to make some general statement about all of them, sometimes you have to sacrifice the really strong structure you have and work with something in the middle.
In the case of loop spaces, to take an example I do know something a little about, then one can always consider the various Sobolev spaces of loops, say $L^s M$.  Each of those is a Hilbert manifold and its tangent bundle is thus a Hilbert bundle and can be given a strong metric.  However, the "natural" structure group of $T L^s M$ is $LO_n$, loops on the orthogonal group.  This only acts orthogonally on the "usual" Hilbert completion of $L\mathbb{R}^n$.  So any other orthogonal structure requires changing the structure group.  Admittedly, the space of all the choices is contractible, but that's homotopy theory and by passing to Hilbertian manifolds one is essentially declaring that one wants to do analysis so one would have to keep track of all the homotopies involved and keep taking them into account.  (To make the point a little clearer, it's the difference between knowing that a solution exists and actually going out and finding it.  If you really need to know the solution, knowing that it exists gives you a little hope but doesn't really help you actually write it down.)
I would also like to say that the usual classification of orthogonal structures into just "weak" and "strong" is a little simplistic.  There is almost always some addition structure in the background (usually related to the structure group) and taking it into account can give a much finer picture.  I have such a finer classification in my paper How to Construct a Dirac Operator in Infinite Dimensions together with examples of the different types.
So to return to the actual question.  Let me see if I can simplify it a little.  We can work locally, and in the actual question you are only concerned with what happens over the submanifold.  So we have some open subset of a model space, $U$, and a Hilbert space, $H$, two trivial bundles over $U$ modelled on $H$, say $E_1$ and $E_2$, and an inclusion of bundles $E_1 \to E_2$ such that the image of each fibre is closed.  Then we want to know if $E_2^\top$ is locally trivial.  We can, if we choose, impose some codimensionality conditions on the inclusions.
Adjointing, we have a smooth map $\theta : U \to \operatorname{Incl}(H,H)$ where $\operatorname{Incl}(H,H)$ is the space of closed linear embeddings of $H$ in itself.
Now we see where the crux of the matter lies.  What topology do we have on $\operatorname{Incl}(H,H)$?  We have lots of choices.  The two most popular are the strong and weak topologies.  Without making further assumptions, we can only assume that we have the weak topology!  Where this distinction comes into play is that the weak topology is very badly behaved.  With the weak topology, $\operatorname{Incl}(H,H)$ is not an ANR so there's no nice extension results.  With the strong topology, it's a CW-complex and so lots of nice things follow - in particular, orthogonal complementation will be continuous and the complement will be a locally trivial bundle.
Imposing finite codimensionality doesn't help either.  That's a bit like saying that you know that you end up in Fredholm operators.  Again, in the strong case then that's okay since the index is well-defined and so the complement will have constant dimension and be locally trivial.  With the weak topology, the index is not continuous as, with the weak topology, the space of Fredholm operators is contractible.
So there's where to find your counterexample: find such an inclusion $E_1 \to E_2$ such that the adjoint map is only continuous into the weak topology, not the strong one.  A good source of such examples is with the obvious representation of a Lie group $G$ on $L^2(G)$.  I expect that with a bit of bundle-crunching, this could be turned into an example.
It is just possible that the bits of your question that I threw out would save you (namely that the bundles were tangent bundles) but I doubt it since I expect that you could take an example as I outlined above and consider that as the inclusion of manifolds, then the tangent bundles of that inclusion would have the same properties of the inclusion itself.  That is, if you can find a counterexample, $E_1 \to E_2$, to my version of your question then viewing $Y = E_1$ and $X = E_2$ then I think you get a counterexample to your original setting.<|endoftext|>
TITLE: Alternative proof of unique factorization for ideals in a Dedekind ring
QUESTION [9 upvotes]: I'm writing some commutative algebra notes, but I'm facing a difficulty in organizing the order of the topics. I'd like to have the topics about factorization before speaking of integral closure. This is fine, as long as I talk of UFD and primary decomposition.
The problem is that a topic worth mentioning is the factorization theory for ideals in a Dedekind ring. Now, there are a few ways to define a Dedekind ring, but I guess one of the most natural is a Noetherian domain, integrally closed, of dimension 1.
At this point I haven't yet introduced the concept of dimension, nor integral closure. It is easy not to speak of dimension, and just say that every prime ideal is maximal. I'm also fine in writing out explicitly what integrally closed means.
The real problem is to get a proof of unique factorization for ideals without using anything about integral closure, apart from direct arguments. For instance, I'd be fine in saying: "...so this element satisfies this monic equation, hence it is in A." Less so in saying "...so this element lies in a ring which is finitely generated as an A-module, hence it is integral. Since it is in the field of fractions of A, is must belong to A."
The only missing step in proving that ideals in a Dedekind ring satisfy unique factorization is the fact that primary ideals are prime powers.

Is there a direct proof of this fact which does not rely on anything about integrally closed domains, apart from the definition?

I should make clear that other standard techniques are available at this point: localization, Noetherian and Artinian stuff, primary decomposition, symbolic powers and so on.
I should also say that changing the order of the topics woule a major headache. I have thought out for long the order, and this is the only point where I get things in the wrong order. If possible, I would like to leave it as it is.
Edit (added in response to KConrad comment).
The steps which are easy are the following. Since $A$ is Noetherian, a primary decomposition exists. Since every prime is maximal, there are no embedded primes, so all primary components are unique. Finally, using again that every prime is maximal, all primary components are coprime, so intersections become products. So the only step where one uses integrality is the proof that the primary ideals are actually prime powers.
For the definition, there is no need to speak about integral closure, let alone proving that the integral closure is a ring. The integral domain $A$ is said to be integrally closed if every element $x$ of the quotient field of $A$, which satisfies a monic equation $x^n + a_{n-1} x^{n-1} + \cdots + a_0$ with coefficients in $A$, is itself in $A$.
As for the examples, the compromise for now is to list some number rings, with the promise that it will be shown in a later section that these are actually Dedekind rings. Of course I'm not happy with this solution. But I'm also not happy with putting an aside on integral closure in the middle of a section about factorization and primary decomposition; even less so because there IS a later section on integral closure.
I cannot even reverse the two, because in the section of integral closure I want to be able to speak about the integral closures of $\mathbb{Z}$, so I need the factorization theory for Dedekind rings.

REPLY [2 votes]: You could define a Dedekind ring as a noetherian domain s.t. the localization at any nonzero prime ideal is a discrete valuation ring (see the beginning of Serre's Local fields).
From there it is easy to show unique factorization.
It is of course a cheat since the equivalence between the two definitions relies on the "integrally closed" property.

REPLY [2 votes]: Here is a route I took when teaching this material to bright high school students at PROMYS. (Of course, not using this language.) Let $R$ be a subring of $\mathbb{C}$, of finite rank over $\mathbb{Z}$. I was actually only doing the particular case of $\mathbb{Z}[\sqrt{-D}]$, for some positive integer $D$, but you could presumably be more general with your audience. 
It is easy to show that ideal classes form a semi-group, and that this semi-group is finite (using Minkowski's theorem). Moreover, the proof is constructive; they can compute the class semi-group in practice without difficulty. It is also easy to show that, if the class semi-group is a group, then unique factorization into prime ideals holds. 
I then had them compute lots of examples, and see that the class semigroup often was a group. You can then discuss those examples without mentioning integral closure at all. When you do get to integral closure, you can have them check their list of examples and see that the class semi-group is a group precisely when the ring is integrally closed. Hopefully, this will make the notion seem better motivated. I never actually got to proving that "all ideals invertible" is equivalent to "integrally closed", but I don't see why I couldn't have if I had more time.
In your setting of general commutative algebra, my proposal is to define the class semi-group; show that one dimensional, Noetherian and class semi-group is a group implies unique factorization into ideals; and compute class semi-groups, using Minkowski's theorem, for the number fields which you wish to exhibit.<|endoftext|>
TITLE: If $2^x $and $3^x$ are integers, must $x$ be as well?
QUESTION [137 upvotes]: I'm fascinated by this open problem (if it is indeed still that) and every few years I try to check up on its status. Some background: Let $x$ be a positive real number.

If $n^x$ is an integer for every $n \in \mathbb{N}$ then $x$ must be an integer. This is a fun little puzzle.
If $2^x$, $3^x$ and $5^x$ are integers then $x$ must be an integer. This requires fairly sophisticated tools, and can be derived from the results in e.g. Lang, Algebraic values of meromorphic functions. II., Topology 5, 1966.
Finally, if all you know is that $2^x$ and $3^x$ are integers, then as far as I know it is not known if $x$ is forced to be an integer (unbelievable, isn't it?). Although of course one can never be certain, I am quite sure this was still the case as recently as 2003. 

So the question is, is that still an open problem, and is there any sort of relevant progress that may provide some hope?

REPLY [50 votes]: This question came up recently on the NMBRTHRY mailing list and I can't resist paraphrasing a comment I made there. So, perhaps surprisingly, this question has links to automorphic forms! For if $x$ is a complex number and $||.||^x$ is the associated Grossencharacter of the ideles of $\mathbf{Q}$, that is, the map $\mathbf{A}_\mathbf{Q}^\times/\mathbf{Q}^\times\to\mathbf{C}^\times$ sending an idele to the $x$'th power of its norm, then the assumption that $p^x$ is an integer for all primes $p$ (which is clearly equivalent to the assumption that $n^x$ is an integer for all $n$) implies that the grossencharacter is arithmetic. Now a standard conjecture in the theory of automorphic representations is that an automorphic representation is arithmetic iff it's algebraic, and this conjecture is a theorem for tori, so the theorem in this case says that $||.||^x$ is algebraic which is precisely the statement that $x$ is an integer!
So for tori over general number fields it's a theorem of Waldschmidt that arithmetic implies algebraic for automorphic forms. So in practice we get a vast generalisation of the first question above, where integers can be replaced by algebraic integers and where we can add finite order characters and so on.
As an example, one sees that if $x$ is complex and if there's a number field $E$ in the complexes such that $n^x$ is an integer in $E$, for all $n$, then again the grossencharacter is arithmetic, so algebraic, and hence $x$ must be an integer. I don't know if there's any low-level proof of this (but it follows from standard transcendence theorems). As other examples $n$ can be replaced by the algebraic integers in a number field and so on.<|endoftext|>
TITLE: Prescribing Gaussian curvature
QUESTION [11 upvotes]: Let $K(r)$ be the piecewise function
                                            

I want to solve the PDE $$\Delta u + K(|x|) e^{2u} = 0$$ for radially symmetric $u$ with boundary condition $u = 0$ at infinity.  I'm content for a solution in two dimensions, so switching to polar coordinates this is the ODE $$\tfrac 1 r u' + u'' + K(r) e^{2u} = 0.$$
The motivation is that the Riemannian metric $g_{ij} = e^{2u} \delta_{ij}$ has curvature profile $K$.  This should be a relatively simple exercise in geometric PDE, and is surely explained in the literature.  Could you please point me toward a reference which explains how to solve this equation explicitly?

REPLY [8 votes]: I'm not sure that this will help, but let me suggest thinking about the following:  You are looking for a metric of the form $g = e^{2u(r)}(dr^2 + r^2\ d\theta^2)$ where $u(r)$ is to be chosen so that the curvature of $g$ is a certain function $K(r)$ and so that $u$ tends to zero as $r\to\infty$.    Now, I wouldn't have called this problem "specifying the curvature profile" just because $r$ won't represent the $g$-distance from the origin when you are done.  Instead, the $g$-distance $s= h(r)$ from the origin will be given by solving $ds = e^{u(r)}\ dr$ with $s(0)=0$, and I would have called $K\bigl(h^{-1}(s)\bigr)$ the 'curvature profile'.
Are you sure that you wouldn't have rather had the metric in the form $g = ds^2 + f(s)^2\ d\theta^2$ where $f(0)=0$ and $f'(0)=1$ and then choose $f$ so that it satisfies the equation 
$$
f''(s) + K(s)\ f(s) = 0
$$
where $K$ is your given function?  
If this is really your problem (and I'm not saying it has to be, but...), then you can, indeed, solve for $f$ explicitly, in a sense, but its definition will be piecewise, of course.  You'll have $f(s) = \sin s$ for $0\le s \le 1$, but on the intervals $1\le s\le 3$ and $3\le s\le 4$, $f$ will be given in terms of translated Airy functions (different ones on the different intervals), and then, for $s\ge 4$, you'll have $f$ be a linear expression in $s$.  Of course, determining the constants at the breakpoints so that $f$ is $C^2$ there is probably not going to be doable in any fully explicit fashion.<|endoftext|>
TITLE: Newlander-Nirenberg for surfaces
QUESTION [8 upvotes]: Quite a long ago, I tried to work out explicitly the content of the Newlander-Nirenberg theorem. My aim was trying to understand wether a direct proof could work in the simplest possible case, namely that of surfaces. The result is that the most explicit statement I could get is a PDE I was not able to solve.
Assume a quasi-complex structure $J$ is given on the surface $S$; we want to prove that this is induced by a complex one (in this case there are no compatibility conditions). This can be easily transformed in the problem of local existence for a second order PDE, as follows.
We look for local charts on $S$ which are holomorphic (with respect to the quasi-complex structure on $S$). Two such charts are then automatically compatible. So the problem is local.
Fix a small open set $U \subset S$ and identify it with a neighboorhood of $0 \in \mathbb{R}^2$ via a differentiable chart. Locally we can write $J = \left(\begin{matrix}[a | b] \\ [c | d ]\end{matrix}\right)$ for some functions $a, \cdots, d$ (pretend it is a two by two matrix).
A chart is given by a complex valued function $f = u + iv$. The condition that the differential is $\mathbb{C}$-linear can be verified on a basis of the tangent space; moreover if it is true for a vector v, it remains true for Jv, which is linearly independent. Here we have used that $J^2 = -1$.
So we need only to check it for the vector $\partial_x$. Since $J \partial_x = a \partial_x + c \partial_y$, the condition says
$-v_x = a u_x + c u_y$
$u_x = a v_x + c v_y$
Hence we need to solve this system, with $f = u + i v$ non singular ($f$ will be then locally invertible). Since $a$ and $c$ do not vanish simultaneously, we can assume $c(0) \neq 0$, hence $c \neq 0$ on $U$ provided $U$ is small.
We can then solve for $u_y$ and get the equivalent system
$u_x = a v_x + c v_y$
$-u_y = \frac{1 + a^2}{c}v_x +a v_y$
Moreover the Jacobian $J_f = u_x v_y + u_y v_x = \frac{1}{c}(v_x^2 + (a v_x + c v_y)^2)$, so $f$ is nonsigular if $v$ is. By Poincaré's lemma, the system admits a local solution if and only if
$\frac{\partial}{\partial_y} \left( a v_x + c v_y \right) - \frac{\partial}{\partial_x} \left( \frac{1 + a^2}{c}v_x +a v_y \right) = 0$.
Hence we are looking for a local solution of the last equation with $(v_x(0), v_y(0)) \neq (0, 0)$.
So my question is:

Is there a simple way to prove local existence for a nonsingular solution of the last displayed equation?

I should make clear that I'm not looking for a proof of Newlander-Nirenberg; of this there are plenty. I am more interested in seeing what Newlander-Nirenberg becomes in terms of PDE in the simplest possible case, and then see that the PDE thus obtained is solvable. According to the answer of Andy, the equation which comes out is the Beltrami equation, so I will have a look at it. Still, I'm curious if any standard PDE technique can solve the equation I derived in the most stupid way above.

REPLY [7 votes]: I'm not sure if the following is elementary enough, but it does only use standard PDE machinery (plus some basic Riemannian geometry).  It's also nice in that it suggests an approach to proving the uniformization theorem (via metrics of constant curvature).
Say you have an almost-complex structure on the unit disk.  Your goal is to find a conformal isomorphism of this disk (or at least some neighborhood of the origin) with an open subset of the complex plane.  To do it, first choose a metric $g$ on the disk which is compatible with the given complex structure.  Let $K$ be the curvature of this metric.  If you can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then you'll be done, since the exponential map with respect to $\tilde{g}$ will be an isometry, hence also a conformal isomorphism. 
So, multiply $g$ by an arbitrary positive function $e^f$, and compute the curvature of the new metric. You'll find that it's given by the formula:
$\tilde{K} = e^{-2f}(K - \Delta f) $ 
where $\Delta$ is the Laplacian with respect to the metric $g$.  Setting the left hand side equal to zero, you have reduced to solving the Laplace equation, which can be done locally using standard PDE techniques.
As for a more "direct" approach...
The equation you wrote down should reduce to solving the Laplace equation as well, using the notion of conjugate harmonic functions.  However, solving your equation will inevitably be a bit more subtle due to your requirement that the solution have nonvanishing differential at the origin.  There is a proof along the lines you're suggesting in Taylor's PDE book, chapter 5, section 11, and I think there's a similar one in Jost's "Postmodern analysis".  Basically the idea is to rescale your coordinate system so that the metric is nearly flat, in which case you should have a conformal map that is close to the identity map in a high enough sobolev space, and therefore has a nonvanishing derivative at the origin.<|endoftext|>
TITLE: A Model Structure on Symmetric Monoidal Categories
QUESTION [6 upvotes]: The recent article found here revisits Thomason's proof that symmetric monoidal categories model all connective spectra, but stops short of showing that there is a full closed model structure on this category (as does, it seems, Thomason's original paper.)  Is there such a thing?
My guess is some lifting similar to how the model structure on small categories is derived would work, but I'm not sure if there are any complications.

REPLY [9 votes]: One basic problem is that the category of symmetric monoidal categories isn't complete.  Its completion, in a basic sense, is the category of multicategories, on which it seems reasonable to conjecture there is a model category structure whose homotopy category "is" the connective part of stable homotopy -- we hope to prove this soon.  See Elmendorf and Mandell, "Permutative categories, multicategories, and algebraic K-theory", which just appeared in Algebraic and Geometric Topology.<|endoftext|>
TITLE: Path integrals, localisation
QUESTION [12 upvotes]: Physicists use the "Atiyah-Bott formula" for path "integrals" (for instance the supersymmetric proof of the Atiyah-Singer index theorem. Is there some way to make atleast some of these ideas rigorous? As in is there a rigorous localisation formula for infinite dimensional integration? (i.e. using the Weiner measure when applicable?)

REPLY [2 votes]: There is the work of Johan Martens on equivariant localization of nonabelian group actions on non-compact spaces ( http://arxiv.org/abs/math/0609841 ), but I think that the general problem as you posed it is still open, seeing that there is an AIM workshop partly on this question next week: http://www.aimath.org/ARCC/workshops/localization.html
edit: Charlie Frohman is right, I didn't read the AIM workshop page carefully enough.
As for the original question, I have only encountered localization techniques in physics when calculating instanton contributions to the path integral, which gives equivariant localization on finite dimensional spaces (although badly behaved in other ways). The supersymmetric "proof" of the Atiyah-Singer index theorem can be made quite rigorous (see the 1983 paper by Getzler for example), without any recourse to infinite dimensional spaces. Most of the articles I read from the physics literature skipped over the infinite dimensional case, with some reference to a nonrenormalization theorem.
To my latest knowledge (around middle of 2008) there is no general method for equivariant localization in infinite dimensional spaces. Also for most high-energy physics purposes, Wiener measures are not the right approach for doing Feynman path integrals.<|endoftext|>
TITLE: Triangulating surfaces
QUESTION [49 upvotes]: I've had a few undergraduate students ask me for references for the classical fact (due to Rado) that closed topological surfaces can be triangulated.  I know two sources for this, namely Ahlfors's book on Riemann surfaces and Moise's book "Geometric topology in dimensions 2 and 3".  Both of these strike me as being a bit much for a bright undergraduate.  Question : in the 30+ years since Moise's book, has anyone written a more accessible account?

REPLY [3 votes]: This is not an answer that points to a more recent and more accessible account of the triangulability of surfaces, but rather a way to make the account in the first chapter of Ahlfors' & Sario's book more accessible, if sufficient time is available. It should be noted that the proof given by Ahlfors & Sario works for all (connected, 2nd countable) surfaces: compact or noncompact; with or without boundary. I will describe what I found to be the difficulties with Ahlfors' & Sario's presentation and how one can mitigate these difficulties, especially if the learning of this material is by self-study, as was the case for me, and not in the context of a university course. Disclaimer: I am a mathematician but not a topologist.
I found there were three main difficulties, all stemming from Ahlfors' & Sario's terse style of writing. The first difficulty is the absence of any references for background material. I found that the classic, self-contained book, Elements of the Topology of Plane Sets of Points (2nd ed.), by M.H.A. Newman, and the first two sections of the third chapter of the book, Algebraic Topology, by E. Spanier (for the basics of the theory of abstract simplicial complexes), provide sufficient background. The second difficulty is that almost every sentence resembles the statement of a lemma whose proof is left to the reader. The third difficulty is an intentionally omitted proof of a rather difficult result, "46C".
Regarding the second and third difficulties: after filling in the missing details, I decided to write them up in the form of a list of notes (as opposed to an article). I then created a website, on which I posted these notes. Included in them, is a proof of the result "46C" that relies heavily on the material in the cited book by Newman. Although I did not strive for either optimum mathematical efficiency or elegance, perhaps my notes will be useful to others, in making Ahlfors' & Sario's account more accessible.
While I was at it, I also posted some details for a proof of Schoenflies' Theorem via Complex Analysis; these are details for the presentation in the book, Boundary Behaviour of Conformal Maps, by C. Pommerenke. Note that this proof assumes the Riemann Mapping Theorem, proofs of which are more widely available. (These notes were written before I was aware of Newman's book, which happens to also contain a proof of Schoenflies' Theorem -- a proof that is purely topological in nature. As mentioned in Allen Hatcher's answer, an historical account of proofs of Schoenflies' Theorem, including a new one at the time, appeared in the paper and its errata. A preprint of the paper is freely available here.)
A careful accounting of all the prerequisite material for Ahlfors' & Sario's approach, reveals that it is a very substantial amount. Although such a proof of the triangulability of surfaces can be made accessible to undergraduates, it is not clear whether an undergraduate who embarks on such a project of self-study will still be an undergraduate upon completion of the project.<|endoftext|>
TITLE: On the failure of the infinite dimensional Brouwer Theorem
QUESTION [5 upvotes]: Let $K$ be the closed unit ball of some infinite dimensional Banach
space, and let $H$ be an autohomeomorphism of $K$, having fixed
points. Can $H/2$ be fixed point free ?
Also, let ${\mathcal{F}}$ := { $S\in\mbox{C}(K,K), \mbox{Fix}(S)\neq\textrm{Ø } $}.
Let $T$ in $\mbox{C}(K,K)$  such that $TS\in\mathcal{F}$ for all $S\in\mathcal{F}$
. Must $T$ be necessarily compact ?

REPLY [3 votes]: Indeed, $H/2$ can be fixed point free. An example for this can be constructed as follows: take $X = c_0(\mathbb{Z},\mathbb{R})$, the Banach space of bounded, null-convergent, real sequences indexed by the integers $\mathbb{Z}$. Now take an equicontinuous sequence of homeomorphisms $f_i:[-1,1] \to [-1,1], i\in \mathbb{Z}$ with the property that $f_i(0) \to 0$ as $i\to \infty$ and so that the sequences of inverses is equicontinuous as well.
We can now construct a homeomorphism of $H:K\to K$ by
$$ (H(x))_{i+1} = f_i(x_i) .$$
The map can be extended to the space of all twosided sequences with values in $[-1,1]$ and there it is obviously a bijection. Also sequences in $K$ are mapped to sequences in $K$ by the assumption of equicontinuity and the convergence $f_i(0) \to 0$. Similarly, the preimage of a sequence in $K$ is in $K$. Finally, equicontinuity gives continuity of $H$ and its inverse.
Now the fixed points of $H$ are those sequences $x=(x_i)_{i\in \mathbb{Z}}$ such that 
$$ x_{i+1} = f_i(x_i) \quad \forall i\in \mathbb{Z}. \tag{1}$$
I.e., the fixed points in $K$ are those solutions of the nonautonomous dynamical system (1) which have the property that they converge to $0$.
The trick is now to construct a nonautonomous system of the described form which has a solution converging to zero and so that the system 
$$ x_{i+1} = \frac{1}{2} \ f_i(x_i) . \tag{2}$$
does not have this property. For this it suffices to give the system (2) a globally attractive fixed point away from $0$. In this way no solution of (2) lies in $X$ and so $H/2$ is fixed point free.
A concrete example for (1) is as follows. Take $f_i=\mathrm{id}$ for $i< 0$. For $i\geq 0$ define $f_i$ by the linear interpolation of the following points:
$$ (-1,-1) \ , \quad (-1/4, -1/2) \ , \quad (2^{-(i+2)},0)\ , \quad (2^{-(i+1)},2^{-(i+2)}) ,\, \quad (1,1). $$
A solution $x$ of (1) is then given by $x_i = 1/2, i\leq 0$, $x_i = 2^{-i}, i>0$. And we have a fixed point of $H$ as $x \in K$.
The maps defining system (2) are interpolations of the points
$$ (-1,-1/2) \ , \quad (-1/4, -1/4) \ , \quad (2^{-(i+2)},0)\ , \quad (2^{-(i+1)},2^{-(i+3)}) \, \quad (1,1/2). $$
Note the nice fixed point at $(-1/4,-1/4)$ and it is not hard to see that all solutions become negative in finite positive time as the solution $x$ with the condition $x(0) = 1$ is $ ...,1,1/2,5/24,17/224,101/4480,...$ and at this stage the solution becomes negative. Also solutions are strictly decreasing if they live in the (invariant) interval $(-1/4,0)$. Thus $H/2$ has no fixed point as there are no solutions of $(2)$ that converge to $0$.
(BTW: I was worried there for a moment that I am contradicting the Banach fixed point theorem, but this is not the case, luckily. The Lipschitz constants of the $f_i$ converge to $2$, so that $H/2$ is not a contraction.)
Re question 2: The identity map would be a $T$ that works. This is not compact. If there are other maps that work I do not know.<|endoftext|>
TITLE: Realizations and pinnings (épinglages) of reductive groups
QUESTION [9 upvotes]: Let $G$ be a reductive group over an (say, algebraically closed) field $k$. Springer (in his book on algebraic groups) calls for a chosen maximal torus $T$ in $G$ a family $(u_\alpha) _{\alpha \in \Phi(G,T)}$ of immersions $u _\alpha:\mathbf{G}_a \rightarrow G$ such that
(i) $t u_\alpha(c) t^{-1} = u_\alpha( \alpha(t) c)$ for all $c \in \mathbf{G}_a$ and $t \in T$,
(ii) $n_\alpha := u_\alpha(1) u_{-\alpha}(-1) u_\alpha(1)$ lies in $\mathrm{N}_G(T) \setminus T$,
(iii) $u_\alpha(x) u_{-\alpha}(-x^{-1}) u_\alpha(x) = \alpha^\vee(x) n_\alpha$ for all $x \in k^\times$,
a realization of (G,T) (or $\Phi(G,T)$) in $G$. We then have $\mathrm{Im}(u_\alpha) = U_\alpha$.
In the book of Conrad-Gabber-Prasad on pseudo-reductive groups a pinning of $G$ is defined as a tuple $(T,\Phi^+,(\varphi_\alpha)_ {\alpha \in \Delta})$ where $T$ is a maximal torus, $\Phi^+$ is a positive system for $\Phi(G,T)$, $\Delta$ is the corresponding basis and $\varphi _{\alpha}: (\mathrm{SL} _2, \mathrm{SL} _2 \cap \mathrm{D} _2) \rightarrow (G _\alpha, G _\alpha \cap T)$ are central isogenies such that $\varphi _\alpha( \mathrm{diag}(x,x^{-1}) ) = \alpha^\vee(x)$ for all $x \in k^\times$, where $G _\alpha = \langle U _\alpha,U _{-\alpha} \rangle$.
My question is: are these two notions somehow equivalent? If a pinning is given, then by defining 
$u_\alpha(x) = \varphi_\alpha\begin{pmatrix} 1 & x \\\\ 1 & 0 \end{pmatrix}$,    $u_{-\alpha}(x) = \varphi_\alpha\begin{pmatrix} 1 & 0 \\\\ x & 1 \end{pmatrix}$
I get closed immersions satisfying the properties above, but unfortunately, as I have $\varphi_\alpha$ only for $\alpha \in \Delta$ this does not yet define a realization. How can I define the $u_\alpha$ for $\alpha \notin \Delta \cup -\Delta$? What about the other direction?
Moreover (as C-G-P also mentions) in SGA3, exposé XXIII, there is defined the notion of épinglages and Conrad mentions that these carry the same information as the pinnings above. Can somebody make this precise? Moreover in SGA, it is mentioned that an épinglage induces monomorphisms $p_\alpha: \mathbf{G}_a \rightarrow G$ for $\alpha \in \Delta \cup -\Delta$. I suspect that these are the morphisms I defined above, but again, can I get a realization from this?
A further problem is the following: For a given realization and a total order on $\Phi(G,T)$ Springer defines structure constants which appear in the expression of the commutator $\lbrack u_\alpha(x), u_\beta(y) \rbrack $ in terms of $u_\gamma$ for linearly independent $\alpha, \beta \in \Phi$. Springer shows that for root systems NOT of type $G_2$ a realization with integral structure constants exist. Demazure also calculates these commutators in SGA3, exposé XXII, for the $p_\alpha$ mentioned above in case of rank 2 root systems. Here, I was surprised that the structure constants seem to be independent of the pinning chosen. Is this now a rank 2 phenomenon that is also true for realizations or does this mean that pinnings/épinglages are more restrictive than realizations? 
I hope, somebody can help me here.

REPLY [17 votes]: OK, here's the deal.  
I.  First, the setup for the benefit of those who don't have books lying at their side. Let $(G,T)$ be a split connected reductive group over a field $k$, and choose $a \in \Phi(G,T)$ (e.g., a simple positive root relative to a choice of positive system of roots).  Let $G_a$ be the $k$-subgroup generated by the root groups $U_a$ and $U_{-a}$.  (Recall that $U_a$ is uniquely characterized as being a nontrivial smooth connected unipotent $k$-subgroup normalized by $T$ and on which $T$ acts through the character $a$.) This is abstractly $k$-isomorphic to ${\rm{SL}}_ 2$ or ${\rm{PGL}}_ 2$, and $G_a \cap T$ is a split maximal torus.
So there is a central isogeny $\phi:{\rm{SL}}_ 2 \rightarrow G_a$ (either isomorphism or with kernel $\mu_2$), and since ${\rm{PGL}}_ 2(k)$ is the automorphism group of ${\rm{SL}}_ 2$ and of ${\rm{PGL}}_ 2$ over $k$ there is precisely this ambiguity in $\phi$ (via precomposition with its action on ${\rm{SL}}_ 2$).  The burning question is:  to what extent can we use $T$ and $a$ to nail down $\phi$ uniquely?
The action of $G_a \cap T$ on $U_a$ is via the nontrivial character $a$, and among the two $k$-isomorphisms $\mathbf{G}_ m \simeq G_a \cap T$ the composition with this character is $t \mapsto t^{\pm 2}$ in the ${\rm{SL}}_ 2$-case and $t \mapsto t^{\pm 1}$ in the ${\rm{PGL}}_ 2$-case. Fix the unique such isomorphism making the exponent positive.  
Now back to the central isogeny $\phi:{\rm{SL}}_ 2 \rightarrow G_a$. By conjugacy of split maximal tori (when they exist!) we can compose with a $G_a(k)$-conjugation if necessary so that $\phi$ carries the diagonal torus $D$ onto $G_a \cap T$.  Recall that we used $a$ to make a preferred isomorphism of $G_a \cap T$ with $\mathbf{G}_ m$.  The diagonal torus $D$ also has a preferred identification with $\mathbf{G}_ m$, namely  $t \mapsto {\rm{diag}}(t, 1/t)$.  Thus, $\phi: D \rightarrow G_a \cap T$ is an endomorphism of $\mathbf{G}_ m$ with degree 1 or 2, so it is $t \mapsto t^{\pm 2}$ (${\rm{PGL}}_ 2$-case) or $t \mapsto t^{\pm 1}$ (${\rm{SL}}_ 2$-case).  Since the standard Weyl element of ${\rm{SL}}_ 2$ induces inversion on the diagonal torus, by composing with it if necessary we can arrange that $\phi$ between these tori uses the positive exponent.  That is exactly the condition that $\phi$ carries the standard upper triangular unipotent subgroup $U^+$ onto $U_a$ (rather than onto $U_{-a}$). 
II. So far, so good: we have used just $T$ and the choice of $a \in \Phi(G,T)$ to construct a central isogeny $\phi_a:{\rm{SL}}_ 2 \rightarrow G_a$ carrying $U^+$ onto $U_a$ and $D$ onto $G_a \cap T$, with the latter described uniquely in terms of canonically determined identifications with $\mathbf{G}_ m$ (as $t \mapsto t$ or $t \mapsto t^2$). The remaining ambiguity is precomposition with the action of elements of ${\rm{PGL}}_ 2(k)$ that restrict to the identity on $D$, which is to say the action of $k$-points of the diagonal torus $\overline{D}$ of ${\rm{PGL}}_ 2$. Such action restrict to one on $U^+$ that identifies $\overline{D}(k)$ with the $k$-automorphism group of $U^+$ (namely, $k^{\times}$ with its natural identification with ${\rm{Aut}}_ k(\mathbf{G}_ a)$). 
Summary: to nail down $\phi$ uniquely it is equivalent to specify an isomorphism of $U^+$ with $\mathbf{G}_ a$.  But $\phi$ carries $U^+$ isomorphically onto $U_a$.  So it is the same to choose an isomorphism of $U_a$ with $\mathbf{G}_ a$.  Finally, the Lie functor clearly defines a bijection $${\rm{Isom}}_ k(\mathbf{G}_ a, U_ a) \simeq
{\rm{Isom}}(k, {\rm{Lie}}(U_ a)).$$
So we have shown that to specify a pinning in the sense of Definition A.4.12 of C-G-P is precisely the same as to specify a pinning in the sense of SGA3 Exp. XXIII, 1.1.
III. Can we improve a pinning to provide unambiguous $\phi_c$'s for all roots $c$?  No, there is a discrepancy of units which cannot be eliminated (or at least not without a tremendous amount of work, the value of which is unclear, especially over $\mathbf{Z}$), and if we insist on working over $k$ and not $\mathbf{Z}$ then there are further problems caused by degeneration of commutation relations among positive root groups in special cases in small nonzero characteristics (e.g., ${\rm{G}}_ 2$ in characteristic 3 and ${\rm{F}}_ 4$ in characteristic 2).
As we saw above, to nail down each $\phi_c$ it is equivalent to do any of the following 3 things:  fix an isomorphism $\mathbf{G}_ a \simeq U_c$, fix a basis of ${\rm{Lie}}(U_ c)$, or fix a central isogeny ${\rm{SL}}_ 2 \rightarrow G_c$ carrying $D$ onto $G_c \cap T$ via $t \mapsto t$ or $t \mapsto t^2$ according to the canonical isomorphisms of $\mathbf{G}_ m$ with these two tori (the case of $G_c \cap T$ being determined by $c$).  This latter viewpoint provides $\phi_{-c}$ for free once $\phi_c$ has been defined (compose with conjugation by the standard Weyl element), so the problem is to really define $\phi_c$ for $c \in \Phi^+$. 
Consider the unipotent radical $U$ of the Borel corresponding to $\Phi^+$, so $U$ is directly spanned (in any order) by the $U_c$'s for positive $c$.  If we choose an enumeration of $\Phi^+$ to get an isomorphism $\prod_c U_c \simeq U$ of varieties via multiplication, then for simple $c$ we have a preferred isomorphism of $U_c$ with $\mathbf{G}_ a$ and one can ask if the isomorphism $\mathbf{G}_ a \simeq U_c$ can be determined for the other positive $c$ so that the group law on $U$ takes on an especially simple form. This amounts to working out the commutation relations among the $U_a$'s for $a \in \Delta$ (when projected in $U_c$ for various $c$), and by $T$-equivariance such relations will involve monomials in the coordinates of the $U_a$'s along with some coefficients in $k^{\times}$ (and some coefficients of 0).  These are the confusing "structure constants".  Chevalley developed a procedure over $\mathbf{Z}$ to make the choices so that the structure constants come out to be positive integers (when nonzero), but there remained some ambiguity of signs since 
$${\rm{Aut}}_ {\mathbf{Z}}(\mathbf{G}_ a) = \mathbf{Z}^{\times} = \{\pm 1\}.$$
Working entirely over $k$, there are likewise $k^{\times}$-scalings that cannot quite be removed.  I am told that Tits developed a way to eliminate all sign confusion, but honestly I don't know a reason why it is worth the heavy effort to do that.  For most purposes the pinning as above is entirely sufficient, and this is "justified" by the fact that all ambiguity from $(T/Z_G)(k)$-action is eliminated in the Isomorphism Theorem when improved to include pinnings (see Theorem A.4.13 in C-G-P). 
IV.  What about the realizations in the sense of Springer's book?  All he's doing is making use of the concrete description of the group law on ${\rm{SL}}_ 2$ to describe a central isogeny $\phi_c$ for every positive root $c$.  (His conditions relate $c$ with $-c$, which amounts to the link between $\phi_c$ and $\phi_{-c}$ in the preceding discussion.)  As long as he restricts to $\alpha \in \pm \Delta$ then he's just defined a pinning in the above sense.  But he goes further to basically do what is described in II but without saying so explicitly.  He then has to confront the puzzle of the structure constants.  (It is a real puzzle, since in the theory over $\mathbf{Z}$, which logically lies beyond the scope of his book, the structure constants are not always $0$ and $\pm 1$; in some rare cases one gets coefficients of $\pm 2$ or $\pm 3$, which implies that if one insists on working over fields and not over $\mathbf{Z}$ then life in characteristic 2 and 3 will look a bit funny in some cases.)  The only conceptual way I know of to overcome the puzzle of the structure constants is to work over $\mathbf{Z}$ and to follow either SGA3 or Chevalley in this respect.  For the former, one has to really do the whole theory of reductive groups over a base that is not necessarily a field.  For the latter, perhaps the (unpublished?) Yale notes of Steinberg are the best reference.<|endoftext|>
TITLE: How to solve $f(f(x)) = \cos(x)$?
QUESTION [116 upvotes]: I found the following equation on some web page I cannot remember, and found it interesting:
$$f(f(x))=\cos(x)$$
Out of curiosity I tried to solve it, but realized that I do not have a clue how to approach such an iterative equation except for trial and error. I also realized that the solution might not be unique, from the solution of a simpler problem
$$f(f(x)) = x$$
which has, for example, solutions $f(x) = x$ and $f(x) = \frac{x+1}{x-1}$.
Is there a general solution strategy to equations of this kind? Can you perhaps point me to some literature about these kind of equations? And what is the solution for $f(f(x))=\cos(x)$ ?

REPLY [7 votes]: This does not directly answer the question, which concerned iterative roots on the real axis. However, I have also seen complex solutions mentioned in some of the discussions, so it seems relevant to mention the following.
Baker (in "The iteration of entire transcendental functions and the solution of the functional equation f{f(z)} = F(z)") proves the following.
Theorem 1. If $F(z)$ is an entire function of finite order bounded on some continuous curve $\Gamma$ which extends to infinity, then the functional equation $f\{f(z)\} = F(z)$ has no entire solution.
Recall that finite order means that $\log_+ \log_+ |F(z)| = O(\log|z|)$ as $z\to\infty$. Of course, $\cos$ has order $1$, and is bounded on the real axis. So the equation $f(f(z)) = \cos(z)$ has no analytic solution defined on the entire complex plane. 
Baker's argument is beautifully simple: Firstly, it is well-known that the composition of two functions of positive order cannot have finite order, so any such solution $f$ would have order zero. Now, either $f$ is itself bounded on $\Gamma$, or $f(\Gamma)$ is unbounded and $f$ is bounded on $f(\Gamma)$. In either case, $f$ is bounded on an unbounded connected set. But it is well-known that this is impossible for functions of order less than $1/2$.<|endoftext|>
TITLE: A question about ordinal definable real numbers
QUESTION [43 upvotes]: If ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice) is consistent, does it remain consistent
when the following statement is added to it as a new axiom? 
"There exists a denumerably infinite and ordinal definable set of real numbers, not all of whose elements
are ordinal definable"
If the answer to the above question is negative, then it must be provable in ZFC that every denumerably
infinite and ordinal definable set of real numbers is hereditarily ordinal definable. This is because
every real number can be regarded as a set of finite ordinal numbers and every finite ordinal number is
ordinal definable.
                                                                        Garabed Gulbenkian

REPLY [23 votes]: The original problem solves in the positive: there is a model of ZFC in which there exists a countable OD (well, even lightface $\Pi^1_2$, which is the best possible) set of reals $X$ containing no OD elements. The model (by the way, as conjectured by Ali Enayat at http://cs.nyu.edu/pipermail/fom/2010-July/014944.html) is a $\mathbf P^{<\omega}$-generic extension of $L$, where $\mathbf P$ is Jensen's minimal $\Pi^1_2$ real singleton forcing and $\mathbf P^{<\omega}$ is the finite-support product of $\omega$ copies of $\mathbf P$. 
A few details. Jensen's forcing is defined in $L$ so that $\mathbf P =\bigcup_{\xi<\omega_1} \mathbf P_\xi$, where each $\mathbf P_\xi$ is a ctble set of perfect trees in $2^{<\omega}$, generic over the outcome $\mathbf P_{<\xi}=\bigcup_{\eta<\xi}\mathbf P_\eta$ of all earlier steps in such a way that any $\mathbf P_{<\xi}$-name $c$ for a real ($c$ belongs to a minimal countable transitive model of a fragment of ZFC, containing $\mathbf P_{<\xi}$), which $\mathbf P_{<\xi}$ forces to be different from the generic real itself, is pushed by $\mathbf P_{\xi}$ (the next layer) not to belong to any $[T]$ where $T$ is a tree in $\mathbf P_{\xi}$. The effect is that the generic real itself is the only $\mathbf P$-generic real in the extension, while examination of the complexity shows that it is a $\Pi^1_2$ singleton. 
Now let $\mathbf P^{<\omega}$ be the finite-support product of $\omega$ copies of $\mathbf P$. It adds a ctble sequence of $\mathbf P$-generic reals $x_n$. A version of the argument above shows that still the reals $x_n$ are the only $\mathbf P$-generic reals in the extension and the set $\{x_n:n<\omega\}$ is $\Pi^1_2$. Finally the routine technique of finite-support-product extensions ensures that $x_n$ are not OD in the extension.
Addendum. For detailed proofs of the above claims, see this manuscript.
Jindra Zapletal informed me that he got a model where a $\mathsf E_0$-equivalence class $X=[x]_{E_0}$ of a certain Silver generic real is OD and contains no OD elements, but in that model $X$ does not seem to be analytically definable, let alone $\Pi^1_2$. The model involves a combination of several forcing notions and some modern ideas in descriptive set theory recently presented in Canonical Ramsey Theory on Polish Spaces. Thus whether a $\mathsf E_0$-class of a non-OD real can be $\Pi^1_2$ is probably still open.
Further Kanovei's addendum of Aug 23. 
It looks like a clone of Jensen's forcing on the base of Silver's (or $\mathsf E_0$-large Sacks) forcing instead of the simple Sacks one leads to a lightface $\Pi^1_2$ generic $\mathsf E_0$-class with no OD elements. The advantage of Silver's forcing here is that it seems to produce a Jensen-type forcing closed under the 0-1 flip at any digit, so that the corresponding extension contains a $\mathsf E_0$-class of generic reals instead of a generic singleton. I am working on details, hopefully it pans out.
Further Kanovei's addendum of Aug 25. 
Yes it works, so there is a generic extension $L[x]$ of $L$ by a real in which the 
$\mathsf E_0$-class $[x]_{\mathsf E_0}$ is a lightface $\Pi^1_2$ (countable) set with no OD elements. I'll send it to Axriv in a few days.
Further Kanovei's addendum of Aug 29.   arXiv:1408.6642<|endoftext|>
TITLE: Can you find linear recurrence relation for dimensions of invariant tensors?
QUESTION [10 upvotes]: Let $V$ be a finite dimensional highest weight representation of a (semi)-simple Lie algebra. For each $n\ge 0$ take $a_n$ to be the dimension of the space of invariant tensors in $\otimes^n V$.
In certain cases there is a formula for $a_n$. For example, for $V$ the two dimensional representation of $sl(2)$ we get $a_n=0$ if $n$ is odd and for $n$ even we get the ubiquitous Catalan numbers. In general I don't expect a formula but the sequence does satisfy a linear recurrence relation with polynomial coefficients (known as D-finite).
For example, for the seven dimensional representation of $G_2$ this sequence starts:
1, 0, 1, 1, 4, 10, 35, 120, 455, 1792, 7413, 31780, 140833, 641928, 3000361, 14338702, 69902535, 346939792, 1750071307, 8958993507, 46484716684, 244187539270, 1297395375129, 6965930587924
for more background see http://www.oeis.org/A059710 
This satisfies the recurrence
$(n+5)(n+6)a_n=2(n-1)(2n+5)a_{n-1}+(n-1)(19n+18)a_{n-2}+ 14(n-1)(n-2)a_{n-3}$
Question How does one find these recurrence relations?
Then I also have a more challenging follow-up question. The space of invariant tensors in $\otimes^n V$ also has an action of the symmetric group $S_n$ and so a Frobenius character which is a symmetric function of degree $n$.
Question How does one calculate these symmetric functions?
I know these can be calculated using plethysms individually. I am hoping for something along the lines of the first question.
Further remarks David's answer solves the problem theoretically but I want to make some remarks about the practicalities. This is in case anyone wants to experiment and also because I believe there is a more efficient method.
The $sl(2)$ example can easily be extended. For the $n$-dimensional representation $a_k$ is the coefficient of $ut^k$ in
$$\frac{u-u^{-1}}{1-t\left(\frac{u^n-u^{-n}}{u-u^{-1}}\right)}$$
For the case $n=3$ see http://www.oeis.org/A005043 and
http://www.oeis.org/A099323
I am not aware of any references for $n\ge 4$. I don't know if these are algebraic.
The limitation of this method is that there is a sum over the Weyl group. This means it is impractical to implement this method for $E_8$. For the adjoint representation of $E_8$ the start of the sequence is
1 0 1 1 5 16 79 421 2674 19244 156612 1423028 14320350
(found using LiE)

REPLY [8 votes]: Finding the recurrence (and proving it is correct) can be done by the standard techniques for extracting the diagonal of a rational power series. 
Let $\beta_1$, $\beta_2$, ..., $\beta_N$ be the weights of $V$. Let $\rho$ be half the sum of the positive roots and $\Delta = \sum (-1)^{\ell(w)} e^{w(\rho)}$ be the Weyl denominator. Then
$$\sum_{n=0}^{\infty} t^n \chi \left( V^{\otimes n} \right) = \frac{1}{1- \sum_{i=1}^N t e^{\beta_i}}$$
and
$$\sum_{n=0}^{\infty} t^n \dim \left(  V^{\otimes n} \right)^{\mathfrak{g}} = \mbox{Coefficient of}\ e^{\rho}\ \mbox{in} \ \left( \Delta \frac{1}{1- \sum_{i=1}^N t e^{\beta_i}} \right).$$
For example, if $\mathfrak{g}=\mathfrak{sl}_2$ and $V$ is the two dimensional irrep, the right hand side is
$$ \mbox{Coefficient of}\ u \ \mbox{in} \left(  \frac{(u-u^{-1})}{1-tu^{-1} - tu} \right)$$
which can be seen without too much trouble to be the generating function for Catalan numbers.
The diagonal of a rational generating function is $D$-finite by a result of Lipshitz. The particular recurrence can be found by Sister Celine's method (see theorems 10 and 11). I found these references in Stanley, Enumerative Combinatorics Vol. II, solution to exercise 6.61. Stanley warns that there is a gap in Zeilberger's argument, but hopefully his algorithm is right.<|endoftext|>
TITLE: solving $f(f(x))=g(x)$
QUESTION [102 upvotes]: This question is of course inspired by the question How to solve f(f(x))=cosx
and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a bounded interval. [EDIT: actually he can do rather better than this, solving the equation away from a bounded interval (with positive measure)].
I've always found such questions ("solve $f(f(x))=g(x)$") rather vague because I always suspect that solutions are highly non-unique, but here are two precise questions which presumably are both very well-known:
Q1) Say $g:\mathbf{R}\to\mathbf{R}$ is an arbitrary function. Is there always a function $f:\mathbf{R}\to\mathbf{R}$ such that $f(f(x))=g(x)$ for all $x\in\mathbf{R}$?
Q2) If $g$ is as above but also assumed continuous, is there always a continuous $f$ as above?
The reason I'm asking is that these questions are surely standard, and perhaps even easy, but I feel like I know essentially nothing about them. Apologies in advance if there is a well-known counterexample to everything. Of course Q1 has nothing to do with the real numbers; there is a version of Q1 for every cardinal and it's really a question in combinatorics.
EDIT: Sergei Ivanov has answered both of these questions, and Gabriel Benamy has raised another, which I shall append to this one because I only asked it under an hour ago:
Q3) if $g$ is now a continuous function $\mathbf{C}\to\mathbf{C}$, is there always continuous $f$ with $f(f(x))=g(x)$ for all $x\in\mathbf{C}$?
EDIT: in the comments under his answer Sergei does this one too, and even gives an example of a continuous $g$ for which no $f$, continuous or not, can exist.
Related MO questions: f(f(x))=exp(x) and other functions just in the middle between linear and exponential, and Does the exponential function has a square root.

REPLY [6 votes]: The following is inspired by the chapter Partition Polynomials in Riordan's Combinatorial Identities where $g(x)$ is analytic and $g(0)=0$. Find the Taylor series of $f(x)$ at zero by evaluating the derivatives of $f(f(x))=g(x)$ at $0$ in succession.
Set $f(0)=0$, giving $f(f(0))=g(0)=0$.
The first derivative gives $f'(x) f'(f(x))=g'(x)$, therefore $f'(0)=\sqrt{g'(0)}$ and $f'(0)=-\sqrt{g'(0)}$. Set $f'(0)=\sqrt{g'(0)}$ for this example. 
The second derivative $f'(x)^2 f''(f(x))+f'(f(x)) f''(x)=g''(x)$ produces $f''(0)=\frac{g''(0)}{g'(0)+\sqrt{g'(0)}}$.
The first few term of the Taylor series are
$f(x)=\sqrt{g'(0)}x+\frac{ g''(0)}{2
   \left(g'(0)+\sqrt{g'(0)}\right)}x^2$ $+\frac{
   \left(-3 g''(0)^2+g^{(3)}(0) g'(0)^{3/2}+2
   g^{(3)}(0) g'(0)+g^{(3)}(0) \sqrt{g'(0)}\right)}{6
   \left(\sqrt{g'(0)}+1\right)^2 g'(0)
   \left(g'(0)+1\right)}x^3+O(4)$.
This solution is based on $g(x)$ having a fixed point. Technically the question stated is whether there is always a solution and is focused on counter examples instead of a broad general answer as I have provided. But it does raise the question of whether there is a connection between $g(x)$ not having a fixed point and the counter examples that the other authors have provided.<|endoftext|>
TITLE: Is there a canonical notion of "mod-l automorphic representation"?
QUESTION [10 upvotes]: As the title says.
In particular, I am interested in the story for a general reductive group $G$, say defined over $\mathbb{Q}$.  I can imagine that mod-$\ell$ (algebraic) automorphic representation should correspond to conjugacy classes of homomorphisms from the motivic Galois group into $\widehat{G}(\overline{\mathbb{F}_\ell})$, but this does not seem like a very workable definition.  Is there a more pragmatic approach?

REPLY [5 votes]: If $G$ is a connected, linear, reductive group over $\mathbb Q$, let $q_0$ be the dimension so denoted in Borel--Wallach, i.e. $q_0 = (d - l_0)/2,$ where $d = \text{dim } G_{\infty} - 
\text{ dim } A_{\infty}K_{\infty}$ and $l_0 = \text{rank } G_{\infty} - 
\text{ rank } A_{\infty}K_{\infty},$  where $G_{\infty} = G(\mathbb R)$, $K_{\infty}$ is a maximal compact subgroup of $G_{\infty}$,  and $A_{\infty} = A(\mathbb R)$ with $A$ being the maximal $\mathbb Q$-split torus in the centre of $G$.  (So $q_0$ is the first interesting dimension of homology in the locally symmetric arithmetic quotients attached to $G$.)
Then the inductive limit of the cohomology groups $H^{q_0}(G(\mathbb Q)\backslash
G(\mathbb A)/A_{\infty}K_{\infty}K_f,\mathbb F_{\ell})$, as $K_f$ runs over all compact open subgroups of $G(\mathbb A_f)$, is an admissible smooth representation of $G(\mathbb G_f)$,
which is (I believe) a reasonable substitute for a space of mod $\ell$ automorphic forms.  
In the case when $G$ satifies the conditions of Gross's "Algebraic modular forms" paper,
$q_0 = 0$, the locally symmetric quotients are just finite sets, and this is precisely the space of mod $\ell$ algebraic modular forms.
This doesn't deal with Kevin's problems; this space will (conjecturally) only contain
automorphic forms whose systems of Hecke eigenvalues correspond to odd Galois representations.  Nevertheless, it is a space, and one can (and does) try to work with it.<|endoftext|>
TITLE: Why are the sporadic simple groups HUGE?
QUESTION [48 upvotes]: I'm merely a grad student right now, but I don't think an exploration of the sporadic groups is standard fare for graduate algebra, so I'd like to ask the experts on MO. I did a little reading on them and would like some intuition about some things.
For example, the order of the monster group is over $8\times 10^{53}$, yet it is simple, so it has no
normal subgroups...how? What is so special about the prime factorization of its order? Why is it $2^{46}$ and not $2^{47}$? Why is it not possible to extend it to obtain that additional power of 2 without creating a normal subgroup? Some of the properties seem really arbitrary, and yet must be very fundamental to the algebra of groups.
I don't think I'm the only person curious about this, but I hesitated posting due to my relative inexperience.

REPLY [19 votes]: Indeed the question is too vague for a precise answer, but nevertheless somehat natural ;-)
I want to give some more details and clearifications to the "hierarchy", that has been broached by Carnahan above. The "generic" simple groups are the Lie type groups of arbitrary size and the alternating groups. Furthermore, the main induction step of the classification theorem was, that the centralizer of an involution of a simple group (an order-2-element/involution is the "only" thing we have "a-priori" in an arbitrary simple group by Feit-Thomson) is close (!) to simple. So there is the chance of sporadic group branching off from a Lie-Type or alternating group and inductively proceed for some steps until it terminates. 
This inductive process of constructing a much larger simple group from it's involution centralizer being a prescribed (already large) simple group in extremely rare (!) situations could be thought of some sort of answer to your question. It is by the way one reason for the incredible length of the classification result (a tremendous case-by-case argument)...and for my personal view on the meta-debate above, that sporadics are more sporadic (not more unnatural!) than others, as much to us as to species 8472  ;-) ;-)
Most examples go only one step (and still are very large!), e.g. almost all so-called pariahs:

$J_1,J_3\leftarrow A_5$
$Ly \leftarrow A_{11}$
$ON \leftarrow SL_3(4)$
$Ru \leftarrow\;^2SO_5(8)$  "twisted" Lie-type (alike the unitaries over finite fields)
$J_4\leftarrow M_{22}\leftarrow\ldots$ branches off already one induction step beyond Lie (see below)

Note that most of these cases already appear as involution centralizers of Lie-type groups, which is somewhat miraculous and was often the reason to study this particular class and find only distinct "irradic" different choices. E.g. $^2G(3^{2n+1})\leftarrow SL_2(3^n)$ and the only other possible case $SL_2(4)\cong SL_2(5)\cong A_5$ lead Janko 1965 to the first new sporadic $J_1$ in almost a century.
On the other hand there is a VERY remarkable string of induction steps to the Monster and with modifications to the other sporadic groups "involved" in it. It goes roughly as
$M\leftarrow Co_1 \leftarrow M_{24}\leftarrow SL_3(4)$ 
and heavily relies on the already mentioned Golay-Code resp. Steiner System $S(24,8,5)$ - beautiful, very sporadic and purely combinatoric objects! Along the induction steps, the combinatorical objects with these groups as automorphism groups can be extended as well, very roughly like 
Griess-Algebra $\leftarrow$ Leech-Lattice $\leftarrow$ Steiner-System $\leftarrow$ Projective-Plane
One striking numerical reason for this construction and the very exotic behaviour to work exactly for $24$ dimensions (also responsible for the $2^{24}$-factor mentioned above) is: 
$1^2+2^2+\ldots+23^2+24^2=70^2$
This is provably impossible for larger numbers (by hard number theory) and is the striking numerical coincidence used in the side-by-side construction of the Golay code, the Steiner System and the $24$-dimensional Leech lattice, which is the most dense sphere packing of all dimensions and the reason e.g. the kissing number is known for this dimension! 
Hope that gives some intuition and "personality" for the various sporadics ;-) ;-)<|endoftext|>
TITLE: Definition of Chow groups over Spec Z
QUESTION [13 upvotes]: Usually (eg, intro. in M. Rost's 'Cycle modules with coefficients'), for a variety, $X$, over a field one can define the Chow group of p-cycles, $CH_p (X)$, as 
$$CH_p (X) = coker\; \left[\bigoplus_{x\in X_{p+1}} k(x)^\times \rightarrow \bigoplus_{x\in X_p} \mathbb{Z}\; \right]$$.
What about for an arithmetic scheme, eg when $X$ is, say, normal, separated, of finite type  and flat over $Spec \; \mathbb{Z} $?  Does something go wrong with the above definition?
Peter Arndt had posed part of this question already, but it seems without an answer.

REPLY [3 votes]: So with a lot of extra care about dimension/codimension it seems to be possible to define Chow groups over Spec Z if I understand the above answers correctly.
I may point out that in the book by Elman, Karpenko and Merkurjev "Algebraic and Geometry Theory of Quadratic Forms" (even though the title does not suggest so) they very carefully work out Chow groups, even some version of higher Chow groups. They begin by treating Chow groups over general excellent schemes (something you do not have written so explicitly in Fulton), so quite general and only later impose additional assumptions, like equidimensionality, being over a field, and all that. So maybe it is worth having a look at that.
On the other hand, they get a pullback along non-flat morphisms only with the typical more restrictive conditions. This however is crucial for turning Chow groups into the Chow ring.
So I think the construction of the intersection product [which uses the pullback along the embedding of the diagonal X -> X x X] is another very very critical matter over Z (but according to one of the other answers it can be done, that sounds very interesting).
Last but not least, just maybe another perspective, if one writes down the classical intersection multiplicity of two cycles, that can be done by first multiplying both cycles of complementary codim [so for this we need a ring structure, but let's just assume somebody can give such a structure even over Z, just to find out where we would actually be going], then the product lies in CH^n(X), n being the dimension of our scheme. Now to turn that into the classical intersection multiplicity one could pushforward this cycle along the structural map to the base field,
$X$ --> $Spec$ $(k)$
over a field(!) and $CH{\_0}(Spec k) = \mathbb{Z}$ and we get our intersection number. Voilà.
But if we are proper over Spec Z, we could at best pushforward
$X$ --> $Spec$ Z
but $CH{\_0}(Spec(\mathbb{Z}) = 0$, so nothing very interesting seems to result here.
[this argument however only makes sense if the dimension shifts in this Spec Z setup would be carried over analogously, which maybe is also stupid here for the reason that Spec Z is one-dimensional and Spec k zero-dimensional. I am just saying all this, because best and supercool would of course be somebody with a Spec *F*$_1$ having
CH_0(Spec *F*$_1$)= ? (....something, probably rather R than Z)
and that could then be our Spec Z intersection number by giving *F*$_1$ the role of a "virtual base field" and I guess some people say this should link to the Arakelovian one.... but well, that's very speculative]
So I think some people's expectation goes in the direction that the "interesting" way of doing intersection theory over Spec Z needs such a final *F*$_1$-twist.
Note maybe that the classical analgoue would be
P1 <-> Spec Z + (infinite place)
but CH_0(P1) = Z, whereas CH_0(Z) = 0, so we kind of miss something if we just use classical Chow groups over Spec Z. For other questions, classical methods work well even for Z without needing *F*$_1$ or so, for example the étale fundamental groups of both P1 and Spec Z are trivial. But for Chow theory some additional tricks seem to be required.
At least that is my impression. Of course this arithmetic aspect of intersection theory over Spec Z is a kind of different story and it also makes perfect sense to talk about classical Chow groups over Spec Z, so there is certainly nothing wrong in having CH_0(Z) = 0, just maybe for some sorts of questions of arithmetical content, this type of Chow theory may not be the right approach.<|endoftext|>
TITLE: Drawing 3-configurations of points and lines with straight lines
QUESTION [12 upvotes]: It is well-known that the black-and-white coloring of the Heawood graph on 14 vertices determines a combinatorial 3-configuration with 7 "points" and 7 "lines", known as Fano plane. Similarly, any cubic bipartite graph of girth at least 6 with a given black-and-white coloring can be regarded as the Levi graph or incidence graph of a 3-configuration.  The Fano plane can be drawn in the Euclidean plane with 6 straight lines and one curved line but not with all lines straight. There are many combinatorial 3-configurations, such as Pappus or Desargues configurations that can be realized as geometric configurations of points and lines in the Euclidean plane. Call such configuration realizable. It is easy to see that if a combinatorial configuration is realizable then its dual configuration is realizable. (The combinatorial dual is obtained by interchanging black and white colors in the coloring of its Levi graph). This means that the property of realizability is, in fact, a property of bipartite graphs and the Heawood graph is not realizable.
I would like to know what is known about the status of the following complexity decision problem.

Input: Cubic connected bipartite graph G of girth at least 6.
Question: Is G realizable?

I am aware of recent book "Configurations of Points and Lines" by Branko Grunbaum, the book by
Juergen Bokowski: "Computational Oriented Matroids" and the book "Computational Synthetic Geometry" by
Bokowski and Sturmfels. I am not sure if any of them gives the final answer to this problem.

REPLY [2 votes]: Let's try this again. If there weren't a degree constraint on the graph, then you could adapt the proof of Mnev's universality theorem (see my previous answer) to show that the problem was equivalent to the existential theory of the reals. So one approach to try is to find gadgets to reduce arbitrary degree to degree three (like SAT → 3-SAT). I'm almost positive you can do this for some constant degree. I'm not at all sure whether you can get it down to degree 3, though.<|endoftext|>
TITLE: Is there a source for a diagrammatic description of the induction functor C->Z(C)?
QUESTION [8 upvotes]: Suppose that C is a fusion category (over the complex numbers) and that Z(C) is its Drinfel'd center.  By definition an object in Z(C) consists of an object V in C together with a collection of half-braidings $V \otimes W \rightarrow W \otimes V$ for every object W in C satisfying some naturality conditions.  Hence there is a restriction functor R:Z(C)->C given by forgetting the half-braiding.  Adjoint to this is an induction functor I:C->Z(C).
A Theorem of Etingof-Nikshych-Ostrik says that $R(I(V)) = \bigoplus_X X \otimes V \otimes X^{\*}$.  In particular, we see that I(V) is $\bigoplus_X X \otimes V \otimes X^{\*}$ (where X ranges over simple objects up to isomorphism) together with some particular choice of half-braidings.  I'm pretty sure I know what those half-braidings are.  In particular there's a nice picture (the X,Y summand of the half-braiding with W is a sum over diagrams with a trivalent vertex connecting W to X* and Y* and another trivalent vertex connecting X and Y to W, where the two vertices range over dual bases).
What I really would like is a reference that explains this so that I don't have to write it up myself.  The only description I know is in ENO's "On Fusion Categories" where it's written in terms of weak Hopf algebras.
The motivation is removing any mention of weak Hopf algebras from the construction in Section 5 (about cyclotomicity of certain Drinfel'd centers) in Scott and my Noncyclotomic Fusion Categories.  It turns out that the diagram description above can be slightly modified (in a way suggested to me by Ben Webster) in order to give a description of I(V) where V is an object in a non-split fusion category over an arbitrary field.

REPLY [2 votes]: In the meanwhile a reference has appeared: Theorem 2.3 in http://arxiv.org/abs/1004.1533<|endoftext|>
TITLE: infinite permutations
QUESTION [16 upvotes]: This question is related to this one: Continued fractions using all natural integers. Suppose we have the set of natural numbers $N$ with order and we perform permutation on it. So we obtain the same elements with different order. Suppose we describe such permutations by usual notation when (1,2,3,4,5...) means identity permutation. Then lets say that permutation denoted by (1,3,2,4,5,6...) ( from the 4th place there is list of natural numbers in usual order) is finite because it only mixes numbers 1,2,3 -> 1,3,2 and for remaining elements it is identity permutation. As I find here there is definition of such objects, namely a few possibilities as states the answer of Qiaochu Yuan. 
Questions:

Are  infinite permutation decomposable into cycles? Transpositions?
Is possible to find such permutation of natural numbers that it cannot be a limit of finite permutations?

REPLY [7 votes]: Every permutation decomposes $\mathbb N$ into orbits. You can arrange these into a cycle decomposition if you allow infinite cycles.
I'm not comfortable with all infinite products, particularly ones which do not define permutations at all times. 
One notion of an infinite product of transpositions is a limit under pointwise convergence. That is, say $\pi$ is an infinite product of transpositions if there is a sequence of permutations $e=\pi_0, \pi_1, \pi_2...$ which converges pointwise to $\pi$ so that $\pi_{n+1} \pi_n^{-1}$ is a transposition.  
Every permutation is an infinite product of transpositions: Define $\pi_n = \bigg(\pi(n) ~~ \pi_{n-1}(n)\bigg)\pi_{n-1}.$ Then  $\pi_n$ agrees with $\pi$ on $\{1,...,n\}$, so $\pi_0, \pi_1, \pi_2, ... \to \pi$. 
Therefore, every permutation is a pointwise limit of what you called finite permutations.<|endoftext|>
TITLE: Books on reductive groups using scheme theory
QUESTION [9 upvotes]: Prof. Conrad mentioned in a recent answer that most of the (introductory?) books on reductive groups do not make use of scheme theory.  Do any books using scheme theory actually exist?  Further, are there any books that use the functor of points approach?  Demazure-Gabriel's second book would have covered general group schemes in this way, but it was never written, and it's not clear whether or not it would have covered reductive groups anyway.  There is a lot of material in SGA 3 using more modern machinery to study group schemes, but I'm not aware of any significant treatment of reductive groups in that book (although I haven't read very much of it).  
Correction: Prof. Conrad has noted that SGA 3 does contain a significant treatment of reductive groups using modern machinery.

REPLY [29 votes]: Personally, I find the "classical" books (Borel, Humphreys, Springer) unpleasant to read because they work in the wrong category, namely, that of reduced algebraic group schemes rather than all algebraic group schemes. In that category, the isomorphism theorems in group theory fail, so you never know what is true. For example, the map $H/H\cap N\rightarrow HN/N$ needn't be an isomorphism (take $G=GL_{p}$, $H=SL_{p}$, $N=\mathbb{G}_{m}$ embedded diagonally). Moreover, since the terminology they use goes back to Weil's Foundations, there are strange statements like "the kernel of a homomorphism of algebraic groups defined over $k$ need not be defined over $k$". Also I don't agree with Brian that if you don't know descent theory, EGA, etc. then you don't "know scheme theory well enough to be asking for a scheme-theoretic treatment'. 
Which explains why I've been working on a book whose goal is to allow people to learn the theory of algebraic group schemes (including the structure of reductive algebraic group schemes) without first reading the classical books and with only the minimum of prerequisites (for what's currently available, see my website under course notes).  In a sense, my aim is to complete what Waterhouse started with his book.
So my answer to the question is, no, there is no such book, but I'm working on it....<|endoftext|>
TITLE: Orientation of a smooth manifold using sheaves
QUESTION [9 upvotes]: Is there any way to define the orientation of an orientable smooth manifold using sheaves (when our smooth manifold is viewed as a locally ringed space) without our definition being overly contrived?

REPLY [9 votes]: In the study of (finite-dimensional?) paracompact and locally compact (?) spaces there is Verdier's topological duality theorem, expressed in terms of a dualizing complex (which is built up from a sheafification process using duals of compactly-supported cohomologies of open subspaces, or something like that).  It is pure topology, having nothing to do with ringed spaces (just like the orientation sheaf!). In the special case of smooth (paracompact) manifolds, this recovers the orientation sheaf up to a shift on each connected component. It is analogous to the fact that the super-abstract dualizing complexes in Grothendieck duality for (quasi-)coherent cohomology collapses to the old friend "top-degree relative differentials" (up to shifting) in the smooth case. 
But that's all just fancy mumbo-jumbo which puts the orientation sheaf into a broader perspective (like many duality theories for cohomologies). This does not qualify as a good way to initially "define" the orientation sheaf, much as appealing to Grothendieck duality would be a strange (and even circular, from some viewpoints) way to "define" top-degree relative differentials in the smooth case.  To get a real theorem out, we have to put some content in.
It seems more illuminating at a basic level to understand how the orientation sheaf is constructed using punctured neighborhoods along the lines of Emerton's comment or the oriented double cover as in David Roberts' answer, and how one can remove some orientability hypotheses in some classical results on "constant coefficient" cohomology by instead allowing coefficients in the locally constant sheaf given by the orientation sheaf.  And likewise to understand why the constant sheaf associated to $\mathbf{Z}(1) = \ker(\exp)$ has $n$th tensor power that serves as an orientation sheaf on a complex manifold of dimension $n$ (and so the natural isomorphism $\mathbf{C}(1) = \mathbf{C}$ via multiplication explains the absence of needing to choose orientations for various cohomological calculations on complex manifolds (very relevant if one is to have a hope to translating things into algebraic geometry).<|endoftext|>
TITLE: A historical question: Hurwitz, Luroth, Clebsch, and the connectedness of $\mathcal{M}_g$
QUESTION [22 upvotes]: The connectedness of the moduli space $\mathcal{M}_g$ of complex algebraic curves of genus $g$ can be proven by showing that it is dominated by a Hurwitz space of simply branched d-fold covers of the line, which in turn can be shown to be connected by proving the transitivity of the the natural action of the braid group on n-tuples of transpositions in $S_n$ with product 1, which generate $S_n$:  in this action, a generator $\sigma_i$ of the braid group acts as
$$(g_1, ... g_n) \to (g_1, ... g_{i+1}, g_i^{g_{i+1}}, g_{i+2}, ..., g_n)$$
This argument is often referred to as "a theorem of Clebsch (1872 or 1873), Luroth (1871), and Hurwitz (1891)."  Does anyone know the history of this argument more precisely, and in particular which parts are due to Luroth, which to Clebsch, and which to Hurwitz?

REPLY [8 votes]: I was able to find the resources online (6 years after this question was asked):

1871 https://eudml.org/doc/156527 Lüroth - 4 pages 
1873 https://eudml.org/doc/156610 Clebsch - 16 pages
1891 https://eudml.org/doc/157563 Hurwitz - 61 pages

I think it's pretty clear Luroth was first, but Hurwitz developed this material much further.  

I get thrown off because we way Riemann surface is just a polygon with edges glued together, and that's pretty much the picture of moduli space painted here.<|endoftext|>
TITLE: Lagrange four-squares theorem: efficient algorithm with units modulo a prime?
QUESTION [10 upvotes]: I'm looking at algorithms to construct short paths in a particular Cayley graph defined in terms of quadratic residues. This has led me to consider a variant on Lagrange's four-squares theorem. 
The Four Squares Theorem is simply that for any $n \in \mathbb N$, there exist $w,x,y,z \in \mathbb N$ such that
$$
n = w^2 + x^2 + y^2 + z^2 .
$$
Furthermore, using algorithms presented by Rabin and Shallit (which seem to be state-of-the-art), such decompositions of $n$ can be found in $\mathrm{O}(\log^4 n)$ random time, or about $\mathrm{O}(\log^2 n)$ random time if you don't mind depending on the ERH or allowing a finite but unknown number of instances with less-well-bounded running time.
I am considering a Cayley graph $G_N$ defined on the integers modulo $N$, where two residues are adjacent if their difference is a "quadratic unit" (a multiplicative unit which is also quadratic residue) or the negation of one (so that the graph is undirected). Paths starting at zero in this graph correspond to decompositions of residues as sums of squares.
It can be shown that four squares do not always suffice; for instance, consider $N = 24$, where $G_N$ is the 24-cycle, corresponding to the fact that 1 is the only quadratic unit mod 24. However, finding decompositions of residues into "squares" can be helpful in finding paths in the graphs $G_N$. The only caveat is that only squares which are relatively prime to the modulus are useable.
So, the question: let $p$ be prime, and $n \in \mathbb Z_p ( := \mathbb Z / p \mathbb Z)$. Under what conditions can we efficiently discover multiplicative units $w,x,y,z \in \mathbb Z_p^\ast$ such that $n = w^2 + x^2 + y^2 + z^2$? Is there a simple modification of Rabin and Shallit's algorithms which is helpful?
Edit: In retrospect, I should emphasize that my question is about efficiently finding such a decomposition, and for $p > 3$. Obviously for $p = 3$, only $n = 1$ has a solution. Less obviously, one may show that the equation is always solvable for $n \in \mathbb Z_p^\ast$, for any $p > 3$ prime.

REPLY [7 votes]: There is also an unconditional deterministic polynomial-time algorithm to find $x,y,z,w \in \mathbf{F}_p^\times$ such that $x^2+y^2+z^2+w^2=n$, given any $n \in \mathbf{Z}$ and any prime $p \ge 7$.
First, given $a,b,c \in \mathbf{F}_p^\times$, Theorem 1.10 of Christiaan van de Woestijne's thesis lets one find an $\mathbf{F}_p$-point $P$ on the smooth conic $C \colon ax^2+by^2=cz^2$ in $\mathbf{P}^2$ over $\mathbf{F}_p$.  The usual trick of drawing lines through $P$ and taking the second point of intersection with $C$ lets one parametrize $C(\mathbf{F}_p)$.  At most $6$ points of $C(\mathbf{F}_p)$ have one of $x,y,z$ equal to $0$, so by trying at most $7$ lines, one finds a point on the affine curve $ax^2+by^2=c$ with $x,y \in \mathbf{F}_p^\times$.
Now, to solve $x^2+y^2+z^2+w^2=n$, choose $c \in \mathbf{F}_p \setminus \{0,n\}$, and apply the previous sentence to find $x,y,z,w \in \mathbf{F}_p^\times$ satisfying $x^2+y^2=c$ and $z^2+w^2=n-c$.<|endoftext|>
TITLE: Difference between measures and distributions
QUESTION [19 upvotes]: On the one hand, Wikipedia suggests that every distribution defines a Radon measure:

http://en.wikipedia.org/wiki/Distribution_(mathematics)#Functions_as_distributions

On the other hand, Terry Tao and LK suggest not:

http://www.math.ucla.edu/~tao/preprints/distribution.pdf
When can a function be recovered from a distribution?

Can someone please clarify this for me?

REPLY [3 votes]: This is a summary of what I've learned about this question based on the answers of the other commenters.
[*] Any positive distribution defines a positive Radon measure.
I had naively assumed a result for distributions like The Hahn Decomposition Theorem[1] for measures, i.e. I assumed that a distribution could be expressed as the difference of two positive distributions.  If it could be, then applying Theorem [*] would yield the result that any distribution is a signed measure.
However, this is not the case.  The derivative of the delta function, i.e. δ', satisfies
δ'(f) = -f'(0).  This is not a measure.  I can't find any way of proving it's not the difference of two positive distributions, other than by contradiction using the above result.
[1] http://en.wikipedia.org/wiki/Hahn_decomposition_theorem<|endoftext|>
TITLE: Is there lore about how endofunctors of Cat interact with the formation of presheaf categories?
QUESTION [11 upvotes]: This is a request for references about a peculiar categorical
construction I've run into in some work I've been doing, and about which I'd
like to learn as much as I can.
Let $\mathrm{Cat}$ be the category of small categories, and let
$\mathrm{PSh}(C)$ be the category of presheaves of sets on a category $C$.
Suppose we
are given a "reasonable" endofunctor $\Xi\colon \mathrm{Cat}\to
\mathrm{Cat}$.  I want to consider a certain "intertwining" functor
$$
V\colon \Xi\mathrm{PSh}(C) \to \mathrm{PSh}(\Xi C)
$$
defined by the formula
$$
(VX)(\gamma) = \mathrm{Hom}_{\Xi\mathrm{Psh}(C)}(A\gamma, X),
$$
where $X$ is an object of $\Xi\mathrm{PSh}(C)$, $\gamma$ is an object
of $\Xi C$, and $A\colon \Xi C\to 
\Xi\mathrm{PSh}(C)$ is the functor obtained by 
applying $\Xi$ to the Yoneda functor $C\to \mathrm{PSh}(C)$.
Note: it's unreasonable to expect for a randomly chosen $\Xi$ that the
category $\Xi \mathrm{PSh}(C)$ is even defined, since
$\mathrm{PSh}(C)$ is a large 
category, and $\Xi$ is given as a functor on small categories.  And even if it is defined, it's unreasonable to expect that $V$ is
well-defined, since $(VX)(\gamma)$ may not be a set.  But here are
some reasonable examples:

Let $\Xi C= C\times C$.  Then $V\colon \mathrm{PSh}(C)\times
  \mathrm{PSh}(C)\to \mathrm{PSh}(C\times C)$ is the "external
product" functor, which takes a pair of presheaves $(X_1,X_2)$ on $C$ to
the presheaf $(c_1,c_2) \mapsto X_1(c_1)\times X_2(c_2)$ on $C^2$.
You can generalize this by considering $\Xi C= \mathrm{Func}(S,C)$,
where $S$ is a fixed small category.
Let $\Xi C = C^{\mathrm{op}}$.  Then $V\colon
  \mathrm{PSh}(C)^{\mathrm{op}} \to \mathrm{PSh}(C^{\mathrm{op}})$ is
a sort of "dualizing" functor, which sends a presheaf $X$ on $C$
to the presheaf $c\mapsto \mathrm{Hom}_{\mathrm{PSh}(C)}(X, Fc)$ on
$C^\mathrm{op}$; here $F\colon C\to \mathrm{PSh}(C)$ represents the
Yoneda functor.
Let $\Xi C=\mathrm{gpd} C$, the maximal subgroupoid of $C$.
Then $V\colon \mathrm{gpd}\\,\mathrm{PSh}(C)\to
  \mathrm{PSh}(\mathrm{gpd}C)$ is such that $(VX)(c)$ is the set of
isomorphisms between $X$ and the presheaf represented by $c$.

The sorts of questions I have include the following.

What makes a functor $\Xi$ reasonable?  Is it enough if it's
accessible?
I think $V$ should be the left Kan extension of the Yoneda
functor $B\colon \Xi C\to \mathrm{PSh}(\Xi C)$ along $A$.  Is this
true?  When can I expect  to have $VA\approx B$?
How does $V$ of a composite $\Xi \Psi$ relate to the composite of the
$V$s of each term?
Given a functor $f\colon C\to D$, you get a bunch of functors
between the associated presheaf categories.  How does $V$ interact
with such functors?

There's really only one or two examples of $\Xi$ that really I need to
understand this for, and I don't want to spend time working out a general theory of this thing.  It would be most convenient if someone can
point me to a reference which talks about this construction.  Even one
that deals with particular instances of it would be helpful.

REPLY [12 votes]: This is really just a comment, but it's too long to fit. 
Many people have come up against the problem that PSh isn't an endofunctor of Cat, because even if C is small, PSh(C) usually isn't.  There's a standard way to solve this problem, as follows.  

Replace Cat (small categories) with CAT (locally small categories)
Replace PSh (presheaves) with psh (small presheaves, i.e. small colimits of representables)

Then psh is genuinely an endofunctor of CAT.  If C is small then psh(C) = PSh(C).  But if C is not small then psh(C) is a proper subcategory of PSh(C).
In fact, psh is not only an endofunctor of CAT, but a monad.  It's free small-cocompletion.  That is, it takes a category and freely adjoins colimits.
The unit of this monad is the Yoneda embedding.  Given this, and given that the Yoneda embedding plays a part in your considerations, I wonder whether the multiplication of the monad plays a part too.<|endoftext|>
TITLE: Way to memorize relations between the Sobolev spaces?
QUESTION [46 upvotes]: Consider the Sobolev spaces $W^{k,p}(\Omega)$ with a bounded domain $\Omega$ in n-dimensional Euclidean space. When facing the different embedding theorems for the first time, one can certainly feel lost. Are there certain tricks to memorize the (continuous and compact) embeddings between the different $W^{k,p}(\Omega)$ or into $C^{r,\alpha}(\bar{\Omega})$ ?

REPLY [2 votes]: The way I remember is that the embedded space must have more smoothness, and the additional smoothness that is required gets larger when the difference in $1/p$ gets larger. Then you have to divide the smoothness by $n$, so there will be fraction lines everywhere. So a necessary condition for $W^{s,p}\hookrightarrow W^{r,q}$ is
$$
\frac{s-r}{n} \geq \frac1p - \frac1q.
$$<|endoftext|>
TITLE: Conjugacy classes in the absolute galois group
QUESTION [7 upvotes]: We consider $G_{\mathbb Q} = Gal(\mathbb {\bar Q}/\mathbb Q)$. The Frobenius elements corresponding to each prime are well-studied. But these are really not elements; these are only defined as some conjugacy classes(upto inertia, etc..)
Question: Are these the only conjugacy classes in the absolute Galois group? If there are others, please give examples or methods to construct them.
The conjugacy classes are of course defined algebraically; this question is not asking for results of the form that the Frobenii form a dense set.

REPLY [8 votes]: I agree with the answer of Kevin Buzzard -- you better look only at quotients $Gal(K/Q)$ unramified outside some (finite) set $S$ to make the question make sense as it is.  
But regardless, you can ask about the conjugacy classes in the absolute Galois group, and what they "mean".  An answer was given many years ago by Ax (I guess in one of his Annals papers, 1968 or 1969), and there are elaborations and new results in the thesis of James Gray, now published in the J. of Symbolic Logic as "Coding complete theories in Galois groups" (his thesis is also available freely online).
The reason why these "other conjugacy classes" (which are essentially all of them) are difficult to construct is that the associated fixed fields are the algebraic (over $Q$) subfields of pseudofinite fields of characteristic zero.  In Theorem 1.27 of his thesis, Gray states that there is a [natural, homeomorphic in the appropriate topology] bijection between the set of conjugacy classes in $Gal(\bar Q / Q)$ and the Stone space of completions of $ACFA_0$ -- the theory of algebraically closed fields of characteristic zero with generic automorphism.  Given such a completion of $ACFA_0$, realized by a model $(K, \sigma)$ where $K$ is algebraically closed, and $\sigma$ is a generic automorphism (see MacIntyre, "Generic automorphisms of fields" for the definition of generic used here), the fixed field $K^\sigma$ is a pseudofinite field of characteristic zero. 
Now, while two conjugacy classes in $Gal(\bar Q / Q)$ are quite simple to describe -- the trivial conjugacy class and the conjugacy class of order 2 elements -- other conjugacy classes contain elements of infinite order.  The associated fixed fields (in this infinite-order case) in $\bar Q$ are psuedofinite fields, which are difficult to get your hands on.  Probably the best way is to take an ultraproduct (for a non-principal ultrafilter on the set of prime numbers) of finite fields, and take the algebraic elements within.  This is certainly nonconstructive, relying heavily on Zorn's lemma.  
Still, such fields have arithmetic significance.  The model theory related to $ACFA_0$ has had a great deal of impact on number theory lately, and Fried-Jarden also touch on related matters in their "Field Arithmetic" book.<|endoftext|>
TITLE: Is there a version of inclusion/exclusion for vector spaces?
QUESTION [31 upvotes]: I am asking for a way to compute the rank of the 'join' of a bunch of subspaces whose pairwise intersections might be non-zero. So in the case n=2 this is just $\dim(A_1+A_2) = \dim(A_1) + \dim(A_2) - \dim(A_1\cap A_2)$. For general $n$, I don't know a good formula.

REPLY [5 votes]: I just wanted to point out, for other readers, the excellent paper by Gian-Carlo Rota "On the foundations of combinatorial theory I. Theory of Möbius function"
http://www.maths.ed.ac.uk/~aar/papers/rota1.pdf
The paper somehow sets up the proper framework to generalize the inclusion-exclusion principle (which is basically a special case of Möbius inversion).
(I wanted to make a comment following Shor's answer, but it seems I don't have enough reputation credit ^^ ...)

REPLY [2 votes]: The question immediately reminded me of this answer to the famous post of "Examples of common false beliefs in mathematics". Obviously, it does not answer the question, but I believe it adds something informative and worthy.<|endoftext|>
TITLE: Automorphism group objects
QUESTION [9 upvotes]: Consider a monoidal category C with operation $\otimes$, unit object $1$, and diagonal map $\delta:A \to A \otimes A$ for all $A \in C$ (with naturality conditions on the diagonal map).
We can define a notion of "group object" in the category, where a group object is an object G of C along with a map from $1$ to G (playing the role of the identity element), a map from G to G (playing the role of the inverse map) and a map $G \otimes G \to G$ (playing the role of the multiplication). Then, we put compatibility conditions on these operations that are commutative diagrams corresponding to associativity, identity elements and inverses. Note that we need to use the diagonal morphism to formulate the condition on inverses.
Wikipedia defines group objects only in the case where the monoidal operation is the categorical product, so if necessary, we can restrict attention to just those cases.
We can define group object morphisms, etc. Some examples of group objects are: topological groups (category of topological spaces with continuous maps and Cartesian product), Lie groups (category of differential manifolds with smooth maps and manifold product), abelian groups (category of groups with group homomorphisms and direct product, by the Eckmann-Hilton principle), and groups (category of sets).
We can then proceed to define a "group object action" of a group object G on an object A by the analogues of the two conditions for a group action. My questions:

Under what situations does there exist a group object that plays the role played by the symmetric group on a set? i.e., A group object $\operatorname{Sym}(X)$ for each X in C with an action on X such that any group object action of G on X corresponds to a group object homomorphism $G \to \operatorname{Sym}(X)$ with the desired compatibility conditions?
How does this group object $\operatorname{Sym}(X)$, if it exists, relate to the automorphism group $\operatorname{Aut}_C(X)$ (which is an actual group, not a group object)?
We can also define a "group object action by automorphisms" as the action of one group object on another satisfying the additional condition of being group automorphisms. (We need to use the diagonal morphism to formulate this compatibility condition). Is there some object called the "automorphism group object of a group object" such that any group object action corresponds to a group object morphism to the automorphism group object?
How does this automorphism group object relate to the automorphism group in the group object category?

REPLY [2 votes]: I think that it might be worth refining your conditions to allow for some other interesting examples.  Rather than have a diagonal map defined for each object, instead have the diagonal map as part of the group object.  This would include vector spaces in your list of examples.  The group objects would then by Hopf algebras.
The following example may be of interest:
Let H be a Hopf algebra, this may act on an algebra A, we just ask that the algebra multiplication map of A is a morphism of H-modules.  Equivalently A is an algebra object in the monoidal category of H-modules; A is called a H-module algebra.
Example: Let V be a representation of a Lie algebra g.  Let A=SV be the free commutative algebra on a vs V.  Then the universal enveloping algebra Ug acts on SV with g acting as derivations.
One may define a category of Hopf algebras acting on a fixed algebra A.  This category has a terminal object, PA.  This is very much reminiscent of what you might call Sym(A).  PA contains the automorphism group algebra of A.
I'm afraid that I can't relate this example directly to your questions, but it's a good generalisation of the automorphism group, perhaps it might help.<|endoftext|>
TITLE: How to teach introductory statistic course to students with little math background? 
QUESTION [9 upvotes]: Next semester I will teach an elementary statistic course for the first time (which I am actually quite excited about). A brief description can be found here. I am told to expect very little math background from the students. The class size is about 40-50, meeting twice a week  , each class lasts about one and half hour. I do have a graduate student helping with grading and recitation. 
My first question is about textbook. Please let me  know of suitable texts which are:
1) Fun to teach and learn from, especially for students with less background. I would like  a book with lot of interesting, engaging, probably more exotic examples ( so less  diseases and more, say, gambling! ).
2) Relatively cheap (preferably less than 100 USD). In this economy we don't want to make students pay too much. Most texts I came across seem quite expensive. (Unless if the text is really a treasure, then one can worry less about price.) 
The second question is about the best way to teach such a course. My instinct is that a heavy lecture style would not work very well, especially since the class is one and half hour long. Do you know ways (preferably with references, especially visual references) to teach this materials without too much lecturing? 
Thanks in advance. 
(I did not make this community wiki since I want to reward the best answer, and also I am not sure there will be many equally good answers given how much I wanted. Let me know if you disagree, I am open to change on this issue.)   
EDIT: The student body will be quite varied. I am told there will be psychology majors and some engineering students, but the majority would prefer examples to proofs.

REPLY [3 votes]: I was amazed at how much was done in the book by Freedman, Pisani, Purves, and Adhikari without using any math beyond what an equation of a line looks like, and that in only one small part of the course.<|endoftext|>
TITLE: Closed hyperbolic manifold with right-angled fundamental domain
QUESTION [13 upvotes]: What is an example (as simple as possible, please!) of a closed hyperbolic three-manifold with a right-angled polyhedron as fundamental domain? 

If we allow cusps then the Whitehead link or the Borromean rings are good answers (fundamental domains have not too many sides and the gluings can be understood).  If we allow orbifolds then the (4,0) filling on all components of the Borromean rings is a good answer.  (This is carefully explained in the NotKnot video.)  I tried to answer the original question, using SnapPea (see also SnapPy), to build a cyclic, four-fold, manifold cover of the Borromean orbifold.  This was not successful.  It is "obvious" that the resulting manifold has a right-angled fundamental domain, but the domain is huge (two octagons, eight hexagons, 16 pentagons) and it is hard (for me) to describe the face pairings or check that I haven't messed up in some way. 
More generally, I guess that one could use Andreev's theorem to build as "simple as possible" right-angled polyhedra and then look for low index torsion free subgroups of the resulting reflection groups.  However I don't know how large an index we'd have to sacrifice or even if the resulting manifold will have a right-angled domain... 
Edit: I've accepted bb's answer below, because the first paper cited gives the required construction.  However, I didn't understand this until I asked an expert off-line, who puckishly told me that this was equivalent to the four color theorem! Here is the construction: 
Suppose that $R$ is a right-angled three-dimensional hyperbolic polyhedron.  Note that the edges of $R$ form a cubic graph in $\partial R$.  Let $G_R$ be the subgroup of isometries of $H^3$ generated by reflections in the sides of $R$.  Now, by the four-color theorem there is a four coloring of the faces of $R$ (so no two adjacent faces have the same color).  This defines surjective homomorphism from $G_R$ to $(Z/2)^4$.  Let $\delta = (1,1,1,1) \in (Z/2)^4$ and let $\Delta$ be the preimage in $G_R$ of the subgroup generated by $\delta$.  Note that $\Delta$ has index eight in $G_R$ and is torsion-free.  Finally, a fundamental domain for $\Delta$ is a obtained by gluing eight copies of $R$ around a vertex.  
Other cryptic tidbits I was fed:  There is such a hyperbolic manifold in dimension four, coming from the 120-cell.  This is the only example known.  There are no such examples in dimensions five and higher.  See Bowditch and Mess, referring to Vinberg and Nikulin.
Further edit: In fact the argument using the four-color theorem can be found in the second paper of Vesnin, as cited by bb.

REPLY [10 votes]: A. Yu. Vesnin has some articles on these Lobell manifolds.  The first one describes how to construct arbitrarily many non-isometric closed hyperbolic manifolds from one right-angled polyhedron.

"Three-dimensional hyperbolic manifolds with a common fundamental polyhedron"   Math. Notes  49  (1991),  no. 5-6, 575--577 
"Three-dimensional hyperbolic manifolds of Löbell type" Siberian Math. J. 28 (1987), no. 5, 731--733<|endoftext|>
TITLE: Small simplicial complexes with torsion in their homology?
QUESTION [35 upvotes]: Fix a prime $p$.  What is the smallest integer $n$ so that there is a simplicial complex on $n$ vertices with $p$-torsion in its homology?
For example, when $p=2$, there is a complex with 6 vertices (the minimal triangulation of the real projective plane) with 2-torsion in its homology.  I'm pretty sure that it's the smallest possible: with 5 or fewer vertices, there should be no torsion at all.  When $p=3$, there is a complex with 9 vertices (a triangulation of the mod 3 Moore space, for instance) with 3-torsion.  Is there one with 8 vertices?  With $p=5$, there is a complex with 11 vertices, found by randomly testing such complexes on my computer.
We can refine this: fix $p$ and also a positive integer $d$.  What's the smallest $n$ so that there is a simplicial complex $K$ on $n$ vertices with $p$-torsion in $H_d(K)$?  Or we can turn it around: for fixed $n$, what kinds of torsion can there be in a simplicial complex on $n$ vertices?
(A paper by Soulé ("Perfect forms and the Vandiver conjecture") quotes a result by Gabber which leads to a bound on the size of the torsion for a fixed number $n$ of vertices; however, this bound is far from optimal, at least for small $n$.)

REPLY [15 votes]: Andrew Newman just posted a preprint to the arXiv, showing that for every prime $p$ and $d \ge 2$, you can get $p$-torsion in homology $H_{d-1}(K)$ with only $O(\log^{1/d} p)$ vertices. (The implied constant depends on $d$, but not on $p$.) This is best possible, up to a constant factor.
His construction starts with something similar to what Speyer describes above, which gets you a complex with $O( \log p)$ vertices. Then he applies the probabilistic method, taking a certain carefully chosen random quotient of the complex, gluing together vertices in a random way. This doesn't affect torsion in homology. The problem is that it might not result in a simplicial complex. But Newman uses the Lovász local lemma to show that with positive probability, it does. Hence there exists a vertex identification that works.<|endoftext|>
TITLE: Does the image of a p-adic Galois representation always lie in a finite extension?
QUESTION [28 upvotes]: I have been looking at Serre's conjecture and noticed that there are two conventions in the literature for a p-adic representation $\rho:\mbox{Gal}(\bar{\mathbb Q}/\mathbb Q)\to \mbox{GL}(n,V).$  In some references (eg Serre's book on $\ell$-adic representations), $V$ is a vector space over a finite extension of $\mathbb Q_p$.  However, in more recent papers (eg Buzzard, Diamond, Jarvis) $V$ is a vector space over $\bar{\mathbb Q_p}$.  It is easy to show that the former definition is a special case of the latter, but I suspect, and would like to prove that they are actually the same.  That is, I would like to show that the image of any any continuous Galois representation over $\bar{\mathbb{Q}_p}$ actually lies in a finite extension of $\mathbb Q_p$.  
Is this the case?  
I think that a proof should use the fact that $G_{\mathbb Q}$ is compact and that $\bar{\mathbb Q}_p$ is the union of finite extensions.  I have tried to mimic the proof that $\bar{\mathbb Q}_p$ is not complete, but have not been able to find an appropriate Cauchy sequence in an arbitrary compact subgroup of GL($n,V$).
(This is my first question, so please feel free to edit if appropriate.  Thanks!)

REPLY [19 votes]: I tried to cut and paste here an argument from a.tex file, but it came out looking like a complete mess, so I'll give a link to a webpage link here.
Concerning the comments by Kevin and David about proofs using the Baire category theorem, I think the proof I posted above (due to Warren Sinnott) should be viewed in a different light.  Consider the theorem that the alg. closure of $\mathbf Q_p$ is not complete.  There are a couple of different proofs of it. (Note Jen said a proof of that noncompleteness theorem is what she was trying to adapt to prove the compactness theorem for the matrix groups, so I suspect the proof in the link above is the direction she was trying to go in, whether or not other proofs of the compactness theorem may be considered more slick.) I'll briefly describe two such proofs.

In the $p$-adic book by Koblitz, he explicitly constructs an infinite series $\sum c_ip^i$ with $c_i$ in $\overline{\mathbf Q}_p$ of  absolute value 1 and increasing degree over $\mathbf Q_p$, and then use the increasing-degree condition on the coefficients to show the series can't converge in $\overline{\mathbf Q}_p$, although it's Cauchy since the general term tends to 0.  (This is essentially what takes place in the compactness proof at the link I posted above, but in a multiplicative setting: form a product of matrices tending to the identity whose entries have higher and higher degree over $\mathbf Q_p$.  The compactness hypotheses imply the product converges in $GL_n(\overline{\mathbf Q}_p)$ and then we get a contradiction.  The same argument shows any compact additive subgroup of $\overline{\mathbf Q}_p$ is inside a finite extension of $\mathbf Q_p$.)

In the ultrametric analysis book by Schikhof, there is a proof that $\overline{\mathbf Q}_p$ is not complete which uses the Baire category theorem: the elements of $\overline{\mathbf Q}_p$ with degree up to $n$, as $n$ varies, provide a countable cover of $\overline{\mathbf Q}_p$ by closed subsets which each turn out to have no interior point, while of course their union $\overline{\mathbf Q}_p$ has many interior points. The closed set formulation of the Baire category theorem is that a countable union of closed subsets which each have no interior does not have an interior either.  Thus we have a contradiction, so $\overline{\mathbf Q}_p$ is not complete.


I don't think these two strategies for proving a space is incomplete are the same, at least psychologically: in the first one you explicitly construct a non-convergent Cauchy sequence and in the second one you show a general property of complete spaces doesn't hold.  For the same reason, I think the Baire and non-Baire proofs of this compactness theorem are pretty different proofs.<|endoftext|>
TITLE: Books you would like to see translated into English
QUESTION [78 upvotes]: I have recently been told of a proposal to produce an English translation 
of Landau's Handbuch der Lehre von der Verteilung der Primzahlen, and
this prompts me to ask a more general question: 

Which foreign-language
  books would you most like to see translated into English?

These could be classics of historical interest, books you would like your 
students to read, books you would like to teach from, or books of use in 
your own research.

REPLY [3 votes]: I hope I did not miss that in one of the previous answers, but a book worthy of an English translation is "Théorie des distributions" by Laurent Schwartz. The later edition combined Vol I and II as a single book. One could also consider translating Vol III and IV which are his two long papers on vector-valued distributions. About the latter, see this related MO question: English translation of Schwartz's papers on vector-valued distributions<|endoftext|>
TITLE: components of E[p], E universal in char p.
QUESTION [15 upvotes]: I have just realised that a group scheme I've known and loved for years is probably a bit wackier than I'd realised.
In this question, in Charles Rezk's answer, I erroneously claim that his construction of the space representing Drinfeld $\Gamma_1(p)$ structures on elliptic curves must be flawed, because the global properties of $Y_1(p)$ that I know from Katz-Mazur seemed to contradict global properties that his construction appeared to me to have. We took the conversation to email and I also started writing down my thoughts more carefully to check there were no problems with them. I found a problem with them---hence this question.
Let $p$ be prime, let $N\geq4$ be an integer prime to $p$, and consider the fine moduli space $Y_1(N)$ over an algebraically closed field $k$ of characteristic $p$. The $N$ isn't important, it just saves me having to use the language of stacks. Let $Y^o$ denote the open affine of $Y_1(N)$ obtained by removing the supersingular points. Over $Y^o$ we have an elliptic curve $E$ (obtained from the universal family over $Y_1(N)$).
In brief: here's the question. The $p$-torsion $E[p]$ of $E$---it's a group scheme and its identity component is non-reduced. But (regarded as an abstract scheme) does it have a component which is reduced? I think it might! This goes against my intuition.
Now let me go more carefully. Let's consider the scheme $E[p]$ of $p$-torsion points. This is finite flat over $Y^o$ and hence as an an abstract scheme over $k$ it's going to be some sort of 1-dimensional gadget. It also sits in the middle of an exact sequence of group schemes over $Y^o$:
$0\to K\to E[p]\to H\to 0$
with $K=ker(F)$, $F$ the relative Frobenius map (an isogeny of degree $p$). Now at every point in $Y^o$, the fibre of $K$ is isomorphic to $\mu_p$ and the fibre of $E[p]$ is isomorphic to $\mu_p\times\mathbf{Z}/p\mathbf{Z}$. In particular all components of all fibres are isomorphic and non-reduced. Now here is where my argument in the thread in the question linked to above must become incorrect. I wanted to furthermore claim that 
(a) $K$ (as an abstract curve) is non-reduced, and then
(b) hence (because $K$ is the identity component of $E[p]$ and "all components of a group are isomorphic as sets") all components of $E[p]$ are non-reduced.
I now think that (b) is nonsense. In fact I know (b) is nonsense in the sense that $\mu_p$ over $\mathbf{Q}$ has only two components and they look rather different when $p$ is odd, but in some sense I feel here that the difference is more striking. In fact I now strongly suspect that $E[p]$ as an abstract scheme has two components, one being $K$ and the other being a regular scheme (an Igusa curve) mapping down in an inseparable way onto $Y^o$ (so the component isn't smooth over $Y^o$ but abstractly it's a smooth curve).
If someone wants a proper question, then there is one: am I right? The identity component of $E[p]$ is surely non-reduced---but does $E[p]$ have any regular components? I know how to prove this but it will be a deformation theory argument and I've got to go to bed :-/ If so then I think it's the first example I've seen, or at least internalised, of a group scheme where the behaviour of a non-identity component is in some sense a lot better than the behaviour of the identity component. I say "in some sense" because somehow it's the map down to $k$ that is better-behaved, rather than the map down to $Y^o$. Someone please tell me I'm not talking nonsense ;-)

REPLY [10 votes]: Speaking of "connected components" is a delicate thing since you really mean in a relative sense, and more specifically the etale quotient $H$ can have its open and closed non-identity part with very nontrivial $\pi_1$-action (so more subtle than on geometric fibers over the base).  But even if the $\pi_1$-action is trivial over whatever base, there are generally no no nontrivial sections through the non-identity part, so you can't do translation arguments, so there's no reason to expect intuition about "homogeneity" to have any relevance.  Likewise for any property which isn't local for whatever topology the thing admits local sections.  (In this case the fppf topology, for which regularity is not a local property, and ditto for reducedness.)
In this case Katz-Mazur (or better: Kummer!) did all of the deformation theory work.   If we pass to the complete local ring at a geometric point of the base curve then (by the Serre-Tate deformation theorem) you're really asking a question about something over the universal deformation ring $R = k[[x]]$ of the $p$-divisible group 
$$\mu_{p^{\infty}} \times (\mathbf{Q}_ p)/\mathbf{Z}_ p$$
over which the universal $p$-divisible group $\Gamma$ has finite flat $p$-torsion $G = \Gamma[p]$ with a connected etale sequence that is described explicitly in Katz-Mazur.  So there you can stare at the non-identity factors, and if those are regular then you're done. And if not regular somewhere then likewise for the global case over the modular curve.
If you look at (8.7.1.1) in KM (with $i \ne 0$) and then the proof of Prop. 8.10.5 there, or instead think about Kummer theory for group schemes, you'll see that there is a unit $q$ unique up to $p$-power unit multiple which "classifies" (up to isomorphism) the extension structure on $G$ (as $p$-torsion extension of $\mathbf{Z}/p \mathbf{Z}$ by $\mu_p$).  We can scale so $q$ is a 1-unit.  
Now I claim that $q-1$ has ord equal to 1 in the deformation ring $k[[x]]$.  Indeed, otherwise it would say that every first-order deformation of the $p$-divisible group has split $p$-torsion, which we know is nonsense (since we can use the unit $1+x$ to build a deformation violating that). 
So then  we can change $x$ so $q = 1 + x$ and the equations of the non-identity components are $T^p - (1+x)^i$ for $1 \le i \le p-1$ (from K-M, or thinking on our own).  I claim the quotient in each case is a discrete valuation ring.  Since $(1+x)^i = 1 + ix + x^2(\dots)$, by change of $x$ it is always the same as $T^p - (1+y)$ over $k[[y]]$, and writing it as $(T-1)^p - y$ since in characteristic $p$ we see it is Eisenstein, so we're done. 
Since function fields of the modular curves you had in mind are not perfect, perhaps a more amusing example for you of funny behavior is to give a reduced group scheme over a field which is not smooth, and a non-reduced group scheme whose underlying reduced scheme is not a subgroup scheme (affine groups of finite type, but ground field must be imperfect of course).<|endoftext|>
TITLE: Why are local systems and representations of the fundamental group equivalent
QUESTION [44 upvotes]: My question: Let X be a sufficiently 'nice' topological space. Then there is an equivalence between representations of the fundamental group of X and local systems on X, i.e. sheaves on X locally isomorphic to a constant sheaf. Does anyone know of a self contained, detailed treatment of this suitable for my background? I've looked at the first few pages of Delignes "Équations différentielles à points singuliers réguliers" (which my advisor suggested I take a look at) but here it just says that the equivalence is "well known", giving no reference. Neither googling ("local systems representations fundamental group") (nothing usable comes up), wiki nor the nLab entry (not detailed anough and more interested in generalisation) on local systems were of much help to me. I apologise in case the equivalence should obvious once one knows about universal covering spaces/deck transformations. I haven't learned those yet. If so, please let me know. 
Why I care: I am trying to read Simpsons "Higgs bundles and local systems",. Publ. Math. I. H. E. S. 75 (1992) 5–95". Simpson assumes this equivalence but gives no references. 
If there is any way this question could be improved upon, please let me know. 
Feel free to retag.

REPLY [4 votes]: I'd strongly recommend having a look at Atiyah & Bott's paper The Yang-Mills Equations over Riemann Surfaces.  They explain this pretty well, and embed it in a beautiful larger picture.  
The basic idea is that a locally constant sheaf can be viewed as the horizontal sections of a bundle with respect to a flat connection.   The holonomy of a flat connection along a curve only depends on the homotopy class of the curve, hence gives a representation of the fundamental group.<|endoftext|>
TITLE: Groupoid structure on G/H?
QUESTION [9 upvotes]: Let $G$ be a group and let $H$ be a subgroup. If $H$ is normal in $G$, then $G/H$ has a group structure. But in general, can there be a groupoid structure on $G/H$(left cosets or right cosets) that generalizes the normal case?

REPLY [2 votes]: Another way of putting this is that there is an equivalences of categories between actions of a group (or groupoid) $G$ and covering morphisms of the group (or groupoid) $G$. See for example 
Higgins, P.J. Notes on categories and groupoids, Mathematical Studies, Volume~32.
Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of
  Categories, No. 7 (2005) pp 1-195.
This links the notion of covering morphisms of groupoids  nicely with the theory of covering spaces.  There one finds that for "nice" spaces $X$ the fundamental groupoid $\pi_1$  gives an equivalence of categories  from covering maps of $X$ to covering morphisms of $\pi_1 X$. Thus a map of spaces is modelled by a morphism of groupoids, and this is convenient. For an exposition in these terms, see my book "Topology and groupoids", or earlier editions (1968, 1988).<|endoftext|>
TITLE: Approximation Property
QUESTION [5 upvotes]: It seems to be a folk result that l_infinity has the approximation property, even the bounded approximation property, and also, I think, even the so-called propery pi (approximation property) of Lindenstrauss. 
This is alluded to in a few texts, but I cannot seem to find the proof, which is presumably obvious. 
Does anyone have a reference or an easy solution?
Cheers,
R. Fry

REPLY [11 votes]: Partition the measure space into finitely many sets and consider the span of their indicator functions.  This space is isometrically isomorphic to $\ell_\infty^n$ for appropriate $n$ and hence is the range of a norm one projection.  Index the partitions by refinement to get a net of norm one finite rank projections that converge strongly to the identity.<|endoftext|>
TITLE: What is the infinite-dimensional-manifold structure on the space of smooth paths mod thin homotopy?
QUESTION [12 upvotes]: This question is motivated by the recent paper An invitation to higher gauge theory by Baez and Huerta, and the 2007 paper Parallel Transport and Functors by Schreiber and Waldorf.
Let $M$ be a smooth, finite-dimensional manifold.  A lazy path in $M$ is a smooth function $\gamma: [0,1]\to M$ such that all derivatives of $\gamma$ vanish at $0,1$.  A homotopy of lazy paths $\gamma_0,\gamma_1: [0,1] \to M$ is a smooth function $\Gamma: [0,1]^2 \to M$ such that all derivatives of $\Gamma(s,t)$ vanish near $t=0,1$, and such that $\Gamma(s,t) = \gamma_s(t)$ for $s = 0,1$.  A homotopy of lazy paths is lazy if additionally we have that for each $t$, all the $s$ derivatives of $\Gamma(s,t)$ vanish near $s = 0,1$.  A homotopy (of possibly non-lazy paths) is thin if $\text{rank}\\, d\Gamma < 2$ everywhere.  Note that (non-lazy) thin homotopies include all reparameterizations, and so any (possibly non-lazy) piecewise-smooth path is thinly homotopic to a lazy path.  Note also that lazy paths concatenate smoothly, and the concatenation of lazy paths is associative up to thin homotopy.  Note also that if $\gamma^{-1}(t) = \gamma(1-t)$, then the concatenation $\gamma^{-1}\gamma$ is thinly homotopic to a constant path.  Note also that lazy thin homotopies concatenate, and so define an equivalence relation, and if two paths are thinly homotopic, then they are lazily-thinly homotopic.  So define $\mathcal P^1(M)$ to be the set of all lazy thin homotopy equivalence classes of lazy paths in $M$.  It is a groupoid with base $M$.
The idea is to consider $\mathcal P^1(M) \rightrightarrows M$ as not just a groupoid but an infinite-dimensional Lie groupoid.  I think I understand the smooth structure on $\mathcal P^1(M)$: a curve in $\mathcal P^1(M)$ should be precisely a (non-thin) homotopy of lazy thin paths, up to thin homotopy.  It's not entirely clear to me that this defines a smooth structure.  But it probably works in some formalism.
But if I really want to think of $\mathcal P^1(M) \rightrightarrows M$ as a Lie groupoid, then I should treat $\mathcal P^1(M)$ not just as a smooth space, but actually as an (infinite-dimensional) manifold, and there are various things to check about the maps (the source and target maps should be surjective submersions, etc.).  And it's not clear to me how to write down a smooth manifold structure on $\mathcal P^1(M)$.
Here's what I'd like.  Given a point in $\mathcal P^1(M)$, I'd like a neighborhood of it and a "diffeomorphism" between that neighborhood and some (Fréchet, maybe?) vector space, and I'd like it to be clear that the gluings are smooth.  I can make a start: it's clear that the space of lazy paths in a finite-dimensional vector space is a vector space, and that thin homotopies respect addition, so that $\mathcal P^1(\mathbb R^n)$ is a vector space.  It's not clear to me how to put a topology on it, and it's not clear that I can approximate $\mathcal P^1(M)$ by chopping $M$ into trivializable pieces, take $\mathcal P^1$ of each piece, and try to glue back together — thin homotopies can take a path in one trivializable patch and make it wrap around $M$ in a complicated way, providing it wraps back, for example.
Hence the question:

What is the manifold structure on $\mathcal P^1(M)$?

REPLY [10 votes]: Okay, you asked for it!

Question: What is the manifold structure on $P^1(M)$?

 

Answer: There isn't one.


Update: The biggest failing is actually that the obvious model space is not a vector space.  The space of paths mod thin homotopy in $\mathbb{R}^n$ does not inherit a well-defined addition from the space of paths in $\mathbb{R}^n$.  Full details at the nLab page http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid.
(Update added here, rather than at the end, as it's the most direct answer to the specific question; the rest should be viewed as extra for those interested in more than just whether or not this space is a smooth manifold.)

It is, as you say, a smooth space.  This is formal: whatever category of generalised smooth spaces you like, take the quotient of $P(M)$ by thin homotopies.  All the proposed categories of generalised smooth spaces admit quotients, so the quotient exists and is a smooth space.  Depending on your choice of category, the description of this smooth space may vary.  For example, its Frolicher structure and its diffeological structure are very different.
But it is not "locally linear" in any sense.  The basic problem is that, as you say, within an equivalence class you have paths wrapping all the way around the manifold.  This destroys any hope of local linearity.
As for the proposed local model, you hit the nail on the head when you say:

It's not clear to me how to put a topology on it,

Absolutely!  Topologising these spaces can lead to quite strange behaviour.  You want a LCTVS structure, else you haven't a hope of even starting, and that can distort the topology from what you expect.  For example, if you take piecewise-smooth paths (with no quotient) then the LCTVS topology on that is the $C^0$-topology!  Indeed, simply taking so-called "lazy paths" could be fraught with difficulties (I notice that you define "lazy" slightly differently to how I've seen it done before with sitting instances).  Is that space a manifold?  (I know the answer to this one, but if you don't then you should start with that one as it is a much easier question and will hone your skills a little.)
If you really want a manifold, the solution is to go one step further.  Rather than quotienting out by thin homotopies, make your "thing" into a 2-structure and put the thin homotopies in at the 2-level.  Keep all paths at the 1-level.  Then each level has a manifold structure and by mapping into a 1-structure you effectively quotient out by the 2-structure but never actually have to consider the quotient itself.
To coin a phrase:

Quotients are horrible, it's a shame so many people think otherwise.

Lastly, that's not to say that there is no way of making $P^1(M)$ into a manifold.  There may well be.  But if there is, it'll be so convoluted and contrived that it won't look anything like the quotient of $P(M)$.  A cautionary tale here is the case of all paths in a manifold, $C^\infty(\mathbb{R},M)$.  That can be made into a manifold, but it has uncountably many components, for example, so looks absolutely horrid.
Okay, not quite lastly.  There's lots of details here that have been glossed over.  If you are really interested in working out the smooth space structure of this particular space then I (and I suspect Urs and Konrad) would be very interested in seeing it done and helping out.  But MO isn't the place for that.  Hop on over to the nLab, create a spin-off of http://ncatlab.org/nlab/show/path+groupoid, and start working.
Further Reading

Constructing smooth manifolds of loop spaces, Proc. London Math. Soc. 99 (2009) pp195–216 (doi:10.1112/plms/pdn058, arXiv:math/0612096).  The point of this is to figure out exactly when the "standard method" (alluded to by Tim) works.  The distinction between "loop" and "path" is irrelevant.
The Smooth Structure of the Space of Piecewise-Smooth Loops, Glasgow Mathematical Journal, 59 (2017) pp27-59. (arXiv:0803.0611, doi:10.1017/S0017089516000033).  Why you should be very, very nervous whenever anyone says "consider piecewise-smooth maps"; and take as a cautionary tale as to the inadvisability of going beyond smooth maps in general.
Work of David Roberts on the nLab.  This is where I got the 2-idea that I mentioned above.
Other relevant nLab pages: http://ncatlab.org/nlab/show/generalized+smooth+space, http://ncatlab.org/nlab/show/smooth+loop+space and further.
Of course, the magnificent book by Kriegl and Michor.  (I'm going to create a separate MO account for that book; its role will be to post an answer on relevant questions simply saying "Read Me".)


In response to Konrad's comment below, I've started an nlab page to work out the smooth structure of this space.  The initial content considers the linear structure of the space of paths in some Euclidean space modded out by thin homotopy.  The page is http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid.<|endoftext|>
TITLE: Graphs where every two vertices have odd number of mutual neighbours
QUESTION [10 upvotes]: There was a rather cute question last week about graphs where every pair of distinct vertices has an odd number of mutual neighbours.
The question was to show that such a graph must have an odd number of vertices, and it can be accomplished with a nice algebraic graph theory argument.
But let's up the ante a bit: can we actually characterize the graphs with this property? 
Here are some examples in the family:

complete graphs of odd order
anything obtained by gluing together a bunch of odd complete graphs at a single vertex
a graph of the form A - B - C where A and C have the "even" version of this property (every pair of vertices have even number common neighbours) B is an odd complete graph, and A is completely joined to B, B completely joined to C.

Is this the lot?

REPLY [4 votes]: Take a Steiner triple system on $v$ points.  Let $X$ be the graph with the $v(v-1)/6$ triples
as its vertices, two triples adjacent if the have exactly one point in common.  We need
$v\equiv1,3$ modulo 6.  Then two adjacent triples have exactly $(v+3)/2$ common neighbours, and two disjoint triples have exactly 9 common neighbours.  If we take $v\equiv3,7$ modulo
12 we get examples.
Of course I am just constructing strongly regular graphs with $\lambda$ and $\mu$ odd.
The are strongly regular graphs with this property besides the ones listed, for example generalized quadrangles with $s$ and $t$ even.  Further examples appear in Andries Brouwer's on-line tables (http://www.win.tue.nl/~aeb/graphs/srg/srgtab.html), or Gordon Royle's
(http://units.maths.uwa.edu.au/~gordon/remote/srgs/).  
This suggest that a classification might be difficult.<|endoftext|>
TITLE: Morphisms of supermanifolds
QUESTION [5 upvotes]: I am confused regarding supermanifolds. Suppose I consider R^(1,2) (1 "bosonic", 2 "fermionic"), This map (x,a,b) -> (x+ab, a,b) (a,b are fermionic) is supposed to be a morphism of this supermanifold. But I thought a morphism should be a continuous map from R->R together with a sheaf map of the sheaf of supercommutative algebra of smooth functions. How is this (x-> x+ab) giving me a continuous map from R->R?

REPLY [7 votes]: Unlike many schemes, but similar to ordinary manifolds, a map of super-manifolds $$(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$$ determines and is completely determined by the map of superalgebras obtained by looking at global sections:
$$\mathcal{O}_Y(Y) \to \mathcal{O}_X(X)$$
In the example at hand this is the graded ring map:
$$ x \mapsto x' + a'b'$$
$$ a \mapsto a'$$
$$ b \mapsto b'$$
This map induces a map of rings after we mod out by nilpotents:
$$C^\infty(Y) = \mathcal{O}_Y/Nil \to \mathcal{O}_X / Nil = C^\infty(X)$$
This map in turn induces a smooth map $X \to Y$ (in fact it is equivalent to such a map). In this case, after modding out by nilpotents we get the map $x \mapsto x'$, i.e. the identity on the underlying manifold $\mathbb{R}$.

REPLY [4 votes]: The ring of functions on your supermanifold is $C^\infty(\mathbb{R}) \otimes \mathbb{C}[a,b]$, where $a$ and $b$ are odd.  The even part is then $C^\infty(\mathbb{R}) \oplus C^\infty(\mathbb{R})ab$, where $(ab)^2 = 0$, so there is an even nilpotent direction.  You might want to view it as a thickening in a perpendicular direction.  The map in question is the identity on the reduced quotient $C^\infty(\mathbb{R})$, and this yields the identity map of manifolds (which is continuous).  The $(x \mapsto x+ab)$ can be viewed as an infinitesimal shearing on the even part.<|endoftext|>
TITLE: Can a corollary follow a conjecture?
QUESTION [7 upvotes]: It is typical to find a corollary that following theorems, but is it right to use the word corollary for a statement following a conjecture, where the statement is true only if the unproven conjecture is true?

REPLY [4 votes]: The correct term for such an item is CONJOLLARY.
;)<|endoftext|>
TITLE: What's the origin of the naming convention for the standard basis of $\mathfrak{sl}_2(\mathbb{C}) $?
QUESTION [15 upvotes]: $\mathfrak{sl}_2(\mathbb{C})$ is usually given a basis $H, X, Y$ satisfying $[H, X] = 2X, [H, Y] = -2Y, [X, Y] = H$.  What is the origin of the use of the letter $H$?  (It certainly doesn't stand for "Cartan.")  My guess, based on similarities between these commutator relations and ones I have seen mentioned when people talk about physics, is that $H$ stands for "Hamiltonian."  Is this true?  Even if it's not, is there a connection?

REPLY [16 votes]: The letters $\mathrm X$ and $\mathrm Y$ are already used by Cayley in what Dieudonné (in MR) calls the first description of all finite-dimensional irreducible $\mathfrak{sl}_2$-modules: A Second Memoir upon Quantics (1856, §§29–31). He apparently has no name for $\mathrm{XY-YX}$. Same in e.g. Faà di Bruno (1876, §§113–114).
$\mathrm H$ seems to stand for Hauptmatrix, as introduced by Weyl (1922, p. 125; see also 1931, pp. 114, 122) to describe Cartan subalgebras, roots (Multiplikatoren) and root spaces (Länder): 

Kommt in $\mathfrak g$ eine „Hauptmatrix“ $H$ vor, in der alle Elemente außerhalb der Hauptdiagonale verschwinden, während in der Hauptdiagonale die Zahlen $\alpha_1,\alpha_2,\dots,\alpha_n$ stehen, so bilde man die Differenzen $\alpha_i - \alpha_k$ und teile mit Bezug auf $H$ das Schema einer beliebigen Matrix in „Länder“ ein, indem man jedem Feld $(ik)$ des Schemas ($i$ der Zeilen-, $k$ der Kolonnenindex) die Zahl $\alpha_i - \alpha_k$ als „Multiplikator“ zuordnet (...)

$\mathfrak{sl}_2$-triples with your bracket relations appear in Killing (1888, p. 281), denoted $(X_{r-1}, X_r, X_{r-2})$; Cartan (1894, p. 116), denoted $\mathrm{(Y, X,X')}$; Weyl (1925, p. 276), denoted $(h_\alpha,e_\alpha,e_{-\alpha})$, with the $h_\alpha$ again called “Diagonal- oder Hauptmatrizen”; Dynkin (1952, §8.1), denoted $(f,e_+,e_-)$; Chevalley (1955, p. 28; 1955, p. 96), denoted $(D,N,N')$, $(H_r,X_r,X_{-r})$, and finally the desired $\mathrm{(H,X,Y)}$. 
(Standardization was slow: Lie (1876, p. 53; 1890, p. 353) used $(X_1,X_2,X_3)=(\mathrm{-Y,\smash{\frac12}H,X})$, and similarly rescaled bases and bracket relations still appear in Pauli (1927, p. 614), Born-Jordan (1930, p. 135), Casimir-van der Waerden (1931, p. 46; 1935, p. 4), Bauer (1933, p. 126), Harish-Chandra (1950, p. 301; 1952, p. 337), Jacobson (1951, p. 107; 1958, p. 825), Séminaire “Sophus Lie” (1955, p. 10-01), Kostant (1959, p. 977), etc. Settling on the “Chevalley” basis $\mathrm{(H,X,Y)}$ over Lie’s seems ultimately motivated by the smaller $\mathbf Z$-form $\mathfrak g_\mathrm{sc}\subset\mathfrak g_\mathrm{ad}$ it spans, cf. Borel (1970, §2.7).)<|endoftext|>
TITLE: Steinberg Representations of Finite Groups of Lie Type
QUESTION [10 upvotes]: Let G be a finite group of Lie type. Assume G is also of universal type. Is the Steinberg representation of G generic, i.e., does the Steinberg representation admit a Whittaker model? 
A Whittaker model for a representation of G is defined in a similar fashion as in the case of GL(2, F) in Bump's "Automorphic Forms and Representations." I am interested in the genericity of the Steinberg representation of a group of matrices over a finite field.

REPLY [11 votes]: I think that for finite groups of Lie type, the analogue of "having a Whittaker model" is that the representation occurs in a Gelfand-Graev representation: these are the representations obtained by inducing a "regular" character from the unipotent subgroup of a rational Borel. Such representations are multiplicity free and so constitute a "model" (in the sense I think people say "Whittaker model"). Now when the center of $G$ is connected, all regular characters are conjugate under the action of the maximal torus of the Borel, so the Gelfand-Graev representation is unique (otherwise there is a family of such representations). In their famous paper, Deligne and Lusztig decompose the Gelfand-Graev representation in this case and show that there is exactly one constituent in each "geometric conjugacy class" of irreducible representations (which can be thought of as a semisimple conjugacy class in the dual group). The Steinberg representation is then the representative in the conjugacy class of the identity element -- that is the representative among the "unipotent representations".
To focus more on the actual question (!) the character of the Steinberg representation is explicitly known, and it is easy to check from this that its restriction to $U$ is the regular representation, so it certainly occurs in the Gelfand-Graev representation.<|endoftext|>
TITLE: Why should I prefer bundles to (surjective) submersions?
QUESTION [27 upvotes]: I hope this question isn't too open-ended for MO --- it's not my favorite type of question, but I do think there could be a good answer.  I will happily CW the question if commenters want, but I also want answerers to pick up points for good answers, so...
Let $X,Y$ be smooth manifolds.  A smooth map $f: Y \to X$ is a bundle if there exists a smooth manifold $F$ and a covering $U_i$ of $X$ such that for each $U_i$, there is a diffeomorphism $\phi_i : F\times U_i \overset\sim\to f^{-1}(U_i)$ that intertwines the projections to $U_i$.  This isn't my favorite type of definition, because it demands existence of structure without any uniqueness, but I don't want to define $F,U_i,\phi_i$ as part of the data of the bundle, as then I'd have the wrong notion of morphism of bundles.
A definition I'm much happier with is of a submersion $f: Y \to X$, which is a smooth map such that for each $y\in Y$, the differential map ${\rm d}f|_y : {\rm T}_y Y \to {\rm T}_{f(y)}X$ is surjective.  I'm under the impression that submersions have all sorts of nice properties.  For example, preimages of points are embedded submanifolds (maybe preimages of embedded submanifolds are embedded submanifolds?).
So, I know various ways that submersions are nice.  Any bundle is in particular a submersion, and the converse is true for proper submersions (a map is proper if the preimage of any compact set is compact), but of course in general there are many submersions that are not bundles (take any open subset of $\mathbb R^n$, for example, and project to a coordinate $\mathbb R^m$ with $m\leq n$).  But in the work I've done, I haven't ever really needed more from a bundle than that it be a submersion.  Then again, I tend to do very local things, thinking about formal neighborhoods of points and the like.
So, I'm wondering for some applications where I really need to use a bundle --- where some important fact is not true for general submersions (or, surjective submersions with connected fibers, say).

REPLY [5 votes]: A reason for working with submersions rather than bundles is that submersions of open manifolds have been classified up to regular homotopy: A. Phillips, Submersions of open manifolds, Topology 6, 1967, 171-206. MR0208611).  See also D. Spring, The golden age of immersion theory in topology, Bull. Amer. Math. Soc. 42, 2005, 163-180.<|endoftext|>
TITLE: How to optimize student happiness in group work?
QUESTION [5 upvotes]: There are $n$ students in a class, and they must be divided into, say, $k$ groups.  Each student ranks the other students in order of preference of working together.  Is there a way to generally optimize student happiness (where happiness is based on working with preferred teammates). We could assume for simplicity that happiness is correlated in a simple (say linear) way with preference rank of group members.
When will there be a unique optimal grouping?
What if the happiness is not linearly correlated to preference rank?

REPLY [9 votes]: This is a generalization of the stable roommate problem (which is the same thing where $k = n/2$, ie, groups of 2). In general, there exist groups in which under any pair of groups contain members who would both like to switch teams. 
From wikipedia: 

For a minimal counterexample, consider 4 people A, B, C and D where all prefer each other to D, and A prefers B over C, B prefers C over A, and C prefers A over B (so each of A,B,C is the most favorite of someone). In any solution, one of A,B,C must be paired with D and the other 2 with each other, yet D's partner and the one for whom D's partner is most favorite would each prefer to be with each other.<|endoftext|>
TITLE: The simplicial Nerve
QUESTION [7 upvotes]: The nerve functor $N:Cat\to SSet$ from the category of small categories to simplicial sets can be obtained as follows: The left Kan extension of the functor $F$ which sends $[n]$ to the category $\bullet\to\ldots\to\bullet$ along the Yoneda embedding $\Delta\to SSet$ gives an adjunction
$$
     h:SSet \leftrightarrows Cat:N
$$
with $N_n(\mathcal{C})=Hom_{Cat}(F(n),\mathcal{C})$.
There exists also a simplicial nerve functor $\mathfrak{N}$ invented by Cordier, I think. Its construction can be found in Lurie's HTT. It is a functor from the category of small simplicial categories (small categories enriched over simplicial sets) $Cat_\Delta$ to $SSet$. It seems to me that $\mathfrak{N}$ can be constructed using a left Kan extension as above too, but of another functor $F$, Lurie calls ${\mathfrak{C}}[-]$ which I can not typeset correctly. Is this true?
Can $\mathfrak{N}$ be constructed as follows also? Maybe it's the same thing:
There exists an enriched version of left Kan extensions. The categories $SSet$ and $Cat_\Delta$ are canonically enriched over $SSet$. The Yoneda embedding is a simplicial functor if $\Delta$ is considered as a discrete simplicial category. Maybe ${\mathfrak{C}}[-]$ is a simplicial functor too and perhaps it can be defined more naturally in this way, I don't know. Then one would get an enriched adjunction
$$
     sh:SSet \leftrightarrows Cat_\Delta:R
$$
and I wonder if $R$ is $\mathfrak{N}$ then.

REPLY [7 votes]: About the first question: Yes, the simplicial nerve is an instance of the same general construction which gives you the usual nerve (and e.g. also the Quillen equivalences between simplicial sets and topological spaces, between models for the $\mathbb{A}^1$-homotopy category given by simplicial presheaves and sheaves respectively and much more): 
A cosimplicial object in a cocomplete category E gives, via Kan extension, rise to an adjunction between $E$ and simplicial sets, where the left adjoint goes from simplicial sets to $E$ and comes from the universal property of the Yoneda embedding (namely that functors from $C$ to $E$, $E$ cocomplete, are the same as colimit perserving functors from $Set^{C^{op}}$ to $E$). This is (part of) Proposition 3.1.5 in Hovey's book on model categories.
Lurie uses the same pattern; it is enough to give a cosimplicial object in simplicial categories, which is done in his definition 1.1.5.1.<|endoftext|>
TITLE: Stable Tables on Fluctuating Floors
QUESTION [8 upvotes]: If a four-legged, rectangular table is rickety, it can nearly always be stabilised just by turning it a little. This is very useful in everyday life! Of course it relies on the floor being the source of the ricketiness; if the table's legs are different lengths, it doesn't work (this is why I said 'nearly always').
Here is a quick 'proof': Label the feet A, B, C, D in order, and integrate the function F(A) + F(C) - F(B) - F(D) while the table is turned through 180° (here F is the height of the floor). The result is ∫ F - ∫ F = 0 (these integrals are over 360°), so the function must be zero somewhere; at this point, the table is stable.
This proof can easily be adapted for any cyclic quadrilateral ABCD.
The proof only works if the slope S of the floor is small enough that S2 can be ignored. If not, and if the stable solution is not horizontal, then the feet will not lie on the projection of the circle that we integrated over. So a more complicated argument is required:
Suppose ABCD is a square, and suppose that |F(x)-F(y)| <= S|x-y| everywhere, where S = sin-1(1/√3). Fix a vertical line V. Then ABCD can always be positioned with its centre on V, so that each corner touches the floor.
To see this, consider the two diagonals AC and BD separately. Choose a starting position such that

(i) their endpoints lie on the floor, 
(ii) their midpoints are on V,
(iii) they are perpendicular to each other.

We are not requiring that their centres coincide (so we will have to dismantle the table to achieve this). We may suppose that in the starting position, the centre of AB lies above the centre of BD. Now rotate them so that constraints (i)-(iii) remain satisfied, until AC occupies the starting position of BD and vice versa. (The constraint on the slope ensures that we can do this continuously.) Now the centre of AB lies below the centre of BD, so the two centres must have coincided at some time.
(This proof generalises to rectangles, but the maximum slope depends on the rectangle. It does not generalise to cyclic quadrilaterals.)
My first question is this: Is the condition on the slope (or a less stringent slope condition) necessary? Or can we always stabilise the table, whatever the slope of the floor?
My second question (if the answer to the first question is that we can always stabilise the table): Given any partition of $\mathbb R$3 into {L,U} with L bounded above and U bounded below, and a vertical line V, can we always find a unit square whose corners lie in the closures of both L and U, and whose centre lies on V?
Update The paper referenced below by Q.Q.J. answers my first question: a rectangular table can always be stabilised, if the floor function F is continuous. But it can only be stabilised by a smooth rotation if the floor satisfies the slope condition. (I was wrong about different rectangles requiring different slope conditions.)

REPLY [10 votes]: "Mathematical table turning revisited" by Baritompa, L"owen, Polster, and Ross
I am no expert on what is or isn't possible but there are at least two different groups who have looked at this type of problem and this article contains a number of references that are probably relevant to you.<|endoftext|>
TITLE: Can every finite graph be represented by an arithmetic sequence of natural numbers?
QUESTION [5 upvotes]: (This is a follow-up to my previous questions Natural models of graphs?.)
Erdös in The Representation of a Graph by Set Intersections (1966) states:

Theorem. Let $G$ be an arbitrary
  graph. Then there is a set $S$ and a
  family of subsets $S_1, S_2, ...$ of
  $S$ which can be put into one-to-one
  correspondence with the vertices of
  $G$ in such a way that $x_i$ and $x_j$ are joined by an
  edge of $G$ iff $i \neq j$
  and $S_i \cap S_j \neq \emptyset$.

If we identify $S$ with a set of prime numbers and each $S_i$ with the product of its members we get the following:

Corollary. Let $G$ be an arbitrary finite
  graph. Then there is a sequence of natural numbers $(n_1, n_2, ..., n_k)$ 
  which can be put into one-to-one
  correspondence with the vertices of
  $G$ in such a way that  $x_i$ and $x_j$ are joined by an edge iff $i \neq j$ and GCD$(n_i, n_j) > 1$.

We can choose the prime numbers (the elements of $S$, from which the $n_i$ are built) arbitrarily, and so the question arises, whether they can always be choosen in such a way, that the set $(n_1, n_2, ..., n_k)$ is an arithmetic sequence. 
Of course every complete graph on $k$ nodes can be represented by an arithmetic sequence: just take some consecutive sequence of even numbers. Green-Tao's Theorem guarantees that also every empty graph on $k$ nodes can be represented by an arithmetic sequence $(p_1, p_2, ..., p_k)$ of primes.

Question: Can every graph on $k$ nodes be represented by an arithmetic sequence
  of natural numbers such that  $n_i$ and $n_j$ are joined by an edge iff $n_i \neq n_j$ and GCD$(n_i, n_j) > 1$

This would be one kind of natural model of a graph, that I was looking for, originally.
Maybe some references?
Added: Due to Kevin's concise answer and Thomas' comment, I'd like to add the following question:

Question: If not every graph on $k$ nodes can be represented by an arithmetic sequence
  of natural numbers such that  $n_i$ and $n_j$ are joined by an edge iff $n_i \neq n_j$ and GCD$(n_i, n_j) > 1$: Are there interesting classes of graphs with this property?

REPLY [11 votes]: OK so take the unique tree on 3 vertices. Claim: you can't encode this with an arithmetic progression (AP). For if the AP is $a,a+d,a+2d$ then (because we have two edges) either vertices 1 and 2 are joined, or vertices 2 and 3 are joined (or both). Hence there is some $p>1$ such that either $p$ divides both $a$ and $a+d$, or $p$ divides both $a+d$ and $a+2d$. In either case, $p$ then divides $d$, so it divides $a$, so it divides everything, so the graph is complete.<|endoftext|>
TITLE: Why would one expect a derived equivalence of categories to hold?
QUESTION [26 upvotes]: This question is perhaps somewhat soft, but I'm hoping that someone could provide a useful heuristic. My interest in this question mainly concerns various derived equivalences arising in geometric representation theory.
Background
For example, Bezrukavnikov, Mirkovic, and Rumynin have proved the following: Let $G$ be a reductive algebraic group over an algebraically closed field of positive characteristic. Then there is an equivalence between the bounded derived category of modules for the sheaf $\cal D$ of crystalline (divided-power) differential operators on the flag variety, and the bounded derived category of modules with certain central character for the enveloping algebra $\cal U$ of Lie($G$). What is interesting is that it is not true that this equivalence holds on the non-derived level: The category of $\cal D$-modules is not equivalent to the category of $\cal U$-modules with the appropriate central character. This is true in characteristic 0 (this is the Beilinson-Bernstein correspondence), but something is broken in positive characteristic: there are certain "bad" sheaves that are $\cal D$-modules which make the correspondence not hold.
There are other results in geometric representation theory of this form. For example, Arkhipov, Bezrukavnikov, and Ginzburg have proved that there is an equivalence (in characteristic 0) between the bounded derived category of a certain block of representations for the quantum group associated to $G$, and the bounded derived category of $G \times \mathbb C^*$-equivariant sheaves on the cotangent bundle of the flag variety of $G$. Again, this equivalence does not hold on the non-derived level.
In general, there are a number of results in geometric representation theory that hold on the derived level, but not the non-derived level.
Question
Here's my question: Why would one be led to expect that a derived equivalence holds, when the non-derived equivalence does not? It seems as though the passage to the derived level in some sense fixes something that was broken on the non-derived level; how does it do that?

REPLY [18 votes]: I asked a similar question to Daniel Huybrechts some time ago, in the form of "If I have a derived equivalence between two varieties, what is this telling me about the relation between the two varieties?"  
His answer was that this should be crudely regarded as saying that each of the varieties is the moduli space for a (sufficiently interesting) moduli problem on the other variety.  This is definitely the 'Fourier-Mukai' perspective, that says to understand a derived equivalence, see where the skyscraper sheaves of points go under the equivalence.
It doesn't answer your question in general, but it makes a useful heuristic in some geometric cases.<|endoftext|>
TITLE: Mumford Conjecture
QUESTION [6 upvotes]: The Mumford Conjecture (now a theorem) says basically what is the (tautological subring)* of the rational cohomology ring of the stable moduli space of curves. Meaning that we know the ring structure (corresponding to the tautological ring)* of the cohomology of the moduli space of Riemann surfaces when the genus is very very big. 
Though the problem could have been stated as an algebraic geometry one, It was worked out (as far as I know), roughly speaking by arguments coming from algebraic topology. 
However, what are the implications (in general) of such a conjecture? In particular in Algebraic geometry and Algebraic topology.
*corrected, Thks.

REPLY [7 votes]: One application that I know of the Mumford conjecture is Teleman's proof of Givental's conjecture in this paper. Givental's conjecture states that when the quantum cohomology of a smooth projective variety (or compact symplectic manifold) is (generically) semisimple (meaning that the algebra is semisimple for generic values of the deformation parameter), then the higher genus Gromov-Witten invariants are uniquely and explicitly determined by the quantum cohomology. Since quantum cohomology consists of genus 0 information, another way to say this is that the higher genus Gromov-Witten invariants are uniquely and explicitly determined by the genus 0 Gromov-Witten invariants, when the quantum cohomology is semisimple.
The proof, extremely roughly, uses the Mumford conjecture in the following way: semisimplicity allows you to go back and forth between low genus and high genus without loss of information. The Mumford conjecture tells you what things look like in very high genus, so if you want to know what something in some arbitrary genus looks like, you kick it up to very high genus, identify it by the Mumford conjecture, and then kick it back down to its original genus.
---Begin large parenthetical remark---
You might then ask, which smooth projective varieties (or compact symplectic manifolds) have semisimple quantum cohomology? Smooth projective toric varieties form a class of examples. Arend Bayer showed that semisimple quantum cohomology is preserved under blowing up points. There are other examples...
There is an interesting conjecture of Dubrovin which states that a variety has semisimple quantum cohomology if and only if its derived category has a full exceptional collection. (This conjecture is based heavily on mirror symmetry philosophy...) I don't know the status of this conjecture. But I think it is true, for example, that having a full exceptional collection is preserved under blowing up points. (Proved by Orlov? Bondal?)
---End large parenthetical remark---
It is also apparently possible to view the Mumford conjecture as a special case of the cobordism hypothesis. Take a look at Jacob Lurie's paper on TFTs, particularly section 2.5.<|endoftext|>
TITLE: Is there a presentation of the cohomology of the moduli stack of torsion sheaves on an elliptic curve?
QUESTION [5 upvotes]: Let $E$ be your favorite elliptic curve, and let $Tor^m$ be the moduli stack of torsion sheaves of degree $m$ on $E$.  This sounds horrible, but it's not so bad; it's a global quotient of a smooth Quot scheme (the space of rank m subbundles of degree -m in $\mathcal{O}_E^{\oplus m}$) modulo the obvious action of GL(m).
For various nefarious reasons of my own, I would like to have an explicit presentation of the cohomology ring of this stack (i.e. the equivariant cohomology of this Quot scheme for GL(m)).  


Does anyone know of such a presentation?


EDIT: Since Torsten asked in comments, I'd be perfectly happy just to understand the rational cohomology, though obviously integral would be even better.  I would also mention that I really want the ring structure, not just the vector spaces.

REPLY [8 votes]: It seems that one can obtain the additive structure of rational cohomology
without too much effort (in no way have I checked this carefully so caveat
lector applies). As Allen noticed, for rational cohomology it is enough to
compute $T$-equivariant cohomology and then take $\Sigma_m$-invariants (if this is to
work also for integral cohomology a more careful analysis would have to be made  I think). Now, the space $Tor^m_C$ of length $m$ quotients of $\mathcal O^m_C$ (I use $C$
as everything I say will work for any smooth and proper curve $C$) is smooth and  proper so we may use the Bialinski-Birula analysis (choosing a general $\rho\colon \mathrm{G}_m \to T$) and we first look at the fixed point locus of $T$.                                                                                     
Now, for every sequence $(k_1,\ldots,k_m)$ of non-negative integers with
$k_1+\ldots+k_m=m$ gives a map $S^{k_1}C\times\cdots\times S^{k_m}C \to Tor^m_C$, where $S^kC$ is the
symmetric product interpreted as a Hilber scheme, and the map takes $(\mathcal
I_1,\ldots,\mathcal I_m)$ to $\bigoplus_i\mathcal I_i \hookrightarrow \mathcal O^m_C$. It is
clear that this lands in the $T$-fixed locus and almost equally clear that this
is the whole $T$-fixed locus (any $T$-invariant submodule must be the direct sum of its
weight-spaces).
We can now use $\rho$ to get a stratification parametrised by the sequences
$(k_1,\ldots,k_m)$. Concretely, the tangent space of $Tor^m_C$ to a point of
$S^{k_1}C\times\cdots\times S^{k_m}C$ has character $(k_1\alpha_1+\cdots+k_m\alpha_m)\beta$, where
$\beta=\sum_i\alpha_i^{-1}$ (and we think of characters as elements of the group ring
of the character group of $T$ and the $\alpha_i$ are the natural basis elements of
the character group). This shows that $\rho$ can for instance be chosen to be $t \mapsto
(1,t,t^2,\ldots,t^{m-1})$. In any case the stratum corresponding to $(k_1,\ldots,k_m)$ has
as character for its tangent space the characters on the tangent space on which
$T'$ is non-negative and its normal bundle consists of those on which $T'$ is
non-negative (hence with the above choice, the character on the tangent space is
$\sum_{i\geq j}k_i\alpha_i\alpha_j^{-1}$ and in particular the dimension of the stratum is
$\sum_iik_i$). We now have that each stratum is a vector bundle of the
corresponding fixed point locus so in particular the equivariant cohomology of
it is the equivariant cohomology of the fixed point locus and in particular is
free over the cohomology ring of $T$. Furthermore, if we build up the cohomology
using the stratification (and the Gysin isomorphism), at each stage a long exact
sequence splits once we have shown that the top Chern class of the normal bundle
is a non-zero divisor in the cohomology of the equivariant cohomology of the
component of the fixed point locus (using the Atiyah-Bott criterion).
However, the non-zero divisor condition seems more or less automatic (and that
fact should be well-known): The normal bundle $\mathcal N$ of the
$(k_1,\ldots,k_m)$-part, $F$ say, of the fixed point locus splits up as direct sum
$\bigoplus_\alpha \mathcal N_\alpha$, where $T$ acts by the character $\alpha$ on $\mathcal N_\alpha$. Then
the equivariant total Chern class of $\mathcal N_\alpha$ inside of
$H^\ast_T(F)=H^\ast(F)\bigotimes H^\ast_T(pt)$ is the Chern polynomial of $\mathcal N_\alpha$
as ordinary vector bundle evaluated at $c_1(\alpha)=\alpha\in H^2_T(pt)$. Hence, if we quasi-order
the characters of $T$ by using $\beta \mapsto -\rho(\beta)$ (``quasi'' as many characters get the
same size), then as $-\rho(\alpha)>0$ (because $\alpha$ appears in the normal bundle) we get
that $1\otimes\alpha^{n_\alpha}\in H^\ast_T(F)$, where $n_\alpha$ is the rank of $\mathcal N_\alpha$, is the term of
$c_n(\mathcal N_\alpha)$ of largest order. Hence, we get that for the top Chern class
of $\mathcal N$ which is the product $\prod_\alpha c_{n_\alpha}(\mathcal N_\alpha)$ its term of
largest order has $1$ as $H^\ast(F)$-coefficient and hence is a non-zero divisor.
As the cohomology of $S^nC$ is torsion free we get that all the involved
cohomology is also torsion free and everything works over the integers but as
I've said to go integrally from $T$-equivariant cohomology to
$\mathrm{GL}_m$-equivariant cohomology is probably non-trivial.
If one wants to get a hold on the multiplicative structure one could use that
the fact that the Atiyah-Bott criterion works implies that that $H^\ast_T(Tor^m_C)$
injects into the equivariant cohomology of the fixed point locus. The algebra
structure of the cohomology of $S^nC$ is clear (at least rationally) so we get
an embedding into something with known multiplicative structure. The tricky
thing may be to determine the image. We do get a lower bound for the image by
looking at the ring generated by the Chern classes of of the tautological bundle
but I have no idea how close that would get us to the actual image. (It is a
well-known technique anyway used in for instance equivariant Schubert calculus
so there could be known tricks.)
There is another source of elements, namely we have the map $Tor^m_C \to S^mC$. This map becomes 
even better if one passes to the $\Sigma_m$-invariants.
[Later] Upon further thought I realise that the relation between $T$- and $G=GL_m$-equivariant cohomology
is simpler than I thought. The point is (and more knowledgeable people certainly know this) that the map $EG\times_TX \to EG\times_GX$ is a $G/T$ bundle and $G$ being special $H^*_T(X)$ is free as a $H^*_G(X)$-module (with $1$ being one of the basis elements). That means that $H^*_G(X) \to H^*_T(X)$
is injective but more precisely $H^*_T(X)/H^*_G(X)$ is torsion free with $\Sigma_m$-action without invariants so that $H^*_G(X)$ is the ring of $\Sigma_m$-invariants of $H^*_T(X)$.<|endoftext|>
TITLE: Estimating flat norm distance from a planar disc
QUESTION [8 upvotes]: Let $D\subset\mathbb R^2\subset\mathbb R^n$ be a unit planar disc in $\mathbb R^n$. Let $S$ be an orientable two-dimensional surface in $\mathbb R^n$ such that $\partial S=\partial D$. Of course, we have $area(S)\ge area(D)$. Assume that $area(S)< area(D)+\delta$ where $\delta>0$ is small.
Then $S$ is close to $D$ in the following sense: there is a 3-dimensional surface $F$ filling the gap between $S$ and $D$ such that $volume(F)<\varepsilon(\delta)$ where $\varepsilon(\delta)\to 0$ as $\delta\to 0$ ($n$ is fixed). "Filling the gap" means that $\partial F=S-D$.
This fact immediately follows from the compactness theorem for flat norms. But this proof is indirect and does not answer the following questions (I am especially interested in the second one):
1) Are there explicit upper bounds for $\varepsilon(\delta)$? How do they depend on $\delta$ and $n$?
2) Can $\varepsilon(\delta)$ be independent of $n$? Or, equivalently, does the above fact hold true in the Hilbert space?
In the unlikely event that 2-dimensional surfaces are somehow special, what about the same questions about $m$-dimensional surfaces, for a fixed $m$?
Remarks: "Surfaces" here are Lipschitz surfaces or rectifiable currents or whatever you prefer to see in this context. Rather than talking about the filling surface $F$, one could equivalently say that the integral flat norm of $S-D$ is less than $\varepsilon(\delta)$.

REPLY [4 votes]: There is Almgren's isoperimetric inequality:

Let $\Sigma$ be a $k$-surface in $\mathbb R^n$. Assume $vol _k \Sigma \le vol_k S^k$. Then one can fill $\Sigma$ by a $(k+1)$-surface with volume $\le vol_{k+1} B^{k+1}$. (Here the "surfaces" might have singularities.)

I will use it to show that there is an estimate $\epsilon(\delta)$ which does not depend on $n$.
Take $r$-nbhd $Z_r$ of $D$.
Note* that one can give an explicite estimate of $r$, independent of $n$ so that total area of $S$ outside of $Z_r$ is very small. Moving a bit $r$, one can make the length of intersection curve $\gamma=\partial Z_r\cap S$ sufficiently small. 
Use Almgren to fill $\gamma$ by a surface;
it breaks $S$ into two pieces $S=S_1+S_2$; 

the surface $S_1$ lies in $Z_r$ and $\partial S_1=\partial D$, 
the surface $S_2$ has small area and $\partial S_2=0$. 

Fill both $S_1$ and $S_2$ separately:

taking all segments from point on $S_1$ to its projection on $D$ gives a filling of $S_1-D$ 
fill $S_2$ using Almgren again.


(*)There is a map $\mathbb R^n\to D$ which decrease distances by some factor $k=k(r)<1$ outside of $Z_r$ and $k(r)$ can be found explicitly.
So if an essential piece of $S$ is outside of $Z_r$ then the area of $S$ is essentially bigger that $area(D)$.
Say, take $f(x)=$ "sum of maximal and minimal distance to the points in $D$".
This function is convex and it is constant on $D$. 
Take Sharafutdinov retruction for the level sets of this function.<|endoftext|>
TITLE: Algebraic de Rham cohomology vs. analytic de Rham cohomology
QUESTION [21 upvotes]: Let $X$ be a nice variety over $\mathbb{C}$, where nice probably means smooth and proper.
I want to know: How can we show that the hypercohomology of the algebraic de Rham complex agrees with the hypercohomology of the analytic de Rham complex (equivalently the cohomology of the constant sheaf $\mathbb{C}$ in the analytic topology)? Does this follow immediately from GAGA? If not, how do you prove it?
I think that this does not follow immediately from GAGA because, while the sheaves $\Omega_X^i$ are coherent, the de Rham $d$ is not a map of coherent sheaves (it is not multiplicative). Am I correct in my thinking?

REPLY [8 votes]: Various people have answered the question, and also brought up some of the subtleties in applying GAGA. So I won't  rehash all that. So let me just suggest the additional reference: 
Deligne, Équations différentielles à points singuliers réguliers
especially Chapter II, section 6. These issues are dealt with carefully in a more
general setting of de Rham cohomology with coefficients in a regular integrable connection.
The result is no doubt true for a regular holonomic D-module, and it would be nice if
someone wrote this down carefully. But perhaps I'm straying too far from the original topic.<|endoftext|>
TITLE: What (if anything) happened to Viennot's theory of Heaps of pieces?
QUESTION [26 upvotes]: In 1986 G.X. Viennot published "Heaps of pieces, I : Basic definitions and combinatorial lemmas" where he developed the theory of heaps of pieces, from the abstract: a geometric interpretation of Cartier-Foata's commutation monoid. This theory unifies and simplifies many other works in Combinatorics : bijective proofs in matrix algebra (MacMahon Master theorem, inversion matrix formula, Jacobi identity, Cayley-Hamilton theorem), combinatorial theory for general (formal) orthogonal polynomials, reciprocal of Rogers-Ramanujan identities, graph theory (matching and chromatic polynomials). 
In the references the subsequent articles "Heaps of pieces, 4 and 5" are listed as "in preparation" where the applications of the theory to solving the directed animal problem and statistical physics are supposed to be developed. I know that these parts of the theory have appeared in literature but I am sort of puzzled as to why the series of papers was not continued (searching for Heaps of pieces II or III or IV doesn't give results). Is there any survey of the full theory somewhere else?
Also, since I didn't feel like asking this in a separate question, is there any paper that proves classical theorems of dimers (Kasteleyn's theorem, Aztec diamond etc.) using Viennot's theory?

REPLY [3 votes]: There has been recent work by Tyler Helmuth which applies Viennot's theory to statistical mechanics and in particular finding new expression for the two point function of Ising-like systems.
See his article: https://arxiv.org/abs/1209.3996<|endoftext|>
TITLE: What tensor product of chain complexes satisfies the usual universal property?
QUESTION [8 upvotes]: Recall that a chain complex is a (finite) diagram of the form
$$ V = \{ \dots \to V_3 \overset{d_3}\to V_2 \overset{d_2}\to V_1 \overset{d_1}\to V_0 \to 0 \} $$
where the $V_n$ are (finite-dimensional) vector spaces and for each $n$, $d_n \circ d_{n+1} = 0$.  If $V$ and $W$ are chain complexes, a chain map $f: V \to W$ is a map $f_n : V_n \to W_n$ for each $n$ such that all the obvious squares commute — "$[d,f]=0$" — and the pair (chain complexes, chain maps) defines a category.  In fact, it is a 2-category: the 2-morphisms between $f,g : V \rightrightarrows W$ are the chain homotopies, i.e. a system of maps $h_n: V_n \to W_{n+1}$ such that "$[d,h] = f-g$".  The category of chain complexes has a biproduct (both a product and a coproduct) $\oplus$ given by the pointwise direct sum.
I thought I knew what the tensor product of chain complexes was.  Namely, if $V$ and $W$ are chains, then the usual thing is to define
$$ (V\otimes W)\_n = \bigoplus_{k=0}^n V_k \otimes  W_{n-k} $$
and the chain maps are the sums of the obvious tensor products of differentials, decorated with signs.
But now I'm not sure why this is the tensor product picked.  Namely, if I have a linear category, I think that a tensor product $V \otimes W$ should satisfy the following universal property: for any $X$, $\hom(V \otimes W,X)$ should be naturally isomorphic to the space of bilinear maps $V \times W \to X$.  Now, I've never really known how to write down the word "bilinear" in a general category, without refering to individual points.  But I think I do know what the "set" $V \times W$ is when $V$ and $W$ are chains — it's the set underlying $V \oplus W$ — and then I think I do know what bilinear maps should be.
In any case, then it's clear that the usual tensor product is not this.  For example, if $V,W$ have no non-zero terms above degree $n$, then the bilinear maps $V \times W \to X$ I think cannot be interesting above degree $n$, whereas the above $\otimes$ has terms in degree $2n$.
In any case, in HDA6, Baez and Crans consider two-term chain complexes $V_1 \to V_0$ (they argue that these are the same as "2-vector-spaces"), and then construct a different tensor product, given by:
$$ V\otimes W = \{ (V_1 \otimes W_1) \oplus (V_1 \otimes W_0) \oplus (V_0\otimes W_1) \to (V_0 \otimes W_0) \} $$
where the differential is the sum of the obvious tensor products of differentials and identity maps.  They then assert that this tensor product satisfies the correct universal property, although they leave the details to the reader.
This leads naturally to:

Question: What is the precise universal property that $\otimes$ ought to have, and what "product" of chain maps satisfies this universal property?

REPLY [3 votes]: Several people have mentioned that $\otimes N$ should left adjoint to a suitable $Hom(N,-)$. There are numerous instances of this sort of construction (including one using joins of augmented simplicial sets mentioned in another question w.r.t Lurie's HTT). There is a good discussion of the ideas in MacLane's Homology and in Hilton and Stammbach. The sign conventions are, however, often not fully explained.  (They work, but why? is the feeling one gets.)
The point in Baez and Crans is more that there are variants that have good more or less geometric interpretations and they give sensible answers. (There is also work by Andy Tonks on tensor products of crossed complexes that makes this clear.)
The discussion in older sources such as Spanier is also good for explaining why the signs are important.
You ask about bilinearity. This is really a strange idea.  It can be handled in more generality than the simple form you mention but it can often be best interpreted as coming from the evaluation map $ Hom(A,B)\times A \to B$ in which $(f,x)$ gets sent to $f(x)$. If you look at the adjunction that was mentioned between $\otimes$ and $Hom$, then from $\phi : C\to Hom(A,B)$ you do a naive thing and get $C\times A \to Hom(A,B)\times A\to B$ and that will be bilinear, ugh! Thus the representability of bilinear maps is the same universal property as the one everyone has been mentioning.<|endoftext|>
TITLE: Can epi/mono for natural transformations be checked pointwise?
QUESTION [26 upvotes]: Let $\mathcal C$ be a category.  Recall that a morphism $f : X \to Y$ is epi if $$\circ f: \hom(Y,Z) \to \hom(X,Z)$$ is injective for each object $Z \in \mathcal C$.  ($f$ is mono if $f\circ : \hom(Z,X) \to \hom(Z,Y)$ is injective.)
Let $\mathcal C,\mathcal D$ be categories.  Then $\hom(\mathcal C,\mathcal D)$, the collectional of all functors $\mathcal C \to \mathcal D$, is naturally a category, where the morphisms are natural transformations: if $F,G: \mathcal C \to \mathcal D$ are functors, a natural transformation $\alpha: F \Rightarrow G$ assigns a morphism $\alpha(x) : F(x) \to G(x)$ in $\mathcal D$ for each object $x \in \mathcal C$, and if $f: x \to y$ is a morphism in $\mathcal C$, then $\alpha(y) \circ F(f) = G(f) \circ \alpha(x)$ as morphisms in $\mathcal D$.

Given a natural transformation, can I check whether it is epi (or mono) by checking pointwise?  I.e.: is a natural transformation $\alpha$ epi (mono) iff $\alpha(x)$ is epi (mono) for each $x$?
If not, is there an implication in one direction between whether a natural transformation is epi and whether it is pointwise-epi?

A more general question, one that I never really learned, is what types of properties of a functor are "pointwise" in that they hold for the functor if they hold for the functor evaluated at each object.  E.g.: is the (co)product of functors the pointwise (co)product?

REPLY [6 votes]: In all the preceding answers the category $D$ is required to have pushouts / pullbacks in order to epi / mono being equivalent to pointwise epi / mono in functor categories $D^C$. 
The discussion in On a corollary in Mitchell's book draw my attention to another important class of categories that usually doesn't have pushouts or pullbacks and where epi / mono is also equivalent to pointwise epi / mono in $D^C$: That is when $D$ is exact. 
A category is exact, if 

it has a zero object 
kernels and cokernels exist 
every monomorphism is a kernel and every epimorphism is a cokernel 
every morphism can be written as a composition of an epimorphism followed by a monomorphism. 

As a reference see Barry Mitchell: "Theory of Categories", II.11 Functor Categories.<|endoftext|>
TITLE: Sections of etale morphisms
QUESTION [5 upvotes]: We all know that smooth morphisms have sections etale locally. However, the following similar statement is not obvious for me:
If X->Y->Z, X is etale over Y, Y is finite and surjective over Z, then a section of X->Y exists etale locally on Z, i.e. there exists an etale cover U of Z such that X_U->Y_U has a section. Where _U means pullback on U.
I think it is supposed to be easy.
Can anyone explain this to me? Thanks.

REPLY [2 votes]: Brian Conrad says: "First reduce to the case when Y is also finitely presented over Z. Then you can use limit arguments, so it becomes an easy application of basic facts about strictly henselian local rings and finite algebras over them. It will be instructive for you to think about it some more for yourself in view of these hints.
I should have also added that you must be assuming $X\to Y$ is surjective, or else it clearly isn't true. That condition enters into the argument when doing analysis of the situation with strictly henselian local rings."<|endoftext|>
TITLE: Google question: In a country in which people only want boys
QUESTION [95 upvotes]: Hi all!
Google published recently questions that are asked to candidates on interviews. One of them caused very very hot debates in our company and we're unsure where the truth is. The question is:

In a country in which people only want
  boys every family continues to have
  children until they have a boy. If
  they have a girl, they have another
  child. If they have a boy, they stop.
  What is the proportion of boys to
  girls in the country?

Despite that the official answer is 50/50 I feel that something wrong with it. Starting to solve the problem for myself I got that part of girls can be calculated with following series:
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\left (1-\frac{1}{n+1}\right )$$
This leads to an answer: there will be ~61% of girls.
The official solution is:

This one caused quite the debate, but
  we figured it out following these
  steps:

Imagine you have 10 couples who have 10 babies. 5 will be girls. 5
  will be boys. (Total babies made: 10,
  with 5 boys and 5 girls)
The 5 couples who had girls will have 5 babies. Half (2.5) will be
  girls. Half (2.5) will be boys. Add
  2.5 boys to the 5 already born and 2.5 girls to the 5 already born. (Total
  babies made: 15, with 7.5 boys and 7.5
  girls.)
The 2.5 couples that had girls will have 2.5 babies. Half (1.25) will
  be boys and half (1.25) will be girls.
  Add 1.25 boys to the 7.5 boys already
  born and 1.25 girls to the 7.5 already
  born. (Total babies: 17.5 with 8.75
  boys and 8.75 girls).
And so on, maintianing a 50/50 population.


Where the truth is?

REPLY [4 votes]: Caveat: This is not an entirely serious answer.
There has been some (heated?) discussion as to the sensitivity of the various answer to the particular model. I thought, for my own amusement, that I would
do some Monte-Carlo experiments with a "plausible" model involving Pilgrims
traveling to the New World. However, in thinking about possible models, I came up with the following issue.
Suppose we assume that:

Once Pilgrims marry, they stay married, and do not re-marry if their spouse dies.
No living Pilgrim has a direct living ancestor older than their grandparents.

Under these assumptions, it follows that if there were N male Pilgrims in the first settlement, then, at any given time, there are at most 3N male Pilgrims. Moreover, the probability of the settlement dying out over several generations (because all the children are girls) is non-zero. By Kolmogorov's zero-one law (overkill), it follows that almost surely the Settlement will die out, and not become the kick-ass country it may well have been.<|endoftext|>
TITLE:   Is there a known way to formalise notion that certain theorems are essential ones?   
QUESTION [7 upvotes]: Suppose You ask a question beginning from "Why some structure is..." or "Why some object has property..." and several
answers arises. Which criteria do You
use to qualify which answer is correct?

For example here You may find interesting picture (gzipped postscript file) of proofs formalized in Mizar system.. Mizar library of formalized theorems is really huge. On the picture You may see, that theorems arises from other and are used in proofs of another ones, forming big graph of structure of theorems formalized in Mizar so far. If I may read something from this graph, there is no theorem which will have more that 3 or 4 incoming edges what means there is no theorem which is used in more that 3 or 4 proofs. Of course there are some with 5 incoming edges, but in fact there is many theorems which have more or less equal number of incoming edges, which may mean that most theorems are equivalently important. Maybe it should be measured by tree deepness? Maybe there is something like Google page rank algorithm for theorems? 
Probably we would like to have such relation: "theorems recognized as important should be influential, or foundational for broad area of theory". 

I understand that one may believe that this is a real state of matter,
but are there any strict results
based on real data in this matter?

By real data I mean at best proof theory analysis, or even citation analysis, but not someone opinion (which of course may be enlightening and inspirational). I would like to learn something about structure of deductive theories, and not about "real practice". It is the same as in real life: we try to measure risks, and income rate not based on someone opinion but on facts. Could we know the facts here? 
It seems from Mizar graph ( which is the only one accessible for me in this area) I could not find any object which will correspond to our intuition of importance of theorems. Maybe this is effect of present Mizar state of affair, and in bigger/other system, some theorems begin central one? Are there any conditions to state such position? 

What about other proof assistants, as
Isabelle or COQ. Is there any similar
graph from other systems suitable for
such analysis?

REPLY [26 votes]: Although your question is vague in certain ways, one robust answer to it is provided by the subject known as Reverse Mathematics. The nature of this answer is different from what you had suggested or solicited, in that it is not based on any observed data of mathematical practice, but rather is based on the provable logical relations among the classical theorems of mathematics. Thus, it is a mathematical answer, rather than an engineering answer.
The project of Reverse Mathematics is to reverse the usual process of mathematics, by proving the axioms from the theorems, rather than the theorems from the axioms. Thus, one comes to know exactly which axioms are required for which theorems. These reversals have now been carried out for an enormous number of the classical theorems of mathematics, and a rich subject is developing. (Harvey Friedman and Steve Simpson among others are prominent researchers in this area.)
The main, perhaps surprising conclusion of the project of Reverse Mathematics is that it turns out that almost every theorem of classical mathematics is provably equivalent, over a very weak base theory, to one of five possibilities. That is, most of the theorems of classical mathematics turn out to be equivalent to each other in five large equivalence classes. 
For example, 

Provable in and equivalent to the theory RCA0 (and each other) are: basic properties of the natural/rational numbers, the Baire Cateogory theorem, the Intermediate Value theorem, the Banach-Steinhaus theorem, the existence of the algebraic closure of a countable field, etc. etc. etc.
Equivalent to WKL0 (and each other) are the Heine Borel theorem, the Brouer fixed-point theorem, the Hahn-Banach theorem, the Jordan curve theorem, the uniqueness of algebraic closures, etc. , etc. etc.
Equivalent to ACA0 (and each other) are the Bolzano-Weierstraus theorem, Ascoli's theorem, sequential completeness of the reals, existence of transcendental basis for countable fields, Konig's lemma, etc., etc.
Equivalent to ATR0 (and each other) are the comparability of countable well orderings, Ulm's theorem, Lusin's separation theorem, Determinacy for open sets, etc. 
Equivalent to Π11 comprehension (and each other) are the Cantor-Bendixion theorem and the theorem that every Abelian group is the direct sum of a divisible group and a reduced group, etc. 

The naturality and canonical nature of these five axiom systems is proved by the fact that they are equivalent to so many different classical theorems of mathematics. At the same time, these results prove that those theorems themselves are natural and essential in the sense of the title of your question.
The overall lesson of Reverse mathematics is the fact that there are not actually so many different theorems, in a strictly logical sense, since these theorems all turn out to be logically equivalent to each other in those five categories. In this sense, there are essentially only five theorems, and these are all essential. But their essential nature is mutable, in the sense that any of them could be replaced by any other within the same class.
I take this as a robust answer to the question that you asked (and perhaps it fulfills your remark that you thought ideally the answer would come from proof theory). The essential nature of those five classes of theorems is not proved by looking at their citation statistics in the google page-rank style, however, but by considering their logical structure and the fact that they are logically equivalent to each other over a very weak base theory. 
Finally, let me say that of course, the Reverse Mathematicians have by now discovered various exceptions to the five classes, and it is now no longer fully true to say that ALL of the known reversals fit so neatly into those categories. The exceptional theorems are often very interesting cases which do not fit into the otherwise canonical categories.<|endoftext|>
TITLE: When is the Galois representation on the étale cohomology unramified/Hodge-Tate/de Rham/crystalline/semistable?
QUESTION [13 upvotes]: Let $X/K$ be a variety over a global field $K$. When (and why) is the Galois representation $H^i_{et}(X \times_K \bar{K}, \mathbf{Q}_\ell)$ unramified at a place $v$ of $K$?
I guess this is true if $X$ has a model smooth (or regular?) over $v$ by using base change theorems.
Same question for Hodge-Tate/de Rham/crystalline/semistable.

REPLY [2 votes]: By asking "when is it ramified," you might be also be asking whether there are any other conditions under which it's ramified, i.e. a converse to Emerton's answer. For abelian varieties, the converse is true, and this is known as N\'eron-Ogg-Shafarevich. There are $p$-adic Hodge theory versions as well.
However, in general, a variety with Galois representation unramified at $v$ need not have good reduction at $v$. In all cases where this has been proven, one relates the Galois representation in question to the representations of auxiliary varieties that do have good reduction at $v$. As a simple example, there are smooth projective curves with bad reduction at $v$ but whose Jacobian has good reduction at $v$.
One might conjecture that any Galois representation coming from a variety actually comes in some fashion from varieties with good reduction over $v$. More precisely, any motive over a global field $K$ whose associated $\ell$-adic Galois representation is unramified at a finite place $v$ (whose residue characteristic $p$ is not $\ell$) is the base change of a motive (of smooth proper varieties) over $\mathcal{O}_{K,v}$, the local ring at $v$. One can make a similar conjecture for $\ell=p$ if one assumes the representation is crystalline. One can even make a modification by defining a category of motives using semistable varieties. I don't know if this is written down anywhere, but I and other graduate students I know have wondered this.<|endoftext|>
TITLE: decomposition of representations of a product group
QUESTION [15 upvotes]: Suppose $G_i$ are finite groups for $i=1,2$ and G is the direct product of $G_i$. If V is a finite dimensional irreducible representation of $G$, then it is well known that $V$ is a tensor product of $V_i$,$i=1,2$ and each $V_i$ is an irreducible representation of $G_i$.
The question I have is when $V$ is given, is there a canonical way to construct $V_i$ from $V$?

REPLY [21 votes]: I agree with David Speyer's answer, and furthermore there is no canonical way to construct $V_i$ from $V$.  This is a subtle and oft-overlooked point in representation theory, in my opinion.  Many texts prove that an irrep of $G_1 \times G_2$ is isomorphic to a tensor product of an irrep of $G_1$ with an irrep of $G_2$.  The typical slick proof relies on character theory -- kind of a cheat, in my view, since it only says something about isomorphism classes.
Here's a categorical explanation of the theorem:  Let $G_1$ and $G_2$ be finite groups, and let $\pi$ be an irrep of $G_1 \times G_2$ on a complex vector space $V$.
Then, for every pair $(\rho_1, W_1)$, $(\rho_2, W_2)$ of representations of $G_1, G_2$, one gets a complex vector space:
$$H_\pi(\rho_1, \rho_2) := Hom_G(\rho_1 \boxtimes \rho_2, \pi).$$
In fact, this extends to a contravariant functor:
$$H_\pi:  Rep_{G_1} \times Rep_{G_2} \rightarrow Vec.$$
Here we use categories of finite-dimensional complex representations and vector spaces.
This is also functorial in $\pi$, yielding a functor:
$$H:  Rep_G \rightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ],$$
where the right side of the arrow denotes the category of functors (for categories enriched in $Vec$).  What this demonstrates is that the canonical thing is to take representations of $G$ to objects of an appropriate functor category related to $Rep_{G_1}$ and $Rep_{G_2}$.  By Yoneda's lemma (for categories enriched in $Vec$), there is an embedding of categories:
$$Rep_{G_1} \times Rep_{G_2} \hookrightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ].$$
It turns out -- and this is where some finiteness is important, and a proof necessarily uses some counting, character theory, or the like -- that for any irrep $\pi$ of $Rep_G$, the functor $H_\pi \in [Rep_{G_1} \times Rep_{G_2}, Vec]$ is representable.  It is not uniquely representable, but it is uniquely representable up to natural isomorphism.
Practically, what this means is that given an irrep $\pi$ of $G$, there exists an isomorphism $\iota: \pi \rightarrow \pi_1 \boxtimes \pi_2$ for some irreps $\pi_1, \pi_2$ of $G_1, G_2$, respectively.  The pair $(\pi_1, \pi_2)$ is not unique, but the triple $(\iota, \pi_1, \pi_2)$ is unique up to unique isomorphism.  This is usually good enough.<|endoftext|>
TITLE: What geometric properties do properties of ell-adic Galois representations imply? 
QUESTION [10 upvotes]: This is the converse question to an earlier question. More precisely,
Let $X/K$ be a curve(or variety) over a global field $K$. We consider the Galois representation obtained by the absolute Galois group $G_K$ acting on $H_{et}^i(X_{/\bar K}, \mathbb{Q}_\ell)$.
Do properties of this representation, such as "unramified at a place $v$", semistable, de Rham, crystalline, Hodge-Tate, and so on and so forth, imply some geometric properties about $X_{/K}$? (I must confess that I know the proper definition of only the first property in this list, but I nevertheless put in the whole list from the original question for good measure.)
If so, please give examples.

REPLY [10 votes]: The converse is false. See the lecture notes by Chandan Singh Dalawat at http://arxiv.org/abs/math/0605326, which give some examples of varieties over finite extensions of $\mathbb{Q}_p$ whose $\ell$-adic cohomology is unramified for $\ell \ne p$ and crystalline at $\ell = p$, but the variety does not have good reduction.<|endoftext|>
TITLE: Is there a relative version of Tannakian reconstruction?
QUESTION [7 upvotes]: According to some form of Tannakian reconstruction, given a finite tensor category with a fiber functor to the category of vector spaces, one determines a Hopf algebra by considering tensor endomorphisms of the fiber functor.  As far as I know, a similar procedure is used to reconstruct a group from its symmetric tensor category of representations.
I am curious about what happens if one is given a finite tensor category $\mathcal{C}$ and a tensor functor $\mathcal{C} \to Rep(G)$ for $G$ a finite group.  It follows that there should exist a Hopf algebra $H$ (by the previous reconstruction business applied to the composition of this tensor functor with the forgetful functor $Rep(G) \to \mathrm{Vect}$) and homomorphism $\mathbb{C}[G] \to H$.  

Under what conditions will $H$ be a
  semidirect product of $G$ with some Hopf algebra?

REPLY [7 votes]: Akhil,
Let $\mathcal{C}$ be a tensor category, and let $(A,\mu) \in \mathcal{C}-Alg$ be an algebra in $\mathcal{C}$.  So $\mu:A\otimes A\to A$ is a morphism in $\mathcal{C}$ and $\otimes$ here means the $\otimes$ in $\mathcal{C}$ (of course there isn't another one around at this point, but I mean to emphasize it's not just the $\otimes$ of Vect).  
In this context, it makes sense to talk about $A$-modules in $\mathcal{C}$, whose definition you can guess.  These form a k-linear abelian category $D$ with a forgetful functor to $\mathcal{C}$ which forgets the $A$ action.
Now if $\mathcal{C}$ has a fiber functor F, then $\mathcal{C}$ is realized as the Hopf algebra $End(F)$, as you said (well $End(F)^{op}$ I think , but nevermind).  The algebra $A$ can be pushed forward by $F$ to an ordinary algebra $F(A)$ in Vect.  However, $D$ is not the category of $A$-modules, but a well-known proposition tells that $D$ is the category of $A\rtimes H$-modules, the semi-direct product you asked about.
Notice that no part of the discussion so far asked for any symmetric structure on $\mathcal{C}$, and also $A$ is only an algebra.  To define a bialgebra in $\mathcal{C}$, however, one needs $\mathcal{C}$ to be braided, because the compatibility between $\Delta$ and $\mu$ will use the braiding.  In your case braiding just means symmetry.  I never worked this out in detail, but I imagine that if $A$ is actually a bi-algebra in $\mathcal{C}$, then $D$ gets endowed with a monoidal structure, and that $D$ is the category of $A\rtimes H$-modules, where $A\rtimes H$ is a bi-algebra.  Likewise if $A$ is Hopf in $\mathcal{C}$, then $D$ is tensor and $A\rtimes H$ is a Hopf algebra, and $D\cong A \rtimes H$-mod.

Thus ends the part where I'm pretty sure I'm not saying anything too incorrect.  Below I will try to answer your actual question.  I would not trust it though until somebody smarter agrees with it.

So, now your question becomes (let's revert to considering algebras at the top and not Hopf algebras, since it should be clear how to extend):  Given a functor $F:D\to \mathcal{C}$, when is $F$ the forgetful functor corresponding to some algebra $A \in \mathcal{C}$?  I think for this, it will be enough to assume (in addition, of course, to assuming that $F$ is faithful and exact) that $D$ has a projective generator $M$ (although maybe this is guaranteed by a lesser assumption?).  This is definitely necessary to be able to realize $D$ as some category of modules of a ring, as you desire, and I imagine that you then let $A=\underline{Hom}(M,M)$ (meaning $\mathcal{C}$-internal homs, which are distinct from $Hom_\mathcal{C}(M,M)$!), which will be an algebra in $\mathcal{C}$, and you can plug into the above.
You should definitely read Ostrik's http://arxiv.org/abs/math/0111139 and other papers by Etingof, Nikshych, and Ostrik about fusion and finite tensor categories.<|endoftext|>
TITLE: solving equations in the braid group
QUESTION [10 upvotes]: Is there a systematic way to solve equations in the braid groups?  In particular, if $B_3$ is the braid group on three strands with the presentation $\{ a,b\ | \ aba = bab \}$, how do I find $x$ so that $xaxbx^{-1}b^{-1}x^{-1} = bax^{-1}b^{-1}x$ (if such an $x$ exists)?

REPLY [5 votes]: Your equation is $xaxbXBXXbxAB = 1$; thus the substitution $x = b$ answers your particular question.  May I ask where this equation comes from?<|endoftext|>
TITLE: Can we decompose Diff(MxN)?
QUESTION [16 upvotes]: If you have two manifolds $M^m$ and $N^n$, how does one / can one decompose the diffeomorphisms $\text{Diff}(M\times N)$ in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$? Is there anything we can say about the structure of this group?  I have looked in some of my textbooks, but I haven't found any actual discussion of the manifolds $\text{Diff}(M)$ of a manifold $M$ other than to say they are "poorly understood."
Can anyone point me to a source that discusses the manifold $\text{Diff}(M)$?  My background is in physics, and understanding the structure of these kinds of groups is important for some of the the things we do, but I haven't seen any discussion of this in my differential geometry textbooks.

REPLY [10 votes]: When $N$ is the circle there's a sort of answer. In fact there's a whole chapter in the book 
Burghelea, Dan; Lashof, Richard; Rothenberg, Melvin: Groups of automorphisms of manifolds. With an appendix ("The topological category'') by E. Pedersen. Lecture Notes in Mathematics, Vol. 473. Springer-Verlag, Berlin-New York, 1975.
dedicated to showing that after looping once, the space $\text{Diff}(M\times S^1)$ splits up to homotopy as (the loops of) 
$$
\text{Diff}(M\times I) \times B\text{Diff}(M\times I) \times \eta(M) ,
$$
where the middle term is a non-connective one-fold delooping of $\text{Diff}(M\times I)$ and
$\eta(M)$ is the mysterious "nil-term" (when writing $\text{Diff}(W)$ for a manifold $W$ with boundary, the convention is that the diffeomorphisms are to preserve the boundary pointwise).
In particular, once gets a decomposition on the level of homotopy groups. (This theorem is an analog of the Bass-Heller-Swan type result which says $K(R[t]) \simeq K(R) \times BK(R) \times \eta(R)$.)
One can say something about the homotopy type the nil-term in the concordance stable range, roughly, $\dim M/3$.
Furthermore, $\text{Diff}(M\times I)$ sits in a fibration sequence
$$
\text{Diff}(M\times I) \to C(M) \to \text{Diff}(M)
$$
where $C(M)$ is the topological group of concordances of $M$. After inverting 2, this
sequence is homotopically trivial and $\pi_k(\text{Diff}(M\times I))$ can be identified with the invariant part of the $\Bbb Z_2$-action on $\pi_k(C(M))$  induced by conjugating a concordance with the diffeomorphism which turns $M\times I$ upside down ($(x,t) \mapsto (x,1-t)$).  Lastly, $\pi_k(C(M))$ can be studied via algebraic $K$-theory methods when $k$ is within the concordance stable range.<|endoftext|>
TITLE: Symmetric Groups and Poisson Processes
QUESTION [8 upvotes]: Consider the number of fixed points in a permutation chosen uniformly at random from the symmetric group on $n$ elements - this gives a probability distribution.  For $k < n$, the $k$-th moments of this distribution are the same as the Poisson distribution with $\lambda = 1$.  
Since you can't pick uniformly randomly from an infinite set, it doesn't make sense to ask if you get exactly the Poisson distribution in the limit of the symmetric group on a countably infinite set.  But can you get there in another way?  Is there a known way to take limits in some category of distributions and groups that leads to the Poisson distribution for the fixed points of some countably infinite group of permutations?  Alternately, is there a known structure that can be put on $|S_\omega|$ that gives rise to a Poisson process that is analogous to the finite case?
Also, are there other infinite families of finite permutation groups that give rise to a similarly elegant characterization of the distribution of fixed points?  Say by taking one of the families of finite simple groups and applying the same finite extension to each of them? 
(Please excuse me if I accidentally posted this twice - I didn't see the first attempt go through.)

REPLY [4 votes]: This isn't a problem I've looked at before, but I've been thinking about it since reading your post, and there does seem to be an interesting limit. The following looks like it should all work out, but I haven't gone through all the details yet.
Embed the set {1,...,n} into the unit interval I=[0,1] by θ(i) ≡ i/n. Then look at θ(A) for the fixed points A ⊂ {1,..,n} of a random permutation. Increasing n to infinity, the distribution of θ(A) should converge weakly to the Poisson point process on I with intensity being the standard Lebesgue measure.
That is only the fixed points though. You can look at the limiting distribution of the set of all points contained in orbits of bounded size (≤ m, say), and at the action of the random permutation on this set. This should have a well defined limit, and you can then take the limit m → ∞ to to give a random countable subset of I and a random permutation on this, such that all orbits are finite.
Let In={1/n,2/n,...,n/n} ⊂ I, and consider a random permutation π of this. The probability that any point P ∈ In is an element of an orbit of size r is (1-1/n)(1-2/n)...(1-(r-1)/n)(1/n). The expected number of points lying in such orbits is (1-1/n)...(1-(r-1)/n) and the expected number of orbits of size r is (1-1/n)...(1-(r-1)/n)/r, which tends to 1/r as n → ∞.
Now represent orbits of size r of the permutation π by sequences of distinct elements of the interval I, (x1,...,xr), where π(xi) = xi+1 and π(xr) = x1. We'll identify cyclic permutations of such sequences, as they represent the same orbit and the same action of π on the orbit. So (x1,...,xr) = (xr,x1,...,xr-1). Let I(r) be the space of such sequences up to identification of cyclic permutations.
Then, every permutation π of In gives a set of fixed points A1 ⊂ I(1)=I, orbits of size 2, A2 ⊂ I(2), orbits of size 3, A3 ⊂ I(3), etc.
Then, I propose that A1, A2, A3,... will jointly converge to the following limit; (α1,α2,α3,...) are independent Poisson random measures on I(1),I(2),I(3),... respectively such that the intensity measure of αr is uniform on I(r) with total weight 1/r.
Taking the union of all these orbits gives a random countable subset of I and a random permutation of this.
I used the unit interval as the space in which to embed the finite sets {1,...,n} but, really, any finite measure space (E,ℰ,μ) with no atoms will do. Just embed the finite subsets uniformly over the space (i.e., at random).<|endoftext|>
TITLE: Algebraic topology beyond the basics: any texts bridging the gap?
QUESTION [65 upvotes]: Peter May said famously that algebraic topology is a subject poorly served by its textbooks. Sadly, I have to agree. Although we have a freightcar full of excellent first-year algebraic topology texts - both geometric ones like Allen Hatcher's and algebraic-focused ones like the one by Rotman and more recently, the beautiful text by tom Dieck (which I'll be reviewing for MAA Online in 2 weeks, watch out for that!) - there are almost no texts which bring the reader even close to the frontiers of the subject. 
GEOMETRIC topology has quite a few books that present its modern essentials to graduate student readers - the books by Thurston, Kirby and Vassiliev come to mind - but the vast majority of algebraic topology texts are mired in material that was old when Ronald Reagan was President of The United States. This is partly due to the youth of the subject, but I think it's more due to the sheer vastness of the subject now. Writing a cutting edge algebraic topology textbook - TEXTBOOK, not MONOGRAPH - is a little like trying to write one on algebra or analysis. The fields are so gigantic and growing, the task seems insurmountable. 
There are only 2 "standard" advanced textbooks in algebraic topology and both of them are over 30 years old now: Robert Switzer's Algebraic Topology: Homology And Homotopy and George Whitehead's Elements of Homotopy Theory. Homotopy theory in particular has undergone a complete transformation and explosive expansion since Whitehead wrote his book. (That being said, the fact this classic is out of print is a crime.) There is a recent beautiful textbook that's a very good addition to the literature, Davis and Kirk's Lectures in Algebraic Topology - but most of the material in that book is pre-1980 and focuses on the geometric aspects of the subject. 
We need a book that surveys the subject as it currently stands and prepares advanced students for the research literature and specialized monographs as well as makes the subject accessible to the nonexpert mathematician who wants to learn the state of the art but not drown in it. The man most qualified to write that text is the man to uttered the words I began this post with. His beautiful concise course is a classic for good reason; we so rarely have an expert give us his "take" on a field. It's too difficult for a first course, even for the best students, but it's "must have" supplementary reading. I wish Dr. May - perhaps when he retires - will find the time to write a truly comprehensive text on the subject he has had such a profound effect on. Anyone have any news on this front of future advanced texts in topology? 
I'll close this box and throw it open to the floor by sharing what may be the first such textbook available as a massive set of online notes. I just discovered it tonight; it's by Garth Warner of The University Of Washington and available free for download at his website. I don't know if it's the answer, but it sure looks like a huge step in the right direction. Enjoy. And please comment here. 
  http://www.math.washington.edu/~warner/TTHT_Warner.pdf

REPLY [11 votes]: I have written up some detailed lecture notes
Introduction to Stable homotopy theory
Prelude -- Classical homotopy theory (pdf, 99 pages)
Part 1 -- Stable homotopy theory

Part 1.1 -- Sequential spectra (pdf, 67 pages)
Part 1.2 -- Structured spectra (pdf, 80 pages)

Part 2 -- Adams spectral sequences (pdf, 56 pages)
This introduces and then proceeds systematically via model categories. Full details and proofs are given. 
To view the web version you need the Firefox browser<|endoftext|>
TITLE: Axiom of Infinity needed in Cantor-Bernstein?
QUESTION [9 upvotes]: Can one prove the Cantor-Bernstein (or Schröder-Bernstein) theorem without using the Axiom of Infinity?

REPLY [10 votes]: Actually, the usual proof of the Cantor-Schröder-Bernstein Theorem does not use the Axiom of Infinity (nor the Axiom of Powersets). 
By the usual proof, I mean the one found on Wikipedia, for example. Using the notation from that proof, the main point of contention is whether we can form the sets $C_n$ and $C = \bigcup_{n=0}^\infty C_n$. These sets exist by comprehension:
$C_0 = \{x \in A: \forall y \in B\,(x \neq g(y))\}$
$C_n = \{x \in A: \exists s\,(s:\{0,\dots,n\}\to A \land s(0) \in C_0 \land (\forall i < n)(s(i+1) = g(f(s(i)))) \land s(n) = x\}$,
where abbreviations such as $s:\{0,\dots,n\}\to A$ should be replaced by the equivalent (bounded) formulas in the language of set theory. The definition of $C_n$ is uniform in $n$, and so
$C = \{x \in A: \exists n\,(\mathrm{FinOrd}(n) \land x \in C_n)\}$,
where $x \in C_n$ should be replaced by the above definition and $\mathrm{FinOrd}(n)$ is an abbreviation for "$n$ is zero or a successor ordinal and every element of $n$ is zero or a successor ordinal." An alternate definition of $C$ is
$C = \{x \in A: \forall D\,(C_0 \subseteq D \subseteq A \land g[f[D]] \subseteq D \to x \in D)\}$,
which shows that $C$ is $\Delta_1$-definable. The rest of the proof uses induction on finite ordinals, but since the definition of the sets $C_n$ is uniform these are special cases of transfinite induction. 
In conclusion, it looks like the proof could work in Kripke-Platek Set Theory — which has neither the Axiom of Infinity nor the Axiom of Powersets — provided that the two definitions of $C$ given above are provably equivalent. I haven't tried to check whether the two definitions are provably equivalent but, in any case, the proof can be carried out in Kripke-Platek Set Theory with $\Sigma_1$-Comprehension.<|endoftext|>
TITLE: A special integral polynomial
QUESTION [5 upvotes]: Given $n \in \mathbf{N}$,is always possible to construct a monic polynomial in $\mathbf{Z}[x]$ of degree $2n$, whose roots are in $\mathbf{C} \setminus \mathbf{R}$ and whose Galois group over $\mathbf{Q}$ is $S_{2n}$?
I have an approximate idea of how to solve the problem for the Galois group (I immagine something related to the Hilbert irreducibility theorem), but I have no idea for the condition on the roots.
Furthermore, is it possible to give an explicit example?

REPLY [9 votes]: An easy way to ensure that a polynomial $g$ of degree $m$ over $\mathbf{Z}$ has
Galois group $S_m$ is to take primes $p_1$, $p_2$ and $p_3$
with $g$ irreducible modulo $p_1$, a linear times an irreducible
modulo $p_2$ and a bunch of distinct linears times an
irreducible quadratic modulo $p_3$. Then the Galois group
must be doubly transitive and have a transposition, so it's $S_m$.
Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real
roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by
close rationals won't create any real roots. So replace the $x^k$
coefficient of $f$ by a sufficient close rational $a_k/b_k$
where $a_k$ and $b_k$ are congruent modulo $p_1 p_2 p_3$
respectively to the $x^k$ coefficient of $g$ and to $1$. Then
the new polynomial has rational coefficients, no real roots
and Galois group $S_{2n}$. You can easily convert it to one
with these properties and integer coefficients should you wish.<|endoftext|>
TITLE: Is the Jaccard distance a distance?
QUESTION [52 upvotes]: Wikipedia defines the Jaccard distance between sets A and B as $$J_\delta(A,B)=1-\frac{|A\cap B|}{|A\cup B|}.$$ There's also a book claiming that this is a metric. However, I couldn't find any explanation of why $J_\delta$ obeys the triangle inequality. The naive approach of writing the inequality with seven variables (e.g., $x_{001}$ thru $x_{111}$, where $x_{101}$ is the number of elements in $(A\cap C) \backslash B$) and trying to reduce it seems hopeless for pen and paper. In fact it also seems hopeless for Mathematica, which is trying to find a counterexample for 20 minutes and is still running. (It's supposed to say if there isn't any.)
Is there a simple argument showing that this is a distance? Somehow, it feels like the problem shouldn't be difficult and I'm missing something.

REPLY [2 votes]: I found a very brief and easy-understanding proof in the paper by Sven Kosub, "A note on the triangle inequality for the Jaccard distance".<|endoftext|>
TITLE: Question about Ext
QUESTION [8 upvotes]: I heard that $Ext(M,N)$ is naturally isomorphic to $Ext(M^*\otimes N,1)$ where 1 is the trivial representation and $M,N$ some representations of a group $G$.
Can anyone explain why?
Is there an explicit construction of a map from one to the other or does it just follow from some general considerations about derived functors?
Thanks.

REPLY [2 votes]: This is probably typographic, but the original question writes

Ext(M,N)  is naturally isomorphic to Ext(M*⊗N,1)

when what is true (see other answers) is that Ext(M,N) is isom.
to Ext(M $\otimes$ N*,1).
By the way, it is often useful to view Ext(M,N)
as Ext(1,M* $\otimes$ N) instead, since this later ext-group
coincides with the group cohomology H(G,M* $\otimes$ N);
here H*(G,-) = derived functor of "G-fixed-points".<|endoftext|>
TITLE: What is the manner of inconsistency of Girard's paradox in Martin Lof type theory
QUESTION [25 upvotes]: I am aware that assigning the type of Type to be Type (rather than stratifying to a hierarchy of types) leads to an inconsistency.  But does this inconsistency allow the construction of a well-typed term with no normal form, or does it actually allow a proof of False?  Are these two questions equivalent?

REPLY [6 votes]: It leads to a well-typed term, having no normal form, which is assigned the type False.  You can find the term given explicitly in A simplification of Girard's paradox<|endoftext|>
TITLE: Polynomial with the primes as coefficients irreducible?
QUESTION [48 upvotes]: If $p_n$ is the $n$'th prime, let $A_n(x) = x^n + p_1x^{n-1}+\cdots + p_{n-1}x+p_n$. Is $A_n$ then irreducible in $\mathbb{Z}[x]$ for any natural number $n$?
I checked the first couple of hundred cases using Maple, and unless I made an error in the code those were all irreducible.
I have thought about this for a long time now, and asked many others, with no answer yet.

REPLY [12 votes]: Since I don't have enough "reputation" to comment on Bjorn's answer, I will write this in an answer. The remark about the location of the zeros of $A_n$ goes back at least to Kakeya, On the Limits of the Roots of an Algebraic Equation with Positive Coefficients, Tôhoku Math. J., vol. 2, 140-142, 1912. It also appears in the nice book by E. Landau, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, Springer 1916, (Hilfssatz p.20).<|endoftext|>
TITLE: Theorems with unexpected conclusions
QUESTION [91 upvotes]: I am interested in theorems with unexpected conclusions. I don't mean
an unintuitive result (like the existence of a space-filling curve), but
rather a result whose conclusion seems disconnected from the
hypotheses. My favorite is the following. Let $f(n)$ be the number of
ways to write the nonnegative integer $n$ as a sum of powers of 2, if
no power of 2 can be used more than twice. For instance, $f(6)=3$
since we can write 6 as $4+2=4+1+1=2+2+1+1$. We have
$(f(0),f(1),\dots) = $ $(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,\dots)$. The
conclusion is that the numbers $f(n)/f(n+1)$ run through all the
reduced positive rational numbers exactly once each. See A002487 in
the On-Line Encyclopedia of Integer Sequences for more
information. What are other nice examples of "unexpected conclusions"?

REPLY [3 votes]: Eckmann-Hilton argument. I mean, WHY?<|endoftext|>
TITLE: Must a surface obtained by exponentiating a plane in a tangent space of a Riemannian manifold be geodesically convex?
QUESTION [10 upvotes]: Perhaps this is basic knowledge in Riemannian geometry, but I can't seem to figure out the answer.  Here is the precise statement of my question.
Let $M$ be a Riemannian manifold, $p$ a point in $M$.  Let $R$ be small enough that $exp_p$ restricts to a diffeomorphism on the ball $B_R(0)$ of radius $R$ centered at the origin, and let $U_R$ be the intersection of $B_R(0)$ and any two dimensional plane through the origin in $T_p M$.  Question: does there exist $R$ such that $exp_p(U_R)$ is geodesically convex, in the sense that for every two points of $exp_p(U_R)$ the unique geodesic segment connecting them lies entirely in $exp_p(U_R)$?  
It would be really convenient for me if the answer is yes.  If so, I am curious to know if the statement is still true if $P$ is replaced by a subspace of higher dimension, but I only need the result for planes.  
Thanks!

REPLY [3 votes]: I don't think that's true. If the dimension of $M$ is bigger than that of $P$, then a necessary condition is that every tangent hyperplane $P\subset T_pM$ develops locally into a totally geodesic submanifold of $M$. This is not true for arbitrary manifolds. Some examples of manifolds in which this is true include those of constant sectional curvature. Symmetric spaces being a special subclass which verifies this conditions. Someone more awake can probably come up with a sufficient criterion.  
I can't think of a very good example right now, but I am pretty sure that if take the 3-dimensional Riemannian Schwarzschild solution, start from a $r$-orthogonal plane outside of the apparent horizon, you'd get a counterexample. 
If the dimension of $P$ is the same as dimension of $M$, however, then you should be okay as long as you make $U$ small enough. 
Edit: Ah, for $P$ 2 dimensional and $M$ 3, by applying the Codazzi equations one sees that a necessary condition for locally developing the hyperplanes to totally geodesic submanifolds is that $Ric(X,Y) = 0$ whenever $g(X,Y) = 0$. This is obviously a very strong condition that is not satisfied by most manifolds.<|endoftext|>
TITLE: Clifford Algebra in Dirac Equation
QUESTION [9 upvotes]: I am wondering if there is any mathematical (or physical, besides the fact that classical quantum mechanics uses complex numbers) justification for why the complexified (1,3) Clifford algebra is used in Dirac's equation.  A (the?) key point of special relativity is that spacetime is a real 4-d vector space with an inner product of signature (1,3).  But by complexifying the signature becomes irrelevant-- all complex Clifford algebras in a given dimension are isomorphic where for real Clifford algebras, even signatures (p,q) and (q,p) are not isomorphic in general (as a side question: can there be a physical significance to this fact?  or do only the spin group and the even subalgebra of the Clifford algebra, which are the same for (p,q) and (q,p), matter?  I only hear about spinor bundles and spin structures, never pinor bundles or pin structures).
Thanks and I hope this isn't too physicsy of a question!

REPLY [4 votes]: I am slightly confused by this question.  The fact that one can formulate the Dirac equation in either (3,1) or (1,3) signature, which have non-isomorphic Clifford algebras and hence Clifford modules of different type (real for (1,3) and quaternionic for (3,1) in my naming conventions), does not mean that one is complexifying the Clifford algebra in formulating the Dirac equation.
Unitarity of the time evolution -- a physical requirement independent of choices -- requires that $i D$, where $D$ is the Dirac operator, be hermitian, and in turn this forces a certain hermiticity condition on the "gamma" matrices, in essence choosing a real form of the complex Clifford algebra.  It is the spinor representation which can be taken to be complex since after all wave functions live in the tensor product of the Clifford module with a complex Hilbert space.  This is not the same thing as complexifying the Clifford algebra, though.
So to summarise, in the Dirac equation the Clifford algebra is real, but the pinor representation can be taken to be complex.<|endoftext|>
TITLE: Can we reconstruct positive weight invariants in algebraic topology using algebraic geometry?
QUESTION [18 upvotes]: I can't really say that I understand what a weight is, but the qualitative distinction between weight zero and positive weight has come up a couple times in MathOverflow questions:

The étale fundamental group of a pointed connected complex scheme has a canonical map to the profinite completion of the topological fundamental group, and for regular varieties, this seems to be an isomorphism.  However, in the case of a nodal rational curve (see this question), one finds that the étale fundamental group is not profinite, and has an honest isomorphism with the topological fundamental group.  Similarly, the degree 1 étale cohoomology of the nodal curve with coefficients in $\mathbb{Z}$ is just $\mathbb{Z}$ as expected from topology, where one typically expects étale cohomology with torsion-free coefficients to break badly in positive degree.  Emerton explained in this blog comment that the good behavior of étale cohomology and the étale fundamental group in these cases is due to the fact that the contribution resides in motivic weight zero, and the singularity is responsible for promoting it to cohomological degree 1.
Peter McNamara asked this question about how well formal loops detect topological loops, and Bhargav suggested in a rather fantastic answer that the formal loop functor only detects weight zero loops (arising from removing a divisor).  In particular, he pointed out that maps from $\operatorname{Spec}\mathbb{C}((t))$ only detect the part of the fundamental group of a smooth complex curve of positive genus that comes from the missing points.

I have a pre-question, namely, how does one tell the weight of a geometric structure, such as a contribution to cohomology, or the fact that removing a divisor yields a weight zero loop?
My main question is: Are there algebraic (e.g., not using the complex topology) tools that always yield the correct invariants in positive weight, such as cohomology with coefficients in $\mathbb{Z}$ and the fundamental group of a pointed connected complex scheme?
I've heard a claim that motivic cohomology has a Betti realization that yields the right cohomology, but I don't know enough about that to understand how.  Any hints/references?
With regard to the second example above, I've seen some other types of loops in algebraic geometry, but I don't really know enough to assess them well.  First, there are derived loops, which you get by generalizing to Top-valued functors on schemes, defining $S^1$ to be the sheaf associated to the constant circle-valued functor (in some derived-étale topology), and considering the topological space $X(S^1)$ or the output of a Hom functor.  As far as I can tell, derived loops are only good at detecting infinitesimal things (e.g., for $E$ an elliptic curve, $LE$ is just $E \times \operatorname{Spec}\operatorname{Sym}\mathbb{C}[-1]$, which has the same complex points as $E$).  Second, there is also some kind of formal desuspension operation in stable motivic homotopy that I don't understand at all.  One kind of loop has something to do with gluing 0 to 1 in the affine line, and the other involves the line minus a point.  I'm having some trouble seeing a good fundamental group come out of either of these constructions, but perhaps there is some miracle that pops out of all of the localizing.

REPLY [11 votes]: I don't think I understand the question, but for the pre-question, I recommend Deligne's ICM address on weights. The point is that the cohomology of algebraic varieties "knows its place." If the variety is smooth and complete, the weight matches (by definition) the degree of the cohomology. By resolution of singularities, it is possible to arrange for all the cohomology to come from smooth complete varieties. But this process of relating them goes through long exact sequences / cones which can shift the degree that the vector spaces end up in.
For example, if X is a smooth complete curve, then its 2nd cohomology is weight 2 and is called Z(1). If you delete a point, its top cohomology is now zero. You have "subtracted" Z(1) from degree 2. If you delete another point, the same thing should happen, but you can't subtract again, and the proper generalization is coning, resulting in Z(1) appearing in degree 1. But Z(1) is still weight 2, so it is now out of place in cohomological degree 1. Dually, if you glue two points together, you should kill off the difference of their homology classes. But they were already homologous, so instead you create a Z in degree 1. This is weight 0 sitting in degree 1. Singularities raise the degree; holes lower the degree.<|endoftext|>
TITLE: Geometric group theory and analysis
QUESTION [9 upvotes]: Geometric group theory is mainly concerned with topological and geometric properties of groups, spaces on which they act etc., so the ideas employed in GGT are mainly algebraic/geometric/topological. Is there any subfield of GGT where methods from analysis find applications? I once heard that analytical tools, e. g. geodesic flows, are used in studying ends of groups, but that's all I know.

REPLY [5 votes]: Not sure why this is not being mentioned but one prominent use of analysis in geometric group theory is in studying boundaries of groups (towards Cannon's conjecture and beyond). See for example discussion here and here.<|endoftext|>
TITLE: What are some fundamental "sources" for the appearance of pi in mathematics?
QUESTION [17 upvotes]: I thought it might be fun to ask this question as a way of celebrating Pi Day.  One way in which people popularize pi is that they say that even though it's defined in terms of properties of a circle, it shows up in contexts such as the normal distribution which have (apparently) nothing to do with circles, and this is supposed to say something about the mysterious and amazing character of mathematics.  So I'm curious what MO have to say about this phenomenon.  If I had to venture a guess, it would be something about Fourier analysis, but I know there are many people here who could make such a statement much more precise.
Answers should ideally be given in general terms, but feel free to illustrate your generalities with interesting examples.

REPLY [2 votes]: Here is another devil's advocacy: We (mathematicians) are not familiar with many transcendental numbers such as $\pi$ or $e$. When in a formula appears some transcendental number that can be expressed as a function of $\pi$, we are happy with that and we have our "solution" for the problem. But if it is not related to $\pi$ (or $e$, or etc...), we just think "OK, we don't have a solution yet." 
Another way of saying pretty much the same thing is the fact that since we know very well $\pi$ and its properties, as soon as it is "in the neighborhood" of our problem, we see it, but if it is not here (and neither $e$ or another famous constant), we do not know where exactly to look to find an acceptable solution.
My 2-cent guess is "the better we know a number (through formulas for instance), the more often it will appear in new formula in the future."<|endoftext|>
TITLE: Faithful representations and tensor powers
QUESTION [20 upvotes]: The following result was mentionned earlier in this thread, I searched a bit in the related threads and couldn't find a proof. I would really like to see a proof of it:
Let $G$ be a finite group and $\rho : G \rightarrow GL(\mathbb{C}, n)$ a faithful representation of $G$. Then every irreducible representation of $G$ is contained in some tensor power of $\rho$.

REPLY [4 votes]: This was a homework problem for a course that I am a TA for. The solution that I had in mind involved using a Vandermonde determinant argument (See Theorem 19.10 in the book by James and Liebeck). But, I was surprised by the following beautiful solution that was submitted by multiple students:
Let $V$ be a faithful representation and $W$ an irreducible representation of $G$. Let $a=\dim(V)$ and $b=\dim(W)$, and let their respective characters be $\chi$ and $\psi$. Then, for all $g\in G$, we have $|\psi(g)|\leq b$, whereas, due to faithfulness, for all $g\in G\setminus\{e\}$, we have $|\chi(g)|\leq a-\varepsilon$ for some $\varepsilon>0$. Then, we have:
\begin{align*}
|\langle\chi^n,\psi\rangle|&=\frac{1}{|G|}\left|\sum_{g}\chi(g)^n\overline{\psi(g
)}\,\right|\\
&\geq \frac{1}{|G|}\left(a^nb - \sum_{g\neq e}\left|\,\chi(g)^n\overline{\psi(g)}\,\right| \right)\\
&\geq \frac{1}{|G|}\big(a^nb-(|G|-1)(a-\varepsilon)^nb\big),
\end{align*}
and as $n\rightarrow \infty$, the above expression becomes positive, showing that the inner product of $\psi$ is non-zero with some power of $\chi$, and thus, $W$ is a sub-representation of some tensor power of $V$, completing the proof!<|endoftext|>
TITLE: Is the universal covering of an open subset of $\mathbb{R}^n$ diffeomorphic to an open subset of  $\mathbb{R}^n$ ?
QUESTION [30 upvotes]: Is the universal covering of a connected open subset $U$ of ℝn diffeomorphic to an open subset of ℝn  (standard differentiable structure)?
If not true in general, is there any condition on $U$ which guarantees a positive answer?

REPLY [25 votes]: The answer is no, and there is a counter-example in dimension $4$. 
A theorem of Whitney and Massey states that the total space of a disc-bundle over a non-orientable surface $\Sigma$ embeds in $S^4$ if and only if the normal Euler class of the disc bundle is one of the integers:
$$\{2\chi -4, 2\chi, \cdots, 4-2\chi\}$$
where $\chi$ is the Euler characteristic of the surface. 
So for example, if $\Sigma = \mathbb RP^2$, $\chi = 1$.  So normal Euler classes $-2$ and $+2$ appear for embeddings $\mathbb RP^2 \to S^4$.   These come from the standard embeddings of $\mathbb RP^2$ in $S^4$. 
The universal cover of this total space is the pull-back of that bundle along the covering map $S^2 \to \mathbb RP^2$.  But the total space of this bundle is orientable so it can't embed in $S^4$ as it's Euler class is not zero -- the pull-back bundle must be isomorpic to the tangent bundle of $S^2$.  And this does not embed in $S^4$.<|endoftext|>
TITLE: restriction of a representation of GL(n) to GL(n-1)
QUESTION [6 upvotes]: Let $R$ be real numbers and consider an irreducible unitary representation (\pi,V) of $GL_n(R)$ in some Hilbert space $V$, now $GL_{n-1}(R)$ embeds in $GL_n(R)$ on the left upper diagonal block. 
Now I wanna ask properties of the restrictions of $V$ to $GL_{n-1}$. This representation is not irreducible, so we may ask is it possible to say something about the closed invariant subspaces? Does this restriction have a multiplicity free decomposition?
Note that we can ask the same question for other irreducible Banach or Frechet representation of $GL_n$, or replacing real numbers by complex or p-adic numbers.

REPLY [8 votes]: Though Peter's answer addresses the finite-dimensional representation theory, I believe that the question asks about the unitary representations on Hilbert spaces, and more general irreps on Banach and Frechet spaces.
This question has been the subject of much recent work by Avraham Aizenbud, Dmitry Gourevitch, Steve Rallis, Gerard Schiffmann, and Eitan Sayag.  In particular, Aizenbud and Gourevitch prove the following in their paper "Multiplicity One Theorem for $(GL_{n+1}(R), GL_n(R))$":
Let $F = R$ or $F = C$.  Let $\pi$ and $\tau$ be irreducible admissible smooth Fr\'echet representations of $GL_{n+1}(F)$ and $GL_n(F)$, respectively.  Then
$$dim \left( Hom_{GL_n(F)}(\pi, \tau) \right) \leq 1.$$
This paper is on the ArXiv, and now published in Selecta, according to Aizenbud's webpage.
Zhu and Binyong have also proved this, I believe.  The result has also been proven for irreducible smooth repreesentations of $GL_{n+1}(F)$ and $GL_n(F)$, when $F$ is a $p$-adic field by Aizenbud-Gourevitch-Rallis-Schiffmann.
Considering the smooth Fr\'echet case should suffice for the case of unitary representations on Hilbert spaces, I believe, by considering the subspace of smooth vectors and Garding's theorem.  I'd guess it would also work for Banach space representations, but I'm not an expert on these analytic things.
It's important to note that semisimplicity may be lost when one restricts smooth representations in these settings -- so their theorem says something about occurrences of quotients after restriction.  It's important to be careful about the meaning of "multiplicity-free" in these situations.<|endoftext|>
TITLE: How do you relate the number of independent vector fields on spheres and Bott Periodicity for real K-Theory?
QUESTION [17 upvotes]: The theory of Clifford algebras gives us an explicit lower bound for the number of linearly independent vector fields on the $n$-sphere, and Adams proved that this is actually always the best possible: there are never more linearly independent vector fields.
More precisely, this gives the following number: if $n+1 = 16^a 2^b c$ with $c$ odd, $0 \leqslant b \leqslant 3$, we get $\rho(n) = 2^b + 8a$ and there are exactly $\rho(n) - 1$ linearly independent vector fields on $S^n$. This lower bound comes by construction of vector fields from Clifford module structures on $\mathbb{R}^{n+1}$, and figuring these out isn't too hard, it follows from the classification of real Clifford algebras with negative definite quadratic form. This is detailed for example in Fibre Bundles by Husemöller; the material comes from the paper Clifford Modules by Atiyah, Bott, Shapiro. This classification hinges on a particular mod 8 periodicity for real Clifford algebras.
Question: How does this description of vector fields on spheres relate to Bott Periodicity in the real case (either for real $K$-Theory, in the form $KO^{n+8} \cong KO^{n}$, or for the homotopy groups of the infinite orthogonal group, $\pi_{n+8}(O) \cong \pi_n(O)$)? 
In particular, I'm inclined to think there should be a rather direct relationship: after all, $K$-theory is talking about vector bundles, sections of which are vector fields! Surely the formula for the number of vector fields on spheres should have a concrete interpretation in terms of $K$-theory? The (underlying) mod $8$ periodicities must be linked!
In addition, the result of periodicity mod 8 for Clifford algebras is also often called Bott Periodicity; what is the deeper relationship here? This other post mentions that the periodicity for Clifford algebras relates to the periodicity for complex K-Theory and so it mentions BU and not BO.

REPLY [9 votes]: Another source for this material: Matt Ando's notes from a course Haynes Miller gave on this subject are available from Miller's webpage: Part I, 
Part II.
(I couldn't get the links to appear correctly in the comment box, but I copied the exact same html code here and it works just fine.  Anyone know why?)<|endoftext|>
TITLE: Lifting matrices mod 2 to integers.
QUESTION [21 upvotes]: The following question was motivated by my research.
Consider a $n\times n$ matrix whose elements are $0$'s or $1$'s such that the determinant is odd. The question is: is it possible to assign signs to matrix elements such that the determinant of the matrix will be equal to $1$? I do not know an answer even to a weaker question: is it possible to replace some of the $1$'s in the matrix with odd integers so that the determinant will be equal to $1$?
Remark: it is known that a natural reduction mod $N$ map $SL_n(\mathbb Z) \to SL_n(\mathbb Z/ N\mathbb Z)$ is surjective for any $n,N$.

REPLY [2 votes]: I want to address the weaker question in a more general setting:
Given any matrix over $\mathbb{Z}$ with determinant $1$ mod $m$. Is it possible to add multiples of $m$ to each entry to get a matrix with determinant one?
Call a matrix transformable, iff this is possible. Given any $A$ matrix over $\mathbb{Z}$. Consider the image of the induced map $\mathbb{Z}^n\rightarrow \mathbb{Z}^n$. It is a submodule of $\mathbb{Z}^n$.
For any such submodule with rank $k$ it is always possible to find a basis of $b_1,\ldots,b_n$ of $\mathbb{Z}^n$ and numbers $r_1,\ldots,r_k$, such that $r_i|r_{i+1}$ for $i=1\ldots k-1$ and $r_1b_1,\ldots r_kb_k$ is a basis of the given submodule. The proof of this is roughly the same as the proof of the structure theorem of finitely generated abelian groups.
This tells us, that we can write our matrix $A$ in the form $A=BDC$, where $B$ and $C$ are invertible and $D$ is a diagonal matrix, with the entries $r_1,\ldots ,r_n$ from above. As the determinant is not zero, the image must have full rank and hence $k=n$.
It is easy to see, that left (right) multiplication with invertible matrices doesn't change the transformability. So we might assume, that $A$ has the given form. We will reduce inductively the number of non-one diagonal entries of $A$ without changing the determinant modulo $m$. This number is the length of $\mathbb{Z}^n/Im(A)=\bigoplus_{i=1}^n\mathbb{Z}/r_i$.
Suppose there is a non-one diagonal entry $r_i$. Then there has to be a second one $r_{i'}$, as the product of all diagonal entries is $1$ mod $m$. Otherwise this is the last non-one diagonal entry and it has to be $1$ mod $m$ and we can transform the matrix into the identity matrix.
Now add $m$ to $r_i$ and it will become coprime to the other non-one entry $r_{i'}$, as $r_i|r_{i'}\;,\;gcd(m,r_i)=1=gcd(m,r_i')$. Hence we get $\mathbb{Z}/(r_i+m)\oplus\mathbb{Z}/r_{i'}\cong\mathbb{Z}/((r_i+m)r_{i'})$
Call the resulting matrix $A'$ and observe that the length of $\mathbb{Z}^n/Im(A')$ is one smaller, than the length of $\mathbb{Z}^n/Im(A)$. Then we can again find the normal form for $A'$ and repeat this process until we end up with the identity matrix. Hence every matrix with determinant $1$ mod $m$ is transformable.<|endoftext|>
TITLE: Category theoretic interpretation of matroids?
QUESTION [44 upvotes]: First time poster, long time lurker here.  I have a really basic question that has been bugging me for sometime.  Specifically, I'm not exactly sure what the 'correct' category theoretic definition of a matroid should be.  The only definition I know involves heavy use of set-theory, and is kind of clumsy:
Given a set $E$, a matroid $\mathcal{I} \subseteq 2^E$ is a non-empty collection of subsets which satisfy the following axioms:

(Heredity) If $X \in \mathcal{I}$ and $X' \subset X$, then $X' \in \mathcal{I}$.

(Exchange) If $X, Y \in \mathcal{I}$ and $|X| > |Y|$, then there exists some $b \in X \backslash Y$ such that $Y \cup \{ b \} \in \mathcal{I}$.


Given that both categories and matroids were introduced around the same time and both were studied by MacLane, it stands to reason that someone ought to have thought about this before.  Also it is obvious from the Heredity axiom that each matroid is a category, since the containment relation is reflexive and transitive.  The second property is a bit more difficult to model, as I am not sure how to get rid of the ugly element / cardinality operators.
In the optimal solution, it would be nice to get rid of the set $E$ entirely, and instead view the specific interpretation of the abstract matroid as a functor from $\mathcal{I} \to 2^E$, the power-set lattice.  This would also suggest a functorial interpretation of the graph theoretic and linear algebra applications of matroids.  I strongly suspect that someone has already done this, but am having great difficulty locating any references.  (Of course I may also be totally wrong headed here too...)

REPLY [5 votes]: The following related article was recently published:
Heunen, C. & Patta, V., The Category of Matroids, Appl Categor Struct (2018) 26: 205. https://doi.org/10.1007/s10485-017-9490-2
In section 9, the authors give a categorical characterization of matroids based on the "greedy algorithm characterization"; very roughly speaking, the optimality of the greedy algorithm for all possible weight functions is encoded as the property that the limits of certain diagrams are all the same. The paper also begins with a nice discussion of the various properties of the category of matroids and strong maps mentioned in Tony Huynh's answer.<|endoftext|>
TITLE: Combinatorics of lattice walks with forbidden points
QUESTION [7 upvotes]: If I'm not in error, the number of walks on a 2-dimensional integer lattice of length 2n steps from the origin to a point (x,y) has a nice closed form:
${2n \choose n + (x + y)/2}{2n \choose n + (x - y)/2}$
Question: Is there a similar closed form for the number of walks from the origin to (x,y), where a chosen "forbidden" point ($x_f$, $y_f$) is disallowed from ever appearing anywhere in a walk?
More generally, note that without the forbidden point, for a fixed length, a uniformly randomly chosen walk of that length is asymptotically Gaussian distributed. This sort of gives us a way of seeing how the $\ell_2$ norm on the euclidean plane arises as we "zoom out" while looking at the lattice and let the lattice size go to zero; that is, since the Gaussian distribution gives equal probabilities for equal $\ell_2$ distances from the origin.
A further, vaguer question is: how does forbidding a point --- removing it from the lattice --- affect the "effective" notion of distance that arises in the analogous way from the combinatorics of paths?

REPLY [3 votes]: First, let me confirm your formula for walks $(0,0) \xrightarrow{m}(x,y)$ with $x+y\equiv m ~\mod 2$, $p_m(x,y)$, which you stated for $m=2n$.
Choose $(m+x+y)/2$ out of $m$ steps $s_i$ to be $(+1/2,+1/2)$ versus $(-1/2,-1/2)$.
Choose $(m+x-y)/2$ out of $m$ steps $t_i$ to be $(+1/2,-1/2)$ versus $(-1/2,+1/2)$.
Then letting $w_i = s_i + t_i$, $(w_i)$ are the steps of a walk $(0,0) \xrightarrow{m}(x,y)$ with steps of $(\pm 1,0)$ or  $(0,\pm 1)$. 

Second, let me fill in some details for the generating function approach.
Any walk $(0,0) \xrightarrow m (x,y)$ either avoids $(x_f,y_f)$, or it can be broken into $3$ pieces: 
$(0,0) \xrightarrow t (x_f,y_f)$ which first visits $(x_f,y_f)$ at the end,
$(x_f,y_f) \xrightarrow {2u} (x_f,y_f)$ of size ${2u \choose u}^2$.
$(x_f,y_f) \xrightarrow {m-t-2u} (x,y)$ which last visits $(x_f,y_f)$ at the start.
The first and third pieces have the same form reversed in time. Let $q_n(a,b)$ be the number of paths  $(0,0)\xrightarrow n (a,b)$ which only visit (a,b) at the end. 
$$\sum_n p_n(a,b)x^n = \bigg(\sum_n q_n(a,b)x^n\bigg)\bigg(\sum_n \sideset{}{^{^{^2}}}{2n \choose n} x^{2n}\bigg)$$
$$\sum_n q_n(a,b)x^n = \sum_{n} p_n(a,b)x^n \bigg/\sum_n \sideset{}{^{^{^2}}}{2n \choose n} x^{2n}$$
So, a generating function for walks $(0,0) \to (x,y)$ which visit $(x_f,y_f)$ is 
$$\bigg(\sum_n q_n(x_f,y_f)x^n\bigg)\bigg( \sum_nq_n(x-x_f,y-y_f)x^n\bigg)\bigg(\sum_n \sideset{}{^{^{^2}}}{2n \choose n} x^{2n}\bigg)$$
$$=\bigg(\sum_n p_n(x_f,y_f)x^n\bigg)\bigg( \sum_nq_n(x-x_f,y-y_f)x^n\bigg)$$
$$=\bigg(\sum_n p_n(x_f,y_f)x^n\bigg)\bigg( \sum_np_n(x-x_f,y-y_f)x^n\bigg)\bigg/\sum_n \sideset{}{^{^{^2}}}{2n \choose n} x^{2n}$$
You want to subtract this from $\sum_n p_n(x,y)x^n$.
I don't see a closed form expression for the coefficients.<|endoftext|>
TITLE: Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds
QUESTION [6 upvotes]: On a compact Riemannian oriented manifold $M$,for each singular $k$-chain $\sigma$ (with real coefficients), $\sigma$ induces a linear functional on the $\mathbb{R}$-vector space of differential k-forms, by integration of the form over $\sigma$. At the same time the metric induces an inner product on that space, by $<\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.
I'm just playing around here with possible relationships between basic stuff I was learning about, but it seemed to me like an obvious way to compare singular chains and forms is to compare the induced functionals (kind of in the spirit of Poincare duality and Stokes' theorem, which pair classes of closed forms with classes of cycles. However, the restriction to only considering singular chains may not make sense in this context...since you just need to integrate k-forms on them, not take boundaries or anything...). For example, when can a chain $\sigma$ and a form $\omega$ have the same functional?
I believe one can show that for any $\sigma$ there is a unique corresponding "dual" form so to speak, say $D(\sigma)$, with the same functional. Since $M$ is compact, there exists a countable orthonormal basis $e_j$ for the space of k-forms, and every element is determined by its inner product with the basis elements (its coordinates). So if we have some chain $\sigma$, we take $< D(\sigma),e_j>$ to be $\displaystyle\int_{\sigma}{e_j}$, and then $\displaystyle D(\sigma)= \sum_{j}{\left(\int_{\sigma}{e_j}\right)e_j}$, and one checks using the basis again that by construction this form has the same functional as $\sigma$. Furthermore, the functional of a form completely determines it, so a priori the dual form must be unique (and it doesn't matter that we chose a basis).
So my question is obviously first of all, does the above make sense? I don't recall seeing it yet. Another obvious question is the other direction - given a form, does it have a dual  singular chain? (Or if one broadens from considering just singular chains?) If not, what can one say about the set of forms that do have duals, relative to the whole space of forms? (e.g. it might be dense.)
EDIT: Thanks, Petya, I guess I need to restrict to smooth singular chains. Also thanks Gonçalo for pointing out that I appear to really be talking about currents - I will take a look at the book! My remaining question: first of all, it seems to me like in the context of a compact Riemannian manifold, the space of k-currents is naturally identified with the space of k-forms via the inner product. So in this space, is the set of currents given by integration over a smooth k-submanifold a proper subspace? I understand that the point of currents is that in the general case they are broader, but in the compact case it seems like maybe that doesn't happen, and I'm having difficulty understanding the statements about the mass norm that seem to concern this question.

REPLY [5 votes]: Let me make sure I understand you correctly.  Given a compact Riemannian manifold of dimension $n$, any smooth $k$-chain induces a functional on the space of $k$-forms, and any $n-k$-form induces a functional on the space of $k$-forms.  You're asking what the relationship is between these two functionals.
Answer: There is no overlap between these two kinds of functionals.  Why?  Suppose a $k$-chain $\sigma$ induces the same functional as a $n-k$-form $\omega$.  Clearly, $\omega$ must be nonzero on some open set $U$ that doesn't intersect any part of $\sigma$.  Then if $\alpha$ is a test $k$-form, then $\sigma(\alpha)$ doesn't depend on the values of $\alpha$ on $U$, but $\omega(\alpha)$ certainly does.
Now, if you look at these functionals at the level of homology/cohomology, then as you probably know, the relationship is Poincare duality.  That is, $[\sigma]$ and $[\omega]$ induces the same functionals on the $k$-th cohomology iff they are Poincare dual.
Here's what currents have to do with the story.  $k$-currents are essentially defined to be all linear functionals on the space of smooth $k$-forms (with appropriate topology).  Then the two types of functionals you describe, the $\sigma$'s and the $\omega$'s are now specific examples of $k$-currents.  If you know much about distributions, then you pretty much already know about currents--you just define everything via integration by parts.  The boundary operator on currents generalizes both the boundary operator on chains and the $d^*$ operator on forms.  Not surprisingly, the homology of currents gives you the (real) homology of $M$.  The reason I immediately said that the $\sigma$'s and $\omega$'s have no overlap is that, in analogy with plain old distributions, the $\omega$'s are like smooth functions, while the $\sigma$'s are like singular measures supported on sets of measure zero.
I don't know much history, but I think that your idea was what lead de Rham to invent currents.<|endoftext|>
TITLE: Growth of the "cube of square root" function
QUESTION [11 upvotes]: Hello all, this question is a variant (and probably a more difficult one)
of a (promptly answered ) question that I asked here, at Is it true that all the "irrational power" functions are almost polynomial ?.
For $n\geq 1$, let $f(n)$ denote the "integer part" 
(largest integer below) $n^{\frac{3}{2}}$, and let
$g(n)=f(n+2)-2f(n+1)+f(n)$. 
   Question : Is it true that $g(n)$ is always in $\lbrace -1,0,1\rbrace$ 
(excepting the initial value $g(1)=2$) ? I checked this up to
$n=100000$.
It is not too hard to check that if $t$ is large enough compared to $r$ (say
$t\geq \frac{3r^2+1}{4}$), 
$f(t^2+r)$ is exactly $t^3+\frac{3rt}{2}$ (or  $t^3+\frac{3rt-1}{2}$ if $t$
and $r$ are both odd ) and similarly 
$f(t^2-r)$ is exactly $t^3-\frac{3rt}{2}$ (or  $t^3-\frac{3rt+1}{2}$ if $t$
and $r$ are both odd ). So we already know that the answer is "yes" for
most of the numbers.

REPLY [8 votes]: Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.
If $f(n)=n^{3/2}$ without the floor, then $g(n)\sim \frac{3}{4\sqrt n}$, so it is positive and tends to 0. When you replace $n^{3/2}$ by its floor, $g(n)$ changes by at most 2, hence the only chance for failure is to have $g(n)=2$ when the fractional parts of $n^{3/2}$ and $(n+2)^{3/2}$ are very small and the fractional part of $(n+1)^{3/2}$ is very close to 1 (the difference is less than $\frac{const}{\sqrt{n}}$).
Let $a,b,c$ denote the nearest integers to $n^{3/2}$, $(n+1)^{3/2}$ and $(n+2)^{3/2}$. Then $c-2b+a=0$ because it is an integer very close to $(n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}$. Denote $m=c-b=b-a$. Then $(n+1)^{3/2}-n^{3/2}<m$ and $(n+2)^{3/2}-(n+1)^{3/2}>m$. Observe that
$$
 \frac{3}{2}\sqrt{n}<(n+1)^{3/2} - n^{3/2} < \frac{3}{2}\sqrt{n+1}
$$
(the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore
$$
\frac{3}{2}\sqrt{n} < m < \frac{3}{2}\sqrt{n+2}
$$
or, equivalently,
$$
 n < \frac49 m^2 < n+2.
$$
If $m$ is a multiple of 3, this inequality implies that $n+1=\frac49 m^2=(\frac23m)^2$, then $(n+1)^{3/2}=(\frac23m)^3$ is an integer and not slightly smaller than an integer as it should be. If $m$ is not divisible by 3, then
$$
 n+1 = \frac49 m^2 + r
$$
where $r$ is a fraction with denominator 9 and $|r|<1$. From Taylor expansion
$$
 f(x+r) = f(x) +r f'(x) +\frac12 r^2 f''(x+r_1), \ \ 0<r_1<r,
$$
for $f(x)=x^{3/2}$, we have
$$
 (n+1)^{3/2} = (\frac49 m^2 + r)^{3/2} = \frac8{27}m^3 + mr + \delta
$$
where $0<\delta<\frac1{4m}$.
This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.<|endoftext|>
TITLE: discrete stochastic process: exponentially correlated Bernoulli?
QUESTION [7 upvotes]: There is a question that was asked on stackoverflow that at first sounds simple but I think it's a lot harder than it sounds.
Suppose we have a stationary random process that generates a sequence of random variables x[i] where each individual random variable has a Bernoulli distribution with probability p, but the correlation between any two of the random variables x[m] and x[n] is α|m-n|.
How is it possible to generate such a process? The textbook examples of a Bernoulli process (right distribution, but independent variables) and a discrete-time IID Gaussian process passed through a low-pass filter (right correlation, but wrong distribution) are very simple by themselves, but cannot be combined in this way... can they? Or am I missing something obvious? If you take a Bernoulli process and pass it through a low-pass filter, you no longer have a discrete-valued process.
(I can't create tags, so please retag as appropriate... stochastic-process?)

REPLY [2 votes]: The above solution is very nice, but relies on the very special structure of the desired process. In a much more general framework, I think that one could use a perfect simulation algorithm as described in:
Processes with long memory: Regenerative construction and perfect simulation,
Francis Comets, Roberto Fernández, and Pablo A. Ferrari,
Ann. Appl. Probab. 12, Number 3 (2002), 921-943.<|endoftext|>
TITLE: What out-of-print books would you like to see re-printed?
QUESTION [45 upvotes]: It's excellent news that the LMS are to re-publish Cassels & Fröhlich.  There are many other excellent mathematics books which are just about impossible (or at least very expensive) to get hold of, though this problem seems to be getting a bit better with some texts being printed on demand.

Which book(s) would you most like to see re-published?

A couple of comments:
Perhaps nobody under 30 actually reads real books made from trees any more, but personally I find it more convenient to refer to a paper copy, to the extent that  I will happily buy a copy of something which is available free on-line (like SGA 1 and 2, or Milne's Arithmetic Duality Theorems).
And of course there can be legal issues with re-publishing works - EGA & SGA seem to be a case in point at the moment.
Here are two to start off with:

Manin, Cubic forms
Grothendieck et al., Dix exposés sur la cohomologie des schémas

(not including Cassels & Fröhlich because I picked up a copy on Amazon a couple of years ago :-) )

REPLY [2 votes]: Élie Cartan, Œuvres complètes. See how expensive and rare used copies are.<|endoftext|>
TITLE: Induction of tensor product vs. tensor product of inductions
QUESTION [17 upvotes]: This is a pure curiosity question and may turn out completely devoid of substance.
Let $G$ be a finite group and $H$ a subgroup, and let $V$ and $W$ be two representations of $H$ (representations are finite-dimensional per definitionem, at least per my definitions). With $\otimes$ denoting inner tensor product, how are the two representations $\mathrm{Ind}^G_H\left(V\otimes W\right)$ and $\mathrm{Ind}^G_HV\otimes \mathrm{Ind}^G_HW$ are related to each other? There is a fairly obvious map of representations from the latter to the former, but it is neither injective nor surjective in general. I am wondering whether we can say anything about the decompositions of the two representations into irreducibles.

REPLY [3 votes]: Just for the case someone is interested, here is another answer I've found. Probably it's equivalent to Robin's and Marty's answers, with the only difference that it's more abstract-nonsense than Robin's (so if you don't consider this as a virtue in itself, you don't have to read on) and shorter than Marty's (in particular, I don't switch to dual representations).
We will always abbreviate $\mathrm{Ind}^G_H$ by $\mathrm{Ind}$ and $\mathrm{Res}^G_H$ by $\mathrm{Res}$. Then, the push-pull formula states that $\mathrm{Ind}\left(U\otimes\mathrm{Res} T\right)\cong \mathrm{Ind}U\otimes T$ for any representation $T$ of $G$. Applying this to $T=\mathrm{Ind}V$, we get $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$. Now, we can see the representation $\mathrm{Res}\mathrm{Ind}V$ as the left $k\left[H\right]$-module $k\left[G\right]\otimes _{k\left[H\right]}V$. Then, there is a canonical $H$-equivariant injection $V\to \mathrm{Res}\mathrm{Ind}V$ given by $v\mapsto 1\otimes v$, and there is a canonical $H$-equivariant projection $\mathrm{Res}\mathrm{Ind}V\to V$ given by $g\otimes v\mapsto gv$ for $g\in H$ and $g\otimes v\mapsto 0$ for $g\not\in H$. This projection splits the injection, and therefore the representation $V$ is canonically a direct summand of the representation $\mathrm{Res}\mathrm{Ind}V$. Hence, $\mathrm{Ind}\left(U\otimes V\right)$ is canonically a direct summand of $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$.
"Canonically" means "canonically with respect to $U$ and $V$ and kind-of canonically with respect to $G$ and $H$" here. "Kind-of canonically with respect to $G$ and $H$" means that it's functorial with respect to maps which preserve both "lying in $H$" and "not lying in $H$", and I think we can't do better. As opposed to Robin's and Marty's proof, we don't need to rely on some randomly chosen system of representatives of cosets or double cosets.<|endoftext|>
TITLE: Decomposition theorem and blow-ups
QUESTION [5 upvotes]: Yet another question of the form 'How to apply the decomposition theorem?' The example that I am considering ought to have a simple answer, but I'm getting confused and I would appreciate if someone could point out where I'm going astray. The confusing point can be stated briefly,  at the end of observation 3. But I'll give an explanation of what I do understand, hoping that this will be helpful to other people and make clear what I am missing.
Let $Y$ be a quasi-projective 3-fold with a unique singular point $0 \in Y$ and suppose that the blow-up at $0$ is a resolution $p: X \rightarrow Y$ and the exceptional divisor $p^{-1}(0)=S$ is a smooth projective surface.The goal is to understand the summands of $Rp_{\ast}IC_{X} \simeq \bigoplus_{i}
{}^{\mathfrak{p}}\mathcal{H}^{i}(Rp_{\ast}IC_{X})[-i]$, where the decomposition is into perverse cohomology sheaves given by the decomposition theorem.
Observations: 

By base change, the fact that $p$ is an isomorphism over the open set $U=Y\setminus 0$ 
implies that $Rp_{\ast} IC_{X}$ restricted to $U$ is just $IC_{U}$, so ${}^{\mathfrak{p}}\mathcal{H}^{0}(Rp_{\ast}IC_{X})\simeq IC_{Y} \oplus E$ for some sky-scraper 
$E$ at $0$. Further, the other perverse cohomology sheaves ${}^{\mathfrak{p}}\mathcal{H}^{i}(Rp_{\ast}IC_{X})[-i]$ must be supported at $0$ and so consist of shifted sky-scrapers. Thus we just have to understand the stalk of $Rp_{\ast} IC_{X}$ at $0$.
By base change and the fact that $p^{-1}(0)=S$ a smooth projective surface, we have that the stalk of $Rp_{\ast} IC_{X}$ at $0$ is $H^{0}(S,\mathbb{Q})[3]\oplus H^{1}(S,\mathbb{Q})[2]\oplus H^{2}(S,\mathbb{Q})[1]\oplus H^{3}(S,\mathbb{Q})\oplus H^{4}(S,\mathbb{Q})[-1]$. Since $IC_{Y}$
is concentrated in degrees $-3,-2,-1$ with respect to the standard t-structure, 
$E \simeq H^{3}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}$ and 

${}^{\mathfrak{p}}\mathcal{H}^{1}(Rp_{\ast}IC_{X})[-1] \simeq H^{4}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}[-1]$. 
Further, there are no higher perverse cohomology sheaves, for degree reasons.

By Verdier duality and self-duality of $Rp_{\ast}IC_{X}$, the only other perverse cohomology
sheaf in the decomposition is ${}^{\mathfrak{p}}\mathcal{H}^{-1}(Rp_{\ast}IC_{X})[1]$, which
must be dual to 

${}^{\mathfrak{p}}\mathcal{H}^{1}(Rp_{\ast}IC_{X})[-1] \simeq  H^{4}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}[-1]$. 
I would think that it should look like
$H^{0}(S,\mathbb{Q})\otimes \mathbb{Q}_{0}[1]$, but then the degree in which $H^{0}(S)$ sits is off by $-2$ from what appears in observation 2. The only sky-scraper sitting in degree $-1$
is $H^{2}(S)$, but that isn't dual to $H^{4}(S)$ in general.
Presumably I am having a problem with applying Verdier duality, but I don't see where the problem lies. Any comments are very welcome.

REPLY [4 votes]: In this example we have $p : X \to Y$ and we may assume, wlog, that $X$ is isomorphic to the total space of the normal bundle to the surface, and $p$ is the contraction of the zero section.
Then, by the Deligne construction, $IC(Y) = \tau_{\le -1} j_* \mathbb{Q}[3]$, where $j : Y^0 \hookrightarrow Y$ is the inclusion of the smooth locus (which is isomorphic to $X^0$ the complement of the zero section in $X$).
In order to work this out, we can use the Leray-Hirsch spectral sequence
$E_2^{p,q} = H^p(S) \otimes H^q(\mathbb{C}^*) \Rightarrow H^{p+q}(X^0)$
this converges at $E_3$ and we get that the degree 0, 1 and 2 parts of the cohomology of $X^0$ is given by the primitive classes in $H^i(S)$ for $i = 0, 1, 2$. Note that this is everything in degrees 0 and 1, but in degree two the primitive classes form a codimension one subspace $P_2 \subset H^2(S)$.
The Deligne construction above, gives us that $IC(Y)_0 = H^0(S)[3]
 \oplus H^1(S)[2] \oplus P_2[1]$.
(This is a general fact: whenever you take a cone over a smooth projective variety, the stalk of the intersection cohomology complex at 0 is given by the primitive classes with respect to the ample bundle used to embed the variety. This follows by exactly the same arguments given above.)
Then the decomposition theorem gives
$p_* \mathbb{Q} = \mathbb{Q}_0[1] \oplus ( IC(Y) \oplus H^3(S) ) \oplus H^4(S)[-1]$.
EDIT: fixed typos pointed out by Chris.<|endoftext|>
TITLE: Seiberg-Witten theory on 4-manifolds with boundary
QUESTION [11 upvotes]: What generalizations of Seiberg-Witten theory to 4-manifolds with boundary do exist?
I would be especially interested in theories which "behave good" under gluing along the boundary (comparable to TQFTs).

REPLY [18 votes]: Hi Fabian! Kronheimer and Mrowka's book Monopoles and three-manifolds lays out comprehensively the construction of a Seiberg-Witten TQFT, called monopole Floer homology. 
It is conjectured to be  isomorphic to the Heegaard Floer homology TQFT of Ozsváth-Szabó. 
There are also beautiful constructions due to Froyshov and to Manolescu which do not apply in quite so  much generality.
The structure of monopole Floer homology is as follows. The TQFT is a functor on the cobordism category $COB_{3+1}$ whose objects are connected, smooth, oriented 3-manifolds. In fact, the TQFT consists of a trio of functors, denoted $\widehat{HM}_{\bullet}$, $\overline{HM_\bullet}$ and $\check{HM}_{\bullet}$ (the last of these is such a sophisticated invariant that you have to download a special LaTeX package just to typeset it properly).  These are $\mathbb{Z}[U]$-modules; there's a story about gradings that's too long to be worth summarising here. There are natural transformations which, for any connected 3-manifold $Y$, define the maps in a long exact sequence
$$ \cdots\to \widehat{HM} _{\bullet}(Y) \to  \overline{HM_\bullet}(Y)\to \check{HM}_{\bullet}(Y) \to \widehat{HM}_{\bullet}(Y) \to  \cdots 
$$
Why this structure? Well, the theory is based on the Chern-Simons-Dirac functional $CSD$ on a global Coulomb gauge slice through a space of (connection, spinor) pairs. $CSD$ is a $U(1)$-equivariant functional, and $\check{HM}_{\bullet}$ is, philosophically, its $U(1)$-equivariant semi-infinite Morse homology. $\overline{HM}_\bullet$ is the part coming from the restriction of $CSD$ to the $U(1)$-fixed-points, and $\widehat{HM}_\bullet$ is the equivariant homology relative to the fixed point set.
Now here's a subtlety for the TQFT enthusiasts out there to get your teeth into (axiomatize, explain...)! The invariant of a closed 4-manifold $X$ in any of the three theories is... zero. The famous SW invariant of a 4-manifold with $b_+>0$ comes about via a secondary operation, not part of the TQFT itself. Delete two balls from $X$ to get a cobordism from $S^3$ to itself.  When $b_+(X)>0$, there are generically no reducible SW monopoles on this cobordism, and this implies that the TQFT-map $\widehat{HM}_\bullet(S^3) \to \widehat{HM}_\bullet(S^3)$ lifts canonically to a map $\widehat{HM}_\bullet(S^3) \to \check{HM}_\bullet(S^3)$; it is this lift that carries the SW invariant.<|endoftext|>
TITLE: What are the recommended books for an introductory study of elliptic curves?
QUESTION [22 upvotes]: I am currently doing a self study on algebraic geometry but my ultimate goal is to study more on elliptic curves. Which are the most recommended textbooks I can use to study? I need something not so technical for a junior graduate student but at the same time I would wish to get a book with authority on elliptic curves. Thanks

REPLY [3 votes]: If you are also interested in exercises on elliptic curves with solutions I recommend 
the book "Computational Commutative Algebra and Algebraic Geometry: Course and exercises with detailed solutions" by Yengui, see https://www.amazon.com/dp/1096374447?ref_=pe_3052080_397514860<|endoftext|>
TITLE: Can a continuous, nowhere differentiable function have specified "shape" at every point?
QUESTION [7 upvotes]: I'm a bit embarrassed to admit that:
a) This is a rather unmotivated question.
b) I can't remember whether or not I've asked this before, but searching doesn't seem to turn anything up so ...
Consider some "shape" function $\phi: \mathbf{R} \to \mathbf{R}$. Then given some function $f: \mathbf{R} \to \mathbf{R}$, one can ask whether the "difference quotient",
$\lim_{y\to x} \frac{f(y)-f(x)}{\phi(y-x)}$,
exists at various points $x$. Letting $\phi(x) = x$ corresponds to taking normal derivatives, and intuitively when the limit exists this means that near $x$, the function $f$ "looks like" $\phi$ does near 0. 
However, if the ratio $\phi(x)/x$ is not bounded above or away from 0 as $x\to 0$ (I'm mostly thinking of the case when it is neither, so that $\phi$ is "wildly oscillating" in some sense), then anywhere the above limit exists and is nonzero, the function $f$ is necessarily non-differentiable. 
My question: If $\phi$ is some wildly oscillating function as described above (pick your favorite), can there be an $f$ for which this limit exists everywhere?
(Edit: I suppose I really want $\phi$ and $f$ to be continuous functions.)

REPLY [6 votes]: Assume WLOG that $\phi(x)>0$ when $x>0$. Since the limit described exists for all $x$ in the source of $f$. We get for any $x$ the bound:
$f(x+\delta)-f(x) \leq C\phi(\delta)$
for $0 < \delta < \delta_0$ for some $C,\delta_0>0$ which may depend on $x$.
diving by $\delta$ we get by the assumptions on $\phi$ that
$\underline{\lim}_{\delta \to 0} ( \frac{f(x+\delta) - f(x)}{\delta}) \leq 0$
This is one the four derivatives of $f$, and proposition 2 chapter 5 in Real Analysis by H.L. Royden states that if $f$ is continuous then it is (non-strictly) decreasing. Similar for increasing. So $f$ is constant.<|endoftext|>
TITLE: Positivity in stack geometry
QUESTION [10 upvotes]: I was wondering how much of the theory say of Lazarsfeld books can be carried to algebraic stacks (if this has been done).
Do we have a sensible notion of an ample (big, nef) line bundle? Of an ample vector bundle? How many of the usual results carry over? Do we have multiplier ideal sheaves? Are the usual vanishing theorems valid in this settings? And so on.
I hope the question even makes sense. I am complete beginner with stacks, so it may even turn out that the relevant objects cannot be defined and the question is too naive.
EDIT: It seems from the answers below that there is some tentative notion of positivity for line bundles. But I'd be more interested in knowing whether something has been done for vector bundles, or multiplier ideal sheaves, and whether vanishing theorems other than Kodaira are known to hold in this context.
Another interesting question (but here I'm really wildly speculating) would be if there are notions of plurisubharmonic functions over (complex) differentiable stacks, and associated analytic multiplier ideals.
Please note that a negative answer would be of equal interest to me ('no, we don't know yet how to generalize these objects...')

REPLY [7 votes]: Whatever you mean by ample line bundle on a DM (say) stack, you cannot of course require that some power of the bundle embeds the stack in projective space.  You could ask that some power of the line bundle embeds the stack into a weighted projective stack, but this imposes restrictions on the kinds of stacks you will be talking about.  This is studied in a preprint by Abramovich and Hassett where they call such stacks cyclotomic.  If you define ampleness in terms of some other positivity (like Kleiman's criterion, Nakai-Moishezon, etc), then you will have many of the same theorems as in the case of varieties - because more or less this positivity will just be "pulled back" from the coarse moduli space.  So the answer depends on the situation you are in and the kinds of properties in which you are interested.<|endoftext|>
TITLE: Flatness of relative canonical bundle
QUESTION [9 upvotes]: I was wondering if there is any general theorem, which guarantees the flatness of $\omega_{X/B}$ over $B$ for a flat morphism $f : X \to B$ of schemes of finite type over $\mathbb{C}$ with equidimensional fibers. I am specially interested in statements which apply to the case of non-reduced $B$. Here by $\omega_{X/B}$ I mean $h^{-n}(f^! \mathcal{O}_B)$ where $n$ is the relative dimension of $f$ and $h^{-n}$ means the cohomology of the complex at the $-n$-th position.

REPLY [8 votes]: It sounds like you may want Exercise 9.7 in Hartshorne's "Residues and Duality".  I paraphrase the statement:

Exercise 9.7 (RD):
Let $f: X \to B$ be a flat morphism of finite type of locally Noetherian preschemes.  Then, $f^!(\mathcal{O}_B)$ has a unique non-zero cohomology sheaf, which is flat over $B$, iff all the fibers of $f$ are Cohen-Macaulay schemes of the right dimension.  Moreover $f^!(\mathcal{O}_B)$ is isomorphic to (a shift of) an invertible sheaf (on $X$) iff all the fibers of $f$ are Gorenstein schemes of the right dimension.

In particular, this addresses the case you mention in your comment ($f$ Gorenstein), since then $f^!(\mathcal{O}_Y)$ is locally free on $X$ and, since $f$ is flat, certainly flat over $B$.
[Aside: I believe I learned this reference from Brian's book "Grothendieck Duality and Base Change", which I think also contains a proof of this.]<|endoftext|>
TITLE: What are surprising examples of Model Categories?
QUESTION [54 upvotes]: Background
Model categories are an axiomization of the machinery underlying the study of topological spaces up to homotopy equivalence.  They consist of a category $C$, together with three distinguished classes of morphism: Weak Equivalences, Fibrations, and Cofibrations.  There are then a series of axioms this structure must satisfy, to guarantee that the classes behave analogously to the topological maps they are named after.  The axioms can be found here.
(As far as I know...) The main practical advantage of this machinery is that it gives a rather concrete realization of the localization category $C/\sim$ where the Weak Equivalences have been inverted, which generalizes the homotopy category of topological spaces.  The main conceptual advantage is that it is a first step towards formalizing the concept of "a category enriched over topological spaces".
A discussion of examples and intuition can be found at this question.
The Question
The examples found in the answers to Ilya's question, as well as in the introductory papers I have read, all have a model category structure that could be expected.  They are all examples along the lines of topological spaces, derived categories, or simplicial objects, which are all conceptually rooted in homotopy theory and so their model structures aren't really surprising.
I am hoping for an example or two which would elicit disbelief from someone who just learned the axioms for a model category. Along the lines of someone who just learned what a category being briefly skeptical that any poset defines a category, or that '$n$-cobordisms' defines a category.

REPLY [45 votes]: The category of sets admits precisely nine model category structures, no more no less. 
I learned this fact from Tom Goodwillie's comments on a different MO question. It always shocks people when I mention it to them, so I guess it is surprising. I am not sure which is more surprising, that you can actually compute all the model structures or that there are exactly nine of them. Working out the details is such a fun exercise that I don't want to spoil the fun here.<|endoftext|>
TITLE: Does a birational involution of C^n always have a fixed point?
QUESTION [7 upvotes]: The title describes completely the question.
For n=1 it is an easy exercise. For n=2 the statement is still true, and depends on the classification of birational involutions of P^2 (see e.g. http://arxiv.org/abs/math.AG/9907028). For n=3 and above I could not find any useful hint in the literature.

REPLY [3 votes]: The answer is no. There are plenty of counterexamples, for example the map given by "a-fortiori", which is $(x,y)\mapsto (x+1/y,-y)$, which can be generalised to any dimension.
The good question is probably to look at birational involutions of $\mathbb{P}^n$.<|endoftext|>
TITLE: What is the explanation for the special form of representations of three string braid group constructed using quantum groups information supplied
QUESTION [6 upvotes]: It is well-known that representations of quantised enveloping algebras give representations of braid groups. For the examples that I know explicitly the representations of the three string braid group take a specific form. Is there an explanation of this? The examples I know are the simplest examples so what can I expect in general?
More specifically: Fix a quantised enveloping algebra $U$. Let $V$ and $W$ be highest weight finite dimensional representations. Then the three string braid group acts on $Hom_U(\otimes^3V,W)$.
The specific form that appears is the following. Let $P$ be the $n\times n$ matrix with $P_{ij}=1$ if $i+j=n+1$ and $P_{ij}=0$ otherwise. Then we can write $\sigma_1$ and $\sigma_2$ with the following properties  

$\sigma_1$ is lower triangular   
$\sigma_2=P\sigma_1P$
$\sigma_i^{-1}=\overline{\sigma}_i$ which means apply the involution $q\mapsto q^{-1}$ to each entry

The simplest example is
$$\sigma_1=\left(\begin{array}{cc} q & 0 \\\ 1 & -q^{-1}\end{array}\right)$$ 
I get the feeling this has something to do with canonical bases.
A specific question is: Take $V$ to be the spin representation of $Spin(2n+1)$. Then do these representations have this form and if so how do I find it?
[In fact, I have representations of this specific form which I conjecture are these representations]
Further comment Assume the eigenvalues of $\sigma_i$ are distinct. This condition holds for the spin representation. Then if this basis exists it is unique. Consider a change of basis matrix $A$ which preserves this structure. Then $A$ commutes with $\sigma_1$ so is lower triangular. Then $A$ also commutes with $P$ so is diagonal. Then the final condition requires $A$ to be a scalar matrix. 
The problem is existence. The Tuba-Wenzl paper shows such a basis exists in small examples.

REPLY [5 votes]: The R-matrix is always upper triangular, for any basis of the tensor product which is compatible with the weight spaces on the two factors; depending on your convention, the R-matrix only decreases weight in the first factor and increases it in the second.
The second element in your list is essentially by definition: $\sigma_1$ and $\sigma_2$ involve doing the exact same thing to the first two or last two factors, so they are conjugate by, say, the map that cyclically permutes the factors.
I believe the last condition is what's usually called "unitarity" of the R-matrix, but I should probably just wait for Noah to show up and give a better answer for that one with references and such.<|endoftext|>
TITLE: Relation between $KO$ and $K$
QUESTION [7 upvotes]: What can be said about the relation between the complex and the real K-theory of a CW complex? An $n$-dimensional complex vector bundle is an $2n$-dimensional real vector bundle but not vice versa. You get a map
$$
K(X)\to KO(X).
$$
What can be said about that?

REPLY [7 votes]: There is a long exact sequence of (reduced) K-groups
$$
K^{n-1}(X) \to KO^{n+1}(X) \to^\eta KO^n(X) \to^c K^n(X) \to^f KO^{n+2}(X) \to \cdots
$$
The map $c$ is induced by complexification, sending a real vector bundle $\xi$ to the associated complex vector bundle.  The map $\eta$ is multiplication by the Hopf element (the Mobius band bundle) in
$$
KO^{-1}(S^0) = KO^0(S^1) = \pi_1(BO).
$$
The map $f$ is induced by the forgetful map from complex vector bundles to real vector bundles, together with the Bott periodicity isomorphism:
$$
K^n(X) \cong K^{n+2} \to KO^{n+2}(X)
$$
The maps $c$, $f$, and Bott periodicity can be used to produce the splitting of $KO(X)$ off $K(X)$ after inverting the prime 2 that Andrew mentioned.<|endoftext|>
TITLE: Theorem versus Proposition
QUESTION [62 upvotes]: As a non-native English speaker (and writer) I always had the problem of understanding the distinction between a 'Theorem' and a 'Proposition'. When writing papers, I tend to name only the main result(s) 'Theorem', any auxiliary result leading to this Theorem a 'Lemma' (and, sometimes, small observations that are necessary to prove a Lemma are labeled as 'Claim'). I avoid using the term 'Proposition'.
However, sometimes a paper consists of a number important results (which by all means earn to be called 'Theorem') that are combined to obtain a certain main result. Hence, another term such as 'Proposition' might come in handy, yet I don't know whether it suits either the main or the intermediate results.
So, my question is: When to use 'Theorem' and when to use 'Proposition' in a paper?

REPLY [2 votes]: Edmund Landau had theorems only, and otherwise nothing (when it comes to naming special statements which required a proof; there should be also definitions and axioms on the top of the theorems).
Irving Kaplansky liked the Landau's style, thus he did the same.
(I am missing many of my books; in the case of Landau, there was a 2-volume differential and integral calculus textbook CLASSIC; in the case of Kaplansky, I think it was a monograph on ring theory).
You may always emphasize important results in several ways, without putting down the other results. One may put a name of a theorem in parenthesis, just after Theorem nn or one may create a subsection which identify the theorem in the section title, e.g. Section 2.4 The fundamental Theorem of Algebra (while, inside the section, this theorem could be Theorem 72, with a sentence of an explanation preceding that theorem).<|endoftext|>
TITLE: motivation of surgery
QUESTION [8 upvotes]: an $n$-surgery on m dim manifold M is to cut out $S^n\times D^{m-n}$and replace it by $D^{n+1}\times S^{m-n-1}$.
I want to know how this is invented?
I do know that the effect of passing a critical point of index $n$ in $m$-manifold is equivalent to attach an $n+1$-handle $D^{n+1}\times D^{m-n-1}$.Now the boundary of $D^n\times D^{m-n}$ is $S^n\times D^{m-n}\cup D^{n+1}\times S^{m-n-1}$,i think there must be some close relation between the special form of $n$-surgery and handle.can someone help make this clear?

REPLY [2 votes]: This question has already been answered, but there's a tiny piece of intuition which I'd like to make explicit:
If you're thinking about a manifold in the PL world, surgery might look a bit arbitrary- why cut out and glue in those pieces and not others? Surgery's natural setting is the smooth world, where you're equipping a manifold with a Morse function $f\colon\, M\to \mathbb{R}$, and using information about critical points of $f$ to encode $M$.
It's actually a bit more involved than you might think it might be, but when you pass a critical point of $f$ you add a handle to $M$, and the boundary changes by surgery. For details, see answers to this question.
So really, surgery isn't an a-priori construction which somebody pulled from a hat- it is rather an operation which stems naturally and inevitably from Morse theory.<|endoftext|>
TITLE: What is the relationship between the Bell numbers, the Bell polynomials, and the partition numbers?
QUESTION [9 upvotes]: A friend of mine and I were wondering what relationship exists between the partition numbers $p_{n}$ and the Bell numbers $B_{n}$ (and also possibly the Bell polynomials $B_{n,k}(x_1,x_2,\dots,x_{n-k+1})$ )? Since integers are the decategorification of finite sets, it is appealing to think that there should be some relationship between the generating functions of the Bell numbers and the partition numbers. My friend pointed out the the equivalence classes of partitions of sets correspond to the conjugacy classes of the symmetric group $S_{n}$, but it was not clear how to convert that into a relationship between the  two types of partitions.
One definite historical flavored question arose in our minds: why is there such a disconnect between the study of integer partitions -- which leads to things like the Ramanujan-Rademacher formula, and involves lots of clever q-series manipulations, and the study of set partitions, which involves things like Stirling numbers, Bell numbers, Bell polynomials and whatnot. Her book about integer partitions apparently didn't mention the Stirling numbers at all.

REPLY [3 votes]: This doesn't answer the question, but I would add another example to the Bell number side: The Bell number of a graph is the number of set partitions of the vertices into independent sets, so the Bell number of the empty graph on $n$ vertices is  $B_n$. 
Winston Yang proved that the Bell number of a tree on $n+1$ vertices is $B_n$, and the Bell number of the  $1$-skeleton of a simplicial complex called a $k$-tree with $n+k$-vertices is also $B_n$.<|endoftext|>
TITLE: Spectral theory for self-adjoint field operators on a symmetric Fock space
QUESTION [12 upvotes]: Background
Suppose we have a finite-dimensional Hilbert space $H = \mathbb{C}^s$ (for a natural number s) and we construct the symmetric (or bosonic) Fock space built from it: $$F(H):= \mathbb{C} \oplus H \oplus S(H \otimes H) \oplus S(H \otimes H \otimes H) \oplus \ldots$$
where S is the symmetrising operator.
Vectors in F are sequences of vectors $\psi = (\psi_0, \psi_1,\psi_2,\ldots)$ such that
$\psi_0 \in \mathbb{C}$, $\psi_1 \in H$, $\psi_2 \in S(H \otimes H)$ etc such that
$\sum_{n=0}^\infty ||\psi_n||_n^2 < \infty$ where || ||n denotes the appropriate norm.
For any vector f $\in$ H we can define a pair of unbounded densely defined operators $a^\dagger(f)$ and $a(f)$ acting on F. These are called the "creation and annihilation operators". They are mutually adjoint and satisfy a commutation relation of the form: $$a(f) a^\dagger(g) - a^\dagger(g) a(f) = \langle f, g\rangle $$
where $\langle f, g\rangle $ is the inner-product of f, g $\in$ H.
The best reference for all this is M. Reed, B. Simon, "Methods of Mathematical Physics, Vol 2", section X.7 p207-212. This is partially available on Google books here: http://books.google.co.uk/books?id=Kz7s7bgVe8gC&lpg=PA141&dq=reed%20and%20simon%20x.7&client=firefox-a&pg=PA210#v=onepage&q=&f=false
The sum $\phi(f) = a(f) + a^\dagger(f)$ is self-adjoint (more properly the closure of their sum is self-adjoint) and is called the Segal quantisation of f (up to a factor of $\sqrt{2}$).

Since $\phi(f)$ is self-adjoint we can apply the spectrum theorem to it. The question is, what is its spectral decomposition? Or more loosely, what are its eigenvalues and eigenvectors? or what can we tell from about its spectral decomposition?

REPLY [4 votes]: One convenient way to do analysis on the symmetric Fock space is to use its isomorphism to
the Bargmann (reproducuing Kernel Hilbert) space (sometimes called the Bargmann-Fock pace) 
of analytic functions on $\mathbb C^s$ (with respect to the Gaussian measure) defined in the classical paper:
Bargman V. On a Hilbert space of analytic functions and associated integral transform I, 
Pure Appl. Math. 14(1961), 187-214. 
An introduction to the Bargmann space may be found in chapter 4 of the
  book by Uri Neretin 
On the Bargmann space the creation and anihilation operators 
are just the multiplication $a_j = z_j$ and the derivation $a^*_j = d/dZ_j$ 
and consequently, the theory of several complex variables can be used for the analysis on this space, 
for example the trace of (a trace class) operator can be represented as an integral on its symbol.
Remark: The isomorphism between the symmetric Fock and Bargmann spaces is not proved in the Book. It can be found for example in the references of the following article:
Regarding the question about $a(f)+a^*(f)$, it is proportional to  the position operator of quantum mechanics. 
This is an unbounded operator, its spectrum is the whole real line, but it does not have 
eigenfunvectors within the Fock space (Loosly speaking, they are Dirac delta functions), however one can find a series of vectors which approximate arbitrarily closely its eigenvectors. Using the corresponding projectors, one can approximate the spectral decomposition of this operator.
The case of the momentum operator $i(a(f)-a^*(f))$ is used more frequently, a possible choice of the approximate eigenvectors is by means of wave packets.<|endoftext|>
TITLE: Is this a known compactification of the natural numbers?
QUESTION [12 upvotes]: Given two infinite sets $A$, $B$ of natural numbers, write $A\preceq B$ if $B\setminus A$ is a finite set. Define the equivalence relation $A\sim B$ if $A\preceq B$ and $B\preceq A$, and let $\partial\mathbb{N}$ be the set of equivalence classes of infinite sets under this equivalence relation. Write $[A]$ for the equivalence class of $A$.
Now define a topology on the disjoint union $\overline{\mathbb{N}}=\mathbb{N}\cup\partial\mathbb{N}$ as follows: A set $U\subseteq\overline{\mathbb{N}}$ is open if and only if, for every $[A]\in U\cap\partial\mathbb{N}$, $[B]\in U$ whenever $B\prec A$, and moreover $A'\subset U$ for some $A'\sim A$.

Is this a known topology? Does it have a name?

It is not hard to see that $\overline{\mathbb{N}}$ is compact (hence the question title): For any neighbourhood of $[\mathbb{N}]$ is the entire space minus a finite subset of $\mathbb{N}$. In particular, it is very much a non-Hausdorff space. Also, $\overline{\mathbb{N}}$ contains $\mathbb{N}$ as an open subset with the discrete topology on the latter. The subspace $\partial\mathbb{N}$ is indeed the boundary of $\mathbb{N}$ in this topology (hence my chosen notation), and it is an Alexandrov space.
I came up with this topology while thinking about sequences, subsequences, and their limits. It seems rather natural, so I don't think I am the first one to ever think of it.

REPLY [5 votes]: I was about to answer the same thing as Joel, but here is a more topological perspective. I will show that the lattice of open subsets of $\overline{\mathbb{N}}$ is very similar to the lattice of open subsets of the Stone-Čech compactification $\beta\mathbb{N}$. So, from the localic point of view, there is but a small difference between your space $\overline{\mathbb{N}}$ and $\beta\mathbb{N}$.
View $\beta\mathbb{N}$ as the set of ultrafilters on $\mathbb{N}$ and let $\mathbb{N}^*$ be the remainder $\beta\mathbb{N}\setminus\mathbb{N}$. (Prinicipal ultrafilters are identified with the corresponding point of $\mathbb{N}$, so $\mathbb{N}^*$ is the subspace of nonprincipal ultrafilters.) Recall that the clopen subsets of $\beta\mathbb{N}$ are precisely those of the form
$$\langle A \rangle = \{ \mathcal{U} \in \beta\mathbb{N} : A \in \mathcal{U} \}$$
for $A \subseteq \mathbb{N}$. Note that $A \preceq B$ iff $\langle A \rangle \cap\mathbb{N}^* \subseteq \langle B \rangle\cap\mathbb{N}^*$, so the points of $\partial\mathbb{N}$ can be identified with the clopen subsets $[A] = \langle A \rangle \cap \mathbb{N}^*$ of the remainder $\mathbb{N}^*$ (including the empty set). Given an open set $U \subseteq \beta\mathbb{N}$, the set
$$U' = (U \cap \mathbb{N}) \cup \{ [A] : [A] \subseteq U \}$$
is open in $\overline{\mathbb{N}}$. Conversely, given an open set $V \subseteq \overline{\mathbb{N}}$, your conditions ensure that
$$V' = (V \cap \mathbb{N}) \cup \bigcup \{ [A] : [A] \in V \}$$
is open in $\beta\mathbb{N}$. This correspondence is not perfect since $A \cup B \cup [A] \cup [B] = A \cup B \cup [A \cup B]$, but 
$$A \cup B \cup \{[C] : C \preceq A \lor C \preceq B\} \quad\mbox{and}\quad A \cup B \cup \{[C] : C \preceq A \cup B\}$$
are not always the same. However, these are the only errors that occur, i.e. the translation is perfect for open subsets of $\overline{\mathbb{N}}$ whose part in $\partial\mathbb{N}$ is upward directed in the ${\preceq}$ ordering.
Although your space $\overline{\mathbb{N}}$ is interesting, this approximate translation suggests most of its applications could be transferred to work over the well studied space $\beta\mathbb{N}$ instead.

Here is yet another perspective which suggests that there may be more to $\overline{\mathbb{N}}$ after all. The soberification of $\partial\mathbb{N}$, which I will denote $\mathrm{Fil}_{\mathbb{N}}$, is the space of all nonprincipal filters on $\mathbb{N}$, with the topology generated by the basic open sets
$$[A] = \{ \mathcal{F} \in \mathrm{Fil}_{\mathbb{N}} : A \in \mathcal{F} \}.$$
The points of $\partial\mathbb{N}$ can be identified with the filters $\mathcal{F}_A = \{ B : A \preceq B \}$. Note that the space $\mathbb{N}^*$ is also a subspace of $\mathrm{Fil}_{\mathbb{N}}$, which explains the connection found above.
For the pointless topology aficcionados, the space $\mathrm{Fil}_{\mathbb{N}}$ is obtained by imposing the trivial Grothendieck topology on the preorder $(\mathcal{P}\mathbb{N},{\preceq})$ viewed as a category (or, equivalently, the quotient partial order $\mathcal{P}\mathbb{N}/\mathrm{fin}$ as suggested by Joel). The subspace $\mathbb{N}^*$ is similarly obtained by imposing the finite cover (aka coherent) Grothendieck topology on $(\mathcal{P}\mathbb{N},{\preceq})$.<|endoftext|>
TITLE: Does a locally free sheaf over a product pushforward to a locally free sheaf?
QUESTION [7 upvotes]: Suppose $X$ and $Y$ are two (smooth, affine) algebraic varieties.  Let $\mathcal{F}$ be a locally free coherent sheaf over $X \times Y$, and let $\mathcal{G}$ be the pushforward of $\mathcal{F}$ to $X$.  Is it true that $\mathcal{G}$ is a locally free quasicoherent sheaf?

REPLY [2 votes]: Hi. I don't know how to make a comment; this is on Torsten's example with 
elliptic curve minus zero point. When you take a line bundle corresponding to 
a point (for simplicity not a two torsion point) and add the line bundle corresponding to minus of that point then this rank=2 vector bundle restricted to the elliptic curve minus zero is (even globally) free. 
Indeed, working on the elliptic curve twist with the degree one line bundle L
corresponding to the zero point and take the canonical inclusion of the structure
sheaf on both factors, the quotient must be isomorphic to L squared. 
This is just to understand Bass' theorem on this nice example.<|endoftext|>
TITLE: Fixed Point Property in Algebraic Geometry
QUESTION [13 upvotes]: I am wondering about the following problem: for which (say smooth, complex, connected) algebraic varieties $X$ does the statement any regular map $X\to X$ has a fixed point hold?
MathSciNet search does not reveal anything in this topic.
This is true for $\mathbb{P}^n$ (because its cohomology is $\mathbb{Z}$ in even dimensions
and $0$ otherwise, and the pullback of an effective cycle is effective, so all summands
in the Lefschetz fixed point formula are nonnegative, and the 0-th is positive -- is this a correct argument?). Is it true for varieties with cohomology generated by algebraic cycles (i.e. $h^{p,q}(X)=0$ unless $p=q$ and satisfying Hodge conjecture), for example for Grassmannians, toric varieties, etc.? This is not at all clear that the traces of $f$ on cohomology will be nonnegative.
Probably you have lots of counterexamples. What about positive results?

REPLY [12 votes]: By demand I expand a little on my answer. The holomorphic Lefschetz fixed point formula (aka the Woods-Hole formula) considers an endomorphism $f\colon M \to M$ of a smooth and compact complex manifold $M$ (or proper
smooth algebraic variety) with only isolated fixed points which are also assumed to be non-degenerate (i.e., the
tangent map of $f$ at a fixed point does not have eigenvalue $1$. Then the alternating trace of the action of $f$ on $H^*(M,\mathcal O_M)$ is equal to a sum over the fixed points $p$ of $1/det(1-T_p(f)$. In particular if there
are no fixed points the alternating trace is equal to $0$. However, in the case when also  $H^i(M,\mathcal O_M)=0$ for $i>0$ then the alternating trace is equal to $1$ so the assumption that there are no fixed points gives a contradiction. Note, that the dimension of $H^i(M,\mathcal O_M)$ is just $h^{0,i}$ so that an assumption that the Hodge numbers vanish off the diagonal gives the required vanishing by a good margin. Furthermore, the vanishing of just $h^{0,i}$ for $i>0$ is much much weaker, it is for instance a birational condition whereas blowing up a smooth curve of genus $>0$ in a variety of dimension at least 3 always give
off diagonal Hodge numbers.
As for the question of whether a birational involution of $\mathbb C^n$ always has a fixed point this seems trickier. It is true that the involution can be made to act regularly on a smooth and proper model and hence by the above has a fixed point. It is not clear however that the fixed point will map to a point of $\mathbb C^n$ as $\mathbb C^n$ is not proper.<|endoftext|>
TITLE: Do the Baumslag-Solitar groups occur in nature?
QUESTION [29 upvotes]: The
Baumslag-Solitar groups
$BS(m,n) = \langle b,s\mid s^{-1}b^ms = b^n\rangle$, with $mn\neq 0$,
are important examples (more often, counter-examples) in group theory.
They are residually finite if, and only if, either  $m$ and $n$
are equal in absolute value, or one of $m$ and $n$ has absolute value equal to
$1$.  In that case, they are Hopfian, and also when $m$ and $n$ have the same
prime divisors.  Otherwise, they are non-Hopfian.
For $m=n$, we get examples of one-relator groups with non-trivial center.
The group  $BS(2,2)$ is an example in which the Howson property fails.
I can only recall having seen these groups defined by means of a presentation.  I would like to know whether these groups (apart from the obvious special cases such as the metabelian ones) can be realized by some other fairly elementary and concrete mechanism.  My question is:

Does $BS(m,n)$ occur "in nature"?

(For example, the Sanov matrices
$\left(\begin{matrix} 1&2 \\ 0&1 \end{matrix}\right)$
and
$\left(\begin{matrix} 1&0 \\ 2&1\end{matrix}\right)$
generate a free subgroup of $SL(2,\mathbb{Z})$, so I would say that free groups occur "in nature".)
Obviously, since the Baumslag-Solitar groups are often non-Hopfian, they
cannot be constructed as groups of matrices.  But, perhaps there is some other concrete realization of these groups.
If there isn't a general construction, it would still be useful to get a concrete realization for the non-Hopfian group $BS(2,3)$.

REPLY [28 votes]: For $m,n \gt 0, m\ne n$, $BS(m,n)$ acts on $\mathbb H^2$ by isometries with a common ideal fixed point. In particular, you can represent it by the action on the upper half plane of
$S = \bigg(\begin{matrix}\sqrt{m/n} & 0 \\\  0 &\sqrt{n/m}\end{matrix}\bigg),$
 $B = \bigg(\begin{matrix}1 & \alpha \\\  0 &1\end{matrix}\bigg)$
where $\alpha$ is arbitrary.
There are elements of $BS(m,n)$ which act trivially, but you can lift this to a free action on a topological  $T\times \mathbb R$ where $T$ is a $n+m$-regular directed tree with outdegree $m$ and indegree $n$ so that for any path $P \subset T$, $P\times \mathbb R$ is a copy of $\mathbb H^2$ so that the sets $\{p\}\times \mathbb R$ are concentric horocycles (like horizontal lines in the upper half plane). 
This is related to the tilings of the hyperbolic plane by horobricks, e.g., there are tiles with arbitrarily small diameter which tile $\mathbb H^2$ which are really fundamental domains of $BS(m,n)$, and analogously there are polyhedra of arbitrary Dehn invariant which tile $\mathbb H^3$.<|endoftext|>
TITLE: How do they verify a verifier of formalized proofs?
QUESTION [56 upvotes]: In an unrelated thread Sam Nead intrigued me by mentioning a formalized proof of the Jordan curve theorem. I then found that there are at least two, made on two different systems. This is quite an achievement, but is it of any use for a mathematician like me? (No this is not what I am asking, the actual question is at the end.)
I'd like to trust the theorems I use. To this end, I can read a proof myself (the preferred way, but sometimes hard to do) or believe experts (a slippery road). If I knew little about topology but occasionally needed the Jordan theorem, a machine-verified proof could give me a better option (and even if I am willing to trust experts, I could ensure that there are no hidden assumptions obvious to experts but unknown to me).
But how to make sure that a machine verified the proof correctly? The verifying program is too complex to be trusted. A solution is of course that this smart program generates a long, unreadable proof that can be verified by a dumb program (so dumb that an amateur programmer could write or check it). I mean a program that performs only primitive syntax operations like "plug assertions 15 and 28 into scheme 9". This "dumb" part should be independent of the "smart" part.
Given such a system, I could check axioms, definitions and the statement of the theorem, feed the dumb program (whose operation I can comprehend) with these formulations and the long proof, and see if it succeeds. That would convince me that the proof is indeed verified.
However I found no traces of this "dumb" part of the system. And I understand that designing one may be hard. Because the language used by the system should be both human-friendly (so that a human can verify that the definitions are correct) and computer-friendly (so that a dumb program can parse it). And the definitions should be chosen carefully - I don't want to dig through a particular construction of the reals from rationals to make sure that this is indeed the reals that I know.
Sorry for this philosophy, here is the question at last. Is there such a "dumb" system around? If yes, do formalization projects use it? If not, do they recognize the need and put the effort into developing it? Or do they have other means to make their systems trustable?
UPDATE: Thank you all for interesting answers. Let me clarify that the main focus is interoperability with a human mathematician (who is not necessarily an expert in logic). It seems that this is close to interoperability between systems - if formal languages accepted by core checkers are indeed simple, then it should be easy to automatically translate between them.
For example, suppose that one wants to stay within symbolic logic based on simple substitutions and axioms from some logic book. It seems easy to write down these logical axioms plus ZF axioms, basic properties (axioms) of the reals and the plane, some definitions from topology, and finally the statement of the Jordan curve theorem. If the syntax is reasonable, it should be easy to write a program  verifying that another stream of bytes represents a deduction of the stated theorem from the listed axioms. Can systems like Mizar, Coq, etc, generate input for such a program? Can they produce proofs verifiable by cores of other systems?

REPLY [14 votes]: I think the question you are asking - about how can we trust a formal proof - is a very important question.  I have spent considerable effort developing software to specifically address this question.  You touch on various things I have concentrated on.
It is true that various systems prominent in the formalisation of mathematics - including the HOL systems (HOL4, ProofPower HOL, HOL Light), Isabelle and Coq - are built according to the "LCF style", which means that all deduction must go via a relatively small kernel of trusted source code (implementing the primitive inference rules), and that this greatly reduces the risks of producing unsound proofs on these systems.  It would also not be an exaggeration to say that almost everyone working in formal proof is happy with this situation.  Indeed, probably most (but not those working on the above systems) feel that resorting to the LCF style is overkill and an unnecessary drain on user execution time and on development effort.
However, there are 3 major problems with this status quo:
A) Most "LCF-style" systems do not implement the LCF-style kernel idea as purely as may be expected.  Some systems have back doors to creating "proved theorems", such as importing the statements (but not the proofs) of previously proved results from disk, and trust that the user will not abuse this.  Also, to reduce execution time, most systems implement various derivable inference rules as primitives, multiplying up the size of the trusted source code.  Also, the kernels of most systems typically incorporate large amounts of supporting code (e.g. for organising theories) and are not particularly clearly implemented, and so are difficult to review.  It should be noted that HOL Light does not suffer from any of these problems.
B) The trusted aspects of an LCF-style system is NOT limited to the design/implementation of its LCF-style kernel.  Like in all systems, it at least also includes the design of the concrete syntax and the implementation of the pretty printer, since, in practice, the user will only view results displayed in concrete syntax via the pretty printer.  However, each system has problems with its concrete syntax and/or pretty printer that allows misleading information to be displayed to the user (e.g. by using irregular variables names, or names that are overloaded).  I have found many ways of appearing to prove "false" in these systems!  Also, depending on how the system is used, the parser is arguably also a trusted component.
C) The importance of the human process of checking that the intended result has in fact been proved (I call this "proof auditing") is generally greatly underestimated, and in practice often not even carried out at all.  As you rightly point out, the axioms and definitions used in a formal proof need to be carefully checked, as well as the statement of the theorem itself.  I have known subtle mistakes in definitions to render real formal proofs on real projects completely invalid.
I have developed an open source theorem prover called HOL Zero, that addresses issues A-C above and is designed for use in proof auditing and generally to be as trustworthy as possible.  It has a watertight inference kernel, a well-designed concrete syntax and pretty printer, and the source code aims to be as clear and well-commented as possible.  However, it should be noted that it does not have the advanced automatic and/or interactive proof facilities of the existing systems I mention above, and is not suited to developing large formal proofs.  HOL Zero can be downloaded from here (it needs OCaml and a Unix-like operating system):
http://www.proof-technologies.com/holzero
The concept of checking one system using another is not only philosophically reassuring, but also of pressing need (due to the above issues A-C).  As you say, what is needed is the ability to port proof objects between systems (the so called "de Bruijn criterion").  Strictly speaking, the de Bruijn criterion is as you state your requirement - the ability to capture a proof as an object (e.g. a text file) - rather than the LCF style (but let's not get too philosophical about the equivalence of these approaches).  Anyway, there are some practical issues here:

The "dumb program" that you refer to needs to be surprisingly sophisticated, otherwise it loses much of its purpose.  If it is just an LCF-style kernel, the data it outputs will be too slow to review for large projects.  As you say, it needs to be human-friendly - a decent pretty printer is a practical necessity.  Also, to make proof exporting (see next item) work in practice, it needs to work at least at a slightly higher level than the kernel, and so some supporting theory is required to be built.  So yes, a dumb program is required, which should be as easy to understand as possible, but it is more challenging to build than a few lines of code.
Capturing proof objects in a suitably efficient way is a non-trivial exercise.  The work that others mention above successfully port things like the base system of HOL Light, but are completely incapable of handling something the scale of Hale's HOL Light proof of the Jordan Curve Theorem, let alone Hale's Flyspeck Project (using HOL Light to check his non-formal proof of the Kepler Conjecture).
Some sort of neat and trivial correspondence of equivalent theory between systems is useful.  This is better than importing language statements as huge expressions, in terms of some highly-complex embedding of one notation inside another, which would either greatly increase the complexity of the checking system or make its human usage difficult.

HOL Zero is primarily aimed at the "dumb program" proof checker role.  The idea is it will import and replay proofs that have been exported from other HOL systems.  I have implemented a proof exporting mechanism which, unlike others' mechanisms, handles with ease large proofs such as Hale's Jordan Curve proof and the (almost-complete) Flyspeck Project.  I am currently working on a proof importing mechanism for HOL Zero.  (Note that a former proof importing mechanism I developed worked on an old version of HOL Zero and successfully ported Hale's Jordan Curve proof and Harrison's proof of the consistency of the HOL Light kernel.)

REPLY [4 votes]: One simple suggestion no-one seems to have mentioned is to have the verifier prove itself correct.
Obviously, this cannot really give any assurance that the verifier is correct, since if the verifier is incorrect its proofs are worthless. However, on heuristic grounds, this should give some confidence. The reason is that errors in assertions about the verifier's proofs (which will be checked) should be uncorrelated with the errors in the verifier itself.
Of course, no physical process can be expected to prove anything with complete accuracy, so these kind of heuristics are the best one can hope for anyway.<|endoftext|>
TITLE: Proving that a group is free
QUESTION [31 upvotes]: I've got a group $G$ that I'm trying to prove is free.  I already know that $G$ is torsion-free.  Moreover, I can "almost" prove what I want : I can find a finite index subgroup $G'$ of $G$ that is definitely free.
This leads me to the following question.  Can anyone give me an example of a torsion-free group $G$ that is not free but contains a free subgroup of finite index?  I've tried pretty hard to find groups like this, but i can't seem to avoid introducing torsion.  Thanks!

REPLY [3 votes]: If a torsion free group is quasi-isometric to a (nontrival) free product, then it is free product.(Gromov). 
And we know that a finite index subgroup of G is quasi-isometry to G.
So G is also free.<|endoftext|>
TITLE: Convergence of spectral sequences of cohomological type
QUESTION [9 upvotes]: Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the homological situation.
An exact couple consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that
$$
\begin{array}{rcl}A&\xrightarrow{i}&A\newline {\scriptsize k}\nwarrow&&\swarrow{\scriptsize j}\newline&E&\end{array}
$$
is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=j\circ k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $a\in \mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees
$$
\begin{array}{rcl}A^{(r)}&\xrightarrow{(1,-1)}&A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&&\swarrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&E^{(r)}&\end{array}
$$
induces a spectral sequence $E_{p,q}^{r+a}=E_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}\circ k^{(r)}$. Here ${\scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.
Here is the canonical example. Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them:
$$
\begin{array}{lcccccr}
\to H_{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H_{p+q-1}(C_{p-1})&\xrightarrow{i}&H_{p+q-1}(C_p)&\xrightarrow{j}&H_{p+q-1}(C_p,C_{p-1})\to\newline
\to H_{p+q}(C_{p-1},C_{p-2})&\xrightarrow{k}&H_{p+q-1}(C_{p-2})&\xrightarrow{i}&H_{p+q-1}(C_{p-1})&\xrightarrow{j}&H_{p+q-1}(C_{p-1},C_{p-2})\to\newline
\to H_{p+q}(C_{p-2},C_{p-3})&\xrightarrow{k}&H_{p+q-1}(C_{p-3})&\xrightarrow{i}&H_{p+q-1}(C_{p-2})&\xrightarrow{j}&H_{p+q-1}(C_{p-2},C_{p-3})\to
\end{array}
$$
There is an exact couple as above with $a=1$ and $E_{p,q}=H_{p+q}(C_p,C_{p-1})$ and $A_{p,q}=H_{p+q}(C_p)$. Please note, that all the bigrades are correct. To follow $d^1:E_{p,q}^1\to E_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.
Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.

The spectral sequence $E^1(C_\bullet)$ of homological type
  converges to $H_{p+q}(C)$ if it is bounded (=only
  finitely many non-zero entries on
  every fixed diagonal $p+q$). One has
  $$ E^\infty_{p,q}=i(H_{p+q}(C_p))/i(H_{p+q}(C_{p-1})) .$$
  ($i$ denotes the image in the colimit $H_{p+q}(C)$.)

Proof. Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple
$$
\begin{array}{c}
\to E^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}_{p+r,q-r}\newline
\xrightarrow{j^{r}}E^{r}_{p,q}\xrightarrow{k^{r}}\newline
A^{r}_{p-1,q}\xrightarrow{i^{r}}A^{r}_{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to.
\end{array}
$$
The first and the last term are $0$ because of the bounding assumption. We have $A_{p,q}^{r}=i(A^1_{p-r,q+r})$ and since $C_n$ is zero for $n<0$, the last three terms are $0$. Exactness implies the result.
Now the cohomological situation.
Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H_n(\hom(C,\mathbb{Z}))$, the homology of the dualized complex as one does in topology.
Again we have many long exact sequences:
$$
\begin{array}{lcccccr}
\to H^{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H^{p+q}(C_p)&\xrightarrow{i}&H^{p+q}(C_{p-1})&\xrightarrow{j}&H^{p+q+1}(C_p,C_{p-1})\to\newline
\to H^{p+q}(C_{p+1},C_p)&\xrightarrow{k}&H^{p+q}(C_{p+1})&\xrightarrow{i}&H^{p+q}(C_{p})&\xrightarrow{j}&H^{p+q+1}(C_{p+1},C_{p})\to\newline
\to H^{p+q}(C_{p+2},C_{p+1})&\xrightarrow{k}&H^{p+q}(C_{p+2})&\xrightarrow{i}&H^{p+q}(C_{p+1})&\xrightarrow{j}&H^{p+q+1}(C_{p+2},C_{p+1})\to
\end{array}
$$
One can again follow $d^1:E_{p,q}^1\to E_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C_p)$ and $E^{p,q}=H^{p+q}(C_p,C_{p-1})$ and $a=1$.
An exact couple with bidegrees
$$
\begin{array}{rcl}A_{(r)}&\xrightarrow{(-1,1)}&A_{(r)}\newline {\scriptsize (0,0)}\nwarrow&&\swarrow{\scriptsize (a+r,-(a-1+r))}\newline&E_{(r)}&\end{array}
$$
induces a spectral sequence $E_{r+a}=E_{(r)}$ of cohomological type with differentials $d_{r+a}=j_{(r)}\circ k_{(r)}$. Here ${\scriptsize something}_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.
Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:

The spectral sequence $E_1(C_\bullet)$ of cohomological type
  converges to $H^{p+q}(C)$ if it is bounded. One has
  $$ E_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C_{p})) .$$

Proof? Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple
$$
\begin{array}{c}
\to E_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline
\xrightarrow{j_{r}}E_{r}^{p,q}\xrightarrow{k_{r}}\newline
A_{r}^{p,q}\xrightarrow{i_{r}}A_{r}^{p-1,q+1}\xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}\to.
\end{array}
$$
The last term is $0$ because of the bounding assumption. We have $A_{r}^{p,q}=i(A^{p+r,q-r}_1)$ and since $C_n$ is zero for $n<0$, the second and the third term are $0$. Exactness implies that
$$
 E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))\to i(H^{p+q}(C_{p+r-1}))).
$$
I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:

How does Lemma 1.2. of Hatcher's text works in the cohomological case?

REPLY [11 votes]: The lemma you refer to has two halves.  The first half covers the case of homology and the second half covers the case of cohomology.  Proofs are given for both halves (though the last sentence of the proof in the cohomology case requires a moment's thought to convince oneself of).  The way to derive the cohomology spectral sequence from the second half is explained on the page following the lemma. Perhaps there is something that seems unclear in what is written there?  Please feel free to ask me about this, though this might be better done in email.<|endoftext|>
TITLE: Is there a non-trivial topological group structure of $\mathbb{Z}$?
QUESTION [9 upvotes]: More specificaly, is there a haussdorf non-discrete topology on $\mathbb{Z}$ that makes it a topological group with the usual addition operation?

REPLY [4 votes]: There is a huge number of such topologies. Let $G$ be any discrete abelian group. By $\widehat G$ we denote the family (in fact, a group) of its characters, that is of homomorphisms from $G$ to the unit circle group $\Bbb T$. By Pontrjagin duality theory, the family $\widehat G$ separate points of $G$ (that is for each two distinct elements $g,h\in G$ there exists a character $\chi\in\widehat G$ such that $\chi(a)\ne\chi(b)$). Thus a diagonal product $\Delta\{\chi:\chi\in\widehat G\}:G\to\prod \{\Bbb T_\chi: \chi\in\widehat G\}$ is an injective homomorphism of $G$ into a compact group. In endows $G$ with a (totally bounded) group topology (called Bohr topology on the group $G$). In particular, if $G$ is infinite then its Bohr topology is indiscrete. On the other hand, according to this answer, $G$ admits continuum many (pairwise transversal) group topologies, none of which is totally bounded. The case of $G=\Bbb Z$ is espectially simple because it admits an injective homomorphism into each topological group with an element of infinite order. As a more concrete example, in this MSE answer I proposed for each $\kappa\le\frak c$ an injective homomorphism of the group $\Bbb Z$ onto a dense subroup of a compact group $\Bbb T^k$.<|endoftext|>
TITLE: What algorithm in algebraic geometry should I work on implementing?
QUESTION [50 upvotes]: This summer my wife and one of my friends (who are both programmers and undergraduate math majors, but have not learned any algebraic geometry) want to learn some algebraic geometry from me, and I want to learn how to program from them, so we were planning on working on some computational algebraic geometry together. While there are several books which we could work through, I thought it might be more fun and productive if we had the goal of developing a usable new algorithm, or at least implementing an algorithm which no one has implemented before. I do not have any ideas, but I thought that some mathoverflowers might have had an idea for an algorithm they would like to see implemented but have never had the time to work through the details.  Keep in mind that my wife and friend will have to learn any mathematics past a first course in topology and abstract algebra as we go. 
So does anyone have any ideas for an algorithm they would like to use which is within the reach of my "team" to implement within a summer?  We are planning on working on this stuff between 2 and 3 hours a day for about 3 months.

REPLY [4 votes]: Steve, do you know about Macaulay 2? It's a computer algebra system designed for commutative algebra and algebraic geometry. 
Moreover it has a fairly easy to use language and an easy way to create packages. 
Probably if you would join the Google Group "Macaulay 2" and asked this same question you would get some offers.<|endoftext|>
TITLE: Density results for equality of Galois/automorphic representations
QUESTION [5 upvotes]: I have two "vague questions" which are the following: if you have two $n$-dimensional $\ell$-adic Galois representations of a number field $K$ (with the standard ramification conditions) that have the same char. polynomial in a set of primes of density $1$, then they are isomorphic. The same is true if I replace Galois representations by automorphic representations and I ask the local representations to be the same in a set of density $1$. So the question is: "what is the best bound for such result"? (assume $n=2$ and $K= \mathbb Q$ if you want).
So for example, is it true that given $\epsilon >0$ there exist two representations that are not isomorphic but whose Frobenius/local components are the same in a set of density $1-\epsilon$? (at least in the case $n=1$?)
The second somehow related question (but has nothing to do, just in spirit) is if I have a restricted tensor product of local automorphic representations (say $K=\mathbb Q$, $n=2$) then there is no reason for it to be modular, but is there any result saying "there exists $\delta <1$ such that if you change the local component appropiately in a set of primes of density $\delta$ then it is an automorphic form"?
Of course, if one starts with a Galois representation and construct a restricted tensor product via the local langlands correspondence, one would like to say that such representation $\Pi$ is modular (under the appropiate assumptions on the Galois representation), so my question can be viewed as "how far (in terms of density) are we from proving modularity in general"

REPLY [9 votes]: Firstly, at the beginning of the question you are missing irreducibility/cuspidality assumptions. If $\rho_1$ and $\rho_2$ are $\ell$-adic Galois reps with the same char poly in a set of primes of density 1, then you can only deduce their semisimplifications are isomorphic. A counterexample to your statement would be given by $\rho_1=1+\omega$ ($\omega$ the cyclotomic character) and $\rho_2$ some non-split extension coming from Kummer theory (taking $\ell$-poower roots of some prime number $p$ for example). Of course you knew that already.
On the automorphic side you make "the same slip", well, a related slip. If $\chi$ and $\psi$ are two Grossencharacters and their ratio is the norm character at some place $v$ (or possibly even infinitely many, or even all $v$), then the induction of $\chi+\psi$ from the Borel to $GL_2(\mathbf{A})$ is reducible, and any Jordan-Hoelder factor (which by definition means a tensor product of J-H factors of the local inductions) ramified at all but finitely many places is an automorphic representation. So now you can build two non-cuspidal automorphic representations which are isomorphic at all but one place (and such that the local components don't even have the same dimension at the bad place) easily. Of course for cuspidal representations you have strong multiplicity 1 theorems.
Passing remark: it is a source of some confusion to me as to why these errors are "similar" but on the other hand they don't "biject". The problem is that on the Kummer theory side, if you allow ramification at two primes, you get a whole projective line of non-isomorphic Galois representations (all with the same semisimplificiation). On the automorphic side the amount of control you have is much more combinatorial (you can change a finite set of places from trivial to Steinberg and that's it). So $\pi$ and $\rho$s don't match up (so Toby Gee and I only conjecture that given any (EDIT: algebraic) $\pi$ one expects a semi-simple $\rho$, and nothing more, and for $GL_n$ this observation is no doubt much older).
OK so after these pedantic remarks, that you no doubt knew anyway, but could have avoided if you'd put "irreducible" and "cuspidal" in the appropriate places, we move on to the far more interesting question of whether one can do better than multiplicity 1. And your hunch is correct. First you should do the following exercise on the Galois side: if $\rho_1$ and $\rho_2$ are two irreducible 2-dimensional representations of a finite group $G$, and their traces agree on a set $S$ in $G$ of density greater than 7/8, then $\rho_1$ and $\rho_2$ are isomorphic. Big hint: orthogonality relations. Next you should convince yourself that 7/8 is optimal in this result Big hint: D_8 x D_8. (EDIT: slightly simpler is D_8 x C_2---here D_8 has 8 elements). So now for silly Artin representation reasons you can't expect to beat 7/8 (like you can't expect to beat 1/2 in the GL_1 case as in Jared's comment).
So now the great news is that Dinakar Ramakrishnan proved an analogue on the automorphic side! At least in some cases. See the appendix to "$l$-adic representations associated to modular forms over imaginary quadratic fields. II." by Richard Taylor (Inventiones 116).
As for your second question though, it's completely hopeless. Even for $GL_1$ your hope is (provably) way out, so for $GL_2$ one can perhaps construct "Eisenstein" counterexamples (induced from non-automorphic characters of $GL_1$). The first objection is that $\pi_p$ can be ramified for infinitely many $p$, making you dead in the water. But even if $\pi_p$ is unramified for all $p$ you've still got no chance. Let me stick to $GL_1/\mathbf{Q}$. The point is that every Grossencharacter for $GL_1/\mathbf{Q}$ is the product of a finite order character $\chi$ (a Dirichlet character) and $||.||^s$, so you can recursively construct characters of $\mathbf{Q}_p^\times$ each of which is totally at odds with everything that came before. For example lets send $Frob_2$ to 1. Now let's write down all the solutions to $2^s.\zeta=1$, with $\zeta$ a root of unity and $s$ a complex number. There are only countably many values of $s$ in this list. So choose $s$ not in the list and send $Frob_3$ to $3^s$. Now knock off countably many more $s$ and send $Frob_5$ to $5^s$ for $s$ in neither list. We are making a collection of representations here that are pairwise completely incompatible! This is only one of the many obstructions. For example, for cuspidal automorphic representations there are Weil bound obstructions (Ramanujan conjecture) and arithmeticity obstructions in the holomorphic case---all deep theorems or conjectures about automorphic forms in some cases that can easily be violated if we're allowed to build $\pi$ locally. These arguments trivially show that the set of automorphic $\pi$s have density zero (and "a very small zero" at that) amongst all the $\pi$s. In fact here's a much cleaner objection for $GL_2$: if you stick to cuspidal automorphic representations with a fixed central character then there are only countably many! But you have uncountably many choices at each local place! So it's completely hopeless I think.<|endoftext|>
TITLE: Forms over finite fields and Chevalley's theorem
QUESTION [7 upvotes]: Chevalley's theorem says that if $k$ is a finite field and $f(X_1,...,X_n)$ is a form (homogeneous polynomial) of degree $d < n$, then the equation $f(X_1,...,X_n) = 0$ has a non-trivial solution in $k^n$. 
It is known that this result is optimal, in the sense that for each $n$ there exists a form $f(X_1,...,X_n)$ - coming from a norm - of degree $d = n$ which has only the trivial zero. See Brian Conrad's answer below.
I am interested in each form with this property, i.e. each form $f(X_1,...,X_n)$ of degree $d = n$ which has only the trivial zero. What are the known examples/classes of such forms? It would be very nice if we could describe/classify them all...
I am in particular interested in the case of quartic forms.

REPLY [3 votes]: Perhaps it is obvious to most readers, but about a year ago I spent several days trying to determine for which pairs (d,n) there existed an anisotropic degree d form in n variables over a finite field $\mathbb{F}_q$.  The question was motivated by Exercise 10.16 in Ireland and Rosen's classic number theory text: "Show by explicit calculation that every cubic form in two variables over $\mathbb{F}_2$ has a nontrivial zero."  
As many students have discovered over the years, this is false: e.g. take 
$f(x_1,x_2) = x_1^3 + x_1^2 x_2 + x_2^3$.
I knew about the existence and anisotropy of norm hypersurfaces for all $n = d$.  But what about $n < d$?  I confess that I spent some time proving this result in several special cases and even dragged a postdoc into it.  Here is a copy of the sheepish email I sent out (in particular to Michael Rosen) later on:

If K is a field, and f(x_1,...,x_n) is an anisotropic form of degree d in n variables, then f(x_1,...,x_{n-1},0) is an anisotropic
  form of degree d in n-1 variables. 
So let K be any field which admits field extensions of every positive degree d.  Then for all d there is an anisotropic norm form
  N in d variables of degree d.  For any n < d, setting (d-n) of the variables equal to 0 gives an anisotropic form of degree d in
  n variables.  In particular, this proves "the converse of Chevalley-Warning".
So, not so fascinating after all, then.
I think it is still nontrivial to ask what happens if the hypersurface f is required to be geometrically irreducible.  For instance, despite the fact that (q,3,3) is anisotropic, every geometrically irreducible cubic curve over a finite field has a rational point.

AS's question about classifying anisotropic hypersurfaces with $d = n$ is interesting.  It may also be interesting to look at the case $d < n$.  It is certainly not clear to me that all such anistropic hypersurfaces come from intersecting a norm hypersurface of larger dimension with a linear subspace.
I also want to add that the following generalization seemed less trivial to me (and I still don't know the answer): Chevalley-Warning is also true for sytems of polynomial equations $f_1(x_1,\ldots,x_n) = \ldots = f_r(x_1,\ldots,x_n)$ so long as the sum of the degrees of the $f_i$'s is strictly less than $n$.  What kind of counterexamples can we construct here when $d = d_1 + \ldots + d_r \geq n$?<|endoftext|>
TITLE: Fundamental groups of noncompact surfaces
QUESTION [60 upvotes]: I got fantastic answers to my previous question (about modern references for the fact that surfaces can be triangulated), so I thought I'd ask a related question.  A basic fact about surface topology is that if $S$ is a noncompact connected surface, then $\pi_1(S)$ is a free group (possibly trivial or $\mathbb{Z}$).  I've had a lot of people ask me for references for this fact.  I know of two such references:
1) In section 44A of Ahlfors's book on Riemann surfaces, he gives a very complicated combinatorial proof of this fact.
2) This isn't a reference, but a high-powered 2-line proof.  Introducing a conformal structure, the uniformization theorem shows that the universal cover of $S$ is contractible.  In other words, $S$ is a $K(\pi,1)$ for $\pi=\pi_1(S)$.  Next, since $S$ is a noncompact $2$-manifold, its integral homology groups vanish in dimensions greater than or equal to $2$.  We conclude that $\pi_1(S)$ is a group of cohomological dimension $1$, so a deep theorem of Stallings and Swan says that $\pi_1(S)$ is free.
There should be a proof of this that you can present in a first course in topology!  Does anyone know a reference for one?

REPLY [2 votes]: A new approach to spines is available via mass transport theory and Kantorovich duality. This is developed in my PhD thesis. 
The idea is elementary: consider the retract $x\mapsto x/|x|$ from the closed unit ball $B:=\{x \in \mathbb{R}^{N} | ~~ |x| \leq 1\}$ to its boundary sphere $\partial B$. The retract has locus-of-discontinuity $Z=\{pt\}$ equal to a point, namely $x=0$. Observe that the inclusion $\{pt\} \hookrightarrow B$ is a homotopy-equivalence. Claim: this is general principle which follows from Kantorovich duality and the optimal transport theory.
For example, let $S$ be closed hyperbolic surface with metric $d$, and $C\hookrightarrow S$ an embedded Cantor set. Let $X:=S-C$ be the Cantor-punctured surface, and let $\sigma$ be the Hausdorff measure on $X$. Similarly let $\tau$ be the Hausdorff measure of $C$ viewed as a subset of $(S,d)$. Now consider the function $c: X \times C \to (0,\infty)$ defined by the rule $$c(x,y_0):= [\int_C d(x,y)^{-2} d\tau(y) ] - \frac{1}{2} d(x,y_0)^{-2}.$$ We view $c(x,y_0)$ as the cost of transporting a unit mass from the source $x\in X$ to target $y_0\in C$. If $\int_X \sigma > \int_C \tau$, then there exist semicoupling measures $\pi$ on $X\times C$ with the property $$proj_X \# \pi \leq \sigma, ~~~\text{and}~~~~proj_C \# \pi = \tau.$$ In otherwords, $\pi$ is a transference plan from the abundant source $\sigma$ to the prescribed target $tau$. (Such measures are called "Semicouplings"). It is standard result of optimal transport that there exists a unique $c$-optimal semicoupling $\pi_*$ which minimizes the total cost $$c[\pi]:=\int_{X\times C} c(x,y) d\pi(x,y).$$ 
Now imagine we rescale the target measure $\tau\mapsto \lambda \tau$ for scalar $\lambda>0$. If $\lambda \int_C \tau$ is sufficiently close to $\int_X \sigma$, then the $c$-optimal semicoupling $\pi_*$ will have a "locus-of-discontinuity" $Z \hookrightarrow X$ such that $Z$ is a strong-deformation retract of $X$ and $Z$ will be codimension-one (i.e., the "singularity" is the spine). 
More specifically, the $c$-optimal semicouplings $\pi_*$ are characterized by the existence of a $c$-concave potentials $\psi: C \to \mathbb{R}$ satisfying $(\psi^c)^c=\psi$. This is Kantorovich's duality theory. The $c$-optimal transport has the form $x\mapsto \partial^c \psi^c (x)$ for every $x\in X$. Here $\partial^c \psi^c(x)$ is a subset of $C$, namely  the $c$-subdifferential of $\psi^c$ at $x\in X$. The "locus-of-discontinuity" is more precisely described as the set of $x\in X$ where $\# \partial^c \psi^c (x) \geq 2$, i.e. where the $c$-convex potential $\psi^c$ is not uniquely differentiable. The locus-of-discontinuity $Z$, where $\psi^c$ is finite and not uniquely differentiable, is a closed lipschitz subvariety of $X$. And Kantorovich duality shows $Z \hookrightarrow X$ is a deformation-retract. The existence of this retract is probably not obvious, unless you are well-studied in the mass-transport theory...
But all the details are in my thesis, including applications to spines for the Teichmueller spaces and symmetric spaces of arithmetic-groups. I'd be happy to share the details, since my supervisor has absolutely zero interest in topological applications, and is quite plainly indifferent to algebraic topology.<|endoftext|>
TITLE: Dualizing sheaf of reducible variety?
QUESTION [9 upvotes]: Sorry for my poor English.
Let $X$ be a reducible projective variety. 
My question is:

How can I compute the dualizing sheaf of $X$ and express it in an explicit way?
Is there a method to get dualizing sheaf of whole reducible variety $X$ from the information of dualizing sheaves of its irreducible components?

Currently I'm not concern the general case, but I want a few accessible concrete examples such as:

reducible hypersurfaces,
union of toric varieties glued at isomorphic orbits(Alexeev calls it stable toric variety).

The reason why I concern is to understand the limit in moduli spaces of stable pairs.

REPLY [14 votes]: With regards part 2.  
Let's assume that you have two components $X_1$ and $X_2$ (or even unions of components) such that $X_1 \cup X_2 = X$=.  Let $I_1$ and $I_2$ denote the ideal sheaves of $X_1$ and $X_2$ in $X$.
Set $Z$ to be the scheme $X_1 \cap X_2$, in other words, the ideal sheaf of $Z$ is $I_1 + I_2$. 
It is easy to see you have a short exact sequence
$$0 \to I_1 \cap I_2 \to I_1 \oplus I_2 \to (I_1 + I_2) \to 0$$
where the third map sends $(a,b)$ to $a-b$.
The nine-lemma should imply that you have a short exact sequence
$$0 \to O_X \to O_{X_1} \oplus O_{X_2} \to O_Z \to 0$$
If you Hom this sequence into the dualizing complex of $X$, you get a triangle
$$\omega_Z^. \to \omega_{X_1}^. \oplus \omega_{X_2}^. \to \omega_{X}^. \to \omega_Z^.[1]$$
You can then take cohomology and, depending on how things intersect (and what you understand about the intersection), possibly answer your question.  
If $X_1$ and $X_2$ are hypersurfaces with no common components (which should imply everything in sight is Cohen-Macualay) then these dualizing complexes are all just sheaves (with various shifts), and you just get a short exact sequence
$$0 \to \omega_{X_1} \oplus \omega_{X_2} \to \omega_{X} \to \omega_{Z} \to 0$$
Technically speaking, I should also probably push all these sheaves forward onto $X$ via inclusion maps.<|endoftext|>
TITLE: Rado graph containing infinitely many isomorphic subgraphs 
QUESTION [5 upvotes]: The Rado graph contains every finite graph as an induced subgraph. It surely contains some finite graphs infinitely often as an induced subgraph, e.g. $K_2$. Does it contain all finite graphs infinitely often as an induced subgraph? Or can an example of a graph be given that is not contained infinitely often?

REPLY [3 votes]: I realise that the question is almost three years old, but maybe that's not that long in Maths. 
I am not sure who the following is originally due to, but as far as I am aware, it's the standard to show that the Rado graph contains every countable graph. The idea is to start with whatever graph you want and construct the Rado graph around it.
Let $G=G_0$ be any countable graph. For $i>0$ define a new Graph $G_i$ as follows: 

The vertices $V(G_{i})$ of $G_i$ are all of $V(G_{i-1})$ plus an extra vertex $v_A$ for every finite subset $A$ of $V(G_{i-1})$.
All edges of $V(G_{i-1})$ are also edges of $V(G_{i})$
Add an edge between $a\in V(G_{i-1})$ and $v_{A}\in V(G_{i})$ whenever $a\in A$.

Let $R$ be the union of the graphs $G_i$ over all $i>0$. Then show that $R$ is the Rado graph. Clearly, $R$ contains $G=G_0$.
One of the nice things about this construction is that you can show without much difficulty that any automorphism of $G$ extends to an automorphism of $R$. So not only does the Rado graph contain every countable graph $G$, but it contains "special" copies of $G$ with the above extension property.<|endoftext|>
TITLE: Maximum bipartite graph (1,n) "matching"
QUESTION [6 upvotes]: Last month I discovered a nice question on stackoverflow and thought the 1,n matching problem could be solved via introducing a 1,k tree matching. Look here for my question, but as Moron pointed out this is not known to be solvable in polynomial time.
Now I discovered mathoverflow and would like to know if one could reduce some possibilities or if one could introduce some restrictions so that the 1,n matching could be solved in polytime. Do you have an idea?
Background: I would like to implement a heuristic for that problem. At the moment I am using an algorithm based on the GPA heuristic for weighted maximum matchings in general graphs as described here (broken link).
Here is my formulation of the problem:
Instead of the 'normal' "one-to-one" matching in a bipartite graph I want to calculate the maximum "one-to-many" matching for a bipartite graph G. (Should I define it or is this clear?)  For the initial problem 1:n the n was arbitrary. Then I fixed the n to k 'static' choices via trees instead of edges, but this was hard, too.
Now I would like to find a similar graph or different formulation which is solvable in polytime (or at least there should be a high accurate and 'fast' algorithm). E.g. couldn't I apply the polytime matching algorithm from general graphs and apply it to a slightly changed graph G'? I think about the following graph G':
     (source)
Clearly there are matchings (red) in the general graph G' which could result in strange/wrong matchings for the corresponding bipartite matching in the bipartite graph G. But maybe one could further tweak the graph G' or the edge weights of G'?
Or how is this related to hypergraphs (e.g. if edge would have 3 nodes)

REPLY [3 votes]: This problem seems pretty hard.  Let $G$ be a bipartite graph with bipartition $(A,B)$.  One easy case to consider is if each vertex in $A$ has degree $k$ and we seek a maximum $(1, k)$ matching.  For each $a \in A$, if we let $N(a)$ be the set of neighbours of $a$, then we seek a maximum size subfamily $\mathcal{S}$ of
{ $N(a) : a \in A$ },
such that any two members of $\mathcal{S}$ are disjoint.  I believe this problem is polynomially equivalent to maximum-clique, so even in this easy case your problem is still hard.  However, maximum-clique is polynomially solvable for perfect graphs, so perhaps this is a partial answer to your question.  
Another way to go is to look for good approximation algorithms, say by adapting approximation algorithms for maximum-clique. 
Edit. A general comment is that if we restrict to a class of graphs with bounded tree-width, then your problem is indeed polynomial.  This works for many NP-hard problems.<|endoftext|>
TITLE: An elementary number theoretic infinite series
QUESTION [13 upvotes]: For a positive integer $k$, let $d(k)$ be the number of divisors of $k$. So $d(1)=1$, $d(p)=2$ if $p$ is a prime, $d(6)=4$, and $d(12)=6$. 
What are the precise asymptotics of $\sum_{k=1}^n 1/(k d(k))$?
Background:
1) This came up on the side in the polymath5 project.
2) There, Tim Gowers wrote: If nobody knows the answer, maybe that’s one for MathOverflow, where I imagine a few minutes would be enough. 
3) Asked: 14:17 Jerusalem time. (The first accurate answer: 17:44 Jerusalem time.) 
4) Looking only at primes or only at integers with a typical number of divisors suggested a $\log\log n$ behavior, but looking at semiprimes indicates the sum is larger. I don't know how much larger.  
5) I couldn't find an answer on the web. If there is an easy way searching for an answer that I missed this will be interesting too.
Follow up:
Great answers! Thanks. What about the sum $\sum_{k=1}^n 1/(kd^2(k))$ ?

REPLY [13 votes]: The idea (from the Selberg-Delange) method to doing this problem is the following steps:
1) Let $F(s) = \sum_{n\ge 1} \frac{1}{n^s d(n)} = \prod_{p} \left(1 + \sum_{k=1}^{\infty} \frac{1}{(k+1) p^{ks}} \right)$.  The latter is by multiplicativity of $d(n)$.
2) If we look, instead at $G(s) = \prod_p \left( 1 + \frac{1}{2 p^s} \right)$ we can see that $F(s)/G(s)$ has a non-zero limit as $s \rightarrow 1$ from above.  $G(s)$ corresponds in our original sum to restricting $n$ to be square-free.
3) $G(s)^2$ almost looks like $\zeta(s)$.  Show that $G(s)^2/\zeta(s)$ also has a non-zero limit at $s \rightarrow 1$.
4) You then use some Tauberian theorems to show that since $H_n \sim \log n$ (which is the sum associated with $\zeta(s)$ then the corresponding sum for $G(s)$ (i.e. over the square-free $n$) is $\sim to \sqrt{\log n}$.<|endoftext|>
TITLE: Example of the completion of a noetherian domain at a prime that is not a domain
QUESTION [16 upvotes]: Let $R$ be a Noetherian domain, and let $\mathfrak{p}$ be a prime ideal; consider the completion $\hat R_{\mathfrak{p}}$ of $R$ at $\mathfrak{p}$ (the inverse limit of the system of quotients $R/\mathfrak{p}^n$). If $R$ is a PID, it is easy to see that $\hat R_{\mathfrak{p}}$ is a domain. 
Someone asked in sci.math if $\hat R_{\mathfrak{p}}$ would always be a domain. I thought it would, but looking at Eisenbud's "Commutative Algebra", I found a reference to a theorem of Larfeldt and Lech that says that if $A$ is any finite-dimensional algebra over a field $k$, then there is a Noetherian local integral domain $R$ with maximal ideal $\mathfrak{m}$ such that $\hat{R_{\mathfrak{M}}}\cong A[[x_1,\ldots,x_n]]$ for some $n$; and so this completion will not be a domain if $A$ is not a domain. I would like to know an example directly, if possible.
Does someone know an easy example of a noetherian domain $R$ and a prime ideal $\mathfrak{p}$ such that $\hat R_{\mathfrak{p}}$ is not a domain? Thanks in advance.

REPLY [2 votes]: At least for algebraic local domains R, we have 1-1 correspondence between the minimal primes in the completion $\cap{R}$ and the maximal ideals in the integral closure of R in its quotient field. This will produce a lot of examples including curves near a point whose neighbourhood in the curve can't be covered by a single parameterization.
    In fact, various connections come to the fore here. For several ways to characterize the number of minimal primes in the completion of a local ring of an irreducible plane curve, one can look at S. S. Abhyankar's  Chavounet prize winning paper "Historical Ramblings in Algebraic Geometry and related algebra".
Several connections are worth mentioning, eg. concepts such as Henselization and Hensel's lemma, Zariski's Main theorem (as a special case, it mentions that the completion of a normal algebraic local domain is again a normal domain), links associated to the singularities of algebraic curves, various reciprocity laws from Kummer to Artin in Algebraic Number Theory etc.<|endoftext|>
TITLE: What is enough to conclude that something is a CW complex?
QUESTION [9 upvotes]: This question was something I considered when looking into CW-structures on Grassmannians, but I found no general treatment of this in the literature:
Question: Assume that $X$ is an $n-1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Assume also that $X'$ is compact Hausdorff. Is $X'$ homeomorphic to a CW complex given by attaching a single $n$ cell to $X$?
Remark: Maybe I am missing some obvious counter example!

REPLY [6 votes]: Let $X$ be the closed subspace of $R^2$ which is the union of $0\times [-1,1]$ and the point $(1, \sin(1))$.  Let $e$ be the graph in the plane of $f(x)=\sin(1/x)$ for $x\in (0,1)$ (the "topologists sine curve"), and let $X'=X\cup e$, viewed as a subspace of $R^2$.  
I believe that $X'$ is closed, and so is compact Hausdorff, $e$ is an open subset of $X'$ homeomorphic to the open interval, and $X$ is a CW-complex.  But $X'$ is not obtained by attaching a $1$-cell to $X$:  you can't produce a map $\Phi\colon[0,1]\to X'$ which restricts to a homeomorphism $(0,1)\to e$.
This is not quite a counterexample to your problem, because you wanted the new cell $e$ to have greater dimension than the cells of $X$.  But it seems like you could use this kind of idea to make a counterexample.
Added. Let $S^2$ be the unit sphere in $R^3$, pick a point $p$ on $S^2$, and consider the function $f\colon (S^2-\{p\})\to R$ given by $f(x)=\sin(1/|x-p|)$.  Let $X'$ be the closure of the graph of $f$ inside $S^2\times [-1,1]$; this should consist of $X=\{p\}\times [-1,1]$ (a CW complex) and $e=$ graph of $f$ (homeomorphic to an open $2$-disk).  This would seem to be the counterexample you want.<|endoftext|>
TITLE: Is a left invertible element of a group ring also right invertible?
QUESTION [14 upvotes]: Given a group $G$ we may consider its group ring $\mathbb C[G]$ consisting of all finitely supported functions $f\colon G\to\mathbb C$ with pointwise addition and convolution. Take $f,g\in\mathbb C[G]$ such that $f*g=1$. Does this imply that $g*f=1$?
If $G$ is abelian, its group ring is commutative, so the assertion holds. In the non-abelian case we have $f*g(x)=\sum_y f(xy^{-1})g(y)$, while $g*f(x)=\sum_y f(y^{-1}x)g(y)$, and this doesn't seem very helpful.
If $G$ is finite, $\dim_{\mathbb C} \mathbb C[G]= |G|<\infty$, and we may consider a linear operator $T\colon \mathbb C[G]\to\mathbb C[G]$ defined by $T(h) = f*h$. It is obviously surjective, and hence also injective. Now, the assertion follows from $T(g*f)=f=T(1)$.
What about infinite non-abelian groups? Is a general proof or a counterexample known?

REPLY [20 votes]: A ring is called Dedekind-finite if that property holds. Semisimple rings are Dedekind finite, so this covers $\mathbb CG$ for a finite group $G$; this is easy to do by hand. It is a theorem of Kaplansky that this also holds $KG$ for arbitrary groups $G$ and arbitrary fields $K$ of characteristic zero. See [Kaplansky, Irving. Fields and rings. The University of Chicago Press, Chicago, Ill.-London 1969 ix+198 pp. MR0269449] It is open, I think, for general fields.<|endoftext|>
TITLE: K-theory as a generalized cohomology theory
QUESTION [11 upvotes]: Which of the statements is wrong:

a generalized cohomology theory (on well behaved topological spaces) is determined by its values on a point
reduced complex $K$-theory $\tilde K$ and reduced real $K$-theory $\widetilde{KO}$ are generalized cohomology theories (on well behaved topological spaces)
$\tilde K(*)= \widetilde{KO} (*)=0$

But certainly $\tilde K\neq \widetilde{KO}$.

REPLY [17 votes]: 1 is doubly wrong.  First, you need to distinguished generalized cohomology theories and reduced generalized cohomology theories.  If you want to work with the latter, you should replace "a point" in 1 by "$S^0$", and then the corrected version of 3 no longer holds.  But even this new version 1' is false; a generalized cohomology theory is not determined by its coefficients, unless they are concentrated in a single degree (example: complex K-theory vs. integer cohomology made even periodic).<|endoftext|>
TITLE: Closed monoidal structure on the derived category of sheaves
QUESTION [5 upvotes]: Given a topological space X, i'd like to find Der X - the derived category of sheaves of abelian groups on X - to be a closed monoidal category. Hom should be cohomological and the internal-hom should be triangulated.

Is this possible in full generality? (Unbounded complexes, no restrictions on X)
Consider a sheaf of rings R or equivalently a ring of sheaves. This gives us two things: An abelian category of left R modules that we can derive; let's call this one Der R; A monoid R in Der X whose category of modules we denote dMod R. Is Der R = dMod R'? Ff not: how do they relate?
Given monoids R,S,T in Der X do we get the usual adjunctions in two variables between their categories of bimodules?
Given rings R,S,T in Sh X do we get the usual adjunctions in two variables between their derived categories of bimodules?

Now for the question: What is the right setting to do this? As i understand it, there's no suitable model structure that gives 4 in full generality.

REPLY [6 votes]: The most direct answer to question (2) is no, even in the case of sheaves on a point.  In this case we simply have a ring R and we are asking whether the derived category of R-modules is equivalent to the category of R-module objects in the derived category of abelian groups.  For example, let $R = Z/p^2$.  Then in the derived category of R-modules, we have nonzero Ext groups
$$Ext^n_{Z/p^2}(Z/p,Z/p) = Hom_{D(Z/p^2)}(Z/p,Z/p[n])$$
for all positive degrees $n$.  However, because no object in the derived category of Z-modules has any nontrivial Ext-groups in high degrees, we have for $n$ large
$$Hom_{dMod(Z/p^2)}(Z/p,Z/p[n]) = 0.$$
The problem is that dMod(R) doesn't really remember that R is an actual ring, but instead only remembers that it can be lifted to a chain complex with a chain-homotopy associative multiplication map.  These kinds of ring objects are too weak to do really serious homological algebra and define a proper derived category.
The answer to question 3 is correspondingly no, because we can't really construct proper derived functors of the tensor product given only a monoid in Der(X).<|endoftext|>
TITLE: How should I visualise RP^n?
QUESTION [26 upvotes]: So I did some algebraic topology at university, including homotopy theory and basic simplicial homology, as well as some differential geometry; and now I'm coming back to the subject for fun via Hatcher's textbook. A problem I had in the past and still have now is how to understand projective space RP^n - I just can't visualise it or think about it in any concrete way. Any ideas?
edit: Essentially RP^n is always the example I don't understand. So when for example Hatcher says that S^n is of a CW complex with two cells e^0 and e^n, I can picture what's going on because I know what spheres look like and I can imagine the attachment in some concrete-ish way. But when he says "we see that RP^n is obtained from RP^{n-1} by attaching an n-cell [...] it follows by induction on n that RP^n has a cell complex structure e^0 U e^1 U ... e^n" - my brain just gives up.

REPLY [5 votes]: A Point in $RP^n$ corresponds to a pair of antipodal points on $S^n$ - so just practice visualizing two antipodal points on a sphere every time you say Point.  Such an approach is clearly equivalent to other definitions, but I find it good for "seeing."  The standard cell structure is then defined by a Point being in the interior of the $i$-cell if and only if the first $n-i$ coordinates of the corresponding pair of points vanish.
You can also visualize Lens spaces in this way: a single point in a Lens space is seen as a collection of some $k$ points on the unit sphere in some ${\mathbb C}^n$, which are all related by multiplication by some $k$th rood of unity.<|endoftext|>
TITLE: Semistable filtered vector spaces, a Tannakian category.
QUESTION [15 upvotes]: Let $k$ be a field (char = 0, perhaps).  Let $(V,F)$ be a pair, where $V$ is a finite-dimensional $k$-vector space, and $F$ is a filtration of $V$, indexed by rational numbers, satisfying: 

$F^i V \supset F^j V$ when $i < j$.
$F^i V = V$ for $i << 0$.  $F^i V = \{ 0 \}$ for $i >> 0$.
$F^i V = \bigcap_{j < i} F^j V$.

We define:
$$F^{i+} V = \bigcup_{j > i} F^j V.$$
The slope of $(V,F)$ (when $V \neq \{ 0 \}$) is the rational number:
$$M(V,F) = \frac{1}{dim(V)} \sum_{i \in Q} i \cdot dim(F^i V / F^{i+} V).$$
The pair $(V,F)$ is called semistable if $M(W, F_W) \leq M(V, F)$ for every subspace $W \subset V$, with the subspace filtration $F_W$.  
A paper of Faltings and Wustholz constructs an additive category with tensor products, whose objects are semistable pairs $(V,F)$.  A paper of Fujimori, "On Systems of Linear Inequalities", Bull. Soc. Math. France, seems to imply that the full subcategory of slope-zero objects (together with the zero object) is Tannakian (the abelian category axioms require semistability), with fibre functor to the category of $k$-vector spaces (though Fujimori considers quite a bit more).
Does anyone know another good reference for the properties of this Tannakian category?  Can you describe the associated affine group scheme over $k$?  I'm particularly interested, when $k$ is a finite field or a local field.
Update:  I think the slope-zero requirement is too strong (though it is assumed in Fujimori).  It seems to exclude almost all the semistable pairs $(V,F)$, if my linear algebra is correct.  Anyone want to explain this to me too?

REPLY [2 votes]: I am not sure if you are still interested in this, one year later.
Check out the papers by Yves Andre (see this and this) as well as the articles of Burt Totaro and Laurent Fargues listed in the second.<|endoftext|>
TITLE: Sheaves over simplicial sets
QUESTION [12 upvotes]: Is there a good way to define a sheaf over a simplicial set - i.e. as a functor from the diagram of the simplicial set to wherever the sheaf takes its values - in a way that while defined on simplex by simplex corresponds in some natural manner to what a sheaf over the geometric realization of the simplicial set would look like?
Edited to add: I'm actually interested in the question in a concrete fashion rather than an abstract one - I'm trying to figure out whether there might be some interesting interpretation of sheaves over the nerve of a category generated by a network; similar to current work by Robert Ghrist that takes a network and views it as a graph, and thus as a topological space (1-dim simplicial complex), and manages to find useful interpretations of sheaves on this particular space in terms of network analysis.
Hence, what I'm really looking for is an interesting definition for, say, the nerve of the category generated by a finite directed graph, or so...
Edited to add: In off-channels, fpqc has clarified his argument in the answer I've accepted. Specifically, $N(C)$ for a category has inherent direction data that is lost in $|N(C)|$. This messes up attempts to formulate an idea of sheaves over $N(C)$ in a way that stays faithful to the definition of sheaves over $|N(C)|$.

REPLY [10 votes]: Clearly looking at sheaves on the geometric realisation gives something too far
removed from the simplicial picture. This is essentially because there are too
many sheaves on a simplex have (most of which are unrelated to simplicial
ideas). What one could do is to consider such sheaves which are constructible
with respect to the skeleton filtration, i.e., are constant on each open
simplex. This can be described inductively using Artin gluing. I think it
amounts to the following for a simplicial set $F$.
For each simplex $c\in F_n$ we have a set $T_c$, the constant value of the sheaf
$T$ on the interior of the simplex corresponding to $c$.
For each surjective map $f\colon [n] \to [m]$ in $\Delta$ the corresponding (degeneracy) map
on geometric simplices maps the interior of $\Delta_n$ into (onto in fact) the interior of $\Delta_m$ and
hence we have a bijection $T_{f(c)} \to T_c$. These bijections are transitive with
respect to compositions of $f$'s.
For each injective map $f\colon [m] \to [n]$ in $\Delta$ the corresponding map on geometric
simplices maps $\Delta_m$ onto a closed subset of $\Delta_n$. If $j\colon \Delta^o_n \hookrightarrow \Delta_n$ is the
inclusion of the interior we get an adjunction map $T \to j_\ast j^\ast T$ and
$j_\ast j^\ast T=T_c$ where $T_c$ also denotes the constanct sheaf with value $T_c$. If
$f'\colon \Delta^o_m \hookrightarrow \Delta_n$ is the inclusion of the interior composed with $f$ we can
restrict the adjunction map to get a map $T_{f(c)}=f'^\ast T \to f'T_c$ and taking
global sections we get an actual map $T_{f(c)} \to T_c$. These maps are transitive
with respect compositions of $f$'s.
We have a compatibility between maps coming from surjections and
injections. Unless something very funny is going on this compatibility should be
that we wind up with a function on the comma category $\Delta/F$ which takes
surjections $[n] \to [m]$ to isomorphisms.
There is the stronger condition on the sheaf $F$, namely that it is constant on
each star of each simplex. This means on the one hand that it is locally
constant on the geometric realisation, on the other hand that $T_{f(c)} \to T_c$
is always an isomorpism.
[Added] Some comments intended to give some kind of relation with the answer
provided by fpqc. My suggested answer is not homotopy invariant in the sense
that a weak (or even homotopy) equivalence of simplicial sets does not induce an
equivalence on the category of sheaves. This is so however if one, as per above,
adds the condition that all the maps $T_{f(c)} \to T_c$ are isomorphisms. However,
that condition is not so good as many maps that are not weak equivalences
induces category equivalences (it is enough that the map induce isomorphisms on
$\pi_0$ and $\pi_1$). This is a well-known phenomenon and has to do with the fact
the $T_c$ are just sets. One could go further and assume that the $T_c$ are
topological spaces and the maps $T_{f(c)} \to T_c$ continuous. Of course adding
the condition that these maps be homeomorphisms shouldn't be right thing to do,
instead one should demand that they be homotopy (or weak) equivalences. Again,
this shouldn't be quite it because of the transitivity conditions. We should not
have that the composite $T_{g(f(c))} \to T_{f(c)} \to T_c$ should be equal to
$T_{g(f(c))} \to T_c$ but rather homotopic to it. Once we have opened that can of
worms we should impose higher homotopies between repeated composites. This can
no doubt be (has been) done but there seems to be an easier way out. In the
first step away from set-valued $T_c$ we have the possibility of they being
instead categories. In that case the higher homotopy conditions is that we
should have a pseudofunctor $\Delta/F \to \mathcal{C}\mathrm{at}$. Even they are
somewhat unpleasant and it is much better to pass to the associated fibred
category $\mathcal{T} \to \Delta/F$. In the general case, and admitting that $\Delta/F$ is
essentially the same things as $F$ itself, we should therefore look at
(Serre) fibrations $X \to |F|$ or if we want to stay completely simplicial, Kan
fibrations $X \to F$. This gives another notion of (very flabby) sheaf which now
should be homotopy invariant (though that should probably be in the sense of
homotopy equivalence of $\Delta$-enriched categories).<|endoftext|>
TITLE: Number of unique determinants for an NxN (0,1)-matrix
QUESTION [18 upvotes]: I'm interested in bounds for the number of unique determinants of NxN (0,1)-matrices. Obviously some of these matrices will be singular and therefore will trivially have zero determinant. While it might also be interesting to ask what number of NxN (0,1)-matrices are singular or non-singular, I'd like to ignore singular matrices altogether in this question.
To get a better grasp on the problem I wrote a computer program to search for the values given an input N. The output is below:
1x1: 2 possible determinants
2x2: 3 ...
3x3: 5 ...
4x4: 9 ...
5x5: 19 ...
Because the program is simply designed to just a brute force over every possible matrix the computation time grows with respect to $O(2^{N^2})$. Computing 6x6 looks like it is going to take me close to a month and 7x7 is beyond hope without access to a cluster. I don't feel like this limited amount of output is enough to make a solid conjecture.
I have a practical application in mind, but I'd also like to get the bounds to satiate my curiosity.

REPLY [4 votes]: From Hadamard's bound the largest possible determinant of an $n\times n$ (0,1) matrix is $h_n=2^{-n}(n+1)^{(n+1)/2}$.   The data at http://www.indiana.edu/~maxdet/spectrum.html suggest several conjectures:

The spectrum is "dense" up to a certain point, after which it becomes "sparse".  An integer in the "dense" part is almost certain to be the determinant of some $n\times n$ (0,1) matrix; and integer in the "sparse" part is almost certain not to be.  The point at which the spectrum becomes sparse is asymptotically some constant times $h_n$.  The data suggest that the constant is near 0.5.  I think this is basically the conjecture made by Gerhard Paseman in his answer.
A stronger statement is as follows: let $g_n$ be the position of the first "gap", that is, the first positive integer that is not the determinant of some $n\times n$ (0,1) matrix.  Then asymptotically $g_n$ is some constant times $h_n$, and again the constant appears to be close to 0.5.  If $D_n$ denotes the set of positive $n\times n$ determinants, and if the above ideas are on the right track, then it seems likely that asymptotically $g_n/|D_n|$ is 1.

I don't have even a heuristic explanation as to why such conjectures ought to be true, and one can always worry about how much one should try to conclude from data that, in fact, go only as high as $n=21$.  As mentioned in some of the comments to other answers, $D_n$ is really only known for $n\le 10$ and $n=12$.  The sets $D_n$ for these $n$ are given at the above link, as are conjectures for all $n$ up to 16.  Data for $17\le n\le21$ have not yet been added to the site.  (Note that the $n$ on the web site refers to $(-1,1)$ matrices, so one should subtract 1 from it if one is talking about $(0,1)$ matrices.)  I do have high confidence that the listed sets $D_n$ up to $n=16$ are not missing any values, even the ones that have not been proved complete.<|endoftext|>
TITLE: Relations between two particular elements of SL_2(Z)?
QUESTION [5 upvotes]: Let $S_4 = \left(\begin{array}{cc}0&-1 \\\ 1&0 \end{array}\right) \textrm{ and } S_6 = \left(\begin{array}{cc} 1&-1 \\\ 1&0\end{array}\right)$. Serre proves in his book on trees that  $SL_2(\mathbb{Z}) \cong \mathbb{Z}/4 *_{\mathbb{Z}/2} \mathbb{Z}/6$, and $S_4$ and $S_6$ are the elements corresponding to the generators of $\mathbb Z/4$ and $\mathbb Z/6$ (I'm not sure if this is related to my question). Then let $a = S_4 S_6$ and $b = S_4 S_6^2$. I believe every element of $SL_2(\mathbb Z)$ can be written as $S_6^d w S_6^e$, where $w$ is a word in $a$ and $b$ but not $a^{-1}$ or $b^{-1}$.
I wrote a program (for other purposes) that seems to show that there aren't any relations between $a$ and  $b$ that have length 15 or less and don't involve $a^{-1}$ or $b^{-1}$. I'm not certain that the program is right, but if it is, one might make a naive guess that these two elements generate a free group. This makes me suspicious. 
1) Does $SL_2(\mathbb Z)$ contain a free group (of rank > 1)? If it does, is there an easy way to determine whether the subgroup generated by $a$ and $b$ is free?
2) A slightly less naive guess is that $a$ and $b$ generate a free monoid in $SL_2(\mathbb Z)$. Is there a reason why  $SL_2(\mathbb Z)$ can't contain a free monoid, or an example showing that it does?
EDIT: Thanks for the quick replies. As Robin and Jack pointed out, $a$ and $b$ generate SL(2,Z), so clearly don't generate a free group. Also, there are free subgroups that are easy to write down. I'm still curious about #2, though.

REPLY [6 votes]: For a broader perspective on proving that groups of matrices and what not are free, I highly recommend reading chapter II.B of Pierre de la Harpe's book "Topics in Geometric Group Theory", which gives a beautiful account with numerous references of known facts, especially for subgroups of SL_2.<|endoftext|>
TITLE: does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index?
QUESTION [8 upvotes]: I thought that I read a paper making this claim a few months ago, but now I can't find it.  If the answer is yes, is there a nice way to go from the presentation of the right-angled coxeter group to a presentation of its right-angled artin subgroup?  Thanks.

REPLY [13 votes]: As James points out, the paper of Davis and Januskiewicz proves the inverse.  To see that the answer to your question is 'no', consider the right-angled Coxeter group whose nerve graph is a pentagon.  That is, it's the group with presentation
$\langle a_1,\ldots, a_5 \mid a_i^2=1, [a_i,a_{i+1}]=1\rangle$
where the indices are considered mod 5.
This group acts properly discontinuously and cocompactly on the hyperbolic plane, and it's not hard to see that it has a finite-index subgroup which is the fundamental group of a closed hyperbolic surface.  Every finite-index subgroup of a right-angled Artin group is either free or contains a copy of $\mathbb{Z}^2$, but the fundamental group of a closed hyperbolic surface has no finite-index subgroups of this form.<|endoftext|>
TITLE: Is there a machinery describing all the irreducible representations ?
QUESTION [11 upvotes]: Suppose we have a finite dimensional Lie algebra $g$， Is there a machinery to describe all the irreducible representation of $g$. 
Consider toy example: $sl_{2}$ or $sl_{3}$, how do we describe all the irreducible representations of them. 
Further, consider quantum case, Is there a machinery way(like algorithm)describing all the irreducible representations of $U_{q}(sl_{2})$
EDIT: What I am looking for is an "mechanical" and canonical machinery describing all the irreducible representations(of course, not only finite dimensional representations,not only unitary representations)
EDIT2: What I am looking for is some reference to describe them in explicitly(such as $sl_{3}$）

REPLY [29 votes]: The problem of classifying irreducible $sl_2(\mathbb C)$-representations is essentially
untractable as it contains a wild subproblem. Indeed, the action of the Casimir
element $C$ on any irreducible representation is by a complex scalar (by a
theorem of Quillen I believe). If we consider the case when $C$ acts by zero, by
a result of Beilinson-Bernstein the category of $sl_2$-representations with
$C=0$ is equivalent to the category of quasi-coherent $\mathcal D_{\mathbf
  P^1}$-modules. In this $1$-dimensional case every irreducible $\mathcal
D_{\mathbf P^1}$-module is holonomic. If we restrict ourselves to irreducible
regular holonomic modules we have two possibilities. One case is that they are
supported at a single point and then the point is a complete invariant. In the
other case they are classified by a finite collection of points of $\mathbf P^1$
and equivalence classes of irreducible representation of the fundamental group
of the complement of the points which map the monodromy elements of the points
non-trivially. In particular we can consider the case of three points in which
case the fundamental group is free on two generators (they and the inverse of
their product being the three monodromy elements). The irreducible
representations where one of the monodromy elements act trivially correspond to
removing the corresponding point and thinking of the representation as a
reprentation of the fundamental group of that complement. 
Hence, we can embed the category of finite-dimensional representations of the
free group on two elements as a full subcategory closed under kernels and
cokernels of the category of $sl_2(\mathbb C)$-modules. This makes the latter
category wild in the technical sense. However, the irreducible representations
of the free group on two letters are also more or less unclassifiable.
There is no contradiction between this and the result of Block. His result gives
essentially a classification of irreducibles in terms of equivalence classes of
irreducible polynomials in a twisted polynomial ring over $\mathbb C$. So the
consequence is that such polynomials are essentially unclassifiable.
[Added] Intractable depends on your point of view. As an algebraic geometer I agree with Mumford
making (lighthearted) fun of representation theorists that think that wild problems are intractable. After
all we have a perfectly sensible moduli space (in the case of irreducible representations) or moduli stack (in the general case). One should not try to "understand" the points of an algebraic variety but instead try to understand the variety geometrically. Today, I think that this view point has been absorbed to a large degree by representation theory.<|endoftext|>
TITLE: circle action on sphere
QUESTION [5 upvotes]: surely $S^1$ can act on $S^n$ as a rotation.I want to know if there is some other way that a circle can act on sphere.

REPLY [14 votes]: Classification of circle actions on (standard and on exotic) spheres is a classical activity in transformation groups, see e.g. the article of Schultz in the collection "Group actions on manifolds", proceedings from 1983 conference, or earlier account in Bredon's book "Introduction to compact transformation groups". 
One standard example is this. Take a smooth compact contractible $(n+1)$-manifold $C$ and consider $C\times D^k$ where circle acts trivially on $C$ and linearly
on the $k$-disk $D^k$, $k>0$. Now if $n+k>4$, the boundary of $C\times D^k$ is diffeomorphic to the standard sphere (after the corners of $C\times D^k$ are rounded). But the fixed point set of the action is the original homology sphere that bounds $C$. I think it is clear that the fixed point sets of linear actions are standard spheres.
Incidentally, in dimensions $k>4$ any homology $k$ sphere bounds a contractible manifold after possibly changing its smooth structure; also any homology $4$-sphere bounds a smooth contractible manifold, while any homology $3$-sphere bounds a topological contractible manifold, but not necessarily smooth ones. References to the above results can by found e.g. in my paper pp.8-9.<|endoftext|>
TITLE: Why are tensors a generalization of scalars, vectors, and matrices?
QUESTION [30 upvotes]: Take two vector spaces $V$ and $W$ over a field $F$. One may form the tensor product $V\otimes W$ and it fulfills an universal property. Elements of $V\otimes W$ are called tensors and they are linear combinations of elementary tensors $v\otimes w$, the elementary tensors generate $V\otimes W$.
People from physics think of a tensor as a generalization of scalars, vectors, and matrices, I think and I have seen them tensoring matrices with matrices as entries with matrices and so on.
What does this mean and what has it to do with the definition from above? What "is" a tensor?

REPLY [3 votes]: The following discussion is for finite-dimensional vector spaces only:
The tenor product of two vector spaces $V$ and $W$ arises, because you want to work with bilinear, rather than just linear, maps and functions of $V \times W$. The key observation is that the space of all possible bilinear scalar-valued functions or vector-valued maps is itself a vector space that is spanned by simple bilinear functions constructed by multiplying a linear function on $V$ by a linear function on $W$.
Let $T$ denote the vector space of all scalar-valued bilinear functions on $V \times W$ and observe that there is a natural map $V^* \times W^* \rightarrow T$. Moreover, it is easy to see that $V^* \times W^* $ generates all of $T$ in the sense that its image does not lie in any proper subspace of $T$. So this leads to the notation $T = V^* \otimes W^*$
The next observation is that there is a natural bilinear map $V\times W \rightarrow T^* $ that corresponds to evaluation of the given $(v,w)$ with an element of $T$. Again, the image "spans" all of $T^* $ in the sense that it does not lie in any proper subspace. Moreover, the linear duality between $T$ and $T^* $ corresponds exactly to the evaluation of a bilinear function on an element in $V\times W$. So it is reasonable to denote $T^*$ by $V \otimes W$.
So, morally speaking, $V\otimes W$ is the "smallest" vector space such that there is a bilinear injective map $V \times W \rightarrow V\otimes W$. Of course, this statement can be made precise by defining $V\otimes W$ as a universal object.
Finally, it is not difficult to prove the rather cool (and for me not so obvious) observation that the space of linear maps $V \rightarrow W$ is naturally isomorphic to $W\otimes V^* $. In this sense, tensors generalize the matrices viewed as linear transformations.<|endoftext|>
TITLE: Is a inverse limit of compact spaces again compact ?
QUESTION [17 upvotes]: Then one can construct a model for the inverse limit by taking all the compatible sequences. 
This is a subspace of a product of compact spaces. This product is compact by Tychonoff. If all the spaces are Hausdorff, then this is even a closed subspace. 
However, if the spaces are not Hausdorff, it needn't be a closed subspace. If you take a two point space with the trivial topology as $X_n$ and constant structure maps, you will get as the inverse limit the space of all constant sequences, which is not a closed subspace of the infinite product, as the infinite product also has the trivial topology. 
But the space is again compact. So I am wondering, whether there is a generalization of the proof of Tychonoff's theorem, that applies directly to inverse limits.

REPLY [11 votes]: Although the question is quite old I just ran into it. You might still want to (if you haven´t done so yet) look at the article:
A.H. Stone, Inverse limits of compact spaces, General Topology and its Applications 10 Issue 2 (1979) pp 203–211 https://doi.org/10.1016/0016-660X(79)90008-4.
It contains some positive results.<|endoftext|>
TITLE: reduction of CM elliptic curves
QUESTION [7 upvotes]: Can someone indicate how to prove the following equivalences for a CM elliptic curve $E$:
(i) $p$ is inert in End($E$)
(ii) $E_p$ is supersingular
(iii) The trace of the Frobenius at $p$ is $0$ [correction: $\equiv 0 \pmod{p}$]
Also, are there generalisations of this to abelian varieties?

REPLY [4 votes]: Let $k$ be a finite field of char. $p$ and $E$ an elliptic curve over $k$. Denote by $R_E$ the endomorphism ring of $E$ over $k$. This ring has a distinguished element given by the purely inseparable $k$-isogeny $\pi_E:E\rightarrow E$ given by the Frobenius endomorphism relative to $k$. Its degree is $|k|$. We know that $\pi_E$ satisfies the Weil polynomial $f(x)=x^2-a_Ex+|k|$, where $a_E$ is the error term $|k|+1-|E(k)|$ (it is not hard to deduce this from the fact that $|E(k)|$ is equal to the degree of the separable $k$-isogeny $1-\pi_E$).
Assume now that $F$ is an imaginary quadratic field that embeds inside $R_E\otimes\mathbf{Q}$. If $E$ is supersingular, then there is a finite extension $k'$ of $k$ such that $R_E':={\rm End}_{k'}(E\otimes_k k')$ becomes a maximal order of "the" quaternion over $\mathbf{Q}$ ramified precisely at $p$ and infinity (in fact $k'$ of degree $2$ suffices "most times"). Considering the embedding $F\rightarrow R_E\otimes\mathbf{Q}\subset R_E'\otimes\mathbf{Q}$ you see that $F$ cannot be split at $p$ (and at infinity) for otherwise it would not embed in such a quaternion algebra (Vigneras' book has all of this basics).
If $E$ is ordinary then $R_E\otimes\mathbf{Q}$ is isomorphic to the quadratic field obtained by joining to $\mathbf{Q}$ a root of $f(x)$, therefore so is $F$ and since $p\nmid a_E$ we see easily that the discriminant of $f(x)$ is not divisible by $p$ and it is a square mod $p$, thus $p$ splits in $F$.
Back to your original question, the reduction mod $p$ of a CM ellipitic curve induce a nice, injective reduction map at the endomorphisms level. Therefore you should get what you wanted!
(Waterhouse thesis "Abelian varieties over finite fields" is a great source to learn these things. The Bourbaki talk by Tate on abelian varieties over finite fields is a must to learn these things)<|endoftext|>
TITLE: What are the worst notations, in your opinion?
QUESTION [56 upvotes]: With which notation do you feel uncomfortable?

REPLY [17 votes]: The term "symplectic group" used to mean the group $U(n,{\mathbb H})$. It's as if people called $U(n)$ and $GL(n,{\mathbb R})$ by some single name.<|endoftext|>
TITLE: Is there a known formula for the number of SSYT of given shape with partition type?
QUESTION [6 upvotes]: Let $s_{\lambda}$ and $m_{\lambda}$ be the Schur and monomial symmetric functions indexed by an integer partition $\lambda$ ($\ell(\lambda)$ is the number of parts of $\lambda$ and $m_i(\lambda)$ is the multiplicity of part $i$). By the hook-content formula we have:
$$
s_{\lambda}(1^n) = \prod_{u\in \lambda} \frac{n+c(u)}{h(u)},
$$
where $c(u)$ and $h(u)$ are the content and hook length of the cell $u\in \lambda$.
Using  $s_{\lambda} = \sum_{\mu} K_{\lambda \mu} m_{\mu}$ where $K_{\lambda \mu}$ is the Kostka number, the number of semistandard Young tableaux of shape $\lambda$ and type $\mu$. Then we get $\sum_{\mu} K_{\lambda \mu} m_{\mu}(1^n)=\prod_{u\in \lambda} \frac{n+c(u)}{h(u)}$. This counts semistandard Young tableaux of shape $\lambda$ and any type.
Does the sum $\sum_{\mu}K_{\lambda,\mu}$ have a known formula for $\ell(\lambda)\geq 2$? This would be the number of semistandard Young tableaux of shape $\lambda$ with partition type.

REPLY [9 votes]: Let $k(\lambda)=\sum_\mu K_{\lambda\mu}$. Then we have the generating function
$\sum_\lambda k(\lambda)s_\lambda = \prod_{n\geq 1} (1-h_n)^{-1}$.<|endoftext|>
TITLE: Finding all paths on undirected graph
QUESTION [18 upvotes]: I have an undirected, unweighted graph, and I'm trying to come up with an algorithm that, given 2 unique nodes on the graph, will find all paths connecting the two nodes, not including cycles. Here's an illustration of what I'd like to do: Graph example
Does this algorithm have a name? Can it be done in polynomial time?
Thanks,
Jesse

REPLY [4 votes]: Let $G=(V,E)$ be a graph.
$FindPaths(p,f)$ prints all paths which end in $f$ and can be obtained by adding nodes to path $p$. $p$ is for path, $f$ is for final (node).
Def $FindPaths(p,f)$:
Let $x$ be the last node of $p$.
For each edge $xy$ for some $y$ in $E$
$\ \ \ \ $If $y$ is not in $p$
$\ \ \ \ $$\ \ \ \ $If $y=f$
$\ \ \ \ $$\ \ \ \ $$\ \ \ \ $Print $p-y$
$\ \ \ \ $$\ \ \ \ $Else
$\ \ \ \ $$\ \ \ \ $$\ \ \ \ $$FindPaths(p-y,f)$
If $s$ is the start node and $t$ is the ending node, run $FindPaths(s,t)$.
You can represent path and edges as strings. To check if a node is in a path $p$ you just have to check whether the string contains the character that represents the node. To get the final node of a path use the function to get the last character of a string.
EDIT: My answer is not math research level, but introduction to programming.<|endoftext|>
TITLE: Counting Eulerian Orientation in a 4-regular undirected graph
QUESTION [7 upvotes]: We would like to know how hard it is to count Eulerian orientation in an undirected 4-regular graph. For a given edge orientation to be Eulerian, we mean that every vertex has 2 in-edges and 2 out-edges.
It is known that counting Eulerian orientation in undirected graphs are #P-complete. We have tried to construct some gadget to reduce the general case to 4-regular case, but did not succeed. Any idea about that?
Thank you.

REPLY [10 votes]: Let $G$ be a planar graph. Consider a medial graph $H=H(G)$, which is always $4$-regular.  Often, problems about $G$ can be translated into the language of $H$ and vice versa.  Closer to your question, the number of Eulerian orientations of $H$ is "almost"  an evaluation of the Tutte polynomial:
$$(\ast) \qquad \sum_{O} 2^{\alpha(O)} = 2\cdot T_G(3,3),$$
where the summation is over all Eulerian orientations $O$ of $H$, and $\alpha(O)$ is the number of saddle vertices (i.e. where the orientation is in-out-in-out in cyclic order).  This is due to Las Vergnas (JCTB 45, 1988).  My former student Mike Korn and I generalized this here. 
Of course, evaluations of the Tutte polynomial of planar graphs, including at ($3,3)$, are pretty much all #P-hard (with a few known exceptions), see D.J.A. Welsh, Complexity: knots, colourings and counting book (1993).  Now, there is a bijective proof of $(\ast)$, which maps orientations $O$ into certain subsets of edges of $G$.  It is possible that when you map the number of orientations without weight you still get a hard-to-compute stat. sum, which will prove what you want.<|endoftext|>
TITLE: Binary Quadratic Forms in Characteristic 2
QUESTION [19 upvotes]: One of the reasons why the classical theory of binary quadratic forms
is hardly known anymore is that it is roughly equivalent to the theory 
of ideals in quadratic orders. There is a well known correspondence
which sends the $SL_2({\mathbb Z})$-equivalence class of a form 
$$ (A,B,C) = Ax^2 + Bxy + Cy^2 $$
with discriminant 
$$ \Delta = B^2 - 4AC $$
to the equivalence class of the ideal
$$ \Big(A, \frac{B - \sqrt{\Delta}}2\Big). $$
One then checks that this gives a group isomorphism between the 
primitive forms with discriminant $\Delta$ and the class group 
of ideals coprime to the conductor of the order with discriminant $\Delta$.
If we go from forms with integral coefficients to forms with 
coefficients in a polynomial ring $k[T]$ for some field $k$, 
then everything transforms nicely as long as $k$ has 
characteristic $\ne 2$. In characteristic $2$, there is a well 
known theory of function fields (instead of $Y^2 = f(T)$,
consider equations $Y^2 + Y h(T) = f(T)$), but I have no idea
what the right objects on the forms side are. There should be 
decent objects, since, after all, calculations in the Jacobian
of hyperelliptic curves are performed essentially by composition
and reduction of forms. But I don't thinkg that $Ax^2 + Bxy + Cy^2$
will work; for example, these guys all have square discriminant.
Question: which objects (forms or something else?) correspond to 
ideals in quadratic function fields with characteristic $2$? 
Added: Thank you for the (three at this point) excellent answers.
I am now almost convinced that binary forms over $F_2[T]$ actually
do work; the reason why I thought they would not was not so much the
square discriminant but the fact that the action of SL$_2$ fixed the
middle coefficient $B$ of $(A,B,C)$, so I didn't get much of a 
reduction. But now I see that perhaps this is not so bad after all,
and that some pair $(B,?)$ will be the analog of the discriminant in 
the characteristic 2 case.
I knew about Kneser's work on composition, but was convinced (by a
lengthy article of Towber in Adv. Math.) that classical Gauss composition 
of forms over PID's, which I was interested in, is quite far away from
composition of quadratic spaces over general rings. Special thanks to
Torsten for showing how to go from quadratic spaces back to forms!

REPLY [12 votes]: Classes of binary quadratic forms over any commutative ring R (with no conditions on characteristic, PID, etc.) correspond to certain modules over quadratic extensions of R.  This theory is completely general also in the forms it covers, i.e. covers non-primitive and discriminant zero forms.  We can then restrict our attention to the cases of interest. 
For example, when R is an integral domain and the binary quadratic form has non-zero discriminant, the modules associated to the forms are exactly the ideal classes of the quadratic extension. 
The construction is as follows, simplied for your case of interest.  From a form $ax^2+bxy+cy^2$, we form a quadratic $R$-algebra $Q:=R[\tau]/(\tau^2+b\tau+ac)$ and a module $M =Rx\oplus Ry$ with $\tau x=-cy-bx$ and $\tau y=ax$.  Given a quadratic algebra and ideal, we can pick an $R$ basis $x,y$ of the ideal and then shift a generator $\tau$ of the quadratic algebra as necessary so that $\tau y$ is a multiple of $x$.  Then you can simply read off the inverse map.
What I've given above is the case when the binary quadratic form, quadratic algebra, and module are free over $R$.  For general $R$, they will only be locally free, and the above gives a construction locally on free patches.  There are also global, coordinate-free descriptions of the correspondence between binary quadratic forms and modules over quadratic algebras.  For the details I refer you to my paper
"Gauss composition over an arbitrary base" http://www.sciencedirect.com/science/article/pii/S0001870810003257<|endoftext|>
TITLE: Number of invertible {0,1} real matrices?
QUESTION [17 upvotes]: This question is inspired from here, where it was asked what possible determinants an $n \times n$ matrix with entries in {0,1} can have over $\mathbb{R}$.  
My question is: how many such matrices have non-zero determinant?  
If we instead view the matrix as over $\mathbb{F}_2$ instead of $\mathbb{R}$, then the answer is 
$(2^n-1)(2^n-2)(2^n-2^2) \dots (2^n-2^{n-1}).$
This formula generalizes to all finite fields $\mathbb{F}_q$, which leads us to the more general question of how many $n \times n$ matrices with entries in { $0, \dots, q-1$ } have non-zero determinant over $\mathbb{R}$?

REPLY [2 votes]: Lurking around MO, I found a question which is related to the second part of my question.  Namely, Greg Martin and Erick B. Wong prove that assuming that the entries of an $n \times n$ matrix are chosen randomly with respect to a uniform distribution from the set $\{-k,  \dots, k\}$, then the probability that the resulting matrix will be singular is $\ll k^{-2 + \epsilon}$.  
See this MO question (where the above paragraph is plagarized from) and also here for the link to the Martin & Wong paper.<|endoftext|>
TITLE: Uniqueness of Chern/Stiefel-Whitney Classes
QUESTION [11 upvotes]: This question is closely related to  this previous question.
Chern and Stiefel-Whitney classes can be defined on bundles over arbitrary base spaces.  (In Hatcher's Vector Bundles notes, he uses the Leray-Hirsch Theorem, which appears to require paracompactness of the base space.  The construction in Milnor-Stasheff works in general, as does the argument given by Charles Resk in answer to the above question.  A posteriori, this actually shows that Hatcher's construction works in general too, since he really just needs $w_1$ and $c_1$ to be defined everywhere.)   
The proof of uniqueness (as discussed in Milnor and Stasheff, or in Hatcher's Vector Bundles notes, or in the answers to the above question) relies on the splitting principle, and hence (it seems to me) requires the existence of a metric on the bundle in question.  More precisely, if we have two sequences of characteristic classes satisfying the axioms for, say, Chern classes, and we want to check that they agree agree on some bundle $E\to B$, the method is to pull back $E$ along some map $f: B'\to B$ (with $f^*$ injective on cohomology) so that $f^*E$ splits as a sum of lines.  Producing the splitting seems to require a metric on $E$ (or at least on $f^*E$).
If $B$ is not paracompact, bundles over $B$ may not admit a metric (and may admit a classifying map into the universal bundle over the Grassmannian), so my question is:
Are Chern and/or Stiefel-Whitney classes unique for arbitrary bundles?  If not, do $w_1$ and $c_1$ at least determine the higher-dimensional classes?

REPLY [3 votes]: You do not need to have metric to have splitting principle. Let $p: E \to B$ be a $n$-dimensional vector bundle and as usual let $E_0$ consist of nonzero vectors. Then consider the $\bar{E}= E_0/ \sim$ where $\sim$ identifies any two vectors in the fiber if they lie on the same line passing through origin(which just requires vector space structure)(Milnor and Stasheff describe this as defining fiber of new bundle as the quotient vector space $F/(\mathbb{C}v), v \neq 0$ pg157). Then $\bar{p}: \bar{E} \to B$ defines the associated projective bundle. We obtain splitting map by repeating this $n$-times.  
But the problem with non-paracompact space $B$ is that it may not admit a classifying map $f : B \to Gr_n$ such that $f^{\ast}\gamma^n \cong E$(bundle isomorphism) where $p : E \to B$ is $n$-plane bundle. Since for a given cover there is no locally finite refinement, in the local trivializations of vector bundle $p: E \to B$ we cannot use the Urysohn lemma properly to get a bundle map $f' :E \to E(\gamma^n)$ which restricts to linear map in each fiber as Milnor-Stasheff do in the proof of lemma 5.3. So we cannot prove theorem 5.6(Milnor-Stasheff) for non-paracompact spaces and cannot define $w_1,c_1$(characteristic classes) as they define in page 69. 
In the case of long line there is a paper showing that its tangent bundle is nontrivial without using any characteristic classes.<|endoftext|>
TITLE: How does this geometric description of the structure of PSL(2, Z) actually work?
QUESTION [7 upvotes]: There is a beautiful way to see that the congruence subgroup $\Gamma(2)$ is free on two generators: the action of $\Gamma(2)$ on $\mathbb{H}$ is free and properly discontinuous, and there is a modular function $\lambda$ with respect to $\Gamma(2)$ coming from Legendre normal form such that $\mathbb{H}/\Gamma(2) \xrightarrow{\lambda} \mathbb{C} - \{ 0, 1 \}$ is an isomorphism.  (Details.)  It follows that $\mathbb{H}$ is the universal cover of $\mathbb{C} - \{0, 1 \}$, hence that $\Gamma(2)$ is isomorphic to the fundamental group of $\mathbb{C} - \{0, 1 \}$.
However, the action of $\Gamma(1) \simeq PSL_2(\mathbb{Z})$ on $\mathbb{H}$ is not properly discontinuous free; there are problems, which maybe I should call "ramification," at the points $i, e^{ \frac{\pi i}{3} }, e^{ \frac{2\pi i}{3} }$.  This is supposed to be responsible for the fact that $\Gamma(1)$ is not free, but is instead the free product of a cyclic group of order $2$ and a cyclic group of order $3$, where the former somehow comes from the behavior at $i$ and the latter somehow comes from the behavior at the sixth roots of unity.  That's what I've been told, anyway, but I don't know how the argument actually goes.  What general context does it fit into?  (Is "monodromy" a keyword here?)

REPLY [10 votes]: The key word here is "Bass-Serre Theory" -- using the action on the hyperbolic plane, you can easily cook up a nice action of $PSL_2(\mathbb{Z})$ on a tree.  This is all described nicely in Serre's book "Trees".
EDIT: Let me give a few more details.  It turns out that a group $G$ splits as a free produce of two subgroups $G_1$ and $G_2$ if and only if $G$ acts on a tree $T$ (nicely, meaning that it doesn't flip any edges) with quotient a single edge $e$ (not a loop) such that the following holds.  Let $e'$ be a lift of $e$ to $T$ and let $x$ and $y$ be the vertices of $e'$.  Then the stabilizers of $x$ and $y$ are $G_1$ and $G_2$ and the stabilizer of $e'$ is trivial.
If you stare at the fundamental domain for the action of $PSL_2(\mathbb{Z})$ on the upper half plane, then you will see an appropriate tree staring back at you.  There is a picture of this in Serre's book.
EDIT 2: This point of view also explains why finite-index subgroups $\Gamma$ of $PSL_2(\mathbb{Z})$ tend to be free.  If you restrict the action on the tree $T$ to $\Gamma$, then unless $\Gamma$ contains some conjugate of the order 2 or order 3 elements stabilizing the vertices, then $\Gamma$ will act freely.  This means that the quotient $T/\Gamma$ will have fundamental group $\Gamma$.  Since $T/\Gamma$ is a graph, this implies that $\Gamma$ is free.<|endoftext|>
TITLE: Set theory in practice
QUESTION [15 upvotes]: This is more of a philosophy/foundation question.
I usually come across things like "the set of all men", or for example sets of symbols, i.e. sets of non-mathematical objects.
This confuses me, because as I understand it, the only kind of objects that exists in set theory are sets. It doesn't make sense to speak of other objects unless we have formalized them in terms of sets. So what to do with something like the set of all men? Are we working with a different set theory, a naive one? Or is it that we are omitting the formalization, because it is straightforward (e.g. assign a number to every man)?

REPLY [22 votes]: You don't have to define your objects as sets, in fact, you should avoid such unnatural definitions. I don't think a number theorist would be happy to see a proof referring to elements of a natural number or using the identity $1=\{0\}$. Such proofs are not acceptable because they won't survive even the slightest change in the foundations.
Similarly, if you develop Euclidean geometry, you don't define a point as a two-element set whose first element is a Dedekind cut and the second one is another weird set. You rather begin with axioms (either Euclidean ones or some axiomatization of the real line) and build the geometry on these. The set theory comes in if you want to show that your theory is consistent (as long as ZF is), and you do that by building a model within ZF.
In your example with men, your actually create a mathematical model of whatever you want to study, in the same way as physics does. There are always translation steps from real world to mathematics and back, they just happen to be trivial in this case. So it's not a problem that men are (modelled by) sets.
Only if you like to believe that mathematical objects do exist in some metaphysical sense, you will have a problem with the counter-intuitive claim that everything is a set. But you can just remove this axiom and stay agnostic about whether everything is a set or not. You will not lose anything - the only essential use of this axiom is that you are able to define the notion of equality rather than having it built into logic. And this is hardly of any importance outside the logic itself.<|endoftext|>
TITLE: PD3 groups and PD4 complexes
QUESTION [5 upvotes]: I am interested at the moment in what groups can occur as the fundamental group of a 4-manifold (or more generally, a 4-dimensional CW complex) with prescribed conditions on the intersection form.  I have what I am hoping is a basic homotopy theory question:
A (orientable) PD-$n$ group is a group $G$ such that the Eilenberg-Maclane space $K(G,1)$ admits "Poincare duality", i.e. there is an $n$-dimensional integer homology class in $K(G,1)$ (thought of as the "fundamental class") such that cap product with it yields an isomorphism between the corresponding cohomology and homology groups (like for closed oriented manifolds). This is more general than saying that $K(G,1)$ admits the structure of an orientable closed manifold of dimension $n$.
Let $G$ be a PD-3 group. Is there any reason why $G$ cannot be the fundamental group of an orientable PD4 complex $X$ with vanishing second homotopy group, $\pi_2(X)=0$?

REPLY [2 votes]: Suppose that an infinite group $G$ is the fundamental group of a non-aspherical compact $PD(4)$ complex $X$ with $\pi_2(X)=0$. Then $G$ has to have at least 2 ends (since $H_3(\tilde{X})\ne 0$, where $\tilde{X}$ is the universal cover of $X$). Since $PD(3)$ groups are 1-ended, they are never fundamental groups of $PD(4)$ complexes $X$ with $\pi_2(X)=0$. 
Furthermore, conjecturally, if $G$ is torsion-free and is isomorphic to the fundamental group of a  compact $PD(4)$ complex $X$ with $\pi_2(X)=0$, then $G$ splits a free product of infinite cyclic groups and $PD(4)$ groups.<|endoftext|>
TITLE: Continuous function from $[0,1]$ to $[0,1]$
QUESTION [10 upvotes]: Does there exist a continuous function $f:[0,1]\rightarrow [0,1]$ such that $f$ takes every value in $[0,1]$ an infinite number of times?

REPLY [25 votes]: Yes.  In fact, there exists such an $f$ taking every value uncountably many times. 
Take a continuous surjection $g: [0, 1] \to [0, 1]^2$.  (Such things exist: they're space-filling curves.)  Then the composite $f$ of $g$ with first projection $[0, 1]^2 \to [0, 1]$ has the required property.

REPLY [17 votes]: Take a projection of a well-known Peano curve, which is a surjective continuous mapping $[0,1] \to [0,1]^2$ to a factor.<|endoftext|>
TITLE: Cohomology rings of $ GL_n(C)$, $SL_n(C)$
QUESTION [7 upvotes]: Can anyone provide me with the reference for the following fact
(idea of the proof will be appreciated too):
Cohomology ring with $\mathbb Q$-coefficients of a group $G$ (I don't know precisely what the assumptions are: reductive complex algebraic group or maybe complex Lie group G with some restrictions. The cases I'm interested in are $GL_n(\mathbb C)$ and $SL_n(\mathbb C)$) is the exterior algebra on the generators of odd degrees, with the number of generators equal to the rank of $G$.
This fact is attributed to H.Hopf, but I wasn't able to find a reference. 
Thanks.

REPLY [11 votes]: If $G$ is a connected Lie group (or just a connected loop space with finite homology) then $H^*(G,\mathbf{Q})$ is a Hopf algebra. Graded connected Hopf algebras over $\mathbf{Q}$ are always tensor products of exterior algebras in odd degrees with polynomial algebras in even degrees. Since polynomial algebras are infinite, they can't occur. The reference to Hopf is probably H. Hopf, Über die algebraische Anzahl von Fixpunkten, Math. Z. 29 (1929), 493–524. For the classification of Hopf algebras, see also A. Borel, Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de
groupes de Lie compacts, Ann. of Math. (2) 57 (1953), 115–207.<|endoftext|>
TITLE: j-invariant of a supersingular elliptic curve
QUESTION [6 upvotes]: Let E be a supersingular curve over a finite field. Why is the j-invariant always in F_p^2?

REPLY [7 votes]: (Note: the following argument uses the fact that an isogeny of elliptic curves is inseparable iff it factors through the Frobenius isogeny.  This is a result in Silverman's book, for instance.)
Let $E$ be an elliptic curve over an algebraically closed field $k$ of positive characteristic $p$.  Recall that $[p]: E \rightarrow E$ is always an inseparable isogeny.  Therefore, by the above, it factors through $F: E \rightarrow E^p$.  Moreover $E$ is supersingular iff $E[p](k) = 0$ iff $[p]$ is purely inseparable, iff the dual isogeny to Frobenius $V: E^p \rightarrow E$ (the "Verschiebung") is also inseparable.  But again, this means that $V$ factors through the Frobenius isogeny for $E^p$ -- i.e., $E^p \rightarrow E^{p^2}$ -- and since both have degree $p$ this means that $E$ is isomorphic to $E^{p^2}$.  Thus on $j$-invaraiants we have $j(E)^{p^2} = j(E)$, done.

REPLY [5 votes]: In characteristic $p$, every map $E_1 \to E_2$ factors as a power of the Frobenius $\varphi_r \colon E_1 \to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} \to E_2$, and we find $r$ by looking at the inseparable degree of our map (if the map is separable, then $r=0$, as Pete pointed out).
Now, in the case of interest, if $E$ is supersingular, $\widehat{\varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being purely inseparable). But then $\widehat{\varphi} \colon E^{(p)} \to E$ factors as $E^{(p)} \to E^{(p^2)} \to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism.
It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$.<|endoftext|>
TITLE: Homotopy type of set of self homotopy-equivalences of a surface
QUESTION [14 upvotes]: Let $\Sigma$ be an oriented topological surface.  For simplicity, assume that the genus of $\Sigma$ is at least $2$.  There are a number of classical results on the homotopy types of various groups of self-maps of $\Sigma$:
1) Earle and Eells proved that the components of $\text{Diff}(\Sigma)$ are contractible.
2) Hamstrom proved that the components of $\text{Homeo}(\Sigma)$ are contractible.
3) Peter Scott proved that the components of $\text{Homeo}^{\text{PL}}(\Sigma)$ are contractible, where by $\text{Homeo}^{\text{PL}}(\Sigma)$ we mean the group of PL self-homeomorphisms of $\Sigma$.
Of course, in 1 and 3 we are fixing a $C^{\infty}$ or $\text{PL}$ structure on $\Sigma$, respectively.
Another standard fact is that every self homotopy-equivalence of $\Sigma$ is homotopic to a homeomorphism.  This leads me to my question.  Denote by $\text{HE}(\Sigma)$ the set of self homotopy-equivalences of $\Sigma$.  Are the components of $\text{HE}(\Sigma)$ contractible?
Another related question is as follows.  There is a beautiful alternate proof of the above theorem of Earle and Eells due to Earle and McMullen (see their paper "Quasiconformal Isotopies"; their proof uses complex analysis).  The proofs of Hamstrom's theorem and Scott's theorem are very complicated -- are there any alternate approaches to them in the literature?

REPLY [15 votes]: A couple comments. For the result about diffeomorphism groups there is a very nice alternative proof due to A. Gramain in the Annales Scient. E.N.S. v.6 (1973), pp. 53-66, that uses no analysis, just basic differential topology. Another approach, which I'm not sure is written down anywhere in detail, is to take the proof of the corresponding result for Haken 3-manifolds (due independently to Ivanov and myself) and scale it back from 3-manifolds to surfaces. This too uses no analysis, just basic topology.  I don't recall Scott's method for the PL case, but it might be similar to this.
For the result about homotopy equivalences it's best to look first at the space of homotopy equivalences that fix a basepoint. This has contractible components for any $K(\pi,1)$ space, by an elementary obstruction theory argument. (Just deform families of maps to the identity map, cell by cell, which is possible since the obstructions lie in the higher homotopy groups of the space.) By evaluating arbitrary homotopy equivalences at the basepoint one gets a fibration where the total space is the space of all homotopy equivalences, the base space is the $K(\pi,1)$ space, and the fiber is the earlier space of basepoint-preserving homotopy equivalences.  Looking at the long exact sequence of homotopy groups for this fibration then gives the desired result.  Triviality of the center of $\pi$ comes in at this point to show that the boundary map from $\pi_1$ of the base to $\pi_0$ of the fiber is injective.<|endoftext|>
TITLE: Existence of Fermi coordinates on a Riemannian manifold
QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold, $p$ a point on the manifold and $v \in T_p M$.  Let $\gamma$ be the geodesic starting at $p$ in the direction $v$.  There exists a time $t_f$ such that there exists a Fermi coordinate system adapted to $\gamma$ up to time $t_f$.  
My question: Does there exist a lower bound for $t_f$ in terms of the 2-jet of $g$ at $p$?
That is, I have solid estimates on $g$ up to its second derivatives at $p$:  $$\|g\| + \|\nabla g\| + \|\nabla^2 g\| \le h \qquad (1)$$ for some $h$.  I would like to show that there exists $f(h)$ such that $$t_f \ge f(h)$$ for all Riemannian metrics satisfying (1) at $p$.
Edit (Mar 19):  Taking the helpful advice of Anton, Deane, TK and Willie into account, I've reworded the question:
Let $U = B(0,r)$ be the closed Euclidean ball of radius $r$ in $\mathbb R^n$.  Write $$\lambda = \inf_{x \in U} \inf_{\|v\|=1} \langle v,v \rangle_{g(x)}$$ as the minimum eigenvalue of the metric in $U$, and suppose that $$\frac{1}{\lambda} + \|g\|_{C^2(U)} \le h.$$  This is a more refined version of (1) above.  Since $$\ddot \gamma^k = -\Gamma^k_{ij} \dot \gamma^i \dot \gamma^j,$$ our estimate gives a control on the acceleration of a geodesic $\gamma$ in $U$, so there exists a minimum self-intersection time $t_i$ depending on $h$ and $r$ (i.e., if $t, t' \le t_i$ then $\gamma(t) \ne \gamma(t')$).  
Does this imply the existence of a uniform lower bound on $t_f$ (depending only on $r$ and $h$)?  If so, can we relax the control on the second derivative of $g$?
More succinctly:  Are existence and non-self-intersection of a geodesic the only obstructions to the existence of Fermi coordinates?

REPLY [3 votes]: Let me start with definitions:  Given a submanifold $N$ in a Riemannian manifold $(M,g)$, one can consider exp as a map from the normal bundle of $N$ to $M$.  If $N$ is compact (possibly with boundary), then for small values of $t$, exp is a diffeomorphism from the normal disk bundle of radius $t$ to a neighborhood of $N$.  (I believe that this diffeomorphism is more or less what is meant by "Fermi coordinates" or "Gaussian coordinates."  Note that when $N$ is a point, these are just normal coordinates.)  I'll define the injectivity radius of $N$ to be the sup over all such $t$ that give you a diffeomorphism.  This is reasonable terminology, since I believe that (just as in the case when $N$ is a point) injectivity of exp restricted on the normal disk bundle of radius $t$ implies that exp is a diffeo (i.e. that there are no "focal points" or "conjugate points").
Proposition:  Suppose that the sectional curvature of $(M,g)$ is bounded above by $K$, and that the injectivity radius of each point in $M$ is bigger than $R_0$.  Then for any minimizing geodesic $\gamma$, the injectivity radius of $\gamma$ is at least as big as $R_1=\min(R_0,\pi/2\sqrt{K})$.
Proof:  Non-injectivity of exp means that you have two geodesics shorter than $R_1$ that are normal to $\gamma$ and intersect each other.  This gives you a geodesic triangle in $M$.  Consider a comparison triangle with the corresponding lengths in the sphere of curvature $K$.  Since two of the sides are shorter than $\pi/2\sqrt{K}$, the comparison triangle must have an acute angle on the edge corresponding to $\gamma$.  But this violates Toponogov's Theorem since our original triangle has right angles on the $\gamma$ edge.
The proof above is pretty much applicable to your situation.  You can prove a lower bound on the injectivity radii of points that are not too close to the boundary of your chart.  Similarly, if the endpoints of your geodesic are far away enough from the boundary, then your geodesic is minimizing.  Finally, the bounds on your metric certainly give you an upper bound on curvature.  (Moreover, you do not really need the full-strength version of Toponogov.)<|endoftext|>
TITLE: Sylow Subgroups
QUESTION [20 upvotes]: I had been looking lately at Sylow subgroups of some specific groups and it got me to wondering about why Sylow subgroups exist.  I'm very familiar with the proof of the theorems (something that everyone learns at the beginning of their abstract algebra course) -- incidentally my favorite proof is the one by Wielandt -- but the statement of the three Sylow theorems still seems somewhat miraculous.  What got Sylow to imagine that they were true (especially the first -- the existence of a sylow subgroup)?  Even the simpler case of Cauchy's theorem about the existence of an element of order $p$ in a finite subgroup whose order is a multiple of $p$ although easy to prove (with the right trick) also seems a bit amazing.   I believe that sometimes the hardest part of a proving a theorem is believing that it might be true.  So what can buttress the belief for the existence of Sylow subgroups?

REPLY [3 votes]: An extension of the Vipul's ideas can be found in the article (couldn't find a link to the pdf with google)
Subgroup complexes by Peter Webb,
pp. 349-365 in: ed. P. Fong, The Arcata Conference on Representations of Finite Groups, AMS Proceedings of Symposia in Pure Mathematics 47 (1987).
But as Mariano already commented, the analogy to the maximal unipotent subgroups of the general linear group was probably not Sylow's motivation. As commented before, he was maybe looking for maximal $p$-subgroups (i.e., maximal with respect to be a $p$-subgroup).
This is also the leitmotif of my favorite proof of the Sylow theorems given by Michael Aschbacher in his book Finite Group Theory. It is based on Cauchy’s theorem (best proved using J. H. McKay’s trick to let $Z_p$ act on the set of all $(x_1, \dots, x_p) \in G^p$ whose product is $1$ by rotating the entries) and goes essentially like this:
The group $G$ acts on the set $\mathrm{Syl}_p(G)$ of its maximal $p$-subgroups by conjugation. Let $\Omega$ be a (nontrivial) orbit with $S\in\Omega$. If $P$ is a fixed point of the action restricted to $S$ then $S$ normalizes $P$ and $PS=SP$ is a $p$-group. Hence $P=S$ by maximality of both $P$ and $S$, and $S$ has a unique fixed point. As $S$ is a $p$-group, all its orbits have order $1$ or a multiple of $p$, in particular $|\mathrm{Syl}_p(G)| = 1 \bmod p$. All orbits of $G$ are disjoint unions of orbits of $S$ proving $\Omega = 1 \bmod p$ and $\Omega' = 0 \bmod p$ for all other orbits $\Omega'$ of $G$. This implies that $\Omega = \mathrm{Syl}_p(G)$, as $\Omega$ was an arbitrary nontrivial orbit of $G$, showing that the action of $G$ is transitive. The stabilizer of $S$ in $G$ is its normalizer $N_G(S)$, and as the action is transitive $|G:N_G(S)| = |\mathrm{Syl}_p(G)| = 1 \bmod p$. It remains to show that $p$ does not divide $|N_G(S):S|=|N_G(S)/S|$. Otherwise, by Cauchy’s theorem there exists a nontrivial $p$-subgroup of $N_G(S)/S$ whose preimage under the projection $N_G(S) \to N_G(S)/S$ is $p$-subgroup properly containing $S$ contradicting the maximality of $S$.<|endoftext|>
TITLE: Linear disjointness of subfields of a centrally finite division algbera
QUESTION [6 upvotes]: I am looking for papers or books which discuss this problem. Thank you for reading:
Let K and L be two subfields of a non-commutative division algebra D with the center Z. Suppose that K and L contain Z and that D is finite dimensional over Z. Let V be the tensor product of K and L over Z, viewed as a vector space over Z. Consider the linear map g from V to D defined by g(x \otimes y) = xy, for every x in K and y in L. My question is this:
Under what condition(s) on K and L the map g is injective?
What is clear is that the intersection of K and L has to be Z. But that is not sufficient in general. Would this necessary condition be also sufficient if K and L were maximal subfields?

REPLY [2 votes]: If K and L are maximal subfields, then most of the time one would expect the map g to be injective, but not always. I will sketch how to construct examples of this form for which g is not injective.
We will take D to be a cyclic division algebra of index greater than two. Explicitly, let E/F be a cyclic Galois extension of degree n, with Galois group generated by $\sigma$. Let $E[x]_{\sigma}$ be the twisted polynomial ring, with multiplication defined by $xa=(\sigma a)x$.
Pick $t\in F^\times$ and let D be the quotient of $E[x]_{\sigma}$ by the central element xn-t. Let K be obvious copy of E in D, and let L be the conjugate of K by an element of the form $l_0+l_2x^2+\cdots+l_{n-1}x^{n-1}$ with all $l_i$ non-zero (yes, there is deliberately no linear term. This is the crux of the construction). I claim that this gives the desired counterexample.
To show this, it suffices to tensor with E, and show that the product map from $(K\otimes E)\otimes(L\otimes E)$ to $D\otimes E=M_n(F)$ is not injective. K embeds via D into Mn(F) via $a\mapsto diag(a,\sigma a,\ldots,\sigma^{n-1}a)$ and hence $K\otimes E$ is the space of diagonal matrices. Pick the obvious basis k1,...,kn of this space, and let l1,...,ln be the conjugate basis of $L\otimes E$.
Then (exercise), one can find indices i and j such that $k_il_j=0$. This yields the desired non-injectivity.<|endoftext|>
TITLE: Mathematical symbols, their pronunciations, and what they denote: Does a comprehensive ordered list exist?
QUESTION [14 upvotes]: Often, certain symbols in mathematics denote different things in different fields.  Is there any sort of ordered list that will tell you what a certain symbol means in alphabetical order by the symbol's alias in LaTeX, perhaps with the way to pronounce it out loud?  
I'm thinking of something like this Wikipedia page but more comprehensive and usefully ordered by LaTeX alias (The one on wikipedia has very few symbols, and I am familiar with all of them already).  The problem is that when you want to find the meaning of a symbol, there is no way to search on google (because google has no support at all for searching for symbols).  Oftentimes, I'm forced to ask someone around the department what it means or how to say it out loud.  
For example, I'm trying to find the meaning of the symbol $\uplus$, but I have no way of finding out what it means.  Also, for the longest time, I couldn't figure out what to call $f_!$ or $f^!$.  How should I know that they're called "f lower shriek" or "f upper shriek".  
So for the question: Does any such list exist for either pronunciation, meaning, or both (aside from the one on Wikipedia that I just noted)?

REPLY [6 votes]: Relatedly, the Detexify utility can help you find the LaTeX name for a symbol you can draw.<|endoftext|>
TITLE: Is model structure on CatSet unique? 
QUESTION [9 upvotes]: On the  category CatSet of usual set based categories, 
there is a "folk" model structure, as described on the first page of

Model structures for homotopy of internal categories
by T. Everaert, R.W. Kieboom and T. Van der Linden. Namely: in
CatSet, ws are weak equivalences, cs are functors injective on
objects, fs are functors with the lifting property for isomorphisms.
wfs are then precisely the full faithful functors surjective on
objects.
Is there's any nice sense in which this model 
category structure on CatSet is unique?

REPLY [13 votes]: As has been pointed out above, there are many possible model structures on Cat with different weak equivalences. This is the only proper model structure on Cat for which the weak equivalences are the categorical equivalences.<|endoftext|>
TITLE: Integer points (very naive question)
QUESTION [15 upvotes]: Well, I don't have any notion of arithmetic geometry, but I would like to understand what arithmetic geometers mean when they say "integer point of a variety/scheme $X$" (like e.g. in "integer points of an elliptic curve").
Is an integer point just defined as a morphism from $Spec\mathbb{Z}$ into $X$?
Suppose $X$ is given a structure of variety over, say, $Spec\mathbb{C}$; does the notion of integer point interact with this structure?
How is all of this related to finding the integer solutions of a concrete polynomial equation (perhaps with integer coefficients)?
Now a very, very, naif question. Suppose you have the plane $\mathbb{A}^{2}_{\mathbb{C}}$ and draw two lines on it: $X=\{x=0\}$ and $X'=\{x=\pi\}$ (with the obvious reduced induced closed subscheme structure). Well, $X$ and $X'$ are clearly isomorphic as schemes over $\mathbb{C}$ (hence as schemes). But, if having integer points is somehow related to finding integer solutions to equations, how do you explain that $X$ (as embedded in $\mathbb{A}^{2}$) has plenty of points with integer coordinates, while $X'$ has none? 
This to mean: how can the intrinsic notion of a morphism from $Spec\mathbb{Z}$ be possibly related to finding solutions to concrete (coordinate dependent) equations?

REPLY [22 votes]: To make sense of the notion of integer points, your scheme should be defined over $\mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(\mathbb{Z})$, since every scheme has one such map. The right notion is the following.
Let $X$ be a scheme over $\mathbb{C}$; so by definition we have a structure map $X \to \mathop{Spec} \mathbb{C}$. Then we say that $X$ is defined over $\mathbb{Z}$ is there exists a scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ such that $X$ is the base change of $X_{\mathbb{Z}}$ to $\mathbb{C}$, i. e. $X \cong X_{\mathbb{Z}} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$.
Now for such a scheme an integral point is a map $\mathop{Spec}\mathbb{Z} \to X_{\mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $\mathbb{Z}$.
With this definition, the line $\{ x = 0 \}$ is defined over $\mathbb{Z}$, but the line $\{ x = \pi \}$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore.
EDIT: Abstractly of course the two lines are isomorphic over $\mathbb{C}$, so the line $r = \{x = \pi \}$ actually has a model over $\mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $\mathbb{A}^2$, that is, there will be no map $r_{\mathbb{Z}} \to \mathbb{A}^2_\mathbb{Z}$ whose base change is the inclusion of $r$ into $\mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $\mathbb{A}^2$ should be generated by polynomials with integer coefficients.
As for your second question, there can be different models, that is, nonisomorphic schemes over $\mathbb{Z}$ which become isomorphic after base change to $\mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model.
For instance take the two conics $\{ x^2 + y^2 = 2 \}$ and $\{ x^2 + y^2 = 3 \}$. Both have an obvious choice of a model, given by the inclusion in $\mathbb{A}^2$; moreover they are isomorphic over $\mathbb{C}$. But the integral points on the first one are $(\pm 1, \pm 1)$, while the second has none.
Finally you consider the possibility that the structure over $\mathbb{C}$ is not relevant. This is false: the base change $X_\mathbb{Z} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$ is endowed with a natural map to $\mathop{Spec} \mathbb{C}$, and we ask for the isomorphism with $X$ to be over $\mathbb{C}$.

REPLY [12 votes]: Is an integer point just defined as a morphism from Spec ℤ to X?

Yes, but in order for there to be any, your scheme has to be genuinely "over ℤ", that is, equipped with a surjective map to Spec ℤ.  To see what a difference this makes, consider affine space $\mathbb{A}^1_{\mathbb{Q}} = \textrm{Spec} \mathbb{Q}[t]$.  A morphism from Spec ℤ to $\mathbb{A}^1_\mathbb{Q}$ is a homomorphism $\mathbb{Q}[t] \to \mathbb{Z}$, but there aren't any (the image of such a homomorphism would have to contain a copy of ℚ).
So to talk about "integer points" of $\mathbb{A}^1_\mathbb{Q}$, you need to choose a model over ℤ, that is, a scheme $\mathcal{X}$ over Spec ℤ with an isomorphism $\mathcal{X} \times_{\textrm{Spec} \mathbb{Z}} \textrm{Spec} \mathbb{Q} \cong \mathbb{A}^1_\mathbb{Q}$; then you can talk about integer points of $\mathcal{X}$.  A likely candidate is $\mathbb{A}^1_\mathbb{Z} = \textrm{Spec} \mathbb{Z}[t]$, which does indeed have a ℤ-valued point for each element of ℤ, as you might expect.  Now you can call a point of $\mathbb{A}^1_\mathbb{Q}$ an integer point if it extends to a ℤ-valued point of $\mathcal{X}$.  The notion of "integer point" you get totally depends on the model you choose; it's not intrinsic to the original scheme over ℚ.
What makes this a bit more confusing is that we're used to dealing with projective varieties; and, for proper schemes $\mathcal{X}$ over ℤ, every ℚ-valued point extends to a ℤ-valued point; this is an easy exercise using the valuative criterion of properness.  For projective varieties this is fairly obvious: for example, any ℚ-valued point of $\mathbb{P}^1_\mathbb{Q}$ can be represented as $(x:y)$ with x, y integers, so extends to a ℤ-valued point of $\mathbb{P}^1_\mathbb{Z}$.
To see how these two notions fit together, imagine $\mathbb{A}^1_\mathbb{Z}$ embedded in $\mathbb{P}^1_\mathbb{Z}$ as usual; the complement is a divisor D corresponding to the ℤ-point $(1:0)$.  (I'm assuming that you are familiar with what schemes over ℤ look like – spend a while studying the nice pictures in Mumford's Red Book if not.)  A point $(x:y) \in \mathbb{P}^1_\mathbb{Q}(\mathbb{Q})$ lies in $\mathbb{A}^1_\mathbb{Q}(\mathbb{Q})$ if and only if $y \neq 0$.  Choosing $x$ and $y$ to be integers, we extend this point to a point $(x:y) \in \mathbb{P}^1_\mathbb{Z}(\mathbb{Z})$.  Now $(x:y)$ comes from an integer point of $\mathbb{A}^1_\mathbb{Z}$ if and only if the 1-dimensional subscheme of $\mathbb{P}^1_\mathbb{Z}$ corresponding to the point $(x:y)$ doesn't meet the divisor D.  For any prime p, if $p \mid y$ then $(x:y)$ intersects D in the point $(1:0) \in \mathbb{P}^1_{\mathbb{F}_p}$.  So $(x:y)$ lies in $\mathbb{A}^1_\mathbb{Z}$ if any only if no prime divides $y$, and so $y = \pm 1$, which is exactly as you would hope.<|endoftext|>
TITLE: Choice function for Borel sets?
QUESTION [8 upvotes]: Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. We don't want to invoke the axiom of choice. Clearly we require $\emptyset\notin S$.
For example, if $S$ is the set of non-empty open sets for the usual topology, then we can fix an enumeration of the rationals and for every open $A$ pick the first rational (in this particular enumeration) lying in $A$. 
If $S$ is the set of non-empty closed sets, then for any $A\in S$ we can consider the least $n$ such that $\[-n,n\]\cap A$ is non empty and then pick the infimum of this non empty compact set.
The question is the following: can you define a choice function for, say, $S=F_{\sigma}\setminus \{\emptyset\}$ or $G_{\delta}\setminus\{\emptyset\}$ or maybe for the higher levels of the Borel hierarchy? Is it possible to prove that such choice function exists for such $S$ without using the axiom of choice?

REPLY [6 votes]: As Simon pointed out, you can't do this without some use of choice. However, you can "almost" do it: the trick is to work with codes for Borel sets instead of actual Borel sets (when you can).
Let $S \subseteq \mathbb{R}\times\mathbb{R}$ be a universal analytic set, i.e. $S$ is analytic and for every analytic set $A \subseteq \mathbb{R}$ there is a $x$ such that $A = S_x = \{ y \in \mathbb{R} : (x,y) \in S \}$. By the Jankov–von Neumann Uniformization Theorem (which is provable in ZF), the set $S$ has a uniformizing (partial) function $f:\mathbb{R}\to\mathbb{R}$, i.e. $\mathrm{dom}(f) = \{ x \in \mathbb{R} : S_x \neq \varnothing \}$ and $\{(x,f(x)) : x \in \mathbb{R}\} \subseteq S$. Since every Borel set is analytic, this $f$ gives you what you want provided you know codes $x$ such that the $S_x$ are Borel sets you're interested in. Picking a unique code for each Borel set is a difficult task which requires some choice. However, you can invoke the Axiom of Choice once to get such a function that gives unique codes and keep working with that function until the end of time, thereby avoiding repeated uses of choice.
See, for example, Kechris's Classical Descriptive Set Theory (II.18) for details on the Jankov–von Neumann Uniformization Theorem, and many other useful uniformization theorems (Kuratowski–Ryll Nardzewski, Kondô–Novikov, etc) that can be used in a variety of contexts.<|endoftext|>
TITLE: Does the Baker-Campbell-Hausdorff formula hold for vector fields on a (compact) manifold?
QUESTION [13 upvotes]: Consider a compact manifold M. For a vector field X on M, let $\phi_X$ denote the diffeomorphism of M given by the time 1 flow of X.
If X and Y are two vector fields, is $\phi_X \circ \phi_Y$ necessarily of the form $\phi_Z$ for some vector field Z?
Since $X\mapsto \phi_X$ can be thought of as the exponential map from the Lie algebra of vector fields to the group of diffeomorphisms, an obvious candidate is that Z should be given by the Baker-Campbell-Hausdorff formula $B(X, Y) = X+Y+\frac{1}{2}[X,Y]+\cdots$. But does this hold in this infinite-dimensional setting? If so, in which sense does the series converge to Z?
Also, I'm interested in the case where M is a symplectic manifold and we consider only symplectic vector fields (ie. vector fields for which the contraction with the symplectic form is a closed 1-form). Locally, X and Y are the Hamiltonian vector fields associated to smooth functions f and g, so I assume that asking whether B(X, Y) makes sense/is symplectic corresponds to asking whether B(f, g) makes sense/defines a smooth function (where, of course, we use the Poisson bracket in the expansion of B(f, g)). The right-hand side of B(f,g) consists of lots of iterated directional derivatives of f and g in the Xf and Xg directions; it is not clear to me that the coefficients in the BCH formula make the series converge (uniformly, say) for any choice of f and g.

REPLY [2 votes]: The BCH formula holds in a finite-dimensional Lie algebra for $X,Y$ in some open neighborhood of $0$.  Therefore, the coefficients of BCH are bounded by $a^n$ for some $0&lt;a&lt;\infty$.  So suppose that $X,Y$ are vector fields for which degree-$n$ bracket monomials (things like $[X,[X,Y]]$, with $n$ many $X$s and $Y$s total) don't grow too fast in your topology — not faster than $b^n$ for some $0 &lt; b &lt; \infty$.  Then for $s &lt; b/a$, ${\rm BCH}(sX,sY)$ converges, in your topology, if your topology is complete.
Now, I'm not much of one for infinite-dimensional topological vector spaces, so I can't say which topologies are great to use.<|endoftext|>
TITLE: Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group
QUESTION [7 upvotes]: I am looking for a reference for the following well-known fact: For any subdiagram $\Delta_0$ of of the Dynkin diagram $\Delta=D(G)$
of an adjoint simple group $G$ over an algebraically closed field $k$, there exists a reductive subgroup of maximal rank $G_0\subset G$
with Dynkin diagram $\Delta_0$.
To be more precise, I am looking for a reference for a proof of the following well-known lemma:
Lemma 1.  Let $G$ be an adjoint, connected, simple algebraic group with Dynkin diagram $\Delta=D(G)$
over an algebraically closed field $k$ of any characteristic.
Let $\Delta_0$ be a  subdiagram of $\Delta$
(that is, a subset $\Pi_0$ of the set $\Pi$ of vertices of $\Delta$,
together with all the edges of $\Delta$ connecting pairs of vertices of $\Pi_0$).
Then there exists a connected reductive $k$-subgroup of maximal rank  $G_0$ of  $G$
such that the corresponding adjoint semisimple group $G_0^{ad}$
has  Dynkin diagram $\Delta_0$.
I know a simple proof of Lemma 1, but I would prefer to give a reference rather than a proof.
The proof goes as follows. Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system,
then our $\Pi$ is a basis of $R$.  Let $S$ be the subgroup of $T$ orthogonal to $\Pi_0$,
then it is a subtorus of $T$ (because $G$ is adjoint). 
Set $G_0=C_G(S)$, the centralizer of $S$ in $G$.
Then $G_0$ is a connected reductive subgroup of $G$.
It is easy to see that (the adjoint group of) $G_0$ has Dynkin diagram $\Delta_0$.
Note that Lemma 1 is a special case of the following Lemma 2,
for which I would also be happy to have a reference.
Lemma 2.   Let $G$ be an adjoint, connected, simple algebraic group 
over an algebraically closed field $k$ of any characteristic.
Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system.
Let $R_0$ be a closed symmetric subset of $R$.
Then there exists a connected reductive $k$-subgroup of maximal rank  $G_0$ of $G$
with root system $R_0$.
I will be grateful to any references, comments, etc. (also to a proof of Lemma 2).
Mikhail Borovoi

REPLY [10 votes]: Dear Mikhail,
If I understand correctly, your Lemma 2 is implied by
SGA 3 Exposé 22, Théorème 5.4.7.
Everything is on a general base S (that you may take as your algebraically closed field). The kind of subgroup you want is called "de type (R)" (see Définition 5.2.1) and a subset of R that corresponds to such a group is also called "de type (R)". Now the theorem above exactly says that when a subset of R is closed, it is "de type (R)" which exactly means that there is a corresponding connected subgroup of G. By the way, Théorème 5.4.7 does not assume the subset to be symmetric, and you get things like Borel subgroups if you take only "half" of the roots. In the symmetric case, the group is reductive by Proposition 5.10.1.
Hope this helps.
Baptiste Calmès<|endoftext|>
TITLE: Cayley graphs of finitely generated groups
QUESTION [26 upvotes]: Let $\approx$ be the binary relation on the class of finitely generated groups 
such that $G \approx H$ iff $G$ and $H$ have isomorphic (unlabeled nondirected)
Cayley graphs with respect to suitably chosen finite generating sets. Is $\approx$ an equivalence relation?

REPLY [18 votes]: The answer is no, as expected. The following proof is "joint work" with L. Scheele.
Consider $G=\mathbb{Z}$, $K=D_\infty$ and $H:=\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Then $G \approx K$ and $K\approx H$, but $G \not\approx H.$
Indeed, the Cayley graph associated to $\{-1,1\}$ for G and the Cayley graph associated to $\{s,t\}$ where $D_\infty=\langle s,t: s^2=t^2=1 \rangle$ are clearly isometric.
Similarly, the Cayley graph associated to $\{s,st,ts\}$ for $K$ and the graph associated to $\{(0,1),(-1,0),(1,0)\}$ for $H$ are isometric.
However, let $S$ be some symmetric generating set for $G$. Then $S_k$, the set of vertices which have distance precisely $k\geq 1$ from the identity, has even cardinality because $S_k$ is invariant under the mapping $x \mapsto -x$ and doesn't contain 0.
Now let $T$ be some symmetric generating set for $H$. Let $k_0$ be the distance of $(0,1)$ from the identity in the associated graph. Then $T_{k_0}$, the set of vertices which have distance precisely $k_0$ from the identity, has odd cardinality. Indeed, it is invariant under the mapping $(x,y) \mapsto (-x,-y)$ since $T$ is assumed to be symmetric. But $(0,1)$ is a fixed point.<|endoftext|>
TITLE: Occurrence of the trivial representation in restrictions of Lie group representations
QUESTION [12 upvotes]: Suppose $G$ is a semisimple group, and $V_{\lambda}$ is an irreducible finite-dimensional representation of highest weight $\lambda$.  Suppose $H \subset G$ is a semisimple subgroup.  What is the multiplicity of the trivial representation in $V_{\lambda}|_{H}$?  Is there a simple way to read this off from $\lambda$ and the Dynkin diagrams of $G$ and $H$?

REPLY [3 votes]: I want to refer you to the following article by B. Cahen, treating the case of compact Lie groups, where an approach based on the realization of the irreducible representations of G according to the Borel-Weil theorem as reproducing kernel Hilbert spaces within the framework of Berezin quantization is adopted. The article describes a multiplicity formula (equation - 3.3) of an irreducible representation of a compact group G in the decomposition of a reproducing kernel Hilbert space on which G acts unitarily whose elements are functions on a manifold M.
When applied to the case in the question, the reproducing kernel Hilbert space can be taken as the space of holomorphic sectionsof the line bundle associated with lambda (according to the Borel-Weil theorem) on the flag manifold G/T. The multiplicity formula contains a double integral, the first is over the maximal torus of H which acts as a projector onto the required representation space and the second is over the flag manifold G/T which acts as a multiplicity counter. 
In this method no explicit sum on the Weyl group is needed, but it is replaced by the task of the double integration. The integration on the maximal torus is quite easy. The integration on the flag manifold looks like a formidable task but at least for the case of weight multiplicity computation it can be replaced by a computation of the rank of
a Hermitian matrix (proposition 3.6). In the article some low dimensional examples are included.
This method does not give nice closed formulas like the combinatorial methods, but may be it has a reduced complexity in some special cases. I think that at least the construction of the reproducing kernel in the affine coordinates of the big cell of the flag manifold can be implemented relatively efficiently by polynomial multiplication using 
a multidimensional fast Fourier transform.<|endoftext|>
TITLE: Triangles, squares, and discontinuous complex functions 
QUESTION [6 upvotes]: Is there some onto function $f:$ $\mathbb{C}$ $\rightarrow$ $\mathbb{C}$
such that for each triangle $T$ (with its interior), $f(T)$ is a
square (with interior, too) ?
I would have the same question for triangles and squares without interior, respectively.

REPLY [10 votes]: With interior: yes. Fix a sequence of squares $Q_1\subset Q_2\subset\dots$ whose union is the entire plane. Then arrange a map $g:\mathbb R\to\mathbb R^2$ such that, for every nontrivial segment $[a,b]\subset\mathbb R$, its image is one of the squares $Q_i$. To do that, construct countably many disjoint Cantor sets so that every nontrivial interval contains at least one of them. Then  send every Cantor set $K$ bijectively onto $Q_n$ where $n$ is the minimum number such that $K\cap [-n,n]\ne\emptyset$. Send the complements of these Cantor sets to a fixed point inside $Q_1$. Then define $f(x,y)=g(y)$.
(This is a detailed version of gowers' answer.)
UPDATE
Without interior: no. Take any triangle $T$ and consider its image $Q$ with vertices $ABCD$. There is a side $I$ of $T$ whose image has infinitely many points on (at least) two sides of $Q$. If these are opposite sides, say $AB$ and $CD$, the image of any triangle containing $I$ must stay within the strip bounded by the lines $AB$ and $CD$. And if these are two adjacent sides of $Q$, say $AB$ and $AD$, the image of any triangle containing $I$ stays within the quarter of the plane bounded by the rays $AB$ and $AD$. In both cases, the images of the triangles containing $I$ do not cover the plane, hence the map is not onto.

REPLY [7 votes]: I think this works but haven't checked. I'm pretty sure that for any open set it's easy to find a map that takes any open subset of that set to all of $\mathbb{C}$, or to all of a square, or to whatever single set you feel like. So now for each n choose a map that takes every open subset of the annulus $\{z: n < |z| \leq n+1\}$ to the square that consists of all points with real and imaginary parts less than or equal to n. Now, given any triangle, there will be a maximum n such that it belongs to the nth annulus, and it will intersect that annulus in an open set and therefore map to a square.<|endoftext|>
TITLE: Montague's Reflection Principle and Compactness Theorem
QUESTION [17 upvotes]: Here's a question I can't answer by myself: The Reflection Principle in Set Theory states for each formula $\phi(v_{1},...,v_{n})$ and for each set M there exists a set N which extends M such that the following holds
$\phi^{N} (x_{1},...,x_{n})$ iff $\phi (x_{1},...,x_{n})$ for all $x_{1},...x_{n} \in N$
Thus if $\sigma$ is a true sentence then the RFP yields a model of it and as a consequence any finite set of axioms of ZFC has a model (as a consequence ZFC is not finitely axiomatizable by Gödel's Second Incompleteness Theorem)
But why can't I just use now the Compactness Theorem (stating that each infinte set of formulas such that each finite subset has a model, has a model itself) to obtain a model of ZFC (which is actually impossible)??

REPLY [11 votes]: As Sridhar already explained, Lévy–Montague Reflection is a theorem scheme and not a single theorem which resolves the apparent contradiction, but here are a few additional cool facts.
First, note that ZFC is not finitely axiomatizable (otherwise we would indeed have a contradiction) but there is a recursive listing of the axioms of ZFC. Let's fix such a listing $\phi_0$,$\phi_1$,$\phi_2$,... If $M$ is a model of ZFC, then either $M$ is an $\omega$-model (i.e. the finite ordinals of $M$ are truly finite) or it is not (i.e. $M$ has some nonstandard finite ordinals). Let's see what happens in each case.
Suppose first that $M$ is an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... exists in $M$ and, by Lévy–Montague, people living in $M$ believe that $\{\phi_0,\ldots,\phi_n\}$ has a model for each $n < \omega$. Since people living in $M$ also believe in the Compactness Theorem, they also believe that there is a model of ZFC. This is surprising, but note that the hypothesis that $M$ is an $\omega$-model is essential since without it we there is no reason for $M$'s notion of finite to agree with ours.  This is where your initial reasoning strayed, you naturally assumed that every model of ZFC was an $\omega$-model.
Suppose now that $M$ is not an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... makes sense in $M$, but since $M$ has nonstandard finite ordinals this listing continues beyond the true $\omega$ and people who live in $M$ believe that these nonstandard $\phi_N$'s are real axioms of ZFC! By Lévy–Montague, $M$ believes that $\{\phi_0,\ldots,\phi_n\}$ has a model for every standard $n$, but since Lévy–Montague Reflection doesn't say anything about nonstandard axioms, there may be some nonstandard finite ordinal $N$ in $M$ such that people living in $M$ do not believe that the nonstandard finite set $\{\phi_0,\ldots,\phi_N\}$ has a model.
Now here is a funny thing that was pointed out by Joel David Hamkins in answer to another question. Suppose $M$ is a model of ZFC + ¬Con(ZFC). Since people in $M$ believe that their finite ordinals are wellordered, there must be a first finite ordinal $N$ in $M$ such that $\{\phi_0,\ldots,\phi_N\}$ has no model in $M$. This $N$ must be nonstandard finite ordinal, and so must its predecessor $N-1$. By minimality of $N$, people in $M$ believe that $\{\phi_0,\ldots,\phi_{N-1}\}$ does have a model. Let $M'$ be such a model. Note that $M' \models \phi_n$ for every standard axiom $\phi_n$ since $n < N-1$. Therefore, although people living in $M$ certainly don't believe it, this $M'$ is in fact a model of ZFC!!!
Thus, Lévy–Montague Reflection does imply that every model of ZFC contains another model of ZFC, but the models are not necessarily aware of that fact...<|endoftext|>
TITLE: Is there a good reference for the relationship between the Yangian and formal based loop group?
QUESTION [9 upvotes]: For every finite dimensional semi-simple Lie group $\mathfrak{g}$, we have a loop algebra $\mathfrak{g}[t,t^{-1}]$.  This loop algebra has a natural invariant inner product by taking the residue at zero of the Killing form applied to two elements (i.e. $t^k\mathfrak{g}$ and $t^{-k-1}\mathfrak{g}$ are paired by the Killing form.)
This Lie algebra actually has a Manin triple structure with respect to this inner product:  the subalgebras $\mathfrak{g}[t]$ and $t^{-1}\mathfrak{g}[t^{-1}]$ are both isotropic, and non-degenerately paired by this form.  This makes $\mathfrak{g}[t]$ into a Lie bialgebra, by getting the cobracket from the bracket on $t^{-1}\mathfrak{g}[t^{-1}]$.  
Now, as we all know, Lie bialgebras can be quantized: in this case, the result is a quite popular Hopf algebra called the Yangian.  By the usual yoga of quantization of Lie bialgebras, the dual Hopf algebra to the Yangian quantizes the universal enveloping $t^{-1}\mathfrak{g}[t^{-1}]$, so if you take a different associated graded of the Yangian, you must get the Hopf algebra of functions on the group with Lie algebra $t^{-1}\mathfrak{g}[t^{-1}]$, which is $L_<G$, the based formal loop group.

Now, all of these things also have explicit descriptions in terms of equations, and it seems as though this story must be worked out explicitly somewhere, but I've had little luck locating it.  Does anyone know where?  Or is this story just wrong, and that's why I can't find it?

EDIT: The comment below mostly answers this question.  I would be interested if anyone out there has written something more explicit than the Etingof and Kazhdan paper, but it's the sort of thing I was looking for.  If it were to be left in the form of an answer, I would probably accept it (hint, hint).

REPLY [4 votes]: I will suggest the article Quantization of Lie bialgebras, III by Pavel Etingof and David Kazhdan.  It discusses both the Yangian and the dual Yangian as examples in the context of quantization of Lie bialgebras of functions on a curve with punctures.
Also, so far as I know, no-one ever explicitly derives the relations of the Yangian from the conditions that it be a Hopf algebra quantization of the Lie bialgebra, but Drinfeld explains it in his article called Quantum Groups (Proc. ICM Berkeley, 1986).  If you assume the coproduct takes a certain form on the Lie algebra (the most obvious choice given that it is a quantization of that Lie bialgebra) then impose the condition that the Hopf algebra coproduct be an algebra homomorphism then you can derive the relations of the Yangian in the first presentation term by term, using the fact that it is a homogeneous (graded) deformation.  
Personally I tried to do this but I found I had to assume little things along the way, like that the RHS of equation (13) in that paper is a symmetric sum of the orthonormal basis elements of the Lie algebra.  I couldn't really understand why I had to do this, but I guess it's probably obvious to people who are smarter than me.  At the end you end up with an algebra that you can then prove is isomorphic to the second presentation he gives, and then get a PBW theorem based on this presentation and verify that it is a quantization of the Lie bialgebra after all (I still working on understanding this part).  Then given the uniqueness, which is meant to follow from cohomological arguments described in Section 9 of that same Quantum Groups paper, you don't need to justify any assumptions you make along the way.  It seems that understanding the cohomology is key.<|endoftext|>
TITLE: How to start Game theory?
QUESTION [41 upvotes]: I recently got interested in Game Theory but I don't know where should I start.
Can anyone recommend any references and textbooks?
And what are the prerequisites of Game Thoery?
Thanks in advance.

REPLY [7 votes]: There is a new (well, the English translation is) book that treats both noncooperative and cooperative (but not combinatorial) game theory on a high level, is extremely well written, mathematically rigorous and fairly comprehensive: Game Theory by Michael Maschler, Eilon Solan, and Shmuel Zamir. For someone who knows some undergraduate real analysis and linear algebra, the book should be self contained (with a few exceptions, where reference literature is recommended in the book). The book doesn't contain everything (there is very little on refinements), but it contains enough to get one near the frontier of research fast.<|endoftext|>
TITLE: Contractible manifold with boundary - is it a disc?
QUESTION [13 upvotes]: I'm sure this is standard but I don't know where to look. Let $M$ be a contractible compact smooth $n$-manifold with boundary. Does it have to be homeomorphic to $D^n$? What about diffeomorphic?
[UPDATE: the answer is well-known to be negative as many people kindly pointed out. But actually I assume more about the manifold, namely the following:]
There is a Riemannian metric on $M$ such that every two points are connected by a unique shortest path. So $M$ can be contracted to a point $p\in M$ by sending every point along a shortest path to $p$. These paths can bend along the boundary and can merge because of this. But they are relatively nice (namely $C^{1,1}$) curves and their first derivatives depend continuously on their endpoints. Given all this, can one conclude that $M$ is a disc?
ADDED: These curves are of course gradient curves of a function (the distance to $p$) which is $C^1$ and has no critical points in the interior of $M$, except at $p$.

REPLY [3 votes]: Given a function $\psi:\mathbb R\to \mathbb R$,
set 
$$\Psi=\psi\circ\mathrm{dist}_ {\partial M},\ \ \ \ \ f=\Psi\cdot(R-\mathrm{dist}_ p)$$
for some fixed $R>\mathrm{diam}\ M$.
Further,
$$d\,f =
(R-\mathrm{dist}_ p)\cdot  d\,\Psi-\Psi\cdot d\,\mathrm{dist}_ p$$
Thus, we may choose smooth increasing  $\psi$, 
such that $\psi(0)=0$ 
and it is constant outside of little nbhd of $0$ so that 
$\Psi$ is smooth.
(It is possible since the function $\mathrm{dist}_ {\partial M}$ is smooth and has no critical points in a small neighborhood of $\partial M$.)
Note that $d\,\Psi$ is positive muliple of $d\,\mathrm{dist}_ {\partial M}$.
Thus $d_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $\partial M$, which can not happen.
Now we can apply Morse theory for $f$...<|endoftext|>
TITLE: Modular Arithmetic in LaTeX
QUESTION [19 upvotes]: This question may end up [closed], but I'm going to ask and let the people decide.  It's certainly the kind of question that I'd ask people at tea, and it's not one whose answer I've been able find with Google.
TeX, I have heard, is Turing complete.  In theory, this means that we can do modular arithmetic with LaTeX programs.  I'd like to know how this can be done in practice.
Background:  I've been using the \foreach command in TikZ to draw NxN arrays of nodes, indexed by pairs of integers (m,n).  I'd like to be able to use modular arithmetic and an ifthenelse statement to put different decorations on the nodes, depending on the value of (m+n) mod p.  Obviously, one can just do this by hand.  But that's not the world I want to live in.

REPLY [5 votes]: I've had to implement a lot of code in TeX while writing a package for drawing spectral sequences (sseq.sty). pgf makes computations in TeX easier, but if you really have a significant amount of code, I recommend you take a look at luatex/lualatex, which is included in many distributions and merges TeX with an interpreter for the easy-and-fast language Lua, which you can learn in a couple of hours. It's much more readable and speeds up the typesetting enormously.
TeX may be Turing complete (LaTeX as well, by the way), but so is a Turing machine, and you wouldn't want to implement modular arithmetic on a Turing machine, would you?<|endoftext|>
TITLE: Does 2^m = 3^n + r have finitely many solutions for every r?
QUESTION [9 upvotes]: Is it true that for every integer $r$, the equation $2^m = 3^n + r$ has at most a finite number of integer solutions? I understand that this is a special case of Pillai's conjecture, which is unsolved.
If the statement is true, then can we verify the finiteness of the solution set using modular arithmetic? To be precise, is the following proposition true?
$$\forall r,\ \exists M,\ \exists N,\  \forall m,n \ge N,\ \ 2^m \not\equiv 3^n + r \pmod{M}$$
I have verified the proposition for $0 \le r \le 12$, and found the least possible modulus $M(r)$ for each $r$ in this interval. Note that $M(r) = 2$ if $r$ is even.
$M(1) = 8$, $M(3) = 3$, $M(5) = 1088$, $M(7) = 1632$, $M(9) = 3$, $M(11) = 8$.

REPLY [13 votes]: Yes, it is true that this kind of equation ax+by=c, where a,b,c are non-zero and fixed and x,y are allowed to only have prime factors in a finite set, has only finitely many solutions. This is a special case of Siegel's theorem on integral points on curves. 
Your second question may be unknown in the generality you pose. It is interesting that it holds. A remark: if there is a solution to $2^m = 3^n + r$, then $2^{m+k\phi(M)} \equiv 3^{n+k\phi(M)} + r (\mod M)$ for all $k,M$ if $(M,6)=1$, so if $M$ exists in this case, then $(M,6)>1$. If there is no solution to the equation $2^m = 3^n + r$, then the existence of $M$ (with $N=0$) is a special case of a conjecture of Skolem.
T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der
Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad.
Oslo, 12 (1937), 1–16.
Another comment. There are no solutions when $r=11$ but $M=8$ doesn't work since $2^2 \equiv 3^2 + 11 \mod 8$. $M(11)=205$. (Edit: $M(11)=8$ is OK. I misunderstood the definition, see comments)

REPLY [5 votes]: I have no comment on your methods, and I know very little about this, but that case of Pillai's conjecture appears to have been solved in the 80's by Stroeker and Tijdeman [Edit: see below]. Here's a paper by Bennett from 2001 that shows more: http://www.math.ubc.ca/~bennett/B-CJM-Pillai.pdf. In particular, the number of solutions is at most 2 for each fixed $r$.  More generally, Bennett shows that for fixed integers $a\geq2$, $b\geq2$, and $r\neq0$, there are at most 2 solutions $(m,n)$ to the equation $a^m=b^n+r$.  The more general form of Pillai's conjecture allows $a$ and $b$ to vary and appears to still be unsolved.
Edit: What Stroeker and Tijdeman actually did was sharpen the result by showing that except when $r$ is in $\{-1,5,13\}$, your equation has at most one solution, and that in the exceptional cases it has two.  The finiteness of the set of solutions $(m,n)$ to the equation $a^m=b^n+r$ had long been known, and Pallai himself gave some quantitative results on this using Siegel's Theorem.  For finiteness alone without quantification, Bennett cites this 1918 Polya paper.  My source for all of this is Bennett's paper.<|endoftext|>
TITLE: Jordan Hölder decomposition for group objects
QUESTION [10 upvotes]: Is there some generalization of the Jordan-Hölder decomposition for group objects in a category $\mathcal{C}$?
If $\mathcal{C}$ is the category Sch$(S)$ of schemes over a base scheme $S$ then (I think) this is true, also probably for other categories of "spaces" like Top or Diff it should be true, but I don't have any idea for general categories.

REPLY [6 votes]: You can start by looking at the paper by P.J. Hilton and W. Ledermann, "On the Jordan-Hölder theorem in homological monoids", where three axioms are needed to establish the decomposition, and the third one is essentially guaranteeing the second isomorphism theorem.
In "Mal'cev, protomodular, homological and semi-abelian categories" by F. Borceux, D. Bourn, there is a chapter devoted to homological categories (which are  pointed, regular and protomodular), these are the categories where certain lemmas of homological algebra hold true (five lemma, nine lemma, snake lemma, Noether isomorphism theorems etc.). The fact that Jordan-Holder holds for these categories is proven in "Jordan-Holder, Modularity  and Distributivity in Non-Commutative Algebra" by F. Borceux, M. Grandis.<|endoftext|>
TITLE: Infinity groupoid objects
QUESTION [9 upvotes]: I was wondering if there is a model-theoretic way of defining the infinity category of infinity-groupoid objects in a category $C$ (more generally, if $C$ is an infinity category itself, but, right now a 1-category is enough). Is there a model structure on $C^{\Delta^{op}}$ such that those objects which are fibrant and cofibrant correspond to "internal Kan-complexes" in the correct way? So e.g. I want the C-enriched nerve of an actual groupoid object of C to be fibrant and cofibrant in this model structure. If you don't know the answer in general, for now I am mostly interested in the case that $C$ is the 1-category of topological spaces (here I DO NOT want to think of $C$ as being the same thing as infinity-groupoids or simplicial sets, I actually care about the topology).
More generally, if $C$ is an infinity-category associated to a model category $D$, does this correspond to the Reedy-model structure on $D^{\Delta^{op}}$? 
EDIT: I should really be asking for a SIMPLICIAL model category structure.

REPLY [2 votes]: There is not a model category on simplicial smooth manifolds which gives the data you're looking for, but there is a structure of category of fibrant objects such that nerves of groupoids are fibrant.  The fibrations are Kan maps (a la Henriques, but with the additional requirement that the maps on vertices be submersions), and the weak equivalences are maps which induce isomorphisms on all simplicial homotopy groups (again, as Henriques defines them).  Nerves of groupoids are fibrant in this sense, hypercovers are trivial fibrations, and morphisms qua principal bibundles are equivalent to morphisms via spans where the source leg is a hypercover.  
This category of fibrant objects structure doesn't extend to a model structure on simplicial manifolds because there aren't any positive dimensional cofibrant objects (e.g. hypercovers are trivial fibrations, but given any positive dimensional manifold, you can build a hypercover on it which doesn't admit a section).  
If you still want to work in a model category capturing this, the natural thing is to use Yoneda and pass to the local model structure on simplicial presheaves.  You can show that this is a fully faithful embedding on the level of homotopy categories.<|endoftext|>
TITLE: Does every smooth manifold of infinite topological type admit a complete Riemannian metric?
QUESTION [6 upvotes]: To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.
I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.

REPLY [3 votes]: We can even prove more: every smooth manifold can be equipped with a Riemannian metric of bounded geometry (positive injectivity radius, bounded curvature tensor and all derivatives of the curvature tensor are also bounded).
This is Theorem 2' in "R. E. Greene, Complete metrics of bounded curvature on noncompact manifolds, Archiv der Mathematik 31 (1978), no. 1, 89-95".<|endoftext|>
TITLE: Are there countable index subrings of the reals?
QUESTION [18 upvotes]: Does ${\mathbb R}$ have proper, countable index subrings?  By countable I mean finite or countably infinite.  By subring I mean any additive subgroup which is closed under multiplication (I don't care if it contains $1$.)  By index, I mean index as an additive subgroup.
Given some real number $x$, when is it possible to find a countable index subring of ${\mathbb R}$ which does not contain $x$?

REPLY [12 votes]: Simon Thomas's approach answers Question 2 too.  The answer is that for any nonzero real number $x$ there exists such a subring (possibly without 1) not containing $x$.
Proof: Let $K$ be the Puiseux series field $\overline{\mathbf{Q}}((t^{\mathbf{Q}}))$, let $A$ be its valuation ring, and let $\mathfrak{m}$ be its maximal ideal.
If $x \in \mathbf{R}$ is not algebraic over $\mathbf{Q}$, then choose an identification $\mathbf{C} \simeq K$ sending $x$ to the transcendental element $1/t$, and use the subring $\mathbf{R} \cap A$.
If $x \in \mathbf{R}^\times$ is algebraic over $\mathbf{Q}$, then choose an identification $\mathbf{C} \simeq K$ again, and use $\mathbf{R} \cap \mathfrak{m}$ (a subring of $\mathbf{R}$ without $1$). $\square$
Remark: If one insists on using subrings with $1$, then the answer is that such a subring not containing $x$ exists if and only if $x \notin \mathbf{Z}$.  
Proof: Repeat the argument above, but in the case where $x$ is algebraic (and outside $\mathbf{Z}$), use $\mathbf{R} \cap (\mathbf{Z} + \mathfrak{m})$.<|endoftext|>
TITLE: Stark's conjecture and p-adic L-functions
QUESTION [18 upvotes]: Not long back I asked a question about the existence of p-adic L-functions for number fields that are not totally real; and I was told that when the number field concerned has a nontrivial totally real or CM subfield, then there is a construction due to various people including Coates-Sinnott and Katz.
But my favourite number field at the moment is K = $\mathbb{Q}(\sqrt[3]{2})$, and sadly K contains no totally real or CM subfield, so for trivial reasons $L(n, \chi) = 0$ for every Groessencharacter $\chi$ of $K$ and every $n \le 0$. So in this case the above constructions just give zero. When I learnt this, I thought "that can't be the whole story, what about higher derivatives at 0"? Asking around, I was told about Stark's conjectures, which apparently predict that the leading term at $s = 0$ of the L-function of any GC of K should be the product of an explicit transcendental regulator and an algebraic number (which, if I've understood this right, should lie in the field $\mathbb{Q}$(values of $\chi$).)
My question is this: assuming Stark's conjecture, can we construct a distribution on the Galois group of the maximal unramified-outside-p abelian extension of K whose evaluation at any locally constant character of this group gives the algebraic part of the leading term at 0 of the L-series of the corresponding Groessencharacter?

REPLY [7 votes]: Conjecturally, the answer is yes, but the amount of work required is not trivial at all. The general set-up is roughly as follows: the special values of $L$-functions (in your case, for Tate motives) are predicted by the Tamagawa Number Conjecture, and by the Equivariant Tamagawa Number Conjecture (ETNC) when one wishes to incorporate the action of some Galois group (as you do). The ETNC links the leading term at 0 of the $L$-function to the determinants of some cohomological complexes. However, in order to do this coherently for any locally constant character (again, as you want to), one needs rather strong hypotheses on the complexes involved: namely, they need to be semisimple at $\rho$ if one wishes to interpolate at $\rho$. Here, the bad news start: showing that a complex is semisimple at $\rho$ amounts to Leopoldt's conjecture for Tate motives so is in general very hard. But things should be fine in your favourite case. In this way, you can construct (leading terms of) $p$-adic $L$-functions for Tate motives. 
If you are ready to admit all conjectures, all of this has been known for quite some time, see for instance B.Perrin-Riou Fonctions $L$ $p$-adiques des représentations $p$-adiques section 4.2 Conjecture CP(M). For a more recent and more flexible formulation, I recommend D.Burns and O.Venjakob On the leading terms  of Zeta Isomorphism... section 3.2.2 and 3.2.3.<|endoftext|>
TITLE: Analog to the Chinese Remainder Theorem in groups other than Z_n.
QUESTION [6 upvotes]: The idea hit me when I was in my Elliptic Curve Cryptography class. $Z_n \leftrightarrow Z_{f_1} \times Z_{f_2} \times ...$ where $f_1 \times f_2 \times ... = n$ and $\{f_1, f_2, ...\}$ are pairwise coprime. Applications of this Chinese Remainder Theorem not only include computational speedups (in the case of decryption in RSA) but also stronger cryptographic attacks against $Z_n$ (for example, Pollard Rho factoring exploits this structure). Can we extend these applications into other areas? (Admittedly I don't know many computationaly examples where this could be useful, but can imagine that Mathematica/Maple would find some uses).
So the real question: is this property just a "coincidence" or are there analogs in other  groups? If there are, is there some group theory analog that applies equally well to every group? If there are not, what underlying structure in the natural numbers makes this possible?

REPLY [2 votes]: I did a course titled something similar in my undergraduate, and while it didn't teach the following applications, H. Cohen's A Course in Computational Algebraic Number Theory (which I read right after the course, and you should too) does.
As you mention, one can use the Pollard rho algorithm to find a factor $p$ of $N$, in time $O(\sqrt p)$. There are two other basic algorithms that use CRT implicitly, both in Cohen's book:
1) Pollard's $p-1$ (and its generalizations, such as Williams' $p+1$): Compute $gcd(a^{n!}-1, N)$. If $p-1 | n!$, then the gcd will be divisible by $p$, and one can factor. This uses CRT implicitly in the following way: we can compute the $gcd(a^{n!}-1,N)$ using only mod-$N$ operations - but we find $p$ because of the existence of CRT. If we accidently get $N$ as the gcd, we can still factor using another application of CRT. Read the above referenced book for details.
2) The much faster Elliptic Curve Method: Initialize an "elliptic curve" $E$ mod $N$ and a point $P$ on it. Compute $(n!)P$. I write "elliptic curve" because we aren't really defining an elliptic curve - $\mathbb{Z}/(N)$ is not a field! But, using CRT, we treat it as the combination fields. We hope that the order of $P$ on $E/\mathbb{F}_p$ divides $n!$, and mod any other prime dividing $N$, the order does not divide $n!$. In this case $(n!)P$ will "have" $p$ in its denominator, but not the other primes, allowing us to recover $p$. This, again, is using CRT much in the same way that Pollard's Rho does. We compute things only mod $N$ - but we get things that are structurely inherit, such as $p$.
3) A bit of a different kind of computational application of CRT is D. J. Bernstein's "Doubly focused enumeration of locally square polynomial values." (Pages 69--76 in High primes and misdemeanours: lectures in honour of the 60th birthday of Hugh Cowie Williams, edited by Alf van der Poorten, Andreas Stein. Fields Institute Communications 41, American Mathematical Society, 2004. ISBN 0-8218-3353-7).
The author uses CRT explicitly in order to enumerate over numbers satisfying certain congruence properties. It is not cryptographic, but computationally interesting and simple to understand, not to mention record braking.<|endoftext|>
TITLE: Grothendieck's Tohoku Paper and Combinatorial Topology
QUESTION [21 upvotes]: I've read some discussions of Grothendieck's famous Tohoku Paper, and I understand that one reason it was a landmark paper was that it introduced abelian categories and gave us sheaf cohomology as a certain type of derived functor. However, I've heard from various sources (Manin's Homological Algebra, my prof, and the 2 part AMS Notices articles) that one of the famous aspects of this paper is that it "reinterpreted" the basics of combinatorial topology.  Does anybody know what this means?  Slightly more specific: how was combinatorial topology understood at that time and how did the Tohoku paper force a reinterpretation of the conventional concepts? I know this question might be a bit ill posed, but I'd like to know if somebody understands what this means and can explain it at a first year graduate student level.
Thanks,
Ben

REPLY [19 votes]: One important piece of information in the Tohoku paper is that for X a finite CW complex and G a finite group acting on X, there is an isomorphism $H^* (X/G; K) \simeq H^* (X; K)^G$, where $K$ is a field of characteristic zero.  I don't know of a way to see this without sheaf cohomology.  Note that the action need not be free.  
(MacDonald's paper on the cohomology of symmetric products references this result to the Tohoku paper, specifically Theorem 5.3.1 and the Corollary to Proposition 5.2.3.)
This is an example of how reformulating algebraic topological concepts sheaf-theoretically can be very helpful, and maybe that's the sense in which the paper "reinterpreted combinatorial topology."<|endoftext|>
TITLE: Are there interesting problems involving arbitrarily long time series of small matrices?
QUESTION [5 upvotes]: Are there well-known or interesting applied problems (especially of the real-time signal processing sort) where arbitrarily long time series of small (say $d \equiv \dim \le 30$ for a nominal bound, and preferably sparse) matrices arise naturally?
I am especially interested in problems that can be mapped onto a setup in which for each event of a reasonably nice point process on $\mathbb{R}$ (the simplest two such processes would be a Poisson or discrete-time process) there is an associated pair $(j,k) \in \{1,\dots,d\}^2$. In this case time-windowed sums $N_{jk}(t)$ of the various pairs can be formed in an obvious way (although there may be plenty of subtlety or freedom in the windowing itself): these supply such a matrix time series. 
Each such pair $(j,k)$ could be regarded as a transition from server $j$ to another (possibly identical) server $k$ in a closed queue with $d$ servers and infinitely many clients. It is not hard to see that in the setting of communication networks, this framework amounts to a very general form of traffic analysis. Such an application should not be considered for an answer: it's already been covered.
A slightly more restrictive but simpler example is where the pairs $(j,k)$ are inherited from a cadlag random walk on the root lattice 
$A_{d-1} :=\left \{x \in \mathbb{Z}^d : \sum_{j=1}^d x_j = 0\right \}$. 
Examples of this sort would also be of considerable interest to me.

REPLY [2 votes]: A very significant application in the context of communications engineering is the modelling of multiple-input-multiple-output (MIMO) communications channels. 
These channels are typically modeled by complex $n \times m$ matrices where $n$ is the number of receive antennas and $m$ is the number transmit antennas. The $(i,j)$ entry in the  matrix describes the channel between the $i$th transmit antenna and the $j$th receive antenna.  In most applications $n$ and $m$ are reasonably small, less than 16.  Also, in most real world applications the channel (and hence the matrix) changes over time. This gives you your time series of matrices. In some situations the matrix will even be sparse because some transmit antennas might not see some receive antennas.
There is a seriously huge amount of literature on the MIMO channel and a large amount of it deals with the static case, i.e. for the sake of simplicity it is assumed that the channel doesn't change with time. However there are also many papers that deal with the time varying case.  For example:
Chen and Su, "MIMO Channel Estimation in Correlated Fading Environments"
I unfortunately am not an expert in MIMO, but I do know some people who are and could ask them for more details if you were interested.<|endoftext|>
TITLE: What is Shimura referring to by "an incorrect formula given by Minkowski... known to most experts."
QUESTION [7 upvotes]: In the article 'André Weil As I Knew Him' in the April 1999 issue of Notices of the AMS, Shimura recounts how in 1996 André Weil (then 90 years old) didn't remember a mistake of Minkowski. Specifically, quoting from mid-paragraph:
"... In fact, to check that point, I asked him whether Minkowski was reliable. He said, 'I think so.' At that point I realized that his recollection was faulty, since Minkowski gave an incorrect formula, as Siegel pointed out, and that was known to most experts. ..."
My question is:
Where specifically in Siegel and Minkowski's published works can I find the aforementioned formula (of Minkowski) and observation (of Siegel)?

REPLY [12 votes]: I believe he is referring to the result in Minkowski's dissertation that gives a formula for the mass (in the number/lattice theory sense) over a genus, that is, the sum of reciprocals of the group orders of all inequivalent quadratic forms in a genus. (wikipedia)
Siegel found and corrected the error in Minkowski's formula and also made many more generalizations in this Annals paper. Namely, there is an incorrect power of 2 in the formula.

REPLY [6 votes]: I do not know where in the original sources, but the topic under discussion is the mass formula for integral quadratic forms. The accepted source with correct information is Conway and Sloane,
Low Dimensional Lattices. IV. The Mass Formula
Proceedings of the Royal Society of London, A 419, 259-286 (1988).
In the actual publication, the tables are sprinkled throughout, and I found it difficult to read the text. I have some sort of preprint around here where the tables are all at the end, easier to find what you want. But it still takes some real patience, and in fact some imagination, to use properly.  
I can recommend the book by the same Conway and Sloane, called Sphere Packings, Lattices, and Groups.<|endoftext|>
TITLE: Can algebraic number fields be generalized in a similar way to function fields in 1 variable over a finite field?
QUESTION [6 upvotes]: Global fields consist of finite extensions of $\mathbb{Q}$ (algebraic number fields) and finite extensions of $\mathbb{F}_q(x)$ (function fields in 1 variable over a finite field). The latter are isomorphic to the category of curves over $\mathbb{F}_q(x)$, and they can be generalized to function fields in $n$ variables over $\mathbb{F}_q(x)$, which are isomorphic to the category of varieties over $\mathbb{F}_q(x)$. Is there an analogous generalization of algebraic number fields?

REPLY [5 votes]: The function fields (in one or more variables) over $\mathbb{F}_q$ are precisely the infinite, finitely generated fields of characteristic $p$.  Thus an at least reasonable characteristic $0$ analogue is given by the (necessarily infinite!) finitely generated fields of characteristic $0$.  In other words, function fields in finitely many (possibly zero) variables over a number field $K$.
Indeed there has been much work on generalizing arithmetic geometric statements over global fields to arithmetic geometric statements over arbitrary infinite, finitely generated fields.  The one which springs most readily to my mind is the following generalization of the Mordell-Weil theorem due to Lang and Neron: the group of rational points on any abelian variety over any finitely generated field is a finitely generated abelian group.<|endoftext|>
TITLE: Homotopy pullbacks of simplicial spaces, and Bousfield-Friedlander
QUESTION [6 upvotes]: Let $X_\bullet \longrightarrow Y_\bullet \longleftarrow Z_\bullet$ be a diagram of simplicial spaces (=bisimplicial sets, if you like).
On p. 14-9 of these notes there is an example which shows that if $Y_\bullet$ and $Z_\bullet$ are levelwise connected then the homotopy pullback of the geometric realisation of the diagram is the geometric realisation of the levelwise homotopy pullback.
The theorem is proved using the Bousfield-Friedlander theorem, which only requires that $\pi_0Z_\bullet \to \pi_0Y_\bullet$ is a Kan fibration and that $Z_\bullet$ and $Y_\bullet$ are $\pi_*$-Kan. Being levelwise connected implies both of these conditions, but is not necessary.
Can the conditions of the Bousfield-Friedlander theorem be relaxed? How about if $Z_\bullet \to Y_\bullet$ is something like a $\pi_*$-Kan fibration", though I'm not sure of the precise definition this should have?

REPLY [7 votes]: The following represents what I know about this; I don't know of a published reference.
Given a map $f:Z_\bullet\to Y_\bullet$ of simplicial "spaces" (to make things easy, assume spaces are simplicial sets), let's call it a realization quasi-fibration (RQF) if for every $U_\bullet\to Y_\bullet$, the homotopy pullback of the geometric realizations is weakly equivalent to the realizations of the levelwise homotopy pullbacks.  The Bousfield-Friendlander theorem gives a sufficient condition for $f$ to be a  RQF, in terms of the dreaded $\pi_*$-Kan condition.
Some facts:

The pullback of an RQF $f$ along any $U_\bullet\to Y_\bullet$ is itself an RQF.
Let $F[n]_{\bullet}$ be the simplicial space which is free on a point in degree $n$.  Then $f$ is an RQF if and only if its pullback along all $g: F[n]_\bullet \to Y_\bullet$, for all $n$, is an RQF.

These two facts are consequences of something that is sometimes called "descent"; basically, the facts that homotopy colimits distribute over homotopy pullbacks, and compatible homotopy pullbacks assembled by a homotopy colimit result in a homotopy pullback.
So the above gives exact criteria for $f$ to be an RQF.  Whether the pullback of an  RQF $f$ along a map $g$ is again an RQF only depends on the homotopy class of $g$.  So if $f:Z_\bullet\to Y_\bullet$ is any map, let 
$\pi_0Y$ be the simplicial set whose $k$-simplices are $\pi_0(Y_k)$, which is to say all homotopy classes of maps $F[k]_\bullet\to Y_\bullet$.  Let $RQF(f)\subseteq \pi_0Y$ be the sub-simplicial set whose $k$-simplices correspond to $g:F[k]_\bullet\to Y_\bullet$ such that the pullback of $f$ along $g$ is an RQF.  
So the criterion is: $f$ is an RQF iff $RQF(f)=\pi_0Y$.
It turns out that since geometric realization always preserves products, any map $Z_\bullet \to point_\bullet$ is an RQF.  Thus $RQF(f)$ contains all $0$-simplices of $\pi_0Y$.  Thus, if all $Y_k$ are connected, $f$ is an RQF, which implies the result you describe.<|endoftext|>
TITLE: Union of closed subschemes with the structure sheaf over it
QUESTION [11 upvotes]: Elementary commutative algebra fact: for two proper ideals I and J of a commutative ring R, we have $V(IJ)=V(I\cap J)=V(I)\cup V(J)$.
Closed subschemes are related to sheaves of ideals. There is operation of intersection and product between sheaves of ideals, which is similar to the affine case.
I see in many places that the structure sheaf over $V(I)\cup V(J)$ are defined to be $R/I\cap J$ rather than $R/IJ$. Why should the structure sheaf be define in that way?

REPLY [5 votes]: You probably agree that for the intersection, things are easy: the most reasonable definition of $V(I)$ and $V(J)$ is as the fibred product  $Spec\,R/I\times_{Spec\,R} Spec\,R/J=Spec\,R/I\otimes_R R/J=Spec\,R/(I+J).$
Now one natural way to view the union $V(I)\cup V(J)$ is as the result of glueing $V(I)$ and $V(J)$ along their intersection, that is, as the pushout of the diagram $V(I)\leftarrow V(I+J)\rightarrow V(J)$. And it is known (see e.g. Ferrand, Conducteur, descente et pincement) that such a pushout is representable in the category of schemes by the affine scheme whose function ring is the fibred product. Then you can do the following exercise: check that the 'diagonal' map $R\to (R/I)\times_{R/(I+J)}(R/J)$ is surjective with kernel $I\cap J$. In other words the sought-for fibred product is $R/(I\cap J)$. We end up with the conclusion that this pushout viewpoint leads naturally to the definition of the scheme-theoretic structure on the union being $Spec\,R/(I\cap J)$.<|endoftext|>
TITLE: What are Mean Values of Ideal Densities in Galois Extensions?
QUESTION [13 upvotes]: In an unfinished (and as of now unpublished) article intended for the encyclopedia of mathematics, Arnold Scholz wrote:
"Classifying extensions according to the Galois group
 of their normal closure provides us with a new point
 of view. Not only the minimal discriminants but also 
 the mean values of the ideal densities differ considerably,
 and have the following values for discriminants with large
 prime factors: 

$\sqrt{\zeta(2)}$ for quadratic extensions;
$\sqrt[3]{\zeta(3)^2}$ and $\sqrt{\zeta(2)}\sqrt[3]{\zeta(3)}$
for a cubic extension according as it is cyclic or noncyclic;
$\sqrt{\zeta(2)}\sqrt{\zeta(4)}$ and $\sqrt{\zeta(2)}^3$
for cyclic and biquadratic quartic extensions, respectively."

I'd like to know what Scholz is talking about here. Ideal density might be some limit of the form  "number of ideals with norm $\le x$" / $x$, and mean value should denote some average over number fields. But what exactly is Scholz doing here?  
Edit. Apparently (this is suggested by some remarks he made elsewhere), Scholz called the expression
$$ \prod_p \phi(p^n)/\Phi_K(p) $$
the ideal density of a number field $K$, where $\phi$ and $\Phi_K$ denote Euler's phi function in the rationals and in $K$, respectively, and where $n$ denotes the degree
of $K$. This expression occurs in the product formula for the zeta function. I still don't know where to go from here.
As for Robin's remark on the density of fields ordered by discriminants, Scholz claimed, in a letter to Hasse dated  Sept. 27, 1938, the following: The Dirichlet series
$$ G(s) = \sum_{Gal(K)=G} D_K^{-s}, $$
where the sum is over all quartic fields whose normal closure has Galois group $G$, have abscissa of convergence $\alpha(D)=1$, $\alpha(Z) = \alpha(V) = \frac{1}{2}$ and
probably $\alpha(S)=1$, $\alpha(A)=\frac{1}{2}$, where $D$, $Z$, $V$, $A$, $S$ denote
the dihedral, cyclic, four, alternating and symmetric group.  Moreover,
$$ \lim_{s \to 1/2} \frac{Z(s)}{V(s)} = 0, $$
where $Z(s)$ and $V(s)$ are the Dirichlet series defined above for $G=Z$ and $G=V$. This is all correct, as we know now, but how could Scholz have discovered (and, for $G = D$, $Z$, $V$, proved) these results in the 1930s?

REPLY [2 votes]: For explaining Scholz's ideas on ideal densities, consider the Dedekind zeta
function of a number field $K$. If we collect all Euler factors for the
prime ideals above a prime number $p$, then a short calculation shows that
this product equals 
$$ E_p =  \frac{1 - \frac1p}{(1 -p^{-f_1}) \cdots (1 - p^{-f_g})}. $$
If $\Phi$ denotes Euler's phi function in $K$ and $\phi$ the usual
phi function for integers, then this equation may be rewritten in the form
$$ \lim_{s \to 1} \frac{\zeta_K(s)}{\zeta(s)} 
            = \frac{2^t\pi^shR}{v \sqrt{|D|}} 
            = \prod_p \frac{\phi(p^n)}{\Phi(p)}. $$
The mean ideal density of a family $F$ of number fields  $K_1$, $K_2$, 
$K_3$, \ldots is defined as
$$ {\rm MID}(F) = \lim_{n \to \infty}  
    \Big( \lim_{s \to 1} \frac{\zeta_{K_1}(s)}{\zeta(s)}
     \cdots \lim_{s \to 1} \frac{\zeta_{K_n}(s)}{\zeta(s)} \Big)^{\frac 1n} $$
if this limit exists. 
Let us now compute the mean ideal density of quadratic number fields
ordered by discriminants. Here 
$$ \Phi_K(p) = \begin{cases}
               (p-1)^2 & \text{ if $p$ splits},  \\
               p(p-1)  & \text{ if $p$ ramifies},   \\
                p^2-1  & \text{ if $p$ is inert},
              \end{cases} $$
hence the product of the Euler factors is
$$ E_p = \frac{\phi(p^2)}{\Phi_K(p)}
            = \begin{cases}
         \frac{p}{p-1} > 1 & \text{ if $p$ splits},  \\
         \qquad \quad    1 & \text{ if $p$ ramifies},   \\
         \frac{p}{p+1} < 1 & \text{ if $p$ is inert}
              \end{cases} $$
and we find
$$ \lim_{s \to 1} \frac{\zeta_K(s)}{\zeta(s)}
      = \prod_p \frac{p}{p-1} \prod_q \frac{q}{q+1} , $$
where $p$ runs through the split primes and $q$ through those that are inert.
Using some handwaving we expect that half the primes split and the other
half remains inert, so heuristically we should get
$$ {\rm MID}(F) = \lim_{n \to \infty}
   \Big( \prod_{p_1}  \frac{p_1}{p_1-1} \prod_{q_1} \frac{q_1}{q_1+1}
    \cdots  
  \prod_{p_n} \frac{p_n}{p_n-1} \prod_{q_n} \frac{q_n}{q_n+1}\Big)^{\frac1n}, $$
where $p_j$ are primes splitting in $K_j$ etc. Changing limits we find
$$ \lim_{s \to 1} \lim_{n \to \infty}
   \sqrt[n]{\frac{\zeta_1(s)}{\zeta(s)} \cdots \frac{\zeta_n(s)}{\zeta(s)}}
   = \Big(\prod_p \frac{p}{p-1} \prod_q \frac{q}{q+1} \Big)^{\frac12} ,$$
where the products now are over all primes $p$ and $q$. Now 
$$ \Big(\prod_p \frac{p}{p-1} \prod_p \frac{p}{p+1} \Big)^{\frac12} 
   = \Big(\prod_p \frac{p^2}{p^2-1} \Big)^{\frac12}
   = \Big(\prod_p \frac1{1-\frac1{p^2}} \Big)^{\frac12}
   = \sqrt{\zeta(2)} $$
in agreement with Scholz's claim.
This prediction does not agree well with numerical experiments, and a
closer examination reveals that this is due to our omitting the ramified
primes from our considerations. In fact, for a given prime number $p$
about $\frac{X}{p+1}$ of the discriminants below $X$ are divisible by $p$,
and in the remaining fields half the primes split and the rest are inert.
If we take this behaviour into consideration it turns out that we expect 
the mean ideal density for quadratic number fields to be
$$ {\rm MID(F)} = \sqrt[3]{\frac{2^2}{2^2-1}} \sqrt{\zeta(2)} , $$
which agrees very well with numerical experiments.
"Computing" the mean ideal densities for other families of number fields
is straightforward. I do not know how difficult it is to make these
calculations rigorous, or what to do with them.<|endoftext|>
TITLE: Triangulations coming from a poset. Or: What conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?
QUESTION [7 upvotes]: Every partially ordered set gives a triangulation of (the geometric realisation of) its order complex. (The n-simplices of the order complex are the chains $x_0\leq x_1\leq\cdots\leq x_n$.) However, there are triangulations of topological spaces that do not arise this way.
Is there a name for triangulations having this special property of "coming from a poset?"
EDIT: Apparently, the following formulation of my question is cleaner: what conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?

REPLY [13 votes]: Here are necessary and sufficient conditions for an abstract, finite simplicial complex $\mathcal{S}$ to be the order complex of some partially ordered set.
(i) $\mathcal{S}$ has no missing faces of cardinality $\geq 3$; and
(ii) The graph given by the edges (=$1$-dimensional simplices) of $\mathcal{S}$ is a comparability.
[Definitions. 
(a) A missing face of $\mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that 
$M \not \in \mathcal{S}$, but all proper subsets $P\subseteq M$ satisfy $P\in \mathcal{S}$.
 (b) A graph (=undirected graph with no loops nor multiple edges) is a comparability if its
edges can be transitively oriented, meaning that whenever edges $\{p, r_1\}, \{r_1, r_2\},\ldots, \{r_{u−1}, r_u\}, \{r_u, q\}$ are oriented as $(p, r_1), (r_1, r_2),\ldots, (r_{u−1}, r_u), (r_u, q)$, then there exists an edge $\{p, q\}$ oriented as $(p, q)$.]
This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in

M. M. Bayer, Barycentric subdivisions. Pacific J. Math. 135 (1988), no. 1, pp. 1-16. 

As Bayer points out, the result was first observed in 

R. Stanley, Balanced Cohen-Macaulay complexes, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.

@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities must be used. For this, first check that the barycentric subdivision of any simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$. 
I hope this helps.<|endoftext|>
TITLE: Invariant Vector Fields for Homogenous Spaces
QUESTION [8 upvotes]: As we all know, the space of invariant vector fields on a Lie group can be identified with the tangent space at the identity (or any other point for that matter). My question is: How does this generalize to homogeneous spaces? My guess would be that one can equate the space with the tangent space at any point point. However, it is not clear to me why everything should carry over smoothly to this more general setting.

REPLY [8 votes]: The short answer is that many (most? all?) homogeneous spaces do NOT have such a nice description.  In particular, at any point $p\in M$, the set of the $G$ or $H$ invariant vectors is a strict subset of $T_pM$.  (Here, and below, I'm assuming $G$ and $H$ are compact - it's all I'm really familiar with)
Here's the quick counterexample when looking for $G$ invariance:  Take any $H\subseteq G$ with rank$(G)$ = rank$(H)$.  Then the Euler characteristic of $G/H$ is greater than $0$.  It follows that EVERY vector field has a 0, and hence, the only $G$-invariant vector field is trivial (since $G$ acts transitively).
For (much more) detail, including how it often fails even with rank$(H)<$rank$(G)$...
Suppose $M = G/H$ (and hence, $M$ is also compact), that is, I'm thinking of $M$ as the RIGHT cosets of $G$.  For notation, set $e\in G$ as the identity and let $\mathfrak{g}$ and $\mathfrak{h}$ denote the Lie algebras to $G$ and $H$.  
Since $G$ is compact, it has a biinvariant metric.  The biinvariant metric on $G$ is equivalent to an $Ad(G)$ invariant inner product on $\mathfrak{g}$.  This gives an orthogonal splitting (as vector spaces) of $\mathfrak{g}$ as $\mathfrak{h}\oplus \mathfrak{m}$.  (The vector subspace $\mathfrak{m}$ is not typically an algebra)
Now, the tangent space $T_{eH}M$ is naturally be identified with $\mathfrak{m}$.  In fact, this identification is $H$ invariant (where $H$ acts on $\mathfrak{m}$ via $Ad(H)$ and $H$ acts on $M$ by left multiplication: $h * (gH) = (hg)H$).
Now, when you say "invariant vector fields", you must first specify $G$ invariant or $H$ invariant.  Lets focus of $G$ first, then $H$.  
Typically, there are not enough $G$-invariant vector fields to span the tangent space at any point.  This is because not only does $G$ act transitively on $M$, but it acts very multiply transitive (an $H$s worth of elements move any element to any other given element).  Thus, a $G$-invariant vector field on $M$ is actually equivalent to an $H$ invariant vector in $T_{eH}M$.
So, for example, viewing $S^{n} = SO(n+1)/SO(n)$, You'd find that $SO(n)$ acts transitively on the unit vectors in $\mathfrak{m}$, and hence, there are no $SO(n+1)$ invariant vector fields on $S^{n}$.
Now, on to $H$ invariant vector fields.  I haven't thought much about this, but even in this case, there are (often?  always?) not enough.  For example, viewing $S^2 = SO(3)/SO(2)$, we see that $H$ is a circle.  In this case, viewing the north pole of $S^2$ as $eSO(2)$, the only $H$ invariant vector fields are the velocity vectors of rotation through the north-south axis - i.e., the flows are lines of lattitude.
Then we see that at generic points, the set of all vectors tangent to an $H$ invariant vector is a 1 dimensional vector space (and hence, doesn't span the whole tangent space), and the situation at the north and south pole is even worse:  the only $H$ invariant vectors are trivial.
Even if you come up with a case where at some point, there were enough $H$ invariant vector fields, you can not translate them around all of $M$, since $H$ does NOT act transitively on $M$.<|endoftext|>
TITLE: Sarrus determinant rule: references, extensions
QUESTION [7 upvotes]: SEEKING REFERENCES FOR SARRUS' RULE AND EXTENSIONS 
An undergraduate came to me with an identity for 4x4 determinants that is actually correct:
$\det(A)=h(A)+h(RA)+h(R^{2}A)$ 
where R cyclically permutes the last three rows of the matrix A. I wont define h here but,  except for the signs of the terms, it is the usual incorrect extension of Sarrus Rule that is familiar to anyone who has has taught linear algebra (His identity  is not the Laplace expansion, as it has three 4x4 matrices, rather than four 3x3 matrices. )
Is something like this known? Mathscinet lists a paper Monaco and Monaco that might be relevant, but my library couldnt  get it. I haven't even found the original reference for Sarrus' rule itself.
Eric Schmutz

REPLY [2 votes]: I've posted a diagram version of 4x4 Sarrus here: http://www.theoremoftheday.org/GeometryAndTrigonometry/Sarrus/Sarrus4x4.pdf<|endoftext|>
TITLE: Chinese Remainder Theorem for rings: why not for modules?
QUESTION [12 upvotes]: This is a followup to Analog to the Chinese Remainder Theorem in groups other than Z_n. . I shouldn't have used the comments to ask a new question, in fact...
Here is the statement of the Chinese Remainder Theorem, as it occurs in most books and websites:
(1) Let $R$ be a commutative ring with unity, and $I_1$, $I_2$, ..., $I_n$ be finitely many ideals of $R$ such that ($I_i+I_j=R$ for any two distinct elements $i$ and $j$ of $\left\lbrace 1,2,...,n\right\rbrace$). Then, $I_1I_2...I_n=I_1\cap I_2\cap ...\cap I_n$, and the canonical ring homomorphism $R/\left(I_1I_2...I_n\right)\to R/I_1 \times R/I_2 \times ... \times R/I_n$ is an isomorphism.
But there seems to be another, even more general form of (1) which doesn't get even half of the attention:
(2) Let $R$ be a commutative ring with unity, and $I_1$, $I_2$, ..., $I_n$ be finitely many ideals of $R$ such that ($I_i+I_j=R$ for any two distinct elements $i$ and $j$ of $\left\lbrace 1,2,...,n\right\rbrace$). Let $A$ be an $R$-module. Then, $I_1I_2...I_n\cdot A=I_1A\cap I_2A\cap ...\cap I_nA$, and the canonical $R$-module homomorphism $A/\left(I_1I_2...I_n\cdot A\right)\to A/I_1A \times A/I_2A \times ... \times A/I_nA$ is an isomorphism.
I am wondering: is (2) a trivial corollary of (1)? Because otherwise I don't see any reason why (2) shouldn't appear in literature as "the" Chinese Remainder Theorem, with (1) being but a corollary. Or is (2) wrong? The only way I see to get (2) from (1) is to apply (1) to the ring $R\oplus A$ (with multiplication on $R\oplus 0$ inherited from $R$, multiplication between $R\oplus 0$ and $0\oplus A$ given by the $R$-module structure on $A$, and multiplication on $0\oplus A$ given by $0$), which seems quite artificial to me. Am I missing something very obvious?

REPLY [16 votes]: The second result you're talking about is also sometimes called the Chinese remainder theorem, and can be derived from the Chinese remainder theorem for rings by "tensoring the CRT isomorphism" with $A$. Explicitly, (1) gives
$R/\prod_{k=1}^n I_k\cong\prod_{k=1}^n R/I_k$
via the natural map. This is an isomorphism of rings as well as an isomorphism of $R$-modules. Therefore, upon tensoring with $A$, it becomes

$A/\big(\prod_{k=1}^nI_k\big)A\simeq \prod_{k=1}^n A/I_kA$
via the natural map, using the canonical isomorphism $R/I\otimes_RA\cong A/IA$ as well as the fact that tensor product commutes with finite direct products. It follows that the kernel of $A\rightarrow\prod_{k=1}^n A/I_kA$, which is clearly $\bigcap_{k=1}^n I_kA$, is equal to $\big(\prod_{k=1}^n I_k\big)A$. So, you've derived (2) from (1). Keep in mind that (2) is an isomorphism of $R$-modules, while (1) is an isomorphism of rings (as well as $R$-modules).<|endoftext|>
TITLE: Dwork's use of p-adic analysis in algebraic geometry
QUESTION [17 upvotes]: Using p-adic analysis, Dwork was the first to prove the rationality of the zeta function of a variety over a finite field. From what I have seen, in algebraic geometry, this method is not used much and Grothendieck's methods are used instead. Is this because it is felt that Dwork's method is not general or powerful enough; for example, Deligne proved the Riemann Hypothesis for these zeta functions with Grothendieck's methods, is it felt that Dwork's method can't do this?

REPLY [11 votes]: I agree with Emerton's answer (and had similar thoughts, but since he is a leading expert in this field, it's better if the answer comes from him).  I would say that if anything the opposite is true: nowadays people (especially those of an arithmetic bent) are more interested in $p$-adic cohomology than $\ell$-adic cohomology and the former is viewed as richer and more difficult than the latter.  Thus the importance of Dwork's work could scarcely be higher.
Flipping to the other side of the Weil conjectures, I think it is also not quite fair to say that Deligne proved the Riemann hypothesis "with Grothendieck's methods".  I know you mean that he used Grothendieck's methods ($\ell$-adic cohomology) rather than Dwork's methods ($p$-adic analysis) but it doesn't do justice to the range of new ideas that Deligne brought to the table (as well as the ideas that were left on Grothendieck's table, to Grothendieck's eternal consternation).  
Flipping back again, note that the Weil conjectures (and even parts of Deligne's significant generalization "Weil II") have since been proven completely by $p$-adic methods, c.f.:

Kedlaya, Kiran S.(1-MIT)
  Fourier transforms and $p$-adic `Weil II'. (English summary)
  Compos. Math. 142 (2006), no. 6, 1426--1450. 
http://arxiv.org/abs/math/0210149

This is another sign that the "Dwork school" is going strong in contemporary research.<|endoftext|>
TITLE: Did Apollonius invent co-ordinate geometry?
QUESTION [8 upvotes]: When I read descriptions of Apollonius' treatise on conics, some of them say that he invented co-ordinate geometry, some say that he kind of did and others are silent on the matter. Or is it the case that there is no simple answer?

REPLY [19 votes]: Let V be the vertex of a parabola, F its focus, X a point on its symmetry axis, and A a point on the parabola such that AX is orthogonal to VX. It was well within the power of the Greeks to prove relations such as $VX:XA = XA:4VF$. If you introduce coordinate axes, set $x = VX$, $y = XA$ and $p = VF$, you get $y^2 = 4px$, the modern form of the equation of a parabola.
Everything now depends on what "invention of coordinate geometry" means to you. I do not think that the Greeks' work on conics should be confused with coordinate geometry since they did not regard the lengths occurring above as coordinates. It's just that parts of their results are very easily translated into modern language. 
In a similar vein, Eudoxos and Archimedes already were close to modern ideas behind integration, but they did not invent calculus. Euclid, despite Heath's claim to the contrary, did not state and prove unique factorization. And Euler, although he knew the product formula for sums of four squares, did not invent quaternions (Blaschke once claimed he did).
In any case, we are much more careful now with sweeping claims such as "Appolonius knew coordinate geometry" than historians were, say, 100 years ago.<|endoftext|>
TITLE: Pontrjagin numbers and exotic spheres
QUESTION [10 upvotes]: Hi everyone, im reading Milnor's article "On manifolds homeomorphic to the 7-sphere", in which he constructs the first example of an exotic structure, id like to know if there's a particular reason why Pontrjagin classes become so relevant in defining the invariant that "detects" exotic smoothness. More generally, is there a way to interpret the Pontrjagin classes that connects them with the way the smooth structure of the manifold behaves? Or are they just happy coincidences and the most "geometric" meaning you can get from them is the obstruction of finding n-k-2 linearly independent sections on the tangent bundle?
References are much appreciated! 
Thanx in advance

REPLY [12 votes]: The Pontryagin classes of the tangent bundle are not easy to interpret (witness the Novikov conjecture), but one geometric datum that can be extracted from them is the rational cobordism class of a manifold. According to Thom, the Pontryagin numbers $p_1^{a_1}\cdots p_k^{a_k}[X]$ of a closed, oriented, smooth $4n$-manifold $X$ vanish iff there's a compact, oriented $(4n+1)$-dimensional manifold bounding a disjoint union of copies of $X$. From Thom's theorem follows Hirzebruch's, expressing the (cobordism-invariant) signature $\sigma(X)$ as a certain Pontryagin number $L(p_1,\dots,p_n)[X]$. When $n=2$, 
$$\sigma(X)=(-p_1^2+7p_2)[X]/45.$$ 
One striking thing about this formula is that $\sigma(X)$ is an integer, while $(-p_1^2+7p_2)[X]/45$ is a priori only a rational number. Milnor's observation is that an integrality theorem for a characteristic class of closed $4n$-manifolds gives rise to an invariant of those $(4n-1)$-manifolds which bound $4n$-manifolds. This is a useful principle, a variant of which also underlies Chern-Simons theory. 
We can define an invariant for homotopy 7-spheres $S$ by taking $\kappa(S)=45\sigma(Y)+p_1^2[Y,\partial Y] \mod 7$; if $Y'$ is another such bounding manifold, the difference between their invariants will be $45 \sigma(X)+p_1^2(X)$ for the closed manifold $X= (-Y') \cup_S Y$, hence a multiple of 7 by the signature theorem. (Milnor prefers the invariant $\lambda=2\kappa \mod 7$.) 
In his later work with Kervaire ("Groups of homotopy spheres I"), Milnor identifies two different reasons why a homotopy-sphere may not be (h-cobordant to) a standard sphere: (i) it may not bound a parallelizable manifold; or (ii) it may bound a parallelizable manifold, but not one which is also contractible. A homotopy 7-sphere which bounds a parallelizable 8-manifold is in fact standard, but this is not true of homotopy 8-spheres. The invariant $\lambda$ of homotopy 7-spheres is an obstruction to bounding a parallelizable 8-manifold; not a complete invariant, since Kervaire-Milnor show that there are exactly 28 h-cobordism classes.<|endoftext|>
TITLE: Why can't the Klein bottle embed in $\mathbb{R}^3$?
QUESTION [26 upvotes]: Using Alexander duality, you can show that the Klein bottle does not embed in $\mathbb{R}^3$. (See for example Hatcher's book Chapter 3 page 256.) Is there a more elementary proof, that say could be understood by an undergraduate who doesn't know homology yet?

REPLY [24 votes]: If you are willing to assume that the embedded surface $S$ is polyhedral, you can prove that it is orientable by an elementary argument similar to the proof of polygonal Jordan Theorem. Of course the proof is translation of a homology/transversality/separation argument.
Fix a direction (nonzero vector) which is not parallel to any of the faces. For every point $p$ in the complement of $S$, consider the ray starting at $p$ and goint to the chosen direction. If this ray does not intersect edges of $S$, count the number of intersection points of the ray and the surface. If this number is even, you say that $p$ is black, otherwise $p$ is white. If the ray intersects an edge of $S$, you paint $p$ the same color as some nearby point whose ray does not intersect edges. It is easy to see that the color does not change along any path in the complement of $S$ (it suffices to consider only polygonal paths avoiding points whose rays contain vertices of $S$).
Now take points $p$ and $q$ near the surface such that the segment $pq$ is parallel to the chosen direction. Then they are of different colors. But if the surface is non-orientable, you can go from $p$ to $q$ along a Mobius strip contained in the surface. This contradicts the above fact about paths in the complement of $S$.<|endoftext|>
TITLE: If F is left adjoint to G, when does FG preserve limits? When do counits interchange with limits?
QUESTION [13 upvotes]: Motivation
Suppose that $F\colon X\to A$ is left adjoint to $G\colon A\to X$, and let
$\varepsilon\colon FG\stackrel{.}{\to}I_A$ be the counit of the adjunction.
Suppose also that $A$ is $J$-complete (for some category $J$), so that
$\operatorname{Lim}$ is a functor $C^J\to C$, where for an arrow
$\alpha\colon T_1\stackrel{.}{\to} T_2$ of $C^J$,
$\operatorname{Lim}(\alpha)$ is the unique arrow of $A$ for which the
following diagram is commutative:
$$
\begin{matrix}
\operatorname{Lim}(T_1)& \stackrel{\text{limiting cone}}{\longrightarrow} & T_1\\
| & & |\\
 \operatorname{Lim}(\alpha) & &  \alpha\\
\downarrow & & \downarrow \\
\operatorname{Lim}(T_2)& \stackrel{\text{limiting cone}}{\longrightarrow} & T_2
\end{matrix}
$$
Let $T\colon J\to A$ be a functor.  We have the natural transformation
$\varepsilon T\colon FGT\stackrel{.}{\to} T$, and
$\operatorname{Lim}(\varepsilon T)$ is the dotted line making the
following diagram commutative:
$$
\begin{matrix}
\operatorname{Lim}(FGT)& \stackrel{\text{limiting cone}}{\longrightarrow} & FGT\\
| & & |\\
 \operatorname{Lim}(\varepsilon T) & &  \varepsilon T\\
\downarrow & & \downarrow \\
\operatorname{Lim}(T)& \stackrel{\text{limiting cone}}{\longrightarrow} & T
\end{matrix}
$$
If $FG$ preserves $J$-limits, and
$\tau\colon \operatorname{Lim}(T)\stackrel{.}{\to}T$ is the lower limiting cone,
then $FG\tau\colon FG\operatorname{Lim}(T)\stackrel{.}{\to}FGT$ is the upper
limiting cone, and the above diagram becomes
$$
\begin{matrix}
FG\operatorname{Lim}(T)& \stackrel{FG\tau}{\longrightarrow} & FGT\\
| & & |\\
 \operatorname{Lim}(\varepsilon T) & &  \varepsilon T\\
\downarrow & & \downarrow \\
\operatorname{Lim}(T)& \stackrel{\tau}{\longrightarrow} & T
\end{matrix}
$$
Since the naturality of $\varepsilon$ implies that for all $j\in
\operatorname{obj}(J)$ the diagram
$$
\begin{matrix}
FG\operatorname{Lim}(T)& \stackrel{FG\tau_j}{\longrightarrow} & FGT(j)\\
| & & |\\
\varepsilon_{\mathrm{Lim}T}& &  \varepsilon_{T(j)}\\
\downarrow & & \downarrow \\
\operatorname{Lim}(T)& \stackrel{\tau_j}{\longrightarrow} & T(j)
\end{matrix}
$$
is commutative, it follows that $\varepsilon_{\mathrm{Lim}T}$
can replace $\operatorname{Lim}(\varepsilon T)$ in the last but one
diagram while keeping it commutative. By uniqueness, we get
the nice equation
$$
\varepsilon_{\mathrm{Lim}T} = \operatorname{Lim}(\varepsilon T).
$$
Note that it seems that all depends on $FG$ preserving $J$ limits.
Question
If $F\colon X\to A$ is left adjoint to $G\colon A\to X$ and $A$ has $J$-limits,
when does $FG$ preserve $J$-limits?
This is obviously true when $F$ preserves limits (for example, when
there is also a left adjoint to $F$), but are there other interesting
situations? 
Background
For solving an exercise from Mac Lane, I used some
results from A. Gleason, ''Universally locally connected
refinements,'' Illinois J. Math, vol. 7 (1963), pp. 521--531.  In that
paper, Gleason constructs a right adjoint to the inclusion functor
$\mathbf{L\ conn}\subset \mathbf{Top}$ ($\mathbf{L\ conn}=$ locally
connected spaces with continuous maps), and proves that the counit
of the product of two topological spaces is the product of the
counits (Theorem C). This made me curious when do counits
and limits interchange.

REPLY [2 votes]: Have you looked at the paper by B. Eckmann and P. J. Hilton entitled "Commuting Limits with Colimits" in the "Journal of Algebra", 11, 116-144 (1969)?<|endoftext|>
TITLE: Generating Classical Groups over Finite Local Rings
QUESTION [5 upvotes]: I am interested in classical groups (in particular $SL_n$, $Sp_{2n}$, $SO_n^{+}$) over finite
rings of the form $$R_k=\mathbb{F}_q[t]/(t^k)$$ for some prime power $q$ (where $q$ is odd in the 
orthogonal case) and $k \in \mathbb{N}$. Over a finite field, the maximal subgroups of the 
classical groups are known, and so it is for example known that any two semisimple elements of 
orders $q^n+1$ and $q^n-1$ generate the group $Sp_{2n}(\mathbb{F}_q)$ (and similar results for 
the other classical groups). I am wondering whether 
it is true that any two semisimple elements of orders $q^{n(k-1)}(q^n+1)$ and $q^{n(k-1)}(q^n-1)$ generate 
$Sp_{2n}(R_k)$. Is anyting like this known? Are there lists of maximal subgroups for classical
groups over finite rings? I would also be interested in similar results over $\mathbb{Z}_p/(p^k)$.
EDIT:  The answer to the question as asked is usually vacuously yes.  Instead, asm meant actually to ask about the group generated by two tori (not just two semisimple elements) of the specified orders, specifically they said (in a now deleted answer):
"I misphrased my question. What I meant to ask is whether any two maximal tori of order $q^{n(k-1)}(q^n+1)$ and $q^{n(k-1)}(q^n-1)$ generate $Sp_{2n}(R_k)$. Of course these tori are far from cyclic. I guess my head was still in the cyclic case $Sp_{2n}(\mathbb{F}_q)$."

REPLY [3 votes]: The answer is YES (at least if $n,k \ge 2$), because elements of the orders you prescribe do not exist!  The problem is that the orders of elements in the kernel of $\operatorname{Sp}_{2n}(R_k) \to \operatorname{Sp}_{2n}(R_1) = \operatorname{Sp}_{2n}(\mathbf{F}_q)$ are quite small.
Let $A$ and $B$ be your two elements.  I am assuming that by semisimple, you mean in particular that the images of $A$ and $B$ in $\operatorname{Sp}_{2n}(R_1)$ should be semisimple.  You are requiring $A^{q^n+1}$ and $B^{q^n-1}$ to have order $q^{n(k-1)}$.  By the semisimplicity, these powers would have to be of the form $I+tM$.  But then they are killed by raising to the $q^r$ power already for $r:=n(k-1)-1$, since $q^r \ge 2^{2(k-1)-1} \ge k$ and hence
$$(I+tM)^{q^r} = I + t^{q^r} M^{q^r} = I$$
in $\operatorname{Sp}_{2n}(R_k)$.  This contradicts your specifications.<|endoftext|>
TITLE: Is the category commutative monoids cartesian closed?
QUESTION [7 upvotes]: Is the category commutative monoids cartesian closed?
I understand it breaks down at the point at which the evaluation map cannot be confirmed, however this is not enough to show it is not cartesian closed.
Thanks!

REPLY [21 votes]: Here's a simple observation: the category of commutative monoids has an object that is both initial and terminal (just like the category of groups or abelian groups), so it cannot be cartesian closed. Indeed, if a cartesian closed category $\mathcal{C}$ has the property that $0 \cong 1$, then
$$\mathcal{C}(X, Y) \cong \mathcal{C}(1, Y^X) \cong \mathcal{C}(0, Y^X) \cong 1$$
and so $\mathcal{C}$ is equivalent to the trivial category with one object and one morphism.<|endoftext|>
TITLE: Question on a theorem of Eisenbud's and Harris' "The geometry of schemes"
QUESTION [6 upvotes]: My problem is perhaps a general lack of understanding but it occurred in a special case of a theorem in Eisenbud's and Harris' "The geometry of schemes" (Theorem VI-29). Let $K$ be a field and $n\in\mathbb{N}$.

Given a closed subscheme $X$ of $\mathbb{P}^n$ with Hilbert Polynomial $P$. The tangent space $T_{[X]}$ to the Hilbert scheme $\mathcal{H}_{P}$ at the point corresponding to $X$ is the space of global sections of the normal sheaf $\mathcal{N}_{\mathbb{P}^n/X}$.

I don't know what "is" precisely means here. $T_{[X]}$ has a natural $K([X])$-vector space structure and I would like to have an isomorphism of $K([X])$-vector spaces but my problem begins much earlier in the definition of $\mathcal{H}om$-sheaves:
Let $J$ be the ideal sheaf of $X$. The presheaf
$$
U \mapsto \mathcal{H}om_{\mathcal{O}_X|_U}(J/J^2|_U,\mathcal{O}_X|_U)
$$
is already a sheaf of $\mathcal{O}_X$-modules (this does not depend on the special arguments here). This sheaf $\mathcal{N}_{\mathbb{P}^n/X}$ is called the normal sheaf of $X$ in $\mathbb{P}^n$. The global sections
$$
\mathcal{N}_{\mathbb{P}^n/X}(X)=\mathcal{H}om_{\mathcal{O}_X}(J/J^2|_X,\mathcal{O}_X)
$$
of this normal sheaf is an $\mathcal{O}_X(X)$-module. Is there an isomorphism of $\mathcal{N}_{\mathbb{P}^n/X}(X)$ to some $hom$ of rings or modules where I can work with? What is a possible $K([X])$-vector space structure on it?

REPLY [4 votes]: Dear roger123, 
This is largely a response to your question aksed as a comment below Charles Siegel's answer, but it won't fit in the comment box.  Since $\mathcal N_{\mathbb P^n/X}$ is a sheaf of modules over the sheaf of rings $\mathcal O_{\mathbb P^n}$ (the structure sheaf of projective space),
its global sections $\mathcal N_{\mathbb P^n/X}(\mathbb P^n)$ are a module over the ring $\mathcal O_{\mathbb P^n}(\mathbb P^n)$, which in turn are just $k$ (the ground field).
In short, the global sections of $\mathcal N_{\mathbb P^n/X}$ form a $k$-vector space.
Maybe you are being confused by the fact that $\mathcal N_{\mathbb P^n/X}$ is a sheaf on $\mathbb P^n$ that is supported on $X$, so that people often simultaneously regard it as a sheaf on either $X$ or $\mathbb P^n$. This is okay, because if $U$ is any open in $\mathbb P^n$ and $\mathcal F$ is a sheaf supported on $X$, then the sections over an open subset $U$ of $\mathbb P^n$
(when it is regarded as a sheaf on $\mathbb P^n$) will coincide with the sections over 
$U\cap X$ (when it is regarded as a sheaf on $X$).  
In particular, one has the equation $\mathcal N_{\mathbb P^n/X}(X) =
\mathcal N_{\mathbb P^n/X}(\mathbb P^n)$ (an abuse of notation if taken literally; however
one is supposed to regard $\mathcal N_{\mathbb P^n/X}$ as a sheaf on $X$ on the left-hand side,
and as a sheaf on $\mathbb P^n$ on the right-hand side).
One more thing: If the residue field of the Hilbert scheme at the point $P$ is $k(P)$,
then $P$ is a $k(P)$-valued point of the Hilbert scheme,
and  so the corresponding closed subscheme $X$ lies in $\mathbb P^n_{k(P)}$.  Thus, in the
above discussion, $k$ can (and should) be taken to be $k(P)$.  Thus the above discussion explains
why $\mathcal N_{\mathbb P^n/X}(X)$ is a $k(P)$-vector space, as it should be.<|endoftext|>
TITLE: Finding questions between functional analysis and set theory
QUESTION [13 upvotes]: Are there some good questions on functional analysis whose solution depends on tools in set theory? My major is mathematical logic, I think tools in set theory, especially infinity combinatorics and forcing, should be used to solve some questions in functional analysis. For functional analysis, I just have read the main part of Conway's textbook. In this book, I have not found such questions.

REPLY [7 votes]: Chris Phillips and Nik Weaver wrote a paper called The Calkin Algebra has Outer Automorphisms, where they showed that the Continuum Hypothesis implies that the Calkin algebra $\mathcal{B(H)/K(H)}$ has outer automorphisms.  See also this paper of Farah, McKenney, and Schimmerling, and references therein; they show that it is relatively consistent with ZFC that the Calkin algebra has only inner automorphisms, and hence the question of existence of outer automorphisms is independent of ZFC.<|endoftext|>
TITLE: Brownian motion and spheres
QUESTION [17 upvotes]: Consider a Brownian motion on $[0;1]$. A (finite) discrete approximation of this Brownian motion consists of $N$ iid Gaussian random variables $\Delta W_i$ of variance $\frac{1}{N}$:
$$ W\left(\frac{k}{N}\right) = \sum_{i=1}^k \Delta W_i. $$
The vector $V_N = (\Delta W_1, \ldots, \Delta W_N) \in \mathbb{R}^N$ has a norm approximately equal to $1$ since the random variable $\|V_N\|^2$ has a variance equal to $\frac{C}{N}$ and $\frac{C}{N} \to 0$ (basic concentration of measure results can make this statement more precise). This is why (?) one can approximately say that in order to sample a Brownian path, it suffices to sample a point uniformly on the unit sphere of $\mathbb{R}^N$.
Question: letting $N$ go to $\infty$, how can one formalize (if possible and/or correct) the idea that a Brownian path on $[0;1]$ is like a point uniformly chosen on the unit sphere of an infinite dimensional Banach space ?

REPLY [3 votes]: H.P. McKean wrote a fun paper on this topic. He argues that one can think of Wiener measure as uniform measure on an infinite-dimensional sphere with radius "$\sqrt{\infty}$".
See http://www.jstor.org/pss/2959482
for a copy of the paper.<|endoftext|>
TITLE: Avoiding Minkowski's theorem in algebraic number theory.
QUESTION [25 upvotes]: For any course in algebraic number theory, one must prove the finiteness of class number and also Dirichlet's unit theorem. The standard proof uses Minkowski's theorem. Is there a way to avoid it?
The reasons I am asking this question are the following.
$1$. Minkowski lived long after Dirichlet and Dedekind(esp Dirichlet). So the original proof cannot likely have used Minkowski's theorem as such. If the original proof did use Minkowski's theorem, then it was of course found by someone else, most probably Dirichlet, and it is unfair to use the name Minkowski's theorem.
$2$. Even more importantly, the finiteness of classnumber and some version of unit theorem is true(at least I hope so) for all global fields. And there of course one cannot talk of Minkowski's theorem.
The objection I have for Minkowski's theorem is that it seems to be ad hoc, coming out of nowhere. And it seems that not much work is going on nowadays in the subject of geometry of numbers.
So it will be really nice to have a method which would feel more natural and is perhaps more general.

REPLY [23 votes]: For a course on algebraic number theory, you certainly can prove the finiteness of the class group without Minkowski's theorem.  For example, if you look in Ireland-Rosen's book you will find a proof there which they attribute to Hurwitz.  It gives a worse constant (which depends on a choice of $\mathbf Z$-basis for the ring of integers of the number field; changing the basis can shrink the constant, but it's still generally worse than Minkowski's) but it is computable and you can use it to show, say, that $\mathbf Z[\sqrt{-5}]$ has class number 2. 
As for the history of the proof of the unit theorem, it was proved by Dirichlet using the pigeonhole principle.  If you think about it, Minkowski's convex body theorem is a kind of pigeonhole principle (covering the convex body by translates of a fundamental domain for the lattice and look for an overlap).  You can find a proof along these lines in Koch's book on algebraic number theory, published by the AMS.  Incidentally, Dirichlet himself proved the unit theorem for rings of the form $\mathbf Z[\alpha]$; the unit theorem is true for orders as much as for the full ring of integers (think about Pell's equation  $x^2 - dy^2 = 1$ and the ring $\mathbf Z[\sqrt{d}]$, which need not be the integers of $\mathbf Q(\sqrt{d})$), even though some books only focus on the case of a full ring of integers.  Dirichlet didn't have the general conception of a full ring of integers.
One result which Minkowski was able to prove with his convex body theorem that had not previously been resolved by other techniques was Kronecker's conjecture (based on the analogy between number fields and Riemann surfaces, with $\mathbf Q$ being like the projective line over $\mathbf C$) that every number field other than $\mathbf Q$ is ramified at some prime.<|endoftext|>
TITLE: Right actions of operads and monads
QUESTION [10 upvotes]: Given an operad $A$, there is an associated monad $M_A$ given by $M_A(X) = \coprod_n A(n) \otimes X^{\otimes n}$ such that being an $A$-algebra and being an algebra over the monad $M_A$ is the same thing (the categories are equivalent). This is very classical. However, there is also the notion of a right action of an operad on $X$, given by compatible maps
$$
X^{\otimes n} \otimes A(k_1) \otimes \cdots \otimes A(k_n) \to X^{\otimes \sum k_i}
$$
(one could also replace $X^{\otimes n}$ by a more general sequence of spaces $X(n)$).
Also, a right module over a monad $M \in End(C)$ is a functor $F\colon C \to D$ to some other category with a right action map $F \circ M \to F$ which is unital and associative.
Is there any relationship between these two concepts? Is there a purely monadic way of describing a right action of a monad $M$, which specializes to the right action of an operad $A$ if $M=M_A$?

REPLY [3 votes]: I think I found a construction that looks right in many circumstances. In the simplest case of a monad $M$ on sets, define a right action of $M$ on an object $X$ in a concrete category C as a natural transformation of functors in sets $Y$,
$$
X \times M(Y) \to M(X \times Y)
$$
where $X \times S$ for a set $S$ denotes the $S$-fold coproduct of $X$ in the category C. There is a unitality and an associativity condition.
For example, if $C$ is the category of commutative algebras and $M$ is the free commutative monoid monad, then a right $M$-action on $X$ is precisely a bialgebra structure on $X$. This isn't entirely obvious. You can enrich this definition and do it over categories other than sets.
I know this is very brief, but I don't know if there are people still following this... So if you want to know more, have comments, or have seen something like that before, let me know.<|endoftext|>
TITLE: A Poisson Geometry Version of the Fukaya Category
QUESTION [11 upvotes]: Is there any possibility of a Poisson Geometry version of the Fukaya category? Given a Poisson manifold Y, objects could be manifolds with isolated singularities X which have the property that TX is contained in NX maximally. The naive example would be something like the Poisson structure on R^2 which is (x^2 + y^2)(d/dx^d/dy). Branes would in this case be curves with some nodal singularity at the origin. 
The morphisms could still be from holomorphic disks with respect to the standard complex structure. In principal, it seems like in this example the Fukaya category could be defined in the standard way (although maybe there is something more subtle one should do with the morphisms?). For a brane L passing through the origin there should be some interesting multiplicative structure in the algebra A(L) owing to the fact that the brane is required to remain fixed at the origin.... I would hope that the Hochschild cohomology could be related to the Poisson cohomology of the manifold though I haven't studied my example yet. What obstructions arise when trying to construct this category?

REPLY [4 votes]: How about looking at the source-simply connected symplectic groupoid of an integrable Poisson manifold. and then forming ITS Fukaya category?    This symplectic groupoid is naturally attached to the Poisson manifold, so any invariant of it is a Poisson invariant.   If M has the zero Poisson structure, the symplectic groupoid is T*M.  If M is the dual g* of a Lie algebra, the groupoid is the cotangent bundle of the simply connected Lie group G.  If M is symplectic,  the groupoid is the fundamental groupoid of M (just $M x M^{opp}$ if M is simply connected).<|endoftext|>
TITLE: Reference for the existence of a Shapovalov-type form on the tensor product of integrable modules
QUESTION [7 upvotes]: Shapovalov and Jantzen showed us how to construct a nice inner product on finite dimensional representations of a semi-simple Lie algebra, by simply giving the highest weight vector inner product 1 with itself and making the upper and lower halves adjoint.
The result I need is an extension of this to tensor products.  Roughly, I would like a statement like:

There is a unique system of $U_q(\mathfrak{g})$-invariant Hermitian inner products on all tensor products $V_{\lambda_1}\otimes \cdots \otimes V_{\lambda_\ell}$ such that

On $V_\lambda$, it is the Shapovalov form.
The action of $E_i$ and $F_i$ are biadjoint (up to some powers of $q$). 
For any $j<\ell$, the natural map $V_{\lambda_1}\otimes\cdots\otimes V_{\lambda_j}\hookrightarrow V_{\lambda_1}\otimes \cdots \otimes V_{\lambda_\ell}$ is an isometric embedding. 


I said to myself, "Self, it would be silly to post this question on MathOverflow.  You are in a math library, just feet away from Lusztig's book. Surely it is in there."  However, I've had no luck finding it in Lusztig's book, which is sadly lacking in index.  Is this actually written down anywhere?
EDIT: Jim asks for more motivation.  I feel like this is the sort of question where motivation will not be very helpful in actually finding an answer, but there's no harm in saying a little (and it will allow me to put off real work).  
One of the foundational principles of categorification is that things with nice categorifications have nice inner products (since Grothendieck groups have a nice inner product given by Euler characteristic of the Ext's between objects).  I'm working right now on categorifying tensor products of representations, so it would be rather convenient for me to find some earlier references that used this form.

REPLY [6 votes]: I know a couple of ways to get a Shapovalov type form on a tensor product. The details of what I say depends on the exact conventions you use for quantum groups. I will follow Chari and Pressley's book. 
The first method is to alter the adjoint slightly. If you choose a * involution that is also a coalgebra automorphism, you can just take the form on a tensor product to be the product of the form on each factor, and the result is contravariant with respect to *. There is a unique such involution up to some fairly trivial modifications (like multiplying $E_i$ by $z$ and $F_i$ by $z^{-1}$). It is given by: 
$$
*E_i = F_i K_i, \quad *F_i=K_i^{-1}E_i, \quad *K_i=K_i,
$$
The resulting forms are Hermitian if $q$ is taken to be real, and will certainly satisfy your conditions 1) ad 3). Since the $K_i$s only act on weight vectors as powers of $q$, it almost satisfies 2).
The second method is in case you really want * to interchange $E_i$ with exactly $F_i$. This is roughly contained in this http://www.ams.org/mathscinet-getitem?mr=1470857  paper by Wenzl, which I actually originally looked at when it was suggested in an answer to one of your previous questions.
It is absolutely essential that a * involution be an algebra-antiautomorphism. However, if it is a coalgebra anti-automorphism instead of a coalgebra automorphism there is a work around to get a form on a tensor product. There is again an essentially unique such involution, given by
$$
*E_i=F_i, \quad *F_i=E_i, \quad *K_i=K_i^{-1}, \quad *q=q^{-1}.
$$
Note that $q$ is inverted, so for this form one should think of $q$ as being a complex number of the unit circle. By the same argument as you use to get the Shapovalov form, then is a unique sesquilinear *-contravariant form on each irreducible representation $V_\lambda$, up to overall rescaling. 
To get a form on $V_\lambda \otimes V_\mu$, one should define
$$(v_1 \otimes w_1, v_2 \otimes w_2)$$
to be the product of the form on each factor applied to $v_1 \otimes w_1$ and $R( v_2 \otimes w_2)$, where $R$ is the universal $R$ matrix. It is then straightforward to see that the result is *-contravariant, using the fact that $R \Delta(a) R^{-1} =\Delta^{op}(a).$
If you want to work with a larger tensor product, I believe you replace $R$ by the unique endomorphism $E$ on $\otimes_k V_{\lambda_k}$ such that $w_0 \circ E$ is the braid group element $T_{w_0}$ which reverses the order of the tensor factors, using the minimal possible number of positive crossings. Here $w_0$ is the symmetric group element that reverses the order of the the tensor factors.  
The resulting form is *-contravariant, but is not Hermitian. In Wenzl's paper he discusses how to fix this. 
Now 1) and 2) on your wish list hold. As for 3): It is clear from standard formulas for the $R$-matrix (e.g. Chari-Pressley Theorem 8.3.9) that $R$ acts on a vector of the form $b_\lambda \otimes c \in V_\lambda \otimes V_\mu$ as multiplication by $q^{(\lambda, wt(c))}$. Thus if you embed $V_\mu$ into $V_\lambda \otimes V_\mu$ as $w \rightarrow b_\lambda \otimes w$, the result is isometric up to an overall scaling by a power of $q$. This extends to the type of embedding you want (up to scaling by powers of $q$), only with the order reversed. I don't seem to understand what happen when you embed $V_\lambda$ is $V_\lambda \otimes V_\mu$, which confuses me, and I don't see your exact embeddings.<|endoftext|>
TITLE: When your paper makes a borderline case for a top journal
QUESTION [25 upvotes]: Say you write a paper that you feel makes a borderline case for publication in a prestigious journal (Annals, Acta, Inventiones, JAMS, etc).  What are the advantages and disadvantages of submitting the paper to the prestigious journal over a less prestigious journal where your paper is very likely to be accepted? My novice take is that there is little downside since if the paper is rejected you can always then submit it to a less prestigious journal. Are there factors I'm not considering? Of course, in most cases, the paper is Arxiv'ed early on so one's claim to the result is never hurt by a lengthy publication process. 
I'd be interested to also hear about the implications on job applications. How much does it hurt a job applicant to have a very good result Arxiv'ed and submitted versus accepted? (Presumably if the validity of the result was a make or break factor for a job the hiring group could take a look at the preprint or talk to an expert in the field. Maybe that is an unrealistic expectation).

REPLY [27 votes]: This requires a conditional response.  First, who is submitting?

Tenured professor:  Not much repurcussion either way.
Worth the wait if you think it deserves publication.
Graduate student:  Some downside to waiting a long time
and not having the thing published come application time, but this can happen
at a lesser journal, too (for grads, the paper is often
submitted only shortly before they are graduating). 
And postdoc hiring is not
as publications-based as tenure-track hiring.  Letters
of recommendation mean more.
Postdoc.  Here, submitting with a high probability of
a rejection after a long wait can be a major gambit.
Mitigating factor:  does the postdoc have other worthy
publications?  If this will be the flagship result,
it's hard not to think that a slightly lesser journal
would have a higher expected yield (in terms of jobs).
Note that it is true that some journals may reject
your paper quickly -- then you can turn around and
submit somewhere else -- but those papers are not really
the marginally-great ones being asked about.

In the end, you are left with a hard decision.
I don't think there's a formula which can help. 
In this case, you either go for broke or you don't.

Collaboration:  decide based on the member with the
most to lose.<|endoftext|>
TITLE: Non-degenerate multilinear forms
QUESTION [12 upvotes]: Is there a standard notion of non-degeneracy for multilinear forms?
My motivation is simple curiosity, by the way!

REPLY [6 votes]: Basic idea: a bilinear form $B$ is degenerate if there are two nonzero vectors $v$ and $w$ so that not only is
$$B(v,w)=0,$$
but also either vector is enough to kill $B$ without help from the other:
$$B(v,-) = 0 \mbox{   and   } B(-,w)=0.$$
Another way to say it: even if we perturb $v$ and $w$, $B(v \otimes w)=0$ to first order.  From this point of view, we see that a bilinear form is degenerate iff it is an element of the variety dual to the Segre embedding of $\mathbb{P}V \times \mathbb{P}W$ in $\mathbb{P}(V \otimes W)$.
This motivation generalizes gracefully to the following definition from Gelfand, Kapranov, and Zelevinsky's book Discriminants, resultants, and multidimensional determinants:
A $p$-linear form $T$ is said to be degenerate if either of the following equivalent conditions holds:

there exist nonzero vectors $\beta_i$ so that, for any $1 \leq j \leq p$,
$$ T \left( \beta_1, \ldots , \beta_{j-1} , x_j ,\beta_{j+1}, \ldots , \beta_{p} \right) = 0 \mbox{ for all $x_{j}$;} $$
there exist nonzero vectors $\beta_{i}$ so that $T$ vanishes at $\otimes \beta_{i}$ along with every partial derivative with respect to an entry of some $\beta_{i}$:
$$ T \mbox{ and } \frac{\partial T}{\partial \beta^{(j)}_{i}} \mbox{ vanish at $\otimes \beta_{i}$.}$$

In certain favorable cases (when the dimensions of the vector spaces $V_i$ are not too different) the dual to the Segre is a hypersurface; in other words, there is a single polynomial---the hyperdeterminant---which vanishes exactly at the degenerate multilinear forms.  This polynomial possesses many magical properties and is much subtler than determinants of bilinear forms.
I can attest that this definition is at least useful, if not standard, since it came up in a substantial way in an elementary question about coin flipping: http://arxiv.org/abs/1009.4188 .<|endoftext|>
TITLE: Can one "soup-up" the LAW OF THE MEAN in the following way?
QUESTION [5 upvotes]: Let J be a closed interval of real numbers whose length is finite and positive. Let f be a real valued
function defined on J which has a continuous second derivative at all points of J.
QUESTION: If P1,P2,P3 are any three pairwise distinct and non-collinear points on the graph of f(J), does
there always exist at least one point p on J such that the absolute value of the curvature of this graph
at the point (p,f(p)) is greater than or equal to 1/r-where r is the radius of the circle through the three
points P1,P2,P3?

REPLY [4 votes]: The answer is yes. Suppose the contrary and rescale the picture so that $r=1$. We may assume that $P_1$ and $P_3$ are the endpoints of the graph. There must be points on the graph that are outside the circle - otherwise the curvature at $P_2$ is at least 1. WLOG assume that there are points below the circle. The lower half of the circle is the graph of a function $f_0$ defined on an interval of length 2. Let $q$ be a point where $f_0-f$ attains its maximum, then $f'(q)=f_0'(q)$. We may assume that $f'(q)=f_0'(q)\ge 0$, otherwise reflect the picture through the $y$-axis.
I claim that $f(t)<f_0(t)$ for all $t\ge q$ such that both $f(t)$ and $f_0(t)$ are defined, contrary to the fact that $P_2$ lies on the circle. To prove this, consider arc-length parametrizations $\gamma(t)=(x(t),y(t))$ and $\gamma_0(t)=(x_0(t),y_0(t))$ of the two graphs, where $\gamma(0)=(q,f(q))$ and $\gamma_0(0)=(q,f_0(q))$. Then
$$
 \gamma'(0)=\gamma_0'(0)=(\cos\alpha,\sin\alpha)
$$
where $0\le\alpha<\pi/2$. Since the curvature of $\gamma$ is less than 1, we have
$$
 \angle(\gamma'(t),\gamma'(0)) < t
$$
for all $t>0$. Therefore $x'(t)>\cos(\alpha+t)$ and $y'(t)<\sin(\alpha+t)$ for $0<t<\pi/2-\alpha$. And for $\gamma_0$ these inequalities turn to equalities. The integration yields that $x(t)\ge x_0(t)$ and $y(t)< y_0(t)$ for all $t\in[0,\pi/2-\alpha]$. Since $f_0$ is increasing after $q$, these inequalities imply that $\gamma(t)$ is below the half-circle (or has already left the domain where $f_0$ is defined).
Since $\gamma_0(\pi/2-\alpha)$ is the rightmost point of the circle, the inequality $x(t)\ge x_0(t)$ for $t=\pi/2-\alpha$ means that the $x$-coordinate of $\gamma(\pi/2-\alpha)$ is already outside the domain of $f_0$ and one does not need to care about larger values of $t$.<|endoftext|>
TITLE: Bringing Number and Graph Theory Together: A Conjecture on Prime Numbers
QUESTION [33 upvotes]: Some MOers have been skeptic whether something like natural number graphs can be defined coherently such that every finite graph is isomorphic to such a graph. (See my previous questions [1],  [2], [3], [4])
Without attempting to give a general definition of natural number graphs, I invite you to consider the following

DEFINITION
A natural number $d$ may be called demi-prime
  iff there is a prime number $p$ such
  that $d = (p+1)/2$. The demi-primes' distribution is exactly
  like the primes, only shrinked by the
  factor $2$:
$$2, 3, 4, 6, 7, 9, 10, 12, 15, 16, 19, 21, 22, 24, 27, 30, 31, 34, 36, 37, 40, 42, 45, 49, ...$$
Let D($k,n$) be the set which consists
  of the $k$-th up to the $(k+n-1)$-th
  demi-prime number. 

After some - mildly exhaustive - calculations I feel quite confident to make the following

CONJECTURE
For every finite graph $G$ there is a $k$ and a bijection $d$ from the vertex set
  $V(G)$ to D($k,|G|$) such that $x,y$ are adjacent if and only if $d(x),d(y)$ are coprime.

I managed to show this rigorously for all graphs of order $n\leq $ 5 by brut force calculation, having to take into account all (demi-)primes $d$ up to the 1,265,487th one for graphs of order 5. For graphs of order 4, the first 1,233 primes did suffice, for graphs of order 3 the first 18 ones. 
Looking at some generated statistics for $n \leq$ 9 reveals interesting facts(1)(2), correlations, and lack of correlations, and let it seem probable (at least to me) that the above conjecture also holds for graphs of order $n >$ 5. 
Having boiled down my initial intuition to a concrete predicate, I would like to pose the following

QUESTION
Has anyone a clue how to prove or disprove the above conjecture?

My impression is that the question is about the randomness of prime numbers: Are they distributed and their corresponding demi-primes composed randomly enough to mimick – via D($k,n$) and coprimeness – all (random) graphs? 

(1) E.g., there is one graph of order 5 - quite unimpressive in graph theoretic terms - that is very hard to find compared to all the others: it takes 1,265,487 primes to find this guy, opposed to only 21,239 primes for the second hardest one. (Lesson learned: Never stop searching too early!) It's – to whom it is of interest –  $K_2 \cup K_3$:
0 1 0 0 0
1 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0

(2) Added: This table shows the position of the smallest prime (among all primes) needed to mimick the named graphs of order $n$. All values not shown are greater than $\approx 2,000,000$
order    |   3     4      5      6       7      8
-------------------------------------------------
empty    |  14    45     89     89      89   3874 
complete |   5    64    336   1040   10864  96515 
path     |   1     6   3063  21814  
cycle    |   5   112  21235  49957

REPLY [2 votes]: On the general theme of Gödel codings of graphs, see my work on Riffs and Rotes, for example, here.<|endoftext|>
TITLE: What is the status of the Gauss Circle Problem?
QUESTION [62 upvotes]: For $r > 0$, let $L(r) = \# \{ (x,y) \in \mathbb{Z}^2 \ | \ x^2 + y^2 \leq r^2\}$ be the number of lattice points lying on or inside the standard circle of radius $r$.  It is easy to see that $L(r) \sim \pi r^2$ as $r \rightarrow \infty$.  The Gauss circle problem is to give the best possible error bounds: put
$E(r) = |L(r) - \pi r^2|$.
Gauss himself gave the elementary bound $E(r) = O(r)$.  In 1916 Hardy and Landau showed that it is not the case that $E(r) = O(r^{\frac{1}{2}})$.  It is now believed that this is "almost" true: i.e.:
Gauss Circle Conjecture: For every $\epsilon > 0$, $E(r) = O_{\epsilon}(r^{\frac{1}{2}+\epsilon})$.
So far as I know the best published result is a 1993 theorem of Huxley, who shows one may take $\epsilon > \frac{19}{146}$.
(For a little more information, see here)
In early 2007 I was teaching an elementary number theory class when I noticed that Cappell and Shaneson had uploaded a preprint to the arxiv claiming to prove the Gauss Circle Conjecture:

http://arxiv.org/abs/math/0702613

Two more versions were uploaded, the last in July of 2007.
It is now a little more than three years later, and so far as I know the paper has neither been published nor retracted.  This seems like a strange state of affairs for an important classical problem.  Can someone say what the status of the Gauss Circle Problem is today?  Is the argument of Cappell and Shaneson correct?  Or is there a known flaw?

REPLY [13 votes]: This recent arxiv posting by Shaneson claims that one may take $\epsilon > 1153/9750$, improving on Huxley's bound.  It also includes the passage

In 2007 Cappell and the author posted a paper on the arXiv claiming to obtain
  the estimate [in my notation -- PLC] $O(r^{1/2+\epsilon})$.  Unfortunately we have not been able to produce an error free version. The present paper shares with that paper the Proposition in section 5 and there is also something there akin to what immediately follows the Proposition.

I guess that's that.<|endoftext|>
TITLE: Generalizing cosine rule to symmetric spaces
QUESTION [6 upvotes]: The sine and cosine rules for triangles in Euclidean, spherical and hyperbolic spaces can be understood as invariants for triples of lines. These invariants are given in terms of the distance (both lengthwise and angular) between pairs of lines. There is also a converse statement. Suppose we are living in a complete Riemannian manifold of constant curvature. If a certain sine/cosine rule is satsifies by triples of lines, then we can determine the curvature.
I'm wondering if we can generalize these sine and cosine rules to arbitrary symmetric spaces. That is, give a triple of geodesics (or parallel submanifolds, if we consider higher dimensions), are there similar invariants? Perhaps these invariants will be given in terms of representations of the coset of symmetries take a geodesic to antoher. 
It would also be great if these invariants can characterize the symmetric space we are living in, just like in the case of the constant curvature case.

REPLY [6 votes]: There exist generalizations of the trigonometry laws to symmetric spaces. The following article by Ortega and Santander works out the trigonometry laws for the case of real symmetric spaces of constant curvature and the following one the case of rank-1 Hermitian symmetric spaces. The following article by Aslaksen and Huynh treats more general cases of n-point trigonometrey of symmetric spaces.<|endoftext|>
TITLE: On order of subgroups in abelian groups
QUESTION [8 upvotes]: I wonder whether any of you guys has already read the homonymous note by R. Beals in the December 2009 issue of the Monthly.
If so, would you be so kind as to let me know about the main ideas in Beal's approach? As you know, the whole point of his note is to present a solution to the following exercise in Herstein's Topics in Algebra:
Let G be an abelian group having subgroups of order m and n. Prove that G also possesses a subgroup of order lcm(m, n).
The funny thing about this proposal is that in subsequent editions of his book, Prof. Yitz would proclaim that he himself didn't have a solution using the authorized tools. Besides, he even went on to saying: "I've had more correspondence about this problem than about any other point in the whole book.".
Being aware of some of the history behind this little pearl, I'd really like to know what it is that Beals came up with. Is his approach crystal-clear? Is it somehow related to the standard attack of proving it first for the case gcd(m, n)=1?
Thanks in advance for you insightful replies.
P.S. The local library is the only access that I have to the literature. Unfortunately, they don't subscribe to any of the MAA periodicals. 
1 Beals, Robert. "On Orders of Subgroups in Abelian Groups: An Elementary Solution of an Exercise of Herstein." The American Mathematical Monthly, Vol. 116, No. 10 (Dec., 2009), pp. 923-926; https://maa.tandfonline.com/doi/abs/10.4169/000298909X477032 https://www.jstor.org/stable/40391251

REPLY [9 votes]: The question is to prove that if $H$ and $K$ are subgroups of a
finite Abelian group or orders $m$ and $n$ then $G$ has a subgroup of order
$\mathrm{lcm}(m,n)$.
Beals starts by doing the case where $H$ and $K$ are cyclic. He proves
that $H$ is an internal direct product of cyclic groups of prime power orders.
Then he proves that a product of cyclic subgroups of coprime orders is
cyclic of the right order. The cyclic case is proved by breaking
up $H$ and $K$ as products of cyclic prime power groups, taking
the larger one for each prime and multiplying them up.
The general case follows roughly the same line. Proving that a product
of subgroups of coprime orders has the right order is straightforward.
But decomposing a subgroup into prime power factors using
results earlier in Herstein is more involved. Beals uses Theorem 2.5.1
in Herstein that $|HK| = ~|H||K|/|H\cap K|$. Then Beals finishes
the proof in the same way as the cyclic case.<|endoftext|>
TITLE: Deformations for complex space germs
QUESTION [5 upvotes]: Is there a space such that it doesn't have any deformation but its space of first order infinitesimal deformations is non trivial?

REPLY [5 votes]: Such germs of spaces exist. See Section 7.6 of Ravi Vakil's paper Murphy's law in Algebraic Geometry.<|endoftext|>
TITLE: Colimits in the category of smooth manifolds
QUESTION [21 upvotes]: In the category of smooth real manifolds, do all small colimits exist? In other words, is this category small-cocomplete? I can see that computing push-outs in the category of topological spaces of smooth manifolds need not be manifolds, but this is not a proof.

REPLY [17 votes]: I'd like to recast Reid's (excellent) answer slightly.  The essence of it is the following principle:

To show that a limit or colimit doesn't exist in some category, embed your category in one where limits or colimits do exist and find some diagram in the original category whose colimit in the larger category does not lie in the image of the embedding.

The point is, it's usually much easier to show that an object $X$ of $\mathcal{D}$ is not an object of $\mathcal{C}$ than it is to show that $\mathcal{C}$ has nothing that looks like $X$.  For a simpler analogy, think of the difference between proving that $(0,1)$ is not complete versus proving that $(0,1) \subseteq \mathbb{R}$ is not closed.  The essence is the same, but the latter always seems to me to be a lot easier to grasp.
Back to the principle.  As stated, it's not quite strong enough.  You need a condition on the embedding:

Make sure that your embedding preserves those limits or colimits that already exist.

Again, by analogy: to prove that a metric space $X$ is not complete, we need a continuous map from $X$ to a complete space with non-closed image.  An arbitrary map won't do.
Back to the case in hand.  As the functor $M \mapsto C^\infty(M,\mathbb{R})$ is a (contravariantly) representable embedding, it preserves colimits and so is suitable for the argument to go through.
However, it does not preserve limits so if you asked the corresponding question about limits, you'd need a different embedding.  It turns out, though, that there is a complete and cocomplete category in which the category of manifolds embeds preserving all limits and colimits.  That is the category of Hausdorff Froelicher spaces.  Froelicher spaces may feel a little more topological than algebras so for those who, like myself, prefer topology to algebra, here's a recasting of Reid's answer using (Hausdorff) Froelicher spaces.
The key thing is that a Froelicher space is completely determined by either the smooth functions from it to $\mathbb{R}$ or the smooth curves in it (i.e. smooth functions from $\mathbb{R}$).
We take the same colimit: the pushout of

$$
\begin{matrix}
\{0\} &\to& \mathbb{R}\\
\downarrow \\
\mathbb{R}
\end{matrix}
$$

We shall show that it is the union of the $x$ and $y$ axes in $\mathbb{R}^2$, which is clearly not a manifold.
Let us write the colimit as $X$.  First, we define a smooth function $F \colon X \to \mathbb{R}^2$.  It is the obvious one: it sends the first copy of $\mathbb{R}$ to the $x$-axis and the second copy to the $y$-axis.  As these two functions agree on $\{0\}$, this is a well-defined smooth function.
We want to show that this is an initial map.  One sufficient (but not necessary) condition for this is that every smooth function $f \colon X \to \mathbb{R}$ factors through $F$.
As Reid says, a smooth function $f \colon X \to \mathbb{R}$ consists of two smooth functions $f_1, f_2 \colon \mathbb{R} \to \mathbb{R}$ satisfying $f_1(0) = f_2(0)$.  Let $g \colon \mathbb{R}^2 \to \mathbb{R}$ be the function $g(x,y) = f_1(x) + f_2(y) - f_1(0)$.  This is smooth and we have $g(x,0) = f_1(x) + f_2(0) - f_1(0) = f_1(x)$ and, similarly, $g(0,y) = f_2(y)$.  Thus $g \circ F = f$ and so every function $X \to \mathbb{R}$ factors through the inclusion $X \to \mathbb{R}^2$.  Hence the inclusion $X \to \mathbb{R}^2$ is initial.  Thus we can identify $X$ with its image, that being the union of the two axes.
As I said, this is merely a recasting of Reid's answer.  I post it partly to make it more topological in feel, but mainly to expose the general principle which Reid uses.<|endoftext|>
TITLE: Decomposing a tensor product
QUESTION [9 upvotes]: I guess this question only requires standard knowledge, but I'm a bit rusty with highest weight theory. I'm trying to catch up, but maybe I don't need the theory in full generality.
Background
Let $V$ be a Euclidean space of dimension $n$, and consider the representations of the group $G = O(V) \cong O(n)$. If I'm not wrong we can decompose
$$
\operatorname{Sym}^2 V = W \oplus Z,
$$
where $Z$ is the trivial one-dimensional representation and $W$ is an irreducible representation of $G$.
More background
This is how I can see the above decomposition. I don't know if it is of any use for the question itself, so feel free to skip it.
It is enough to decompose $\operatorname{Sym}^2 V^{*}$, that is, degree $2$ homogeneous polynomials on $V$. Degree $n$ homogeneous polynomials are a representation of $G$ via
$$
A.f(x) = f(Ax),
$$
where $A$ is an orthogonal matrix and $f \in \operatorname{Sym}^n V^{*}$.
Now for any $n$ we have a $G$-morphism $f : \operatorname{Sym}^n V^{*} \to \operatorname{Sym}^{n + 2} V^{*}$ which is given by multiplication by $x^2 = x_1^2 + \cdots + x_n^2$. This exhibits $\operatorname{Sym}^n V^{*}$ as a subrepresentation of $\operatorname{Sym}^{n + 2} V^{*}$.
The complement is easily found. The invariant scalar product on $\operatorname{Sym}^n V^{*}$ is given by $(f, g) = f(D)g(x)$, where $D$ is the derivation operator. From this one finds that the adjoint of $f$ is the laplacian $\Delta$.
So one can decompose
$$
\operatorname{Sym}^{n + 2} V^{*} = \operatorname{Sym}^{n} V^{*} \oplus \mathcal{H}_n,
$$
where $\mathcal{H}_n$ is the space of harmonic homogeneous polynomials of degree $n$. If I recall well, $\mathcal{H}_n$ is irreducible.
The decomposition which interests me is then
$$
\operatorname{Sym}^{2} V^{*} = \mathcal{H}_2 \oplus \mathcal{H}_0.
$$
Problem
I'd like to understand how to decompose the tensor products of $W$; in particular

What is the decomposition of $W \otimes W$ and $W \otimes W \otimes W$ into irreducible representations of $G$?

REPLY [4 votes]: For $SO(n)$ a calculation using LiE gives:
(using partition notation so $W$ is [2])
and assuming $n$ is not small
For $W\otimes W$, [4],[3,1],[2,2],[2],[1,1],[]
(all with multiplicity one)  
and for $W\otimes W\otimes W$,
1.[6] 2.[5,1] 3.[4,2] 1.[3,3] 1.[4,1,1] 2.[3,2,1] 1.[2,2,2] 3.[4] 6.[3,1] 2.[2,2] 3.[2,1,1] 6.[2] 3.[1,1] 1.[]
The same works for $Sp(n)$ by taking conjugate partitions.  
There is also a relationship with $SL(n)$.
This is taking your question at face value. If it is understanding you're after instead then the best approach is to use crystal graphs.
The notation I have used denotes a representation by a partition. I have put $m.$ in front to denote multiplicity is $m$. A partition is $[a_1,a_2,a_3,...]$ where $a_i\ge a_j$ if $i$ less than $j$. To convert to a highest weight vector add the appropriate number of $0$s to the end.
Then take $[a_1-a_2,a_2-a_3,a_3-a_4,...]$. This gives a dominant integral weight. The fundamental weights are the partitions $[1,,,,1]$. If this has length $k$ this corresponds to the $k$-th exterior power of the vector representation (provided $2k-1$ less than $n$).
In particular the trivial representation is $[]$, the vector representation $V$ is $[1]$, the exterior square of $V$ is $[1,1]$, the symmetric square is $[2]+[]$.
For the $k$-th tensor power of $W$ you will see partitions of $2k-2p$ for $0\le p\le k$ only and it remains to determine the multiplicities (possibly $0$). For $SL(n)$ just take the partitions of $2k$ (with their multiplicities) and ignore the rest.<|endoftext|>
TITLE: Is there a dense subset of the real plane with all pairwise distances rational?
QUESTION [89 upvotes]: I heard the following two questions recently from Carl Mummert, who encouraged me to spread them around. Part of his motivation for the questions was to give the subject of computable model theory some traction on complete metric spaces, by considering the countable objects as stand-ins for the full spaces, to the extent that they are able to do so. 
Question 1. Is there a countable subset $D$ of the real plane $\mathbb{R}^2$ that is dense and has the property that the Euclidean distance $d(x,y)$ is a rational number for all $x,y\in D$? 
The one dimensional analogue of this question has an easy affirmative answer, since the rationals $\mathbb{Q}$ sits densely in $\mathbb{R}$ and the distance between any two rationals is rational. 
Question 2. More generally, does every separable complete metric space have a countable dense set $D$ with all distances between elements of $D$ rational? [Edit: Tom Leinster has pointed out that if the space has only two points, at irrational distance, this fails. So let us consider the case of connected spaces, generalizing the situation of Question 1.]
If one is willing to change to an equivalent metric (giving rise to the same topology), then the answer to Question 1 is Yes, since the rational plane $\mathbb{Q}\times\mathbb{Q}$ is dense in the real plane $\mathbb{R}\times\mathbb{R}$, and has all rational distances under the Manhattan metric, which gives rise to the same topology. Is the answer to the correspondingly weakened version of Question 2 also affirmative, if one is willing to change the metric?
Note that one cannot find an equivalent metric such that all distances in $\mathbb{R}^2$ become rational, since omitted values in the distance function lead to disconnectivity in the space. This is why the questions only seek to find a dense subset with the rational condition.
The question seems related to the question of whether it is possible to find large non-linear arrangements of points in the plane with all pair-wise distances being integers. For example, this is true of the integers $\mathbb{Z}$ sitting inside $\mathbb{R}$, but can one find a 2 dimensional analogue of this? Clearly, some small arrangements (triangles, etc.) are possible, but I am given to understand that there is a finite upper bound on the size of such arrangements. What is the precise statement about this that is known?

REPLY [8 votes]: Victor Klee and Stan Wagon write about this and other fun problems in their book:
Old and New Unsolved Problems in Plane Geometry and Number Theory<|endoftext|>
TITLE: Rational numbers as an extension of the field with one element?
QUESTION [8 upvotes]: Greetings.
I would love to have a field $\mathbb F$ which is a subfield of the field of rational numbers $\mathbb Q$, and such that the Galois group $Gal (\mathbb Q / \mathbb F)$ has preferably infinitely many elements.
While there is no such field $\mathbb F$, since $\mathbb Q$ has no proper subfields at all, I've recently heard of this field $\mathbb F_1$ with one element concept.
As far as I understand there is no definition which would be set in stone for this object, at least not yet. My question to those who know the subject: does any of the currently studied definitions of $\mathbb F_1$ allow for realization of $\mathbb F_1$ as a "subfield" of $\mathbb Q$ in some sense?

REPLY [15 votes]: Imo the best theory today for the field with one element is Borger's proposal to consider Lambda-rings and use their Lambda-structure as a substitute for descent from the integers (or rationals) to the field F1 with one element.
Some examples of this philosophy are contained in the nice short paper by Borger and Bart de Smit (arXiv:0801.2352) 'Galois theory and integral models of Lambda-rings'.
Lambda-rings finite etale over the rationals Q are finite discrete sets equipped with a continuous action of the monoid Gal(Qbar/Q) x N' where N' are the positive integers under multiplication. This suggest that the Galois monoid Gal(Qbar/F1) = Gal(Qbar/Q) x N'.
Likewise, Lambda-rings over Q having an integral Lambda-model correspond to finite sets with a continuous action of the monoid Zhat, that is the set of profinite integers as a topological monoid under multiplication. This suggests that the absolute Galois monoid of F1, that is Gal(F1bar/F1) = Zhat.<|endoftext|>
TITLE: "adjoint" =?= "inverse of composite endofunctor is uniform bi-composition"
QUESTION [5 upvotes]: Understanding adjoints has always been (and continues to be) a bit of a struggle for me.
Today I stumbled upon a property of adjoint functors which seemed extremely intuitive to me.  I was wondering why this property isn't mentioned more often in introductory category theory literature, and whether or not it completely characterizes adjunctions.
If two functors $F:C\to D$ and $U:D\to C$ are adjoint $F\dashv U$, then for every $f:F(Y)\to X$ in $D$ there exists an $\hat f:Y\to U(X)$ in $C$ such that
$$ U(f)\circ \eta_Y = \hat f$$
$$ \epsilon_X\circ F(\hat f)=f$$
If we substitute the top equation into the bottom, we get
$$ \epsilon_X\circ F(U(f)\circ \eta_Y)=f$$
and by functoriality we get
$$ \epsilon_X\circ F(U(f))\circ F(\eta_Y)=f$$
$$ \epsilon_X\circ (F\circ U)(f)\circ F(\eta_Y)=f$$
What the last equation says is that we can recover any morphism $f$ from the action of the "round trip endofunctor" $F\circ U$ on it by pre-composing with $\epsilon_X$ and post-composing with $F(\eta_Y)$.  These two morphisms are determined only by the domain and codomain of $f$ -- we only needed to know $X$ and $F(Y)$ in order to pick the two morphisms.  We would have picked the same two morphisms for some $g\neq f$ as long as $g:F(Y)\to X$.
So, I believe it is correct to say that "if the domain of a morphism is within the range of a functor which has a right adjoint, then it can be recovered from the action of the composite endofunctor on it by pre-composition with some morphism and post-composition with some other morphism, where the choice of these two morphisms is completely determined by the domain and codomain of the original morphism".  There is, of course, an equivalent statement for morphisms with a codomain in the range of a functor with a left adjoint.
So, my three questions are: (1) is this correct, (2) if so, why isn't it used to explain adjunctions to beginners (I certainly would have caught on quicker!) and (3) does the condition completely characterize adjoint functors?
Thanks,

REPLY [4 votes]: (1) Yes.  (2) Well, it doesn't give me any additional intuition.  You didn't say why it helps you understand, so I can't judge what the advantage of it might be.  I think this is really just a complicated way of giving the "bijection of hom-sets" condition.
(3) No, you need something more.  For instance, let $r:B\to A$ be a surjection with section $s$, let $C$ have two objects $x$ and $y$ with $C(x,y)=B$, $C(y,x)=\emptyset$, and $C(x,x)=C(y,y)=1$ (only identities), let $D$ be similar using $A$ instead, and let $F:C\to D$ and $U:D\to C$ be the identity on objects and with action on arrows given by $r$ and $s$ respectively.  Pick $\varepsilon$ and $\eta$ to be identities.  Then every morphism in $D$ can be recovered, as you describe, but the components of $\eta$ are not natural, and the dual condition fails.
The "unknown (google)" comment above explained why if you additionally require the dual condition, plus naturality of $\eta$ and $\varepsilon$, then you do get an adjunction.  (Although it's not clear to me from the condition you stated whether you wanted to require the morphism playing the role of $F(\eta)$ to actually be $F$ of something, which is also necessary for this argument to work.)<|endoftext|>
TITLE: Motivation for strong law of large numbers
QUESTION [21 upvotes]: I always find the strong law of large numbers hard to motivate to students, especially non-mathematicians.  The weak law (giving convergence in probability) is so much easier to prove; why is it worth so much trouble to upgrade the conclusion to almost sure convergence?
I think it comes down to not having a good sense of why, practically speaking, a.s. convergence is better than convergence i.p.  Sure, I can prove that one implies the other and not conversely, but the counterexamples feel contrived.  I understand the advantages of a.s. convergence on a technical level, but not on the level of everyday life.
So my question: how would you explain to, say, an engineer, the significance of having a.s. convergence as opposed to i.p.?  Is there a "real-life" example of bad behavior that we're ruling out?

REPLY [7 votes]: Suppose you are in the context of collecting data and estimating the mean.
Imagine a situation where SLLN does not hold. It means that with positive probability accumulating new data is useless.<|endoftext|>
TITLE: Why is a topology made up of 'open' sets?
QUESTION [307 upvotes]: I'm ashamed to admit it, but I don't think I've ever been able to genuinely motivate the definition of a topological space in an undergraduate course. Clearly, the definition distills the essence of many examples, but it's never been obvious to me how it came about, compared, for example, to the rather intuitive definition of a metric space. In some ways, the sparseness of the definition is startling as it tries to capture, apparently successfully, the barest notion of 'space' imaginable.
I can try to make this question more precise if necessary, but I'd prefer to leave it slightly vague, and hope that someone who has discussed this successfully in a first course, perhaps using a better understanding of history, might be able to help me out.
Added 24 March: 
I'm grateful to everyone for their thoughtful answers so far. I'll have to think over them a bit before I can get a sense of the 'right' answer for myself. In the meanwhile, I thought I'd emphasize again the obvious fact that the standard concise definition has been tremendously successful. For example, when you classify two-manifolds with it, you get equivalence classes that agree exactly with intuition. Then in as divergent a direction as the study of equations over finite fields, there is the etale topology*, which explains very clearly surprising and intricate patterns in the behaviour of solution sets. 
*If someone objects that the etale topology goes beyond the usual definition, I would argue that the logical essence is the same. It is notable that the standard definition admits such a generalization so naturally, whereas some of the others do not. (At least not in any obvious way.) 
For those who haven't encountered one before, a Grothendieck topology just replaces subsets of a set $X$ by  maps $$Y\rightarrow X.$$ The collection of maps that defines the topology on $X$ is required to satisfy some obvious axioms generalizing the usual ones.
Added 25 March: 
I hope people aren't too annoyed if I admit I don't quite see a satisfactory answer yet. But thank you for all your efforts. Even though Sigfpe's answer is undoubtedly interesting, invoking the notion of measurement, even a fuzzy one, just doesn't seem to be the best approach. As Qiaochu has pointed out, a topological space is genuinely supposed to be more general than a metric space. If we leave aside the pedagogical issue for a moment and speak as working mathematicians, a general concept is most naturally justified in terms of its consequences. As pointed out earlier, topologies that have no trace of a metric interpretation have been consequential indeed.
When topologies were naturally generalized by Grothendieck, a good deal of emphasis was put on the notion of an open covering, and not just the open sets themselves. I wonder if this was true for Hausdorff as well. (Thanks for the historical information, Donu!) We can see the reason as we visualize a two-manifold. Any sufficiently fine open covering captures a combinatorial skeleton of the space by way of the intersections. Note that this is not true for a closed covering. In fact, I'm not sure what a sensible condition might be on a closed covering of a reasonable space that would allow us to compute homology with it. (Other than just saying they have to be the simplices of a triangulation. Which also reminds me to point out that homology can be computed for ordinary objects without any notion of topology.)
To summarize, a topology relates to analysis with its emphasis on functions and their continuity, and to metric geometry, with its measurements and distances. However, it also interpolates between these and something like combinatorial geometry, where continuous functions and measurements play very minor roles indeed.
For myself, I'm still confused.
Another afterthought: I see what I was trying to say above is that open sets in topology provide an abstract framework for describing local properties of functions. However, an open cover is also able to encode global properties of spaces. It seems the finite intersection property is important for this, but I'm not able to say for sure. And then, when I try to return to the pedagogical question with all this, I'm totally at a loss. There are very few basic concepts that trouble me as much in the classroom.

REPLY [5 votes]: I have long understood that the initial ideas of  topology arose from the notion of "neighbourhood" and were then found to be equivalent to the definition in terms of open sets. One advantage of the neighbourhood concept was that the definition of continuity using that is  nearer to the $\varepsilon$-$\delta$ definition used in analysis. 
The neighbourhood definition is more easily motivated than that in terms of open sets, but one then shows the equivalence. However one finds difficulties with the neighbourhood definition in defining, say,  identification spaces, and this illustrates nicely a feature of mathematics, that equivalent concepts may have their best uses in different areas. Horses for courses! 
Einstein wrote in 1915: 
"Concepts which have proved useful for ordering things easily assume so great an authority over us, that we forget their terrestrial origin and accept them as unalterable facts. They then become labelled as conceptual necessities, a priori situations, etc. The road of scientific progress is frequently blocked for long periods by such errors. It is therefore not just an idle game to exercise our ability to analyse familiar concepts, and to demonstrate the conditions on which their justification and usefulness depend, and the way in which these developed, little by little... "
Thus Grothendieck in his 1984 "Esquisse d'un programme" Section 5, argues that the notion of topological space is motivated from analysis rather than geometry, and the latter requires spaces with more structure, in particular what he calls stratified spaces. I have found filtered spaces important in basic homotopical algebraic topology.<|endoftext|>
TITLE: Reference request for a proof of Ramanujan's tau conjecture
QUESTION [9 upvotes]: In the Wikipedia article it states that Ramanujan's tau conjecture was shown to be a consequence of Riemann's hypothesis for varieties over finite fields by the efforts of
Michio Kuga, Mikio Sato, Goro Shimura, Yasutaka Ihara, and Pierre Deligne. Do their papers consist of the only published proof of this result? And is this proof of a similar level of difficulty to Deligne's proof of Riemann's hypothesis?

REPLY [6 votes]: Following the discussion at meta.MO, I'm going to post a good answer from the comments (made by JT) as a "community wiki" answer.  I should mention that the Rogawski article mentioned by Tommaso says almost nothing about the proof of Ramanujan's conjecture, but it seems to be a very nice introduction to Jacquet-Langlands.
Deligne reduced Ramanujan's conjecture about the growth of tau to the Weil conjectures (in particular, the Riemann hypothesis) applied to a Kuga-Sato variety, in his paper Formes modulaires et representations l-adiques, Seminaire Bourbaki 355.  I believe Jay Pottharst has made an English translation available.
Deligne then proved the Weil conjectures in his paper La conjecture de Weil. I.
As far as I know, all known proofs of this conjecture involve the use of cohomology of varieties over finite fields in an essential way.
Added by Emerton: One point to make is that the Weil conjectures (in their basic form,
saying that the eigenvalues of Frobenius on the $i$th etale cohomology of a variety over
$\mathbb F_q$ have absolute value $q^{i/2}$) apply only to smooth proper varieties.  On the other hand, the Kuga-Sato variety is the symmeteric power of the universal elliptic curve over a modular curve, which is not projective.  Thus one has to pass to a smooth compactification in order to apply the Weil conjectures, and then hope that this does not mess anything up in the rest of the argument.  A certain amount of Deligne's effort in
his Bourbaki seminar is devoted to dealing with this issue.  If you don't worry about this
(i.e. you accept that it all works out okay) then the proof is essentially just Eichler--Shimura theory (i.e. the relation between modular forms and cohomology of modular curves), but done with etale cohomology, combined with the Eichler--Shimura congruence relation that connects the $p$th Hecke operator to Frobenius mod $p$. (The latter was treated in the
following question.)<|endoftext|>
TITLE: How good is Kamenetsky's formula for the number of digits in n-factorial?
QUESTION [26 upvotes]: In Number of digits in n!, now closed, there was a mention of Dmitry Kamenetsky's formula, $[\bigl(\log(2\pi n)/2+n(\log n-\log e)\bigr)/\log 10]+1$, for the number of decimal digits in $n$-factorial. Here, $[x]$ is the integer part of $x$. The formula appears at A034886 in the Online Encyclopedia of Integer Sequences, http://oeis.org/A034886. My question is whether this formula is exact for all $n$, or is it occasionally off. No proof of exactness is given at the OEIS, no paper of Kamenetsky appears in Math Reviews. 
In the other thread, I mentioned the discussion in the Usenet newsgroup sci.math in January-February, Subject: Number of digits in factorial. Although neither proof nor counterexample was found, I'd recommend looking over that discussion before starting in on this question. 
EDIT 11 Aug 2011: I note that the question also came up at m.se: question 8323, 30 Oct 2010.

REPLY [69 votes]: A counterexample is $n_1 := 6561101970383$, with
$$
\log_{10} \left( (n_1/e)^{n_1} \sqrt{2\pi n_1} \right)
= 81244041273652.999999999999995102483 - \phantom; ,
$$
but
$$
\log_{10} (n_1!)
= 81244041273653.000000000000000618508 + \phantom;.
$$
If I computed correctly, $n_1$ is the first counterexample, and the only one up to $10^{13}$.  The computation should reach $10^{15}$ sometime next week, with a probability of about $1 - \exp(-\frac16) \sim 15\%$ of finding an $n_2$.
The computation (in gp/pari) took about 40 CPU hours here, compressed to 4 hours by running in parallel on 10 of the 12 heads of alhambra.math.harvard.edu .  This was not done by calculating $\log_{10} (n!)$ to enough precision for every $n \leq 10^{13}$, which would have taken hundreds of times longer.  The problem of finding nearly integral values of $\log_{10} (n!)$ is a special case of the "table maker's dilemma" (Wikipedia attributes this felicitous coinage to William Kahan); in this case, the linear-approximation technique suggested by Lefèvre at the bottom of page 15 of his slides takes time $\tilde O(N^{2/3})$ to find all examples with $n < N$.  That's what's running on alhambra now.
Along the way a few more terms of sequence A177901 turned up:
$252544447$,
$1430841730$,
$5042264463$,
$31774693500$,
$40752166709$,
$46787073630$,
$129532358256$,
$421559495894$,
$2418277169072$,
$6105111564681$,
and then $n_1 = 6561101970383$, which might even turn out to be the last term up to $10^{15}$ because $\log_{10} (n_1!)$ is so close to an integer (about $9$ times closer than necessary for our purpose).  [EDIT It's the last term $<10^{14}$ but not $10^{15}$, see below.]
The term $252544447$ was reported on math.se #8323 by Byron Schmuland [EDIT and a few months earlier by David Cantrell on sci.math], though it has not been posted to OEIS yet.  The further ones seem to be new, and I'll post them on OEIS soon.
Kamenetsky was right to suggest that the approximation should fail sometimes: in base 10, we expect $n$ to be a counterexample with probability about $1/cn$ with $c = 12 \log 10$, so on average each range $[N, 10^{12}N]$ should have about one.  Thus it is not surprising that the first one (past $n=1!$) turns out to have $13$ digits.  This heuristic is also the source of the estimate $1-\exp(-\frac16)$ for the probability of another counterexample in $ [10^{13}, 10^{15}]$.
UPDATE The calculation has now passed $10^{14}$, finding no new counterexample.  It did, however, find a new term for the OEIS sequence a bit beyond $10^{14}$: $n=125291661119688$, with $\log_{10}(n!)$ close to but just below the nearest integer $1711938609606982$ (where a counterexample must be a bit above), and also not quite as close as $1/(12n)$ — the difference is about $1/(8.4n)$.
While I'm at it: I should have mentioned that the gp/pari computation also found (in a minute or two) all the terms in $[10^4,10^8]$ listed by OEIS, which lends the new results some credibility; and I thank Gerry Myerson for drawing my attention to this question with his edit of about two weeks ago.<|endoftext|>
TITLE: Informal online seminars or reading groups via videoconferencing?
QUESTION [36 upvotes]: Does the following exist, and if not, does anyone besides me wish it did? A web site where a mathematician (say) could find other mathematicians who want to study the same book or paper, and arrange to meet via videoconference, and run their own informal seminar around that topic, and then disband when they're done.
The reason I ask is that I have plenty of material I'd like to learn this way. I have an interesting job in industry, with occasional mathematical challenges, but there are more topics I would like to learn outside of work, and that I didn't get a chance to cover in grad school. I live near two major universities, and they do offer some interesting courses I could audit, but I'd rather focus on the exact material I want to learn.
One reason such a site might be successful (or, more precisely, popular) is that I expect academic mathematicians and grad students would find great use for it as well, since it can be hard to branch out to new areas on your own, if your own department doesn't overlap enough. If you have a deep professional network in the math community, you are probably better off, but not everyone is in that position for one reason or another.
I expect such a site, if it doesn't exist today, wouldn't be hard to create. If the videoconferencing portion was done outside the site using Skype or another service, then the site would just have to match up the groups, and maybe offer other features to support a running seminar. It would basically be a specialized social networking site, or maybe even an application within an existing site like Facebook.
UPDATE 1: The more I think about it, the more I think meetup.com is a good fit too, since it already has the calendaring and rsvp-ing features. The downside is that each seminar would probably have to be a separate meetup, and meetup.com is sort of tied to geography.
UPDATE 2: I have spent a lot of time looking for good, free, non-sleazy video streaming services, and I ended up at livestream.com. They have a couple key features. They offer a free plan with a reasonable business model behind it (the "freemium" model with a paid tier and a free tier that shows ads to viewers, though I haven't seen what these ads look like yet). They have Mac and Windows client software that lets you use your webcam OR stream your desktop, the latter of which I think is very important as a replacement for a physical blackboard or whiteboard. The other piece I have been researching is drawing on the screen with a pen. The best solution I have found is slightly sub-optimal: this pen or this one. They are similar in that they are standard ink pens with a wireless transmitter, and you clip a receiver to the paper and then write normally. The notes are either saved as images (the primary use case), or if you have a recent version of Windows, it can activate the pen features of the OS and you can draw directly into OneNote and other similar programs. This latter is what I want -- the seminar speaker can write on a paid of paper, and the remote attendees will see the writing rendered in real time on their screen. The problems are several: the pens cost at least $50-$60, I haven't tested them, reviews are scarce and Mac reviews are even scarcer. Most digital pen enthusiasm is focused on the Livescribe pen, which is strictly for saving notes and audio for later, and does not transmit in real time. Lastly, I also found this chat software that supports audio and a whiteboard, and which is free. By the way, these pens all look really great for students, especially the Livescribe, because it syncs the audio with the notes, so you can play it back and know what was being said when something was written. And you can convert a session to Flash and upload it as a "pencast". I think pencasts could really be wonderful for mathematics presentations that are recorded offline and then uploaded for an audience to learn from asynchronously.

REPLY [2 votes]: look @ this website http://www.twiddla.com/ . you can use it with your tablet and also has latex capability as well<|endoftext|>
TITLE: Where do stable Kronecker coefficients live "in nature"?
QUESTION [15 upvotes]: Background:
For a partition $\lambda$, let $\lambda[N] = (N - |\lambda|, \lambda_1, \lambda_2, \lambda_3, \dots)$, also let $\chi_\lambda$ be the corresponding irreducible character of the symmetric group $S_{|\lambda|}$. Even if $\lambda[N]$ is not a partition, we can make sense of $\chi_{\lambda[N]}$ as a class function of $S_N$ using a determinant, and Murnaghan proved that there exist coefficients $G^\nu_{\lambda, \mu}$ (the stable, or reduced, Kronecker coefficients) such that
$\chi_{\lambda[N]} \chi_{\mu[N]} = \sum_\nu G^\nu_{\lambda, \mu} \chi_{\nu[N]}$
for all $N \ge 0$. 
Question:
In a sense, the construction of $\lambda[N]$ is a bit ad hoc. Is there a more representation-theoretic way to define these coefficients? In particular, if $|\lambda| + |\mu| = |\nu|$, then $G^\nu_{\lambda, \mu}$ coincides with the corresponding Littlewood-Richardson coefficient, so I am hopeful there is some connection. I am looking for an answer which addresses the following point: as I have defined these coefficients, it seems that the most accessible way to work with these coefficients is to use combinatorics. If I wanted to use tools from say, invariant theory or algebraic geometry, what is a more natural context for these coefficients to appear?

REPLY [7 votes]: Just for the record, I would like to confirm that David Jordan's feeling was right. Deligne's category $Rep(S_t)$ is monoidal, depends on a parameter $t$ and is semisimple for $t\not\in\mathbb N$. In that case, its simple objects $X_\lambda$ are parameterized by partions $\lambda$ of arbitrary size and the multipicity of $X_\nu$ in $X_\lambda\otimes X_\mu$ is precisely the stable Kronecker coefficient $G_{\lambda,\mu}^\nu$.
The idea is as follows: For $t=N\in\mathbb N$ there is a monoidal functor $F:Rep(S_t)\to Rep(S_N)$ (the latter is the ordinary category of $S_N$-representations). If $N$ is big enough (with respect to a fixed $\lambda$), the simple object $X_\lambda$ is also defined for $t=N$. Deligne shows that $F(X_\lambda)$ is the irreducible $S_N$-representation $V_{\lambda[N]}$ corresponding the partition $\lambda[N]$. Therefore, $X_\lambda\otimes X_\mu$ has the same decomposition as $F(X_\lambda\otimes X_\mu)=V_{\lambda[N]}\otimes V_{\mu[N]}$ yielding the result. For details see Inna Aizenbud's paper Deligne categories and reduced Kronecker coefficients.<|endoftext|>
TITLE: bibl. q.s on Dwork's "p-adic cycles", Mazur's "p-adic variations":
QUESTION [5 upvotes]: Matthew Emerton mentioned recently the relevance of Dwork's "p-adic cycles". As I wonder if I should read that, reviews of it are ambiguous, I'd be happy on remarks and possible further bibl. hints. And I'd enjoy hints where one finds surveys etc. related to Mazur's "Theme of p-adic variation" and how that develops.

REPLY [6 votes]: You could read Mazur's article in the $p$-adic monodromy volume.  And also Katz's Travaux de Dwork, as well as his two articles on Serre--Tate theory (LNM 828?), and the accompanying article of Deligne and Illusie on K3 surfaces.   You could also read Gross's Tameness Criterion paper in Duke from the late 80s, which uses Dwork's ideas and related $p$-adic techniques.  And there is Nygaard's article on the Tate conjecture for K3's over finite fields.
Dwork is difficult, and I don't recommend reading him in a vacuum or for casual entertainment.  But his ideas and insights are very deep, and very original.  (His actual techniques are very involved, and I am not sure that I would recommend learning them before you learned some more standard ideas from $p$-adic geometry, such as are explained in the above references.)<|endoftext|>
TITLE: Category = Groupoid x Poset?
QUESTION [5 upvotes]: Is it possible to split a given category $C$ up into its groupoid of isomorphisms and a category that resembles a poset?
"Splitting up" should be that $C$ can be expressed as some kind of extension of a groupoid $G$ by a poset $P$ (or "directed category" $P$ the only epimorphisms in $P$ are the identities, all isomorphisms in $P$ are identities).

REPLY [4 votes]: In the finite, one-object case, Krohn-Rhodes theory gives a way to decompose a semigroup as a wreath product of finite simple groups and aperiodic semigroups, which are in some sense "as non-grouplike as possible", though in a different way from posets. There is an extension of the theory to categories due to Wells. I don't know much about this, but it is similar to the sort of decomposition you're asking for.<|endoftext|>
TITLE: Divisors on Proj(UFD)
QUESTION [6 upvotes]: Hello to all,
I have been perusing Harthorne for some time, and I noticed something: it is well known that the class group on $\mathbb{P}^n_k$ is $\mathbb{Z}$. But as I look at Harthorne's proof it seems to mee that it works in much greater generality. Namely if I consider any projective scheme $X=Proj(A)$, where $A$ is a graded $UFD$ in such a way that there exists an irreducible element of degree $1$, then the exact same reasoning shows that the class group of $X$ is also $\mathbb{Z}$, and generated by the prime divisor $(a)$. Is this true ?

REPLY [3 votes]: Well, if you read on to Chapter 2, exercise 6.3, then it is stated that:
$$Cl(A) \cong Cl(X)/\mathbb Z[H]$$ 
here $[H]$ represents the hyperplane section. So the answer is yes.
There is a less well-known but very nice generalization. Suppose that $X$ is smooth. Let $R=A_m$ be the local ring of A at the irrelevant ideal. Then one has a (graded) isomorphism of $\mathbb Q$- vector spaces:
$$CH(X)_{\mathbb Q}/[H]CH(X)_{\mathbb Q} \cong A_*(R)_{\mathbb Q}$$
Here $CH(X)$ is the Chow ring of $X$ and $A_*(R)$ is the total Chow group of  $R$. Details can be found in this paper by Kurano.<|endoftext|>
TITLE: Conditions for "bootstrapping" a smooth DM stack?
QUESTION [12 upvotes]: In the preprint "Smooth toric DM stacks", Fantechi, Mann and Nironi define the stacks of their title, and show that each of these can be obtained through the following sequence of steps:
1) start with a scheme (the coarse moduli scheme) with at worst finite quotient singularities, and take the associated canonical stack;
2) use a root stack construction to possibly add some extra stack structure to divisors (given by an integer for each divisor);
3) finally add a gerbe.
Not all smooth DM stacks can be obtained this way, e.g. for $n>3$ take the global quotient $\mathbb{C}^n/S_n$, where the symmetric group $S_n$ acts by permuting the factors of $\mathbb{C}^n$.  The coarse moduli scheme is smooth here, and there is no gerbe, but the stack doesn't seem to arise as a root stack.
Are there conditions known for reasonable (finite type over a field,...) smooth DM stacks under which the stack can be obtained by the "bootstrapping" procedure described above (or a similar one)?

REPLY [6 votes]: Notation: Let $x$ be a point of a smooth separated finite type DM stack $\mathcal X$ over a field. Suppose
• $G$ is the stabilizer of $x$,
• $V$ is the tangent space of $x$ (which comes equipped with an action of $G$),
• $G^\textrm{triv}\subseteq G$ is the subgroup which acts trivially on $V$,
• $H = G/G^\textrm{triv}$ (note $H$ acts on $V$),
• $K$ is the subgroup of $H$ generated by pseudoreflections on $V$, and
• $K'$ the commutator subgroup of $K$.

$\mathcal X$ can be expressed as you described (in an étale neighborhood of $x$) if and only if $K'$ is trivial.

I'll now unpack that answer. Any smooth separated finite type DM stack over a field can be (canonically!) obtained from its coarse space with the following steps (this is basically the main Theorem of my paper with Matt Satriano, A "bottom up" characterization of smooth Deligne-Mumford stacks):

take the canonical stack of the coarse space,
do a root stack construction along the ramification divisor of the coarse space map, rooting each component of the ramification divisor by the degree of ramification,
take the canonical stack again (the root stack may not be smooth any more, but it will have quotient singularities!), and
add a gerbe.

Your question is "when can we skip step 3?" That is, when is the root stack from step 2 already smooth?
Using the notation above, and looking formally locally around $x$ (so we can assume $\mathcal X=[V/G]$; you can describe it étale locally too, but it's clearer this way), the above steps are:

$\bigl[(V/K)/(H/K)\bigr]$ is the canonical stack of the coarse space $V/H = V/G$,
$\bigl[(V/K')/(H/K')\bigr]$ is a root stack of $\bigl[(V/K)/(H/K)\bigr]$,
$[V/H]$ is the canonical stack of $\bigl[(V/K')/(H/K')\bigr]$, and
$\mathcal X = [V/G]$ is a $G^\textrm{triv}$-gerbe over $[V/H]$.

Note that step 1 is the familiar way of building the canonical stack of a space with quotient singularities (in this case $V/H$) by expressing it as a quotient by a finite group somehow, and then quotienting out the subgroup generated by pseudoreflections, with the Chevalley-Shephard-Todd theorem ensuring that you don't lose smoothness. This description tells us that the canonical stack is any description of the space as a quotient where the group acts without pseudoreflections.
Note that in step 3, we're quotienting out by $K'$, which has no pseudoreflections since it's a commutator subgroup (all its elements must therefore act with determinant 1, and there are no pseudoreflections of determinant 1). So it makes sense that this step is a canonical stack.
It's pretty clear that step 4 is a (trivial) gerbe.
Seeing that step 2 is a root stack is more complicated; you can find the details in the section titled "A local description of Theorem 1" in the paper linked above.<|endoftext|>
TITLE: A question on liftings of supersingular elliptic curves over the prime fields
QUESTION [8 upvotes]: Let $p$ be a fixed prime number $>3$. The motivation for asking the question below is the coincidence of the following two numbers. Namely, the number $h_p^{(1)}$ of supersingular $j$-invariants belonging to the prime field $F_p$, and half of the number $\Sigma_p$ of isomorphism classes classes of complex elliptic curves $E$ so that the imaginary quadratic order $\mathbf{Z}[\sqrt{-p}]$ embeds in the ring $\textrm{End}_\mathbf{C}(E)$.
Explicitly, if $h(\sqrt{-p})$ denotes the class number of $Q(\sqrt{-p})$, then
$h_p^{(1)}=\Sigma_p/2=h(\sqrt{-p})/2$, when $p\equiv 1\bmod 4$;
$h_p^{(1)}=\Sigma_p/2=2h(\sqrt{-p})$, when $p\equiv 3\bmod 8$;
$h_p^{(1)}=\Sigma_p/2=h(\sqrt{-p})$, when $p\equiv 7\bmod 8$.
Proving that $\Sigma_p/2$ is equal to the function of $h(\sqrt{-p})$ above is elementary. On the other hand, the fact that $h_p^{(1)}$ is also given by the same function of $h(\sqrt{-p})$ is harder, and a possible proof follows from Eichler's trace formula relating traces of Hecke operators on $S_2(\Gamma_0(p),\mathbf{C})$ to those of Brandt matrices (cf. Gross, Heights and the Special Values of L-series, formula (1.10)).
My question is: is there a direct way of showing that $h_p^{(1)}=\Sigma_p/2$?
My naive thought about it is that one could try to study the reduction mod $p$ of suitable models of the elliptic curves $E$ with complex multiplication by $\sqrt{-p}$ and see what kind of (supersingular?) elliptic curves do arise, hoping that the map on $j$-invariants is 2-to-1. One could ask even more: for any supersingular elliptic curve $E$ over $F_p$ with Frobenius satisfying $X^2+p$, can we find a lifting to an elliptic curve $E'$ over an extension of $Q_p$ that depends functorially on $E$? (The functoriality requirement is not perhaps an "a-priori" nonsense since the endomorphism ring of $E$ over $F_p$ is just an imaginary quadratic order (containing $\sqrt{-p}$)). I do not know enough about lifting of elliptic curves to say how reasonable these questions are. If some of you has more structured and elaborated thoughts on the matter, then I would like to hear them! Thanks in advance.

REPLY [8 votes]: This question is addressed as part of the proof of Theorem 14.18 (attributed to Deuring) given by Cox in "Primes of the Form $x^2 +ny^2$," see pp. 321-322.  Cox counts curves rather than $j$-invariants, but for $p > 3$ there are exactly $p-1$ elliptic curves for each of the $p$ possible $j$-invariants of an elliptic curve over $\mathbb{F}_p$ (see Ex. 14.19).  He first shows that the number of elliptic curves with trace $a\ne 0$ is exactly
$$
\frac{p-1}{2}H(a^2-4p),
$$
where $H(D)$ is the Hurwitz class number of the quadratic order with discriminant $D<0$ (NB: many authors negate $D$).  Cox then notes that the total number of elliptic curves over $F_p$ is
$$
p(p-1) = N + \sum_{0<|a|\le 2\sqrt{p}}\frac{p-1}{2}H(a^2-4p),
$$
where $N$ counts the number of supersingular elliptic curves over $\mathbb{F}_p$ (so $N=(p-1)h_p^{(1)}$ in your notation).  He then applies the class number formula
$$
2p = \sum_{0\le|a|\le2\sqrt{p}}H(a^2-4p),
$$
to obtain $N=(p-1)/2H(-4p)$.  Dividing by $p-1$ gives $h_p^{(1)} = H(-4p)/2$, which covers all three of the cases you list above.
This argument may not be as direct as you would like, but it is fairly simple.<|endoftext|>
TITLE: what notions are "geometric" (characterized by geometric fibers)?
QUESTION [5 upvotes]: Sorry The title might not be suggestive enough.
The question is about things like the following: A reductive group scheme is defined to be a (really nice) group scheme whose geometric fibers are reductive groups. So in some sense, "reductiveness" is some kind of "geometric" notion.
So whatelse properties of schemes can be checked only on geometric fibers? What I know, for example, given a scheme over a field $k$, it is projective iff it is projective over $\bar{k}$. But can this be extended to any base scheme?
In particular, is there any reference that collects such results? And "WHY" should this work? for example, WHY geometrically-reductive group schemes turn out to be the right generalization of reductive algebraic groups?
Sorry this question might be a little to vague, and thank you in advance.

REPLY [2 votes]: I'm not sure I'm answering your question, but let me say some (vague and general) things and hopefully they'll be helpful.
Algebraic geometry over algebraically closed fields is really nice, essentially because of the Nullstellensatz: points are actually points, and functions are actually functions on these points.  So it's often easy (or classical) to define a property of a variety over an algebraically closed field.
But then Grothendieck comes along and sez, hey, we should say things relatively, over an arbitrary base scheme.  And the procedure for that is this: if you have a class of varieties C over algebraically closed fields, you extend it to the relative situation by saying that a map of schemes X --> S is in C if it is flat (which, experience and several nice theorems tell us, amounts to saying that the fibers are continuously parametrized by S) and each geometric fiber is in C.
There are two things that this buys you right off the bat: first, C is closed under base change, and second, C satisfies fpqc descent.  Both are great indications that you have a good in-families notion; for instance they are certainly necessary if you want to make a good moduli functor out of C.  If you didn't use geometric points, you couldn't guarantee fpqc descent, only Zariski descent.
But it seems like you were more interested in the converse: why should, if we have a good in-families notion, it be sufficient to check it on geometric fibers?  Well actually, here "geometric" has nothing to do with it: you should always be able to check on fibers, because you want to be talking about a family of elements of C parametrized by the base.  It's just that over non-algebraically closed fields it's probably harder to say what it means to lie in C -- if you don't do it geometrically, you potentially lose stability under base change and descent.<|endoftext|>
TITLE: Holomorphic vector fields acting on Dolbeault cohomology
QUESTION [20 upvotes]: The question.
Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L_u$ on forms preserves the (p,q) decomposition and also that it commutes with $\bar{\partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?
Some context.
To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L_v = d \circ i_v + i_v \circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)
Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $\alpha$ be a $\bar{\partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L_u \alpha = d(i_u \alpha)$ is also of type (p,q). So, 
$$
L_u\alpha = \bar{\partial}\left((i_u\alpha)^{p, q-1}\right) + \partial\left((i_u \alpha)^{p-1, q}\right)
$$
and the other contributions $\bar{\partial}((i_u\alpha)^{p-1,q}$) and $\partial((i_u\alpha)^{p,q-1})$ vanish. Now the fact that $\bar\partial((i_u\alpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $\beta$ such that 
$$
(i_u\alpha)^{p-1,q}+ \bar\partial \beta
$$
is closed. Hence
$$
\partial \left((i_u\alpha)^{p-1,q}\right) = \bar\partial \partial \beta
$$
and so
$$
L_u\alpha = \bar \partial \left( (i_u \alpha)^{p,q-1} + \partial \beta\right)
$$
which proves the action of $u$ on $H^{p,q}(X)$ is trivial.

REPLY [2 votes]: Klemyatin proved that this action is trivial if the corresponding
${\Bbb C}$-flow is compatible with some metric (hence can be extended
to a compact torus action),
https://arxiv.org/abs/1909.04075,
(N. Klemyatin, Dolbeault cohomology of compact complex manifolds with an action of a complex Lie group, J. Geom. Phys. 157 (2020), 103823.)
Also, this action is trivial whenever the Hodge to de Rham spectral sequence degenerates in $E_1$. This happens on all complex surfaces, on Vaisman manifolds, Oeljeklaus-Toma manifolds, and on some other interesting classes of non-Kahler complex manifolds.
Examples when the action is non-trivial are given by Akhiezer in this paper:
Akhiezer, Dmitri
Group actions on the Dolbeault cohomology of homogeneous manifolds.
Math. Z. 226 (1997), no. 4, 607–621.<|endoftext|>
TITLE: Ambiguous definition of "nerve of an open covering" on wikipedia?
QUESTION [7 upvotes]: Let $(U_i)_{i\in I}$ be an open covering of a topological space $X$.
At http://en.wikipedia.org/wiki/Nerve_of_an_open_covering,
the nerve of the open covering is defined as follows:

the nerve $N$ is the set of finite subsets of $I$ defined as follows:

the empty set belongs to $N$;
a finite set $J\subset I$ belongs to $N$ if and only if the intersection of the $U_i$ whose subindices are in $J$ is non-empty.


On the other hand, http://en.wikipedia.org/wiki/Nerve_(category_theory) states:

If $X$ is a topological space with open cover $U_i$, the nerve of the cover is obtained from the above definitions by replacing the cover with the category obtained by regarding the cover as a partially ordered set with relation that of set inclusion.

Here, "the above definitions" refers to the usual construction of the nerve of a category: A vertex for each object, and a $k$-simplex for each $k$-tuple of composable morphisms.
My question is: Does this categorical construction really yield the previously defined nerve of the open covering?
For instance, cover the inverval by two intersecting invervals non of them containing the other one. Then it seems to me that the first construction yields two vertices connected by an edge, while the second construction yields to bare vertices.
What am i missing?
If the second definition is indeed wrong, what is the right way to obtain the nerve of an open covering as a special case of the nerve of a category?

REPLY [5 votes]: I think the second construction is not correct. If you replace the cover with the category whose objects are all intersections of elements of your original cover, then the two notions agree.<|endoftext|>
TITLE: Uniqueness in Composition of Polynomials
QUESTION [9 upvotes]: The following situation came up in my research:
Suppose two functions $f$ and $g$ map $[0,\infty)$ to (a subset of) itself. The function $f$ is linear and $g$ is quadratic, but $g$ is one-to-one on the interval $[0,\infty)$.
My conjecture/desired property: Any permutation of compositions of these two functions yields a unique polynomial.
The only progress I've made is the easy step of looking at degrees to argue that if two permutations are the same then they must have the same number of $g$'s.
For example, in my specific case $f(x)=x+1$ and $g(x)=x^2+x=x(x+1)$. Looking at length-3 compositions with two $g$'s gives the following different polynomials:
$f\circ g \circ g (x) = 1 + x + 2 x^2 + 2 x^3 + x^4$
$g\circ f \circ g (x) = 2 + 3 x + 4 x^2 + 2 x^3 + x^4$
$g\circ g \circ f (x) = 6 + 15 x + 14 x^2 + 6 x^3 + x^4$
I feel like this is probably easier than I'm making it, so please feel free to just give a reference for dealing with the semigroup(?) of polynomial composition.

REPLY [2 votes]: This is a bit more than what you want, but in MR1438634 (98c:12003), it is shown that 
f(x)=x+1 and g(x)=x^3 generate a free subgroup of the group of homeomorphisms of R. 
What you need is a semi-group and sometimes it is easier to find them. There is a bit more general discussion of this question and others in pages 30-31 and 37 of Topics in Geometric Group theory, by Pierre de la Harpe.<|endoftext|>
TITLE:  Smooth manifolds that don't admit a partition of unity
QUESTION [7 upvotes]: When I first starting studying differential geometry, I asked my lecturer a question about smooth manifolds that didn't admit a partition of unity. He promptly told not to worry about such objects as they were only studied by the extremely eccentric. I would like to know if this is true, ie, does anyone study manifolds that don't admit a partition of unity (not whether such people are eccentric).

REPLY [13 votes]: The answer to your stated question ("Does anyone study non-paracompact manifolds?") is certainly yes.  Here are a few papers which do just this:

Gauld, David.
  Manifolds at and beyond the limit of metrisability. (English summary) Proceedings of the Kirbyfest (Berkeley, CA, 1998), 125--133 (electronic),
  Geom. Topol. Monogr., 2, Geom. Topol. Publ., Coventry, 1999. 
http://www.emis.de/journals/GT/ftp/main/m2/m2-7.pdf

Among other things, Gauld references that there are two paracompact and two nonparacompact 1-manifolds, and $\aleph_0$ paracompact and $2^{\aleph_1}$ non-paracompact 2-manifolds.  (That's a lot!)

Foliations and foliated vector bundles, 1969 MIT lecture notes
http://www.foliations.org/surveys/FoliationLectNotes_Milnor.pdf

Milnor entertains non-paracompact manifolds.  In particular he constructs a (necessarily non-paracompact) surface with uncountable fundamental group.  Milnor also says: "The main object of this exercise is to imbue the reader with suitable respect for non-paracompact manifolds."

Balogh, Zoltan; Gruenhage, Gary.
  Two more perfectly normal non-metrizable manifolds. (English summary)
  Topology Appl. 151 (2005), no. 1-3, 260--272. 

The existence of perfectly normal, non-metrizable (hence non-paracompact) manifolds is shown to depend upon one's set-theoretic assumptions.  
And so forth.  I could find 10 more papers without much effort.  I'm not sure I could find 100.  (A MathSciNet search with "manifold" and ("nonmetrizable" or "non-paracompact") in the anywhere fields doesn't return many hits.)  So some serious mathematicians take non-paracompact manifolds seriously enough to write some papers about them.  On the other hand, although one could use more complimentary language than "extremely eccentric", your lecturer's take on non-paracompact manifolds seems to be an accurate reflection of how most geometric topologists feel: they seem mostly to be used as a source of counterexamples and to be of interest to general and set-theoretic topologists.<|endoftext|>
TITLE: Bousfield-Kan spectral sequence with local coefficients
QUESTION [5 upvotes]: Let $F : \mathcal{D} \to \mathbf{Top}$ be a diagram of topological spaces. A local system of coefficients $M$ on $\mathrm{colim}_\mathcal{D} F$ pulls back to a local system $M_d$ on $F(d)$ for each $d \in \mathcal{D}$, and also a local system $M_h$ on $\mathrm{hocolim}_\mathcal{D} F$.
Is there a Bousfield-Kan type spectral sequence of the form
$$E^2_{s,t} = \mathrm{colim}^s_{\mathcal{D}} H_t(F(d);M_d) \Rightarrow H_{s+t}(\mathrm{hocolim}_\mathcal{D} F;M_h)$$
and if so where can one find it in the literature? I would also be content to know if this is not possible.

REPLY [8 votes]: Let LOC be the category in which an object is a space plus a local system on it, and a morphism is a map of spaces covered by a map of coefficient systems in the obvious sense. There's an obvious functor $C$ from LOC to CH, the category of chain complexes; one can speak of the hocolim of a diagram of chain complexes; and $C$ commutes with hocolim up to natural equivalence. Your setup yields a functor $\mathcal D\to LOC$, and then your problem  becomes the algebraic problem of making a spectral sequence going from $E^2_{s.t}=colim_{\mathcal D}^s H_t(F)$ to $H_{s+t}(hocolim_{\mathcal D} F)$ when $F$ is a functor $\mathcal D\to CH$. For this you can, as Tyler suggests, use a simplicial model for $hocolim_{\mathcal D}F$ related to the nerve of $\mathcal D$ and thus get a two-quadrant double chain complex for which one of the two standard filtrations yields a spectral sequence of the desired kind. ($colim^i_{\mathcal D}$ means the $ith$ derived functor of the functor $colim_{\mathcal D}$ from $\mathcal D$-diagrams of abelian groups to abelian groups; it is the same as $ith$ homology of hocolim of the diagram of (abelian groups viewed as) chain complexes.)<|endoftext|>
TITLE: Is the average first return time of a partitioned ergodic transformation just the number of elements in the partition?
QUESTION [5 upvotes]: For some reason my thinking is very fuzzy today, so I apologize for the following rather silly question below...
Let $T$ be an ergodic transformation of $(X,\Omega, \mathbb{P})$ and let $X$ be partitioned into $n < \infty$ disjoint sets $R_j$ of positive measure. For $x \in R_k$ define $\tau(x) := \inf \{\ell>0:T^\ell x \in R_k\}$. The Kac lemma (see, e.g. http://arxiv.org/abs/math/0505625) gives that $\int_{R_k} \tau(x) \ d\mathbb{P}(x) = 1$. 
Now $\int_X \tau(x) \ d\mathbb{P}(x) = \sum_k \int_{R_k} \tau(x) \ d\mathbb{P}(x) = n$, or equivalently $\mathbb{E}\tau = n$.
Can anyone provide a sanity check on the above assertion that the expected return time is just the size of the partition? I've never seen this explicitly stated as a corollary of the Kac lemma, which seems odd.

REPLY [3 votes]: I found a 2003 paper of Choe containing your sanity check, called "A universal law of logarithm of the recurrence time".  See the first few lines of section 3 on page 888.  The "$K_n$" used there is essentially your $\tau$, but corresponding to a partition $P_n$ in a sequence of partitions.  See definition 1.7 on page 885 for what the notation means.<|endoftext|>
TITLE: Algebraic proof of 4-colour theorem?
QUESTION [47 upvotes]: 4-colour Theorem. Every planar graph is 4-colourable.
This theorem of course has a well-known history.  It was first proven by Appel and Haken in 1976, but their proof was met with skepticism because it heavily relied on the use of computers.  The situation was partially remedied 20 years later, when Robertson, Sanders, Seymour, and Thomas published a new proof of the theorem.  This new proof still relied on computer analysis, but to such a lower extent that their proof was actually verifiable. Finally, in 2005, Gonthier and Werner used the Coq proof assistant to formalize a proof, so I suppose only the most die hard skeptics remain.
My question stems from reading An Update on the Four-Color Theorem by Robin Thomas.  The paper describes several interesting reformulations of the 4-colour theorem.  Here is one:
Note that the cross-product on vectors in $\mathbb{R}^3$ is not an associative operation.  We therefore define a bracketing of a cross-product $v_1 \times \dots v_n$ to be a set of brackets which makes the product well-defined.
Theorem. Let $i, j, k$ be the standard unit vectors in $\mathbb{R}^3$.  For any two different bracketings of the product $v_1 \times \dots \times v_n$, there is an assignment of $i,j,k$ to $v_1, \dots, v_n$ such that the two products are equal and non-zero.
The surprising fact is that this innocent looking theorem implies the 4-colour theorem.
Question. Is anyone working on an algebraic proof of the 4-colour theorem (say by trying to prove the above theorem)?  If so, what techniques are involved?  What partial progress has been made?  Or do most people consider the effort/reward ratio of such an endeavor to be too high?
I think it would be interesting to have an algebraic proof, even a very long one, particularly if the algebraic proof does not use computers. Given its connection to many other areas (Temperley-Lieb Algebras), the problem seems to be amenable to other forms of attack.

REPLY [6 votes]: There are several reformulations by Matiyasevich, like this with polynomial, this with binomial coefficients modulo 7, also some others but they are less `algebraic', whatever it means.<|endoftext|>
TITLE: Do the empty set AND the entire set really need to be open?
QUESTION [5 upvotes]: My question is motivated by the previous discussion 'Why is a topology made of open sets?'. While the axioms for arbitrary unions and finite intersections are without doubt essential to the concept of a topological space, the 1st axiom (the set itself and the empty set are open) seems rather technical. So, do we really need these conditions in order to build most (if not all) of the point-set-topology without significant changes? In other words, if we leave out the 1st axiom, can point-set-topology stil remain as useful and powerful in terms of what we actually need to do analysis and geometry?
EDIT: Expanded the topic name in accordance with my question.

REPLY [2 votes]: To address the question in a somewhat less-categorical way, I would point out that you can in fact do all of topology without using the expression "open set", by instead refering to the filter of neighborhoods of every point --- then a function $f:X\rightarrow Y$ is continuous iff for all $x\in X$ and every neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)\subset V$.  You'll remember this as the "epsilon-delta"-style of continuity criterion from calculus, only without mentioning $&epsilon;$ or $&delta;$.  The viewpoints are equivalent in the sense that one can define an open set as being a neighborhood of all its points, or contrariwise define a neighborhood of $x$ as including some open set containing $x$.
There's another alternative, to study presentations of topologies; usually a base for a topology is given.  These have the advantage of being available as raw data, because the closure requirements for a base are much less stringent from a set-theoretic point of view than those of the whole system of open sets; one consequence is that you can study the topological space $(X,\langle B\rangle)$ in any universe including $X$, $B$ and $(X,B)$.  This notion of a presented topology then lets you compare how the properties of a space-given-the-base change after a forcing extension of the universe --- if you're into that sort of thing.
The moral of this story is that open sets are not really the object of study in topology, and changing what you mean by "open set" is mostly going to distract you; continuous functions are one object of study --- as others have pointed out --- and there are many ways to define those.<|endoftext|>
TITLE: Bounds on a partition theorem with ambivalent colors
QUESTION [5 upvotes]: I've been running into the following type of partition problem.

Given positive integers h, r, k, and a real number ε ∈ (0,1), find n such that if every (unordered) r-tuple from an n element set X is assigned a set of at least εk 'valid' colors out of a total of k possible colors, then you can find H ⊆ X of size h and a single color which is 'valid' for all r-tuples from H.

Lower bounds on the smallest such n can be obtained from lower bounds for Ramsey's Theorem. If k is sufficiently large, then partition the set of colors into [1/ε] pairwise disjoint sets of approximately equal size to emulate a proper [1/ε]-coloring of r-tuples. A simple pigeonhole argument shows that this is essentially sharp when r = 1 and k is large enough, i.e. one color must be 'valid' for at least nε points.
Is the Ramsey bound more or less sharp for r > 1 or are there better lower bounds? The interesting case is when k is large since the proposed Ramsey lower bound is (surprisingly?) independent of k.

REPLY [3 votes]: Here is a generalization of domotorp's answer to arbitrary h > r > 1.
Independently for each color i ∈ {1,2,...,k}, pick a random Hi from a family H of r-hypergraphs that don't contain any complete r-hypergraph of size h. Declare color i to be 'valid' for the r-tuple t = {t1,...,tr} iff t ∈ Hi. Let Yt be the number of 'valid' colors for t. Note that Yt is binomial with parameters (k, p) for some 0 < p ≤ 1/2 which is independent of k and also independent of t when H is closed under isomorphism. Hoeffding's Inequality then gives
Prob[Yt ≤ εk] ≤ exp(-2k(p-ε)2)
for 0 < ε < p. So the probability that Yt ≥ εk for all i is positive whenever n ≤ exp(2k(p-ε)2/r) (not optimal). 
This is not enough since p implicitly depends on n. However, for fixed h > r > 1, p can be bounded away from 0. This can be seen by using for H the family of r-partite hypergraphs as domotorp did, but different choices of H give better bounds.<|endoftext|>
TITLE: Topological results from geometry
QUESTION [5 upvotes]: Hi people,
I'm interested in results, such as the Gauß-Bonnet theorem, Fàry-Milnor theorem or classification theorems for manifolds, which give topological properties from geometric considerations. Can anyone recommend some good texts? In particular I'd like to see a nice proof of Fàry-Milnor and of the theorem of turning tangents (total curvature of an imbedded plane curve is $2\pi$).
thanks

REPLY [5 votes]: This topic being quite large, I cannot insist enough to recommand you to take a look to Marcel Berger's Panoramic view of Riemannian geometry. The Bonnet-Myers theorem, the sphere theorems (for the recent development on this one, I think the web page of Simon Brendle contains a survey) are two celebrated examples of the topological consequences of geometric properties in the setting or Riemannian geometry.<|endoftext|>
TITLE: Drawing (graphs) by numbers: a minimality question
QUESTION [5 upvotes]: Every simple graph $G$ can be represented ("drawn") by numbers in the following way:

Assign to each vertex $v_i$ a number $n_i$ such that all $n_i$, $n_j$ are coprime whenever $i\neq j$. Let $V$ be the set of numbers thus assigned. 
Assign to each maximal clique $C_j$ a unique prime number $p_j$ which is coprime to every number in $V$.
Assign to each vertex $v_i$ the product $N_i$ of its number $n_i$ and the prime numbers $p_k$ of the maximal cliques it belongs to.


Then $v_i$, $v_j$ are adjacent iff $N_i$
  and $N_j$ are not coprime,

i.e. there is a (maximal) clique they both belong to. Edit: It's enough to assign $n_i = 1$ when $v_i$ is not isolated and does not share all of its cliques with another vertex.
Being free in assigning the numbers $n_i$ and $p_j$ lets arise a lot of possibilites, but also the following question:

QUESTION
Can the numbers be assigned  systematically such that the greatest $N_i$
  is minimal (among all that do the job) — and if so: how?

It is obvious that the $n_i$ in the first step have to be primes for the greatest $N_i$ to be minimal. I have taken the more general approach for other - partly answered  - questions like "Can the numbers be assigned such that the set $\lbrace N_i \rbrace_{i=1,..,n}$ fulfills such-and-such conditions?"

REPLY [3 votes]: As an alternative to my earlier computational answer for particular graphs G, here is a worst-case description of the asymptotic growth of the minimum size of the integers Nj 
Let h be the sum of |V(G)| and the number of maximal cliques in G (bounded above by |E(G)|, which is saturated for bipartite graphs, whose edges are the maximal cliques).
Let c be the maximum number of maximal cliques to which a vertex v ∈ V(G) may belong. Because ph ≤ h (ln(h) + ln ln(h)), we can bound
$$ \large \Big(\min_{\small\text{weightings}} \; \max_{v \in V(G)}\;  N_v\Big) \;\;\in\;\; \mathrm O\Big(h^c \log^c(h)\Big). $$
This bound is asymptotically saturated by placing the largest prime weights on the vertex in the largest number of maximal cliques, and those maximal cliques of which it is a part, which is obviously a bad thing to do. But e.g. in Cayley graphs G, every vertex belongs to the same number of maximal cliques, this asymptotic growth cannot be avoided, as there will exist vertices v for which Nv will consist exclusively of a product of primes pt, for t bounded below by a constant fraction of h.
One can construct bipartite Cayley graphs in which the degree of each vertex is a constant fraction of n = |V(G)|. We then have h = α n for some 0 < α < 1; and h = |V(G)| + |E(G)| ∈ O(n2), so that
$$ \large \Big(\min_{\small\text{weightings}} \; \max_{v \in V(G)}\;  N_v\Big) \;\;\in\;\; \mathrm O\Big(n^{2\alpha n} \log^{\alpha n}(n)\Big) $$
for such graphs. Thus there exist graphs for which the coefficients Nj grow much more quickly than e.g. the factorial function.<|endoftext|>
TITLE: Simply connectedness of algebraic group
QUESTION [5 upvotes]: $G$ is a semisimple algebraic group over $k$, if $G_{\bar k}$ is simply connected when we do base change to $\bar k$, can we descent the simply connectedness to $G$?
Here, simply connectedness means no nontrivial connected central isogeny onto $G$.
Can we say that simply connected algebraic group is geometrically connected? If then we can give an affirmative answer by considering the universal cover of $G$.
Welcome for any answer under further assumption that $\text{char }k=0$.

REPLY [9 votes]: To amplify Brian Conrad's semi-answer, I need a more precise definition of
"simply connected" at the outset.   In characteristic 0 some of the classical
ways of thinking about this concept can be carried over to the algebraic 
setting, but in prime characteristic the most common definition starts with
a connected semisimple group.   Over an algebraically closed field,
the algebraic criterion for such a group to be simply connected
is that the character group of a maximal torus be the
full weight lattice.   
Here the "fundamental group" of the adjoint group in the compact case
is re-interpreted as the quotient of the weight lattice by the root lattice,
which may also be regarded as the (scheme-theoretic) center of the simply
connected group. 
There may be no quotable source earlier than the 1956-58 Chevalley
seminar.  The classification work of Tits and others then descends to arbitrary
fields of definition.  In SGA 3, Expose 22 (by Demazure), Definition 4.3.3 defines "simply
connected" in terms of the behavior of fibers relative to this 
criterion using the root datum language.<|endoftext|>
TITLE: What is the etymology for the term conductor?
QUESTION [12 upvotes]: This is related to the previous question of how to define a conductor of an elliptic curve or a Galois representation.

What motivated the use of the word "conductor" in the first place?

A friend of mine once pointed out the amusing idea that one can think of the conductor of an elliptic curves as "someone" driving a train which lets you off at the level of the associated modular form.
A similar statement can be made concerning Szpiro's conjecture, which provides asymptotic bounds on several invariants of an elliptic curve in terms of its conductor. Here one might think of the conductor as "someone" who controls this symphony of invariants consisting of the minimal discriminant, the real period, the modular degree, and the order of the Shafarevich-Tate group (assuming BSD).
Was there some statement of this sort which motivated Artin's original definition of the conductor?

Does anyone have a reference for the first appearance of the word conductor in this context?

I apologize if this question is inappropriate for MO.

REPLY [14 votes]: "Es steht alles schon bei Dedekind", as Emmy Noether was fond of saying. In fact,

R. Dedekind, Über die Anzahl der Idealklassen in den verschiedenen 
Ordnungen eines endlichen Körpers, Gauss Festschrift 1877

defined the "Führer" of an order in a number field. [BTW: in German, Führer
does not actually mean a strong leader but rather someone who guides you (as in 
tourist guide). But of course . . . ]
Class groups of orders in quadratic number fields are ring class groups, which generalize immediately to ray class groups (Weber); from there the word spread to complex multiplication and class field theory.<|endoftext|>
TITLE: What are some examples of narrowly missed discoveries in the history of mathematics? 
QUESTION [42 upvotes]: What are the examples of some mathematicians coming very close to a very promising theory or a correct proof of a big conjecture but not making or missing the last step?

REPLY [4 votes]: One example of a missed opportunity, in my opinion, is the Aleksandrov-Zeeman theorem, which states (in one of its different forms) that any bijection of $d$-dimensional Minkowski space-time onto itself ($d>2$) which sends light ray segments into light ray segments (= Einstein's postulate of constancy of the speed of light in Special Relativity) is the composite of one or more of the following:

A space-time translation; 
A spatial rotation; 
A scale transformation;
A time reflection;
A space reflection;
And a Lorentz boost.

Particularly, this singles out Lorentz boosts as the only transformation law between inertial space-time frames obeying Einstein's postulate. Einstein did show that Lorentz boosts satisfy the latter, but he did not prove whether Lorentz boosts are unique in the above sense, neither did any of his predecessors (Lorentz, Poincaré). Surprisingly, the answer came only about half a century later, through the work of A.D. Aleksandrov in the 50's (assuming the bijection acts linearly) and independently by E.C. Zeeman in 1964 (Causality Implies the Lorentz Group, J. Math. Phys. 5 (1964) 490-493). Aleksandrov dropped the hypothesis of linearity from his argument in 1967 (A Contribution to Chronogeometry, Canad. J. Math. 19 (1967) 1119-1128).
Aleksandrov's proof was topological, based on the concept nowadays called "Alexandrov topology", which is just the order topology derived from the chronology relation in Minkowski space-time. Zeeman's proof, however, was fairly elementary and classical, relying only on bits of analytic geometry. Both the theorem and Zeeman's proof were perfectly within the grasp of the likes of Felix Klein at the time Einstein's work was published, and perfectly within the spirit of Klein's Erlanger Programm as he noticed himself by dismissing special relativity as a simple special case of his programme and calling it the "invariant theory of the Lorentz group" in his address Über die geometrischen Grundlagen der Lorentzgruppe (Jahresbericht der Deutschen Mathematiker-Vereinigung Bd. 19 (1910), pages 533-552 of his collected works). This indicates that he could have proven Zeeman's theorem at least as early as 1910, if only he had the interest.<|endoftext|>
TITLE: Applications of Brouwer's fixed point theorem
QUESTION [30 upvotes]: I'm presenting Brouwer's fixed point theorem to an audience that knows some point-set topology. Does anyone have any zippy / enlightening / cool applications or consequences of it? So far, I have:

Physical realizations: stuff involving maps of my city, crumpled pieces of graph paper, a stationary gin molecule in a cocktail shaker, etc.
That every n*n real matrix with all-positive entries has a positive eigenvalue.

Thanks! :)

REPLY [3 votes]: The proof of the fundamental theorem of algebra using the Brouwer fixed point theorem is given here:
 B.H. Arnold, "A topological proof of the fundamental theorem of algebra" Amer. Math. Monthly , 56 (1949) pp. 465–466<|endoftext|>
TITLE: Is every integral epimorphism of commutative rings surjective?
QUESTION [20 upvotes]: That's the question. Recall that a morphism $f\colon A\to B$ of commutative rings is integral if every element in $B$ is the root of a monic polynomial with coefficients in the image of $A$ and that $f$ is an epimorphism if and only if the multiplication map $$B\otimes_A B\to B$$ is an isomorphism.
If we make the additional assumption that $B$ is finitely generated as an $A$-algebra, then it is true. This can be proven by Nakayama's lemma, for example.
This came up not so long ago when I was trying to show that the Witt vector functor (of finite length) preserves separatedness of algebraic spaces. In this application I was able to reduce things to the finitely generated case and could therefore use the weaker statement above, but I still wonder about the general case.

REPLY [7 votes]: If $A$ is noetherian, then every integral epimorphism $f \colon A \to B$ is surjective. This has been proven by Ferrand: Prop. 3.8 in Monomorphismes de schémas noethérien", Exp. 7 in Séminaire Samuel, Algèbre commutative, 2, 1967-1968. For schemes, this is phrased as:
Theorem: Let $f \colon X \to Y$ be a morphism of schemes such that $Y$ is locally noetherian. Then $f$ is a closed immersion if and only if $f$ is a universally closed monomorphism.
(It can be seen any universally closed injective morphism of schemes is affine and integral [EGA IV, 18.12.10], so the scheme version is equivalent to the affine version.)
In Ferrand's theorem, one can also replace $X$ and $Y$ with algebraic spaces: In fact, the question is local on $Y$ so we can assume that $Y$ is a scheme. Since $f$ has affine fibers and is universally closed, it follows that $f$ is affine by arXiv:0904.0227 Thm 8.5.
PS. How come that you needed this question for non-finite morphisms for your application to Witt vectors? If the diagonal of an algebraic space is a universally closed monomorphism then (since the diagonal of an algebraic space always is locally of finite type), it is a proper monomorphism, hence a closed immersion ([EGA IV 18.12.6]). In particular, it follows that if $f \colon X \to Y$ is universally closed and surjective and $X$ is separated, then so is $Y$. Perhaps this was what you meant by "reduce things to the finitely generated case"?<|endoftext|>
TITLE: How do you axiomatize topology via nets?
QUESTION [28 upvotes]: Let $X$ be a set and let ${\mathcal N}$ be a collection of nets on $X.$
I've been told by several different people that ${\mathcal N}$ is the collection of convergent nets on $X$ with respect to some topology if and only if it satisfies some axioms.  I've also been told these axioms are not very pretty.
Once or twice I've tried to figure out what these axioms might be but never came up with anything very satisfying.  Of course one could just recode the usual axioms regarding open sets as statements about nets and then claim to have done the job.  But, come on, that's nothing to be proud of.
Has anyone seen topology axiomatized this way?  Does anyone remember the rules?

REPLY [6 votes]: (too long for a comment to Pete's answer)
Garrett Birkhoff was my Ph.D. advisor.  Let me provide a few remarks
of a historical nature.
From a 25-year-old Garrett Birkhoff we have: Abstract 355, "A new
definition of limit" Bull. Amer. Math. Soc. 41 (1935) 636.
(Received September 5, 1935)
According to the report of the meeting (Bull. Amer. Math. Soc. 42
(1936) 3) the paper was delivered at the AMS meeting in New York on
October 26, 1935.
In the abstract we find what would nowadays be called convergence
of a filter base.  (See also Definition 4 in Birkhoff's 1937 paper.)
Birkhoff remarked to me once that Bourbaki never acknowledged his
(Birkhoff's) priority.
It seems that some time after Birkhoff's talk, his father (G. D. Birkhoff)
remarked that it reminded him of a paper of Moore and Smith.  So
young Garrett read Moore and Smith, and in the end adopted their
system for the subsequent paper, calling it "Moore-Smith convergence
in general topology".  Since that Annals of Mathematics paper was
received April 27, 1936, one can only imagine young Garrett working
furiously for 6 months converting his previous filter-base material
into the Moore-Smith setting!<|endoftext|>
TITLE: How far can the analogy between a Cayley graph and a symmetric space be pushed?
QUESTION [6 upvotes]: If $G$ is a finitely group and $S$ a finite symmetric set of generators, the associated Cayley graph, then $x \mapsto x^{-1}$ gives rise to a geodesic symmetry $i$ at the identity:
If $g=s_1^{e_1}\cdots s_k^{e_k}$ with $e_i \in \{\pm 1\}$, then let $i(g):=s_1^{-e_1}\cdots s_k^{-e_k}$.`
Translating $i$ via $G$, there is a symmetry at every vertex.
For points in the interior of edges, there is likewise a (local) symmetry.
This prompts my question: What concepts of locally symmetric Riemannian spaces can be applied to the study of Cayley graphs?

REPLY [7 votes]: The Cheeger constants for graphs and Riemannian locally symmetric spaces are closely related. Via inequalities of Buser and Cheeger, these are also related to eigenvalues of the laplacians for each. This analogy led to the first construction of expander graphs, by Margulis, via Property (T). More recently, this analogy has been exploited by several people, notably Marc Lackenby, to study finite-sheeted coverings using Cayley graphs of finite quotients as a finite simplicial approximation.
The point, roughly, is the following. Let $\Gamma$ be a group with generating set $S$, and suppose $\Gamma = \pi_1(M)$ for some Riemannian manifold $M$. Then any finite quotient $F$ under a homomorphism $\phi$ has a generating set $\phi(S)$, so we can form the corresponding Cayley graph $\mathcal{G}(F, \phi(S))$. Properties of $\mathcal{G}(F, \phi(S))$ like girth, spectrum, expansion constants, Cheeger constant, and so forth are closely related to the analogous concept for the finite-sheeted covering $M_\phi$ of $M$ corresponding to the subgroup $\mathrm{kernel}(\phi)$ of $\Gamma$. This analogy is most potent when you consider a family {$\mathcal{G}(F_j, \phi_j(S))$} of Cayley graphs corresponding to a family $F_j$ of finite quotients of $\Gamma$.
References for all these concepts are the books On Property ($\tau$) by Lubotzky and Zuk (unpublished, but on Lubotzky's website), Discrete Groups, Expanding Graphs and Invariant Measures by Lubotzky, Elementary Number Theory, Group Theory and Ramanujan graphs by Davidoff, Sarnak, and Valette, and Marc Lackenby's paper Expanders, ranks and graphs of groups, Israel J. Math. 146 (2005) 357-370.<|endoftext|>
TITLE: Why do gerbes live in H^2?
QUESTION [18 upvotes]: Line bundles on a scheme $X$ live in $H^1(X,O_X^*)$, where $O_{X}^{*}$  is the sheaf of invertible functions. If $X$ is noetherian separated, then we can think of this $H^1$ to be Čech cohmology w.r.t. an open affine cover of $X$. We can think of a line bundle as a principal $\mathbb{G}_{m}$ -bundle,  where $\mathbb{G}_{m}$  is the multiplicative group scheme, i.e. the result of patching together, via $\mathbb{G}_{m}$ -valued transition functions, local pieces that look like $ U \times \mathbb{G}_{m} $ for $U$ open in $X$.
I apologize if the question sounds trivial to people who have a serious knowledge of stack theory. 
First of all, let's take the following definition of "gerbe" (can find in Wikipedia): a gerbe on $X$ is a stack $G$ of groupoids over $X$ which is locally non-empty (each point in $X$ has an open neighbourhood $U$ over which the section category $G(U)$ of the gerbe is not empty) and transitive (for any two objects $a$ and $b$ of $G(U)$ for any open set $U$, there is an open set $V$ inside $U$ such that the restrictions of $a$ and $b$ to $V$ are connected by at least one morphism). And in this context, I think we should add that, locally over $X$, $G$ should be isomorphic to $U\times B \mathbb{G}_{m}$ ($B \mathbb{G}_{m}$ is the classifying stack of $\mathbb{G}_{m}$).
I was told that $\mathbb{G}_{m}$ -gerbes over $X$ up to equivalence correspond to cohomology classes in $H^2(X,\mathbb{G} _{m})$. 
I would like to understand in concrete (Čech) terms why this bijection should take place. In other words: why the process of patching classifying spaces (edit: rather, classifying stacks) of a group involves passing to the second cohomology group?

REPLY [13 votes]: A nice point of view is to consider principal bundles with structure group $B\mathbb{C}^\times$. One can probably take any abelian Lie group instead of $\mathbb{C}^\times$. Principal $B\mathbb{C}^\times$-bundles are one way to give a precise meaning to what you are calling "patching together classifying spaces". 
The point is that there is an isomorphism
$$
H^1(X,B\mathbb{C}^\times) = H^2(X,\mathbb{C}^\times),
$$
which explains the relation to the second cohomology group. It is basically saying that principal $B\mathbb{C}^\times$-bundles are the same as $\mathbb{C}^\times$-gerbes. 
This isomorphism is induced from the exact sequence
$$
1 \to \mathbb{C}^\times \to E\mathbb{C}^\times \to B\mathbb{C}^\times \to 1
$$
of groups, and the fact that the sheaf of $E\mathbb{C}^\times$-valued functions on a paracompact space $X$ is soft. All this is very nicely explained in Gajer's Inventiones paper "Geometry of Deligne cohomology".<|endoftext|>
TITLE: Good example of a non-continuous function all of whose partial derivatives exist
QUESTION [7 upvotes]: What's a good example to illustrate the fact that a function all of whose partial derivatives exist may not be continuous?

REPLY [3 votes]: I was travelling when this came up.
With the same results as Georges Elencwajg's comment and Marco's recent answer, take
$$ f(x,y) = \frac{2 x^2 y}{x^4 + y^2} $$ and set to $0$ at the origin $(0,0).$  Along any line  through the origin $ x = a t, \; y = b t$ the limit is 0, as 
$$ | f(a t, b t) | \leq a^2 | t / b | .  $$ However, along  the parabola $y = x^2$ the value is 1, and along the parabola $y = - x^2$ the value is $-1.$
To get "directional derivative" 0 in every direction through the origin switch to
$$ g(x,y) = \frac{2 x^3 y}{x^6 + y^2} $$ as
$$ | g(a t, b t) | \; \leq \; t^2 \; | a^3 / b |   $$ when $b \neq 0,$ but then if $y = \pm x^3$...the directional derivative  generalizes to the Gateaux derivative in other settings.<|endoftext|>
TITLE: Is there an analogue of curvature in algebraic geometry?
QUESTION [96 upvotes]: I am not an expert, but there seems to be an enormous technical difference between algebraic geometry and differential/metric geometry stemming from the fact that there is apparently no such thing as curvature in the former context while curvature is everywhere in the latter (indeed, it is hard to produce nontrivial results in Riemannian geometry that DON'T involve curvature).  Of course it seems unreasonable to just port definitions of curvature into an algebraic context, but maybe there are constructions that play the same role in algebraic geometry that curvature does in other kinds of geometry.  Here are two specific ways the notion of curvature shows up which I can imagine making sense in more general contexts.
Algebraic Chern - Weil Theory?  In differential geometry one uses the curvature of a connection on a vector bundle to produce explicit cohomology classes.  Does this have an algebraic analogue?
Algebraic Curvature Bounds?  One supremely important theme in modern geometry involves proving theorems that depend only on the large scale geometry of a space, and the main strategy is to compare the space to a simpler one with the same large scale properties.  This reminds me a little bit of tropical geometry wherein one replaces an algebraic variety with a simple combinatorial proxy, but from what little I know the analogy seems to stop there.
Any thoughts?  I hope this question is not too vague, but it seems worthwhile and part of the problem is that I can't formulate a precise question along these lines.  Thanks in advance!

REPLY [47 votes]: An algebraic analog of Chern-Weil theory (explicitly taking symmetric polynomials of curvature) is given by the Atiyah class.
Given a vector bundle $E$ on a smooth variety we can consider the short exact sequence 
$$ 0\to End(E) \to A(E) \to T_X\to 0$$
where $T_X$ is the tangent sheaf and $A(E)$ is the "Atiyah algebroid" --- differential operators of order at most one acting on sections of $E$, whose symbol is a scalar first order diffop (hence the map to the tangent sheaf). A (holomorphic or algebraic) connection is precisely a splitting of this sequence, and a flat connection is a Lie algebra splitting. Now algebraically such splittings will often not exist (having a holomorphic connection forces your characteristic classes to have type $(p,0)$ rather than the $(p,p)$ you want..) but nonetheless we can define the extension class, which is the Atiyah class 
$$a_E\in H^1(X, End(E)\otimes \Omega^1_X).$$
This is the analog of the curvature form in the Riemannian world -- we now can take symmetric polynomials in the $End(E)$ factor to get the characteristic classes of $E$ in $H^p(X,\Omega_X^p)$ as desired.
This answer and Mariano's agree of course in the sense that Atiyah classes can be interpreted via Hochschild and cyclic (co)homology and generalized to arbitrary coherent sheaves (or complexes) on varieties (or stacks) (let me stick to characteristic zero to be safe). Namely the Atiyah class of the tangent sheaf can be used to define a Lie algebra structure (or more precisely $L_\infty$) on the shifted tangent sheaf $T_X[-1]$, and Hochschild cohomology is its enveloping algebra. This Lie algebra acts as endomorphisms of any coherent sheaf (which is another way to say Hochschild cohomology is endomorphisms of the identity functor on the derived category), and one can take characters for these modules, recovering the characteristic classes defined concretely above.
(In fact the notion of characters is insanely general... for example an object of any category - with reasonable finiteness - defines a class (or "Chern character") in the Hochschild homology of that category, which is cyclic and so descends to cyclic homology. An example of this is the category of representations of a finite group, whose HH is class functions, recovering usual characters, or coherent sheaves on a variety, recovering usual Chern character. or one can go more general.)<|endoftext|>
TITLE: Connected components of the orthogonal group O(2n) in characteristic 2.
QUESTION [10 upvotes]: I am looking for a reference for the following fact:
The orthogonal group $O_{2n}$ over an algebraically closed field of characteristic 2
has exactly two connected components.
To be more precise, let $O_q$ denote the orthogonal group of the quadratic form $q(x)=x_1 x_2 +x_3 x_4+\cdots +x_{2n-1}x_{2n}$
over an algebraically closed field $k$.
In characteristic $p\neq 2$ the determinant takes two values on $O_q$, 1 and $-1$,
and therefore the subgroup $SO_q:=O_q\cap SL_{2n}$ is of index 2 in $O_q$; it is known that $O_q\cap SL_{2n}$ is connected.
In characteristic 2 the determinant takes only one value 1 on $O_q$ (because $-1=1$), and therefore $O_q\cap SL_{2n}=O_q$.
Still there is a homomorphism $D\colon O_q\to \mathbf{Z}/2\mathbf{Z}$ given by a polynomial $D$ called the Dickson invariant,
see J.A.~Dieudonn\'e, Pseudo-discriminant and Dickson invariant, Pacific. J. Math. 5 (1955), 907--910.
This homomorphism $D$ indeed takes both values 0 and 1 on $O_q$, and therefore its kernel ker $D$
is a closed subgroup of index 2 in $O_q$. I would like to know that ker $D$ is connected.
In other words, I am looking for a reference to the assertion that the orthogonal group $O_q$ has at most two connected components.
This is proved in Brian Conrad's handout "Properties of orthogonal groups" to his course Math 252 "Algebraic groups",
see http://math.stanford.edu/~conrad/252Page/handouts/O(q).pdf . Is there any other reference for this fact?
I will be grateful to any references, comments, etc.
Mikhail Borovoi

REPLY [2 votes]: An alternative presentation of Dickson's invariant is given in §4.1.2 of the article On the adjoint quotient of Chevalley groups over arbitrary base schemes, JIMJ 2010, by P.-E. Chaput and myself. There we show how the determinant $O(2n)\to\mathbb{Z}/2\mathbb{Z}$ in characteristic 0 (or say different from 2) extends to a morphism of group schemes over $\mathbb{Z}$, reducing to Dickson's invariant modulo the prime 2.<|endoftext|>
TITLE: Example Wanted: When Does Cech Cohomology Fail to be the same as Derived Functor Cohomology?
QUESTION [32 upvotes]: I want to know exactly how derived functor cohomology and Cech cohomology can fail to be the same.
I started worrying about this from this answer to an MO question, and Brian Conrad's comments to another MO question. Let $\mathcal{F}$ be a sheaf of abelian groups on a space X. (Here I want to be a little vague about what a "space" means. I'm thinking of either a scheme or a topological space). Then Cech cohomology of X with respect to a cover $U \to X$ can be defined as cohomology of the complex
$$ \mathcal{F}(U) \to \mathcal{F}(U^{[2]})  \to \mathcal{F}(U^{[3]}) \to \cdots $$
Where $U^{[ n ]} = U \times_X U \times_X \cdots \times_X U$. The total Cech cohomology of $X$, $\check H^{ * }(X, \mathcal{F}) $, is then given by taking the colimit over all covers $U$ of X. Now if the following condition is satisfied:

Condition 1: For sufficiently many covers $U$, the sheaf $\mathcal{F}|_{U^{[ n ]}}$ is an acyclic sheaf for each n

then this cohomology will agree with the derived functor version of sheaf cohomology. We have,
$$\check H^{ * }(X, \mathcal{F}) \cong H^*(X; \mathcal{F}).$$
I am told, however, that even if $\mathcal{F}|_U$ is acyclic this doesn't imply that it is acyclic on the intersections. It is still okay if this condition fails for some covers as long as it is satisfied for enough covers. However I am also told that there are spaces for which there is no cover satisfying condition 1.
Instead you can replace your covers by hypercovers. Basically this is an augmented simplicial object $$V_\bullet \to X$$ which you use instead of the simplicial object $U^{[ \bullet +1 ]} \to X$. There are some conditions which a simplicial object must satisfy in order to be a hypercover, but I don't want to get into it here. You can then define cohomology with respect to a hypercover analogously to Cech cohomology with respect to a cover, and then take a colimit. This seems to always reproduce derived functor sheaf cohomology.
So my question is when is this really necessary?

Question 1: What is the easiest example of a scheme and a sheaf of abelian groups (specifically representable ones such as $\mathbb{G}_m$) for which Cech cohomology of that sheaf and derived functor cohomology disagree?
Question 2: What is the easiest example of a (Hausdorff) topological space and a reasonable sheaf for which Cech cohomology and derived functor cohomology disagree?


I also want to be a little flexible about what a "cover" is supposed to be. I definitely want to allow interesting Grothendieck topologies, and would be interested in knowing if passing to a different Grothendieck topology changes the answer. It changes both the notion of sheaf and the notion of Cech cohomology, so I don't really know what to expect.
Also, I edited question 1 slightly from the original version, which just asked about quasi-coherent sheaves. Brian Conrad kindly pointed out to me that for any quasi-coherent sheaf the Cech cohomology and the sheaf cohomology will agree (at least with reasonable assumptions on our scheme, like quasi-compact quasi-separated?) and that the really interesting case is for more general sheaves of groups.

REPLY [6 votes]: In the paper Pathologies in cohomology of non-paracompact
Hausdorff spaces, Stefan Schröer constructs a Hausdorff space which is not paracompact, and for which sheaf cohomology with values in the sheaf of germs of $S^1$-valued functions does not agree with the Čech cohomology (for example - the same is true for other sheaves).
The space is constructed by taking the countably infinite join of disks $D^2$, with the CW-structure consisting of two 0-cells, two 1-cells and a single 2-cell, using one of the 0-cells as a basepoint. Then he takes a coarser topology, whereby the open sets are open sets from the CW-topology, but only those that either don't contain the basepoint, or contain all but finitely many of the closed disks.
With this topology the space is not paracompact, but is $\sigma$-compact, Lindelöf, metacompact.... and contractible! Sheaf cohomology is non-trivial however.
An interesting point to note is that this is not a k-space, and the k-ification of this space is the original CW-complex.<|endoftext|>
TITLE: How to localize a model category with respect to a class of maps created by a left Quillen functor
QUESTION [12 upvotes]: Let $M$ and $N$ be "nice" model categories.  I'm happy to have "nice" mean combinatorial model category.  Consider a Quillen pair
$$ L: M\rightleftarrows N: R.$$
I want the following result:

There exists a set of maps $S$ in $M$, such that $L$ and $R$ descend to a Quillen pair
  $$ L: S^{-1}M \rightleftarrows N: R,$$
  with the property that a map $f:X\to Y$ between cofibrant objects in $M$ is a weak equivalence in $S^{-1}M$ if and only if $L(f)$ is a weak equivalence in $N$.

Here $S^{-1}M$ denotes the model category with the same underlying category $M$ obtained by localizing $M$ with respect to the set of maps $S$.
This result seems to encode a standard technique; in fact, the very first example of a localized model category (Bousfield's localization of spaces with respect to a homology theory) can be viewed (in retrospect) of a special case of this.
I think I could prove this result if I need to.  But I would rather have a reference.  I've looked in the usual places, but I can't seem to find anything exactly like it.  I expected to find something like this in one of Dugger's papers on presentable model categories, but I don't find it there; he proves that under an additional condition, you can get a Quillen equivalence $S^{-1}M\rightleftarrows N$, but his construction of the set $S$ does not apply in my case.

REPLY [11 votes]: I suspect that you already have one, but here is a proof. I will assume that $M$ and $N$ are combinatorial and that $M$ is left proper (otherwise, I don't think that the literature contains a general construction of the left Bousfield localizations of $M$ by any small set of maps). Everything needed for a quick proof is available in Appendix A of
J. Lurie, Higher topos theory, Annals of Mathematics Studies, vol. 170, Princeton University Press, 2009.
First, there exists a cofibrant resolution functor $Q$ in $M$ which is accessible: the one obtained by the small object argument (as accessible functors are closed under colimits, it is sufficient to know that $Hom_M(X,-)$ is an accessible functor for any object $X$ in $M$, which is true, as $N$ is combinatorial). Let $W$ be the class of maps $f$ of $M$ such that $L(Q(f))$ is a weak equivalence in $N$. As $N$ is combinatorial, the class of weak equivalences of $N$ is accessible see Corollary A.2.6.9 in loc. cit. Therefore, by virtue of Corollary A.2.6.5 in loc. cit, the class $W$ is accessible. To Prove what you want, it is sufficient to check that $M$, $W$ and $C=${cofibrations of $M$} satisfy the conditions of Proposition A.2.6.8 in loc. cit. The only non trivial part is the fact that the class $C\cap W$ satisfies all the usual stability properties for a class of trivial cofibrations (namely: stability by pushout, transfinite composition). That is where we use the left properness. For instance, if $A\to B$ is in $W$ and if $A\to A'$ is a cofibration of $M$, we would like the map $A'\to B'=A'\amalg_A B$ to be in $W$ as well. This is clear, by definition, if $A$, $A'$ and $B$ are cofibrant. For the general case, as $M$ is left proper, $B'$ is (weakly equivalent to) the homotopy pushout $A'\amalg^h_A B$, and as left derived functors of left Quillen functors preserve homotopy pushouts, we may assume after all that $A$, $A'$ and $B$ are cofibrant (by considering the adequate cofibrant resolution to construct the homotopy pushout in a canonical way), and we are done. The case of transfinite composition is similar.<|endoftext|>
TITLE: Most striking applications of category theory?
QUESTION [98 upvotes]: What are the most striking applications of category theory? I'm trying to motivate deeper study of category theory and I have only come across the following significant examples:

Joyal's Combinatorial Species
Grothendieck's Galois Theory
Programming (unification as computing a coequalizer, Tatsuya Hagino's categorical construction of functional programming)

I am sure that these only touch on the surface so I would be most grateful to hear of more examples, thank you!
edit: To try and be more precise, "application" in the context of this question means that it makes use of slightly deeper results from category theory in a natural way. So we are not just trying to make a list of 'maths that uses category theory' but some of the results which exemplify it best, and might not have been possible without it.

REPLY [9 votes]: David Spivak has found applications of category theory in many areas outside of pure mathematics, and many are recorded in his book “Category Theory for the Sciences.” He's also done important work regarding the foundations of databases and schema, and it uses non-trivial results from category theory. A whole collection of literature in that direction can be found on his webpage, and probably satisfies the OP's desire that the application use "slightly deeper results" from category theory.
The rest of this answer is a giant list of examples of how you can use category theoretic thinking in pretty much every science. Many of these examples are taken from Spivak's writings. The list here comes from a general-audience talk I gave at my university (lecture notes here) back in 2015, and I figured I may as well post what I came up with somewhere, in case it helps others who need examples of category theory. You should probably take this list with a grain of salt: for many of the items, it would take some work to formalize the relationship to category theory. 
The talk tried to highlight the value of category theory in several stages:

Objects and the relationships between them.
The use of functors to build bridges between different categories.
Breaking an object up into simple pieces; understanding how to build complicated structure from these pieces. Limits and colimits.
Localization: shifting view so that two objects you previously viewed as different are now viewed as the same.
Replacing an object by one which is easier to work with but has the same fundamental properties you are trying to study.
Mapping an object to a small bit of information about the object. Showing that two are different because they differ on this bit. Trying to find a complete set of invariants so you know precisely when two are the same.

Let's start with examples of (1), i.e. of categories themselves:

Classical mechanics can be viewed as studying the state of the world around us as time goes on. So it works just like the example above, except an object is the whole state of the universe at time $t$.
States of the economy as time goes on.
Crystallography: Objects are arrangements of atoms in a molecule, morphism is a symmetry.
Databases: an object can be a table, a morphism can be a shared column (called a foreign key).
Going a bit more meta, an experiment is like a category. Objects could be observables and a relationship could tell us if they're correlated. Spivak writes: "Reusable methodologies can be formalized, and that doing so is inherently valuable. Category theory also provides a language for experimental design patterns, introducing formality while remaining flexible." 
Even more meta, the collection of all experiments is a category. Objects are experiments and we say two are related if they got the same conclusion (perhaps just on one question of interest across all experiments).
In material science, objects could be materials and we could draw $A\to B$ if A is an ingredient or part of B, so water $\to$ concrete. A different way to view it as a category would be to draw $A\to B$ if $A$ is less electrically conductive than $B$, so concrete $\to$ water. 
Robert Rosen introduced in the 90s a category of morphogenetic networks to study morphogenetic problems. Objects are elements and their different states, morphisms come from neighborhoods.

Example stolen from Spivak: Category theory can serve as a mathematical model for mathematical modeling. Our minds simultaneously keep several models of the world, often in conflict. The value of a model can therefore be measured by how well it fits with other models. What is true will be present across all models, so we should study the relationship between models.
Now for some examples of functors, (2) in my list of stages of the talk...

If A  be the set of amino acids and Str(A) the set of all strings formed from A. The process of translation  gives a functor turning a list of RNA triplets into a polypeptide. 
Quantum field theory was categorified by Atiyah in the late 1980s, with much success (at least in producing interesting mathematics). In this domain, an object is a reasonable space, called a manifold, and a morphism is a manifold connecting two manifolds, like a cylinder connects two circles. Such connecting manifolds are called cobordisms. Topological quantum field theory is the study of functors Cob $\to$ Vect that assign a vector space to each manifold and a linear transformation of vector spaces to each cobordism.
Suppose you are interested in different algorithms to buy a car. If you fix your preferences then this ordering makes them a category. Consider the price function that tells you the cost of a car, and lands in $\mathbb{R}_{>0}$. In order for this to be a functor it must respect the ordering: is it true that better cars cost more and worse cars cost less? In other words, does the model from category theory match reality? There seems to be debate about this among economists.
Suppose you're running an experiment and in all cases so far have observed 4 traits. You've created a mental model for what's going on, but then you observe several cases where only the first 3 traits are true. You shift to a new mental model and that process of shifting your point of view is a functor.
An experiment can be thought of as a functor from the category of pairs (Experimenter, Variables) to the category of measurements of the variables under observation. Viewing it this way makes it explicit that the experimenter can affect the outcome, something well-known in psychology and sociology.

Turning to (3), let's think about a natural human tendency: to break things that are hard to understand down into simple pieces, and then try to cobble those pieces together again to understand the original hard thing.

Chemistry breaks down to the study of atoms and the molecules they make up.
Physics breaks the world down even further, into strings (in the sense of string theory).
Molecular Biology studies the cell. Robert Rosen introduced a categorical presentation of (M,R)-systems, which model the activities of a cell. This is a category of automata (sequential machines). 
Geoscience breaks materials down into their simplest constituent pieces.
Neuroscience tries to understand mental processes via the simplest pieces: neurons.
Computer Science breaks computation down into 0s and 1s, at the end of the day.
Economics and game theory try to isolate a single cause and effect relationship by holding all other variables constant ("decision making on the margin")
Political science and the action of individuals.
Understanding how materials are built up of their constituent parts. For example, a tendon is made of collagen fibers. Each collagen fiber is made of collagen fibrils (what matters is how these simple pieces are reassembled). A collagen fibril is made up of tropocollagen collagen molecules, i.e. twisted strands of collagen molecules, and you can keep breaking things down this way.

A related example is spider silk, which Spivak has studied.
The process of putting those simple pieces back together again into an understanding of the original problem is an example of a colimit.

The current state of any evolutive system is a colimit of previous states. Here, "evolutive system" means a subcategory of time, i.e. for each time $t$ there is a category $K_t$ (the state of the system at time $t$), and for each period $[t,t']$ there is a functor $K_t \to K_{t'}$. So, an evolutive system is itself an example of a functor from time (viewed as a poset) to $Cat$.
Emergent Phenomena like the behavior of an ant colony, or of people starting to clap at the end of a performance, or of birds flocking - all examples of colimits.
modeling the biological tendency toward homeostasis is again a movement through time, of a collection of individuals following local rules, so it's a colimit.
Suppose you have different temperature reading devices measuring a terrain, perhaps with some overlapping areas. You can patch them together to get a maximally accurate reading by taking the colimit. This is simply a categorification of some kind of weighted averaging operation (weighted by knowledge of the devices).
Consider outer space. Different astronomers record observations using telescopes. We can patch together different observations of space as a colimit. Objects here using pixels and the set of wavelengths in the visible light spectrum (written in nanometers). 
The set of laws of the land; are there inconsistencies? Do they assemble properly? This is why we have lawyers.
The individuals making up society, and realizing society as the sum of its parts, i.e. at the object built up from all these individuals. When something happens and individuals are effected, the net effects on the colimit can be studied this way.

Turning to (4), localization...

Adding more isomorphisms to any of the examples above, e.g. in economics deciding which features of a snapshot in time matter and which don't, and saying two periods are "the same" if they are the same on those features.
viewing two different driving routes as the same if they take the same time.
viewing two assignments or exercises or exam problems as equivalent if they are the same difficulty and test the same concept.
viewing two products as the same if they cost the same and if I don't know/care about any differences in quality.
In linguistics, they study phonemes (and morphemes, graphemes, and lexemes, but I won't talk about those), which abstract the types of sounds we hear in speech. The point is to blur away details that cannot serve to differentiate meaning. This is an example of a localization.
I could go on, and probably did, but it's not written in my lecture notes from 2015.

Finally, we turn to (5), replacing an object by one which is easier to work with but has the same fundamental properties, and (6) is a special case of (5).

Information theory asks: what is the least amount of information required to describe something?
Macroeconomics tries to predict behavior at time $t$ based on behavior at time $t'$ just based on the macro environments at those times. It'd be great if you knew which indicators really mattered so you could make predictions like that.
Biological classification divides the set of organisms into distinct classes, called taxa. The result is a phylogenetic tree, a partial order on the set of taxa. This is reducing biological information to the information present in the phylogenic tree. Note that the ranking of taxa into kingdom, phylum, etc., can be understood as morphisms of orders. I think I learned this example from Baez.
Reducing the information of a human heart to an EKG read-out. 
In all the examples of categories, I can think of ways to discard extraneous information, giving plenty of examples of localization (4), replacement (5), and compression/invariants (6).<|endoftext|>
TITLE: Algebraic (semi-) Riemannian geometry ?
QUESTION [28 upvotes]: I hope these are not to vague questions for MO. 
Is there an analog of the concept of a Riemannian metric, in algebraic geometry? 
Of course, transporting things literally from the differential geometric context, we have to forget about the notion of positive definiteness, cause a bare field has no ordering. So perhaps we're looking to an algebro geometric analog of semi- Riemannian geometry.
Suppose to consider a pair $(X,g)$, where $X$ is a (perhaps smooth) variety and $g$ is a nondegenerate section  of the second symmetric power of the tangent bundle (or sheaf) of $X$.
What can be said about this structure? Can some results of DG be reproduced in this context? Is there a literature about this things?

REPLY [4 votes]: If a holomorphic Riemannian metric $g=g_{ij}(z) dz^i dz^j$ on a compact Kaehler manifold $X$ is everywhere nondegenerate, then the metric has a holomorphic Levi-Civita connection, so the Atiyah class of the tangent bundle of $X$ is zero. Therefore a finite etale cover of $X$ is a complex torus, and the metric pulls back to be translation invariant. Hence $X$ is a quotient of such a torus by a finite group acting as affine isometries without fixed points. On the other hand, if you allow degeneracies of the holomorphic Riemannian metric, I suppose anything could happen. If you allow $X$ to be a compact complex manifold, perhaps not Kaehler, then you might look at the papers of Sorin Dumitrescu where you find a low dimensional classification.<|endoftext|>
TITLE: Torsors in Algebraic Geometry?
QUESTION [29 upvotes]: I think I am confused about some terminology in algebraic geometry, specifically the meaning of the term "torsor". Suppose that I fix a scheme S. I want to work with torsors over S. Let $\mu$ be a sheaf of abelian groups over S. Then my understanding is that a $\mu$-torsor, what ever that is, should be classified by the cohomology gorup $H^1(X; \mu) \cong \check H^1(X; \mu)$. 
Now suppose that $\mu$ is representable in the category of schemes over S, i.e. there is a group object $$\mathbb{G} \to S$$
in the category of schemes over $S$, and maps (over S) to $\mathbb{G}$ is the same as $\mu$. Lots of interesting example arise this way. 
I also thought that in this case a torsor over S can be defined as a scheme $P \to S$ over S with an action of the group $\mathbb{G}$ such that locally in S it is trivial. I.e. there exists a cover $U \to S$ such that 
$$ P \times_S U \cong \mathbb{G} \times_S U $$
as spaces over S with a $\mathbb{G}$-action (or rather as spaces over U with a $\mathbb{G} \times_S U$-action). 
The part that confuses me is that these two notions don't seem to agree. Here is an example that I think shows the difference. Let $S= \mathbb{A}^1$ be the affine line (over a field k) and let $x_1$ and $x_2$ by two distinct points in $S$. Consider the subscheme $Y = x_1 \cup x_2$, and let $C_Y$ be the complement of Y in S. Let $A$ be your favorite finite abelian group which we consider as a constant sheaf over S. Then we have an exact sequence of sheaves over S,
$$0 \to A_{C_Y} \to A \to i_*A \to 0$$
Where $i_*A(U) = A(U \cap Y)$. I believe the first two are representable by schemes over S, namely $$C_Y \times A \cup S \times 0$$
and  $S \times A$, where we are viewing the finite set $A$ as a scheme over $k$ (and these products are scheme-theoretic products of schemes over $spec \; k$). 
In any event, the long exact sequence in sheaf cohomology shows that
$$H^1(S; A_{C_Y}) \cong \check H^1(S; A_{C_Y}) \cong A$$
and it is easy to build an explicit C$\check{\text{e}}$ch cocycle using the covering given by the two opens consisting of the subschemes $U_i = S \setminus x_i$, for $i = 1,2$. 
Now the problem comes when I try to glue these together to get a representable object over S, i.e. a torsor in the second sense. Then I am looking at the coequalizer of
$$C_Y  \times A \rightrightarrows (C_Y\cup C_Y) \times A$$
where the first map is the inclusion and the second is the usual inclusion together with addition by a given fixed element $a \in A$. This seems to just gives back the trivial "torsor" $C_Y \times A$. 
Am I doing this calculation wrong, or is there really a difference between these two notions of torsor?

REPLY [19 votes]: As remarked by Brian Conrad above, there is an excellent explanation of all this in Milne's book Étale cohomology, Section III.4.  There wouldn't be much point in reproducing the details here, but the main issues are:

You need to decide whether a torsor is going to be a scheme over S which locally looks like a trivial torsor, or merely a sheaf of sets over S which locally looks like a trivial torsor.  What people mean by "torsor" can be either of these things.  As Milne says, "The question of which sheaf torsors arise from schemes is, in general, quite delicate".  If you go for the sheaf definition, then isomorphism classes of torsors are indeed classified by $\check H^1(S,G)$.  Beware that if G is not commutative then you need to define $\check H^1(S,G)$ appropriately as a pointed set.
You need to decide which topology all this is happening in; the usual definition of torsor uses the flat topology, though if G is smooth over S then you can use the étale topology instead.
Depending on what topology you're using, and what S and G look like, there may be issues about whether $\check H^1(S,G)$ and $H^1(S,G)$ are isomorphic.<|endoftext|>
TITLE: Fundamental group of a compact space form.
QUESTION [15 upvotes]: Space forms are complete (connected) Riemannian manifolds of constant sectional curvature. 
These fall into three classes: Euclidean, with universal covering isometric to $\mathbb{R}^n$, 
spherical, with universal covering isometric to $S^n$, and hyperbolic, with universal covering isometric to $\mathbb{H}^n$.
Does there exist compact space forms of the same dimension from two different classes having isomorphic fundamental groups?
This cannot happen for $n = 2$, since the Gauss-Bonnet theorem 
$$
\int_M K{ \ }\mathrm{vol}_M = 2{\pi}\chi(M)
$$
shows that the Euler characteristic $\chi(M)$ is positive, zero, or negative when the space form $M$ is Euclidean, spherical or hyperbolic respectively. But for (closed) surfaces the fundamental group determines the Euler characteristic.
It is essential for the question to require compactness, otherwise there are trivial examples. Dividing out $\mathbb{R}^2$ by the group of isometries generated by 
$(x,y) \mapsto (x+1,y)$ yields a Euclidean space form with fundamental group isomorphic to $\mathbb{Z}$, while dividing out $\mathbb{H}^2$ by the group of isometries generated by 
the hyperbolic isometry $(x,y) \mapsto (x+1,y)$ yields a hyperbolic space form also with fundamental group isomorphic to $\mathbb{Z}$.
The standard reference on space forms is Spaces of Constant Curvature by Joseph A. Wolf. 
The classification of Euclidean space forms is given in Chapter 3, and of spherical ones in Chapter 7. Wolf does not treat hyperbolic space forms, possibly because not much was known about them in 1967. Unfortunately, the fundamental groups are infinite for the compact Euclidean space forms, and finite for the spherical space forms (which are necessarily compact, being quotients of $S^n$). So a hypothetical example has to involve a hyperbolic space form.
An example might drop out of the theory of three-manifolds. In dimension three the space forms belong to three of the eight Thurston model geometries. A pair yielding an example would have to be one Euclidean and one hyperbolic, since it follows from Perelman's geometrization theorem that the spherical ones are precisely those with finite fundamental group.

REPLY [5 votes]: For the record, no compact 3-manifold can ever admit two different geometries; this isn't just for space forms. You should be able to show this using quasi-isometric techniques, as in Sergei Ivanov's answer above (although I haven't thought about this in a while). One nice corollary is that you can determine the geometry just from the fundamental group without a lot of technical conditions (this is copied from Wikipedia with minimal editing, since it is laid out so nicely there).
Assume $M$ is a compact 3-manifold which admits one of the 8 geometries. Then:

If $\pi_1(M)$ is finite then the geometry on $M$ is spherical.
If $\pi_1(M)$ is virtually cyclic but not finite then the geometry on $M$ is $S^2×\mathbb{R}$.
If $\pi_1(M)$ is virtually abelian but not virtually cyclic then the geometry on $M$ is Euclidean.
If $\pi_1(M)$ is virtually nilpotent but not virtually abelian then the geometry on $M$ is Nil geometry.
If $\pi_1(M)$ is virtually solvable but not virtually nilpotent then the geometry on $M$ is Sol geometry.
If $\pi_1(M)$ virtually splits as a semidirect product with $\mathbb{Z}$ but is not virtually solvable then the geometry on $M$ is the universal cover of $SL_2(\mathbb{R})$.
If $\pi_1(M)$ has an infinite normal cyclic subgroup but not of the above form and is not virtually solvable then the geometry on $M$ is $\mathbb{H}^2\times\mathbb{R}$.
If $\pi_1(M)$ has no infinite normal cyclic subgroup and is not virtually solvable then the geometry on $M$ is hyperbolic.

This is false for finite-volume non-compact manifolds, for example the complement of a trefoil knot, which admits both $\widetilde{SL_2(\mathbb{R})}$ and $\mathbb{H}^2\times\mathbb{R}$ geometries.<|endoftext|>
TITLE: How has "what every mathematician should know" changed?
QUESTION [91 upvotes]: So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school.
To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different.
My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students.
Some other guesses:

Today, people are expected to have a lot more of a quick idea of a larger number of subjects, and less of an in-depth understanding of "Big Proofs" in areas outside their subdomain of expertise. Basically, the Great Books or Great Proofs approach to learning may be declining. The rapid increase in availability of books, journals, and information via the Internet (along with the existence of tools such as Math Overflow) may be making it more profitable to know a bit of everything rather than master big theorems outside one's area of specialization.
Also, probably a thorough grasp of multiple languages may be becoming less necessary, particularly for people who are using English as their primary research language. Two reasons: first, a lot of materials earlier available only in non-English languages are now available as English translations, and second, translation tools are much more widely available and easy-to-use, reducing the gains from mastery of multiple languages.

These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated.
NOTE: This might be too soft for Math Overflow! Moderators, please feel free to close it if so.

REPLY [9 votes]: In earlier times, it seems that great emphasis was put on technical calculation, such as checking for convergence, handling logarithms, and so on (this has been pointed out in an earlier answer).
As for today, the only thing I am convinced every mathematician should know is to formulate correct proofs and be able to come up without hesitation with correct and readable proofs for simple statements such as


If $G,H$ are groups, and $f:G\to H$ is a group homomorphism, then $\text{ker}(f) =\{g\in G:f(g) = 1_H\}$ is a subgroup of $G$.


Mathematicians should all be able to handle quantifiers with care -- for instance see the difference between continuity uniform continuity.<|endoftext|>
TITLE: The Join of Simplicial Sets ~Finale~
QUESTION [8 upvotes]: Background

Let $X$ and $S$ be simplicial sets, i.e. presheaves on $\Delta$, the so-called topologist's simplex category, which is the category of finite nonempty ordinals with morphisms given by order preserving maps.
How can we derive the structure of the face and degeneracy maps of the join from either of the two equivalent formulas for it below:
The Day Convolution, which extends the monoidal product to the presheaf category:
$$(X\star S)_{n}:=\int^{[c],[c^\prime] \in \Delta_a}X_{c}\times S_{c^\prime}\times Hom_{\Delta_a}([n],[c]\boxplus[c^\prime])$$
Where $\Delta_a$ is the augmented simplex category, and $\boxplus$ denotes the ordinal sum.  The augmented simplex category is the category of all finite ordinals (note that this includes the empty ordinal, written $[-1]:=\emptyset$.
The join formula (for $J$ a finite nonempty linearly ordered set):
$$(X\star S)(J)=\coprod_{I\cup I=J}X(I) \times S(I')$$
Where $\forall (i \in I \text{ and } i' \in I'),$ $i < i'$, which implies that $I$ and $I'$ are disjoint.
Then we would like to derive the following formulas for the face maps (and implicitly the degeneracy maps):

The $i$-th face map $d_i : (S\star T)_n \to (S\star T)_{n-1}$ is defined on $S_n$ and $T_n$ using the $i$-th face map on $S$ and $T$. Given $\sigma \in S_j\text{ and }\tau\in T_k$ , we have:
$$d_i (\sigma, \tau) = (d_i \sigma,\tau)\text{ if } i \leq j, j ≠ 0.$$
$$d_i (\sigma, \tau) = (\sigma,d_{i-j-1} \tau) \text{ if } i > j, k ≠ 0.$$
$$d_0(\sigma, \tau) = \tau \in T_{n-1} \subseteq (S\star T)_{n-1} \text{ if } j = 0$$
$$d_n(\sigma, \tau) = \sigma \in S_{n-1} \subset (S\star T)_{n-1}\text{ if } k = 0$$

We note that the special cases at the bottom come directly from the inclusion of augmentation in the formula for the join.
Edit: Another note here: I got these formulas from a different source, so the indexing may be off by a factor of -1.
Question

How can we derive the concrete formulas for the face and degeneracy maps from the definition of the join (I don't want a geometric explanation.  There should be a precise algebraic or combinatorial reason why this is the case.)?
Less importantly, how can we show that the two definitions of the join are in fact equivalent?
Edit:
Ideally, an answer would show how to induce one of the maps by a universal property.
Note also that in the second formula, we allow $I$ or $I'$ to be empty, and we extend the definition of a simplicial set to an augmented simplicial set  such that $X([-1])=*$, i.e. the set with one element.
A further note about the first formula for the join: $\boxplus$ denotes the ordinal sum. That is, $[n]\boxplus [m]\cong [n+m+1]$. However, it is important to notice that there is no natural isomorphism $[n]\boxplus [m]\to [m]\boxplus [n]$. That is, there is no way to construct this morphism in a way that is natural in both coordinates of the bifunctor. This is important to note, because without it, it's not clear that the ordinal sum is asymmetrical.

REPLY [12 votes]: (A note: I am going to regard simplicial sets as also defined on the empty ordinal as well, with $X(\emptyset) = *$, which is required for the join formula.  This is implicit in your first definition and will remove the need for two extra cases for $d_i$ at the end.)
Regarding the "minor" question.  The short explanation is that this follows by decomposing the Hom-set according to the preimage of $c$ and $c'$ in $n$, and observing that each decomposition of $n$ provides an initial choice.
In more category-theoretic language, one way to rewrite the convolution is using the "over" category:
$$
(X \star S)_n = \int^{[n] \to [c] \boxplus [c']} X_c \times S_{c'}
$$
where now the coend is taken over the comma category $n \downarrow \boxplus$ whose objects are triples $([c],[c'],f)$ of a pair of objects of $\Delta$ and a morphism from $[n]$ to their ordinal sum.  We note that this comma category decomposes as a disjoint union of categories: each $([c],[c'],f)$ determines a decomposition $[n] = f^{-1} [c] \cup f^{-1} [c']$ into a disjoint union, and morphisms preserve such a decomposition.  Therefore,
$$
[n] \downarrow \boxplus \simeq \coprod_{[n] = I \cup I'} (I \downarrow \Delta) \times (I' \downarrow \Delta)
$$
This makes the coend decompose:
$$
(X \star S)_n = \coprod_{[n] = I \cup I'} \int^{I \to [c], I' \to [c']} X_c \times S_{c'}
$$
However, the comma category $(I \downarrow \Delta)\times(I' \downarrow \Delta)$ has an initial object: $I \times I'$ itself.  Thus, the coend degenerates down to simply being the value:
$$
(X \star S)_n = \coprod_{[n] = I \cup I'} X_{|I|} \times S_{|I'|}
$$
This is slightly different notation for the second definition of the join that you gave.
Now, as for the boundary formulas.
The definition of $d_i$ is as follows.  For each $0 \leq i \leq n$, there is a unique map $d^i:[n-1] \to [n]$ in $\Delta$ whose image is $[n] \setminus \{i\}$: $d^i(x) = x$ for $x < i$, and $d^i(x) = x+1$ for $x \geq i$.  The induced map $(X \star S)_n \to (X \star S)_{n-1}$ is the map induced by applying the contravariant functor to $d^i$.
Since $(X \star S)_n$ is a disjoint union of sets, it suffices to show that the formula is correct on $X(I) \times S(I')$ for all decompositions of $[n]$ into $I \cup I'$, where $|I| = j+1$ and $|I'| = k+1$.  There are two possibilities: either $i \in I$ when $0 \leq i \leq j$, or $i \in I'$ when $j < i \leq n$.
In either case, the map $[n-1] \to [n] = I \cup I'$ induces, by taking preimages, a unique ordered decomposition $[n-1] = J \cup J'$ of $[n-1]$.  If $i \in I$, then $J$ has size $|I| - 1$ and $J'$ is mapped isomorphically to $I'$ by $d^i$.  In this case, the map $d^i$ is isomorphic to the map $d^i \boxplus id$ on $[j-1] \boxplus [k] \to [j] \boxplus [k]$.  If $i \in I'$, we have the reverse possibility, with $d^i$ isomorphic to $id \boxplus d^{i-j-1}$ (the upper index necessary because inserting the identity at the beginning adds $j+1$ elements to the ordered set at the beginning).
In the case $0 \leq i \leq j$, the induced map
$$
d_i: X(I) \times S(I') \to \coprod_{[n-1] = K \cup K'} X(K) \times S(K')
$$
is therefore the map $X(d^i) \times id: X(I) \times S(I') \to X(J) \times S(J')$, followed by the inclusion into the coproduct.  In the case $j < i \leq n$, the map is $id \times S(d^{i-n-1})$ followed by inclusion.
This recovers the formula for $d_i$ that you have written down, up to inserting copies of a point $*$ as in the remark at the beginning.<|endoftext|>
TITLE: Prime numbers that lead to relatively prime
QUESTION [5 upvotes]: It might be well-known (and sorry if it is), but a quick search did not return the answer.
Consider prime numbers $p \neq q$.
Are $\displaystyle \frac{p^q-1}{p-1}$ and $\displaystyle  \frac{q^p-1}{q-1}$ relatively prime?

REPLY [18 votes]: The answer is no.  As the Wikipedia article in my comment states, the counterexample $p = 17, q = 3313$ was found by Stephens in 1971, but the stronger question of whether one can ever divide the other is a famous open problem because its solution would greatly simplify a step in the proof of the Feit-Thompson theorem.<|endoftext|>
TITLE: Intuition behind the Eichler-Shimura relation?
QUESTION [40 upvotes]: The modular curve $X_0(N)$ has good reduction at all primes $p$ not dividing $N$.  At such a prime, the Eichler-Shimura relation expresses the Hecke operator $T_p$ (as an element of the ring of correspondences on the points of $X_0(N)$ in $\overline{ \mathbb{F}_p }$) in terms of the geometric Frobenius map.  This is already weird enough; the definitions of the Hecke operators with which I'm familiar give no indication that such a relation should be true, and the sources I've read so far give no intuition as to why such a relation should be true.  (In fact, I still don't feel very comfortable with the Hecke operators themselves; I like the definition given in Milne the best so far, but any enlightening alternate definitions are welcome if they shed light on the question.)
What's much weirder, to me anyway, is an important corollary of the Eichler-Shimura relation, which says that given a cusp eigenform $f$ of weight $2$ with respect to $\Gamma_0(N)$ it is possible to construct an elliptic curve $E_f$ whose L-function is (more or less) the Mellin transform of $f$.  
There are several reasons this is weird to me, but here's a specific one.  The modular form $f$ satisfies a functional equation more or less by definition.  Its Mellin transform therefore satisfies a corresponding functional equation (part of which has been absorbed into the fact that $f$ is written in terms of its Fourier coefficients), again more or less by definition.  The zeta function of an elliptic curve, however, satisfies a functional equation because (or so I'm told) of Poincaré duality in the étale cohomology.  So:
What on Earth does Poincaré duality have to do with modular symmetry?
(Ignore the above; I seem to have mixed up the local and global functional equations again.)  
One of the many things that's weird about the above is that the L-function of an elliptic curve naturally has an Euler product, but for modular forms the Euler product for the Mellin transform comes about because of certain properties of the Hecke operators (which, again, I don't really understand conceptually).  What do these properties have to do with multiplying local zeta functions together?  

I guess I should also clarify what I mean by "intuition":
For the first part of the question, if something in the definition of the Hecke operators suggests that they should be related to the Frobenius map if certain natural things were true, and the proof of Eichler-Shimura (which I haven't really looked at yet...) consists of verifying these natural things are true, that would be great intuition.  I would appreciate an answer telling me whether or not this was the case in terms of "first principles."
For the second part of the question, intuition might more naturally come from a more sophisticated perspective.  Here I would appreciate an answer about the "big picture" instead, giving some vague high-level sense of how this all fits into more general things people know about automorphic forms and so forth.

REPLY [27 votes]: Let me highlight some issues that Emerton doesn't:
1) you seem to hint that you don't know that modular forms can be viewed as a product of a bunch of local terms. So there is an adelic story, where GL_2(R)/GL_2(Z) gets replaced by GL_2(adeles)/GL_2(Q), and in that story Hecke operators are entirely local objects, each with its own local Euler factor on an eigenform etc etc. Maybe if you knew this story some stuff would be clearer---e.g. a modular form really does have a "local factor at p" for p a prime, but it's an infinite-dimensional representation of GL_2(Q_p)!
2) the "big picture" story of Eichler-Shimura, as I'm sure I've typed into this forum before somewhere, is that Langlands, a long time ago, made conjectures about how the cohomology of a very general class of Shimura varieties can be completely explained using automorphic forms, but the conjectures in full generality are very difficult to explain and have many subtleties (coming from endoscopy, non-compactness at infinity, multiplicity issues and so on). For modular curves the conjectures boil down to the statement that, vaguely speaking, the Tate module of the Jacobian of a compact modular curve X_0(N) should break up into 2-dimensional pieces each explained by an eigenform of weight 2 and level N. But there is a lot of stuff secretly built in there: you're using X_0(N) instead of Y_0(N), you're assuming "multiplicity 1" holds for cusp forms on GL_2, which is a theorem of Jacquet and Langlands, and so on. Once you take all this on board, the precise relation between the pieces and the modular forms is given to you by Eichler-Shimura. If you look at it in this way you can start to guess what e.g. the H^3 of the Siegel modular variety parametrising princ polarized abelian surfaces might look like, but your guess might be wrong, because now endoscopy and failure of multiplicity 1 and issues involving compactification really start rearing their ugly heads in a less trivial way. Somehow there are hundreds of pages of Corvallis devoted to this sort of thing, and one thing Clozel, Harris, Taylor and their collaborators have done in the last few years is to completely sort these issues out in the case of unitary groups when the Shimura variety is compact. Langlands' conjectures are stronger than, but imply, a generalised Eichler-Shimura correspondence in this setting, and probably Matt's approach involving considering correspondences in char p will also give some insights into what the precise statement should be. But my feeling is that it is genuinely hard to explain the big picture without a lot of work the moment one leaves GL_2. If you want to read about another case other than classical modular forms, perhaps the next easiest case is Hilbert modular forms. Good luck!
3) A nice place to read about Eichler-Shimura for GL_2/Q is the appendix by Conrad to "lectures on Serre's Conjectures" by Ribet-Stein.<|endoftext|>
TITLE: Transformation formulae for classical theta functions
QUESTION [9 upvotes]: I am looking for a reference for the transformation formulae
for the classical theta-functions
$$\theta_4(\tau)=\sum_{n=-\infty}^\infty (-1)^n q^{n^2}$$
and
$$\theta_2(\tau)=\sum_{n=-\infty}^\infty q^{(2n+1)^2/4}$$
under the congruence group $\Gamma_0(4)$.
Here $\tau$ lies in the upper-half plane and $q^x$ denotes
$\exp(2\pi i x\tau)$. More precisely I want the exact automorphy
factors for each $A\in\Gamma_0(4)$ (some eighth root of
unity times $\sqrt{c\tau+d}$). I know these can easily
be deduced from those for the basic theta-function
$$\theta_3(\tau)=\sum_{n=-\infty}^\infty q^{n^2}$$
for which a nice reference for the automorphy factors is Koblitz's Introduction
to Elliptic Curves and Modular Forms. However

a citation would be useful to me,

I want to check my calculation and

a reference may give the formulae in a more convenient form than I have.


Thanks in advance.
EDIT I have now found a convenient reference: Rademacher's
Topics in Analytic Number Theory.
FURTHER EDIT Rademacher actually gives full transformation formula
for the two-variable classical Jacobi theta functions under arbitrary
matrices in $\mathrm{SL}_2(\mathbb{Z})$. From these we can deduce
for $A\in\Gamma_1(4)$ that
$$\frac{\theta_2(A\tau)}{\theta_3(A\tau)}
=i^b\frac{\theta_2(\tau)}{\theta_3(\tau)}$$
and
$$\frac{\theta_4(A\tau)}{\theta_3(A\tau)}
=i^{-c/4}\frac{\theta_4(\tau)}{\theta_3(\tau)}$$
in the usual notation. Once noticed, these relations are easy to prove
from scratch.
Thanks to all who replied to this question.

REPLY [2 votes]: K. Chandrasekharan "elliptic functions" chapter 5  discuss also 2 variales transformation but theta-{2,4} becomes {1,2} in his notation<|endoftext|>
TITLE: How to show that x-y is Lebesgue-Lebesgue measurable
QUESTION [14 upvotes]: Which is the cleanest way to show that the difference, $d:R^n\times R^n\rightarrow R^n$, $d(x,y)= x-y$, is Lebesgue-Lebesgue measurable? (i.e. foreach A lebesgue measurable in $R^n$, $d^{-1}(A)$ is Lebesgue measurable in $R^n\times R^n$).
Thanks in advance.

REPLY [19 votes]: Nicolo is asking about functions where the inverse image
of a Lebesgue measurable set is Lebesgue measurable. This
is stronger than the usual definition of measurability
where it is required only the inverse image of each Borel
set must be Lebesgue measurable. Continuous functions need not
be measurable by this stronger criterion. If $B$ has zero
Lebesgue measure and $A=f^{-1}(B)$ has nonzero measure then each
subset of $B$ is Lebesgue measurable but its inverse image may
be non-measurable. A simple example is given by $f:x\mapsto (x,0)$
from $\mathbb{R}$ to $\mathbb{R}^2$. Taking $A$ to be a
non-measurable subset of $\mathbb{R}$ and $B=f(A)$ we see this
$f$ is not Lebesgue-Lebesgue measurable. More interesting examples
occur on the real line when there are continuous homeomorphisms
from $\mathbb{R}$ to itself taking Cantor sets of positive measure
to Cantor sets of zero measure.
To return to Nicolo's example. Each surjective linear map
from $\mathbb{R}^m\to\mathbb{R}^n$ is Lebesgue-Lebesgue measurable
as it can be decomposed as a composition of linear bijections
and the projection map $\mathbb{R}^m\to\mathbb{R}^n$ mapping onto
the first $n$ coordinates (both these types of maps can be seen to
be Lebesgue-Lebesgue measurable). By definition, the class
Lebesgue-Lebesgue measurable maps is closed under composition
(unlike the class of Lebesgue-measurable maps!).<|endoftext|>
TITLE: Approximation of a normal distribution function
QUESTION [8 upvotes]: I am reviewing and documenting a software application (part of a supply chain system) which implements an approximation of a normal distribution function; the original documentation mentions the same/similar formula quoted here
$$\phi(x) = {1\over \sqrt{2\pi}}\int_{-\infty}^x e^{-{1\over 2} x^2} \ dx$$
This is approximated with what looks like an asymptotic series like here
however the expression is slightly different:
$$\phi(x) \simeq {1\over x\sqrt{2\pi}} \Bigl( 1 - {1\over x^2} + {3\over x^4} - {15\over x^6} + {105 \over x^8}\Bigr)$$
The original document quotes a "Cambridge statistical tables" book which I don't have; also, the software uses a precalculated table for values of $x$ between $[-8.2, 8.2]$ and uses the approximation function for values of $x$ outside that interval.
I need to find some reference that explains why this particular formula was chosen, and how accurate the approximation really is.
(The documentation claims that an error less then the $7$th decimal is "not bad")
Also, the constant $1/\sqrt{2\pi}$ does it have a name? All I was able to find was that "this expression ensures that the total area under the curve $\Phi(x)$ is equal to one"
Does it matter if the constant has only $10$ decimals when the double type in Java (IEEE 754) has a precision of approximately $16$ decimals?

REPLY [4 votes]: For the normal CDF:
$$
\Phi(x) \approx \frac{1}{1+e^{-1.65451*x}}
$$
worked very well for me.<|endoftext|>
TITLE: Definition of forgetful functor
QUESTION [9 upvotes]: I was wondering if it is possible to make a formal definition of what it means for a functor to be forgetful. That is, using only the terminology of categories. I have seen so many examples of forgetful functors, but they have always been specific examples, and the fact that they were forgetful was "clear simply from what we untuitively mean by forgetful.
So, is there a formal definition?

REPLY [13 votes]: Here is a proof that every functor is "forgetful." Let $F: \mathcal{C} \to \mathcal{D}$ be a functor. Then $\mathcal{C}$ is equivalent (in fact, isomorphic) to the category of pairs $(x, y) \in \mathcal{C} \times \mathcal{D}$ such that $F(x) = y$, where morphisms are pairs $(f, F(f)): (x, y) \to (x', y')$. Under this equivalence, $F$ is the functor to $\mathcal{D}$ that "forgets $\mathcal{C}$." This seems like a silly construction, but it is simply the restriction of the projection functor $\mathcal{C} \times \mathcal{D} \to \mathcal{D}$ to the graph of $F$, and if we believe that the former functor is forgetful, then surely its restriction to a subcategory should be forgetful.
The adjective "forgetful" is only meaningful when it is placed into the context of some specified extra structure (or the various generalizations of "structure" suggested by Baez-Dolan-Bartels) that one is supposed to be forgetting. For instance, if the structure one cares about is "algebras over some arbitrary monad," then Beck's monadicity theorem gives necessary and sufficient conditions for a functor to be "forgetful" in this sense. Without specifying such a context, "forgetful" is meaningless.<|endoftext|>
TITLE: Do Abel summation and zeta summation always coincide?
QUESTION [25 upvotes]: This is a more focused version of Summation methods for divergent series.

Let $a_n$ be a sequence of real
  numbers such that $\lim_{x \to 1^{-}}
> \sum a_n x^n$ and $\lim_{s \to 0^{+}}
> \sum a_n n^{-s}$ both exist. (In
  particular, we assume that both of the
  sums in question converge in the
  appropriate region.) Need the limits
  be equal?


I've thought about this before, and am doing so again thanks to the linked question. Here are some things I've tried:
It is true if $a_n$ is periodic with average $0$.
If it is true for $a_n$, then it is true for the sequence $b_{kn+r}=a_n$, $b_m=0$ if $m \not \equiv r \mod k$.
It is true for $a_n=1$ if $n$ is an even square, $-1$ if $n$ is an odd square and $0$ otherwise. I tried to prove in general that, if it is true for $a_n$ then it is true for $b_{n^2}=a_n$, $b_m=0$ for $m$ not a square, but couldn't.
It appears to be true for $a_n = (-1)^n \log n$, although I didn't check all the details. 
I do not know any explicit sequence $a_n$ which obeys the hypotheses of the question and is not $(C, \alpha)$-summable for some $\alpha$. So it is possible that this is really a theorem about higher Cesaro summability. But I suspect such sequences do exist.
A natural generalization is: Let $a_n$, $\lambda_n$ and $\mu_n$ be three sequences of real numbers, with $\lambda_n$ and $\mu_n$ approaching $\infty$. If
$\lim_{s \to 0^{+}} \sum a_n e^{-\lambda_n s}$ and $\lim_{s \to 0^{+}} \sum a_n e^{-\mu_n s}$ both exist, need they be equal?
As far as I can tell, Wiener's generalized Tauberian theorem does not apply.

REPLY [8 votes]: G. H. Hardy, DIVERGENT SERIES, page 76...
"It is easy to give examples of series summable $(A,\log n)$ but not summable $(A)$:
we shall see, for example, that $\sum n^{-1-ci}$, where $c>0$, is such a series."
Before that, on page 73, we see if a series is both summable $(A,\log n)$ and summable $(A)$, then the two sums are equal.<|endoftext|>
TITLE: Existence of probability measure defined on all subsets
QUESTION [15 upvotes]: Let S be an uncountable set. Does there exist a probability measure which is defined on all subsets of S, with P({x}) = 0 for any element x of S ?
If I remove the condition P({x}) = 0, then I can trivially get a measure defined on all subsets as follows:
Fix some a in S. For any subset U of S, define P(U ) = 1 if a is in U and 0 otherwise.
But what happens if I am not allowed to put nonzero probability on individual points ?

REPLY [9 votes]: Joel's answer is the correct one, but in some cases one only needs a finitely additive probability measure rather than a countably additive one, and in this case one can use an non-principal ultrafilter to create such a measure, which would give every set in the ultrafilter a measure of 1 and all the other sets a measure of zero.  Indeed, one important way to think about ultrafilters is as a {0,1}-valued finitely additive probability measure on a set.<|endoftext|>
TITLE: what mistakes did the Italian algebraic geometers actually make?
QUESTION [129 upvotes]: It's "well-known" that the 19th century Italian school of algebraic geometry made great progress but also started to flounder due to lack of rigour, possibly in part due to the fact that foundations (comm alg etc) were only just being laid, and possibly (as far as I know) due to the fact that in the 19th century not everyone had come round to the axiomatic way of doing things (perhaps in those days one could use geometric plausibility arguments and they would not be shouted down as non-rigorous and hence invalid? I have no real idea about how maths was done then).
But someone asked me for an explicit example of a false result "proved" by this school, and I was at a loss. Can anyone point me to an explicit example? Preferably a published paper that contained arguments which were at the time at least partially accepted by the community as being OK but in fact have holes in? Actually, to be honest I'd probably prefer some sort of English historical summary of such things, but I do have access to (living and rigorous) Italian algebraic geometers if necessary ;-)
EDIT: A few people have posted solutions which hang upon the Italian-ness or otherwise of the person making the mathematical mistake. It was not my intention to bring the Italian-ness or otherwise of mathematicians into the question! Let me clarify the underlying issue: a friend of mine, interested in logic, asked me about (a) Grothendieck's point of view of set theory and (b) a precise way that one could formulate the statement that he "made algebraic geometry rigorous". My question stemmed from a desire to answer his.

REPLY [30 votes]: A beautiful survey article on the Italian school, with a discussion of several errors of all kinds by Severi, can be found in 

The Legacy of Niels Henrik Abel (Oslo 2002), Springer-Verlag 2004:
Brigaglia, Ciliberto, Pedrini,  The Italian school of algebraic geometry and Abel's  legacy, 295--347<|endoftext|>
TITLE: Values of cusp forms at q = 1 ?
QUESTION [5 upvotes]: Take a cusp form $f$ and let $f(q) = q + a_2q^2 + q_3q^3 + \ldots$" denote its $q$-expansion (assume that the $a_k$ are integers, and that $f$ comes from an elliptic curve $E$). Of course the series $f(1) = 1 + a_2 + a_3 + \ldots$ diverges, but I wonder whether there is any work on evaluating $f(1)$ via some regularization method (I do not even know whether  $\lim_{q \to 1-0} f(q)$ exists or not). I am kind of hoping that $f(1)$ is connected with the order of the Tate-Shafarevich group of the associated elliptic curve $E$ and perhaps the order of its torsion group (the actual expression would have to be invariant under isogenies). 
Does this ring a bell with anyone?

REPLY [2 votes]: I'll give a slightly uncertain answer, based somewhat on my recollection of conversations with Zagier a month ago about similar questions.
If we were to imitate Euler, we might consider $f(1)$ as
$$f(1) = \sum_{n \geq 1} a_n = \sum_{n \geq 1} a_n n^{-0} = L(f,0).$$
So the analytic continuation of the L-function suggests that $f(1)$ should be identified with the value of the L-function at zero.  By the functional equation, this relates to the L-function at the right edge of the critical strip.
So, for a cusp form of weight two, arising from an elliptic curve $E$ over $Q$, the value $L(f,0)$ is related to $L(E,2)$.  An interpretation of this L-value, conjectured by Zagier, was proven by Goncharov and Levin, in "Zagier's conjecture on $L(E,2)$", Invent. Math. 132 (1998).
As for the analytic question, you are considering the "value" of a cusp form $f$ on the real axis, which bounds the upper half-plane.  Almost by definition, there is a Sato hyperfunction $f_{bdr}$ on the real axis, which describes this boundary behavior of the holomorphic function $f$ on the upper half-plane.  I am not sure if the following is published, but I have the impression that there might be a preprint now or soon which proves the following result:
At every (positive? I don't recall) rational number $q$, the hyperfunction $f_{bdr}$ is $C^\infty$ at $q$.  Its value at $1$ is $L(f,0)$ as described above.
I think that saying "a hyperfunction is $C^\infty$ at $q$" means that the hyperfunction can be expressed as the distributional derivative of a continuous function -- $f = g^{(k)}$ for some $k \geq 0$ -- and $g$ happens to be $C^\infty$ at $q$.  But I'm not much of an analyst.
I think that the value $f(1)$ also exists as $\lim_{z \rightarrow 1} f(z)$ limit, if $z$ approaches $1$ via a geodesic in the upper half-plane.
I don't think you'll see Sha or the torsion directly, as these appear at the central value $L(f,1)$.  On the other hand, I do think you'll find $L(f,-n)$ for all $n \geq 0$ (or equivalently, $L(f,2+n)$ ), by looking at the derivatives $f^{(n)}(1)$ of the boundary hyperfunction of $f$ at $1$.<|endoftext|>
TITLE: Galois group of a product of irreducible polynomials
QUESTION [6 upvotes]: Hello
Suppose given a polynomial $P=Q_1\cdots Q_k$ of degree $n$, where each $Q_i$ is irreducible. Suppose also that I know the Galois group $G_i$ (over the rationals) of each irreducible factor $Q_i$.
Is there an easy correlation between the Galois group of $P$, and the $G_i$?

REPLY [18 votes]: The Galois group of $P$ will be a subdirect product of
the $G_i$, that is a subgroup of $G_1\times\cdots\times G_k$
projecting surjectively onto each of the $G_i$.

REPLY [17 votes]: If you mean: does knowing the Gi tell you the Galois group of P, then no.
Examples:
$P = (X^2+1)(X^2-2)$ has Galois group $C_2 \times C_2$, and both factors have Galois group $C_2$; this works because the splitting fields of the two factors intersect only in $\mathbb{Q}$.
But $P = (X^2 + X + 1)(X^2+3)$ has Galois group $C_2$, although both factors again have Galois group $C_2$.  Here both factors, though they're coprime, define the same extension $\mathbb{Q}(\sqrt{-3})$.
I've just seen Robin's answer, so to relate to that: in the first example, the Galois group of P is the whole of $G_1 \times G_2$.  In the second example, it is the diagonal subgroup of $G_1 \times G_2$, which is smaller although still projects surjectively onto each factor.<|endoftext|>
TITLE: Limit law for the number of local maxima in a square lattice of IID random variables
QUESTION [6 upvotes]: For $i, j \in \{ 1, \ldots, n \}$, let $X_{i,j}$ be a real-valued random variable uniformly distributed on the interval $[0,1]$.  The $X_{i,j}$ are independent.
Let $A_{i,j}$ be the indicator random variable of the event that $X_{i,j}$ is a local maximum, i. e. it is the largest of the five random variables $X_{i,j}, X_{i,j+1}, X_{i,j-1}, X_{i-1,j}, X_{i+1, j}$.   For the sake of not having to think about boundary conditions, interpret all coordinates modulo $n$.
Then it's clear that $E A_{i,j} = 1/5$, by symmetry. It's also not hard to see that:

$E (A_{i,j} A_{i,j+1}) = 0$, since we can't have local maxima both at $(i,j)$ and at $(i,j+1)$.  (The case where $X_{i,j} = X_{i,j+1}$ can be ignored since it occurs with probability zero.
I believe $E (A_{i,j} A_{i,j+2}) = 2/45$ and $E(A_{i,j} A_{i+1,j+1}) = 1/20$.  (In any case, these are clearly constants, and their exact values don't matter.)
$E(A_{i,j} A_{k,l}) = 1/25$ for all choices of $i, j, k, l$ other than the ones I already listed and those that clearly are related to them by symmetry.  That is, $A_{i,j}$ and $A_{k,l}$ are independent unless the rook-neighborhoods of $(i,j)$ and $(k,l)$ overlap.

From this we can compute the mean and variance of $M_n = \sum_{i=1}^n \sum_{j=1}^n A_{i,j}$, the total number of maxima.  Of course $EM = n^2/5$.  The variance is a bit harder to compute and I haven't actually written out the computation, but it ought to be asymptotic to $c^2 n^2$ for some positive $c$.  (It is the sum of $\Theta(n^2)$ covariances each of order 1.)
Let $\tilde{M}_n = (M_n-n^2/5)/(cn)$.  Then $\tilde{M}_n$ has mean $0$ and variance $1$.  Is it true that the sequence $\{ \tilde{M_n} \}_{n=1}^\infty$ converges in distribution to the standard normal?  Intuitively this seems like it should be true -- we're adding up a bunch of small, almost-independent contributions.  If it's true, how can this be proven?
This problem came from last year's qualifying exam in probability at Penn; it's been making the rounds around here over the past few days but nobody seems to remember how to do it.

REPLY [4 votes]: This is an archetypical example where classical Bernstein's method of "sections" or "blocks" works. Among popular probability books, Kai Lai Chung's "A Course in Probability Theory" contains a complete proof of CLT for m-dependent r.v.'s  indexed by $\mathbb{N}$. Here the indexation is 2-dimensional, but the idea is absolutely the same: divide the $N\times N$ array of r.v.'s into independent blocks of size that grows with $N$, with small "corridors" separating them. The corridors are needed to make the sums over blocks independent. 
The sums over blocks are i.i.d. and satisfy CLT for arrays, and the correction from the "corridors" is relatively small and does not spoil the CLT.<|endoftext|>
TITLE: Is an invertible biset necessarily a bitorsor?
QUESTION [10 upvotes]: Question
Let $G$ be a group, and let $X$ be a $G$-biset that is (weakly) invertible with respect to the contracted product. Is $X$ necessarily a bitorsor?
Background
By $G$-biset, I mean a set equipped with commuting left and right $G$-actions. There is a standard tensor product on the category of $G$-bisets called the contracted product; it is defined by $X \times_G Y = X \times Y / (x \cdot g, y) \sim (x, g \cdot y)$, where $G$ acts on the left by its left action on $X$, and on the right by its right action on $Y$. The unit object is the group $G$, where $G$ acts by left and right multiplication on itself.
A left $G$-torsor is a left $G$-set $X$ such that the map $G \times X \to X \times X$, $(g, x) \mapsto (g \cdot x, x)$ is a bijection. A right $G$-torsor is defined analogously. A $G$-bitorsor is a $G$-biset that is both a left and a right $G$-torsor. A $G$-bitorsor $X$ is necessarily invertible with respect to the contracted product; its inverse is the opposite $G$-bitorsor $X^{\operatorname{op}}$. This bitorsor has the same objects as $X$, but $g \in G$ acts on the left (resp. the right) by the right (resp. left) action of $g^{-1}$.
It follows from a simple counting argument that when $G$ is a finite group, any invertible $G$-biset is a $G$-bitorsor. Is this true for arbitrary groups (and more generally, in an arbitrary topos)? What about if we replace "invertible" by "right- (or left-) invertible"?
I can show, at least in the punctual topos (and I think it's true in general), that if $X^{\operatorname{op}}$ is an inverse to $X$, then $X$ must be a $G$-bitorsor. So the question is whether a $G$-biset can have an inverse not of this form.
The reason I'm interested in this question is that I want to understand how to generalize bitorsors to higher categorical settings. A possible generalization would be an invertible profunctor, but this is only a good definition if the answer to my question is affirmative.

REPLY [10 votes]: A torsor is a faithful transitive $G$-set. If the left $G$-action on $X$ is not faithful, the left $G$-action on $X\times_G Y$ will not be faithful. If the left $G$-action on $Y$ is not transitive, the left $G$-action on $X\times_G Y$ will not be transitive. By symmetry, it follows that a $G$-biset with a left and right inverse is a $G$-bitorsor.<|endoftext|>
TITLE: Is every smooth function Lebesgue-Lebesgue measurable?
QUESTION [32 upvotes]: This is motivated by pure curiosity (triggered by this question). A map $f:\mathbb R^n\to\mathbb R^m$ is said to be Lebesgue-Lebesgue measurable if the pre-image of any Lebesgue-measurable subset of $\mathbb R^m$ is Lebesgue-measurable in $\mathbb R^n$. This class of maps is terribly inconvenient to deal with but it might be useful sometimes. And maybe it is not that bad in the case $m=1$, especially if the answer to the following question is affirmative.
Question: Is every $C^1$ function $f:\mathbb R^n\to\mathbb R$ Lebesgue-Lebesgue measurable? If not, what about $C^\infty$ functions?
I could not figure out the answer even for $n=1$. However, there are some immediate observations (please correct me if I am wrong):

Since the map is already Borel measurable, the desired condition is equivalent to the following: if $A\subset\mathbb R$ has zero measure, then $f^{-1}(A)$ is measurable.
If $df\ne 0$ almost everywhere, then $f$ is Lebesgue-Lebesgue measurable (because locally it is a coordinate projection, up to a $C^1$ diffeomorphism). So the question is essentially about how weird $f$ can be on the set where $df=0$.
If the answer is affirmative for $C^1$, it is also affirmative for Lipschitz functions (by an approximation theorem).
The answer is negative for $C^0$, already for $n=1$. An example is a continuous bijection $\mathbb R\to\mathbb R$ that sends a Cantor-like set $K$ of positive measure to the standard (zero-measure) Cantor set. There is a non-measurable subset of $K$ but its image is measurable since it is a subset of a zero-measure set.

REPLY [20 votes]: It seems that your example of bijection  that sends one Cantor set with positive measure to another Cantor set with zero measure can be made $C^\infty$. 
Am I missing something?<|endoftext|>
TITLE: Does a finite dimensional algebra having a Cartan matrix with determinant 1 imply finite global dimension (possibly with more hypotheses)?
QUESTION [6 upvotes]: Let $A$ be your favorite finite dimensional algebra, and $P_i$ be a sets of representatives for the indecomposible projectives (or PIMs, if you like).  Then we have the Cartan matrix $C$ of the algebra, whose entries are $\dim Hom(P_i, P_j)$.  You can think of this as the matrix of the Euler form on the Grothendieck group $K^0(A-pmod)$ of projective $A$-modules.
Now, if the algebra $A$ has finite global dimension, then we can define the classes of the simple heads of these $L_i$ in $K^0(A-pmod)$ as integer linear combinations of the $P_i$'s, and $[P_i]$ and $[L_i]$ are dual bases in the Euler form.  That is, the matrix $C$ is integer valued and has integer-valued inverse, i.e. it has determinant 1.

To what degree is the converse of this true?  Is there a weaker hypothesis than finite global dimension itself such that $det(C)=1$ and that hypothesis will imply finite global dimension?

The application I have in mind for this is a little more complex.  I'd like to consider a graded version of this question.  So, let $A$ be a graded algebra such that each degree is finite dimensional (and let say the appearing gradings are bounded below).  The the graded version of $C$ is well-defined in $\mathbb{Z}((q))$, and similarly, if every simple has a resolution by projectives where only finitely many projectives generated in a given degree appear, this implies that this matrix has an inverse in $\mathbb{Z}((q))$, that is determinant with leading coefficient 1.

The same question as above: can I use a hypothesis like the graded Cartan matrix having determinant with integral leading coefficient to conclude the existence of such a projective resolution?

REPLY [6 votes]: In general, no. See [Burgess, W. D.; Fuller, K. R.; Voss, E. R.; Zimmermann-Huisgen, B. The Cartan matrix as an indicator of finite global dimension for Artinian rings. Proc. Amer. Math. Soc.  95  (1985), no. 2, 157--165. MR0801315]
It does work for artin algebras of Loewy length at most two, though, and various other families, like quasi-stratified algebras or left serial (that's the main result of the paper mentioned above)<|endoftext|>
TITLE: Is the sum of 2 Lebesgue measurable sets measurable?
QUESTION [21 upvotes]: Is the sum of two measurable set measurable? I think it is not...

REPLY [6 votes]: Note that the problem is trivial if you talk about subsets of the plane $\mathbb R\times \mathbb R$.  Let $A\subseteq \mathbb R$ be non-measurable, then $A\times \{0\}$ and $\{0\}\times \mathbb R$ both have Lebesgue measure 0 in the plane, but their sum $A\times \mathbb R$ is not measurable.<|endoftext|>
TITLE: Can I detect the point of impact without looking at it?
QUESTION [6 upvotes]: I'm going to postpone the motivation for this question because the question itself involves no complicated maths and may well have a very simple solution so I don't want to put anyone off with high falutin' symbols.
Here's the question.  I have a smooth curve $c \colon (0,1) \to \mathbb{R}^2$ which does not intersect the $x$-axis.  As $t \to 0$, this curve approaches the $x$-axis.  I want to find out if it has a point of impact.  At my disposal, I have any smooth function $f \colon \mathbb{R}^2 \to \mathbb{R}$ which is constant along the $x$-axis.  I know that for any such $f$, $f \circ c \colon (0,1) \to \mathbb{R}$ extends to a smooth function $[0,1) \to \mathbb{R}$ (that is, all one-sided derivatives exist at the origin and are the limits of the corresponding derivatives as we approach the origin) with value $f(0,0)$ at $0$.  So:

Is there some $f$ (satisfying the condition) such that the composition $f \circ c$ tells me that $c$ approaches a particular point on the $x$-axis?

If the answer to that is "yes", then my follow-up question is about the derivatives of $c$ at the point of approach.
Here are some comments and partial results:

I'm allowed to use any information about the compositions $f \circ c$: their values, their derivatives, and so forth.
The $y$-value of $c$ easily extends to a smooth function $[0,1) \to \mathbb{R}$ since the second projection $\mathbb{R}^2 \to \mathbb{R}$ is one of our detectors.  Moreover, it extends taking the value $0$ at $t = 0$.
I can show that the $x$-value of $c$ is bounded as $t$ approaches $0$.  If it weren't, I could stick bump functions along the image of $c$ with disjoint support and that were $0$ on the $x$-axis.  As they have disjoint support, their sum, call it $f$, is a smooth function and is at the disposal of my detection agency.  So there is some sequence $(t_n) \to 0$ such that $f \circ c(t_n) = 1$, but $f(0,0) = 0$ so this violates my condition.
I can show that if $c$ approaches the $x$-axis with any sort of speed then I can detect its point of impact (and all derivatives).  To do this, I use the function $g \colon (x,y) \mapsto x y$.  So long as some derivative of the (extended) $y$-value of $c$ is non-zero at $0$, I can use this to find out the $x$-value by differentiating $g \circ c$ that many times.
If $c$ approaches the $x$-axis infinitely slowly, has a point of impact, and the $x$-value extends to a smooth function $[0,1) \to \mathbb{R}$ (so then $c$ extends to a smooth function $[0,1) \to \mathbb{R}^2$) then I cannot detect the actual point of impact.  This is because I can use the chain rule to find the value of any derivative of $f \circ c$ and each term vanishes because either it involves a derivative of the $y$-value (assumed zero) or it involves a pure $x$-derivative of $f$ (zero by assumption on $f$ as we're then on the $x$-axis).

So it seems to me that it's a reasonable conjecture that I can't show that $c$ has a point of impact, if it approaches infinitely slowly.  However, the above is not a proof of that fact.

Motivation I'm trying to finish off the details of an example of a Froelicher space (http://ncatlab.org/nlab/show/Froelicher+space).  The space in question, let's call it $X$, is the quotient of the plane by the $x$-axis.  By the rules for taking quotients, the smooth functions on this space are simply the smooth functions on $\mathbb{R}^2$ which are constant along the $x$-axis.  I want to work out the smooth curves.  Let $c \colon \mathbb{R} \to X$ be a smooth curve.  It's straightforward to show that $c$ lifts to a smooth curve $U_c \to \mathbb{R}^2$, where $U_c$ is the complement of the preimage of the collapsed point.  As $U_c$ is (easily shown to be) open, it decomposes as a union of intervals.  On each interval, $c$ is a smooth curve in $\mathbb{R}^2$ which approaches the $x$-axis at the end-points.  So the question is as to what can be said as $c$ gets near one of these end-points.  That's the source of this question.

Edit Added in response to Bjorn's answer.  The underlying question is:

What are the smooth functions $c \colon \mathbb{R} \to \mathbb{R}^2/\mathbb{R}$?

(Blah, blah, Froelicher space structure, blah, blah)
Thus the point of the question is not "I have a curve, what is it?" but "Which curves can I get?".  However, I figured that the question "What are the smooth curves in $\mathbb{R}^2/\mathbb{R}$?" wouldn't get much interest, but something about extending smooth curves in $\mathbb{R}^2$ might!
Also If, as I suspect, I can get curves with no definite "point of impact", can I limit how bad these curves must be in some way?  Can I put some bound on their ($x$-)derivatives, or at least limit how fast they go to infinity?

REPLY [4 votes]: Andrew's comments showed me that in my first answer I was misunderstanding several aspects of his question.  Since I am still not entirely sure that I am capturing the spirit of the problem, let be begin this answer by stating in my own words (in very dry mathematical terms) what I interpret the question(s) to be, so that he or someone else can correct me if necessary.

Let $\mathcal{F}$ be the set of smooth functions $f \colon \mathbf{R}^2 \to \mathbf{R}$ whose restriction to the $x$-axis is constant.  Let $\mathcal{C}$ be the set of smooth functions $c \colon (0,1) \to \mathbf{R}$ such that for every $f \in \mathcal{F}$,

the function $f \circ c \colon (0,1) \to \mathbf{R}$ extends to a smooth function $[0,1) \to \mathbf{R}$ (in the sense that all one-sided derivatives exist at the origin and equal the one-sided limits of the corresponding derivatives), and 
$\lim_{t \to 0^+} f(c(t)) = f(0,0)$.

Question 1: If $c \in \mathcal{C}$, must $\lim_{t \to 0^+} c(t)$ exist?  (Actually, Andrew is also asking more generally what one can say about $c$ if the limit does not exist.)
Question 2: Is there a rule (function) $\mathcal{C} \to \mathcal{F}^r$ for some positive integer $r$, say taking $c$ to $(f_{1,c},\ldots,f_{r,c})$, such that $f_{1,c}$ is independent of $c$, and $f_{2,c}$ depends only on $f_{1,c} \circ c$, and $f_{3,c}$ depends only on $f_{1,c} \circ c$ and $f_{2,c} \circ c$, and so on, together with a rule (function) $R$ that takes as input a sequence of smooth functions $(g_1,\ldots,g_r)$ from $(0,1)$ to $\mathbf{R}$ and outputs a point in $\mathbf{R}^2$ such that for every $c \in \mathcal{C}$, we have $R(f_{1,c}\circ c,\ldots,f_{r,c} \circ c) = \lim_{t \to 0^+} c(t)$ whenever the latter exists?
Question 3: Same as Question 2, but with $\lim_{t \to 0^+} c'(t)$ in place of $\lim_{t \to 0^+} c(t)$.

Answer to Question 1: No.  A negative answer was essentially given already by Andrew himself.  Namely, let $c(t) := (\sin 1/t,e^{-1/t})$.  This $c(t)$ has the following properties: the $x$-coordinate is bounded, the derivatives of the $x$-coordinate grow at most polynomially in $1/t$ as $t \to 0^+$, and the $y$-coordinate and derivatives of the $y$-coordinate decay to $0$ exponentially as $t \to 0^+$.  As explained by Andrew, for any $f \in \mathcal{F}$, the chain rule shows that $f(c(t))$ extends to a smooth function $[0,1) \to \mathbf{R}$ whose value at $0$ is $f(0,0)$ and whose higher derivatives at $0$ are all $0$. $\square$
Answer to Questions 2 and 3: Yes.  In fact, we can do it with $r=2$, and with both $f_{1,c}$ and $f_{2,c}$ independent of $c$.  Namely, use $f_1(x,y)=y$ and $f_2(x,y)=e^x y$.  From $f_1 \circ c$ and $f_2 \circ c$, we may recover not only the $y$-coordinate of $c$ as $f_1 \circ c$, but also the $x$-coordinate of $c$ as $\log (f_2(c(t))/f_1(c(t)))$.  So the whole function $c(t)$, and hence any property of $c(t)$, can be detected. $\square$<|endoftext|>
TITLE: Chow Ring of Moduli Space of Abelian Varieties
QUESTION [6 upvotes]: Is there a good reference for the structure of the Chow ring of $\mathcal{A}_g$, the moduli space of complex principally polarized abelian varieties? More generally, references for the intersection theory, enumerative geometry, and vector bundles on $\mathcal{A}_g$ would be nice.

REPLY [3 votes]: Van der Geer has written a paper computing what he calls the tautological subring of the chow ring of $\mathcal{A}_g$.  He also computes the tautological ring for a smooth toroidal compactification.
G. van der Geer, Cycles on the Moduli Space of Abelian Varieties, in "Moduli of Curves and Abelian Varieties (The Dutch Intercity Seminar on Moduli)", p. 65-89 (Carel Faber and Eduard Looijenga, editors), Aspects of Mathematics, Vieweg, Wiesbaden 1999.
It is available on the van der Geer's website here
Regarding intersection theory,  Erdenberger, Grushevsky, and Hulek have been working on this for the toroidal compactifications, mostly for small values of $g$.  For example, see the following references.
C. Erdenberger,  S. Grushevsky,  K. Hulek,  Intersection theory of toroidal compactifications of $\mathcal{A}_4$.  Bull. London Math. Soc.  38  (2006),  no. 3, 396--400.
C. Erdenberger, S. Grushevsky, K. Hulek,  Some intersection numbers of divisors on toroidal compactifications of $\mathcal{A}_g$.  J. Algebraic Geom.  19  (2010),  no. 1, 99--132.
S. Grushevsky,  Geometry of $\mathcal{A}_g$ and its compactifications.  Algebraic geometry---Seattle 2005. Part 1,  193--234, Proc. Sympos. Pure Math., 80, Part 1, Amer. Math. Soc., Providence, RI,  2009.<|endoftext|>
TITLE: Fields of definition of a variety
QUESTION [23 upvotes]: Let $K$ and $L$ be two subfields of some field. If a variety is defined over both $K$ and $L$, does it follow that the variety can be defined over their intersection?

REPLY [19 votes]: No, not even for genus $0$ curves, if "$X$ is defined over $K$" means that the variety $X$, initially defined over an extension $F$ of $K$, is to be isomorphic to the base extension of some variety over $K$.  (Pete in his answer was implicitly assuming that he was allowed to base extend to an algebraic closure of $F$ before going down to $K$.)
For finite extensions $F \supseteq E$ of $\mathbb{Q}_p$, if $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $E$ if and only if $[F:E]$ is odd (because $\operatorname{Br}(E)[2] \to \operatorname{Br}(F)[2]$ is multiplication by $[F:E]$ from $\frac{1}{2} \mathbb{Z}/\mathbb{Z}$ to itself).
So if $F$ is an $A_4$-extension of $k:=\mathbb{Q}_2$ (such an extension exists), and $K$ and $L$ are the subfields of $F$ fixed by two $3$-cycles in $A_4$, and $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $K$, and similarly over $L$, but not over $K \cap L = \mathbb{Q}_2$.<|endoftext|>
TITLE: How many unit squares can you pack into a rectangle with nearly integer side lengths?
QUESTION [20 upvotes]: Earlier today, somebody asked what looks like a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:  
      (source)
If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of $1$. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?

REPLY [4 votes]: https://erich-friedman.github.io/mathmagic/0610.html has some answers for small numbers of squares.<|endoftext|>
TITLE: Doing geometry using Feynman Path Integral?
QUESTION [37 upvotes]: I have often heard in the folk-lore that Feynman Path Integral can be used to compute geometric invariants of a space. 
Coming from a background of studying Quantum Field Theory from the books like that of Weinberg, I have myself used Feynman Path Integrals to compute scattering of particles. 
Earlier I had done courses in Riemannian Geometry and these days I am also doing courses in Algebraic Topology and hence I think it would be very educative if I can see how exactly the calculation of topological invariants that one does here are related to Feynman's ideas. 
It would be helpful if someone can give me references which explain (hopefully starting with simple examples!) how one can use path integrals in geometry.

REPLY [3 votes]: Try this article: 
Braiding Statistics and Link Invariants of Bosonic/Fermionic Topological Quantum Matter in 2+1 and 3+1 dimensions by 
Pavel Putrov, Juven Wang, Shing-Tung Yau
Annals of Physics 384C, 254-287 (2017)
https://arxiv.org/abs/1612.09298
It explains many interesting link invariants and their relations to braiding statistics of anyons in 3d spacetime and anyonic strings in 4d spacetime in Quantum Matter (condensed matter and lattice models) and Quantum Field Theories:<|endoftext|>
TITLE: Mathematics of path integral: state of the art
QUESTION [65 upvotes]: I was told that one of the most efficient tools (e.g. in terms of computations relevant to physics, but also in terms of guessing heuristically mathematical facts) that physicists use is the so called "Feynman path integral", which, as far as I understand, means "integrating" a functional (action) on some infinite-dimentional space of configurations (fields) of a system.
Unfortunately, it seems that, except for some few instances like Gaussian-type integrals, the quotation marks cannot be eliminated in the term "integration", cause a mathematically sound integration theory on infinite-dimensional spaces — I was told — has not been invented yet.
I would like to know the state of the art of the attempts to make this "path integral" into a well-defined mathematical entity.
Difficulties of analytical nature are certainly present, but I read somewhere that perhaps the true nature of path integral would be hidden in some combinatorial or higher-categorical structures which are not yet understood...
Edit: I should be more precise about the kind of answer that I expected to this question. I was not asking about reference for books/articles in which the path integral is treated at length and in detail. I'd have just liked to have some "fresh", (relatively) concise and not too-specialistic account of the situation; something like: "Essentially the problems are due to this and this, and there have been approaches X, Y, Z that focus on A, B, C; some progress have been made in ... but problems remain in ...".

REPLY [10 votes]: First, there are several rigorous definitions of integration in infinite dimensional spaces, like the Bochner integral in Banach spaces (see Wikipedia), or see the book by Parthasarathy: "Probability measures on metric spaces" (this includes the Gaussian probability measures used by constructive QFT already mentioned).
These cannot be used to make the Feynman path integral into a rigorous defined mathematical entity with finite values, i.e. the problem is to get an integral that spits out finite numbers in physically interesting models.
For starters, there cannot be a translationally invariant measure (on the Borel sigma algebra) other than the one that assigns infinite volume to every open set in an infinite dimensional metric space (hint: a ball of radius r contains infinitly many pairwise disjunct copies of the ball of radius r/2). So the path integral, as it is written by physicists, certainly has no interpretation via a translationally invariant measure, contrary to what the notation usually employed may suggest.
While there currently is no mathematically rigorous definition of a Feynman path integral applicable to an interesting subset of physical models, here are some books that give some hints at the current state of the affair:
Huang and Yan: "Introduction to Infinite Dimensional Stochastic Analysis" (this contains a description of the Feynman path integral from the viewpoint of "white noise analysis"),
Sergio Albeverio, Raphael Hoegh-Krohn; Sonia Mazzucchi:"Mathematical theory of Feynman path integrals. An introduction",
Pierre Cartier, Cecile DeWitt-Morette: "Functional integration: action and symmetries".
BTW: This is in a certain sense a "one million dollar" question because one of the millenium problems of the Clay Mathematics Institute is a rigorous construction of Yang-Mills theories.<|endoftext|>
TITLE: For which fields K is every subring of K...?
QUESTION [9 upvotes]: This question was inspired by 
How to prove that the subrings of the rational numbers are noetherian?
which some people found too routine to be of interest.  So I have decided to liven things up a bit with the following questions.  In the interest of full disclosure, I have not thought seriously about these questions, and I think that I probably could answer at least some of them myself, but I do think they are interesting and, if I may say so, educational.  
Find all (commutative!) fields $K$ such that every (unital!) subring $R$ of $K$ is:
a) a principal ideal domain.
b) a Dedekind domain.
c) a Noetherian domain.  
I mean here to be asking three different questions, one for each condition.  Evidently the classes of such fields are nondecreasing from a) to b) and from b) to c).  
If you would like to answer the question with a), b) or c) replaced by some other standard property of commutative rings -- especially if it yields a different class of fields than in the first three questions -- please feel free.
Addendum: How about
d) a Dedekind domain if it is integrally closed?
e) a PID if it is integrally closed?

REPLY [9 votes]: Let me put together the previous two answers (plus epsilon) to give an answer to all three questions.
Step 1: By Gilmer's theorem, a field $K$ has all its subrings Noetherian iff:
(i) It is a finite extension of $\mathbb{Q}$, or
(ii) It is an algebraic extension of $\mathbb{F}_p$ or a finite extension of $\mathbb{F}_p(t)$.
Step 2: Suppose $K$ is a number field which is not $\mathbb{Q}$.  We may write $K = \mathbb{Q}[\alpha]$ for some algebraic integer $\alpha$.  Then $R = \mathbb{Z}[2\alpha]$ is a non-integrally closed subring of $K$ so is not a Dedekind domain.  So the only field of characteristic $0$ which has every subring a Dedekind domain is $\mathbb{Q}$, in which case (by the previous question) every subring is a PID.
Step 3: Suppose $K$ has characteristic $p > 0$.  If $K$ is algebraic over $\mathbb{F}_p$, then every subring is a field, hence also Dedekind and a PID.  If $K$ is a finite extension of $\mathbb{F}_p(t)$ then it admits a subring of the form $\mathbb{F}_p[t^2,t^3]$, which is not integrally closed.  
So the fields for which every subring is a Dedekind ring are $\mathbb{Q}$ and the algebraic extensions of $\mathbb{F}_p$.  For all such fields, every subring is in fact a PID.<|endoftext|>
TITLE: Computations of the Link homology categorifying the second colored Jones polynomial
QUESTION [7 upvotes]: Has anybody done computations of such a theory?  Is there a place I could look up and see what the answers are for low crossing knots?

REPLY [2 votes]: Hi Charlie, 
I did some calculations, but they are hard. Already for the unknot one gets an interesting, but infinite complex with cohomologies in all degrees!
We are just finishing a paper on this which hopefully will appear at the end of next week. 
Are there any specific knots you are interested in?
Catharina<|endoftext|>
TITLE: What is meant by smooth orbifold?
QUESTION [29 upvotes]: There seems to be some confusion over what the tangent space to a singular point of an orbifold is.
On the one hand there is the obvious notion that smooth structures on orbifolds lift to smooth $G$-invariant structures on $\mathbb R^n$ ($G$ being the finite group so that the orbifold is locally (about some specific point $x$) the quotient of $\mathbb R^n$ by the action of $G$).  One might be tempted to consider cone points as differentiable spaces (that is, subsets of some $\mathbb R^k$, inheriting their differential structure by restriction), however, we are told, for example, that $\mathbb R^2/\mathbb Z_3$ and $\mathbb R^2/\mathbb Z_4$ are distinct as orbifolds, so it is not the case that cone points can be modeled merely with cone-like subsets of some $\mathbb R^k$.  The definition in which 'smooth' means 'lifts to $G$-invariant smooth' distinguishes these two cones, as the set of functions with 3-fold symmetry and the set of functions with 4-fold symmetry, in $\mathbb R^2$, are distinct.  The third item in Satake's seminal paper [On a Generalization of the Notion of Manifold] corroborates this, giving $C^\infty$ forms of degree $p$ at a singularity $x$ as those $C^\infty$ $p$-forms in $\mathbb R^n$ which are invariant under $G_x$.  If we require the same property of vectors, that is, that they lift to $G$-invariant vectors in $\mathbb R^n$, then we have that the dimension of the tangent space of an orbifold is the dimension of the invariant subspace upstairs.  In particular the dimension tends to drop at the singular points.  For example, the dimension of the tangent space at the singularity in $\mathbb R^2/\mathbb Z_3$ is 0.  This notion of vector agrees with the notion of vector as derivation on the germ of smooth functions.  In this case smooth functions in $\mathbb R^2$ which have 3-fold symmetry necessarily have vanishing derivatives at the origin.
On the other hand, one finds descriptions of smooth orbifolds as objects which have tangent bundle-like structures, which are locally $\mathbb R^n/G$.  It is not clear what this means as far as smooth structures go, but the explanation above of $\mathbb R^2/\mathbb Z_3$ having a 0 dimensional tangent space at the cone point seems to contradict the notion that the tangent-like space at the singularity in $\mathbb R^2/\mathbb Z_3$ is $\mathbb R^2/\mathbb Z_3$, whatever that means.
It is also said that manifolds with boundary can be viewed as orbifolds, which have isotropy group reflection by $\mathbb Z_2$ along their boundaries.  It would be nice to include the note that the differentiable structures are different.  Specifically, smooth manifolds with boundaries have tangent spaces along their boundaries which are the same dimension as the manifold.  In contrast, the same topological space as an orbifold with $\mathbb Z_2$ structure group along the boundary should have a tangent space which is one dimension less than the dimension at a generic point, if the definition of tangent space follows Satake's guideline.  Indeed, smooth functions in $\mathbb R^n$ which locally have symmetry by reflection through a codimension 1 hyperplane, have vanishing partial derivatives in the normal direction.
I am asking for concurrence or correction and clarification, since I am still not certain I have the correct notion of tangent space to an orbifold, although I'm fairly confident in the first given here.

REPLY [6 votes]: Edit: Nov. 26, 2015
Another example about how diffeology represents the smooth structure of orbifolds: the Seifert Orbifolds, as space of fibers of a "Seifert fibered manifold":
http://math.huji.ac.il/~piz/documents/DBlog-Rmk-SeifO.pdf
Edit: Nov. 3, 2015
I had recently to explain to a student the difference of an Orbifold (or a V-Manifold) according to Satake's original definition, with "local uniformizing systems" and "defining family", and the Diffeology version of these objects. I wrote this short text (published in my bolg)
http://math.huji.ac.il/~piz/documents/DBlog-Rmk-SO.pdf,
where I tried to make clear and simple how we can understand the smooth structure of an orbifold through the frame of diffeology. I think/hope it can help.
It doesn't prevent to read the original paper on diffeological orbifold
http://www.ams.org/journals/tran/2010-362-06/S0002-9947-10-05006-3/home.html
-------------------------------- Dec. 29, 2010
I don't want to discourage you, but I don't think that the tangent space is a good concept for quotient-like spaces, especially for orbifolds. More adapted would be differential forms. Of course, it depends on what we are willing to do with that. But if you consider "symplectic geometry" you can definitely avoid tangent spaces and focus on forms. Moment Maps, for example, don't need any kind of tangent spaces to be defined and computed in even more general situations than orbifolds. 
First focus on what you want to prove or find, then choose, accordingly, the right tool to use and surprisingly some generalizations coming just from habits of manifolds become obsolete, or drive you in the wrong direction. Let me say straight, what tangent spaces are useful for? Because of the Cauchy-Lipschitz theorem on Hausdorff manifolds: they generate (locally and under some conditions) a one parameter group of diffeomorphisms. So if you are interested in diffeomorphisms of the orbifolds, go directly to the group, and analyze it. For example (taking the diffeological approach) you can check immediately that any diffeomorphism may exchange only points with equivalent structure groups, etc. etc. If you start with vector field you have already a long way before reaching this starting point.<|endoftext|>
TITLE: An S-unit equation, with S an infinite and sparse set of primes.
QUESTION [6 upvotes]: Say we have three infinite sequences $\{a_i\},\{b_i\},\{c_i\}$ of natural numbers, satisfying the equations $$a_1+b_1=c_1,\dots, a_n+b_n=c_n,\dots $$
Assume further that $gcd(a_i,b_i,c_i)=1$ for each $i$ and that $(a_i,b_i,c_i)\neq (a_j,b_j,c_j)$ for all $i,j$. Now let's define $S$ as the set of primes $p$ which divide $a_ib_ic_i$ for at least one $i$. From the S-unit theorem we know that $S$ has to be infinite. 
Now the question is: Can $S$ be sparse? This can be taken to mean Dirichlet density zero, for example.
I haven't thought much about this but there are reasons to believe the answer is yes, indeed if there are infinitely many Mersenne primes $q=2^p-1$ then the equations $1+q=2^p$ give such a sparse $S$. However, I am looking for an unconditional result.

REPLY [2 votes]: An even simpler answer: take triples of the form $(2,n!-1,n!+1)$ for a sufficiently sparse sequence of values of $n$.<|endoftext|>
TITLE: Why is a topology made up of 'open' sets? Part II
QUESTION [16 upvotes]: Because the display was getting quite cluttered, I thought I'd post a second part to this question separately. I hope the Gods of Math Overflow don't take too much offense. I'll  go now into some details with which I wished intially to avoid prejudicing the replies.
Here are three natural areas of progression in the study of topology:
(1) point-set topology;
(3) sheaves and their cohomology;
and a very important 
(2) middle ground 
that I won't give a name to, involving matters like the classification of two-manifolds. I don't quite agree that the metric space intuition is sufficient even for a first course, simply because we always look ahead to the way the material will help or hinder students' future understanding. 
Now, it is in (3) that a topology in the abstract sense plays a very active role, as the relation to global invariants emerges prominently. At this point, the definition needs no other motivation. However, in an introductory course on (1), I have yet to incorporate successfully interesting material from (3). When dealing with (2), curiously enough, the definition of a topology plays a very passive role. Many arguments are strongly intuitive. However, it is clear that one needs to have already absorbed (1) for the material in (2) to feel really comfortable. Of course there are exceptionallly intuitive people who can work fluently on (2) without a firm foundation in point-set topology. But for most ordinary folks (like me), the rigor of (1) is needed if only as a psychological crutch. Note here that  most of the natural spaces that come up in (2) are in fact metrizable. However, getting bogged down with worrying about  the metric would be a definite hindrance in working through the operations that come up constantly: stretching, bending, and perhaps most conspicuously, gluing. A quotient metric is a rather tricky notion, while a quotient topology is obvious.
With this future work in mind, how best then should one get this background in (1)? When I was an undergraduate, in fact, there was plenty of motivation in  course (1). We used the first part of Munkres' book, and had exercises dug out of 'Counterexamples in Topology,' involving many strange spaces that have one property but not another. It was great fun. You may wonder then, what exactly I'm worried about. It's that I felt later that this kind of motivation had not been quite right. It wasn't entirely wrong either. Certainly we became very confident in working with the definitions, and that was good. However, and this is a big 'however,' when I moved on to (2), I had a distinct sense that the motivational material and exercises used in (1) were actually preventing me from learning the new notions efficiently. It took me quite a long time to make the transition,  bugged by a persistent longing for the axioms and the exotic examples. A number of conversations I've had over the years indicate that my experience was far from unique. At any rate, this experience makes the issue for me quite different from a course on, say, linear algebra. No doubt the motivational material for diagonalization used in most courses is hardly convincing. Substantive examples come later, sometimes much later. But most of the operators and eigenvectors in standard textbooks are good toy models for a wide range of serious objects.
One obvious remedy would be to incorporate material from (2) into (1). I found this very hard, mostly because of the point already made above, that the role of (1) in (2) is quite implicit. When I first posed the question, I was hoping someone had figured out a good way of doing something like this. Incidentally, sheaves came around much later than the definition of a topology, so the historical question remains as well. How were the standard properties decided upon?
Let me make clear that I am not arguing that everyone has to go through the progression just outlined. Obviously, some people will take (1) elsewhere, in a way that the issues become quite different. Perhaps many people will never need more than metric spaces (or normed spaces, for that matter). But (1)-->(2)-->(3) is certainly common enough (perhaps especially for arithmetic geometers) to call for some reasonable methodology.
Meanwhile, I also appreciate Andrew Stacey's point (possibly even more than he does!). The long paragraphs above notwithstanding, the pedagogical question isn't something I lose sleep over. But it would be nice to have a few concrete and systematic ideas to use. They would certainly help me to understand the subject better!
Added:
Perhaps I should rephrase the question: How should we teach point set topology so as to facilitate  the transition to the topology of natural spaces?
Somehow, this way of putting it seems much vaguer to me. 'How to teach X?' is such a broad question I would never be able to answer it myself in a finite amount of time. It seems to invite a diffuse discussion of everythng under the sun. That's why I had preferred to focus on one rather precise mathematical facet of that question. But I don't feel too strongly about it either way.

REPLY [6 votes]: Your question reminds me of Grothendieck's Esquisse d'un Programme, specifically the notion of "tame topology". Was this on your mind when posing the question? I am thinking of the part that starts like this:
After some ten years, I would now say, with hindsight, that “general topology” was developed (during the thirties and forties) by analysts and in order to meet the needs of analysis, not for topology per se, i.e. the study of the topological properties of the various geometrical shapes. That the foundations of topology are inadequate is manifest from the very beginning, in the form of “false problems” (at least from the point of view of the topological intuition of shapes) such as the “invariance of domains”, even if the solution to this problem by Brouwer led him to introduce new geometrical ideas. Even now, just as in the heroic times when one anxiously witnessed for the first time curves cheerfully filling squares and cubes, when one tries to do topological geometry in the technical context of topological spaces, one is confronted at each step with spurious difficulties related to wild phenomena. 
To me the notion of a "false problem" is a very succinct way of expressing the problems associated with going from (1) to (2) in your formulation, at least if I have understood your question correctly. Namely, in a first course in point-set topology, one spends a lot of time learning about topologist's sine curves, long lines and Sorgenfrey planes, and the problem is that all of these are cooked up to solve/exemplify problems that are "false", i.e. not very natural/interesting to someone who actually does geometry in his day-to-day life. 
If we then believe Grothendieck, the problem is not one of didactics at all: the difficulties in passing from (1) to (2) are an inevitable byproduct of the fact that point-set topology is not adequate as a framework for geometry. Maybe this is true. In a first topology course, one does need to spend time on pathological behavior, simply because anyone who does topology should know that a topological space with no extra conditions can be weird and not very geometric. I think this will be true as long as topological spaces are considered the basic framework for topology. Then one does have to unlearn the instinct to think about pathologies when actually doing topology. On the other hand, of course you can't tell students in a first topology course that the pathological spaces they are learning about will not actually be so useful later in life; no one would learn the material after hearing something like that. I suspect that I have just restated what you said in a clumsier manner.<|endoftext|>
TITLE: Is there any disadvantage from non-academic job turn to academic job in math
QUESTION [7 upvotes]: If you get your PhD in math , and then work for 1 or 2 years in a non-academic institution and then turn to apply for postdoc or tenure-track position in math like usual, is there any disadvantage (I mean for your application for postdoc or tenure-track position)?
An appendix: I just want to make sure whether or not I can't or it is difficult to get reference letter, take conference or give talks (in the future) because you are not in academic institution. This is the most important for someone (like me) who will returen to academic job.(but for some reasons he can't now)

REPLY [10 votes]: It's indeed possible [I spent many years in industry building math software, and now I'm a tenured prof, albeit in computer science and software engineering department rather than math; although my research involves building mechanized mathematics systems...].  I started my academic career having previously 'published' 0 academic papers!  I was, however, already well-known within the computer algebra community, and my work was known [so I was able to get many good academic reference letters].  The reason for me to report this is that it is important to be able to convince the academic community that you really have something to contribute, else why would they hire you?  So, if you intend to move back to academia, either write papers or make sure that somehow the community 'knows' you and appreciates your work.
From my experience, I would say that the hardest part is to go from having well-defined goals with precise deadlines, often driven by external pressures, to writing research papers with no deadline.  Getting up to speed on producing papers while being on the 'tenure clock' was most unpleasant.  And, of course, at the beginning teaching courses can (and likely will) swallow all available time.  Unless you're in an enlightened department (I wasn't) where untenured faculty are given a lighter teaching load to allow them the time to get settled into their research career.
If at all possible, get a post-doc in between a non-academic job and a tenure-track position.  This will give you the time needed to 'switch gears'.  I probably would not have done that myself (the salary cut was already substantial enough as it is, I didn't want to make it even worse).  It depends on your personal situation.<|endoftext|>
TITLE: The difference between $l^1(G)$ and the reduced group $C^*$ algebra $C_r^*(G)$
QUESTION [14 upvotes]: Let $G$ be a group and $l^2(G)$ the Hilbert space on $G$. The complex group algebra $CG$ can be imbedded in $B(l^2(G))$, the set of all bounded linear operators, by left translation. The reduced group $C^*$ algebra $C_r^*(G)$ is the operator norm completion of $CG$. It's clear that $l^1(G)$ lies in $C_r^*(G)$. 
Q1. Are there some typical elements in $C_r^*(G)\setminus l^1(G)$? 
Q2. For a normed algebra $A$, let $C$ be the completion of the subalgebra $\{ab-ba|a,b\in A\}$ and $T(A)=A/C$. It's not hard to see $T(l^1(G))$ is the $l^1$ completion of $C[conj(G)]$, the set of all finitely supported functions on the conjugacy classes $conj(G)$. How to describle $T(C_r^*(G))$? In particular, if $g$ is an element of infinite order, can $g$ have the same image in the quotient algebra as $0$ or a finite order element in $G$? Thank you very much.

REPLY [6 votes]: Q1 seems to be related to symmetry of $\ell^1(G)$ (a Banach $^*$-algebra $A$ is symmetric if, for every $a\in A$, the spectrum of $a^*a$ is contained in $\mathbb{R}^+$). If $\ell^1(G)$ is not symmetric, then for an element  $a^*a$ with non-positive $\ell^1$-spectrum, since the $C^*$-spectrum is clearly positive, the resolvent of $a^*a$ will define elements in $C^*_r(G)\backslash\ell^1(G)$. For examples of groups with non-symmetric algebras, see papers by J. Boidol, H. Leptin and D. Poguntke from the 1980's.<|endoftext|>
TITLE: What is the insight of Quillen's proof that all projective modules over a polynomial ring are free?
QUESTION [68 upvotes]: One of the more misleadingly difficult theorems in mathematics is that all finitely generated projective modules over a polynomial ring are free.  It involves some of the most basic notions in commutative algebra, and really sounds as though it should be easy (the graded case, for example, is easy), but it's not.  The question at least goes as far back as Serre's FAC, but it wasn't proved until 1976, by Quillen in Projective modules over polynomial rings EDIT: and also independently by Suslin.
I decided that this is the sort of fact that I should know a rough outline of how to prove, but the paper was not very helpful.  Usually when someone kills off a famous conjecture in 5 pages, it's because they've developed some fantastic new piece of machinery people didn't have before.  And, indeed, Quillen is famous for inventing some fancy and wonderful machinery, and the paper is only 5 pages long, but as far as I can tell, none of that fancy machinery actually appears in the proof.

So, what was it that Quillen saw, that Serre missed?

REPLY [58 votes]: Here is a summary of what I learned from a nice expository account by Eisenbud (written in French), can be found as number 27 here.1
First, one studies a more general problem: Let $A$ be a Noetherian ring, $M$ a finite presented projective $A[T]$-module. When is $M$ extended from $A$, meaning there is $A$-module $N$ such that $M = A[T]\otimes_AN$?
The proof can be broken down to 2 punches:
Theorem 1 (Horrocks) If $A$ is local and there is a monic $f \in A[T]$ such that $M_f$ is free over $A_f$, then $M$ is $A$-free
(this statement is much more elementary than what was stated in Quillen's paper).
Theorem 2 (Quillen) If for each maximal ideal $m \subset A$, $M_m$ is extended from $A_m[T]$, then  $M$ is extended from $A$
(on $A$, locally extended implies globally extended).
So the proof of Serre's conjecture goes as follows: Let $A=k[x_1,\dotsc,x_{n-1}]$, $T=x_n$, $M$ projective over $A[T]$. Induction (invert all monic polynomials in $k[T]$ to reduce the dimension) + further localizing at maximal ideals of $A$ +  Theorem 1 show that $M$ is locally extended. Theorem 2 shows that $M$ is actually extended from $A$, so by induction must be free.
Eisenbud's note also provides a very elementary proof of Horrocks's Theorem, basically using linear algebra, due to Swan and Lindel (Horrocks's original proof was quite a bit more geometric).
As Lieven wrote, the key contribution by Quillen was Theorem 2: patching. Actually the proof is fairly natural, there is only one candidate for $N$, namely $N=M/TM$, so let $M'=A[T]\otimes_AN$ and build an isomorphism $M \to M'$ from the known isomorphism locally.
It is hard to answer your question: what did Serre miss (-:? I don't know what he tried. Anyone knows?
1Eisenbud, David, Solution du problème de Serre par Quillen–Suslin, Semin. d’Algebre, Paul Dubreil, Paris 1975-76 (29eme Annee), Lect. Notes Math. 586, 9-19 (1977). DOI: 10.1007/BFb008711, PDF on the MSRI website, ZBL0352.13005, MR568878.<|endoftext|>
TITLE: Definition of L-function attached to automorphic representation
QUESTION [12 upvotes]: Suppose $\pi$ is an irreducible automorphic representation of a reductive connected algebraic group $G$ over $\mathbb{A}_K$, here $K$ is a number field, $\mathbb{A}_K$ denotes its adeles. We have a restricted tensor product decomposition of $\pi=\otimes\pi_v$, where $\pi_v$ is an irreducible admissible representation for $G(K_v)$, and for all but finitely many $v$, $\pi_v$ is unramified.
We know how to define local L-factors at $v$ is $\pi_v$ is unramified, and we also know how to define local L-factors at archimedean places because of Langlands classification. So the question is how to define L-factors at ramified places?
As far as I know, at least for $GL_n$, we can define it as the gcd of some family of integrals via integral representation of L-function.

REPLY [9 votes]: This question was posted a while back but I just saw it. Here are some thoughts. In practice there are a couple of methods to construct L functions for local ramified representations. The first one is the Langlands-Shahidi method which works for "generic representations" of quasi-split groups and the second one is the method of integral representations (Rankin-Selberg method, Shimura's integral and the doubling method, etc). It is probably a bit painful to give a meaningful description of these two methodologies in such a limited space, so instead let me refer you to a couple of places where you can see accessible accounts of the two approaches.  A good reference for the basics of the Langlands-Shahidi method is the beautiful monograph "Analytic properties of automorphic L functions" by Gelbart and Shahidi. The same reference has a nice introduction to the method of integral representations. Dan Bump has written two very informative survey papers on the Rankin-Selberg method. Cogdell's ICTP lectures on the Rankin-Selberg method are lovely. Another good book to look at is the AMS book by Cogdell, Kim, Murty.  
As it stands there is no Langlands-Shahidi method for non-generic representations. What is missing from the picture is a good supply of easy to use unique models, like the Whittaker model in the generic setting. For orthogonal groups, however, recent progress by Waldspurger and others on the Gross-Prasad conjectures gives one the hope that maybe one can now develop a Langlands-Shahidi method, although there are serious obstacles to deal with. 
Most of the integral representations known to mankind too are closely linked with unique models (Whittaker, Bessel, etc). Sakellaridis has a theory that "explains" (some) integral representations in terms of spherical subgroups of reductive groups.<|endoftext|>
TITLE: Do Jones-Wenzl idempotents lift to anything interesting in the Hecke algebra?
QUESTION [12 upvotes]: Background
Inside the Temperley-Lieb algebra $TL_n$ (with loop value $\delta=-[2]$ and standard generators $e_1,\ldots,e_{n-1}$), the Jones-Wenzl idempotent is the unique non-zero element $f^{(n)}$ satisfying
$$ f^{(n)}f^{(n)} = f^{(n)} \quad \textrm{and} \quad e_i\;f^{(n)} = 0 = f^{(n)}e_i \quad \textrm{for each } i.$$
Consider the Iwahori-Hecke algebra $\mathcal{H}_n$, $n\ge3$, normalized so that $(T_i-q)(T_i+q^{-1})=0$, where $q$ is generic.  Let $\mathcal{I}$ be the two-sided cellular ideal generated by canonical basis element
$$C_{121} = T_1T_2T_1-qT_1T_2-qT_2T_1+q^2T_1+q^2T_2-q^3.$$
The assignment $\mathcal{H}_n \rightarrow TL_n$ given by $T_i \mapsto e_i + q$ is a surjective $\mathbb{C}(q)$-algebra homomorphism with kernel $\mathcal{I}$.
We can lift the generators $e_i$ in $TL_n$ to the Kazhdan-Lusztig elements $C_i=T_i-q \in \mathcal{H}\_n$.  In fact, we have $C_{121} = C_1C_2C_1 - C_1$, hence the relation down below.  Rescaling a bit, $E=-\frac{1}{[3]!}C_{121}$ is an idempotent, corresponding to the partition $(1,1,1)$.  Actually, all of the primitive idempotents in $\mathcal{H}_n$ that correspond to Young diagrams with more than two rows live in the ideal $\mathcal{I}$.
Now, any preimage of $f^{(n)}$ in the Hecke algebra (call it $F^{(n)}$) satisfies
$$F^{(n)}F^{(n)} \equiv F^{(n)} \quad \textrm{and} \quad C_iF^{(n)} \equiv 0 \equiv F^{(n)}C_i \quad (\operatorname{mod} \mathcal{I})$$
Question

Can we choose $F^{(n)}$ to be an idempotent in $\mathcal{H}_n$?

When $n=2$, the map is an isomorphism and we have no choice.
$$F^{(2)} = \frac{1}{[2]}(T_1+q^{-1}),$$
which projects onto the $q$-eigenspace for $T_1$.  In other words, it is the idempotent corresponding to the partition $(2)$.

REPLY [8 votes]: Yes. No problem. This is the $q$-analogue of the symmetriser. In terms of $T_i$ we have
$$ \frac{1}{[n]!}\sum_{\pi\in S_n} q^{\ell(\pi)} T_\pi$$
where for $T_\pi$ we take a reduced word for $\pi$ and $\ell(\pi)$ is the length of a reduced word.
There is another presentation for the Hecke algebra which I am used to writing as
generators $u_i$ and defining relations
$$u_i=[2]u_i$$
$$u_iu_j=u_ju_i\qquad\text{for $|i-j|>1$}$$
$$u_iu_{i+1}u_i-u_i=u_{i+1}u_iu_{i+1}-u_{i+1}$$
Strictly speaking this is the subring of the Hecke ring which is invariant under the bar involution. This algebra is defined over $\mathbb{Z}[\delta]$. The Hecke algebra is the algebra over $\mathrm{Z}[q,q^{-1}]$ obtained by the specialisation $\delta\mapsto q+q^{-1}$.
Then we have $u_i=-C_i$.
The Temperley-Lieb algebra is the quotient by $u_iu_{i\pm 1}u_i=u_i$. 
This has an involution given by $u_i\leftrightarrow \delta-u_i$.
To define the idempotents we need to divide by $[n]!$.
Define $R_i(k)=1-\frac{[k]}{[k+1]}u_i$. Then these satisfy the Yang-Baxter equation
$$R_i(r)R_{i+1}(r+s)R_i(s)=R_{i+1}(s)R_i(r+s)R_{i+1}(r)$$
Using this you can write everything (well a lot, at least) explicitly.
For example the idempotents can be written
$$1\qquad R_1(1)\qquad R_1(1)R_2(2)R_1(1)\qquad R_1(1)R_2(2)R_1(1)R_3(3)R_2(2)R_1(1)$$
and so on. You get another sequence of idempotents by applying the involution. Under the involution we have $R_i(k)\leftrightarrow R_i(-k)$. The three string idempotent is
$$\left(\frac{u_1}{\delta}\right)\left(1-\delta u_2\right)\left(\frac{u_1}{\delta}\right)$$
which gives the relation for the Temperley-Lieb algebra.<|endoftext|>
TITLE: Dimension of subalgebras of a matrix algebra
QUESTION [17 upvotes]: If $n$ is given and $A$ is a subalgebra of $M_n(\mathbb C)$, the algebra of $n \times n$ matrices with entries in the field of complex numbers, then what are the possible values of dimension of $A$ as a vector space over $\mathbb C$?

REPLY [4 votes]: Soit $E$ un $\mathbb C$-espace vectoriel de dimension $n$.
J'ai démontré entre autres les deux résultats suivants dans un article à paraître dans la revue française Quadrature :

On suppose que $k$ vérifie les inégalités $k \ge 2$ et 
$k^{2}\le n$. Soit $\mathcal{A}$ une sous-algèbre de $\mathcal{L}(E)$ qui
vérifie la relation $n^{2}-kn+k^{2}-k+1 < \dim \mathcal{A} < n^{2}-kn+n.$
Alors, $\mathcal{A}$ vérifie la relation $\dim \mathcal{A}=n^{2}-kn+k^{2}.$
Soient $n$ un entier naturel et $p$ un entier de l'intervalle $[0,n^{2}].$ On
suppose $p$ écrit sous la forme $p=n(n-k)+t,\ 0\le t \le n-1$. Alors il existe
une sous-algèbre de dimension $p$ dans $\mathcal M_n (\mathbb C )$ si
et seulement s'il existe une sous-algèbre de dimension $t$ dans $\mathcal M_k(\mathbb C)$.<|endoftext|>
TITLE: About the Gauss map of a surface in euclidean 3 space
QUESTION [7 upvotes]: Regarding the sphere as complex projective line (take $(0,0,1)$ as the infinite point), the Gauss map of a smooth surface in the 3 dimensional space pulls a complex line bundle back on the surface. 
My question is, what the bundle is? 
(In the trivial case, if the surface is sphere itself, the bundle is just the tautological line bundle.)
Does the chern class (Of course the first one) of this bundle depend on the embedding of the surface? 
(The Jacobian determinant of Gauss map is just the Gauss curvature, hence is intrinsic. Also its degree is the Euler $\chi$, so I ask for more...)
If yes, how much does the chern class/bundle reflect the geometry of embedding?
There may be something to make the question meaningless, such as there is no cannonical way to identify a sephere with the projective line... But as a beginner in learning geometry, I am still curious to it...
edit Thank Sergei for your answer, thank you

REPLY [4 votes]: Let me add something: As Sergei said, the pull-back bundle (without any geometric structure) will be the same. Nethertheless, there is more structure: Regrading your smooth embedded surface as a Riemann surface $M$ (induced from the metric or first fundamental form), you have the canonical bundle $K\to M$ on it (the bundle of complex linear 1-forms). This bundle is in a natural way a holomorphic bundle, holomorphic sections are exactly the closed complex linear forms. Now, on every Riemann surface, there are exactly $2^{2g}$ complex holomorphic line bundles $S\to M$ which satisfy $S^2=K$ holomorphically. These bundles are called spin-bundles.
It turns out that your pull-back bundle is a spin bundle of the Riemann surface $M.$ Moreover, the type of the spin-bundle tells you something about the way your surface is embedded. For example, the spin-bundle of an embedding has no global holomorphic section. Moreover two immersions are homotopic, iff the spin bundles are the same. There is a nice paper of Pinkall
("Regular homotopy classes of immersed surfaces") were all these questions are answered.<|endoftext|>
TITLE: Is Lusztig's conjecture solved?
QUESTION [12 upvotes]: What I said is Lusztig's conjecture about representation of quantum group at root of unity and representation of Lie algebra at positive characters.
It seems that Andersen-Jantzen-Soergel ever wrote a book on this conjecture. 
Is it solved? Any recent development? I am looking for reference talking about it. 
Thank you

REPLY [20 votes]: Geordie Williamson has found counterexamples to Lusztig's conjecture: 
http://people.mpim-bonn.mpg.de/geordie/Torsion.pdf
Apparently, any bound on the largest prime that breaks Lusztig's conjecture must be at least proportional to $n\log n$, and this is probably still far too optimistic; I think it's now believed that you'll need an exponential bound.<|endoftext|>
TITLE: When does a CW-complex of dimension 2 embedd in $R^4$ ?
QUESTION [13 upvotes]: Let X be a finite CW-complex of dimension two having just one 0-cell
(+ finitely many 1-cells + finitely many 2-cells).
Is it true that X can be embedded in $R^4$ ?
If true, is it due to Stallings ?
More generally, let Y be a finite CW-complex of dimension two
of which the 1-skeleton embeds in $R^2$;
is is true that Y can be embedded in $R^4$ ?
(Reminder : any finite CW-complex of dimesnion two embeds in $R^5$.)
Any comment, or even better a reference, would help.

REPLY [2 votes]: Shapiro's obstruction:
A. Shapriro, "Obstructions to the imbedding of a complex in Euclidean space, I. The first obstruction," Ann. of Math., 66 No. 2 (1957), 256--269.<|endoftext|>
TITLE: question about kahler cone of a compact kahler manifold
QUESTION [5 upvotes]: Hi to all!
I'm studying complex geometry from Huybrechts book "Complex Geometry"
and i have problems with an exercise, please can anyone help me?
I define the kahler cone of a compact kahler manifold X as the set 
$K_X \subseteq H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ 
of kahler classes. I have to prove that $K_X$ doesn't contain any line
of the form $\alpha + t \beta$ with $\alpha , \beta\in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ 
and $\beta\neq 0$ (i identify classes with representatives).
This is what i thought: i know that a form  $\omega \in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$
that is positive definite (locally of the form $\frac{i}{2}\sum_{i,j} h_{ij}(x)dz^i\wedge d \overline{z}^{j}$ and $(h_{ij}(x))$ is a positive definite hermitian matrix $\forall x\in X$) is the kahler form associated to a kahler structure. Supposing $\alpha$ a kahler class i want to show that there is a $t\in\mathbb{R}$ such that $\alpha + t \beta$ is not a kahler class. Since $\beta\neq0$ i can find a $t\in\mathbb{R}$ such that $\alpha + t \beta$ is not positive definite any more, now i want to prove that there is no form $\omega \in H^{(1,1)}(X)\cap H^2(X,\mathbb{R})$ such that $\omega=d\lambda$ with $\lambda$ a real 1-form and $\omega=\overline{\partial}\mu$ with $\mu$ a complex (1,0)-form (what i'd like to prove is: correcting representatives of cohomology classes with an exact form i don't get a kahler class). From $\partial\overline{\partial}$-lemma and a little work i know that $\omega=i\partial\overline{\partial}f$ with f a real function. And now (and here i can't go on) i want to prove that i can't have a function f such that $\alpha + t \beta+i\partial\overline{\partial}f$ is positive definite. 
Please, if i made mistakes, or you know how to go on, or another way to solve this, tell me.
Thank you in advance.

REPLY [4 votes]: Here is another try:
WLOG, we assume $\alpha$ is kahler, fix it as a metric on $M$.
Assume $\alpha+t\beta$ is kahler for every $t$. So
$\int(\alpha+t\beta)\wedge \alpha^{n-1}=\int
\alpha^n+t\int\beta\wedge\alpha^{n-1}>0$ for every $t$. It then
follows $\int\beta\wedge\alpha^{n-1}=0$. In a same manner, by considering $\int(\alpha+t_1\beta)\wedge(\alpha+t_2\beta)\wedge\alpha^{n-2}$, we have
$\int \beta^2\wedge\alpha^{n-2}=0$. By Lefschetz decomposition, we
can write $\beta=\beta_1+c\alpha$, where $\beta_1$ is a primitive
cohomology class. Then $\beta_1\wedge\alpha^{n-1}=0$. By the fact
$\int\beta\wedge\alpha^{n-1}=0$, we conclude $c=0$ and $\beta$
itself is primitive class. Then it is a contradiction that $\int
\beta^2\wedge\alpha^{n-2}=0$ unless $\beta=0$ by Hodge-Riemann
bilinear relation for primitive classes.<|endoftext|>
TITLE: "Physical" construction of nonconstant meromorphic functions on compact Riemann surfaces?
QUESTION [17 upvotes]: Miranda's book on Riemann surfaces ignores the analytical details of proving that compact Riemann surfaces admit nonconstant meromorphic functions, preferring instead to work out the algebraic consequences of (a stronger version of) that assumption.  Shafarevich's book on algebraic geometry has this to say:
A harmonic function on a Riemann surface can be conceived as a description of a stationary state of some physical system: a distribution of temperatures, for instance, in case the Riemann surface is a homogeneous heat conductor.  Klein (following Riemann) had a very concrete picture in his mind:

"This is easily done by covering the Riemann surface with tin foil... Suppose the poles of a galvanic battery of a given voltage are placed at the points $A_1$ and $A_2$.  A current arises whose potential $u$ is single-valued, continuous, and satisfies the equation $\Delta u = 0$ across the entire surface, except for the points $A_1$ and $A_2$, which are discontinuity points of the function."
Does anyone know of a good reference on Riemann surfaces where a complete proof along these physical lines (Shafarevich mentions the theory of elliptic PDEs) is written down?  How hard is it to make this appealing physical picture rigorous?  (The proof given in Weyl seems too computational and a little old-fashioned.  Presumably there are now slick conceptual approaches.)

REPLY [2 votes]: The answers to this previous question are relevant. To fill in some gaps, that question is about building functions on a disc with specified Laplacian. As Tim Perutz says, you want to generalize that to building functions on Riemmann surface whose Laplacian is a specified $2$-form of integral zero.<|endoftext|>
TITLE: Is there an "adjacency matrix" for weighted directed graphs that captures the weights?
QUESTION [9 upvotes]: I am currently writing up some notes on the max-plus algebra $\mathbb{R}_{\max}$ (for those that have never seen the term "max-plus algebra", it is just $\mathbb{R}$ with addition replaced by $\max$ and multiplication replaced by addition. For some reason, authors whose main interest is control-theoretic applications never seem to use the term "tropical", and I have been reading from such authors). There is a nice result which says the following: 
$\textbf{Theorem.}$ Let $G$ be a directed graph on $n$ vertices such that each arc $(i,j)$ in $G$ has a real weight $w(i,j)$. Define the $n \times n$ matrix $A$ by $(A)_{ij} = w(i,j)$ if $(i,j)$ is an arc, and $(A)_{ij} = -\infty$ otherwise. Then for each $k > 0$, the maximum weight of a path of length $k$ from vertex $i$ to vertex $j$ is given by $(A^{\otimes k})_{ij}$ (here, $A^{\otimes k}$ is just the $k$th power of $A$, computed using the $\mathbb{R}_{\max}$ operations).
This result is certainly analagous to the standard result that the $ij$-entry of the $k$th power of the adjacency matrix gives the number of walks of length $k$ from vertex $i$ to vertex $j$. When writing up my notes I found myself claiming that the above theorem provides some evidence that $\mathbb{R}_{\max}$ is in fact a natural setting in which to study weighted digraphs, since there is no natural definition of an ``adjacency matrix'' of a weighted digraph (in the usual setting of $\mathbb{R}^{n \times n}$) that gives useful information about the weights. This seemed like too strong of a claim, especially since I am no expert in networks or combinatorial optimization. This leads to the question:
$\textbf{Question.}$ Is there a standard matrix (in $\mathbb{R}^{n \times n}$) associated with a weighted digraph that is analogous to the adjacency matrix and captures in a useful way the weights of the arcs?
$\textbf{Clarification:}$ By "analogous to the adjacency matrix" I mean a matrix that is defined simply in terms of the graph (vertices, arcs, and weights). I imagine there are all sorts of matrices associated to weighted digraphs so that computers can be used to analyze networks. But I am  not interested in, say, a matrix that requires a complicated algorithm to compute its entries.

REPLY [5 votes]: $\textbf{Question.}$ Is there a standard matrix (in $\mathbb{R}^{n \times n}$) associated with a weighted digraph that is analogous to the adjacency matrix and captures in a useful way the weights of the arcs?

Yes, and in fact it is essentially the matrix that you define in the theorem that you state. (Typically one sets A[i,j] = $\infty$, since this matrix is used to help find shortest paths in a graph.) This is generally known (at least in the algorithms and data structures community) as the "weighted adjacency matrix." 
I don't know what else you would want from a matrix that is supposed to represent a weighted digraph.
Concerning Qiaochu's comment: In fact there is an algorithm for computing the "max-plus matrix product" that uses precisely this trick, and relies on the existence of fast matrix multiplication over rings. 

Theorem [Alon, Galil, Margalit]: If $n \times n$ matrix multiplication over the integers can be done in $O(n^{\omega})$ arithmetic operations, then the max-plus matrix product of two $n \times n$ matrices with entries in the range $[-M,M]$ can be computed in about $O(M n^{\omega})$ bit operations.

Given a matrix $A$ with weights, the idea is to make a matrix $B[i,j] = (n+1)^{A(i,j)}$, compute $B \cdot B$ (over the integers) and check the high-order terms to find the largest $k$ such that $A(i,k) + A(k,j)$ is maximized.<|endoftext|>
TITLE: Computing stable reduction of finite covers of curves in practice
QUESTION [7 upvotes]: The general theory is described in various places, but I'll be following (sketchily) the description of this process appearing in section 1 of Bouw and Wewers' "Reduction of covers and Hurwitz spaces".
Background
Let $R$ be a complete DVR, and $K$ its function field. Say we have a $G$-Galois map of (smooth projective) curves over $K$, $f:Y_K \rightarrow X_K$. Assume also that the order of $G$ is not divisible by the characteristic of the residue field of $R$. After replacing $K$ by a finite extension we may assume the ramification points are $K$-rational, and and the smooth stably marked curve $(Y_K, D)$ (where $D$ is the ramification divisor) can be defined over $R$: $(Y_R,D_R)$. There is some variation between different papers as to what "stably marked curve" means, but I think I mean minimal semi-stable, which happens to be stable (am I wrong? correct me if I am.) If we quotient $Y_R$ by the action of $G$ we should get a semi-stable curve, which we shall denote: $X_R$. This may no longer be a minimal semi-stable model of $X_K$ (but it's definitely a semi-stable model of it).
If I understand the theory correctly, if we assume that $K$ is such that we have an $R$-model of $X_K$ which is semi-stable and such that the branch points specialize to different points, then it must be $X_R$ as constructed above.
Question
In order to understand this better, I wish to have some concrete computations under my belt. Let's try a simple yet interesting example:
Let $R:= \mathbb{C}[[t]]$, $X_{\mathbb{C}((t))}:=\mathbb{P}_{\mathbb{C}((t))}^1$ (with parameter $x$), and let $f$ and $Y _ { \mathbb{C} ((t))}$ be given affinely by $y^2=x(x-t)$. (So $f$ is the projection to $x$, and $Y _ { \mathbb{C} ((t))}$ is a $\mathbb{P}_{\mathbb{C}((t))}^1$ with parameter $y/x$. In other words the function field of $X$ is $\mathbb{C}((t))(x)$ and the function field of $Y$ is $Quot(\mathbb{C}((t))[x,y]/(y^2-x(x-t)))$, which, in turn, is equal to $\mathbb{C}((t))(y/x)$.)
If we let $X_{\mathbb{C}[[t]]}:=\mathbb{P}_{\mathbb{C}[[t]]}^1$, then this is clearly a semi-stable curve, and the branch points (in $X _ { \mathbb{C}((t))}$), which were 0 and t, specialize to the same point. But I want to guarantee that this would be the quotient of the stably marked curve on top. According to the last paragraph in the background section, I would get this guarantee if the branch points (interpreted in $X_{\mathbb{C}[[t]]}$) would specialize to different points. So instead choose $X_{\mathbb{C}[[t]]}$ to be the blow up of $\mathbb{P} _ { \mathbb{C} [[t]]}^1$ at $t=x=0$. If we work affinely, this would be: $\mathbb{C}[[t]][x,z]/(xz-t)$. The question now is: how do I find the stable reduction upstairs, and the map between them? How do I finish this example?

REPLY [4 votes]: If the normalization of $X_R$ in the function field of $Y_K$ has no vertical ramification, then I think so. You might have a look here, § 2 and 3.<|endoftext|>
TITLE: Seeing math when viewing abstracts on arxiv.org
QUESTION [12 upvotes]: One very enjoyable feature of mathoverflow is that the math just works. You enter it using the usual LaTeX, and then the jsMath magic does the rest. One of the most frequent activities for many mathematicians is checking abstracts on arXiv. So you open an abstract and you see all these dollar signs which you have to mentally decipher.
Question: Is there a way to set up jsMath, Firefox, etc., so that you can see formulas when viewing abstracts on arxiv.org?
This may be a meta question, but who reads meta threads, right? And there is a high concentration of experts here who may be able to answer this. That would improve quality of life for many mathematicians, I am sure.
Notes on the comments:

Yes, people should make an effort and write abstracts on arXiv without using math. symbols, preferably. But the easiest thing to do, really, is to reuse your article abstract for the arXiv abstract, and that's what people do. And you can easily use some math symbols in your article abstract.
Is it a question for arXiv people and not for MO? Maybe. But this is a very active community of people who encounter this problem on a daily basis. I hoping there is someone clever enough that solved this problem for themselves already.

REPLY [5 votes]: You may be interested in looking at the following greasemonkey script:
http://www.gold-saucer.org/mathml/greasemonkey/
It doesn't look as though it's been worked on recently but that doesn't stop it being usable. The basic idea is to overlay a simplified latex->MathML converter on top of a webpage.  Since it is a greasemonkey script, it is entirely user-controlled and can be applied to any webpage, arxiv or otherwise.
Of course, there is always the problem of non-standard macros in abstracts and I completely agree with the sentiment that abstracts should be 100% legible without requiring extra parsing.
(In addition to posting this answer, I am add my vote to the list of "to close" as I agree with the various sentiments on that line in the comment thread.  That I'm answering the question is not paradoxical since this isn't actually an answer to the question that was asked.  I'm not convinced that I would want to see the question to which this is the answer also posted on MO, but think that the easiest way to prevent it being asked is to answer it here.  I'm community-wikifying this answer not because I want others to be able to edit it, but to underline my opinion by forgoing any reputation for this answer (not that I assume it would actually get any!).)<|endoftext|>
TITLE: Relation between Hecke Operator and Hecke Algebra
QUESTION [20 upvotes]: In the study of number theory (and in other branches of mathematics) presence of Hecke Algebra and Hecke Operator is very prominent.
One of the many ways to define the Hecke Operator $T(p)$ is in terms of double coset operator corresponding to the matrix $ \begin{bmatrix} 1 & 0  \\ 0 & p \end{bmatrix}$ . 
On the other hand Hecke Algebra $\mathcal{H}(G,K)$ associated to a group $G$ of td-type ( topological group, such that every neighborhood of unity contains a compact open subgroup), where $K$ is a compact open subgroup of $G$ is defined as the space of locally constant compactly supported $K$ bi-invariant functions on $G$. Convolution product makes it an associative algebra. 
I was told that  the hecke algebra $\mathcal{H}(Gl(2,\mathbb{Q}_p) , Gl(2,\mathbb(Z)_p))$ corresponds to the classical algebra of hecke operators attached to $p$ via Satake Isomorphism Theorem. Using Satake Isomorphism theorem I can show $\mathcal{H}(Gl(2,\mathbb{Q}_p) ,Gl(2,\mathbb(Z)_p))$ is commutative and finitely generated over $\mathbb{C}$. 
So my question is how one uses Satake Isomorphism Theorem (or otherwise) to see this? 
And secondly in general what is the relation between hecke operators and hecke algebra?

REPLY [24 votes]: The fact that Hecke operators (double coset stuff coming from $SL_2(\mathbf{Z})$ acting on modular forms) and Hecke algebras (locally constant functions on $GL_2(\mathbf{Q}_p)$) are related has nothing really to do with the Satake isomorphism. The crucial observation is that instead of thinking of modular forms as functions on the upper half plane, you can think of them as functions on $GL_2(\mathbf{R})$ which transform in a certain way under a subgroup of $GL_2(\mathbf{Z})$, and then as functions on $GL_2(\mathbf{A})$ ($\mathbf{A}$ the adeles) which are left invariant under $GL_2(\mathbf{Q})$ and right invariant under some compact open subgroup of $GL_2(\widehat{\mathbf{Z}})$. 
Now there's just some general algebra yoga which says that if $H$ is a subgroup of $G$ and $f$ is a function on $G/H$, and $g\in G$ such that the $HgH$ is a finite union of cosets $g_iH$, then you can define a Hecke operator $T=[HgH]$ acting on the functions on $G/H$, by $Tf(g)=\sum_i f(gg_i)$; the lemma is that this is still $H$-invariant.
Next you do the tedious but entirely elementary check that if you consider modular forms not as functions on the upper half plane but as functions on $GL_2(\mathbf{A})$, then the classical Hecke operators have interpretations as operators $T=[HgH]$ as above, with $T_p$ corresponding to the function supported at $p$ and with $g=(p,0;0,1)$. Because the action is "all going on locally" you may as well compute the double coset space locally, that is, if $H=H^pH_p$ with $H_p$ a compact open subgroup of $GL_2(\mathbf{Q}_p)$, then you can do all your coset decompositions and actions locally at $p$.
Now finally you have your link, because you can think of $T$ as being the characteristic function of the double coset space $HgH$ which is precisely the sort of Hecke operator in your Hecke algebra of locally constant functions. Furthermore the sum $f(gg_i)$ is just an explicit way of writing convolution, so everything is consistent.
I don't know a book that explains how to get from the classical to the adelic point of view in a nice low-level way, but I am sure there will be some out there by now. Oh---maybe Bump?<|endoftext|>
TITLE: What  does «generic» mean in algebraic geometry?
QUESTION [13 upvotes]: As a beginner, when I read some books in algebraic geometry such as the book complex projective variety by Mumford,I found a lot of "generic" object.
Could any one tell me how to understand "generic"?

REPLY [4 votes]: Generic often refers to true in a Zariski open dense set, i.e. true outside some "small" set of proper codimension (in the Zariski topology). It is something like the algebro-geometric version of "almost everywhere" or "residual" in analysis and topology.
There is a general fact in algebraic geometry, already mentioned in previous answers, that whenever a constructible set (i.e., one obtained from closed sets using a finite number of boolean operations, at least one when has suitable noetherian hypotheses) contains the "generic point" of an irreducible scheme, it is generic in the sense of containing a Zariski open (and thus dense, under irreducibility hypotheses) set.  One simple example of this phenomenon is the following: let $M$ be a matrix with coefficients in $k(T)$ for $k$ a field.  Then the rank of $M$ ("at the generic point") is equal to the rank of almost all the specializations $M(t)$ obtained by substituting $T \to t, t \in k$ (i.e. in a "generic set").<|endoftext|>
TITLE: Category of copresheaves over commutative monoids
QUESTION [8 upvotes]: Let C be a symmetric monoidal category. Let Comm(C) be the category of commutative monoids in C. Consider the topos X = CoPSh(Comm(C)) of covariant functors from Comm(C) to the category Set of sets.
Which extra data do we have to specify on the topos X such that we can recover (up to some notion of equivalence) the essential structure of the underlying site? For example, every representable copresheaf F = Hom(A, _) has a category of modules attached to it, which is a kind of extra data.
[I know I am a bit unprecise here with what I mean by essential, so making this precise could also be part of the answer to my question.]
One could also rephrase this question as follows: Given a topos X that satisfies Giraud's axioms, one can extract a site such that X is the Grothendieck topos over this site. Which extra data do we need to impose on X such that we can recover X as a Grothendieck topos over a site that is (the dual) to commutative monoids in a symmetric monoidal category.
When I write down this question, I have the following example in mind: Let C be the category of abelian groups. Then X is the topos of presheaves on the category of affine schemes, which gives rise to algebraic geometry. X possesses a commutative ring object, namely the affine line A1 and one has the stack of categories of quasi-coherent sheaves over objects of X. Taking the idea of (Grothendieck) topoi seriously, one should be able to forget about C and just consider the topos X (i.e. without a fixed base site). Of course, one has to remember (at least) A1. This allows to recover the stack of categories of quasi-coherent modules. 
Added for clarification:
But what if C is not the category of abelian groups? In this case, X = CoPSh(Comm(C)) also carries a stack QCoh of categories of quasi-coherent modules as follows: Let F be an object of X, i.e. F is a covariant functor from Comm(C) to Set. An object M of the category QCoh(F) maps a morphism a: Hom(A, _) -> F to an A-module M(a) in C together with
natural isomorphisms M(b) = M(a) ⊗A B for all morphisms a -> b in Comm(C). [It is here where the category C itself comes in.]
What is the minimal amount of data we need on X so that X is equivalent to sheaves on a site S such that the dual of S is of the form Comm(C') with C' giving rise to a somewhat equivalent stack of categories of quasi-coherent modules.

REPLY [2 votes]: I'm not sure if this is part of your question, but I just want to observe that there's no hope of recovering C from Comm(C), let alone from X = CoPSh(Comm(C)).  
For example, take any commutative monoid and regard it as a discrete symmetric monoidal category C.  In other words, the objects of the category are the elements of the monoid, the only morphisms are identities, and the tensor product in the category is the multiplication in the monoid.  Then there is precisely one commutative monoid in C (namely, the unit object), and it has no endomorphisms.  Hence Comm(C) is the terminal category.  This is true for whatever commutative monoid you started with.  So C can't be recovered from Comm(C).  
Or were you only trying to recover Comm(C), not C, from X?<|endoftext|>
TITLE: Ackermann-related function
QUESTION [12 upvotes]: While doing some research, I came up with a problem of proving that 
$ f(a,b,c)=\begin{cases}1 &\text{ if }A(a,b)=c\\ \\\\ 0 &\text{ otherwise }\end{cases} $
is primitively recursive ($A$ is the Ackermann's function).
Any references, ideas or proofs?
(This may not be a good MO question, but since the participants in problem-solving sites listed in MO posting FAQ failed to solve it - I posted it there before - I was hoping for a solution here.)

REPLY [15 votes]: The key to do this is to realize the difference between computation and verification. Although computing value A(a,b) of the Ackermann function cannot be done primitive recursively, verifying whether a proposed number c is the correct value of A(a,b) can be done primitive recursively. (Note that computation and verification is also what distinguishes P and NP.)
In this case, the fact that this can be done hinges on the strong monotonicity properties of Ackermann's function. Indeed, if A(a,b) = c then all the previous values of A needed to compute A(a,b) are bounded by c. Therefore, the search for a valid computation verifying that A(a,b) = c can be bounded by a primitive recursive function B(a,b,c). Knowing this function B we can take a proposed value c for A(a,b), compute the bound B(a,b,c) and search up to this bound for a valid computation verifying that c is indeed equal to A(a,b). If no such computation is found we return 1, otherwise we return 0.
To make this a little more specific, let's say that computation for A(a,b) = c is a (coded) finite sequence of triples (ai,bi,ci) which ends with the triple (a,b,c). Each such triple codes the fact that A(ai,bi) = ci. The computation is valid when each such triple follows from previous triples and the rules for computing the Ackermann function. (Checking this is obviously primitive recursive.) For example a valid computation for A(1,1) = 3 is the sequence (0,1,2), (0,2,3), (1,0,2), (1,1,3). 
Using the rules for computing Ackermann function and our specific method for coding finite sequences, we can compute an explicit bound B(a,b,c) on a valid computation verifying A(a,b) = c. The verification that B is primitive recursive should be straightforward if our coding of finite sequences is reasonable. Therefore, verifying A(a,b) = c can be done by a primitive recursive computation.
(The details of the last paragraph are best carried out in the privacy of one's office.)

REPLY [8 votes]: Here's a sketch of an argument which I expect could be made into a proof.
The key fact is that the Ackermann function fails to be primitive recursive only because it grows so quickly.  More formally:
Claim.  There exists a Turing machine T and a primitive recursive function f(a, b, c) (which is an increasing function of c) such that on input (a, b), T computes A(a, b) in at most f(a, b, A(a, b)) steps.
"Proof".  Starting with the expression "A(a, b)", repeatedly expand terms of the form A(x, y) with the recursive definition, but do not simplify any of the resulting additions.  The length of the string increases at every step, if we agree that the symbol "+" is "longer" than "A".  The resulting string is a formal sum of positive integers, so its length is bounded by (a multiple of) A(a, b); hence the number of steps is also bounded above in terms of A(a, b), and we may perform each step in time polynomial in A(a, b).
Now, we can simulate a given Turing machine for a fixed number of steps using a primitive recursive function.  We may therefore compute the graph of the Ackermann function with a primitive recursive function as follows:  Given a, b, c,

Compute f(a, b, c).
Simulate the Turing machine T on input (a, b) for f(a, b, c) steps.
If T has halted, then return whether c is equal to the output of T.  If T has not halted, then c < A(a, b) so return false.<|endoftext|>
TITLE: How do you recover the structure of the upper half plane from its description as a coset space?
QUESTION [11 upvotes]: This is maybe a dumb question.  $SL_2(\mathbb{R})$ has a natural action on the upper half plane $\mathbb{H}$ which is transitive with stabilizer isomorphic to $SO_2(\mathbb{R})$.  For this reason, people sometimes write $\mathbb{H}$ as the coset space $SL_2(\mathbb{R})/SO_2(\mathbb{R})$.  
Now, it's clear how this description recovers the topology of $\mathbb{H}$: it's just the quotient topology.  But can you recover either the Riemann surface structure or the hyperbolic metric on $\mathbb{H}$ from this description?  How much of the structure of $SL_2(\mathbb{R})$ and $SO_2(\mathbb{R})$ do you need to do this, if it's possible?

REPLY [4 votes]: Edit: I should have put a short version of the answer in the beginning, so here is how the various structures are recovered.  To get a smooth manifold structure on the quotient, you use the fact that $SL_2(\mathbb{R})$ is a real Lie group and $SO_2(\mathbb{R})$ is a closed subgroup.  To get a hyperbolic structure, you use the fact that $SL_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (n,1) for some n (giving a transitive action on hyperbolic n-space).  To get a complex structure, you use the fact that $SL_2(\mathbb{R})$ is isomorphic to an orthogonal group of signature (2,m) for some m (giving an action on a hermitian symmetric space).
As others have noted, you can get a bijection on points using the Iwasawa decomposition, and you can get a hyperbolic structure using the exceptional isomorphism $PSL_2(\mathbb{R}) \cong SO_{2,1}^+(\mathbb{R})$.  First, I'd like to clean up the Iwasawa treatment a bit.  Any element of $SL_2(\mathbb{R})$ can be uniquely decomposed as BK, where K is a rotation and B is upper triangular with positive diagonal.  Any rotation K fixes i, so we should consider what elements B do.  A bit of fiddling shows that $\begin{pmatrix} \sqrt{y} & x/\sqrt{y} \\ 0 & 1/\sqrt{y} \end{pmatrix} \cdot i = x+iy$.
We can view the exceptional isomorphism in another way that makes the complex structure more apparent, by viewing the hyperbolic plane as the Grassmannian $O_{2,1}(\mathbb{R})/(O_2(\mathbb{R}) \times O_1(\mathbb{R}))$.  From the standpoint of special relativity, this is the space of timelike lines through the origin in $\mathbb{R}^{2,1}$.  Taking a quotient of the total space of these lines (minus origin) by positive rescaling, we find that this space is isomorphic to the space of pairs of antipodal points of norm -1.  In particular, we have an isomorphism of the Grassmannian with the quotient of the hyperboloid with two sheets (i.e., solutions of the equation $x^2 + y^2 - z^2 = -1$) by the antipodal automorphism.
One way to explain the origin of the complex structure is by the fact that all Grassmannians of the form $O(2,n)/(O(2) \times O(n))$ are hermitian symmetric spaces, and the hyperbolic plane is just the case $n=1$.  The 2 in $O(2)$ is essential, because the orthogonal group action is what yields the ninety degree rotation in the tangent space of any point, and this is what endows the quotient with an almost complex structure.  If you want to see more about hermitian symmetric spaces than the Wikipedia blurb, I recommend looking in chapter 1 of Milne's introduction to Shimura varieties.
Finally, I'd like to point out Deligne's description of the upper half plane as a moduli space of structured elliptic curves.  Points on H parametrize elliptic curves with an oriented basis of first homology (as mentioned a few times in our class).  If you want to say it is a fine moduli space, you need a functor that it represents, and it is unfortunately a bit complicated.  The functor takes as input the category of complex analytic spaces, and for any such space S, it gives the set of isomorphism classes of elliptic curves over S (i.e., diagrams $E \underset{\pi}{\leftrightarrows} S$ of complex analytic spaces, where $\pi$ is smooth and proper with one-dimensional genus one fibers and the leftward map is a section) equipped with an isomorphism $R^1\pi_*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z} \times \mathbb{Z}}$ that induces the canonical identity $R^2\pi_*\underline{\mathbb{Z}} \cong \underline{\mathbb{Z}}$ on exterior squares.  Here, the underscore indicates a constant sheaf.  The functor also takes morphisms to "the evident diagrams".  To be honest, I have never seen a complete proof that this functor is represented by the complex upper half plane, although it seems to be more a question of doing lots of writing than an honest theoretical problem.  You can probably do it using the fact that H is a classifying space of polarized Hodge structures, as Kevin Buzzard mentioned in the comments.<|endoftext|>
TITLE: The Symmetry of a Soccer Ball
QUESTION [14 upvotes]: Let $P$ be a polyhedron which satisfies the following three conditions:

$P$ is built out of regular hexagons and regular pentagons.
Three faces meet at each vertex.
$P$ is topologically a sphere.

An easy Euler characteristic argument tells you that $P$ has exactly twelve pentagonal faces.
An example of a polyhedron like this is a truncated icosahedron (soccer ball for those of us in the States, football for everyone else). In this case, the pentagonal faces are arranged with some nice symmetry, and the polyhedron has icosahedral symmetry.
Another (trivial) example is the regular dodecahedron, which again has icosahedral symmetry.
Here's my question: Is this symmetry forced? What, if anything, can be said in general about the symmetry of a polyhedron which satisfies the above three conditions? 
Edit: Since the discussion below points out that there are precisely two polyhedra which satisfy the above conditions, a suitable evolution of the question, which has already begun to be discussed below, is this: What symmetry groups can a polyhedron have if one or more of the above conditions are relaxed?

REPLY [3 votes]: If you are willing to relax the trivalency requirement (and not require convexity, which was not one of the stated constraints), you can make all sorts of polyhedra, using only regular pentagons and hexagons, which have all sorts of symmetry. 
For example, start with two truncated icosahedra, and remove one hexagon from each of them. Now you can glue them together along the removed hexagons (matching up pentagonal and hexagonal faces), forming a new polyhedron with 3 and 4-valent vertices and considerably less symmetry than what you started with. Continuing, you could make long rod-shaped polyhedra, or you could make some spiky polyhedron by replacing multiple hexagons with other truncated icosahedra.<|endoftext|>
TITLE: Problems concerning subspaces of $M_n(\mathbb{C})$
QUESTION [19 upvotes]: Let $M_n(\mathbb{C})$ denote the n times n matrices over the complex number field. N be a subspace of $M_n(\mathbb{C})$.

If all the matrices in N are non-invertible , what is the maximum the dimension of N can be?
If all the matrices in N commute with each other, what is the maximum the dimension of N can be?
If all the matrices in N are nilpotent, what is the maximum the dimension of N can be?
If all the non-zero matrices in N are invertible, what is the maximum the dimension of N can be?

REPLY [6 votes]: For problem 4 over the real field, the answer is the Radon-Hurwitz function at $n$. See for instance Petrovic, "On nonsingular matrices and Bott periodicity." The Radon-Hurwitz function is defined to be $\rho(n)=8a+2^b$, where the largest power of 2 dividing $n$ is $2^{4a+b}$, $a\geq 0$, $0\leq b\leq 3$.<|endoftext|>
TITLE: Decomposition of Tate-Shafarevich groups in field extensions
QUESTION [12 upvotes]: Suppose $E/\mathbb{Q}$ is an elliptic curve with rank zero.  According to the conjecture of Birch and Swinnerton-Dyer, the special value $L(1,E_{/\mathbb{Q}})$ should be equal (up to some harmless factors) to the order of the Tate-Shafarevich group Sha$(E/\mathbb{Q})$.  Now, suppose $\chi$ is a Dirichlet character, and $K/\mathbb{Q}$ is the cyclic extension cut out by $\chi$.  My question is:
What is the precise (conjectural) relation between the individual special values $L(1,E\times\chi^i)$ and Sha$(E/K)$?
More precisely, we have $L(s,E_{/K})=\prod_{i=0}^{\mathrm{ord}(\chi)} L(s,E\times \chi^i)$.  If $E/K$ has rank zero, is there a decomposition of Sha$(E/K)$ with respect to an action of $\mathrm{Gal}(K/\mathbb{Q})$ such that the individual pieces of this decomposition have orders given by the individual values $L(1,E \times \chi^i)$?

REPLY [8 votes]: Fix a prime p which doesn't divide the degree of K over ${\mathbb Q}$, and let ${\mathcal O}$ denote the ring of integers of ${\mathbb Q}_p(\chi)$ i.e. an extension of ${\mathbb Q}_p$ containing the values of $\chi$.  Then the group algebra ${\mathcal O}[G]$ decomposes into a direct sum of 1-dimensional pieces over ${\mathcal O}$, one for each power of $\chi$. 
Then $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ being an ${\mathcal O}[G]$-module inherits such a decomposition.  Concretely, the $\chi^i$-component of $Sha(E/K)[p^\infty] \otimes {\mathcal O}$ is the subset where $G$ acts by $\chi^i$.  
This $\chi^i$-component is then a reasonable candidate to compare to the $p$-adic valuation of the algebraic part of $L(E,\chi^i,1)$.  
Some further comments:
-Note that extending scalars to ${\mathcal O}$ increases the size of the modules so this has to be taken into account.
-The component corresponding to the trivial character is the invariants under $G$, and when $G$ has size prime-to-p this is simply $Sha(E/{\mathbb Q})[p^\infty] \otimes {\mathcal O}$ (which is good).
-To make a precise relationship between the $L$-value and Sha, you need to take into account the other terms in BSD.  Namely:
*The torsion-term should work out exactly as above (decomposing into $\chi$-components).  
*The periods have to be considered (which was ignored above in my vague phrase "the algebraic part of").  
*The Tamagawa numbers give me pause -- possibly there is an analogous $\chi$-decomposition, but I don't see it now.  
*Lastly, if K is ramified over ${\mathbb Q}$ then the discriminant of K appears in the BSD quotient (in the denominator which increases the size of Sha).  To handle this, I imagine what should be done is that rather then considering the L-value alone, consider the L-value times the Gauss sum of the character.  (By the conductor-discriminant formula this should give exactly the extra powers of p needed.)

REPLY [6 votes]: There is the Stickelberg element $\Theta$ considered by Mazur and Tate which gives more information in this direction. It is conjectured to be in the Fitting ideal and hence in the annihilator of the Selmer group. 
To simplify the notation suppose $K/\mathbb{Q}$ is of odd prime degree $d$. Then evaluating a non-trivial character $\chi$ on $\Theta$ gives
$$\overline{\chi}(\Theta) = \frac{G(\overline{\chi})\cdot L(E,\chi,1)}{\Omega} \qquad\text{ in }\qquad\mathbb{Q}[\chi],$$
I believe. Here $G(\chi)$ is the Gauss sum and $\Omega$ is the Néron period. The element $\overline{\chi}(\Theta) \in \mathbb{Q}[\chi]=\mathbb{Q}[\zeta_d]$ is of norm
$$\frac{\sqrt{\Delta_K}\cdot L(E/K,1)}{\Omega^d}\cdot \frac{\Omega}{L(E/\mathbb{Q},1)},$$
which has a conjectural BSD-expression.
Suppose that the Tamagawa numbers of $E/K$ are trivial and that $E(K)$ is the trivial group. Then $\Theta\in\mathbb{Q}\bigl[\textrm{Gal}(K/\mathbb{Q}]\bigr]$ annihilates the Tate-Shafarevich group of $E/K$. In other words one should look at the prime decomposition of the fractional ideal generated by $\overline{\chi}(\Theta)$ of $\mathbb{Z}[\zeta_d]$ to extract information about the size of the $\chi$-parts of Sha. So that will link the number $L(E,\chi,1)$ to Sha, but only up to units in $\mathbb{Z}[\zeta_d]$. It seems difficult to pin down the correct unit.
The presence of Tamagawa numbers and torsion points in $E(K)$ will make this less precise and maybe there is no easy description.<|endoftext|>
TITLE: Homotopy commutativity of the cup product
QUESTION [13 upvotes]: Where can I find explicit formulas for the higher homotopies, which exhibit the cup product (in singular simplicial cohomology, say) as homotopy commutative on the cochain level?  Same question in Cech cohomology.

REPLY [12 votes]: These operations in the singular setting were fully and explicitly developed and generalized beautifully by McClure and Smith (who also credit Benson and Milgram) in their paper "Multivariable cochain operations and little $n$-cubes":
http://arxiv.org/pdf/math.QA/0106024.pdf<|endoftext|>
TITLE: Homotopy colimits/limits using model categories
QUESTION [17 upvotes]: A homotopy (limits and) colimit of a diagram $D$ topological spaces can be explicitly described as a geometric realization of simplicial replacement for $D$. 
However, a homotopy colimit can also be described as a derived functor of limit. A model category structure can be placed on the category $\mathrm{Top}^I$, where $I$ is a small index category, where weak equivalences and fibrations are objectwise, so that $\mathrm{colim} : \mathrm{Top}^I \leftrightarrow \mathrm{Top} : c$ form a Quillen pair, where $c$ is the diagonal functor taking an object $A$ to the constant diagram at $A$. Then the homotopy colimit can be described as a derived functor for $\mathrm{colim}$: take a cofibrant replacement $QD$ for a diagram $D$, then compute $\mathrm{colim}(QD) = \mathrm{hocolim}(D)$. It turns out that two cofibrant replacements will give weakly equivalent homotopy colimits. As such, you would suspect that this choice is not really important.
This leaves two questions: firstly, is it necessary in most cases to construct homotopy colimits explicitly, or are its properties as a homotopical functor enough? Secondly, do any problems arise from the fact that homotopy colimit is well-defined only up to weak equivalence (through the derived functor angle)? Do cases ever arise where a more canonical definition is required? 
Context: I am reading through Goodwillie's "Calculus II: Analytic Functors." There the explicit simplicial construction is used, and in particular it is needed that certain maps from holim(D) are fibrations (Definition 1.1a, for example). However, being a fibration is not invariant under weak equivalence. Does this reflect that properties of this particular choice of holim are needed, or that the paper itself is too rigid? Can these arguments be made with a non-canonical choice of holim? 
I apologize ahead of time for the vague question: I've been trying to read up in this subject area for a few months now, and this has been a stumbling block.

REPLY [14 votes]: In the context of Goodwillie's paper, he's got an explicit natural transformation $f:holim_I(X)\to holim_J(X|_J)$, where $X:I\to Top$ is a functor to spaces, and $J\subset I$ is a subcategory of $I$.  With the construction of holim he's using, this map is always a fibration.
What if you tried to use a different construction of holim?  Then maybe you get a map $f'$ which  is not a fibration anymore.  In that case, you could still have taken the homotopy fiber of $f'$, and this would be a notion which is invariant under weak equivalence.  That is, you could (functorially) replace $f'$ with a fibration via the path construction, and take the fiber of that.
Of course, the homotopy fiber is exactly the thing he wants here.  In fact, he's manufactured the situation exactly so that the homotopy fiber he wants is just the fiber of this map. 
(It's worthwhile to note that in his setting, the category $I$ (which is a cube) has an initial object $\varnothing$.  This means that the evident map $holim_I(X)\to X(\varnothing)$ is a weak equivalence.  In other words, $holim_I(X)$ is really just $X$ evaluated at $\varnothing$, but modified so that it maps to (and fibers over) $holim_J X|_J$.)<|endoftext|>
TITLE: Approximately holomorphic functions
QUESTION [6 upvotes]: In real analysis one can define something known as the approximative derivative of a function. See here eg Roughly speaking one asks that the limit of the difference quotient exists as long as h goes to zero while only taking values in some subset that is sufficiently dense. 
Does anyone know if this concept has been studied for complex-valued valued functions of a complex variable? The basic definition should go through without problem so it should make sense to speak of an approximately holomorphic function as one that has an approximate complex derivative at every point of some open set. It would be interesting how much of the classical complex analysis one could generalize. Even knowing whether there exists functions that are approximately holomorphic but not holomorphic in the normal sense would be interesting.

REPLY [6 votes]: Men'shov proved in 1936 that if $f\colon D\to\mathbb C$ is continuous and approximately differentiable outside of a countable set, then it is holomorphic in $D$ 1 (Russian original with French summary). In the same paper he gives an example, attributed to Lusin, which shows that continuity cannot be dropped entirely even if the function is approximately differentiable at every point. Let $\varphi\colon\mathbb C\to\mathbb C$ be an entire function that tends to $0$ as $z\to\infty$ within any sector $\{|\arg z|<\pi-\epsilon\}$ (approximation theory can be used to create such examples). Then $f(z)=z\varphi(1/z)$ has an approximate derivative everywhere in $\mathbb C$, but of course it is not differentiable or even continuous at $0$. 
Men'shov asked if continuity can be replaced with boundedness. This was answered affirmatively by Telyakovskii 50 years later 2. In fact, boundedness can be weakened to logarithmic integrability. 
Brodovich 3 proved that an injective function with an approximate derivative at every point is holomorphic. (MR review omits the injectivity assumption). A survey of results in this area was written by Dolzhenko 4, but it predates the work of Brodovich.<|endoftext|>
TITLE: Free actions of finite groups on products of even-dimensional spheres
QUESTION [5 upvotes]: Suppose a finite 2-group G acts freely on X = $\prod_{i=1}^k$ *S*$^{2n_i}$, a product of k even-dimensional spheres, k > 2. Is it possible for G to be non-abelian? What if we additionally assume that spheres in the product are equidimensional?
Some comments: The equality 2k = $\chi(X)$ = |G|$\chi(X/G)$ coming from the covering X $\to$ X/G ensures that a finite group acting freely on X is a 2-group and also answers the question for k = 1, 2. (Actually, for k = 1 this gives a proof of a classical theorem: the only group which can act freely on an even-dimensional sphere is the cyclic group of order 2. Does anyone know who is this result originally due to? Sorry for a question inside the question; perhaps someone can comment on this one.) Hence if one would like to construct a sort of "minimal" example, it should involve an action of either the quaternion group Q8 or the dihedral group Dih4 for k = 3. I thought about this a little bit, but I feel like I'm not comfortable enough with non-abelian groups.
I've stumbled across some papers where authors characterize arbitrary finite groups acting freely on X in terms of existence of particular representations, but explicit examples are given only in the abelian case.

REPLY [5 votes]: [Adem, Alejandro; Davis, James F. Topics in transformation groups.  Handbook of geometric topology,  1--54, North-Holland, Amsterdam, 2002. MR1886667] mentions the fact that a finite $2$-group such that every element of order $2$ is central acts freely on $(S^{|G|/2-1})^k$, where $k$ is the rank of $G$, that is, the rank of the biggest subgroup of the form $(\mathbb Z_2)^{(r)}$: «The action is built by inducing up sign representations on $k$ elements of order $2$ which span the unique central elementary abelian subgroup of $G$ and then taking their product».
Using GAP I find a smallest non abelian example $G$ of order 32, which it describes as $(C_4\times C_2):C_4$.<|endoftext|>
TITLE: Different way to view action of fundamental group on higher homotopy groups
QUESTION [31 upvotes]: There are a couple of ways to define an action of $\pi_1(X)$ on $\pi_n(X)$. When $n = 1$, there is the natural action via conjugation of loops. However, the picture seems to blur a bit when looking at the action on higher $\pi_n$. All of them have the flavor of the conjugation map, but are more geometric than algebraic, and in some cases work is needed to show the map is well defined. Here are a couple I have seen: 
There is a homotopy equivalence $f : S^n \to S^n \vee I$. taking the basepoint of $S^n$ to the endpoint of the unit interval "far away" from $S^n$. Given a path $\alpha$ from $x_0$ to $x_1$, one can get a basepoint changing homomorphism $\pi_n(X,x_0) \to \pi_n(X,x_1)$ by taking $g : S^n \to X$ and mapping it to $(g \vee \alpha) \circ f$. If $\alpha$ is a loop this gives an action of $\pi_1$
Another way to proceed may be to look at elements of $\pi_n(X,x_0)$ as homotopy classes of maps $I^n \to X$ that send $\partial I^n$ to $x_0$. Then a base change homomorphism could be obtained by using a path $\alpha$ to define a map $I^n \cup (\partial I^n \times I) \to X$, which can be filled in to a map $I^{n+1} \to X$. Then the action would be to take the face opposite the original $I^n \subset I^{n+1}$. 
These both define the same standard action of $\pi_1$ on $\pi_n$, but lose the algebraic flavor of the group action and instead have this stronger geometric feel, which can make working with the action a bit cumbersome. Are there other ways of looking at this action that are more algebraic? 
Perhaps, can something be done wherein $\pi_0(Y)$ acts on $\pi_n(Y)$, where $Y$ is some sufficiently nice space like $\Omega X$, and does this coincide with the above defined actions? Is this a useful way of viewing the action?

REPLY [6 votes]: I doubt this would be considered less geometric than the actions in your question, but if $(X,x_0)$ has a universal cover with covering map $p:(\tilde X, \tilde x_0)\rightarrow (X,X_0)$, then $p$ induces isomorphisms $p_*:\pi_n(\tilde X, \tilde x_0)\rightarrow \pi_n(X,x_0)$ and  $\tilde p: D\rightarrow \pi_1(X,x_0)$ where $D$ is the group of deck transformations (covering transformations) of $(\tilde X,\tilde x_0)$.  
Now for $\alpha\in \pi_1(X,x_0)$ and $\rho\in\pi_n(X,x_0)$ let $\tilde \alpha$ be the preimage of $\alpha$ under $\tilde p$ and $\tilde\rho$ be the preimage of $\rho$ under $p_*$.  
Then $\sigma=(\tilde\alpha)_*(\tilde\rho)$ is in $\pi_n(\tilde X, \tilde x_0)$ and we get
$\alpha\cdot \rho$ as $p_*(\sigma)$.
Now that I think about it, this is at least as geometric as the actions in your question, but I like the picture better.  In the picture, the universal cover not only unrolls the elements of $\pi_1$, but it also unrolls the action of the elements of $\pi_1$ on the maps of spheres.<|endoftext|>
TITLE: Countable paracompactness, normality and locally countable open covers
QUESTION [6 upvotes]: (repost from the topology Q&A board)
I have a (T_1), Normal, countably paracompact space X. I would like to know if every locally countable open cover of X (i.e. an open cover such that every x in X has a neighbourhood which intersects only countably many members of the cover) has a locally finite refinement.
My suspicion is that the answer is a resounding no, but every time I try to construct a counterexample it starts to seem more plausible. 
If the answer does turn out to be yes I'd love to know if it generalises from aleph_0 to arbitrary cardinals.

REPLY [4 votes]: In Caryn Navy's thesis under Mary Ellen Rudin she constructed several spaces that are normal, countably paracompact and paralindelöf (every cover has a locally countable refinement) but not paracompact.
All such spaces provide counterexamples (we can refine a cover without a locally finite refinement to a locally countable one and then we cannot continue...)
I'm not sure (as I do not have access to the PhD-thesis in question, and I only know it from references like http://www1.elsevier.com/homepage/sac/opit/10/article.pdf) whether these examples are all under extra set-theoretic assumptions (like MA + non-CH) or whether there are absolute ones as well.<|endoftext|>
TITLE: When do the Reedy and injective model category structures agree?
QUESTION [10 upvotes]: Let $R$ be a Reedy category and consider the category $\mathcal{P}(R) = \mathbf{sSet}^{R^{\mathrm{op}}}$ of simplicial presheaves on $R$.  When are the Reedy and injective model structures on $\mathcal{P}(R)$ the same?  Are there useful sufficient conditions on $R$?
I know this is the case when all the morphisms of $R$ raise degree (I hope that's the right direction), and also when $R = \Delta$.  I am specifically interested about the case where $R$ is $\Theta_n$, or a product of several copies of $\Theta_n$.

REPLY [9 votes]: I complete the answer of Charles Rezk by some precise references: you may find the answer to your questions in my book Les préfaisceaux commes modèles des types d'homotopie, Astérisque 308 (2006). In chapter 8 of loc cit, I gave an axiomatic approach to this. Indeed, if $R$ is a skeletal category in the sense of Definition 8.1.1 in loc cit, and in which the automorphisms are trivial, then the class of monomorphisms in $\mathcal{P}(R)$ is generated by the set of boundary inclusions $\partial x\to x$, where $x$ runs over the family of representable precheaves; see Proposition 8.1.37. Now, these nice categories are closed under finite cartesian product (see Remark 8.1.7), from which you deduce easily that the Reedy model structure on the category of simplicial presheaves on $R$ coincides with the injective one. This applies of course to $R=\Theta_n$.
However, as I wrote this in French, I suspect some people might be happy to have a reference in English: Section 4 in this paper of Samuel B. Isaacson.<|endoftext|>
TITLE: Infinite projective space
QUESTION [14 upvotes]: Is infinite (say complex) projective space a scheme? More generally, can schemes have infinite cardinal dimension? It seems that infinite dimensional projective space is not a manifold, since it is not locally Euclidean for any R^n.
Related question. If inifinite projective space is a scheme, then take a nonclosed point. Taking the closure of this nonclosed point, can we get infinite dimensional subschemes? Sorry for I'm quite foreign to schemes.

REPLY [13 votes]: Starting with the affine case, if you try to define infinite dimensional affine space as Spec of k{x1,x2,...], then you realise that this is not a vector space of countable dimension, but something much larger. If you want a vector space over k of countable dimension, then this will not be a scheme, but instead will be an ind-scheme. A similar description should hold in the projective case.
Edit: Regarding why I am saying that Spec(k[x1,x2,...]) is too big: A (k-)point of Spec(k[x1,x2,...]) is an infinite sequence a1,a2,... of elements of k. If I wanted a vector space of countable dimension, then I should be asking for sequences a1,a2,... of elements of k, only finitely many of which are non-zero. This latter space is the inductive limit of affine n-space as n tends to infinity.

REPLY [4 votes]: Since rings can have infinite Krull dimension, affine schemes can have infinite dimension.

REPLY [3 votes]: You can define $Proj S$ for any graded ring $S$ and this is certainly a scheme; this is in Hartshorne II.2.  Infinite projective space is $Proj S$ where $S = k[x_0, x_1, ....]$ and $k$ is the base field.
Regarding your second question, if you take any homogeneous element $f \in S$, then the vanishing of this should define a closed subscheme of codimension 1 (in particular still infinite dimensional).
Maybe I should say $Proj S$ is the algebraic analogue of infinite projective space.  As a topological space it is very different from $\mathbb{C}^{\infty} - 0$/scaling.  But this is even true in the finite dimensional case (Zariski topology is not the same as topology considered as a real or complex manifold).<|endoftext|>
TITLE: How much complex geometry does the zeta-function of a variety know
QUESTION [8 upvotes]: From Weil conjecture we know the relation between the zeta-function and the cohomology of the variety, however it appears that there are certainly more information containing in the zeta-function, and the question remains whether they can be used to compute some more geometric invariants of the variety, such as the Chern classes. For instance, can one spot a Calabi-Yau manifold just by looking at the zeta-function? Is the zeta-function a birational invariant, or stronger?
And consider the roots and poles of the zeta-function. Their absolute values are determined by the Riemann Hypothesis, nevertheless the "phases" still appear to be very mysterious. Are that any good explanations for them, e.g. in the case of elliptic curves?

REPLY [19 votes]: Although the question is phrased a bit sloppily, there is a standard interpretation: Given a smooth complex proper variety $X$, choose a smooth proper model over a finitely generated ring $R$. Then one can reduce modulo maximal ideals of $R$ to get a variety $X_m$ over a finite field, and ask what information about $X$ it retains. As has been remarked, the zeta function of $X_m$ gives back the Betti numbers of $X$. 
I believe Batyrev shows in this paper that the zeta function of a Calabi-Yau is a birational invariant and deduces from this the birational invariance of Betti numbers for Calabi-Yau's. And then, Tetsushi Ito showed here that knowledge of the zeta function at all but finitely many primes contains info about the Hodge numbers. (He did this for smooth proper varieties over a number field, but a formulation in the 'general' situation should be possible.) 
For an algebraic surface, once you have the Hodge numbers, you can get the Chern numbers back by combining the fact that
$c_2=\chi_{top},$
the topological  Euler characteristic, and Noether's formula:
$\chi(O_X)=(c_1^2+c_2)/12.$
I guess this formula also shows that if you know a priori that $m$ is a maximal ideal of ordinary reduction for both $H^1$ and $H^2$ of the surface, then you can recover the Chern numbers from the zeta functions, since $H^1(O_X)$ and $H^2(O_X)$ can then be read off from the number of Frobenius eigenvalues of slope 0 and of weights one and two.
You might be amused to know that the homeomorphism class of a simply-connected smooth projective surface can be recovered from the isomorphism class of $X_m$. (One needs to formulate this statement also a bit more carefully, but in an obvious way.) However, not from the zeta function. If you compare $P^1\times P^1$ and $P^2$ blown up at one point, the zeta functions are the same but even the rational homotopy types are different, as can be seen from the cup product in rational cohomology. See this paper.
Added: Although people can see from the paper, I should have mentioned that Ito even deduces the birational invariance of the Hodge (and hence Betti) numbers for smooth minimal projective varieties, that is, varieties whose canonical classes are nef. Regarding the last example, I might also point out that this is a situation where the real homotopy types are the same.
Added again: I'm sorry to return repeatedly to this question, but someone reminded me that Ito in fact does not need the zeta function at 'all but finitely many primes.' He only needs, in fact, the number of points in the residue field itself, not in any extension.<|endoftext|>
TITLE: Completion of the rationals to the reals as an inverse limit construction?
QUESTION [10 upvotes]: There is of course the standard construction of the reals by considering the set of sequences that are Cauchy with respect to the standard metric and taking the quotient by sequences that converge to 0, but in many other cases of the completion of a ring or group, we can complete by taking an inverse limit of factor rings/groups corresponding to some predetermined ideals/subgroups.  For example, the $\mathfrak{p}$-adic integers are constructed using precisely this type of inverse limit construction, where the ideals are the powers of $\mathfrak{p}$, where $\mathfrak{p}$ is the ideal $(p)$ for $p$ a prime.
Is there any way to construct the reals using a similar inverse limit construction?  
It seems like this might be related to the so-called "infinite prime", but I don't know if there is really a way to complete with respect to an "infinite prime", which doesn't seem like it is strictly a prime integer or ideal, so that doesn't really seem to help..
(If anyone can think of better tags, please go ahead and change the ones I've used.  I wasn't really sure how to classify this).

REPLY [2 votes]: In general, a completion of a normed field is naturally a colimit, since you can make a poset of normed fields generated by equivalence classes of sets of Cauchy sequences.  I don't think there is an inverse limit construction of the reals, but it may be interesting to analyze the failure modes of some attempts. 
First, we should take a closer look at the p-adic construction.  We start by restricting our view to a valuation ring, namely the subring of elements of absolute value at most one.  Then, we take an inverse limit over quotients by powers of the maximal ideal.  Finally, we take the fraction field of the resulting complete ring.
The first step in the construction is already problematic with the archimedean valuation, because it is not an ultrametric, i.e., the set of rationals of norm at most 1 is not closed under addition.  We can boldly press on, noting that we still have a continuous multiplicative monoid structure, and it has a maximal ideal $M = \{ x \in \mathbb{Q} : |x| < 1 \}$.  Unfortunately, positive powers of M are equal to M itself, so the inverse limit of the quotients is just the monoid $\{-1,0,1\}$, which is not the closed unit disk.  We could also try the inverse limit of quotients by the directed system of ideals $I_\alpha = \{x \in \mathbb{Q} : |x| < \alpha \}$, but all of the quotients are totally disconnected.  This is bad, because (if I'm not mistaken) the inverse limit of a directed system of totally disconnected spaces is totally disconnected, while the interval $[-1,1]$ (often called the "ring of integers of $\mathbb{R}$") is what we really want, and its connected component(s) are far from singletons.<|endoftext|>
TITLE: The space of Lie group homomorphisms
QUESTION [10 upvotes]: Let $\ \mathrm{Hom}(H,G)\ $ be the space of Lie group homomorphisms between compact connected Lie groups $H$, $G$.  What is known about homology (or homotopy) groups of $\mathrm{Hom}(H,G)$? 
UPDATE: $G$ acts on $\mathrm{Hom}(H,G)$ by conjugation, and the orbits are 
a)  connected, as $G$ is connected,  
b) closed, as easily follows from compactness of $G$, and 
c) open for it is an classical result that any nearby representations of $H$ into $G$ are conjugate; this uses compactness of $H$, and a proof can be found in the book by Conner-Floyd, "Differentiable periodic maps", Chapter VIII, Lemma 38.1.
Thus $\mathrm{Hom}(H,G)$ is the disjoint union of the $G$-orbits, and the $G$-orbit that contains a representation $r$ is homeomorphic to $G/Z_G(r(H))$, where $Z_G(r(H))$ is the centralizer of $r(H)$ in $G$.
What I do not yet understand is how to see whether $\mathrm{Hom}(H,G)$ has infinitely many connected components. 
UPDATE: the topology on $\mathrm{Hom}(H,G)$ is that of uniform convergence.

REPLY [6 votes]: The general criterion is: 

Given connected compact Lie groups $G,H$, $\mathrm{Hom}(G,H)$ has finitely many components if and only if $H$ is semisimple or $G=1$. 

Proof.
If $G=1$ there is nothing to prove.
Assume that H is not semisimple and $G\neq 1$. Then there exists a 1-dimensional torus $T$ in $H$ and a closed normal subgroup $N$ of codimension 1 such that $TN=H$. Define $T=T\cap N$, so that $Z$ is finite and define $T'=T/Z$. Then $\mathrm{Hom}(T',G)$ can be identified to the set of elements in $\mathfrak{g}$ whose exponential is 1; we have already seen that it has infinitely many components.
We have a continuous restriction map $\mathrm{Hom}(H,G)\to \mathrm{Hom}(T,G)$. Its image contains $\mathrm{Hom}(T',G)$ (identified to those homomorphisms trivial on $N$), and $\mathrm{Hom}(T',G)$ is clopen in $\mathrm{Hom}(T,G)$ (indeed, let $Z$ have exponent $n$; then 1 is an isolated point in the set $K$ of elements $g\in G$ such that $g^n=1$ and $\mathrm{Hom}(T',G)$ is the fiber of $(1,\dots,1)$ for the mapping $\mathrm{Hom}(T,G)\to K^Z$ mapping $f$ to $z\mapsto f(z)$). Thus $\mathrm{Hom}(H,G)$ has infinitely many components (the proof even shows that it has a continuous map onto an infinite discrete set).

Conversely, suppose $H$ semisimple.
Consider the map $L:\mathrm{Hom}(H,G)\to\mathcal{L}(\mathfrak{h},\mathfrak{g})$ mapping $f$ to its tangent map between Lie algebras ($\mathcal{L}(\mathfrak{h},\mathfrak{g})$ denoting the whole space of linear maps). Then $L$ is injective. Since $L(f)$ is locally conjugate to $f$ by the exponential map, $L$ is continuous, and actually is a homeomorphism to its image (because if $L(f_n)$ tends to $L(f)$, then in a given compact neighborhood of 1 we have $f_n$ tending to $f$ uniformly, and this implies that $f_n$ tends to $f$ uniformly on all of $H$).
If $H$ is simply connected, then the image of $f$ is equal to the set of Lie algebra homomorphisms $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$, which is Zariski closed, and hence has finitely many components in the ordinary topology. In general (still assuming $H$ semisimple), let $\tilde{H}$ be its universal covering and $Z$ the (finite) kernel of $\tilde{H}\to H$. Recalling that the exponential is surjective, write $Z=\exp(Z')$ for some finite subset $Z'$ of $\mathfrak{h}$; then the image of $L$ is the set of elements $u$ in $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ such that for all $z\in Z'$ we have $u(\exp(z))=1$. Let $\mathfrak{g}_1$ be the set of elements $x$ in $\mathfrak{g}$ such that $\exp(x)=1$: then we have the restatement: the image of $L$ is
$$\{f\in \mathrm{Hom}(\mathfrak{h},\mathfrak{g}): f(Z')\subset \mathfrak{g}_1\}.$$
Fix a faithful continuous linear representation of $G$ (in $\mathrm{SO}(n)$ for some $n$), so that we have a faithful representation of $\mathfrak{g}$ (by antisymmetric matrices): then $\mathfrak{g}_1$ is the set of elements in $\mathfrak{g}$ whose eigenvalues are in $2i\pi\mathbf{Z}$. Note that as soon as $G\neq 1$, it has infinitely many connected components (since all these eigenvalues can be achieved, using a 1-parameter subgroup whose image in $G$ is a 1-dimensional torus).
Nevertheless, observe that (still assuming $H$ compact semisimple) $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ is a compact subset of the vector space $\mathcal{L}(\mathfrak{h},\mathfrak{g})$. Indeed, $\mathfrak{h}$ admits a basis $(e_j)$ such that each $e_j$ belongs to some subalgebra $\mathfrak{h}_j$ isomorphic to $\mathfrak{so}(3)$, such that some 2-dimensional complex representation of $\mathfrak{h}_i$ maps $e_j$ to the diagonal matrix $(i,-i)$ (where $i^2=-1$). It follows from representation theory of $\mathfrak{sl}_2$ that for every $n$-dimensional complex representation $\rho$ of $\mathfrak{h}_j$ (and hence of $\mathfrak{h}$) maps $e_j$ to an element with eigenvalues in $\{-i(n-1),\dots,i(n-1)\}$. Thus the trace of $-\rho(e_j)^2$ is $\le (n-1)^2n$. Since $-\mathrm{trace}(XY)$ is a definite positive symmetric bilinear form on antisymmetric matrices, this shows that $\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ is bounded, hence compact.
In particular, there exists $n_0$ such that eigenvalues of $f(z)$ for all $z\in Z'$ and $f\in\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ are in the interval $[-2i\pi n_0,2i\pi n_0]\subset i\mathbf{R}$. Thus the image by $L$ of $\mathrm{Hom}(H,G)$ is the set of $f\in\mathrm{Hom}(\mathfrak{h},\mathfrak{g})$ such that for every $z\in Z'$ we have $\prod_{k=-n_0}^{n_0} (f(z)-2i\pi k)=0$. This is a Zariski-closed subset, and hence has finitely many connected components in the ordinary topology, and hence so does the homeomorphic space $\mathrm{Hom}(H,G)$.<|endoftext|>
TITLE: Topologizing free abelian groups
QUESTION [12 upvotes]: For any set $S$ one can consider the free abelian group $\mathbb{Z}[S]$ generated by this set. Now suppose, there is a topology on $S$ given. Is it possible to find a topology on $\mathbb{Z}[S]$ in such a way, that:
i) The map $S\rightarrow \mathbb{Z}[S]$ is a homeomorphism onto the image.
ii) The addition and the inverse map are continuous
And if it is possible, is this topology unique?

REPLY [9 votes]: I don't know if such a topology is unique, but it exists if and only if $S$ is completely regular. This includes locally compact hausdorff spaces and CW complexes.
With Freyd's Adjoint Functor Theorem, it can be shown that the forgetful functor from abelian top. groups to top. spaces has a left adjoint. This is essentially the same proof as in the discrete case. Explicitely, $\mathbb{Z}[S]$, the free abelian top. group over the top. space $S$, is the usual free abelian group endowed with the weak topology for all homomorphisms $\mathbb{Z}[S] \to A$, such that $S \to \mathbb{Z}[S] \to A$ is continuous. Here, $A$ is an arbitrary abelian top. group. In order to show that this topology exists, we may assume that $A$ is, as a group, a quotient of $\mathbb{Z}[S]$, so that these $A$ form a set. But the description of the topology does not change and even without Freyd's Theorem it is easy to see that $\mathbb{Z}[S]$ thus becomes an abelian top. group satisfying the desired universal property.
Now I claim that the three assertions

$S \to \mathbb{Z}[S]$ is a homeomorphism onto its image.
$S$ is a subspace of an abelian top. group.
$S$ is completely regular.

are actually equivalent!
1) implies 2), that's clear. Now assume 2), thus $S \subseteq A$ for some top. abelian group. Extend the inclusion $S \to A$ to a continuous homomorphism $\mathbb{Z}[S] \to A$. Every open subset of $S$ can be extended to an open subset of $A$. Pull it back to $\mathbb{Z}[S]$. This is an open subset of $\mathbb{Z}[S]$ which restricts to the given oben subset of $S$. This proves 1).
2) implies 3), this follows from the fact that every topological group is completely regular and subspaces of completely regular spaces are obviously completely regular.
Finally assume 3), i.e. $S$ carries the initial topology with respect to all continuous functions $S \to \mathbb{R}$. Endow $\mathbb{R}^{(S)}$ with the initial topology with respect to all homomorphisms $\mathbb{R}^{(S)} \to \mathbb{R}$, such that the restriction $S \to \mathbb{R}$ is continuous. Then $\mathbb{R}^{(S)}$ is an abelian topological group and $S \to \mathbb{R}^{(S)}$ is an embedding, thus 2).
I also believe that (but cannot prove)

If $S$ is hausdorff and completely regular, $\mathbb{Z}[S]$ is hausdorff.

In another comment, it was suggested to endow $\mathbb{Z}[S]$ with the final topology with respect to $S \to \mathbb{Z}[S]$. But this does not even yield a translation invariant topology: If $S=\{a,b\}$ with the only nontrivial open subset $\{a\}$, then $\{a\}$ is open in $\mathbb{Z}[S]$, but $\{b\}$ is not.

But maybe, if $S$ is a completely regular space, the topology of $\mathbb{Z}[S]$ used above is the final topology?<|endoftext|>
TITLE: Whenever I read "centraliser of maximal split torus", I think of...
QUESTION [7 upvotes]: Inspired by this question
I'd like to ask something more specific: 
In the theory of connected reductive groups over fields, one often reads about the centraliser of a maximal split torus. Here is one example: let $k$ be a field and $D$ a skewfield containing $k$ such that $D$ is a finite-dimensional central simple $k$-algebra. Then for any $n \ge 1$, in the $k$-group ${\rm{SL}}_ n(D)$ the $k$-subgroup $T$ consisting of diagonal matrices with entries in ${\rm{GL}}_ 1$ is a maximal split $k$-torus, while the centraliser $S$ of $T$ consists of the diagonal matrices with entries in $D^{\times}$ (viewed as a $k$-group in the usual way). 
Is this example typical or is it too simple-minded to capture the mysteries of these centralisers?

REPLY [5 votes]: The centralizer of a max. split torus is (as Loren noted) the anisotropic
kernel of G.  Maybe the following additional example is useful: Let k be a 
field and Q a non-degenerate quadratic form over k, and let G = SO(Q) (let's 
avoid char. 2 for simplicity...) 
Then [Witt's Theorem] Q can be decomposed into an (orthogonal) sum
Q = Q_an + Q_hyp
where Q_an is an anisotropic quadratic form (has no non-trivial zeros), and
where the quadratic from Q_hyp is hyperbolic ("looks like a quadratic form over
an alg. closed field").
The choice of a hyperbolic basis for Q_hyp is (almost) the same as a choice
of maximal split torus.
And the derived group of the centralizer of that maximal split torus is
the anisotropic group SO(Q_an). [For detail on all this see e.g. [Borel, 
Linear Algebraic Groups 23.4] I'm sure there is an analogous reference
in [Springer, LAG] but my copy of that book is elsewhere at the moment].
Of course, this is similar in spirit to your division algebra example.
For a more elaborate source of examples, see the references Jim cites.<|endoftext|>
TITLE: How do we recognize an integer inside the rationals?
QUESTION [33 upvotes]: My question is fairly simple, and may at first glance seem a bit silly, but stick with me.  If we are given the rationals, and we pick an element, how do we recognize whether or not what we picked is an integer?
Some obvious answers that we might think of are:
A. Write it in lowest terms, and check the denominator is 1.
B. Check that the p-adic valuation is non-negative, for all p.
C. Decide whether the number is positive (or negative) and add 1 to itself (or -1 to itself) until it is bigger than the rational you picked.  (If these multiples ever equaled your rational, then you picked an integers.)
Each of these methods has pluses and minuses.  For example, in option A we presuppose we know how to write an arbitrary rational number q as a quotient of integers and reduce.  In C, we have issues with stopping times. etc..
To provide some context for my question: We know, due to the work of Davis, Putnam, Robinson, and Matijasevic, that the positive existential theory of $\mathbb{Z}$ is undecidable.  The same question for $\mathbb{Q}$ is not entirely answered.  One approach to this new question is to show that that, using very few quantifiers, one can describe the set of integers inside the rationals; and then reduce to the integer case.  For example, see Bjorn Poonen's paper "Characterizing integers among rational numbers with a universal-existential formula."  There, he finds a way to describe the p-valuation of a rational number (i.e. he finds a way to encode option B in the language of quantifiers and polynomials on the rationals).  I'm wondering if there are other characterizations of the integers which would follow suit.

REPLY [2 votes]: The following paper by Stan Wagon and Dan Flath might be of some interest to you:
How to pick out the integers in the rationals: An application of number theory to logic, American Mathematical Monthly, 98 (1991) 812-823.
(I would have included a hyperlink if I was able to find one - unfortunately, my Internet browsing skills failed me this time.)<|endoftext|>
TITLE: A list of machineries for computing cohomology
QUESTION [25 upvotes]: This is not a question, but I just hope to hear more from everyone here on it.
A list of ready-to-use machineries to compute the de Rham / Cech cohomology of a manifold / variety. As far as I know, I have never seen this being made explicit.
What I have in mind at the moment:
"Basic" methods:
*) The definition: for example Simplicial cohomology makes the problem into one of pure linear algebra which can then be done by hand or by many computer program packages at the moment. For singular cohomology this is not really reasonable though.
*) The Axioms: Things Such as the Mayer–Vietoris sequence or the LES of a Pair. These two methods allow you to compute the cohomology of most cell complexes that you are likely to encounter early in your education. More detailed study of the maps in the sequences can get you even farther.
"Advanced" methods:
*) Spectral sequences. Leray-Serre seems to be the most commonly used, since many interesting spaces can be written in terms of fibrations.
*) Morse theory. Surprisingly effective for many difficult problems, especially if one can construct a good energy function, such that the critical sets and flows are simpler.
*) Weil conjecture. After Deligne's proof, one can go in the opposite direction and find Betti numbers by point-counting. Unfortunately it can not give the torsions as far as I know.
For the last two methods, I find Atiyah-Bott's celebrated paper on the moduli space of bundles an excellent demonstration.
Now I am looking forward to your inputs. How many important methods are missing here?

REPLY [6 votes]: 1) If you want to compute the rational or real cohomology of something, you can try to use rational homotopy theory. Rational homotopy theory says that the rational singular chain complex is (as a dga) chain equivalent to a very small chain complex, the minimal model, whose generators are in correspondence to the generators of the rational homotopy groups. So, if you have knowledge of the rational homotopy groups, you can try this. It works also quite well to study the cohomology of the (free) loop space of a space, because you can compute the minimal model of the (free) loop space of a space if you know the minimal model of the space. 
2) If you can show that your space is a $BG$, its cohomology equals the group cohomology of $G$ which is computable in some cases. But I suppose that for manifolds the direct usage of this method is not very efficient since the fundamental group of all acyclic manifolds is infinite and group cohomologies of infinite groups can be very difficult to compute by algebraic means. 
3) For a compact connected Lie group you can use the theorem that the De Rham cohomology equals the equivariant forms (see chapter V.12 in Bredon). 
4) Even if you have no strict Lie group structure, but only a multiplication which fulfills the axioms up to homotopy, i. e. an H-space, you can make use of this structure. With field coefficients, you have the structure of a Hopf algebra on the cohomology/homology and there are various structure theorems. E.g. an easy application of this method is that an H-space which is a finite CW-complex with non-trivial homology has zero Euler-characteristic.<|endoftext|>
TITLE: Has decidability got something to do with primes?
QUESTION [55 upvotes]: Note: I have modified the question to make it clearer and more relevant. That makes some of references to the old version no longer hold. I hope the victims won't be furious over this.

Motivation:
Recently Pace Nielsen asked the question "How do we recognize an integer inside the rationals?". That reminds me of this question I had in the past but did not have chance to ask since I did not know of MO.
There seems to be a few evidence which suggest some possible relationship between decidability and prime numbers:
1) Tameness and wildness of structures
One of the slogan of modern model theory is " Model theory = the geography of tame mathematics". Tame structure are structures in which a model of arithmetic can not be defined and hence we do not have incompleteness theorem. A structure which is not tame is wild.
The following structures are tame:

Algebraic closed fields. Proved by Tarski.
Real closed fields e.g $\mathbb{R}$. Proved by Tarski.
p-adic closed fields e.g $ \mathbb{Q}_p$. Proved by Ax and Kochen.

Tame structures often behave nicely. Tame structures often admits quantifier elimination i.e. every formula are equivalent to some quantifier free formula, thus the definable sets has simple description. Tame structures are decidable i.e there is a program which tell us which statements are true in these structure.   
The following structures are wilds;

Natural number (Godel incompleteness theorem)
Rational number ( Julia Robinson)

Wild structure behaves badly (interestingly). There is no program telling us which statements are true in these structures.
One of the difference between the tame structure and wild structure is the presence of prime in the later. The suggestion is strongest for the case of p-adic field, we can see this as getting rid of all except for one prime.
2) The use of prime number in proof of incompleteness theorem
The proof of the incompleteness theorems has some fancy parts and some boring parts. The fancy parts involves Godel's Fixed point lemma and other things. The boring parts involves the proof that proofs can be coded using natural number. I am kind of convinced that the boring part is in fact deeper. I remember that at some place in the proof we need to use the Chinese Remainder theorem, and thus invoke something related to primes.
3) Decidability of Presburger arithmetic and Skolem arithmetic ( extracted from the answer of Grant Olney Passmore)
Presburger arithmetic is arithmetic of natural number with only addition. Skolem arithmetic is arithmetic of natural number with only multiplication.
Wishful thinking: The condition that primes or something alike definable in the theory will implies incompleteness. Conversely If a theory is incomplete, the incompleteness come from something like primes.

Questions:
(following suggestion by  François G. Dorais)
Forward direction: Consider a bounded system of arithmetic, suppose the primes are definable in the system. Does it implies incompleteness.
Backward direction: Consider a bounded system of arithmetic, suppose the system can prove incompleteness theorem, is primes definable in the system? is the enumeration of prime definable? is the prime factoring function definable?

Status of the answer:
For the forward direction: A weak theory of prime does not implies incompleteness. For more details, see the answer of Grant Olney Passmore and answer of Neel Krishnaswami
For backward direction: The incompleteness does not necessary come from prime. It is not yet clear whether it must come from something alike prime. For more details, see the answer of Joel David Hamkins.

Since perhaps this is as much information I can get, I accept the first answer by Joel David Hamkins. Great thanks to Grant Olney Passmore and Neel Krishnaswami who also point out important aspects.
Recently, Francois G. Dorais also post a new and interesting answer.

REPLY [6 votes]: I know this is old, but there are still two unmentioned results that can shed some light.
The first is closely related to recursion (computability) theory and it follows from the diagonal lemma (see https://en.wikipedia.org/wiki/Diagonal_lemma for the basics). The assertion is that in order to prove the existence of an unprovable sentence in theory $T$, the theory must be able to represent all primitive recursive functions.
The diagonal lemma intuitively says that in all such $T$ there exists a sentence that is the fixed point of a function that assigns predicates to the Gödel numerals of sentences in $T$.
A different angle was pursued by Lawvere in "Diagonal Arguments and Cartesian Closed Categories". This approach is category-theoretic, but it yielded similar results. Lawvere proved that Tarski's undefinability theorem, Gödel incompleteness theorem, Cantor's powerset theorem, and Russel's paradox all follow from a fixed-point theorem in cartesian closed categories.
The basic requirements for the fixed-point theorem is that:

$T$ must have a model that is a cartesian closed category (CCC).
$T$ must prove the existence of an object $A$ and a map $f:A\longrightarrow Y^A$ that is weakly point-surjective in the CCC (see Lawvere's paper for details, it roughly concerns recovering truth values of maps to function spaces).

In conclusion, it seems that prime numbers are not specifically required for diagonal arguments. In general, decidability appears to be a more general concept that is independent of prime numbers. There is, however, a lot of information to be retrieved by analyzing the links between primitive recursive functions together with Gödel numberings, on the one hand, and CCCs together with weakly point-surjective morphism, on the other.<|endoftext|>
TITLE: Images and Monomorphisms of Schemes
QUESTION [14 upvotes]: If $X$ is an object in an arbitrary category, there is a natural definition of a subobject of $X$ as a monomorphism into $X$ (or really an equivalence class of monomorphisms).  If $X$ is a scheme, however, the term 'subscheme' is conventionally reserved only for locally closed immersions (as in EGA I.4.1.2).  There are certainly many monomorphisms of schemes that in this sense aren't subschemes, for example the inclusion of Spec of a local ring such as $Spec K[x]_{(x)} \to Spec K[x]$.
When we restrict 'subscheme' to mean 'locally closed immersion', defining images of schemes becomes problematic.  A sensible definition, in any category, of the image of a morphism is the minimal subobject through which it factors.  Using the above definition of subscheme, there are perfectly well-behaved examples of morphisms of schemes that don't have images in this sense.  For example, consider the morphism $\mathbb A^2_K \to \mathbb A^2_K $ induced by the ring homomorphism $(x,y) \mapsto (x,xy)$; the set-theoretic image is the union of the origin and the complement of the $y$-axis, and there is no minimal locally closed set containing this.  
There is, however, always a minimal closed immersion through which a given morphism factors, and so if one defines the scheme-theoretic image in this sense, it always exists.  My question is that, if we let our notion of 'subscheme' include all monomorphisms, would the resulting notion of 'scheme-theoretic image' always exist?  In other words, is there always a minimal monomorphism of schemes through which a given morphism factors?  Say, in the above example? If I hand you a constructible subset of a scheme, can you only find a monomorphism onto that set if it's locally closed?
As a 'softer' question, can someone explain why we don't want to call general monomorphisms subschemes?  In particular, suppose I have a morphism that is a submersion onto a locally-but-not-globally closed subscheme.  It seems much more sensible to call that locally closed subscheme the image, rather than its global closure.

REPLY [3 votes]: For any subset $|Y| \subseteq |X|$, there exists a monomorphism $\iota \colon Y \hookrightarrow X$ supported on $Y$. However, these do not have the property that any map landing in $|Y|$ factors through $Y$.
In fact, I will show (using results from the Samuel seminar on epimorphisms of rings [Sam], [Laz]) that in the example $\mathbb A^2 \to \mathbb A^2$ given by $(x,y) \mapsto (x,xy)$ that you give, there is no minimal monomorphism it factors through.

Lemma 1. For any subset $|Y| \subseteq |X|$, there exists a monomorphism $\iota \colon Y \hookrightarrow X$ supported on $Y$.

Proof. Indeed, let $Y = \coprod_{y \in |Y|} \operatorname{Spec} \kappa(y)$. Then the natural map $\iota \colon Y \to X$ is a monomorphism, because $Y \times_X Y$ is just $Y$. (See [ML, Exc. III.4.4] for this criterion for monomorphism.) $\square$
Now we focus on the example $\mathbb A^2 \to \mathbb A^2$ given by $(x,y) \mapsto (x,xy)$. Write $f \colon Z \to X$ for this map, and note that $f$ is an isomorphism over $D(x) \subseteq X$. Assume $\iota \colon Y \to X$ is a minimal immersion such that there exists a factorisation of $f$ as
$$Z \stackrel g\to Y \stackrel \iota \to X.$$
For any irreducible scheme $S$, write $\eta_S$ for its generic point. We denote the origin of $\mathbb A^2$ by $0$.

Lemma 2. We must have $|Y| = |\!\operatorname{im}(f)|$ or $|Y| = |\!\operatorname{im}(f)| \cup \{\eta_{V(x)}\}$, and $Y$ is integral.

Proof. Clearly $|Y|$ contains $|\!\operatorname{im}(f)|$. If this inclusion were strict, then $|Y|$ contains some point $y$ that is not in the image of $f$. If $y$ is closed in $X$, then the open immersion $Y \setminus \{y\} \to Y$ gives a strictly smaller monomorphism that $f$ factors through, contradicting the choice of $Y$. Thus, $y$ has to be the generic point of $V(x)$. This proves the first statement.
For the second, note that the scheme-theoretic image $\operatorname{im}(g)$ of $g$ is $Y$. Indeed, if it weren't, then replacing $Y$ by $\operatorname{im}(g)$ would give a smaller monomorphism factoring $f$, contradicting minimality of $Y$. But the scheme-theoretic image of an integral scheme is integral, proving the second statement. $\square$
We now apply the following two results from the Samuel seminar on epimorphisms of rings [Sam]:

Theorem. Let $\iota \colon Y \to X$ be a quasi-compact birational monomorphism of integral schemes, with $X$ normal and locally Noetherian. Then $\iota$ is flat.

Proof. See [Sam, Lec. 7, Cor. 3.6]. $\square$
If $U \subseteq Y$ is an affine open neighbourhood of $0$ and $R = \Gamma(U,\mathcal O_U)$, then we get a flat epimorphism $\phi \colon k[x,y] \to R$ of $k$-algebras (not necessarily of finite type).

Theorem. Let $f \colon A \to B$ be a flat epimorphism of rings, and assume $A$ is normal and $\operatorname{Cl}(A)$ is torsion. Then $f$ is a localisation, i.e. $B = S^{-1}A$ for $S \subseteq A$ a multiplicative subset.

Proof. See [Laz, Prop. IV.4.5]. $\square$
Thus, $R = S^{-1}k[x,y]$ for some multiplicative set $S \subseteq k[x,y]$. This implies that $V = X \setminus U$ is a union of (possibly infinitely many) divisors. Moreover, $V \cap D(x) \subseteq D(x)$ has finitely many components since $D(x)$ is Noetherian (and $\iota$ is an isomorphism over $D(x)$). But then $V \cap V(x)$ is either finite or all of $V(x)$. This contradicts the fact that $U \cap V(x)$ equals $\{0\}$ or $\{0,\eta_{V(x)}\}$ (depending whether $\eta_{V(x)} \in Y$). $\square$

References.
[Laz]  D. Lazard, Autour de la platitude, Bull. Soc. Math. Fr. 97, 81-128 (1969). ZBL0174.33301.
[ML] S. Mac Lane, Categories for the working mathematician. Graduate Texts in Mathematics 5. New York-Heidelberg-Berlin: Springer-Verlag (1971). ZBL0232.18001.
[Sam]  P. Samuel et al., Séminaire d’algèbre commutative (1967/68): Les épimorphismes d’anneaux. Paris: École Normale Supérieure de Jeunes Filles (1968). ZBL0159.00101.<|endoftext|>
TITLE: In a locally CAT(k) space, does uniqueness of geodesics imply the lack of conjugate points?
QUESTION [10 upvotes]: A complete, simply connected Riemannian manifold has no conjugate points if and only if every geodesic is length-minimizing. I just realized that I don't know whether the same is true for a locally CAT(k) space (i.e. a geodesic space with curvature bounded above in the Alexandrov sense).
Thanks to Alexander and Bishop, there is a developed "geodesic analysis" in these spaces, including Jacobi fields and conjugate points. And there is a Cartan-Hadamard theorem for spaces without conjugate points: if it is simply connected, then every pair of points is connected by a unique geodesic.
In the Riemannian case, the converse statement follows from the basic fact that a geodesic beyond a conjugate point is no longer minimizing. This is proved by constructing a length-decreasing variation, or something similar, from a vanishing Jacobi field. Unfortunately, this argument uses a lower curvature bound. Well, not quite that, because it also works in Finsler geometry, but anyway it fails for CAT(k): on a bouquet of two spheres there are geodesics that remain minimizing beyond a conjugate point.
However this does not disprove the converse Cartan-Hadamard theorem. Hence the question:
Let $X$ be a space with curvature locally bounded above. Let's not talk about monsters: the space is complete, locally compact, all geodesics are extensible (otherwise one can play dirty tricks with a boundary). Suppose that every geodesic in $X$ is minimizing. Or even better: every pair of points is connected by a unique geodesic. Does this imply that the geodesics have no conjugate points?
UPDATE. Thanks to Henry Wilton, I've found that there is no standard definition of a conjugate point. In fact, some definitions are designed so as to imply the affirmative answer to my question immediately. When I asked the question, I meant the following (maybe not the best possible) definition.
Fix a point $p\in X$ and consider the space $X_p$ of geodesic segments emanating from $p$. The segments are parametrized by $[0,1]$ proportionally to arc length. The space $X_p$ is regarded with the $C^0$ metric. The exponential map $\exp_p:X_p\to X$ is defined by $\exp_p(\gamma)=\gamma(1)$. A point $q=\gamma(1)$ is conjugate to $p$ along $\gamma$ iff $\exp_p$ is not bi-Lipschitz near $\gamma$.

REPLY [5 votes]: Consider a surface of revolution with an equator $\ell$ of lenght $2{\cdot}\pi$ such that its Gauss curvature $$K=1/\left(1+\sqrt[5]{\mathrm{dist}_ \ell}\right).$$
Choose $z\in \ell$ and let $\Sigma=B_{\pi/2}(z)$.
Clearly $\Sigma$ is a $\mathrm{CAT}(1)$-space it has just one pair of conjugate points (say $p$ and $q$ --- the ends of $\Sigma\cap\ell$) and it has unique geodesics between each pair.
It remains to make geodesics extensible. 
To do this, we take $\Lambda=(S^1\times [0,\infty), d)$ with flat metric and concave boundary $\partial \Lambda=\partial\Sigma$.
Then we glue $\Lambda$ and $\Sigma$ along the boundary.
The metric on $\Lambda$ is completely described by curvature $k(u)$ of its boundary [$u\in \partial \Lambda=\partial \Sigma$].
We only need to choose a function $k$ which is 

on one had is large enough so that the glued surface still has unique geodesics between each pair (in particular $k(p)=k(q)=\infty$).
on the other hand is $\int_{S^1} k<\infty$, so that glued space is locally compact.

I believe it is possible...<|endoftext|>
TITLE: Is there a general projection formula for morphisms of ringed topoi?
QUESTION [5 upvotes]: What's the general projection formula in algebraic geometry, for instance on the level of derived categories of ringed topoi? And what's the reference? I guess it might be in SGA 4, but couldn't find it. 
Two examples:

Zariski site, $D^b_{qcoh}$ on schemes. Let $f:X\to Y$ be a proper map of noetherian schemes (maybe there are some other mild conditions), and let $F\in D^b_{qcoh}(X)$ and $G\in D^b_{qcoh}(Y).$ Then $(Rf_*F)\otimes^L G=Rf_*(F\otimes^L Lf^*G)$. 
Etale site, say ringed by a torsion ring like $Z/n.$ Let $f:X\to Y$ be a (seperated; but this condition can be removed. See for instancde Laszlo and Olsson, The six operations on Artin stacks...) map of schemes of finite type over some base $S$ ($S$ may need to satisfy some assumptions, in order for the classical results in SGA 4/4.5 or Gabber's new results on finiteness of $f_*$ and dualizing complexes to work; but let's be sloppy). Let $F\in D^-_c(X,Z/n)$ and $G\in D^-_c(Y,Z/n).$ Then $Rf_!F\otimes^L G=Rf_!(F\otimes^L f^*G).$ 

We used $f_!$ in example 2 in order to allow $F$ and $G$ to be in $D^-_c$ rather than $D^b_c.$ If one restricts to $D^b_c,$ is it also true for $f_*?$

REPLY [4 votes]: In the context of sheaves of $\mathcal O_X$-modules,
there is the following reference: Prop. 3.9.4 in Lipman's
Notes on derived functors and Grothendieck duality.
A closely related result is in Neeman's paper The Grothendieck duality theorem ...; see Prop. 5.3.
I'm not sure that analogous results should be expected to hold in arbitrary generality;
for example, both references place a restriction on the base scheme, and require quasi-coherence assumptions.  (In some sense, one has to reduce to the locally free case,
where the statement is obvious.  Quasi-coherent sheaves then admit locally free resolutions.
The proofs of the cited results apply some form of this argument in rather subtle and sophisticated ways.)<|endoftext|>
TITLE: Universal group?
QUESTION [24 upvotes]: I can construct a finitely presented group $G$ with the following property (which I use to construct something else).

Given a finitely preseted group $\Gamma$, there is a subgroup $G'\le G$ of finite index
such that  $$\Gamma=G'/\langle\mathrm{Tor}\, G'\rangle ,$$ where $\mathrm{Tor}\, G'\subset G'$ is the set of all elements of finite order.

I think to call such group $G$ universal.
Questions:

Was it already constructed?
Does it already has a name? Is there any closely related terminology?


P.S.

The group which I construct is in fact hyperbolic.
The construction is simple, but it takes 2--3 pages. Let me know if you see a short way to do it.
Here, the term "universal group" was used in very similar context (thanks to D. Panov for the reference).
Thanks to all your comments, we call them "telescopic" actions now.

REPLY [5 votes]: Аnswered to move the question to answered status.
Thank you all for your comments they were helpful for me and Dima.<|endoftext|>
TITLE: Slick proof related to choosing points from an interval in order
QUESTION [30 upvotes]: Choose a point anywhere in the unit interval $[0, 1]$.  Now choose a second point from the same interval so that there is one point in each half, $[0, \frac12]$ and $[\frac12, 1]$.  Now choose a third point so that there is one point in each third of the interval, and so on.  How long can you choose points like this?
In other words, what is the largest $n$ so that there exists a (finite) sequence $(a_i)_{i=1}^n$ so that for all $k$, among the first $k$ points $(a_i)_{i=1}^k$, there is at least one in each interval $[0, \frac1k], [\frac1k, \frac2k], \ldots, [\frac{k-1}k,1]$?
My question
Is there a slick proof that $n$ is bounded?  (Note that by compactness, the existence of such a sequence for all $n$ is equivalent to the existence of an infinite sequence with the same property.)
Backstory
I was recently re-reading old columns of Martin Gardner's, and he describes this problem in "Bulgarian solitaire and other seemingly endless tasks".  (Original column 1983, available in his collections "The last recreations: ..." and "The colossal book of mathematics".)  He says that the problem first appeared in "One hundred problems in elementary mathematics" by Hugo Steinhaus.  There, he shows that there is a sequence of length 14, but there is none of length 75.  The first published proof of the longest sequence, which has length 17, was by Elwyn Berlekamp and Ron Graham in their paper "Irregularities in the distributions of finite sequences" in the Journal of Number Theory.  This was followed by a shorter proof by Mieczysƚaw Warmus in "A supplementary note on the irregularities of distributions" in the same journal.
Now, these proofs mostly use case analysis to some degree or other.  They are of varying complexities, oddly with Warmus's proof of the optimal $n$ being the shortest.  It's also not too hard to write a computer check oneself that finds the optimal $n$.  However, I feel that because of the elegant nature of the problem, there should be some "nice" proof -- not of optimality, but simply that such a sequence can't continue forever.  Can anyone find one?
Technical note:  The problem is usually stated with half-open intervals.  I made them closed so that compactness worked.  (Edit:  The possibility that this changes the answer did occur to me.  I assume it doesn't, and I'll check by computer soon.  I am fine with answers for any kind of intervals -- open, closed, half-open.)

REPLY [10 votes]: The proof given by Steinhaus (following A. Schinzel) is slick indeed. I'll paraphrase what he does:
At step number $n$, the interval (0,1) is divided into $n$ intervals $I_0^{(n)}=(0,\frac{1}{n}), \ldots, I_{n-1}^{(n)}=(\frac{n-1}{n},1)$. Then $\lfloor nP \rfloor$ gives you the number of the interval to which a point $P$ in the sequence belongs in the $n.$ step.
He proceeds by proving the following claim:
If $P_1 \in (\frac{7}{35},\frac{8}{35})$ and $P_2 \in (\frac{9}{35}, \frac{10}{35})$, then there is some $n \in \mathbb{N}$ big enough such that $\lfloor (n-1)P_1 \rfloor < \lfloor nP_1 \rfloor$ , but $\lfloor (n-1)P_2 \rfloor = \lfloor nP_2 \rfloor$.
Let's see how this claim implies that there is no infinite such sequence. Eventually, in your sequence at the $k.$ step there will be two points $P_1,P_2$ as in the claim. Let $n$ be as in the claim, in particular $n>k$. 
Let $k_1:=\lfloor (n-1)P_1\rfloor$ and $k_2:=\lfloor (n-1)P_2\rfloor$. Then $P_i$ belongs to the $k_i$. interval in the $(n-1).$st step and there are $k_2-k_1-1$ points of $a_1, \ldots, a_{n-1}$ of the sequence in-between. But in the $n.$ step, $P_1$ will move to the $k_1+1$.st interval, while $P_2$ will remain in the $k_2$.st interval, and so there will be a ''collision'' of the in-between points.<|endoftext|>
TITLE: Are monads monadic?
QUESTION [13 upvotes]: Is there some sort of monad whose algebras are monads? How about if we are internal to a bicategory B? Are internal monads in B monadic? Certainly not always, as otherwise free T-multicategories a la Leinster would always exist. What is known?

REPLY [10 votes]: An important reference on free monoids (and by extension, free monads) is:

G.M. Kelly, A unified treatment of transfinite constructions for free algebras, free monoids, colimits, associated sheaves, and so on, Bulletin of the Australian Mathematical Society 22 (1980), 1--83.

It's 83 pages long and famously impenetrable.  But those who've braved it say that it contains just about every free monoid construction that anyone's ever thought of.  
So if you want to construct a free monoid/monad, the question is whether it's easier to (a) read Kelly's paper, or (b) do it yourself and cite Kelly anyway (on the assumption that your construction is probably in there, somewhere).<|endoftext|>
TITLE: Lax and Colax Monads
QUESTION [5 upvotes]: Is there much known about the theory of lax and colax monads on a bicategory? Here, I really mean lax or colax, not weak. I'm aware of some literature about weak monads. I'm interested in distributive laws of lax and colax monads and their relationships with algebras. I've managed to prove some things, but, I don't want to reinvent the wheel.

REPLY [2 votes]: This isn't technically an answer, but depending on your examples, you might want to think about lax/colax monads on (pseudo) double categories instead.  Part of the problem with lax monads on bicategories is that there is no tricategory of bicategories and lax functors, whereas there is a 2-category of pseudo double categories and lax functors, so that all of the "formal theory of monads" can be applied directly to lax monads on double categories.  Most of the lax functors and lax monads that I've seen on bicategories have "actually" lived on double categories, except that people tend to forget about the extra direction of arrows and think only about the bicategory.<|endoftext|>
TITLE: What's the "correct" smooth structure on the category of manifolds?
QUESTION [8 upvotes]: As will become clear, this is in some sense a follow up on my earlier question Why should I prefer bundles to (surjective) submersions?.  As with that one, I hope that it's not too open-ended or discussion-y.  If y'all feel it is too discussion-y, I will happily close it.
Let $\rm Man$ be the category of smooth (finite-dimensional) manifolds.  I can think of (at least) two natural "smooth structures" on $\rm Man$, which I will outline.  My question is whether one of these is the "right" one, or if there is a better one.
I should mention first of all that there many subtly different definitions of "smooth structure" — see e.g. n-Lab: smooth space and n-Lab: generalized smooth space and the many references therein — and I don't know enough to know which to prefer.  Moreover, I haven't checked that my proposals match any of those definitions.
In any case, the definition of "smooth structure" that I'm happiest with is one where I only have to tell you what all the smooth curves are (and these should satisfy some compatibility condition).  So that's what I'll do, but I'm not sure if they do satisfy the compatibility conditions.  Without further ado, here are two proposals:

A smooth curve in $\rm Man$ is a fiber bundle $P \to \mathbb R$.
A smooth curve in $\rm Man$ is a submersion $Y \to \mathbb R$.

Then given a manifold $M$, we can make it into a category by declaring that it has only identity morphisms.  Then I believe that the smooth functors $M \to {\rm Man}$ under definition 1 are precisely the fiber bundles over $M$, whereas in definition 2 they are precisely the submersions over $M$.
(Each of these claims requires checking.  In the first case, it's clear that bundles pull back, so all bundles are smooth functors, and so it suffices to check that if a surjective submersion to the disk is trivializable over any curve, then it is trivializable.  In the second case, it's clear that if a smooth map restricts to a submersion over each curve, then it is a submersion, so any smooth functor in a submersion, and so one must check that submersions pull back along curves.)
I can see arguments in support of either of these.  On the one hand, bundles are cool, so it would be nice if they were simply "smooth functors".  On the other hand, we should not ask for smooth functions (i.e. 0-functors) to be necessarily "locally trivializable", as then they'd necessarily be constant. Maybe the correct answer is definition 2, and that bundles are "locally constant smooth functors", or something.
Anyway, thoughts?  Or am I missing some other good definition?
Addendum
In the comments, folks have asked for applications, which is very reasonable.  The answer is that I would really like to have a good grasp of words like "smooth functor", at least in the special case of "smooth functor to $\rm Man$".  Of course, Waldorf and Shreiber have explained these words in certain cases in terms of local gluing data (charts), but I expect that a more universal definition would come directly from a good notion of "smooth structure" on a category directly.
Here's an example.  Once we have a smooth structure on $\rm Man$, we can presumably talk about smooth structures on subcategories, like the category of $G$-torsors for $G$ your favorite group.  Indeed, for the two definitions above, I think the natural smooth structure on $G\text{-tor}$ coincide: either we want fiber bundles where all the fibers are $G$-torsors, or submersions where all the fibers are $G$-torsors, and in either case we should expect that the $G$ action is smooth.  So then we could say something like: "A principle $G$-bundle on $M$ is (i.e. there is a natural equivalence of categories) a smooth functor $M \to G\text{-tor}$", where $M \rightrightarrows M$ is the (smooth) category whose objects are $M$ and with only trivial morphisms.  (Any category object internal to $\rm Man$ automatically has a smooth structure.)  And if I understood the path groupoid mod thin homotopy $\mathcal P^1(M) \rightrightarrows M$ as a smooth category, then I would hope that the smooth functors $\mathcal P^1(M) \to G\text{-tor}$ would be the same as principle $G$-bundles on $M$ with connection.  Functors from the groupoid of paths mod "thick" homotopy should of course be bundles with flat connections.  Again, Schreiber and Waldorf have already defined these things categorically, but their definition is reasonably long, because they don't have smooth structures on $\rm Man$ that are strong enough to let them take advantage of general smooth-space yoga.
Here's another example.  When I draw a bordism between manifolds, what am I actually drawing?  I would like to say that I'm drawing something close to a "smooth map $[0,1] \to \rm Man$".  I'm not quite, by my definitions — if you look at the pair of pants, for instance, at the "crotch" it is not a submersion to the interval.  So I guess there's at least one more possible definition of "smooth curve in $\rm Man$":

A smooth curve in $\rm Man$ is a smooth map $X \to \mathbb R$.

But this, I think, won't be as friendly a definition as those above: I bet that it does not satisfy the compatibility axioms that your favorite notion of "smooth space" demands.

REPLY [4 votes]: I think that the most interesting part of your question is the part you put in parentheses!

(and these should satisfy some compatibility condition)

What are your compatibility conditions?  That is everything here.  If you specify the correct conditions, you may find that all your definitions collapse to just one.
I have an issue with Konrad's answer (which I doubt very much that he will be surprised to hear me express!).  Whenever I heard words like "Grothendieck topology" or "sheaves" or encoding similar ideas then I feel that something's been lost.  I don't like the idea that "smooth" is just "really nice continuity".  "Smooth" sits alongside continuity and can be expressed in a different way which is extremely simple: takes smooth curves to smooth curves.
Of course, I would say that, as everyone by now presumably knows that I prefer Frolicher spaces to the other variants (like Chen spaces or diffeological spaces, see generalized spaces for links).  It is interesting that Chen's third definition (by my count) was stronger than his eventual sheaf condition and was more along the lines of "a map is smooth if enough tests say that it is smooth".
But Frolicher spaces have a problem, which is that it is extremely difficult to prise them away from being a set-based theory.  The compatibility condition is so strong that it forces an underlying set.  I'd really like to figure out how to make this extension, and I know that Urs would as well.  If I could just encourage you and Konrad over to the nLab to play around with these ideas to see how they could work ...
If you want to study The Smooth, The Whole Smooth, and Nothing But The Smooth, then you should do so and not flirt continually with continuity.  The stronger compatibility condition means that more stuff is smooth than you first thought (witness my recent question on this) and that makes it interesting!  The unexpected happens, so study it!
This isn't much of an answer so far, it's more of a commentary on your question which (as is usual for me) is too long for an actual comment.  So let me end with an actual answer (which I freely confess that I stole from a rabbi):

That is such a great question, why on earth would you want an answer?<|endoftext|>
TITLE: A reference: the splitting principle for exterior powers of coherent sheaves?
QUESTION [6 upvotes]: It's well known that if E is a vector bundle with Chern roots $a_1,\ldots, a_r$,
then the Chern roots of the $p$th exterior power of E consist of all sums of $k$ distinct $a_i$'s.  I would like to say the same is true if E is just a torsion-free coherent sheaf on $P^n$.  It seems non-obvious, though, maybe because an exterior power isn't generally an additive functor.  
Presumably this is either false or also well known, but I can't find a reference.

REPLY [3 votes]: My guess would be that the formula you want does not extend to the case of coherent sheaves. As indicated in Mariano and David answers (which has unfortunately been deleted), the best hope to compute is via a resolution $\mathcal F$ of $E$ by vector bundles. In general, for 2 perfect complexes $\mathcal F, \mathcal G$ of vector bundles, there is a formula for the localized chern classes 
$$ch_{Y\cap Z}(\mathcal F \otimes \mathcal G) = ch_Y(\mathcal F)ch_Z(\mathcal G)$$, with $Y,Z$ being the respective support. Unfortunately, this only gives the right formula for the "derived tensor product".
So to mess up the formula, one can pick $E$ such that $Tor^i(E,E)$ are non-trivial. I think an ideal sheaf of codimension at least 2 would be your best bet for computation purpose.<|endoftext|>
TITLE: Why is it called *spectral* triple?
QUESTION [8 upvotes]: I know the definition a spectral triple and that it is some kind of non-commutative generalisation of (the ring of functions on) a compact spin manifold.
But, why is it called spectral triple?

REPLY [10 votes]: Well, it uses the spectral properties of the Dirac operator $D$ in the spectral triple quite extensively.
Also, in the article where he (essentially) introduces the notion of spectral triples ( http://www.alainconnes.org/docs/reality.pdf ) Alain Connes writes about the naming:
"We shall need for that purpose to adapt the tools of the differential and integral calculus
to our new framework. This will be done by building a long dictionary which relates the usual
calculus (done with local differentiation of functions) with the new calculus which will be done
with operators in Hilbert space and spectral analysis, commutators.... The first two lines of the
dictionary give the usual interpretation of variable quantities in quantum mechanics as operators in
Hilbert space. For this reason and many others (which include integrality results) the new calculus
can be called the quantized calculus’ but the reader who has seen the word “quantized” overused
so many times may as well drop it and use “spectral calculus” instead."<|endoftext|>
TITLE: Is there a general statement about structures on spheres relating to division algebras?
QUESTION [13 upvotes]: It is classical to take a division algebra over $\mathbb{R}$ and defining an H-space structure on the unit spheres by restricting and normalizing.
There are commutative division algebras of dimension 1 and 2 leading to commutative products on $S^0$ and $S^1$ identifying them as Eilenberg-MacLane spaces - Or if we forget some structure as an $E_{\infty}$-spaces.
The associative division algebras $\mathbb{H}$ defines an associative product on $S^3$, which is also a Lie-group, but forgetting some structure it is an $A_\infty$-space.
There division algebra $\mathbb{O}$ defines an $A_2$ structure on $S^7$, which is not $A_\infty$ (is it $A_3$?).
As is well known it is possible to prove that no other spheres has $A_2$ structure.
Question: Is there a hierarchy of structures below $A_2$ yet related such that $S^{15}$ has this structure, but $S^{31}$ does not?
Remark: A hierarchy below $A_2$ could be that $A_2=D_\infty$ for some definition of structures $D_n$, analagous to $E_1$ being $A_\infty$. 
Question: Is there an even more general definition of "lower" structures and a statement about all spheres (including possibly non-trivial structures on even-dimensional spheres)?

REPLY [4 votes]: I like your spelling of hierarchy! $S^7$ is not $A_3$ -- if it were, you could construct the projective space $\mathbb{O}P^3$, but that's impossible (some decomposition of Steenrod operations argument). You'll find this and answers to your other question in Baez's article on the octonions in the Bulletin of the AMS. There are higher dimensional algebras in the sequence $\mathbb R$, $\mathbb C$, $\mathbb H$, $\mathbb O$, which you can get by the Cayley-Dickson construction, but they are not normed, so you can't take the unit sphere. The 16-dimensional guy is "power-commutative", meaning that powers of an element x commute with x (not obvious if you're not associative), this starts failing in dimension 32 if I remember correctly. So there's some sort of hierarchy of structure on the algebras themselves, if not on the spheres.
Some of the spheres have an H-space structure or a loop space structure after completing at a prime p ($S^{2p-3}$, for example). Maybe this goes a bit in the direction of your question about some kind of multiplication on higher spheres.<|endoftext|>
TITLE: Writing papers in pre-LaTeX era?
QUESTION [75 upvotes]: I wonder how people wrote papers in the pre-LaTeX era? I mean, when typewriters and simple computers were (60th-70th?). Did they indeed put formulas by hand in the already printed articles?

REPLY [6 votes]: In the 1960s I was writing at North American Aviation memos and then a book explaining application of Kalman Filters to celestial navigation for the Apollo Lunar Mission which had complex matrices, etc.  The typists at North American Aviation told me these equations could not be typed.   I bought a set of Type-it symbols and a used IBM Executive typewriter which had half spacing for superscripts and subscripts.  I enlisted my wife and she typed
my memos and book, mainly to show it could be done.  The typists ended up being very grateful to me, because I showed upper management that if they would provide some of the typists with the same Executive Typewriters and boxes of Type-it symbols they could save money by not having these memos, etc. sent to printers!<|endoftext|>
TITLE: What is a symplectic form intuitively?
QUESTION [45 upvotes]: Hi,
to completely describe a classical mechanical system, you need to do three things:
-Specify a manifold $X$, the phase space. Intuitively this is the space of all possible states of your system.
-Specify a hamilton function $H:X\rightarrow \mathbb{R}$, intuitivly it assigns to each state its energy.
-Specify a symplectic form $\omega$ on $X$. What is $\omega$ intuitively? What kind of information about physics does it capture?

REPLY [15 votes]: Incidentally, I more or less disagree that symplectic geometry captures what I would consider "classical mechanics".  The reason is that in all the examples that I think deserve to be called "classical mechanics", I actually have a configuration space $N$, and your symplectic manifold is $X = {\rm T}^*N$ the cotangent bundle.  Then, of course, the symplectic form is precisely (part of) the cotangent structure.
This is not to say that symplectic geometry isn't interesting — it's led to great mathematics, and certainly captures some of "classical mechanics".  From the physics perspective, what I think makes it most interesting is that it shows that there are strange symmetries between mechanical systems, when you have a symplectomorphism ${\rm T}^*N \to {\rm T}^*N'$ that does not arise from a diffeomorphism $N \to N'$.
But physics is not invariant under all symplectomorphisms.  Otherwise, how would I know which coordinates are "position" and which are "momentum"?  And I do believe that I know this, although maybe I'm wrong.  You and I should get together and compare if our Darboux coordinates differ only by a map $N \to N'$, or by some more interesting symplectomorphism.<|endoftext|>
TITLE: Pairing used in Lefschetz duality
QUESTION [13 upvotes]: I am thinking about the precise formulation of the Lefschetz duality for the relative cohomology.  If I understand this Wikipedia article correctly, there is an isomorphism between $H^k(M, \partial M)$ and $H_{n-k}(M)$ and hence (I suppose) a non-degenerate pairing $H^k(M, \partial M) \times H^{n-k}(M) \rightarrow \mathbb{R}$.  However, I have trouble visualizing this pairing.  Let $[(\alpha, \theta)] \in H^k(M, \partial M)$ and $[\beta] \in H^{n - k}(M)$, is it then true that
$$
  \left< [(\alpha, \theta)], [\beta] \right> =
\int_M \alpha \wedge \beta + \int_{\partial M}\theta \wedge \beta_{|\partial M}
$$
or am I missing something?  If unrelated to Lefschetz duality, does this pairing ever appear in topology?
I can understand how to define a pairing on the homology by counting intersections, but I really don't see how this works for cohomology.  Also, a reference on Lefschetz cohomology or just analysis/topology on manifolds with boundary would be greatly appreciated!

REPLY [4 votes]: Yes, your formula is right. For the intuitive understanding just compute it for 1- and 2- dimensional half-spaces.
See Bott & Tu, Differential forms in Algebraic topology, $\S 5$, Poincaré duality.
I give only sketch of proof for your question.

First of all you need pairing between $H_c^k(M, \partial M)$ and  $H^{n-k}(M)$.
Just consider $M= [0,+\infty)$, find $H_c^k(M), H_c^k(M,\partial M), H^k(M), H^k(M,\partial M) $  and check that you have non-generating pairing.
By induction, expand previous statement to $\mathbb R_{+}^n = \{(x_1,x_2\dots,x_n)|x_1\geqslant 0\}$ (read Bott & Tu $\S 4$ and do the same things).
Prove that there is Mayer-Vietoris sequence for $H_c^k(M,\partial M)$ similar to Mayer-Vietoris sequence for $H_c^k(M)$. 
Prove duality the same way as in $\S 5$ (check the commutativity of diagram and apply 5-lemma).

That's all, I performed these actions without any troubles.<|endoftext|>
TITLE: What is the time complexity of computing sin(x) to t bits of precision?
QUESTION [43 upvotes]: Short version of the question:  Presumably, it's poly$(t)$.  But what polynomial, and could you provide a reference?
Long version of the question:
  I'm sort of surprised to be asking this, because it's such an extremely basic sounding question.  Here are some variants on it:

How much time does it take to compute $\pi$ to $t$ bits of precision?
How much time does it take to compute $\sin(x)$ to $t$ bits of precision?
How much time does it take to compute $e^x$ to $t$ bits of precision?
How much time does it take to compute $\mathrm{erf}(x)$ to $t$ bits of precision?
How much time and how many random bits does it take to generate a (discrete) random variable $X$ such that there is a coupling of $X$ with a standard Gaussian $Z \sim N(0,1)$ for which $|X - Z| < \delta$ except with probability at most $\epsilon$?

In my area of theory of computer science, no one seems to pay much attention to such questions; an algorithm description might typically read 
"Generate two Gaussian random variables $Z$ and $Z'$ and check if $\sin(Z \cdot Z') > 1/\pi$"
or some such thing. But technically, one should worry about the time complexity here.
One colleague of mine who's more of an expert on these things assured me that all such "calculator functions" take time at most poly$(t)$.  I well believe that this is true.  But again, what polynomial (out of curiosity, at least), and what is the reference?  
I kind of assumed that the answers would be in every single numerical analysis textbook, but I couldn't find them there.  It seems (perhaps reasonably) that numerical analysis cares mainly about getting the answers to within a fixed precision like 32 or 64 bits or whatever.  But presumably somebody has thought about getting the results to arbitrary precision, since you can type
Digits := 5000; erf(1.0);
into Maple and it'll give you an answer right away.  But it seemed hard to find.  After much searching, I hit upon the key phrase "unrestricted algorithm" which led me to the paper "An unrestricted algorithm for the exponential function", Clenshaw-Olver-1980.  It's pretty hard to read, analyzing the time complexity for $e^x$ in terms of eight (??!) parameters, but its equation (4.55) seems to give some answers: perhaps  $\tilde{O}(t^2)$ assuming $|x|$ is constant?  
And really, all that work for little old $e^x$?  As for erf$(x)$, I found the paper "The functions erf and erfc computed with arbitrary precision" by Chevillard in 2009.  It was easier to read, but it would still take me some time to extract the answer; my first impression was $\tilde{O}(t^{3/2})$.  But again, surely this question was not first investigated in 2009, was it?!
(By the way, question #5 is the one for which I really want to know the answer, but I can probably work it out from the answer to question #4.)

REPLY [2 votes]: Knuth gives several algorithms for sampling from a normal distribution using elementary functions, given samples from a uniform distribution.  As above, the expected complexity is $O(M(n) \log{n})$.  Chapter 7 of Numerical Recipes is a brief survey of related methods.  It might help to consult MPFR, since they cannot assume $O(1)$ floating-point operations.  GSL has various implementations.<|endoftext|>
TITLE: How do you approach your child's math education?
QUESTION [61 upvotes]: My son is one year old, so it is perhaps a bit too early to worry about his mathematical education, but I do. I would like to hear from mathematicians that have older children: What do you wish you'd have known early? What do you think you did particularly well? What do you think would be particularly bad? Is there a book (for children or parents) that you recommend?
(This a community wiki, so please give one advice per answer, as usual.)
Background
I ask here because I believe that the challenges a mathematician faces in educating a child are special. For example, at least some websites and books address the parents' fear of not knowing how to solve homework, which keeps them from becoming involved. On the contrary, I fear I might get too involved and either bore my son or make him think he likes math when in fact his skills are elsewhere.
Christos Papadimitriou said in an interview that, even though his father was teaching math in high-school, they never discussed math. I wonder if that means his father didn't teach him how to count and I wonder if it's a good strategy. (It certainly turned out well in one case.)
Timothy Gowers (in Mathematics, a very short introduction) says that it was inappropriate to explain to his son, who was six, the concept 'zero' using the group axioms. (Or something to this effect, I don't have the book near to check.) That was surprising to me, because I wouldn't have thought that I need to restrain myself from mentioning abstract concepts. (Update. Here's the quote: "[The non-abstract] way of thinking makes it hard to answer questions such as the one asked by my son John (when six): how can nought times nought be nought, since nought times nought means that you have no noughts? A good answer, though not one that was suitable at the time, is that it can be deduced from the [field axioms] as follows. [...]")
There is a somewhat related Mathoverflow question. This one is different, because I'm looking for advice (rather than statistics/anecdotes) and because my goal is to give my son a good math education (rather than to make him a mathematician). I also found an online book that seems to give particularly good generic advice. Here I'm looking more for advice geared towards parents that are mathematicians.
In short, I'm looking for specific advice on how a mathematician should approach his/her child's math education, especially for the 1 to 10 age range.

REPLY [8 votes]: I have formulated a theorem regarding the involvement of mathematicians in mathematics education.  It says: the personal educational experience of a mathematician is of no value in drawing general conclusions in mathematics education.  I agree that dealing with one's own children is different from "a general conclusion" but even so I'd be careful about sentences that begin "When I was in school..."<|endoftext|>
TITLE: Attaching maps for Grassmann manifolds
QUESTION [14 upvotes]: The Grassmannian $G_n(\mathbb{R}^k)$ of n-planes in $\mathbb{R}^k$ has a CW-complex structure coming from the Schubert cell decomposition.
What is known about the attaching maps in this CW-complex structure?
I understand that a lot of work has been done to try to understand the answer to this question using things like Schubert calculus, Young diagrams, Steenrod operations, etc. I'd like to see some kind of collection of known results about the attaching maps and the specific techniques used to obtain those results.
I'm also interested in the case of the complex Grassmannians.

REPLY [6 votes]: The degrees of the attaching maps (and hence the integral chain complex) for the real Grassmannians have been determined in 

L. Casian and Y. Kodama. On the cohomology of real Grassmann manifolds. arXiv:1309.5520 (Link to arXiv)

The integral chain complex is determined by an explicit combinatorial procedure involving the Young diagram representations of Schubert cells. There are other papers by the authors in which they determine the incidence graphs of Grassmannians. Maybe the techniques used there can also help to obtain more precise information on the attaching maps.

Concerning the attaching maps for the complex Grassmannians, there is some discussion in the following paper: 

C. Lenart. The combinatorics of Steenrod operations on the cohomology of Grassmannians. Adv. Math. 136 (1998), no. 2, 251–283. (link to paper on ScienceDirect)

In the final section, you find the specific example $Gr_2(\mathbb{C}^5)$ and remarks on how to detect non-trivial cell attachment using the action of Steenrod operations.<|endoftext|>
TITLE: GAGA and Chern classes
QUESTION [23 upvotes]: My question is as follows.
Do the Chern classes as defined by Grothendieck for smooth projective varieties coincide with the Chern classes as defined with the aid of invariant polynomials and connections on complex vector bundles (when the ground field is $\mathbf{C}$)?
I suppose GAGA is involved here. Could anybody give me a reference where this is shown as detailed as possible? Or is the above not true?
Some background on my question:
Let $X$ be a smooth projective variety over an algebraically closed field $k$. For any integer $r$, let $A^r X$ be the group of cycles of codimension $r$ rationally equivalent to zero. Let $AX=\bigoplus A^r X$ be the Chow ring.
Grothendieck proved the following theorem on Chern classes.
There is a unique "theory of Chern classes", which assigns to each locally free coherent sheaf $\mathcal{E}$ on $X$ an $i$-th Chern class $c_i(\mathcal{E})\in A^i(X)$ and satisfies the following properties:
C0. It holds that $c_0(\mathcal{E}) = 1$. 
C1. For an invertible sheaf $\mathcal{O}_X(D)$ on $X$, we have that $c_1(\mathcal{O}_X(D)) = [D]$ in $A^1(X)$. 
C2. For a morphism of smooth quasi-projective varieties $f:X\longrightarrow Y$ and any positive integer $i$, we have that $f^\ast(c_i(\mathcal{E}))  =c_i(f^\ast(\mathcal{E}))$.
C3. If $$0\longrightarrow \mathcal{E}^\prime \longrightarrow \mathcal{E} \longrightarrow \mathcal{E}^{\prime\prime} \longrightarrow 0$$ is an exact sequence of vector bundles on $X$, then $c_t(\mathcal{E}) = c_t(\mathcal{E}^\prime)c_t(\mathcal{E}^{\prime\prime})$ in $A(X)[t]$.
So that's how it works in algebraic geometry. Now let me sketch the complex analytic case.
Let $E\longrightarrow X$ be a complex vector bundle. We are going to associate certain cohomology classes in $H^{even}(X)$ to $E$. The outline of this construction is as follows.
Step 1. We choose a connection $\nabla^E$ on $E$;
Step 2. We construct closed even graded differential forms with the aid of $\nabla^E$; 
Step 3. We show that the cohomology classes of these differential forms are independent of $\nabla^E$. 
Let us sketch this construction. Let $k= \textrm{rank}(E)$. Let us fix an invariant polynomial $P$ on $\mathfrak{gl}_k(\mathbf{C})$, i.e.   $P$ is invariant under conjugation by $\textrm{GL}_k(\mathbf{C})$. 
Let us fix a connection $\nabla^E$ on $E$. We denote denote its curvature by $R^E = (\nabla^E)^2$. One shows that $$R^E \in \mathcal{C}^\infty(X,\Lambda^2(T^\ast X)\otimes \textrm{End}(E)).$$ That is, $R^E$ is a $2$-form on $X$ with values in $\textrm{End}(E)$. Define $$P(E,\nabla^E) = P(-R^E/{2i\pi}).$$ (This is well-defined.)
The Chern-Weil theorem now says that:
The even graded form $P(E,\nabla^E)$ is a smooth complex differential form which is closed.  The cohomology class of $P(E,\nabla^E)$ is independent of the chosen connection $\nabla^E$ on $E$. 
Choosing $P$ suitably, we get the Chern classes of $E$ (by definition). These are cohomology classes. In order for one to show the equivalence of these "theories" one is forced to take the leap from the Chow ring to the cohomology ring.
How does one choose $P$? You just take $P(B) = \det(1+B)$ for a matrix.
Motivation: If one shows the equivalence of these two theories one gets "two ways" of "computing" the Chern character.

REPLY [23 votes]: See this question.
Also, read Milnor-Stasheff or Hatcher's book "Vector Bundles and K-Theory".  In particular, Milnor-Stasheff and Hatcher prove that there is a unique "theory of Chern classes" for complex vector bundles over topological spaces satisfying axioms totally analogous to the C0, C1, C2, C3. Milnor-Stasheff constructs Chern classes using the Thom isomorphism theorem; Hatcher constructs them using the Leray-Hirsch theorem. Hatcher's construction (in topology) is essentially the same as Grothendieck's construction (in algebraic geometry). I think if you study the two constructions (Hatcher's and Grothendieck's) carefully, their equivalence should follow fairly easily. I did this once a while ago.
I don't think you need any GAGA theorem. I think you just need the fact that there is an analytification functor.
Appendix C of Milnor-Stasheff then proves the equivalence with the Chern-Weil theory.<|endoftext|>
TITLE: How can I tell whether a Poisson structure is symplectic "algebraically"?
QUESTION [7 upvotes]: My primary motivation for asking this question comes from the discussion taking place in the comments to What is a symplectic form intuitively?.
Let $M$ be a smooth finite-dimensional manifold, and $A = \cal C^\infty(M)$ its algebra of smooth functions.  A derivation on $A$ is a linear map $\{\}: A \to A$ such that $\{fg\} = f\{g\} + \{f\}g$ (multiplication in $A$).  Recall that all derivations factor through the de Rham differential, and so: Theorem: Derivations are the same as vector fields.
A biderivation is a linear map $\{,\}: A\otimes A \to A$ such that $\{f,-\}$ and $\{-,f\}$ are derivations for each $f\in A$.  By the same argument as above, biderivations are the same as sections of the tensor square bundle ${\rm T}^{\otimes 2}M$.  Antisymmetric biderivations are the same as sections of the exterior square bundle ${\rm T}^{\wedge 2}M$.  A Poisson structure is an antisymmetric biderivation such that $\{,\}$ satisfies the Jacobi identity.
Recall that sections of ${\rm T}^{\otimes 2}M$ are the same as vector-bundle maps ${\rm T}^*M \to {\rm T}M$.  A symplectic structure on $M$ is a Poisson structure such that the corresponding bundle map is an isomorphism.  Then its inverse map makes sense as an antisymmetric section of ${\rm T^*}^{\otimes 2}M$, i.e. a differential 2-form, and the Jacobi identity translates into this 2-form being closed.  So this definition agrees with the one you may be used to of "closed nondegenerate 2-form".

Question: Is there a "purely algebraic" way to test whether a Poisson structure is symplectic?  I.e. a way that refers only to the algebra $A$ and not the manifold $M$?

For example, it is necessary but not sufficient that $\{f,-\} = 0$ implies that $f$ be locally constant, where I guess "locally constant" means "in the kernel of every derivation".  The easiest way that I know to see that it is necessary is to use Darboux theorem to make $f$ locally a coordinate wherever its derivative doesn't vanish; it is not sufficient because, for example, the rank of the Poisson structure can drop at points.
Please add tags as you see fit.

REPLY [3 votes]: In the purely algebraic setting, Daniel Farkas proved in his beautiful paper [Farkas, Daniel R. Characterizations of Poisson algebras. Comm. Algebra  23  (1995), no. 12, 4669--4686. MR1352562] that a Poisson-simple linear Poisson algebra over an algebraically closed field is a regular symplectic domain, a partial converse of the much easier fact that a commutative regular affine domain which is symplectic is Poisson-simple. There are examples of non symplectic Poisson-simple polynomial algebras, though.<|endoftext|>
TITLE: What is the local structure of a Lie groupoid?
QUESTION [6 upvotes]: A manifold is locally $\mathbb R^n$.  An orbifold is locally $\mathbb R^n/\{\text{finite group}\}$.  Is there a similar way to think about the local structure of a Lie groupoid $G_1 \rightrightarrows G_0$?
For example, the Lie algebroid determines a distribution on $G_0$, and I think that it is locally integrable?  What extra structure "local" structure of the groupoid is there (e.g. this distribution loses the data about the automorphisms of a point).
Finally, is the right notion of "local structure" well-behaved under equivalences of groupoids?  If it is, then I really should change the title of the question to "What is the local structure of a smooth stack?".

REPLY [6 votes]: This is roughly what I know about how to do devissage on algebraic stacks. It may or may not apply to differentiable stacks. 

Given an Artin stack X (say of finite type over the base, a field if you like), there is an nonempty open substack U in X over which the inertia is flat. This means the automorphism groups (a family of Lie groups) is varying "smoothly" over points in U. 
When the inertia is flat over X, one can take the "rigidification" of X with respect to the full inertia. This means we modulo the automorphisms, and get an algebraic space Y (non-stacky stack, which might be singular, so I don't want to call it a manifold).
The morphism X --> Y is called a gerbe, namely fibers are classifying spaces of the automorphism groups. Then one can use many techniques (like Leray spectral sequence etc.) to deduce/compute something (like cohomology) on X from Y and the fibers.<|endoftext|>
TITLE: Finding all roots of a polynomial
QUESTION [19 upvotes]: Is it possible, for an arbitrary polynomial in one variable with integer coefficients, to determine the roots of the polynomial in the Complex Field to arbitrary accuracy? When I was looking into this, I found some papers on homotopy continuation that seem to solve this problem (for the Real solutions at least), is that correct? Or are there restrictions on whether homotopy continuation will work? Does the solution region need to be bounded?

REPLY [5 votes]: I give a course on matrix analysis, and I like to mention the following thing. It is equivalent to finding the roots of polynomials, or to finding the eigenvalues of matrices, because to a polynomial $P$, you can associate its companion matrix $C_P$, and to a square matrix $M$, you can associate its characteristic polynomial $\chi_M$.
Now, because the calculation of $\chi_M$ is costly, and direct solving methods are ill-conditionned, it is usually a bad idea to use polynomial solving in order to compute eigenvalues.
Instead, the efficient idea is to apply the QR method to the companion matrix $C_P$ when you want to calculate the roots of $P$. This way has several advantages:

the method does converge. Proven.
it is fast. Actually, each iteration produces a Hessenberg matrix (only one non-zero sub-diagonal) and this ensures that the cost of an iteration is only $O(n^2)$, instead of $O(n^3)$.
it is stable. Because an iteration acts as a unitary conjugation. Therefore the round-off errors only add up, but are not amplified.

I agree that because of the round-off errors, this produces on a computer approximations of limited (but outstanding) precision. What you can do is to stop after a few hundred iterations, and then apply Newton's method or something similar, starting from one of the approximate eigenevalues.<|endoftext|>
TITLE: Visualizing a complex plane cubic together with the real plane
QUESTION [9 upvotes]: In Alain Roberts "Elliptic curves: notes from postgraduate lectures given in Lausanne 1971/72" page 11 (available on google books unless you already tried to read another chapter), there is a hand drawn picture of a real 2-dimensional torus and a real plane, which topologically represent the way a complex cubic (with two real components) and the real projective plane sit in the complex projective plane. Taking the picture on face value, one should be able to project an open subset of the complex projective plane to $\mathbb{R}^3$, so that there is some real line $L$ that passes through the "doughnut" defined by the image of the complex cubic.
I tried to reproduce this picture on a computer, using the map $\mathbb{CP}^2\to\mathbb{R}^7$ given by
$(z_1:z_2:z_3)\mapsto(z_2\overline{z_3},z_3\overline{z_1},z_1\overline{z_2},|z_1|^2-|z_2|^2)/(|z_1|^2+|z_2|^2+|z_3|^2)$,
projecting to various $\mathbb{R}^3$s, and looking for $L$ by trial and error; all in vain. Which brings me to....
Questions:

Is there such a line (the map I used does not send the real projective plane to a plane, so it does not have to be the case even if Roberts picture is correct) ?
Is there an algorithm to find such a line ? 
Is there a "better" way to project an open part of the complex projective plane to $\mathbb{R}^3$ ?

REPLY [3 votes]: I found this article:
"Visualizing Elliptic Curves"
by Donu Arapura
it is available at the following URL:
http://www.math.purdue.edu/~dvb/graph/elliptic.pdf
In it he discusses a projection that sends sends the real
part of $x$ to $x_1$ and the real part of $y$ to $x_3$ thus
it would seem to preserve the entire real plane and any line
in it. So this might be useful to you.<|endoftext|>
TITLE: Does central limit theorem hold for general weakly dependent variables?
QUESTION [5 upvotes]: Say I have $X_{ij}$, $j \le i$ with the property that $X_{ij}$ are centered and identically distributed and $E(X_{ij} X_{ij'}) = o(\exp(-i)))$. Then does $\sum_j X_{ij}$ have Gaussian domain of attraction? 
As a related question, if $X_1, X_2, X_3$ are identically distributed and centered and $E(X_i X_j) = c$, what bound can I get for $E(X_1 X_2 X_3)$ in terms of $c$?

REPLY [6 votes]: Not necessarily. One has to impose more restrictive mixing and moment conditions. A classical book is:
Ibragimov I.A., Linnik Yu.V. Independent and stationary sequences of random variables
There is a long-standing question asked by Ibragimov: is $\phi$-mixing and finiteness of second moment sufficient for CLT to hold for a stationary sequence?
Also, there are various concepts of dependence. For example, if your r.v.'s are associated (i.e. satisfy FKG inequalities) and the covariance decays as you describe, then CLT holds. 
UPD. As for the second part of your question: you cannot estimate higher-order moments in terms of lower-order ones unless the joint distributions have some special structure.

REPLY [3 votes]: Your double subscripts are extraneous.  Let's consider a simpler situation, where we have a single family of random variables $\{X_i\}$.
As Yuri Bakhtin says above, your condition is not sufficient for a CLT to hold.  Here is a simpler situation, however:  suppose that $X_i$ and $X_j$ satisfy finite-range dependence.  That is, there exists a positive integer $R$ such that if $|i-j| \ge R$, then $X_i$ and $X_j$ are independent.  We will prove a law of large numbers for $\{X_i\}$.  If you're interested, you can push it farther to prove a central limit theorem.  Suppose that $X_i$ has mean $\mu$ for each $i$.
Let $S_N = \tfrac{1}{N} \sum_{i=1}^N X_i$ as usual.  Without loss of generality, we may consider indices only divisible by $R$:  $S_{RN} = \tfrac{1}{RN} \sum_{i=1}^{RN} X_i$.  Let $$S_{RN}^{(k)} = \tfrac{1}{N} \sum_{j=0}^{N-1} X_{Rj+k}$$ for $k= 1, \dots, R$, so that $$S_{RN} = \tfrac{1}{R} \left( S_{RN}^{(1)} + \dots + S_{RN}^{(R)} \right).$$Each sum $S_{RN}^{(k)}$ is comprised of independent random variables, so the classical law of large numbers applies and $S_{RN}^{(k)} \to \mu$ both in probability and almost surely.  Consequently, $S_{RN} \to \mu$.  
Obviously, this argument breaks down when $R = \infty$.  In that case, the problem is no longer trivial and you will have to be more cautious with your assumptions.<|endoftext|>
TITLE: Who thought that the Alexander polynomial was the only knot invariant of its kind?
QUESTION [15 upvotes]: I apologize that this is vague, but I'm trying to understand a little bit of the historical context in which the zoo of quantum invariants emerged.
For some reason, I have in my head the folklore:
The discovery in the 80s by Jones of his new knot polynomial was a shock because people thought that the Alexander polynomial was the only knot invariant of its kind (involving a skein relation, taking values in a polynomial ring, ??).  Before Jones, there were independent discoveries of invariants that each boiled down to the Alexander polynomial, possibly after some normalization.
Is there any truth to this?  Where is this written?

REPLY [2 votes]: You can find some historical remarks on polynomial invariants in chapter 9 of the book
"Knots and links" by Cromwell, he also gives a lot of references.<|endoftext|>
TITLE: The Closure-Complement-Intersection Problem
QUESTION [21 upvotes]: Background
Let $A$ be a subset of a topological space $X$.  An old problem asks, by applying various combinations of closure and complement operations, how many distinct subsets of $X$ can you describe?
The answer is 14, which follows from the observations that $Cl(Cl(A))=Cl(A)$, $\neg(\neg (A)=A$ and the slightly harder fact that
$$ Cl(\neg (Cl(\neg (Cl(\neg (Cl(A))))=Cl(\neg (Cl(A))$$
where $Cl(A)$ is the closure of $A$ and $\neg (A)$ is the complement.  This makes every expression in $Cl$ and $\neg$ equivalent to one of 14 possible expressions, and all that remains is to produce a specific choice of $A$ which makes all 14 possiblities distinct.
This problem goes by the name Kuratowski's closure-complement problem, since it was first stated and solved by Kuratowski in 1922.
The Problem
A very similar problem recently came up in a discussion that was based on a topological model for modal logic (though the logical connection is unrelated to the basic question).  The idea was to take a subset $A$ of $\mathbb{R}$, and to consider all possible expressions on $A$ consisting of closure, complement, and intersection.  To be clear, we are allowed to take the complement or closure of any subset we have already constructed, and we are allowed to intersect any two subsets we have already constructed.
The question: Is this collection of subsets always finite?
A potentially harder question: If there are multiple starting subsets $A_1$, $A_2$, $A_3$... (finite in number), is this collection of subsets always finite?
The first question is essentially the Kuratowski question, with the added operation of intersection.  There is also the closely related (but slightly stronger) question of whether there are a finite number of formally distinct expressions on an indeterminant subset $A$ (or a collection of subsets $A_1$, $A_2$...).
Some Thoughts
My guess to both questions is yes, but the trick is showing it.  I can take an example of a set $A$ which realizes all 14 possibilities from Kuratowski's problem, and show that the collection of distinct subsets I can construct from it is finite.  However, just because such a set captures all the interesting phenomena which can happen when closing and complementing, doesn't mean this example is missing a property that is only important when intersecting.
It also seems difficult to approach this problem formally.  The problem is that there are many non-trivial intersections of the 14 expressions coming from Kuratowski's theorem.  Then, each of these intersections could potentially have its own new set of 14 possible expressions using closures and complements.  In examples, the intersections don't contribute the full number of 14 new sets, but its hard to show this aside from case-by-case analysis.

REPLY [6 votes]: As François mentioned, in 1922 Kuratowski gave the first published example of a space and seed set $x_0$ such that the sequence $\lbrace x_i\rbrace$ is infinite, where $x_{i+1}=x_i\cap cl(cl(x_i)-x_i)$.
The following equations give a more detailed picture of why the sequence is infinite:
space = $\lbrace1,2,3,\ldots\rbrace$
topology = $\lbrace {\rm space}, \lbrace \rbrace ,\lbrace 1\rbrace ,\lbrace 1,2\rbrace ,\lbrace 1,2,3\rbrace ,\ldots\rbrace $
$x_0=\lbrace 2,4,6,\ldots\rbrace $
$cl(x_0)=\lbrace 2,3,4,\ldots\rbrace $
$y=cl(x_0)-x_0=\lbrace 3,5,7,\ldots\rbrace $
$cl(y)=\lbrace 3,4,5,\ldots\rbrace $
$x_0\cap cl(y)=\lbrace 4,6,8,\ldots\rbrace $
$\ldots$<|endoftext|>
TITLE: What is the relationship between being normal and being regular?
QUESTION [14 upvotes]: On a scheme, being normal means that each stalk of the structure sheaf is a integrally closed domain.
Being regular means that each stalk of the structure sheaf is a regular local ring.
As for a local ring, being regular or being integrally closed does not imply another.
What is their connection with each other and classical/usual intuition of being smooth(being regular on stalk of each closed points)?
Moreover, is there a smooth/regular variety which is not normal?

REPLY [26 votes]: Dear 7-adic, yes there is an implication between the two notions.
For a local ring, regular implies normal. Actually Auslander and Buchsbaum proved in 1959 that a regular local ring is a UFD and it is an easy result that a UFD (local or not) is integrally closed. Serre then gave a completely different proof. He proved that regular is equivalent to having finite global (=homological) dimension . This finiteness means that any module over the ring has a finite projective resolution. I have heard it claimed that this was the beginning of the acknowledgment of the importance of homological algebra in commutative algebra.
An example.The cone $z^2=xy$ in affine 3-space (over a field, say) is normal but not regular: its very equation suggests that we don't have the UFD property and this intuition can be converted into a rigorous proof. 
Normality is a weak form of regularity. The two concepts coincide in dimension one but not in higher dimensions: the quadratic cone above shows this in dimension two. 
Finally, smoothness is even stronger: it is a relative concept meaning regular and remaining regular after base change.

REPLY [6 votes]: You seem a bit confused.  A regular* local ring is a UFD hence integrally closed.  In other words, regular implies normal.  See for instance
http://www.math.iitb.ac.in/atm/caag1/jayanthan.pdf
for a relatively elementary algebraic treatment.  
*: I had previously included Noetherian here, but after checking on this I see I was being overly careful: it is part of the definition of a regular local ring that it be Noetherian.<|endoftext|>
TITLE: Applications of string topology structure
QUESTION [26 upvotes]: Chas and Sullivan constructed in 1999 a Batalin-Vilkovisky algebra structure on the shifted homology of the loop space of a manifold: $\mathbb{H}_*(LM) := H_{*+d}(LM;\mathbb{Q})$. This structure includes a product which combines the intersection product and Pontryagin product and a BV operater $\Delta: \mathbb{H}_*(LM) \to \mathbb{H}_{*+1}(LM)$.
I was wondering about the applications of this structure. Has it even been used to prove theorems in other parts of mathematics? A more concrete question is the following: Usually, considering a more complicated structure on topological invariants of a space allows you to prove certain non-existince results. For example, the cup product in cohomology allows you to distinguish between $ S^2 \vee S^1 \vee S^1 $  and 
$T^2$. Is there an example of this type for string topology?

REPLY [8 votes]: As a shameless plug, I may say that in my thesis we do show that string topology, interpreted in a broader context, is NOT a homotopy invariant. What we do is the following : instead of looking at loops in $M$ we think of them as arcs in $M\times M$ with its boundary in the diagonal $M$ that sits inside $M\times M$. Now we look at the space of such arcs $\mathcal{S}(M)$, which, when they intersect the diagonal at intermediate stages, do so transversely. One can then define a suitable coalgebra structure which is NOT a homotopy invariant. In particular, this structure distinguishes the Lens spaces $L(7,1)$ from $L(7,2)$, which are homotopy equivalent but NOT homeomorphic. 
Of course, this new structure is not related to the loop product or the BV operator as per the question asked. Moreover, this structure is defined on a much smaller space then $LM$. However, if you take the point of view that string topology is broadly the study of loops in a manifold then this is a new and interesting algebraic structure.

REPLY [2 votes]: The string topology of a manifold is isomorphic to the Hamiltonian Floer Homology of the cotangent bundle of the manifold.<|endoftext|>
TITLE: How to find ICM talks?
QUESTION [83 upvotes]: I am very interested in reading some and skimming through the list of invited talks at the International Congress of Mathematicians. Since the proceedings contain talks supposedly by top experts in each area, even the list of invited talks would hopefully provide some picture of how mathematics changed throughout the last century or so. 
I looked it up but wikipedia only provides links to the proceedings of ICM since 1998. So it excludes many talks I really want to read, like those by Serre, Grothendieck, Auslander, Quillen, etc.  
Does anyone know how to find the rest of the ICM proceedings, hopefully online? Thanks. 
UPDATE: The whole collection of all the proceedings of the ICMs is available here!

REPLY [60 votes]: Update: (Oct. 2018) For the first time, all ICM 2018 lectures (plenary, invited and special) as well as panels and special events are presented (by good-quality videos) on the ICM 2018 You tube channel.
Update:(Dec 2017) The ICM launched a new website. All previous ICM proceedings are available here. However, I cannot find the old page with access to individual papers and search options.
Just recently the International Mathematical Union (IMU) put online all the previous proceedings on the ICM's! Here: (update:broken). This webpage is based on joint work by R. Keith Dennis (Ithaca) and Ulf Rehmann (Bielefeld). So you can read for free all the articles (including Hilbert's famous problem paper; Martin Grötschel demonstrated it at the opening ceremony of ICM 2010). I don't know when the proceedings of ICM2010 will be added. UPDATE: the 2010 articles have now been added! Update The ICM2014 papers are available here (you can download each of 4 volumes; the IMU site does not contain these papers yet.)
As for the talks themselves, there is a page with all the ICM 2010 plenary talks here; links for videos from earlier ICM's (plenary talks and other events) can be found here.
Update (August, 6 2014) Many (46 for now) of the ICM 2014 proceedings contributions are already available on arXiv, via this search. (I got it from Peter Woit's blog.) Videos of lectures can be found here.
Update (May 2014): Starting 1992 there is also every four years the European Congress of Mathematics (ECM) that the European Mathematical Society (EMS) is running. 
The proceedings of the first three ECMs are now freely open. These volumes are available here.
(Digitising the proceedings of the first three ECMs, published by Birkhäuser was a task carried out by the EMS Electronic Publishing Committee.)
Starting with the 4ECM (2004), the Proceedings are published by the EMS Publishing House. The EMS decided to make them freely available online too. I expect that this will happen soon and I will keep you posted. Further update (June 2014) The ECM Proceedings are now available here! I was told that in a few months, the EMS will put also the 6ECM volume.<|endoftext|>
TITLE: How would You encourage graduate students to learn algebraic geometry and/or complex analysis?
QUESTION [6 upvotes]: Hello,
I am the 3rd year undegraduate student of mathematics.
After I obtain a bachelor degree I want to study maths at graduate level, especially algebraic geometry and complex analysis.
This fields of mathematics are well-represented at my univeristy, so at the first glance this plan looks fine.
Unfortunately almost all of my collegues are not interested in this fields (they think that AG and CA are too technically involved; most of them are interested mainly in functional analysis and topology).
It will have such unpleasant effect on my studies, that most probably the most (or all) of courses in AG and CA in the upcoming year won't start at my university.
So I'll end up learning alone, from books. 
It's not that it's a big problem to learn from the book, it's not a problem at all. But I think such learning have no comparison with the regular course, where I could discuss problems and see another approches of other (more gifted) students.
To avoid such situation I may try to convince my collegues to study CA and AG, but I don't have many arguments since I'm still an ignorant in this fields.
And here arise my questions:


How would You encourage graduate students to learn algebraic geometry?
How would You encourage graduate students to learn complex analysis?

REPLY [5 votes]: This feels really like an "Ask Professor Nescio" question. 
Let me ask you a question: if you feel like you cannot learn what you want to learn, why are you staying at the same university? I see from your profile that you are studying a Jagielloni, and you are Polish, so I understand somewhat that, if you want to remain in Poland, you feel that you should stay. But mathematics being the international field as it is, I would recommend going to a different university in a different country. (For example, in the US there seems to be no lack of graduate students who share your interest. I'm sure if you ask around a bit you can find out about other places in Europe.) 
Now, with that said, if you decide that you want to stay:

Don't over sell it. Being too pushy will have a negative effect on the other students. 
Don't be evangelical. You should not tell them why they should be interested or why they ought to study the subject with you. That'll have the opposite effect. 
From your descriptions you need to first dispel the myth that complex analysis and algebraic geometry is too technical. I guess the best thing to do is to introduce them to partial differential equations. (That's a joke.) But you need to be able to show them some examples of how sometimes, things becomes much more clear when viewed in the right framework. Show them a nice theorem or two with relatively simple proofs. A nice forum for this could be an informal seminar organized by the students for themselves: try to run a seminar where each student presents a result (not due to himself) that he finds interesting (don't just hijack it for your ulterior motives). When it is your turn talk about something really pretty from algebraic geometry. It may win you some converts. 
Find out what your fellow students like to do. You said functional analysis and topology. Anything else? You need to sell to your audience. For the topologists, at least some introductory complex analysis and algebraic geometry should be that hard to sell: tell them about (Hirzebruch-)Riemann-Roch! Tell them about the works of Kodaira! Complex analysis in one-variable is basically just topology anyway. (Can't help you with the functional analysts there.)
An extension of the above: convince your fellow students that those subjects are useful for them. So a good idea is to find some theorems in their field that was first proven, or has nice interpretations, using the tools of complex analysis or algebraic geometry. 

If all else fail, and you cannot get another person to study with you, you can always ask questions here or on sci.math.<|endoftext|>
TITLE: A question on group action on categories 
QUESTION [8 upvotes]: Let $Gr$ be the affine Grassmannian of $G=G((t))/G[[t]]$, and let $Perv(Gr)$ be the category of perverse sheaves on $Gr$. We have action of $G((t))$ on the left-hand side of $Perv(Gr)$, also we have action of the tensor category $Rep(G^\vee)$ on the right-hand side, through geometric Satake correspondence. 
It is clear that we have the action of $G((t))$ on $Perv(Gr)$ as functors.
Follow Gaitsgory's paper "the notion of category over stack", he claimed that an action of algebraic group scheme $H$ on a category $\mathcal{C}$ is actually equivalent to a category $\mathcal{C}$ with the action of the tensor category "Rep(H)".
Go back to the original set-up, that means we also have an action of $G^\vee$ on $Perv(Gr)$.
I think, on this $Perv(Gr)$, the actions of $G((t))$ and $G^\vee$ should have different meaning, but why they gave them the same name?
Am I confused?

REPLY [8 votes]: The $G((t))$ action and the $Rep(G^\vee)$ [or equivalently of $G^\vee$ itself after deequivariantization, as Victor explains] are of quite different natures -- the former is a "smooth" action, and the latter an "algebraic" or "analytic" actions (the adjectives smooth and analytic come from analogy with p-adic rep theory). 
i.e. there are many kinds of notion of group action, and they are (to me) most conveniently summarized by describing the corresponding notion of group algebra which acts. An algebraic action of a group on a category is an action of the "quasicoherent group algebra" of G, ie the monoidal category of quasicoherent sheaves wrt convolution. (though I'd feel much safer if we said all this in a derived context, makes me uneasy otherwise).
A smooth action is an action of the monoidal category of D-modules on G, the "smooth group algebra" -- analog of smooth functions on a p-adic group. Such an action is the same as an algebraic action, which is infinitesimally trivialized. Such examples are studied in Chapter 7 of Beilinson-Drinfeld's Hecke manuscript and the appendix to the long paper by Gaitsgory-Frenkel, in particular.
PS the "equivariantization" dictionary between categories over BH and categories with H action is a nice simple case of descent --- you describe things over BH as things over a point with descent data, that descent data is given by the map H --> pt, 
the two maps H x H---> H, and so on. When you assemble this together (most efficiently using the Barr-Beck theorem) you get the desired dictionary. (Of course if you want to consider categories as forming a 2-category you'd need a 2-categorical version of Barr-Beck, but for most practical purposes I know of you can get by with the current
Lurie [$(\infty,$]1-categorical version.<|endoftext|>
TITLE: Question on transversal slice of Lie group
QUESTION [6 upvotes]: Assume we have action of Lie group $G$ on a manifold $X$. Fix some orbit $\mathcal{O}$, it is known there exist transversal slice $S$ with respect to this orbit. Fix some point $x$ in $\mathcal{O}$, and let $G_x$ be the stabilizer of $x$.
My question is, can we find a transversal slice which is $G_x$-stable?
How about in the algebraic situation?

REPLY [2 votes]: Assume the base field is of characteristic zero. If $G$ is an affine algebraic group with reductive connected component which acts by morphisms on an affine variety $X$, then Luna has shown that there exists a slice étale at $x$ for each closed orbit $G\cdot x$ in $X$. This means there exists a $G_x$-invariant locally closed affine subvariety $S$ of $X$ containing $x$ such that the morphism $\psi: G*_{G_x}S\to X$, $[g,s]\mapsto gs$ is excellent. In particular, the image of $\psi$ is a saturated open subset $V$ of $X$ and 
$\psi: G*_{G_x}S\to V$ is étale. Here $G*_{G_x}S$ denotes the homogeneous fiber bundle 
$(G\times S)//G_x$. A good, easily available reference are the notes by 
Drézet.<|endoftext|>
TITLE: Conjugacy classes intersecting subgroups of finite groups
QUESTION [11 upvotes]: I figured out the first part of this years ago, but completely forget how I did it.  I looked at the second, but don't think I figured it out.
This I am sure is true, but don't remember why.  Suppose that G is a finite group of size $n$, and H is a normal subgroup with |G/H| = $k$.  Then at least $\frac{1}{k}$ of the conjugacy classes of G are within H.
This I don't know the answer to.  If H is allowed to be an arbitrary subgroup, must H intersect at least $\frac{1}{k}$ of the conjugacy classes?
An example of a simple consequence of the first statement is that if we look at $S_n$ and $A_n$ is that at least half the partitions of $n$ have an even number of even parts.

REPLY [4 votes]: If we let $k(G)$ denote the number of conjugacy classes of the finite group $G$,
then for any subgroup $H$ of $G$ (normal or not), a Theorem of P.X. Gallagher
states that $[G:H]^{-1}k(H) \leq k(G) \leq [G:H]k(H)$ (this has probably been discovered
and rediscovered many times). I find that the easiest way to see it is using irreducible
complex characters. Of course, $k(G)$ is also the number of complex irreducible characters of $G$,
and likewise for $H$. For each irreducible character $\chi$ of $G$, there is an irreducible 
character $\mu$ of $H$ such that $\mu$ occurs with non-zero multiplicity in the restriction of $\chi$ to $H$. By Frobenius reciprocity, $\chi$ is an irreducible constituent of the character of $G$
induced from the character $\mu$ of $H$. On the other hand, using Frobenius reciprocity again, each irreducible constituent of $\mu$ induced to $G$ must have degree at least $\mu(1).$ Thus there are (even including multiplicities) at most $[G:H]$ irreducible constituents of $\mu$ induced to $G$
Since this is true for each irreducible character of $H$, and since each irreducible character of $G$ must appear with non-zero multiplicity in at least one such character, we have 
$k(G) \leq [G:H]k(H).$ Going in the other direction, if $\chi$ is an irreducible character of $G$
and $\mu$ is an irreducible constituent of te restriction of $\chi$ to $H$,  then we have 
$\chi(1) \leq [G:H]\mu(1)$, since $\chi$ occurs as a constituent of $\mu$ induced to $G$.
Thus $\mu(1) \geq \frac{\chi(1)}{[G:H]}$. Hence there are at most $[G:H]$ distinct irreducible
constituents of the restriction of $\chi$ to $H$. Since each irreducible character of $H$ occurs
as a constituent of some such irreducible character of $G$ (consider, for example, the restriction
of the regular character), we have $k(H) \leq [G:H] k(G).$<|endoftext|>
TITLE: A ring on which all finitely generated projectives modules are free but not all projectives are free?
QUESTION [11 upvotes]: Dear all, 
Sorry if the question is naive: any nice example of such a ring or, better, of a class of such rings?

REPLY [29 votes]: Cher Michel, these rings are uncommon.
1) Over a local ring ALL projective modules are free : this is a celebrated theorem due to Kaplansky.
2) If  $R$ is commutative noetherian and $Spec(R)$ is connected, every NON-finitely generated projective module is free. This is due to Bass in his article "Big projective modules are free" which you can download for free here 
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.ijm/1255637479
And now for the good news:  the rings you are after are uncommon but they exist. Bass in the article just quoted shows that the ring $R=\mathcal C([0,1])$ of continuous functions on the unit interval has all its finitely generated projective modules free. Nevertheless the ideal consisting of functions vanishing in a neighbourhood of zero (depending on the function) is projective, not finitely generated and not free. Bass attributes the result to Kaplansky.<|endoftext|>
TITLE: Surprising Analogue of Q
QUESTION [21 upvotes]: I was describing Manish Kumar's work a few weeks ago to a fellow graduate student, and she stumped me with a big-picture question I couldn't answer.
Manish Kumar proved that the commutator subgroup of $\pi_1(\mathbb{A}^1_K)$, where $K$ is a characteristic $p$, algebraically closed field, is pro-finite free. He proved this, in fact, for any smooth affine curve over $K$.
(He proved this for algebraically closed fields of char $p$ which are uncountable in his thesis: http://www.math.msu.edu/~mkumar/Publication/thesis.pdf; and without the cardinality restriction in: http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf)
As I explained to my colleague, this is a geometric analogue of Shafarevich's conjecture, that $Gal(\mathbb{Q}^{ab})$ is pro-finite free. Indeed $Gal(\mathbb{Q}^{ab})=\pi_1^c(Spec(\mathbb{Q}))$ ($c$ stands for the commutator subgroup). But why is $\mathbb{A}^1_K$, for $K$ an algebraically closed characteristic $p$ field (or indeed, any smooth affine curve over $K$), an analogue of $Spec(\mathbb{Q})$? Usually $\mathbb{A}^1_K$ (for $K$ an algebraically closed field) is an analogue of $Spec(\mathbb{Z})$. I came up with some partial explanations, but no full heuristic. Can you think of one?

REPLY [6 votes]: One can see that the commutator subgroup of the topological fundamental group of a complex
curve is free for elementary reasons, but this is a pretty weak analogy. 
In fact I don't have a good heuristic of why it should be true, other than the fact that is.
I was Manish's adviser, and I was pretty surprised by the result when he proved it.<|endoftext|>
TITLE: Interesting results in algebraic geometry accessible to 3rd year undergraduates
QUESTION [55 upvotes]: On another thread I asked how I could encourage my final year undergraduate colleagues to take an algebraic geometry or complex analysis courses during their graduate studies.
Willie Wong proposed me following idea - to show them some interesting results in this fields with relatively simply proofs and some consequences in other fields.
Thus by 'interesting' result in algebraic geometry I here mean the result which may convince 3rd year undergraduate student to study algebraic geometry.
In fact I'm supposed to give some talk during the seminar dedicated to final year undergraduates, and I can propose my own topic, so I thought that it could work.
But my problem is that I'm just wanna-be student of algebraic geometry and I don't have enough insight and knowledge to find a topic which 'could work'.
Also I doubt whether it is possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field.
So in short my first question is as above:

Is it possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field?

To be more precise - the talk should take a one or two meetings, 90 minutes each one.
The audience will be, as I said 3rd year undergraduate students, all of them after two semester course in algebra, some of them after one or two semesters in commutative algebra. All taking the course in one complex variable, and after one semester introductory course in differential geometry.
Some of them may be interested in number theory.
The second, related question is as follows:

If You think that the answer to the previuos question is positive, please try to give an example of idea/theorem/result which would be accessible in such time for such audience, and which You find interesting enough to make them consider possibility of studying algebraic geometry. Try to think about the results which shows connections of AG with some other fields of mathematics.

REPLY [28 votes]: If you are willing to cheat over some details, you could present a "proof" of the statement that every smooth cubic surface in $\mathbb{CP}^3$ contains exactly 27 lines. The fact that lines on such a surface can be counted, and that the result turns out not depend on the cubic seemed to me like magic before I learned algebraic geometry.
I think this topic is good for a number of reasons. The statement is elementary and suprising. The proof is fairly elementary as well (granting some things), but is far from trivial and shows many interesting ideas typical of algebraic geometry, like the use of parameter spaces. Moreover, it is not so elementary that you will be able to give a complete proof in one lecture, so interested students will want to know more. Finally the most delicate concepts from an algebraic geometry point of view are those of smoothness and dimension, but you can do with the analog concept in differential geometry, thereby providing a link with something they already know.
I have written all the steps just to be sure, and I don't think there is anything too complicated for your audience. Of course, depending on the time, you will want to skip over some details and present them as black boxes.
The proof
First, you can define smoothness in terms of differential geometry, and make the claim that an algebraic variety is smooth in a point if and only if the Jacobian of the defining equations has maximal rank. Remark that one implication always holds in the differentiable case, but there may be smooth manifolds defined by equations which are singular. Polynomials are rigid enough to prevent this. You could actually prove this fact for hypersurfaces, which is the case you are interested in. So, students will be able to understand what a smooth subic surface is.
The first step of the proof is the fact that every cubic surface, smooth or not, contains lines. This is the least elementary part, and you may want to skip this until the end of the talk.
Once you have a line $l$ consider planes $H$ containing $l$. The intersection of $H$ with the cubic is a plane cubic. Factoring the polynomial, it must be the union of $l$ with a smooth conic, or three lines. Moreover you can show that in the second case the three lines are distinct and do not meet in one point, otherwise the cubic would not be smooth there. This is completely elementary: you just prove that if the three lines do not form a triangle, all partial derivatives of the defining equation for your cubic surface vanish at a point.
The third step is to give a bijection between the set of planes containing $l$ and $\mathbb{P}^1$. A simple (but a bit long) computation tells you when the residual conic in the plane is smooth. Namely the vanishing of the determinant of the residual conic is an equation of degree $5$ on this $\mathbb{P}^1$. One can show that this equation has distinct roots, again because the cubic is smooth.
We conclude that for a given line $l$ there are exactly $5$ planes through $l$ on which the cubic decomposes as the union of three lines. Said differently, every line meets exactly $10$ other lines.
Finally, this implies the total number of lines is $27$. Indeed take any of these planes containing $3$ lines, call them $r$, $s$ and $t$. Any other line on the surface meets the plane, hence it meets one between $r$, $s$ and $t$. There are no triple of lines meeting in a point, because such a point would be singular. Hence each of $r$, $s$ and $t$ meets $8$ other lines, giving a total of $3 \times 8 + 3 =27$ lines.
The black box
It remains the black box that you can find at least one line on the surface. Here you will have to be sketchy, but hopefully this will wet your students appetite to see more about algebraic geometry.
First, you define the Grassmannian as a set, and make the claim that it is an algebraic variety. This is particularly easy for the Grassmannian of lines in $\mathbb{P}^3 = \mathbb{P}(V)$, since it is defined by the unique quadratic equation $\alpha \wedge \alpha = 0$ inside $\mathbb{P}(\bigwedge^2 V)$. Similarly you show that the parameter space for cubic surfaces is again an algebraic variety itself, namely a projective space $\mathbb{P}^{19}$.
Then you define the incidence variety of couples $(l, X)$ of a line and a cubic such that $l \subset X$; this is a subvariety of the product. The argument now uses the theorem on the dimension of the fibres of an algebraic morphism. Of course you cannot prove it, but it is rather plausible. Make the remark that such a neat statement can possibly hold only because polynomials are rigid enough. For the notion of dimension, you can use the dimension as complex manifolds, or half the real dimension.
Show that the set of cubics containing a given line is itself a projective space $\mathbb{P}^{15}$. Since the Grassmannian has complex dimension $4$ (this is plausible, being a hypersurface in $\mathbb{P}^5$), the incidence variety has dimension $19$, like the parameter space for cubics. By the theorem on the dimension of the fibres, to show that the projection is surjective you only need to exhibit a cubic which has a fibre of dimension $0$, that is, a cubic with a finite number of lines.
This is easy to do explicitly. Note that it is easier to do with a singular cubic, so to obtain a result about smooth cubics it will be easier to work with a parameter space containing singular ones.<|endoftext|>
TITLE: How to compute the Picard-Lefschetz monodromy matrix of a non-semistable fiber?
QUESTION [13 upvotes]: Let $f:X\to B$ be a family of curves of genus $g$ over a smooth curve $B$. Let $F_0$ be a singular fiber. 
If $F_0$ is a semistable fiber, the monodromy matrix can be gotten by the classical Picard-Lefschetz formula. 
If $F_0$ is non-semistable, I don't know how to compute its monodromy matrix. For example, in Namikawa and Ueno's paper[1], they can compute the Picard-Lefschetz monodromy matrix for each type of singular fiber of genus 2.  It's not clear to me how they did that. 
[1] Namikawa, Y. and Ueno, K., The complete classification of fibres in pencils of curves of genus two, Manuscripta math., Vol. 9 (1973), 143-186.

REPLY [5 votes]: One approach (I don't know how effective it is in the genus 2 case you asked about)
is to explicitly apply the semi-stable reduction theorem, and so reduce to the semi-stable
case.  
To achieve semi-stable reduction, you have to alternately blow-up singular points in
the special fibre, and then make ramified base-changes.  The latter operation just extracts
a root of the monodromy operator (i.e. if $\gamma$ is a generator of $\pi_1$ of the punctured
$t$-disk, and we set $t = s^n$, then $\gamma = \tau^n,$ where $\tau$ is a generator of
$\pi_1$ of the punctured $s$-disk), so it is easy to see how the monodromy matrix
changes.  And blowing up a point in the special fibre doesn't change the monodromy action
around the puncture at all.
So using this process, one can relate the original (unknown) monodromy matrix
to the corresponding matrix in the semi-stable context, where it is known thanks to the
Picard--Lefshcetz formula.<|endoftext|>
TITLE: Why is Proj of any graded ring isomorphic to Proj of a graded ring generated in degree one?
QUESTION [16 upvotes]: I have seen it stated that Proj of any graded ring $A$, finitely generated as an $A_0$-algebra, is isomorphic to Proj of a graded ring $B$ such that $B_0 = A_0$ and $B$ is generated as a $B_0$-algebra by $B_1$.
Could someone either supply a reference for or a sketch a proof of this statement?
Note: An obvious approach to this question is to make $B$ a Veronese subring of $A$.  However, when I try this approach, I end up getting a terrible combinatorics problem that I do not know how to approach.

REPLY [17 votes]: Bourbaki Commutative Algebra Chapter 3:
Let $A$ be a non-negatively graded algebra.  Assume that $A$ is finitely generated over $A_0$.  There exists $e \geq 1$ such that $A^{me} = A_0[A^{me}_1]$ for any $m \geq 1$.
Here $A^{e} = \oplus_{n \in\mathbb Z} A^{e}_n$, where $A^{e}_n := A_{ne}$.
The desired result follows because replacing $A$ with $A^{me}$ does not change Proj.<|endoftext|>
TITLE: "Fourier-Mukai" functors for Fukaya categories?
QUESTION [11 upvotes]: I just skimmed a bit of this fresh-off-the-press paper on homological mirror symmetry for general type varieties.
One thing that intrigued me was statement (ii) of Conjecture 3.3. It suggests that, just as there are Fourier-Mukai functors $DCoh(X) \to DCoh(Y)$ associated to objects of $DCoh(X \times Y)$, there are also "Fourier-Mukai" functors $DFuk(M) \to DFuk(N)$ associated to objects of $DFuk(M \times N)$. How does this work, or how might this work? The paper does not seem to explain it.

REPLY [13 votes]: I can't speak for these authors, but what I understand by a "Fourier-Mukai" transform between Fukaya categories is the functor between extended Fukaya categories associated with a Lagrangian correspondence. I expect these will appear a good deal in the next few years, in symplectic topology and its applications to low-dim topology and mirror symmetry (cf. these papers of Auroux and Abouzaid-Smith).
A Lagrangian correspondence from $X$ to $Y$ is an embedded Lagrangian submanifold of $X\times Y$ with symplectic form $(-\omega_X)\oplus\omega_Y$. One can take the graph of a symplectomorphism, for instance, or the vanishing moment-map locus $\mu^{-1}(0)$ as a correspondence from a Hamiltonian $G$-manifold $M$ to the quotient $\mu^{-1}(0)/G$. According to Wehrheim-Woodward, a generalised Lagrangian in $X$ is a sequence of symplectic manifolds $X_0=pt., X_1, X_2,\dots,X_d=X$ and Lagrangian correspondences $L_{i,i+1}$ from $X_i$ to $X_{i+1}$. Generalised Lagrangians (subject to the usual sorts of restrictions and decorations) form objects in the extended Fukaya category $F^{\sharp}(X)$, whose $A_\infty$-structure is under construction by Ma'u-Wehrheim-Woodward. If it happens that two adjacent Lagrangian correspondences have a smooth, embedded composition, say $L' = L_{i+1,i+2}\circ L_{i,i+1}$, then deleting $X_{i+1}$ from the chain and substituting $L'$ for its two factors results in an isomorphic object; see arxiv:0905.1368. 
The geometric mechanism behind this is the idea of "pseudo-holomorphic quilts", see arXiv:0905.1369, arxiv:math.SG.0606061.
Whilst $F^\sharp(X)$ seems even less tractable than the Fukaya category $F(X)$, we expect that in many cases the embedding $F(X)\to F^\sharp(X)$ induces a quasi-isomorphism of module categories.
A Lagrangian correspondence from $X$ to $Y$ induces a "Fourier-Mukai" $A_\infty$-functor between extended Fukaya categories. Even better, we expect that in this way one gets an $A_\infty$-functor $F(X_-\times Y)\to \hom(F^{\sharp}(X),F^{\sharp}(Y))$. 
Interpolated paragraph, in response to Kevin's query: The definition of the $A_\infty$-functor associated with a correspondence $C$ from $X$ to $Y$ is very simple, at least on objects. An object of $F^\sharp(X)$ is a chain of Lagrangian correspondences, beginning at the 1-point manifold and ending at $X$. We just tack $C$ to the end of this sequence. The clever thing about this formalism is that when there's a nice geometric way to pass Lagrangians through a correspondence (for instance taking the preimage of $L\subset \mu^{-1}(0)/G$ in $\mu^{-1}(0)\subset M$) the geometric and formal approaches give quasi-isomorphic objects in $F^\sharp(Y)$. (Usually you can't pass a Lagrangian submanifold through a correspondence without making a terrible mess - hence the formal approach.)
For me, all this is exciting because we can at last compute Floer cohomology using its naturality properties, rather than by direct attack on the equations.
[In the new paper that inspired your question, the authors suggest that their F-M kernels should be coisotropic branes in the sense of Kapustin-Orlov. Floer cohomology for such branes is supposed to be some weird mixture of pseudo-holomorphic discs and Dolbeault cohomology over a holomorphic foliation - but there is no concrete proposal on the table
and for the moment this is just an intriguing idea. For the purposes of homological mirror symmetry, idempotent endomorphisms in the Fukaya category apparently provide the enlargement that K-O observed was necessary.]<|endoftext|>
TITLE: Is every homogeneous G-variety of the form G/H?
QUESTION [6 upvotes]: Let $G$ be an algebraic group over an algebraically closed field $k$. Then G/H is a quasi-projective homogeneous G-variety for any closed subgroup $H$. Now, several times I have seen something like "Let $X$ be a homogeneous $G$-variety, i.e. $X = G/H$ for a closed subgroup $H$ of $G$" and I wonder if this "i.e." is correct. This would imply that any homogeneous $G$-variety is already quasi-projective. I think this is true, when $\mathrm{char}(k) = 0$, because then the canonical abstract isomorphism $\pi:X \rightarrow G/G _x$ is separable and thus an isomorphism of varieties for any $x \in X$ (is this correct?). But what about $\mathrm{char}(k) > 0$? Are there counter-examples or is any (quasi-projective) homogeneous $G$-variety up to isomorphism of the form $G/H$?

REPLY [3 votes]: An aside: BCnrd's argument above shows that, over any perfect field $k$, $X$ is isomorphic to $G/H$ if and only if $X$ has a $k$-point. Otherwise, consider, e.g., conics with no point: they are homogeneous under a twist of $G=SO_3$ but are not $G/H$. I don't know what happens for more general base schemes.<|endoftext|>
TITLE: candidate for rigorous _mathematical_ definition of "canonical"?
QUESTION [7 upvotes]: In this question: What is the definition of "canonical"?
, people gave interesting "philosophical" takes on what the word "canonical" means. Moreover I percieved an underlying opinion that there was no formal mathematical definition. 
Whilst looking for something else entirely, I just ran into Bill Messing's post
http://www.cs.nyu.edu/pipermail/fom/2007-December/012359.html
on the FOM (Foundations of Mathematics) mailing list. I'll just quote the last paragraph:
"It is my impression that there is very little FOM discussion of either
Hilbert's epsilon symbol or of Bourbaki formulation of set theory.  In
particular the chapitre IV Structures of Bourbaki.  For reasons,
altogether mysterious to me, the second edition (1970) of this book
supressed the appendix of the first edition (1958).  This appendix gave
what is, as far as I know, the only rigorous mathematical discussion of
the definition of the word "canonical".  Given the fact that Chevalley
was, early in his career, a close friend of Herbrand and also very
interested in logic, I have guessed that it was Chevalley who was the
author of this appendix.  But I have never asked any of the current or
past members of Bourbaki whom I know whether this is correct."
It's a 4-day weekend here in the UK and I'm very unlikely to get to a library to find out what this suppressed appendix says. Wouldn't surprise me if someone could find this appendix on the web somewhere though! Is there really a mathematical definition of "canonical"??
NOTE: if anyone has more "philosophical" definitions of the word, they can put them in the other thread. I am hoping for something different here.

REPLY [12 votes]: Although the Bourbaki formulation of set theory is very seldom used in foundations, the existence of a definable Hilbert $\varepsilon$ operator has been well studied by set theorists but under a different name. The hypothesis that there is a definable well-ordering of the universe of sets is denoted V = OD (or V = HOD); this hypothesis is equivalent to the existence of a definable Hilbert $\varepsilon$ operator.
More precisely, an ordinal definable set is a set $x$ which is the unique solution to a formula $\phi(x,\alpha)$ where $\alpha$ is an ordinal parameter. Using the reflection principle and syntactic tricks, one can show that there is a single formula $\theta(x,\alpha)$ such that for every ordinal $\alpha$ there is a unique $x$ satisfying $\theta(x,\alpha)$ and every ordinal definable set is the unique solution of $\theta(x,\alpha)$ for some ordinal $\alpha$. Therefore, the (proper class) function $T$ defined by $T(\alpha) = x$ iff $\theta(x,\alpha)$ enumerates all ordinal definable sets.
The axiom V = OD is the sentence $\forall x \exists \alpha \theta(x,\alpha)$. If this statement is true, then given any formula $\phi(x,y,z,\ldots)$, one can define a Hilbert $\varepsilon$ operator $\varepsilon x \phi(x,y,z,\ldots)$ to be $T(\alpha)$ where $\alpha$ is the first ordinal $\alpha$ such that $\phi(T(\alpha),y,z,\ldots)$ (when there is one).
The statement V = OD is independent of ZFC. It implies the axiom of choice, but the axiom of choice does not imply V = OD; V = OD is implied by the axiom of constructibility V = L.

When I wrote the above (which is actually a reply to Messing) I was expecting that Bourbaki would define canonical in terms of their $\tau$ operator (Bourbaki's $\varepsilon$ operator). However, I was happily surprised when reading the 'état 9' that Thomas Sauvaget found, they make the correct observation that $\varepsilon$ operators do not generally give canonical objects.
A term is said to be 'canonically associated' to structures of a given species if (1) it makes no mention of objects other than 'constants' associated to such structures and (2) it is invariant under transport of structure. Thus, in the species of two element fields the terms 0 and 1 are canonically associated to the field F, but $\varepsilon x(x \in F)$ is not since there is no reason to believe that it is invariant under transport of structures. They also remark that $\varepsilon x(x \in F)$ is actually invariant under automorphisms, so the weaker requirement of invariance under automorphisms does not suffice for being canonical.

To translate 'canonically associated' in modern terms:
1) This condition amounts to saying that the 'term' is definable without parameters, without any choices involved. (Note that the language is not necessarily first-order.)
2) This amounts to 'functoriality' (in the loose sense) of the term over the core groupoid of the concrete category associated to the given species of structures.
So this seems to capture most of the points brought up in the answers to the earlier question.<|endoftext|>
TITLE: When the splitting fields of shifted generic polynomials are linearly disjoint?
QUESTION [9 upvotes]: Let me start by rigorously pose my question. 
Let $K$ be an algebraically closed field of characteristic $2$, let $n$ be an even integer number, let $f(X) = X^n + T_1 X^{n-1} + \cdots + T_n$, be the generic polynomial, that is, $T = (T_1, \ldots, T_n)$ is a tuple of algebraically independent variables over $K$. 
Let $\Omega = $ { $\omega_1, \ldots, \omega_m$} be a finite subset of $K$, let $f_i(X) = f(X) - \omega_i$, and let $F_i$ be the splitting field of $f_i$ over $K(T)$ ($i=1,\ldots, m$).
Question: For which $\Omega$ the splitting fields $F_1, \ldots, F_m$ are linearly disjoint over $K(T)$?
Remarks:

If the characteristic of $K$ is NOT $2$, or if $n$ is odd, then the splitting fields are linearly disjoint for arbitrary $\Omega$. Thus, I pose the question the specific case of $p=2$ and $n$ even.
The answer cannot be ALWAYS, as in the previous remark. Indeed, one can show that if $p=n=2$, $m=4$, and $\omega_1 + \omega_2 + \omega_3 + \omega_4 = 0$, then the splitting fields are not linearly disjoint. In fact, if $p=n=2$, the answer is that the splitting fields are linearly disjoint if and only if the sum of any even number of elements of $\Omega$ does not vanish. 
How one proves 1 + 2: The linear disjointness of the splitting fields can be reduced to the linear independent of the discriminant as elements in $H^1(K,\mathbb{Z}/2\mathbb{Z})$. If $p\nmid n$, then one can use ramification theory to achieve this, if $p\neq 2$ but divides $n$, one can calculate this by hand using the formula given by the determinant of the Sylvester matrix. If $p=2$, I know of no formula for the discriminant in terms of the coefficients. However when $p=n=2$ situation is simple enough to do calculations and hence get 2. 

Motivation: The linear disjointness of the splitting fields allows one to calculate a Galois group of a composite of polynomials, which in turn yields arithmetic features of the ring of polynomials over large finite fields. Let me not elaborate on that here

REPLY [2 votes]: Lior, you say:

If $p=2$, I know of no formula for the
discriminant in terms of the
coefficients.

However, this link says that the Sylvester matrix works the same over every field, whether $p=2$ or not. The article on PlanetMath about determinants seems to support this, not including a requirement of $p\neq2$. Using the determinant Sylvester matrix you get a formula in terms of the polynomial's coefficients like you wanted; I don't see why counting a determinant over a field with $p=2$ should be any different than in any other field, of course other than the fact that your definition of '$+$' is different. It doesn't seem you need anything other than to know how to differentiate polynomials over your field.
On the other hand, the Wikipedia article on the determinant also says that the determinant of quadratic forms cannot be generalized for fields of characteristic $p=2$. Is this what you were thinking of?
I have a feeling an answer to your question might be in the book "Generic Polynomials. Constructive Aspects of the Inverse Galois Problem" By Jensen, Ledet, Yui.
You might also find inspiration in the paper "Computing Galois Groups of Polynomials through ODEs", by Cormier, Singer, Trager, Ulmer, 2000, wherein they construct a linear differential operator that takes a polynomial and spits out information about its Galois group over the field as well as its algebraic closure; you might be able to come up with some form of chain rule for composite polynomials, solving your original problem. The paper was submitted to Journal of Symbolic Computation.<|endoftext|>
TITLE: Is there an analog of class field theory over an arbitrary infinite field of algebraic numbers?
QUESTION [11 upvotes]: Recently, I found a paper by Schilling http://www.jstor.org/pss/2371426, which mentions that for certain infinite field of algebraic numbers there is an analog of class field theory. By infinite field of algebraic number we mean an infinite extension of $\mathbb{Q}$. The paper cite a previous paper by Moriya which was the origin of the idea. I could not read the later since it is in German. Since the first paper is quite old (1937), I believe there must have been a lot of development in the mean time.
My question: Do we have an analog of class field theory over an arbitrary infinite field of algebraic number?
An even more general question: Do we have an analog of class field theory over an arbitrary field. This seems a bit greedy, but since we know that an algebraic closed field of characteristic 0 is totally characterized by its trancendence degree so if the answer to the previous question is positive the answer to this is perhaps not too far. Am I making sense?

REPLY [8 votes]: Iwasawa theory studies abelian extensions of fields $K$ where $K$ is a $\mathbb{Z}_p$-extension of $\mathbb{Q}$, that is the Galois group of $K/\mathbb{Q}$ is $\mathbb{Z}_p$. The corresponding Galois groups (of extensions of $K$) and class groups (of really subfields) of $K$, suitably interpreted, become $\mathbb{Z}_p$-modules and there are interesting relations between these modules and $p$-adic $L$-functions. It is a vast subject. Washington's book, Introduction to Cyclotomic Fields, is a good entry point.<|endoftext|>
TITLE: Is there a notion of "Morse index" for geodesics in a manifold with indefinite metric that is well-behaved under cutting and gluing?
QUESTION [5 upvotes]: More generally, I'm interested in the situation of Lagrangian mechanics.  And actually my question is local, so you can work on $\mathbb R^n$ if you like.  I will begin with some background on Lagrangian mechanics, and then recall the notion of Morse index in the positive-definite case.  My final question will be whether there is a similar story in the indefinite case.
Background on Lagrangian mechanics
Let $\mathcal N$ be a smooth manifold.  A Lagrangian on $\mathcal N$ is a function $L: {\rm T}\mathcal N \to \mathbb R$, where ${\rm T}\mathcal N$ is the tangent bundle to $\mathcal N$.  I will always suppose that the Lagrangian is nondegenerate, in the sense that when restricted to any fiber ${\rm T}_q\mathcal N$, the second derivative $\frac{\partial^2 L}{\partial v^2}$ is everywhere invertible (for $v\in {\rm T}_q\mathcal N$, $\frac{\partial^2 L}{\partial v^2}(v)$ makes sense as a map ${\rm T}_v({\rm T}_q\mathcal N) \to {\rm T}_v^*({\rm T}_q\mathcal N)$).  In this case, the Euler-Lagrange equations
$$ \frac{\partial L}{\partial q}\bigl(\dot\gamma(t),\gamma(t)\bigr) = \frac{\rm d}{{\rm d}t}\left[ \frac{\partial L}{\partial v}\bigl(\dot\gamma(t),\gamma(t)\bigr)\right]$$
define a nondegenerate second order differential equation for $\gamma$ a parameterized path in $\mathcal N$.
An important example is when $\mathcal N$ is equipped with a (semi-)Riemannian metric, in which case the Euler-Lagrange equations specify that $\gamma$ is an arc-length-parameterized geodesic.
By nondegeneracy, a solution to the Euler-Lagrange equations is determined by its initial conditions, a point $\bigl(\dot\gamma(0),\gamma(0)\bigr) \in {\rm T}\mathcal N$.  Thus, the Euler-Lagrange equations determine a smooth function $\text{flow}: {\rm T}\mathcal N \times \mathbb R \to \mathcal N \times \mathcal N \times \mathbb R$, which sends an initial condition and a duration to the triple (initial location, final location, duration).  Actually, the function is only defined on an open neighborhood of the zero section of ${\rm T}\mathcal N \times \mathbb R$.
I will use the shorthand path to mean a smooth function $[0,T] \to \mathcal N$, i.e. a parameterized path of finite duration.  A path is classical if it solves the Euler-Lagrange equations, so that classical paths are in bijection with points in (that open neighborhood of) ${\rm T}\mathcal N \times \mathbb R$.  A classical path is nonfocal if near it the function $\text{flow}: {\rm T}\mathcal N \times \mathbb R \to \mathcal N \times \mathcal N \times \mathbb R$ is a local diffeomorphism.  Thus a choice of nonfocal classical path of duration $T$ determines a function $\gamma: \mathcal O \times [0,T] \to \mathcal N$, where $\mathcal O \subseteq \mathcal N \times \mathcal N$, such that for each $(q_0,q_1) \in \mathcal O$, the path $\gamma(q_0,q_1,-)$ is classical with $\gamma(q_0,q_1,0) = q_0$ and $\gamma(q_0,q_1,T) = q_1$.  Given a nonfocal classical path $\gamma$, the corresponding Hamilton principal function $S_\gamma: \mathcal O \to \mathbb R$ is given by:
$$S_\gamma(q_0,q_1) = \int_{t=0}^T L\left( \frac{\partial \gamma}{\partial t}(q_0,q_1,t), \gamma(q_0,q_1,t)\right){\rm d}t$$
Finally, given a classical path $\gamma: [0,T] \to \mathcal N$, there is a well-defined second-order linear differential operator $D: \gamma^*{\rm T}\mathcal N \to \gamma^*{\rm T}^*\mathcal N$, given by:
$$ D_\gamma[\xi] = -\frac{\rm d}{{\rm d}t} \left(\frac{\partial^2 L}{\partial v^2} \frac{{\rm d}\xi}{{\rm d}t}\right) - \frac{\rm d}{{\rm d}t}\left( \frac{\partial^2 L}{\partial q \partial v}\xi\right) + \frac{\partial^2 L}{\partial v \partial q} \frac{{\rm d}\xi}{{\rm d}t} + \frac{\partial^2 L}{\partial q \partial q}\xi$$
The second derivatives of $L$ are evaluated at $(\dot\gamma(t),\gamma(t))$ and act as "matrices"; in particular, $\frac{\partial^2 L}{\partial v \partial q}$ and $\frac{\partial^2 L}{\partial q \partial v}$ are transpose to each other.  These individual matrices require local coordinates to be defined, but $D_\gamma$ is well-defined all together if $\gamma$ is classical.
Then $\gamma$ is nonfocal iff $D_\gamma$ has trivial kernel among the space of sections of $\gamma^*{\rm T}\mathcal N \to [0,T]$ that vanish at $0,T$.
Of course, really what's going on is that for $(q_0,q_1) \in \mathcal N \times \mathcal N$, the space of paths of duration $T$ that start at $q_0$ and end at $q_1$ is an infinite-dimensional manifold.  Using the Lagrangian $L$ we can define an action function on this manifold.  The Euler-Lagrange equations assert that a path $\gamma$ is a critical point for this function, $S_\gamma$ is the value of the function, and the operator $D_\gamma$ is the Hessian at that point.
The Morse index of a classical path
Recall the following fact.  Let $\mathcal V$ be a vector space and $D: \mathcal V\otimes \mathcal V \to \mathbb R$ a symmetric bilinear form on $\mathcal V$.  Then any subspace  $\mathcal V_- \subseteq \mathcal V$ that is maximal with respect to the property that $D|_{\mathcal V_-}$ is negative-definite has the same dimension as any other such subspace.  This dimension is the Morse index $\eta$ of the operator $D$ acting on $\mathcal V$.
Recall that there is a canonical pairing between sections of $\gamma^*{\rm T}\mathcal N$ and sections of $\gamma^*{\rm T}^*\mathcal N$ (pairing the vectors and covectors for each $t\in [0,T]$ gives a function on $[0,T]$, which we then integrate).  By composing with this pairing, we can think of the operator $D_\gamma$ as a bilinear form on $\gamma^*{\rm T}\mathcal N$, and it is symmetric on the space of sections of $\gamma^*{\rm T}\mathcal N$ that vanish at the endpoints $0,T$.
Given a nonfocal classical path $\gamma$, its Morse index $\eta(\gamma)$ is the Morse index of $D_\gamma$ acting on such endpoint-zero sections.
Let $L$ be a Lagrangian on $\mathcal N$, and assume moreover that the matrix $\frac{\partial^2 L}{\partial v^2}$ is not just everywhere invertible but actually everywhere positive-definite (this is a coordinate-independent statement, even though the value of the matrix depends on coordinates).  Then the Morse index of any nonfocal classical path is finite.  (And conversely: if $\frac{\partial^2 L}{\partial v^2}$ is not positive-definite along $\gamma$, then the Morse index as defined above is infinite.)
Moreover, suppose that $\gamma: [0,T] \to \mathcal N$ is classical and nonfocal and choose $T' \in [0,T]$ such that the obvious restrictions $\gamma_1: [0,T'] \to \mathcal N$ and $\gamma_2: [T',T] \to \mathcal N$ are both nonfocal.  Let $S_{\gamma_1}$ and $S_{\gamma_2}$ be the corresponding Hamilton-principle functions.  Then $q = \gamma(T')$ is a nondegenerate critical point for $S_{\gamma_1}(\gamma(0),-) + S_{\gamma_2}(-,\gamma(T))$.  Define its Morse index of $\eta(q)$ to be the number of negative eigenvalues of the Hessian of $S_{\gamma_1}(\gamma(0),-) + S_{\gamma_2}(-,\gamma(T))$ at $q$.  Then the following is a fact:

$\eta(\gamma) = \eta(\gamma_1) + \eta(q) + \eta(\gamma_2)$

My question
Is there a similar story when $\frac{\partial^2 L}{\partial v^2}$ is invertible but indefinite?  More precisely:

Suppose that you are given a nondegenerate (but not convex on fibers) Lagrangian $L$ on a manifold $\mathcal N$.  Is there a way to assign a (finite) number $\eta(\gamma)$ to each classical nonfocal path $\gamma$ such that $\eta(\gamma) = \eta(\gamma_1) + \eta(q) + \eta(\gamma_2)$, where $\gamma,\gamma_1,\gamma_2$ are as above, and $\eta(q)$ is the usual Morse index of $S_{\gamma_1}(\gamma(0),-) + S_{\gamma_2}(-,\gamma(T))$?

The starting idea would be to recall the fact that in the positive-definite case, $\eta(\gamma)$ counts with multiplicity the number of times $T' \in [0,T]$ such that the restriction $\gamma|_{[0,T']}$ is focal.  (The multiplicity is given by the rank of the differential of the flow map.)  This counting still makes sense in the indefinite case.  So perhaps it can be used, and the signature of $S_{\gamma_1}(\gamma(0),-) + S_{\gamma_2}(-,\gamma(T))$ can be added in by hand?

REPLY [2 votes]: Yes, there is: The Maslov index and a generalized Morse index theorem for non-positive definite metrics.<|endoftext|>
TITLE: A plausible inequality
QUESTION [8 upvotes]: I come across the following problem in my study. 
Let $x_i, y_i\in \mathbb{R}, i=1,2,\cdots,n$ with $\sum\limits_{i=1}^nx_i^2=\sum\limits_{i=1}^ny_i^2=1$, and $a_1\ge a_2\ge \cdots \ge a_n>0 $. Is it true $$\left(\frac{\sum\limits_{i=1}^na_i(x_i^2-y_i^2)}{a_1-a_n}\right)^2\le 1-\left(\sum\limits_{i=1}^nx_iy_i\right)^2~~?$$
Has anyone seen this inequality before, or can you give a counterexample?

REPLY [5 votes]: The following proof is a bit heavy-handed; I'm sure you it can be simplified.  Assume $a_1=1, a_n=0$ as suggested above and write:
Write $$F(x,y) = \sum_{i=1}^n a_i(x_i^2-y_i^2)$$, $$G(x,y)=F(x,y)^2+\langle x,y\rangle^2.$$
Let $(x,y)\in S^{n-1}\times S^{n-1}$ be a point where $G$ is maximized, where we may assume that for each $i$ at least one of $x_i,y_i$ is non-zero.  It is clear that $-\sum_i y_i^2 \leq F(x,y) \leq \sum_i x_i^2$ so we may also assume $\langle x,y\rangle \neq 0$.
By the method of Lagrange multipliers there exist $\xi,\eta$ such that for all $i$
$$  4a_i x_i F+2\langle x,y\rangle y_i=2\xi  x_i$$ and
$$ -4a_i y_i F+2\langle x,y\rangle x_i=2\eta y_i.$$
Multiplying the first equation by $x_i$, the second by $y_i$, adding the two and summing over $i$ gives
$ 4G = 2(\xi+\eta)$.  Multiplying the second equation by $y_i$, the first by $x_i$ and adding gives $$ \langle x,y\rangle (y_i^2+x_i^2) = (\xi+\eta)x_i y_i = 2G\cdot x_i y_i. $$  By assumption one of $x_i,y_i$ is non-zero.  Dividing by the square of that number we see that the quadratic $$\langle x,y\rangle t^2 - 2G t + \langle x,y\rangle = 0$$ has a real root.  Evaluating the discriminant it follows that $$ G^2 \leq \langle x,y\rangle^2 \leq 1.$$<|endoftext|>
TITLE: What is a noncommutative fiber bundle?
QUESTION [16 upvotes]: Given a spectral triple (A,H,D) in the sense of Connes, what would be the right notion of a fiber bundle or a principal fiber bundle on it? An example of this type is the Connes' cosphere algebra S*A, which is the noncommutative analogue of the cosphere bundle of a 
Riemannian manifold. It is defined as the image in the Calkin algebra Q(H) of the C*-algebra generated by the compact opertators in H and all dialations of A by the continuous 1-parameter family of unitaries generated by the the derivation [|D|,.] on B(H). Connes showed that when 
the spectral triple is the canonical Dirac triple of a compact spin^c Riemannian manifold, the cosphere algebra is the C*-algebra of continuous functions on the cosphere bundle of the manifold. 
The question is: can we put this into a more general framework by finding the right notion of a fiber bundle on a spectral triple?

REPLY [2 votes]: It might not be the context you want (although if you think about this, it might be very close to what you want) but for some people a non-commutative bundle is a Banach bundle with non-commutative fibres. unlike a fibre bundle, the fibres are not necessarily isomorphic. Also bundles related to categories of Morita equivalence bimodules might be related at to the context I'm referring to because Connes' differential calculus is built into Morita equiv bimodules. The latter is a way to generalise the "tangent bundle" over the space described by A. I'm not an expert. Anyway, this approach seems to be working.<|endoftext|>
TITLE: Terminology: Algebras where long strings of products are 0?
QUESTION [9 upvotes]: I'd like a name for an augmented algebra $A = \langle 1\rangle \oplus A_+$ for which there is an $N$ so that any product of more than $N$ elements in the augmentation ideal is $0$, i.e., $(A_+)^N = 0$.  Is there a name?  It seems related to nilpotence, and it implies that all elements in $A_+$ are nilpotent, but is stronger than that.  There is a uniform bound on the degree of nilpotence, but that's not enough either, as the example of the exterior algebra in infinitely many variables over $\mathbb{Z}/2$ shows.
MathWorld defines a nilpotent algebra or nilalgebra to be one where every element is nilpotent.  (They are therefore not considering unital algebras, contrary to an earlier discussion here.)  Is this standard?  Is there a better term?

REPLY [3 votes]: It seems to me that Jordan and Dotsenko are giving different answers from one another, and I agree with Dotsenko’s.   The condition Thurston has stated is the definition of $A_{+}$ being nilpotent.   “Locally nilpotent” is a weaker condition.  There are many examples of nonunital rings $A_{+}$ that are locally nilpotent (meaning for any finite set $a_1, \ldots, a_k \in A_{+}$ there exists $n \in \mathbb{N}$ such that $a_{i_1} \cdots a_{i_n} = 0$ provided every $i_j \in \lbrace 1, \ldots, k\rbrace$) but not nilpotent (meaning there exists $n \in \mathbb{N}$ such that $a_1 \cdots a_n = 0$ provided every $a_i \in A_{+}$, which is the condition Thurston stated).   Even better: two nice (and quite different) examples of a locally nilpotent prime nonunital ring can be found in E. I. Zelmanov, “An example of a finitely generated primitive ring,” Sibirsk. Mat. Zh. 20 (1979), no. 2, 423, 461, and J. Ram, “On the semisimplicity of skew polynomial rings,” Proc. Amer. Math. Soc. 90 (1984), no. 3, 347–351.   (Of course, if one merely wants an example where $A_{+}$ is locally nilpotent but not nilpotent—and so does not satisfy Thurston’s condition—one could take something like $A_{+} = \bigoplus_{i=2}^{\infty} 2\mathbb{Z}/2^i\mathbb{Z}$.)
N.B. Mathematical Reviews incorrectly lists the title of Zelmanov’s paper as “An example of a finitely generated primary ring.”   It’s listed correctly in Zentralblatt.   Possibly the problem lies in the translation from the original Russian; the condition primitive in the English translation of the paper (Siberian Math. J. 20 (1979), no. 2, 303–304) is what we would today call prime.<|endoftext|>
TITLE: What's the cell structure of K(Z/nZ, 1)?  Does it let me sum this divergent series?  What about other finite groups?
QUESTION [19 upvotes]: The Eilenberg-Maclane space $K(\mathbb{Z}/2\mathbb{Z}, 1)$ has a particularly simple cell structure: it has exactly one cell of each dimension.  This means that its "Euler characteristic" should be equal to
$$1 - 1 + 1 - 1 \pm ...,$$
or Grandi's series.  Now, we "know" (for example by analytic continuation) that this sum is morally equal to $\frac{1}{2}$.  One way to see this is to think of $K(\mathbb{Z}/2\mathbb{Z}, 1)$ as infinite projective space, e.g. the quotient of the infinite sphere $S^{\infty}$ by antipodes.  Since $S^{\infty}$ is contractible, the "orbifold Euler characteristic" of the quotient by the action of a group of order two should be $\frac{1}{2}$.  
More generally, following John Baez $K(G, 1)$ for a finite group $G$ should be "the same" (I'm really unclear about what notion of sameness is being used here) as $G$ thought of as a one-object category, which has groupoid cardinality $\frac{1}{|G|}$.  In particular, $K(\mathbb{Z}/n\mathbb{Z}, 1)$ should have groupoid cardinality $\frac{1}{n}$.  I suspect that $K(\mathbb{Z}/n\mathbb{Z}, 1)$ has $1, n-1, (n-1)^2, ...$ cells of each dimension, hence orbifold Euler characteristic
$$\frac{1}{n} = 1 - (n-1) + (n-1)^2 \mp ....$$
Unfortunately, I don't actually know how to construct Eilenberg-Maclane spaces...
Question 1a:  How do I construct $K(\mathbb{Z}/n\mathbb{Z}, 1)$, and does it have the cell structure I think it has?  (I've been told that one can write down the cell structure of $K(G, 1)$ for a finitely presented group $G$ explicitly, but I would really appreciate a reference for this construction.)  
Question 2:  $K(\mathbb{Z}/2\mathbb{Z}, 1)$ turns out to be "the same" as the set of all finite subsets of $(0, 1)$, suitably interpreted; the finite subsets of size $n$ form the cell of dimension $n$.  Jim Propp and other people who think about combinatorial Euler characteristic would write this as $\chi(2^{(0, 1)}) = 2^{\chi(0, 1)}$.  Is it true more generally that $K(\mathbb{Z}/n\mathbb{Z}, 1)$ is "the same" as the set of all functions $(0, 1) \to [n]$, suitably interpreted?
Question 3:  What notion of "sameness" makes the above things I said actually true?  
Question 4:  Let $G$ be a finite group and let $K(G, 1)$ be constructed using the standard construction I asked about in Question 1.  If $c_n$ denotes the number of cells of dimension $n$, let $f_G(z) = \sum_{n \ge 0} c_n z^n$.  Can $f_G$ always be analytically continued to $z = -1$ so that $f_G(-1) = \frac{1}{|G|}$?

REPLY [11 votes]: The answer to something like question 4 is "yes".  It not only has an analytic continuation to $-1$, it has a unique such.  In fact, it's a rational function.  Something similar is true not only for finite groups but for "most" finite categories.  See this paper, especially Example 2.4.
(I say "something like question 4" because I'm taking $c_n$ to be the number of nondegenerate $n$-simplices in the nerve of $G$, which I think isn't quite what you have in mind.)<|endoftext|>
TITLE: Query: Year of Death of Three Mathematicians
QUESTION [5 upvotes]: I need the year of death of the following mathematicians all of whom are written up in R.C.Archibald's book Mathematical Table Makers.
Carl Burrau, b. 1867, d. ???? - Danish, astronomer and actuary
Herbert Bristol Dwight, b. 1885, d. ???? - American, tables of integrals
Alexander John Thompson, b. 1885, d. ????, British, statistician, BAASMTC
Thanks for any insight.
Cheers, Scott

REPLY [6 votes]: Just a comment on Šiljak's source about Dwight: the footnote (in Russian) says "Год смерти этого автора точно неизвестен и оценен как 1975", i.e. the year of death of Dwight was unknown by the author of that website and "estimated" as 1975.<|endoftext|>
TITLE: Homomorphisms of Topological Groups which are Automatically Fiber Bundles?
QUESTION [6 upvotes]: Suppose I have a surjective homomorphism of topological groups $f:E \to G$. Let K be the kernel of f. The topological group K acts on E in an obvious way. When is this a fiber bundle over G? (It will then be a K-principal bundle over G).  
I'm writing a paper where I make a claim about when this holds. I thought I had a reference but when I went looking for it, my claim was not in the reference. 
I don't want to consider examples that are too pathological so lets assume everything is Hausdorff and paracompact. (However if people are familiar with the more general setting, I'd be curious about that too!).
Clearly a necessary condition is that $G \cong E/K$, so let's assume this is the case. By homogeneity it is enough to show that f admits a local section in a neighborhood of the identity element of G. So my question is equivalent to asking if there are conditions I can impose on E and G which will ensure that f admit local sections near the identity of G. 
I know by work of G. Segal ("Cohomology of Topological Groups" Symposia Math. Vol IV 1970 pg 377, in the appendix) that if G is abelian and locally contractible then the sequence $$G \to EG \to BG$$
is of this kind. 
I want to know: 

Does this hold when K is locally contractible?
What if K is globally contractible?
Are there any simple (but not tautological) conditions I can put on K, E, or G to make this hold?

I'd also like to know some reasonable examples where this fails to be a principal bundle (if there are any).

REPLY [3 votes]: The following statement follows from results of Palais (see theorem 2.3.3 in "On the existence of slices for actions of non-compact Lie groups"). When $G$ is a Lie group, any principal $G$-bundle (in the sense of Husemoller's book, which does not assume local triviality) whose total space is completely regular is actually locally trivial.
It follows that when $H$ is a Lie group, and a closed subgroup of a Hausdorff topological group $G$, then $G\to G/H$ is a locally trivial principal $H$-bundle.
In a slightly different direction, it is a theorem of Skljarenko (theorem 15 of "On the topological structure of locally bicompact groups and their quotient spaces") that when $G$ is a locally compact Hausdorff group, and $H$ is a closed subgroup of $G$, then $G\to G/H$ is a Hurewicz fibration. In particular, if $G/H$ is (relatively) locally contractible then $G\to G/H$ is a locally trivial principal $H$-bundle. Again, by a result of Skljarenko, it is also enough to assume that $G/H$ is finite dimensional to conclude local triviality. See also Mostert's "Sections in principal fibre spaces".<|endoftext|>
TITLE: Flatness in Algebraic Geometry vs. Fibration in Topology
QUESTION [36 upvotes]: I am currently trying to get my head around flatness in algebraic geometry. In particular, I'm trying to relate the notion of flatness in algebraic geometry to the notion of fibration in algebraic topology, because they do formally seem quite similar. I'm guessing that the answers to my questions are "well-known", but I am struggling to find anything decent in the literature. Any help/references will be most useful.
The set up is this: Let $E,B$ be smooth projective complex algebraic varieties, and let $\pi:E \to B$ be a surjective flat map such that the fibres $E_b:=\pi^{-1}(b), b \in B$ are smooth projective complex algebraic varieties.
I am aware that each fibre has the same Hilbert polynomial, so cohomologically they are quite similar. But each fibre can certainly be non-isomorphic as algebraic varieties (e.g. moduli spaces). However:

Using GAGA type methods, we can consider $E$ and $B$ as complex manifolds. Is it true that $(\pi,E,B)$ is a fibration? That is, satisfies the homotopy lifting property with respect to any topological space?
Again considering $E,B$ and each fibre $E_b$ as a complex manifold, is it true that each fibre $E_b$ is homotopy equivalent to each other? What about homeomorphic/diffeomorphic?

Thanks,
Dan

REPLY [36 votes]: A surjective flat (equals faithfully flat) map with smooth fibres is in fact a smooth morphism, and hence induces a submersion on the underlying manifolds obtained by passing
to complex points.  Since the fibres are projective, it is furthermore proper (in the sense of algebraic geometry) [see the note added at the end; this is not a logical deduction from the given condition on the fibres, but nevertheless seems to be a reasonable reinterpretation of that condition], and hence proper (in the sense of topology).  A theorem
of Ehresmann states that any proper submersion of smooth manifolds is a fibre bundle.
In particular, it is a fibration in the sense of homotopy theory, and the fibres are 
diffeomorphic (thus also homeomorphic, homotopic, ... ). 
Note: Your specific question is really about smooth morphisms (these are flat morphisms
with smooth fibres, although there are other definitions too, which are equivalent
under mild hypotheses on the schemes involved, and in particular, are equivalent for maps
of varieties over a field).  One point about the notion of flat map is that it allows one
to consider cases in which the fibres over certain points degenerate, but still vary continuously (in some sense).  It may well be a special feature of algebraic geometry
(and closely related theories such as complex analytic geometry) that one can have such a reasonable notion, a feature related to the fact that one can
work in a reasonable manner with singular spaces in algebraic geometry, because the singularities are so mild compared to what can occur in (say) differential topology.
[Added: I should add that I took a slight liberty with the question, in that I interpreted the condition that the fibres are projective stronger than is literally justified, in so far as I replaced it with the condition that the map is proper.  As is implicit in Chris Schommer-Pries's comment below, we can find non-proper smooth surjections whose fibres are projective varieties: e.g. if, as in his example, we consider the covering of  $\mathbb P^1$ by two copies of $\mathbb A^1$ in the usual way, then the fibres consists of either one or two points (one point for $0$ and the point at $\infty$, two points for all the others), and any finite set of points is certainly a projective variety.
Nevertheless, my interpretation of the question seems to have been helpful; hopefully, with the addition of this remark, it is not too misleading.]<|endoftext|>
TITLE: Inverting the Weyl Character Formula
QUESTION [9 upvotes]: The Weyl Character formula tells us how to write the character of a representation as a linear combination of integral weights. Since characters are invariant under the action of the Weyl group, $W$, we can write a character as a linear combination of $W$-symmetrized dominant integral weights. It is know that the representation ring of a Lie algebra is isomorphic to $\mathbb{C}\[P\]^W$ as a vector space, where $P$ is the weight lattice of some Cartan subalgebra.
So we have two bases for the same vector space: the $W$-symmetrized dominant integral weights and the the character basis. The Weyl character formula tells us how to go from the former to the latter. My question is: is there much known about the matrix of going from the latter to the former? I've gone through a few low rank examples, and many of the coefficients are coming out to be zero. Does anyone know of a reference for this question in general?
Addendum: Jim makes a good point. The Weyl character formula isn't really needed. Perhaps we should just say that the matrix from the weight basis to the character basis is precisely the matrix of weight multiplicities. From this point of view it is clear that the matrix will be "upper triangular" (since weight multiplicities are zero above the highest weight). Thus the inverse should also be upper triangular. So my modified question is there any way to interpret the coefficients of the inverse matrix as counting anything interesting? As the matrix is upper triangular, we can certainly give recursive formulas for the coefficients. Does anyone have any other insight?

REPLY [4 votes]: Thanks Bruce Westbury for reminding me that I wrote something about this in my younger days. Here's what I make of it today, which provides a "closed formula" of sorts, a bit along the line of Allen Knutson's answer.
I'll deviate from the OP in writing $\Lambda$ for the weight lattice, $\Lambda^+$ for the set of dominant weights, and $X^\lambda$ for the basis elements of $\mathbf{Z}[\Lambda]$ with $\lambda\in\Lambda$, I'll use the dot-action $w\cdot\lambda=w(\rho+\lambda)-\rho$ and a related operator $J:\mathbf{Z}[\Lambda]\to\mathbf{Z}[\Lambda]$ sending $P\mapsto\sum_{w\in W}(-1)^{l(w)}w(X^\rho P)X^{-\rho}$ in general, and in particular $X^\lambda\mapsto\sum_{w\in W}(-1)^{l(w)}X^{w\cdot\lambda}$. Weyl's character formula says that the character $\chi_\lambda$ of $V_\lambda$ satisfies
$$
   \chi_\lambda.J(1)=J(X^\lambda)\quad\text{for all $\lambda\in\Lambda^+$}
$$
The left hand side is in fact also equal to $J(\chi_\lambda)$, since for any $P\in\mathbf{Z}[\Lambda]^W$ one has
$$
  J(P)=\sum_{w\in W}(-1)^{l(w)}w(X^\rho P)X^{-\rho}
  = P \sum_{w\in W}(-1)^{l(w)}w(X^\rho)X^{-\rho} = P.J(1),
$$
the second equality by $W$-invariance of $P$. Thus $\chi_\lambda$ and $X^\lambda$ have the same image by $J$.
Every $P\in\mathbf{Z}[\Lambda]$ is equivalent modulo $\ker(J)$ to a unique $P'\in\mathbf{Z}[\Lambda^+]$, i.e., a polynomial supported on the dominant weights. Concretely, define a $\mathbf{Z}$-linear operator $\alpha:\mathbf{Z}[\Lambda]\to\mathbf{Z}[\Lambda^+]$ by $\alpha(X^\mu)=0$ if $X^\mu\in\ker(J)$, which happens if $r\cdot\mu=\mu$ for some reflection $r\in W$, and otherwise $\alpha(X^\mu)=(-1)^{l(w)}X^{w\cdot\mu}$ where $w\in W$ is the unique element with $w\cdot\mu\in\Lambda^+$. Then $J(P)=J(\alpha(P))$ for all $P\in\mathbf{Z}[\Lambda]$. It follows from the above that $\alpha(\chi_\lambda)=X^\lambda$ for all $\lambda\in\Lambda^+$. In other words if $\chi:\mathbf{Z}[\Lambda^+]\to\mathbf{Z}[\Lambda]$ is the "character" map that linearly extends $X^\lambda\mapsto\chi_\lambda$, then $\alpha$ restricted to $\mathbf{Z}[\Lambda]^W$ defines the inverse "decomposition" map.
Now the decomposition of an orbit sum $m_\lambda=\sum_{\mu\in W(\lambda)}X^\mu$ is given by $\alpha(m_\lambda)=\sum_{\mu\in W(\lambda)}\alpha(X^\mu)$; since each $\mu$ gives at most one term, this shows that its decomposition involves at most as many irreducible factors as the size the $W$-orbit of $\lambda$.
If $\lambda$ is very large, it may be convenient to write $m_\lambda=\frac1s\sum_{w\in W}X^{w^{-1}(\lambda)}$ where $s$ is the size of the stabiliser in $W$ of $\lambda$, and using $\alpha(X^\mu)=\alpha((-1)^{l(w)}X^{w\cdot\mu})$ for any $\mu,w$, obtain 
$$
  \alpha(m_\lambda)
  =\frac1s\alpha\left(\sum_{w\in W}(-1)^{l(w)}X^{w\cdot(w^{-1}(\lambda))}\right)
  =\frac1s\alpha(X^\lambda J(1)).
$$
If $\lambda$ is strictly dominant one has $s=1$, and if moreover $\lambda$ is far enough off the walls that $X^\lambda J(1)$ is entirely supported on dominant weights, then the right hand side simply becomes $X^\lambda J(1)$, which has $|W|$ distinct terms.
This coincides with the expression that Allen Knutson guessed. However the requirement is rather stronger than he suggested: in type $A_n$ for instance, dominant weights can be represented by weakly decreasing $n+1$-tuples of integers, with $\rho=(n,n-1,\ldots,1,0)$, and the "off the walls" condition means that the successive entries for $\lambda$ decrease by at least $n+1$. In other words, in this case the simplified formula holds only if $\lambda-(n+1)\rho$ is dominant.<|endoftext|>
TITLE: Symplectic Steinberg group
QUESTION [8 upvotes]: I have several questions about Steinberg group and K2 for symplectic group:

Can I extend the definition of Steinberg symbols to symplectic case? Will they generate the center of Steinberg group?
Does the center of symplectic Steinberg group coincide with K2 (the kernel of $\mathrm{SpSt}\rightarrow\mathrm{Sp}$ as usual)?
Is there an analogue for Matsumoto's theorem?

I tryed to read "Sur les sous-groupes arithmetiques des groupes semi-simples deployes" by Hideya Matsumoto and all I got to know about symplectic case is that there is some problems with long roots in Cl. Also, is it written in english anywhere about non-Al K-theory? There is "The Classical groups and K-theory" by Hahn and O'Meara, but it tells about SL and about unitary groups only.

REPLY [2 votes]: Recently, centrality and Suslin's local-global principle for symplectic K_2 are studied by Andrei Lavrenov, student of Nikolai Vavilov.<|endoftext|>
TITLE: distribution of degree of minimum polynomial for eigenvalues of random matrix with elements in finite field
QUESTION [7 upvotes]: This is an attempt to extend the current full fledged random matrix theory to fields of positive characteristics. So here is a possible setup for the problem: Let $A_{n,p}$ be an $n \times n$ matrix with entries iid taking values uniformly in $F_p$. Then one should be able to find its eigenvalues together with multiplicities, which might lie in some finite extension of the field $F_p$. To ensure diagonalizability, one might even take $A_{n,p}$ to be symmetric or antisymmetric (I am not so sure if that guarantees diagonalizability in $F_p$ but I have no counterexamples either). Now the question is if we associate to each eigenvalue $\lambda$ the degree of its minimal polynomial $d(\lambda)$, then does the distribution of $d(\lambda)$ as $n$ goes to infinite converge to some law upon normalization (say maybe Gaussian)? I am very curious whether others have studied this problem before. Maybe it's completely trivial.

REPLY [12 votes]: The survey article
Jason Fulman, Random matrix theory over finite fields, Bulletin of the AMS 39 (2002), 51-85
and the references therein should answer your questions to the extent that the answers are currently known.  See in particular Example 3 in Section 2.2.  Roughly, the distribution of the degrees of the factors of the characteristic polynomial of a random matrix over a finite field is close to the distribution of the degrees of the factors of a random polynomial over the same finite field, which is close to the distribution of the cycle lengths of a random element of a symmetric group.<|endoftext|>
TITLE: What was Gödel's real achievement?
QUESTION [47 upvotes]: When I first heard of the existence of Gödel's theorem, I was amazed not just at the theorem but at the fact that the question could be made precise enough to answer: how on earth, even in principle, could one show that it was impossible to prove something in a given system? That doesn't bother me now, and that is not my question.
It seems to me that Gödel's theorem is a combination of at least three amazing achievements, namely these.

Formalizing the notions of proof, model, etc. so that the question could be considered rigorously.
Daring to think that there might be true but unprovable statements in Peano arithmetic.
Thinking of the idea of Gödel numbering and getting the proof to work.

One might think that 3 constitutes two separate achievements, but I think that actually getting the proof to work, though pretty good going, is somehow a technicality once you have had the idea that in principle a proof along those lines might be possible. (I'm not saying I could have done it, but Gödel would have been deeply immersed in these ideas.)
My guess is that pretty well all the credit for 2 and 3 goes to Gödel (except that the idea of diagonal proofs was not invented by him). My question is how much he contributed to 1 as well. Had it occurred to anyone else that it might be possible to think about such questions rigorously, or did an entirely new way of thinking appear pretty well out of nowhere? Popular accounts suggest the latter, but common sense would suggest the former, at least to some extent.

REPLY [4 votes]: If I remember correctly, the notion of a model is already present in Hilbert-Bernays where finite structures are used for proving absolute consistency of some theories, and is probably older. Again, if I remember correctly, Frege, Russel, and Hilbert did have formal systems and the notion of a formal proof. Skolem's construction of a term model (which is now famous as Skolem's Paradox because the set of real numbers of the model turns out to be countable) is in his 1922 paper, where the Godel's completeness theorem is from 1929. In other words, it seems that Skolem did already have all the tools necessary for proving completeness in 1922. It seems that Hilbert had even stated the question of completeness for first-order theories before this date and Godel has learned about this problem in Carnap's logic course in 1928.
Hilbert's 10-th problem from his famous 23 problems asks for an algorithm to decide existence of solutions for Diophantine equations. I think there were attempts after this for understanding what is an algorithm. There were many definitions which came out before Turing's definition which were equivalent to his definition, although they were not philosophically satisfactory, at least Godel did not accept any of them as capturing the intuitive notion of computability before Turing's definition.
Godel's collected works can shed more light on these issues.
EDIT: Also
Solomon Feferman, "Gödel on finitism, constructivity and Hilbert’s program"
http://math.stanford.edu/~feferman/papers/bernays.pdf

Hilbert and Ackermann posed the fundamental problem of the completeness of the first-order predicate calculus in their logic text of 1928; Gödel settled that question in the affirmative in his dissertation a year later. [page 2]
Hilbert introduced first order logic and raised the question of completeness much earlier, in his lectures of 1917-18. According to Awodey and Carus (2001), Gödel learned of this completeness problem in his logic course with Carnap in 1928 (the one logic course that he ever took!). [page 2, footnote]

Martin Davis, "What did Gödel believe and when did he believe it", BSL, 2005

Godel has emphasized the important role that his philosophical views had
played in his discoveries. Thus, in a letter to Hao Wang of December 7,
1967, explaining why Skolem and others had not obtained the completeness
theorem for predicate calculus, ... [page 1]<|endoftext|>
TITLE: $2^{\omega_1}$ separable?
QUESTION [5 upvotes]: I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof?

REPLY [3 votes]: This is indeed the Hewitt-Marczewski-Pondiczery theorem. My proof, following Engelking, is here. It's in fact not that hard, the fact for a product of copies of 2 point discrete spaces already implies the general theorem pretty quickly.<|endoftext|>
TITLE: Why do Littlewood-Paley projections behave like iid random variables
QUESTION [13 upvotes]: I have read more than once that the Littlewood-Paley (LP) projections of a function (i.e. decomposing a function into parts with frequency localization in different octaves) behave in some sense like iid random variables.
I am also aware of some facts (like inequalities for square functions vs. Khinchine Inequality) which "look similar". 
Is there any precise way of stating this similarity?
And why do we have this similarity?
Can we somehow interpret the LP projections as something like independent random variables?
A related question concerns systems of functions of the form
$$
    f_k(\cdot):=f(n_k \cdot )\quad {k\geq 1} ,
$$
with $(n_k)_{k\geq 1}$ a lacunary sequence. Also in this case (under suitable assumptions) the functions $f_k$ behave like iid random variables in the sense that they satisfy the Central Limit Theorem and the LIL.

REPLY [5 votes]: If one replaces the real line with the Walsh ring $F_2[t](\frac{1}{t})$ (or equivalently, replaces the Fourier transform by the Fourier-Walsh transform), then Littlewood-Paley projections become precisely the same thing as martingale differences.  See for instance the lecture notes of Pererya and Ward at
http://www.math.unm.edu/~crisp/papers/princeton.pdf
or my own lecture notes at
http://www.math.ucla.edu/~tao/254a.1.01w/notes5.ps
Very roughly speaking, the difference between the two is the difference between a sine wave and a square wave - and the latter, when viewed in binary, depicts the fluctuation of a random bit.  In contrast, a sine wave of frequency comparable to $2^k$ (and more generally, a Littlewood-Paley projection to that range of frequencies) depends primarily, but not exclusively, of the $k^{th}$ bit in the binary expansion of the domain variable -  and one can view the binary bits of the domain variable as independent random variables.<|endoftext|>
TITLE: Cohomology of sheaves in different Grothendieck topologies
QUESTION [29 upvotes]: Suppose I have a sheaf $\mathcal{F}$ on the (small) étale site over $X$. By restriction, $\mathcal{F}$ is also a sheaf on $X$ (with the Zariski topology). When is it that the sheaf cohomologies (i.e. derived functors of the global section functors) agree in these two sites?
For example, in SGA 4 (Chapter VII, p355), Grothendieck proves that the above cohomologies agree for quasicoherent sheaves. I do not really understand the proof however, and would appreciate any elaboration on this. It uses the Leray spectral sequence, and it seems that the key ingredient is that the higher direct images $R^q f_*(\mathcal{F})$ are $0$, where $f_*$ is the direct image functor induced by the inclusion of the Zariski site in the étale one).
Question: Are there any precise conditions on $\mathcal{F}$ that guarantee that the two cohomologies agree? Is the above condition on $R^q f_*(\mathcal{F})$ necessary/sufficient, and when does it hold?
I should give an example of failure: constant sheaves seem like the obvious choice, and here the cohomologies don't agree even in the case of a point (Zariski cohomology is zero, étale cohomology reduces to Galois cohomology).
Maybe an easier question would be obtained by instead starting with a sheaf $\mathcal{G}$ on $X$ (with the Zariski topology).
Subquestion: Under what conditions can you extend $\mathcal{G}$ to a sheaf on the (small) étale site? (I'm thinking about extending by inverse image, i.e. for an étale $f \colon U \to X$, setting $\mathcal{G}(U) =f^*(\mathcal{G})(U)$, but maybe there are other possibilities)
(Possibly) Easier question: Supposing that $\mathcal{G}$ extends, are there any precise conditions on $\mathcal{G}$ that guarantee that the Zariski cohomology and the étale cohomology of $\mathcal{G}$ agree?
Here again I have in mind the case of quasicoherent sheaves, where everything works out nicely provided that you extend in the right way: starting with a quasicoherent sheaf on the Zariski topology, you need to make sure your sheaf ends up being quasicoherent for the étale topology. You need to take some tensor products to ensure this, as Scott points out in the comments.
(Of course, I have specialised this question to a particular example of sites/toposes, ie Zariski and étale. If you know how this works in more general cases, please share that too!)

REPLY [18 votes]: As you pointed out, your question (for general sheaves of abelian groups)
follows from the vanishing of the derived functors $R^if_*{\mathcal F}$ for $i>0$, 
and that's all one can say.
A good example is the Beilinson-Lichtenbaum conjectures, which states that for a certain
complex of sheaves $\mathbb Z(n)$, the etale and Zariski cohomology groups agree up to degree $n+1$, and then differ. The same example also show that many deep questions in arithmetic geometry can be formulated as a comparison result between the two cohomology groups,
which you could view as a meta-proof that there is no good answer.<|endoftext|>
TITLE: Invariant quadratic forms of irreducible representations
QUESTION [7 upvotes]: Let $G$ be a finite group, and $k$ be a field of characteristic zero (not necessarily algebraically closed!). Let $\rho : G \to \mathrm{End}_k \left(k^n\right)$ be a irreducible representation of $G$ over $k$. Consider the vector space
$S=\left\lbrace H\in \mathrm{End}_k\left(k^n\right) \mid \rho\left(g\right)^T H\rho\left(g\right)=H\text{ for any }g\in G\right\rbrace$
$=\left\lbrace \sum\limits_{g\in G}\rho\left(g\right)^T H\rho\left(g\right)\mid H\in \mathrm{End}_k\left(k^n\right)\right\rbrace$
and its subspace
$T=\left\lbrace H\in S\mid H\text{ is a symmetric matrix}\right\rbrace$.
It is easy to show that, if we denote our representation of $G$ on $k^n$ by $V$, then the elements of $S$ uniquely correspond to homomorphisms of representations $V\to V^{\ast}$ (namely, $H\in S$ corresponds to the homomorphism $v\mapsto\left(w\mapsto v^THw\right)$), while the elements of $T$ uniquely correspond to $G$-invariant quadratic forms on $V$ (namely, $H\in T$ corresponds to the quadratic form $v\mapsto v^THv$).
(1) In the case when $k=\mathbb C$, Schur's lemma yields $\dim S\leq 1$, with equality if and only if $V\cong V^{\ast}$ (which holds if and only if $V$ is a real or quaternionic representation). Thus, $\dim T\leq 1$, and it is known that this is an equality if and only if $V$ is a real representation. (Except of the equality parts, this all pertains to the more general case when $k$ is algebraically closed of zero characteristic).
(2) In the case when $k=\mathbb R$, it is easily seen that $T\neq 0$ (that's the famous nondegenerate unitary form, which in the case $k=\mathbb R$ is a quadratic form), and I think I can show (using the spectral theorem) that $\dim T=1$. As for $S$, it can have dimension $>1$.
(3) I am wondering what can be said about other fields $k$; for instance, $k=\mathbb Q$. If $k\subseteq\mathbb R$, do we still have $\dim T=1$ as in the $\mathbb R$ case? In fact, $T\neq 0$ can be shown in the same way.

REPLY [4 votes]: There are certainly examples over $k=\mathbb{Q}$ where $\dim T\ge2$.
Let's take the cyclic group $G$ of order $5$ and the representation
space
$$V=\{(a_0,\ldots,a_4)\in\mathbb{Q}^5:a_0+\cdots +a_4=0\}$$
where $G$ acts by cyclic permutation. Two linearly independent
elements of $T$ are given by
$$\left(\begin{array}{rrrrr}
2&-1&0&0&-1\\\
-1&2&-1&0&0\\\
0&-1&2&-1&0\\\
0&0&-1&2&-1\\\
-1&0&0&-1&2\end{array}\right)
$$
and
$$\left(\begin{array}{rrrrr}
2&0&-1&-1&0\\\
0&2&0&-1&-1\\\
-1&0&2&0&-1\\\
-1&-1&0&2&0\\\
0&-1&-1&0&2\end{array}\right)
$$
(these define quadratic forms on $V$ since they annihilate the all-one
vector).
The point here is that this representation splits into two over
$\mathbb{R}$. I think the dimension of $T$ in general will be
the number of irreducible representations it splits into over
$\mathbb{R}$.<|endoftext|>
TITLE: Did Gelfand's theory of commutative Banach algebras influence algebraic geometers?
QUESTION [72 upvotes]: Guillemin and Sternberg wrote the following in 1987 in a short article called "Some remarks on I.M. Gelfand's works" accompanying Gelfand's Collected Papers, Volume I:

The theory of commutative normed rings [i.e., (complex) Banach algebras], created by Gelfand in the late 1930s, has become today one of the most active areas of functional analysis.  The key idea in Gelfand's theory -- that maximal ideals are the underlying "points" of a commutative normed ring -- not only revolutionized harmonic analyis but had an enormous impact in algebraic geometry.  (One need only look at the development of the concept of the spectrum of a commutative ring and the concept of scheme in the algebraic geometry of the 1960s and 1970s to see how far beyond the borders of functional analysis Gelfand's ideas penetrated.)

I was skeptical when reading this, which led to the following:

Basic Question: Did Gelfand's theory of commutative Banach algebras have an enormous impact, or any direct influence whatsoever, in algebraic geometry?

I elaborate on the question at the end, after some background and context for my skepticism.
In the late 1930s, Gelfand proved the special case of the Mazur-Gelfand Theorem that says that a Banach division algebra is $\mathbb{C}$.  In the commutative case this applies to quotients by maximal ideals, and Gelfand used this fact to consider elements of a (complex, unital) commutative Banach algebra as functions on the maximal ideal space.  He gave the maximal ideal space the coarsest topology that makes these functions continuous, which turns out to be a compact Hausdorff topology.  The resulting continuous homomorphism from a commutative Banach algebra $A$ with maximal ideal space $\mathfrak{M}$ to the Banach algebra $C(\mathfrak{M})$ of continuous complex-valued functions on $\mathfrak{M}$ with sup norm is now often called the Gelfand transform (sometimes denoted $\Gamma$, short for Гельфанд).  It is very useful.
However, it is my understanding that Gelfand wasn't the first to consider elements of a ring as functions on a space of ideals.  Hilbert proved that an affine variety can be considered as the set of maximal ideals of its coordinate ring, and thus gave a way to view abstract finitely generated commutative complex algebras without nilpotents as algebras of functions.  On the Wikipedia page for scheme I find that Noether and Krull pushed these ideas to some extent in the 1920s and 1930s, respectively, but I don't know a source for this.  Another related result is Stone's representation theorem from 1936, and a good summary of this circle of ideas can be found in Varadarajan's Euler book.
Unfortunately, knowing who did what first won't answer my question.  I have not been able to find any good source indicating whether algebraic geometers were influenced by Gelfand's theory, or conversely.  

Elaborated Question:  Were algebraic geometers (say from roughly the 1940s to the 1970s) influenced by Gelfand's theory of commutative Banach algebras as indicated by Guillemin and Sternberg, and if so can anyone provide documentation?  Conversely, was Gelfand's theory influenced by algebraic geometry (from before roughly 1938), and if so can anyone provide documentation?

REPLY [18 votes]: In conversations and discussions, David Mumford mentioned several times that Gelfand's result about commutative Banach algebras had great influence for the development of schemes. He also mentioned that local coordinate systems on manifolds had influence.<|endoftext|>
TITLE: Errata for Emil Artin's 'The Gamma Function'?
QUESTION [9 upvotes]: In the English translation of The Gamma Function by Emil Artin (1964 - Holt, Rinehart and Winston) there appears to be a mistake in the formula given for the gamma function on page 24:
$$\Gamma(x) = \sqrt{2\pi}x^{x-1/2}e^{-x+\mu(x)}$$
$$\mu(x)=\sum_{n=0}^\infty(x+n+\frac{1}{2})\text{log}(1+\frac{1}{x+n})-1=\frac{\theta}{12x},\ \ \ \ \ 0 < \theta < 1$$
and on page 22 where this is derived, it is noted that '$\theta$ is a number independent of $x$ between 0 and 1'.
This sounds incorrect, as $\theta$ does depend on $x$, but since the wording is a little ambiguous it may just be an unclear translation. The original German might have meant that $0< \theta(x) < 1$ for any $x$. That the variable $x$ is suppressed from $\theta$ could be just confusing notation, or someone's misunderstanding (possibly mine.)
The preface does mention that a (different) formula had to be corrected for the English reprint.
I would like to know if there are mistakes in this book, and if so, whether they exist in the German edition. Is there an available list of errata?

REPLY [4 votes]: It seems clear that $\theta$ can indeed be chosen to be a number independent of $x$ as stated, to get Stirling's formulas for the gamma function when
$x$ is large.    The wording, at least in English, is not too helpful
in this section.
But I'm less clear about where in the formula on page 24 there is supposed to be
a mistake.   Here as in any mathematics book (especially a translation) one has to be
wary about misprints or errors.   Probably there is no publicly available list
of errata for this small monograph published originally in 1931 in German and
later republished in 1964 in an English translation by 
Michael Butler.    This English version is included in the 2007 AMS softcover
book Exposition by Emil Artin: A Selection edited by Michael Rosen.  (There is an older 1965 book
The Collected Papers of Emil Artin published by Addison-Wesley and edited
by Lang & Tate.  This contains Artin's research papers, in the original
German or English.)    As Zavosh observes, the 1964 preface by Edwin Hewitt reprinted here does indicate one formula corrected in the translation:
" ... a small error following formula (59) (this edition) was corrected..."
However, the formula seems to be the one actually numbered (5.9).   Caveat lector.<|endoftext|>
TITLE: Maps inducing zero on homotopy groups but are not null-homotopic
QUESTION [21 upvotes]: Today my fellow grad student asked me a question, given a map f from X to Y, assume $f_*(\pi_i(X))=0$ in Y, when is f null-homotopic?
I search the literature a little bit, D.W.Kahn 
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102995805
And M.Sternstein has worked on this, and Sternstein even got a necessary and sufficient condition, for suitable spaces. 
http://www.jstor.org/stable/pdfplus/2037939.pdf
However, his condition is a little complicated for me as a beginner. Right now I just wanted a counter example of a such a map. Kahn in his paper said one can have many such examples using Eilenberg Maclance spaces. Well, we can certainly show a lot of map between E-M spaces induce zero map on homopoty groups just by pure group theoretic reasons, but I can not think of a easy example when you can show that map, if it exists, is not null-homotopic. Could someone give me some hint?
or, maybe even some examples arising from manifolds?

REPLY [14 votes]: I realize this is a very old question.  Nonetheless, here is a large, easy,  and well-known class of examples.
Let $X$ be a path connected CW-complex, and consider the diagonal map $\delta_X\colon X\to X\wedge X$ into the smash product (i.e., the composite of $X\xrightarrow{\text{diag}} X\times X \xrightarrow{\text{quot}} X\wedge X$).  
This map is often non-null: for instance, 
$$
\widetilde{H}^*(X)\otimes \widetilde{H}^*(X)\xrightarrow{\text{Kunneth}} \widetilde{H}^*(X\wedge X)\xrightarrow{\delta_X} \widetilde{H}^*(X)
$$
is exactly the cup-product, so any $X$ with non-trivial cup-product in positive degrees must have non-null $\delta_X$.
On the other hand, $\delta_X$ is always trivial on homotopy groups: any $f\colon S^k\to X$ fits in the commutative square
$$\begin{array}{ccc}
S^k & \xrightarrow{\delta_{S^k}} & S^k\wedge S^k
\\
\downarrow & & \downarrow
\\
X & \xrightarrow{\delta_X} & X\wedge X
\end{array}
$$
and $\delta_{S^k}\sim *$ for $k\geq 1$.  In fact, this idea proves that any composite $\Sigma Y\to X\xrightarrow{\delta_X} X\wedge X$ is null, or equivalently that $\Omega(\delta_X)\colon \Omega X\to \Omega(X\wedge X)$ is null-homotopic for any $X$.<|endoftext|>
TITLE: Approximately Invert x^x
QUESTION [6 upvotes]: What is the best asymptotic approximation of the inverse $x=g(y)$ of $y = x^x$ for large $x$?  [Clearly, if $x>e$, then $f(x) > e^x$ implies $g(x) < \log x$.]

REPLY [16 votes]: I don't know how accurate you want to be, but a quick and dirty approximate inversion of $x\log x$ is $x/\log x$. So if $y=x^x$ then $\log y\approx x\log x$, so $x\approx\log y/\log\log y$. But perhaps you want something better than this.<|endoftext|>
TITLE: Details for the action of the braid group B_3 on modular forms
QUESTION [19 upvotes]: I'm reading Terry Gannon's Moonshine Beyond the Monster, and in section 2.4.3 he hints at (but does not explicitly describe) a way to extend the action of $SL_2(\mathbb{Z})$ on modular forms to an action of the braid group $B_3$.  Here is what he says about this action: 
First, we lift modular forms $f : \mathbb{H} \to \mathbb{C}$ to functions $\phi_f : SL_2(\mathbb{R}) \to \mathbb{C}$ as follows: let
$$\phi_f \left( \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right] \right) = f \left( \frac{ai + b}{ci + d} \right) (ci + d)^{-k}.$$
Thinking of $f$ as a function on $SL_2(\mathbb{R})$ invariant under $SO_2(\mathbb{R})$, we have now exchanged invariance under $SO_2(\mathbb{R})$ for invariance under $SL_2(\mathbb{Z})$.  ($SO_2(\mathbb{R})$ now acts by the character corresponding to $k$.)  In moduli space terms, an element $g \in SL_2(\mathbb{R})$ can be identified with the elliptic curve $\mathbb{C}/\Lambda$ where $\Lambda$ has basis the first and second columns (say) of $g$, and $\phi_f$ is a function on this space invariant under change of basis but covariant under rotation.
Second, $SL_2(\mathbb{R})$ admits a universal cover $\widetilde{SL_2(\mathbb{R})}$ in which the universal central extension $B_3$ of $SL_2(\mathbb{Z})$ sits as a discrete subgroup.  Unfortunately, Gannon doesn't give an explicit description of this universal cover (presumably because it's somewhat complicated).
Question:  What is a good explicit description of this universal cover and of how $B_3$ sits in it (hence of how it acts on modular forms)?  In particular, does it have a moduli-theoretic interpretation related to the description of $B_3$ as the fundamental group of the space $C_3$ of unordered triplets of distinct points in $\mathbb{C}$?  (These triplets $(a, b, c$) can, of course, be identified with elliptic curves $y^2 = 4(x - a)(x - b)(x - c)$.)

REPLY [4 votes]: This is only a partial answer to your question, concerning how to visualise the universal cover of $\mathrm{SL}(2,\mathbb{R})$.
A nice picture of this group (together with a proof that it is not a matrix group) is given in Graeme Segal's lectures in the book Lectures on Lie groups and Lie algebras.  Alas, the Google Books preview does not cover Segal's lectures and I don't have the book here with me in order to scan the nice picture he draws.
The bi-invariant metric on $\mathrm{SL}(2,\mathbb{R})$ is lorentzian and has constant negative sectional curvature.  Its universal cover is the lorentzian analogue of hyperbolic space and in the Physics literature it goes by the name of (three-dimensional) anti de Sitter spacetime $\mathrm{AdS}_3$ and has been studied a lot.<|endoftext|>
TITLE: How do you pronounce "Hartshorne"?
QUESTION [15 upvotes]: What is the "correct" pronunciation of Robin Hartshorne's last name? Mostly I hear it pronounced "Har-shorn" although I've also heard "Harts-orn" and maybe a few other variations.

REPLY [26 votes]: He prefers it be pronounced as in Hart's  Horn. I asked him a few years ago, our brief common ground being assisting Marvin Jay Greenberg with revisions for the fourth edition of his book on Euclidean and non-Euclidean geometry. That is not to say that I have ever heard anyone else say it that way. But then few people get my name right.<|endoftext|>
TITLE: Sum of difference moduli vs. sum of modulus differences
QUESTION [10 upvotes]: This is a failed attempt of mine at creating a contest problem; the failure is in the fact that I wasn't able to solve it myself.
Let $x_1$, $x_2$, ..., $x_n$ be $n$ reals. For any integer $k$, define a real $f_k\left(x_1,x_2,...,x_n\right)$ as the sum
$\sum\limits_{T\subseteq\left\lbrace 1,2,...,n\right\rbrace ;\\ \ \left|T\right|=k} \left|\sum\limits_{t\in T}x_t - \sum\limits_{t\in\left\lbrace 1,2,...,n\right\rbrace \setminus T} x_t\right|$.
We mostly care about the case of $n$ even and $k=\frac n 2$; in this case, $f_k\left(x_1,x_2,...,x_n\right)$ is a kind of measure for the dispersion of the reals $x_1$, $x_2$, ..., $x_n$ (more precisely, of their $\frac n 2$-element sums).
Now my conjecture is that if $n$ is even and $k=\frac n 2$, then
$f_k\left(x_1,x_2,...,x_n\right)\geq f_k\left(\left|x_1\right|,\left|x_2\right|,...,\left|x_n\right|\right)$
for any reals $x_1$, $x_2$, ..., $x_n$.
I think I have casebashed this for $n=4$ and maybe $n=6$; I don't remember anymore - it's too long ago. Sorry. I still have no idea what to do in the general case, although my attempts at big-$n$ counterexamples weren't of much success either.

REPLY [5 votes]: Hi, Darij!
This is actually quite simple (and also much more appropriate for AoPS than for MO). The idea is to show that for every $t$, the expression $\sum_T|t+D_T|$, where $D_T$ is your difference, goes down if you replace all $x_k$ by their absolute values ($t=0$ is your claim). The base $n=2$ is rather trivial and boils down to the inequality $|t-a|+|t+a|=2\max(|t|,|a|)\ge 2\max(|t|,|b|)=|t-b|+|t+b|$ when $|a|\ge |b|$. Now, assume that we know the statement for $n$ and want to show it for $n+2$. The trick is to choose a random pair of indices $i,j$ and notice that the full sum is just the average over such choices of $\sum_T(|t+x_i-x_j+D_T|+|t+x_j-x_i+D_T|)$ where $T$ runs over all $n/2$ element subsets of the set of remaining indices. Now, applying the statement with $n=2$ for fixed $T$, we see that we can replace $x_i$ and $x_j$ with their absolute values and our sum (with fixed $i,j$) will go down in each term. After that, we replace everything else by the absolute values using the induction assumption and, again, the sum will go down. But now we are completely done: we showed that for each fixed $i,j$, the sum goes down when we replace everything by the absolute value, so it is true after averaging as well.
The whole thing is just a textbook case of the "inventor's paradox". Strange that you haven't figured it out...<|endoftext|>
TITLE: Good books on theory of distributions
QUESTION [27 upvotes]: Hi all. 
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.

REPLY [2 votes]: "Mathematics for the Physical Sciences", Laurent Schwartz, Dover 2008 is a simplified English language book that covers some (maybe even much) of Schwartz's theory of distributions. Very readable, helpful and interesting (also $19.95). The title sounds more general than it actually is--really is focused on distributions, and their applications.  Schwartz says in the preface: 'This work is concerned with the mathematical methods of physics'.<|endoftext|>
TITLE: Is it decidable whether a given set generates the whole group?
QUESTION [7 upvotes]: Upon thinking about this question, I have a feeling that there is an interesting general problem like that, but I cannot verbalise it. Here is an approximation.
The question is: given a finitely generated group $G$ and a finite set $S\subset G$, we want to find out whether the subgroup generated by $S$ is the whole group. Is there an algorithm deciding this?
One has to specify how $G$ and $S$ are presented to a machine. As for $S$, its elements are given as words in some generating set for $G$. The question how to represent a group is more delicate. Obviously not by a set of relations, since the identity problem is undecidable. There are two ways I see.
1) Let's say that $G$ is nice if there is an algorithm deciding the above problem for this group only. For example, $\mathbb Z^n$ is nice, [edit:] free groups are nice, and probably many others. Are all groups nice? What about some reasonable classes, like lattices in Lie groups?
2) This one is the best approximation of my intuitive picture of a question. Suppose we have an algorithm $A$ deciding the identity problem in $G$. (This algorithm tells us whether two elements of $G$, presented as words, are equal.) Or maybe it's an oracle rather than algorithm, I don't feel the difference. Can we decide the above problem using $A$?
Update. It turns out that even $F_6\times F_6$ is not nice in the above sense (see John Stillwell's answer). This kills the second part too. The only question that remains unanswered is the one with the least motivation behind it: Is "generating problem" solvable in lattices of Lie groups?

REPLY [5 votes]: Let me mention that in the context of computational group theory, it is more natural to consider the group membership problem: given the generators, say as matrices $A_i \in SL(n,\Bbb Z)$ you want to decide whether matrix $M$ lies in the group generated by $A_i$.  Clearly, we can apply the group membership to standard generators to decide whether $A_i$ they generate some given group $G$.  Unfortunately, this is in fact a much harder problem - since $SL(4,\Bbb Z)$ contains a product of two copies of $F_2$, this is undecidable.  Interestingly, one can decide in polynomial time whether the group the $A_i$'s generate is finite or infinite, as well as whether it is virtually solvable: see here and here.<|endoftext|>
TITLE: What is a reference for an explicit, logic-based, statement of duality in category theory (in ''complicated'' situations)?  And what are the prerequisites for a beginner in logic?
QUESTION [6 upvotes]: Background
In the course of reading Mac Lane linearly (currently in Chapter VI),
I have seen again and again that duality can make life much easier. My
problem is that I have almost no background in logic, and duality is a
theorem in logic about category theory.
When I first read about duality in Chapter II of Mac Lane in the
context of the elementary theory of a single category, everything was
pretty clear even without knowing any logic.  However, when I got to
the chapter on adjunctions, involving two categories and functors
between them, a bijection of hom-sets, and two natural
transformations, I got confused to the point that I wasn't even sure
how to use duality (let alone, why it is correct).
At this stage, I made a rather long pause and read the first three
chapters of Ebbinghaus, Flum, and Thomas' ''Mathematical logic'' (so, I
have read about the syntax and semantics of first-order logic).  From
this, I built my own (hopefully correct) ''poor man's proof of
duality'' up to the situation of a single adjunction.   This has both
clarified the validity of duality for formulas involving adjunctions,
and helped me understand how to use duality in such situations.
But a single adjunction is far from the most ''complicated'' situation
one meets.  There are composition of adjunctions, pointwise limits in
functor categories, and many other situations in which I  am still
not totally convinced that I understand duality (both theoretically
and practically).
For example, in one answer to a recent question
on pointwise limits in functor categories, it was stated that the
reference for limits is Mac Lane, while the reference for colimits
is Mac Lane--Moerdijk.  I really wanted to comment that the
assertion on colimits is just the dual of the one on limits, but then
I realized that I am not totally sure.  I would be most grateful for some solid source that
I can consult whenever I have doubts in what I get after doing the intuitive
things (reverse arrows but not functors, etc.).
Questions

What is a good reference for an explicit, logic-based, statement of a duality
theorem of category theory in ''complicated situations?''
What are the prerequisites in logic?  For example, up to which
point of Ebbinghaus--Flum--Thomas should I read?

REPLY [9 votes]: I would go even farther than the comments above, at least in the specific case you mention about computing (co)limits objectwise in a functor category.  Once you know the statement for limits, deducing the statement for colimits is not even a syntactic transformation to the proof, and needs no arguments from formal logic at all.  Instead, to prove that colimits are computed objectwise in the functor category [I, C], simply use the fact that limits are computed objectwise in the functor category [Iop, Cop] = [I, C]op, and the fact that colimits in a category are the same as limits in the opposite category.
I think this is the common case in this kind of argument, but perhaps someone can come up with an example where the argument really needs to be repeated in the dual situation.<|endoftext|>
TITLE: Detailed proof of cup product equivalent to intersection
QUESTION [15 upvotes]: Consider a smooth, closed, compact finite-dim manifold. We have Poincare Duality to relate the cocycles and cycles.

I would like to know where I can find
  a reference for a proof that the cup
  product of the Cohomology Ring is
  given by the intersection of the
  corresponding cycles.

Griffiths and Harris talk about intersection number, and discuss this result in chapter 0, Hatcher's book doesn't mention this explicitly as far as I can tell, Katz' little book on enumerative geometry alludes to this, Fulton's book on Young Tableaux dodges this, etc.
I am preparing to give a talk on Schubert Cells and Schubert calculus, and I realized that I have not checked the details of this proof.
Thanks in advance!

REPLY [12 votes]: Bott and Tu do this completely, in the de Rham theoretic setting of course.
Here's an alternate proof I have used when I teach this material, which I find slightly more clean and direct than using Thom classes in de Rham theory (which require choice of  tubular neighborhood theorem, etc) and works over the integers.
Definition: Given a collection $S = \{W_i\}$ of submanifolds of a manifold $X$, define the smooth chain complex transverse to $S$, denoted ${C^S}_*(X)$, by using the subgroups of the singular chain groups in which the basis chains $\Delta^n \to X$ are smooth and transverse to all of the $W_i$.
Lemma: The inclusion ${C^S}_*(X) \to C_*(X)$ is a quasi-isomorophism, for any such collection $S$.
Now if $W \in S$ then "count of intersection with $W$" gives a perfectly well-defined element $\tau_W$ of
${\rm Hom}(C^S_*(X), A)$ and thus by this quasi-isomorphism a well-defined cocycle if the $W$ is proper and has no boundary.  It is immediate that this cocycle evaluates on cycles which are represented by closed submanifolds through intersection count.
There are two approaches to show that cup product agrees with intersection on cohomology.  Briefly, one is to take $W, V$ over $M$ and consider the special case of $W \times M$ and $M \times V$ over $M \times M$.  There some work with the K"unneth theorem
leads to direct analysis in this case.  But this case is "universal" - cup products in $M$ are pulled back from ``external'' cup products over $M \times M$.  A second proof given in https://arxiv.org/abs/2106.05986 uses a variant of the theory, where one fixes a triangulation or cubulation, and assumes $W, V$ transverse to those.  There we explicitly see that these products do not agree at the cochain level (they can't since intersection is commutative, but non-commutativity of cup product is reflected in Steenrod operations), but Friedman, Medina and I show a vector field flow leads to a cobounding of the difference.<|endoftext|>
TITLE: Is there a formal notion of what we do when we 'Let X be ...'?
QUESTION [10 upvotes]: This is likely an elementary question to logicians or theoretical computer scientists, but I'm less than adequately informed on either topic and don't know where to find the answer. Please excuse the various vague notions that appear here, without their appropriate formal setting. My question is exactly about what I should have said instead of what follows:
When we make a definition, of either a property, or the explicit value of a symbol, it seems that we are somehow changing the language. Prescribing meaning to a word might be viewed as a kind of transformation of the formal language akin to taking a quotient, where we impose further relations on a set of generators.
I don't know how to describe a 'semantic' object, but am assuming an ad-hoc definition could be as a class of words under an equivalence relation supplied by an underlying logic. If the complexity of such an object is the size of the smallest word in the language that describes it, then making a definition lowers the complexity of some objects (and doesn't change the rest.) The obvious example is that if I add the word group to my language, then saying G is a group is a lot shorter than listing its properties.
It seems that lowering complexity is a main point of making definitions. Further, that one reason mathematical theory-building works, is a compression effect through which I am able to use less resources to describe more complex objects, at the cost of the energy it takes to cram definitions from a textbook.
Likely there is some theory out there that describes this process, but I've not been able to google it. I would appreciate being pointed towards the right source, even if it's a wikipedia link. Specifically:
Where can I find a theory of formal logic or complexity theory that studies the process of adding definitions to a mathematical language, viewed as a transformation that changes complexity?

REPLY [9 votes]: Kieffer, Avigad, & Frideman, 2008 A language for mathematical knowledge management, which I mentioned in the Proof formalization thread, discusses DZFC, an extension of ZFC with definitions of terms and partial terms.  Theorem 1 proves conservativity over ZFC, with respect to which the paper says:

the usual method of eliminating defined function symbols and 
  relation symbols by replacing them by their definiens can result in an exponential 
  increase in length.

Which is roughly in line with some analogous results for other formalisms.  Neel mentioned the doubly expontential blow-up for normalisation in the simply typed lambda calculus, and more drastic blowups are possible with higher type systems.
It's unclear to me what notion of complexity is sought, but the expansion in size of terms under expansion will typically be the same as the time complexity of the expansion algorithm.<|endoftext|>
TITLE: Verb form of 'homotopy'?  'Homotope'?
QUESTION [10 upvotes]: Is there a transitive verb, in common use, which means 'deform via a homotopy'?  I used to think 'homotope' was the answer, but it produces surprisingly few relevant matches on Google, so now I have my doubts.  I want to be able to say things like "By Lemma 17 we can homotope the k-cells (rel boundary) into the subspace Y".  I am well aware that I could rearrange the sentence to say something like "...there is a homotopy such that...", but that's not what I want to do.  I want a simple, one-word verb with this meaning.  I suppose 'deform' might work, but it's not as specific as I would like.
Another form of this question: Does the use of 'homotope' in the sample sentence of the previous paragraph sound strange?  Sound standard and idiomatic?

REPLY [8 votes]: Since all of the responses thus far have been in the form of comments rather than answers, I'll summarize the results in this answer.
Of the first seven people to comment, all thought that "homotope" was a bit informal.  Four commenters thought it was fine to use in print, while three thought it should be avoided in formal papers.
I did a full-text arXiv search, and found 408 occurrences of "homotoped" and 362 occurrences of "homotope".  (I looked at only the mathematics sections, not physics etc.)<|endoftext|>
TITLE: In what category is the sum of real numbers a coproduct?
QUESTION [13 upvotes]: (if any?)
I understand that in the natural numbers, the sum of two numbers can be readily thought of as the disjoint union of two finite sets.
John Baez even spent a week talking about how you can extend this idea to thinking about the integers here: TWF 102. This led into a discussion of the homotopy groups of spheres.
But then you have to pass to a certain colimit to get the rationals, and take a certain completion to get the reals. It all gets very complicated.
One place where we rely on a correspondence between a sum of real numbers and a certain coproduct is in measure theory-- perhaps in analogy to the relation between finite sets and natural numbers, we should think of some measure space as the categorification of the real numbers. But this sounds unpromising-- what space would be in any sense a canonical categorification? Moreover, what I was really hoping for originally was a precise sense in which the $\sigma$-additivity of a measure states that it preserves coproducts or something, so I was hoping there might be more to it than sigma-algebras.

REPLY [15 votes]: Isomorphism classes of finitely generated right Hilbert modules over a II_1 factor are in a bijective
correspondence with the nonnegative reals.
The correspondence sends every module to its dimension.
Moreover, the dimension function is additive with respect to the coproduct of modules.
I believe you can obtain all reals using some form of supersymmetry (Quillen construction?),
but then the addition of reals will no longer correspond to the coproduct of objects.<|endoftext|>
TITLE: the central issues in complex geometry
QUESTION [5 upvotes]: I want to know about people in researching complex (maybe differential) geometry are careing about what currently ? For example ,$L^2$ estimate inspired by Lars Hormander is a very useful tool,and how does this theory be developed currently ? As myself , i like this method very much ,but i don't know which is the next important problem be solved by this method . How far will this method go ?  As well , just like the holomorphic morse inequalities , when it is proved by Demailly in 1985,twenty years passed , it seems that during thest twenty years there are no important results comes out in complex geometry ? I'm a beginner in complex geometry, i think that i can't scratch the direction of complex geometry ? I don't know people are careing about what in complex geometry ?Only doing some little questions or leave this field? So this is just the purpose of this question i asked , i want to communicate with all who are interested in this field.

REPLY [4 votes]: There also problems along the lines of proving L2 extension theorems for vector valued forms (there are such theorems already like the Ohsawa-Takegoshi extension theorem however, that applies to (n,1) forms. We want more general results). Besides, there is the problem of deformation invariance of plurigenera over Kahler manifolds (Siu proved it for the projective case).<|endoftext|>
TITLE: When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group?
QUESTION [12 upvotes]: I should admit the question below does not have a serious motivation. But still I found it somehow natural.
Let $G$ be a finite group of order $n$ with $h$ conjugacy classes. If $c_1,\ldots,c_h$ are the orders of the conjugacy classes of $G$, then clearly
$n=c_1+c_2+\ldots+c_h$.
Let now $\pi_1,\ldots,\pi_h$ be the pairwise non-isomorphic, irreducible complex representations of $G$. It is well known that another partition of $n$ of length $h$ is given by the squares of the degrees $d_i$'s of the $\pi_i$'s:
$n=d_1^2+d_2^2+\ldots+d_h^2$.
Question: Assume that, up to reordering, the two partitions of $n$ described above are the same. Then what can we say about $G$? Is $G$ forced to be abelian?

REPLY [6 votes]: This is not an answer, but more of an extended comment. It seems that the first one to raise this question was E. Bannai Association schemes and fusion algebras. (An introduction), J. Algebr. Comb. 2, No. 4, 327-344 (1993). The motivation was to find examples of when a certain matrix attached to a group association scheme is unitary (which is the case for "naturally occuring" association schemes). This happens precisely when the irreducible characters and conjugacy classes of the finite group can be paired up such that the size of the conjugacy class is the square of the corresponding character degree.
On p. 341 Bannai mentions that M. Kiyota has proved (using the CFSG) that this property does not hold for any non-abelian simple group. This shouldn't be too hard; as Jim Humphreys has noted, the case of simple groups of Lie type follows from the existence of the Steinberg character, whose degree equals the order a Sylow $p$-subgroup, hence the square of its degree cannot divide the order of the group. For simple alternating groups, it is easy to find conjugacy classes which are not squares, and for the sporadic groups one can explicitly check whether the condition holds in each case.
Bannai also mentions that Kiyota "conjectures that $G$ must be nilpotent if the condition is satisfied", so Kevin Buzzard's question has been asked before.
An interesting development that gives some support to the conjecture that these groups are nilpotent is this paper:
Andrus, Ivan; Hegedűs, Pál; Okuyama, Tetsuro, Transposable character tables and group duality., Math. Proc. Camb. Philos. Soc. 157, No. 1, 31-44 (2014). ZBL1330.20007,
where it is shown that every transposable group is nilpotent. Transposable groups are finite groups $G$ whose character table has the property that its transpose (viewed as a matrix) is, up to multiplication by suitable diagonal integer matrices on each side, equal to the character table of some other finite group $G^T$. Moreover, $G$ is called self-dual if it is isomorphic to $G^T$ (there are other, equivalent, definitions). By definition, self-dual groups satisfy the condition that conjugacy class sizes are the squares of corresponding character degrees (i.e., that the two partitions in the OP agree). Thus, at least self-dual groups are nilpotent. However, I think there exist non-self-dual groups satisfying the condition that the two partitions agree, so in general Kiyota's conjecture seems to be open.<|endoftext|>
TITLE: Crystalline cohomology of abelian varieties
QUESTION [19 upvotes]: I am trying to learn a little bit about crystalline cohomology (I am interested in applications to ordinariness).  Whenever I try to read anything about it, I quickly encounter divided power structures, period rings and the de Rham-Witt complex.  Before looking into these things, it would be nice to have an idea of what the cohomology that you construct at the end looks like.
The l-adic cohomology of abelian varieties has a simple description in terms of the Tate module.  My question is: is there something similar for crystalline cohomology of abelian varieties?
More precisely, let $X$ be an abelian scheme over $\mathbb{Z}_p$.  Is there a concrete description of $H^1(X_0/\mathbb{Z}_p)$?  (or just $H^1(X_0/\mathbb{Z}_p) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$?)  I think that this should consist of three things: a $\mathbb{Z}_p$-module $M$, a filtration on $M \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ (which in the case of an abelian variety has only one term which is neither 0 nor everything) and a Frobenius-linear morphism $M \to M$.
I believe that the answer has something to do with Dieudonné modules, but I don't know what they are either.

REPLY [12 votes]: To add a bit more to Brian's comment: the crystalline cohomology of an abelian variety (over a finite field of characteristic p, say) is canonically isomorphic to the Dieudonné module of the p-divisible group of the abelian variety (which is a finite free module over the Witt vectors of the field with a semi-linear Frobenius). If you start with an abelian scheme over the Witt vectors of this field then the crystalline cohomology of the special fibre is canonically isomorphic to the algebraic de Rham cohomology of the thing upstairs, hence receives a Hodge filtration also.
A good starting place to understand Dieudonné modules is Demazure's 'Lectures on p-divisible groups', which appears in Springer LNM. In particular, he gives a nice description of the analogy with Tate modules (and the relation between the various Frobenii that appear). For a general picture of crystalline cohomology, and the various structures that can be placed on it, I would look at Illusie's survey in the Motives volumes (this is a little out of date now, but gives a good description of the basic theory).<|endoftext|>
TITLE: What finite group schemes can act freely on a rational function field in one variable?
QUESTION [15 upvotes]: Suppose that $G$ is a finite group scheme over a field $k$ (we may want to assume that $k$ is perfect). How does one tell whether there exists a free action of $G$ on the function field $k(t)$ in one variable? By this I mean that there exists an action $G \times_{\mathop{\rm Spec}k}\mathop{\rm Spec}k(t) \to \mathop{\rm Spec}k(t)$, making $\mathop{\rm Spec}k(t)$ into a $G$-torsor over a scheme (necessarily of the form $\mathop{\rm Spec} k(s)$, where $s \in k(t)$, by Lüroth's theorem).
The question is a very natural one when one studies essential dimension of group schemes: see http://www.math.ubc.ca/~reichst/lens-notes6-27-8.pdf for a nice survey of the topic of essential dimension, and http://front.math.ucdavis.edu/1001.3988 for the essential dimension of group schemes. When $G$ is smooth over $k$, then it is easy to see that the action extends to an action on $\mathbb{P}^1$, so $G$ must be a subgroup of ${\rm PGL}_{2,k}$; but when $G$ is not smooth it is not all clear to us that this must happen. The sheaf of automorphisms of $k(t)$ over $k$ is enormous in positive characteristic, and we find it very hard to see what group schemes it contains.
For example, how about twisted forms of the group scheme $\mu_p$, where $p$ is the characteristic of the field? I would conjecture that most of them can't act freely on $k(t)$, but we can't prove it.

REPLY [3 votes]: I guess by looking at it algebraically one can at least rule out the forms of $\mu_p$. Let $H$ be the function algebra of the group scheme, $H^\ast$ its dual. $H^\ast$ is a cocommutative Hopf algebra. Algebraically, you ask whether $H^\ast$ can act on $k(t)$ so that
$k(t)>k(s)$ is Hopf-Galois. If I remember correctly, there is a theorem (see chapter 8 of Montogmery' Hopf Algebra actions) that this is equivalent to the semidirect product $k(t)*H^\ast$ being simple. This will necessarily require $H^\ast$ to be semisimple. Now I read your $\mu_p$ as a form of the cyclic group. Thus, your $H^\ast$ fails to be semisimple by Maschke's theorem.
Sorry, if I misunderstood or misquoted something.<|endoftext|>
TITLE: Mathematics as a hobby
QUESTION [43 upvotes]: I would like to know if practicing mathematics, constituting a hobby for some of you who are neither academics nor (advanced) mathematics, is an important part of your career. How do you go and learn a new mathematical field on your own?
Do you just pick up a book and go over all proofs and do all exercises on your own? Is there any technique would you like to share?

REPLY [11 votes]: I do mathematics as a hobby and have found various blogs and MathOverflow and Stack Exchange site Mathematics very helpful in continuing my education and revealing my limitations.
I have always enjoyed maths and have done well, but in the end that wasn't a good enough reason for me to make a career. I became interested in too many other things as well.
So what do I do?
First, I do puzzles and problems - I enjoy solving things and always have done it.
Second, I read textbooks - the demise of bookshops in favour of online resources is a bit of a menace here, because in a bookshop I could browse more easily for something interesting which appeared to be within my range and opened up an area I might be a little unfamiliar with.
Third, I do intentional study - for example to understand the classification of finite simple groups, or (as far as possible) the proof of Fermat's Last Theorem, or the Riemann hypothesis, or PvNP. But (to give a benefit of online resources) I do download a number of the papers linked in posts on this site. I did some algebraic geometry when I was younger and am now trying out Ravi Vakil's notes to get myself up to speed - but as a hobbyist I don't always have the time to consolidate what I've read.
What I do find is that I miss some of the informal knowledge (the stuff people talk about, but don't write down), so I don't always link things together as quickly as I might. And also I find that my intuition is not what it was - I think the exercises in good textbooks feed intuition by giving a sense of what is possible and what is not, and help to direct imagination in fruitful ways. I am not doing enough work to sustain my intuition at a high level.
However posting answers on sites like this does force me to commit myself in public, and I find that a learning experience, with lots of helpful (if occasionally sharp) comments and feedback. Since such self-learning is not in line with the research goal on MathOverflow, I indulge myself rather more on Stack Exchange, and tend to scan here for insights on things I might be reading at the moment.

REPLY [2 votes]: Here's a couple of interesting articles about math as a hobby:

Mathematical learning (and math as a hobby) by Rishidev Chaudhuri
Mathematics as a hobby (math as a hobby) ? Yes – Really!<|endoftext|>
TITLE: How to motivate the skein relations?
QUESTION [13 upvotes]: I am teaching an advanced undergraduate class on topology. We are doing introductory knot theory at the moment. One of my students asked how do we know to use this skein relation to compute all these wonderful polynomial invariants of knots. 
I was not trained as a knot theorist, so I was at loss. Intuitively  it does not seem to be that powerful, because it is not clear (to me at least) that you can always pick a recursive relation involving simpler knots. Could somebody help me motivating my students here? In other words,
Why are the skein relations so useful for computing polynomial invariants of knots, and when did people realize that is the case? 
Thanks.

REPLY [13 votes]: One of the earliest appearances of the ingredients for a skein relation can be found in Romilly Allen's 1904 book on Celtic Knotting. There he explains that designers of Celtic knot patterns first start with a regular weave and then systematically replace some of the crossings by "horizontal" and "vertical" smoothings. These days, it is easy to find working demonstrations of this process on the internet. The skein triple consists in an unoriented crossing and its two possible smoothings. This is the skein triple underlying the bracket polynomial model for the Jones polynomial and directly related to Temperley-Lieb algebra. Of course neither Romilly Allen nor the Celtic knotters had any intention to calculate polynomial invariants of knots! That came much later. Another important point about this simplest skein relation is that it is related (by shading the link diagram) to the contraction and deletion operations on the edges of a plane graph. This is why the bracket polynomial and the Jones polynomial are related to the Dichromatic (Tutte) polynomial and to signed generalizations of the Tutte polynomial. And then finally, the two smoothings of a crossing are related to one another in that they form the 'top' and 'bottom' of a saddle surface that joins them. This is the key geometric idea behind Khovanov homology.

REPLY [3 votes]: Here is a sketch of how the skein relations appear in the approach to knot invariants based on braided monoidal categories coming e.g. from representations of quantum groups. 
Suppose $V$ is a dualizable object in a braided monoidal category $(C, \otimes)$. This means, among other things, that $V^{\otimes n}$ acquires an action of the braid group $B_n$, and moreover we can meaningfully take traces of the action of a braid in a way which produces an invariant of the knot obtained by closing the braid up. Famously, the Jones polynomial arises in this way when $C$ is taken to be the braided monoidal category of representations of the quantum group $U_q(\mathfrak{sl}_2)$ and $V$ is taken to be the standard $2$-dimensional representation. 
Now, one way to describe the skein relation satisfied by the Jones polynomial is that it is a linear relation between three endomorphisms of $V^{\otimes 2}$, namely

two parallel strands, describing the identity,
a crossing, describing the braiding $b_{V, V} : V^{\otimes 2} \to V^{\otimes 2}$, and
the other crossing, describing the inverse $b_{V, V}^{-1}$.  

Now, why might there be a linear relation between these three elements? Certainly a sufficient condition is if $\text{End}(V^{\otimes 2})$ is at most $2$-dimensional. And this actually happens in the Jones polynomial case. (The point is that the representation theory of $U_q(\mathfrak{sl}_2)$ is sufficiently similar to that of $\mathfrak{sl}_2$ that it remains true that $V^{\otimes 2}$ is a direct sum of two nonisomorphic irreducibles.)<|endoftext|>
TITLE: Conjugating a subgroup of a group into a proper subgroup of itself
QUESTION [11 upvotes]: The following question came up in the class I'm teaching right now.  There definitely exist groups $G$ with subgroups $H$ such that there exists some $g \in G$ such that $g H g^{-1}$ is a proper subgroup of $H$.  For instance, let $G$ be the (big) permutation group of $\mathbb{Z}$ (by the big permutation group, I mean that elements of $G$ can move infinitely many elements of $\mathbb{Z}$).  Let $H \subset G$ be the big permutation group of $\mathbb{N}$ and let $g \in G$ be the permutation that takes $n \in \mathbb{Z}$ to $n+1 \in \mathbb{Z}$.  Then $g H g^{-1}$ is a proper subgroup of $H$.
The same sort of trick produces many examples like this.  However, a feature of all of them is that $G$ is "big" in some way -- for instance, $G$ is not finitely presentable.  By Higman's embedding theorem, you can embed such a $G$ into a finitely presentable group, so there exist examples where $G$ is finitely presentable.  However, in all the examples I can come up with, the subgroup $H$ is not finitely presentable.  I'm pretty sure that there exist examples in which $H$ is finitely presentable.  Does anyone know of one?  Even better, are there examples in which both $G$ and $H$ are of finite type (ie have compact $K(\pi,1)$'s)?

REPLY [18 votes]: There are very simple examples with $H\cong\mathbb{Z}$. For instance
let $G$ be the affine linear group over $\mathbb{Q}$ consisting
of all maps $x\mapsto ax+b$ where $a\in\mathbb{Q}^*$ and $b\in\mathbb{Q}$.
Let $H$ be the set of maps $x\mapsto x+b$ with $b\in\mathbb{Z}$.
Then $x\mapsto 2x$ conjugates $H$ into the proper subgroup $2H$.
In this example we could restrict $a$ to be a power of $2$
and $b$ to be a dyadic rational. Then the new $G$ has generators $a:x\mapsto 2x$
and $b:x\mapsto x+1$ and is defined by the single relation
$h^2=ghg^{-1}$.<|endoftext|>
TITLE: Path integrals outside QFT
QUESTION [16 upvotes]: The main application of Feynman path integrals (and the primary motivation behind them) is in Quantum Field Theory - currently this is something standard for physicists, if even the mathematical theory of functional integration is not (yet) rigorous.
My question is: what are the applications of path integrals outside QFT? By "outside QFT" I mean non-QFT physics as well as various branches of mathematics.
(a similar question is Doing geometry using Feynman Path Integral?, but it concerns only one possible application)

REPLY [2 votes]: These articles explain many interesting link invariants and their relations to braiding statistics of anyons in 3d spacetime and anyonic strings in 4d spacetime in 

Entangled Quantum Matter (condensed matter and lattice models). 

Try these articles:
Braiding Statistics and Link Invariants of Bosonic/Fermionic Topological Quantum Matter in 2+1 and 3+1 dimensions by Pavel Putrov, Juven Wang, Shing-Tung Yau
Annals of Physics 384C, 254-287 (2017) - doi.org/10.1016/j.aop.2017.06.019
and
Tunneling Topological Vacua via Extended Operators: (Spin-)TQFT Spectra and Boundary Deconfinement in Various Dimensions
Juven Wang, Kantaro Ohmori, Pavel Putrov, Yunqin Zheng, Zheyan Wan, Meng Guo, Hai Lin, Peng Gao, Shing-Tung Yau
Prog. Theor. Exp. Phys. 053A01 (2018) - doi.org/10.1093/ptep/pty051
The careful uses of path integral are performed in order to identify the phases of matters beyond the Ginzburg-Landau theory. These new phases include: 

symmetry protected topological [SPT] orders, 
interacting topological superconductors/insulators, 
invertible topological orders 
intrinsic topological orders.<|endoftext|>
TITLE: Conjugate prior of the Dirichlet distribution?
QUESTION [16 upvotes]: What is the conjugate prior distribution of the Dirichlet distribution?

REPLY [21 votes]: Neil sent me an email asking:
===
I read your post at http://www.stat.columbia.edu/~cook/movabletype/archives/2009/04/conjugate_prior.html and I was wondering if you could expand on how to update the Dirichlet conjugate prior that you provided in your paper:
S. Lefkimmiatis, P. Maragos, and G. Papandreou,
Bayesian Inference on Multiscale Models for Poisson Intensity Estimation: Applications to Photon-Limited Image Denoising,
IEEE Transactions on Image Processing, vol. 18, no. 8, pp. 1724-1741, Aug. 2009
In other words, given in your paper's notation the prior hyper-parameters (vector $\mathbf{v}$, and scalar $\eta$), and $N$ Dirichlet observations (vectors $\mathbf{\theta}_n, n=1,\dots,N$), how do you update $\mathbf{v}$ and $\eta$?
===
Here is my response:
Conjugate pairs are so convenient because there is a standard and simple way to incorporate new data by just modifying the parameters of the prior density. One just multiplies the likelihood with its conjugate prior; the result has the same parametric form as the prior, and the new parameters can be readily "read-off" by comparing the likelihood-prior product with the prior parametric form. This is described in detail in all standard texts in Bayesian statistics such as Gelman et al. (2003) or Bernardo and Smith (2000).
In the case of the Dirichlet and its conjugate prior described in our paper and using its notation, after observing $N$ Dirichlet vectors $\mathbf{\theta}_n$, $n=1,\dots,N$, where each vector $\mathbf{\theta}_n$ is $D$ dimensional with elements $\theta_n[t]$, $t=1,\dots,D$, the $D+1$ hyper-parameters should be updated as follows:

$\eta_N = \eta_0 + N$
$v_N[t] = v_0[t] - \sum_{n=1}^N \ln \theta_n[t], \quad t=1,\dots,D$, where $\eta_0$,  $\mathbf{v}_0$ and $\eta_N$, $\mathbf{v}_N$ are the initial and updated model parameters, respectively.

You can verify this in a few lines of equations by following the previously described general rule.
Hope this helps!<|endoftext|>
TITLE: Automorphisms of supergroups of non-coHopfian groups
QUESTION [8 upvotes]: In this question, I asked whether there existed groups $G$ with finitely presentable subgroups $H$ such that $gHg^{-1}$ is a proper subgroup of $H$ for some $g \in G$.  Robin Chapman pointed out that the group of affine automorphisms of $\mathbb{Q}$ contains examples where $H \cong \mathbb{Z}$.
This leads me to the following more general question.  A group $\Gamma$ is "coHopfian" if any injection $\Gamma \hookrightarrow \Gamma$ is an isomorphism.  To put it another way, $\Gamma$ does not contain any proper subgroup isomorphic to itself.  The canonical example of a non-coHopfian group is a free group $F_n$ on $n$ letters.  Chapman's example exploits the fact that $F_1 \cong \mathbb{Z}$ contains proper subgroups $k \mathbb{Z}$ isomorphic to $\mathbb{Z}$.
Now let $\Gamma$ be a non-coHopfian group and let $\Gamma' \subset \Gamma$ be a proper subgroup with $\Gamma' \cong \Gamma$.  Question : does there exist a group $\Gamma''$ such that $\Gamma \subset \Gamma''$ and an automorphism $\phi$ of $\Gamma''$ such that $\phi(\Gamma) = \Gamma'$?  How about if we restrict ourselves to the cases where $\Gamma$ and $\Gamma''$ are finitely presentable?  I expect that the answer is "no", and I'd be interested in conditions that would assure that it is "yes".
If such a $\Gamma''$ existed, then we could construct an example answering my linked-to question above by taking $G$ to be the semidirect product of $\Gamma''$ and $\mathbb{Z}$ with $\mathbb{Z}$ acting on $\Gamma''$ via $\phi$.  This question thus can be viewed as asking whether Chapman's answer really used something special about $\mathbb{Z}$.

REPLY [11 votes]: Let $\alpha: \Gamma\to\Gamma$ be an injection sending $\Gamma$ to $\Gamma'$.  Then the $\Gamma''$ you're looking for is the infinite amalgamated product
$\cdots *_{\Gamma}\Gamma *_{\Gamma}*\Gamma*_{\Gamma}\cdots$
where, at each stage, $\Gamma$ maps to the left by the identity and to the right by $\alpha$.  Now the 'shift' automorphism has the property that you want, and the semidirect product with $\mathbb{Z}$ that you suggest is just the ascending HNN extension of $\Gamma$ via $\alpha$.

REPLY [4 votes]: Effectively you have a group $\Gamma$ and a monomorphism
$\phi:\Gamma\to\Gamma$ which is not a surjection. Take the direct limit
of the sequence $(\Gamma_n)$ where each $\Gamma_n=\Gamma$
and each map from $\Gamma_n$ to $\Gamma_{n+1}$ is $\phi$.
I think this direct limit is the group $\Gamma''$ you want.<|endoftext|>
TITLE: When does a conditional expectation preserve some trace?
QUESTION [8 upvotes]: In developing a theory of index for inclusions of finite von Neumann algebras, several authors ([Kosaki, 1986], [Fidaleo & Isola,1996], etc.) define the index of a conditional expectation of a von Neumann algebra M onto a vN-subalgebra N (here, a conditional expectation is a normal, faithful N-N bimodule map fixing the subalgebra pointwise).  An inclusion is said to have finite index if there exists a conditional expectation that has finite index.  However, in the case where M is finite we might be interested in restricting ourselves to the conditional expectations that preserve some trace on M.
This leads us to the question: For a given (normal, faithful, finite) trace on M, Umegaki gives us a unique trace preserving conditional expectation E:M->N.  Are there any nice necessary and sufficient conditions for a conditional expectation to arise in this manner?  What if we allow the trace to be semifinite?
Since subfactors give rise to more than one conditional expectation, it is certainly not the case that all conditional expectations come from traces. A necessary condition is that E(xy)=E(yx) whenever x or y is an element of the relative commutant $N^\prime \cap M$.  This is not sufficient, however.

REPLY [3 votes]: I would expect a general answer to be difficult, because the set of traces on your von Neumann algebra will depend a lot on the centre of the algebra. 
In the case of a factor, the question becomes how to tell if a given expectation is the one that commutes with the trace. Not checking all my facts very carefully, I think that in this case the necessary condition is also sufficient: that is, if $E(xy)=E(yx)$ whenever x is in $N^\prime \cap M$, then $E$ commutes with the trace. This follows from the fact that this condition is equivalent to the modular group of the expectation (see Combes-Delaroche, 1975) being trivial; and in the case of a factor, the modular group characterizes the expectation (Remark 4.12.b in Combes-Delaroche).<|endoftext|>
TITLE: What's the name of this 2D cellular automaton?
QUESTION [5 upvotes]: Does this 2D cellular automaton have a known name and history?

n colors (numbered 1 to n), assigned randomly at the start.
For each generation, every cell that has at least one neighbour cell with a color that is one higher changes its color to that "next higher" color. Additionally, the "lowest" color is considered "next higher" to the highest one.
Emergent behaviour shows up best around n=16, disappears for much higher or much lower n

I have it implemented on my website so you can see it in action.
I saw this ages ago and always remembered it as a great example of emergent behaviour, but can't remember what it was called, and couldn't find it on Wikipedia or Wolfram Mathworld.

REPLY [10 votes]: Cyclic cellular automaton<|endoftext|>
TITLE: Is there an explicit example of a complex number which is not a period?
QUESTION [33 upvotes]: Kontsevich and Zagier proposed a definition of a period (see for example http://en.wikipedia.org/wiki/Ring_of_periods ). The set of periods is countable, so not all of $\mathbb{C}$. I heard a rumour today that there is now a known explicit complex number which is not a period; does anyone know if this is true, or have more details?

REPLY [26 votes]: Such a number is constructed in this article of M.Yoshinaga where it is proved that periods can be effectively approximated by elementary rational Cauchy sequences.<|endoftext|>
TITLE: Presentation of the monoid of surfaces
QUESTION [11 upvotes]: In the following every surface is assumed to be connected.
I've read that the commutative monoid of homeomorphism classes of closed surfaces is generated by $P$ (projective plane) and $T$ (torus) subject to the only(!) relation $P^3=PT$. Here the product is given by the connected sum. Now what about the commutative monoid of homeomorphism classes of compact surfaces (with boundary)? Does it also have a nice presentation? I think it is generated by the $P[k]$ and the $T[k]$, where $[k]$ means that $k$ holes have been inserted, and $k$ runs through the natural numbers. What relations do we need? And how do you prove that no others are needed? 
Another question, which is rather informal: Do you think that it's worth to read the proofs of these classical classifications? I know the importance of the results, but I suspect that the proofs are just technical.

REPLY [12 votes]: I agree with Tom and Ryan that it is worthwhile to learn proofs of the classification of surfaces.  I think that the result get a bit of a bad rap since the "standard" combinatorial proof that everyone used to learn (which appears in Seifert-Threlfell's book and Massey's book) is complicated and unenlightening.  However, there are now a number of nicer proofs available.  Here are a few of my favorites.
1) If you like Morse theory, there is a nice proof in Hirsch's book on differential topology.
2) There is a slick combinatorial proof in Armstrong's book "Basic Topology".  I believe that this is the source for the proof mentioned above by Tom Church that Benson Farb likes to give.
3) In Fomenko-Matveev's book "Algorithmic and Computer Methods for Three-Manifolds", there is a nice proof using handle decompositions.
4) There is finally John Conway's "ZIP proof", which was written up by Francis and Weeks in their paper "Conway's ZIP Proof".
All of these proofs assume that the surface has been equipped with either a triangulation (for numbers 2-4) or a smooth structure (for 1).  For nice approaches to this, see the answers to my question here.  However, when you are first approaching these types of results, I would recommend just assuming that the surfaces can be triangulated or smoothed.<|endoftext|>
TITLE: Request: A Serre fibration that is not a Dold fibration
QUESTION [13 upvotes]: A Serre fibration has the homotopy lifting property with respect to the maps $[0,1]^n \times \{0\} \to [0,1]^{n+1}$. A Dold fibration $E \to B$ has the weak covering homotopy property: lifts with respect to maps $Y\times \{0\} \to Y \times [0,1]$ such that the lift agrees with the map $Y \to E$ up to a vertical homotopy (see the nLab page for more details. All Hurewicz fibrations are Dold fibrations, but not conversely, and not all Dold fibrations are Serre fibrations. I'm sure I read that not all Serre fibrations are Dold fibrations, but I don't have a counterexample.
My request is thus: an example of a Serre fibration that is not a Dold fibration.
Edit: I have found that a slight variant on this question was asked by Ronnie Brown in Proc. Camb. Phil.Soc. in October 1966, under the caveat that the base is path-connected and the base and the fibre have the homotopy type of a CW complex.

REPLY [14 votes]: You've already answered your own question, but here is another example. 
Let $f: \mathbb{Q}^\delta \to \mathbb{Q}$ be the obvious map from the rational numbers with the discrete topology to the rational numbers with the usual topology. Let $M_f$ be the mapping cylinder. Then the projection 
$$p:M_f \to [0,1]$$
 is a Serre fibration. This follows because any map of a disc into $\mathbb{Q}$ factors through f. Hence as far as discs are concerned, $M_f$ might as well be $\mathbb{Q}^\delta \times [0,1]$.  
However this projection  is not a Dold fibration. It is easy to construct a diagram using $Y = \mathbb{Q}$ which will have no weak homotopy lift. Indeed consider the projection map $\mathbb{Q} \times [0,1] \to [0,1]$ with the obvious initial lift. Any other initial lift vertically homotopic to this one in fact coincides with this one, hence it is easy to see that there is no weak lift of this map. 
By replacing $\mathbb{Q}^\delta$ and $\mathbb{Q}$ with their cones, we get a similar example where now the base, total space, and fibers are contractible. Hence they are path-connected and also have the homotopy type of CW complexes. So this also answers Ronnie Browns question (but surely the answer to that has been known for some time).<|endoftext|>
TITLE: Semiring of algebraic vector bundles on projective space
QUESTION [11 upvotes]: Let $K$ be a field and $n \geq 1$. Then the set of isomorphism classes of vector bundles over $\mathbb{P}^n_K$ is a semiring (i.e. almost a ring, but no additive inverses are possible). By introducing additive inverses and quotienting out exact sequences, we get the $K$-theory of $\mathbb{P}^n_K$, which is known to be $\mathbb{Z}^{n+1}$. But is it also possible to compute exactly the semiring?
For $n=1$, there is a result by Dedekind-Weber (1892) which proves that the semiring is $\mathbb{N}[x,x^{-1}]$, where $x=\mathcal{O}(1)$ (related topic). Some months ago, I was told that the structure is far more complicated for $n>1$. Can anybody elaborate this or even give a presentation of the semiring?
If necessary, you may assume $K = \mathbb{C}$.

REPLY [10 votes]: This semiring carries an enourmous amount of information about vector bundles on $\mathbb{P}^n$, including stuff we don't yet know. For example, you can read from it whether there are indecomposable vector bundles of any given rank; and for small rank we know very little about it (this is discussed, for example, in C. Okonek, M. Schneider, H. Spindler, "Vector bundles on complex projective spaces" , Birkhäuser (1987); I don't have access to the book here, and can't give you a more precise reference). I doubt you can can even get close to what you want.<|endoftext|>
TITLE: Coend computation
QUESTION [17 upvotes]: Let

$F:A^{\mbox{op}} \to \mbox{Set}$

and define

$G_a:A\times A^{\mbox{op}} \to \mbox{Set}$
$G_a(b,c) = \mbox{hom}(a,b) \times F(c)$.

I think the coend of $G_a$,

$\int^AG_a$,

ought to be $F(a)$--it's certainly true when A is discrete, since then hom is a delta function.  But my colimit-fu isn't good enough to actually compute the thing and verify it's true.  Can someone walk me through the computation, please?

REPLY [28 votes]: Hi Mike.  This is what's often called the Density Formula, or (at the n-Lab) the coYoneda Lemma (I think), or (by Australian ninja category theorists) simply the Yoneda Lemma.  (But Australian ninja category theorists call everything the Yoneda Lemma.)  In any case, it's a kind of dual to the ordinary Yoneda Lemma.
But you asked to be walked through it.  First: yes, it is $F(a)$.  Another way of writing your coend
$$
\int^A G_a
$$
is as
$$
\int^{b \in A} G_a(b, b) = \int^b \mathrm{hom}(a,b) \times F(b).
$$
I claim this is canonically isomorphic to $F(a)$.  I'll prove this by showing that for an arbitrary set $S$, the homset $\mathrm{hom}(\mathrm{this}, S)$ is canonically isomorphic to $\mathrm{hom}(F(a), S)$.  The claim will then follow from the ordinary Yoneda Lemma.
So, let $S$ be a set.  Then
$$
\begin{align}
\mathrm{Set}(\int^b \mathrm{hom}(a, b) \times F(b), S) &
\cong 
\int_b \mathrm{Set}(\mathrm{hom}(a, b) \times F(b), S) \\
&\cong 
\int_b \mathrm{Set}(\mathrm{hom}(a, b), \mathrm{Set}(F(b), S)) \\
&\cong 
\mathrm{Nat}(\hom(a, -), \mathrm{Set}(F(-), S)) \\
&\cong 
\mathrm{Set}(F(a), S)
\end{align}
$$
I don't know how much of this you'll want explaining, so I'll just say it briefly for now.  If you want further explanation, just ask.  The first isomorphism is kinda the definition of colimit.  The second is the usual exponential transpose/currying operation.  The third is maybe the most important: it's a fundamental fact about ends that if $F, G: C \to D$ are functors then
$$
\mathrm{Nat}(F, G) = \int_c D(F(c), G(c)).
$$
The fourth and final isomorphism is the ordinary Yoneda Lemma applied to the functor $\mathrm{Set}(F(-), S)$.<|endoftext|>
TITLE: Are G_infinity algebras B_infinity? Vice versa?
QUESTION [20 upvotes]: What is the relationship between $G_\infty$ (homotopy Gerstenhaber) and $B_\infty$ algebras?
In Getzler & Jones "Operads, homotopy algebra, and iterated integrals for double loop spaces" (a paper I don't well understand) a $B_\infty$ algebra is defined to be a graded vector space $V$ together with a dg-bialgebra structure on $BV = \oplus_{i \geq 0} (V[1])^{\otimes i}$, that is a square-zero, degree one coderivation $\delta$ of the canonical coalgebra structure (stopping here, we have defined an $A_\infty$ algebra) and an associative multiplication $m:BV \otimes BV \to BV$ that is a morphism of coalgebras and such that $\delta$ is a derivation of $m$. 
A $G_\infty$ algebra is more complicated.  The $G_\infty$ operad is a dg-operad whose underlying graded operad is free and such that its cohomology is the operad controlling Gerstenhaber algebras.  I believe that the operad of chains on the little 2-discs operad is a model for the $G_\infty$ operad.  Yes?
It is now known (the famous Deligne conjecture) that the Hochschild cochain complex of an associative algebra carries the structure of a $G_\infty$ algebra.  It also carries the structure of a $B_\infty$ algebra.  Some articles discuss the $G_\infty$ structure while others discuss the $B_\infty$ structure.  So I wonder: How are these structures related in this case?  In general?

REPLY [9 votes]: There is a nice summary of the relationship between B infinity and G infinity in the first chapter of the book "Operads in Algebra, Topology and Physics" by Markl, Stasheff and Schnider. The short answer is G infinity is the minimal model for the homology of the little disks operad (the G operad). B infinity is an operad of operations on the Hochschild complex. Many of the proofs of Deligne's conjecture involve constructing a map between these two operads.<|endoftext|>
TITLE: Realizability of irreducible representations of dihedral groups
QUESTION [12 upvotes]: Consider a dihedral group of degree n and order 2n. Its two-dimensional irreducible representations can be realized over the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$, with the usual action by rotations and reflections. Also, any splitting field of characteristic zero for this group must contain $\mathbb{Q}(\cos(2\pi/n))$, because this is the field generated by the characters of the irreducible representations. (Here, splitting field for a finite group means a field over which all the irreducible representations are realized).
When n is a multiple of 4, these two fields are the same; otherwise, they are not. My question: when n is not a multiple of 4, what conditions would ensure that the smaller subfield $\mathbb{Q}(\cos(2\pi/n))$ is a splitting field for the dihedral group? I think we can restrict attention to n odd.
For instance, when $n = 3$, the dihedral group of degree 3, order 6, has $\cos(2\pi/3) = -1/2$, $\sin(2\pi/3) = \sqrt{3}/2$. So, any splitting field must contain $\mathbb{Q}(1/2) = \mathbb{Q}$. Also, $\mathbb{Q}(1/2,\sqrt{3}/2) = \mathbb{Q}(\sqrt{3})$ is a splitting field since it contains the usual representation given by rotations and reflections.
However, $\mathbb{Q}$ is also a splitting field. To see this, we think of the group as the symmetric group of degree three and take the standard representation. So in this case, we see that $\mathbb{Q}(\cos(2\pi/n))$ works as splitting field even though it is smaller than the field $\mathbb{Q}(\cos(2\pi/n),\sin(2\pi/n))$.
NOTE: It is not true in general that if the characters all take values in a field, the representation can be realized over that field. The standard counterexample is the quaternion group, whose characters all take rational values but whose irreducible representations are realized only when we go to $\mathbb{Q}(i)$. However, some weaker variant of the result may be true for the groups that we are interested in here, namely, the dihedral groups.

REPLY [4 votes]: Since, as Pete Clark pointed out, it is one way to answer the original question, 
it may be worth explaining with Schur index theory, rather than invariant theory, why a finite 
irreducible complex linear group $G$ in dimension $n$ which contains even a single pseudoreflection is conjugate within ${\rm GL}(n,\mathbb{C})$ to a subgroup of ${\rm GL}(n,F)$, where $F$ is the
field generated by the traces of the elements of $G$. Let $\chi$ be the character of the given representation of $G$. Let $x$ be a (genuine, ie non-scalar) pseudo-reflection of $G$. Then
${\rm Res}^{G}_{\langle x \rangle}(\chi) = (n-1)1 + \lambda$ for a non-trivial character $\lambda$
of $\langle x \rangle$. 
Recall that the Schur index $m_{F}(\chi)$ is the smallest positive multiplicity with which $\chi$
occurs as a constituent of a character afforded by a representation of $G$ over $F$. A key property of $m_{F}(\chi)$, from the general theory, is that $m_{F}(\chi)$ even divides the multiplicity
with which $\chi$ occurs in the character afforded by a representation of $G$ realised over $F$.
Now certainly the regular representation of $G$ can be realized over $F$, and $\chi$ occurs with multiplicity $n = \chi(1)$ in the regular character. Hence $m_{F}(\chi)$ divides $n$. But by Frobenius reciprocity, $\chi$ occurs with multiplicity $n-1$ in the permutation representation 
of $G$ induced from the trivial representation of $\langle x \rangle$, and this representation is certainly also realised over the field $F$. Hence $m_{F}(\chi)$ also divides $n-1$, so we conclude
that $m_{F}(\chi) = 1$, that is, the given representation may be realised over $F$.<|endoftext|>
TITLE: Biographic Data/Stories about André Néron
QUESTION [26 upvotes]: Tomorrow, April 6, 2010, André Néron will have been dead for 25 years.
In spite of the weight of his work on abelian varieties, I've only been able to ascertain the following information:

His birth and death dates (born November 30, 1922)
In 1943 he graduated from the École Normale. He got his doctorate in 1951 and his advisor was Châtelet (though not at any particular school, just somewhere in Paris according to German Wikipedia) and his only student was Colliot-Thélène at Orsay (shared with Swinnerton-Dyer)
He was employed at Poitiers. It also seems that in 1953 when he was inducted into the Société Mathématique de France he was listed as being at Orsay. In years 59-60 he was at the IAS and in 1954 he was an invited speaker at the ICM in Amsterdam.

That's pretty much what I found. And all that does is hint that there's a really fantastic story in there!
What happened during the war?
How did he get back to mathematics?
Who else did he work with (not just publish papers with, MathSciNet suggests his only collaborators were Serge Lang and Pierre Samuel)?
Why so few students? Was he difficult to get along with or was it just a sign of the times?
Why did he die at age 62?
If you have any data or stories about him, please leave them here, one answer per.

REPLY [11 votes]: It seems that Colliot-Thélène was asked this question three times: by me, by Olivier, and by Kevin Buzzard.  Here is the reply that he gave to Kevin and forwarded to me:

Le patron d'André Néron était Albert Châtelet.
Néron eut un autre étudiant en thèse, Gérard Ligozat,
  qui après plusieurs travaux sur les courbes modulaires
   quitta le département de mathématiques.
André Néron parlait le langage de la géométrie algébrique
  de Weil à une époque où l'école de Grothendieck était devenu
  dominante. Les jeunes fringants allaient naturellement voir
  du côté des schémas.
La notion de patron et d'élève dans les années 1970
  en France était souple. Après une excellente scolarité,
  on obtenait un poste quasi-permanent soit à l'Université
  soit au CNRS sans avoir publié une ligne dans une revue.
  Ensuite on faisait une thèse si on en avait envie. Voir
  son patron une fois par an était souvent suffisant.
Et le système, sur la durée, a marché aussi bien
  que dans notre époque de publish ou perish.
André Néron mourut d'un cancer en 1985.

Kevin explicitly gave Colliot-Thélène the option of responding in French, which seems appropriate given the subject matter.  

English translation by François G. Dorais:

André Néron's advisor was Albert Châtelet.
Néron had one more thesis student, Gérard Ligozat, who left the Department of Mathematics after much work on modular curves.
André Néron spoke the language of Algebraic Geometry in the style of Weil at a time when the Grothendieck school had become dominant. The dashing youth naturally leaned toward [Grothendieck's] schemes.
The notion of advisor and student was flexible in France during the 1970's. After excellent scholarly work, one could obtain a quasi-permanent job either at the University or at the CNRS, without having published a single line in a journal. Then, if desired, one could do a thesis. Seeing one's advisor once per year was often enough.
And this system, while it lasted, worked just as well as the publish or perish [system] of our times.
André Néron died of cancer in 1985.

(I took a few very minor liberties for readability, but the translation is mostly literal.)<|endoftext|>
TITLE: Why are free groups residually finite?
QUESTION [40 upvotes]: Why is it that every nontrivial word in a free group (it's easy to reduce to the case of, say, two generators) has a nontrivial image in some finite group?  Equivalently, why is the natural map from a group to its profinite completion injective if the group is free?
Apparently, this follows from a result of Malcev's that finitely generated matrix groups over an arbitrary commutative ring are residually finite, but is there a more easily accessible proof if we only want the result for free groups?

REPLY [2 votes]: A proof based on algebraic topology can also be found in Thomas Koberda's lecture notes, RAAGS and their subgroups; in fact, he proves more generally that a finitely generated free group is residually $p$ for any prime. The proof is rather laconic, so I gives some details below.
A finitely generated free group $F$ is the fundamental group of a finite bouquet of circles $X_0$. By induction, we build a sequence of regular coverings $$\cdots \to X_2 \to X_1 \to X_0,$$ where $\mathrm{Aut}(X_{i+1} \to X_i) \simeq H_1(X_i, \mathbb{Z}_p)$; in order to construct $X_{i+1}$, it is sufficient to take the covering of $X_i$ associated to the subgroup of $\pi_1(X_i)$ consisting of the kernel of $$\pi_1(X_i) \twoheadrightarrow \pi_1(X_i)^{\mathrm{ab}} \simeq H_1(X_i,\mathbb{Z}) \twoheadrightarrow H_1(X_i, \mathbb{Z}_p).$$
In particular, notice that $H_1(X_i,\mathbb{Z}_p)$ is a finitely generated abelian torsion group, so it is a finite $p$-group. Therefore, any covering $X_{i+1} \to X_i$ of our sequence has a degree a power of $p$. To conclude, it is sufficient to notice that any loop in $X_0$ does not lift into $X_i$ as a loop for some large enough $i$. 
Because the action $\mathrm{Aut}(X_{i+1} \to X_i) \curvearrowright X_{i+1}$ is free, we deduce
Claim 1: A loop $\gamma \subset X_i$ lifts into $X_{i+1}$ as a loop iff $\gamma=1$ in $H_1(X_0, \mathbb{Z}_p)$.
Claim 2: A loop minimising the length among the homotopically nontrivial loops of a graph $Y$ defines a nontrivial element of $H_1(Y,\mathbb{Z})$.
Now let $l(X_i)$ denote the minimal length of a homotopically nontrivial loop in $X_i$. Of course, we have $l(X_0)=1$.
Claim 3: $l(X_{i}) \geq i+1$ for all $i \geq 0$.
Let $\gamma$ be a loop in $X_{i+1}$ and let $p$ denote the covering map $X_{i+1} \to X_i$. Then $p(\gamma)$ is a loop in $X_i$ that lifts into $X_{i+1}$ as a loop, so $\mathrm{lg}(p(\gamma)) \geq l(X_i)+1$ according to claims 1 and 2 (if $p$ is large enough). Hence $$\mathrm{lg}(\gamma) \geq \mathrm{lg}(p(\gamma)) \geq l(X_i)+1,$$ and finally $l(X_{i+1}) \geq l(X_i)+1$. Now, claim 3 follows easily.
We deduce from claim 3 that, if a homotopically nontrivial loop $\gamma \subset X_0$ loops into $X_i$ as a loop $\gamma'$, then $$\mathrm{lg}(\gamma)=\mathrm{lg}(\gamma') \geq l(X_i) \geq i+1;$$ therefore, for large enough $j$, $\gamma$ cannot lifts into $X_j$ as a loop.<|endoftext|>
TITLE: Is there a theory on two sequences of measures weakly asymptotic to each other?
QUESTION [5 upvotes]: Suppose $(P_n)_{n\ge 1}$ is a sequence of probability measures on a metric space $E$. Everybody knows what weak convergence of $P_n$ to a measure $P$ is. 
Instead, let $(Q_n)_{n\ge 1}$ be another sequence of probability measures, and I want to ask the following question: when are $Q_n$ and $P_n$ asymptotically close in the sense of weak convergence? Of course, I can say that it means that $d(P_n,Q_n)\to 0$ where $d$ is one of the metrics compatible with weak convergence of measures. However, is there a useful theory helping to prove $d(P_n,Q_n)\to 0$?  For the usual case i.e. weak convergence $P_n\to P$, there is a number of technologies based on tightness, and here we may have a situation where none of the two sequences is tight, but they still "converge to each other".
This could be useful if one of the two sequences is much simpler or better understood than the other one, and I am mostly interested in situations where the metric space $E$ is a functional space like $C$ or $D$.
UPD. It turns out that "merging" is the keyword and there is some literature on the issue
(thanks to Mark Meckes for pointing to one of these papers), but it looks like there is still no tool to prove merging in functional spaces.
It seems that the most recent paper related to this topic is 
Davydov & Rotar: On asymptotic proximity of distributions. J. Theoret. Probab. 22 (2009), no. 1, 82--98.

REPLY [4 votes]: D'Aristotile, Diaconis, and Freedman called this "merging" of probability measures, and studied relationships among different notions of it in this paper.<|endoftext|>
TITLE: A graph on irrationals where p is adjacent to q if p^q or q^p is rational.
QUESTION [9 upvotes]: When I was in high school I learned about an elementary proof that there exist irrational numbers $p$ and $q$ such that $p^q$ is rational. Put $p = q = \sqrt{2}$; if $p^q$ is rational, we are done. Otherwise take $p = \sqrt{2}^\sqrt{2}$and $q = \sqrt{2}$.  
I am sure that by now much more is known about this phenomenon. Maybe the answer can be described in terms of graph theory. Define a graph $G$ on the positive irrationals. Two irrationals $p$ and $q$ are adjacent in $G$ if and only if at least one of the two powers $p^q$ or $q^p$ is a rational number that is not an integer power of another rational. A similar graph $G^*$ could be defined on the positive real numbers. I would like to know what can be said about $G$ and $G^*$. For instance, what is the cardinality of the edge set $E(G)\ ?$ What is the maximum degree of $G\ ?$
Suggestions for proper tagging are most welcome.

REPLY [14 votes]: The answers are that $|E(G)|=2^{\aleph_0}$ and that every vertex of $G$ has degree $\aleph_0$.  
Proof:
The positive real solutions to $x^y=2$ form a curve of cardinality $2^{\aleph_0}$, and at most $\aleph_0$ of these have $x$ or $y$ rational, so $|E(G)|=2^{\aleph_0}$.  (It cannot be larger, because this is also the number of pairs of vertices.)  
For $g \in G$ with $g>1$, the set of positive real numbers $x$ such that $g^x$ is a prime number contains at most one rational number since if there were two, then we would obtain an equation $p^q=p'$ with $p,p'$ distinct primes and $q$ rational, which is impossible by unique factorization.  Thus $g$ has infinite degree.  On the other hand, the degree of $g$ is at most countable since $g^x = q$ and $x^g=q$ have at most one solution $x$ each for each $q \in \mathbf{Q}$.  Finally, the same arguments apply when $g<1$, with reciprocals of primes in place of primes.
Remark: The same statements hold for $G^*$ except that one should exclude the vertex $1$.<|endoftext|>
TITLE: How does eigenvalues and eigenvectors change if the original matrix changes slightly?
QUESTION [5 upvotes]: Hi,
We have a square symmetric matrix M with all the elements 0 or 1, and the eigenvalues and vectors of M are computed.
Now if we change one arbitrary element of M (from 0 to 1 or 1 to 0), what will happen to eigenvalues (and eigenvectors)? Does there exist a clear relation or not?
Thank you for your help
Hoda

REPLY [2 votes]: As Matthew Daws pointed out with his link above, Terry Tao answers your specific question.  However, you might be interested in the larger framework of random matrix theory, which is includes the study of eigenvalue distributions of large random matrices.  This subject makes precise Douglas Zare's comment:I would guess that there will be some properties of the eigenvalues which hold for at least 98% of the alterations of a large matrices.
By putting special probability distributions on the space of $N \times N$ matrices, one can do everything explicitly using orthogonal polynomials, Riemann-Hilbert theory, and a whole slew of other exact tools.  For example, the Gaussian Unitary Ensemble (GUE) is the class of complex $N \times N$ matrices with i.i.d. Gaussian entries constrained so that the matrix is Hermitian.  Here, the distribution of largest eigenvalue follows the Tracy-Widom law, the cumulative distribution of eigenvalues is Wigner's semicircle law, and one can calculate explicitly pretty much any other quantity of interest.
Just as the central limit theorem holds for a much wider class of i.i.d. random variables than just Gaussians, much current work in the analysis of random matrix theory is to show that these properties are universal, and don't rely on the specific Gaussian structure.<|endoftext|>
TITLE: What is torsion in differential geometry intuitively?
QUESTION [199 upvotes]: Hi,
given a connection on the tangent space of a manifold, one can define its torsion:
$$T(X,Y):=\triangledown_X Y - \triangledown_Y X - [X,Y]$$
What is the geometric picture behind this definition—what does torsion measure intuitively?

REPLY [3 votes]: Define a path as a set of instructions to move, as given in the reference frame of the starting point, i.e. one step forwards, one step to your right, one step back and one step left, all given with respect to the initial reference frame. Of course, in order to follow the instructions, we have to parallel transport the reference frame with the path as we move.
In a flat space, when we parallel transport our reference frame through a closed path (closed in the flat map we parallel transported), we end up at the initial point and the reference frame ends in its initial orientation. If we do this in a torsionless, curved space, we probably will end up somewhere else, unless the path is infinitesimal. Yet, in this case, the reference frame will probably be changed with respect to the initial frame. In Riemannian geometries, the reference frame will simply be rotated with respect to the original but relative angles  and the length of the basic vectors is left invariant. In non-Riemannian geometries relative angles and lengths of the basis vectors of the reference frame can vary.
What happens when you go through an infinitesimal closed path if torsion is not zero?
The set of instructions close, but we end up in a different point of the space. 
Torsion is the measure by which an infinitesimal closed path does not result in a closed loop.<|endoftext|>
TITLE: Is it known that the ring of periods is not a field?
QUESTION [73 upvotes]: I have just learned here that we know numbers that are not periods; is it known meanwhile that the ring of periods is not a field? I know that it is conjectured that $1/\pi$ is not a period, but the existence of a period whose inverse is not a period seems to be still open. Is this correct?
More generally: is it believed that the unit group of the ring of periods is bigger than the nonzero algebraic numbers?

REPLY [24 votes]: I think the questions were about unconditional proofs or counter examples. I don't have an answer to any of those questions but I think it is still interesting to understand how the yoga of motives suggests natural answers to theses questions. Even though this may seem trivial to people familiar with the subject.   
Let's work in the setting of Voevodsky's $\otimes$-triangulated categories $DM^{eff}(\mathbb{Q}):= DM_{gm}^{eff}(Spec(\mathbb{Q});\mathbb{Q}) \subset DM_{gm}(Spec(\mathbb{Q});\mathbb{Q}) =: DM(\mathbb{Q})$. Remember that the latter is obtained by formally inverting $\mathbb{Q}(1)$ and that it is a rigid $\otimes$-triangulated category. 
Question 1: Is the ring of (effective) periods a field?
Following Beilinson's "Remarks on Grothendieck's standard conjecctures", let's assume
Motivic conjecture: There exists a non degenerate t-structure on Voevodsky's category $DM(\mathbb{Q}) := DM(Spec(\mathbb{Q});\mathbb{Q})$ and such that the Betti realization function $\omega_B: DM(\mathbb{Q}) \to D^bMod_f(\mathbb{Q})$ is a $t$-exact $\otimes$-functor.
This is an extremely strong conjecture as it implies the standard conjectures in characteristic 0.
Under this conjecture, the heart of the motvitic $t$-structure is a tannakian category $MM(\mathbb{Q})$. We have Betti and Rham realization functors $\omega_B,\omega_{dR}: MM(\mathbb{Q}) \rightrightarrows Mod_f(\mathbb{Q})$. And we can define 
$$
  Per := Isom^\otimes(\omega_{dR},\omega_{B})
$$
This is a fpqc-torsor under the motivic Galois group $G_B := Aut^\otimes(\omega_B)$. 
Define the algebra of motivic periods as the ring of regular functions on the Betti/de Rham torsor:
$$
  P_{mot} := \mathcal{O}(Per)
$$ 
Integration of differential forms (or more generally the Riemann-Hilbert correspondance) defines an $\mathbb{C}$-point 
$$
  Spec(\mathbb{C}) \longrightarrow Per
$$
The image of the corresponding morphism $P_{mot} \to \mathbb{C}$ is the ring of periods $P$.
Note: I think this whole part is actually known unconditionnally in the setting of Nori's motives (see arXiv:1105.0865v4).   
Period conjecture: The morphism $P_{mot} \to P$ is an isomorphism.
Now based on these tiny little conjectures we can say 
Prop: $P_{mot}$ is not a field so $P$ isn't either.
Proof: Indeed in his comment G-torsor whose ring of regular functions is a field.   @quasi-coherent explained how faithfull flatness would imply that if $P_{mot}$ were a field then it would be algebraic over $\mathbb{Q}$ which contradicts the fact that $2\pi i$ belongs to the image of $P_{mot}\to \mathbb{C}$.
Question 2 Is it true that $(P_{mot}^{eff})^\times = \overline{\mathbb{Q}}^\times$? 
This post is getting too long already so I'll try and write down the rest later but the basic idea is that invertible effective motives are Artin motives. This can be proved in terms of weights or niveau (level).<|endoftext|>
TITLE: Algebraic cycles of dimension 2 on the square of a generic abelian surface
QUESTION [9 upvotes]: I would like to know, what is known on algebraic cycles of dimension 2
modulo algebraic or rational equivalence
on the square of a generic abelian surface.
First, let $A$ be a generic abelian surface (generic abelian variety of dimension 2)
over $\mathbb{C}$.
Then the group of cycles of dimension 1 (divisors)
up to rational equivalence is known, it its the Picard group of $A$,
see e.g. the answers to this question
and Fulton, Intersection Theory, Chapter 19.
Still I have a stupid question: Can one "write down" all the (positive?) divisors on $A$?
The real question concerns the square $A^2=A\times_{\mathbb{C}} A$ of a generic abelian surface $A$.
This is an abelian variety of dimension 4.
I am interested in algebraic cycles of dimension 2 on $A^2$.
I think I know the group of algebraic cycles of dimension 2 on $A^2$
modulo homological equivalence and modulo torsion, it is  $\mathbb{Z}^6$
(because the space of invariants of $\mathrm{Sp}_{4,\mathbb{Q}}$
in $\wedge^4(\mathbb{Q}^4\oplus\mathbb{Q}^4)$ is of dimension 6).
What is known about the group of algebraic cycles of dimension 2 on $A^2$ modulo rational or algebraic equivalence?
In particular, what is known about the Griffiths group?
Again a stupid question: Can one "write down" all the cycles of dimension 2 on $A^2$ (in some sense)?

REPLY [2 votes]: Not an answer per se, but you might be interested in http://arxiv.org/abs/1003.3183, where similar questions are investigated.<|endoftext|>
TITLE: Is K(R-Mod) compactly generated when R is an artin algebra?
QUESTION [5 upvotes]: I wonder if the triangulated category K(R-Mod) is compactly generated when R is an artin algebra? R-Mod denotes all left R-modules. I understand this would be true if R has finite representation type since R-modules then are direct sums of finitely generated ones, but I am interested in the general case. Could it be that a generating set are finitely generated R-modules and shifts of them. (This would not be true for general rings, e.g. Neeman showed that K(Z-mod) is not compactly generated.)
Thanks.

REPLY [2 votes]: The answer is in general no - $K(R\text{-}\mathrm{Mod})$ can fail to be well generated even when $R$ is artinian. As you mention $K(R\text{-}\mathrm{Mod})$ is compactly generated if $R$ is of finite representation type. It turns out that the converse holds. This is a result of Jan Šťovíček which occurs as Proposition 2.6 in  this paper. The precise result is:
Proposition Let $R$ be a ring. The following are equivalent:
(i) $K(R\text{-}\mathrm{Mod})$ is well generated;
(ii) $K(R\text{-}\mathrm{Mod})$ is compactly generated;
(iii) $R$ is left pure semisimple.
In particular, when $R$ is artinian this occurs precisely when $R$ has finite representation type.<|endoftext|>
TITLE: Characterizing invertible matrices with {0,1} entries
QUESTION [8 upvotes]: Related to the question link text I was asking myself some time ago the following. Can one precisely describe the invertible n\times n matrices with{0, 1} entries? For example, is anything special about the graph asociated to this matrix?

REPLY [3 votes]: At least for symmetric matrices, such graphs have been studied under the name singular graphs. See, for instance, this paper.<|endoftext|>
TITLE: conformally embedding complex tori into R^3
QUESTION [9 upvotes]: Let $L$ be a lattice in $\mathbb{C}$ with two fundamental periods, so that $\mathbb{C}/L$ is topologically a torus. Let $p:\mathbb{C}/L \mapsto \mathbb{R}^3$ be an embedding ($C^1$, say). Call $p$ conformal if pulling back the standard metric on $\mathbb{R}^3$ along $p$ yields a metric in the equivalence class of metrics on $\mathbb{C}/L$ (i.e. a multiple of the identity matrix).

Is there an explicit formula for such a p in the case of L an oblique lattice?


Background
The existence of such $C^1$ embeddings is implied by the Nash embedding theorem (fix a metric on $\mathbb{C}/L$, pick any short embedding, apply Nash iteration to make it isometric and hence conformal).
For orthogonal lattices, the solution is simple: Parametrise the standard torus of radii $r_1$, $r_2$ in the usual way. Make the ansatz $\pi(\theta, \phi) = (f(\theta), h(\phi))$, pull back the standard metric on $\mathbb{C}/L$ and solve the resulting system of ODEs. This relates $r_1/r_2$ to the ratio of the magnitudes of the periods. This shows that no standard torus can be the image of $p$ in the original question (oblique lattice), although that is geometrically clear anyway.

REPLY [11 votes]: You should have a look in Pinkall's Hopf Tori paper. You take the preimage of a curve in $S^2$ under the Hopf fibration. The lattice of the torus and hence the conformal class is then given by the generators $1\in C$ and $L+i/2 A$ (if I remember right), where $L$ is the length and $A$ is the enclosed area of the curve.<|endoftext|>
TITLE: Decidability of conjugacy problem for finitely generated subgroups of free groups
QUESTION [10 upvotes]: The conjugacy problem for a free group $F_n$ on $n$ letters has an easy solution.  Each element of $F_n$ is conjugate to a unique and easily computable "cyclically reduced element" (this means that if you arrange the word around a circle, then there are no cancellations), so two elements of $F_n$ are conjugate if and only if they have the same cyclically reduced conjugates.
I've been trying unsuccessfully to generalize this to solve the following problem.  Let $\{x_1,\ldots,x_k\}$ and $\{y_1,\ldots,y_{k'}\}$ be two finite sets of elements of $F_n$.  Let $G_x$ and $G_y$ be the subgroups of $F_n$ generated by the $x_i$ and the $y_i$, respectively.  Is there an algorithm to decide if $G_x$ and $G_y$ are conjugate?  Does anyone know how to do this?  Thank you very much!

REPLY [7 votes]: There is an algorithm to do this.  I would have thought that it was classical, but in any case an algorithm is given in: I. Kapovich and A. Myasnikov "Stallings foldings and the subgroup structure of free groups", J. Algebra 248 (2002), no 2,   pp. 608-668.  In the online version I found here, it is Corollary 7.8 on page 18.<|endoftext|>
TITLE: Basis for modular forms of half-integral weight
QUESTION [8 upvotes]: Given a character $\chi$ and $k$ odd how can one compute a basis for the space of modular forms $M_\frac{k}{2}(\Gamma_0(4),\chi)$. By compute a basis I mean, finding the beginning of the Fourier expansions. I am looking for computer programs, which can do that for me.
I have heard of the package SAGE, which seems to do the job for integral weight modular forms. There is even the function http://www.sagemath.org/doc/reference/sage/modular/modform/half_integral.html but the examples all have q-expansions starting with q, so I guess this is not really a basis for the space of all modular forms but only cusp forms.
MAGMA does not seem to include this functionality, either.
So, are there any packages which can do this? Since I have not found a package, I have some doubts that there is really an algorithm working in general. If there is no algorithm known to handle this, what methods are available in order to compute a basis "by hand"?
Thanks.

REPLY [5 votes]: "MAGMA does not seem to include this functionality, either."

Basis(HalfIntegralWeightForms(DirichletGroup(4).1^2,11/2));

[
    1 - 88*q^3 - 330*q^4 - 4224*q^7 - 7524*q^8 - 30600*q^11 + O(q^12),
q + 4*q^3 + 56*q^4 + 132*q^5 + 224*q^6 + 512*q^7 + 912*q^8 + 1525*q^9 + 
2752*q^10 + 4044*q^11 + O(q^12),

q^2 + 6*q^3 + 20*q^4 + 56*q^5 + 130*q^6 + 256*q^7 + 472*q^8 + 800*q^9 + 
1266*q^10 + 1970*q^11 + O(q^12)

]

Basis(HalfIntegralWeightForms(DirichletGroup(112).1^2,3/2)); 

[
    1 + 2*q^16 + 2*q^28 + O(q^30),
q - q^21 + 2*q^29 + O(q^30),

...
]
http://magma.maths.usyd.edu.au/calc<|endoftext|>
TITLE: What is the minimum N for which there exist N points in the plane that cannot be covered by any number of non-overlapping closed unit discs?
QUESTION [19 upvotes]: This problem was posed in March 2010 at G4G9 in a talk by the Japanese mathematician Hirokazu "Iwahiro" Iwasawa. He claims there is a simple proof that N > 10, though he did not share it with the audience, since it proving it is apparently an enlightening exercise in its own right.

REPLY [7 votes]: In the answer to Open problems in Euclidean geometry? , 
Alexey Ustinov brings into attention to a 2012 article. 

Greg Aloupis, Robert A. Hearn, Hirokazu Iwasawa, Ryuhei Uehara, Covering Points with Disjoint Unit Disks, 24th Canadian Conference on Computational Geometry (2012)

The abstract of that article confirms that it's concerned of the same problem, and gives improved bounds.

We consider the following problem. How many points must be placed in the plane so that no collection of disjoint unit disks can cover them? The answer, k, is already known to satisfy 11 ≤ k ≤ 53. Here, we improve the lower bound to 13 and the upper bound to 50. We also provide a set of 45 points that apparently cannot be covered, although this has been determined via computer search.

The article also claims that the lower bound of 11 was published in 2008<|endoftext|>
TITLE: What is the oriented Fano plane?
QUESTION [35 upvotes]: One way to remember the multiplication table of the octonions is to use the following diagram (which I got from John Baez's online paper): if $(e_i,e_j,e_k)$ is one of the lines listed according to the cyclic order indicated in the diagram, then $e_ie_j=e_k$ and $e_je_i=-e_k$ in $\mathbb O$.

If we forget the cyclic orientation of the lines, this is of course a well-known depiction of the Fano plane $P^2(\mathbb F_2)$, which is an example of many different structures: it is a Steiner triple system, a quasigroup, &c.
What kind of object is this oriented Fano plane?
NB1: Naive googling informs of the concept of Mendelsohn triple systems and of transitive triple systems, both of which are enrichments of the notion of Steiner triple systems with orderings on the blocks. The oriented Fano plane above is not an example of these concepts, though.
NB2: One way to reconstruct the orientation is as follows: it is (up to projective linear automorphisms) the unique way to cyclically orient the lines in the plane in such a way that for each point $x$, the set of three points which follow $x$ in the three lines that go through it is itself a line. In fact, it is the only Steiner triple system which can be oriented with this property.

REPLY [15 votes]: Here is one answer: It is an oriented line over $\mathbb{F}_7$.
An affine line over $\mathbb{F}_7$ is a set of 7 points with a simply transitive action of $\mathbb{Z}/7\mathbb{Z}$, but no distinguished origin.  Here, we don't have a distinguished origin and we also don't remember the precise translation action, but we have a distinguished notion of addition by a square (think of what this would mean for real numbers).  In other words, it is a set with seven elements, equipped with an unordered triple of simply transitive actions of $\mathbb{Z}/7\mathbb{Z}$, such that translation by 1 under one of the actions is equivalent to translation by the square classes $2$ and $4$ under the other two actions.
If you take any pair of points $(x,y)$ in the above picture and subtract their indices, the orientation of the arrow between them is $x \to y$ if and only if $y-x$ is a square mod 7.  Furthermore, a triple of points $(x,y,z)$ with directed arrows $x \to y \to z$ is collinear if and only if $\frac{z-y}{y-x} = 2$.  Even though the numerator and denominator are only well-defined up to multiplication by squares, the quotient is a well-defined element of $\mathbb{F}_7^\times$, since each of the three translation actions yield the same answer.  These two data let us reconstruct the diagram from the oriented line structure.
There is a group-theoretic interpretation of this object.  The oriented hypergraph you've given has automorphism group of order 21, generated by the permutations $(1234567)$ (one of the translation actions) and $(235)(476)$ (changes translation action by conjugating).  This can be identified with the quotient $B^+(\mathbb{F}_7)/\mathbb{F}_7^\times$, where $B^+(\mathbb{F}_7)$ is the group of upper triangular matrices with entries in $\mathbb{F}_7$ and invertible square determinant, and $\mathbb{F}_7^\times$ is the subgroup of scalar multiples of the identity.  This group is the stabilizer of infinity under the transitive action of the simple group of order 168 on the projective line $\mathbb{P}^1(\mathbb{F}_7)$.  In this sense, we can view the simple group as the automorphism group of an oriented projective line, since it is the subgroup of $PGL_2(\mathbb{F}_7)$ whose matrices have square determinant.
Unfortunately, I do not know a natural notion of orientation on an $\mathbb{F}_2$-structure.  I tried something involving torsors over $\mathbb{F}_8^\times$ and the Frobenius, but it became a mess.<|endoftext|>
TITLE: Is there a ground between Set Theory and Group Theory/Algebra?
QUESTION [12 upvotes]: It is well known that there are strong links between Set Theory and Topology/Real Analysis.
For instance, the study of Suslin's Problem turns out to be a set theoretic problem, even though it started in topology: namely, whether $\mathbb{R}$ is the only complete dense unbounded linearly ordered set that satisfies the c.c.c.
Another instance is when we see that what's behind extending Lebesgue Measure is really the theory of large cardinals, with the introduction of measurable cardinals. Also another example of a real analysis problem that ends up in Set Theory is whether every set of reals is measurable. So the links are clear between Set Theory and Topology/Real analysis.
My question is this: are there links, as strong as the ones I roughly described in the last paragraph, between Set Theory and Abstract Algebra? The only example I know of is the Set Theoretic solution to the famous Whitehead Problem by Shelah (namely that if $V=L$ then every Whitehead group is free and if MA+$\neg$CH then there is a Whitehead group which is not free).
Can we hope to discover more of these type of links between Set Theory and Abstract Algebra? In contrast, Model Theory seems to be strongly grounded in Abstract algebra. I have seen that Shelah has some papers about uncountable free Abelian groups but he seems to be the only one investigating some areas of Abstract Algebra with the help of Set Theory. So again is there hope for links?

REPLY [6 votes]: It turns out that when it comes to infinite groups/modules, some algebraic concepts are deeply connected to the underlying set theory (for example, the notion of freeness, the structure of Ext, etc).
A good reference for this subject is the book "Almost free modules"  by Eklof and Mekler. This book introduces the works of Shelah, Gobel, Eklof and many other important contributors in this field. This research has also led to some interesting developments in "pure" set theory, such as the introduction of black-boxes by Shelah (some diamond-like combinatorial principles which can be proved in ZFC alone, and allow the construction of many interesting algebraic objects).<|endoftext|>
TITLE: Coend computation continued
QUESTION [9 upvotes]: This is a follow-up question to this coend computation.  Here's an attempt at a slightly simpler computation:

$\int^{a \in A} \mbox{hom}_A(a,a)$

This should be similar to the trace operator.  In attempting to follow the derivation

$\begin{array}{l}\mbox{Set}(\int^{b \in B}\mbox{hom}(a, b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b), \mbox{Set}(F(b), S)) \\ \cong \mbox{Nat}(\mbox{hom}(a,-), \mbox{Set}(F(-), S) \\ \cong \mbox{Set}(F(a), S),\end{array}$ 

I get

$\begin{array}{l}\mbox{Set}(\int^{a \in A} \mbox{hom}_A(a,a), S) \\ \cong \int_{a \in A} \mbox{Set}(\mbox{hom}_A(a,a), S) \\ \cong \mbox{Nat}(\mbox{hom}_A(-,-), S)\end{array}$

So here I guess we have the set of natural transformations from the hom functor to the constant functor $S$.  For any first parameter $a$, we have the set of natural transformations from hom$(a,-)$ to $S(a,-)$, which by Yoneda's lemma is isomorphic to $S(a,a) = S$.  So I think it goes

$\begin{array}{l}\cong \displaystyle \prod_a \mbox{Nat}(\mbox{hom}_A(a,-), S(a,-)) \\ \cong \prod_a S\\ \cong S^{Ob(A)} \\ \cong \mbox{Set}(\mbox{Ob}(A), S).\end{array}$

So $\int^{a \in A} \mbox{hom}_A(a,a) \cong \mbox{Ob}(A).$ Is that right?

REPLY [8 votes]: I agree with Reid's answer, but I want to add a bit more.
Putting Reid's calculation into a more general setting, if $A$ is any category then 
$$
\int^{a \in A} \mathrm{hom}_A (a, a)
= (\mathrm{endomorphisms\ in\ } A)/\sim
$$
where $\sim$ is the (rather nontrivial) equivalence relation generated by $gh \sim hg$ whenever $g$ and $h$ are arrows for which these composites are defined.  
You can see confirmation there that your instinct about traces was right.  If we wanted to define a 'trace map' on the endomorphisms in $A$, it should presumably satisfy $\mathrm{tr}(gh) = \mathrm{tr}(hg)$, i.e. it should factor through $\int^a \mathrm{hom}(a, a)$.
In fact, Simon Willerton has done work on 2-traces in which exactly this coend appears.  See for instance these slides, especially the last one.
You can see in that slide something about the dual formula, the end
$$
\int_{a \in A} \mathrm{hom}_A(a, a).
$$
By the "fundamental fact" I mentioned before, this is the set
$$
{}[A, A](\mathrm{id}, \mathrm{id}) = \mathrm{Nat}(\mathrm{id}, \mathrm{id})
$$
of natural transformations from the identity functor on $A$ to itself.  Evidently this is a monoid, and it's known as the centre of $A$.  For example, when $G$ is a group construed as a one-object category, it's the centre in the sense of group theory.  So your set might, I suppose, be called the co-centre of $A$.<|endoftext|>
TITLE: How does categoricity interact with the underlying set theory?
QUESTION [11 upvotes]: Here's the setup: you have a first-order theory T, in a countable language L for simplicity. Let k be a cardinal and suppose T is k-categorical. This means that, for any two models
M,N |= T
of cardinality k, there is an isomorphism f : M --> N.
Supposing all this happens inside of ZFC, let's say I change the underlying model of ZFC, e.g by restricting to the constructible sets, or by forcing new sets in. I would like to understand what happens to the k-categoricity of T.
I'll assume the set theory doesn't change so drastically that we lose L or T. Then, a priori, a bunch of things may happen:
(i) We may lose all isomorphisms between a pair of models M,N of cardinality k;
(ii) Some models that used to be of cardinality k may no longer have bijections with k;
(iii) k may become a different cardinal, meaning new cardinals may appear below it, or others may disappear by the introduction of new bijections;
(iv) some models M, or k itself, may disappear as sets, leading to a new set being seen as "the new k".
Overall, nearly every aspect of the phrase "T is k-categorical" may be affected. How likely is it to still be true? Do some among (i)-(iv) not matter, or is there some cancellation of effects? (Say, maybe all isomorphisms M-->N disappear, but so do all bijections between N and k?)

REPLY [9 votes]: In contrast to absoluteness of categoricity in First Order Logic, there are many interesting non-absoluteness phenomena at low infinite cardinals outside of first order. Perhaps the most important is the following:
(Under wGCH at $\kappa$, i.e. $2^\kappa<2^{\kappa^+}$) For every abstract elementary class $(\mathfrak K,\prec_{\mathfrak K})$ with $LS(\mathfrak K)\leq \kappa$, if $\mathfrak K$ is categorical in $\kappa$ and fails to have the amalgamation property at $\kappa$ ($AP_\kappa$), then $\mathfrak K$ is not categorical in $\kappa^+$ (indeed, it has the maximum number of models of size $\kappa^+$, $2^{\kappa^+}$).
In contrast, Martin's Axiom provides a completely different picture:
There exists an AEC (axiomatizable in $L_{\omega_1,\omega}(Q)$) $\mathfrak K_r$ with $LS(\mathfrak K_r)=\aleph_0$, $\mathfrak K_r$ is categorical in $\aleph_0$ and fails $AP_{\aleph_0}$ that is categorical in $\aleph_1$ in the presence of $MA_{\aleph_1}$.
The theorem and the example are due to Shelah (but have had improved presentations due to various other authors - Grossberg and Baldwin most prominent). Notice that here categoricity of a certain class in $\aleph_1$ is NOT absolute. Many open questions remain.<|endoftext|>
TITLE: Prime decomposition for knots in manifolds
QUESTION [6 upvotes]: A knot in S^3 is uniquely decomposed into a connected sum of prime knots. What is known for knots in other three-manifolds?

REPLY [9 votes]: This was addressed in Miyazaki, Conjugation and the prime decomposition of knots in closed, oriented 3-manifolds, using the definition of connected sum suggested by Ryan Budney in the comments.
The main results are that a prime decomposition of K exists iff a meridian of K is not null-homotopic in the complement of K, and if a prime decomposition of K does not contain a particular knot R in S1×S2 then it is the unique decomposition of K, whereas knots with this summand R can admit several prime decompositions.<|endoftext|>
TITLE: What role does Cauchy's determinant identity play in combinatorics?
QUESTION [19 upvotes]: When studying representation theory, special functions or various other topics one is very likely to encounter the following identity at some point:
$$\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}$$
This goes under the name of Cauchy's determinant identity and has various generalizations and analogous statements. There is also a lot of different proofs using either analysis or algebra. In my case I have always seen it introduced (or motivated) as an identity that plays an important role in combinatorics, but I realized that I haven't really seen this identity in a combinatorial context before. In this question I'm asking for a combinatorial interpretation of the above identity. A bonus to someone who can give such an interpretation to Borchardt's variation:
$$\det \left(\frac{1}{(x _i+y _j)^2}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)} \cdot \text{per}\left(\frac{1}{x _i +y _j}\right) _{1 \le i,j \le n}$$
(This seems a little too ambitious though, and I would be happy to accept an answer of just the first question)

REPLY [16 votes]: See also pp. 397--398 of Enumerative Combinatorics, vol. 2. Cauchy's determinant is given in a slightly different but equivalent form. It is explained there that the evaluation of the determinant is equivalent to the fundamental identity $\prod(1-x_iy_j)^{-1} =\sum_\lambda s_\lambda(x)s_\lambda(y)$ in the theory of symmetric functions.<|endoftext|>
TITLE: Radii and centers in Banach spaces
QUESTION [12 upvotes]: Suppose I have a Banach space $V$ and a set $A \subseteq V$ such that for all $\epsilon > 0$ there exists $v$ such that $A \subseteq \overline{B}(v, r + \epsilon)$. Does there exist $c$ such that $A \subseteq \overline{B}(c, r)$?
The answer is clearly yes for finite dimensional normed spaces: Define $T_\epsilon = \bigcap_{a \in A} \overline{B}(a, r + \epsilon)$. The $T_\epsilon$ form a chain of closed sets and for $\epsilon > 0$ are non-empty, so have the finite intersection property. Thus when $V$ is finite dimensional they have non-empty intersection, and any element of the intersection works as $c$.
For more general Banach spaces I feel like you should be able to choose a cauchy sequence $x_n$ such that $x_n \in T_{\epsilon_n}$ with $\epsilon_n \to 0$, but I can't seem to make it work.
Note that an arbitrary choice of $x_n \in T_{\epsilon_n}$ can't be guaranteed to be Cauchy: If $V$ is $l^\infty$ and $A = \{ x : x_0 = 0, ||x|| \leq 1 \}$ then diam$(T_\epsilon) \geq 2$ because you can choose $c_0$ arbitrarily in $[-1, 1]$
Note also that the assumption of $V$ a Banach space is essential: If $V$ is not Banach and $c$ is an element of the completion which is not in $V$ then $A = \overline{B}(c, 1) \cap V$ has no center.

REPLY [4 votes]: I think that http://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654848-2/S0002-9947-1982-0654848-2.pdf [together with its references] provides us with several counterexamples [as well as with some remarkable examples], in the infinite-dimensional framework.<|endoftext|>
TITLE: Construction of the petit Zariski topos out of the gros topos of a scheme
QUESTION [17 upvotes]: Let S be a scheme. Let (Sch/S) be a small category of schemes over S (including essentially all finitely presented schemes affine over S). Let E = (Sch/S)zar denote the gros Zariski topos with its local ring object A1.
Is there a nice way to construct the petit Zariski topos X = Szar out of the locally ringed topos E? (By nice I mean, for example, that there is a universal property that the locally ringed topos X possesses with respect to E.)
There are variations of this question in which I am also interested: For example, one can replace E by the gros étale (or fppf or fpqc) topos (Sch/S)ét and ask for the construction  of Szar out of (Sch/S)ét. Or one can replace X by the petit étale (or fppf or fpqc) topos Sét and ask for the construction of it out of E = (Sch/S)zar.

REPLY [5 votes]: Many of these toposes admit descriptions as internal classifying toposes, hence indeed enjoy useful universal properties. Here is a selection of such descriptions:
Constructing the big Zariski topos from the little Zariski topos. The big Zariski topos of a scheme $S$ is the externalization of the result of constructing, internally to the little Zariski topos of $S$, the classifying topos of local $\mathcal{O}_S$-algebras which are local over $\mathcal{O}_S$.
(One might hope that it would simply be the internal big Zariski topos of $\mathcal{O}_S$, that is the classifying topos of local $\mathcal{O}_S$-algebras where the structure morphism needn't be local. This alternative description is true if $S$ is of dimension $0$.)
Constructing the little Zariski topos from the big Zariski topos. Additionally to $\mathbf{A}^1$, which is the functor $T \mapsto \Gamma(T,\mathcal{O}_T)$, the big Zariski topos contains an additional local ring object: The functor $\flat \mathbf{A}^1$, which maps an $S$-scheme $(f : T \to S)$ to $\Gamma(T,f^{-1}\mathcal{O}_S)$. There is a local ring homomorphism $\flat \mathbf{A}^1 \to \mathbf{A}^1$, and the little Zariski topos can be characterized as the largest subtopos of the big Zariski topos where this morphism is an isomorphism. (This fits with the comments as follows. If $S = \operatorname{Spec}(A)$ is affine, we have the string of ring homomorphisms $\underline{A} \to \flat \mathbf{A}^1 \to \mathbf{A}^1$ starting in the constant sheaf $\underline{A}$. The map $\underline{A} \to \flat \mathbf{A}^1$ is always a localization, and the composition is iff $\flat \mathbf{A}^1 \to \mathbf{A}^1$ is an isomorphism.)
Constructing the big étale topos from the big Zariski topos. The big étale topos of a scheme $S$ is the largest subtopos of the big Zariski topos of $S$ where $\mathbf{A}^1$ is separably closed. This fact is essentially a restatement of Gavin Wraith's theorem on what the big étale topos classifies.
Constructing the big infinitesimal topos from the little Zariski topos. A back-of-the-envelope computation indicates that the recent result of Matthias Hutzler on what the infinitesimal topos of an affine scheme classifies can be relativized to the non-affine case as follows. The big infinitesimal topos of a scheme $S$ is the externalization of constructing, internally to the little Zariski topos of $S$, the classifying topos of local and local-over-$\mathcal{O}_S$ $\mathcal{O}_S$-algebras equipped with a nilpotent ideal.
Some details can be found in Sections 12 and 21 of these notes.
A word of warning: When I say "the largest subtopos where foo", I refer to the largest element in the poset of subtoposes which validate foo from their internal language. (By general abstract nonsense (more or less the existence of classifying toposes), such a largest element always exists in case foo is a set-indexed conjunction of geometric implications.) In particular, I'm not referring to "the subtopos of those objects $Y$ of $E$ such that $\mathbf{A}^1$ restricted to $Y$ enjoys foo" (as in the comments). Indeed, this category is in general not a subtopos (typically it doesn't contain the terminal object). Maybe I interpreted that phrase too literally.<|endoftext|>
TITLE: Additive reduction of elliptic curves
QUESTION [5 upvotes]: Suppose $E/ \mathbf{Q}$ is an elliptic curve with additive reduction at a prime $p$.  Is there an easy way to tell if $E$ is a quadratic twist of an elliptic curve $E'/\mathbf{Q}$ with good reduction at $p$?  I have asked one or two experts about this, without a satisfying answer...

REPLY [5 votes]: In other words, given an elliptic curve $E/\mathbb{Q}_p$ with additive reduction, you wish to know whether there is a quadratic extension $F/\mathbb{Q}_p$ such that $E/F$ has good reduction. Of course the $j$-invariant must be integral. 
 Let $\Phi_p$ be the Serre-Tate group which is the Galois group of the minimal extension of $\mathbb{Q}_p^{unr}$ such that $E$ acquires good reduction. In Serre's paper "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques", page 312, there is a table of what $\Phi_p$ is. You wish to know when $\Phi_p$ is cyclic of order 2.,
If $p\neq 2,3$ then there exists a quadratic extension $F$ that makes $E$ having good reduction if and only if the Kodaira type is $I_0^{*}$, i.e. if and only if the order of $v_p(\Delta)$ in $\mathbb{Z}/12\mathbb{Z}$ is 2. (as in Robin Chapman's answer). If the equation is minimal, this is equivalent to $v_p(\Delta) = 6$.
If $p=2$ or 3, then it is still true that we must have $v_p(\Delta)\cdot 2\equiv 0\pmod{12}$ to guarantee that you can twist to good reduction. This excludes certain Kodaira types. To find the complete answer one would have to analyse this better.<|endoftext|>
TITLE: Is there analogue of Peter-Weyl theorem for non-compact or quantum group
QUESTION [7 upvotes]: From the Peter-Weyl theorem in wikipedia, this theorem applies for compact group. I wonder whether there is a non-compact version for this theorem.
I suspect it because the proof of the Peter-Weyl theorem heavily depends on the compactness of Lie group. It is related to the spectral decomposition of compact operators. 
Thanks Mariano pointing out the Peter-Weyl theorem does not hold for non-compact group. But I really wants to know is: is there any Peter-Weyl analogue decomposition for non-compact group, say decompose to integral representations but not finite dimensional representations?
Another related questions is about the definition of quantized flag variety. In the work of Lunts and Rosenberg on localization for quantum group, they tried to establish the quantum analogue of Beilinson-Bernstein localization theorem. They defined the quantized flag variety in the framework of noncommutative algebraic geometry. They used the Peter-Weyl philosophy for quantum group to define the coordinate ring of quantized base affine space as the direct sum of all simple $U_{q}(g)$-modules with highest weight $\lambda$(positive).(Denoted by $R_{+}$)
Then one can define category of quasi coherent sheaves on "quantized flag variety" as proj-category of graded $R_{+}$. 
What I want to ask is there any other way to define quantized flag variety? In the classical case, It is well known that flag variety can be define as $G/B$, say $G$ is general linear group and $B$ is Borel subgroup. Is there any analogue for quantum case? Is there a definition like $G_{q}$ as "quantum linear group" and $B_{q}$ as quantum analogue of Borel subgroup?
However,I suspected, because the quantum flag variety is essentially not a real space

REPLY [6 votes]: Marty's answer discusses the Plancherel formula, and in a comment on his answer, I mentioned Harish-Chandra's work on the Plancherel formula in the case of reductive Lie groups. Yemon Choi's answer also mentions the case of semisimple Lie groups as being easier than the general case.  The point of this answer is to elaborate slightly on my comment, and to point out that, while the semi-simple case might be easier, it is a very substantial piece of mathematics; indeed, it is essentially Harish-Chandra's life's work.
When Harish-Chandra began his work, it was known (thanks to Mautner?) that semisimple Lie groups were type I, and hence that for such a group $G$, the space $L^2(G)$ admits a
well-defined direct integral decomposition into irreducibles (typically infinite
dimensional, if $G$ is not compact).  However, this is a far cry from knowing the precise
form of the decomposition.  
The most fundamental, and difficult, question, turns out to
be whether there are any atoms in the Plancherel measure, i.e. whether $L^2(G)$ contains
any non-zero irreducible subspaces, i.e. whether the group $G$ admits discrete series.
This was solved by Harish-Chandra, who established his famous criterion: $G$ admits discrete series if and only if it contains a compact Cartan subgroup.  He also gave a complete enumeration of the discrete series representations up to isomorphism, and described
their characters via formulas analogous to the Weyl character formula.
Harish-Chandra then want on to describe
the Plancherel measure on $L^2(G)$ inductively in terms of direct integrals
of parabolic inductions of discrete series represenations of Levi subgroups of $G$.
(It is the appearance of Levi subgroups, which are always reductive but typically
never semisimple, that also forces one to generalize from the semisimple to the reductive case.)
After completing the theory of the Plancherel measure for reductive Lie groups, he then went on to develop the analogous theory for $p$-adic reductive groups.  However, in this case, one still doesn't have a complete enumeration of the discrete series representations in general: there are certain "atomic" discrete series representations, called "supercuspidal", which have no analog for Lie groups, and which aren't yet classified in general (i.e. for all $p$-adic reductive groups). 
Harish-Chandra's work, as well as standing on its own as an amazing edifice, was a central
inspiration for Langlands in his development of the Langlands program, and remains 
at the core of the Langlands proram today.
For a very nice introduction to Harish-Chandra's work, and the surrounding circle of ideas,
one can read this article by Rebecca Herb.<|endoftext|>
TITLE: Special Holonomy Groups for Lorentzian Manifolds
QUESTION [12 upvotes]: Let $X$ be a Riemannian manifold. If $X$ is simply connected, irreducible, and not a symmetric space then we know that the possible holonomy groups of the metric on $X$ are: 
1) $O(n)$ General Riemannian manifolds
2) $SO(n)$ Orientable manifolds
3) $U(n)$ Kahler manifolds 
4) $Sp(n)Sp(1)$ Quaternionic Kahler manifolds
5) $SU(n)$ Calabi-Yau manifolds
6) $Sp(n)$ Hyperkahler manifolds
7) $G_{2}$ (if $X$ has dimension 7) $G_{2}$ manifolds
8) $Spin(7)$ (if X has dimension 8)  $Spin(7)$ manifolds
Of these, cases 5-8 are important in string theory.  The reduced holonomy implies that these manifolds have vanishing Ricci curvature and hence are automatically solutions to Einstein's equations of general relativity.
However, physics does not take place on a Riemannian manifold, but rather on a Lorentzian one.  Thus my question is: what is known about special holonomy manifolds for metrics of general signature? (I am most interested in the (1,n) case!)  In particular I would like to know if there is a classification of allowed holonomy groups and further if there are interesting examples that I should be aware of.

REPLY [16 votes]: I think it would be more accurate to say that the real reason why
Calabi-Yau, hyperkähler, $G_2$ and $\mathrm{Spin}(7)$ manifolds are of
interest in string theory is not their Ricci-flatness, but the fact
that they admit parallel spinor fields.  Of course, in
positive-definite signature, existence of parallel spinor fields
implies Ricci-flatness, but the converse is still open for compact
riemannian manifolds, as discussed in this
question, and known to fail for noncompact manifolds as pointed
out in an answer to that question.
The similar question for lorentzian manifolds has a bit of history.
First of all, the holonomy principle states that a spin manifold
admits parallel spinor fields if and only if (the spin lift of) its
holonomy group is contained in the stabilizer subgroup of a nonzero
spinor.  Some low-dimensional (i.e., $\leq 11$, the cases relevant to
string and M-theories) investigations (by Robert Bryant and myself,
independently) suggested that these subgroups are either of two types:
subgroups $G < \mathrm{Spin}(n) < \mathrm{Spin}(1,n)$, whence $G$ is
the ones corresponding to the cases 5-8 in the question, or else $G =
H \ltimes \mathbb{R}^{n-1}$, where $H < \mathrm{Spin}(n-1)$ is one of
the groups in cases 5-8 in the question.  Thomas Leistner showed that
this persisted in the general case and, as Igor pointed out in his
answer, arrived at a classification of possible lorentzian holonomy
groups.  Anton Galaev then constructed metrics with all the possible
holonomy groups, showing that they all arise.  Their work is reviewed
in their
paper (MR2436228).
The basic difficulty in the indefinite-signature case is that the de
Rham decomposition theorem is modified.  Recall that the de Rham
decomposition theorem states that if $(M,g)$ is a complete, connected
and simply connected positive-definite riemannian manifold and if the
holonomy group acts reducibly, then the manifold is a riemannian
product, whence it is enough to restrict to irreducible holonomy
representations.  This is by no means a trivial problem, but is
tractable.
In contrast, in the indefinite signature situation, there is a
modification of this theorem due to Wu, which says that it is not
enough for the holonomy representation to be reducible, it has to be
nondegenerately reducible.  This means that it is fully reducible
and the direct sums in the decomposition are orthogonal with respect
to the metric.  This means that it is therefore not enough to restrict
oneself to irreducible holonomy representations.  For example,
Bérard-Bergery and Ikemakhen proved that the only lorentzian holonomy
group acting irreducibly is $\mathrm{SO}_0(1,n)$ itself: namely, the
generic holonomy group.
It should be pointed out that in indefinite signature, the
integrability condition for the existence of parallel spinor fields is
not Ricci-flatness.  Instead, it's that the image of the Ricci
operator $S: TM \to TM$, defined by $g(S(X),Y) = r(X,Y)$, with $r$ the
Ricci curvature, be isotropic.  Hence if one is interested in
supersymmetric solutions of supergravity theories (without fluxes) one
is interested in Ricci-flat lorentzian manifolds (of the relevant
dimension) admitting parallel spinor fields.  It is now not enough to
reduce the holonomy to the isotropy of a spinor, but the
Ricci-flatness equation must be imposed additionally.<|endoftext|>
TITLE: Motivation for equivariant sheaves?
QUESTION [9 upvotes]: Hello everyone;
i'm looking for a motivation for equivariant sheaves (see http://ncatlab.org/nlab/show/equivariant+sheaf) ~ Why are we interested in them?
More explicitely: Can I think of G-equivariant sheaves on a space X as a quotient of the category of sheaves (by some action? in a more general sense) by G?

REPLY [16 votes]: If you know that the sections of a vector bundle form a standard example of a sheaf, then the corresponding example of a $G$-equivariant sheaf on a space $X$ with $G$-action is a vector bundle $V$ over $X$ with a $G$-action compatible with the projection (i.e. making the projection $G$-equivariant, i.e. intertwining the actions). Such actions on vector bundles over homogeneous spaces were considered in representation theory by Borel, Weil, Bott and Kostant ("homogeneous vector bundles"; and later many generalizations to sheaves by Beilinson-Bernstein, Schmid, Miličić etc.). David Mumford introduced $G$-equivariant structures on sheaves under the name $G$-linearization for the purposes of geometric invariant theory. 
While for a function on a $G$-space the appropriate notion is the $G$-invariance, for sheaves the invariance is useful only up to a coherent isomorphism, what spelled out yields the definition of the $G$-equivariant sheaf. This is an example of a categorification (recall that functions form a set and sheaves form a category). Using the Yoneda embedding one can indeed consider the $G$-equivariant objects as objects in some fibered category of objects on $X$ with an action on each hom-space (see the lectures by Vistoli). 
While for a function to be invariant is a property, for a sheaf the $G$-equivariance entails the additional coherence data, so it is a structure. 
Category of $G$-equivariant sheaves on $X$ is not a quotient of the category of usual sheaves on $X$, but rather equivalent to the category of sheaves on the geometric quotient $G/X$, in the case when the action of $G$ on $X$ is principal; or in general if we replace the geometric quotient by the appropriate stack $[G/X]$.<|endoftext|>
TITLE: Why is complex projective space triangulable?
QUESTION [21 upvotes]: In an exercise in his algebraic topology book, Munkres asserts that $\mathbf{C}P^n$ is triangulable (i.e., there is a simplicial complex $K$ and a homeomorphism $|K| \rightarrow \mathbf{C}P^n$).  Can anyone provide a reference or a proof?

REPLY [3 votes]: Although I am more than a decade late to the discussion, here is an alternate approach to an explicit triangulation of $\mathbb{C}P^n$.
(1) A simplicial set $X$ is a natural generalization of a simplicial complex with locally ordered vertices.  $X$ has a realization $|X|$ which is a CW complex, with a cell for each non-degenerate simplex $\sigma \in X$.  However, $X$ may or may not be regular as a CW complex.
(2) If $X$ is a simplicial set, then so is its $n$th symmetric power $X^n/S_n$, and miraculously $|X^n/S_n| \cong |X|^n/S_n$ (since geometric realization preserves finite limits).  In particular we can let $|X| \cong S^2 \cong \mathbb{C}P^1$ by collapsing the boundary of a triangle to a point.   Then $|X^n/S_n| \cong \mathbb{C}P^n$ is a canonical simplicial set model of $\mathbb{C}P^n$.
(3) Finally, there is a canonical way to subdivide a simplicial set to make a simplicial complex; see arXiv:2001.04339. It is a modified second barycentric subdivision and is not all that efficient of a triangulation, but it works.
(4) You can make the construction a little more efficient by letting $|X| \cong S^2$ be the boundary of a tetrahedron.  Then I believe that $|X^n/S_n|$ is a regular cellulation of $\mathbb{C}P^n$ with simplices, so that its first barycentric subdivision is already an honest simplicial complex.<|endoftext|>
TITLE: Is a bialgebra pairing of Hopf algebras automatically a Hopf pairing?
QUESTION [8 upvotes]: The following question came up in the course on Quantum Groups here at UC Berkeley.  (If you care, I have been TeXing uneditted lecture notes.)
Let $X,Y$ be (infinite-dimensional) Hopf algebras over a ground field $\mathbb F$.  A linear map $\langle,\rangle : X\otimes Y \to \mathbb F$ is a bialgebra pairing if $\langle x,y_1y_2 \rangle = \langle \Delta x,y_1\otimes y_2\rangle$ and $\langle x_1x_2,y\rangle = \langle x_1\otimes x_2,\Delta y\rangle$ for all $x,x_1,x_2 \in X$ and $y,y_1,y_2 \in Y$.  (You must pick a convention of how to define the pairing $\langle,\rangle : X^{\otimes 2} \otimes Y^{\otimes 2} \to \mathbb F$.)  And we also demand that $\langle 1,- \rangle = \epsilon_Y$ and $\langle -,1\rangle = \epsilon_X$, but this might follow from the previous conditions.  (See edit.)
A bialgebra pairing is Hopf if it also respects the antipode: $\langle S(x),y \rangle = \langle x,S(y)\rangle$.  A pairing $\langle,\rangle : X\otimes Y \to \mathbb F$ is nondegenerate if each of the the induced maps $X \to Y^*$ and $Y \to X^*$ has trivial kernel.

Question: Is a (nondegenerate) bialgebra pairing of Hopf algebras necessarily Hopf?  (Does it depend on whether the pairing is nondegenerate?)

My intuition is that regardless of the nondegeneracy, the answer is "Yes": my motivation is that a bialgebra homomorphism between Hopf algebras automatically respects the antipode.  But we were unable to make this into a proof in the infinite-dimensional case.
Edit: If $\langle,\rangle: X\otimes Y \to \mathbb F$ is nondegenerate, then it is true that as soon as it satisfies $\langle x,y_1y_2 \rangle = \langle \Delta x,y_1\otimes y_2\rangle$ and $\langle x_1x_2,y\rangle = \langle x_1\otimes x_2,\Delta y\rangle$, so that the induced maps $X \to Y^*$ and $Y \to X^*$ are (possibly non-unital) algebra homomorphisms, then it also satisfies $\langle 1,- \rangle = \epsilon_Y$ and $\langle -,1\rangle = \epsilon_X$, so that the algebra homomorphism are actually unital.  But I think that this does require that the pairing be nondegenerate.  At least, I don't see how to prove it without the nondegeneracy assumption.  So probably the nondegeneracy is required for the statement about antipodes as well.

REPLY [4 votes]: Hi
I was in class when the question came up, and I remember the discussion on whether a bialgebra pairing of $A$ and $B$ always induces a bialgebra map $A\to B^\circ$. You guys tricked me back then :): it is, in fact, true. In other words, a bialgebra pairing is always (regardless of (non)degeneracy) exactly the same thing as a bialgebra map from one of the two bialgebras to the finite dual of the other. 
I'm going to view the elements of $a$ as linear maps on $B$ via the pairing. We want to show that the map from $A$ to the dual of $B$ induced by the pairing actually lands in $B^\circ$. 
This is pretty clear using the following characterization of the finite dual: $B^\circ$ is precisely the set of $f\in B^*$ for which one can find finitely many $g_i,h_i\in B^*$ such that $f(xy)=\sum_ig_i(x)h_i(y)$ (for all $x,y\in B$). It follows immediately from the bialgebra pairing conditions that any $a\in A$ satisfies this property: if, say $\Delta(a)=\sum_i a'_i\otimes a''_i$, then take $g_i=a'_i$ and $h_i=a''_i$.<|endoftext|>
TITLE: Generalization of Finch Cheney's 5 Card Trick
QUESTION [8 upvotes]: The Original Fitch Cheney puzzle goes like this:

A Volunteer from the crowd chooses any
  five cards at random from a deck, and
  hands them to you so that nobody else
  can see them.  You glance at them
  briefly, and hand one card bakc, which
  the volunteer then places face down on
  the table to one side.  You quickly
  place the remaining four cards face up
  on the table, in a row from left to
  right. Your confederate, who has not
  been privy to any of the proceedings
  so far, arrives on the scene, inspects
  the faces of the  four cards, and
  promptly names the hidden fifth card.

The solution to this is:

One of the 4 suits must be represented 2 times in your set of 5 cards due to the pigeon hole principle.  Consider the 13 cards of that suit A=1,... K=13 to be arranged in a clockwise circle.  We can see that the cards are at most 6 away from each other, meaning counting clockwise one of them lies at most 6 vertices past the other.  Use the "higher" one  as the hidden card and place the "lower" one as the first card face up on the table.
Now, using an established value for all 52 cards in the deck, the remaining 3 cards can be placed in 6 orders, Low-Middle-High=1, LHM=2, MLH=3, MHL=4, HLM=5, and HML=6, and give each one of these a value of +1,..., +6 from the first card.
For example, Jc, 2c, 3h, 4d, 2s are handed to you.  Choosing the lower club, (2-Jack(11) Mod 13 is 4, so the Jack is the lower one and the 2 is +4 from the jack.  Hide the 2c and place the Jc in the first spot.  Then using MHL, our +4 value, we arrange the remaining 3 cards with the 3 first, than the 4, now the 2.  This implies that the hidden card is a club, and it is +4 from the Jc in the first spot, so we know it is the 2c.

The 124 card solution is discussed in the January 2001 issue of Emissary or Michael Kleber’s article in The Mathematical Intelligencer, Winter 2002 (which I don't have access to).
It can be shown that with 5 cards there is a strategy to do the trick on a deck of size up to 124 cards (n!+(n-1)).  
My question is this: With the audience choosing n cards out of a deck of size (n!+(n-1)) and one card being hidden, how many unique strategies could the magician and the assistant use?

REPLY [6 votes]: Kleber's paper will certainly point you in the right direction if you can find it. (I have a printed out copy, and I don't remember where on the web I got it. Sorry.)
In it, he suggests thinking about a strategy as a pairing up of the $(n!+n-1)\_{n-1}$ messages with the ${{n!+n-1}\choose n} = (n!+n-1)\_{n-1}$ hands the audience can give you. Of course, you can only pair a message with one of the n! hands that contain its cards, and you can only pair a hand with one of the n! messages you can make with it. So this is just like finding a perfect matching on an n!-regular bipartite graph, and by Hall's Marriage Theorem this is possible.
The bipartite graph you draw here is quite symmetric: aside from being n!-regular, it's also vertex transitive, so any of the n! edges you could choose from a particular vertex v are part of some perfect matching (since one of them is). This is where Kleber's claim that there are "at least n!" strategies come from.
We can get a much better lower bound if this paper is to be believed. Here it says there are at least $\left(\frac{(n!-1)^{n!-1}}{n!^{n!-2}}\right)^{{n!+n-1}\choose n}$ strategies, which for n=2 isn't very enlightening (it says there's at least one, when we know there are really two), but for n>2 puts our previous estimate to shame: with a three-card hand (and thus an eight-card deck), we get at least 2.54x10^21 strategies, a number so fantastically larger than our previous estimate of 6 that I'm still a little bit in shock. (It is, at least, many orders of magnitude lower than the naive upper bound of 6^56, where for each of the 3-subsets of 8 we choose one of the possible messages it could send without worrying about overlap with other hands.)
Anyway, I haven't read the linked paper, but the abstract suggests we might not get a much better lower bound than this. To improve, one might look for results on vertex transitive k-regular bipartite graphs, but I haven't found any.<|endoftext|>
TITLE: What is the equivariant cohomology of a group acting on itself by conjugation?
QUESTION [36 upvotes]: This question makes sense for any topological group $G$, but I'd particularly like to know the answer for $G$ a compact, connected Lie group.
$G$ acts on itself by conjugation. One has the equivariant singular cohomology
$H^{\ast}_G(G) = H^{\ast}( (G\times EG)/G )$, with integer coefficients, say.

What is $H^{\ast}_G(G)$? 

Concretely, $H^{\ast}_G(G)$ is the target of the Serre spectral sequence for the fibration $$G \hookrightarrow (G\times EG)/G \to BG.$$
When $G$ is path-connected, so that $BG$ is simply connected, this spectral sequence has $E_2^{p,q}=H^p(BG; H^q(G))$ (trivial local system). Does the spectral sequence always degenerate at $E_2$? It does when $G$ is abelian, because then the conjugation action is trivial.

REPLY [6 votes]: For any interested latecomers who somehow discover this question in the future, I've found a very low-tech answer, bootstrapping from the low-tech answer to Is a Lie group equivariantly formal under conjugation by a maximal torus?.
Anyway, once you believe that the conjugation action upon a compact Lie group $G$ of its maximal torus $T$ is equivariantly formal, it follows that the action of $G$ by conjugation is as well. This argument will actually work for any reasonably good space $M$ on which $G$ acts and the restricted $T$-action is equivariantly formal, because one has $$H_G(M) = H_T(M)^W = (H(M) \otimes H(BT))^W = H(M) \otimes H(BT)^W = H(M) \otimes H(BG)$$ as $H(BT)$-modules, where $W = N_G(T)/T$ is the Weyl group of $G$.
If, say, you don't buy that the action of $W$ on $H(M)$ is trivial, a longer proof goes like this.
$\require{AMScd}$
The homotopy quotient $M_G$ is a further quotient of $M_T$, and the projection $EG \times M \to EG$ then induces a commutative diagram
\begin{CD}
M @= M\\
@VVV @VVV\\
M_T @>>> M_G\\
@VVV @VVV\\
BT @>>> BG
\end{CD}
where the upper vertical maps are fiber inclusions.
The projection $BT = EG/T \to EG/G = BG$ induces an inclusion $H(BG) \cong H(BT)^W \hookrightarrow H(BT)$ in cohomology, and there are induced maps both in cohomology and on the Serre spectral sequences for the equivariant cohomologies, starting with this $E_2$ page:
\begin{CD}
H(M) @= H(M)\\
@AAA @AAA\\
H(M) \otimes H(BT) @<<< H(M) \otimes H(BG)\\
@AAA @AAA\\
H(BT) @<<< H(BG)
\end{CD}
Because the top and bottom horizontal maps are injective, so is the middle one, so the differentials for the spectral sequence converging to $H_G(M)$ are restrictions of those for $H_T(M)$. But the differentials for $H_T(M)$ are all zero, by equivariant formality, so the spectral sequence for $H_G(M)$ collapses as well.<|endoftext|>
TITLE: Smooth dg algebras (and perfect dg modules and compact dg modules)
QUESTION [16 upvotes]: Katzarkov-Kontsevich-Pantev define a smooth dg ($\mathbb{C}$-)algebra $A$ to be a dg algebra which is a perfect $A \otimes A^{op}$-module. They say that an $A$-module $M$ is perfect if the functor $Hom(M,-)$ from ($A$-modules) to ($\mathbb{C}$-modules) preserves small homotopy colimits. (Definition 2.23 of KKP) *
Kontsevich-Soibelman define smooth dg (or $A_\infty$) algebras in the same way; but they define perfect $A$-modules to be ones which are quasi-isomorphic to a direct summand of an extension of a sequence of modules each of which is quasi-isomorphic to a shift of $A$. (Definition 8.1.1 of KS) **
First question: I have trouble wrapping my head around what it means for a functor to preserve small homotopy colimits. I don't understand the definition from Kontsevich-Soibelman, either. What are the "moral" meanings of these conditions?
Second question: Presumably these two conditions are equivalent. How do you prove this?
Third question: If $A$ is a commutative $(\mathbb{C}$-)algebra, then presumably smoothness of $A$ as a dg algebra is equivalent to smoothness of $\operatorname{Spec} A$ as a ($\mathbb{C}$-)scheme. (Maybe you need $A$ to be finite type?) How do you prove this?  Furthermore, Example 8.1.4 of KS gives some more examples of dg algebras that are allegedly smooth (for instance free algebras $k\langle x_1, \dots, x_n \rangle$). Again, I don't know how to prove that these are in fact smooth, and Kontsevich-Soibelman don't seem to provide proofs; or maybe I overlooked something.

*The answers below indicate that this is the standard definition of compact (also known as small) module.
**The answers below indicate that this is the standard definition of perfect module.

REPLY [4 votes]: As David explained above we have two notions: perfect objects and compact(small) objects.
They coincide for derived categories of DG modules over a DG algebra and, more generaly, 
over a DG category. Proof can be found, for example, in Bernhard Keller paper
link text Section 5.3.
For answer on the third question you can look paper of Rouquier
link text and Lemma 7.2 there which says that if $A$ is a finite dimensional $k$-algebra or a commutative $k$-algebra essentially of finite type such  that for given $V$ a simple A-module, $Z(End_A(V ))$ is a separable extension of $k$. Then, $pdim_{A^{en}} A = gldim A$.<|endoftext|>
TITLE: How many pairs (M, N) of sets of size n have M + N = {0, 1, ..., n^2-1}?
QUESTION [14 upvotes]: How many pairs (M, N) of sets of size n have M + N = {0, 1, ..., n^2-1}?
Manfred Schroeder, in Number Theory in Science and Communication, 4th edition, asks (p. 27): find all pairs of sets $(M,N)$, each of size 10, such that
$$ M + N = \{ m + n : m \in M, n \in N \} $$
is the set $\{ 0, \ldots, 99 \}$.  One example is, of course, the pair $M = \{ 0, 1, \ldots 9 \}$ and $N = \{ 0, 10, \ldots, 90 \}$.  Of course there's nothing special about 10 here, and one can instead look for pairs $(M,N)$ such that $|M| = |N| = n$ and $M + N = \{0, 1, \ldots, n^2-1\}$.
This reminded me of the following problem: are there two six-sided dice, other than the standard ones, such that the probabilities of obtaining $2, 3, \ldots, 12$ are the same as on the standard dice?  The solution I know is as follows: this is equivalent to finding polynomials $f(z), g(z)$ with positive coefficients, with $f(1) = g(1) = 6$, such that
$$ f(z) g(z) = z^2 + 2z^3 + 3z^4 + 4z^5 + 5z^6 + 6z^7 + 5z^8 + 4z^9 + 3z^{10} + 2z^{11} + z^{12}$$
Then $f$ and $g$ are the generating functions (polynomials) of the two dice, and $f \cdot g$ the generating function of the possible sums, counted with multiplicity.  
Now, this polynomial of degree $12$ factors as
$$ z^2 (z+1)^2 (z^2+z+1)^2 (z^2-z+1)^2 $$
and the only way we can group these factors together to get $f(z)$ and $g(z)$ as desired is to take
$$f(z) = z(z+1)(z^2+z+1) = 1+2z^2+2z^3+z^4, g(z) = z(z+1)(z^2+z+1)(z^2-z+1)^2 = z+z^3+z^4+z^5+z^6+z^8$$
That is, two dice labelled $(1,2,2,3,3,4)$ and $(1,3,4,5,6,8)$ have the same probabilities of each outcome as the standard dice.  (See Sicherman dice in Wikipedia.)
So I tried to do something analogous for Schroeder's problem: factor $1+z+z^2+\cdots+z^{99} = (z^{100}-1)/(z-1)$ into two factors $f(z), g(z)$ with $f(1) = g(1) = 10$ and all coefficients positive.  (Factoring $z^{100}-1$ isn't too hard, with the help of the theory of cyclotomic polynomials.)  But this method of solution doesn't seem to work.  How do we put the factors back together?  For example we can take
$$ f(z) = 1 + z + z^2 + z^3 + z^4 + z^{50} + z^{51} + z^{52} + z^{53} + z^{54}, g(z) = 1 + z^5 + z^{10} + z^{15} + z^{20} + z^{25} + z^{30} + z^{35} + z^{40} + z^{45} $$
which I found by experiment.  But I can't see how to find all the ``good'' factorizations, since a lot of the irreducible factors of $1+z+z^2 + \cdots + z^{99}$ have negative signs.  Presumably someone a bit more comfortable with cyclotomic polynomials can answer Schroeder's question, or the (probably not much simpler) question of finding the number of pairs of sets $(M,N)$ having this splitting property.

REPLY [5 votes]: Let $a$, $b$ be divisors of $n$.  Convince yourself that the sets
$M = \{0,1,\ldots,a-1\} + ab \cdot \{0,1,\ldots,n/a-1\}$
$N = \{0,a,\ldots,(b-1)a\} + bn \cdot \{0,1,\ldots,n/b-1\}$
have the desired property.  These pairs $(M,N)$ are all distinct as $a$, $b$ vary over divisors other than $1$, $n$.  On the other hand the pairs coming from choosing $a=n$ or $b=1$ or $(a,b)=(1,n)$ all coincide with the obvious ("base $n$") pair, while the pair coming from $(a,n)$ coincides with the pair coming from $(1,a)$.  So if $\sigma$ is the number of divisors of $n$, this gives you at least $\sigma^2 - 3 \sigma + 3$ such pairs.  For $n=10$ you can check (e.g. using the generating function method that you described) that this gives all the possibilities.  
In general there will be more, because you can iterate the construction: you have pairs of the form
$M = \{0,1,\ldots,d_1-1\} + d_1 d_2 \{0,1,\ldots,d_3-1\} + d_1 d_2 d_3 d_4 \{0,1,\ldots,d_5-1\} + \cdots  $
$N = d_1 \{ 0,1,\ldots,d_2-1\} + d_1 d_2 d_3 \{0,1,\ldots,d_4-1\} + \cdots $
for divisors $d_1,d_2,\ldots$ of $n$ with  $d_2 d_4 \cdots d_{2k} = n$ and similarly for the product of the odd-indexed $d$'s.  Now collisions between the various pairs should happen precisely whenever any of the $d_i$'s happens to be $1$, and so distinct pairs should come from distinct sequences $d_1,\ldots,d_{\ell}$ of divisors of $n$ other than $1$ such that the odd terms multiply to $n$ and the even terms also multiply to $n$. For instance we find that we've correctly re-counted the $\sigma^2-3\sigma +3$ possibilities that we found and counted before: the sequences of $d$'s of length at most $4$ are
$n,n$
$a,n,n/a$
$a,b,n/a,n/b$
as $a,b$ range over nontrivial divisors of $n$. But of course if $n$ has more than two prime factors, you get longer sequences as well.  I'll leave it to someone else to do the enumeration of the sequences.
I think I've convinced myself that this construction gives you everything, but it might be a pain to write down.  (Suppose $M$ is the set containing $1$.  Take $d_1$ to be the smallest nonzero integer contained in $N$.  Then the smallest integers in $N$ have to consist of multiples of $d_1$ up to some $d_1(d_2-1)$, and the next integer contained in $M$ is $d_1 d_2$.  But then  $d_1 d_2 + 1,\ldots,d_1 d_2 + (d_1-1)$ each have to be in one or the other of the sets, and in fact they have to be in $M$ or else you could form the sum $d_1 (d_2+1)$ in two ways.  And so forth.)<|endoftext|>
TITLE: Quantum group as (relative) Drinfeld double?
QUESTION [26 upvotes]: The most elementary construction I know of quantum groups associated to a finite dimensional simple Hopf algebra is to construct an algebra with generators $E_i$ and $F_i$ corresponding to the simple positive roots, and invertible $K_j$'s generating a copy of the weight lattice.  Then one has a flurry of relations between them, and a coproduct defined on the generators by explicit formulas.  These are not mortally complicated, but are still rather involved.  Then come explicit checks of coassociativity, and compatibility between multiplication and comultiplication.  Finally, one has the $R$-matrix which is an infinite sum with rather non-obvious normalizations.  Enter more computations to verify $R$-matrix axioms.
I recall learning about a nice way to construct the quantum group, which in addition to requiring less formulas has the advantage of making it clear conceptually why it's braided.

I'm hoping someone can either point me to a reference for the complete picture, or perhaps fill in some of the details, since I only remember the rough outline.  That, precisely, is my question.

I include the remarks below in hopes it will jog someone's memory.
You start with the tensor category $Vect_\Lambda$ of $\Lambda$-graded vector spaces, where $\Lambda$ is the weight lattice.  We have a pairing $\langle,\rangle:\Lambda\times\Lambda\to \mathbb{Z}$, and we define a braiding $\sigma_{\mu,\nu}:\mu \otimes \nu \to \nu\otimes\mu$ to be $q^{\langle \mu,\nu \rangle}$.  Here $q$ is either a complex number or a formal variable.  We may need to pick some roots of $q$ if we regard it as a number; I don't remember (and am not too worried about that detail).  Also, here we denoted by $\mu$ and $\nu$ the one dimensional vector space supported at $\mu$ and $\nu$ respectively, and we used the fact that both $\mu\otimes\nu$ and $\nu\otimes\mu$ are as objects just $\mu+\nu$.
Okay, so now we're supposed to build an algebra in this category, generated by the $E_i$'s, which generators we regard as living in their respective gradings, corresponding to the simple roots.  Here's where things start to get fuzzy.  Do we take only the simples as I said, or do we take all the $E_\alpha$'s, for all roots $\alpha$?  Also, what algebra do we build with the $E_i$'s?  Of course it should be the positive nilpotent part of the quantum group, but since we build it as an algebra in this category, there may be a nicer interpretation of the relations?  Anyways, let's call the algebra we are supposed to build here $U_q(\mathfrak{n}^+)$.  I definitely remember that it's now a bi-algebra in $Vect_\Lambda$, and the coproduct is just $\Delta(E_i)=E_i\otimes 1 + 1\otimes E_i$ (the pesky $K$ that appears there usually has been tucked into the braiding data). Now we take $U_q(\mathfrak{n}^-)$ to be generated by $F_i$'s in negative degree, and we construct a pairing between $U_q(\mathfrak{n}^+)$ and $U_q(\mathfrak{n}^-)$.  The pairing is degenerate, and along the lines of Lusztig's textbook, one finds that the kernel of the pairing is the q-Serre relations in each set of variables $E_i$ and $F_i$.
Finally, once we quotient out the kernel, we take a relative version of Drinfeld's double construction (the details here I also can't remember, but would very much like to), and we get a quasi-triangular Hopf algebra in $Vect_\Lambda$.  As an object in $Vect_\Lambda$ it's just an algebra generated by the $E_i$'s and $F_i$'s, so no torus.  But since we're working in this relative version, we can forget down to vector spaces, and along the way, we get back the torus action, because that was tucked into the data of $Vect_\Lambda$ all along.
So, the construction (a) gives neater formulas for the products, coproducts, and relations (including the $q$-Serre relations),  and (b) makes it clear why there's a braiding on $U_q(\mathfrak{g})$ by building it as the double.
The only problem is that I learned it at a seminar where to my knowledge complete notes were never produced, and while I remember the gist, I don't remember complete details.  Any help?

REPLY [8 votes]: There is a detailed exposition of this in Majid's paper Double-bosonization of braided groups and the construction of $U_q(\mathfrak{g})$, Math Proc Cambridge Phil Soc 125(1). Especially appendix B where quantum group is obtained by a version of Tannaka-Krein duality for braided monoidal categories applied to the category of Yetter-Drinfeld modules of the positive part $U_q(\mathfrak{n})$ living in the category of comodules over a weakly quasitriangular group algebra $k\Lambda$ with respect to a braiding coming from the Cartan datum and a parameter $q$.
Here, the quantum groups are a special case of what Majid calls double bosonization which is the a briaded version of the Drinfeld double (the ordinary Drinfeld double arises by applying reconstruction to Yetter-Drinfeld modules in the symmetric monoidal category of $k$-vector spaces). It is not possible to define the Drinfeld double in a briaded monoidal category. This is why one needs to work with a fibre functor to vector spaces, and bosonizations (reflected by the commutator relations with the $K_i$ become necessary.
The nice thing about this construction is that the category of Yetter-Drinfeld modules is very special, it is the center of the category of $U_q(\mathfrak{n})$-modules. And $U_q(\mathfrak{n})$ is a Nichols algebra.<|endoftext|>
TITLE: What's a natural candidate for an analytic function that interpolates the tower function?
QUESTION [13 upvotes]: I know that there are analytic functions whose composition with itself is the exponential function, the so-called functional square root of the exponential function, with the additional property that it is real on the real line. 
Is a similar property possible for a holomorphic function that interpolates the tower function?
Tower function on the positive integers is defined recursively by $f(n+1) = e^{f(n)}, f(1) = 1$.

REPLY [4 votes]: In addition to the above formulas, we can also use this very old formula, dating back to 1945 (J. Ginsburg, Iterated exponentials, Scripta Math. 11 (1945), 340-353.):
$$f(x)=r+\sum_{n=1}^{\infty} \frac{\left(\ln a \right)^{n-1}\left(\ln \left(a^r \right)\right)^{nx}\left(1-r\right)^n B_n^{x-1}}{n!}$$
Where $B_n^x$ are the Bell numbers of $x$-th order and $r=\frac{W(-\log (a))}{\log (a)}$ ($W(x)$ is the Lambert function). Here: http://arxiv.org/abs/0812.4047 one can read about Bell numbers of higher orders.
The problem is that Bell numbers are only defined for integer order. We can easily generalize that to any real number by induction as follows:
$$A_0^x=1$$
$$A_{n+1}^x=\sum_{k=0}^{x-1} A_n^x\star A_n^k$$
And then
$$B_n^x=A_{n-1}^{x+1}$$
where $f(n)\star g(n)$ is the binomial convolution as described by Donald Knuth:
$$f(n)\star g(n)=\sum_{k=0}^n \binom nkf(n-k)g(k)$$
To obtain the value for any real $x$, we can note that the right part in $A_{n+1}^x=\sum_{k=0}^{x-1} A_n^x\star A_n^k$ is a polynomial of $x$ and $k$ of degree $n-1$ and integer coefficients and we can take indefinite sum of it symbolically following the rule 
$$\sum_x cx^n=\frac{B_{c+1}(x)}{c+1}$$
Where $B_c(x)$ are the Bernoulli polynomials.
Unfortunately this method works only for $a \le e^{1/e}$ in $f(x+1)=a^{f(x)}$.
Here is the plot of the function, for $a=\sqrt{2}$, obtained with this method and $5$ terms:
     (source)<|endoftext|>
TITLE: A question regarding a claim of V. I. Arnold
QUESTION [40 upvotes]: In his Huygens and Barrow, Newton and Hooke, Arnold mentions a notorious teaser that, in his opinion, "modern" mathematicians are not capable of solving quickly. Then, he adds that the exception that proved the rule in this case of his was the German mathematician Gerd Faltings.
My question is whether any of you knows the complete story behind those lines in Arnold's book. I mean, did Arnold pose the problem somewhere (maybe Квант?) and Faltings was the only one that submitted a solution after Arnold's own heart? Is the previous conjecture totally unrelated to the actual development of things?
I thank you in advance for your insightful replies.
P.S. It seems that this teaser of Arnold eventually became a cult thingy in certain branches of the Russian mathematical community. Below you can find a photograph taken by a fellow of mine of one of the walls of IUM's cafeteria (where IUM stands for Independent University of Moscow). As the Hindu mathematician Bhāskara would say (or so the legend has it): BEHOLD!

REPLY [24 votes]: The "inspecting the graph" comment might refer to something like this. 
Consider two smooth curves $y = f(x)$ and $y = g(x)$ that are tangent to $y = x$ at $(0,0)$. For $x$ near 0, define $u(x)$ and $v(x)$ so that $g^{-1}(x) - f^{-1}(x) = u(x)$ and $f(x) - g(x) = v(x)$.  In the picture, $v(x) = BC$ and $u(x) = AD$.  But both curves have slope very close to 1, so $AD \approx BC$, i.e. $u(x) \approx - v(x)$, and $\frac{f(x) - g(x)}{g^{-1}(x) - f^{-1}(x)} \approx 1$.<|endoftext|>
TITLE: Degree constrained edge partitioning
QUESTION [6 upvotes]: I've come up with the following optimization problem in my research. Is this a known problem in graph-theory and/or combinatorial optimization? If not, which of the known problems are the most similar to it? 
Let's have a graph $G=(V,E)$ with real positive or negative weights assigned to its edges: $w: E \rightarrow \Re$. The problem is to remove a set of edges ($E_1$) from $G$ such that the sum of the remaining edges ($E_2$) is minimized, and no vertex in G has degree less than 2 (i.e. no leaf vertices should exist).

REPLY [3 votes]: I'm not sure about published references to this specific problem, but I'm pretty sure it can be solved in polynomial time via a reduction to minimum weight perfect matching, as follows.
Replace each vertex of degree $d$ by a complete bipartite graph $K_{d-2,d}$ where each of the original edges incident to the vertex becomes incident to one of the vertices on the $d$-vertex side of the bipartition. Set all edge weights in this complete bipartite graph to zero. Next, add a complete graph (with zero edge weights) connecting all of the vertices in the whole graph that are on the $(d-2)$-vertex sides of their bipartitions.
For each vertex $v$ in the original graph, a perfect matching in this modified graph has to include at least two of the original graph's edges incident to $v$, because at least two of the vertices on the $d$-vertex side of the complete bipartite graph that replaces $v$ are not matched within that complete bipartite graph. Because all the other edges have cost zero, the cost of the perfect matching is the same as the cost of the solution it leads to in the original graph.
On the other hand, whenever one has a subgraph of the original graph that includes at least two edges at each vertex, it can be completed to a perfect matching at no extra cost by matching the remaining vertices on the $d$-vertex sides of their bipartition to vertices on the $(d-2)$-vertex sides, and then using the complete graph edges to complete the matching among any remaining unused vertices on the $(d-2)$-vertex sides.
Therefore, the cost of the minimum weight perfect matching in this graph is the same as the cost of the optimal solution to your problem.
Added later: something like this appears to be in "An efficient reduction technique for degree-constrained subgraph and bidirected network flow problems", Hal Gabow, STOC 1983, doi:10.1145/800061.808776<|endoftext|>
TITLE: Solving recurrence equation with indexes from negative infinity to positive infinity
QUESTION [7 upvotes]: In many cases, the recurrence equations that people are solving involves index of only non-negative values. Here I have a recurrence equation which arises from transport of light in an infinite 1D chain:
$a_m=\sum _{j=1}^{\infty } \left(T_ja_{m+j}+T_ja_{m-j}\right) + \delta _{m,0}$
where $\delta_{m,0}$  is the Kronecker delta function. i.e.:
$\delta_{i,j} = \begin{cases} & 1 \text{ if } i=j \\ & 0 \text{ if } i \neq j \end{cases}$ 
Here I would like to solve $a_m$, where the index of m is from negative infinity to positive infinity, while $T_j$ is a given sequence, and p is just a given constant.
Defining the generating function $G(z)=\sum _{k=-\infty }^{\infty } a_kz^k$, I found that:
$G(z)=\frac{1}{1-\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)}$
The problem is, how am I going to do series expansion on G? Doing a simple expansion of $\frac{1}{1-\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)}=\sum _{j=0}^{\infty } \left(\sum _{k=1}^{\infty } t_k\left(z^{-k}+z^k\right)\right){}^j$ won't help. Since the power is too difficult to expand out.
And contour integration isn't helping as well, since it is too difficult to compute analytically or numerically too.
Here I would like to ask about direction in obtaining analytical solution, or approximated one.
And in my case, my function G is given by:
$G(z)=\left(1+\frac{3i}{2r^3}\left(r^2\left(\ln \left(1-\frac{e^{i r}}{z}\right)+\ln \left(1-e^{i r}z\right)\right)\right)-i r\left(\text{Li}_2\left(\frac{e^{i r}}{z}\right)+\text{Li}_2\left(e^{i r}z\right)\right)+\text{Li}_3\left(\frac{e^{i r}}{z}\right)+\text{Li}_3\left(e^{i r}z\right)\right){}^{-1}$
p.s.:I have posted the same problem in Voofie.

REPLY [3 votes]: I see your problem more like a linear equation on an infinite dimensional space of doubly infinite sequences than as a recurrence equation, since there are no initial values from which start to build up the solution. In the following I will assume that $\sum_{j=1}^\infty|t_j|<\infty$ and that $\mathbf{a}=(a_m)$ is bounded. The linear operator $T$ defined on $\ell^\infty(\mathbb{Z})$ by
$$
(T\mathbf{a})_m=\sum_{j=1}^\infty t_j(a_{m+j}+a_{m-j})
$$
is bounded with operator norm $\|T\|\le2\sum_{j=1}^\infty|t_j|$. Your equation can be written as
$$
(I-T)\mathbf{a}= \mathbf{b},
$$
where $I$ is the identity operator and $\mathbf{b}\in\ell^\infty(\mathbb{Z})$. If $\sum_{j=1}^\infty|t_j|<\dfrac{1}{2}$, then $I-T$ is invertible and the above equation has a unique solution for all $\mathbf{b}\in\ell^\infty(\mathbb{Z})$, given by
$$
\mathbf{a}=(I-T)^{-1}\mathbf{b}=(I+T+T^2+T^3+\dots)\mathbf{b}.
$$
This is an explicit formula that in practice may be useless, although one can get an approximation by computing a few terms of the sum.
Just to check that this can really give a solution, let's study the particular case in which $t_1=t$ and $t_j=0$ for all $j>1$, and $(\mathbf{b})_m=\delta_{m,0}$. Then we find that
$$
a_m=a_{-m}=\sum_{k=1}^\infty\binom{2k+m}{k+m}t^{2k+m}=t^m{}_2F_1\left(\frac{m+1}{2},\frac{m+2}{2},m+1,4t^2\right),
$$
where ${}_2F_1$ is the hypergeometric function. We see that indeed one must have $|t|<\dfrac{1}{2}$ for this to make sense.
This analysis doesn't mean that there are no other solutions under different conditions, but I think that it will be difficult to avoid the requirement that $\sum_{j=1}^\infty|t_j|<\infty$.<|endoftext|>
TITLE: Is there a "natural" characterization of when X × βN is normal?
QUESTION [6 upvotes]: As per a recent question of mine, $\omega_1 \times \beta \mathbb{N}$ is not normal. I'm wondering whether there's some sort of "natural" condition that describes when a space has a normal product with  $\beta \mathbb{N}$, analagous to Dowker's characterisation that $X$ is countably paracompact iff $X \times [0, 1]$ is normal. 
The context here is that I'm looking for something I can weaken to something sensible, as I have a property implied by $X \times \beta \mathbb{N}$ being normal which I am "surprised" isn't an equivalence ($\omega_1$ has this property) and would like to see if I can show its equivalenct to some slightly weaker topological condition.

REPLY [6 votes]: The space $X\times\beta\mathbb{N}$ is normal if and only if $X$ is normal and $\mathfrak{c}$-paracompact. This follows from results of Morita (Paracompactness and product spaces, MR132525), where he generalizes Dowker's characterization of countable paracompactness.
First note that $X\times\beta\mathbb{N}$ is normal if and only if $X\times K$ is normal for every separable compact Hausdorff space $K$. This is because every separable compact Hausdorff space is a perfect image of $\beta\mathbb{N}$.
Morita's Theorem 2.2 shows that if $X$ is normal and $\mathfrak{c}$-paracompact, then $X \times K$ is normal for every compact Hausdorff space $K$ of weight at most $\mathfrak{c}$. Hence, $X\times K$ is normal for every separable compact Hausdorff space $K$ since these all have weight at most $\mathfrak{c}$.
Morita's Theorem 2.4 shows that a space $X$ is normal and $\mathfrak{c}$-paracompact if (and only if) $X\times[0,1]^{\mathfrak{c}}$ is normal. Since the space $[0,1]^{\mathfrak{c}}$ is a separable compact Hausdorff space, this closes the implication loop.<|endoftext|>
TITLE: Does there exist a holomorphic function which takes given values on the positive integers?
QUESTION [6 upvotes]: Inspired of course by What's a natural candidate for an analytic function that interpolates the tower function?
I am minded to ask what looks to me like a more natural question: given a sequence $a_1,a_2,a_3,\ldots$ of complex numbers, is there always a holomorphic function $f$ defined on the entire complex plane, with $f(n)=a_n$ for $n=1,2,3,\ldots$? No idea what the answer is myself, but wouldn't surprise me if it were well-known and even easy.

REPLY [15 votes]: This is Exercise 6, Page 26, of Knopp's Problem Book in the Theory of Functions, Volume 2: For any sequence of complex numbers $z_n$ with no finite limit point, and for any sequence of complex numbers $w_n$, there is an entire function mapping $z_n$ to $w_n$. The proof goes like this: Use the Weierstrass Factor Theorem to construct a function $W$ with simple zeros at the $z_n$. Use the Mittag-Leffler theorem to construct a function $M$ with simple poles at the $z_n$ with residues $\frac{w_n}{W'(z_n)}$. Then the function $W\cdot M$ does the job.<|endoftext|>
TITLE: Do "surjective" degree zero maps exist?
QUESTION [24 upvotes]: Is there a map $f\colon X \to Y$ of closed, connected, smooth and orientable $n$-dimensional manifolds such that the degree of $f$ is 0 but $f$ is not homotopic to a non-surjective map?
Added: The motivation is: There is a "mild version" of the Nearby Langrangian conjecture stating: any exact Lagrangian manifold $X \to T^*Y$ has non-zero degree when composed with the projection $T^*Y \to Y$. It is known that the map is always surjective. I am looking at a possible inbetween stating that the map cannot be homotoped to a non-surjective map.

REPLY [37 votes]: It is a theorem of H. Hopf that a map between connected, closed, orientable n-manifolds of degree 0 is homotopic to a map that misses a point, when n > 2. See D. B. A. Epstein, The degree of a map. Proc. London Math. Soc. (3) 16 1966 369--383, for a "modern" discussion including the analogous situation in the non-orientable case. The same result holds for n = 2, but is more difficult and is due to Kneser. See Richard Skora, The degree of a map between surfaces. 
Math. Ann. 276 (1987), no. 3, 415--423, for a thorough discussion of the non-orientable case in dimension 2.<|endoftext|>
TITLE: What to expect from attending an ICM?
QUESTION [15 upvotes]: Since ICM 2010 is due to be held in Hyderabad next august, and since I never attended such an event, I am wondering if those MO users who did attend at least one ICM could share some of their wisdom (the normal price registration period ends in may, so I'd like to plan things before then if at all).
What's for sure is that the list of speakers, both plenary and sectional, is mouth-watering to say the least.


is an ICM a good place to learn some new math, including outside one's main speciality?

what about the "meeting new people" aspect, is the setting making discussions easy?



At the secret blogging seminar the topic has been discussed at length, and two kind of quite different answers emerged (intended to mature-enough grad students and young postdoc):
a. the math talks are okay, sometimes great, but forget about meeting new folks
b. some math talks are truly inspiring, and you get a real opportunity to meet experts you didn't know yet
I'd like to have a wider sample of opinions, or at least know which between a and b is the opinion of the majority, many thanks.
(Note to moderators: ideally this question should be made community wiki, but I don't know how to do that.)

REPLY [15 votes]: Regarding 1, ICM's are a place to acquire new ideas.  With rare exception, there will be no "Introduction to ..." type material, just "Motivation for..." and "Teaser to ..." and
"Of course, since we all know ..., the following will be obvious ...", which is true for me after a few pages and perhaps months of study.  However, even the last type may contain an idea I can use, even if I don't know what "we all know".
Regarding 2, there are times, places, and opportunities to meet; see below.
I hope one day to compare the lists of attendees from all the ICMs since the one in Berkeley in 1986.  I know I am in all of them, and I suspect that I share that distinction with at most 10 other people.
The one in 1986 convinced me to go to Berkeley for graduate school;  the others have given me a quick exposure to various fields in mathematics and usually too much to think about.  Your mileage may vary, but I found the following routine helpful to me.
1) Each night, plan your lecture tour for the following day.  You will need to get a list of programme changes (usually needed only for Short Communications sessions), and select your A, B and C list choices for each hour. (Sometimes your A choice and B choice get rescheduled or canceled.) If you have friends attending and want to split up to cover more lectures, plan with them.  At the first few conferences, I had mild sessions of agony because there were more lectures I wanted to attend than slots available.  I now have restricted my interests in the afternoons to three or four sections (currently logic, algebra, computing, and combinatorics), to make my planning easier.
2) When you suffer from lecture burnout, try a tour or even just walk around in the part of the city near the conference hall.  One of my treats from the 2006 ICM was looking at the inspired architecture in that part of Madrid.  Or check out the array of posters.
(Ideally, ask permission before photographing a poster.  In any case, don;'t use the
photo beyond personal use without arrangments with the author of the poster.)
3) If you are very social (I was not), figure out how to work that into your visit.  Lunch breaks and afternoon breaks are the most opportune for this.  I did manage to chat up a few attendees (including Tim Gowers and Greg Kuperberg) during moments in between  lectures.  There will be a few social occasions, and in most cases informal dress is acceptable (but check the invitations).  During the
lecture sessions, there will be limited or no time for questions, so make your question count.  Also, unless you find someone to walk to the same lecture with you, it is not likely that you will have much time to chat between lectures.  
4) Find the information booth and help desk.  If any problems arise, they are your first step towards resolution, even if it is visit related and not conference related.  Also,
avoid the temptation to check your email during the conference.  You can easily get sucked in to spending too much time on the computer.  (You get to choose how much time to devote to computer versus sleep, however.)
Regarding your a and b choices, I would say both are true.  In point of a, if I had made a persistent ass of myself, I might have gotten more than a few words with Tim Gowers, but as he had been awarded the Fields Medal recently, I chose to think his time with me would be very limited and in high demand.  Thus I did not persist, and being in lecture-attending mode most of the time, I often have the perception that there is little to no time for socializing.
Towards point b., when I relax, I find I can talk to a number of people and create opportunities for socializing quite easily, if I put my focus on it.  There will be some inspiring talks, and after hearing a few of these, you can use one or more points from them as icebreakers.  I will thus say it depends on what kind of experience you will want to make of your ICM.
Gerhard "Collectors edition ICM T-shirts for sale" Paseman, 2010.04.08<|endoftext|>
TITLE: Bounding a spectral gap: what proof techniques exist?
QUESTION [9 upvotes]: The following situation is ubiquitous in mathematical physics.  Let 
$\Lambda_N$ 
be a finite-size lattice with linear size 
$N$.  An typical example would be the subset of 
$\mathbb{Z}\times\mathbb{Z}$
given by those pairs of integers 
$(j,k)$ such that 
$j,k \in$ {
$0,\ldots,N-1$}.  On each vertex 
$j$ of the lattice place a copy of the vector space 
$\mathbb{C}^d$.  The total space will be the tensor product of all of these spaces.  Then define a Hamiltonian acting on this total space as follows:
$$ H = \sum_{k \in \Lambda_N} h_k$$
for some Hermitian matrices 
$h_k$ which act like the identity everywhere except on the vector spaces located on site 
$k$ and in the neighborhood surrounding 
$k$.  Typically, one is interested in the case where there is a translational symmetry (except at the boundary) in the definition of the 
$h_k$.  Denote the eigenvalues of 
$H$ in increasing order by 
$\lambda_1 \le \lambda_2 \le \ldots \le \lambda_M$.  

For an arbitrary fixed family of Hamiltonians
  $H$, what proof techniques exist for computing an upper and a lower bound on 
  $\Delta = \lambda_2 - \lambda_1$ 
  as a function of 
  $N$?  In particular, we want to know if 
  $\Delta$ decays to zero as a function of 
  $N$, or if it is lower-bounded by some constant independent of 
  $N$.

The gap 
$\Delta$ is the energy gap between the ground state and the first excited state of an interacting quantum system.  Understanding this quantity tremendously impacts our understanding of the different phases of matter, but it is extremely difficult to compute or even bound for all but the simplest cases (like when all the 
$h_k$ commute).  This difficulty persists even when there is significant additional (physically motivated) structure in the problem, such as considering only 
$h_k$ which are projectors, and where there is a unique zero-energy eigenstate (all others having positive energy for any finite 
$N$).
More general formulations of this question also have applications to expansion properties of graphs, mixing times of Markov chains, and many other things.  I’m happy to hear answers related to these as well, but I’m hoping to find answers that are useful for the structure of local Hamiltonians, as defined above.

REPLY [2 votes]: For VBS quantum antiferromagnets in one dimension see also :
Ian Affleck, Tom Kennedy, Elliott H. Lieb and Hal Tasaki,
Valence bond ground states in isotropic quantum antiferromagnets.
Comm. Math. Phys., Volume 115, Number 3 (1988) 
and
Stefan Knabe, Energy gaps and elementary excitations for certain VBS-quantum antiferromagnets, Journal of Statistical Physics
Volume 52, Numbers 3-4, 1988<|endoftext|>
TITLE: About dense orbits on dynamical systems
QUESTION [5 upvotes]: Preliminars and notation:
Let $M$ be an $n$-dimensional compact manifold, $T\colon M\rightarrow M$ a diffeomorphism and $( x_n)_{n\in\mathbb{Z}}$ a dense orbit under $T$, ($x_n = T^n(x_0)$). Let $p\in M$ be another point and define, for $\delta>0$, $B_n(\delta) = B(p, e^{-n\delta})$. 
Question: Is it true that for every $\delta>0$ the set
$A = $(  $n\in\mathbb{N} | x_n \in B_n(\delta)$  )
has finite cardinality? How can it be proven?
Thank you for the answers!

REPLY [6 votes]: This is called the shrinking target problem, and there is a reasonably large literature on it. For hyperbolic dynamical systems we can usually find quite a few pairs $x$, $p$ such that $A$ is infinite for all $\delta$. Indeed, I believe that there are results showing that in certain cases, for any point $z$ and positive real number $\delta>0$, the set of all $x$ such that $d(T^nx,z)<\exp(-n\delta)$ for infinitely many $n \geq 1$ has positive Hausdorff dimension. A good place to start would be the articles "Ergodic theory of shrinking targets" and "The shrinking target problem for matrix transformations of tori", both by Hill and Velani, but there are many results beyond this.
For illustration, here is a nice example in the case where $T$ is a smooth map of the circle which is not a diffeomorphism. I realise that this falls slightly outside the purview of your question, but it is possible to extend this argument to the case of toral diffemorphisms using the technical device of a Markov partition. (I will not attempt this here because it is very fiddly.) Let $X=\mathbb{R}/\mathbb{Z}$ be the circle, let $T \colon X \to X$ be given by $Tx = 2x \mod 1$, and let $d$ be a metric on $X$ which locally agrees with the standard metric on $\mathbb{R}$. Take $p=0 \in X$ and fix any $\delta>0$. Now, the orbit of $x$ is dense if and only if it enters every interval of the form $(k/2^n,(k+1)/2^n)$, if and only if every possible finite string of 0's and 1's occurs somewhere in the tail of its binary expansion.
On the other hand, we have $d(T^nx,0)<2^{-\delta n}$ as long as the binary expansion of $x$ contains a string of zeroes starting at position $n$ and having length $\lceil \delta n \rceil$. I think that it is not difficult to see that we can construct an infinite binary expansion, and hence a point $x$, such that this condition is met for infinitely many $n$, whilst simultaneously meeting the condition that the orbit of $x$ is dense. In particular we can construct a point $x$ for which $A$ is infinite, even for all $\delta$ simultaneously if you like.<|endoftext|>
TITLE: Does the set of open sets in a topological space have a topology itself?
QUESTION [8 upvotes]: If X is a topological space, and A consists of all of X's open sets, can we define a natural topology on A (using the topology of X)?

REPLY [15 votes]: Of course there are many answers to your question. The interesting thing to ask is if there is a "best" or "right" answer. In many respects the "correct" topology for the lattice of open sets is the Scott topology. In case $X$ is locally compact, the Scott topology coincides with the compact-open topology of the continuous function space $C(X,\Sigma)$, where $\Sigma$ is the Sierpinski space (where we identify open sets with their characteristic functions into $\Sigma$).
There are several reasons why the Scott topology is the "right" one. One of them is that the following are equivalent for a space $X$:

$X$ is an exponentiable space in the category of topological spaces ($Y^X$ exists for all $Y$).
The exponential $\Sigma^X$ exists.
The topology of $X$ is a continuous lattice.
The lattice of open sets of $X$ equipped with the Scott topology is the exponential $\Sigma^X$.

I recommend the following paper by Martin Escardó and Reinhold Heckmann in which they explain many things related to topology of the lattice of open sets (and function spaces in general):

M.H. Escardo and R. Heckmann. Topologies on spaces of continuous functions. Topology Proceedings, volume 26, number 2, pp. 545-564, 2001-2002.<|endoftext|>
TITLE: Is there an introduction to probability theory from a structuralist/categorical perspective?
QUESTION [222 upvotes]: The title really is the question, but allow me to explain.
I am a pure mathematician working outside of probability theory, but the concepts and techniques of probability theory (in the sense of Kolmogorov, i.e., probability measures) are appealing and potentially useful to me. It seems to me that, perhaps more than most other areas of mathematics, there are many, many nice introductory (as well as not so introductory) texts on this subject.
However, I haven't found any that are written from what it is arguably the dominant school of thought of contemporary mainstream mathematics, i.e., from a structuralist (think Bourbaki) sensibility.  E.g., when I started writing notes on the texts I was reading, I soon found that I was asking questions and setting things up in a somewhat different way.  Here are some basic questions I couldn't stop from asking myself:
[0) Define a Borel space to be a set $X$ equipped with a $\sigma$-algebra of subsets of $X$.  This is already not universally done (explicitly) in standard texts, but from a structuralist approach one should gain some understanding of such spaces before one considers the richer structure of a probability space.]
1) What is the category of Borel spaces, i.e., what are the morphisms?  Does it have products, coproducts, initial/final objects, etc?  As a significant example here I found the notion of the product Borel space -- which is exactly what you think if you know about the product topology -- but seemed underemphasized in the standard treatments.
2) What is the category of probability spaces, or is this not a fruitful concept (and why?)?  For instance, a subspace of a probability space is, apparently, not a probability space: is that a problem?  Is the right notion of morphism of probability spaces a measure-preserving function?  
3) What are the functorial properties of probability measures?  E.g., what are basic results on pushing them forward, pulling them back, passing to products and quotients, etc.  Here again I will mention that product of an arbitrary family of probability spaces -- which is a very useful-looking concept! -- seems not to be treated in most texts.  Not that it's hard to do: see e.g.
http://math.uga.edu/~pete/saeki.pdf
I am not a category theorist, and my taste for how much categorical language to use is probably towards the middle of the spectrum: that is, I like to use a very small categorical vocabulary (morphisms, functors, products, coproducts, etc.) as often as seems relevant (which is very often!).  It would be a somewhat different question to develop a truly categorical take on probability theory.  There is definitely some nice mathematics here, e.g. I recall an arxiv article (unfortunately I cannot put my hands on it at this moment) which discussed independence of events in terms of tensor categories in a very persuasive way.  So answers which are more explicitly categorical are also welcome, although I wish to be clear that I'm not asking for a categorification of probability theory per se (at least, not so far as I am aware!).

REPLY [5 votes]: Bart Jacobs has a new textbook out now. It is on his webpage.  It is called "Structured Probabilistic Reasoning"  I believe this is going to be an important reference in the next five years.<|endoftext|>
TITLE: Representations of \pi_1, G-bundles, Classifying Spaces
QUESTION [13 upvotes]: This question is inspired by a statement of Atiyah's in "Geometry and Physics of Knots" on page 24 (chapter 3 - Non-abelian moduli spaces). 
Here he says that for a Riemann surface $\Sigma$ the first cohomology $H^1(\Sigma,U(1))$, where $U(1)$ is just complex numbers of norm 1, parametrizes homomorphisms $$\pi_{1}(\Sigma)\to U(1).$$ 
This is fine by me, after all by Brown Representability we know $$H^1(\Sigma,G)\cong [\Sigma,BG]=[B\pi_{1}\Sigma,BG]$$ since we know that Riemann surfaces are $K(\pi_{1},1)$s. We then use the handy fact from Hatcher Prop. 1B.9 (pg 90) that shows that... 

For $X$ a connected CW complex and $Y$ a $K(G,1)$ every homomorphism $\pi_1(X,x_0)\to\pi_1(Y,y_0)$ is induced by a map $(X,x_0)\to(Y,y_0)$ that is unique up to homotopy fixing $x_0$.

So group homomorphisms $\pi_{1}(\Sigma)\to U(1)$ correspond on the nose with first cohomology of the $\Sigma$ with coefficients in $U(1)$.
EDIT: To be accurate the Hatcher result shows we have an injection of group homomorphisms into homotopy classes of maps. Does anyone know if every homotopy class of maps is realized by a group homomorphism?
What bothers me is what comes next. Now replace $U(1)$ with $G$ - any compact simply connected Lie group, take $G=SU(n)$ for example. Now Atiyah claims that $H^1(\Sigma,G)$ parametrizes conjugacy classes of homomorphisms $\pi_{1}(\Sigma)\to G$.
Now if the fundamental group or the Lie group were abelian this would be the same statement, but higher genus Riemann surfaces (genus greater than 1) have non-abelian fundamental groups and Atiyah is looking specifically at non-abelian $G$. Also, it seems that the statement of Brown Representability requires abelian coefficient groups, so I am stuck. 
Does anyone know a clean way of proving Atiyah's claim?
EDIT2: I renamed the question to draw in the "right" people. I think the answer has to do with the fact that principal G-bundles over a Riemann surface are determined by maps of $\pi_1(\Sigma)\to G$ (can someone explain why?). This is related to local systems and/or flat connections, which I don't understand well. Thanks!

REPLY [6 votes]: Since it is not always clear what $H^1(X;G)$ means for a non-abelian group, Atiyah might have meant this loosely, perhaps using Cech 1-cocycles to construct flat $G$ bundles, with homologous cycles giving isomorphic bundles. The moduli space of flat bundles is homeomorphic (and real analytically isomorphic) to the space of conjugacy classes of $G$ representations via the holonomy. I suspect this is what Ben is hinting at in his answer above.
But another  interpretation in terms of homotopy classes of maps to $BG$ is as follows. If you give $G$ the discrete topology, then $BG$ is a pointed space with a fixed choice of isomorphism $\pi_1(BG)\cong G$.  So
$Hom(\pi_1(X),G)$ is in bijective correspondence with the pointed homotopy classes of maps $[X,BG]_0$. The free homotopy classes $[X,BG]$ are then in bijective correspondence with the conjugacy classes $Hom(\pi_1(X),G)/conj$, since the action of $G=\pi_1(BG)$ on the pointed homotopy classes corresponds to conjugation, with quotient the free homotopy classes. So if you define $H^1(X;G)=[X,BG]$ (unbased), you get Atiyah's statement.  
In the case of Atiyah's book, though,  more important that the notation for the space of conjugacy classes of reps is the fact that at a representation $r:\pi_1(\Sigma)\to G$, the (usual, twisted by the adjoint rep) cohomology $H^1(\Sigma; g_r)$ ($g$ the lie algebra of $G$)is the Zariski tangent space to the variety of conjugacy classes of representations at $r$, i.e. to the moduli space of flat $G$ connections on $\Sigma$. The cup product
$H^1(\Sigma; g_r)\times H^1(\Sigma; g_r)\to R$ determines a symplectic form on this variety, and a holomorphic structure on $\Sigma$ induces a complex structure on this variety, which reflects itself in the Zariski tangent space as an almost complex structure $J:H^1(\Sigma; g_r)\to H^1(\Sigma; g_r)$ which coincides with the Hodge $*$ operator on harmonic forms.  
Incidentally, all principal $SU(2)$ bundles over a Riemann surface are topologically trivializable.<|endoftext|>
TITLE: The problem of finding the first digit in Graham's number
QUESTION [24 upvotes]: Motivation
In this BBC video about infinity they mention Graham's number. In the second part, Graham mentions that "maybe no one will ever know what [the first] digit is". This made me think: Could it be possible to show that (under some assumptions about the speed of our computers) we can never determine the first digit? 
In logic you have independence results like "We cannot decide if AC is true in ZF". But we cannot hope for this kind of result in this case, since we can easily program a computer to give us the answer. The problem is, that we don't have enough time to wait for the answer! 
In complexity theory you prove things like "no program can solve all problems in this infinite set of problems, fast". But in this case you only have one problem, and it is easy to write a program, that gives you the answer. Just write a program that prints "1" another that prints "2" ... and a program that prints "9". Now you have a program that gives you the answer! The problem is, that you don't know which of the 9 programs that are correct.
Questions
Edit: I have now stated the questions differently. Before I asked about computer programs instead proofs.

Could it be possible to show that any proof of what the first digit in Grahams numbers is, would have length at least $10^{100}$?
Do there exist similar results? That is, do we know a decidable statement P and a proof that any proof or disproof of P must have length $10^{100}$.
Or can we prove, that any proof that a proof or disproof of P must have length at least $n$, must itself have length at least $n$?

I think the answer to 3) is no, at least if all proofs are in the same system. Such a proof would prove that it should have length all least n for any n. 
(Old Questions:

Could it be possible to show that it would take a computer at least say $10^{100}$ steps to determine the first digit in Grahams number?
Do there exist similar results? That is, do we know a decidable statement P and a proof that P cannot be decided in less that say $10^{100}$ steps.
Or can we prove that we need at least $n$ steps to show that a decidable statement cannot be decided in less than $n$ steps?)

(I'm not sure I tagged correctly. Fell free to retag, or suggest better tags.)

REPLY [3 votes]: In general, the first digit of a number $A$ can be determined from the fractional part of $\log_{10} A$. Specifically, if $\log_{10} A = n + \gamma$ where $n = \lfloor \log_{10} A \rfloor$ and $\gamma = \{ \log_{10} A \}$ then $A = 10^\gamma 10^n$ and $1 \leq 10^\gamma < 10$. Therefore the first decimal digit of $A$ is $k$ iff $k \leq 10^\gamma < k+1$, or, equivalently, iff $\log_{10} k \leq \gamma < \log_{10} (k+1)$.
Taking the special case where $A = G$ is Graham's number, we have $G = 3^{3^x}$ for some natural number $x$. Then $\log_{10} G = 3^x \log_{10} 3$. Observe that multiplication by $3^x$ is the same as shifting the base 3 digits of a number $x$ places to the left. Setting $\alpha = \log_{10} 3$ the problem reduces to finding the base 3 digits of $\alpha$ starting $x$ places after the decimal point.
Then the following conjecture would imply that it is impossible to prove anything about the first digit of $G$ in time $o (x)$:
Conjecture: Given a natural number $x$ and a sequence of base 3 digits $z_0, \dots, z_{k-1}$, any proof in ZFC of the statement "The first $k$ base 3 digits of $\alpha$ starting $x$ places after the decimal point are not $z_0 \dots z_{k-1}$" must have length $\Omega\:(n)$.
Analogous conjectures can be asked for any other $b \geq 2$ and replacing $\alpha = \log_{10} 3$ with any other well-known irrational numbers. The reason to expect such a conjecture to hold is that I don't think there is any way to prove anything about $x$th digit of $\alpha$ that is essentially different from calculating the $x$th digit with a general-purpose algorithm, and I don't think there are any such sublinear general-purpose algorithms. Given that there does exist a quasilinear time algorithm for computing the digits of $\alpha$ (see Modern Computer Arithmetic for details), this conjecture would give a fairly sharp estimates $\tilde {\Theta}\:(\log \log G)$ steps for the shortest proof determining the first digit of $G$.
Showing that there are no sublinear-time algorithms for computing the $x$th digits of any natural irrational number seems more difficult than the P vs. NP problem, which is already far beyond what current theoretic computer science can achieve, and the conjecture I stated above is probably more difficult still. However, if we ever get to the point where it is possible to give sharp lower bounds on the most efficient possible algorithms for most problems, then this problem would be potentially solvable.<|endoftext|>
TITLE: In what sense (if any) does the cohomology of profinite groups commute with projective limits?
QUESTION [11 upvotes]: Background:
Let $G$ be a profinite group. If $M$ is a discrete $G$-module, then $M=\varinjlim_U M^U$, where the direct limit is taken with respect to inclusions over all open normal subgroups of $G$, and one naturally has $H^n(G,M)\simeq\varinjlim H^n(G/U,M^U)$, where the cohomology groups on the right can be regarded as the usual abstract cohomology groups of the finite groups $G/U$ (this is sometimes, as in Serre's Local Fields, taken as the definition of $H^n(G,M)$).
More generally if one has a projective system of profinite groups $(G_i,\varphi_{ij})$ and a direct system of abelian groups $(M_i,\psi_{ij})$ such that $M_i$ is a discrete $G_i$-module and the pair $(\varphi_{ij},\psi_{ij})$ is compatible in the sense of group cohomology for all $i,j$, then $\varinjlim M_i$ is canonically a discrete $\varprojlim G_i$-module, the groups $H^n(G_i,M_i)$ form a direct system, and one has $H^n(\varprojlim G_i,\varinjlim M_i)\simeq\varinjlim H^n(G_i,M_i)$. The statement and straightforward proof of this more general result can be found, for instance, in Shatz' book on profinite groups.
Question:
In general, I'm wondering if there are, under appropriate hypotheses, any similar formulae for projective limits of discrete $G$-modules. Now, given a projective system of discrete $G$-modules $(M_i,\psi_{ij})$, it isn't even obvious to me that the limit will again be a discrete $G$-module, and at any rate, while each $M_i$ is discrete, the limit (in its natural topology) will be discrete if and only if it is finite. So, for the sake of specificity, I'll give a particular situation in which I'm interested. If $R$ is a complete, Noetherian local ring with maximal ideal $\mathfrak{m}$ and finite residue field and $M$ is a finite, free $R$-module as well as a discrete $G$-module such that the $G$-action is $R$-linear, then the canonical isomorphism of $R$-modules $M\simeq\varprojlim M/\mathfrak{m}^iM$ is also a $G$-module isomorphism (each $M/\mathfrak{m}^iM$ is a discrete $G$-module with action induced from that of $M$). Moreover, in this case, one can see that the limit is a discrete $G$-module (because it is isomorphic to one as an abstract $G$-module!). There is a natural homomorphism $C^n(G,M)\rightarrow\varprojlim C^n(G,M/\mathfrak{m}^iM)$ where the projective limit is taken with respect to the maps induced by the projections $M/\mathfrak{m}^jM\rightarrow M/\mathfrak{m}^iM$, and this induces similar map on cohomology. I initially thought the map at the level of cochains was trivially surjective, just because of the universal property of projective limits. However, given a ``coherent sequence" of cochains $f_i:G\rightarrow M/\mathfrak{m}^iM$, the property gives me a map $f:G\rightarrow M$ that is continuous when $M$ is regarded in its natural profinite topology, which is, as I noted above, most likely coarser than the discrete topology, so this might not be a cochain. So, what I'd really like to know is whether or not the map on cohomology is an isomorphism.
Why I Care: The reason I'd like to know that the map described above is an isomorphism is to apply it to the particular case of $G=\hat{\mathbb{Z}}$. It is well known (and can be found, for instance, in Serre's Local Fields) that $H^2(\hat{\mathbb{Z}},A)=0$ for $A$ a torsion abelian group. In particular the higher cohomology of a finite $\hat{\mathbb{Z}}$-module vanishes, and I'd like to be able to conclude that the same is true for my $M$ above, being a projective limit of finite abelian groups. 
Thanks!

Keenan

REPLY [3 votes]: Hi Keenan,
You're right that the projective limit of discrete $G$-modules is not necessarily discrete. To take the cohomology of such "topological $G$-modules" you can use continuous cochain cohomology and this continuous cochain cohomology commutes with inverse limits under certain conditions. See section 7 of chapter II of Cohomology of Number Fields by Neukirch, Schmidt & Wingberg.<|endoftext|>
TITLE: Is there a notion of Galois extension for Z / p^2?
QUESTION [8 upvotes]: The above title is in fact a special case of what I want to ask. 
Certainly we have a well defined notion of Galois extension for $ \mathbb{Q}_p $. The intersections of these extensions to the ring of integer of the absolute algebraic closure of $\mathbb{Q}_p$ give us a notion of Galois extensions for $\mathbb{Z}_p $. ( I know that there is a notion of Galois extension for commutative rings, and I believe that it should give us this. Am I correct?)
Let's go further. Let $A_K$ be the ring of integer in a finite Galois extension $K$ of $ \mathbb{Q}_p$. Let $e$ be the ramification degree of $K$ over $\mathbb{Q}_p$. The injection of $ \mathbb{Z}_p$ into $A_K$ will induce an injection of $ \mathbb{Z} / p^n $ into $ A_K / \mathfrak{p}^{en} $. In this picture, there seems to be some desire to say that $ A_K / \mathfrak{p}^{en} $ is the correct notion Galois  of extension of $ \mathbb{Z} / p^n $. But there are problems; taking this notion of Galois extension, if $K$ is has ramification degree $e >1$, the corresponding extension $ A_K /p^e $ is not a field (it is not even an integral domain).
Question 1: Is there any notion of Galois extensions corresponding to what I desire?
Question 2: Can a class field theory (i.e a nice description of absolute abelian Galois extension) of $ \mathbb{Z}/p^n$ be developed in this context? Is there any relationship between this and the local class field theory of $\mathbb{Q}_p$ ( which is the same as that of $\mathbb{Z}_p $)?

REPLY [4 votes]: Let $R$ be a commutative ring and $G$ a finite group. A Galois extension of $R$ with Galois group $G$ is a pair consisting of a morphism $f : R \to S$ and an $R$-linear action of $G$ on $S$ such that the natural map $R \to S^G$ is an isomorphism and the natural map
$$S \otimes_R S \ni s_1 \otimes s_2 \mapsto \prod_{g \in G} s_1 g s_2 \in \prod_{g \in G} S$$
is also an isomorphism. This is just a translation into algebra of the definition of a $G$-torsor over $\text{Spec } R$. With this definition, it is not true that Galois extensions of $\mathbb{Q}_p$ induce Galois extensions of $\mathbb{Z}_p$; I think only the unramified extensions do. 
Galois extensions of $R$ in this sense are classified by conjugacy classes of continuous homomorphisms $\pi_1(\text{Spec } R) \to G$ from the étale fundamental group of $\text{Spec } R$ to $G$. It's known that the étale fundamental group of $\text{Spec } \mathbb{Z}/(p^n)$ is $\hat{\mathbb{Z}}$, same as with $\mathbb{F}_p$. This says more or less that the underlying ring extension of every Galois extension is a finite product of the extensions described in Robin Chapman's answer.<|endoftext|>
TITLE: Hermitian symmetric spaces vs Hermitian homogeneous spaces
QUESTION [11 upvotes]: A Hermitian symmetric space is a connected complex manifold with a hermitian metric on which the group of holomorphic isometries acts transitively, and which satisfies the following extra condition: each point (equivalently, some point) is an isolated fixed point of some (holomorphic, isometric) involution of the space.  There is a beautiful classification theory of Hermitian symmetric spaces, and this is the starting point for the study of Shimura varieties.
The first four conditions (connected, complex, hermitian, homogeneous), which define a Hermitian homogeneous space, all seem natural enough to me.  My question: why the need for the extra condition regarding the involution?
This is deliberately vague because I'm not quite sure what form the most satisfying answer will take.  But here are a couple of more specific versions of the question.  (1) What sort of spaces are we excluding by imposing this condition, and why?  (2) Are there other conditions which might seem more natural to me which are equivalent (or equivalent some of the time)?  For instance if you were to tell me that a Hermitian homogeneous space that's negatively curved and simply-connected always has such an involution, I would be completely satisfied.

REPLY [4 votes]: Boringly-but-honestly, for me "symmetric space" means "G/K", where G is reductive or semi-simple, and (for non-compact symmetric spaces) K is maximal compact. That is, the "definition" is not "geometric", but artifactual. 
The explanation of this viewpoint is yet another one of those stories wherein one finds oneself "a long-distance commuter" from A to B, and then wonders why one doesn't simply live closer to B.<|endoftext|>
TITLE: Mechanically instantiating abstract constructions
QUESTION [6 upvotes]: I am looking for work on the effective inverse of abstraction, aka specialization.
There are two ways in which abstraction helps us: 

Get a better understanding of the structural rules at play in specific situations.
Not wasting time re-inventing infrastructure that already exists.

Historically, we have group theory as an extremely successful abstraction of symmetry and permutations.  More recently, category theory (CT) has catalogued a huge number of universal constructions that pervade mathematics (and universal algebra before that, and ...).  Pedagogically, CT is very useful in sense #1 above.
What I really want is to use the constructions in CT in the sense of #2, but I want to do it as a 'conservative extension' of my base theory.  Let me be a bit more precise:  suppose I have a theory T using a language L (you can assume higher-order-logic as the background logic if that helps) for which I have identified some categorical structure already.  In other words, some objects of T with some morphisms for a category with some 'nice' properties.  These properties allow me to pick out some categorical constructions as now being 'available' (from pushouts to coends to Kan extensions, etc, as appropriate).  Now, what I really want to do is to enrich the theory T with the results of these categorical constructions, but with all categorical language erased.  In other words, I want to retract the construction from CT to T by using L.
For example, given the category Set, one should be able to unwind the definitions (mechanically!) to see that pushouts over the empty set is exactly disjoint union [imagine that disjoint union had not yet been defined!].  Much better examples are given all through Abstract and concrete categories: The joy of cats.  
What I really would like to do is to mechanically obtain as many of the examples from The joy of cats as possible.  [Yes, I know that what that means is that I need to discharge a lot of proof obligations mechanically as well].  This turns out to be significantly harder than it seems (category theorists don't seem to put a lot of effective algorithms in their papers).  So I would like to know what previous work has been done on this, so that I can at least start from there.
EDIT: For example, take this MO question about left adjoints and its nice characterization of when you have a left adjoint for a Functor.  With all of that machinery, how can you 'specialize' all of that to the $\mathrm{CHaus} \rightarrow \mathrm{Top}$ case of the given example and 'extract' Stone-Cech compactification?  Tom Leinster's answer to the same question is also full of examples of exactly what I am after.  As another example, how do I use the Adjoint Functor Theorem so 'extract' an Abelianization procedure from the forgetful Functor from AbGroup to Group, ie how does one get 'construction' given in Andrew Critch's answer to 'drop out' of the abstract-nonsense?

REPLY [3 votes]: I don't know if this answers your question, but here's how I would think about it. Work in a dependent type theory, and suppose we've defined a type of categories, $\mathrm{Cat}$. Suppose we've proved some theorem about categories, $\mathit{Thm}$, with proof term $p\colon\forall C\colon\mathrm{Cat},\mathit{Thm}(C)$. An example of the phenomenon in question would then be to specialize this theorem to a specific category, or to a specialized class of categories. Say we want to specialize to monoids. Then we might have a type of monoids, $\mathrm{Mon}$, and a function $\alpha\colon\mathrm{Mon}\to\mathrm{Cat}$ constructing for every monoid the corresponding one-object category. Then we can specialize $\mathit{Thm}$ to get a new theorem, $\forall M\colon\mathrm{Mon},\mathit{Thm}(\alpha M)$, with proof term $\lambda M\colon\mathrm{Mon}, p(\alpha M)$.
Another way to look at this, it that having a theorem in some context $\Gamma$ corresponds to a display map in the context category, $(\Gamma, p\colon T)\to\Gamma$. Specializing to another context, $\Delta$, is given by a morphism of contexts  $\alpha\colon\Delta\to\Gamma$. Pulling back along $\alpha$ gives you a display map over $\Delta$ with your specialized theorem (obtained by substituting according to $\alpha$).
In any case, the specialization procedure is mechanical, given by some sort of substitution, but then the question arises as to how to best simplify and present the result. Say you've specialized the proposition that every split epimorphism in a category is, in fact, an epimorphism. Specializing to monoids states this result in the language of monoids, but maybe you want afterwards to search for definitions and restate the proposition to be that every monoid element with a right inverse is right cancellable. This was just a simple example, but in general there may of course be many possible simplifications of the specialized terms. You could normalize, but the normalization is rarely the most simple form, especially if everything takes place in a non-empty background context.<|endoftext|>
TITLE: Ring-theoretic characterization of open affines?
QUESTION [47 upvotes]: Background
Recall that, given two commutative rings $A$ and $B$, the set of morphisms of rings $A\to B$ is in bijection with the set of morphisms of schemes $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$.  Furthermore, we know that Spec$(A)$ has a base of open affine subsets, the "basic" or "principal" open affines $D(f)$ for all $f\in A$. Furthermore, $D(f)\cong \mathrm{Spec}(A_f)$ as schemes, and the inclusion $D(f)\hookrightarrow\mathrm{Spec}(A)$ corresponds to the localization map $A\to A_f$.
But answers to a recent MathOverflow question show that open affine subschemes of affine schemes can arise in other ways.

Question
In order to try to make sense of the situation above, I'd like to know the following.

Given a commutative ring $A$, is there a "ring-theoretic" characterization of the ring homomorphisms $A\to B$ that realize $\mathrm{Spec}(B)$ as an open affine subscheme of $\mathrm{Spec}(A)$ (more precisely, those morphisms such that the induced map $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ is an open immersion)?

Of course, "ring-theoretic" is a bit vague. Let's certainly avoid any tautological characterizations.  I would prefer if an answer didn't make any reference to the Zariski topology (for instance, the morphisms $A\to A_f$ make perfect sense without the Zariski topology), but I'm not sure whether that's reasonable.

Update: I received two great answers, thank you both!  I chose the one that was closer to the kind of condition that I had in mind.  But Dan Petersen's answer was also very interesting and unexpected.

REPLY [45 votes]: There is the following characterisation. I don't think it's too tautological. Let $T \subseteq A$ be the set of f such that the induced map $A[f^{-1}] \to B[f^{-1}]$ is an isomorphism. Then $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ is an open immersion if and only if the image of $T$ in $B$ generates the unit ideal.

REPLY [36 votes]: Theorem 1: Let $R$ be an integral domain with field of fractions $K$, and $R \to A$ a homomorphism. Then $Spec(A) \to Spec(R)$ is an open immersion if and only if $A=0$ or $R \to K$ factors through $R \to A$ (i.e. $A$ is birational over $R$) and $A$ is flat and of finite type over $R$.
Proof: Assume $Spec(A) \to Spec(R)$ is an open immersion and $A \neq 0$. It is known that open immersions are flat and of finite type. Thus the same is true vor $R \to A$. Now $R \to K$ is injective, thus also $A \to A \otimes_R K$. In particular, $A \otimes_R K \neq 0$. Open immersions are stable under base change, so that $Spec(A \otimes_R K) \to Spec(K)$ is an open immersion. But since $Spec(K)$ has only one element and $Spec(A \otimes_R K)$ is non-empty, it has to be an isomorphism, i.e. $K \to A \otimes_R K$ is an isomorphism. Now $R \to A \to A \otimes_R K \cong K$ is the desired factorization.
Of course, the converse is not as trivial. It is proven in the paper

Susumu Oda, On finitely generated birational flat extensions of integral domains
  Annales mathématiques Blaise Pascal, 11 no. 1 (2004), p. 35-40

It is available online. In the section "Added in Proof." you can find some theorems concerning the general case without integral domains. In particular, it is remarked that in E.G.A. it is shown that
Theorem 2: $Spec(A) \to Spec(R)$ is an open immersion if and only if $R \to A$ is flat, of finite presentation and an epimorphism in the category of rings.
More generally, in EGA IV, 17.9.1 it is proven that a morphism of schemes is an open immersion if and only if it is flat, a (categorical) monomorphism and locally of finite presentation.
There are several descriptions of epimorphisms of rings (they don't have to be surjective), see this MO-question.<|endoftext|>
TITLE: approximate a probability distribution by moment matching
QUESTION [10 upvotes]: Suppose we want to approximate a real-valued random variable $X$ by a discrete random variable $Z$ with finitely many atoms. Suppose all moments of $X$ is finite. We want to match the moments of $X$ up to the $m^{\rm th}$ order: 
(1)   $\mathbb{E}[X^k] = \mathbb{E}[Z^k]$ for $k = 1, \ldots m$.
Here is a positive result, which is a simple consequence of convex analysis (Caratheodory's theorem): there exists $Z$ with at most $m+1$ atoms such that (1) holds.
Here are my questions:
1) Is there a converse result about this? Say $X$ has an absolutely continuous distribution supported on $\mathbb{R}$ (e.g. Gaussian). When $m$ is large, given that $Z$ has only $m$ atoms, can we conclude that we cannot approximate all $2m$ moments of $X$ well, i.e., can we lower bound the error
$\max_{1 \leq k \leq 2m}|\mathbb{E}[X^k] - \mathbb{E}[Z^k]|$? My intuition is the following: for a Gaussian $X$, $\mathbb{E}[X^k]$ grows like $k^{\frac{k}{2}}$ superexponentially. When we find a $Z$ who matches all moments of $X$ up to $m$, it cannot catch up with higher-order moments $X$; if $Z$ matches all moments from $m+1$ up to $2m$, then its low-order moments will be quite different from $X$.
2) Is there an efficient algorithm to compute the location and weights of the approximating discrete distribution? Does there exist a table to record these for approximating common distribution (e.g. Gaussian) for each fixed $m$? It could be very handy...   
3) I heard from folklore that when (1) holds, the total variation distance between their distributions can be upper bounded by, say, $e^{-m}$ or $1/m!$. Of course, this won't be true for a discrete $Z$. But let's say $X$ and $Z$ both has smooth and bounded density on $\mathbb{R}$. Could this be true? Now two characteristic functions matches at $0$ up to $m^{\rm th}$ derivatives. They should be pretty close?

REPLY [3 votes]: In the context of 3), what I have heard from folklore is that when (1) holds, the Kolmogorov distance (not total variation) is bounded by something like $1/\sqrt{m}$.  This bound follows if (1) holds only approximately, and exact equality in (1) suggests that a much stronger bound holds but does not formally imply it, even under the assumption of smooth bounded densities.
See the introduction of this paper, and observe that a lot more work is necessary to prove the main results.<|endoftext|>
TITLE: isomorphism of abelian varieties
QUESTION [12 upvotes]: Let $A, B, C$ and $D$ be abelian varieties (over $\mathbb{C}$) such that $A \times B \cong C \times D$, and $A \cong C$.  From the irreducibility of abelian varieties, we can say that $B$ and $D$ are isogeneous.  But do we actually have $B \cong D$?

REPLY [20 votes]: This is false even for elliptic curves over $\mathbb{C}$. This was proved by T. Shioda in "Some remarks on abelian varieties"  J. Fac. Sci. Univ. Tokyo Sect. IA Math.  24  (1977), no. 1, 11-21, http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6164/1/jfs240102.pdf.<|endoftext|>
TITLE:  Frobenius - Schur indicator and irreducible representations over R
QUESTION [5 upvotes]: from now on let $\mathbb{C}$ denote the complex number field, $G$ a finite group , and 
$T$ be a irreducible representation of G over $\mathbb{C}$ whose character is $\chi$.
Frobenius - Schur indicator is given by 
$ \mu = \frac{1}{|G|} \sum_{g\in G} \chi(g^2) $  
if $\mu$ =1 ,then $\chi$ can be realized over the real number field $\mathbb{R}$.
if $\mu$ =0 , then $\chi$ is not real.
if $\mu$ =-1 , then $\chi$ is real, but $T$ cannot be realized over $\mathbb{R}$.
also we know there are three different division algebra (finite dimension) over $\mathbb{R}$ : $\mathbb{R}$ , $\mathbb{C}$ and $\mathbb{H}$ (Hamilton quternions) , so the ring of endomorphisms commuting with the group action can be isomorphic only to either the real numbers, or the complex numbers, or the quaternions.
So I believe there are some connections between the three possible values of $\mu$ 
and the three kind of division algebra over $\mathbb{R}$. But I do not know why. Can anyone explain the connections between them? 
additionally, these three cases corresponde to where there is a symmetrc/skew symmetric $G$
-invariant bilinear form in V . Though I can go through the proofs, but I don't konw why the indicator can be controled by the existence of a bilinear form of a particular kind.
Is there something essential lying behind?
So far we have 
indicator--bilinear form--division algebra.
Also, for a given field $F$, and a division algebra $D$ over $F$, can we assign $D$ a “number invariant” $\mu_D$ such that $\mu_D$ completely determines $D$ (under isomorphism) ?

REPLY [5 votes]: I've always thought of this formula as an amusing accident. You say "I can go through the proofs", so I don't know if what I'll say helps any more than that, but here goes.

If $M$ is any endomorphism, it's easy to show $Trace(Sym^2(M)) = (Trace(M)^2 + Trace(M^2))/2$. Proof: check it for diagonal $M$, obtain for diagonalizable, extend by continuity to all. The same technique gives $Trace(Alt^2(M)) = (Trace(M)^2-Trace(M^2))/2$.
Basic character theory fact: the average of $Trace(g|_V)$ is $\dim V^G$.

If $V$ is a real representation, and $G$ is compact (yours being finite, apparently), then we can average a bilinear positive definite symmetric inner product to get a $G$-invariant one, which gives an isomorphism $V \equiv V^*$. That passes to the complexification $V_{\mathbb C}$. Hence there is an invariant vector (the inner product) in $ S ym^2(V)$.
If $V$ is a quaternionic representation, and $G$ is compact, then I admit I forget how to get an antisymmetric form on the complex representation $Forget(V)$. Maybe one can average a quaternionic-Hermitian form to get a $G$-invariant one?
Moreover one wants the converses of these: having the symmetric or antisymmetric inner products lets one realize a complex irrep as a complexification or $Forget()$ of a quaternionic rep. I don't remember this being too hard.
Then come a couple of minor miracles. Since $V$ is irreducible, Schur's lemma says there is at most a $1$-d space of invariant maps $V \to V^*$, i.e. $G$-invariant vectors in $V \otimes V$.  Since $V\otimes V$ is a $G\times Z_2$-rep, any $G$-invariant is either in $Sym^2(V)$ or $Alt^2(V)$, not some weird mix. Hence there are only three cases: no invariant at all, $\dim  (Sym^2(V))^G = 1$ and $\dim (Alt^2(V))^G = 0$, or vice versa. We can figure out which occurs by looking at $\dim (Sym^2(V))^G - \dim (Alt^2(V))^G = 0,1,-1$. By facts 1 & 2 above, that's the F-S indicator.<|endoftext|>
TITLE: Finite morphisms between algebraic varieties are flat?
QUESTION [7 upvotes]: Let $f: X\to Y$ be a finite (surjective) morphism between two algebraic varieties. I know when $X$ and $Y$ are non-singular and $\dim Y =1$, $f$ is flat. But in general, is it true that $f$ is a flat morphism?

REPLY [14 votes]: If $X$ and $Y$ are both regular, then this is true. In fact, it's true more generally if $Y$ is regular and $X$ is Cohen-Macaulay (Eisenbud, Commutative Algebra, Corollary 18.17). In general it's certainly false.<|endoftext|>
TITLE: When is the set of zero divisors equal to the union of the minimal primes in a reduced ring?
QUESTION [13 upvotes]: It is straightforward to show, for example, that the set of zero divisors of a (commutative unital) reduced Noetherian ring is precisely the union of its minimal primes.  When else can we say that the set of zero divisors is equal to the union of the minimal primes?  Are there other useful cases where this is true?  Is there a structure theory for such rings?  I'm primarily looking for conditions that do not assume that the ring is Noetherian.
Edit: I messed up this question and received a correct answer for the wrong assumptions, so I have accepted the answer.  The reducedness assumption was for the example, not for the general question.  
The correct question is in there, but I accidentally changed the title, so I apologize for the confusion.

REPLY [22 votes]: The answer is: always, and argument is pretty simple:
Let $R$ be a reduced commutative unital ring.
If $a\in R$ is a zero divisor, then $ab=0,$ for some $b\neq 0.$ Hence
$b\not \in 0 = \text{nil}(R)= \bigcap \text{Spec}(R) =\bigcap \text{Specmin}(R),$ where $\text{Specmin}(R)$ states for family of all minimal prime ideals of $R.$
So there exists a minimal prime ideal $\mathfrak{p}\triangleleft R$ s.t. $b\not \in \mathfrak{p}.$ But $\mathfrak{p}$ is prime, and $ab\in \mathfrak{p},$ thus $a\in \mathfrak{p}.$
Hence the set of zero divisors is contained in union of minimal prime ideals. 
On the other hand it is well-known fact that minimal prime ideals in commutative unital rings consist of zero divisors, so the set of zero divisors in reduced commutative unital ring is exactly the union of minimal prime ideals.<|endoftext|>
TITLE: Are fundamental groups of aspherical manifolds Hopfian?
QUESTION [25 upvotes]: A group $G$ is Hopfian if every epimorphism $G\to G$ is an isomorphism. A smooth manifold is aspherical if its universal cover is contractible. Are all fundamental groups of aspherical closed smooth manifolds Hopfian?
Perhaps the manifold structure is irrelevant and makes examples harder to construct, so here is another variant that may be more sensible. Let $X$ be a finite CW-complex which is $K(\pi,1)$. If it helps, assume that its top homology is nontrivial. Is $\pi=\pi_1(X)$ Hopfian?
Motivation. Long ago I proved a theorem which is completely useless but sounds very nice: if a manifold $M$ has certain homotopy property, then the Riemannian volume, as a function of a Riemannian metric on $M$, is lower semi-continuous in the Gromov-Hausdorff topology. (And before you laugh at this conclusion, let me mention that it fails for $M=S^3$.)
The required homotopy property is the following: every continuous map $f:M\to M$ which induces an epimorphism of the fundamental group has nonzero (geometric) degree.
This does not sound that nice, and I tried to prove that some known classes of manifolds satisfy it. My best hope was that all essential (as in Gromov's "Filling Riemannian manifolds") manifolds do. I could not neither prove nor disprove this and the best approximation was that having a nonzero-degree map $M\to T^n$ or $M\to RP^n$ is sufficient. I never returned to the problem again but it is still interesting to me.
An affirmative answer to the title question would solve the problem for aspherical manifolds. A negative one would not, and in this case the next question may help (although it is probably stupid because I know nothing about the area):
Question 2. Let $G$ be a finitely presented group and $f:G\to G$ an epimorphism. It it true that $f$ induces epimorphism in (co)homology (over $\mathbb Z$, $\mathbb Q$ or $\mathbb Z/2$)?

REPLY [12 votes]: A couple of remarks on this question: 
It is true for closed aspherical manifolds of dim $\leq 3$, since they have residually finite fundamental group. 
If one could find an aspherical 4-manifold $W$ with boundary, such that $\pi_1 W$ is non-Hopfian,
and a degree >1 map $f:W\to W$ which induces a self-covering $f: \partial W \to \partial W$,
then one might be able to use a reflection trick of Davis to find a closed aspherical
manifold with the same property. Since there exist 3-manifolds which are not co-Hopfian
(in fact, which are finite-sheeted covers of themselves), this approach might work.
I attempted this approach by thickening up the Baumslag-Solitar group presentation complex to a 4-manifold with boundary, but the boundary wasn't co-Hopfian, so I couldn't get the self-mapping of the 2-skeleton to extend.<|endoftext|>
TITLE: Estimate for the order of the outer automorphism group of a finite simple group
QUESTION [17 upvotes]: It is known (given CFSG) that all non-abelian finite simple groups have small outer automorphism groups.  However, it's quite tedious to list all the possibilities.  Does anyone know a reference for a statement of the following form?
Let G be a non-abelian finite simple group.  Then |Out(G)| < f(|G|)  (where f is something straightforward that gives a good idea asymptotically of the worst case).
Also, a related question: How good a bound can be obtained in this case without using the classification?  Can one do much better than the bounds one might obtain for the outer automorphism group of an arbitrary finite group?

REPLY [3 votes]: It is not hard to prove $|\mathrm{Out}(T)|\leqslant \log_p|T|$ when $T=L(q)$ is a simple group of Lie type of characteristic $p$. (One uses formulas for $|\mathrm{Out}(T)|=dfg$ and for $|T|$ for different Chevalley groups. The proof is slightly tedious but straightforward.) The bound $|\mathrm{Out}(T)|\leqslant\lfloor \log_8(\frac{16|T|}{15})\rfloor$ stated above (although sharper when $p=2$) has exceptions such as
$\textsf{A}_1(9)$, $^2\textsf{A}_2(5)$, $^2\textsf{A}_2(8)$. For $T=\textsf{A}_n(q), {}^2\textsf{A}_n(q)$ one can show $|\mathrm{Out}(T)|=\frac{2\log_p|T|}{n}$ is true.<|endoftext|>
TITLE: Involutions of $S^2$
QUESTION [17 upvotes]: are there some complete results on the involutions of 2 sphere?
at least I have three involutions:
(let $\mathbb{Z}_2=\{1,g\}$,and $S^2=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=1\}$)
1.$g(x,y,z)=(-x,-y,-z)$(antipodal map) with null fixed point set,and orbit space $\mathbb{R}P^2$
actully,for free involution on $S^n$ with $n\leq3$,the orbit space is homeomorphic to real projective space (Livesay 1960)
2.$g(x,y,z)=(-x,-y,z)$ (rotation $\pi$ rad around $z$ axis)  with fixed point set $S^0$(the north pole and south pole) and orbit space $S^2$.
3.$g(x,y,z)=(x,y,-z)$(reflection along $z=0$) with fixed point set $S^1$ (the equator)and orbit space $D^2$
i want to know if there are some other involutions over 2-sphere.
here we take two involutions as equivalent if there are conjugate in the homeomorphism group of $S^2$

REPLY [3 votes]: Yes, your three examples are the only three up to conjugation by a homeomorphism.  In the orientation-preserving case, this is a theorem of Kerekjarto (Sur les groupes compacts de transformations topologiques des surfaces, 1941). A streamlined modern proof, including the orientation-reversing case, is offered by Boris Kolev (Note sur les sous-groupes compacts d'homeomorphismes de la sphere, 2006). Both of these references are in French.
From what I gather in skimming Kolev (but not poring through the proofs), the outline of the proof has two parts.  First, one shows that each fixed point has an invariant disc neighborhood.  Then, one uses the fact that any involution of a 2-disc is conjugate to the cone of an involution on a circle, the latter of which must conjugate to a linear one.<|endoftext|>
TITLE: Is there a necessary and sufficient condition for the tangent bundle of a fiber bundle to be trivial?
QUESTION [10 upvotes]: My question is very basic (I don't know too much of differential geometry):
given a fiber bundle, is there a necessary and sufficient condition for its tangent bundle to be trivial?
I have some ideas, but submitted to some conditions on the cohomology ring of the bundle.
(I apologize if it is trivial.)

REPLY [10 votes]: I just wanted to elaborate on Benoît Kloeckner's answer, so if you like what I say, please upvote his answer.
By a frame, I mean a basis of the tangent space at a point on a smooth
manifold $M$. The space $F$ of all possible frames, called the frame
bundle, is a principal $GL(n)$-bundle over the manifold, $n$ is the
dimension of the manifold. A point in $F$ is given by $(x, e)$, where
$x \in M$, $e = (e_1, \dots, e_n)$, and $e_i \in T_xM$. Associated
with each point is the dual frame $\omega^1, \dots, \omega^n \in
T_x^*M$. Let $\pi: F \rightarrow M$, $\pi(x,e) = x$, denote the
natural projection.
There is a natural set of $n$ $1$-forms $\hat\omega^1, \dots,
\hat\omega^n$ on $F$, which are called either "tautological" or
"semi-basic" and act as follows: If $v \in T_{(x,e)}F$, then
$
\langle \hat\omega^i,v\rangle = \langle\omega^i,\pi_* v \rangle,
$
where $\omega^1, \dots, \omega^n \in T^*_xM$ form a dual basis to the basis
$e_1, \dots, e_n \in T_xM$. These forms have the universal property
that given any section $s: M \rightarrow F$, $s^*\bar\omega^i$ are
$1$-forms on $M$ dual to the moving frame given by the $e_i$.
You can check that any connection $\nabla$ on $T_*M$ determines a set
of global $1$-forms $\hat\omega^i_j$ on $F$, such given any section
$s = (s_1, \dots, s_n): M \rightarrow F$, $\nabla s_j =
s_is^*\hat\omega^i_j$. Therefore, a connection on $F$ gives a set of global
$1$-forms $\hat\omega^1, \dots, \hat\omega^n, \hat\omega^1_1, \dots, \hat\omega^n_n$
that trivialize $T^*F$. The dual vector fields
trivialize $T_*F$.
Since there always exists a connection on $T_*M $, this shows that $F$
has a parallelizable tangent bundle. The same argument can be extended
to any principal $G$-bundle of tangent frames. As observed by
Hoeckner, the case $G = O(n)$ corresponds to a Riemannian structure.
This, of course, does not answer the original question, but it is a
important case where the answer is yes. These global $1$-forms are
extremely useful in many contexts; the work of Robert Bryant
illustrates this.<|endoftext|>
TITLE: Why are the Dynkin diagrams E6, E7 and E8 always drawn the way they are drawn?
QUESTION [5 upvotes]: The Dynkin diagrams of type ADE are ubiquitous in mathematics as solutions of various classification problems. The diagram E6 is usually drawn by five dots in a row with a sixth dot above the third, see for example here. There would be many other ways to draw   the diagram E6, for example the sixth dot below the five dots, or just a capital 
E.
Is there a reason for drawing that diagram in that particular shape beside or is that just a confirmed habit?

REPLY [6 votes]: The question as stated is not really helpful, but it's worth pointing out that the ADE
and other graphs/diagrams evolved over a couple of decades in different countries.   The
graphs, which encode at first the Coxeter data for finite (or more general) reflection
groups, go back at least to Coxeter's 1934 paper.   In Witt's 1941 paper these now
familiar graphs reappear when the reflection groups are unified with the classification
of root systems for semisimple Lie algebras over $\mathbb{C}$.  Here the vertices correspond to simple roots, not just simple reflections, so the notation has to incorporate some length
information.  Dynkin's fundamental 1952 papers used the resulting "Dynkin diagrams" with
extra labels 0, 1, 2 at vertices to classify efficiently the nilpotent orbits.    Along the
way a number of different choices were made about adding edges and arrows or making some
vertices darker to distinguish lengths.    Bourbaki's 1968 treatise is by now the easiest
standard source to follow.
The graph itself is drawn in a typographically convenient way and can vary as Scott notes
(note especially Coxeter's type E pictures).    An even more arbitrary convention governs
the numbering of vertices or simple roots.   Here again the evolved version in Bourbaki is
well established, though some authors like Carter depart a bit from that numbering.    The
history of ideas is quite interesting, but at the end of the day a convention is just a
convention.<|endoftext|>
TITLE: Twin categories in representation of Lie algebra
QUESTION [9 upvotes]: Let $\mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n^+}$ be a triangular decomposition of semisimple Lie algebra. Let $\mathcal{Z}$ be the central of universal envoloping Lie algebra of $\mathfrak{g}$.
Let $\mathcal{C}$ be the category of representations of $\mathfrak{g}$, on which $\mathfrak{n ^+}$ and $\mathfrak{h}$ acts locally finite, and $\mathcal{Z}$ acts semisimplely. 
Let $\mathcal{D}$ be the another category of representations of $\mathfrak{g}$, on which $\mathfrak{n ^+}$ and $\mathcal{Z}$ acts also locally finitely, and $\mathfrak{h}$ acts semisimply. If I don't make mistake, $\mathcal{D}$ should be called category $\mathcal{O}$.
Claim. $\mathcal{C}$ is equivalent to $\mathcal{D}$.
Do I formulate the problem correctly? 
About the proof of this theorem, where is it written?

REPLY [5 votes]: Following David's suggestion, I'll point out that this is a theorem of Soergel.  That paper is in French, but an English account of a related (and more general) construction is given in the paper of Soergel and Milicic.  While David's explanation is quite nice, the references below show this isn't an intrinsically geometrical result; Soergel gives completely algebraic proofs.<|endoftext|>
TITLE: Is the functor of open subschemes representable?
QUESTION [6 upvotes]: First some simple observations in order to motivate the question:
The functor $Set^{op} \to Set, X \to \{\text{subsets of }X\}, f \to (U \to f^{-1}(U))$ is representable. The representing object is $\{0,1\}$ with the universal subset $\{0\}$. Also the functor $Top^{op} \to Set, X \to \{\text{open subsets of }X\}$ is representable: Endow $\{0,1\}$ with the topology such that $\{0\}$ is the unique nontrivial open subset (Sierpinski space), then it is again the representing object.
But what about schemes. Is the functor $Sch^{op} \to Set, X \to \{\text{open subschemes of }X\}$ representable? Of course, we could also talk about open subsets of $X$. My first idea was to endow the Sierpinski space above with a scheme structure, using DVR, but this does not work properly.

REPLY [10 votes]: It is representable by an fpqc sheaf (SGA I.VIII.4.4).  Also, since the inverse limit of affine schemes has the inverse limit topology (EGA IV.8), the sheaf is locally of finite presentation.  Since the open subsets of an infinitesimal thickening of a scheme are the same as those of the original scheme, this functor is even formally étale over $\mathrm{Spec} \:\mathbf{Z}$.  It's therefore just about as close as it could be to being an algebraic space without actually being one.
What goes wrong is algebraization.  If $X$ is an algebraic space and you have a complete local ring $R$ then $\mathrm{Hom}(\mathrm{Spec}\:R,X) = \varprojlim_n  \mathrm{Hom}(\mathrm{Spec}\:R_n,X)$ where the $R_n$ are the quotients of $R$ by powers of the maximal ideal.  It's pretty clear that this equality doesn't hold for the sheaf of open subsets.<|endoftext|>
TITLE: Formal-group interpretation for Lin's theorem?
QUESTION [22 upvotes]: Background
For compact Lie groups, Atiyah and Segal proved a strong relationship between Borel-equivariant K-theory, defined in terms of the K-theory of $X \times_G EG$, and the equivariant K-theory of X defined in terms of equivariant vector bundles.  Roughly, for "nice" spaces X the K-theory of $X \times_G EG$ is a completion of the equivariant K-theory of X, and in particular the K-theory of BG is a completion of the complex representation ring of G.
The Segal conjecture is an analogous result proven in subsequent years (by many authors, with Carlsson completing the proof).  It's less well-known outside the subject, and obtained by roughly replacing "vector bundles" with "covering spaces" - the original conjecture is that for a finite group $G$, the abelian group of stable classes of maps $\varinjlim[S^n \wedge BG, S^n]$ has as limit to the Burnside ring of finite $G$-sets.  There are further statements describing $\varinjlim [S^{n+k} \wedge BG, S^n]$ in terms of a completion of certain equivariant stable homotopy groups.  It's notable for the fact that it's not really a computational result - we describe two objects as being isomorphic, without any knowledge of what the resulting groups on either side really are.
There are a number of steps in this proof, and over the years most of them have been recast and reinterpreted in a number of ways.  However, the initial steps in the proof are computational.  Lin proved this conjecture for the case $G = \mathbb{Z}/2$, and Gunawardena proved it for the case $G = \mathbb{Z}/p$ for odd primes $p$.  Lin's original proof involved some very difficult computations in the Lambda algebra and a simplified proof was ultimately written up by Lin-Davis-Mahowald-Adams.  It amounts to a computation of certain Ext or Tor groups over the Steenrod algebra - namely, if $H^* \mathbb{RP}^\infty = \mathbb{Z}/2[x]$ has its standard module structure over the Steenrod algebra, then $Ext^{**}(\mathbb{Z}/2[x^{\pm 1}],\mathbb{Z}/2)$ degenerates down to a single nonzero group.
Bordism theory
A lot of the contemporary work in stable homotopy theory uses the relationship between stable homotopy theory and the moduli of formal groups, rather than the Adams-spectral-sequence calculations that are used in the above proofs.  The analogous calculation would be the following.
Let L be the Lazard ring carrying the universal formal group law, with 2-series $[2](t)$ whose zeros are the "2-torsion" of the formal group law.  Then there is an L-algebra
$$
Q = t^{-1} L[[t]]/[2](t)
$$
whose functor of points would be described (up to completion issues) as taking a ring R to the set of formal group laws on R equipped with a nowhere-zero 2-torsion point.  This comes equipped with natural descent data for change-of-coordinates on the formal group law, and so it describes a sheaf on the moduli stack of formal group laws $\mathcal{M}$.
A student of Doug Ravenel's (Binhua Mao) proved in his thesis that the analogous Ext-computation is valid in the formal-group setting: namely, if one computes the Ext-groups 
$$Ext_{\mathcal M}(\mathcal{O}, Q \otimes \omega^{\otimes t})$$
where $\omega$ is the sheaf of invariant 1-forms on $\mathcal{M}$, it converges to a completion of 
$$Ext_{\mathcal M}({\mathcal O}, \omega^{\otimes t}).$$
(The result was stated in different language, and I am still ignoring completion issues.)
However, as I understand the proof (and I don't claim that I really do!) it essentially uses a reduction to the Adams spectral sequence case by using a filtration that reduces to the group scheme of automorphisms of the additive formal group law, and this is very closely connected to the Steenrod algebra.  I would regard it as still being computationally focused, and I don't really have a grip on why one might expect it to be true without carrying the motivation from homotopy theory all the way through.
Question (finally)
Is there is a more conceptual interpretation of this computation in terms of the geometry of the moduli of formal groups?

REPLY [9 votes]: I can't really answer this.   I'll just think out loud for a bit.
Let $R$ be a complete local $\mathbb{F}_p$ algebra.  The additive formal group $G$ is a formal scheme $Spf(\mathbb{F}_p[[x]])$.  An $R$-point of $G(R)$ is an element of $\mathfrak{m}_R$.
Pick $t\in G(R)=\mathfrak{m}_R$, and consider $f(x)=x(x-t)(x-2t)\cdots (x-(p-1)t)=x^p-tx$.  Let $F_t=Spec(R[[x]]/(x^p-tx))=Spec(R[x]/(x^p-tx))$.  Then $F_t$ is a finite subgroup scheme of $G$.  If we base change $F_t$ to $\tilde{F}_t$ over $R[t^{-1}]$, then $\tilde{F}_t$ becomes an etale group scheme.
The universal example of such an $F_t$ lives over $B=\mathbb{F}_p[[t]]$.  The scheme $S$ of automorphisms of $G$ (i.e., the dual Steenrod algebra) acts on $B$, and the action lifts to $B[t^{-1}]=\mathbb{F}_p((t))$.  If $\omega$ is the module of invariant differentials on $G$ (isomorphic as a module with $S$-action to $tB/t^2B$, then there's a map
$$Res_{t=0}: B[t^{-1}] \otimes_{\mathbb{F}_p} \omega\to \mathbb{F}_p,$$
which is a map of $S$-modules.  Lin's theorem asserts that this map induces isomorphisms in $Ext_S^*(\omega^i,{-})$.
So Lin's theorem is something about residues.   
You have this residue map is other cases, for instance if we replace $G$ with a Lubin-Tate deformation.  Neil Strickland has thought about this: in his Formal Schemes and Formal Groups, he spells out some of the relationship between the residues and the Segal conjecture.<|endoftext|>
TITLE: Derangements with repetition
QUESTION [6 upvotes]: Hi all,
given (a1,...,an) formed by distinct letters, it's a well known problem to count the number of permutations with no fixed element.
I've been trying to solve a generalization of this problem, when we allow repetition of the letters.
I was able only to partially solve the problem when we have only repetition of a single letter.
If we have n letters and only one of them is repeated p times, then the number O(n,p) of permutations with no fixed element is given by the following recursive relation:
$O(n,0)=O(n,1)=\mbox{Derangement}(n)$
$O(p+1,p)=\dots=O(2p-1,p)=0, O(2p,p)=p!$
$O(n,p)={n-p\choose p} p!\sum_{k=0}^p{p \choose k}O(n-p-k,p-k)$
Does anybody know where this problem have been studied before? Does anybody know a general solution for this problem?
Thanks in advance.

REPLY [16 votes]: The formula based on Inclusion-Exclusion for the usual number $D(n)$ of derangements of $n$ objects can be generalized. The result is the following.
Fix $k\geq 1$. Let $\mathbb{N}=\lbrace 0,1,2,\dots\rbrace$. For $\alpha=(\alpha_1,\dots,\alpha_k)\in\mathbb{N}^k$, let $D(\alpha)$ be the number of fixed-point free permutations of the multiset with $\alpha_1$ 1's, $\alpha_2$ 2's, etc. Write $x^\alpha = x_1^{\alpha_1}\cdots x_k^{\alpha_k}$. Then
  $$ \sum_{\alpha\in\mathbb{N}^k} D(\alpha)x^\alpha =
       \frac{1}{(1+x_1)\cdots (1+x_k)\left(1-\frac{x_1}{1+x_1}-\cdots -
         \frac{x_k}{1+x_k}\right)} $$
  $$  = \frac{1}{1-\sum_S
        (|S|-1)\prod_{i\in S}x_i}, $$ 
where $S$ ranges over all nonempty subsets of $\lbrace 1,2,\dots,n\rbrace$. 
This result appears as Exercise 4.5.5 in Goulden and Jackson, Combinatorial Enumeration. It can be used to obtain a lot of information about $D(\alpha)$.<|endoftext|>
TITLE: Rational homotopy theory of a punctured manifold
QUESTION [13 upvotes]: Let $M$ be a smooth simply connected manifold and let $N$ be $M$ minus a point. Is it possible to construct an explicit Sullivan model for $N$ (i.e. a commutative differential graded algebra (cdga) which is connected to the algebra of $\mathbf{Q}$-polynomial forms on $N$ by a chain of cdga quasi-isomorphisms) starting from the minimal Sullivan model for $M$?
[upd: in principle the above question is a very particular case of the one discussed in the paper Algebraic models of Poincar\'e embeddings by P. Lambrechts and D. Stanley, AGT 5, 2005. That paper discusses general polyhedra that satisfy some connectivity/codimension assumptions, which are certainly true when the polyhedron is a point. But the general construction involves some non-canonical choices and I was wondering if there is a cleaner and more ``canonical'' construction when the polyhedron to be thrown away is simply a point.]

REPLY [8 votes]: This is an old question but  I hope the following is still of interest.
If $M^n$ is a closed simply connected manifold then the inclusion $M^n\backslash \{ pt\}\hookrightarrow M^n$ is $(n-1)$-connected which means that the induced map of minimal models is an isomorphism through dimension $(n-2)$. Next note that  $M^n\backslash \{ pt\}$ has zero homology in degrees above $n-2$. It's a general fact that given a minimal model up to dimension $k$ of a space whose cohomology vanishes in degrees above $k$ the rest of the minimal model is determined uniquely (and constructively) from the model up to degree $k$.  This provides an easy recipe for computing the minimal model of 
$M^n\backslash \{ pt\}$ which can be more explicitly described as follows.
If $(\Lambda V, d)$ is a minimal model of $M^n$ then consider the following dga $(A,d)=(\Lambda V\oplus\Lambda \langle z\rangle/(z^2), d)$ with $\deg z=n-1, V\cdot z=0$ and $dz=[M]$ - the fundamental class of $M$. This is a model (non-minimal and even a non-free one!) of $M^n\backslash \{ pt\}$.   In practice it's easier to directly compute the minimal model of $A$ by the general procedure outlined above.
Here are a couple of examples.
Let $M^4=\mathbb {CP}^2$. Its minimal model is $(\Lambda \langle x,y\rangle,d)$ with deg x=2, deg y=5, dx=0, dy=x^3. Up to degree $n-2=2$ this is simply given by $\Lambda\langle x\rangle$ with dx=0. Next we need another generator to make $H^4=0$ (which is currently generated by $[M]=x^2$) so we add $z$ of deg $3$ such that $dz=x^2$. Now the model  $(\Lambda \langle x,z\rangle,d)$ with deg x=2, deg z=3, dx=0, dy=x^2 already has $H^i=0$ for $i\ge 4$ so we don't need to add anything else. The resulting model is easily recognized as the model of $\mathbb S^2$ which is of course not surprising since $\mathbb{CP}^2\backslash\{pt\}$ is a Hopf disk bundle over $\mathbb{CP}^1$.
A more interesting example: Let $M=\mathbb S^3\times\mathbb S^5$. Its minimal model is generated by $x,y$ with $\deg x=3, \deg y=5$ and $dx=0,dy=0$. Applying our recipe the model of  $\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$  will be the same through dimension 6. Next, 
we need to kill off cohomology in degree 8 which is currently generated by $[M]=xy$. So we need another generator $z$ of degree 7 with $dz=xy$. However, adding such generator introduces more cohomology in degrees 10 and 12 generated by $xz$ and $yz$. So we need two more generators $a$ and $b$ with $da=xz, db=yz$. However, adding those introduces yet more cohomology and we need to keep adding more generators. This will continue forever because  $\mathbb S^3\times\mathbb S^5\backslash \{ pt\}$ is rationally hyperbolic.
Lastly, let me mention that operations such as cell attachments (or in this case cell deletions) are usually easier handled by Quillen Lie algebra models which are better suited to work with cofibrations (while Sullivan models are better suited for fibrations). In this particular case it's especially easy. If $(\mathbb L_V,d)$ is a minimal Quillen Lie model of $M^n$ then the model of  $M^n\backslash \{ pt\}$ is obtained by simply removing a single generator from $V$ corresponding to the fundamental class of $M$.<|endoftext|>
TITLE: When can we cancel vector bundles from tensor products?
QUESTION [30 upvotes]: Let $E,F,G$ be algebraic vector bundles over $\mathbb P_{\mathbb C}^n$. My general question is:
Assume $E\otimes G \cong F\otimes G$, under what conditions can one conclude that $E\cong F$? 
Some easy answers (if I am not mistaken): one can when $n=1$ or  when $G$ is a line bundle. At this point I am mostly interested in the case when $E$ is a direct sum of line bundles, but any comments/reference/solutions/analogues about other cases would be appreciated.

REPLY [14 votes]: According to this preprint, over a connected proper algebraic variety $X$ there is a universal reductive group $G$ such that isomorphism classes of vector bundles of rank $n$ are in bijection with isomorphism classes of $n$-dimensional representations of $G$. Furthermore the component group is $\pi_1(X)$.
So the cancellation property you seek holds for any simply-connected proper algebraic variety, because it holds for representations of any connected reductive group. We can see this by characters - a representation is uniquely determined by its character, and because the group is connected, there are no zero-divisors in the ring of characters.
This certainly includes $\mathbb P^n$.<|endoftext|>
TITLE: Waring's problem for matrices
QUESTION [9 upvotes]: Probably a well-know question, but I haven't solved it, so I'll ask.
I can show that every matrix in $M_2(\mathbb{R})$ is the sum of two squares of matrices in $M_2(\mathbb{R})$.
If $n>2$, I can also show that every matrix in $M_n(\mathbb{R})$ is the sum of three squares of matrices in $M_n(\mathbb{R})$.
So my question is : Is every matrix in $M_n(\mathbb{R})$ is the sum of two squares of matrices in $M_n(\mathbb{R})$ (n>2)?

REPLY [14 votes]: The answer is YES if $n$ is even.  But if $n$ is odd, then the answer is NO since $-I$ is not a sum of two squares.
See 
Griffin and Krusemeyer, Matrices as sums of squares, Linear and Multilinear Algebra 5 (1977/78), no. 1, 33-44
for the proofs of these facts and generalizations.<|endoftext|>
TITLE: What is the difference between homology and cohomology?
QUESTION [46 upvotes]: In intuitive terms, what is the main difference? We know that homology is essentially the number of $n$-cycles that are not $n$-boundaries in some simplicial complex $X$. This is, more or less, the number of holes in the complex. But what is the geometrical interpretation of cohomology?

REPLY [8 votes]: On a closed, oriented manifold, homology and cohomology are represented by similar objects, but their variance is different and there is an important change in degrees.  For simplicity, consider homology or cohomology classes represented by submanifolds.  Then if $f : M \to N$ is a smooth map between manifolds of dimension $m$ and $n$ respectively and $W$ is a submanifold of $M$ representing a homology class then $f(M)$ represented (really, $f_*([M])$ is) a homology class of $N$, in the same dimension.  On the other hand, if $V$ is a submanifold of $N$, then we can consider $f^{-1}(V)$, which is a manifold if $f$ is transverse to $F$.  Its codimension is the same as that of $V$ (that is, $n - dim(V) = m - dim(f^{-1}(V))$).   
Note that this preimage is generically at least as "nice" as $V$ (smooth is $V$ is, with reasonable singularities if $V$ has, etc.) whereas little can be said about the geometry of $f(W)$.  That's one reason I think that cohomology is of more use in algebraic geometry.
If you want to relate this view of cohomology to the standard one, a codimension $d$ submanifold of $M$ (that is, one of dimension $m-d$) generically intersects a dimension $d$ one in a finite number of points which can be counted with signs.  The former defines a class in $H^d(M)$ while the latter a class in $H_d(M)$, and this intersection count is evaluation of cohomology on homology.  
This point of view is more applicable than it might seem since in a manifold with boundary cohomology classes are similarly defined by submanifolds whose boundary lies on the boundary of the ambient manifold.  Since any finite CW complex is homotopy equivalent to a manifold with boundary, one can view cohomology in this way for finite CW complexes and often infinite ones as well.<|endoftext|>
TITLE: Most 'unintuitive' application of the Axiom of Choice?
QUESTION [85 upvotes]: It is well-known that the axiom of choice is equivalent to many other assumptions, such as the well-ordering principle, Tychonoff's theorem, and the fact that every vector space has a basis.  Even though all these formulations are equivalent, I have heard many people say that they 'believe' the axiom of choice, but they don't 'believe' the well-ordering principle.  
So, my question is what do you consider to be the most unintuitive application of choice?
Here is the sort of answer that I have in mind.  
An infinite number of people are about to play the following game.  In a moment, they will go into a room and each put on a different hat.  On each hat there will be a real number.  Each player will be able to see the real numbers on all the hats (except their own). After all the hats are placed on, the players have to simultaneously shout out what real number they think is on their own hat.  The players win if only a finite number of them guess incorrectly.  Otherwise, they are all executed. They are not allowed to communicate once they enter the room, but beforehand they are allowed to talk and come up with a strategy (with infinite resources).
The very unintuitive fact is that the players have a strategy whereby they can always win.  Indeed, it is hard to come up with a strategy where at least one player is guaranteed to answer correctly, let alone a co-finite set.  Hint: the solution uses the axiom of choice.

REPLY [4 votes]: I know a (perhaps even more counter-intuitive) "game" similar to the one presented in the question. There are 100 people and a room with countable many boxes (numbered by naturals). There is a real number in each box. These 100 people can prepare a strategy and then they are separately going to the room. When one comes to the room he begins to open some boxes. He is allowed to (for example) open infinite number of boxes, then decide which box will be opened next. But on the end there have to remain exactly one closed box. The visitor makes a tip, what number there is in it, and go away. He, of course, can't hint to others. Then all boxes are closed and another visitor comes.
There exists a strategy such that 99 of them gives the right answer.
Hint: The core idea is the same as the one in the problem presented in the question but there is one more step ;-)<|endoftext|>
TITLE: singular homology of a differential manifold
QUESTION [13 upvotes]: Let $M$ be a differentiable manifold, $\Delta$ the closed simplex $[p_0, p_1,...,p_k]$. A differential singular $k$-simplex $\sigma$ of $M$ is a smooth mapping $\sigma:\Delta \to M$.
And we construct a chain complex in the same way we construct the chain complex of singular homology, we gain its homology group.
My question is why this homology group equals the singular homology group? I have tried finding in many books but there is no answer of this question.

REPLY [5 votes]: ons can find a proof for my quesion in "algebraic topology: a first course" by W. Fulton. The proof bases on the fact that in some simple sample like a ball, these two are identical, and the general case would be implied by applying the Mayer-Vertoris principle (in fact, ones can see a manifold as a set of balls that overlap each other, and that principle allows to compute homology groups by reducing on each ball).<|endoftext|>
TITLE: Polish spaces in probability
QUESTION [62 upvotes]: Probabilists often work with Polish spaces, though it is not always very clear where this assumption is needed.
Question: What can go wrong when doing probability on non-Polish spaces?

REPLY [7 votes]: Fun fact: you can find a non-separable Banach space and a Gaussian probability measure on it which gives measure $0$ to every ball of radius $1$. (In particular your intuition about notions like the "support" of a measure goes pretty badly wrong.)
Consider i.i.d. $\xi_n$ and take as your norm $|\xi|^2 = \sup_{k\ge 0}2^{-k}\sum_{n=1}^{2^{k}} |\xi_n|^2$. This is almost surely finite by Borel-Cantelli and almost surely at least $1$ by the law of large numbers. The fact that it gives measure $0$ to every ball of radius $1$ is left as an exercise. This norm isn't even very exotic: if you interpret the $\xi_n$'s as Fourier coefficients, then $B$ is really just the Besov space $B^{1/2}_{2,\infty}$.<|endoftext|>
TITLE: what is the motivation of Shimura variety?
QUESTION [7 upvotes]: Tonight, a friend of mine give me a concise introduction to  Shimura variety . I only get some first impression of it. I think the hodge structure is a generalization of the cohomology ring of Kaehler manifold or algebraic manifold , and i  think of the Shimura variety an anologue of the analytic familly of complex manifolds , just as introduced in Kodaira's book Complex manifolds.And i suspect that there maybe some anologue theorem's as what Kodaira had done by deformation of complex structures .  I'm just doing some imagination unboundedless , do not laugh at me !Heh!

REPLY [21 votes]: The theory of Shimura varieties was begun by Shimura, and further developed by Langlands (who introduced the name), and is now a central part of arithemtic geometry and of the theories
of automorphic forms, Galois representations, and motives.
Shimura varieties are certain moduli of Hodge structures; but that is perhaps not the best point of view to understand why people study them.  Rather, the primary motivation is the following: Shimura varieties are attached to (certain) reductive linear algebraic groups over $\mathbb Q$, and the geometry of the Shimura variety is closely linked to the theory of automorphic forms over the corresponding reductive group.
Thus Shimura varieties make a natural test-case for investigating the conjectural relations between motives and automorphic forms, since they are geometric objects with a direct link to the theory of automorphic forms.   (I'm not sure that it's useful to be more specific about this in this particular answer, but for those to whom it is meaningful: on the one hand, one has an analogue, for any Shimura variety, of Eichler--Shimura theory, relating the cohomology of modular curves to modular forms; and on the other hand, by thinking of cohomology as being etale cohomology, one obtains Galois representations which one would like to show (and in many cases can show) are related to automorphic forms in a manner analogous to the relationship between the Galois representations on the etale cohomology of modular curves and Hecke eigenforms that was established by Deligne.)<|endoftext|>
TITLE: A question on a trace inequality
QUESTION [8 upvotes]: Let $A, B\in M_{n}(\mathbb{R})$ be symmetric positive definite matrices. It is easy to see $Tr(A^2+AB^2A)=Tr(A^2+BA^2B)$. Numerical experiments indicate $$Tr[(A^2+AB^2A)^{-1}]\ge Tr[(A^2+BA^2B)^{-1}],~~(1)$$ but it seems difficult to show it.  
Remark. When $n=2,3$, by direct computation, (1) is true. Here is an expriment done by matlab:
for s=1:1000
as=randn(4,4);
bs=randn(4,4);
ts=as*as';
rs=bs*bs';
ls=trace(inv(ts^2+ts*rs^2*ts)-inv(ts^2+rs*ts^2*rs))

end
{\bf Updated.} What about $A, B\in M_{n}(\mathbb{C})$ be positive definite Hermitian matrices.

REPLY [19 votes]: Note first that $A^2+AB^2A=(A+iAB)(A-iBA)$. The reverse product is $(A-iBA)(A+iAB)=A^2+BA^2B-i(BA^2-A^2B)=X-iC$. Thus, the quantity on the left is $\operatorname{Tr} (X-iC)^{-1}$ and that on the right is $\operatorname{Tr} X^{-1}$. Moreover, the self-adjoint complex matrix $X-iC$ is positive definite (as the product of an invertible operator and its adjoint). Similarly, considering the factorization $A^2+AB^2A=(A-iAB)(A+iBA)$, we can write the quantity on the left as $\operatorname{Tr} (X+iC)^{-1}$. Symmetrizing, we see that it will suffice to show that $(X-iC)^{-1}+(X+iC)^{-1}\ge 2X^{-1}$ in the sense of quadratic forms (then the inequality for traces will hold too). We can multiply by $X^{1/2}$ from both sides to reduce it to $(I-iD)^{-1}+(I+iD)^{-1}\ge 2I$ where $D=X^{-1/2}CX^{-1/2}$ and both operators on the left are positive definite. Diagonalizing the self-adjoint operator $iD$, we see that the inequality reduces to $(1+p)^{-1}+(1-p)^{-1}\ge 2$ for $p\in(-1,1)$.<|endoftext|>
TITLE: Is there an R=T type result for modular forms with additive reduction?
QUESTION [8 upvotes]: Let E be an elliptic curve over the rationals with conductor $Mp^2$ with p>5 and M and p coprime, and let $\rho$ be the Galois representation attached to the p-torsion points of E. Is there a way to consider deformations of $\rho$ to representations that still have "additive reduction at p" with Serre weight 2? Actually in that direction, what is the local condition on p that one has to put down? Assuming that's the case, and assuming this is representable by ring R, is there an R=T result in that direction?

REPLY [7 votes]: If you fix a prime $\ell$, and consider the Galois action of the decomposition group $D_p$ on the (rational) $\ell$-adic Tate module (here "rational" means tensored with $\mathbb Q_{\ell}$),
then (in a standard way, due to Deligne) you can convert this action into a representation
of the Weil--Deligne group, and so in particular of the Weil group.  Restricting to the
inertia group, you get a representation of the inertia group $I_p$, known as the inertial type $\tau$.  It is independent of $\ell$.  (The only reason to detour through the Weil--Deligne group is to deal with possibly infinite image of tame inertia; if the elliptic curve has potentially good reduction, then we can skip this step and just take the representation of $I_p$ on the $\ell$-adic Tate module, which has finite image and is independent of $\ell$.)
[Added: In the above, one should insist that $\ell \neq p$.  If $\ell = p$, then one can also arrive at a Weil-Deligne representation, and hence
inertial type, which is the same as the one obtained as above for $\ell \neq p$, but to do this one must use Fontaine's theory: one forms the $D_{pst}$ of the rational $p$-adic Tate module, 
which then can be converted into a Weil--Deligne representation in a standard way,
and hence gives an inertial type.]
Now one can look at the deformation ring $R_{\rho}^{[0,1],\tau}$ parameterizing lifts
of $\rho$ of which at $p$ are of inertial type $\tau$ and Hodge--Tate weights $0$ and $1$.  (See Kisin's recent
JAMS paper about potentially semi-stable deformation rings.)
[Added: Here $\ell = p$,
i.e. we are looking at $p$-adic deformations of $\rho$ which are potentially semi-stable
at $p$, and whose inertial type, computed via $D_{pst}$ as in the above added remark,
is equal to $\tau$. But note that, by the preceding discussion, these deformations do precisely capture the idea of lifts of $\rho$ having the same "reduction type" as
the original elliptic curve $E$.]
Let's suppose that $E$ really does have potentially good reduction.  Then Kisin's "moduli of finite flat group schemes" paper shows that any lift parameterized by $R^{[0,1],\tau}$ is modular.   This shows that $R^{[0,1],\tau} = {\mathbb T},$ for an appropriately chosen
${\mathbb T}$.   
One thing to note: unlike in the original Taylor--Wiles setting, from this statement one doesn't get quantitive information about adjoint Selmer groups, and one doesn't get any simple interpretation of what this $R = {\mathbb T}$ theorem means on the integral level.
(In other words, Artinian-valued points of $R^{[0,1],\tau}$ have no simple interpretation in terms of a ramification condition at $p$; this is related to the fact that the theory
of $D_{pst}$ only applies rationally, i.e. to ${\mathbb Q}_p$-representations, not integrally,
i.e. not to representations over $\mathbb Z_p$ or over Artin rings.)<|endoftext|>
TITLE: Distinguishing congruence subgroups of the modular group
QUESTION [18 upvotes]: This question is something of a follow-up to
Transformation formulae for classical theta functions .
How does one recognise whether a subgroup of the modular group
$\Gamma=\mathrm{SL}_2(\mathbb{Z})$ is a congruence subgroup?
Now that's too broad a question for me to expect a simple answer
so here's a more specific question. The subgroup $\Gamma_1(4)$
of the modular group is free of rank $2$ and freely generated by
$A=\left(
\begin{array}{cc}
1&1\\\
0&1
\end{array}\right)$
and
$B=\left(
\begin{array}{cc}
1&0\\\
4&1
\end{array}\right)$. If $\zeta$ and $\eta$ are roots of unity there is a
homomorphism $\phi$ from $\Gamma_1(4)$ to the unit circle group
sending $A$ and $B$ to $\zeta$ and $\eta$ resepectively. Then the kernel $K$
of $\phi$ has finite index in $\Gamma_1(4)$. How do we determine whether $K$
is a congruence subgroup, and if so what its level is?
In this example, the answer is yes when $\zeta^4=\eta^4=1$. There are
also examples involving cube roots of unity, and involving eighth
roots of unity where the answer is yes. I am interested in this example
since one can construct a "modular function" $f$, homolomorphic on
the upper half-plane and meromorphic at cusps such that $f(Az)=\phi(A)f(z)$
for all $A\in\Gamma_1(4)$. One can take $f=\theta_2^a\theta_3^b\theta_4^c$
for appropriate rationals $a$, $b$ and $c$.
Finally, a vaguer general question. Given a subgroup $H$ of
$\Gamma$ specified as the kernel of a homomorphism from $\Gamma$ or
$\Gamma_1(4)$ (or something similar) to a reasonably tractable target group,
how does one determine whether $H$ is a congruence subgroup?

REPLY [16 votes]: Although Andy has the correct answer, I thought I would point out that 
there will only be finitely many groups of the type you consider that
are congruence subgroups. Since these are abelian (cyclic) covers of your surface,
they will contain the universal abelian cover (corresponding to the commutator subgroup of 
$\Gamma_1(4)$). By "strong approximation",
this group will map onto all but finitely many congruence quotients of 
$SL_2(\mathbb{Z})$, and therefore so will any group containing it. 
Thus, only finitely many of your groups will be congruence. 
Another way to see this is to notice that abelian covers have Cheeger
constant approaching zero, and therefore first eigenvalue of the 
Laplacian $\lambda_1$ approaching
zero by Buser's inequality. By Selberg's estimate, $\lambda_1\geq 3/16$ for a congruence
subgroup. In principle, one could probably get explicit estimates
on the index of a congruence abelian cover this way.<|endoftext|>
TITLE: Sheaves of Principal parts
QUESTION [12 upvotes]: In EGA IV, Sec. 16, Grothendieck defines the sheaf of principal parts as follows: Let $f:X\rightarrow S$ be a morphism of schemes and $\Delta:X\rightarrow X\times_S X$ the diagonal morphism associated to $f$. $\Delta$ is an immersion, so the corresponding morphism $\Delta^{-1}\mathcal{O}_{X\times_S X}\rightarrow\mathcal{O}_X$ is surjective. Let $\mathcal{I}$ denote its kernel and define the sheaves of principal parts as
$$\mathcal{P}_{X/S}^n:=\Delta^{-1}( \mathcal{O}_{X\times_S X}) / \mathcal{I}^{n+1}$$
In their book on Crystalline cohomology, Berthelot and Ogus define the sheaf of principal parts $\mathcal{P}^n_{X/S}$ as 
$$(\mathcal{O}_X\otimes_{f^{-1}\mathcal{O}_S}\mathcal{O}_X)/\mathbf{I}^{n+1},$$
where $\mathbf{I}$ is the kernel of the multiplication map from the tensor product to $\mathcal{O}_X$.
My question is probably simple, but I don't know how to see it:
Why are those definitions equivalent if $X$ and $S$ are not affine and $n>0$? 
I've not seen the second definition anywhere else, although it seems somewhat nicer than the first one...

REPLY [8 votes]: The statement holds in general if $f : X \to S$ is a morphism of locally ringed spaces. The fibred product of locally ringed spaces can be constructed explicitly without gluing constructions, and also restricts to the fibred product of schemes. See this article (german; shall I translate it?) for details. I will make use of the explicit description given there. Also I use stalks all over the place. Probably this is not the most elegant proof, but it works.
First we construct a homomorphism $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$. For that we compute the stalks at some point $x \in X$ lying over $s \in S$:
$(\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X)_x = \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x},$
$(\Delta^{-1} \mathcal{O}_{X \times_S X})_x = \mathcal{O}_{X \times_S X,\Delta(x)} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x})_{\mathfrak{q}}$,
where $\mathfrak{q}$ is the kernel of the canonical homomorphism
$\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x} \to \kappa(x), a \otimes b \mapsto \overline{ab}.$
Thus we get, at least, homomorphisms between the stalks (namely localizations). In order to get sheaf homomorphisms out of them, the following easy lemma is useful:
(*) Let $F,G$ be sheaves on a topological space $X$ and for every $x \in X$ let $s_x : F_x \to G_x$ be a homomorphism. Suppose that they fit together in the sense that for every open $U$, every section $f \in F(U)$ and every $x \in U$ there is some open neighborhood $x \in W \subseteq U$ and some section $g \in G(W)$ such that $s_y$ maps $f_y$ to $g_y$ for all $y \in W$. Then there is a sheaf homomorphism $s : F \to G$ inducing $s$.
This can be applied in the above situation: Every section in a neighborhood of $x$ in $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X$ induced by an element in $\mathcal{O}_X(U) \otimes_{\mathcal{O}_S(V)} \mathcal{O}_X(U)$ for some neighborhoods $U$ of $x$ and $V$ of $s$ such that $U \subseteq f^{-1}(V)$. This yields a section in $\mathcal{O}_{X \times_S X}$ on the basic-open subset $\Omega(U,U,V;1)=U \times_V U$ and thus a section of $\Delta^{-1} \mathcal{O}_{X \times_S X}$ on $U$. It is easily seen, that this construction yields the natural map on the stalks.
Thus we have a homomorphism $\alpha : \mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$. Now let $J$ be the kernel of the multiplication map $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \mathcal{O}_X$ and $I$ be the kernel of the homomorphism $\Delta^\# : \Delta^{-1} \mathcal{O}_{X \times_S X} \to \mathcal{O}_X$. Then for every $n \geq 1$ our $\alpha$ restricts to a homomorphism
$(\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X)/J^n \to (\Delta^{-1} \mathcal{O}_{X \times_S X})/I^n,$
which is given at $x \in X$ by the natural map
$(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n \to ((\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n)_{\mathfrak{q}}$,
where $\mathfrak{p} \subseteq \mathfrak{q}$ is the kernel of the multiplication map $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x} \to \mathcal{O}_{X,x}$.
We want to show that this map is an isomorphism, i.e. that the localization at $\mathfrak{q}$ is not needed. For that it is enough to show that every element in $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}$, whose image in $\mathcal{O}_{X,x}$ is invertible, is invertible modulo $\mathfrak{p}^n$. Or in other words: Preimages of units are units with respect to the projection
$(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n \to (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^1  \cong \mathcal{O}_{X,x}$.
However, this follows from the observation that the kernel $\mathfrak{p}^1 / \mathfrak{p}^n$ is nilpotent; cf. also this question.
I'm sure that there is also a proof which avoids stalks at all.
EDIT: So here is a direct construction of the homomorphism $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$:
Let $p_1,p_2$ be the projections $X \times_S X \to X$. Then we have for $i=1,2$ the homomorphism
$\mathcal{O}_X \to {p_i}_* \mathcal{O}_{X \times_S X} \to {p_i}_* \Delta_* \Delta^{-1} \mathcal{O}_{X \times_S X} = (p_i \Delta)_* \Delta^{-1} \mathcal{O}_{X \times_S X} = \Delta^{-1} \mathcal{O}_{X \times_S X}$,
and they commute over $f^{-1} \mathcal{O}_S$. Thus we get the desired homomorphism. But I think stalks are convenient when we want to show that this is an isomorphism when modding out the ideals.<|endoftext|>
TITLE: Are there any Hecke operators acting on an elliptic curve with additive reduction that I don't know about?
QUESTION [25 upvotes]: I could have made this question very brief but instead I've maximally gone the other way and explained a huge amount of background. I don't know whether I put off readers or attract them this way. The question is waay down there.
Let $f$ be a cuspidal modular eigenform of level $\Gamma_0(N)\subseteq SL_2(\mathbf{Z})$ (for example $f$ could be the weight 2 modular form attached to an elliptic curve) and let $p$ be a prime. In the theory of modular forms, one Hecke operator at $p$ is singled out, namely $T_p$, sometimes called $U_p$ if $p$ divides $N$, and defined by the double coset attached to the matrix $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$. Now $f$ is an eigenform for $T_p$, and $f$ has an eigenvalue for this operator---a Galois-theoretic interpretation of this eigenvalue is that it is (modulo fixing embeddings of $\overline{\mathbf{Q}}$ in $\overline{\mathbf{Q}}_\ell$ and $\mathbf{C}$) the trace of the geometric Frobenius on the inertial invariants of the $\ell$-adic representation attached to $f$, for $\ell\not=p$ a prime.
Now here is a very naive question that I don't know the answer to, and I really should, and I'm sure it's very well-known to people who do this sort of stuff. Say $N=p^rM$ with $M$ prime to $p$. One can approach the theory of Hecke operators entirely locally. Let $K:=U_0(p^r)$ denote the subgroup of $GL_2(\mathbf{Z}_p)$ consisting of matrices whose bottom left hand entry is $0$ mod $p^r$. Now there is an "abstract Hecke algebra" of locally left- and right-$K$-invariant complex-valued functions on $G:=GL_2(\mathbf{Q}_p)$ with compact support. As a complex vector space this algebra has a basis consisting of the characteristic functions $KgK$ as $KgK$ runs through the double cosets of $K$ in $G$. But this space also has an algebra structure, given by convolution.
If $r=0$ then $K$ is maximal compact, and the structure of this Hecke algebra is well-known and easy. Via the Satake isomorphism, the abstract Hecke algebra is isomorphic to $\mathbf{C}[T,S,S^{-1}]$, with $S$ and $T$ independent commuting polynomial variables. The interpretation is that $T$ is the usual Hecke operator $T_p$ attached to the matrix $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$ and $S$ is the matrix attached to $\left(\begin{array}{cc}p& 0\\ 0&p\end{array}\right)$. One doesn't always see this latter Hecke operator explicitly in elementary developments of the theory because it acts in a very dull way---it acts by scalars on forms of a given weight and level $\Gamma_0(N)$, typically (depending on normalisations) as the scalar $p^{k-2}$ on forms of weight $k$. In particular the "abstract Hecke algebra" doesn't give us any more information than that which classical texts explain, as it's generated by $T_p$, $S_p$ and $S_p^{-1}$.
The next case is $r=1$ and this case I also understand. The abstract Hecke algebra now is non-commutative, "because of oldforms": I don't think the operators attached to $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$ and $\left(\begin{array}{cc}0& p\\ 1&0\end{array}\right)$ (that is, the operators corresponding to these double coset spaces) commute, but if $f$ has level $Mp$ and is old at $p$ then we should be working at level $M$, and if it's new at $p$ then we get two invariants---the $T_p$ (or $U_p$) eigenvalue, which is classical, and the $w$-eigenvalue, which is the local sign for the functional equation. Again both of these numbers are classical and a lot is known about them. I am pretty sure that the abstract Hecke algebra in this case is generated by these operators $T_p$, $w$, and the uninteresting $S_p$ and $S_p^{-1}$, the latter two still acting by scalars on forms of a given weight. Am I right in thinking that these operators generate the local Hecke algebra? I think so.
The next case is $r=2$ and this I am not 100 percent sure I understand. The classical theory gives us $T_p$, $S_p^{\pm1}$ and $w$. Note that on a newform of level $\Gamma_0(Mp^2)$, $T_p$ is zero in this situation, $S_p$ acts by a scalar, and $w$ is some subtle sign which people have clever ways of working out.

Finally then, the question! Let $K$ be the subgroup of matrices in $GL_2(\mathbf{Z}_p)$ consisting of matrices for which the bottom left hand corner is $0$ mod $p^2$. Let $H$ denote the abstract double coset Hecke algebra of compactly supported $K$-bivariant functions on $GL_2(\mathbf{Q}_p)$. 

Is this abstract Hecke algebra generated (as a non-commutative algebra) by the characteristic functions of $KgK$ for $g$ in the set {$\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$, $\left(\begin{array}{cc}0& p^2\\ 1&0\end{array}\right)$, $\left(\begin{array}{cc}p& 0\\ 0&p\end{array}\right)$, $\left(\begin{array}{cc}p^{-1}& 0\\ 0&p^{-1}\end{array}\right)$}? 

In the language I've been using in the waffle above: modular forms of level $p^2$ have an action of the Hecke operators $T_p$, $w$, and the invertible $S_p$. Are there any more, lesser known, Hecke operators that we're missing out on?

REPLY [11 votes]: Just to expand on a comment I made above: I'm not exactly sure what operators generate the Hecke algebra of $\Gamma_0(p^2)$, but the Hecke algebras of the principal congruence subgroups $\Gamma(p^r)$ are easier to handle. 
Let's write $K_n$ for the principal congruence subgroup of level $p^n$ in $G = {\rm GL}_2(\mathbb{Z}_p)$. 

CLAIM: For any $n > 0$, the Hecke algebra $H(G // K_n)$ is generated by the double cosets $K_n x K_n$ for $x$ in the set
$S = \left\{ 
\begin{pmatrix} 1 & 0 \\\ 0 & p \end{pmatrix}, 
\begin{pmatrix} p & 0 \\\ 0 & p \end{pmatrix},
\begin{pmatrix} p^{-1} & 0 \\\ 0 & p^{-1} \end{pmatrix},
\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix},
\begin{pmatrix} 0 & -1 \\\ 1 & 0 \end{pmatrix},
\begin{pmatrix} a & 0 \\\ 0 & 1 \end{pmatrix}
 \right\}$
where a is your favourite generator of $\mathbb{Z}_p^\times$. 
(These correspond, classically, to $T_p$, $S_p$, $S_p^{-1}$, a "twisting operator at p", something close to the Atkin-Lehner $w$, and the diamond operator $\langle a \rangle$.)

Proof: It suffices to show that the subalgebra generated by these operators contains the double coset $[K_n g K_n]$ for any given $g \in G$. Let $X$ be the monoid of elements of the form $\begin{pmatrix} p^r & 0 \\\ 0 & p^s\end{pmatrix}$ with $r \le s \in \mathbb Z$. The Cartan decomposition tells us that any $g \in G$ can be written as $g = k x k'$ for some $k, k' \in K_0$.
We now write 
$K_n\ g\ K_n = K_n\ k\ x\ k'\ K_n$
$ = K_n\ k\ K_n\ x\ K_n\ k'\ K_n$ (using the normality of $K_n$ in $K_0$)
$ = [K_n\ k\ K_n]\ [K_n\ x\ K_n]\ [K_n\ k'\ K_n]$
The first and last terms are obviously in the subalgebra $H(K_0 // K_n)$ of $H(G // K_n)$, which is isomorphic to the group algebra of the finite group $K_0 / K_n = {\rm GL}_2(\mathbb Z / p^n)$. This is clearly generated by the images of the last three elements of $S$, since these are topological generators of ${\rm GL}_2(\mathbb{Z}_p)$. 
Meanwhile, the middle term is in the subalgebra of $H(G // K_n)$ generated by $X$, and it's easy to see that for $x, y \in X$ we have $K_n\ x\ K_n\ y\ K_n = K_n\ xy\ K_n$. Hence this subalgebra is just the monoid algebra of $X$, which is generated by the first three elements of $S$.
Now, as for your original question, the subgroup $U_0(p^2) \subseteq {\rm GL}_2(\mathbb{Z}_p)$ of matrices that are upper triangular modulo $p^2$ contains a conjugate of $K_1$, so its Hecke algebra is isomorphic to a subalgebra of the Hecke algebra of $K_1$. So although I can't give generators for your algebra, I can exhibit it as a subalgebra of something we know generators for.<|endoftext|>
TITLE: Why is the Gamma function shifted from the factorial by 1?
QUESTION [148 upvotes]: I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty  t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty  t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss, but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the multiplication formula is simpler: we have $$\Pi\left(\frac{z}{m}\right) \, \Pi\left(\frac{z-1}{m}\right) \cdots \Pi\left(\frac{z-m+1}{m}\right) = \left(\frac{(2 \pi)^m}{2 \pi m}\right)^{1/2} \, m^{-z} \, \Pi(z)$$
instead of
$$\Gamma\left(\frac{z}{m}\right) \, \Gamma\left(\frac{z-1}{m}\right) \cdots \Gamma\left(\frac{z-m+1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - z} \; \Gamma(z);$$
the infinite product definitions reduce from
$$\begin{align}
\Gamma(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} 
= \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Gamma(z) &= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \\
\end{align}$$
to
$$\begin{align}
\Pi(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{(z+1)\cdots(z+n)} 
= \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Pi(z) &= e^{-\gamma z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}; \\
\end{align}$$
 and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$
I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?

REPLY [5 votes]: I'm going to elaborate on Pietro Majer's answer a bit.
Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$
\begin{align}
\Pr(S_1\in A) & = \left. \int_A s^n e^{-s}\, \frac{ds} s \right/ \Gamma(n), \\[10pt]
\Pr(S_2\in A) & = \left.\int_A s^m e^{-s}\, \frac{ds} s\right/ \Gamma(m),
\end{align}
where $n,m$ are positive real numbers. Then
$$
\Pr(S_1+S_2\in A) = \left.\int_A s^{n+m} e^{-s} \, \frac{ds} s \right/ \Gamma(n+m).
$$
A concrete instance: Suppose the waiting time $T$ until the next phone call arrives at a switchboard as a memoryless probability distribution: for $s,t\ge 0,$ one has $\Pr(T>s+t\mid T>s) = \Pr(T>t).$ That implies that for some $\mu>0$ and all $t>0,\,\,\,$ $\Pr(T>t) = e^{-t/\mu},$ and $\mu$ is the expected value of $T,$ i.e. $\mu$ is the average waiting time.
Then the distribution of the time $T_n$ until the arrival of the $n$th phone call after the present time is given by
\begin{align}
\Pr(T_n \in A) & = \frac 1 {\Gamma(n)} \int_A \left( \frac t \mu \right)^n e^{-t/\mu} \, \frac{(dt/\mu)} {(t/\mu)} \\[10pt]
& = \frac 1 {\Gamma(n)} \int_{A/\mu} u^n e^{-u}\, \frac{du} u.
\end{align}<|endoftext|>
TITLE: What are the Martin's Maximum consequences of Namba forcing?
QUESTION [6 upvotes]: It is known that Namba forcing is stationary-preserving and hence can be used in the setting of Martin's Maximum. Does this result in any striking consequences?

REPLY [2 votes]: Justin Moore asked a similar question a while ago, and I pointed him to my papers with Claverie and Doebler. Even though it doesn't exactly answer your question: Namba-like forcings (in the sense that they are stationary set preserving and make $\omega_2$ $\omega$-cofinal) have many applications in the presence of ${\sf MM}$. The most recent one is my proof with David Asperó that ${\sf MM}^{++}$ implies Woodin's $(*)$, see https://ivv5hpp.uni-muenster.de/u/rds/MM_implies_star.pdf .<|endoftext|>
TITLE: When are fiber bundles reversible?
QUESTION [25 upvotes]: My question, in its most general form is this:

Given a fiber bundle $F\rightarrow E\rightarrow B$, when is there a fiber bundle $B\rightarrow E\rightarrow F$?

Here, F,E, and B can lie in whichever category you wish, but I'm mostly interested in the case where all 3 are smooth closed manifolds.
Now, I realize that the initial answer is "unless E is a product, essentially never", so here is a more focused question (with background).
I've been studying a certain class of free actions of the 3-torus $T^3$ on $S^3\times S^3\times S^3 = (S^3)^3$.  For each of these actions, by quotienting out by various subtori, I can show that the orbit space $E=(S^3)^3/T^3$ simultaneously fits into 2 fiber bundles:
$$S^2\rightarrow E \rightarrow S^2\times S^2$$ and $$S^2\times S^2\rightarrow E\rightarrow S^2$$ where the structure group for both bundles is $S^1$.
(In fact, the class of actions also gives rise to examples where either $S^2\times S^2$ can independently be replaced with $\mathbb{C}P^2\sharp -\mathbb{C}P^2$, the unique nontrivial $S^2$ bundle over $S^2$.)
By computing characteristic classes for (the tangent bundle to) E, I know that for an infinite sublcass of the actions I'm looking at, E is not homotopy equivalent to $S^2\times S^2\times S^2$, and each of the E are pairwise nondiffeomorphic.
I suspect the reason I could find so many E which fit into "reversible" fiber bundles is strongly related with the fact that the fiber and base are so closely related.
And so, I ask

For fixed manifold M, what is the relationship between bundles $X\rightarrow E\rightarrow M$ and $M\rightarrow E'\rightarrow X$ where $X$ is some $M$ bundle over $M$?

And just in case there is no general relationship,

Is there a reason I should have expected there to be a relationship in my examples, even though in general there isn't?

REPLY [2 votes]: In terms of Seifert fiber spaces, there are two examples when you consider torus bundles over $S^1$ among the ones which use periodic mapping classes: These are $(No,1|(1,0))$ and $(No,1|(1,1))$ which are respectively $K\times S^1$ and $K\times_{\tau}S^1$, where $\tau$ is the unique Dehn's twist on the Klein bottle. 
Those fibrations are not unique because also $K\times S^1=(NnI,2|(1,0))$ and $K\times_{\tau}S^1=(NnI,2|(1,1))$. Curiously, if $T=S^1\times S^1$ is the 2-torus, then $T\hookrightarrow K\times S^1\to S^1$ is a "non-trivial" fibering.<|endoftext|>
TITLE: Heegaard genus in hyperbolic 3-manifolds
QUESTION [5 upvotes]: It is well known that for a closed hyperbolic 3-manifold $M$ the rank of $\pi_1(M)$ is bounded above by some universal constant $K$ times the volume of $M$. Using similar methods, i.e. the thick-thin decomposition of $M$, one can also show that the Heegaard genus of $M$ is bounded above by a universal constant times the volume of $M$. (I believe that Thurston showed this first, though I am not sure as to how.)
I am looking to construct a sequence of (closed) hyperbolic 3-manifolds, say {$M_n$} such that the volume grows linearly in the Heegaard genus of $M_n$. That is, I am trying to show that a linear bounded on Heegaard genus in terms of volume is the best one can do. So far, I am having some trouble constructing such an example. 
Does anyone have a good method or reference for constructing such an example? Also, another approach to the problem would be wonderful (short of solving the rank versus Heegaard genus conjecture of hyperbolic 3-manifolds, of course).

REPLY [2 votes]: Agol says: Take a 3-manifold group that maps onto a free group, and take induced covers (the rank and volume will both grow linearly). See this paper of Lackenby.<|endoftext|>
TITLE: Is the classification of supercuspidal representations a wild problem?
QUESTION [7 upvotes]: As explained by Torsten Ekedahl in this reply, the problem of classifying the irreducible (infinite dimensional) representations of $sl_2(\mathbb{C})$ is a wild one. The notion of wild classification problem was discussed in this question. It means that the problem in a sense contains the problem of reducing pairs of matrices over a field to a canonical form by simultaneous conjugation.

Let $F$ be a non-Archimedean local field. Is the classification of supercuspidal representations of $\mathrm{GL}_n(F)$ a wild classification problem?

The work of Bushnell and Kutzko [The admissible dual of $\mathrm{GL}(N)$ via compact open subgroups] gives a classification of the supercuspidal representations of $\mathrm{GL}_n(F)$, and the local Langlands correspondence for $\mathrm{GL}_n$ also gives a parametrisation of the supercuspidal representations. However, it is not obvious to me whether any of these classifications answers the question.

REPLY [2 votes]: I cannot directly answer your question because I am not a specialist of wild classification problems. However I want to make the following remark that could help. 
I was told that the problem of classifying the irreducible (smooth) complex representations of ${\rm GL}_n ({\mathfrak o}_F )$, where ${\mathfrak o}_F$ is the ring of integers of a local field $F$, is wild (at least for $n$ large). 
Moreover to construct and classify supercuspidal representations, Bushnell and kutzko obtain them as compactly induced representations from  irreducible smooth representations of compact mod center subgroups. In particular we obtain a large class of supercuspidals by inducing certain irreducible complex representations of $F^{\times}{\rm GL}_n ({\mathfrak o}_F)$. 
However in Bushnell and Kutzko's theory  not all representations of ${\rm GL}_n ({\mathfrak o}_F )$ are needed, but only very particular ones and one knows how to construct them effectively.  Indeed the whole Bushnell and Kutzko's construction is based on the theory of 'simple characters' which is in principal entirely effective.<|endoftext|>
TITLE: Polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$?
QUESTION [516 upvotes]: Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}$ such that $f\colon\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?

REPLY [36 votes]: There is a new manuscript on the arXiv by Giulio Bresciani, A higher dimensional Hilbert irreducibility theorem, arXiv:2101.01090, which shows that assuming the weak Bombieri--Lang conjecture, there cannot be a polynomial bijection from $\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$.
The author writes that:

Our strategy is essentially the one followed in a "polymath project"
led by T. Tao, see [Tao19], hence this result should be credited to the polymath project as a whole.

[Tao19] https://terrytao.wordpress.com/2019/06/08/ruling-out-polynomial-bijections-over-the-rationals-via-bombieri-lang/<|endoftext|>
TITLE: What is it that makes some diophantine equations interesting, while others are less so
QUESTION [15 upvotes]: The following question is in particular reference to the previous question by Bjorn Poonen. I guess I won't even need to give this link, Polynomial representing all nonnegative integers, since this is perhaps the most famous question of MO until now.
I found this question interesting and natural as a curiosity. But from the interest of people, there seems to be more in it. (recall the exclamation What a nice problem! by  Gil Kalai). May someone explain me why this question particularly interesting from, perhap, a number theorist perspective ? For example,is there some deeper linkage to other results ? 
The same query applies to Fermat polygonal number theorem. Is there anything that these theorems reveal us about the deeper structure of integers? 
More generally, I hope we can address the question: What is it that makes some diophantine equations interesting, while others are less so?
(The change was suggested by Kevin O'Bryant)

REPLY [3 votes]: Your question is probably too general, I simply hope that you'd like to learn some personal experience of people who do their research in diophantine equations or who apply the equations to other areas.
Although my starting research (under- and postgraduate level) was the theory of transcendental numbers, quite tied to diophantine equations (especially the ones related to linear forms in logarithms), I did not try to work on these. (Maybe, because of my father's attempts to prove Fermat's Last Theorem.) But at one occasion I was "introduced" to the Erdős--Moser equation $1^k+2^k+\dots+(m−1)^k=m^k$ ($m$ and $k$ positive integers), Applications of pattern-free continued fractions, and was impressed by the beauty and power of the method which Leo Moser used in 1953 to show that no solution (except $1+2=3$) exists with $m\le 10^{10^6}$. (!) Moser's method could not give much more, and I was happy enough to collaborate on some new ideas and computational achievements of nowadays to significantly extend (after more than 50 years) Moser's bound.
Another favourite diophantine equation is Catalan's equation $x^p-y^q=1$ ($p,q>1$) which was rather recently solved by Preda Mihăilescu. He even managed to avoid linear forms in logs (which are far from beauty because of so many technicalities). There are many steps in the proof, treating some special cases, most of them using completely different methods of diophantine analysis and algebraic number theory. But they are just beautiful! For example, there are two different proofs of the nonsolvability of $x^2-y^q=1$ in integers $x>1$, $y>1$ for a prime $q>3$. The original one, due to Ko Chao ([On the diophantine equation $x^2=y^n+1$, $xy\ne0$, Sci. Sinica 14 (1965) 457--460]; also given in Mordell's "Diophantine equations"), uses the law of quadratic reciprocity in a very elegant way. Another one, due to E.Z. Chein [A note on the equation $x^2=y^q+1$, Proc. Amer. Math. Soc. 
56 (1976) 83--84], is extremely short and elementary.
Summarizing, I would say that natural criteria for considering some diophantine equations (and ignoring other) are the simplicity of the equation (isn't $x^n+y^n=z^n$ simple?) and the beauty and novelty of methods to solve it. Probably, the usefullness of the equation has to be taken into account as well.<|endoftext|>
TITLE: Minimal model which is necessarily singular
QUESTION [12 upvotes]: I was told during a summer school on the MMP a nice example (which I have also mentioned here on MO) that I'm not able to figure out anymore.
The example (due, I think, to Miles Reid) is a smooth compact threefold $X$ such that the number of sections of $\mathcal{O}_X(m K_X)$ grows like $m^3/4$ (if I recall correctly). The nice thing about this example is the following.
Assume $X$ is birational to a smooth variety $Y$ such that $K_Y$ is nef. Then sections of $K_Y$ grow in the same fashion, in particular $K_Y$ is big, so by Kawamata-Viehweg vanishing we have $h^0 (Y, \mathcal{O}_Y(m K_Y)) = \chi(Y, \mathcal{O}_Y(m K_Y)) \sim m^3/4$, and by Riemann-Roch we find $K_Y^3 = 3/2$, which is not possible, since that number must be integer.
So, if one wants to have a minimal model for $X$, one has to allow singular varieties into the picture.

Can anyone tell me how the variety $X$ is obtained (or another example in a similar flavour)?

EDIT: As I wrote in the comments to VA's answer, I'm looking for an elementary example, where $\mathcal{O}_X(m K_X)$ can be computed and compared to the Riemann-Roch expansion, in order to have a completely intersection-theoretic argument. In particular I'd like to avoid using the concepts of canonical and terminal singularities, since I view this example mainly as a motivation to introduce exactly those concepts. It would also be nice if one could directly find a smooth $X$, rather then using Hironaka to resolve a singular variety with fractional $K_Y^3$ (which one secretly knows is terminal).

REPLY [8 votes]: @Andrea: Perhaps this example would be a good motivation (see also this post). I am referring to your comment above to VA's answer.
I think this is example is due to Iitaka, or someone from his school, in any case it predates Miles Reid. Take a $3$-dimensional abelian variety $A$ and mod out by the involution $(−1)$. Resolve the resulting $64$ double points and call the result X. Then it is relatively easy to prove that $X$ is not birational to a smooth projective variety with a nef canonical bundle (the point is that you have to blow down the exceptional divisors over these $64$ double points, so the minimal model will be $A$ mod $(-1)$). I believe that at the time this example was thought of as proof that minimal models did not exist in higher dimensions, but then Reid and Mori realized that it only means that minimal models need not be smooth. 
I think the right way to think about this is that minimal models have no worse than terminal singularities. It turns out that terminal singularities are smooth in codimension 2, so in particular a 2-dimensional terminal singularity is actually smooth. So, one could argue that even minimal models of surfaces have terminal singularities, that is, that's the natural class of singularities for a minimal model. It just so happens that in dimension 2, these singularities are indistinguishable from smooth points.<|endoftext|>
TITLE: Liftability of Enriques Surfaces (from char. p to zero)
QUESTION [7 upvotes]: Let $k$ be an algebraically closed field of characteristic $p > 0$, $X$ a variety over $k$.
We say $X$ lifts to characteristic zero, if there exists a local ring $R$ containing $\mathbb Z$ with residue field $k$ and a flat scheme $\mathcal X$ over $R$, such that
$$ \mathcal X \otimes_R k \simeq X.$$
In words: There exists a family over a ring of mixed characteristic, which has our $X$ for a special fibre.
For most classes of surfaces in Kodaira dimension zero, liftability is known: For K3 surfaces, liftability was established by Deligne, and for abelian surfaces, one can use more general theories developed for abelian varieties. Bi-elliptic and quasi-bi-elliptic surfaces can be dealt with explicitly.
As far as I know, there is nothing in the literature about Enriques Surfaces. For this class, the case p = 2 is the most interesting.
The question seems natural, so it would struck me as strange if it were still open. Does anyone around here know anything about this?
Thanks a lot.

REPLY [8 votes]: This may not be exactly the answer you are looking for: I and Nick
Shepherd-Barron have an unpublished (so far) proof of liftability in
characteristic $2$, the only non-trivial case. To atone for the fact that I
refer to unpublished results I give a quick sketch of proof.
The proof starts by showing that in a family $X/S$ of Enriques surfaces
$\mathrm{Pic}^\tau(X/S)$ is flat (of order $2$) over $S$ and
$\mathrm{Pic}(X/S)/\mathrm{Pic}^\tau(X/S)$ is locally constant. This implies
that the tensor square of any line bundle of an Enriques surface extends along
any formal deformation and hence it is enough to find a formal lifting.
For a surface with $h^2(T_X)=0$ the deformations are unobstructed so we are
OK. There are two types of surfaces with $h^2(T_X)=1$ (which is the only other
possibility); surfaces with $\mathrm{Pic}^\tau(X)=\alpha_2$ and surfaces
$\mathrm{Pic}^\tau(X)=\mathbb Z/2$ having non-trivial vector fields (the latter case
exists). The first case is nicer in that we get a map from deformations of
such surfaces to deformations of $\alpha_2$ and this map is formally smooth. As
we can lift $\alpha_2$ (a formal deformation has base $\mathbb
Z_2[[x,y]]/(xy-2)$) we can also lift the surface, in fact over a base with
absolute ramification of order $2$. The second case we know less
about but as $h^2(T_X)=1$ the base of the deformation is (at most) a
hypersurface singularity and as one can show that it is a very small family one
can show that a versal deformation is flat over $\mathbb Z_2$. We know nothing
about the ramification necessary in this case.<|endoftext|>
TITLE: What is the exterior derivative intuitively?
QUESTION [93 upvotes]: Actually I have several related questions, not worth opening different threads:

What is the exterior derivative intuitively? What is its geometric meaning?
A possible answer I know is, that it is dual to the boundary operator of singular homology. However I would prefer a more direct interpretation.
What is a conceptually nice definition of the exterior derivative?

REPLY [5 votes]: First define the exterior derivative for forms defined on an open set $U \subseteq \mathbb{R^n}$. This uses the notion of integration of a $p$-form over a singular $p$-chain, which needs only the integration of $\mathcal{C}^{\infty}$-functions over compact subsets of $\mathbb{R}^p$ and runs as follows. A singular $p$-cube in $U$ is a $\mathcal{C}^{\infty}$-map $\sigma : I^p \rightarrow U$, where $I := [0,1]$ is the closed unit interval. Let $\Omega^p(U) := H^0(U;\wedge^pT^*U)$ the space of alternating $p$-forms on $U$; then each $\omega \in\Omega^p(U)$ pulls back to a top form $\sigma^*\omega$ $=$ $f dx \in \Omega^p(I^p)$ with $f \in \mathcal{C}^{\infty}(I^p)$ and $dx = dx_1 \wedge \cdots \wedge dx_p$ the canonical volume element of $\mathbb{R}^p$. It thus has an integral
$$
    \int_{\sigma} \omega := \int_{I^p} f dx,
$$
and, in fact, this it is what differential forms are made for:
born to be integrated.
Next define the vector space of $p$-chains to be the free $\mathbb{R}$-vector space on the singular $p$-cubes, so that a $p$-chain $c_p$ is a formal linear combination of singular $p$-cubes:
$$
    c_p = \sum_{i=1}^k \gamma^i \sigma_i \quad,\quad        
          k\in\mathbb{N}, \gamma \in \mathbb{R}.\tag{1}
$$
The integral then extends to $p$-chains by linearity;
$$
    \int_{c_p}\omega := \sum_{i=1}^k \gamma^i \int_{\sigma_i}
                        \omega. 
$$
As a next ingredient we need that any $p$-chain $c_p$ has a boundary $\partial c_p$ which is a $(p-1)$-chain. We first define it on singular $p$-cubes $\sigma$ by
$$
    \partial \sigma := 
    \sum_{j=1}^p (-1)^j
    (\sigma \circ d^j_- - \sigma \circ d^j_+),
$$
where the singular $(p-1)$-cubes $d^j_{\mp}$ in $I^p$ define the $j$-th front and back boundary faces:
$$
    d^j_-(x^1, \dots, x^{p-1}) := 
    (x^1, \dots, x^j, 0, x^{j+1}, \dots, x^{p-1}),
$$
$$
    d^j_+(x^1, \dots, x^{p-1}) := 
    (x^1, \dots, x^j, 1, x^{j+1}, \dots, x^{p-1}).
$$
We extend this boundary operator to $p$-chains by linearity:
$$
    \partial c_p := \sum_{i=1}^k \gamma^i \partial \sigma_i 
$$
with $c_p$ given by (1).
As a last ingredient we need that a point $P \in U$ and a $p$-tuple of vectors $X:= (X_1, \dots, X_p)$ (viewed as tangent vectors at $P$ to $U$) define a singular $p$-chain
$$
    [X]_P : I^p \rightarrow U
$$
via
$$
    [X]_P(x^1, \dots, x^p) := P+\sum_{k=1}^p x^k X_k
$$
as soon as the $X_k$ are so small that all the $P+x^kX_k$ are in
$U$ for all $k$. In fact, these simple linear singular chains are all what is needed of this formalism to define the exterior derivative, to which we proceed next.
After this preliminaries, we now want, given $\omega \in \Omega^p(U)$, define its exterior derivative $d\omega \in \Omega^{p+1}(U)$. We do this pointwise at any point $P \in U$ by exhibiting the value $d\omega_P$, as an alternating $(p+1)$-form, takes on any $(p+1)$-tuple of (tangent) vectors $(X_1, \dots, X_{p+1})$ $\in$ $(\mathbb{R^n})^{p+1}$. We define
$$
    \fbox{$d\omega_P(X_1, \dots, X_{p+1}) :=
    \lim_{t \rightarrow 0}    \dfrac{1}{t^{p+1}}
    \int_{\partial([tX]_P)} \omega.$}
$$
Finally, for the general case of the exterior derivative of a $p$-form $\omega$ on an $n$-dimensional manifold $M$, just take charts $\phi: V \rightarrow U$ with $V$ open in $M$, $U$ open in $\mathbb{R}^n$, with the $V$ covering $M$, and put
$$
    (d\omega)|V := \phi^*d\eta \quad 
    \text{with}\quad \eta := (\phi^{-1})^*(\omega|V) \in \Omega^p(U).
$$
The transformation formula for multivariate integrals then shows that the $(d\omega)|V$ glue well on the overlaps, thus yielding a global well-defined $d\omega$.
Loosely speaking, this defines the exterior derivative as a "volume derivative", a flux density through the boundary of an infinitesimal $(p+1)$-dimensional parallelepiped and so has as a built-in an infinitesimal version of Stokes' Theorem.<|endoftext|>
TITLE: Ultrafilters vs Well-orderings
QUESTION [14 upvotes]: This question was actually asked by John Stillwell in a comment to an answer to this question.  I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it.
Question: Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$?

REPLY [4 votes]: Historically it was proved that the ultrafilter lemma is independent from the axiom of choice by showing that there is a model in which there is an infinite Dedekind-finite set of real numbers, but every filter can be extended to an ultrafilter. Where an infinite Dedekind-finite set is an infinite set which does not have a countably infinite subset.
The existence of infinite Dedekind-finite sets negates not only the axiom of choice, but also the [much] weaker axiom of countable choice. These sets cannot be well-ordered, and since the real numbers have such subset they cannot be well-ordered themselves in such model.
The proof was given by Halpern and Levy in 1964.<|endoftext|>
TITLE: What great mathematics are we missing out on because of language barriers?
QUESTION [17 upvotes]: Primary question: What great mathematics are we missing out on because of language barriers?
Please post interesting results, pursuits, and branches in mathematics that have not been translated to English. Make sure to mention the language(s) they are accessible in!
note: this question (and the further two) are about specific theorems or branches in mathematics, not about bringing up 'general answers'. Please post the (original language) names of specific theorems, or tell about theories/branches in mathematics worked on by specific groups with the name of the group and a simple note how to contact the group (a name + university name, or www address, or email address suffice).
And as a secondary question:
Please post examples of where mathematical thought is more easily expressed or more intuitive in a language other than English, rather than in English.
As a ternary question:
Please post examples of mathematical theories in history that have stalled due to unnatural or awkward spoken-language associated with the theory, while counterparts in another language have flourished.

I know this question is English-centric, but if I were posting this on a German- (Russian-, Spanish-, Japanese-) speaking forum I would make it German-centric; on the other hand English is now the standard into which people translate their papers.
One historical example of a big chunk of maths that never made it to English is Grothendieck's EGA.
A co-question to this question has been asked here:
What are good non-English languages for mathematicians to know?
However, I disagree with the comments that languages other than English are only good for dead maths; pending research in every country happens in the local language, and that is a huge amount of knowledge to wave away.
A similar question but not directly stated in this form is here:
Books you would like to see translated into English
This question is a chance for the native (as well as second-language) speakers of many beautiful languages to tell about mathematical ideas, concepts, theories and results that have not yet transcended the language they were first written in. It is also a place to reflect on expressing specific mathematical concepts in very elegant ways that you think surpass the way we are thinking of them in English. I don't know if this is against the rules, but if not, do not hesitate to post examples in the language you're talking about.

Motivation and background:
The recent question on what books we would like to have translated to English has revived an idea, or question, that I had about the way we learn and propagate ideas in mathematics. What knowledge, and more specifically mathematics (just to be on topic) are we missing out on because we don't know the less-commonly-taught second languages?
One person might say that English is the main language currently in which we are publishing papers. For example Mathematics as a language (Wikipedia)

It is interesting to note that there are very few cultural dependencies or barriers in modern mathematics. 

This might be true, to some extent. We all have seen those bad papers - many of us have commited the crimes - of submitting papers with broken English, explaining delicate concepts with the subtlety of a jackhammer. The lingual density of mathematics is immense and, since the spoken word is much more precise, we enhance it by using small nuances that we exaggerate (e.g. contains/consists of) to fit more information into the language that we speak. Before this paper makes it to an English-speaking journal a lot of mathematics is lost: it might be lost in translation; maybe the author didn't have enough time to translate everything; perhaps the author brought up some interesting adages that didn't work in English; maybe their first language allowed a specific 'slang' that made the concepts much easier to talk about. Most importantly, before a paper is submitted, a big, big amount of work happens - you will not learn of it before the select results are published; for one thing, it is a lot of time; for another, we all know that sometimes the most interesting mathematics stay hidden because they somehow didn't make the cut. Finally, maybe the research group did not publish to English because their work was meant to support other research in their country; or they just didn't want to bother, being happy with reaching their local environment. This is a place to bring up this sort of research.
In this question we are talking about understanding on a level much higher than 'being able to apply the theorems and formulas'. We are also talking about the 'enlightenment'. A lot of - maybe most? - mathematical thought is encoded in the every-day language being used to describe it, which can be more or less elegant. The fitness of this language to the purpose of the concept can make a big difference - compare the Newtonian school of Calculus becoming stalled because they would not want to forgo the 'dot notation' ($\dot{x}$) that is now largely abandoned and limited only to papers in mechanics. Compare Origins of Mathematical Symbols/Names
A concept can be explained in raw, dry definitions and formulas using thousands of sentences, or it can be explained in a swift, elegant way because the language has got just the right logical constructs and subtle interactions between linguistic concepts to express the logical constructs and interactions between the concepts in mathematical theory we are learning. Compare Examples of great mathematical writing
Related but orthogonal questions that might explain the nature of the problem at hand; further reading:
MO: What are some good resources for mathematical translation?
MO: What’s so great about blackboards?
MO: japanese/chinese for mathematicians?
Towards a New Model of Bilingual Mathematics Teaching: the case of China
Mathematics is Not a Universal Language - Tara N. Tevebaugh; Teaching Children Mathematics, Vol. 5, December 1998 - children are having problems understanding simple, basic mathematics across language barriers; what can be said about immensely more complex ideas that we try to juggle? Concepts can be expressed in mathematical notation; the motivation of those concepts cannot be.
Thanks!
P.S. This is one of my first few posts on MO. I welcome any comments that can make my contributions better.

REPLY [17 votes]: Not sure if this quite fits the criteria, but . . .
The Ramanujan-Nagell equation was introduced in 1913 by Ramanujan in the J. Indian Math. Soc. (Vol.5):

Question 464: $2^n - 7$ is a perfect square for the values 
  $3$, $4$, $5$, $7$, $15$ of $n$. Find other values.

The question was asked independently in 1943 by W. Ljunggren in
Norsk Mat. Tidsskr. (Vol.25), and answered by T.Nagell five years
later in the same journal ("Løsning till oppgave nr 2").
Since that journal is in Norwegian, most of the community
did not know of the result until 1961 when Nagell republished it in
Ark. Mat. (Vol.30, "The Diophantine equation $x^2 + 7 = 2^n$").
The result is that there are no solutions other than the five that
Ramanujan listed.<|endoftext|>
TITLE: Why does the naive definition of compactly supported étale cohomology give the wrong answer?
QUESTION [14 upvotes]: Illusie's article about étale cohomology available here (in French) mentions that the standard definition of compactly supported cohomology (and higher direct images with compact support) does not give the right answers in the case of étale cohomology.
He says Grothendieck had the idea of defining them as follows instead:
$$Rf_! = Rg_* \circ j_!$$
where $f = gj$ is a compactification of $f$: $g$ is proper, and $j$ is an open immersion. It takes a bit of work to see that this is well defined (a theorem of Nagata guarantees the existence of this compactification, and the proper base change theorem shows that the result does not depend on which compactification is chosen).
What is the explanation here for the difference between Grothendieck's definition and the usual one, which amounts to $Rf_! = R(g_* j_!)$ ? What is it that allows us to say that either is the "right" one?
It seems to me that a satisfactory explanation should be more than just computing the groups in a special case and appealing to intuition (indeed, the compactly supported cohomology groups are often strange, at first sight, even in the topological situation).
For example, I could imagine that an explanation might involve derived categories: after all, one of the main uses of $Rf_!$ is Verdier duality, which implies many useful properties such as Poincaré duality. This maybe gives some hints as to which definition is preferred.

REPLY [15 votes]: It is important in etale cohomology, as it is topology, to define cohomology
groups with compact support --- we saw this already in the case of curves in
Section 14. They should be dual to the ordinary cohomology groups.
The traditional definition (Greenberg 1967, p162) is that, for a manifold
$U$,
$
H_{c}^{r}(U,\mathbb{Z})=dlim_{Z}H_{Z}^{r}(U,\mathbb{Z})
$
where $Z$ runs over the compact subsets of $U$. More generally (Iversen 1986,
III.1) when $\mathcal{F}$ is a sheaf on a locally compact topological space
$U$, define
$
\Gamma_{c}(U,\mathcal{F})=dlim_{Z}\Gamma_{Z}(U,\mathcal{F})
$
where $Z$ again runs over the compact subsets of $U$, and let $H_{c}%
^{r}(U,-)=R^{r}\Gamma_{c}(U,-)$.
For an algebraic variety $U$ and a sheaf $\mathcal{F}$ on $U_{\mathrm{et}}$,
this suggests defining
$
\Gamma_{c}(U,\mathcal{F})=dlim_{Z}\Gamma_{Z}(U,\mathcal{F}),
$
where $Z$ runs over the complete subvarieties $Z$ of $U$, and setting
$H_{c}^{r}(U,-)=R^{r}\Gamma_{c}(U,-)$. However, this definition leads to
anomolous groups. For example, if $U$ is an affine variety over an
algebraically closed field, then the only complete subvarieties of $U$ are the
finite subvarieties, and for a finite subvariety $Z\subset
U$,
$
H_{Z}^{r}(U,\mathcal{F})=\oplus_{z\in Z}H_{z}^{r}(U,\mathcal{F}).
$
Therefore, if $U$ is smooth of dimension $m$ and $\Lambda$ is the constant
sheaf $\mathbb{Z}/n\mathbb{Z}$, then
$
H_{c}^{r}(U,\Lambda)=dlim H_{Z}^{r}(U,\Lambda)=\oplus_{z\in U}H_{z}%
^{r}(U,\Lambda)=\oplus_{z\in U}\Lambda(-m)$ if $r=2m$, and it is 0 otherwise
These groups are not even finite. We need a different definition...
If $j\colon\ U\rightarrow X$ is a homeomorphism of the topological space $U$
onto an open subset of a locally compact space $X$, then
$
H_{c}^{r}(U,\mathcal{F})=H^{r}(X,j_{!}\mathcal{F})
$
(Iversen 1986, p184).
We make this our definition.
From Section 18 of my notes: Lectures on etale cohomology.<|endoftext|>
TITLE: Is there good intution of the trace map?
QUESTION [6 upvotes]: I have never understood the trace map,not even after reading Geometric Interpretation of Trace. The problem with many answers in the above discussion is the geometric intuition does not apply to other field. 
As I don't want this to be closed, let me make the question more precise. Is there a definition of the trace map which
1) is basis independent, (there was a definition given by Sridhar Ramesh in the old post).
2) explains in an intuitive way why if $L$ is a finite separable extension of $K$, the map $ (x,y) \mapsto Tr(xy) $, where $x,y$ are in $L$, is a non degenerated bilinear form on L?

REPLY [9 votes]: I don't know if this is what you're looking for, but there's a basis-free definition of the trace in general, outside of the algebraic number theory context -- a linear transformation $V\to V$ corresponds in a natural way to an element of $V\otimes V^\ast$, and the trace map is the map $V\otimes V^\ast\to k$ induced by the bilinear map $V\times V^*\to k$ which sends $(v,f)$ to $f(v)$.
But I think the easiest way to see why the trace pairing is nondegenerate for a separable extension is to use bases. Intuitively, nonseparability corresponds to a linear dependency relation among the rows of the matrix because it leads to "repetitions" among the conjugates of some element of the field on top. There's a great (though kind of short) exposition of this stuff in Milne's notes on algebraic number theory: http://jmilne.org/math/CourseNotes/ant.html<|endoftext|>
TITLE: Noetherian rings of infinite Krull dimension?
QUESTION [33 upvotes]: Since Noetherian rings satisfy the ascending chain condition, every such ring must contain infinitely many chains of prime ideals s.t. the heights of these chains are unbounded. 
The only example I know of is the one due to Nagata [1962]: we take a polynomial ring in infinitely many variables over a field, and consider the infinite collection of prime ideals formed by disjoint subsets of the variables. Then we localise the ring by the complement of the union of these prime ideals.  With a little work, we can show that by appropriate choice of the subsets, the localised ring will be Noetherian and of infinite Krull dimension.  Eisenbud (ex. 9.6) provides a good walkthrough.
The question is: what are other examples of Noetherian rings of infinite Krull dimension?

REPLY [15 votes]: If you check the book "Krull Dimension" Memoirs of the American Mathematical Society 133  1973, you will find examples of commutative Noetherian integral domains of arbitrary infinite ordinal Krull dimension<|endoftext|>
TITLE: Can we bound degrees of complex irreps in terms of the average conjugacy class size?
QUESTION [14 upvotes]: This question arises when looking at a certain constant associated to (a certain Banach algebra built out of) a given compact group, and specializing to the case of finite groups, in order to try and do calculations for toy examples. It feels like the answer should be (more) obvious to those who play around with finite groups more than I do, or are at least know some more of the literature.
To be more precise: let $G$ be a finite group; let $d(G)$ be the maximum degree of an irreducible complex representation of $G$; and (with apologies to Banach-space theorists reading this) let $K_G$ denote the order of $G$ divided by the number of conjugacy classes.
Some easy but atypical examples:

if $G$ is abelian, then $K_G=1=d(G)$;
if $G=Aff(p)$ is the affine group of the finite field $F_p$, $p$ a prime, then $$K_{Aff(p)}=\frac{p(p-1)}{p}=p-1=d(Aff(p))$$

Question.
Does there exist a sequence $(G_n)$ of finite groups such that $d(G_n)\to\infty$ while
$$\sup_n K_{G_n} <\infty ? $$
To give some additional motivation: when $d(G)$ is small compared to the order of $G$, we might regard this as saying that $G$ is not too far from being abelian. (In fact, we can be more precise, and say that $G$ has an abelian subgroup of small index, although I can't remember the precise dependency at time of writing.) Naively, then, is it the case that having $K_G$ small compared to the order of $G$ will also imply that $G$ is not too far from being abelian?
Other thoughts.
Since the number of conjugacy classes in $G$ is equal to the number of mutually inequivalent complex irreps of $G$, and since $|G|=\sum_\pi d_\pi^2$, we see that $K_G$ is also equal to the mean square of the degress of complex irreps of $G$. Now it is very easy, given any large positive $N$, to find a sequence $a_1,\dots, a_m$ of strictly positive integers such that
$$ \frac{1}{m}\sum_{i=1}^m a_i^2 \hbox{is small while} \max_i a_i > N $$
so the question is whether we can do so in the context of degrees of complex irreps -- and if not, why not? The example of $Aff(p)$ shows that we can find examples with only one large irrep, but as seen above such groups won't give us a counterexample.

REPLY [7 votes]: As a matter of interest, Bob Guralnick and I proved in "On the commuting probability in finite groups", Journal of Algebra 300 (2006) 509–528, that what you call $K_{G} \to \infty$ as $[G:F(G)] \to \infty$, where $F(G)$ is the largest nilpotent normal subgroup of $G$ (this particular fact does not require the classification of finite simple groups, though several results from that paper do). 
(We were actually working with the reciprocal of your $K_{G}$). Hence, in insisting that $K_{G_n}$ is bounded, you are only allowing finitely many possibilities for the isomorphism type of 
$G_n/F(G_n)$. Now by a Theorem of Ito, the largest degree of an irreducible character of $G$ divides $[G:A]$ whenever $A$ is an Abelian normal subgroup of $G$. These facts together seem to indicate that the ``meat" of your question is contained in the difference between $F(G)$ and the maximal order Abelian normal subgroups of $G$. Of course, while there is a unique maximal nilpotent normal subgroup of any finite group, there is not usually a unique maximal Abelian normal subgroup of a finite group. In the above result that Guralnick and I proved, it is not possible to replace 
$[G:F(G)]$ by the index of an Abelian normal subgroup. Also, note that the largest degree of an irreducible character of $G$ is at most $[G:F(G)]$ times the largest degree of an irreducible character of $F(G)$ (and is at least as great as the largest degree of an irreducible character of 
$F(G)$). Since your question bounds $[G:F(G)]$, it can be made more precise that
your question is really about nilpotent groups.<|endoftext|>
TITLE: Do all Dedekind domains have the "Riemann-Roch property"?
QUESTION [7 upvotes]: Let $R$ be a Dedekind domain with fraction field $K$. 
Say that a Dedekind domain $R$ has the Riemann-Roch property if: for every nonzero prime ideal $\mathfrak{p}$ of $R$, there exists an element $f \in (\bigcap_{\mathfrak{q} \neq \mathfrak{p}} R_{\mathfrak{q}}) \setminus R$, i.e., an element of $K$ which is integral at every prime ideal $\mathfrak{q} \neq \mathfrak{p}$ and is not integral at $\mathfrak{p}$.  

Do all Dedekind domains have the Riemann-Roch property?

Motivation: for any subset $\Sigma \subset \operatorname{MaxSpec}(R)$, put $R_{\Sigma} := \bigcap_{\mathfrak{p} \in \Sigma} R_{\mathfrak{p}}$.  Then the maximal ideals of $R_{\Sigma}$ correspond bijectively to the maximal ideals $\mathfrak{p}$ of $R$ such that $\mathfrak{p} R_{\Sigma} \subsetneq R_{\Sigma}$.  Thus $\operatorname{MaxSpec}(R_{\Sigma})$ may be viewed as containing $\Sigma$.  $R$ has the Riemann-Roch property iff for all $\Sigma$, 
$\operatorname{MaxSpec}(R_{\Sigma}) = \Sigma$.  Equivalently, the mapping $\Sigma \mapsto R_{\Sigma}$ is an injection.  
Remarks: $R$ has the Riemann-Roch property if its class group is torsion: then for every $\mathfrak{p} \in \operatorname{MaxSpec}(R)$ there exists $n \in \mathbb{Z}^+$ and $x \in R$ such that $\mathfrak{p}^n = (x)$, so take $f = \frac{1}{x}$.  Also the coordinate ring $k[C]$ of a nonsingular, integral affine curve $C$ over a field $k$ has the Riemann-Roch property...by the Riemann-Roch theorem.  Unfortunately this already exhausts the most familiar examples of Dedekind domains!

REPLY [7 votes]: Yes. Given a maximal ideal $P$ there exists $x \in K \backslash R_P$. Let $S$ be the finite set of maximal ideals $Q$ so that  $x \notin R_{Q}$. For each $Q \in S$ such that $Q \neq P$ let $y_Q \in Q\backslash P$. The element $f$ given by multiplying $x$ by large positive powers of all the $y_Q$ has the desired property.<|endoftext|>
TITLE: Is there a category-theoretic definition of the arithmetic Grothendieck group
QUESTION [7 upvotes]: Let $X$ be a regular scheme which is flat over $\mathbf{Z}$. The arithmetic Grothendieck group $\hat{K}(X)$ is defined to be the quotient of $\hat{G}(X)$ by $\hat{G}^\prime(X)$. This is actually quite a length definition which I added below for the sake of completeness.
In the classical case, for $X$ any noetherian scheme, the Grothendieck group $K_0(X)$ is defined to be the Grothendieck group of the category of vector bundles on $X$. That is, one applies the notion of a Grothendieck group for an additive subcategory of an abelian category. (In our case the abelian category is the category of coherent sheaves on $X$.) This means just modding out by short exact sequences. 
I would like to know if there is a categorical type of definition for this group. Thus, first one needs to decide what kind of categories we're talking about (objects are pairs in some sense) and then the notion of exact sequence should coincide in some sense with the below definition. 
Probably there is no such thing. I just ask this question in order to understand the arithmetic Grothendieck group better.
Note. Let me sketch the definition of the arithmetic Grothendieck group as given in Faltings. In the above $\hat{G}(X)$ is the direct sum of "the free abelian group generated by all vector bundles which have a hermitian metric on $X_{\mathbf{C}}$ which is invariant under complex conjugation $F$" and the abelian group $\widetilde{A}^\ast(X)$. Here $\widetilde{A}^\ast(X)$ is generated by all $p$-forms $\alpha^p$ such that $F^\ast \alpha^p = (-1)^p \alpha^p$. Furthermore, $\hat{G}^\prime(X)$ is the subgroup generated by elements of the form $E_2 - E_1-E_3 - \widetilde{ch}(E)$, where $E$ is the short exact sequence $$0\rightarrow E_1 \rightarrow E_2 \rightarrow E_3 \rightarrow 0$$ and $\widetilde{ch}(E)$ is the secondary Chern form.

REPLY [2 votes]: The classical group $K_0$ can also be thought of as consisting of equivalence classes of chain complexes of vector bundles, such that the exact sequences represent the zero of $K_0$ --- and furthermore every complex is equivalent to a two-term complex, a.k.a. a bundle morphism; the graded tensor product of complexes also gives a ring structure on $K_0$.  If you like, classical $K_0$ is a categorical interpolation between the Euler Characteristic and the homology of a chain complex.  The present construction looks like a refinement of that idea, where instead of representing a trivial element, an exact sequence is equivalent to a particular (sum of) differential form(s).<|endoftext|>
TITLE: Is there a "primitive-recursively enumerable" set whose complement is not such?
QUESTION [8 upvotes]: Call a subset of $\mathbb{N}$ primitive-recursively enumerable (p-r.e.) if it is empty or an image of a primitive recursive function. I feel like a lot must be known about the poset of such sets ordered by inclusion, but I am unable to dig up references. Concretely, I would like to know whether there exists a p-r.e. set whose complement is not p-r.e.
The answer is affirmative if there is a complete set (in the sense of many-to-one reducibilities) that is enumerated by a primitive recursive function. My hunch is that such a set exists, but cannot come up with one.

REPLY [8 votes]: There is a stronger result: Every r.e. set is primitive r.e. in your sense.
Short proof: Kleene's Normal Form Theorem.
Longer proof:  Let S be an r.e. set, assumed WLOG nonempty; fix a ∈ S, and fix an algorithm e where S is precisely the range of the function computed by e.
Consider the following algorithm:  Given the input pair (n, M), run e on input n for M steps.  If it gives an output by then, output whatever e outputs; otherwise output a.
The functions which set up the initial state of computation, advance a state by one step, and extract the output from a final state, are all p.r.  Thus the above algorithm defines a p.r. function, and it is easy to check that its range is S.
Edit: Cutland's Computability is a decent resource for these questions.

REPLY [7 votes]: This is an interesting question. From B. Rosser, Extensions of some theorems of Gödel and Church:

Corollary I. If a class can be enumerated (allowing repetitions) by a general recursive function, it can be enumerated (allowing repetitions) by a primitive recursive function.

Hence any complete recursively enumerable set (such as K) should work.

REPLY [7 votes]: I claim that a set is primitive recursive enumerable if and only if it is computably enumerable. So the answer to your question is affirmative. 
Clearly any p-r.e. set is c.e., since primitive recursive functions are computable. Conversely, suppose that A is computably enumerable. We want to show A is p-r.e. If A is empty, then we're done. So fix some element a0 in A. Since A is c.e., it is the domain of a computable function f, computed by program e. Consider now the function h(s,n) = n, if s codes the proof of a halting computation of program e on input n, and otherwise h(s,n) = a0. The function h is defined by Δ0 cases, and hence is primitive recursive. Also, the range of h is A, as desired. 
So there are numerous sets A as you desire!<|endoftext|>
TITLE: smooth Gelfand-duality
QUESTION [6 upvotes]: Assume $M$ is a compact smooth manifold (without boundary). What can we say about the spectrum of the $\mathbb{R}$-algebra $A=C^{\infty}(M)$? The elements of $M$ give rise to rational points of $A$, are there other ones? Does the smooth structure of $M$ endow $A$ with additional structure such that $M$ can be completely recovered from $A$? In other words, is there some kind of smooth Gelfand-duality?

REPLY [3 votes]: If I am not mistaken the algebraic data distinguishing $C(M)$ from $C^\infty(M)$ is that $C^\infty(M)$ is equipped with a space of derivations which is a module over the algebra $C^\infty(M)$.
A derivation in this case is an $\mathbb R$-linear map $D$ of $C^\infty(M)$ to itself satisfying Leibniz's product rule: $D(fg) = D(f)g + fD(g)$ for all $f,g\in C^\infty(M)$.
I don't think the Gelfand duality itself is different from what you'd expect. In fact, the point of the Gelfand duality in this case would be to prove that $C(M)$ is the closure of $C^\infty(M)$ under the compact-open topology. The differentiable manifold structure is given by the derivations.<|endoftext|>
TITLE: Where to start reading into p-adic non-abelian Hodge theory? 
QUESTION [8 upvotes]: I'm curious about Faltings' "A p-adic Simpson correspondence ". Do you know more detailed, introductory,  expositions, surveys, texts of seminars on that?
Edit: Annette Werner's survey "Vector Bundles on Curves over C_p" seems to be related.
Edit: The first part of a "new approach for the p-adic Simpson correspondence, closely related to the original approach of Faltings, but also inspired by the work of Ogus and Vologodsky on an analogue in characteristic p>0". An other related article.
Edit: today new in arxiv - "Non-abelian Hodge theory for algebraic curves over characteristic p"

REPLY [3 votes]: This doesn't answer the question, but you might want to check out Martin Olsson's
Towards non--abelian $P$--adic Hodge theory in the good reduction case.
In another (maybe too different to be useful - I'm totally ignorant here) 
direction there's the work of Ogus and Vologodsky
Nonabelian Hodge Theory in Characteristic p.<|endoftext|>
TITLE: Stacks determined by their coarse moduli spaces
QUESTION [9 upvotes]: Is there a non-trivial class $S$ of smooth Deligne-Mumford stacks over a base $B$ with the property that if $\mathcal{X}, \mathcal{Y} \in S$ have isomorphic coarse moduli spaces (assumed to exist) then $\mathcal{X} \cong \mathcal{Y}$? If $B = Spec(k)$, $k$ a field (of characteristic $0$ if necessary), can one take $S$ to be the class of all irreducible, smooth, separated DM stacks with trivial inertia in codimension $\leq 1$?

REPLY [17 votes]: Yes, the class of all smooth, separated DM stacks over a field of characteristic $0$, with trivial inertia in codimension at most $1$, over a field of characteristic $0$, has the propery you want. The point is that every moduli space of such a stack has quotient singularities; and every variety with quotient singularities is the moduli space of a unique such stack. I believe that this was first proved in Angelo Vistoli: Intersection theory on algebraic stacks and on their moduli spaces.  Invent. Math.  97  (1989),  no. 3, 613-670, Proposition 2.8 (uniqueness is not stated there, but it follows from the proof).<|endoftext|>
TITLE: Is there a natural way to view the proof of Hilbert 90?
QUESTION [33 upvotes]: I only know of one proof of Hilbert 90, which is very smart if not magical. See for example http://hilbertthm90.wordpress.com/2008/12/11/hilberts-theorem-90the-math/
Does anyone know of a more intuitive proof or know a good way to view the proof?
I have accepted the answer by Emerton, great thanks as well to David Speyer and Brian Conrad.

REPLY [4 votes]: Another way to conclude that $\tau$ has a fixed point using linear algebra is the following. David Speyer shows that $\tau$ is of order $n$. But, thanks to independence of characters, the minimal polynomial of $\tau$ is $X^n-1$ and thus $\tau$ is a cyclic endomorphism of $L$. Hence 1 is an eigenvalue of $\tau$ with multiplicity one.<|endoftext|>
TITLE: Low dimensional nilpotent Lie algebras
QUESTION [7 upvotes]: This is a reference request question. I would like to know more on the structure of low dimensional nilpotent lie algebras. I heard that up to dimension 6 there are only finitely many isomorphism classes, and every such algebra admits a gradation with only positive degrees (see http://en.wikipedia.org/wiki/Graded_Lie_algebra). 
Do you know of any source where I can find the corresponding proofs?

REPLY [12 votes]: There is indeed lot of works devoted to classification of nilpotent Lie algebras of low dimension (those cited above and many more), with numerous mistakes and omissions. Even worse, all they are using different nomenclature and invariants to classify the algebras, and it is a nontrivial task to compare different lists. Luckily, Willem de Graaf undertook a painstaking task to make an order out of this somewhat messy situation in "Classification of 6-dimensional nilpotent Lie algebras over fields of characteristic not 2", J. Algebra 309 (2007), 640-653 (http://dx.doi.org/10.1016/j.jalgebra.2006.08.006 ); arXiv:math/0511668 . Even better, he provides an algorithm for identifying any given nilpotent Lie algebra with one in his list, and the corresponding code is available as a part of GAP package. He builds on earlier work of Skjelbred-Sund cited above and his own method of identification of Lie algebras by means of Groebner bases.<|endoftext|>
TITLE: Dyer-Lashof based spectral sequence for homotopy classes of maps between infinite loop spaces (spectra).
QUESTION [14 upvotes]: The homology of an infinite loop space, which represents a spectrum, is an algebra over the Dyer-Lashof algebra (see for example Cohen-Lada-May's Springer volume, or for part of the story the more accessible Luminy notes of Bisson-Joyal).  Has anyone used this to construct a spectral sequence converging under some assumptions to $[X,Y]$, the homotopy classes of infinite-loop-maps between $X$ and $Y$, which starts with some kind of derived (Ext/Tor) maps between their homology in the category of algebras over the Dyer-Lashof algebra?  Have any calculations been done with such a spectral sequence?

REPLY [4 votes]: I'm not sure if this is what you want, but Haynes Miller constructs a spectral sequence computing the homology of a connective spectrum $E$ from the homology of $E_0$ as a Hopf algebra over the Dyer-Lashof algebra in the 1978 Pacific Journal of Mathematics paper "A spectral sequence for the homology of an infinite delooping."<|endoftext|>
TITLE: Intuition behind existence of moduli space of stable curves
QUESTION [8 upvotes]: I'm not entirely sure that the title is what I'm looking for. What I'm really asking is for intuition as to why $\bar{\mathcal{M}_g}$ is the compactification of $\mathcal{M}_g$. I'm sure this is covered in the more classic papers (like Deligne and Mumford), but I still find those hard to penetrate.

REPLY [5 votes]: One perspective is geometric. Non-singular closed Riemann surfaces have a unique complete hyperbolic metric by uniformization, with a decomposition into thin and thick regions, where the thin regions are  collar neighborhoods of short geodesics, and the thick regions are compact, with each component having uniformly bounded diameter (depending only on the genus). In the case of a non-compact Riemann surface of finite hyperbolic areaa, one also has cusp components of the thin part.  One also sees that the geometry of a Riemann surface is uniquely determined by the geometry of the thick part, together with which pairs of boundary components bound collar neighborhoods. One may take a limit of a sequence of surfaces with geodesics whose length approaches zero, and the thick parts converge geometrically to a finite area hyperbolic surface, with a pair of cusps for each geodesic whose length has gone to zero. These cusp pairs are naturally identified with noded Riemann surfaces, giving the Deligne-Mumford compactification. The fact that it is compact follows from the compactness of the thick regions in the Gromov-Hausdorff topology. 
It is an interesting result of Masur that the compactification is also given by the metric completion of the Weil-Petersson metric on moduli space.<|endoftext|>
TITLE: Is there any approximated version of Hilbert 90?
QUESTION [5 upvotes]: Suppose $K$ is a local field and $L$ a finite cyclic extension of $K$. By Hilbert 90, we know that if an element $a$ in $ L$ such that $N_{L / K}(a) =1 $ then $ a = b / \sigma(b) $ for some $b$ in $L$ and $ \sigma$ a generator of the Galois group.
My question: Suppose $N_{L / K}(a) \simeq 1 $, is there some $b$ such that  $ a \simeq b /\sigma(b) $?
I guess I could have try to make the question more precise, but there seems to be some merits in leaving it a bit vague.
I have accept the the answer by Paul Broussous, which address the situation when the extension is unramified. It is because that is what I need. I am still curious whether something can be done when the extension is totally ramified?

REPLY [8 votes]: Yes such approximated versions of Hilbert 90 do exist. But you need some technical conditions.
For instance assume that $L/K$ is unramified of degree $d$ and that 
$a\in {\mathfrak o}_{L}^{\times}$. 
Then you condition writes 
$N_{L/K}(a)\equiv 1$  modulo  $\mathfrak{p}_{K}^{n}$,
for some $n>0$ (I assume that this 
is what you mean by $\simeq$). This may be rewritten 
$N_{L/K}(a)=1$ in $U_{L}/U^{n}_L$, where $U$ denotes a 
unit group. So the map 
$$\sigma^u\mapsto a\sigma (a) \cdots \sigma^u (a)$$
 defines a $1$-cocycle of ${\rm Gal}(L/F)$ in $U_L /U^n_L$  (here $\sigma$ denotes the Frobenius substitution). 
So what you want is that this cocycle is split. In fact we have $H^{1}({\rm Gal}(L/K), U_{L}/U^{n}_{L})=1$. This is proved by a standard filtration argument: this is implied by 
$$H^{1}({\rm Gal}(L/K), U_L /U^{1}_L ) = H^{1}({\rm Gal}(k_L /k_K ), k_{L}^{\times})=1$$
and 
$$H^{1}({\rm Gal}(L/K), U_{L}^{i}/U_{L}^{i+1})=H^{1}({\rm Gal}(k_L /k_K ), k_L )=1$$
here $k$ denotes a residue field. You can find the detail of the proof in, I think,  Serre's 'Local fields' or Cassels-Fröhlich's 'Algebréaic Number Theory'.<|endoftext|>
TITLE: Do non-associative objects have a natural notion of representation?
QUESTION [32 upvotes]: A magma is a set $M$ equipped with a binary operation $* : M \times M \to M$.  In abstract algebra we typically begin by studying a special type of magma: groups.  Groups satisfy certain additional axioms that "symmetries of things" should satisfy.  This is made precise in the sense that for any object $A$ in a category $C$, the invertible morphisms $A \to A$ have a group structure again given by composition.  An alternate definition of "group," then, is "one-object category with invertible morphisms," and then the additional axioms satisfied by groups follow from the axioms of a category (which, for now, we will trust as meaningful).  Groups therefore come equipped with a natural notion of representation: a representation of a group $G$ (in the loose sense) is just a functor out of $G$.  Typical choices of target category include $\text{Set}$ and $\text{Hilb}$.
It seems to me, however, that magmas (and their cousins, such as non-associative algebras) don't naturally admit the same interpretation; when you throw away associativity, you lose the connection to composition of functions.  One can think about the above examples as follows: there is a category of groups, and to study the group $G$ we like to study the functor $\text{Hom}(G, -)$, and to study this functor we like to plug in either the groups $S_n$ or the groups $GL_n(\mathbb{C})$, etc. on the right, as these are "natural" to look at.  But in the category of magmas I don't have a clue what the "natural" examples are.  
Question 1:  Do magmas and related objects like non-associative algebras have a "natural" notion of "representation"?
It's not entirely clear to me what "natural" should mean.  One property I might like such a notion to have is an analogue of Cayley's theorem.  
For certain special classes of non-associative object there is sometimes a notion of "natural": for example, among not-necessarily-associative algebras we may single out Lie algebras, and those have a "natural" notion of representation because we want the map from Lie groups to Lie algebras to be functorial.  But this is a very special consideration; I don't know what it is possible to say in general.  
(If you can think of better tags, feel free to retag.)
Edit:  Here is maybe a more focused version of the question.
Question 2:  Does there exist a "nice" sequence $M_n$ of finite magmas such that any finite magma $M$ is determined by the sequence $\text{Hom}(M, M_n)$?  (In particular, $M_n$ shouldn't be an enumeration of all finite magmas!)  One definition of "nice" might be that there exist compatible morphisms $M_n \times M_m \to M_{n+m}$, but it's not clear to me that this is necessarily desirable.
Edit:  Here is maybe another more focused version of the question.
Question 3:  Can the category of magmas be realized as a category of small categories in a way which generalizes the usual realization of the category of groups as a category of small categories?
Edit:  Tom Church brings up a good point in the comments that I didn't address directly.  The motivations I gave above for the "natural" notion of representation of a group or a Lie algebra are in some sense external to their equational description and really come from what we would like groups and Lie algebras to do for us.  So I guess part of what I'm asking for is whether there is a sensible external motivation for studying arbitrary magmas, and whether that motivation leads us to a good definition of representation. 
Edit:  I guess I should also make this explicit.  There are two completely opposite types of answers that I'd accept as a good answer to this question:

One that gives an "external" motivation to the study of arbitrary magmas (similar to how dynamical systems motivate the study of arbitrary unary operations $M \to M$) which suggests a natural notion of representation, as above.  This notion might not look anything like the usual notion of either a group action or a linear representation, and it might not answer Question 3.
One that is "self-contained" in some sense.  Ideally this would consist of an answer to Question 3.  I am imagining some variant of the following construction: to each magma $M$ we associate a category whose objects are the non-negative integers where $\text{Hom}(m, n)$ consists of binary trees with $n$ roots (distinguished left-right order) and $m$ "empty" leaves (same), with the remaining leaves of the tree labeled by elements of $M$.  Composition is given by sticking roots into empty leaves.  I think this is actually a 2-category with 2-morphisms given by collapsing pairs of elements of $M$ with the same parent into their product.  An ideal answer would explain why this construction, or some variant of it, or some other construction entirely, is natural from some higher-categorical perspective and then someone would write about it on the nLab!

REPLY [2 votes]: There is a thing occurring in the overkill context of Lurie's Higher Algebra which seems related to this question.
Namely, in Ch 3.3, Lurie introduces a fairly general notion of "module for an algebra of an operad". More precisely, we fix the following data:

A unital operad $O$ (here "unital" means that for every color $C$ of the operad, there is a unique nullary $C$-valued operation). For example, $O$ could be the operad for unital magmas. I don't know how essential the unitality hypothesis is.
A "fibration of generalized operads" $C \to O$. I believe we're supposed to think of this as a category $C$ with "$O$-monoidal structure", or an $O$-algebra object in $Cat$. For example, if $O$ is the operad for unital magmas, then any monoidal category acquires such a structure by restriction along the map from $O$ to the associative operad.
An $O$-algebra $A$ in $C$. I'm not 100% sure I'm unwinding the definitions correctly (Lurie has several subtly different concepts he denotes by "$Alg$" with various decorations, and I'm not straight on what's what), but I believe that when $O$ is the operad for unital magmas and $C$ is a monoidal category, then $A$ is just a unital magma object in $C$.

Given this data, Lurie defines a category of "$O$-module objects over $A$", denoted 
$$Mod_A^O(C)$$
I don't have a great grasp on this -- part of the issue is that in fact, Lurie doesn't isolate this construction, instead jumping straight to the construction of this category plus a full $O$-monoidal structure (and even packaging it all up in a bigger construction which allows $A$ to vary), which requires his technical "coherence" hypothesis. (Caveat: Lurie doesn't prove that $Mod_A^O(C)$ is a category in general -- without the "coherence" hypothesis, it might just be some simplicial set. But I'll proceed under the assumption that even though coherence is needed to get an $O$-monoidal structure on $Mod_A^O(C)$, it's probably the case that $Mod_A^O(C)$ is at least a category in greater generality.) However, Lurie does say that

When $O$ is the commutative operad, we recover the usual notion of module over a commutative algebra $A$;
When $O$ is the associative operad, we recover the notion of an $(A,A)$-bimodule.

That second point really throws me, and bears repeating -- for Lurie, a "module over an associative algebra object" is neither a left nor a right module, but rather a bimodule (I hasten to add that he does develop the theory of left and right modules separately, without reference to this more general context). For example, Lurie's notion does not recover the notion of a group acting on a set. Lurie explains that the motivation for doing it this way (the idea of which I think he attributes to John Francis) is that he wants to have an $O$-monoidal structure on the category of $A$-modules -- and of course left $A$-modules don't generally have a monoidal structure, but $(A,A)$-bimodules do. I have no idea whether the operad for unital magmas satisfies the technical "coherence" condition guaranteeing that $Mod_A^O(C)$ does in fact have an $O$-monoidal structure; if it doesn't, then the main motivation for introducing this notion evaporates and perhaps it ends up not being useful. But it's still there.
If I have the chance to unwind what this construction yields in the case of the operad for magmas and ordinary 1-categories, I will edit to add more. Or perhaps somebody more familiar with this stuff can chime in.<|endoftext|>
TITLE: Is the pure braid group on three strands generated as a normal subgroup of the braid group by the six-crossing braid?
QUESTION [15 upvotes]: Artin's presentation of braid group on three strands is:
$$ B_3 = \langle l,r : lrl = rlr \rangle $$
where you should think of "$l$" as the positive crossing between the left and middle strands and "$r$" as the positive crossing of the right and middle strands:
     |   |   |         |   |   |
      \ /    |         |    \ /
 l =   \     |     r = |     \
      / \    |         |    / \
     |   |   |         |   |   |

Then there is a surjection $B_3 \to S_3$ given by $l \mapsto (12)$ and $r \mapsto (23)$.  ($S_3$ is the symmetric group on three letters: it is generated by $(12)^2 = 1 = (23)^2$ and the braid relation above.)  The pure braid group $PB_3$ is the kernel of this surjection.
The six-crossing braid $b = lr^{-1}lr^{-1}lr^{-1}$ is an element of the pure braid group.  Let $N$ be the minimal normal subgroup of $B_3$ that contains $b$.  Certainly $N \subseteq PB_3$.

Question: Do we have $N = PB_3$?

Motivation
The motivation for my question comes from a neat trick that Conway showed us years ago.  It leads to a more nuanced question than what I asked that I will pose as its own question if the answer above is "no".  My memory is that at the time Conway did not know the answer to the more nuanced question, which suggests that the answer above cannot be "yes".
Take a long and reasonably thin rectangle of paper, and score it with two cuts, so that you have three strips of paper that are connected at both ends, so that you end up with a (framed, oriented, ...) "theta graph":
 |--------|
 |        |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |        |
 |--------|

Then with some finagling, you can in fact "tie" the six-crossing braid in those three strands, without ripping the paper further, by passing the bottom through itself a few times.  (The trick is that it's easier to unbraid than to braid, so make your paper long enough that you can put $bb^{-1}$ into it, and then unbraid the $b^{-1}$.)  Put another way: you can first put a hair-tie at the end of your ponytail, and then braid your hair.
The harder question is to characterize all braids that you can put on the above "theta graph".  The following facts are essentially obvious:

Any "braiding" of the theta graph is pure.
The set of braidings of the theta graph is a subgroup of $B_3$.
The set of braidings of the theta graph is closed under conjugating by arbitrary braids.

Therefore, the set $T$ of braidings of the theta graph is a normal subgroup of $B_3$, with $N \subseteq T \subseteq PB_3$.  In particular, a positive answer to the question above characterizes $T$.

REPLY [3 votes]: Does anyone have a reference from the literature for the fact that $PB_3=F_2\times F_1$?<|endoftext|>
TITLE: How to classify the algebras C^∞(M)?
QUESTION [16 upvotes]: This continues my question about smooth Gelfand-duality. In the book

Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824

it is shown that $M \mapsto C^\infty(M)$ is a fully faithfull contravariant functor from the category of manifolds (smooth, separable and without boundary) to the category of $\mathbb{R}$-algebras. Isn't this nice? It would be even more nice if there is an algebraic description of the essential image of this functor, so that we have an antiequivalence of categories between manifolds and certain $\mathbb{R}$-algebras. Thus my question is:

Which $\mathbb{R}$-algebras $A$ are isomorphic to $C^{\infty}(M)$ for some manifold $M$?

Of course, you could just formulate that $Spec_r(A)=Hom(A,\mathbb{R})$ with the obvious structure sheaf is a manifold and that the canonical map $A \to C^{\infty}(M)$ is an isomorphism in terms of the ring structure of $A$. But this does not seem to be handy at all. I want some nontrivial purely algebraic formulation. If possible avoiding structure sheaves at all.
Here are some necessary conditions:

If $f \neq g$ in $A$, then there is some $\mathbb{R}$-homomorphism $\phi : A \to \mathbb{R}$ such that $\phi(f) \neq \phi(g)$. In particular, $A$ is reduced.
For every $p \in Spec_r(A)$ with corresponding maximal ideal $m_p$, then the maximal ideal $\overline{m_p}$ of $A_{\mathfrak{m}_p}$ is finitely generated, say by elements $f_1,...,f_n$, and the canonical map
$\mathbb{R}[t_1,...,t_n] / (t_1,...,t_n)^{r+1} \to A_{\mathfrak{m}_p} / \overline{m_p}^{r+1}, t_i \mapsto f_i$
is an isomorphism for all $r \geq 0$.
With the notation above, the canonical map $A/m_p^{r+1} \to A_{\mathfrak{m}_p} / \overline{m_p}^{r+1}$ is an isomorphism.
The function $Spec_r(A) \to \mathbb{N}, p \to \dim_\mathbb{R} \mathfrak{m}_p/{\mathfrak{m}_p}^2$ is locally constant.

Are they sufficient [no, see Michael's answer]? Finally [solved by Dmitri's answer]:

How can we characterize the algebras (at least within all the $C^{\infty}(M)$'s), that come from compact manifolds?

You might admit that "$Spec_r(A)$ is compact with the Gelfand topolgy" is not a satisfactory answer ;-).
Addendum: At first glance, it appears too optimistic to find an algebraic characterization. But many famous problems started like that and involved unexpected methods. I don't claim that this applies to my problem. But at least I invite you to think about it. The properties of the algebras above are just an approximation. Even if we add some of the conditions in the answers (such as $\cap_{r} \overline{m}_p^{r+1} \neq 0$), it would be a great surprise that the conditions are sufficient. But I'm not convinced of the contrary as soon someone provides a counterexample. It is fun trying to deduce some of the differential geometric theorems such as IFT from the properties above (if $A \to B$ is an isomorphism in one tangent space, then it is a local isomorphism). Perhaps a first step is to characterize the local rings $C^{\infty}_p(\mathbb{R}^n)$.

REPLY [2 votes]: This is to expand further on something I wrote in the comments.
Martin wrote:

Here are some necessary conditions: [...] Are they sufficient?

I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:
Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\lim_{r \geq 1} A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.
In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis.
(Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).<|endoftext|>
TITLE: When is a Homology Class Represented by a Submanifold?
QUESTION [41 upvotes]: Possible Duplicate:
Cohomology and fundamental classes 

Given an oriented manifold     $M$ and an oriented submanifold     $\phi:N\to M$ we can obtain a homology class     $\phi_*[N]\in H_*(M)$ where     $[N]$ is the fundamental class of     $N$.  In general, it is not true that every homology class of     $M$ can be represented by a submanifold in this manner, however for some special cases it is.
For example, for     $M$ an oriented (and closed maybe?) 4-manifold every homology class can be represented by a submanifold.  Another example is when     $M$ an Euclidean configuration space.
My questions are:
1) Under what circumstances can every homology class of     $M$ be represented by a submanifold and
2) What are some examples of manifolds who have homology classes not representable in this manner?

REPLY [30 votes]: Here are a few simple answers to the question you asked:

Every class in $H_{n-1}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to S^1$ representing the Poincare dual in $H^1(M;Z)=[M,S^1]$ and take the preimage of a point. In dimensions>2 it can be taken connected.
Similarly, every class in $H_{n-2}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to CP^\infty$ representing the Poincare dual in $H^2(M;Z)=[M,CP^\infty]$, homotop $f$ into a finite skeleton, say $CP^N$, and take the preimage of $CP^{N-1}$.
Transversality says that if you can represent $x\in H_k(M)$ by a map of a smooth manifold (e.g.  elements in the image of the Hurewicz map, or by Thom) , and $2k < n$, then you can represent it by an embedded submanifold (as Andy mentions above). For example, any class in $H_1(M)$ for $dim(M)\ge 3$.  With care you can also make this work for $2k=n$, and there are techniques available in the "metastable" range (no triple points) involving generalizations of Whitney's trick and other ways to replace double points.<|endoftext|>
TITLE: classification of smooth involutions of torus
QUESTION [12 upvotes]: Let $\mathbb{Z}_2=\{1,g\},T^2=\{(e^{i\theta_1},e^{i\theta_2})\}$ and place $T^2$ in $\mathbb{R}^3$ as the locus of the rotation of $2\pi$ rads of the circle$\{(y,z)|(y-2)^2+z^2=1\}$ around $z$ axis.
It is known that there are 5 nonequivalent smooth involutions on torus,and they are:
1.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i(\theta_1+\pi)},e^{i\theta_2})$ (rotation$\pi$ rads around $z$ axis) with null fixed point set and orbit space $T^2$
2.$g(e^{i\theta_1},e^{i\theta_2})=(e^{-i\theta_1},e^{i\theta_2})$(reflection along $x=0$) with fixed point set $S^1\times S^0$ and orbit space an annulus
3.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i\theta_2},e^{i\theta_1})$(switch the two coordinates) with fixed point set the diagonal circle and orbit space Mobius band
4.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i(\theta_1 +\pi)},e^{-i\theta_2})$(restriction of the involution $(x,y,z,\mapsto (-x,-y,-z)$ of $\mathbb{R}^3$ to torus)with null fixed point set and orbit space klein bottle
5.$g(e^{i\theta_1},e^{i\theta_2})=(e^{-i\theta_1},e^{-i\theta_2})$(reflection along $x=0$ plus reflection along $z=0$)
with fixed point set 4 points and orbit space $S^2$
i want to know how to derive the result above.for the free case it seems easy.since the action is free,the orbit space must be a manifold also,and has euler char 0,hence must be torus or klein bottle.
for the nonfree case,the orbit is not manifold,but "orbifold".
and we have Riemann-Hurwitz Formula:
$\chi(O)=\chi(X_O)-\sum_{i=1}^n (1-\frac{1}{q_i})-\frac{1}{2}\sum_{j=1}^m (1-\frac{1}{r_j})$
here$\chi(O)$ is the orbifold euler char and $\chi(X_o)$ is the euler char of the underlying space associated to the orbifold $O$,and $q_i$and $r_j$ denote the angles for sigular points(cone points and reflector corners
can we determine the remaining 3 involutions by using this formula?Thank you!

REPLY [7 votes]: Here is a sketch -- some of the details are a bit hazy:
Suppose that $\iota$ is a smooth involution of $T^2$.  Show that the fixed point set of $\iota$ is a submanifold.  Show that the orbit space of $\iota$ is an orbifold with orbifold Euler characteristic zero. Using the orbifold Euler characteristic you can enumerate all 17 compact, connected, 2-dimensional orbifolds of orbifold Euler characteristic zero.  Now rule out 12 of these for topological reasons. 
The second to last step is a nice exercise that everybody should do once, after learning about the orbifold Euler characteristic.  The non-trivial part in the last step is eliminating $D(2,2;)$ and $P(2,2)$.  Getting rid of the others is easy.<|endoftext|>
TITLE: Making an intuition precise
QUESTION [9 upvotes]: In writing my senior thesis I met the following problem: Sometimes I have some intuition about some mathematical statement. Yet I find it extremely painful trying to put these intuition into precise form on paper. In particular it is very hard to specify the correct condition for statement.
Does anyone have some tips for me in doing so? How do you often do it?
Let me elaborate a bit further. I believe it is an experience that any mature mathematician must have went through. We want to go to D, and we need to go through A, B, C. But A can not be stated clearly until one knows B, B can not be stated clearly until one knows A and C, and C can not be stated clearly until one knows B. But we sort of have a vague picture of A, B, C in our mind. It sounds very stupid, but I don't know where to start.
I wonder if this question is too vague for MO. So please close it if you see fit.

REPLY [7 votes]: My first intuition is always to talk to somebody. This can help, since the speech center is a different part of the brain, works differently, taps different parts of the brain. In that way, even speaking out loud what you want to talk about can supply you with a surge of inspiration. Don't forget that written words are just a way to store the sounds we communicate with.
That said a second opinion is always very good. On the one hand someone smarter might have a ready answer, but on the other hand explaining the material (maybe from ground up, if that's not too much) to someone who doesn't know anything about it yet is apt can give you a lot of inspiration.
Try thinking about the object at hand in terms of real-life words. Ask yourself 'what happens if it does X? What if it does Y?' trying to learn a bit more about the object at hand - just like a physicist could pick up a mechanical contraption, look at it from many angles, and try turning a gear here and there, we have to do that with mathematical objects. Most of this is done on paper, using formulas and words.
Sometimes you can't put together theorems because the notation you came up with isn't intuitive enough - try thinking of a system of notation that limited to your current point of interest will be coherent, complete, and intuitive.
However sometimes some things simply will not 'tick' unless seen or heard: in signal processing you sometimes must listen to (some form of) the signals or functions at hand to understand what's going on. Fourier analysis helps visualize, but it doesn't do justice to the information you can get. In real analysis the picture of a saddle point cannot be replaced by any amount of writing - you just have to see it. Similarly an animation cannot be replaced with a few pictures. Example: I would have never noticed this effect had the display not been animated MO: effect in additive resynthesis
Try visualizing your mathematical objects. Ask a fellow geometer to help you figure out a pretty illustration. Visualizing is important because, again, it taps into different parts of our brain, all which can work for the cause rather than sitting around doing nothing. And they work better when they're looking at something pretty, rather than ugly! If you visualized the data/theorem already, try visualizing differently. Maybe you're thinking of some specific representative of the kind of object your theorem talks about, while the theorem visualizes better on a different one? (example: intermediate value theorems don't visualize too well on straight line functions..)
Another option is it might be a "writer's block". Just forget about it for some time, and get back to it later. Stop thinking about it, have a full day without working on maths, meditate, take a nap, go to the movies, sleep over it, have a good, long, enjoyable, dinner, relax listening to some music.<|endoftext|>
TITLE: Reference request: A theorem by S. Garrison
QUESTION [5 upvotes]: A theorem by S. Garrison states that if $G$ is a finite solvable group and $|cd(G)| = 4$ then $dl(G)\leq |cd(G)|$ (the Taketa inequality, which is conjectured to hold for all finite solvable groups). So far I have been unable to find a proof of this theorem anywhere. The only references I have seen are to Isaacs' book on character theory (where he only mentions that it has been proven by S. Garrison), and to the Ph.d thesis of S. Garrison (which has not been published, so not much help there). oes anyone know where one might find the proof?

REPLY [7 votes]: A new proof was published in:
Isaacs, I. M.; Knutson, Greg. "Irreducible character degrees and normal subgroups."
J. Algebra 199 (1998), no. 1, 302–326.
MR1489366
DOI:10.1006/jabr.1997.7191
This was extended to cd(G)=5 in:
Lewis, Mark L. "Derived lengths of solvable groups having five irreducible character degrees. I."
Algebr. Represent. Theory 4 (2001), no. 5, 469–489.
MR1870501
DOI: 10.1023/A:1012706718244
It mentions that "Because of the length and complexity of his argument, Garrison never published this result." and has some other useful comments.<|endoftext|>
TITLE: Book about fluid dynamics
QUESTION [5 upvotes]: Next Monday, I'll have an interview at Siemens for an internship where I have to know about fluid dynamics/computational fluid dynamics. I'm not a physicist, so does somebody have a suggestion for a good book where I can read about some basics? Thank you very much.

REPLY [4 votes]: Since this is for a job interview, I would focus on the more applied side of computational fluid dynamics (CFD). I really liked CFD Python: 12 steps to Navier-Stokes
A brief description of this course from the author's webpage is:

The course is for beginners. It assumes only basic programming skills —the concepts of iterations, function calls, and so on— and builds immediate hands-on experience via this module and several others that come after.

Hope this helps others in need of a crash course on applied CFD.<|endoftext|>
TITLE: Is there a combinatorial interpretation of the identity $\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m} =4^{-m} \binom{4m+1}{2m}$?
QUESTION [26 upvotes]: I came across the following combinatorial identity in a paper by Victor H. Moll and Dante V. Manna 'a remarkable sequence of integers'. 
$$\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m}
=4^{-m} \binom{4m+1}{2m}.
$$
I gave an elementary proof as follows, yet a combinatorial interpretation seems difficult to a layman like me. So I post it here for discussion.
My elementary proof is through the method of coefficients.
Let $[t^n]f(t)$ be the coefficient of $t^n$ in $f(t)$.
Lemma: $[t^k]\frac{1}{\sqrt{1-t}}=4^{-k} \binom{2k}{k}$.
Proof: 

$$
\begin{aligned}
[t^k]\frac{1}{\sqrt{1-t}}
&=\binom{-1/2}{k} (-1)^k \\\\
&=\binom{1/2+k-1}{k} \\\\
&=\frac{(k-1/2)(k-3/2)\cdots (1/2)}{k!}\\\\
&=\frac{(2k-1)(2k-3)\cdots 1}{k!}2^{-k} \\\\
&=\frac{(2k)(2k-1)(2k-2)(2k-3)\cdots 2\cdot1}{k!\cdot k!} 4^{-k} \\\\
&= 4^{-k} \binom{2k}{k}
\end{aligned}
$$

QED
Moreover, it is easy to see

$$
\begin{aligned}
\binom{2m-k}{m}&=\binom{2m-k}{m-k} \\\\
&=\binom{-(2m-k)+m-k-1}{m-k}(-1)^{m-k}\\\\
&=
\binom{-m-1}{m-k} (-1)^{m-k} \\\\
&=[t^{m-k}]\frac{1}{(1-t)^{m+1}}
\end{aligned}
$$

Proposition:
$$\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m}=
4^{-m} \binom{4m+1}{2m}.$$
Proof: 

$$
\begin{aligned}
\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m} &=
[t^m]\left(\frac{1}{\sqrt{1-t}} \frac{1}{(1-t)^{m+1}}\right) \\\\
&= [t^m]\frac{1}{(1-t)^{m+(3/2)}} \\\\
&=\binom{-m-(3/2)}{m} (-1)^m \\\\
&=\binom{2m+(1/2)}{m} \\\\
&= 2^{-m} \frac{(4m+1)(2m-1)\cdots(2m+3)}{m!} \\\\
&=2^{-m} \frac{(4m+1)(2m-1)\cdots(2m+3)}{m!} \frac{4m(4m-2)\cdots(2m+2)}{4m(4m-2)\cdots(2m+2)} \\\\
&=2^{-2m}\frac{(4m+1)!}{(2m+1)!(2m)!} \\\\
&=2^{-2m} \binom{4m+1}{2m}
\end{aligned}
$$

QED

REPLY [6 votes]: I don't have an answer, but I have spend a couple of hours on it, so here is some of my thoughts on the problem.
In the following I will identify expressions with sets, so $2^n$ corresponds to the set of 01-sequences of length n, $\binom{n}{k}$ is the set of 01-sequence of length n with exactly k 1s, and products and sums corresponds to taking product sets and unions. 
We want to find a bijective function from $\sum_{k=0}^m 2^{2(m-k)} \binom{2k}{k} \binom{2m-k}{m}$ to $\binom{4m+1}{2m}$ (I have multiplied with $4^m$ on both sides). Let me give an example a similar looking equality (you can skip the rest of this paragraph if you want): $\sum_{k=0}^m2^{2(m-k)}\binom{2k}{k}\binom{2m}{2k}=\binom{4m}{2m}$. Take an element in $\binom{4m}{2m}$, and pair the terms, so we have a sequence over {(00),(01),(10),(11)} of length $2m$. We have an even number of 1s in the sequence, so there must be an even number of pairs that contain exactly one 1, and thus and even number of pairs with (00) or (11). Let the number of (00) and (11) in the sequence of pairs be 2k. Now k of these must be (00) and k of them is (11) is the number of 0s and 1s are the same. The 2k terms in the 2m length sequence can be chosen in $\binom{2m}{2k}$ ways, the k (11)s of these 2k terms can chosen in $\binom{2k}{k}$ ways and in the rest of the $2m-2k$ terms we must choose between (10) and (01). This gives a factor $2^{2(m-k)}$, so we have a 1-1 correspondence between $\sum_{k=0}^m2^{2(m-k)}\binom{2k}{k}\binom{2m}{2k}$ and $\binom{4m}{2m}$. 
An important part of such a proof is to find out what k represents. In your equality, it turns out that the k=0 part of the sum is about $\sqrt{\frac{1}{2}}$ of the whole sum. Do anyone know where the $\sqrt{\frac{1}{2}}$ could come from? One way to find out what k is, would be to find an injective function from the k=0 term, $2^{2m}\binom{2m}{m}$, to $\binom{4m+1}{2m}$ and see what the image set looks like. But I haven't been able to find such a function, that is, I cannot find a combinatorial proof that $2^{2m}\binom{2m}{m}\leq \binom{4m+1}{2m}$ (nor that $\binom{4m}{2m}<2^{2m}\binom{2m}{m}$). Perhaps you should try to ask this in you question?<|endoftext|>
TITLE: Nef divisors with few global sections
QUESTION [6 upvotes]: Are there nef divisors D on a complex projective manifold X such that $h^0(X,D)$ is less than or equal to $\dim X$?
Edit: In fact I'm interested in nef line bundles D, not just divisors.

REPLY [2 votes]: Another extremely simple example:
Let $S = \mathrm{Bl}_p \mathbb{P}^2$ be the blowup of $\mathbb{P}^2$ at a point, and look at the divisor $H-E$ on $S$, where $E$ is the exceptional divisor and $H$ is the pullback of a hyperplane class on $\mathbb{P}^2$.  Notice $h^0(H-E)=2 = \dim S$, since sections correspond to lines in $\mathbb{P}^2$ passing through $p$ (up to scalars).  But $(H-E)^2=0$, so $H-E$ is nef.<|endoftext|>
TITLE: Applications of Faber's conjecture
QUESTION [9 upvotes]: Faber's perfect pairing conjecture states that the tautological ring $R^*$ of the moduli space $\mathcal{M}_g$ of curves of genus $g$ behaves as if it were the rational cohomology of a closed, oriented manifold of dimension $g-2$.  Specifically, $R^{g-2}$ is rank one, and multiplication into this degree gives a perfect pairing between $R^k$ and $R^{g-2-k}$.
My understanding is that it is known (through work of Looijenga, Faber, and Pandharipande) that $R^{g-2} = \mathbb{Q}$, but the perfect pairing part hasn't been proven (though it has been verified in low genus cases).  I'd like to know:

Why might Faber have conjectured this to be the case?  What is it about $R^*$ that suggests that it might satisfy Poincare duality?
If true, what sort of applications does this have (to our understanding of $\mathcal{M}_g$, for instance)?

REPLY [6 votes]: Numerical evidence, from computing the cases $g=2,3,\dots$, eventually $g\le 15$, and seeing the symmetry in the numbers $\dim R_g^n$. I recall Carel saying he made the conjecture when $g$ was still pretty low, maybe 6. For any $g$, there is an algorithm computing $\dim R^n_g$ in finite time, that Faber came up with.
That's not so clear. But that's a very mysterious property. The search for the meaning is on.<|endoftext|>
TITLE: When is a blow-up non-singular?
QUESTION [21 upvotes]: Suppose that $X$ is a non-singular variety and $Z \subset X$ is a closed subscheme.  When is the
blow-up $\operatorname{Bl}_{Z}(X)$ non-singular?  
The blow-up of a non-singular variety along a non-singular subvariety is well-known to be non-singular, so the real question is ``what happens when $Z$ is singular?"  The blow-up can be singular as the case when $X = \mathbb{A}^{2}$and $Z$ is defined by the ideal $(x^2, y)$ shows. On the other hand, the example where $Z$ is defined by the ideal $(x,y)^2$ shows that the blow-up can be non-singular.
Edit:  Based on the comments of Karl Schwede and VA, I think that it would also be interesting to find non-trivial examples of appropriate $Z$'s.  I am splitting this off as a  separate question.  In the comments there, the users quim and Karl Schwede say a bit about what can be said about this question using Zariski factorization.

REPLY [2 votes]: A topological remark:
If $E \subset Y$ is a closed analytic subspace of a smooth space $X$, then the boundary
of a tubular neighbourhood is an (odd dimensional, real) sphere bundle over $E$.
Thus if the blowup of $Z \subset X$ is smooth then the boundary of a tubular 
neighbourhood of $Z \subset X$ can be written as a sphere bundle over the exceptional 
divisor $E = P(N_Z X)$.
One might use this to arrive at a contradiction e.g. by calculating cohomology.
Similar ideas were used by Mumford in his study of normal surface singularities.
Mumford: The topology of normal singularities of an algebraic surface and a criterion for simplicity.<|endoftext|>
TITLE: Pairwise intersecting sets of fixed size
QUESTION [24 upvotes]: The Erdős-Ko-Rado theorem talks about how large an intersecting set system (a set of pairwise intersecting sets) can be if the size of the base set is fixed.  I'm interested about intersecting set systems where the base set is not fixed, but the size of the sets is bounded.  I can prove the following lemma (see proof below).
Lemma 1.  For every natural number $k$ there is a natural number $N(k)$ such that for every set $C$ each of whose elements are sets of size at most $k$, if every two element of $C$ has a common member, then there is a kernel $A$ which is a set of size at most $N$ so that every two element of $C$ also has a common member that's in $A$.  
I'd like to know if this lemma is known in some literature, and whether you can give me a simpler proof for it than mine.  
I'd also like to know what bound you can give on $N(k)$.  An exact bound is probably hard and not too interesting, but I'd like to get the order of magnitude, say whether you can make $N(k)$ a polynomial of $k$.  My proof only gives $N(k) = 2^{O(k^2)}$, so anything with a smaller order of magnitude would be nice.  (I know that $N(k)$ has to be $\Omega(k^2)$.  You can show this by chosing a prime $q$ between $k/2-1$ and $k-1$ and then letting $C$ be the set of lines of a finite projective plane of order $q$.)
There's also a strengthening of the lemma, which follows easily from my proof and can be useful.
Lemma 2.  For every natural number $k$ there is a natural number $N^{\ast}(k)$ such that for every set $C$ each of whose elements are sets of size at most $k$, if every two element of $C$ has a common member, then there is a kernel $A$ which is a set of size at most $N^{\ast}$ so that if $Y \in C$ and $X$ is a set that intersects every element of $C$ and $X$ has at most $k$ elements then $X \cap Y \cap A$ is nonempty.
Update: the original phrasing of lemma 2 was wrong, I added the condition that $|X|\le k$.
I'm asking the same questions as above for this stronger version, and also whether it follows easily from the first lemma.  
Proof of lemma 1.  
Fix $k$.  We will use induction on $p$ to show the existence of a set $A_p$ such that the size of $A_p$ is bounded by a constant natural number depending only on $k$ and $p$ (but not $C$), and that for every $X \in C$ either $p \le |X \cap A_p|$ or the intersection $X \cap Y \cap A_p$ is non-empty for every $Y\in C$.  This is enough because $A = A_{p+1}$ satisfies the conditions of the lemma (in fact even $A = A_p$ would work).  The case of $p = 0$ is trivial, because $A_0$ can be the empty set.
Now suppose we have found $A_p$ and we want to construct $A_{p+1}$.  Now sort the elements of $C$ in equivalence classes such that two element is equivalent if their intersection with $A_p$ is equal.  There are at most as many such classes as  subsets of $A_p$ (or even subsets with at most $k$ elements), which is a constant bound because the size of $A_p$ is bounded by a constant.  Now chose a single element from each equivalence class and let $B$ be the set of these elements.  Let $A_{p+1} := A_p \cup \bigcup_{Y\in B} Y$.  
Thus all we have to prove is that for every $X \in C$ either $X \cap A_{p+1}$ has at least $p+1$ elements or it intersects every element of $C$.  From the induction hypothesis we know that $X \cap A_p$ either has at least $p$ elements or intersects with every element of $C$.  If it's the latter, we're done, because $X \cap A_{p+1}$ is a superset of $X \cap A_p$, so let's now assume the former: $X \cap A_p$ has at least $p$ elements.  Now if $X \cap A_{p+1}$ intersects all elements of $C$ then we're done, so we can also assume that there is a $Z \in C$ such that $X \cap A_{p+1} \cap Z$ is empty.  Now consider the class of $Z$ in the equivalence we defined above, that is, all sets $Y$ for which $Y \cap A_p = Z \cap A_p$, and let $Y$ be the representant element we chose from this class for the construction.  This means that $Y \in B$ thus $Y \subset A_{p+1}$.  Now $X$ and $Y$ has a common element, say $x$.  Now it's not possible that $x \in A_p$, because by our second assumption $X \cap Y \cap A_p = X \cap Z \cap A_p \subset X \cap Z \cap A_{p+1}$ is empty.  But then $X \cap A_{p+1}$ has the at least $p$ elements of $X \cap A_p$ from our first assumption (because $A_p \subset A_{p+1}$), and the extra element $x$ which is not in $X \cap A_p$, so it has at least $p + 1$ elements, which completes our proof.

REPLY [3 votes]: I found out that this is a known problem, and was solved in 1973.  The Lovász: Combinatorical problems and exercises actually gives a solution in exercise 13.27.  This gives asymptotically better estimates than anything we could derive here.  
A generalization (where the sets are required to be s-wise intersecting instead of pairwise) can be found in N. Alon, Z. Füredi: On the Kernel of Intersecting Families, Graphs and Combinatorics 3, 91–94 (1987).<|endoftext|>
TITLE: What is state of the art for the Shooting Method?
QUESTION [5 upvotes]: I am interested in examples where the Shooting Method has been used to find solutions to systems of ordinary differential equations that are either 

reasonably large systems, or 
the search algorithm in the shooting parameters is somewhat prohibitive because of the nature of the solutions, or
both of the above.

Any references, descriptions, recent progress, folklore, in the ballpark would be of interest. Feel free to interpret "reasonably large" subjectively if necessary.

REPLY [5 votes]: Another shameless plug ... Coworkers and I used the Evans function formalism, which is a variant of the shooting method to deal with unstable directions (probably the same problem as mentioned by yfarjoun), on a boundary value problem of the form $y'(t) = (\lambda A_1 + A_2(t)) y(t)$ with $y$ specified as $t \to \pm \infty$. This is very similar to a Sturm-Liouville problem except that the differential operator is not self-adjoint. The application we're interested in is to do stability analysis of a travelling wave of a 2d reaction-diffusion equation. The main problem is that $y(t)$ is a fairly big vector with up to about 200 entries. 
For details, please ask or see the paper at http://arxiv.org/abs/0805.1706 and references therein.<|endoftext|>
TITLE: How many groups of size at most n are there? What is the asymptotic growth rate? And what of rings,  fields, graphs, partial orders, etc.? 
QUESTION [32 upvotes]: Question. How many (isomorphism types of) finite groups of size at most n are there? What is the asymptotic growth rate? And the same question for rings, 
fields, graphs, partial orders, etc. 
Motivation. This question arises in the context of a certain finite analogue of Borel equivalence relation theory. I explained in 
this answer that the purpose of Borel equivalence relation theory is to analyze the complexity of various 
naturally occuring equivalence relations in mathematics, such as the isomorphism relations on various types of structures. It turns out that many of the 
most natural equivalence relations arising in mathematics are Borel relations on a standard Borel space, and these fit into a hierarchy under Borel reducibility. Thus, this subject allows us make precise the idea that some classification problems are wild and others tame, by fitting them into a precise hierarchy where they can be compared with one another under reducibility.  Recently, there has 
been some work adapting this research project to other contexts. Last Friday, for example, Sy Friedman gave a 
talk for our seminar on an effective analogue of the Borel theory. Part of his analysis provided
a way to think about very fine distinctions in the relative difficulty even of the various problems of classifying finite structures, 
using methods from complexity theory, such as considering NP equivalence relations under polytime reductions. 
For a part of his application, it turned out that fruitful conclusions could be made when one knows something about 
the asymptotic growth rate of the number of isomorphism classes, for the kinds of objects under consideration.
This is where MathOverflow comes in. I find it likely that there are MO people who know about the number of groups. 
Therefore, please feel free to ignore all the motivation above, and 
kindly tell us all about the values or asymptotics of the 
following functions, where n is a natural number: 

G(n) = the number of groups of size at most n.
R(n) = the number of rings of size at most n.
F(n) = the number of fields of size at most n.
Γ(n) = the number of graphs of size at most n.
P(n) = the number of partial orders of size at most n.

Of course, in each case, I mean the number of isomorphism types of such objects. These particular functions are representative, though of course, there 
are numerous variations. Basically I am interested in the number of isomorphism classes of any kind of natural finite structure, limited by size. 
For example, one could modify Γ for various specific kinds of graphs, or modify P for various kinds of partial orders, such as trees, 
lattices or orders with height or width bounds. And so on. 
Therefore, please answer with other natural classes of finite structures, but I shall plan to accept the answer for my favored 
functions above. In many of these other cases, there are easy answers. For example, the 
number of equivalence relations 
with n points is the intensely studied partition number of n. The number of Boolean 
algebras of size at most n is just log2(n), since all finite Boolean algebras are finite power sets.

REPLY [5 votes]: In a more general vein, spectra of
equational classes have been studied by Ralph
McKenzie and others.  A jumping off point is his
"Locally finite varieties with large free spectra" .
I believe his work on tame congruence theory shed some light
on the growth rates for certain classes of finite algebras.
Gerhard "Ask Me About System Design" Paseman, 2010.04.13<|endoftext|>
TITLE: Which number fields are monogenic? and related questions
QUESTION [39 upvotes]: A number field $K$ is said to be monogenic when $\mathcal{O}_K=\mathbb{Z}[\alpha]$ for some $\alpha\in\mathcal{O}_K$. What is currently known about which $K$ are monogenic? Which are not? From Marcus's Number Fields, I'm familiar with the proof that the cyclotomic fields are monogenic, and for example that $\mathbb{Q}(\sqrt{7},\sqrt{10})$ is not monogenic (it is exercise 30 of chapter 2), but because Marcus eschews anything local, I haven't seen any of the perhaps more natural proofs of these results.
If $K$ is monogenic, is there an effective method of determining those $\alpha\in\mathcal{O}_K$ for which $\mathcal{O}_K=\mathbb{Z}[\alpha]$? 
More generally, what is known about the minimal number of generators of $\mathcal{O}_K$ as a $\mathbb{Z}$-algebra? That is, can we determine, or at least put non-trivial bounds on, the minimal $m$ such that $\mathcal{O}_K=\mathbb{Z}[\alpha_1,\ldots,\alpha_m]$ for some $\alpha_i\in\mathcal{O}_K$? We know that any $\mathcal{O}_K$ has an integral basis of $n=[K:\mathbb{Q}]$ elements, so certainly $m\leq n$ (I'm considering that trivial).

REPLY [6 votes]: I recently ran across the following result in a paper of [Ash-Brakenhoff-Zarrabi][1] (Lemma 3.1), where it is called Dedekind's criterion:
Let $T(x)$ be a monic irreducible polynomial with root $\theta$ and let $K = \mathbb{Q}(\theta)$. Let $\mathcal{O}$ be the ring of integers in $K$ and let $p$ be a prime. We use overlines to denote reduction modulo $p$.
Factor $\bar{T}(x)$ in $\mathbb{F}_p[x]$ as $\prod \bar{t}_i(x)^{e_i}$, and choose lifts of the $\bar{t}_i$ to monic polynomials in $\mathbb{Z}[x]$. Define $g(x) = \prod t_i(x)$, $h(x)=\prod t_i(x)^{e_i-1}$ and $f(x) = (T(x) - g(x) h(x))/p$. Then $p$ divides the index of $\mathbb{Z}[\theta]$ in $\mathcal{O}$ if and only if $GCD(\bar{f}, \bar{g}, \bar{h})$ is not $1$.
$\hspace{1cm}$
[1] A. Ash, J. Brakenhoff, and T. Zarrabi, Equality of Polynomial and Field Discriminants, Experiment.
Math. 16 (2007), 367–374<|endoftext|>
TITLE: Does there exist a number field, unramified over a predetermined finite set of primes of Q, such that the inverse regular Galois problem is correct for that number field?
QUESTION [17 upvotes]: The question is: for any finite group, $G$, and any finite set of primes (of $\mathbb{Z}$), $P$, is there a number field $K$, such that there is a regular $G$-Galois extension of $\mathbb{P}^1_K$, and such that $K$ is unramified over all the primes in $P$ as an extension of $\mathbb{Q}$. Supposedly, of course, the answer is yes because conjecturally there is a $G$-Galois extension of $\mathbb{P}^1_{\mathbb{Q}}$. To clarify: by a regular extension, I mean one that descends from a geometric extension (meaning that if you base change to $\mathbb{C}$ you get a cover of the same degree). Basically this means that I don't allow the action of $G$ to come from an extension of scalars. Also: you should notice that when I talk about a $G$-Galois cover of a $K$-curve, I mean that the cover itself is defined over $K$ and that even over $Spec(K)$ it is $G$-Galois (meaning the $G$-action is defined over $K$).
It is a little known fact that for any finite set of primes $P$ and any finite group $G$, there is a number field $K$ unramified over $P$ and having a $G$-Galois extension (in fact the extension constructed will itself be unramified over $P$.) If you follow the proof carefully, you see that it also proves that for any group, $G$, and any such finite set of primes $P$, there is a $G$-Galois extension of some curve $C$, which descends such that it's still Galois to $K$; where $K$ is unramified over the primes in $P$ as an extension of $\mathbb{Q}$. I'm wondering if this can be extended to $C$ being $\mathbb{P}^1_K$
The proof goes like this:
Imbed $G$ in some $S_m$, and embed this $S_m$ in an $S_n$ such that n is coprime with all the primes in $P$. Start with $\mathbb{Q}[X_1, ..., X_n]$, and mod out by the obvious action of $S_n$. You get a $\mathbb{Q}[\sigma_1, ..., \sigma_n]$ (where the $\sigma_i$'s are the elementary symmetric polynomials). Look at the impositions on the $(a_1, ..., a_n)$ in $\mathbb{Z}^n$: $a_i$ is divisible by all the $p \in P$, for $i=1, ..., n-1$, and $a_n \equiv 1$ modulo $\displaystyle\prod_{p \in P} p$. This is gives a $\displaystyle\prod_{p \in P} p$-adically open set, in which we can find an $(a_1, ..., a_n)$ that would make Hilbert irreducibility work. Meaning that over the $\mathbb{Q}$-rational point $(a_1, ..., a_n)$ the fiber is connected. So we get that the fiber is $Spec$ of the splitting field of $t^n-a_1t^{n-1}+...+(-1)^na_n$. Call this extension of $\mathbb{Q}$, $L$.
Now, in the original proof, one just looks at $L^G$, and gets that since $L$ is unramified over $\mathbb{Q}$ (this can be seen by the impositions on the $a_i$'s), then obviously $L^G$ does also. 
Let's take a different route. Instead of plugging in all of the $a_i$'s, we can plug in all but one. For example plug in $a_1, ..., a_{n-1}$, and thus get an $S_n$ cover of $\mathbb{A}^1_{\mathbb{Q}}$ (all defined over $\mathbb{Q}$) given by $t^n-a_1t^{n-1}+...+(-1)^nx$ (where $x$ is my new name for $X_n$). Let's think of this as an $S_n$ cover of $\mathbb{P}^1_{\mathbb{Q}}$. We don't know the genus of the cover. Let's call this cover $D$. If you mod out $D$ by $G$: $D \rightarrow D/G$, we get a $G$-Galois cover of $D/G$, and all is defined over $L$. So $D/G$ is the $C$ I was talking about. My question is: can we be clever about our choice of $(a_1, ..., a_n)$ so that $D/G$ would be a $\mathbb{P}^1$? Of course completely different approaches are also welcome!

REPLY [10 votes]: Not sure whether this is of help after such a long time, but anyway:
The answer is yes, and this even works over some number field in which all the primes in $P$ are completely split. Namely, denote by $K^P$ the maximal algebraic extension of $\mathbb{Q}$ in which all $p\in P$ split completely.
As noted by Pop on p.3 here: https://www.math.upenn.edu/~pop/Research/files-Res/LF_6Oct2013.pdf ,
the field $K^P$ is a "large" field, meaning that every irreducible curve over $K^P$ with a smooth $K^P$-point has infinitely many such points (I suppose for this particular field, the "large" property actually follows from some variant of Krasner's lemma). The regular inverse Galois problem is known to have a positive answer over all large fields (due to e.g. Pop, Harbater, Jarden etc.), so there is a regular $G$-Galois extension of $\mathbb{P}^1(K^P)$. But this extension must of course be defined over some number field $F\subset K^P$, giving the answer.<|endoftext|>
TITLE: A local ring not a quotient of a regular local ring
QUESTION [11 upvotes]: In his book Commutative Ring Theory, Matsumura proves that if a local ring is equidimensional, and a quotient of a regular local ring, then its completion is equidimensional.
What is an example of a local ring which does not admit such a presentation?

REPLY [15 votes]: A source available online is this paper "Examples of bad Noetherian rings" by Marinari (example 2.1). 
The reason many of these types of construction work is because of the following vague and counter-intuitive phenomemon: 
It is usually easier than we think for a complete local ring to be a completion of a Noetherian ring with certain properties. 
For example, there is this amazing theorem by Heitmann that most complete local ring of depth at least $2$ is a completion of a UFD !  
So back to Marinari's paper, the example is as follows: start with some local Artinian ring $(Q,m)$ such that $Q$ is not Gorenstein. Then $Q[[X]]$ is complete, and one can find a local domain $R$ such that $\hat R=Q[[X]]$. Now if $R$ is a quotient of a regular local ring, then the comletion of $R$ is generically a complete intersection. But $Q[[X]]$ is not even generically Gorenstein, since $Q$ is not.<|endoftext|>
TITLE: How many ways are there to prove flag variety is a projective variety?
QUESTION [13 upvotes]: I am looking for references talking about different ways to prove flag variety $G/B$ is projective variety. Now I have some in mind:

There is a proof in Humphreys Linear algebraic groups, he first prove $G/S$ is a complete algebraic variety hence a projective variety, where $S$ is a Borel subgroup of $G$ of largest possible dimension. Then he used the Borel Fixed Point Theorem to obtain the assertion. 
There is a proof in A.L.Onishchik and E.B.Vinberg Lie groups and algebraic groups, they used Chevalley's theorem to give the unique projective variety structure to the coset variety $G/B$.
There is a sketch of the proof in Tanisaki's D-modules, Perverse sheaves and Representation theory, he identified $G/B$(in particular case, suppose $G=GL_{n}(k)$) with set of flags in $k^{n}$. Then it is clear that one can give the projective variety structure on the set of flags.
One can also prove the flag variety can be embedded to Grassmannian variety, Then using Plucker embedding to give projective variety structure to this Grassmannian.
I think one can prove the line bundle on $G/B$ is positive and then use Kodaira embedding to prove it is a projective variety. But I did not find a proper expository reference.

Are there any other interesting proof? Any related comments are welcome. Thanks!
Edit: $G$ is an algebraic group, $B$ is Borel subgroup and $k$ is algebraic closed and $char k=0$

REPLY [7 votes]: I'm not sure this is an answer, but it got too long to be a comment! The projectivity of $G/B$ seems to me to follow from two facts: a) a homogeneous space $G/H$ is always a quasi-projective variety (which is due to Chevalley), and b) the variety $G/B$ must be complete. 
The first fact clearly has nothing to do with Borel subgroups but it would be maybe interesting to ask about how many proofs we have of it. The idea of the construction is clear: find a representation of $G$ which contains a line $L$ which has $H$ as its stabilizer and then take the orbit of $L$ in $\mathbb P(V)$, but to see that you get a categorical quotient this way takes some more care: you need to use some infinitesimal properties, (which you can tidily say using the Lie algebra of course). 
The second fact perhaps depends on what you're willing to assume, but it must come down to the Borel fixed point theorem in some form or other, since this tells you that if $G/H$ is complete, then any solvable subgroup will be contained in a conjugate of $H$, thus Borel subgroups are the only solvable subgroups with a chance of having an associated homogeneous space which is complete.  
The argument from Humphreys' book uses the strategy of picking first a maximal solvable subgroup of largest dimension so as to show the orbit has to be closed (essentially because the stabilizer is largest so the orbit has smallest, but even then you use the Borel fixed point and the theorem for $GL_n$ if I remember correctly). I don't know how Onishchik and Vinberg get around this (if they do). Then, as the question says, you get the general result from the Borel fixed point theorem.
The proof for $GL_n$ shows that $G/B$ is projective if you take $B$ to be the subgroup of upper triangular matrices, but one still needs to show that this subgroup is a Borel, which again I only know how to do using the Borel fixed point theorem in some form. (And of course you can use the same strategy for other classical groups if you can eyeball a candidate Borel subgroup).
Given that, it's maybe worth pointing out that in all of this you can get away with a weak version of the Borel fixed point theorem: namely if $V$ is a representation of a solvable group $H$, and $X$ is an $H$-stable closed subvariety of $\mathbb P(V)$ then $X$ has an $H$-fixed point. This can be shown just using the Lie-Kolchin theorem (that is, that a representation of a solvable group contains a one-dimensional subrepresentation), which can be proved directly. Since the standard proof of the general Borel fixed point requires you to use something like Zariski's main theorem, this is maybe a noticeable saving. 
All of this leads me to wonde how many proofs do we know for i) the fact that for any closed subgroup $H$ the homogeneous space $G/H$ has to be quasi-projective and ii) the Borel fixed point theorem (or some variant)?<|endoftext|>
TITLE: Smoothness of distance function in Riemannian Manifolds
QUESTION [36 upvotes]: Let $(\mathcal{M},g)$ be a $C^{\infty}$-Riemannian manifold. A basic fact is that $g$ endows the manifold $\mathcal{M}$ with a metric space structure, that is, we can define a distance function $d:\mathcal{M}\times\mathcal{M}\longrightarrow\mathbb{R}$ (the distance between two points will be the infimum of the lengths of the curves which join the points)
which is compatible with the topology of $\mathcal{M}$. Of course $d$ is continuous function, but what can we say about the differentiability of $d$?, is it smooth?. If not, Is there some criterion to know when it is?
Thanks in advance.

REPLY [13 votes]: The purpose of this answer is to provide a proof for the following result which Sergei mentioned in the accepted answer:
Proposition. Let $M$ be a complete Riemannian manifold and $x,y \in M\times M$, $x\neq y$. Then the following are equivalent:

The Riemannian distance $d:M\times M\rightarrow[0,\infty)$ is smooth in a neighbourhood of $(x,y)$.
There is only one length minimising geodesic connecting the points $x$ and $y$  and they are not conjugate along that geodesic.


Proof of Proposition. The Proposition follows from the three Lemmas below which freely use some properties of the so called 
segment domains $\Sigma_x=\{w\in T_xM: d(x,\exp_x(w))=\vert w \vert\}$:

$\exp_x: \mathrm{int} \Sigma_x\rightarrow M$ is a diffeomorphism onto its image [Gallot-Lafontaine, Corollary 3.77 or Petersen, Lemma 5.7.8 and Proposition 5.7.10]
$M= \exp_x(\mathrm{int}\Sigma_x)\cup\exp_x(\partial \Sigma_x)$ and the union is disjoint [Gallot-Lafontaine, Proposition 2.113].
Denote $\partial^1\Sigma_x = \{w\in \partial \Sigma_x: \exp_x(w)=\exp_x(w')$ for some $w'\in \partial \Sigma_x\backslash\{w\}\}$. Then:


If $w\in \partial \Sigma_x\backslash \partial^1\Sigma_x$, then $D\exp_x\vert_w$ is singular. [Gallot-Lafontaine, Scholium 3.78 or Petersen, Lemma 5.78 ]
$\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is dense [Sakai, Remark 4.9], see also [Klingenberg, Theorem 2.1.12 & 14] as well as here.



Lemma 1. $d^2(x,\cdot):M\rightarrow [0,\infty)$ is smooth in a neighbourhood of $y$ if and only if $y\in \exp_x(\mathrm{int} \Sigma_x).$
Proof. On (the open set) $\exp_x(\mathrm{int}\Sigma_x)$ we have $d(x,y)^2 = \vert \exp_x^{-1}(y)\vert^2$ which is clearly a smooth function in $y$. 
For the converse assume that $d(x,\cdot)^2$ is smooth on some open set $U\subset M$ and note that it suffices to prove
$$U \cap \exp_x(\partial \Sigma_x) = \emptyset \tag{1}.$$
Without loss of generality we may assume that $x\notin U$, then also $d(x,\cdot)$ is smooth in $U$ and has a gradient $G\in C^\infty(U;TM)$. Let $\gamma:[0,l]\rightarrow M$ be a length minimising unit-speed geodesic with $\gamma(0)=x$ and $y=\gamma(l)\in U$. Then 
$d(x,\gamma(t))=t$ and differentiation yields $$\langle G_y, \dot \gamma(l)\rangle = 1. \tag{*} $$  Since  $d(x,\cdot)$ is Lipschitz with constant $\le 1$ (triangle inequality) we have $\vert G_y \vert \le 1$. Further $\vert \dot \gamma(l) \vert =1$ and in light of (*) this is only possible if
$
G_y = \dot \gamma(l). 
$
We conclude:
$$\text{Length minimising geodesics which start in $x$ don't intersect in $U$.} \tag{2}$$
Now we can prove (1). Assume to the contrary that $U \cap \exp_x(\partial \Sigma_x) \neq \emptyset$. Then by densitity of $\exp_x(\partial^1 \Sigma_x)$ we also have $U \cap \exp_x(\partial^1  \Sigma_x) \neq \emptyset$ and there are $w,w'\in \partial \Sigma_x$ with $w\neq w'$ and $\exp_x(w) = \exp_x(w')\in U$, which is in contradiction to (2).
Lemma 2. $d^2:M\times M\rightarrow [0,\infty)$ is smooth in a neighbourhood of $(x,y)$ if and only if $y \in \exp_x(\mathrm{int} \Sigma_x)$. 
Proof. If $d^2$ is smooth near $(x,y)$, then $d^2(x,\cdot)$ is smooth near $y$ and the previous Lemma implies that $y\in \exp_x(\mathrm{int}\Sigma_x)$. For the converse define $\Sigma = \bigcup_x \Sigma_x \subset TM$ and note that
$$
\Sigma \text{ is closed }\quad \text{ and } \quad \mathrm{int} \Sigma \cap T_xM = \mathrm{int} \Sigma_x \tag{3}.
$$
Define
$$
F: \mathrm{int} \Sigma \rightarrow M\times M, (x,w) \mapsto (x,\exp_x(w))
$$
and note that 
$$
DF\vert_{(x,w)} = \begin{bmatrix}
\mathrm{id}  & 0\\
\ast& D \exp_x\vert_w
\end{bmatrix}
$$
is invertible for all $(x,w)\in \mathrm{int} \Sigma$. Further $F$ is easily seen to be injective and thus it has a smooth inverse $F^{-1}:F(\mathrm{int} \Sigma) \rightarrow \mathrm{int} \Sigma$. Hence $d^2(x,y)= \vert F^{-1}(x,y)\vert ^2$ is smooth in a neighbourhood of every $(x,y) \in F(\mathrm{int} \Sigma)$, which concludes the proof. 
Lemma 3. $y\in \exp_x(\mathrm{int} \Sigma_x)$ if and only if there exists a unique distance minimising geodesic between $x$ and $y$ and along this geodesic they are not conjugate.
Proof. Let $y=\exp_x(w)$ with $w \in \mathrm{int}\Sigma_x$. Then $t\mapsto \exp_x(tw)$, $0\le t\le 1$ is length minimising (because $w \in \Sigma_x$) and $x$ and $y$ are not conjugate along this geodesic ($D\exp_x\vert_w$ is invertible because $\exp_x$ is a diffeomorphism on $\mathrm{int}\Sigma_x$). If there was another length minimising geodesic from $x$ to $y$, then $y=\exp_x(w')$ for some $w'\in \Sigma_x \backslash \{w\}$. Since $\exp_x(\mathrm{int}\Sigma_x)\cap\exp_x(\partial \Sigma_x)=\emptyset$ we must have $w'\in \mathrm{int} \Sigma_x$, but this is false (since $\exp_x$ is injective on $\mathrm{int} \Sigma_x$).
Conversely assume that there is a unique distance minimising geodesic from $x$ to $y$ and that they are not conjugate along that geodesic. Then $y=\exp_x(w)$ for some $w\in \Sigma_x$. If we had $w\in \partial \Sigma_x$, then either there would be two length minimising geodesic between $x$ and $y$ (corresponding to $w\in \partial^1 \Sigma_x$) or $x$ and $y$ would be conjugate (corresponding to $D \exp_x\vert_w$ being singular).<|endoftext|>
TITLE: How to best distribute points on two concentric circles?
QUESTION [12 upvotes]: An N-subset $\{x_1,\dots,x_N\}$ of a compact set $X\subset \mathbb R^d$ is called a set of Fekete points (named after Michael Fekete) if it maximizes the product $$\prod_{1\le k<j\le N}|x_k-x_j|\qquad (1)$$ among all such $N$-tuples. When $X\subset \mathbb C$, one can express this product in terms of the Vandermonde determinant. In this case Fekete points are of particular interest in approximation theory (as interpolation nodes). Generally, they have no explicit form and must be found numerically. But there are exceptions.
When $X$ is a circle, Fekete points are equally spaced. This is well-known and can be proved like this: we may assume $x_k=\exp(i\theta_k)$ with $0=\theta_1<\theta_2<\dots <\theta_N<2\pi$. For a fixed integer $m$, $1\le m\le \lfloor N/2 \rfloor$, consider the product $\Pi(m)=\prod_{k=1}^N|x_{k+m}-x_{k}|$ with indices taken mod $N$. Since each point $x_k$ appears in just two terms of this product, it is not hard to see that $\Pi(m)$ is maximal when $x_k$ are equally spaced. (Indeed, $\log \sin$ is a concave function.)
Now let $X$ be the union of two concentric circles of radii $r$ and $R$. Assume that  $N=2n$ is even and that exactly $n$ of the points lie on each circle. With this additional assumption we are not quite in the Fekete problem anymore, but we have an obvious candidate for the maximum -

Under the assumptions of the preceding paragraph, is the product (1) maximized by equally spaced, interlaced points? E.g., by 
  $$x_k=\begin{cases} r\exp(2\pi i k/n), \quad k=1,\dots,n \\ 
R\exp(\pi i (2k-1)/n), \quad k=n+1,\dots,2n \end{cases} \qquad (2)$$

Presumably, a solution would consist of two steps. 

Show that the maximum is attained by interlaced points, i.e., when ordered by the 
argument, the points alternate between the two circles. 
Show that for interlaced points, maximum of (1) is attained when they are equally spaced. 

One can hope that step 2 is doable with standard calculus tools, since there should be no other local maxima for interlaced points. But Step 1 seems to require a more imaginative approach. 
Why do I find this question interesting? There is a natural way to distribute N points on a circle (i.e., at equal distances), and this configuration is known to give the solution for many extremal problems, such as the Fekete problem, isoperimetric N-gon problem, etc. There is also a natural way to distribute two sets of N points -- namely, by formula (2), but I don't know of any nontrivial problem for which the optimality of such configuration has been established.

REPLY [2 votes]: (This was too long to fit in a comment.)
For subsets of the complex plane, a slightly different perspective is that the square of the quantity you want to maximize is the absolute value of the discriminant of the monic polynomial whose roots are precisely the points. Up to a factor depending only on the degree, the discriminant of a monic polynomial is the product of the values of the polynomial at the roots of the derivative. Also note that in the configuration you suggested, the derivative of the polynomial is almost the optimal polynomial for points on a circle: maybe you can prove that the derivative of an optimal polynomial is close to an optimal polynomial?<|endoftext|>
TITLE: (0,1)-matrix congruence: is it known?
QUESTION [8 upvotes]: [[UPDATE: This work has now been published at SIAM J Discrete Math.: Formulae for the Alon–Tarsi Conjecture.]]
By equating two formulae (one congruence by Glynn (1) (which has just appeared) and one unpublished formula) for the number of even Latin squares minus the number of odd Latin squares, we find the following result.
For odd primes $p$ we have \[\sum_{A \in B} (-1)^{\sigma_0(A)} \equiv 1 \pmod p\] where $B$ is the set of $(p-1) \times (p-1)$ $\\,(0,1)$-matrices whose determinant is indivisible by $p$ and $\sigma_0(A)$ is the number of zeroes in $A$.  It happens to be true for $p=2$ also (but it does not follow from Glynn's result).

Is this result already known?  If so, it would provide an alternate proof of Glynn's result.

To illustrate, consider when $p=3$.  The (0,1)-matrices whose determinants are indivisible by $p$ are
01 10       01 10 11 11
10 01  and  11 11 01 10

So the sum becomes $+2-4=-2 \equiv 1 \pmod 3$.
It is equivalent to the congruence \[\sum_{A \in C} (-1)^{\sigma_0(A)} \det(A)^{p-1} \equiv 1 \pmod p\] where $C$ is the set of all $(p-1) \times (p-1)$ $\\,(0,1)$-matrices (via Fermat's Little Theorem).
(1) Glynn, D., 2010. The conjectures of Alon-Tarsi and Rota in dimension prime minus one. SIAM J. Discrete Math., 24 (2010), 394-399.

REPLY [4 votes]: Darij's proof was wrong but both his claims (that it is next to trivial and that it requires some long formulae) were actually correct. Here is the demonstration.
To start with, we'll count the sum over degenerate (in $\mathbb Z_p$) matrices instead because the sum over all matrices is clearly $0$. Also, I'll prefer to use $\sigma_1$ instead of $\sigma_0$ as the exponent of $(-1)$ (they have the same parity if $p>2$). Now, let us observe that the degenerate matrices are exactly those that kill some non-zero vector $x\in\mathbb Z_p^{p-1}$. Moreover, the total number of vectors they kill is $-1$ (I'm counting modulo $p$, of course). Thus, we can just cound the sum over all $x\ne 0$ of the "signatures" of matrices that kill $x$. Now, a matrix kills $x$ if and only if each row kills $x$, so this sum is just the sum of signatures of all rows killing $x$ to the power $p-1$. The sum of row signatures over all possible rows is $0$, so we can again replace it with the sul over all rows not killing $x$ to the power $p-1$, which is the same as the sum over all matrices not killing $x$ in any row. Now, if we count modulo $p$, this sum can be written as
$$
\begin{aligned}
&\sum_x\sum_{\varepsilon}\prod_{i,j}(-1)^{\varepsilon_{ij}}\prod_i\left(\sum_j\varepsilon_{ij}x_j\right)^{p-1}
\\
&=\sum_{x,\varepsilon}\sum_\lambda\frac{(p-1)!^{p-1}}{\prod_{i,j}\lambda_{jj}!}\prod_{i,j}(-1)^{\varepsilon_{ij}}\varepsilon_{ij}^{\lambda_{ij}}x_j^{\lambda_{ij}}
\\
&=\sum_{x,\lambda}\frac{(p-1)!^{p-1}}{\prod_{i,j}\lambda_{jj}!}\prod_{i,j}x_j^{\lambda_{ij}}(0^{\lambda_{ij}}-1)
\end{aligned}
$$ 
($\lambda$ here runs over all matrices with non-negative entries such that $\sum_j\lambda_{ij}=p-1$, $x$ runs over non-zero vectors in $\mathbb Z_p$ and $\varepsilon$ runs over all $0,1$ matrices)
Now we use the deep formula $0^0=1$, which tells us that the only $\lambda$ for which the product is not $0$ is the one consisting of all ones. This reduces the sum to 
$$
(p-1)!^{p-1}\sum_x\prod_j x_j^{p-1}=(p-1)!^{p-1}\left(\sum_{a\in\mathbb Z_p}a^{p-1}\right)^{p-1}\equiv (p-1)!^{p-1}(p-1)^{p-1}\equiv 1\mod p
$$
(we can now include the $0$ vector in the sum, it doesn't contribute anything).
The end.<|endoftext|>
TITLE: An example of a series that is not differentially algebraic? 
QUESTION [14 upvotes]: Motivated by this question, I remembered a question I was curious about sometime which I am sure has some easy and nice example for it as well, which I just can't think of for some reason. I want an example of a power series that is not differentially algebraic. A differential algebraic power series is a series $f(t)$ satisfying an equation $P(t,f(t),f'(t),\ldots,f^{(k)}(t))=0$ for some $k$ and some polynomial $P$ in $k+2$ variables. 
Update: examples in the comments below ($\sum t^{n^n}$, $\sum t^{2^n}$) make me ask a refinement (of a sort) for the original question: these examples are reminiscent of all those Liouville-flavoured examples of transcendental numbers, - I wonder if there is a Liouville-flavoured proof, stating that if the polynomial P is of given (multi)degree, some inequality holds that is obviously impossible for the series above?
Update 2: there are quite a few examples now, and I am tempted to accept the $\sum t^{n^n}$ answer since the example itself is easy and it came together with an easy explanation. I wonder what are other general approaches besides the ones that are exhibited in answers here (looking at p-adic norms of coefficients and looking at powers of $t$ with nonzero coefficients).

REPLY [4 votes]: Yet some more examples: the ordinary generating function for the Bell numbers, see Martin Klazar.  Or, Flajolet, Gerhold, Salvy.
I think it would be nice to have some kind of survey of results and methods.  In combinatorics, I frequently encounter generating functions that do not seem to be differentially algebraic, but I have no idea how to prove that.  An example would be the generating functions for walks in the quarter plane with certain step sets, see Bostan, Kauers.  Maybe a little bit more philosophical: is it true that "natural" generating functions are either differentially finite or differentially transcendent?<|endoftext|>
TITLE: Non-Kahler manifolds where the different Laplacians are compatible 
QUESTION [9 upvotes]: On a Kahler manifold, the different Laplacians are compatible: $\Delta_d=2\Delta_{\bar{\partial}}=2\Delta_{\partial}$.
Are there non-Kahler Hermitian manifolds where the above identity holds?

REPLY [3 votes]: I just add some details to Zhao's answer.
Let $(X,\omega)$ be a hermitian manifold and $\tau$ be the operator of type $(1,0)$ and order $0$ defined by
$$
\tau=[\Lambda_\omega,\partial\omega].
$$
Here $[\bullet,\bullet]$ is the (graded) commutator and $\Lambda_\omega$ the formal adjoint of the operator "wedge product with $\omega$". Often $\partial\omega$ is called the torsion of $\omega$ (which is $\equiv 0$ if and only if $\omega$ is Kähler) and $\tau$ the torsion operator. Then, we have the following identities:
$$
\Delta_{\bar\partial}=\Delta_\partial+[\partial,\tau^\star]-[\bar\partial,\bar\tau^\star],
$$
$$
[\partial,\bar\partial^\star]=-[\partial,\bar\tau^\star],\quad[\bar\partial,\partial^\star]=-[\bar\partial,\tau^\star],
$$
and
$$
\Delta_d=\Delta_\partial+\Delta_{\bar\partial}-[\partial,\bar\tau^\star]-[\bar\partial,\tau^\star].
$$
Therefore, $\Delta_\partial$, $\Delta_{\bar\partial}$ and $\frac 12\Delta_d$ no longer coincide, but they differ by linear differential operator of order $1$ only.<|endoftext|>
TITLE: What is the easiest way to classify all possible smooth orientable closed 2-manifolds?
QUESTION [6 upvotes]: If this has been answered already, please let me know and I'll delete the question. 
ADDED: I'd prefer to assume a smooth structure, rather than a triangulation.

REPLY [5 votes]: A very simple and low tech solution to this problem is the very first proof, 
given by Möbius in 1863. He assumes that the surface is smoothly embedded
in $\mathbb{R}^3$, and slices it by a family of parallel planes. Assuming that
the orientation of the planes is general and that they are sufficiently close together, this cuts the surface into simple pieces -- either disks, annuli, or pairs of pants.
It is then quite easy to show that the result of assembling such pieces is always a sphere with handles.<|endoftext|>
TITLE: The Interrelationship Problem Of Modern Mathematics- How To Deal With it In First Year Graduate Courses?
QUESTION [27 upvotes]: I was reading recently online Peter May's complaints (I'm a fan, you can tell, I'm sure) about teaching the third quarter of the graduate algebra sequence at the University of Chicago. This course focuses on homological algebra and attempts to be as up-to-date as possible. May's conundrum stems from the fact that homological algebra is inexorably tied to algebraic topology and as a result, it's difficult to separate the 2 out in the course completely. May questions whether or not this is in fact a good idea; however, since this an algebra course and not a topology course, he feels compelled to work hard to do this.
That being said, he raises a very good pedagogical problem in the teaching of mathematics, particularly at the graduate level where the better schools are trying to prepare students to enter research as quickly as possible. Mathematics is now a very holistic, intertwined discipline: Algebra increasingly permeates virtually all of mathematics, the study of manifolds now requires very sophisticated analytic tools from differential equations and functional analysis, probability theory now partakes of a considerable amount of harmonic analysis, mathematical physics is now a major player in the construction of new mathematical structures-I could go on and on, but you get the idea.
So here's the question: Is the old model of keeping the subdisciplines of mathematics separate in coursework for the sake of focus obsolete? I know a lot of mathematicians in recent decades have begun to draw from various disciplines in constructing the first year graduate sequences of most universities; Columbia is one local example. The question is really are they going far enough? The problem of course is that when you begin weakening those artificial barriers, you run the risk of them collapsing altogether and you ending up with a hodgepodge of theory and methods that seems to have no focus. 
So anyone want to comment on what the solution here might be from their own experiences as both teachers and students? How far should courses go in being interrelated? And does this lead to better prepared graduate students for the research level?

REPLY [23 votes]: I question whether mathematics is really as holistic and intertwined as some people are making it out to be.
Certainly there are a few freaks of nature out there who can understand 20% of the mathematics out there and incorporate ideas from 10 different subfields into their work.  A larger group of us are capable of getting the big picture though maybe not all the specifics of 4 or 5 different subfields, at least to the extent that we know when to reach out to an expert.  Many graduate students, and most of them once one leaves the world of the top ten or twenty departments, are just capable of learning one subfield well enough to write a dissertation narrowly focused on one problem in that subfield, ignoring all the wider connections if indeed there are any.  Most published papers are written by people who have never done serious work outside a single narrow subfield in their entire career, even if the same is not true for the best papers.
A professor or a department may choose to aim its education at the future Fields Medalist (or, somewhat more broadly, the future NSF-or-equivalent-research-grant-receipients), but is this really fair for the other 19 students in the room?<|endoftext|>
TITLE: Proof that pi is transcendental that doesn't use the infinitude of primes
QUESTION [58 upvotes]: I just taught the classical impossible constructions for the first time, and in finding my class a reference for the transcendence of pi, I found a dearth of distinct proofs.  In particular, those that I read all require the existence of infinitely many primes, which strikes me as extraneous.  Is there a known proof that requires only knowledge that I would "expect", namely, integral calculus to get your hands on the actual constant and algebraic properties of polynomials in connection with the assumption that the constant is algebraic?

REPLY [18 votes]: Barry, this post is coming a "little bit" after your class ended, but I want to direct you to a paper on the Lindemann-Weierstrass theorem by Beukers, Bezivin, and Robba in the Amer. Math. Monthly in 1990: https://www.jstor.org/stable/pdf/2324683.pdf.  They give a proof of LW theorem that does not explicitly involve a choice of an auxiliary prime number, and it would be a good exercise to work out what the proof is saying in the special case of the number $\pi$ to see more directly how it proves $\pi$ is transcendental, or more generally how it shows $e^\alpha$ is transcendental when $\alpha$ is a nonzero algebraic number (the Lindemann part of LW) without dealing with the more general setting of a finite set of algebraic $\alpha_i$ (the Weierstrass part of LW). As indicated at the start of the paper by Beukers et al., the proof was inspired by ideas from $p$-adic analysis, but their paper is a simplified version that does not have any explicit dependence on $p$-adic notions.
I don't agree with the post claiming infinitude of primes "might be necessary" to prove transcendence of $\pi$. Just because you are working within an axiomatic framework that would allow you to prove infinitude of primes doesn't mean you ought to be required to do that for what you want, so that whole matter seems irrelevant to answering your question. If I am proving the Eisenstein irreducibility criterion in $\mathbf Z[x]$ you can't seriously tell me that I must go on a detour in the middle of the proof and prove $\mathbf Z[i]$ is a Euclidean domain just because it is logically possible with the axioms I am allowing. In any case, I think the paper I linked to should settle your question in an affirmative way that you were seeking.
UPDATE: The impression that you need infinitely many primes to prove transcendence results is incorrect, not because there are proofs that avoid it like the comparatively recent proof I mention above or older proofs I did not mention, but because even the proofs that mention it don't really need it. For example, look at the two proofs of transcendence of $e$ and $\pi$ here (pp. 3-5) and here. (In case those links change in the future, the proofs are essentially the same as those on pp. 4-6 of Baker's book "Transcendental Number Theory".) The way the proofs work is that an expression $J_p$ is introduced that depends on a prime $p$ (and some fixed data like a hypothetical algebraic relation for the number to be proved transcendental) and $J_p$ is shown to satisfy two properties: (i) a growth estimate $|J_p| \leq c^p$, where $c$ is independent of $p$ and (ii) $J_p$ is a sum of two integers, one being a multiple of $p!$ and the other being a multiple of $(p-1)!$ that for all large prime $p$ is not a multiple of $p!$, so $J_p$ is an integer multiple of $(p-1)$! that is not zero. Therefore $|J_p| \geq (p-1)!$, which contradicts the growth estimate in (i) since we can't have $(m-1)! \leq c^m$ for large $m$. How is the primality of $p$ really used? Its only purpose is to be an arbitrarily large number that is relatively prime to a couple of specific nonzero integers. But there is no need to rely on prime numbers to achieve that: if you want a huge integer $m$ relatively prime to two nonzero integers $a$ and $b$, just take $m = |ab|M + 1$ for a huge integer $M$. So you can work with numbers in such an arithmetic progression in place of a large prime number $p$ in the proof. (Personally I'd also pass to the integer $J_m/(m-1)!$ so the bounds $0 < |J_m/(m-1)!| \leq c^m/(m-1)!$ lead to the "no integers between $0$ and $1$" contradiction once $c^m/(m-1)!< 1$, but that's a matter of taste.) 
Of course the idea of creating numbers relatively prime to particular numbers by taking a multiple of their product and adding $1$ is the main idea in Euclid's proof of the infinitude of the primes, so we aren't bypassing a method that can lead to infinitely many primes but we are bypassing any need for prime numbers in the proof that uses them.
The idea of using a large prime in proofs of the transcendence of $e$ and $\pi$ is due to Hurwitz, who added this detail to a proof by Hilbert that relies on divisibility properties but not on primes.  Mahler discusses this in his Springer Lecture Notes in Math "Lectures on Transcendental Numbers" (1976). The volume is available online here, and while most chapters are behind a paywall the last part called Back Matter is not, and that's the part relevant to us. Mahler presents Hilbert's proof starting on p. 237. On p. 243 he mentions Hurwitz's idea of using a prime instead of a number in a simply constructed arithmetic progression and explains why he thinks this "seemingly simpler" approach of Hurwitz is actually not a good idea.<|endoftext|>
TITLE: Where does the splitting principle come from and does it generalize
QUESTION [10 upvotes]: Basically, I'm aware of "splitting principles" for the following three objects (which are all isomorphic modulo torsion).  
1. The Chow group a la Fulton.
2. The classical Grothendieck group of vector bundles or coherent sheaves.
3. The $\gamma$-graded Grothendieck group.
I was just wondering where the idea of "the splitting principle" comes from. I'm guessing somewhere in topology when one wanted to define Chern classes and show some properties. But I don't know.
And above that, is there some more general way of looking at this? I know there is a theorem that connects higher K-groups with Chow groups in a sense. So I ask, is there a way of deducing the splitting principle for one of the above objects from the other? (It's easy if we want to do this modulo torsion, of course.)

REPLY [5 votes]: In algebraic topology, there are two slightly different ways of thinking about this: one bundle at a time (the more usual way) and universally (via bundles between classifying spaces).  The second approach is surprisingly simple and general, as shown in the very 
brief paper "A note on the splitting principal", http://www.math.uchicago.edu/~may/PAPERS/Split.pdf, #109 on my web page. A key point is to notice that the splitting principle concerns the reduction of the structural group of a pullback of a given (complex) vector bundle from $U(n)$ to its maximal torus $T^n$. Replacing $U(n)$ by other compact Lie groups leads to splitting principals for other kinds of bundles, e.g. symplectic and real with cohomology away from $2$.  Replacing $T^n$ by a maximal $2$-torus gives a splitting principal for real bundles and cohomology at $2$.<|endoftext|>
TITLE: What functor is adjoint to the tensor product of 2-vector spaces?
QUESTION [9 upvotes]: This is a refinement of my (naive, poorly asked) question here.  The reference for my question is Baez and Crans, HDA6.
Background: category objects, etc.
Let $\mathcal V$ be a category.  A category object internal to $\mathcal V$ consists of the following data and properties:

Objects $C_0,C_1 \in \mathcal V$ and morphisms $s,t: C_1 \to C_0$ and $i: C_0 \to C_1$.
Such that $s\circ i = t\circ i = \text{id}_{C_0}$ and the pull-back $C_1 \underset{C_0}{\times} C_1 = C_1 \underset{\displaystyle ^{\searrow\!\!^{\scriptstyle s}} {C_0} ^{^{\scriptstyle t} \!\!\swarrow} }{\times} C_1 $ exists.
A morphism $m: C_1 \underset{C_0}\times C_1 \to C_1$ such that $s\circ m = s_R$ and $t\circ m = t_L$, where $s_R: C_1 \underset{C_0}\times C_1 \to C_0$ is the "$s$" projection from the right factor, and similarly for $t_L$.
And such that the obvious "associativity" square (two ways to get from $C_1 \underset{C_0}\times C_1 \underset{C_0}\times C_1$ to $C_1$) and "identity" triangles (three ways to get from $C_1 = C_1 \underset{C_0}\times C_0 = C_0 \underset{C_0}\times C_1$ to $C_1$) commute.

For example, a category object in $\mathcal V = {\rm SET}$ is a small category.
For this post, I will be interested in $\mathcal V = {\rm VECT}$, the category of vector spaces over your favorite field.  I will call your favorite field "$\mathbb R$".
A category object in ${\rm VECT}$ is a 2-vector space.
2-vector spaces are relatively mild things.  Indeed, it turns out that in $\rm VECT$ the data and properties of 1-2 above uniquely determine a map $m$ satisfying 3-4.
By the general yoga known as "commutativity of internalization", a 2-vector space is the same as a "vector space object in $\rm CAT$".  More precisely, let $\rm CAT$ be the category of small categories.  Then it makes sense to talk about "field objects" — like a category object, a field object consists of some objects, some maps, some pull-backs, and some more maps, and some commuting diagrams.  In particular, by thinking of $\mathbb R$ as a discrete category ($\mathbb R_0 = \mathbb R = \mathbb R_1$ and $s = t = i = {\rm id}$), it is in fact a field object.  Then with some more diagrams, we can talk about "vector space objects over $\mathbb R$" internal to $\rm CAT$, and it is straightforward to check that these are the same as 2-vector spaces.
Background: tensor products
I know of two natural approaches to define "tensor products":

Define a notion of "bilinear map", such that the assignment $X,Y,Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$ is contravariant in the first two spots and covariant in the last.  Then set $X\otimes Y$ to be the object (if it exists) that represents the function $Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$.
Define a notion of "internal hom", i.e. a (nice) functor $\underline{\rm Hom}: \mathcal V^{\rm op} \times \mathcal V \to \mathcal V$.  For each $X\in \mathcal V$, define the functor $-\otimes X$ by declaring that it is left adjoint to $\underline{\rm Hom}(X,-)$.

Approach 1 is the way that tensor products are introduced in grade school.  Approach 2 is I think more standard in the real world.  We can implement each in the case of 2-vector spaces:

The trick for approach 1 is that the notion of "bilinear" depends on more than just the category.  So realize the category of 2-vector spaces as the category of vector spaces objects in $\rm CAT$.  Recall that a morphism of 2-vector spaces is simply a morphism of underlying categories so that some diagrams commute.  Then we can say the following.  Let $X,Y,Z$ be 2-vector spaces.  Then a morphism of underlying categories $X \times Y \to Z$ is bilinear if a bunch of diagrams commute (these diagrams refer to the vector-space-object structures of $X,Y,Z$, and are precisely the diagrams that you learned in grade school).  By reproducing the proofs from vector spaces internal to $\rm SET$, this in fact defines a functor $\otimes$.
Given 2-vector spaces $X,Y$, there is a category whose objects are linear functors $X \to Y$ and whose morphisms are linear natural transformations of functors, and this category has a natural structure as a 2-vector space.  Moreover, the corresponding notion of $\underline{\rm Hom}$ is correctly functorial, and has an adjoint.  So this approach defines a functor $\otimes$.

However:

The two "tensor products" defined in 1-2 above do not agree.

Writing 2-vector spaces $X = (X_1 \rightrightarrows X_0)$ and $Y = (Y_1 \rightrightarrows Y_0)$ as category objects in $\rm VECT$, approach 1 gives $(X\otimes Y)_a = X_a \otimes Y_a$ for $a=0,1$, with the tensor products of the structure maps.  Approach 2 also has $(X\otimes Y)_0 = X_0 \otimes Y_0$, but $(X\otimes Y)_1 \cong $
$$ X_0 \otimes Y_0 \oplus \text{coker}\Bigl( \bigl(\ker(X_1 \overset s \to X_0) \otimes \ker(Y_1 \overset s \to Y_0)\bigr) \overset{t\otimes{\rm id} - {\rm id}\otimes t}\longrightarrow \bigl( X_0 \otimes \ker(Y_1 \overset s \to Y_0) \oplus \ker(X_1 \overset s \to X_0) \otimes Y_0\bigr) \Bigr)$$
The structure maps are: $s$ is the projection onto the first factor, and $t$ is the sum of the same projection and the map ${\rm id}\otimes t + t\otimes {\rm id}$ from the second factor (it is well-defined out of the cokernel).
(There is probably a way to simplify the above description.  The trick is that, as explained in HDA6, the category of 2-vector spaces is equivalent as a category to the category of 2-term chain complexes, and this is the natural "internal Hom" over there.)
Anyway, a dimension count shows that these two "tensor" constructions are inequivalent in general.
Hence:
Question
Does the tensor product of 2-vector spaces given in "approach 1" above — the tensor product defined as representing "bilinear functors" — under this tensor product, does the functor $\otimes X$ have a right adjoint?

REPLY [5 votes]: I'll denote your category of 2-vector spaces by 2Vect.  By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on Δ≤1 where Δ≤1 denotes the full subcategory of Δ on the objects [0] and [1].  Therefore, colimits in 2Vect are computed objectwise under this identification.  So the functor – ⊗ X certainly has a right adjoint Hom(X, –) (by the adjoint functor theorem for locally presentable categories).  This adjunction also respects the Vect enrichment.
To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing Δi for the 2-vector space (Δi)j = HomΔ≤1([j], [i]) • ℝ, we have hom2Vect(Δi, X) = Xi as vector spaces, where hom denotes the Vect-enriched Hom.  So
Hom(X, Y)0 = hom(Δ0, Hom(X, Y)) = hom(Δ0 ⊗ X, Y) = hom(X, Y)
since Δ0 happens to be the unit for this ⊗, but
Hom(X, Y)1 = hom(Δ1 ⊗ X, Y)
will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).<|endoftext|>
TITLE: Why are they called L-functions?
QUESTION [12 upvotes]: I was hoping to see this pop up on the recent big list question about etymology or terms and symbols. Since it has not, and I can't find an answer, I will ask:
What is the reason for the $L$ in $L$-function? I've read that the general use of the term cames from Dirichlet's $L$-functions $L(s,\chi).$ Was there any motivation behind Dirichlet's use or was it just an arbitary choice?
If so, is there any compelling reason that we keep this name other than tradition?

REPLY [4 votes]: Whatever the historical reasons are, I think it is a good thing to use the terminology 'L-function' because of Langlands's amazing contribution to the theory of automorphic forms.
Moreover Langlands functorialities are stated in terms of the 'L-group'.<|endoftext|>
TITLE: Applications of Cauchy's Arm Lemma
QUESTION [6 upvotes]: Cauchy's Arm Lemma is used in the proof of Cauchy's Rigidity Theorem for convex polyhedra. The Lemma states that in the plane or on the sphere that if all but one of the side lengths of two convex polygons $P$ and $P'$ are the same, and the angles formed by the remaining sides of P are less than or equal to those of the remaining sides of $P'$, then the ommitted side length from $P$ is less than or equal to the omitted side length for $P'$, with equality occurring iff the angles are all the same. I know of a couple of extensions of this theorem (a nice presentation of this sort of thing is available in O'Rourke's paper [O'R01]). The Lemma can also be used to show that convex linkages may be (unsurprisingly) straightened.
I'm seeking other applications of this Lemma, particularly those that might be suitable for use as exercises in an advanced undergraduate course, or other applications in the theory of polyhedra.
[O’R01] Joseph O’Rourke, An extension of Cauchy’s arm lemma with application to curve development, Discrete and computational geometry (Tokyo, 2000), Lecture Notes in Comput. Sci., vol. 2098, Springer, Berlin, 2001, pp. 280–291. MR MR2043660 (2004m:52052)

REPLY [2 votes]: If I may add one explicit example as a late-comer, even though it is along the lines with which you are already familiar: Cauchy's arm lemma may be used to prove that the curve that is the intersection of a plane with a convex polyhedron develops on a plane without self-intersection.  This is described in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, p.377ff.<|endoftext|>
TITLE: What should be taught in a 1st course on Riemann Surfaces?
QUESTION [20 upvotes]: I am teaching a topics course on Riemann Surfaces/Algebraic Curves next term.  The course is aimed at 1st and 2nd year US graduate students who have have taken basic coursework in algebra and manifold theory, but may not have had much expose to algebraic geometry.  I will loosely follow the book  Introduction to Algebraic Curves by Griffiths.  In particular, I hope to spend a minimum amount of time developing basic machinery (e.g. sheaf theory) and to start doing concrete geometry (e.g. canonical models of curves of genus up to 4) as soon as possible.  
My question is: what are some good concrete, accessible  geometric topics in Riemann Surface/Curve theory that aren't in the standard textbooks?  
Let's say that the standard textbooks are the book I mentioned and those discussed:
  here.

REPLY [7 votes]: As Gerald said, really understanding some specific surfaces is useful.  And not just the compact ones!  Remember that Riemann's ideas were based on analytic continuation, not algebra.
And if you really mean Riemann surfaces, then Divisors and monodromy (compute some actual monodromy matrices!).
And if you want to cover material which is not in the standard textbooks, cover in depth the relation between Riemann surfaces and differential equations.<|endoftext|>
TITLE: How do you make a good math research poster for a non-mathematical audience?
QUESTION [23 upvotes]: I have the opportunity to prepare a research poster for a non-mathematical, yet scientifically savvy audience, and I want to do it well. I have asked a few mathematicians, and I have heard the following sound advice:

Use interesting graphics.
Elaborate on possible applications to other scientific fields.

Although easier for applied mathematicians, this will be ok as I study subfactors, which have connections to quantum physics and statistical mechanics, and planar algebras, which provide great graphics. But there are practical questions as well:

How can I use LaTeX to make a poster?
How can I avoid using mathematical symbols and technical language?

What makes a good math research poster? What are some good ways to target a non-mathematical audience? Does anyone have examples and/or templates using LaTeX? What other advice would you give a mathematician who has never made a research poster before?
A few asides:

I'm not sure if this is community wiki. I'm more than happy to click the box if requested.
Feel free to retag this question as you see fit.
I will answer my own question after the poster presentation, and I will have all of my materials available online.

REPLY [2 votes]: I agree, somewhat, with Jon Yard. Make individual "pages" with LaTeX and/or beamer. Save the pages as pdf, and arrange using Illustrator. You can preassign the page size in illustrator, and take the file to Kinkos to have it made. Or find out where the cool color copier is in the chemistry, biology, physics, or administrative department is. The cool copier will render a single page glossy poster. 
Since the author of the question has very cool graphics of his work, many of which work better in color, these should be included. 
If you don't want to use beamer, you can import eps files (and I think, pdf) into xfig. I have seen amazing xfig posters.<|endoftext|>
TITLE: the delta system lemma outside set theory
QUESTION [11 upvotes]: The lemma:
Any uncountable set $S$ of finite sets has an uncountable subset $\Delta \subseteq S$ and an $x$ such that $\forall a,b \in \Delta$, if $a \neq b$ then $a \cap b = x$. $\Delta$ is called a $\Delta$-system.
I've seen this lemma used in independence proofs, such as the famous result that the negation of the continuum hypothesis is consistent with ZFC, but it seems like it would be useful in other fields as well. Does anyone have any examples with this lemma outside pure set theory?
See also the finite and generalized infinite versions posted below.

REPLY [3 votes]: Joel gave one generalization; there is another generalization to a different direction:
Let $T$ be a $\xi$-stable (complete) first order theory, and $A$ a set in a model of $T$ with $|A|\le\xi$. If $I$ has cardinality $>\xi$, then $I$ has a subset with cardinality $>\xi$, which is an indiscernible set over $A$. (Here $I$ is a set of finite tuples in some model of $T$ that contains $A$.)
The usual $\Delta$-lemma follows from this because the theory of infinite set with empty language is $\omega$-stable, and an indiscernible set in such a model is just as wanted.
I believe this is from Saharon Shelah's Classification Theory (1990).
Wikipedia links:

Stable theory
Indiscernibles<|endoftext|>
TITLE: What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?
QUESTION [44 upvotes]: Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$, and let $\mathfrak g \text{-rep}$ be its category of finite-dimensional modules.  Then $\mathfrak g\text{-rep}$ comes equipped with a faithful exact functor "forget" to the category of finite-dimensional vector spaces over $\mathbb C$.  Moreover, $\mathfrak g\text{-rep}$ is symmetric monoidal with duals, and the forgetful functor preserves all this structure.  By Tannaka-Krein duality (see in particular the excellent paper André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, 1991), from this data we can reconstruct an affine algebraic group $\mathcal G$ such that $\mathfrak g \text{-rep}$ is equivalent (as a symmetric monoidal category with a faithful exact functor to vector spaces) to the category of finite-dimensional representations of $\mathcal G$.
However, it is not true that every finite-dimensional Lie algebra is the Lie algebra of an algebraic group.  So it is not true that $\mathcal G$ is, say, necessarily the simply-connected connected Lie group with Lie algebra $\mathfrak g$, or some quotient thereof.  So my question is:

Given $\mathfrak g$, what is an elementary description of $\mathcal G$ (that avoids the machinery of Tannaka-Krein)?

For example, perhaps $\mathcal G$ is some Zariski closure of something...?

REPLY [22 votes]: After some thought my pessimism (as expressed in my concurrence with the answer
of Milne) has abated somewhat. If I were bold enough I would conjecture the
following (assuming that the characteristic zero base field is algebraically
closed): Let $\mathfrak g$ be a finite dimensional Lie algebra over $k$ and let
$G$ be the pro-algebraic group whose representation tensor category is
equivalent to the tensor category of finite dimensional $\mathfrak
g$-modules. Then if $S$ is the (pro-)radical of $G$ and $U$ the (pro-)unipotent
radical $U$ and $G/S$ are algebraic groups (unipotent and semi-simple
respectively). Furthmore, the pro-torus $T:=S/U$ has as character group
$\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ considered as an additive
group. Hence the only infinite-dimensional part is $T$ but its character group,
\emph{à priori} only an abstract group, is reasonably well controlled. This is
analogous to the case of  irreducible infinite-dimensional
representations of a semi-simple Lie group where the center of the enveloping
algebra acts by a character and the set of characters as a set is very
large. However it is the set of $k$-points of an algebraic variety which means
that it is under control. The analogy goes further as the category of
$G$-representations (assuming $U$ is finite dimensional) splits up into a direct
product of categories parametrised by cosets of the character group of $T$ with
respect to the subgroup generated by the characters occurring in the action of
$T$ on the Lie algebra of $U$.
Here are some comments on the conjecture (I do not vouch for the complete
veracity of my claims).
We can get a picture of $G/U$ by looking at the irreducible $\mathfrak
g$-representations (as they correspond exactly to the irreducible
$G/U$-representations). All such representations factor through $\mathfrak
g/[\mathfrak{g},\mathfrak{u}]$ which is the product of
$\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and $\mathfrak g/\mathfrak{u}$.
Hence, the irreducible representations are parametrised by pairs of a
$1$-dimensional representation of $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and
an irreducible representation of the semi-simple algebra $\mathfrak
g/\mathfrak{u}$. This gives the prediction that $G/U$ should be the product of a
torus with character group $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and the
simply-connected semi-simple group with Lie algebra $\mathfrak g/\mathfrak{u}$.
As for $U$, the idea is that the category of
unipotent representations (i.e., successive extensions of the trivial
representation) of $\mathfrak g$ is equivalent to the category of
representations of the unipotent group with Lie algebra $\mathfrak g'$, the
maximal unipotent quotient of $\mathfrak g$. Something similar ought to be true
for successive extension of the same irreducible representation and there
shouldn't be too much "intermixing" between different irreducibles.
[Added]
I somewhat rudely hijacked the question by taking up things that maybe weren't that
pertinent to the question so let me give an answer which I think is more on track.
The problem is that one can not always define the algebraisation of an
abstract finite dimensional Lie algebra $\mathfrak g$ even if some
algebraisation exists. As an examples consider a $2$-dimensional Lie algebra
with basis $x,y$ and $[x,y]=y$. This is the Lie algebra of an infinite number of
algebraic groups: Let the $1$-dimensional torus $\mathbb G_m$ act on the
additive group $\mathbb G_a$ by $(t,v) \mapsto t^nv$, where $n\not=0$ and let $G_n$ be the
semi-direct product of this action. These groups all have $\mathfrak g$ as Lie
algebra but the only isomorphisms between them is that $G_n$ is isomorphic to
$G_{-n}$.
What does make sense is to speak of an algebraic hull of an embedding of
$\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak
g$-representation. In that case one may consider the intersection of all
algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak
g$. In terms of Zariski closures (when the base field is $\mathbb C$) it is the Zariski closure of the
exponentials of all elements of $\mathfrak g$ (inside of
$\mathfrak{gl}_m$). From the Tannakian point of view this is the group that
corresponds to the tensor subcategory of the category of $\mathfrak
g$-representations generated by the given representation.
However, if one wants something that is independent of a particular
representation one has to pass to an inverse limit of groups coming from
different representations. This leads to an infinite dimensional monster even in
the case when $\mathfrak g$ is $1$-dimensional.<|endoftext|>
TITLE: How to draw knots with Latex? 
QUESTION [38 upvotes]: I am writing an exam for my students, and the topic is intro knots theory. I have no idea how to put knots into the file, but I know many MO users who can draw amazing diagrams in their papers. 
Can someone please provide some hints on what can be used, preferably with some example codes? I do not need complicated diagrams, just some simple knots and links with few crossings. Thanks in advance.

REPLY [2 votes]: You can draw knots with metapost:
\documentclass{article}
\usepackage{mpgraphics}
\begin{document}
    \begin{mpdisplay}
      for i=0 step 1 until 360:
        pair P;
        P = 2cm* (sind(2*i), cosd(3*i));
        fill fullcircle scaled 4mm shifted P withcolor white;
        draw fullcircle scaled 4mm shifted P;
      endfor;
      picture p;
      p:=nullpicture;
      for i=-180 step 1 until 180:
        pair P;
        P = 2cm* (sind(2*i), cosd(3*i));
        addto p contour fullcircle scaled 4mm shifted P withcolor white;
        addto p doublepath fullcircle scaled 4mm shifted P
          withpen pencircle scaled .5bp;
      endfor;
      clip p to (.5cm,2.5cm) -- (.5cm,1.5cm)--
                (-.5cm,1.5cm) -- (-.5cm,2.5cm)--cycle;
      draw p;
    \end{mpdisplay}
    \end{document}

or use pst-knot package (using postscript).<|endoftext|>
TITLE: Transformations of integer polynomials under combinations of their roots
QUESTION [5 upvotes]: I'm wondering whether the following ideas/questions give rise to an existing body of research. (Accordingly: please suggest appropriate changes to the tags!)
Preamble
We consider polynomials f ∈ ℤ[x] with roots in ℝ, and for each polynomial f, the principal root is the real root with the largest magnitude. In the case of two roots of equal magnitude, we take the positive one.† So, for instance, √5 is the principal root of x2 − 5, the golden ratio φ is the principal root of x2 − x − 1, and −φ is the principal root of x2 + x − 1.
[ † Edit: Previously, I had defined the principal root to be the maximal one; I revised this definition based on remarks by Kevin Buzzard below. This also motivates some revisions to the questions I ask. ]
It's tempting to think that we could represent algebraic numbers by (minimal) polynomials over ℤ for which they are the principal root. We implicitly do this with rational numbers all the time: a/b denotes the real number which is the principal root of bx − a, and can formally be defined in such terms. This is also precisely what we do with algebraic integers: n√b is defined to be the principal root of xn − b, at least for b non-negative.
This approach seems problematic for negative algebraic numbers such as φ−1, whose minimal polynomial is x2 + x − 1, which is the same as for −φ; thus every polynomial for which φ−1 is a root will also have the (larger in magnitude) root −φ. A similar problem arises for −√5, of course. But let us focus on algebraic numbers which are the principal roots of their minimal polynomials.
Questions
For two irredicible polynomials f,g ∈ ℤ[x] with roots in ℝ, let u,v ∈ ℝ be their principal roots.

Are there broad classes of polynomials f and g, including ones of degree 2 or more, of course, for which there is a "simple" formula (e.g. involving no recursive functions more complicated than sums, products, exponents, and "well-known" number sequences) for the minimal polynomials of u + v, uv, or uv−1?
Does there exist such a "simple" formula for some polynomial (not necessarily irredicuble) for which u + v, uv, or uv−1 is the principal root?
If such questions are a proper subject of some body of research or well-studied theory: what is the name of the associated field of mathematics? (E.g. is this a special topic of Galois theory?)

Note that it is unlikely that we can obtain any sort of satisfactory answer for obtaining a minimal polynomial for u − v ; for instance, if u = √2 and v = √3, then the minimal polynomial of all four numbers ± √3 ± √2 have the same minimal polynomial, x4 − 26x2 + 145. For similar reasons, it is unlikely that there is a complete solution for arbitrary sums of principal roots: if u is a principal root of f(x), then −u is a principal root of f(−x), which differs from f in the case that f is not an even function of x. Thus, some differences of principal roots may also be expressed as sums of principal roots.
The difficulties described above and in the preamble suggest that a clean and elegant theory is unlikely; but I'm hoping that there are interesting classes of algebraic numbers which may be treated in this way.

REPLY [3 votes]: If $u$ is a root of $f$ and $v$ is a root of $g$ then $u+v$ is a root of the resultant of $f(x-y)$ and $g(y)$ (for the purpose of calculating the resultant, we take these as polynomials in $y$). The resultant is just a determinant of a matrix whose entries are all coefficients of the polynomials, so it would seem to satisfy your request for a simple formula. 
There are similar formulas for polynomials with $uv$ as a root, or $uv^{-1}$.<|endoftext|>
TITLE: Which curves can be found on Abelian varieties?
QUESTION [20 upvotes]: We know that each genus 2 curve is embedded into its degree 1 Jacobian. 
Under which conditions on $C$, $A$, $g$ and $n$ is it possible for a genus $g$ smooth curve $C$ to be embedded in an Abelian variety $A$ of dimension $n$?
What can be said in the case $n=2$? 
And what if $A$ is a Jacobian (and e.g. $n=2$)?

REPLY [6 votes]: Gian Pietro Pirola in  [Curves on generic Kummer varieties, 
 Duke Math. J. Volume 59, Number 3 (1989), 701-708] proves a rigidity theorem for curves of  genus $g\le q-2$ in the Kummer variety of a $q$-dimensional abelian variety. As a consequence, he shows that a generic abelian variety of dimension $q\ge 3$ does not contain hyperelliptic curves of any genus.<|endoftext|>
TITLE: Typesetting mathematics: how do {\em you} convert text into pdf?
QUESTION [6 upvotes]: Prompted by this question I would like to ask the community how they convert their mathematics into pdf files.  In any given procedure for converting mathematics into pdf I am interested in two issues: first typographical quality of text and of mathematical formulas and second production and placement of figures and labels within figures.  
As a concrete example my current procedure is: latex and bibtex until the references settle down, dvips -o to produce postscript, and then ps2pdf to produce pdf files.  I go through postscript in order to make psfrag labels work.  I've never been fully happy with the output - in particular label placement inside of figures is  difficult. 
As a final issue, in the question referenced, Tilman suggests that pdftex has typographical improvements over latex.  I've looked around on-line and these seem to be margin kerning (hanging punctuation) and glyph scaling (font expansion).  How does one use these features?  Do they make a difference in practice?
EDIT: After a bit of pain, I've managed to switch from my previous procedure (described above) to the much simpler procedure of using pdflatex.  Instead of psfrag I now use Colin Rourke's pinlabel package.  I am very happy with pinlabel -- the fonts are exactly what I expect them to be, and the job of labelling is much easier than it used to be.  It is still possible to align labels inside of a figure, and they virtually always show up where I intended.  
I started using the microtype package, which turns on margin kerning and glyph scaling.  I can see that these change the output, but I honestly can't say that the output is better - I guess my eyes aren't that sensitive.  One thing to watch out for - pdftex 1.20 threw show-stopping errors when typesetting figure captions. I updated to 1.40 and the problem went away. 
Thanks for your suggestions - if other people have other latexing procedures I'd be interested to hear about them.

REPLY [3 votes]: One package that that I like to use (when quasi-wysiwyg LaTeX in figures seems desirable) is: IPE. It is extremely simple to use, extensible too (I think), and works directly with PDF. If you don't feel like programming Tikz, pgf, etc. in detail, then IPE might prove to be very useful (also easy to install, as it comes as a package for Ubuntu). A link to the IPE Wiki is here and the Wikipedia article is here.<|endoftext|>
TITLE: Subset of the plane that intersects every line exactly twice
QUESTION [12 upvotes]: In a comment to this question, Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to be shown that choice is required).
Unfortunately, I haven't been able to track down a reference, so if someone could link me to the original result (or provide a short proof) that would be great.  I have roughly worked out a proof for myself, and I'd like to check it against the literature.

REPLY [3 votes]: Here is a minor variation of the proof.
We show that there is a subset $Z$ of $\mathbb R^2$ which intersects each line exactly twice.
We define the ordinals in such a way that each ordinal $a$ is the set of those ordinals $< a$. For any set $S$ we write $|S|$ for the least ordinal equipotent to $S$, and call it the cardinality of $S$. Let $c$ (for continuum) be the cardinality of $\mathbb R$. For any subset $S$ of $\mathbb R^2$ we denote by $\langle S\rangle$ the set of lines generated by $S$ (a line being generated by $S$ if it has at least two points in common with $S$). Let $d\mapsto L_d$ be a bijection from $c$ onto the set of lines in $\mathbb R^2$.
For each $d\in c$ we define $Z_d\subset\mathbb R^2$, $f(d)\in c$, and $z_d\in\mathbb R^2$, as follows. Let $z_0$ be any point of $L_0$, put $f(0)=0$, and let $Z_0$ be the empty set. Now assume $0< d< c$, and $f(e), z_e$ already defined for $e < d$. Put $Z_d:=$ {$z_e\ |\ e< d$}. As $|\langle Z_d\rangle|< c$ (because $|Z_d|< c$), there is a least $f(d)$ in $c$ such that $L_{f(d)}\notin\langle Z_d\rangle$. Let $z_d$ be any point of the set
$$L_{f(d)}-(Z_d\cup(\cup\langle Z_d\rangle)),$$
which is easily seen to be nonempty.
Let $Z$ be the union of the $Z_d$ and $L$ any line in $\mathbb R^2$. We claim $|L \cap Z|=2$, that is, $Z$ is the sought-for subset of $\mathbb R^2$. To prove $|L\cap Z|\le2\ (*)$ we assume by contradiction that there are $g< h< i$ in $c$ such that $z_g,z_h,z_i\in L\cap Z$. We have $z_g,z_h\in Z_i$ by definition of $Z_i$, and thus $L\in\langle Z_i\rangle$, contradicting the definition of $z_i$. To prove $|L\cap Z|\ge2$ we assume by contradiction $|L\cap Z|< 2$. Put $L=L_d$ and let $g$ be in $c$. The inequality $|L_d\cap Z_g|<2$ implies $L_d\notin\langle Z_g\rangle$, and thus $f(g)\le d$ by minimality of $f(g)$. This shows that $Z$ is contained into the union of the $L_e$ such that $e\le d$. As $|L_e\cap Z|\le2$ by $(*)$, we get $|Z|\le2|d|+2< c$, contradicting the obvious equality $|Z|=c$.
A pdf version is available here (Wayback Machine).<|endoftext|>
TITLE: Can you have a spherical plane?
QUESTION [7 upvotes]: When I was at school I wondered if a surface could locally appear to be a unit sphere, yet `carry on forever'.  More formally, my question is:
Can you place a metric of constant curvature +1 on ${\mathbb R}^2$, such that the identity map to ${\mathbb R}^2$ (with standard Euclidean metric) is uniformly continuous?
It is possible to induce such a metric on ${\mathbb R}^2 - {\mathbb Z}^2$, by identifying each unit square with integer vertices, with a hemisphere on the unit sphere.

REPLY [5 votes]: Intrinsically, no, because one way to understand positive curvature is as a "force" that pulls diverging geodesics closer together. If the curvature is bounded from below away from zero, then any two geodesic arcs leaving a given point will intersect at a finite distance as Ulrich comments above.
But extrinsically, the situation is a lot more interesting and fun. You can embed an intrinsically flat plane in a hyperbolic 3-space and, from within the hyperbolic 3-space, it will look like it has constant positive curvature with respect to geodesic planes (intrinsically, hyperbolic planes) of the hyperbolic 3-space. There are models of the hyperbolic 3-space where the 3-space is represented by the interior of a ball of radius 1, and then one of this embedded flat planes is represented by a 2-sphere internally tangent to the surface of the 3-ball.
See http://en.wikipedia.org/wiki/Horoball, and (for a nice picture) http://en.wikipedia.org/wiki/Horocycle<|endoftext|>
TITLE: Linear algebra and regular orbits
QUESTION [6 upvotes]: If $A$ is an $n\times n$ matrix over a field, and $A^{k} = I$, with $k$ the least positive integer such that this occurs, then must there be some vector $v$ such that $\{v,Av,A^{2}v,\dots,A^{k-1}v\}$ has $k$ distinct elements in it?  In other words:

Must every matrix of finite multiplicative order have a regular orbit?

If A has prime power order, $k = p^{m}$, then $A^{p^{m-1}}-I$ is nonzero, so its kernel is proper, and everything outside of that kernel is a vector in a regular orbit.  Over a finite field of size $q$, the index of a proper subspace is at least $q$, so we can even just choose (on average) $q$ random vectors to find one in a regular orbit.  Over an infinite field, the same idea roughly says any random vector should work, as long as one can make some sort of "uniformly" distributed choice.
If $A$ has order a product of two prime powers, then I am assured this is true by a (special case) of an exercise in Isaacs's Finite Group Theory.  I cannot imagine an argument that does not work for arbitrary orders $k$, but I also cannot find a convincing proof even for the product of two prime powers.  The sum of vectors in regular orbits of the $p$ - parts of $A$ need not themselves be in regular orbits of $A$.  Every matrix (over a finite field) I've tried has a regular orbit.

Assuming this is easy, how does one handle the case where $A$ is an automorphism of a finite group $G$, and the order of $A$ is a product of two prime powers?  In other words:

Prove every automorphism of order $p^{a}q^{b}$ of a finite group has a regular orbit.

Assuming the first question's answer is "yes", then what goes wrong for arbitrary orders?  Isaacs's book gives an example where the general automorphism can fail to have a regular orbit, but it is impossible to compare this until I have at least some idea of why the two-prime case does work.

A related version of this question is: regular orbits are quite important in permutation and (finite) matrix groups and are a standard technique in several important (solved and unsolved) problems in modular representation theory.

Is there sort of a gentle introduction that puts these techniques in context?

For any individual paper is clear that what they say works, but my picture of this area is incredibly disjointed and I suspect that is not true for everyone.  For instance Khukhro has an excellent book on automorphisms of p-groups with few fixed points, and many finite group theory texts have chapters on fixed-point-free automorphisms and the consequences for the group structure of the group being acted upon.  However, I haven't found any "textbook" exposition of regular orbits yet.

REPLY [2 votes]: Thanks to Marty Isaacs for reminding me the exercise was supposed to be easy and for giving a reference to my larger question.  I'll post it here, since I think some people are following it.

If the order of A is paqb then the subgroup generated by A has two minimal subgroups, generated by P=Apa−1qb and Q=Apaqb−1.  If the orbit of g under A is not regular, then the stabilizer is a non-identity subgroup of A, so it contains either P or Q.  Hence either P or Q centralize g.  Hence g centralizes either P or Q (I am always amazed at the power of noticing "centralize" is symmetric), so g is in the union CG(P) ∪ CG(Q).  Both of these subgroups are proper subgroups of G, since A acts faithfully on G itself.  However, G is not the union of two proper subgroups, so there is some g in G − ( CG(P) ∪ CG(Q) ), and such a g represents a regular orbit.

An investigation of which groups G must have regular orbits is in:
Horoševskiĭ, M. V. "Automorphisms of finite groups."
Mat. Sb. (N.S.) 93(135) (1974), 576–587, 630.
(Math. USSR Sbornik 22 (1974) 4, 584–594)
MR347979
DOI: 10.1070/SM1974v022n04ABEH001707
and in particular proves that every automorphism of a nilpotent group or a semisimple
(that is, Fitting-free) group has a regular orbit.  The paper has exercise 3A.8 as a remark after corollary 1 (page 592 in the English translation), and corollary 3.3 as theorem 2.  Its lemma 4 fixes my difficulties with dealing with the orbits prime by prime (don't look at the p-parts of A where the obvious statement has obvious counterexamples, look at non-faithful orbits instead).

I could not generalize Robin Chapman's argument to finite abelian groups, since one
no longer has that finite Z/nZ[t] modules are direct sums of cyclic modules (for
instance, Z[t]/(4,tt-1) has the ideal (2,t+1)/(4,tt-1) of type C2 × C4 with t=A acting as the matrix [1,2;0,1].  This module is non-cyclic and indecomposable.  Of course t has a regular orbit, but I could not simply choose a "generator".  Marty Isaacs has shown me how to use Horoševskiĭ's argument to reduce to the case where G is indecomposable, where presumably it is easier than I think.

In the other direction, keeping G elementary abelian but letting A be an entire group of automorphisms, one is still quite interested in whether there is a regular orbit.  I found this article helpful for getting an idea of how this works:
Fleischmann, Peter. "Finite groups with regular orbits on vector spaces."
J. Algebra 103 (1986), no. 1, 211–215.
MR860700
DOI: 10.1016/0021-8693(86)90180-8.
In particular, nilpotent groups tend to have regular orbits except when p=2 is involved (either in A or G), and the specific problems with p=2 are addressed.  Its methods for abelian groups A give an alternative view of Robin Chapman's answer (basically the paper shows that you can reduce to the algebraically closed/absolutely irreducible case, and then the Bi are all 1x1, and the matrix A is diagonal).<|endoftext|>
TITLE: Question about hypercohomology / spectral sequence of a complex of "almost-acyclic" sheaves
QUESTION [5 upvotes]: I have a very particular situation involving a (non-exact) complex $K$ of coherent sheaves on a nonsingular projective variety $X$, and I need to compute the hypercohomology of the complex.  The associated spectral sequence is highly degenerate.  Of course, in degenerate cases, one hopes that there are techniques for getting at the hypercohomology rapidly.  In the most degenerate case, wherein every sheaf in the complex is acyclic, then the hypercohomology of the complex is the cohomology of the spaces of global sections (with respect to the maps induced by the complex's differential, say, $d$):
$\mathbb{H}^q(X,K^\bullet)\cong H^q_d(\Gamma(X,K^\bullet))$.
The problem with my situation is that the sheaves are not quite acyclic.  I will be more formal now about what I am dealing with.  I have a non-exact complex $(K^\bullet,d)$ of three sheaves:
$K^1\stackrel{d}{\rightarrow}K^2\stackrel{d}{\rightarrow}K^3$.
Here, $K^1$ has no nonzero sheaf cohomology except for $H^1(K^1)$, whilst $K^2$ and $K^3$ have nonzero zero-th and first cohomology and all higher cohomology vanishes.
I would think that there must be a way of dealing with such a specialized, degenerate situation, but I haven't found it yet.
Please feel free to specialize this further, if it helps (e.g. torsion-free instead of only coherent ones, or even vector bundles).  Also, feel free to make this more general: I am only restricting myself to three sheaves because that is precisely the problem I am faced with.  In similar spirit, if giving $K^1$ a nonzero $H^0$ doesn't harm the chances of finding a reasonable solution, then by all means do so.  (This would give the problem at hand a uniform description: hypercohomology of a complex of sheaves in which each sheaf has nonvanishing zero-th and first cohomology, and vanishing cohomology elsewhere.)
If there is a specific reference for where this situation is worked out (as I'm sure it must be and I probably just haven't looked hard enough), then please do pass it along.
Thanks!!

REPLY [7 votes]: Ok, so what will the spectral sequence give you? This is a very easy exercise, but since Altgr is not experienced, here is the solution. The term $E_2^{p,q}$ is the $p^{\rm th}$ cohomology group of the complex $\mathrm{H}^q K^{\bullet}$ (the complex of groups obtained by applying the $q^{\rm th}$ cohomology functor to the complex $K^\bullet$). Then one has equalities
$$\mathbb H^i (K^\bullet) = 0 \quad{\rm for}\ i \neq 2,3,4$$
$$\mathbb H^4 (K^\bullet) = E_2^{3,1}$$
and an exact sequence
$$
0 \to E_2^{2,0} \to \mathbb H^2 (K^\bullet) \to E_2^{1,1} \to E_2^{3,0}
\to \mathbb H^3 (K^\bullet) \to E_2^{2,1} \to 0
$$
The homomorphism $E_2^{1,1} \to E_2^{3,0}$ is the only differential at the $E_2$ level that can possibly be non-zero.
Without further information, there is nothing else one can say, other than this: working with hypercohomology without knowing spectral sequences is like driving nails into a wall without a hammer.<|endoftext|>
TITLE: Two questions about Cohen-Macaulay rings
QUESTION [14 upvotes]: The following questions seem basic, but I can't find them in the literature. I'm also unable to think of a counterexample.
Let $A$ be a local Cohen-Macaulay ring of dimension $d$.

Let $I$ be an ideal generated by $r$ elements. Is is true that the depth of $A/I$ is at least $d-r$?
Let $Q$ be a minimal prime of $A$. Is it true that $A/Q$ is also Cohen-Macaulay?

REPLY [18 votes]: Nice questions! The answers are no in both cases, although the examples are more interesting than one would expect.
1) Even when $A$ is regular, one can always find an ideal $I$ with $3$ generators such that $A/I$ has depth $0$. This is due to a very nice result by Bruns, which says you can construct $3$-generated ideal with all kinds of homological patern. The details are explained in this answer.
2) Let $R=k[X^4,X^3Y,XY^3,Y^4]$. Then $R$ is a domain of dimension $2$ which is not Cohen-Macaulay. So one can write $R=S/Q$, where $R=k[a,b,c,d]$ and $Q$ is a prime ideal of height 2. Take $(f,g)$ to be a regular sequence  in $Q$ and let $I=(f,g)$. Then $A=S/I$ is Cohen-Macaulay (being a complete intersection), but $Q$ is a minimal prime of $A$ and $R=S/Q$ is not CM.<|endoftext|>
TITLE: Stacks in the Zariski topology?
QUESTION [11 upvotes]: I have two naive questions about stacks.

1) Is it possible to define stacks in the Zariski topology? 

Presuming you can:

2) If a stack has a coarse moduli, and the coarse moduli space is a scheme, then does that mean that your stack is a stack in the Zariski topology?

In general, I am trying to understand why a new notion of open cover is necessary if all I am interested in is remembering stabilizers. Certainly this is too simple a mind-set, so feel free to enlighten me.

REPLY [11 votes]: 1.) It's possible to define stacks on ANY category equipped with a Grothendieck topology (such a category with a topology is called a site). In particular, this holds true for the Zariski site. Moreover, there is always a way to define an "Artin stack"- these are those stacks which arise as torsors for a groupoid object in your site. Outside of algebraic geometry, these give rise to notions of topological and differentiable stacks, for instance.
EDIT: As long as groupoid objects exist in your category.
2.) As in Harry's post, any stack which is a stack in a site which is finer than the Zariski topology is also a stack in the Zariski topology.
To address your general question as to "why a new notion of open cover is necessary if all I am interested in is remembering stabilizers", you should learn a bit about Grothendieck topologies. I'll make a couple remarks:
i) If all you cared about were stabilizers, then you wouldn't need to use any covers at all; ordinary fibered categories would do the trick!
Indeed, take a group object in your site acting an object, and take the action groupoid- it is a groupoid object. Look at the pseudo-functor which assigns each object of your site the groupoid of maps into this groupoid object (considering the object as a groupoid with all identity arrows). This remembers the stabilizers for this action.
ii) (subcanonical) Grothendieck topologies are a choice of a type of cover for your objects, in such a way that this object is the colimit of these covers, AND "this is important to remember". This is a little imprecise, so, allow me to elaborate via an example from topology:
Let $U_i$, $i\in I$ be an open cover of a space X. Then, continuous maps from X to another space Y are in bijection with with continuous maps $f_i:U_i \to Y$ which agree on their intersection. This is just saying that X is the colimit of this open cover. Instead, we can view this a property of the presheaf $Hom(blank,Y)$ represented by $Y$ on the category of topological spaces (for you set theorists, choose a Grothendieck universe).
For any $X$ and any open cover $U_i$, $i\in I$ of $X$, (let $Hom(blank,Y)=F$)
the natural map $F(X) \to \varprojlim \left[{\prod{F(U_i)}} \rightrightarrows {\prod{F(U_{ij})}}\right]$
is a bijection.
If $F$ is any presheaf, this is just saying $F$ is a sheaf. Since this is NOT true for an ARBITRARY presheaf $F$, X is no longer the colimit of its open covers in the full category of all presheaves. The same argument holds for all fibred categories- it's only true if we restrict to STACKS (and $X$ then becomes the weak colimit of this cover, but, never mind).
The reason you add the condition for descent for covers, is so that "all maps into your stack from a space are continuous". More precisely, and more generally, it's so that maps from a space, scheme, whatever you site is, into a stack can be determined by mapping out of elements of some covering of your object in a way that glues (for stacks, rather than sheaves, they don't need to AGREE on the intersection, but, agree up to an invertible 2-cell, plus some coherency conditions).
Combining these ideas, if you have a group acting on an object, the pseudo-functor produced by the action groupoid is rarely a stack with respect to your topology, but you can stackify it, and then it will become one and still remember all the stabilizers. I hope this helps!<|endoftext|>
TITLE: Connection between the two-variable case of Hilbert's Tenth Problem and Roth's Theorem.
QUESTION [9 upvotes]: Connection between Hilbert's Tenth Problem and Roth's Theorem.
The following two decision problems seem to be open:

Given a polynomial equation in two variables with integer coefficients, determine whether there are any integer solutions. (The two-variable case of Hilbert's Tenth Problem)
Given a real algebraic number $r$  and given integers $B,N > 0$,  determine whether  the inequalities  $|rx-y|  <  \frac{1}{x^{1+1/N}}$, $0 < x < B$  are solvable in integers $x$ and $y$. (This is the problem of the effective Roth bound -- Roth proved that for any algebraic $r$ and for any $N > 0 $, inequality $|rx-y| <  \frac{1}{x^{1+1/N}}$ has only finitely many  solutions in positive integers $x,y$.)

Now I heard once that an effective algorithm for the Roth bound  would yield an effective algorithm for the two variable case of Hilbert's Tenth Problem.  I can begin to imagine that this might be true for norm-form equations, following the treatment in Schmidt's books, but the general case seems quite opaque. 
Can anyone suggest any references along these lines? Also, does anyone know of  surveys summarizing what has been done to-date on the  of the decidability of the two-variable case of Hilbert's Tenth Problem? Finally, does anyone know any interesting "plausibility arguments" for or against decidability?

REPLY [2 votes]: Here is a plausibility argument for decidability. 

DISCLAIMER: I am in no way an expert and admittedly an optimist.

The way you show that Hilbert's Tenth Problem has a negative solution is by showing that diophantine equations can "cut out" every recursively enumerable subset of $\Bbb Z.$ The negative solution follows from the fact that there are recursively enumerable subsets which are not recursive. For a quick introduction see Mazur's recent expository lecture notes here.

Moral: Diophantine equations can define sets which are too complicated!

For each recursively enumerable subset $S$ of $\Bbb Z$ we can define the diophantine dimension of $S$ as the smallest such $n$ for which there is a diophantine equation in $n$ variables with integer coefficients which cuts out $S$. (Experts: Is there a better name for this integer?)
For example, this popular MO question of Poonen asks roughly to determine if the diophantine dimension of $\Bbb N$ is 2? This seems to be incredibly hard. It seems plausible that the relatively tame subset $\Bbb N$ of $\Bbb Z$ might have diophantine dimension $ > 2$. So I feel free to hope that the even more complicated sets which lead to undecidability are complicated enough that they have diophantine dimension greater than 2.<|endoftext|>
TITLE: What is the Zariski topology good/bad for? 
QUESTION [34 upvotes]: In a comment to this question the quotation "The Zariski Topology is the 'Wrong' topology for Algebraic Geometry" appears.
Well, so some spontaneous questions arise: 
1) What is Zariski topology good for in algebraic geometry? In other words, what can you do with it, without referring directly to some finer Grothendieck topologies?
2) On the other hand, which concepts which are analogs of concepts in -say- analytic geometry or topology, really need finer Grothendieck topologies to be generalized to the algebro geometric setting?
I think one could mention vector bundles for the 1), and principal bundles and projective bundles for 2). And also cohomology for 2), since it seems that the étale topology is best suited for reproducing a cohomology that resembles the singular one in the analytic setting. And also sheaf cohomology for 1).

Edit: I think it would also be nice if in the answers there were some brief comments about the sufficiency (or not) of Zariski topology to capture the essential geometric picture borrowed from analytic geometry and/or topology about (some of) the following contexts:

Patching of morphisms
Inverse function theorem & implicit function theorem
Existence of tubular neighbourhoods (?)
Local reducubility vs. global reducibility (analytic branches of a variety at a point..)
"Infinitesimal" properties given by the local ring at a point (Zariski vs. Hensel)
Vector bundles
Coherent sheaves (and patching thereof...)
Principal bundles (local triviality...)
Projective bundles (as above...)
In general, "locally trivial" fiber bundles (isotriviality vs. triviality...) 
Sheaf cohomology with constant coefficients vs. of coherent sheaves
Cech cohomology (as above...)
Covering spaces and fundamental group
Extracting topological/homotopical information via coverings
(any other topic you find interesting)

REPLY [2 votes]: algebraic geometry is not my field at all, but it seems like this is a bit of an answer, and i don't really know the details but maybe someone can fill them in. The Zariski topology is not good for doing homotopy theory. I have heard this at many seminars, specifically from someone who does motivic homotopy theory. So from that perspective it is not the right topology, but i can only say that i have heard this not that i understand why the Zariski topology is bad. I guess just naively it seems like it would be pretty hard doing homotopy theory on "any" line where your only open sets are finite complements.<|endoftext|>
TITLE: Density of monogenic number fields?
QUESTION [10 upvotes]: Background
Zev Chonoles recently asked the question "which number fields are monogenic?". The answers say that for a specific number field the question is hard. So, I thought, how about looking at all of them.
Question
There are two natural ways, that I can think of, to count number fields:
(1) By discriminant: what is the asymptotic behavior of $$ \frac{\{K:K \text{ is monogenic, } |\Delta_K| < x\}}{\{K :|\Delta_K|<x\}}$$
(2) By minimal polynomial of an element: what is the asymptotic behavior of $$\frac{\{K: K \text{ is monogenic, } \exists \alpha \in K \text{ with } ||m(\alpha)||_{\infty} < x\}}{\{K: \exists \alpha \in K \text{ with } ||m(\alpha)||_{\infty} < x\} }$$
where $m(\alpha)$ is the minimal polynomial of $\alpha$ over $\mathbb{Z}$.

REPLY [5 votes]: A recent article of Bhargava and Shankar, "Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves" (http://arxiv.org/abs/1006.1002), addresses, among many other related questions, the density of monogenic cubic orders, counted by the height (slightly modified) of their invariants $I$, $J$.
This height is almost the same as the height of the discriminant of the polynomial to which they are associated with. So this answers the first case above for cubic fields.
Theorem 4.1 (Bhargava, Shankar):

Let $N_3^{(0)}(X,\delta)$ (resp. $N_3^{(1)}(X,\delta)$) denote
    the number of cubic submonogenized rings $(C,x)$ of index $n$ with
    positive (resp. negative) discriminant such that $H(C,x)\lt X$ and
    $n\lt X^\delta$, where $\delta \leq 1/4$. Then we have
    $$N_3^{(0)}(X,\delta)=\displaystyle\frac{4}{45}X^{5/6\,+\,2\delta/3}+O(X^{5/6});$$
    $$N_3^{(1)}(X,\delta)=\displaystyle\frac{16}{45}X^{5/6\,+\,2\delta/3}+O(X^{5/6}).$$

If I understand correctly, this means that the density for case (1) above (restricted to cubic fields) is in fact 0.
The article goes on to study quartic rings, but the result, Theorem4.8, is a bit more difficult to understand. But if I do, it says (a bit more than) that the density of monogenic quartic fields counted by discriminant is, once more, 0. Note that in the theorem the count is by the invariants of resolvent, which is like the discriminant of the resolvent, which is a constant factor away from the discriminant of the quartic polynomial to which it is associated with.
So, an educated guess is that the answer to (1) above is 0 for all degrees, and then in general (as stated above without degree). On the other hand, the discriminant of an $n$ degree polynomial is a multivariate polynomial, and I suspect that sieve theory can prove that for every constant degree there is a positive density (counted as in (2)) of polynomials that have discriminant with bounded squareful part. This in turn would prove that there is a positive density of polynomials giving orders of bounded index in their respective maximal orders. Taking this educated guess a little further, I suspect the answer to (2) is greater than 0, contrary to (1).<|endoftext|>
TITLE: Is there Domain Invariance for Alexandrov spaces?
QUESTION [16 upvotes]: A colleague asked me this question recently. Every injective continuous map between manifolds of the same (finite) dimension is open - this is Brouwer's Domain Invariance Theorem. Is the same true for complete boundaryless Alexandrov spaces (of curvature bounded below)?
Alexandrov spaces are manifolds almost everywhere, and their singularities have special structure. In dimensions 1 and 2 there are no topological singularities (all Alexandrov spaces are manifolds). Higher dimensional singularities have a sort of inductive description: every point $x$ in an $n$-dimensional Alexandrov space has a neighborhood homeomorphic to the cone over an $(n-1)$-dimensional connected(!) Alexandrov space $\Sigma_x$ which carries a metric of curvature $\ge 1$. The last property implies that $\Sigma_x$ is compact and its fundamental group is finite.
For example, in dimension 3 the only type of singularity is the cone over $RP^2$. In dimension 4 there are cones over lens spaces, $\mathbb R\times Cone(RP^2)$ and maybe other beasts.
There is a similar purely topological question: is Domain Invariance Theorem true for "almost manifolds"? An "almost manifold" is a pseudo-manifold obtained from $n$-simlices by gluing their $(n-1)$-dimensional faces in pairs - that is, there are exactly two $n$-simplices adjacent to each $(n-1)$-simplex, and there are no extra identifications between lower-dimensional faces. I'm not sure that all Alexandrov spaces admit triangulations, but if they do, they are "almost manifolds" (of a special kind).

REPLY [6 votes]: Here I will clarify the cohomological issues in Sergei's answer above. For applications to Alexandrov spaces scroll to the end of the post.
I will use Alexander-Spanier cohomology with compact support and $\mathbb Z_2$ coefficients, and the main reference will be Massey's book "Homology and cohomology theory, an approach based on Alexander-Spanier cochains"; I own a Russian translation with insightful comments by Sklyarenko. As usual, using $\mathbb Z_2$ coefficients allows to ignore orientability issues.
Incidentally, as is explained in Massey's book (or in Spanier's "Algebraic topology" text) for locally contractible, locally compact Hausdorff spaces (e.g. for finite-dimensional Alexandrov spaces) the  Alexander-Spanier cohomology coincide with singular and Cech cohomology. 
Lemma. Let $X$ be a locally compact Hausdorff space that 
contains a closed subset $S$ such that $X-S$ is a connected topological $n$-manifold. 
If $U$ is an open subset of $X-S$, then the homomorphism $H_c^n(X, X-U)\to H_c^n(X,S)$ induced by inclusion is an isomorphism.
Proof. By Theorem 1.4 in Chapter 1 of Massey's book, if $A$ is a closed subset of $X$, then there is an isomorphism $H^n_c(X,A)\cong H^n_c(X-A)$.
Theorem 3.21 in Chapter 3 of Massey's book says that if $U$ is an open subset of a manifold $M$,
then the map $H^n_c(U)\to H^n_c(M)$, which associates to a cocycle with compact support in $U$ the same cocycle with support in $M$, is an isomorphism. 
Look at the inclusion $(X, S)\to (X, X-U)$.
Using the above isomorphism, we can identify the induced map $H^n_c(X, X-U)\to H^n_c(X,S)$ with $H^n_c(U)\to H^n_c(X-S)$, which is an isomorphism, as $U$ is open in the manifold $X-S$. QED
Below we denote by $H^n$ the Alexander-Spanier cohomology with arbitrary support;
they coincide with singular cohomology for nice spaces, such as locally contractible, locally compact Hausdorff spaces. Of course, for compact $X$ cohomology with compact support coincide with the the usual cohomology, so we get:
Corollary. If in the assumptions of the Lemma $X$ is compact, then the map
$H^n(X, X-U)\to H^n(X, S)$ induced by inclusion is an isomorphism. QED
Finally, as in Grove-Peterson's paper from Anton's answer, if $X$ is a compact $n$-dimensional Alexandrov space without boundary, and $S$ is the set of non-manifold points, then $S$ has codimension $2$, so long exact sequence of the pair $(X,S)$ shows that $H^n(X,S)\to H^n(X)$ is an isomorphism by inclusion, and we get the isomorphism $H^n(X, X-U)\to H^n(X)$
which implies $H^n(X-U)=0$ by the exact sequence of the pair $(X, X-U)$ because all cohomology in dimension $>n$ vanish. In summary:

If $X$ is a compact $n$-dimensional Alexandrov space without boundary, and $S$ is the set of non-manifold points, then $H^n(X-U)=0$ for any open subset $U$ of $X-S$.

Now if $x$ is a point of $X-S$, then $X-x$
deformation retracts to some $X-U$, so $H^n(X-x)=0$ which is what's needed for Sergei's answer. 
Remark. In fact, the above assertion that $H^n(X-U)=0$ holds for any open $U$ in $X$, i.e. we need not assume $U\subset X-S$. Indeed, if $V:=U-S$, then the isomorphism $H^n(X, X-V)\to H^n(X)$ factors through
the the homomorphism $H^n(X, X-U)\to H^n(X)$, so the latter is onto, but its cokernel is $H^n(X-U)$, hence $H^n(X-U)=0$.<|endoftext|>
TITLE: Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD?
QUESTION [17 upvotes]: Let $K$ be a number field and let $\mathcal O_K$ be the ring of integers. Following this paper of Cornelissen, Pheidas, and Zahidi, a key ingredient needed to show that Hilbert's tenth problem has a negative solution over $\mathcal O_K$ is an elliptic curve $E$ defined over $K$ with rank$(E(K))=1$.
Recently Mazur and Rubin have shown that such a curve exists assuming the Shafarevich-Tate conjecture for elliptic curves over number fields. They actually use a weaker, but still inaccessible hypothesis (See conjecture $IIIT_2$).
If you wanted to eliminate the need for this hypothesis you would have to write a proof that simultaneously demonstrated that rank$(E(K))=1$ for infinitely many pairs $(K,E)$ where $E$ is an elliptic curve defined over $K.$ This raises (as opposed to begs) the easier question:

Can you show unconditionally that rank$(E(\Bbb Q)) = 1$ for infinitely many elliptic curves $E$ over $\Bbb Q$?

It would appear that Byeon, Jeon, and Kim have done so in this paper (probably need an institutional login). Vatsal obtains a weaker result here that still does the job. Unfortunately both of these results invoke the fact that the BSD rank conjecture is true for elliptic curves over $\Bbb Q$ with analytic rank 1. Which won't help at present working over number fields.

Can anyone do the above WITHOUT invoking the proven part of the BSD rank conjecture or assuming any conjectures?

REPLY [8 votes]: This is just a tiny follow up to Victor Miller's remark above.  Manjul Bharghava has a second paper with Shankar entitled "Ternary cubic forms having bounded invariants, and the existence of a positive proportion of elliptic curves having rank 0", and in there he states the following theorem:
Theorem 5: Assume $Sha(E)$ is finite for all $E$. When all elliptic curves $E/\mathbf{Q}$ are ordered by height, a positive proportion of them have rank $1$. 
This obviously doesn't answer your original question, because Manjul is assuming that Sha is finite.  However, I believe his arguments for this theorem use results toward BSD over $\mathbf{Q}$. 
EDIT: Crud -- I meant to say that I believe his arguments don't use results toward BSD, which was the whole point.  Oops.<|endoftext|>
TITLE: Is there a "groupoid integral" with values in a groupoid?
QUESTION [6 upvotes]: Let $G = \{G_1 \rightrightarrows G_0\}$ be a finite groupoid, i.e. $G_1,G_0$ are both finite sets, and let $A$ be $\mathbb Q$-module.  Regard $A$ as a discrete groupoid $A \rightrightarrows A$, and let $f: G\to A$ be a functor — equivalently, there is a set $G_0/G_1$ of isomorphism classes of $G$, and $f$ is an $A$-valued function on this set.
  Then Baez and Dolan define a groupoid integral:
$$ \int_G f = \sum_{x\in G_0/G_1} \frac{f(x)}{\lvert {\rm Aut}(x)\rvert} $$
where to make the above precise I'm using the fact that if $x,y \in G_0$ are isomorphic, then $\lvert {\rm Aut}(x)\rvert = \lvert {\rm Aut}(y)\rvert$, and choosing an isomorphism $x \to y$ in fact induces an isomorphism ${\rm Aut}(x) \to {\rm Aut}(y)$.  Anyway, the point is that $\int_G f$ actually depends only on the "stack" $G_0 // G_1$, where for our purposes "stack" can mean "groupoid up to equivalence": if $G,G'$ are equivalent groupoids and $f':G' \to A$ is the functor corresponding to $f$ under the equivalence, then $\int_Gf = \int_{G'}f'$.
Suppose now that $A$ is not a $\mathbb Q$-module but some (possibly weak) version of a "$\mathbb Q$-module in groupoids".  (If you want, I have no objection to you thinking of $A$ as a strict object — the weakened version would replace all the axioms for a $\mathbb Q$ vector space with natural isomorphisms that have their own coherency.)  Then it still makes sense to talk about functors $f: G \to A$.

Question: Does there exists an extension of the groupoid integral above to integrals of the form $\int_Gf$ where $f: G \to A$ is a functor but $A$ is not discrete, and that plays well with equivalences of groupoids $G \to G'$ and $A \to A'$?

REPLY [3 votes]: Yes.  It seems you're only looking for an object up to isomorphism, so all you really need is divisibility and addition to yield objects that are well-defined up to isomorphism.  Assuming your notion of rational vector space in groupoids involves a Picard groupoid (i.e., symmetric monoidal category with all objects invertible) with "multiplication by a/b" functors, and a zoo of compatibility isomorphisms under multiplication and addition, then you can use the Baez-Dolan formula.
This is a groupoid version of pushing forward a constant sheaf on a zero dimensional stack (which is well-known under suitable assumptions on stabilizers and the characteristic of the coefficients).<|endoftext|>
TITLE: Upper bound for the size of a $k$-uniform $s$-wise $t$-intersecting set system
QUESTION [7 upvotes]: Given integers $n \geq k \geq t \geq 1$ and an integer $s$, let $m(n,k,s,t)$ denote the maximum size of a family $\mathcal F$ of $k$-subsets of $[n]$, i.e. $\mathcal F \subseteq \binom{[n]}{k}$, such that the intersection of any $s$ members of $\mathcal F$ is at least $t$. 
My question is: what is the best known upper bound on $m(n,k,s,t)$?
Below are what I was able to dig out from the literature.
The classic Erdos-Ko-Rado theorem ([EKR 61] plus a result by [Wilson 1984]) states that $m(n,k,2,t) \leq \binom{n-t}{k-t}$ for $n \geq n_0(k,t) = (k-t+1)(t+1)$. This bound is tight for $n \geq n_0$.
[Ahlswede and Khachatrian 1997] derived tight bounds for the $n &lt; n_0$ case, completely settling the pairwise $t$-intersecting sub-problem.
[Frankl 1974] showed that $m(n,k,s,1) \leq \binom{n-1}{k-1}$, provided that $ks \leq n(s-1)$. This bound is tight. (When $ks > n(s-1)$, the intersection of any $s$ members of $\mathcal F$ is not empty.) Several other papers of Frankl gave some bounds for the non-uniform case, i.e. when members of $\mathcal F$ do not need to be of the same size $k$.
[Tokushige 2007] gave a bound for $m(n,k,3,t)$. Conjecture 1 in that paper specifies a formula for $m(n,k,s,t)$ but not much evidence was given other than that the conjecture holds for $s=2$. I'd be interested to know whether the conjecture holds if we replace $=$ by $\leq$.
There are also some other papers discussing bounds when $s$ and $t$ are small constants.
In summary, I was not able to find any generic upper bound for $m(n,k,s,t)$ (except for the obvious fact that $s$-wise $t$-intersecting systems are also $(s-1)$-wise $t$-intersecting systems, and thus the EKR bound applies).

REPLY [2 votes]: This is not as general as what is asked, but in N. Tokushige, The random walk method for intersecting families, in Horizons of Combinatorics, Bolyai Society Mathematical Studies, 17 (2008) 215-224, it is stated that

For any fixed $p = \frac{k}{n}$ such that $p < \frac{s-1}{s+1}$ and $n$ large enough,
$$m(n,k,s,t) \leq \alpha_{s, p}^t{{n}\choose{k}},$$
where $\alpha_{s,p} \in (p, 1)$ is the unique root of the equation $(1-p)x^s-x+p$.

The proof can be found in N. Tokushige, The maximum size of 4-wise 2-intersecting and 4-wise 2-union families, European J. Combin., 27 (2006) 814-825, where this is used as a tool to prove a strong theorem for specific $t$ and $s$. Upper bounds of the same kind seem to be proved in other places as well to give strong theorems for different fixed $t$ and $s$.
The first paper I linked to also proves that

There exist positive constants $n_1$, $\epsilon$ such that for all $t$ such that $1 \leq t \leq \frac{3^s-2s-1}{2}$ and all $n > n_1$ and $k$ such that $\frac{k}{n} < \frac{1}{3}+\epsilon$,
$$m(n,k,s,t) = \max\left\{{{n-t}\choose{k-t}}, (t+s){{n-t-s}\choose{k-t-s+1}}+{{n-t-s}\choose{k-t-s}}\right\}.$$

Note that ${n-t}\choose{k-t}$ and $(t+s){{n-t-s}\choose{k-t-s+1}}+{{n-t-s}\choose{k-t-s}}$ are exactly the sizes of $\mathcal{F}_0$ and $\mathcal{F}_1$ respectively in $\text{Conjecture}\ 1$ in question. So, this proves the conjecture for a restricted case. Results in the same spirit by the same author can also be found in
N. Tokushige, Multiply-intersecting families revisited, J. Combin. Theory B, 97 (2007) 929-948
and
N. Tokushige, A multiply intersecting Erdos-Ko-Rado theorem - The principal case, Discrete Math. 310 (2010) 453-460.<|endoftext|>
TITLE: Murray-von Neumann classification of local algebras in Haag-Kastler QFT
QUESTION [7 upvotes]: The Haag-Kastler approach to quantum field theory (QFT) is one of the oldest approaches to rigorously define what a QFT is, it deals with nets of operator algebras: You start with a spacetime and assign von Neumann algebras (or $C^*$-algebras, but my question is about the von Neumann algebra situation only) to certain subsets of the spacetime subject to certain axioms.
I am interested in results about the Murray-von Neumann classification of these algebras, i.e. which kind of factors can occur in the central decomposition (the decomposition of the algebra as a direct integral of factors).
To make this more precise, here is an example: One can define vacuum representations on Minkowski spacetime, for details please see 
Haag-Kastler vacuum representation on the nLab.
A net of a vacuum representation is said to satisfy duality, or be a dual net, if one has $\mathcal{M}(\mathcal{O}') = \mathcal{M}'(\mathcal{O})$, put in words: the algebra of the causal complement of a bounded open set $\mathcal{O}$ is the commutant of the algebra associated with $\mathcal{O}$. Then it is a theorem that algebras associated to diamonds can only have factors of type $III_1$ in their central decomposition.

Is the assumption of duality necessary or is causality enough? Is Haag duality enough? (Haag duality means that the duality condition does not have to hold for all algebras associated to bounded open regions, but for diamonds only).
What are the necessary assumptions to deduce that algebras associated to diamonds are a factor of type $III_1$, i.e. have trivial center? What are the necessary assumptions to get that these algebras are hyperfinite?
Are there similar results about the factor decomposition of algebras associated to more general subsets than diamonds of the Minkowski space, like open bounded subsets?
Are there similar results about more general spacetimes, like globally hyberbolic ones?

REPLY [6 votes]: There is a nice overview about algebraic quantum field theory by Halverson and Müger, which covers some of the stuff I mention below and can be found at
http://arxiv.org/abs/math-ph/0602036
Concerning your question(s): Having a factor of type III means that every (non-zero) projection in it is Murray-von Neumann equivalent to the identity. In the Haag-Kastler approach Borchers came up with a slightly weaker notion of "type III-ness", which is sometimes called 'property B'. 
Definition: Let $\mathcal{O} \rightarrow \mathcal{A}(\mathcal{O})$ be a net of von Neumann algebras on some common Hilbert space $H$. $\mathcal{A}$ has property B if for any two double cones $\mathcal{O}_1$ and $\mathcal{O}_2$ such that $\overline{\mathcal{O}_1}\subset \mathcal{O}_2$ and for any non-zero projection $E \in \mathcal{A}(\mathcal{O}_1)$ there is an isometry $V \in \mathcal{A}(\mathcal{O}_2)$, such that $VV^* = E$ and $V^*V= 1$ (in other words: $E$ is equivalent to $1$ in the algebra $\mathcal{A}(\mathcal{O}_2)$).
The point is the following theorem proven by Borchers in 
A remark on a theorem of B. Misra, H.J. Borchers, Communications in Mathematical Physics, Volume 4 (5), page 315-323.
Theorem: Let $\mathcal{O} \rightarrow \mathcal{A}(\mathcal{O})$ be a net of von Neumann algebras satisfying microcausality, the spectrum condition, and weak additivity. Then the net $\mathcal{A}$ satisfies property B.
So, what are these notions? 

microcausality means that the algebras associated to spacelike separated double cones commute
additivity means that for any double cone $\mathcal{O}$ the set of all $\mathcal{A}(\mathcal{O} + x)$ for $x \in T$, where $T$ denotes the translation group generates the associated (universal) $C^*$-algebra (I am not quite sure, what weak means in the statement).
the spectrum condition means there is a subset $T_+ \subset T$ with $T_+ \cap (-T_+) = \{0\}$ and the spectrum of the unitary representation of $T$ is contained in $T_+$.

Since these are all motivated by physics, they are all kind of natural to demand for nets of von Neumann algebras in AQFT.
Anyway there are also some results concerning the type of local algebras. For example, if you take wedge shaped regions $W$ in Minkowski spacetime, then it is shown in 
On the Net of von Neumann algebras associated with a Wedge and Wedge-causal Manifolds, H.J. Borchers, http://www.lqp.uni-goettingen.de/papers/09/12/09120802.html
that the local algebra associated to $W$ is actually of type $III_1$ for nets satisfying similar assumptions motivated by physics.
Furthermore there is the paper 
The Universal Structure of Local Algebras, Buchholz, D., D'Antoni and Fredenhagen, K., Communications in Mathematical Physics 111 (1), page 123-135
in which it is shown that if you assume that your net is derived from a Wightman QFT and satisfies asymptotic scale invariance and  nuclearity, then the local algebras associated to double cones are of the form $\mathfrak{R} \otimes Z(\mathcal{O})$, where $\mathfrak{R}$ is the unique hyperfinite type $III_1$-factor and $Z(\mathcal{O})$ is the center of the local algebra $\mathcal{A}(\mathcal{O})$.
So, this and the fact that in many examples of nets you can check directly that the local algebras are of type $III_1$ is the justification for physicists to assume this to be the case in everything physically relevant.<|endoftext|>
TITLE: Does IP = PSPACE work over other rings?
QUESTION [7 upvotes]: Background: It is possible (see e.g., this) to define a Turing machine over an arbitrary ring.  It reduces to the classical notion when the ring is $\mathbb{Z}_2$; the key difference is that elementary algebraic computations are allowed to be performed in one step (and one has infinite precision).  It is possible to define analogs of classical complexity classes (e.g. $\mathrm{P}_R, \mathrm{NP}_R, \mathrm{BPP}_R$, etc. with respect to a ring $R$).  Sometimes the ring may be required to be ordered, and the machine is allowed to test for ordering (it is always allowed to test for equality).
It should, therefore, be possible to define a class $\mathrm{IP}_R$ of problems that can be solved by a polynomial-time interactive proof system over $R$ with error probability $\leq \frac{1}{3}$, a class $\mathrm{PSPACE}_R$ of problems that can be solved in polynomial space by a (deterministic) Turing machine.  (Probably this has already been done, but I just haven't been able to find it.) When $R = \mathbb{Z}_2$ (i.e. the classical case), it is a known result that $\mathrm{IP}=\mathrm{PSPACE}$.
Question: Does the same work over other rings $R$?
Perhaps there might be a problem, since the only proof I've seen (e.g. in Sipser's Introduction to the Theory of Computation) uses the $\mathrm{PSPACE}$-completeness of $TQBF$, and I don't know whether that would work over an arbitrary ring.

REPLY [2 votes]: This does not directly answer your question, but it is related to issues of relativizing IP and PSPACE.

For a random oracle A, IPA
  ≠ PSPACEA with
  probability 1.

The problems coincide relative to a random oracle with probability either 0 or 1 by Komolgorov's zero-one law. That they are distinct for almost all oracles was proved by Chang-Chor-Goldreich-Hartmanis-Håstad-Ranjan-Rohatgi in "The random oracle hypothesis is false." J. Comput. System Sci.  49  (1994) 1 24--39. Note however that if you pass from IP to IPP, where the error probability is only ≤ 1/2, you do have that IPPA = PSPACEA for a random oracle A, so these are subtle issues.<|endoftext|>
TITLE: What is the definition of the $\uplus$ symbol?
QUESTION [7 upvotes]: Hi,
 I have what I hope is a very simple question related to unfamiliar notation.
 I am looking through a maths paper on a topic related to set theory which contains a symbol, 
$\uplus$, 
and I would like to know how, if at all, it differs from the typical 
$\cup$ 
symbol in terms of its meaning.
 The context leads me to believe that it does not in fact differ at all but since I don't even know the name of the symbol other than the latex id that I looked up, I can't seem to confirm that suspicion.
 Cheers
edit:  it seems that the latex renderer also does not know about this obscure symbol '\cupplus' but '\uplus' does work.
edit2:  thanks for all the replies!  however, since everyone commented rather than providing an answer it seems I cannot grant the coveted 'answer' status to anyone.  the disjoint union makes the most sense.

REPLY [7 votes]: It may also denote that the union is a multi-set (and the both operands are possibly multi-sets). Then the multiplicity of an element in the result is the sum of its multiplicities in the operands, that explains the existence of the + in the symbol.<|endoftext|>
TITLE: Roadmap for studying arithmetic geometry
QUESTION [47 upvotes]: I have read Hartshorne's Algebraic Geometry from chapter 1 to chapter 4, so I'd like to find some suggestions about the next step to study arithmetic  geometry. 

I want to know how to use scheme theory and its cohomology to solve arithmetic problems. 
I also want to learn something about moduli theory.

Would you please recommend some books or papers? Thank you very much!

REPLY [3 votes]: Has not been mentioned yet: James Milne's course notes http://www.jmilne.org/math/CourseNotes/index.html and his books http://www.jmilne.org/math/Books/index.html, especially the one on Arithmetic Duality Theorems.<|endoftext|>
TITLE: Why are modular forms (usually) defined only for congruence subgroups?
QUESTION [15 upvotes]: Modular forms could be defined for arbitrary subgroups of the modular group, and I have read that this is done in some papers, but every definition of a modular form I have seen has been with respect to congruence subgroups.

REPLY [14 votes]: Actually, modular curves for congruence groups have canonical models over number fields, not Q (there exist congruence subgroups other than $\Gamma _0(N)$!). They even have reasonably nice integral models. Moreover, the modular forms are sections of a sheaf defined on the canonical model, and the sheaf extends to the integral model. This has the following consequence: if a modular form has Fourier coefficients $a_n$ in a number field $K$ (so the form is a section of the sheaf over $K$), then the $a_{n}$ have bounded denominators, i.e., lie in $d^{-1}\mathcal{O}_{K}$ for some $d$.
Modular curves for arithmetic groups also have models over number fields, but the last statement definitely fails for forms that don't come from congruence groups. So something goes wrong with the beautiful picture we have for congruence modular curves, but I've never understood exactly what. However, this is another indication that the link to arithmetic is more tenuous in the noncongruence case, and helps explain why number theorists are mainly interested in the congruence case.
[This was written as a comment on Buzzard's answer, but the site wouldn't let me post it (too long).]<|endoftext|>
TITLE: What are some mathematical concepts that were (pretty much) created from scratch and do not owe a debt to previous work?
QUESTION [29 upvotes]: Almost any mathematical concept has antecedents; it builds on, or is related to, previously known concepts. But are there concepts that owe little or nothing to previous work?
The only example I know is Cantor's theory of sets. Nothing like his concrete manipulations of actual infinite objects had been done before.

REPLY [2 votes]: I have to agree with Scott's comment: Every development has its roots. The following three examples are thus only approximations. 
The first is Riemann's work on the "On the Hypotheses which lie at the Bases of Geometry". As a habilitation talk it is almost devoid of any details, but it is not only one of the earliest accounts of geometry in $n$ (or even infinite) dimension, it also gives the ideas of a Riemannian metric and the Riemann curvature tensor! As Riemann said it:

[...] ausser einigen ganz kurzen Andeutungen, welche Herr Geheimer Hofrath
  Gauss in der zweiten Abhandlung über die biquadratischen Reste [...] darüber gegeben hat und einigen philosophischen Untersuchungen Herbart’s, [konnte ich] durchaus
  keine Vorarbeiten benutzen [...].
Translation: expect for a few very short hints, which Privy Councillor Gauss gave in his second work on biquadratic residues, and some philosophical investigations Herbart's, I could not use any previous work. 

Also Gauss's work on the relationship between intrinsic and extrinsic geometry of surfaces, culminating in his Theorema Egregium, might qualify. Of course, there was some previous work on surfaces, but this goes so much deeper that all previous work pales in comparison. 
I also want to mention Grassmann's Die Lineale Ausdehnungslehre, ein neuer Zweig der Mathematik, which already states in the title that is a new branch of mathematics. (Note there are two quite different editions, 1844 and 1862). Essentially he invented linear algebra in this book. Again not completely without precursors, as people solved linear equations before, but to use geometric ideas in $n$ dimension, subspaces, linear independence, exterior algebras etc. was very new. See this this article for an overview of his contributions.<|endoftext|>
TITLE: Is there a simple proof that a group of linear growth is quasi-isometric to Z?
QUESTION [36 upvotes]: I proposed to a master's student to work, from the exercise in Ghys-de la Harpe's book, on the proof that a finitely generated group $G$ that is quasi-isometric to $\mathbb{Z}$ is virtually $\mathbb{Z}$. However I initially had in mind the result that gives the same conclusion from the hypothesis that $G$ has linear growth.
Do you know of any simple (and elementary, in particular without assuming Gromov's theorem on polynomial growth groups) proof that a group of linear growth is quasi-isometric to $\mathbb{Z}$?

REPLY [32 votes]: I believe the substance of the question has been answered, but since it's a  question of what can be done without modern machinery, this may help: I was thinking about the problem of deducing structure of a group from the growth rate in the '70's, when Gromov was still in Russia. When we first met, soon after he arrived, we had much in common that we'd been thinking about, but he had not yet proven the theorem concerning groups of polynomial growth. I was stuck on trying to analyze groups of quadratic growth; I recall that strictly linear growth seemed fairly straightforward using the technology of the time: in particular, Stalling's idea of a minimizing cocyle (from his work analying ends of groups). ( I believe when Gromov and I first met we both knew that groups of linear growth are virtually $\mathbb Z$, and we focused on other things.)  I'm pretty sure there's a fairly elementary complete proof based on Stallings' idea, just thinking of components of the sphere of radius R as defining a cocycle.  I could supply details on demand.

Added in response to Sam Nead's request:

This isn't very different from what Sam Nead and others outlined, but I'll give some  variations of a proof. These are all related:

Consider a Cayley graph for the group.  Between any two vertices $v$ and $w$, ask what is the maximal graph-flow between $v$ and $w$ (Each edge carries a flow of at most 1. A graph flow, in other language, is expressing  the maximum possible multiple of the 0-chain w-v a the boundary of a 1-chain of $L^\infty$ norm 1).  The well-known min-flow / max-cut principle says that the maximal flow equals the minimal cut, that is, the minimum $L^1$ norm of a collection of edges that separates $V$ from $W$, which in other language is the minimum $L^1$ norm of the coboundary of a 0-cochain that is 0 on $v$ and 1 on $w$.
Linear growth implies there is a uniform upper bound for the graph flow, no matter how   distant $v$ and $w$ are, because there are balls with bounded size of boundary about $v$ that don't contain $w$.
Since there are spheres of arbitarily large radius of bounded size, if v and w are far enough apart, there must be a graph flow that is isomorphic in a neighborhood of two such spheres $S_r(v)$ and $S_s(v)$ where $0 < r, s< d(v,w)$. Take the annulus between them, and identify the spheres. A connected component of the resulting graph defines a subgroup of finite index in the group, and it comes equipped with a cyle (the flow) paired non-trivially with the cocycle (the identified spheres of radius $r$ and $s$).
Similarly to (1), you can look at the combinatorial derivative of the distance function from a vertex $v$, that a 1-cocycle on a 2-complex for the group that takes values $\pm 1$ and $0$.  The derivative of the distance function has to repeat on spheres of radius R.  Cut and glue, as before, and get a subgroup of finite index with a non-trivial 1-cocycle, giving a homomorphism to $\mathbb Z$.
Another variation, same general idea, technically harder but perhaps giving a clearer mental image:  you can take a Riemannian manifold with fundamental group $G$, and in its universal cover, do volume-constrained minimization of hypersurfaces: what is the least area $A(V)$ for a hypersurface that bounds a volume $V$?
Such surfaces have constant mean curvature.  There has been a reasonably good existence theorem for solutions for a long time --- the solutions may not be smooth hypersurfaces, although in low dimensions they are, but they still have nice geometric descriptions.
At a local minimum for $A(V)$, the hypersurface is a minimal surface. Even if there are no local minima, one could take limits as $V \to \infty$ to get minimal hypersurfaces.
The images of minimizing hypersurfaces by deck transformations are either disjoint or they coincide, by familiar arguments (cutting and paste to get smaller surfaces doing the job).
Use the separation properties of these to get a map to the infinite dihedral group, a la Stallings.
Automatic group theory. This theory is subsequent to Gromov's proof, but it had incipient forms before -- a number of people, including me and Gromov, had thought about growth patterns for groups.  I think there's a nice pathway  linear growth => automatic =(with polynomial growth)=> virtually Z^n => virtually Z.  I've written enough, so I won't unroll this now.<|endoftext|>
TITLE: Number of trees with n nodes and m leaves
QUESTION [9 upvotes]: Even searching for " 'number of trees' leaves " didn't reveal what I am looking for: an approach for calculating the (approximate) number of trees with exactly n nodes and m leaves. Any hints from MO?

REPLY [5 votes]: The number of labelled trees on $n$ vertices with $m$ leaves is 
$$\binom{n}{m}S(n-2,n-m)(n-m)!$$
where $S(a,b)$ is the Stirling number of the second kind. This can be seen by analysing the multivariate generation which counts trees of any degree sequence given here: https://math.berkeley.edu/~mhaiman/math172-spring10/matrixtree.pdf<|endoftext|>
TITLE: When does adding inverses of morphisms preserve commutativity of a diagram?
QUESTION [5 upvotes]: Here is the essence of a problem I have run in to:  I have a finite poset D with a terminal object.  If I formally invert all of the morphisms, and add these into my diagram, does the new diagram D' still commute?
I think that the resulting diagram will still commute basically because I have done a lot of examples.  Working out a few examples you can see that it basically follows by doing it for the "commutative triangle", and applying this finitely many times. It feels like I should be able to do some kind of messy induction, but I do not really want such a proof cluttering up my work.
Is there a reference I could quote for a result like this? It seems like if it is true it should be a "folk lemma".
Of course, if you have more relaxed criteria for when the result will hold, that would be helpful too.  
Also if you know of a conceptual proof which does not fall back on some messy induction, that would be wonderful!
EDIT:  An example might help to clarify my question. (How do you draw diagrams on MO?)
 a-->b
 ^   ^
 |   |
 c-->d

is my poset.  b is the terminal object.  Now say someone told you that this was actually a subcategory of a larger category, and in that larger category all of the arrows were invertible. Now consider the larger diagram consisting of the 4 original arrows and their inverses.  Is this diagram also commutative?  Yes! It is just one or two lines of formal manipulation.

REPLY [5 votes]: There is an easy conceptual proof using the fact that the category obtained by formally inverting all the arrows in a category C is equivalent to the fundamental groupoid of the nerve NC of C, and that the nerve of a category with a final object is contractible.  Without the assumption of a final object your assertion is false in general, e.g., reverse the arrows from c in your example.
But it should also be easy to prove by induction: for any zigzag of arrows between a and b, the corresponding map in the category with all arrows inverted, when composed with the map from b to the original terminal object, is equal to the map from a to the original terminal object (this is by induction); and so any two maps from a to b in the category with all arrows inverted are equal.  In symbols: let me write $t_x$ for the unique morphism in C from $x$ to the terminal object and $[f]$ for the image of $f$ in the category with all arrows inverted.  Suppose $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1}$ is a typical map in the category with all arrows inverted with domain $a$ and target $b$.  Then the inductive claim is that $[t_b] [f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_a]$, and so $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_b]^{-1} [t_a]$.<|endoftext|>
TITLE: Rational homotopy type of a complement
QUESTION [6 upvotes]: Let $X$ and $X'$ be smooth closed manifolds. Take closed subpolyhedra $D\subset X$ and $D'\subset X'$ (with respect to some triangulations) and let $f:X\to X'$ be a homotopy equivalence such that $f(D)=D'$ and the restriction of $f$ to $D$ is also a homotopy equivalence. Is it possible for the complements $X-D$ and $X'-D'$ to have different rational homotopy types, assuming all spaces ($X,X',D,D'X-D,X'-D'$) simply-connected?
Here is some motivation behind the question: if we replace the rational homotopy type with the integral one and do not require the spaces to be simply connected, then the answer is yes, as shown in a paper by R. Longoni and P. Salvatore http://arxiv.org/abs/math/0401075; a much simpler example is in Ryan's comment below. On the other hand, additively the cohomology of $X-D$ and $X'-D'$ is obviously the same.
upd: in the first version of the question the simple connectedness condition was missing. Apologies for the mix-up.

REPLY [2 votes]: Some of the best results known along these lines are in the paper

Pascal Lambrechts, Don Stanley, Algebraic models of Poincaré embeddings, Algebr. Geom. Topol. 5 (2005) 135-182, doi:10.2140/agt.2005.5.135, arXiv:math/0503605,

which used to be called "Algebraic models of complements...".
Briefly, it seems that even if the complement is simply connected, there are examples of what you seek when the codimension is too small, and when the codimension is large enough the rational homotopy type of a complement is determined.<|endoftext|>
TITLE: "Transitivity" of the Stone-Cech compactification
QUESTION [35 upvotes]: Let $\beta \mathbb{N}$ be the Stone-Cech compactification of the natural numbers $\mathbb{N}$, and let $x, y \in \beta \mathbb{N} \setminus \mathbb{N}$ be two non-principal elements of this compactification (or equivalently, $x$ and $y$ are two non-principal ultrafilters).  I am interested in ways to "model" the ultrafilter $y$ using the ultrafilter $x$.  More precisely, 
Q1.  (Existence) Does there necessarily exist a continuous map $f: \beta \mathbb{N} \to \beta \mathbb{N}$ which maps $x$ to $y$, while mapping $\mathbb{N}$ to $\mathbb{N}$?  To put it another way: does there exist a function $f: \mathbb{N} \to \mathbb{N}$ such that $\lim_{n \to x} f(n) = y$?
Q2.  (Uniqueness)  Suppose there are two continuous maps $f, g: \beta \mathbb{N} \to \beta\mathbb{N}$ with $f(x)=g(x)=y$, which map $\mathbb{N}$ to $\mathbb{N}$.  Is it then true that $f$ and $g$ must then be equal on a neighbourhood of $x$?
I suspect the answer to both questions is either "no" or "undecidable in ZFC", though perhaps there exist "universal" ultrafilters $x$ for which the answers become yes.  But I do not have enough intuition regarding the topology of $\beta \mathbb{N}$ (other than that it is somewhat pathological) to make this more precise.  (The fact that $\beta\mathbb{N}$ is not first countable does seem to indicate that the answers should be negative, though.)

REPLY [12 votes]: The answer to Q1 is even more `no': Kunen showed that there are x and y such that x cannot be mapped to y and y cannot be mapped to x, see this review. This has been strengthened by Rudin and Shelah. It is still open, as far as I know, whether given x there is a y such that neither can be mapped to the other.<|endoftext|>
TITLE: System of eigenvalues of level $N$ weight $k$ mod $l$ 
QUESTION [5 upvotes]: Let $M_k(N)$ be the space of modular forms of level $N$ and weight $k$ whose $q$-coefficients at infinity are rational and $l$-integral.
Put $\tilde M_k(N) = M_k(N) \otimes \mathbb{\bar F}_l$ and let $\mathbb T_k$ be the subring of $\mathrm{End}_{\mathbb{\bar F}_l}(\tilde M_k(N))$ generated by Hecke operators $T_p$ for all prime $p$, prime to $N$. 
What is the number of different systems of eigenvalues for all $T_p$ with $(p,N)=1$? 
My feeling is that this is a number of maximal ideals in $\mathbb T_k$, but can not make it precise.

REPLY [7 votes]: Your intuition is correct. Actually this has nothing to do with modular forms: there is a simple general statement which contains your question:
Let $k$ be an algebraically closed field,
$M$ a finite dimensional $k$-vector space, $(T_i)$ a family of commuting 
linear operators on $M$, and $\tau$ the sub-algebra of $End_R(M)$ they generate.
Then there is a natural bijection between the maximal ideals of $\tau$ and the system of eigenvalues for all the $T_i$ in $M$.
The bijection is easily constructed in one direction: suppose that $T_i \mapsto \lambda_i$
is a system of eigenvalues appearing in $M$ (that is to say, there is a $0 \neq v \in M$ such that $T_i v = \lambda_i v$ for all $i$). Then obviously, $v$ is an eigenvector 
for all $T \in \tau $  and the map $ \tau \rightarrow k$ sending $T \in \tau$ on the unique
$\lambda(T) \in k$ such that $T v = \lambda(T) v$ is a morphism of $k$-algebras.
Its kernel is thus a maximal ideal of $T$. This defines a map "system of eigenvalues" $\rightarrow$ "maximal ideals of $\tau$".
Conversely, if $m$ is a maximal ideal of $\tau$, then $\tau/m=k$ (since $\tau$ is finitely generated and $k$ is algebraically closed) so the map $\lambda : \tau \rightarrow \tau/m=k$ is a morphism of $k$-algebras. You want to prove that there is a $0 \neq v \in M$ such that $T v = \lambda(T)v$ for all $T \in \tau$. It suffices to do it for a family of generators
$T_1,\dots,T_k$ of $\tau$, which we can choose finite. Now since $\lambda (T_1-\lambda(T_1))=0$, $T_1-\lambda(T_1)$ is not invertible in $\tau$, hence is not invertible
in $End_k(M)$ (otherwise its inverse would be by Cayley-Hamilton a polynomial in $T_1$,
hence in $\tau$) hence $V_1 = \ker (T_1 - \lambda(T_1))$ is not $0$. Now $T_2, \dots, T_k$
stabilize $V_1$ since they commute with $T_1$, and you can make the same reasoning for $T_2$ acting on $V_1$, and so on. At the end you get a non-zero subspace of $V$ on which
each $T_i$ acts by the scalar $\lambda(T_i)$, QED.
That's a very pedestrian proof. If you are willing to apply some basic commutative algebra
(e.g. Artinian rings are semi-local) you can do it much faster.<|endoftext|>
TITLE: Nice proof of the triangle inequality for the metric of the hyperbolic plane
QUESTION [11 upvotes]: I am writing something for the journal of the university on the Lorentz space, and I want to prove that the following definition of the hyperbolic distance on the upper sheet of the hyperboloid $v \cdot v=-1$ is a metric:
The hyperbolic distance between $u$ and $v$ is the only positive number $\eta (u,v)$ such that
$$\cosh(\eta(u,v))=-u \cdot v.
$$
The first properties are easy, but all the proofs of the triangle inequality that I have seen seem complicated. In Ratcliffe's Foundations of Hyperbolic Manifolds the lorentzian cross product is used along with many of its properties, and I would like not having to introduce the cross product just for this proof.
I have thought about showing that the hyperbolic angle is the same as the lorentzian arc-length distance, and then showing that the metric given by the arc length satisfies the triangle inequality, but this seems even more tedious. Is there a shorter and nicer proof of this?

REPLY [2 votes]: The inequality we want to prove is
$$
\left|\frac{u-v}{1-\overline{u}v}\right| \le 
\left|\frac{u-a}{1-\overline{u}a}\right| + 
\left|\frac{a-v}{1-\overline{a}v}\right|
$$
for $u,v,a$ in the open unit disk ${\mathbf D}$. Let $f$ be the map
$$ 
f(\zeta)=\frac{\zeta-a}{1-\overline{a}\zeta} \quad (\zeta \in {\mathbf D}).
$$
It is easily seen that $f$ maps ${\mathbf D}$ into itself. Indeed, if $\zeta,a \in {\mathbf D}$, then $(1-|\zeta|^2)(1-|a|^2)> 0$, it follows that
$|\zeta|^2+|a|^2 < 1+ |a|^2 |\zeta|^2$, and after adding $-\overline{a}\zeta-a\overline{\zeta}$ on both sides, we get
$ 
|\zeta-a|^2 < |1-\overline{a}\zeta|^2$, that is 
$f(z)=\frac{\zeta-a}{1-\overline{a}\zeta} \in {\mathbf D}$.
A straightforward computation shows that 
$$
\left|\frac{f(\zeta)-f(\xi)}{1-\overline{f(\zeta)}f(\xi)}\right|
=\left|\frac{\zeta-\xi}{1-\overline{\zeta}\xi}\right|.
$$
Thus, in the inequality we want to prove we may replace $u,v,a$ by $f(u)=:z$,
$f(v)=:w$, $f(a)=0$, respectively, and we are left with proving that
$$
\left|\frac{z-w}{1-\overline{z}w}\right| \le |z|+|w|,
$$
that is,
$$ 
|z-w| \le |z|\,|1-\overline{z}w|+|w|\,|1-\overline{z}w|
$$
for $z,w \in {\mathbf D}$. Since $|1-\overline{z}w|=|1-\overline{w}z|$,
this is equivalent to the inequality
$$
|z-w| \le |z-|z|^2 w|+|w-|w|^2z|.
$$
Denote the points $z$ and $w$ in the complex plane by $Z$ and $W$, and let
$O$ denote the origin. The point $|z|^2w$ is a point $P$ on the line segment
between $O$ and $W$, and $z-|z|^2w$ is the vector from $P$ to $Z$.
It follows that $|z-|z|^2w|$ equals the (Euclidean) length $|PZ|$ of
the line segment between $P$ and $Z$. Analogously, $|w-|w|^2z|=|QW|$ for
some point $Q$ on the line segment between $O$ and $Z$. Let $S$ be the
point of intersection of the line segments $PZ$ and $QW$. We then have
$$
|z-w|  =  |ZW| \le |ZS|+|SW| \le |PZ|+|QW| = |z-|z|^2 w|+|w-|w|^2z|,
$$
which is what we wanted to prove.<|endoftext|>
TITLE: Orbit structures of conjugacy class set and irreducible representation set under automorphism group
QUESTION [23 upvotes]: let G be a finite group. Suppose C is the set of conjugacy classes of G and R is the set of (equivalence classes of) irreducible representations of G over the complex numbers.
The automorphism group of G has a natural action on C and also on R (we can make both of these left actions). My questions:

Under what conditions are C and R equivalent as $\operatorname{Aut}(G)$-sets? This is definitely true, for instance, if every automorphism is inner, if the outer automorphism group of G is cyclic (it then follows from Brauer's permutation lemma) and it is also true if the quotient of the automorphism group by the group of class-preserving automorphisms is cyclic (again by Brauer's permutation lemma). But it also seems to be true in a number of other cases, such as the quaternion group, where the outer automorphism group is a symmetric group of degree three.
A weaker condition: under what conditions are the orbit sizes under $\operatorname{Aut}(G)$ for C and R the same?

REPLY [20 votes]: I think that an example of non-equivalent permutation sets is given by
$G=(\mathbb Z/p\mathbb Z)^n$ for $n>2$ (and $p$ a prime). Then the automorphism
group is $\mathrm{GL}_n(\mathbb Z/p\mathbb Z)$, the conjugacy classes are in
natural bijection with $G$ and the set of irreducible representations are in
bijection with the dual group (or dual $\mathbb Z/p\mathbb Z$-vector space). In
both cases there are only two orbits, one of length $1$ (the identity element
and the trivial representation respectively). The stabilisers for elements in
the non-trivial orbits are not conjugate: Mapping to $\mathrm{PGL}_n(\mathbb
Z/p\mathbb Z)$ map these two kinds of stabilisers two non-conjugate parabolic
subgroups (stabilisers of lines resp. of hyperplanes).